Visualizing the “Greenhouse Effect” – Atmospheric Windows

Guest post by Ira Glickstein

A real greenhouse has windows. So does the Atmospheric “greenhouse effect”. They are similar in that they allow Sunlight in and restrict the outward flow of thermal energy. However, they differ in the mechanism. A real greenhouse primarily restricts heat escape by preventing convection while the “greenhouse effect” heats the Earth because “greenhouse gases” (GHG) absorb outgoing radiative energy and re-emit some of it back towards Earth.

The base graphic is from Wikipedia, with my annotations. There are two main “windows” in the Atmospheric “greenhouse effect”. The first, the Visible Light Window, on the left side of the graphic, allows visible and near-visible light from the Sun to pass through with small losses, and the second, the Longwave Window, on the right, allows the central portion of the longwave radiation band from the Earth to pass through with small losses, while absorbing and re-emitting the left and right portions.

The Visible Light Window
To understand how these Atmospheric windows work, we need to review some basics of so-called “blackbody” radiation. As indicated by the red curve in the graphic, the surface of the Sun is, in effect, at a temperature of 5525ºK (about 9500ºF), and therefore emits radiation with a wavelenth centered around 1/2μ (half a micron which is half a millionth of a meter). Solar light ranges from about 0.1μ to 3μ, covering the ultraviolet (UV), the visible, and the near-infrared (near-IR) bands. Most Sunlight is in the visible band from 0.38μ (which we see as violet) to 0.76μ (which we see as red), which is why our eyes evolved to be sensitive in that range. Sunlight is called “shortwave” radiation because it ranges from fractional microns to a few microns.

As the graphic indicates with the solid red area, about 70 to 75% of the downgoing Solar radiation gets through the Atmosphere, because much of the UV, and some of the visible and near-IR are blocked. (The graphic does not account for the portion of Sunlight that gets through the Atmosphere, and is then reflected back to Space by clouds and other high-albedo surfaces such as ice and white roofs. I will discuss and account for that later in this posting.)

My annotations represent the light that passes through the Visible Light Window as an orange ball with the designation 1/2μ, but please interpret that to include all the visible and near-visible light in the shortwave band.

The Longwave Window
As indicated by the pink, blue, and black curves in the graphic, the Earth is, in effect, at a temperature that ranges between a high of about 310ºK (about 98ºF) and a low of about 210ºK (about -82ºF). The reason for the range is that the temperature varies by season, by day or night, and by latitude. The portion of the Earth at about 310ºK radiates energy towards the Atmosphere at slightly shorter wavelengths than that at about 210ºK, but nearly all Earth-emitted radiation is between 5μ to 30μ, and is centered at about 10μ.

As the graphic indicates with the solid blue area, only 15% to 30% of the upgoing thermal radiation is transmitted through the Atmosphere, because nearly all the radiation in the left portion of the longwave band (from about 5μ to 8μ) and the right portion (from about 13μ to 30μ) is totally absorbed and scattered by GHG, primarily H2O (water vapor) and CO2 (carbon dioxide). Only the radiation near the center (from about 8μ to 13μ) gets a nearly free pass through the Atmosphere.

My annotations represent the thermal radiation from the Earth as a pink pentagon with the designation for the left-hand portion, a blue diamond 10μ for the center portion, and a dark blue hexagon 15μ for the right-hand portion, but please interpret these symbols to include all the radiation in their respective portions of the longwave band.

Sunlight Energy In = Thermal Energy Out
The graphic is an animated depiction of the Atmospheric “greenhouse effect” process.

On the left side:
(1) Sunlight streams through the Atmosphere towards the surface of the Earth.
(2) A portion of the Sunlight is reflected by clouds and other high-albedo surfaces and heads back through the Atmosphere towards Space. The remainder is absorbed by the Surface of the Earth, warming it.
(3) The reflected portion is lost to Space.

On the right side:
(1) The warmed Earth emits longwave radiation towards the Atmosphere. According to the first graphic, above, this consists of thermal energy in all bands ~7μ, ~10μ, and ~15μ.
(2) The ~10μ portion passes through the Atmosphere with litttle loss. The ~7μ portion gets absorbed, primarily by H2O, and the 15μ portion gets absorbed, primarily by CO2 and H2O. The absorbed radiation heats the H2O and CO2 molecules and, at their higher energy states, they collide with the other molecules that make up the air, mostly nitrogen (N2), oxygen (O2), ozone (O3), and argon (A) and heat them by something like conduction. The molecules in the heated air emit radiation in random directions at all bands (~7μ, ~10μ, and ~15μ). The ~10μ photons pass, nearly unimpeded, in whatever direction they happen to be emitted, some going towards Space and some towards Earth. The ~7μ and ~15μ photons go off in all directions until they run into an H2O or CO2 molecule, and repeat the absorption and re-emittance process, or until they emerge from the Atmosphere or hit the surface of the Earth.
(3) The ~10μ photons that got a free-pass from the Earth through the Atmosphere emerge and their energy is lost to Space. The ~10μ photons generated by the heating of the air emerge from the top of the Atmosphere and their energy is lost to Space, or they impact the surface of the Earth and are re-absorbed. The ~7μ and ~15μ generated by the heating of the air also emerge from the top or bottom of the Atmosphere, but there are fewer of them because they keep getting absorbed and re-emitted, each time with some transfered to the central ~10μ portion of the longwave band.

The symbols 1/2μ, , 10μ, and 15μ represent quanties of photon energy, averaged over the day and night and the seasons. Of course, Sunlight is available for only half the day and less of it falls on each square meter of surface near the poles than near the equator. Thermal radiation emitted by the Earth also varies by day and night, season, local cloud cover that blocks Sunlight, local albedo, and other factors. The graphic is designed to provide some insight into the Atmospheric “greenhouse effect”.

Conclusions
Even though estimates of climate sensitivity to doubling of CO2 are most likely way over-estimated by the official climate Team, it is a scientific truth that GHGs, mainly H2O but also CO2 and others, play an important role in warming the Earth via the Atmospheric “greenhouse effect”.

This and my previous posting in this series address ONLY the radiative exchange of energy. Other aspects that control the temperature range at the surface of the Earth are at least as important and they include convection (winds, storms, etc.) and precipitation that transfer a great deal of energy from the surface to the higher levels of the Atmosphere.

I plan to do a subsequent posting that looks into the violet and blue boxes in the above graphic and provides insight into the process the photons and molecules go through.

I am sure WUWT readers will find issues with my Atmospheric Windows description and graphics. I encourage each of you to make comments, all of which I will read, and some to which I will respond, most likely learning a great deal from you in the process. However, please consider that the main point of this posting, like the previous one in this series, is to give insight to those WUWT readers, who, like Einstein (and me :^) need a graphic visual before they understand and really accept any mathematical abstraction.

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About Ira Glickstein, PhD

[Retired] Senior System Engineer (Advanced Avionics and Visionics, Route Planning, Decision Aiding, Five Patents ... at IBM, Lockheed-Martin); Adjunct Associate Professor (System Engineering at University of Maryland, System Science and Computer Science at Binghamton University); PhD in System Science (Binghamton University, 1996); MS in System Science (Binghamton); Bachelors in Electrical Engineering (CCNY)
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489 Responses to Visualizing the “Greenhouse Effect” – Atmospheric Windows

  1. etudiant says:

    Relative humidity in the upper atmosphere (300mb level) has fallen from around 55% in 1950 to about 45% now.
    Should this not have a material impact on the greenhouse effect?
    It is hard to get concerned when such substantial changes don’t even warrant discussion, much less explanation. Or is there some obvious offset that makes this moot?

  2. Gerry says:

    I realize the first graphic is attributed to Wikipedia, but I am used to wavelength graphs with the longer wavelengths (lower frequency) on the left and shorter wavelengths (higher frequency) on the right. Also confusing are the colors which are backwards from the norm.

    Unless 2/3 of a beer has completely ruined me mentally (I’m willing to negotiate that point)…

    Of course, the warmists have manipulated every other convention of science and engineering, so what’s new?

    Gerry

  3. Fred Souder says:

    Ira,
    Are there any experiments which detect the incoming re-emitted radiation from the atmosphere? Did a quick search and only came up with models. It seems like we should be able to detect the radiation coming back at us from the atmosphere if this model is valid.

  4. RJ says:

    Does it heat the earth or does it slow the rate of cooling?

    I do not think you should say its a scientific truth etc. It’s a theory that some experts do not agree with.

    I’m not an expert but I have a problem with a cold body (a small amount of CO2 and water vapour) heating a much larger warmer earth. Logically it does not seem right. If for example a human was put into a container filed with CO2 would they cook or warm up. I do not think so.

    And does energy leave the earth mainly by convection or radiation.

    Slaying the sky dragon addressed these issues. Maybe our understanding of this issue is still not that advanced and the majority viewpoint is flawed.

  5. Russell Duke says:

    Can you please give me the references for CO2 IR re-emittance

  6. mpaul says:

    Ira, I think this explanation is clearly written and understandable and is helpful to people who are new to the topic or who are less technically inclined.

    Most serious skeptics have little disagreement with the basic physics of the green house effect. The big issue (as you point out) is sensitivity. Many of us believe that sensitivity is a function of a large number of interrelated variables and as such lends itself to stochastic modeling rather than deterministic modeling. I’d be interested in your comments on this regard.

    Many on the True Believer side argue that the skeptics don’t ‘understand simple physics’ when the discussion turns to the green house effect. In fact, what the skeptics argue is that the green house effect is anything but simple and that the simplistic modeling approaches so in vogue with the AGW crowd are naive. Having said that, I do like Pielke’s approach of measuring ocean heat content anomaly as the best way to estimate sensitivity.

  7. Gil Dewart says:

    Finally someone has addressed the “atmospheric window” effect. There is a lot more to be said, but for now, “it’s about time”.

  8. Vuk etc. says:

    Most of the above is well understood by many (but not all) on both sides of the argument. The AGWs think that the solar input is constant, but the output is affected by CO2 (hence hockey stick).
    Sceptics think that the solar input is variable so the output has to change accordingly.
    There is also possibility of a half way house:
    Solar input is constant, but the energy storage – release relationship is not a linear function, but kind of thermal hysteresis, where the oceans absorb and store large chunks of energy well above the equilibrium, followed by more than the accumulated excess release. The effect of this is a sequence of the warm periods (roman, medieval & current one) separated by the cooler ones.

  9. Kevin says:

    From the looks of that first chart, ~100% of 7μ and 15μ radiation gets absorbed by the atmosphere on the way back to space, mainly by H20 and CO2. Maybe a tiny trickle bounces its way from molecule to molecule and makes it to space.

    If that’s so, then doubling or even tripling the values of H2O in our atmosphere should have no effect on absorbtion. You can’t absorb more than 100% of the radiation, right?

  10. Bryan says:

    Your graphs are completely misleading.

    They show that the intensity of the Solar Radiation as being equal to the Earth surface IR upward radiation.

    Did you really intend to show this?

    Further the colder atmosphere cannot heat or “warm” the warmer Earth Surface.

    All it can do is to slow to some extent the heat loss from the surface.

  11. Fred Souder says:

    RJ,
    I would guess that the vast majority of energy leaves earth by radiation. It has recently been discovered that our atmosphere is being continuously shed by the solar wind, and replenished by volcanism. This would be the only vehicle for convection that I can think of, and no idea if it is significant.

  12. Fred Souder says:

    Gerry,
    I must be that 2/3 beer. You should have downed the whole draft. Most physics and chemisry texts go from shorter to longer, left to right.

  13. “Other aspects that control the temperature range at the surface of the Earth are at least as important and they include convection (winds, storms, etc.) and precipitation that transfer a great deal of energy from the surface to the higher levels of the Atmosphere.”

    I would suggest CO2 in the atmosphere is much like the “correct” expanation of the Crookes radiometer (the black and white rotating thing in a glass vacuum bulb).

    According to one theory, the black absorbs light so it rotates away from the light, or is it that the black emits IR? Or is it that the black surface is slightly hotter therefore the few air particles exhibit a higher pressure. Or to put it correctly: “Over the years, there have been many attempts to explain how a Crookes radiometer works:” http://en.wikipedia.org/wiki/Crookes_radiometer.

    Just like the Crookes radiometer, carbon dioxide doesn’t just absorb IR, it also emits it. It doesn’t just interact with heat from the surface, it also interacts with the cold of space.

    The so called “science” of global warming can be completely turned on its head: increasing CO2 blocks heat being emitted by the earth –> increased CO2 emits more radiation to the cold sink of space.

    Just like the Crooke radiometer, the “obvious” explanation is neither obvious when you look at the deatils nor (in the case of the “greenhouse” effect which is a convective blocking effect.) is it right.

  14. wayne says:

    Ira, first, I think I will mostly agree with your entire post and that’s a great animated visual, but a question; where in the heck did 5525K for the sun’s temperature come from? Is that a hypothetical effective temperature as if all albedo and reflection has already been removed and falling on a flat Earth? (sounds like more Wikipedia’s hidden ‘clarity’)

  15. Dan in California says:

    Inspection of the infrared absorption windows for CO2 and H2O shows the much greater greenhouse effect of H2O. Yet I’ve often heard that the H2O only appears in the computer models as being forced by CO2, and therefore it doesn’t matter that H2O is a far more effective greenhouse gas than CO2. (People who want to regulate CO2 generation say this) My opinion is that the sensitivity to CO2 is weak, and it is an artifact of the written models that show CO2 effects predominating over H2O. Furthermore, because H2O changes from gas to liquid and back again, the H2O may even have a negative, regulating influence on the CO2 contribution to the greenhouse effect. Certainly the density driven convection caused by cloud formation and evaporation has a strong stirring effect that carries heat aloft. (A key difference between Earth and Venus is Venus’ lack of winds and vertical convection that carry heat upward)

    I would like to see a discussion of these issues, as I think this is key to the AGW enthusiast/ skeptic differences of opinion.
    Thanks,
    Dan

  16. jknapp says:

    Thanks Ira. Interesting post. A couple of questions, which will undoubtedly demonstrate my lack of physics knowledge. I thought that atoms emitted photons based on quanta specific to that atom. As such the H2O and CO2 molecules will absorb and emit only specific wavelengths. That would also be true of the O2 and N2 molecules. But here you are talking about “blackbody radiation” doesn’t that imply that the wavelength emitted will not be specific to the molecule but will, instead, be a function of the temperature. Is there a conflict here or is it a misunderstanding on my part?

    Assuming that my understanding will be resolved, then a further understanding on my part was that the 7u and 15u wavelengths were absorbed very quickly (in terms of distance traveled through the atmosphere. That would it seems to me to effectively “trap” the energy in a fairly narrow region with the vast amount of the energy kinda just hanging around near where it was absorbed neither going up or down very far due to radiation. That then would heat that region resulting in expansion and the physical raising of the molecules. This convective upward flow of the energy would be dominant over the downward portion of the radiative flow. Thus the energy would be lifted toward space. As it got higher more and more would escape outward and the downward radiative portion would serve mainly to keep the convection going with then very little energy actually being returned to the surface.

    Thus you mentioned convection as not being treated here but isn’t it possible that convection is a dominant not secondary driver of energy transmission out.

    Thanks for any enlightenment you can give me.

  17. wayne says:

    etudiant says:
    February 28, 2011 at 2:18 pm

    Relative humidity in the upper atmosphere (300mb level) has fallen from around 55% in 1950 to about 45% now.
    Should this not have a material impact on the greenhouse effect?
    It is hard to get concerned when such substantial changes don’t even warrant discussion, much less explanation. Or is there some obvious offset that makes this moot?
    ——–
    Sadly Dr. Miskolczi’s paper and it’s implications are rarely mentioned here, for that aspect of which you speak is probably why increased co2 has had little effect in reality (or it least it hasn’t had any measureable effect in the last 60 years). The slight warming came from some other reason and that is the real mystery, what other reason.

  18. George E. Smith says:

    Well some people like a wavelength Horizontal scale, and some people like a wave number scale.
    The AGW types, like the frequency scale, because that puts the CO2 frequencies near the peak of the curve; which creates the illusion that CO2 is far more important than it is. CO2 in the 606 to 741 wave number range looks much more impressive than the water bands beyond 1250 cm^-1, where the thermal emissions are only half the flux values as at the CO2 peak.

    Well the width of the water bands in wave numbers is a lot more than double what it is for CO2, so water still wins, although it is not as apparent on their frequency graphs. It may be true that frequency, which relates more directly to photon energy, is the right way to do it; but then Wien’s displacement law is always stated as a product of wavelength and Temperature, and the Planck function is often plotted as a function of the single variable lambda.T.

    I don’t like Ira’s wiki graph, because it creates the incorrect impression that the earth thermal emission spectrum is the same amplitude, as the incoming solar spectrum; which it is not.

    The area under the solar spectrum graph is four times the area under the thermal radiation spectrum, and the spectral width of the thermal spectrum, is 20 times that of the solar spectrum, so the actual amplitude of the thermal spectrum should be about 80 times smaller than it is.

    The solar spectrum goes from 0.25 to 4.0 microns for 98% of the energy, while the thermal spectrum goes from about 5.0 to 80 microns for the same 98% of the energy. Even if you equate those two energies, the thermal spectrum amplitude should be 20 times lower than plotted.

  19. George E. Smith says:

    jknapp; it is a misundertanding on your part. The wavelength specific nature of say CO2 or H2O is a function of the absorption modes of those molecules, which relate to specific mechanical resonances of the molecules themselves.

    BUT most of those excited molecules DO NOT get a chance to re-radiate, that photon, and return to the unexcited state. The time interval between molecular colliwsions is a few nanoseconds; but the lifetimes of the excited states, can be microseconds to milliseconds; so the energy is lost in collisions, before the molecule gets a chance to re-radiate. That thermalization heats the whole atmosphere; the N2, the O2, the Ar, and anything else; and it is that mixture of gases, at somer average temperature in the 250-300 K range, that is the source of the atmospehric emitted thermal continuum spectrum, that is not related to specific molecular energy levels, but only to the Temperature of the gases; whcih determines the dynamics of the collisions, and the resulting chaotic accelerations of the colliding molecules, and the electric charges they contain.

    So the atmosphere does radiate a black body like thermal spectrum, and the presence of the gHG molecules simply means that there will be bands of that continuum emission, that are also captured by the GHG molecules, as well as the emissions from the surface.

    The only function of the GHG molecules is to heat the atmospheric gases; along with all the other mechanisms that are heating it. After that, the GHG molecules serve no function in the climate process whatsoever.

  20. _Jim says:

    A nit pick; the ‘curves’ (from wiki) in that first figure all indicate to have the same peak amplitude (in reality they do not, of course). This is in the same category of drawing sine waves using half-circles from a drawing template which yield a ‘slope’ that is seemingly infinite as (delta y / delta x) -> infinity as x -> 0. (Never mind that the same that amplitude scale seem applicable to both solar and earth intensities … one may click my name and see the 2nd image down to see something drawn to scale relating sun and earth intensity. )

    Using this depiction does not convey graphically the idea that ever-greater amounts of energy ‘escape’ through the atmospheric window (out to space or to the ‘great beyond’) as the temperature rises.

    Please, correct me if I am wrong …

    .

  21. Edward Bancroft says:

    In your animated diagram what happens to the downward re-emitted 7μ radiation when it reaches the earth surface? I believe that what happens is that it may emitted again as 7μ IR radiation, or at a longer wavelength such as a photon in the 10μ band. In the former case it will have a high probability that it will be absorbed and downward emitted again, and in the latter case it will be lost to space.

    Thus the 7μ band sees not an equilibrium of up and down radiation flux, but a continuous stream of loss to 10μ. This would cause more energy to be lost than in your model.

  22. _Jim says:

    George E. Smith February 28, 2011 at 3:23 pm :

    So the atmosphere does radiate a black body like thermal spectrum, and the presence of the gHG molecules simply means that there will be bands of that continuum emission, that are also captured by the GHG molecules, as well as the emissions from the surface.

    I’ll wait to see how this is adjudged.

    (IR Spectroscopy explaining spectral response regarding gas molecule vibrational modes would seem to indicate otherwise.)

    .

  23. Iskandar says:

    This graph, about the atmospheric absorption, is from the 1970’s. Could anyone provide a recent one? Just to compare things.

    Thanks in advance.

  24. Dan in California says:

    jknapp says: February 28, 2011 at 3:01 pm
    Thanks Ira. Interesting post. A couple of questions, which will undoubtedly demonstrate my lack of physics knowledge. I thought that atoms emitted photons based on quanta specific to that atom. As such the H2O and CO2 molecules will absorb and emit only specific wavelengths.
    ————————————————–
    This is true at low pressures, such as inside a fluorescent light bulb or a gas laser. At higher pressures, such as atmospheric, the molecules interact with each other and smear the lines into bands. Think of it as trying to make a perfect baseball pitch with 3 people bumping into you.

  25. DCC says:

    etudiant said:

    Relative humidity in the upper atmosphere (300mb level) has fallen from around 55% in 1950 to about 45% now.

    Can you provide a reference for that?

  26. Philip Peake (aka PJP) says:

    Now this is much better than the bouncing balls :-)

    Now we are getting somewhere.

    To simplify what is going on, we have very little impediment between the Sun’s radiation and the earth. But there is a layer of absorbent material between us and space at the frequencies at which the earth re-radiates its energy.

    This absorbent layer consists mainly of water vapor (H2O), and a little CO2.
    These gases absorb the radiation (heat) and themselves become warm. Like all insulators, all they do is delay the loss of heat. No insulator stops the loss of heat, it just delays it. In this case it is delayed by the many, many absorption re-emission, absorptions of the earths radiated energy.

    The important thing about this is told by the first graph. Look at the words on the right-hand side: “Total absorption and scattering…”.

    What that means is that adding any more H2O or CO2 will have zero effect, because all the radiation from the earth is ALREADY absorbed by these gases.

    The only way that loss of heat can be reduced is to make the insulating layer thicker (increase the depth of the atmosphere). I have never heard any claim that this is what additional CO2 is doing.

    In fact, the only way CO2 can have any significant effect is by the mysterious “forcings”, supposedly pushing more H2O into the atmosphere … but see above .. the absorption by H2O is already 100%, so unless (again) there is a claim that the increased H2O is increasing the depth of the atmosphere, the means by which this insulating layer is becoming more effective escapes me.

  27. Olavi says:

    Thanks for intresting post. Have you considered that Quiet Sun makes atmosphere to shrink and as we know from sound waves, low frequency sound waves go easier through thinner wall as well shorter waves starting to go through if wall is thinner. I believe that IR acts same way. Is there any measuremet of IR behavior between situations when atmosphere is puffed up or shrinked down? Collapse of thermosphere can make diffrencies to athmosphere’s density at least down to 30 km altitude. I believe, it can have measurable effects to atmospheric temperature.

  28. DCC says:

    Fred Souder said:

    It has recently been discovered that our atmosphere is being continuously shed by the solar wind, and replenished by volcanism.

    Could you give us a reference for that?

  29. Alex says:

    If absorption of the re-radiated energy by water vapour is logarithmic akin to that of CO2, and the concentration of water vapour in the atmosphere is genrally already several orders of magnitude higher than CO2, how can an increase in absolute humidity have any more than a miniscule effect? It seems the models rely on this increased humidity feedback to amplify the small incremental warming from CO2 to a significant several degrees C. As we know from the characteristics of CO2 the bulk of the so called “damage” is done within the first 100ppm. Is water vapour any different?

  30. Iskandar says:

    And most importantly, if the backradiation phenomenon does exist, it will change the transmission windows. At maximum absorption, the atmosphere will beome translucent due to back radiation. Which, to my humble opinion, is in contradiction with the first law of thermo.

    Be free to correct me if I am wrong.

  31. kbray in california says:

    Another little thought visualization for CO2 in the atmosphere.

    Let a backyard swimming pool represent the atmosphere.
    Let the air above it represent outer space.
    A black colored ball bobbing in the water represents CO2.
    The black ball is calculated by size in relation to the pool for its appropriate ratio as per CO2 in the atmosphere.

    The sun comes up and warms the water, and the black ball is a heat sink due to its color. The heat absorbing ability of the black ball represents CO2’s greenhouse effect.

    At the end of the day, heat absorbed by the water is released to the air, representing atmospheric heat released to space. Heat absorbed by the ball is released to the water representing CO2’s heat retention/reflection in the atmosphere.

    How much is that tiny ball bobbing in the pool going to maintain the pool temperature?…. I do not think it is even measurable. The dynamics of the entire pool are too great to overcome by this little ball and reach a temperature equilibrium regardless of the little ball’s efforts.

    A black ball representing the man made CO2 would be even more insignificant.

    CO2 is treated in AGW articles like it is a nuclear heat source of magic abilities to retain/conserve heat. I don’t buy it.

    I do not believe that CO2 in tiny amounts can heat the earth as claimed. Mars is almost 100% CO2 and it’s still too cold for us to live there… where is the runaway greenhouse effect on Mars?

    A simple visualization like this displays the obvious that it is physically impossible for CO2 to have any major effect on earth’s temperatures as claimed.

  32. EdH says:

    Interesting post. Along these lines, a recommendation:

    I’m an mild AGW skeptic myself (like Willis; it’s not a religion and you could convince me otherwise with good evidence), but I’ve been following “The Science of Doom” blog recently (found via a link on Climate Audit):

    http://scienceofdoom.com

    The author has been carefully (and I’d say honestly) and at great length going through the physics.

    Worth a good look, IMHO. Currently he’s discussing the better textbooks on the physics of the atmosphere. I bought Taylor’s book a couple of months ago and have just started to go through it (starting my taxes early this year has put it mostly on hold).

    Greenhouse posts:
    “Understanding Atmospheric Radiation and the “Greenhouse” Effect”, parts 1-5 will reward the patient reader.

  33. Jay Davis says:

    Nice visual aid. I’m not a scientist, however I’ve had extensive experience operating in jungle and desert environments. My direct observations have been that in the jungle, where humidity is very high, it stays hot when the sun goes down. Almost the whole night through. In the desert, where humidity is very low, it cools down rapidly when the sun goes down. At altitude it can get damn cold. Now I’m making a WAG here, but I assume that the atmospheric CO2 level is the same in both environments. Therefore, it is the humidity, atmospheric water vapor, that is the moderating element and CO2 does nothing. So in order for CO2 to contribute to global warming, it requires the sun. And I am given to understand that radiation from the sun only acts on water vapor and CO2. The other atmospheric gases, mostly nitrogen and oxygen, do not do anything, they absorb no radiation of any wavelength nor reflect ant radiation of any wavelength. Not being a scientist, for the sake of argument, I’ll accept that. But what I have a very hard time accepting is how a trace gas, CO2, can have any effect on anything.

  34. wayne says:

    Ira, I’m going to give you a big slap on the back. Well done!

    You are the first scientist posting here that has portrayed the whole process (well ignoring some minor details) in such clarity and showing the whole atmosphere, all atoms and molecules, radiating as I always thought was correct. Bravo!

    Now all you have to do is reassure me one last time that you are sure all matter radiates, for that has been my one struggle for nearly 15 months. I sometimes feel I have been led, lied to and swindled out of some of my forty year understanding of physics, constantly being reminded, no told, from all sides that O2, N2 & argon do not and cannot radiate, even when combined as an atmosphere. (never really bought that)

    Now the only thing I don’t seem to know is the proper division of rates of overall radiation from the bulk atmosphere and how much might actually reverse thermalize back to the GHG gases with most molecules just thermalizing again and a small fraction actually radiating. But that is just a minor detail. Thank you again Ira (and davidmhoffer for listening).

  35. richard verney says:

    Solar input (in the sense of the amount of energy being received at the top of the atmosphere from the sun) may be costant although there may yet be processes that we have yet to understand.

    However, the mere fact that the energy from the sun is constant does not mean that the amount of energy received at ground level has remained constant. There only needs to be slight variations in the amount of clouds (their area, shape, reflectivity) to result in changes to the amount of energy received by the oceans and the land. A change in the pattern of the clouds of just 1 or 2% could account for all the observed warming. We have no real historical data of cloudiness.

    The point raised by etudiant February 28, 2011 at 2:18 pm is a good point. I would like to hear some more on this and the effect that this ought from a theoretical point of view had on warming/cooling observed at the surface..

    Measuring the temperature of the oceans is probably the best metric for assessing whether there is any warming. This is becuase the oceans account for about 70% of the surface area of the globe and they contain (ignoring the core) approx 99% of the stored global energy. Further, a change in temperature in the oceans can be converted to joules and thus the heat content is know. Land based air temperature record does not measure energy since it does not also include data for humidity/moisture content.

    i disagree that measuring ocean temperature can be used to assess sensitivity. In my opinion, measuring the heat content of the oceans cannot necessarily be used to assess sensitivity since a slight change in the amount of cloudiness has a significant impact on the amout of solar radiation being received by the oceans. Unless we accurately know the total amount of clouds and their pattern, we will not be able to rule that out as the source of any observed warming (or cooling).

  36. JAE says:

    “A real greenhouse primarily restricts heat escape by preventing convection while the “greenhouse effect” heats the Earth because “greenhouse gases” (GHG) absorb outgoing radiative energy and re-emit some of it back towards Earth.”

    Yeah, that is the theory. But NOBODY has yet been able to show that the “greenhouse effect” actually “heats the Earth,” as stated here. I.e., there is absolutely NO empirical evidence of this, only a simplistic analysis of what PART of the system (the radition) does, WITHOUT ANY CONSIDERATION OF WHAT ELSE IS GOING ON that might affect the putative heating. Does convection increase due to additional backradiation, thereby “erasing” some/all of the heat (the consistency of the lapse rate seems to suggest this is so)? Why are the temperatures in areas which have the highest amount of greenhouse gases at any given time (tropics) never much above 30 C (probably evaporation but does that counter the warming of the GHE? ) Any actual heating from the “atmospheric greenhouse effect” may vanish in the same way that the real greenhouse effect does when the windows to the greenhouse are opened. Radiation is only PART of the system, folks.

  37. James Sexton says:

    “Most Sunlight is in the visible band from 0.38μ (which we see as violet) to 0.76μ (which we see as red), which is why our eyes evolved to be sensitive in that range.”
    ======================================================
    I probably would have come away with a more complete view of your posting had you not included this little statement. I read through your post, and most of it, I agree with.

    But, towards the mentioned statement………….. That’s beautiful, you’ve managed to refute Newton without expressing how or why. But, gleaning what I can from this statement, our eyes evolved to seeing something we couldn’t otherwise see…………..because of the entire specie’s collective inherent knowledge? And thank God……errrr, uhmm…..thank Gaia…..errr…..thank evolution for giving that to us, else we’d be navigating in a manner such as bats!

  38. Katherine says:

    The portion of the Earth at about 310ºK radiates energy towards the Atmosphere at slightly shorter wavelengths than that at about 210ºK, but nearly all Earth-emitted radiation is between 5μ to 30μ, and is centered at about 10μ.

    So, when the Earth warms, it radiates less longwave radiation in the bands absorbed by CO2 and when it cools, it radiates more longwave radiation in the bands absorbed by CO2? Meaning when the Earth warms, more longwave radiation escapes, allowing the planet to cool, and when the Earth cools, more longwave radiation is reflected, allowing the planet to warm? Do I have that right?

  39. RJ says:

    Fred

    “I would guess that the vast majority of energy leaves earth by radiation”.

    It would leave the outer atmosphere by radiation. But Charles R Anderson in STSD Ch 20. In the denser, lower altitude atmosphere, most energy transfer is due to gas molecule collision etc.

    I don’t know but it seems even the experts do not agree on this.

    Most accept the GHGT (even sceptics) but the more I read on this topic the more I’m of the view that the entire GHG theory is flawed. But I’m not an expert.

  40. Myrrh says:

    I’m really at a loss to understand any of this. How on earth does Visible light and near short wave heat the Earth?

  41. kbray in california says:

    Another thought…

    Suppose I bought a container of CO2 for carbonated beverages and boosted the CO2ppm from 390 to 1000ppm in my apartment air.

    Will the winter sunshine boost my apartment temperature by say 10 degrees F due to the greenhouse effect of the CO2 ? Doubtful…

    I’d save a lot on my heating bill if that would work… if it really did work, we’d all be doing it.

  42. Brent Hargreaves says:

    I just tried jotting down, step by step, a sequence of events beginning with evaporation from the sea surface, absorption of latent heat, subsequent condensation at altitude into opaque liquid droplets and the release of selfsame latent heat.

    I soon got bogged down with considerations of cloud opacity, absorbtion and reradiation of IR from above and below by day and by….. (gasp)… and this is purely descriptive stuff; I didn’t reach the next step – attempting the numbers.

    This stuff is com-pli-cated! Could it be that this complexity frightens the bejaysus out of even professional climatographers… with the consequence that they each concentrate on (or hide away in?) some narrow aspect of climate but shy away from contemplating the mind-bogglingly big picture.

    Is the Global Warming scare is a ‘fragmentation failure’? When the house of AGW cards collapses. will they in the postmortem say, “I was just concentrating on my little patch. I thought that others were ensuring that the jigsaw pieces made up a coherent whole!”

    Here’s a ficticious entry to Wikipedia 2061: “With the benefit of hindsight we see that clouds – messy, irregular, fuzzy and inconvenient – were hiding fullsquare in the blindspot of millennium climate science. A key feature of the Earth’s thermostat, the absence of clouds from the IPCC’s analysis allowed the notion of unstable equilibrium – the ‘Tipping Point’ fallacy advanced by faded politician Al Gore – to terrify the public. When, finally, clouds were ‘discovered’ the bottom dropped out of the global warming market. The worldwide celebration of Gore Day is an enduring reminder of those dark days; a warning to future generations to be on one’s guard against end-is-nigh merchants, especially those with expensive haircuts and silver tongues.”

  43. Tim Folkerts says:

    First of all, I like the post in general — it seems to provide a fairly effective visualization.

    Fred Souder says: February 28, 2011 at 2:31 pm
    Are there any experiments which detect the incoming re-emitted radiation from the atmosphere?

    I googled images for “outgoing infrared spectrum” and found this: http://www.skepticalscience.com/images/infrared_spectrum.jpg
    It should the spectrum looking up from somewhere in the arctic on a clear day and the view looking back down from 20 km. The incoming radiation would be the re-emitted IR from the atmosphere that you wanted. It clearly shows the 7 um, 10 um and 15 um bands Ira is talking about.

  44. RJ says:

    Jay

    The other atmospheric gases, mostly nitrogen and oxygen, do not do anything, they absorb no radiation of any wavelength nor reflect ant radiation of any wavelength.

    I’m not sure about this

    I think molecules like oxygen. nitrogen and Argon are warmed by convection. They can exchange this energy by convection as they rise or by radiating IR radiation.

  45. Bill Illis says:

    It is the first time I’ve seen an explanation where the energy represented by the specific frequencies absorbed by GHG molecules are transferred to non-GHGs through collision (and then reemitted by the non-GHGs in the atmospheric windows – almost everyone believes the non-GHGs absorb and emit no IR radiation at all – some bad textbooks mis-educated everyone somewhere along the line).

    But then, to show this has a real greenhouse effect (an extra 150 watts/m2 at the surface), the numbers have to be crunched taking into account the time lag between absorption and effective emission to space.

    The 10 um atmospheric window actually emits at a higher energy level than it should according to Earth’s temperature alone.

    Second, if it was a simple process of 1 CO2 molecule absorbs a 15 um photon and then passes that on to 1 N2 molecule which then promptly emits that to space in the atmospheric window at 10 um, there be no greenhouse effect at all.

    The entire process would only take 0.001 seconds for ALL the photons to escape to space. Once the sun sets, the atmospheric temperature should fall to -220C within a second.

    But the average time the energy represented by a solar photon spends in the Earth system before it is lost to space is 43 hours. It spends time in 5 billion different molecules before it escapes to space on average.

    There is much much more going on here which I would love to see explained to my satisfaction.

  46. Dave Springer says:

    @Ira

    That wikipedia annotation “15%-30% transmitted” is a crock of sht. 100% of outgoing thermal radiation is “transmitted”. There is no other way for energy to leave the atmosphere and if it all doesn’t leave then the earth’s temperature would keep increasing until was as hot as the sun.

    What they mean to say is that 15% to 30% of thermal radiation leaving the surface escapes in a fraction of a second. The rest is delayed to some degree by absorption and re-emission but it too eventually leaves the theater by the only exit possible.

    A theater makes an excellent analogy. At the end of the movie there’s a rush for the exits. A few people make out the door without delay. Then a line forms and things slow down. People start crowding each other and if it’s too slow they start pushing and yelling which tends to speed up the line. The earth’s surface temperature rising in response to the delay caused by absorption and re-emission is like the people leaving the theater pushing and yelling to get the line moving faster.

    Well maybe not a great analogy but I never give up hope that some damn thing will make sense. You’d think this was rocket science by all the confusion and misunderstanding but in reality it’s about as complicated as attic insulation and can
    described so the proverbial bartender can understand it. People seem to just refuse to accept the simple facts of the matter and the more they think they know the bigger and more sciency the words they use in their convoluted alternate explanations.

  47. Stephan says:

    For Leif Svaalgard to ruminate upon..
    http://www.halesowenweather.co.uk/cet_sunshine.htm

  48. etudiant says:

    To DCC
    The data is here: http://www.climate4you.com/
    Click on the Greenhouse Gas section on the left and scroll down to get atmospheric humidity trends since 1948.
    Overall, that site is a fabulous resource, with lots of interesting nooks and crannies.

  49. cal says:

    Ira, I like the description overall but I do not like the Wiki graph of the outgoing infrared. It is technically correct in that it shows the likelihood of a photon of a particular wavelength being emitted by the earth and then passing through the atmosphere without being absorbed. However it might give the impression that it is the spectrum radiated into space. In practice the radiation into space is pretty much the black body spectrum you would expect for a body at about 290K with reductions at particular wavelengths particularly between 5 and 8 micron and between 14 and 18 micron as you describe in your text. Photons of these wavelengths can only be radiated into space by water and CO2 molecules at very high altitude since emissions lower down are absorbed. The lower temperatures at these altitudes acount for the reduced emissions.

    George E Smith. I do not think I agree with your last point assuming I understand it properly. You state “The only function of the GHG molecules is to heat the atmospheric gases; along with all the other mechanisms that are heating it. After that, the GHG molecules serve no function in the climate process whatsoever”

    As I have just explained CO2 and H20 are the main sources of radiation into space from the troposphere and indeed CO2 is the main source of radiation from the the stratosphere which is heated directly by the absorption of UV from the sun mainly by ozone. As I read it you seem to imply that the greenhouse molecules pass all their energy to the O2 and N2 molecules which then radiate into space. However they can’t since there is a law which says that a bad absorber cannot be a good radiator. In reality the energy is partitioned amongst all the molecules but the CO2 and H20 molecules radiate into space and cool the upper troposphere and tropopause. So the greenhouse gasses warm and they cool as well. That is the nature of thermodynamics.

    Further to this point there have been several comments to the effect that you cannot have energy passing from a cooler to a warmer body. This is not true. The rule is that you cannot have a net energy transfer in this direction but there is always energy being radiated from any body above absolute zero. So greenhouse molecules do warm the earth. It is just that they are being warmed by the earth at a greater rate. So the nett flow is upwards.

  50. rbateman says:

    I would ask what happens at night?
    My initial guess is that the 7u/10u/15u process repeats itself under a law of diminishing returns until day arrives.
    What is most interesting is that the 7u bandwidth is only hindered by H20 for outradiating.
    So, only 1/2 of the reflection back to Earth has CO2 involved.
    2nd question: What % is C02 involved in the 15u bandwave, assuming the two worst cases: Maximum humidity and minimum humidity?

  51. kuhnkat says:

    Ira,

    ” The ~7μ and ~15μ generated by the heating of the air also emerge from the top or bottom of the Atmosphere, but there are fewer of them because they keep getting absorbed and re-emitted, each time with some transfered to the central ~10μ portion of the longwave band.”

    Could you give more details as to what shifts the energy into the ~10μ window. Is this due to net out from the surface of the absorbed back radiation or are there other effects??

  52. Theo Goodwin says:

    Really good post. Really good questions in some comments. The questions take us exactly to the crucial issue: what are the physical hypotheses which explain what CO2 does to increase climate sensitivity. There will be several such physical hypotheses, of course. What seems best understood is that changes caused in clouds by the additional CO2 will have an impact on climate sensitivity. However, we also understand that there are no physical hypotheses which explain the expected or imagined change in cloud behavior. There will be some day, after the needed scientific work is done. The other needed physical hypotheses are in worse shape. We do not have so much as good hunches about the phenomena in question, unlike the case of clouds. The science is in its infancy. Any Warmista who denies this surely must be challenged to produce the relevant physical hypotheses. Short of those hypotheses, the claim that someone has scientific reasons for attributing warming to manmade CO2 are as empty as the claim that straw can be spun into gold.

  53. wayne says:

    George E. Smith says:
    February 28, 2011 at 3:23 pm

    The only function of the GHG molecules is to heat the atmospheric gases; along with all the other mechanisms that are heating it. After that, the GHG molecules serve no function in the climate process whatsoever.
    —–
    George, are you positive of including the word “only” in your statement and not “one of”. I read from those words that you don’t think thermalization can run in reverse the way the equatuion reads with symmetry. Like when the atmosphere is warmer than it should be compared to the surface and at dusk or night. I have always viewed GHGs can either heat or cool depending on the exact conditions. Any clarification?

  54. kuhnkat says:

    Bill Illis,

    ” Once the sun sets, the atmospheric temperature should fall to -220C within a second.”

    Even without GHG’s we would still have stored energy in the water and surface of the earth.

  55. RJ says:

    Bill

    almost everyone believes the non-GHGs absorb and emit no IR radiation at all

    Bad textbooks. It seems that way. Because otherwise how would say Nitrogen release energy. Some would be by convection by surely not all. Otherwise it would have to collide with water vapour or CO2. Which might not happen. And then what as it rises higher and higher.

  56. Ken Coffman says:

    I’ve been trying to think of the perfect experiment to demonstrate the effects of “back radiation”. Here’s what I came up with.

    Put a pot on the stove. Fill it with a couple of gallons of water. Put in a thermometer. Heat it up to boiling. Turn off the heat. Every 10 minutes, make a note of the temperature as the water cools back toward room temperature.

    Refill the pot to get the same volume of water. Heat to boiling. Turn off the heat. Slowly move a cookie around the room–put it anywhere you like, but not over the pot because you don’t want to restrict convection in any way (we want to measure the influence of radiation, not convection). The cookie will “back radiate” into the water. Every 10 minutes, make a note of the temperature as the water cools. Voila! The difference between the two temp series as the water cools will be your radiation effect.

    The mass of the water and the mass of the cookie are supposed to represent the relative thermal capacities of the earth’s surface and rarefied CO2.
    It might be more appropriate to use a Cheerio or a speck of dust, I don’t know.

  57. Phil's Dad says:

    Katherine 4:26
    So, when the Earth warms, it radiates less longwave radiation in the bands absorbed by CO2 and when it cools, it radiates more longwave radiation in the bands absorbed by CO2?

    I think you are right except that the longwave radiation that would subsequently move beyond the shorter end of the 10um atmospheric window would be absorbed / re-radiated by H2O instead. What we can say is that, all other things being equal, as the earth warms CO2 becomes a less significant GHG.

    James Sexton 4:20 pm
    Eyes evolved by pure chance, doubtless along with a whole load of other “useless” sensors and probably at all sorts of different sensitivity / frequency ranges. Natural selection favoured those who could see. See?

    (It is interesting to speculate on what other senses could be close to mutating into something useful)

  58. DccMartyn says:

    Ira, can you explain this to me. Approximately 6% of the energy spectrum, hard uv, is absorbed by the upper atmosphere, mostly by Ozone.
    Now this heat must be reradiated, so half must go down. I never see this heat in any of the box figures, but it can’t just disappear.
    So where does this heat go and in what form does it leave or enter the Earths atmosphere?

  59. D. J. Hawkins says:

    kbray in california says:
    February 28, 2011 at 4:07 pm
    Another little thought visualization for CO2 in the atmosphere.

    Let a backyard swimming pool represent the atmosphere.
    Let the air above it represent outer space.
    A black colored ball bobbing in the water represents CO2.
    The black ball is calculated by size in relation to the pool for its appropriate ratio as per CO2 in the atmosphere.

    I think your counter analogy is missing a few elements. During the day, the black ball absorbs “high grade” heat and tranfers it to the water by convection/conduction. At night, “low grade” heat is transferred to the ball from the water by convection/conduction and is radiated out to “space.” All things being equal, add more balls to the pool and you can heat the water to a higher temperature. In point of fact, people actually do this to help their pools warm up in early spring. You just add a pool cover to reduce re-radiation at night.

    Now, I’m not claiming that bumping the CO2 from 350ppm to 1,000ppm is going to roast us alive. Far from it. But if someone could point me to a post on the logarithmic behavior of GHG’s in the atmosphere, I’d appreciate it.

  60. Jim D says:

    The main thing to correct in this explanation is that the CO2 and H2O will not emit in wavelengths that they don’t absorb, so none of the 10 micron photons reaching space will have been emitted by the CO2 or H2O gas. Clouds can emit at all wavelengths, but not these gases. I believe Miskolczi has a similar misconception that the window region will somehow show increasing GHGs, when it can’t because they don’t emit there.
    The other quibble is related to the idea of these photons heating the atmosphere. In fact, the troposphere is mostly heated by convection from the surface with radiation having a net cooling effect, so the atmosphere as a whole emits more photons to space than it gains from the ground.

  61. KevinK says:

    Ira, with respect, a comment;

    You wrote (in part);

    Regarding the incoming (mostly) visible radiation;

    “On the left side:

    ….. (portions omitted for brevity)

    The remainder is absorbed by the Surface of the Earth, warming it.”

    Regarding the outgoing IR radiation;

    “On the right side:

    (1) The warmed Earth emits longwave radiation towards the Atmosphere.”

    Both of these are entirely correct. But you are missing one small detail, if I might be so bold as to supply a minor edit;

    “On the right side:

    (1) The warmed Earth emits longwave radiation towards the Atmosphere.” This subsequently causes the Earth to COOL by an amount EXACTLY equal to the energy contained in that packet of emitted IR radiation.

    And then of course the reemitted radiation from the gas may return to the Earth thereby warming it again. But this warming must be less than the cooling when the IR radiation left. Thus, no “net energy gain” and no ”higher equilibrium temperature”.

    You cannot simultaneously retain energy and also emit it. You can accumulate energy and emit it later thereby de-accumulating it.

    I know it is hard to believe that 30 years of very hard work by climate scientists could have such a major flaw. But sometimes folks get so far into the details of the work that the forest is no longer visible since they have been staring at the tree bark.

    Nice graphics.

    Cheers, Kevin.

  62. HankHenry says:

    If I understand rightly, while to our eye air is transparent, in the longwave earth’s atmosphere is a very murky and foggy kind of thing. I also believe that the greenhouse effect is due to the stairstep kind of transit that longwave radiation coming up from the earth has to go through to reach the surface from which it can be emitted to space. What I have yet to see is how the calculation is done that figures the transit time. To me this would be an important calculation because the longer the transit the greater the heating of the atmosphere should be. Is this something that is dealt with in the field of statistical physics? It would also be interesting to understand in microscopic detail what happens when shortwave radiation hits a solid surface – or for that matter a liquid surface since the earth is 71 % sea.

  63. Dave Springer says:

    Bill Illis says:
    February 28, 2011 at 4:40 pm

    It is the first time I’ve seen an explanation where the energy represented by the specific frequencies absorbed by GHG molecules are transferred to non-GHGs through collision (and then reemitted by the non-GHGs in the atmospheric windows – almost everyone believes the non-GHGs absorb and emit no IR radiation at all – some bad textbooks mis-educated everyone somewhere along the line).

    But then, to show this has a real greenhouse effect (an extra 150 watts/m2 at the surface), the numbers have to be crunched taking into account the time lag between absorption and effective emission to space.

    The 10 um atmospheric window actually emits at a higher energy level than it should according to Earth’s temperature alone.

    Second, if it was a simple process of 1 CO2 molecule absorbs a 15 um photon and then passes that on to 1 N2 molecule which then promptly emits that to space in the atmospheric window at 10 um, there be no greenhouse effect at all.

    The entire process would only take 0.001 seconds for ALL the photons to escape to space. Once the sun sets, the atmospheric temperature should fall to -220C within a second.

    But the average time the energy represented by a solar photon spends in the Earth system before it is lost to space is 43 hours. It spends time in 5 billion different molecules before it escapes to space on average.

    There is much much more going on here which I would love to see explained to my satisfaction.

    Fortunately we don’t have to work through the theoretical quantum physics for every particle in the atmosphere. All we have to do look is look at the surface with a thermal spectrometer from a high altitude and observe the result.

    The following is what you see in the infrared spectrum looking upward and downward:

    http://www.sundogpublishing.com/fig8-2.pdf

    It’s from the 2006 textbook “A First Course in Atmospheric Radiation” by Grant Petty.

    Looking downward from 20km over the arctic ocean you can see how much energy is missing in the CO2 absorption band. In the IR window you can “see” right down to sea level which follows the curve of a 265K blackbody. At CO2 absorption frequency of 15um the curve at 265K suddenly drops down sharply to follow the 225k blackbody curve. Using the dry adiabatic lapse rate of 1 Kelvin per 100 meters this gives us an emission altitude of 4 kilometers or about 12,000 feet. At that point absorption and thermalization (transfer of energy to other atmospheric gases) of 15um upwelling radiation by CO2 ceases and it is transmitted directly out to space and follows a colder blackbody curve.

    Now, if you reduce the surface temperature of 265K by an amount that’s enough to fill in the hole at 15um (and the other smaller holes in CO2’s lesser absorption bands) to make a smooth blackbody curve at a lower temperature the downward difference is the amount of greenhouse effect by all atmospheric CO2.

    CO2 greenhouse effect isn’t linear except for the first 50ppm or so and then it turns into a log curve where additional CO2 has less effect. This is also determined empirically by taking actual measurements of absorption in a laboratory column of gas so it doesn’t rely on theory. I can look up that curve too and have posted a link to it before but don’t have it handy and don’t feel like searching for it again.

    The bottom line when the numbers are crunched on the empirical data is a 1.1C increase in surface temperature per doubling of CO2 starting from a baseline of 280ppm with about half of that already a done deal in going from 280ppm to 390ppm today. That 1.1C is the IPCC low end “sensitivity” estimate which isn’t a scary number at all and in fact is a great number because if that’s all it is then the slight warming, mostly in the winter in the higher latitudes, is a great boon to agriculture especially when the biological effect of higher CO2 on green plant growth rates and water consumption is taken into consideration. It’s actually moving the earth closer to an optimum climate where I define optimum as that climate which dominated the earth’s history and in which green plants evolved. Naturally we would expect plants to be best adapted to the environment in which they spent the most time evolving and adapting over hundreds of millions of years. Even if you don’t believe in macro-evolution just about everyone accepts micro-evolution and that’s all we’re talking about is micro-evolution.

    In order to sex up CO2 greenhouse warming to a point that might make it scary the climate boffins invented, out of whole cloth with no empircal data to back it up, an imaginary “amplification” where a little CO2 warming increases greenhouse warming by water vapor by twice as much. Following that out to its logical end, since warming is warming no matter the source, water vapor warming will cause even more water vapor warming through positive feedback and we get a runaway greenhouse. In the almost sure knowledge that the earth never experienced a runaway greenhouse even with ancient CO2 levels 10 to 20 times greater than today, these anti-science scoundrels insist with a “high level of confidence” that this amplification is real and it’s based on nothing more than faster than expected surface temperature rise in the past few decades which can be TOTALLY explained by multi-decadal cyclic behavior in ocean currents, trade winds, and/or solar magnetic activity causing small global average albedo changes. There isn’t a lick of evidence that water vapor amplification is more than a figment of fevered imaginations. But CO2 greenhouse effect alone IS real and empirically established.

  64. wayne says:

    When looking at http://www.skepticalscience.com/images/infrared_spectrum.jpg link given by Tim Folkerts above, and looking at the spectrum of the surface looking up there appears that the all-atmosphere radiation is somewhere around 5% radiance. You see that 5% especially in the window frequencies between about 8 and 12.5 µm.

    That seems to imply that most of the transfer in Ira’s animation after initial absorption is still via thermalization and re-excitation, both directions and is keeping most of the radiance in the GHG absorption/emission lines. Would you roughly agree with that?

  65. Anthony Zeeman says:

    The mass of the earth is one trillion, that’s 1 followed by 12 zeros, greater than the mass of carbon dioxide in the atmosphere. There is no way that an infinitesimal amount of gas could contain enough energy to significantly change the earth’s temperature. Any heating of the earth is by direct absorption of energy from the sun.

  66. jae says:

    Ken Coffman:

    “I’ve been trying to think of the perfect experiment to demonstrate the effects of “back radiation”. Here’s what I came up with.”

    Well, to repeat myself at least 100 times, here’s another “experiment” to consider:

    Atlanta has at least three times as much GHGs as Phoenix in July (both cities are at virtually the same elevation and latitude, so solar energy is not a factor, at least on clear days). Yet it is virtually always MUCH cooler in Atlanta than Phoenix. EVEN AT NIGHT, I keep wondering, WTF?

    Now, this can be explained “simply” by the heat lost through evaporation of moisture in Atlanta, which cools their world. BUT, the question still remains: does evaporatative cooling negate/minimize/marginalize the radiative effects of GHE? If so, who gives a damn about the putative increase in CO2???

    It is really weird that I have never got a sensible answer to this question, despite asking it for about 4 years now…

  67. Slacko says:

    Bryan says:
    February 28, 2011 at 2:52 pm

    “Your graphs are completely misleading. They show that the intensity of the Solar Radiation as being equal to the Earth surface IR upward radiation.”

    The curve seems to depict atmospheric transmissivity rather than quantitative energy transfer, despite its label.

    “Further the colder atmosphere cannot heat or “warm” the warmer Earth Surface.”

    The colder atmosphere still radiates IR back down, contributing to the total inflow. It’s the net transfer that should be considered.

    “All it can do is to slow to some extent the heat loss from the surface.”

    Yes, that’s how it works. But the warmists would rather sell you a tipping point with catastrophic thermal runaway, when in fact CO2 is already so close to its maximum effectiveness (critical mass if you like) that increasing it will have negligible further effect.

  68. bubbagyro says:

    This is what theoreticians are really good at. I know, I have had to deal with them, as a pragmatic, applied scientists for years. Once they almost talked me out of one of my best inventions, that now saves lives worldwide. Not boasting, just saying. We found out much later, after I put my career on the line, what the key neglected variable was.

    What modelers are good at is simplification, actually oversimplification. The goal is to eliminate as many variables that are unknown, or that compete with the model. We have all seen this in action. Once these are accepted, then the rest plays out like clockwork.

    I found, at a cursory glance, a few major omissions and simplifications. By ignoring these, the model simplifies, and seems reasonable. To keep this short, I will mention the first. It is a type of misdirection, as a magician would use. Theoreticians are like magicians—they must oversimplify in order to describe a chaotic and heterogeneous system.

    Here goes. Did anyone pick up that the “average” incoming radiation is “centered around” 1/2 µ? That is true sometimes, but remember that our sun is a variable M star. What that means is that it has bursts of very short wavelength (very high energy) radiation, like X-rays. Other times, when it gets quiet, the output is switched to lower energy radiation. Including infrared, up to 15000+ nm.

    The sun puts out some X-rays—it also generates some infrared.

    Defining the solar output range to be between 100 nm and 3000 nm is an arbitrary range, based on averages (as Gaussian distributions). There is an infrared component of incoming solar radiation that is not only reflected, but also absorbed by CO2 and other molecules, and then reradiated as lower energy radiation. The higher in the atmosphere this happens, the more is lost back to space.

    CO2 absorbs light primarily in the IR wavelength, but it is not totally transparent to other wavelengths. So the model has another assumption.

    Other energy from the sun, such as plasmas, and particles are very important. Especially protons. Protons can protonate CO2, and other molecules to form formic acid and other species.

    Etudiant mentions water. Water vapor also reacts with CO2 to form carbonic acid, that can go on to do other chemistry, besides being the king at absorbing almost all wavelengths of incoming and outgoing radiation. Water is the boss, not the secondary actor.

    CO2 is generated on the earth’s surface. It becomes more dilute as it diffuses upward. The models, similar to Ira’s, assume it is a narrow band. This is not true. The very dilute CO2 and water even in the stratosphere absorb radiation. The higher that occurs, the more is lost to space. Remember that Mars has 970,000 ppm of CO2 in its atmosphere. But there is not much atmosphere, so there are fewer molecules to absorb. If we halved the CO2 concentration on Mars, it would have a minuscule effect.

    BTW, most organisms on the planet are adapted to “seeing” in the ultraviolet. Humans are pretty rare that they cannot. Many animals “see” also in the infrared, so Ira’s argument in that regard has little impact with me. His purpose, it seems, was to for us to focus on the visible range of light to the exclusion of the other incoming wavelengths, in order to help us apprehend his model.

    The relationship between the earth and sun is involves the relationship of two variable, chaotic and heterogeneous systems. I like to think of the atmosphere like the Jovian or Saturnian atmospheres, where we see banding and concentration of clouds and compounds, flowing countercurrent to one another. Our atmosphere is invisible to us. The situation with ozone teaches us that gases can be more or less concentrated in different areas of the earth, varying in response to unidentified variables, in a fluid, dynamic interaction.

    Notwithstanding these caveats to the models assumptions, Ira does a great job describing the consensus model of the greenhouse effect.

  69. bubbagyro says:

    Oh, no!
    I did it again, misspelled my /unbolding. It was only meant for one word.
    Sorry.

  70. James Sexton says:

    Phil’s Dad says:
    February 28, 2011 at 5:32 pm

    James Sexton 4:20 pm
    Eyes evolved by pure chance, doubtless along with a whole load of other “useless” sensors and probably at all sorts of different sensitivity / frequency ranges. Natural selection favoured those who could see. See?

    (It is interesting to speculate on what other senses could be close to mutating into something useful)
    ======================================================

    lol, It’s all too clear now! Now I can see! Evolution means something was whispering in our soon to be ears that there was something to see out there! Now I’m coming to an understanding! That d.a. Newton……what could he ever logic?

  71. Tim Folkerts says:

    I must disagree with the comments to the effect that N2 & O2 radiate (or absorb) any important amount of IR.

    Look at a somewhat similar and more familiar idea – visible light emitted by a Hydrogen atom (the Bohr model covered in freshman chem & physics around the world). It is easy to observe that H only emits very specific energies of light which are “easily” predicted by the quantum mechanics of the orbiting electrons and the allowed transitions between energy levels of those orbits. Photons of other energies are simply not emitted by H atoms. And photons of other energies are simply not absorbed either. Different atoms and molecules will absorb and emit at their own characteristic visible frequencies due to their own electron orbital energies. If you know the energies of the allowed orbits, you can find the energies of photons that can be absorbed or emitted by other gases.

    For thermal IR (above a few micrometers), the quantum mechanics is different. This is not taught in freshman chemistry. Here the energies are not related to electrons jumping to other orbits (that takes too much energy), but instead it is related to rotations and vibrations of the molecules (which can happen at the energies associated with IR photons). However, it is an observed fact (supported by theory) that monatomic gases (like argon) or symmetric diatomic gases (like N2 and O2) do not have vibration modes or rotation modes that would allow them to absorb (or emit) IR photons. Just like yellow light is not absorbed or emitted by H atoms, IR is not absorbed or emitted by N2 & O2. (There are some unimportant exceptions – for example O2 or N2 with different isotopes can absorb IR weakly for example. But it is my understanding that this is a very small effect, even given that N2 is MUCH more common in the atmosphere.)

  72. old engineer says:

    Thanks for a very clear explanation of the “greenhouse” radiative effect. As someone who is just beginning to learn about the radiative modeling of the atmosphere, it is greatly appreciated. I’m sure that to many of the WUWT regulars this is very elementary, but to those of us who are just learning, the graphics made it very understandable. It goes in my hardcopy file for sure.

    One question. I never see Rayleigh or Mie scattering mentioned, although it is my understanding that, in the atmosphere these equations govern the absorpsion and scattering of electomagnetic radiation from the sun, and the “long wave” radiation of the earth back to space. I am only just aware of Mie theory, having run across it in a project to measure jet engine exhaust particle size.

    Can’t these eqations be used to settle some of the arguements of just how much electromagnetic radiation is absorbed and scattered?

  73. Max Hugoson says:

    PS: Because of Robert Woods 1909 experiment with two miniature “greenhouses”, one with a “rock salt” (i.e., transparent to the longwave IR material) window and the other with a glass window, showing NO MEASURABLE DIFFERENCE in the final equilibrium temperatures in both boxes, we’ve known SINCE 1909 that to ascribe the warming of greenhouses to the allegation of a “one way valve” due to the regular sodium silica glass, is in error.

    Real “Meteorological” textbooks for many years have noted this and used the prefered term: “Atmospheric Effect”.

    I think the WUWT crowd needs to realize this fact and steadfastly refuse to use the terms “Greenhouse gases” or “Greenhouse Effect”.

    Truth needs to win!

    Max

  74. Dave Springer says:

    HankHenry says:
    February 28, 2011 at 6:15 pm

    “If I understand rightly, while to our eye air is transparent, in the longwave earth’s atmosphere is a very murky and foggy kind of thing.”

    Yes.

    “I also believe that the greenhouse effect is due to the stairstep kind of transit that longwave radiation coming up from the earth has to go through to reach the surface from which it can be emitted to space.”

    Yes,

    “What I have yet to see is how the calculation is done that figures the transit time.”

    Don’t need to do that. We lofted infrared spectrometers with high altitude balloons in clear dry air over the arctic ocean and measured the energy level across the spectrum looking downward.

    “To me this would be an important calculation because the longer the transit the greater the heating of the atmosphere should be.”

    Yes.

    “Is this something that is dealt with in the field of statistical physics?”

    Classical (statistical) mechanics and quantum mechanics both. The scale is classical since the wavelength of thermal IR spans billions of molecules at once in the troposphere where all the action takes place. Quantum explanations get truly bizarre. But again there’s no need to use theory when you have empirical observations. It’s like trying to calculate the acceleration of gravity by theory and just measuring how fast a lead ball falls by dropping it from the Leaning Tower of Pisa.

    “It would also be interesting to understand in microscopic detail what happens when shortwave radiation hits a solid surface – or for that matter a liquid surface since the earth is 71 % sea.”

    Interesting yes but again it’s an easy empirical measurement. The ocean absorbs practically every bit of sunlight that reaches the surface and penetrates to a depth of about 300 meters give or take depending on turbidity. At low angles it begins to reflect a significant portion of incident light but low angles only occur when the sunlight is weak to begin with (near dawn and dusk and at very high latitudes).

    Clouds however also cover an average of around 70% of the earth’s surface and they can reflect anywhere between 15% to 85% of the sun’s light directly back out into space. Average albedo of the earth isn’t known to an acceptable margin of error with estimates ranging a few percent above and below 35%. Every percentage point difference is a few degrees warmer or colder surface temperature if the difference is lasting across a number of years. With a marginal exception for seasonal NH/SH difference due to snow cover it’s assumed to be a constant value in the models and different models just pick a number that works out best for fitting hindcast global average temperature with measured past global average temperature. In other words albedo is not just a fudge factor in the climate models but rather a damn big fudge factor that dwarfs any change that anthropogenic CO2 can possibly bring about.

    This is why the hypothesis that galactic cosmic ray intensity changes result in more or fewer clouds and changing albedo. GCRs are modulated by both solar magnetic field, which is largely unpredictable in strength except for generalities associated with 11-year sunspot cycle and is also modulated by unpredictable events like nearby supernovas, and by more predictable very very long slow changes in intensity due to the solar system traversing spiral arms of our galaxy and wandering above and below the galactic plane in cycles lasting tens and hundreds of millions of years.

    Quite frankly it really appears GCR intensity is The Big Kahuna when it comes to global average temperature variations. The latter half of the twentieth century saw the most intense solar magnetic field (by sunspot number proxy) in the past 400 years since sunspot records began. This would have resulted in fewer GCRs, fewer clouds, lower average albedo, and surface warming from the lower albedo.

  75. Dave Springer says:

    Oops. Saw a mistake in my last comment. Sunlight penetrates the ocean to roughly 300 feet not 300 meters. That would be 100 meters (give or take depending on turbidity) before it gets inky black. It sucks to be one of those strange people (Americans and Brits) who learnt English units of measure in our youths and has had the rest of the world pressuring us into going metric. I try. Sort of.

  76. kbray in california says:

    [[[ D. J. Hawkins says:
    February 28, 2011 at 5:52 pm

    ... All things being equal, add more balls to the pool and you can heat the water to a higher temperature. In point of fact, people actually do this to help their pools warm up in early spring. You just add a pool cover to reduce re-radiation at night. ]]]

    If we covered the earth in a pool cover, the heat increase would kill us in short order.
    That would create an environment similar to a real greenhouse which becomes scorching in the summer sun and use water spray, windows, and fans to cool down. They can be deadly for plants and people trapped in them in the summer if they are not managed properly. A pool cover is not an accurate representation of CO2.

    My point with the black balls is that the ratio to the volume of water is miniscule… they cannot have much effect due to the small quantity…. same as CO2 is only a trace gas in the atmosphere. The black balls should actually be drifting under the water to be a more accurate representation of CO2 floating in the atmosphere. You are creating a non-representative condition by covering the surface of the pool with black balls, (and also interfering with surface cooling by radiation/evaporation) and…
    1) That is also creating an overrepresentation of CO2. and..
    2) CO2 does not float at the top of the atmosphere, but is homogenized in.
    I suggest again that there are not enough black balls floating around in the pool to make any measurable change in water (atmospheric) temperature.

    I realize this is not a perfect analogy. CO2 purports to reflect certain wavelengths back to the planet. I visualize CO2 in the atmosphere like a screen door on a submarine…. Some of the energy waves will get reflected back, but there are so many holes in the CO2 cover that after a couple bounces, all will make it out into space. Net gain…. not much. Screen doors on submarines don’t keep the water out. CO2 is a screen door… too many holes in the coverage to do much benefit.

  77. Brian H says:

    Brief note to Myrrh;
    All incoming radiation that reaches the surface heats the Earth. It is absorbed by one form of matter or another, which causes kinetic agitation — i.e., heat. It can’t be re-radiated at those same wavelengths because that would require being as hot as the source, the Sun. It gets emitted at the wavelengths matching the temperature of the matter/material emitting. That’s in the IR band for the temps we experience.

    In deep space, much material is much colder, and emits at even lower wavelengths, down in the radio bands. Other higher frequency radiation also comes in on those bands after being “red-shift” stretched by the relative recession of the sources from our POV. Such as the putative “Big Bang” cosmic background patterns.

  78. Theo Goodwin says:

    bubbagyro writes:

    “CO2 is generated on the earth’s surface. It becomes more dilute as it diffuses upward. The models, similar to Ira’s, assume it is a narrow band. This is not true. The very dilute CO2 and water even in the stratosphere absorb radiation. The higher that occurs, the more is lost to space.”

    Please expand on this. My understanding is that the modelers assume that concentrations of CO2 are the same wherever they occur in the atmosphere – all the way up. In addition, they assume that the behavior of CO2 molecules regarding radiation is the same throughout the atmosphere. It has always seemed to me that these “uniformity” assumptions were just evidence of an unwillingness to do the necessary empirical research. In plain and simple terms, I would like someone to address the linked questions of “where are the CO2 concentrations and how does radiative behavior change depending on where they are?”

  79. Brian H says:

    Correction/refinement of above: “All incoming radiation that reaches the surface and is not reflected …”

  80. James Sexton says:

    bubbagyro says:
    February 28, 2011 at 7:11 pm

    Oh, no!
    I did it again, misspelled my /unbolding. It was only meant for one word.
    Sorry.
    =======================================================
    They are some really nice guys! They fixed it for you!

  81. Dave Springer says:

    “I plan to do a subsequent posting that looks into the violet and blue boxes in the above graphic and provides insight into the process the photons and molecules go through.”

    Please don’t go into quantum electrodynamics. The scales involved don’t warrant it and I guarantee you no good will result from it unless further confusion and unending arguments even between the PhD physicists here is your goal.

  82. KevinK says:

    Oh, one other observation for Ira, by continuously mentioning Albert Einstein’s name in all of your posts you are in fact doing yourself a disservice. You may think that it increases your credibility and more people will automatically accept your arguments since you “drop” Al’s name, but it only works for a small portion of the audience.

    I for one evaluate your posts based solely on their merit. Just because you make nice graphics in the “spirit of Albert Einstein” does not make me give your arguments any more weight. If my garbage collector makes a good argument I consider it with just as much logical analysis as I do yours.

    Just an observation, you might want to go heavy on the LOGIC and light on the “Al Einstein agrees with me” thing.

    Cheers, Kevin.

  83. Dave Springer says:

    You know Ira, there’s a quantum explanation for why you squint your eyes in bright lights. There’s also no good reason to ever try to explain it in quantum terms when a macroscopic explanation is far simpler and completely adequate for any practical purpose. The same goes for the greenhouse effect. Quench that desire. Just say no.

  84. Chad Woodburn says:

    When the up-going long-wave radiation gets reflected back to the earth, how long does it take for it to “bounce off” the earth and return to the level where it was previously reflected back to the earth? (Or does it not work that way?)

    How long does it take for the long-wave radiation to completely dissipate during the night? (If it does not, how much of it is still there when “new” long-wave radiation joins it after sunrise?)

    How long does it take after sunrise for the GHG in the upper atmosphere to again get saturated with long-wave radiation?

    If the long-wave inventory returns to negligible levels by morning, what difference would that make to average global temperatures in contrast to the inventory staying relatively high (or static)?

  85. D. J. Hawkins says:

    kbray in california says:
    February 28, 2011 at 7:47 pm

    …I realize this is not a perfect analogy. CO2 purports to reflect certain wavelengths back to the planet. I visualize CO2 in the atmosphere like a screen door on a submarine…. Some of the energy waves will get reflected back, but there are so many holes in the CO2 cover that after a couple bounces, all will make it out into space. Net gain…. not much. Screen doors on submarines don’t keep the water out. CO2 is a screen door… too many holes in the coverage to do much benefit.

    This was sort of what I was alluding to regarding the behavior of CO2 in the atmosphere. Your black ball can’t be based on the proportion (molar volume) of CO2 in the atmosphere; it has to be proportional to the effect of the CO2 in the atmosphere. You say it’s a “couple bounces.” I say…does anyone know for sure? Or have a defensible WAG?

  86. bubbagyro says:

    Theo Goodwin says:
    February 28, 2011 at 7:51 pm

    I did not want to elaborate, because it is an onion skin, that keeps revealing more layers of variables. More variables than explanatory equations. For example, CO2 is never free. When water vapor is present, depending upon humidity, it is associated reversibly into carbonic acid: CO2 + H2O ⇌ H2CO3. This is a highly radiation absorptive species, probably across most of the bands. The higher the humidity, the more the CO2 is associated. This will absorb incoming radiation at most UV, visible, and IR wavelengths.

    Then there is quenching. (Thanks for mentioning, David). If a molecule is excited by a wavelength, then it can reradiate at higher wavelength (always higher, because lower energy—2nd Law). OR, it has another choice. Another species can “steal” the energy of the π to π* transition state, and excite itself (I’m getting excited). OR, it can do chemistry.

    If a CO2 molecule is excited by IR, let’s say from below, then it can reradiate a higher wavelength IR (direction of that photon is up, down, or sideways, on average). OR, it can have the energy stolen (quenched) by SO2, or CFCs, or chloride, or water, or ozone, or carbon particulates, or dust, or…

    The trouble with the models is that they do not test all of the available permutations of all the species present at all locations and levels of the atmosphere.

    Hence, they simplify, and show us the absorption spectrum of pure CO2 in a vial under ideal conditions.

    Then their is the horrendous problem of diffusion and diffusivity of gases.

    Then we have the states of matter, especially water. Water associates in complex ways, forming structures in the low atmosphere. It is most dense when it is at 4°C, yet expands pronouncedly when it is cooled to 0*C.

    Many variables, few equations. Chaos for models.

  87. etudiant says:
    February 28, 2011 at 2:18 pm
    Relative humidity in the upper atmosphere (300mb level) has fallen from around 55% in 1950 to about 45% now.
    Should this not have a material impact on the greenhouse effect? …

    What is the source of that data and the accuracy of measurement in 1950? Assuming the change is actual, I would think it would impact the “greenhouse effect”, but I do not know which way because relative humidity considers both the temperature and water content. Are there any more expert than me out there with an answer?

  88. kuhnkat says:

    DccMartyn,

    the ozone is broken down by the radiation it absorbs. I am told there is no energy radiated from this interaction.

  89. kuhnkat says:

    Dave Springer,

    You use a common graph of the Arctic. I was wondering why you would use a graph of the second dryest place on earth, the dryest being the Antarctic. These would seem to show a much larger absorption by CO2 when water vapor is the largest effect in most other areas on the surface especially since the poles have a lot less energy to radiate.

  90. rbateman says:

    Dave Springer says:
    February 28, 2011 at 7:36 pm

    I strongly suspect that the GCR’s are a variable, and that variable is modulated by the level of Solar Activity.
    To be very blunt, the level of incoming GCR’s would vary even if the Sun were to remain constant in activity.
    You could have low GCR levels coupled with low Solar Activity or high GCR levels with high Solar Activity, and both cases would produce much the same result.

  91. James Sexton says:

    ” Whence arises this uniformity in all their outward shapes but from the counsel & contrivance of an Author? Whence is it that the eyes of all sorts of living creatures are transparent to the very bottom & the only transparent members in the body, having on the outside an hard transparent skin, & within transparent juyces with a crystalline Lens in the middle & a pupil before the Lens all of them so truly shaped & fitted for vision, that no Artist can mend them? Did blind chance know that there was light & what was its refraction & fit the eys of all creatures after the most curious manner to make use of it? These & such like considerations always have & ever will prevail with man kind to believe that there is a being who made all things & has all things in his power & who is therfore to be feared.” —- Isaac Newton…….(emphasis mine)

  92. Fred Souder says:
    February 28, 2011 at 2:31 pm
    Ira,
    Are there any experiments which detect the incoming re-emitted radiation from the atmosphere? Did a quick search and only came up with models. It seems like we should be able to detect the radiation coming back at us from the atmosphere if this model is valid.

    It is basic Science 101 that any mass with a temperaure above absolute zero emits radiation in all directions at wavelengths that peak according to their “blackbody” temperature. The air and clouds in the Atmosphere are over 200ºK, which is well above absolute zero (0ºK), so they emit radiation accordingly.

    I also did a search and did not find any specific mention of test results, but I am sure they exist and I hope some WUWT reader will provide a link.

    This NASA site (not GISS :^) http://earthobservatory.nasa.gov/Features/Clouds/, is about clouds and says, in part: “When a cloud absorbs longwave radiation emitted by the Earth’s surface, the cloud reemits a portion of the energy to outer space and a portion back toward the surface. ”

    Why believe this site? Well, because they get it that clouds are a net NEGATIVE feedback. That is, if warming, say due to increased CO2, produces more evaporation and therefore more water vapor and thus more clouds, that will feed back and reduce the amount of warming somewhat. Here is what they say: “The balance between the cooling and warming actions of clouds is very close although, overall, averaging the effects of all the clouds around the globe, cooling predominates.”

    So, this site confirms re-emission from clouds to Earth at least.

  93. Kevin says:
    February 28, 2011 at 2:52 pm
    From the looks of that first chart, ~100% of 7μ and 15μ radiation gets absorbed by the atmosphere on the way back to space, mainly by H20 and CO2. Maybe a tiny trickle bounces its way from molecule to molecule and makes it to space.

    If that’s so, then doubling or even tripling the values of H2O in our atmosphere should have no effect on absorbtion. You can’t absorb more than 100% of the radiation, right?

    As you know, the “temperature” of a gas, such as the Atmosphere, is an abstraction of the average speed of the N2, O2, O3, H2O, CO2 and other molecules that make up that gas. When GHGs absorb a photon, they do not hold onto it forever. Rather, in their energized state, they may re-emit it or they may move faster and, by colliding with non-GHGs (such as N2) they will raise the temperature of the gas. Any mass, including a mass of gas, that has a temperature above absolute zero, will emit longwave radiation in random directions and at a variety of wavelengths, peaking at the wavelength corrresponding to the temperature of that gas, as indicated in the first graphic.

    Thus, once the temperature of any portion of the Atmosphere stabilizes, ALL the photons that are absorbed from the Earth are re-emitted in all directions and a variety of wavelengths (~7μ, ~10μ ~and 15μ).

    The ~7μ and ~15μ photons rattle around between GHG molecules until some emerge at the top of the Atmosphere and go off into Space, and others strike the Earth and are absorbed, further warming the Earth. The ~10μ photons get an almost free pass and zip through the GHGs and off to Space or, if they happen to be emitted downward, strike the Earth and get absorbed, further warming the Earth until it stabilizes in temperature, re-emitting the same number of photons as it receives.

    If the density of CO2 molecules is increased, the ~15μ photons will get absorbed and re-emitted after moving a shorter distance, on average, through the Atmosphere. That results in a bit more being re-emitted towards Earth and the others spending more time in the Atmosphere which raise the temperature of both the Earth and the Atmosphere accordingly, until they too stabilize, but at higher average temperatures.

    The key question is about the sensitivity to a doubling of CO2, not to the fact that it is a GHG that, all else being equal, tends to raise the mean Earth temperatures. Is the CO2 sensitivity high (2ºC to 5ºC) as the official climate Team claims, or only (0.25ºC to 1ºC) as some skeptics claim.

    A second key question is the logarithmic parameter of successive doubling. Everyone agrees that subsequent doublings will have less and less effect, in a logarithmic manner, but the specific reduction seem to be a matter of controversy.

  94. wayne says:
    February 28, 2011 at 2:58 pm
    Ira, first, I think I will mostly agree with your entire post and that’s a great animated visual, but a question; where in the heck did 5525K for the sun’s temperature come from? Is that a hypothetical effective temperature as if all albedo and reflection has already been removed and falling on a flat Earth? (sounds like more Wikipedia’s hidden ‘clarity’)

    Thanks for your kind comment. I have seen that value, in various versions (5,500ºK, 6,000ºK, 9,500ºF, 10,000ºF, …) so I am sure it is approximately correct. The source http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png seems to say they assume the Sun is a “blackbody” (which, while not absolutely true is close enough for government work) and they run the numbers to get the corresponding “blackbody” temperature, which is the red curve. The solid red area is what actually gets through the Atmosphere to the surface, which includes cloud-tops and the white roof of Energy Secretary Chu’s home.

  95. Ron Manley says:

    A useful post and an important topic not often given an airing.

    For an alternative view of the long wave effect have a look at:
    http://www.climatedata.info/Forcing/Emissions/introduction.html

    With regard to the question on variability of outgoing radiation have a look at:
    http://www.climatedata.info/Forcing/Forcing/albedo.html

  96. jknapp says:
    February 28, 2011 at 3:01 pm
    Thanks Ira. Interesting post. A couple of questions, which will undoubtedly demonstrate my lack of physics knowledge. I thought that atoms emitted photons based on quanta specific to that atom. As such the H2O and CO2 molecules will absorb and emit only specific wavelengths. That would also be true of the O2 and N2 molecules. But here you are talking about “blackbody radiation” doesn’t that imply that the wavelength emitted will not be specific to the molecule but will, instead, be a function of the temperature. Is there a conflict here or is it a misunderstanding on my part?

    Assuming that my understanding will be resolved, then a further understanding on my part was that the 7u and 15u wavelengths were absorbed very quickly (in terms of distance traveled through the atmosphere. That would it seems to me to effectively “trap” the energy in a fairly narrow region with the vast amount of the energy kinda just hanging around near where it was absorbed neither going up or down very far due to radiation. That then would heat that region resulting in expansion and the physical raising of the molecules. This convective upward flow of the energy would be dominant over the downward portion of the radiative flow. Thus the energy would be lifted toward space. As it got higher more and more would escape outward and the downward radiative portion would serve mainly to keep the convection going with then very little energy actually being returned to the surface.

    Thus you mentioned convection as not being treated here but isn’t it possible that convection is a dominant not secondary driver of energy transmission out.

    Thanks for any enlightenment you can give me.

    Great questions jknapp! I am sure convection, including the rise of warmed air as well as wind and storms is the major driver of energy transmission out. But, for convection to take place, you need the warming due to radiation to preceed it.

    As for the idea that each molecule has a particular wavelength where it absorbs and emits, that is what I used to think. If you look at the source http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png for my first graphic, where I used only the top part, the lower part shows the absorption bands for H2O and CO2 and they are quite wide, spanning around 13μ to 17μ for CO2 and, for H2O, from around 5μ to 8μ plus around 15μ to 70μ. So, clearly, these molecules are not that selective. I do not know about nitrogen (N2). However, since the Sun and the Earth act pretty much like “blackbody” emitters, with a wide range peaking at a characteristic wavelength based on temperature, I now think the Atmosphere is about the same.

  97. derspatz says:

    The Late Great John Daly was writing about this kinda stuff (and of “Fraunhofer bands”) over a decade ago in his excellent book “The Greenhouse Trap – why the Greenhouse Effect will not end Life on Earth” … of which I’ve had a copy of for nearly as long.

    regarDS

  98. HankHenry says:

    Dave Springer

    Thanks for the thorough reply. It is helpful to me. Of course, I’m not completely satisfied with this answer about empirical measurements replacing theoretical ones.

    I was thinking about Einstein’s (and others) investigations of Brownian motion as something to give me a lead into better understanding of the movement of heat through the atmosphere. I am guessing that heat radiating from the surface molecules of a solid also radiate in the same way within the solid. This talk of radiation conflicts with my understanding of heat conduction being vibrational. Although I suppose the two aren’t mutually exclusive; I’ve never thought of it that way.

    From what I’ve read Big Bear observatory has data demonstrating that albedo is variable even on the scale of decades using a technique conceived of by DaVinci. … Fascinating.

    http://www.sciencedaily.com/releases/2001/04/010418072342.htm
    http://www.sciencedaily.com/releases/2004/05/040527233052.htm
    http://www.sciencedaily.com/releases/2009/04/090407105156.htm

    I am just a lay reader of science, and it is important to understand something about models to gauge whether the scientific consensus that I read about is real, or whether there is a large group of scientists that have fashioned their views and taken a stance because an important moral matter is at hand. Do you think it really possible that there is a large number of scientists that have delved into the working of these models and formed their judgments on climate change based on what they understand about the workings and the reliability of models.

  99. Philip Peake (aka PJP) says:
    February 28, 2011 at 4:02 pm
    Now this is much better than the bouncing balls :-)

    Now we are getting somewhere.

    To simplify what is going on, we have very little impediment between the Sun’s radiation and the earth. But there is a layer of absorbent material between us and space at the frequencies at which the earth re-radiates its energy.

    This absorbent layer consists mainly of water vapor (H2O), and a little CO2.
    These gases absorb the radiation (heat) and themselves become warm. Like all insulators, all they do is delay the loss of heat. No insulator stops the loss of heat, it just delays it. In this case it is delayed by the many, many absorption re-emission, absorptions of the earths radiated energy.

    The important thing about this is told by the first graph. Look at the words on the right-hand side: “Total absorption and scattering…”.

    What that means is that adding any more H2O or CO2 will have zero effect, because all the radiation from the earth is ALREADY absorbed by these gases.

    The only way that loss of heat can be reduced is to make the insulating layer thicker (increase the depth of the atmosphere). I have never heard any claim that this is what additional CO2 is doing.

    In fact, the only way CO2 can have any significant effect is by the mysterious “forcings”, supposedly pushing more H2O into the atmosphere … but see above .. the absorption by H2O is already 100%, so unless (again) there is a claim that the increased H2O is increasing the depth of the atmosphere, the means by which this insulating layer is becoming more effective escapes me.

    Thanks Philip Peake for starting me down this trail of bouncing balls with your “trucks” comment way back in Willis Eschenbach’s topic. Do you have any idea how much work you (as “PJP”) , and your co-conspirator “davidmhoffer”, have induced me to undertake? Thanks! It has been a learning pleasure and that is why I am up at 2AM Florida time answering your most recent comment, ignoring the fact I have to lead a bicycle ride this morning at 8AM and do water aerobics at 9AM – Oh, the harried life of the retired engineer, never a day off :^) Thanks also for your kind comments.

    What the 100% absorption means is that 100% of the photons in the appropiate bands are 100% likely to be absorbed by an H2O or CO2 molecule before they travel all the way through the Atmosphere. Indeed, I believe, even at the historical 270 or 280 ppm level of CO2, each photon traveled a small fraction of the height of the Atmosphere before being absorbed and re-emitted. So, with the current level of CO2 around 390 ppm, the photons, on average, will travel a shorter distance before getting absorbed and re-emitted. That means that newly emitted longwave radiation from the Earth’s surface will spend more time in the lower reaches of the Atmosphere and therefore contribute to heating the N2 and O2 and other components of the air more than before. Thus, the air nearest the surface of the Earth will, being a bit warmer, re-emit a bit more longwave radiation in all directions, including towards the surface. Hence, more CO2 = a bit more warming. Fortunately, this is a logarithmic process, so subsequent doublings have less and less effect – but they do have an effect.

  100. izen says:

    Fred Souder says:
    February 28, 2011 at 2:31 pm
    “Ira,
    Are there any experiments which detect the incoming re-emitted radiation from the atmosphere? Did a quick search and only came up with models. It seems like we should be able to detect the radiation coming back at us from the atmosphere if this model is valid.”

    Try this –

    http://www.gewex.org/bsrn.html
    Baseline Surface Radiation Network (BSRN)
    About BSRN
    Because of the important role radiation plays in the climate system, the Baseline Surface Radiation Network (BSRN) was established to provide a worldwide network to continuously measure radiative fluxes at the Earth’s surface.

    Or-

    http://ams.confex.com/ams/Annual2006/techprogram/paper_100737.htm
    Measurements of the Radiative Surface Forcing of Climate

    W.F.J. Evans, North West Research Associates, Bellevue, WA; and E. Puckrin

    The earth’s climate system is warmed by 35 C due to the emission of downward infrared radiation by greenhouse gases in the atmosphere (surface radiative forcing) or by the absorption of upward infrared radiation (radiative trapping). Increases in this emission/absorption are the driving force behind global warming.

  101. kbray in california says:

    [[[ D. J. Hawkins says:
    February 28, 2011 at 8:53 pm

    ... Your black ball can’t be based on the proportion (molar volume) of CO2 in the atmosphere; it has to be proportional to the effect of the CO2 in the atmosphere. You say it’s a “couple bounces.” I say…does anyone know for sure? Or have a defensible WAG? ]]]

    This is not my field of expertise, however, I don’t like treating CO2 molecules like they are magic “heat” beans. Some of the effect attributed to them I regard as speculative or a current theory. I am skeptical about this influence. Every CO2 molecule takes up a measurable space or volume in the airmass. I treat it as a simple molecule and go from there. There are not enough of them floating around to do much in my opinion. Giving CO2 special powers or as you say “proportional to the effect of the CO2″ is magical thinking in my book. I suggest that CO2 might bounce energy around but it eventually escapes into space. Heat has to leave Earth otherwise it would keep building up and we would eventually ignite in flames. Our current atmosphere, land, and sea seem to have an intrinsic control mechanism that swings between certain stable limits. I do not believe that CO2 is a zombie monster out to kill us. It is a minor player in “warming”. If we do accelerate a “warming” somehow, I imagine it will just start the reverse swing sooner than it would have happened because we hit the trigger limit faster than its “natural” oscillation . This is only an opinion based on my sum of life experiences so far. I am happy to learn more or be corrected if it can be proven accurate.

  102. P. van der Meer says:

    Ira Glickstein says in his article above:

    “The absorbed radiation heats the H2O and CO2 molecules and, at their higher energy states, they collide with the other molecules that make up the air, mostly nitrogen (N2), oxygen (O2), ozone (O3), and argon (A) and heat them by something like conduction. The molecules in the heated air emit radiation in random directions at all bands (~7μ, ~10μ, and ~15μ). The ~10μ photons pass, nearly unimpeded, in whatever direction they happen to be emitted “

    No, they don’t because Kirchoff’s law of thermal radiation states that emissivity and absorptivity must be equal to each other. If the surface can emit unhindered in the 10μ band without absorption in the atmosphere than that means that absorptivity of the atmosphere is zero and therefore emissivity must also be zero. The conclusion of this is that the atmosphere cannot possibly emit in the 10μ band.

    And using the argument further, the atmosphere can only loose heat by CO2 and water vapour emitting radiation in their respective bands.

    The model as presented by Ira does not reflect reality!

  103. Richard111 says:

    May I suggest some consideration be given to the Maxell-Boltzmann kinetic energy distribution curves on which subject Tom Vonk made an excellent post here on WUWT a while back.

    http://ibchem.com/IB/ibnotes/full/sta_htm/Maxwell_Boltzmann.htm

    http://dwb.unl.edu/teacher/nsf/c09/c09links/www.kobold.demon.co.uk/kinetics/maxboltz.htm

    To me these effects imply that most long wave radiation absorbed by GHGs in the atmosphere is thermalised and “back radiation” is not related to the MASS of the GHGs in the atmosphere. A simple example is compare average surface temperatures at or near the equator to the average surface temperatures in the desert regions north or south of the equator. True there is more cloud at the equator because of the higher humidity but why doesn’t the temperature rocket up when the sky is clear? I have lived for some years in Singapore and never experienced temperatures there that I experienced in the Namib desert. Yet which region has the most “greenhouse” gases?

  104. Al Tekhasski says:

    Ira Glickstein, PhD said:
    “The ~10μ photons generated by the heating of the air emerge from the top of the Atmosphere”

    I am sorry, but for the sake of accuracy, CO2 in air does not have any states that interact with 10-13um band. Therefore, hot air cannot possibly generate any photons in that range. Therefore the following statement is also wrong, nothing gets “transferred” into 10-um band from “heated air”:

    “The ~7μ and ~15μ generated by the heating of the air also emerge from the top or bottom of the Atmosphere, but there are fewer of them because they keep getting absorbed and re-emitted, each time with some transfered to the central ~10μ portion of the longwave band.”

    As result, the entire explanation of greenhouse effect is wrong. The emission from ~7um and ~15um bands is smaller than at the surface because the overall TEMPERATURE of air is smaller at that height (of emission to space). The temperature gets lower with height because of purely mechanical property of convectively-stirred turbulent atmosphere in the field of gravity, when it HAS to form the vertical temperature gradient called “lapse rate”. No matter what kind of radiation gets diffused (and locally emitted-absorbed) through the air, this mechanics overrides (complements) the process and forms the steady (on average) gradient no matter what. The 10-13um photons just shine through without much interaction with air, all straight from the Earth surface.

    In the animated picture the pink and dark blue areas should not have any 10um components.

  105. John of Kent says:

    “A real greenhouse primarily restricts heat escape by preventing convection while the “greenhouse effect” heats the Earth because “greenhouse gases” (GHG) absorb outgoing radiative energy and re-emit some of it back towards Earth.”

    Has WUWT become a “lukewarmer” site and abandoned Climate Realism?

    This is pure nonsense and I’m surprised that WUWT would print such rubbish. Infra Red Interacting Gasses (IRIGs) CANNOT heat the earths surface. To do so would mean the colder air is warming the hot ground and that goes against the laws of thermodynamics.

    Clearly impossible! (See slaying the Sky Dragon ).

    The “greenhouse effect” on earth actually does work much the same as an actual greenhouse. It is due to the relative slowness of convection that keeps the earth surface from instantly freezing at night.

  106. izen says:

    @-Ira Glickstein, PhD says:
    February 28, 2011 at 10:44 pm
    “However, since the Sun and the Earth act pretty much like “blackbody” emitters, with a wide range peaking at a characteristic wavelength based on temperature, I now think the Atmosphere is about the same.”

    That assumption may not be warrented.
    The atmosphere may not be a ‘blackbody’ radiator, given Kirchoff’s law the emissivity is equal to its absorptivity(?), so the emission spectrum of the atmosphere should mirror its absorption spectrum which as you have so ably described is VERY different from a ‘blackbody’.

    The emissivity of the atmosphere is going to be prefferentialy concerntrated in the H2O and CO2 bands because these are several orders of magnitude ‘better’ at absorbing and emitting at these wavelengths than the N2 AND O2 are at any wavelength.

  107. cal says:

    Ira, a couple more comments.

    The temperature of the sun is in fact an artifact of its radiance. The sun’s actual temperature ranges from several millions of degrees at its core to about 3 million degrees in the very diffuse corona above the “surface”. The photosphere or “surface” is sandwiched between these extremes and behaves as if it were a black body at 5840K. This can be determined approximately by its total power output and peak wavelength and accurately by its absorption lines which are temperature dependent. This does not contradict anything you said but may reassure those who look at the very low temperature and think “how can that be true?”.

    Secondly I have re-read your paper and realised that there is something I do not believe is correct. It relates to the point I made in my earlier comment to George E Smith and was restated rather more strongly by Tim Folkerts when he said “I must disagree with the comments to the effect that N2 & O2 radiate (or absorb) any important amount of IR”.
    You talk about the “7 and 15 micron photons” being absorbed and then some of this energy being passed to other gas molecules and being re-radiated as “10 micron photons” with some of these being lost to space. This implies that there are gas molecules around that can radiate at this wavelength. However if there were then there would not be an atmospheric window. The very reason that this window exists is that there is no absorption and therefore no radiation at these wavelengths.

    Incidentally, I may be wrong but, I get the impression that some posters believe that there are two types of radiation: radiation at specific wavelengths and blackbody radiation. In reality black bodies are radiating and absorbing at specific wavelengths it is just that they do so with 100% efficiency at all wavelengths. A grey body also absorbs and radiates with the same efficiency at all wavelengths but does so with less than 100% efficiency.

    The earth is close to being a grey body for the infrared part of the spectrum where it emits but it is not a grey body in the visible region. That is why the sea is blue! The atmosphere is nothing like a black or grey body and can never behave like one. As I explained in my previous post the combination of the earth’s surface and the atomosphere results in the grey body emission of the surface being modified by the absorption bands of the IR absorbing gases like H2O and CO2 to give dips in the spectrum particularly around 7 micron and 15 microns when viewed by satellites.

  108. Dave Wendt says:

    Why is it that everyone who attempts to create a cartoon or animation to illustrate how the “greenhouse effect” warms the planet, they always feature a big old Sun beating down. No matter which of the theories about the operation of the various component gases that comprise the atmosphere you chose to embrace, you have to recognize that when the Sun is shining on the Earth the main function of the atmosphere is to make the surface of the planet much cooler than it would be without it. If the Earth were just a larger version of the Moon, without an atmosphere, its daytime temp would be close to that of the Moon. The only difference being provided by the differing lengths of the diurnal cycle. If you compare the max, min, and mean temps for the Earth and the Moon it seems to me that, whatever the atmosphere is doing to make the Earth warmer than its theoretical blackbody temp, its doing most of it at night by increasing the min temps. I keep waiting for someone to do one of these neat climate schematics that features a big full Moon and radiative conditions that prevail when the Sun isn’t shining. It seems to me that that is where the “greenhouse effect” dwells, though I still believe that the “flywheel effect” is a more accurately descriptive analogy of what the atmosphere does for the planet.

  109. Myrrh says:

    Brian H says:
    February 28, 2011 at 7:49 pm

    Brief note to Myrrh;
    All incoming radiation that reaches the surface heats the Earth. It is absorbed by one form of matter or another, which causes kinetic agitation – i.e., heat. It can’t be re-radiated at those same wavelengths because that would require being as hot as the source, the Sun. It gets emitted at the wavelengths matching the temperature of the matter/material emitting. That’s in the IR band for the temps we experience.

    Brian, I have a real problem with that explanation in the graphic that it’s only Visible and near shortwave that heat the Earth kinetically, a.k.a. “Solar”. Your “all incoming radiation reaches the surface of the earth” isn’t addressing that, because that’s what I’m saying is happening… Because all radiation includes Thermal IR which isn’t in the chart.

    Ira says that Visible is the largest part of the Sun’s spectrum, wrong, it’s actually the smallest, and compared with the others, very very small. Light is not hot, it is Reflective not Absorptive, and your use of “absorbed by matter of one form of the other” is misleading. The radiation that penetrates to heat the Earth is Thermal IR. And that is not in the Chart.

    So, not only is downwelling Thermal IR taken out of the equation, it is replaced by the very strange notion that Visible light and its two other cool neighbours, by themselves heat the Earth, called “Solar”. I can sit infront of the TV all day in a cold room, it won’t heat me.

    What I want here is conclusive proof that it this narrow band of non-thermal light energy which is the sole cause of the ground and rocks and me heating up from the Sun. That’s basic premise to the AGW Energy Balance claim as shown in the Charts.

    And no, it does not require the source to be as hot as the sun to radiate out Thermal IR, the electric plate on your stove doesn’t have to be even hot enough to produce red light for you to feel the IR it is radiating, but that doesn’t mean that the Sun isn’t radiating our Thermal IR! That’s reaching the Earth at the same time as Solar. And is better going through water vapour, fog and mist, than Visible Light.

    The Earth being heated by kinetic energy alone by Solar without any mention of of the Absorptive energies from the Sun of Thermal IR is nonsense, which is the AGW claim. I’m asking for proof it isn’t nonsense.

  110. The ~10µ photons generated by the heating of the air emerge from the top of the Atmosphere and their energy is lost to Space,

    The ~7µ and ~15µ generated by the heating of the air also emerge from the top or bottom of the Atmosphere, but there are fewer of them because they keep getting absorbed and re-emitted, each time with some transferred to the central ~10µ portion of the longwave band.

    Complete nonsense. The reason that the 10µ photons from the surface are able to reach space is because the atmosphere does not absorb them. For exactly the same reason, the atmosphere also does not emit radiation at the wavelength.

    Jim D does point out one exception, clouds.

    I also agree with Jim D that greenhouse gases have a net cooling effect. When convection from the surface is included in the analysis, it is obvious that increasing the amount of CO2 cools the atmosphere more than it adds heat.

  111. In addition to the problems in the main post, there are numerous errors in the comments.

    George E. Smith says:
    The AGW types, like the frequency scale, because that puts the CO2 frequencies near the peak of the curve

    Actually, the frequency (wavenumber) scale gives a more intuitive plot when discussing the amount of energy. Note, the wikipedia graph uses log of wavelength which completely distorts the plot.

    the actual amplitude of the thermal spectrum should be about 80 times smaller than it is.

    Well, that depends on the x-axis selection. When frequency (wavenumber) is selected, it is the solar spectra that is about 28 times too high. When wavelength is selected, the 300K curve should be about 10 times smaller.

    The only function of the GHG molecules is to heat the atmospheric gases; along with all the other mechanisms that are heating it. After that, the GHG molecules serve no function in the climate process whatsoever.

    True, they do heat the atmosphere .. however, they also cool it. In fact, greenhouse gases provide the only method to cool the atmosphere. Note that some of the energy they release returns to the surface producing the greenhouse effect.

  112. wayne says:
    where in the heck did 5525K for the sun’s temperature come from?

    Scientists have measured the spectra of the Sun and temperatures near 5,777 K produce a similar blackbody spectrum. However, the references use many different values. For instance, the wikipedia plot above references ~5500 K.

    There are several very good reasons why the references don’t agree

    * The real spectra is not very close to any blackbody spectra (slightly different shape)
    * The measurement devices do not agree
    * The radiation temperature of the center of the solar disk is higher than the apparent temperature near the edge
    * The temperature of the outer corona is more than 1,000 hotter than the “surface”
    * The Sun does not have a surface
    * The distance between the Earth and Sun varies (the orbit is an ellipse), and different references handle this differently

    However, what is known fairly well is the number of watts at the top of the atmosphere (TOA).

  113. davidmhoffer says:

    John of Kent says:
    Has WUWT become a “lukewarmer” site and abandoned Climate Realism?
    This is pure nonsense and I’m surprised that WUWT would print such rubbish. Infra Red Interacting Gasses (IRIGs) CANNOT heat the earths surface. To do so would mean the colder air is warming the hot ground and that goes against the laws of thermodynamics.>>>

    Oh my.
    There’s so many things wrong with those few sentences that it would take paragraphs to correct them all. I sometimes wonder why it is that when science is well explained on this site, by highly technical people, who then engage in well informed debate and discussion with other highly technical people, the bulk of whom are skeptics or mild luke warmers at best, there is always some troll jumping in, spouting rage and ridicule, and denouncing the explanation as fraudulent? And why does the criticism almost always include the exact same two statements?

    1. Cold things can’t heat up warm things.
    2. That breaks the laws of thermodynamics.

    Is there some secret troll society somewhere, a cabal of warmists perhaps, determined to mess up every serious discussion of greenhouse effect by shouting those two statements? Are they on a web site someplace worded a dozen different ways so that volunteers can just copy and paste to distract everyone else? Or maybe it is completely automated?

    OK Kent, cold things can’t heat up warm things. Igloos freeze everything inside them to death, that’s why the Inuit are extinct. The laws of thermodynamics don’t allow for net energy transfer. That’s why if you take two very hot things at the exact same temperature and put them side by side, there’s no energy transferring between them so you won’t burn yourself if you stick your hand between them, the laws of thermodynamics say you’ll be just fine. Kent…KENT KKKKEEEENNNNTTTT don’t actually do it!!! Phew. I thought you were going to try it. Would have felt bad about that.

    There’s dozens of threads and hundreds of comments on this site from very well informed skeptics (amongst others) explaining the facts. You would do well to read many of them before popping off with comments that can be falsified by building an igloo.

  114. Several people have said that a cold sky can not heat a warm surface. The truth is that it can. What they miss is that the “cold sky” is much warmer than the colder microwave background. So, while it may be true that a cold body won’t heat a warmer one, it will slow down the heat loss if the cold body wasn’t there.

    In addition, at night the air in the lower 1,000 meters is warmer than the surface.

  115. wayne Job says:

    When it comes to our little blue planet, all this hypothetical analysis of incoming and outgoing radiation in various wave lengths, and how much is reflected back and how much escapes is neatly summed up by by Humphrey Bogarde, It does not amount to a hill of beans. The amount of heat our world receives and its reaction to it gives us our chaotic but wonderfully variable climate. I find it some what incongruous that people find small changes in things like CO2 can cause warming and cooling besides drought and floods, not to mention no snow or massive snow. The entire argument is total nonsense. It is and always will be the sun and solar mechanics that control climate, anything else will be but a bit player and not relevant to our climate.
    We have a heat pump Earth that uses water and water vapour as a control mechanism every thing else is irrelevant. It is time to ask our great sun god Sol to awaken for the energy balance is tipping to the cool side. As our moderating fluid in the oceans cool as it would seem is happening no amount of nit picking as to what IR radiation is going up or down is going to be diddlysquat as far as our well being is concerned.
    The world has suffered a concerted propaganda effort by the by so called experts to brain wash the west into believing that we are evil polluting monsters, AGW never was about global warming, it was about control power and wealth distribution.
    Discussing radiative imbalances is a digression to keep those with cognitive thought busy whilst evil is committed.
    Those here on this site show huge education and intelligence but this subject has nothing to do with global warming the IPCC is totally political with no real science involved. Confound the science of global warming all you like but the politics and the lies will keep overwhelming you in the compliant MSM.

    It is time for a different tack for the science is becoming irrelevant as the brainwashing of the young is total in the schooling system. I have no answer but many on this site are far more intelligent than I and may have an antedote to this modern day grab for control. Cheers Wayne

  116. davidmhoffer says:

    Ira,
    I think the explanation of the atmospheric window that you posted is reasonably sound, but I think the discussion has wandered off the most important aspect of it. Yes, sensitivity and greenhouse effect and what molecules radiate at what frequencies are all parts of the discussion, and necessary for understanding the climate as a whole. But in relation to the atmospheric window, I think these are relatively minor. The atmospheric window has an effect on earth temperature that I think is far more significant than how it interacts with the greenhouse effect. The important aspect of the atmospheric window is how it interacts with ALL upward bound radiance regardless of source.

    Consider the black curve in your diagram, which represents the distribution of LW radiation emitted by a surface at 210 degrees Kelvin, or -63 C. The peak of the black curve is BEYOND the atmospheric window. Guestimating from the drawing, a surface at -63 C would be only emitting about 30% of its radiance in a spectrum that can escape freely to space. The rest is impeded by absorption and re-emission. But the centre curve, at 260 Kelvin, or -13 C, is smack dab in the middle of the sweet spot. For argument’s sake, let’s call that 50% going straight out the window. Then look at that pink line…uh oh….310 K… 27 C….. the percentage going out to space is dropping back down, the hotter it gets, the less surface radiance is in the escape window…25%? Less? Can you hear the tipping point argument sneaking up on you?

    Of course the tipping point argument is ludicrous. Falsified by no other means than the conditions that supposedly would create it have existed multiple times in the geological record, yet no tipping point occured. The question is not if the tipping point is possible, the question is why ISN’T is possible, and the atmospheric window is a big part of that. As another commenter alluded to earluer, the geological record shows that the earth varies between certain upper and lower temperature bounds, and the atmospheric window has much to do with that also.

    The lower end is in my mind the easier of the two to understand, and also easier to demonstrate. Once the earth surface cools below a certain temperature, it loses heat via radiance that is increasingly outside of the atmospheric window. In brief, the colder it get in that temperature range, the more effective the atmosphere is as insulation. You can see this easily in the temperature record. Anyone who lives in a cold climate, say Winnipeg, at 50 degree N latitude will talk with a certain amount of pride about surviving -40 C. Thompson, well north of that they will talk about…. -40 C. Keep going north. In fact, let’s go right to the arctic circle and take a look at the temperature record from DMI:

    http://ocean.dmi.dk/arctic/meant80n.uk.php

    Pick any year you want. Winnipeg in the depths of winter gets a few hours of sunshine every day. The arctic circle gets none for WEEKS. Yet look at the lows. Hardly ever gets past -40. The window is closed to the heat being radiated at that temperature.

    What of the high end? Why no tipping point? That’s more complicated of course. For starters, you can actually see that in fact the atmospheric window does start to close as earth surface warms up. Here’s a snapshot of ERBE:

    http://knowledgedrift.files.wordpress.com/2010/01/erbe-zonal-1986.gif

    The tropics are for the most part slightly positive in terms of net radiation, they absorb more heat than the radiate (which is why its hot there!) and they would in fact ignite if they couldn’t get rid of that heat via convection and other processes. But you can see that the atmospheric window is pretty much closed to the wavelengths being emitted by the warm tropics. Then take a look at the arctic zones. Holy cow grandma, them windas is wide open, look at all them photon torpedos justa leakin’ out ta space, no wonder its so darn cold here, we gots ta build us an igloo right fast or we gonna freeze solid….what’s that grandma? igloos won’t work because they violate the laws of thermodynamics and cold stuff can’t keep warm stuff warm? Ya been datin’ that troll Kent agin’?

    So part of the answer (sorry Kent, had to take that shot) is that convection at the tropics raises warm air upward where it eventually spills toward the poles. As it cools, it is radiating increasingly into the atmospheric window. Cool air from the poles works its way to the equator to replace the rising hot air picking up heat as it goes. For the most part it starts out cold, below the atmospheric window, and warms, radiating to space more efficiently until it gets “too” warm. But the hotter the tropics get, the faster the cycle goes. Check out the GISS or HadCrut broken down by latitude and you’ll find the tropics are VERY stable, even though they are the net retainers of heat for the planet. Any variation in solar or anything else gets pushed to the temperate and arctic zones for mandatory expulsion to space.

    Of course there are other factors too. Those curves represent the RELATIVE distribution of wavelength radiated, they do NOT represent ABOSLUTE magnitude. Since the amount of energy the surface radiates is proportional to the temperature in degrees K raised to the power of FOUR (squared squared!) there’s an awfull lot more photon torpedoes pointed at space from the +27 C tropics than there are at the -63 C poles. So part of the reason for -40 being a sort of bottom end is that the atmospheric window is nearly closed AND there’s less photons being fired upward besides.

    However, at the top end, even at 27 C, there’s still a healthy percentage of that curve radiating within the window, and the total being radiated is MUCH bigger. Grab a calculator and do 210^4 then do 310^4. Even if only 25% of the torpedoes at 310 get fired off within the range of the window, that’s still more than 100% of the torpedos at 210. Does that mean a tipping point is impossible? YES! You’d have to heat the planet to the point where even the POLES are more than 300 degrees K warmer than the top end of that window. Do the net radiation calculation for THAT temperature. OK, NOW we have a violation of the laws of thermodynamics.

    Grandma… GRANDMA… ya can come out now, yu and Kent there done melted that igloo.

  117. richard verney says:

    Dave Springer comments at February 28, 2011 at 7:36 pm are well worth a read, especially the point he makes aboud clouds (a point I made in my earlier post).

    Whilst CGR may be an explanation for changes in clouds/cloud seeding/ cloud patterns, there may be other explanations for these changes. Indeed, in a chaotic system (such as weather/our atmosphere) one would expect there to be changes in clouds and their patterns. Further, given that we are looking at such a small time scale (circa 100 years) it would not be surprising if during such a short period there was a tendancy (ie., a trend) for less cloudiness and hence a slight upward trend in temperature. This might be nothing more than natural variation on the time scales that we are looking at.

    Personally, I consider that an explanation along these lines is a more plausible explanation (by many orders of magnitude) over that of CO2 being responsible for the warming. Given that current computer models do such a poor job with clouds, we cannot rule this out.

  118. I think etudiant has an interesting point concerning the jetstream (300mb level) and the reduction in its relative humidity. Following the reasoning of Steven Wilde a link may be considered concerning the temperature gradients at these heights in the atmosphere and the position (or changes therein) of the high speed vortices around the world driving the weather patterns and ultimately the climate.

  119. Bill Illis says:

    Tim Folkerts says:
    February 28, 2011 at 7:23 pm
    I must disagree with the comments to the effect that N2 & O2 radiate (or absorb) any important amount of IR.

    Look at a somewhat similar and more familiar idea – visible light emitted by a Hydrogen atom (the Bohr model covered in freshman chem & physics around the world). It is easy to observe that H only emits very specific energies of light which are “easily” predicted by the quantum mechanics of the orbiting electrons and the allowed transitions between energy levels of those orbits.
    ————————————

    But the Sun emits mostly like a blackbody.

    It is not being emitted in the Hydrogen frequencies only, it is emitting across the entire spectrum close to what the blackbody curve predicts. If anything the Hydrogen spectra are suppressed (much like the CO2 spectra is suppressed on Earth) and it emits a little more energy in the non-specific-absorption frequencies (more in the visible).

    The absorption and emission spectra bear special attention, but not all the attention.

    http://commons.wikimedia.org/wiki/File:EffectiveTemperature_300dpi_e.png

    http://en.wikivisual.com/index.php/Image:Solar_irradiance_spectrum_1992.gif

  120. Phil. says:

    Ira Glickstein, PhD says:
    February 28, 2011 at 10:02 pm
    Kevin says:
    February 28, 2011 at 2:52 pm
    “From the looks of that first chart, ~100% of 7μ and 15μ radiation gets absorbed by the atmosphere on the way back to space, mainly by H20 and CO2. Maybe a tiny trickle bounces its way from molecule to molecule and makes it to space.”

    If that’s so, then doubling or even tripling the values of H2O in our atmosphere should have no effect on absorbtion. You can’t absorb more than 100% of the radiation, right?

    As you know, the “temperature” of a gas, such as the Atmosphere, is an abstraction of the average speed of the N2, O2, O3, H2O, CO2 and other molecules that make up that gas. When GHGs absorb a photon, they do not hold onto it forever. Rather, in their energized state, they may re-emit it or they may move faster and, by colliding with non-GHGs (such as N2) they will raise the temperature of the gas. Any mass, including a mass of gas, that has a temperature above absolute zero, will emit longwave radiation in random directions and at a variety of wavelengths, peaking at the wavelength corrresponding to the temperature of that gas, as indicated in the first graphic.
    No, gases such as N2, O2 and Ar possess no dipole and are unable to emit in the IR. Gases do not emit as black bodies, they can only emit in certain wavelength bands which correspond to transitions between certain rotational/vibrational states. This is well established physical chemistry.

  121. Ed Zuiderwijk says:

    Unfortunately the text in the figure is incorrect.

    It says: Downgoing solar radiation 70-75% transmitted. The part not transmitted (but reflected) is not downgoing anymore. This may seem pedantic (it probably is) but it informs the howler at the right: Upgoing thermal radiation 15-30% transmitted. This has to be: All, that is 100%, of the upgoing thermal radiation is transmitted (else it would not be upgoing anymore). The atmosphere is in long-term equilibrium which means that all thermally emitted radiation is in balance with the loss at the top of the atmospher to space. The semantic mistake is an indication that the maker of the figure confused the outgoing flux with the local source function.

  122. Dave Springer says:

    Chad Woodburn says:
    February 28, 2011 at 8:38 pm

    “When the up-going long-wave radiation gets reflected back to the earth, how long does it take for it to “bounce off” the earth and return to the level where it was previously reflected back to the earth? (Or does it not work that way?)”

    Maybe instantly maybe 10,000 years. You can’t track energy by naming each photon and following them like airplanes on a radar display.

    The ocean, like all things with a temperature above absolute zero, is constantly radiating energy out to the environment and absorbing energy from the environment at the same time. When an object and its environment are at different temperatures there is a net flow of energy from warmer to colder. The larger the difference the greater the net flow rate. So say the ocean at night is radiating 400 watts per square meter upward but greenhouse gases absorb 100 watts and half of that is radiated downward towards the ocean. So you have the ocean radiating 400w and absorbing 50w at the same time for a net flow of 350w out of the ocean. Now say we have a doubling of greenhouse gases such that they absorb 200w and radiate half of that downward. Now the ocean is radiating 400w and absorbing 100w for a net flow of 300w out of the ocean.

    What this results in is a lower rate of cooling for the ocean. Like coffee in an insulated versus an uninsulated cup.

    The net result of all this is that when the sun warms the ocean during the day greenhouse gases reduce the rate at which that energy can escape. So the ocean gets warmer than it would have absent the greenhouse gases. The warmer ocean radiates heat faster than a cooler ocean due to the larger temperature differential between the cold of outer space and the warm ocean surface. Thus a new equilibrium point is established where energy received during the day is lost at night. In the long haul all energy absorbed by the ocean is radiated back out into space. However there are many other dynamic things happening that keep changing the daily energy flow so that equilibrium becomes a moving target that can be approached but never hit. Some days are cloudy, some are not, so some days the ocean receives more energy and some days less. Evaporation, convection, ocean and wind currents move energy around mechanically from one place to another. But the bottom line remains that greenhouse gases slow down the rate of energy loss to space which causes a higher surface equilibrium temperature the system will target.

  123. Tim Folkerts says:

    Myrrh says: March 1, 2011 at 2:45 am
    “Ira says that Visible is the largest part of the Sun’s spectrum, wrong, it’s actually the smallest, and compared with the others, very very small. Light is not hot, it is Reflective not Absorptive, and your use of “absorbed by matter of one form of the other” is misleading. The radiation that penetrates to heat the Earth is Thermal IR. And that is not in the Chart. ”

    The issue here is the “size of the energy”, not the wavelength. There is a handy table at http://en.wikipedia.org/wiki/Planck%27s_law that shows the energy distribution of sunlight (above the atmosphere).
    * A little over 10% of the ENERGY is in ultraviolet photons or shorter wavelengths.
    * Close to 40% of the ENERGY is visible light photons
    * Over 40% is “near IR” (from 0.7 – 3 um). (So Ira’s claim that visible is the biggest is not quite right, but it is not far off. On the other hand, the atmosphere is still mostly transparent to these wavelengths, so his arguments still hold).
    * Only a few % is above 3 um (“thermal IR”) so it cannot possibly be the main source for heating the earth.

  124. 1DandyTroll says:

    Greenhouses primary design is not to trap heat, but to control the environment, which includes to control the in house atmosphere which, of course, constitutes of gases.

    So what you have in a greenhouse is a triple layer between the plants and the god ol’ Sol. You got the outside atmosphere (that apparently is working as a greenhouse effect), the glass (which apparently the outside atmosphere is functioning as), and the denser inside atmosphere (which apparently the outside atmosphere’s affect is the warming effect of.)

    And greenhouses do not retain the heat very well, no insulation to keep the heat inside really and the much denser atmosphere does a really poor job of trapping the heat since the heat “escapes” so easily.

    A greenhouse is in effect trying to mimic earth’s “summer effect” in a controlled manner. However, other ‘an summer days you usually have to add heat by other means since the heat “evaporates” so quickly to the outside, even with the inside having the denser atmosphere.

    So I still claim that using the “greenhouse effect” as defined as something that gets warmer is just silly. Although if more people sported their greenhouses all through winter the power and gas companies would be ecstatic. :p

  125. kbray in california says:
    February 28, 2011 at 4:07 pm …

    Mars is a lot further from the Sun than is the Earth, which is the main reasion it is too cold to support life as we know it, despite all the CO2 in its Atmosphere. Your analogy of the black balls in the pool does not work because the mechanism is not at all like that of the “greenhouse effect” GHG in the atmosphere. There is more H2O (the main GHG) in the Atmosphere than CO2, but the percentage of H2O is also miniscule, yet, the tiny percentage of GHGs (mostly H2O but also CO2) are responsible for the Earth being at livable temperatures.

  126. Peter Taylor says:

    Ira –

    I have grappled with the ‘basic physics’ for some time – having concluded from studying the flux data (courtesy of NASA), that CO2 can be accounting for at most 20% of the observed warming between 1980-2010. The main driver is very clearly percentage changes in cloud cover – I deal with this at length in my book ‘Chill: a reassessment of global warming theory’ – which calls for a re-analysis of the models. Obviously, if I am right, then there is something seriosuly wrong with the ‘basic physics’.

    But I think you could be focussing your energy on an area of far greater importance than the emission spectra. It took me a long time to realise that the Radiative Forcing (RF) that is at the heart of the model calculations is centred on a notional point toward the top of the troposphere…defined by the temperature at which the earth will reradiate the balancing energy – which is -19 C and about 10km up. This is the point of notional equilibrium – a necessary condition for model initiation (it is very difficult to build in long term cycles and the reality of non-equilibrium when the causes of those cycles, as well as their amplitude and frequency are so poorly known).

    So – the model calculations show that an RF of say 1.6 watts/square metre for current carbon dioxide levels will thus radiate downwards – this is a ‘forcing’ because it has moved the system from its pre-AGW (and fictitious) equlilibrium state.

    This value , and the value for ‘doubling’ CO2 – at about 4 watts/square metre, is quoted everywhere you look – but just try to find the science of RF calculations to back it up!!! It took me a long time to find references in IPCC – I had to go back to 1990. Their special publication on RF is very unhelpful (and hard to find). I still found no academic scientific references. Eventually I realised that the IPCC reference to ‘offline computer codes’ meant the MODTRANS formulae which are produced by the USAF and can be accessed for $300! There may be others, but this unit seems to have the monopoly. I have no idea if there are error margins – but they do all the ‘basic physics’ calculating the heat transfers and photons and all that jazz at any altitude…and this model will allow you to input changes in greenhouse gases.

    It is clear the MODTRANS RF model is LOG…..as many have posted on WUWT.

    There may be nothing wrong with this level of the basic science – But references to this log relationship are VERY sparce in the literature. One colleague of mine persistently asked – ‘how can you say there is a log relationship between concentration of CO2 and its RF, when everything I see in the literature refers to a linear one?’ Eventually, I found him a point in IPCC where the (ln) equation was stated (but not referenced to an academic paper).

    But then the Rf must be translated to a surface temperature – and that is far from basic physics.

    I believe the first equations relating RF to temperature at the surface were courtesy of James Hansen – but I may not have the whole story. It is in THIS relationship that assumptions of linearity come in.

    THUS….modtans calculates the RF at 10 km…..and say for a doubling we have an excess of 4 watts/sq metre of downwelling IR radiation.

    Warmist science then has to translate this value into Temperature at the surface – and compare this to the observed T, and arrive at an equation T at surface = X (RF) where X is the factor that translates RF to T on a degrees C per watt basis. Remember, the RF has a log relation to concentration and Modtrans already assumed this, so the RF does not increase linearly with concentration. In calculating the RF we have to assume – in the absence of any critical scientific review of the privately produced modtrans model – that the asymptotic point has been calculated correctly. Actually, by the time you approach 200ppmv for CO2, you have already reached the break point in the curve, beyond which additional CO2 has much less impact on the RF – and this is close to the glacial value – suggesting that CO2 changes do not drive the glacial cycles (CO2 changes are supposed to amplify T rise during deglaciation, but there is scant evidence for this and the assumption that it did also underlay the IPCC belief – and a great many references in academic papers give a T degrees C per ppmv CO2 without stating over which range of concentrations this is meant to apply.

    So – this X-factor must take account of everything that goes on beneath the calculated RF at 10km…..all the varying cloud and aerosol and water vapour. Values seem to vary between 0.4 and 0.8. This is the assumed ‘linear’ relationship that is always mentioned, whereas the original log relationship gets buried.

    I suspect that given the paucity of knowledge in relation to clouds and aerosols (not to mention cycles)…the original X factors for the equation ranged through values that at the lower end produced no scary warming scenarios for the future doubling (ie at or lower than 1.5 C) to those that were very scary at 3-4.5 C – or even 6 C if you add strong feedbacks from melting ice, permafrost and emissions of methane. Hansen (and others?) was faced with such a range and chose the high ones – perhaps out of a genuine sense of precaution.

    We should now be able to ascribe a real figure to the X-factor. If we can trust the RF calculation in the absence of effective peer-review – and using the observed global rise in T (criticism of urban heat islands and dodgy grid homogenising algorithms not-withstanding), then 0.7 = X (RF), where the RF for CO2 is…..? Do we take the final cumulative figure at the end of the observation period at 1.6? Or the increase since the beginning of the observation period? Taking the former – the RF given by IPCC for the CO2 is 1.6 watts (this is also the more approximate net forcing when all other GHG, cloud and aerosol factors are taken into account).

    Using these values, the X-factor would be 0.43 with the proviso that it is determined over a century and that the system is at equilibrium (Warmists like to have it both ways here, ie that it was at equlibrium to start with, but not at the end, where there is ‘warming in the pipeline’ due to absorption and release of heat by the oceans). If we apply this value to the future predicted total net RF in 2050 of 3.7 watts/sq metre due to the mix of GHGs, then the resultant surface T is 1.59 C – and not at all a scary climate story.

    Thus, observations tend to support a low factor X. Even this assumes that ALL the change in T from 1900 to 2010 was caused by the net downward flux or RF – the forcing, and there was no variation in any of the factors that contribute to that X factor such as clouds and aerosols, or warming and cooling cycles in the ocean heat reservoir – which is unlikely. The defending argument is that any such variables will be compensated for over the time period.

    It is this X factor that needs attention – where is the peer-reviewed science? I assume it IS there, just not so readily available to the non-specialist scientific reviewer.

    And actually, my final conclusion is that there is nothing actually ‘wrong’ with the basic science – it already includes these lower factors and a low prediction for doubling in 2050, but there is an assumption that this low prediction is very unlikely. This is not a problem with the basic science, rather a lack of awareness and bias on the part of the reviewers at IPCC and just about every academic institution that has supported their faulty analysis.

    By the way – the IPCC Summary for Policy Makers has a graphic that has on the left side concentration of CO2 and on the right, the RF in wats/square metre – the right hand axis is shown as linear. This may account for the difficulty the IPCC has in accepting that future increases in CO2 will have very small impacts – they believe that graphic. The relationship is not of course linear, but logarithmic.

  127. Bryan says:

    Robert Clemenzi says:

    ………”Several people have said that a cold sky can not heat a warm surface. The truth is that it can.”……

    A warmer surface will radiate more of every wavelength than it absorbs from the colder surface.
    How then can we say the colder surface “heats” the warmer surface?
    Heat means to “increase the temperature of “.
    In fact it is more correct to say that the colder surface slows the rate of heat loss from the warmer surface.
    In other words it insulates the warmer surface to some extent.
    The igloo example will insulate the person inside to some extent.

  128. Myrrh says:
    February 28, 2011 at 4:31 pm
    I’m really at a loss to understand any of this. How on earth does Visible light and near short wave heat the Earth

    Myrrh, you really need to get outside more and sit in the Sunshine and feel the warmth! That is how visible and near-visible (“shortwave”) light warms he Earth.

    If you don’t or cannot get outside, turn on an old-fashioned incandescent light bulb and hold yourhand near it (not too close, you will get burned). Feel the heat? That is shortwave light because the filament is heated to temperatures similar to the Sun’ surface. You can tell it is shortwave because you can see the light.

  129. Slacko says:

    Tim Folkerts says:
    February 28, 2011 at 7:23 pm

    “However, it is an observed fact (supported by theory) that monatomic gases (like argon) or symmetric diatomic gases (like N2 and O2) do not have vibration modes or rotation modes that would allow them to absorb (or emit) IR photons.
    … IR is not absorbed or emitted by N2 & O2.”

    A hot body that won’t radiate infrared?? That’s gotta be the best-kept secret in physics, if it’s true.

  130. Alan McIntire says:

    In reply to Katherine and to Phil’s dad. When the Earth’s temperature increases
    from 1 to 1 + p, the total radiation increases by a factor of (1 + p) ^4,
    Outgoing radiation increase at all wavelengths, but the PERCENTAGE increase at
    short wavelengths, which are not affected by CO2, increases at a much larger rate.

    So if CO2 results in 40 watts/500 total watts, with a 1% increase in the total to
    505 watts, CO2 will absorb less than 40.4 watts. There’s a negative feedback due to that 4th power increase in radiation with respect to temperature, as you discovered, there’s also a negative feedback with respect to CO2 due to a larger percentage of increased outgoing radiation in wavelengths not affected by CO2.

  131. Brent Hargreaves says:
    February 28, 2011 at 4:34 pm
    … This stuff is com-pli-cated! Could it be that this complexity frightens the bejaysus out of even professional climatographers… with the consequence that they each concentrate on (or hide away in?) some narrow aspect of climate but shy away from contemplating the mind-bogglingly big picture.

    Is the Global Warming scare is a ‘fragmentation failure’?…

    I think you have it summarized quite well. IMHO, the initial Alarmists, like “Chicken Little” who REALLY thought the sky ws falling, were honest, if a bit mistaken. In the case of Global Warming, they really thought that we were heading for a “tipping point” that would lead us over the cliff into “runaway warming”. I believe James Hansen of NASA GISS and Phil Jones of the UK CRU were true believers, at least in the beginning, convinced they were annointed to SAVE THE WORLD AND HUMAN CIVILIZATION AND ANIMAL AND PLANT POPULATIONS.

    Perhaps, as you say, they each focussed on their narrow, fragmented, domain of expertise (Hansen’s PhD is in Astronomy) and missed the “big picture”.

    As a system engineer on military avionics, I dealt with many wonderfully dedicated and intelligent and good natured specialty engineers and scientists and mathematicians who were so deep into their area of expertise that they could “not see the forest for the trees”. Sometimes, when they ventured into the overall system aspect, they made amazing errors in judgment (as I sometimes did when I ventured into their specialties).

    These specialists were so used to being expert in their area (and they were) that they thought they also had perfect insight into all areas, including stuff where their knowledge level was almost childish. As a system engineer, I knew my limitations and that my knowledge of many specialized areas was fairly shallow. Therefore, I was not so sure I was right all the time. When a given specialized domain became of critical importance to the system as a whole, I summoned up my admittedly limited knowledge of physics and math (though, of course, I passed those courses in engineering college) and sat at the feet of these experts, and begged them to explain their stuff to me. I would usually say, “I want to know ‘what time is it’ not ‘how to build a clock'”. Most were quite patient with me since I was modest and fairly quick on the uptake. I am now trying to do the same with climate science and I appreciate those on WUWT who have corrected and instructed me so well.

    Even now, Hansen and Jones and others on the official climate Team may still be able to force and delude themselves to believe this stuff (“Strange is man when he seeks after his gods.”) I give former VP Al Gore a (nearly) free pass because he simply accepted what established scientists told him and ran with it. His idea that the Earth’s core is at “millions of degrees” (high by a factor of a thousand) illustrates his lack of any real scientific knowledge. But he did have some great political allies and and mainstream media connections. Amazing what that man can do with only half a brain :^).

    On the other hand, I believe some percentage of climate scientists are opportunists, in it for the money, who will give scientific support to almost any exaggeration to get more government grants. Then there are those politicians and professional enviromentalists who are in it for opportunistic political reasons. (“Half the air pollution is due to burning fossil fuels and the other have due to the incessant spouting of nonsense by environmentalists” :^)

    That is the problem when the government gets involved in complex stuff that would better be left to the privae sector. (“If the government was in charge of hte Sahara Desert, in six months there would be a shortage of sand.” :^)

    However, some of the alarmist and warmist advocates are honestly convinced that the advance of technolology and transnational corporations and globalization will do us all in. I tend to be an optimist on these issues, as expressed in my free online novel and predictions about life, liberty, and technology in the second half of this century.

  132. izen says:

    @-Myrrh says:
    March 1, 2011 at 2:45 am
    “Ira says that Visible is the largest part of the Sun’s spectrum, wrong, it’s actually the smallest, and compared with the others, very very small. Light is not hot, it is Reflective not Absorptive, and your use of “absorbed by matter of one form of the other” is misleading.”

    The vast majority of the energy direct from the Sun that heats the Earth is in the visible spectrum, NOT the IR. If you doubt that visible light can heat things consider the car on a sunny day. The glass of the windows blocks IR but lets through visible light that is absorbed by the (usually) dark plastic/leather interior. The hot steering wheel and seat are warmed by the visible light not any IR.

    You are correct of course that the surface is ALSO heated by the 7um and 15um IR emissions from the H2O and CO2 in the atmosphere that has been warmed by the IR emissions from the surface….

  133. mkelly says:

    cal says:
    February 28, 2011 at 4:58 pm

    “As I read it you seem to imply that the greenhouse molecules pass all their energy to the O2 and N2 molecules which then radiate into space. However they can’t since there is a law which says that a bad absorber cannot be a good radiator.

    The rule is that you cannot have a net energy transfer in this direction but there is always energy being radiated from any body above absolute zero.”

    OK which is it. N2 and O2 “can’t since there is a law” or “there is always energy being radiated from any body above absolute zero.”

    You can’t have it both ways.

  134. Vince Causey says:

    Ira,

    usefull schematics. However, some posters have pointed out that, according to Kirchoff’s law, if the atmosphere cannot absorb at 10 micro thingies, it can’t emit at that wavelength either, and therefore by implication, cannot this wavelength cannot appear as the result of being warmed by other wavelengths. I don’t pretend to know anything about Kirchoff’s laws, but if they are correct, this does present an issue with your model. Care to comment?

  135. Vince Causey says:

    John of Kent,

    “To do so would mean the colder air is warming the hot ground and that goes against the laws of thermodynamics.”

    Wow. Against the laws of thermodynamics. Which laws of thermodyamics did you have in mind? First law says that energy can be changed from one form to another but cannot be destroyed. That can’t be it then. What about the second law? Perhaps that’s what you meant. Let’s see. Here’s a quote about that from Wiki:

    “The second law of thermodynamics is an expression of the tendency that over time, differences in temperature, pressure, and chemical potential equilibrate in an isolated physical system. From the state of thermodynamic equilibrium, the law deduced the principle of the increase of entropy and explains the phenomenon of irreversibility in nature. The second law declares the impossibility of machines that generate usable energy from the abundant internal energy of nature by processes called perpetual motion of the second kind.”

    Talks about a tendency of temperatures to equilibriate and the impossibility of perpetual motion machines. Ok, if the colder air is supposed to be radiating towards the warmer ground, is that against the second law of thermodynamic?

    Not necessarily. As long as the warmer ground is radiating more back towards the cold air than the cold air is radiating towards the warmer ground, their temperatures would equilibriate (if the ground wasn’t being continually warmed by the sun). Does this mechanism constitute a perpetual motion machine? No, because again, their temperatures would equilibirate over time so no more work could be done.

    By this definition of the second law, GHG back radiation to earth does not violate it. Care to elaborate?

  136. Dave Springer says:

    davidmhoffer says:
    March 1, 2011 at 3:04 am
    John of Kent says:

    “Infra Red Interacting Gasses (IRIGs) CANNOT heat the earths surface. To do so would mean the colder air is warming the hot ground and that goes against the laws of thermodynamics.”

    By the same token insulation in your attic can’t heat your house. But that doesn’t mean attic insulation does nothing at all. It slows down the rate of heat exchange between inside the house and outside the house. That’s what greenhouse gases do except that GHGs are more one-way than attic insulation. They impede very little energy coming from the sun to the earth’s surface but impede a lot of energy going from the earth’s surface to the frigid cold of outer space. This should not be difficult to understand. Arguments to the contrary always have two defining characteristics: they are convoluted and untrue.

  137. pyromancer76 says:

    Anthony, I wish WUWT would follow Max Hugoson’s advice — atmospheric effect. Truth in science.

    Max Hugoson says:
    February 28, 2011 at 7:34 pm
    PS: Because of Robert Woods 1909 experiment with two miniature “greenhouses”, one with a “rock salt” (i.e., transparent to the longwave IR material) window and the other with a glass window, showing NO MEASURABLE DIFFERENCE in the final equilibrium temperatures in both boxes, we’ve known SINCE 1909 that to ascribe the warming of greenhouses to the allegation of a “one way valve” due to the regular sodium silica glass, is in error.

    Real “Meteorological” textbooks for many years have noted this and used the prefered term: “Atmospheric Effect”.

    I think the WUWT crowd needs to realize this fact and steadfastly refuse to use the terms “Greenhouse gases” or “Greenhouse Effect”.

    Truth needs to win!

  138. Tim Folkerts says:
    February 28, 2011 at 4:34 pm
    First of all, I like the post in general — it seems to provide a fairly effective visualization.

    Fred Souder says: February 28, 2011 at 2:31 pm
    Are there any experiments which detect the incoming re-emitted radiation from the atmosphere?

    I googled images for “outgoing infrared spectrum” and found this: http://www.skepticalscience.com/images/infrared_spectrum.jpg
    It should the spectrum looking up from somewhere in the arctic on a clear day and the view looking back down from 20 km. The incoming radiation would be the re-emitted IR from the atmosphere that you wanted. It clearly shows the 7 um, 10 um and 15 um bands Ira is talking about.

    THANKS Tim Folkerts for the link. I knew some WUWT reader would be alert enough to answer Fred Souder’s good question . GREAT LINK, let me repeat it: http://www.skepticalscience.com/images/infrared_spectrum.jpg

    Here it is for all to see:

  139. mkelly says:

    Ira Glickstein

    http://www.scribd.com/doc/34962513/Elsasser1942

    Again I give you this link to a paper on radiative heating of the atmosphere via IR. It may help.

  140. DccMartyn says:
    February 28, 2011 at 5:52 pm
    Ira, can you explain this to me. Approximately 6% of the energy spectrum, hard uv, is absorbed by the upper atmosphere, mostly by Ozone.
    Now this heat must be reradiated, so half must go down. I never see this heat in any of the box figures, but it can’t just disappear.
    So where does this heat go and in what form does it leave or enter the Earths atmosphere?

    Great question DccMartyn. I think that some part of the UV spectrum, like its close cousin visible light, is mostly reflected rather than absorbed and re-emitted, so it would go back out to Space like the visible light that strikes clouds. Other parts are 100% absorbed by oxygen and ozone, or are scattered, see http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png. Some of the UV gets through and gives us happy residents of Florida our nice natural tans.

  141. Steve says:

    “Mars is a lot further from the Sun than is the Earth, which is the main reasion it is too cold to support life as we know it, despite all the CO2 in its Atmosphere.”

    No, Mars isn’t at a “beyond supporting life” distance from the sun. The Martian atmosphere would be warm enough to support life if it wasn’t so thin , at about 1% the density of Earth’s atmosphere. The density of the atmosphere ties in directly with it’s heat capacity. Mars has little to no magnetosphere, so the solar wind stripped much of the atmosphere away over billions of years.

  142. Dave Springer says:

    Enough already with the argument that certain gases that don’t absorb infrared can’t emit it. This is not true for dense gases. The troposphere is a mixture of cold dense gases. Please review Kirchoff’s Laws:

    http://www.physics.rutgers.edu/~matilsky/documents/kirchoff.html

    Radiation Laws

    Kirchoff’s Laws

    First Law: A hot solid, liquid, or dense gas emits radiation at all wavelengths (“a continuous spectrum of radiation”). For example, a perfect blackbody does this. If the light were passed through a prism, you would see the whole rainbow of colors in a continuous band.

    Second Law: A thin hot gas in front of a cooler background emits radiation at a discrete set of isolated wavelengths. These discrete, isolated wavelengths are called the “emission lines” of the spectrum, because if you were to pass the radiation through a prism, you would see isolated lines of different colors. The whole spectrum is called an “emission-line” spectrum. The wavelengths of the emission lines are unique to the type of neutral atom or ionized atom that is producing the emission lines.

    Third Law: A thin cool gas in front of a hotter solid, liquid, or dense-gas background removes the radiation from the background source at special wave lengths. If the resulting radiation were passed through a prism, there would be dark lines superimposed on the continuous band of colors due to the background. These dark lines are called “absorption lines.” The wavelengths of the absorption lines are unique to the type of neutral atom or ionized atom that is producing the emission lines.

    If a certain type of gas produces absorption lines at certain wavelengths when it is in front of a hot background, then when that same type of gas is seen in front of a cooler background, it produces emission lines at the exact same wavelengths.

    Explanation of Kirchoff’s First Law

    Kirchoff’s First Law boils down to blackbody radiation, since solid objects and dense gases emit radiation like blackbodies.

    Explanation of Kirchoff’s Second and Third Laws

    Thin gases don’t emit or absorb radiation like blackbodies. To understand their emission and absorption, we must consider the structure of atoms, as described by Quantum Mechanics.

    The Bohr model of the hydrogen atom: a dense nucleus containing the hydrogen atom’s single proton (and possibly one or more neutrons), surrounded by an electron that can be on one of several different orbits.

    Nitrogen doesn’t absorb infrared radiation but it can certainly gain kinetic energy by excited molecules of CO2 and H2O bumping into nitrogen molecules. The kinetic energy gained by the nitrogen molecules will be lost in a continuous blackbody spectrum characteristic of the temperature of the gas.

    Any argument to the contrary is simply wrong and ignores basic laws of physics formulated 150 years ago.

    What’s hilarious is watching people who have less understanding of classical laws of physics than 19th century physicists start babbling about 20th century quantum mechanics.

  143. Dave Springer says:

    Ira Glickstein, PhD says:
    March 1, 2011 at 7:19 am

    I googled images for “outgoing infrared spectrum” and found this:

    http://www.skepticalscience.com/images/infrared_spectrum.jpg

    “It should the spectrum looking up from somewhere in the arctic on a clear day and the view looking back down from 20 km. The incoming radiation would be the re-emitted IR from the atmosphere that you wanted. It clearly shows the 7 um, 10 um and 15 um bands Ira is talking about.”

    If you’d paid attention to MY comments you’d have found that image months ago. I have posted it several times including last night on THIS thread.

    http://www.sundogpublishing.com/fig8-2.pdf

    It’s from the 2006 textbook “A First Course in Atmospheric Radiation” by Grant Petty.

    I didn’t know you could get a PhD in Obfuscation & Convolution but you’re living proof that it happens.

  144. Max Hugoson says:
    February 28, 2011 at 7:34 pm
    PS: Because of Robert Woods 1909 experiment with two miniature “greenhouses”, one with a “rock salt” (i.e., transparent to the longwave IR material) window and the other with a glass window, showing NO MEASURABLE DIFFERENCE in the final equilibrium temperatures in both boxes, we’ve known SINCE 1909 that to ascribe the warming of greenhouses to the allegation of a “one way valve” due to the regular sodium silica glass, is in error.

    Real “Meteorological” textbooks for many years have noted this and used the prefered term: “Atmospheric Effect”.

    I think the WUWT crowd needs to realize this fact and steadfastly refuse to use the terms “Greenhouse gases” or “Greenhouse Effect”.

    Truth needs to win!

    Max

    Max, you are quite correct, technically. Yes greenhouse (even with ‘scare quotes’ as “greenhouse effect”) is a misnomer for what you prefer to call the “Atmospheric effect”.

    On the other hand, when we speak to the general public, and even to those who are into climate science like the WUWT crowd, we need to use the terms that, for whatever reason, have become established in the language. The popular usage term “Greenhouse effect” is what my PhD advisor calls a “frozen accident” – something that has been accepted and is now dominant, though other alternatives might be equally good or even better. There are many aspects of the genetic DNA code, for example, that, as a computer and programming savy guy, I would change if I could, but Evolution and Natural Selection has chosen those somewhat accidental codes and fixed them so firmly that to change them would be futile. The best is the enemy of the good enough.

    I try to use ‘scare quotes’ and I mention, right at the top of my topics, that the Atmospheric “greenhouse effect” works differently from a physical greenhouse. If I simply said “Atmospheric effect” who, except for you and some subset of Meteorologist specialists, would know what I was talking about?

    Perhaps we can get the agriculturalists to change the name of their transparent boxes to “Hot House” or “Limited Convection Containment Houses” or “Temperature Enhancement Houses” or some other more descriptive term? :^)

    If I am right (see CO2 is Plant Food) we will see captured CO2 from clean coal powerplants pumped into real greenhouses, since elevated levels of CO2, up to 1400 ppm or even higher, aid the growth of most plants. Greenhouses are also usually quite humid. Thus, even the term “Greenbouse gases” (GHG) will apply both to the “Amospheric effect” and the inside of real greenhouses.

  145. JohnWho says:

    Ira Glickstein, PhD says:
    February 28, 2011 at 11:08 pm
    What the 100% absorption means is that 100% of the photons in the appropiate bands are 100% likely to be absorbed by an H2O or CO2 molecule before they travel all the way through the Atmosphere. Indeed, I believe, even at the historical 270 or 280 ppm level of CO2, each photon traveled a small fraction of the height of the Atmosphere before being absorbed and re-emitted. So, with the current level of CO2 around 390 ppm, the photons, on average, will travel a shorter distance before getting absorbed and re-emitted. That means that newly emitted longwave radiation from the Earth’s surface will spend more time in the lower reaches of the Atmosphere and therefore contribute to heating the N2 and O2 and other components of the air more than before. Thus, the air nearest the surface of the Earth will, being a bit warmer, re-emit a bit more longwave radiation in all directions, including towards the surface. Hence, more CO2 = a bit more warming. Fortunately, this is a logarithmic process, so subsequent doublings have less and less effect – but they do have an effect.

    I’m wondering then if the above is true, then what is the minimum altitude (bold section mine) that actually has any discernable effect on either the earth’s surface or the climate?

    Also, while it radiates in all directions, doesn’t whatever becomes warmer rise, moving away from the surface?

    Just wondering.

  146. etudiant says:

    etudiant says:
    February 28, 2011 at 2:18 pm
    Relative humidity in the upper atmosphere (300mb level) has fallen from around 55% in 1950 to about 45% now.
    Should this not have a material impact on the greenhouse effect? …

    What is the source of that data and the accuracy of measurement in 1950? Assuming the change is actual, I would think it would impact the “greenhouse effect”, but I do not know which way because relative humidity considers both the temperature and water content. Are there any more expert than me out there with an answer?

    Dr Glickstein,

    The source of the data is NOAA, here
    http://www.esrl.noaa.gov/psd/cgi-bin/data/timeseries/timeseries1.pl

    I do not know how good the pre satellite data is.
    Presumably it represents balloon measurements.
    The striking thing is the apparent monotonicity of the trend, unlike that at 600mb, which has seen a recovery in recent years.
    It does underscore the tremendous complexity of the atmosphere, that this kind of longer term change happens without us even being much aware of it, much less able to explain it.

  147. mkelly says:

    Dave Springer says:
    March 1, 2011 at 7:53 am
    Nitrogen doesn’t absorb infrared radiation but it can certainly gain kinetic energy by excited molecules of CO2 and H2O bumping into nitrogen molecules. The kinetic energy gained by the nitrogen molecules will be lost in a continuous blackbody spectrum characteristic of the temperature of the gas.

    Agreed. My point to Cal earlier. If N2 and O2 could not shed energy (trapping it) then we would have a problem.

    But that leads to a question: what difference is there between the spectrum of N2 and O2 both at the exact same temperature and radiating. I have not found anything on that.

  148. Ed Scott says:

    The peak wavelength of human body emission is at 9500 nm (9.5 microns) and according to the diagram, passes through the Long-wave Window with small loses.

    It is interesting to note that nearly all Earth-emitted radiation is between 5 to 30 microns and is centered at about 10 microns.

    It seems that most of the thermal energy emitted by Earth and its inhabitants gets a free pass through the Pass-Through Window.

    One of the few articles that recognizes the specific heat capacities of the major atmospheric gases – Nitrogen, Oxygen, Ozone and Argon – as well as the atmospheric trace gases – water vapor and Carbon Dioxide.

  149. Oliver Ramsay says:

    The charts from skepticalscience are very interesting.
    They do not appear to be readings taken at the north pole on January 7th., nor at Tucson on July 9th.
    I would be very interested in reading how they are put together.

  150. Bryan says:

    Dave Springer

    …….”The kinetic energy gained by the nitrogen molecules will be lost in a continuous blackbody spectrum characteristic of the temperature of the gas.”……

    Could you provide a chart showing a continuous black body spectrum for nitrogen gas.
    I dont think such a thing exists.

  151. Philip Peake (aka PJP) says:

    @Ira — sorry about all that extra work :-)

    I agree that adding more CO2 molecules will increase the chances of photons of a particular energy impinging upon another molecule with an energy band exactly matching that of the photon energy (another CO2 or H2O molecule) and so slowing the progress of that “packet” of energy towards space and freedom.

    In short, it will improve the insulation properties of the atmosphere.

    As you say, because of the statistical chance of hitting another molecule, as the concentration of molecules increases, so will the chance of such an encounter, but the effect is, indeed, logarithmic in nature (Log base e, not Log base 10).

    Look ONLY at CO2 and you will find that a doubling does have some headroom to make a measurable difference. However, if you treat CO2 and H2O molecules as effectively the same, which they are in this context, then a doubling of CO2 amounts to a much, much smaller fractional increase when considering all H2O and CO2 molecules combined, and I would be exceedingly surprise if you could actually measure any difference.

  152. Bryan says:

    Ira Glickstein

    The famous experiment by R W Wood proved two things;

    1. Glasshouses(greenhouses) heat up by stopping convection.

    2. The heating radiative effect at atmospheric temperatures is so small as to be negligible for most practical purposes.

    People tend to forget the second part.

  153. Tim Folkerts says:

    Dave Springer says: March 1, 2011 at 7:53 am
    Enough already with the argument that certain gases that don’t absorb infrared can’t emit it.

    At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity.
    http://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation

    The emissivity characterizes the radiation or absorption quality of nonblack bodies.
    http://www.answers.com/topic/emissivity

    The rate of electromagnetic radiation emitted at a given frequency is proportional to the amount of absorption that it would experience by the source. Thus, a surface that absorbs more red light thermally radiates more red light.
    http://en.wikipedia.org/wiki/Thermal_radiation

    A black body absorbs and emits all frequencies of light perfectly. A non-black body will absorb and emit light less well at some (or all) frequencies. But the frequencies where it absorbs poorly are the same as the frequencies where it emits poorly. Thus a material (like the atmosphere) that absorbs some wavelength of light (say 10 um) poorly will also emit that wavelength poorly.

    Since you agree that N2 absorbs 10 um light poorly, then the only conclusion is that it also emits 10 um light poorly.

    In fact, the very image you reference at http://www.sundogpublishing.com/fig8-2.pdf shows that there is almost no radiation coming down at 8-12 um. If the N2 was indeed emitting significant amounts of thermal radiation at those wavelengths, where is it? I see the (nearly) black body radiation above and below those limits due to the H2O and CO2 (and other GHGs) which can (almost perfectly) absorb those wavelengths.

  154. Tim Folkerts says:

    Bill Illis says: March 1, 2011 at 5:14 am
    But the Sun emits mostly like a blackbody.

    It is not being emitted in the Hydrogen frequencies only, it is emitting across the entire spectrum close to what the blackbody curve predicts.

    Quite true. But the circumstances were quite different from what I was describing. The sun is not a single H atom, but a huge collection of ionized H+ and electrons. Ionized H+ cannot have a Balmer spectrum because it has no orbiting electron. (And as you point out, non-ionized H in the cooler outer layers can and does then absorb the Balmer lines, making them dimmer than other wavelengths). Certainly a large enough collection of ionized CO2 or N2 or any other gas could emit a significant continuous spectrum like the sun does. But that does not describe the situation in the earth’s atmosphere.

  155. davidmhoffer says:

    Peter Taylor;
    It took me a long time to realise that the Radiative Forcing (RF) that is at the heart of the model calculations is centred on a notional point toward the top of the troposphere…defined by the temperature at which the earth will reradiate the balancing energy – which is -19 C and about 10km up>>>

    This was one of my very first epiphanies in regard to the IPCC numbers. They keep quoting 3.7w/m2 and 1 degree, but a quick SB Law calculation versus temp at earth surface doesn’t support that. From your post it sounds like you couldn’t find the supporting docs in AR4, but I did. I’d have to dig like crazy because it was a long time ago, but from memory:

    IPCC calculates sensitivity at the “theoretical black body” temperature of earth. I thought it was -15, but could very well have been -19

    They specificaly state that surface temperature changes cannot be directly related to changes in RF. In fact, they even say it is possible that a change in RF results in climate change, but potentially NO surface temp change.

    The implication that CO2 is linear was very carefully worded, but turned out to be a reference to a small change of some sort where treating it as linear was a reasonable approximation. BUT, if you go to the section on ECONOMIC scenarios, that make it VERY difficult to work the numbers backward to CO2 concentrations, but the log functions are very clear in those sections.

    This subterfuge is very misleading, deliberately so, and I’ve remarked on it often. If someone wants to write a detailed expose, I’ll be happy to go scrounge up all the references even if it means reading the whole d*** thing again.

  156. Dave Springer says:
    March 1, 2011 at 8:06 am
    … If [Ira] paid attention to MY comments you’d have found that image months ago. I have posted it several times including last night on THIS thread.

    http://www.sundogpublishing.com/fig8-2.pdf

    It’s from the 2006 textbook “A First Course in Atmospheric Radiation” by Grant Petty.

    I didn’t know you could get a PhD in Obfuscation & Convolution but you’re living proof that it happens.

    Thanks, Dave Springer for your very helpful expert comments and useful links in this and other topic threads on WUWT. I have learned a lot from reading your postings and I hope you continue to contribute your special knowledge to my topic threads.

    My PhD is in System Science with my dissertation being in Hierarchy Theory (an application of Shannon’s Information Theory to Optimal Span – the “magical number 7 +/- 2″ that may usefully be applied to Quantifying the Mythical Man-Month and Management Span of Control).

    Fortunately, I got my PhD late in my career as a system engineer, after I had done useful work in advanced visionics and automation for military aircraft. Thus, being educated beyond my intelligence did not set in until about 15 years ago, when I was within five years of retirement from my real job. However, the PhD came in time for me to qualify as an Associate Professor and I am currently afflicting students with my knowledge of system engineering in an online graduate course.

    Taking your comment to heart, I promise to eschew obsfuscation and modify all my convolutions into pointwise products of Fourier transforms, if you get my drift :^)

  157. Phil. says:

    Slacko says:
    March 1, 2011 at 6:19 am
    Tim Folkerts says:
    February 28, 2011 at 7:23 pm

    “However, it is an observed fact (supported by theory) that monatomic gases (like argon) or symmetric diatomic gases (like N2 and O2) do not have vibration modes or rotation modes that would allow them to absorb (or emit) IR photons.
    … IR is not absorbed or emitted by N2 & O2.”

    A hot body that won’t radiate infrared?? That’s gotta be the best-kept secret in physics, if it’s true.

    A gas is not a hot body and gases that don’t have a dipole, like N2, O2 and Ar, don’t emit IR, this has been known to physics and chemistry for quite some time. See Herzberg G., ‘Molecular spectra and molecular structure. I. Spectra of diatomic molecules’, New York: Van Nostrand Reinhold, 1950, the classic text on the subject.
    mkelly says:
    March 1, 2011 at 8:44 am
    Dave Springer says:
    March 1, 2011 at 7:53 am
    “Nitrogen doesn’t absorb infrared radiation but it can certainly gain kinetic energy by excited molecules of CO2 and H2O bumping into nitrogen molecules. The kinetic energy gained by the nitrogen molecules will be lost in a continuous blackbody spectrum characteristic of the temperature of the gas.”

    Agreed. My point to Cal earlier. If N2 and O2 could not shed energy (trapping it) then we would have a problem.

    But that leads to a question: what difference is there between the spectrum of N2 and O2 both at the exact same temperature and radiating. I have not found anything on that.

    As stated above absolutely wrong, the only way that N2 and O2 molecules can shed energy is by collisional exchange with CO2, H2O etc. molecules which can then radiate.

    Here’s a low resolution spectrum showing N2, O2 and CO2, I’ve left out H2O for simplicity but it occupies the 5-10 micron region at a similar intensity to CO2 (10^-20). Note that the vertical scale is log base 10!

    http://i302.photobucket.com/albums/nn107/Sprintstar400/CO2N2O2.png

  158. cal says:

    mkelly says:
    March 1, 2011 at 8:44 am
    Dave Springer says:
    March 1, 2011 at 7:53 am
    Nitrogen doesn’t absorb infrared radiation but it can certainly gain kinetic energy by excited molecules of CO2 and H2O bumping into nitrogen molecules. The kinetic energy gained by the nitrogen molecules will be lost in a continuous blackbody spectrum characteristic of the temperature of the gas.

    Agreed. My point to Cal earlier. If N2 and O2 could not shed energy (trapping it) then we would have a problem.

    But that leads to a question: what difference is there between the spectrum of N2 and O2 both at the exact same temperature and radiating. I have not found anything on that.

    OK so please describe in detail the mechanism by which N2 and O2 radiate this energy if it is not by the movement of electrons between different energy states (as a consequence of the vibrational modes of the molecule) which causes the characteristic absorption and emission lines.

    I gave the explanation to your supposed paradox in my original comment. The N2 and O2 molecules gain energy and lose energy by collision with CO2, H2O and other greenhouse gases. There is a small amount of radiation as well but it is negligible in the context of the greenhouse gas emissions. Just look at the emission spectra of N2 and O2 and compare it with the IR spectrum that Ira gives. Further all bodies above absolute zero radiate but the emissivities vary dramatically. A black body is 100% efficient a white body (which cannot really exist) is 0% efficient. N2 and O2 are closer to being white bodies at these temperatures.

  159. Brian H says:

    I thought it was still a matter of dispute whether a gas (the molecules in a gas) behaved/radiated according to BB “laws”. There seem to be very positive opinions being asserted above that hinge on that specific issue.

    BTW Ira, I’ve just gone thru the first part of your novel, and I wonder if you still “believe” the scenarios and causalities for climate behavior and fluctuation you posit and “explain” in Chapter 1. The verbiage circles round and round, but boils down to domination by CO2 changes, with some drastic natural and not-so-natural interventions and intervening events forcing it back down to “normal” levels.

    Which, IMO, is alla buncha hooya.

  160. Domenic says:

    to Phil and Cal

    Nope. The full characteristics of N2 and O2 in the long wavelength IR have never been actually tested.

    I have been looking for long wavelength IR real test data for both N2 and O2 , and cannot find it.

    The only ‘graphs’ out there are calculated from models, not from lots of actual test data. With 20+ years in IR measurement and quantification experience, I don’t trust the graphs out there.

    There are no such things as ‘white bodies’ out there. The opposite of a ‘blackbody’ is a ‘perfect mirror’, not a ‘white body’.

    The only materials that exhibit close to ‘perfect mirror’ thermal radiation properties are pure metals. Highly polished gold, aluminum, etc, or their atoms.

    From my direct IR measurement experiences, materials with nitrides (N, N2, etc) and oxides (O, O2, etc) in their surface composition have very high emissivity, closer to blackbodies. Thus they have very small reflective components. Metal oxides are perfect examples. Bare metals are highly reflective. But let the surface oxidize heavily, and their IR reflection capability disappears…and their emittance approaches a blackbody.

    There is something very fishy in the spectral graphs bandied about from calculations regarding the emittance and reflectance of N2 and O2. They are not metals.

    Both N2 and O2 must be tested thoroughly in the long wavelengths and not guessed at.

  161. Disputin says:

    Many thanks, Ira, for a clear explanation, and I understand that you have only attempted to describe the radiative effects in this post.
    My view is that the term ‘Greenhouse Effect’ is dead right because it describes exactly how a greenhouse does not work. A real greenhouse works just as well with polythene, styrene or, classically, rock salt windows because, although the blocking of long wave IR is real it’s irrelevant compared with the greenhouse’s blocking of the convective currents which are the main mechanism for removal of hot air, directly and by evaporation.
    Similarly, although I’m sure there is a radiative effect which would be active in a static atmosphere, the real atmosphere is anything but static, being thoroughly turbulent and full of convective cells. Radiation thus only comes into effect at the top of the troposphere, where the density of all gases is too low to do much blocking.
    In my view the talk of radiative physics is a waste of time since it’s an extremely minor effect.

  162. Oliver Ramsay says:

    Domenic says:
    March 1, 2011 at 11:43 am

    The only materials that exhibit close to ‘perfect mirror’ thermal radiation properties are pure metals. Highly polished gold, aluminum, etc, or their atoms.

    ————————-
    I have read that variations in emissivity between substances (except for metals) are largely due to the topography of the surfaces. I imagine that must also be true for metals, since the differences between polished and unpolished surfaces are so marked.
    I’m suprised that our coarse rubbing with abrasives is able to produce such a strong effect, even though I have seen it with my own eyes. ???

  163. mkelly says:

    Phil. says:
    March 1, 2011 at 10:15 am

    “As stated above absolutely wrong, the only way that N2 and O2 molecules can shed energy is by collisional exchange with CO2, H2O etc. molecules which can then radiate.”

    Thanks for the information Phil. Your photo while nice does not answer the question of N2 and O2 at the same temperature. Since temperature drives the emittion frequency will they both be the same? I math (Weins Law) says yes, but I have found no info on the subject.

    By the way by your statement above you are in the camp that says without GHG we would be a very hot world indeed. Not at -18 C. Once a N2 or O2 molecule was heated via conduction with the ground it would stay heated for a very long time with no H2O or CO2 to reduce its energy. Correct?

  164. George E. Smith says:

    “”””” Scottish Sceptic says:
    February 28, 2011 at 2:56 pm
    “Other aspects that control the temperature range at the surface of the Earth are at least as important and they include convection (winds, storms, etc.) and precipitation that transfer a great deal of energy from the surface to the higher levels of the Atmosphere.”

    I would suggest CO2 in the atmosphere is much like the “correct” expanation of the Crookes radiometer (the black and white rotating thing in a glass vacuum bulb). “””””

    I believe that the original conception of the Crookes Radiometer, was to demonstrate “Radiation pressure” The Kinetic energy of Photons impinging on a surface.

    The idea was that the blackened surface would absorb the photons, setting up a pressure on the black side of the paddles. The shiny side of the paddles however would not absorb the photons but would reflect them, creating twice the momentum change, so the paddles would rotate, with the shiny side pushing the black side before it.

    Of course it rotates the other way instead; and the reason is that the black side is slightly hotter than the shiny side, and it heats the residual gas molecules close to the surface more than the shiny side does thereby creating a higher pressure on the black side. They couldn’t create a high enough vaccuum to get the radiation pressure to dominate the residual gas differential pressure.

  165. Sensor operator says:

    Wow! There are soo many problems and issues being thrown around that it is nearly impossible to address everything. As someone with a degree in physical science (including physics and astronomy) I guarantee you cannot learn the basics from one page, let alone all of the complexities, e.g. spectroscopy.

    I just want to cover four simple items and see if I can provide other insight later:

    1) Steve at 141 is correct: The nearly absent atmosphere is the reason Mars is so cold. Other comparisons: the Moon vs the Earh (basically the same distance from the Sun but no atmosphere on the Moon) and Mercury vs Venus (Venus is further away but with an extemely dense, CO2 heavy atmosphere which causes Venus to be much hotter than Mercury).

    2) On a large scale (planetary), the method of heat transfer is radiation. There may be convection and conduction within the atmosphere (up to about 10 Km as found by E. O. Hulburt in 1931), but between the Earth (including the atmosphere) and “space”, heat is transferred via radiation since space is a vacuum (you need matter for conduction and convection, also shown by Hulburt in 1931).

    3) CO2 and H2O
    Ugh, this comes up soo often it is getting tiresome. So let’s look at two pieces of information.
    3a) Absorption by components
    Please look at the following picture showing the absorption from CO2 and H2O, as well as other constituents. While water is a broad absorber of long wave IR radiation (greater than 10 microns), it is not complete. But look at CO2. This is very important. There is a major absorption feature between ~13-20 microns which is very good at absorbing LWIR radiation. This is important becasue H2O does not completely absorb the radiation in this wavelength region. So when CO2 levels increase, it further impedes LWIR radiation from escaping the atmosphere. This is even more important since according to Planck’s law, a blackbody at the Earth’s surface temperature will radiate in this wavelength region. While the Earth may not be a blackbody, it does radiate in this spectral region.
    3b) Composition of atmosphere relative to CO2 and H2O
    H2O is primarily limited to being located close to the surface of the Earth and is not well mixed into the entire atmospheric column, unlike CO2. However, when we look at things spectrally, it does not matter if the gas we are looking through is localized in a small area or spread out through a larger area/volume (ignoring clouds of course). It will have the same effect. When we measure gases using spectral instruments, we need to take into account the distance from the source to the sensor. We calculate the amount of the gas we see and divide by the distance and get an average over that distance. Same is true when looking down from a satellite towards the Earth. When we consider the entire column we find:
    H2O: ~0.40%
    CO2: ~0.04%
    While H2O is higher than CO2, it is not orders of magnitude larger. The challenge is that the H2O amount is not changing by 100%. Remember from 3a, CO2 is absorbing outgoing radiation where H2O is not and the amount of CO2 is increasing.

    4) Be careful with these plots and simple explanations. Total absorption and opaque are not quite the same, i.e. if 100% is already absorbed than more CO2 cannot be a problem. The graph at the top of the page suggests that the atmosphere is a single step from Earth to space. In reality, models use multiple levels to account for the differences in gas constituents at various heights.

    Last lesson: Be very careful with LWIR. Most people are okay with near IR and short wave IR since this radiation is identical to visible sunlight, which we are accustomed to as humans since our eyes are optimized for that region. In this part of the spectrum, we talk about reflection and absorption. But when we head towards long wave IR, you have absorption/emission, transmission and reflection. It is a somewhat messier region that can be easily confused. Most people familiar with LWIR come from a lab setting using instruments in a controlled setting. But when you start looking at/using the atmosphere, it becomes a much more difficult problem.

    Hopefully NASA’s Glory satellite may help answer more questions when it finally launches.

  166. Mark Nodine says:

    Thanks, Ira, for an interesting post.

    I think there is an important point, however, on which your analysis is misleading. I agree with George E. Smith (February 28, 2011 at 3:23 pm) that the IR absorption occurs by exciting the vibrational modes of GHGs. The reason that N2 and O2 and H2 do not absorb IR is not that they have no dipole (since neither does CO2), but because, being diatomic, they have no asymmetric vibrational modes, which are the ones that absorb in the IR range. George is right in saying that the vast majority of the absorbed light rays are immediately reradiated and do not have time to affect the thermal distribution of the surrounding gases. The reradiating becomes even more preponderant the farther up in the atmosphere you get because of the decreasing density of gases with height. So Ira’s assertion that GHGs exert a large thermal effect on the atmosphere seems to me to be a bit unlikely, except possibly in the lower troposphere. The other thing to consider is that where thermal transfer does take place, any blackbody radiation that occurs will be at the new temperature rather than at the original temperature. Because of adiabatic lapse (think PV=nRT), the temperature of the atmosphere drops rather quickly as a function of height until you get up to the thermosphere, where there is a large rise in temperature, but so rarefied an atmosphere as not to have any significant impact on the atmospheric windows. So even assuming you could model radiation from a layer of atmosphere as if it were a black body (which it is not), the distribution is shifted toward yet longer wavelengths than are radiated by the earth.

    My main point is that my substantial physics training leads me to believe that the main impact of GHGs is vibrational excitation followed by return to the ground state, emitting a photon of the same wavelength but in a random direction, rather than by any significant transfer of internal vibrational energy to thermal energy of surrounding molecules.

  167. Domenic says:

    Hi Oliver,

    I like these discussions because they have been encouraging me to remember my IR skill set.

    Yes, it’s a surface condition effect. And it is quite dramatic.

    But what I suddenly realized, from this discussion is something that I had not thought of before.

    The stars. Outer space.

    Do you realize that it impossible from earth to know whether all the ‘stars’ that are seen in the sky, or from space, are indeed really ‘stars’ and not ‘reflections’ from massive pure metallic bodies, or perfect reflectors, of nearby true stars?

    It is literally impossible to know from a distance. Impossible.

    Now that is wild.

    I have never seen that idea looked at in astrophysics.

    But from a radiational physics point of view, it is entirely possible.

  168. Richard Smith says:

    There is confusion on this thread about the meaning of the term ‘greenhouse effect’. Some of the commenters seem to think that it is merely something that reduces the rate of heat loss (and they draw analogies with blankets and igloos which nobody disputes). But the greenhouse effect according to climate ‘science’ is not the insulation of a blanket or a reduction in the rate of cooling, but a positive addition of heat (33C or more) because of ‘backradiation’. If this theory were true it would mean that the Earth emits more energy than it receives. This is a contravention of the laws of thermodynamics. It is based on the fallacy that if the escape of heating is blocked, then the temperature will continue to rise until a ‘radiative equilibrium’ is reached and the heat bursts through the barrier. It is claimed that the laws of thermodynamics are not broken because the emission at the top of the atmosphere is equal to the incoming solar radiation. But, like the theory of the runaway greenhouse, it is a fallacy. If it were true we could generate huge amounts of energy from a small input simply by placing an infra-red barrier around a radiator – just tap some of the heat off at regular intervals – free energy.

  169. George E. Smith says:

    “”””” _Jim says:
    February 28, 2011 at 3:45 pm
    George E. Smith February 28, 2011 at 3:23 pm :

    So the atmosphere does radiate a black body like thermal spectrum, and the presence of the gHG molecules simply means that there will be bands of that continuum emission, that are also captured by the GHG molecules, as well as the emissions from the surface.

    I’ll wait to see how this is adjudged.

    (IR Spectroscopy explaining spectral response regarding gas molecule vibrational modes would seem to indicate otherwise.) “””””

    Jim needs to take remedial English Reading courses. IR Spectroscopy indicates nothing of the sort.

    The IR Spectroscopy item you cited, refers to ABSORPTION SPECTROSCOPY which as it properly indicates, relates to RESONANCE MODES which are indeed characteristic of specific molecules, and their excited state energy levels.

    No where did I make any statement that is inconsistent with this; it is the reason we talk about the “15 micron” or 666 cm^-1 degenerate bending mode of the linear CO2 molecule. It is degenerate since there are two identical and indistinguishable modes at right angles to each other; each of which can gather its own photon.

    But the question which you chose to challenge, is the quite different THERMAL SPECTRUM EMISSION; which has nothing whatsoever to do with atomic or molecular structure or energy levels. It is a classical physics consequence of the accelerations of electric charge; which according to Maxwell’s equations must (and do) result in the radiation of an EM wave (or photon if you wish; which has a continuous energy spectrum; not a quantized one. Ordinary atoms, once ionized by the removal of one or more electrons, also emit a continuum EM spectrum, which is not quantized, because a newly captured free electron can have any initial energy before being captured by the ion.

    The accelerations of Electric charge in molecular collisions, are entirely unpredictable as the the redistribution of momentum and energy; although one can talk statistically about expected distributions.

    One can learn a thing of two from the Particle Accelerator Physicists. The Stanford two mile long electron linear accelerator is one of if not the most powerful electron accelerators in the world. It is linear, because then the only acceleration the electrons see, is due to the accelerating electric (traveling) waves in the beam line. On the other hand circular track accelerators, have to steer the beam around corners continuously to get them to repass through the accelerator gaps, and gain more energy. To compensate for this increased energy, the bending magnet fields have to increase synchronously with the particle energy, to maintain them in the correct orbits. Since the charged particles are traveling in a circle, they are subject to a continuous acceleration, since acceleration is a change in velocity and velocity is a vector quantity, having a direction as well as a “speed”.

    So particles in a circular accelerator continuously radiate, EM energy as the go around the track. Well pedantically, the path is a number of curved arcs, interleaved with a number of straight sections, and it is only in the arced sections that the particles radiate.

    So the reason why proton beams can be accelerated in a circle, and electrons can’t (beyond certain energies, is that the proton is far more massive than the electron.

    So consider two neutral atoms (or molecules) colliding. Each molecule or atom is electrically neutral, so the electrons will go more or less, where the protons in the nucleus goes (on average), but in the process of collisions and recoil of the particles, the more massive nuclei consisting of Protons, and Neutrons, undergo far lower accelerations, than do the much lighter electrons; so even though and single atom or molecular is electrically neutral, in collision, the electrons undergo much higher accelerations than do the Protons or nuclei; so just as in the particle accelerator, it is the electrons that radiate the EM wave energy, and the much smaller opposing field due to the proton accelerations does not compensate. So one can practically ignore the proton positive charges, and consider only the acceleration of the net electron charge to figure out the radiated EM fields.

    The thermal radiation continuum depends only on the Temperature, and has a BB like spectrum, that does not depend on the atomic or molecular species; only on the Temperature.

    Some of you people need to get your noses out of facebook, and wiki, and hit some Physics books, to learn what you should have learned around the 8th grade.

    Nothing in the IR Spectroscopy paper you cited is inconsistent with that; they talk ONLY about absorption spectra; not emission.

    And Kirchoff’s law applies only in equilibrium, and the atmosphere is never in equilibrium.

  170. George E. Smith says:

    “”””” Tim Folkerts says:
    March 1, 2011 at 9:37 am
    Dave Springer says: March 1, 2011 at 7:53 am
    Enough already with the argument that certain gases that don’t absorb infrared can’t emit it.

    At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity.
    http://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation “””””

    “”””” At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity. “””””

    “””””At thermal equilibrium, ……. “””””

    Can people stop citing Kirchoff’s law, and then TOTALLY ignoring it (spectral) emmissivity equals (spectral) absoptance, ONLY in “THERMAL EQUILIBRIUM ”

    Earth’s atmosphere is NEVER in thermal Equilibrium; it has a continuous energy input from somewhere else, and a continuous energy output to someplace else.

    So Kirchoff’s Law NEVER applies to earth atmospheric situations; EMISSIVITY NEVER EQUALS ABSORPTANCE.

  171. kbray in california says:

    [[[ Ira Glickstein, PhD says:
    March 1, 2011 at 6:01 am

    ...Your analogy of the black balls in the pool does not work because the mechanism is not at all like that of the “greenhouse effect” GHG in the atmosphere. There is more H2O (the main GHG) in the Atmosphere than CO2, but the percentage of H2O is also miniscule, yet, the tiny percentage of GHGs (mostly H2O but also CO2) are responsible for the Earth being at livable temperatures. ]]]

    OK Ira,

    I agree that Earth’s Atmosphere helps make our Earth livable. I think water in all three states, solid, liquid, and gas, clearly plays the major role. The alleged “CO2 effect” is critical that it be accurately quantified, as it is being considered a “problem” and will cost us all dearly (money) if current politicians impose their planned restrictions.

    I want to shrink the “alleged CO2 effect” to a simple provable visual experiment that the average layperson can easily absorb. Some of the material presented in this wonderful blog is quite advanced and beyond the expertise of many lay readers.

    How about this…..

    1) Chill 1 bottle of distilled water and 1 bottle of CO2 injected mineral water in the refrigerator (to keep the CO2 in solution).
    2) Find 2 identical glass or plastic bottles with a screw top that seals.
    3) Place a thermometer upsidedown in each bottle so the bulb won’t be in the liquid.
    4) Fill each bottle half way with identical volumes, one with distilled water, the other with the CO2 impregnated mineral water.
    5) Secure the cap and place in full sun on the window sill.
    6) Keep room temperature just slightly below the thermometer temperature as it changes.
    7) If CO2 really does have a big influence in keeping Earth warmer, at some point in the experiment either during the day, or at sunset, or in the evening, or at night, the air above the mineral water should be warmer.

    Both will have a mix of normal atmosphere and water vapor, but the mineral water will have released additional CO2 into its “atmosphere”, boosting its “greenhouse effect” that will keep retained/reflected heat in its sealed environment.

    Remember to keep the room temperature just slightly below the bottle temperature to keep any heat transfer influence from the room air at a minimum.

    If there is a “CO2 effect” it should show up here on the thermometers.

    How about this one Ira?

  172. izen says:

    @-mkelly says:
    March 1, 2011 at 12:24 pm
    “By the way by your statement above you are in the camp that says without GHG we would be a very hot world indeed. Not at -18 C. Once a N2 or O2 molecule was heated via conduction with the ground it would stay heated for a very long time with no H2O or CO2 to reduce its energy. Correct?”

    If I may reply as someone who would agree with Phil’s spectralcalc picture of the emissivity of N2 and O2 as several orders of magnitude smaller than that of CO2 for present surface temperature. I understand that such values are calculated from quantum mechanical first principles, perhaps those that doubt them can find direct measurements that contradict that. But the research on radiative transfer carried out in connection with heat sensor/seeking systems for military purposes would seem to make it unlikely that any such major error has gone unnoticed.

    Without GHG we would not be hotter because the N2 and O2 atmosphere has a lower emissivity. They could only warm by conduction with the surface to a temperature equal to the surface. Without GHG the surface would lose energy by IR radiation faster because there is no downwelling or back radiation from the atmosphere. Therefore it would be colder, and could not heat up the N2 and O2 molecules to any greater temperature than it had reached.

  173. George E. Smith says:

    “”””” Mark Nodine says:
    March 1, 2011 at 12:35 pm
    Thanks, Ira, for an interesting post.

    I think there is an important point, however, on which your analysis is misleading. I agree with George E. Smith (February 28, 2011 at 3:23 pm) that the IR absorption occurs by exciting the vibrational modes of GHGs. The reason that N2 and O2 and H2 do not absorb IR is not that they have no dipole (since neither does CO2), but because, being diatomic, they have no asymmetric vibrational modes, which are the ones that absorb in the IR range. George is right in saying that the vast majority of the absorbed light rays are immediately reradiated and do not have time to affect the thermal distribution of the surrounding gases. “””””

    Well Mark, you were doing ok until you said this:- “”””” George is right in saying that the vast majority of the absorbed light rays are immediately reradiated and do not have time to affect the thermal distribution of the surrounding gases. “””””

    Because absolutely never in my life have I ever said any such thing; the exact opposite in fact.

    The CO2 (or other GHG molecule) certainly absorbs specific LWIR energies, according to the energy levels of its excited molecular states; ” BUT ” in the lower atmosphere it NEVER (hardly ever) has enough time to re-radiate that absorbed Photon, at the same energy and frequency (wavelength). Molecular collisions happen many thousand times faster, than spontaneous emission at specific (CO2) frequencies. The energy is lost in thermalization collisions, which warm the ordinary gases of the atmosphere. From that time on, the GHG gas molecule has no influence whatsoever on what happens. The (slightly heated) atmospehric gases now radiate a complete thermal spectrum continuum, depending ONLY on the atmospheric Temperature; AND THEN the CO2 or other GHG molecule can re-engage with THAT spectrum, and once again carve out its specific absorption band chunk.

    The proof of this is in the satellite (and calculated) LWIR spectra of the earth. They are BLACK BODY LIKE SPECTRA with the GHG absorption band partly missing (attenuated) by the GHG molecules.

    If the only LWIR emission from the atmopshere was from the GHG molecules, itwould be ONLY at th4e specific GHG resoanance absorption band frequencies; and it is not; it is a full BB like spectrum.

    You have to note that the actual spectrum is the overlaying Temperature BB spectrum, MODIFIED by the spectral radiant emissivity of the radiating material.

    Since a thin gas layer, has few molecules, it has a correspondingly low emissivity.

    A 50 Angstom gold film, also has a correspondingly low emissivity; but of course higher than the gas layer, because it still has a greater atomic or molecular density.

    There is nothing in the phase transition from solid ice, or liquid water, to H2O vapor by sublimation or evaporation; that suddenly turns off the ability of those molecules to emit a BB like thermal continuum radiation spectrum according to Planck formula and other applicable laws (don’t forget the emissivity).

    But I have never said the the atmospheric LWIR spectrum consists of the GHG resonance absorption frequewncies only.

    That condition is approached in the stratosphere, when molecular mean free paths, and intercollision times become long compared to the lifetimes of the excited states. Then the spontaneous emission can occur; but the molecular density is so low, that it cannot be a significant contributor to the total earth LWIR emission spectrum.

  174. George E. Smith says:

    “”””” izen says:
    March 1, 2011 at 1:31 pm
    @-mkelly says:
    March 1, 2011 at 12:24 pm
    “By the way by your statement above you are in the camp that says without GHG we would be a very hot world indeed. Not at -18 C. Once a N2 or O2 molecule was heated via conduction with the ground it would stay heated for a very long time with no H2O or CO2 to reduce its energy. Correct?”

    If I may reply as someone who would agree with Phil’s spectralcalc picture of the emissivity of N2 and O2 as several orders of magnitude smaller than that of CO2 for present surface temperature. I understand that such values are calculated from quantum mechanical first principles, perhaps those that doubt them can find direct measurements that contradict that. “””””

    Those specracalc images that Phil posted, are ABSORPTION SPECTRA; NOT EMISSION SPECTRA.

  175. Domenic says:

    to Izen

    you wrote: “But the research on radiative transfer carried out in connection with heat sensor/seeking systems for military purposes would seem to make it unlikely that any such major error has gone unnoticed.”

    Heat seeking systems are ridiculously simply compared to accurate quantification and complete summation of thermal radiation properties for specific gases.

    Heat seeking sensor systems only need a ‘window’ or two, or three to locate a target.

    Heat seeking sensor systems are qualitative in nature, not quantitative. The military doesn’t care much about the details. They just want to find and lock onto the target.

    If you had any experience with quantifiable IR, you would know this.

    And that is one of the biggest problems. Much research about atmospheric IR has been done by the military in order to locate and track heat signatures. But that data is really of minimal use when absolute quantification is necessary for establishing real world temperature effects of gases in the atmosphere regarding ‘greenhouse effects’.

  176. kbray in california says:

    [[[Ira said:

    “Mars is a lot further from the Sun than is the Earth, which is the main reasion it is too cold to support life as we know it, despite all the CO2 in its Atmosphere.”

    Steve says:
    March 1, 2011 at 7:38 am

    No, Mars isn’t at a “beyond supporting life” distance from the sun. The Martian atmosphere would be warm enough to support life if it wasn’t so thin , at about 1% the density of Earth’s atmosphere. The density of the atmosphere ties in directly with it’s heat capacity. Mars has little to no magnetosphere, so the solar wind stripped much of the atmosphere away over billions of years. ]]]

    kbray asks: How much atmosphere on Mars do you need to produce the “runaway greenhouse effect”(think Hansen) to get Venus-like temperatures ? Mars already seems to have enough CO2 in the atmosphere being measured in the high 90%. Maybe at high concentrations CO2 has a “protective effect” and actually reverses the greenhouse effect. After all, as per Hansen’s ideas, it should be hotter than blazes on Mars shouldn’t it ?… Maybe we can reverse “global warming” by producing MORE CO2 ? /sarc.

  177. George E. Smith says:

    “”””” Dave Springer says:
    March 1, 2011 at 7:53 am
    Enough already with the argument that certain gases that don’t absorb infrared can’t emit it. This is not true for dense gases. The troposphere is a mixture of cold dense gases. Please review Kirchoff’s Laws:

    http://www.physics.rutgers.edu/~matilsky/documents/kirchoff.html

    Radiation Laws

    Kirchoff’s Laws

    First Law: A hot solid, liquid, or dense gas emits radiation at all wavelengths (“a continuous spectrum of radiation”). For example, a perfect blackbody does this. If the light were passed through a prism, you would see the whole rainbow of colors in a continuous band.

    Second Law: A thin hot gas in front of a cooler background emits radiation at a discrete set of isolated wavelengths. These discrete, isolated wavelengths are called the “emission lines” of the spectrum, because if you were to pass the radiation through a prism, you would see isolated lines of different colors. The whole spectrum is called an “emission-line” spectrum. The wavelengths of the emission lines are unique to the type of neutral atom or ionized atom that is producing the emission lines.

    Third Law: A thin cool gas in front of a hotter solid, liquid, or dense-gas background removes the radiation from the background source at special wave lengths. If the resulting radiation were passed through a prism, there would be dark lines superimposed on the continuous band of colors due to the background. These dark lines are called “absorption lines.” The wavelengths of the absorption lines are unique to the type of neutral atom or ionized atom that is producing the emission lines.

    If a certain type of gas produces absorption lines at certain wavelengths when it is in front of a hot background, then when that same type of gas is seen in front of a cooler background, it produces emission lines at the exact same wavelengths.

    Explanation of Kirchoff’s First Law

    Kirchoff’s First Law boils down to blackbody radiation, since solid objects and dense gases emit radiation like blackbodies.

    Explanation of Kirchoff’s Second and Third Laws

    Thin gases don’t emit or absorb radiation like blackbodies. To understand their emission and absorption, we must consider the structure of atoms, as described by Quantum Mechanics. “””””

    Well I found this citation from Dave Springer; about “Kirchoff’s Laws”.

    Would people please read what Rutgers University says about those laws; and please take note of all of the words, in their explanation.

    Notice their reference to THIN HOT GASES and THIN COLD GASES. Note the “”””” THIN “””””

    Those nice Balmer spectra and the like from Mercury vapor lamps or Sodium lamps and the like are observed in VERY LOW DENSITY gases, where the mean time between molecular or atomic collisions is much longer than the lifetimes of the excited states, so that spontaneous emission can occur; only then do you get the bright line spectra as seen in those HOT gases.

    But the earth’s lower troposphere is neither low density nor hot; it is 288 K on average worldwide (so they tell us), and the molecular collision rates are thousands of times faster than the excited state lifetimes, so spontaneous emission almost never occurs in the lower troposphere.

    But the molecular velocities of thermal motion of atoms or molecules; and especially molecules being more massive, are way lower than the vibrational mode velocites of the excited state resonances, so the resulting electron accelerations in intermolecular collisions, are much lower, hence the magnitude of the thermal radiation fields is way down in the mud, which is why it is so hard to observe.

    But the fact remains, that the spectrum which is supplying the GHG resonance energies , is akin to that radiated by an ordinary yard brick or a bottle of water, at room Temperature. That radiation is 10,000 times lower in radiant emittance, than the radiation form an ordinary 100 Watt “heat lamp”, that emits radiation that is absorbed by human tissue and registers itself as what we call heat; and it does so, because we are mostly water, and absorb strongly at the 1 micron wavelength of the heat lamp emissions.

    Humans do NOT perceive 10 micron 288 K radiation as producing heat, because it is only 400 W/m^2, and not 4 million W/m^2 that the heat lamp emits.

    But jsut because we don’t sense it, does not mean the atmospheric gases are not emitting it.

  178. RJ says:

    Izen

    A good question. How would O2 and nitrogen lose energy if there were no GHGs in the atmosphere.

    I assume this means that nitrogen and oxygen molecules can in fact lose energy by conduction (that mostly occurs at lower attitudes) or by radiating IR energy out into space at high attitudes.

  179. Vince Causey says:

    Richard Smith,

    You say “But the greenhouse effect according to climate ‘science’ is not the insulation of a blanket or a reduction in the rate of cooling, but a positive addition of heat (33C or more) because of ‘backradiation’. If this theory were true it would mean that the Earth emits more energy than it receives. This is a contravention of the laws of thermodynamics.”

    Surely, if temperatures are rising, (as posited by the GHG hypothesis) it must mean the exact opposite – namely that the Earth emits LESS energy that it receives. Therefore this is not a contravention of the laws of thermodynamics as you assert.

    You then say “It is based on the fallacy that if the escape of heating is blocked, then the temperature will continue to rise until a ‘radiative equilibrium’ is reached . . ”

    If a proportion of escaping IR is radiated back down to Earth then temperatures must increase. As this happens, then more energy is radiated from the Earth at a flux density that is proportional to the fourth power of the increase in temperature. The idea you suggest where “the heat bursts through the barrier,” although meant to be a derisive dismisal of the GHG mechanism, actually misunderstands the issue. It is a simple application of the Stefan-Boltzman equation for blackbody radiation that gives an increase in radiative emissions from Earth, until it once again equals the radiation coming into the Earth.

    Your final remark “If it were true we could generate huge amounts of energy from a small input simply by placing an infra-red barrier around a radiator – just tap some of the heat off at regular intervals – free energy,” is completely false. There is nothing about a mechanism that radiates energy back to the Earth and resulting in a temperature increase, that lends itself to the notion that this can in any way lead to free energy. If you try to drain energy from the atmosphere all that would happen is the atmosphere would cool down and the Earth would radiate faster and cool down more quickly. I suggest the same thing will happen with your ‘infra-red barrier around a radiator.’

  180. Bryan says:

    George E. Smith

    Did the classical EM radiation laws of Maxwell not run up against the ultraviolet catastrophe and hence the modification by Panck and Einstein introducing the quantised photon?
    I have always understood that continuous spectra are associated with radiating solids.
    For gases that radiate a line spectra is what I’d expect.

    You can flick between absorption and emission spectra for H2O
    http://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&Units=SI&Type=IR-SPEC&Index=1#IR-SPEC

    Same also for CO2
    http://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Units=SI&Type=IR-SPEC&Index=1#IR-SPEC

  181. Tim Folkerts says:

    kbray in california says: March 1, 2011 at 1:23 pm

    How about this…..
    1) Chill 1 bottle of distilled water and 1 bottle of CO2 injected mineral water …

    One problems with this experiment is that it is missing “outer space”. CO2 (or H2O) limits the ability of the IR to “escape” from a warm area (the earth) to a cool area (the rest of space). Without the cold surroundings, you really can’t model the “greenhouse effect”.

    You really need:
    1) a warm object (perhaps with a light shinning onto it)
    2) a cold enclosure around the object (so that the net IR flux will flow from the object to the enclosure). Perhaps a deep freeze.
    3) controllable atmosphere (CO2, humid air, dry air etc) between the two that is significantly warmer than the walls of the enclosure.

    If the greenhouse effect is true, CO2 or H2O would be radiating IR back to the object. The object will cool more slowly (controlling of course for conduction & convection).

    Even this will be tough to see because it takes many meters of CO2 to absorb/emit much IR.

    (hmmm … as a side thought, does anyone know just how far IR of various wavelengths would penetrate thru CO2? Eg how far it takes to attenuate 50% or 99% the intensity of the IR light? For 400 ppm it clearly takes less than the height of the atmosphere, but how much less?)

  182. RJ says:

    Vince

    I still can not see how a small amount of CO2 can heat a much larger warmer earth. Its a bit like a very small heater heating a massive room for one. And secondly I do not accept that a colder body can heat a warmer one.

    If a person was enclosed in a container of CO2 would their temperature rise due to the greenhouse effect. Or would the CO2 just slightly reduce the rate of cooling. I would say it would only reduce the rate of cooling. Is this correct or not a very good example as a comparison.

  183. Richard Smith says:

    Vince Causey said in response to my comment:

    “Surely, if temperatures are rising, (as posited by the GHG hypothesis) it must mean the exact opposite – namely that the Earth emits LESS energy that it receives. Therefore this is not a contravention of the laws of thermodynamics as you assert.”

    No. Look at the Kiehl and Trenberth global energy flows diagram, for example. Earth receives 64wm2 (after deduction of thermals and evapotranspiration from insolation of 161 wm2) which somehow backradiation magnifies to an emission of 396wm2. This is more energy out than in.

    Vince then says:
    “There is nothing about a mechanism that radiates energy back to the Earth and resulting in a temperature increase that lends itself to the notion that this can in any way lead to free energy. If you try to drain energy from the atmosphere all that would happen is the atmosphere would cool down and the Earth would radiate faster and cool down more quickly. I suggest the same thing will happen with your ‘infra-red barrier around a radiator.’”

    Well according to the radiation equilibrium theory, I could get a huge amount of energy with an infra-red barrier. But if that example does not impress you, how about a furnace with an infra-red barrier. A much cheaper way of melting steel. Why is it that nobody has patented this yet?

  184. kbray in california says:

    [[[ Tim Folkerts says:
    March 1, 2011 at 2:38 pm
    kbray in california says: March 1, 2011 at 1:23 pm ]]]

    Tim,
    Have you ever put a sheet or glass in front of an Infra Red heater ? The IR goes right through the glass. The heat will escape through the glass. And the cold? The room is purposely kept colder in the experiment to simulate the colder surroundings of outer space.

    Increased heat should still show up in the mineral water bottle if CO2 has that ability.

    Your model is unlikely to create any meaningful data, it has too many variables and just the staging of your experiment sounds impossible… besides, I am only testing CO2 not the whole shabang. I go by KISS – keep it simple son.

  185. Myrrh says:

    Tim Folkerts says:
    March 1, 2011 at 5:55 am

    Myrrh says: March 1, 2011 at 2:45 am
    “Ira says that Visible is the largest part of the Sun’s spectrum, wrong, it’s actually the smallest, and compared with the others, very very small. Light is not hot, it is Reflective not Absorptive, and your use of “absorbed by matter of one form or the other” is misleading. The radiation that penetrates to heat the Earth is Thermal IR. And that is not in the Chart.”

    The issue here is the “size of the energy”, not the wavelength. There is a handy table at http://en.wikipedia.org/wiki/Planck%27s_law that shows the energy distribution of sunlight (above the atmosphere).
    * A little over 10% of the ENERGY is in ultraviolet photons or shorter wavelengths.
    *Close to 40% of the ENERGY is visible light photons
    *Over 40% is “near IR” (from 0.7-3um). (So Ira’s claim that visible is the biggest is not quite right, but it is not far off. On the other hand, the atmosphere is still mostly transparent to these wavelenghts, so his arguments still hold).
    *Only a few % is above 3 um (“thermal IR”) so it cannot possibly be the main source for heating the earth.

    It doesn’t show any such thing, missing bits. See http://ds9ssl.berkeley.edu/lws_gems/e/espec.htm

    I think I’m going to start calling this AGWplanks law since its use has resulted in that Chart which made incoming Thermal IR from the Sun walk the plank out of our physical Earth and it’s now become practically impossible to find anything rational about this subject. Visible light is a tiny sliver of the electromagnetic spectrum, as you can see from the above link, it may be highly energetic, but it’s not the highest of that either.

    Planck’s law “In physics, Planck’s law describes the spectral radiance of electromagnetic radiation at all wavelengths from a black body at temperature T. As a function of frequency v, Planck’s law is written as: …….Note also that the two functions have different units – the first is radiance per unit frequency interval while the second is radiance per unit wavelength interval.” http:www.worldlingo.com/ma/enwiki/en/Planck’s_law

    All Planck’s law shows as graphs depict is that the hotter an object the higher the peak in Visible light. How does the Radiance peak here mean it has a greater AMOUNT of the ENERGY of that produced by a heated body?

    That’s how steel workers can tell the temperature of steel they’re working on, from the degree of radiance as it goes into colour. It doesn’t mean anything else. It takes very hot to produce white light. The Sun’s heat certainly hot enough to produce white light.., but it’s a peak of radiance into Visible, it hasn’t stopped producing Thermal Infrared, nor has it stopped producing any of the other wavelengths which you haven’t included.

    http://m.plantengineering.com/index.php?id=2831&tx_ttnews%5Btt_news%5D=33209&cHash=db4db9479b

    “Not as widely recognized is the fact that incandescent objects emit a tremendous amount of invisible infrared radiation. For example, the radiance of a steel billet at 1500F is 100,000 times greater in the infrared spectrum than in the visible spectrum.”

    The real “peak amount” then between the Visible and Thermal IR is not in the Visible. Planck’s peak frequency into visible does not equal peak amount given off by a heated object.

    A bog standard lightbulb gives 95% of its energy in Heat, which is Thermal Infrared (Near Infrared included in “Solar”, is not hot, it’s a cold light.) If you drew a graph showing this relationship, where would the peak be?

    ……………………………………..

    izen says:
    March 1, 2011 at 6:45 am

    Re: the same quote of mine, above.

    The vast majority of the energy direct from the Sun that heats the Earth is in the visible spectrum, NOT the IR. If you doubt that visible light can heat things consider the car on a sunny day. The glass of the windows blocks IR but lets through visible light that is absorbed by the (usually) dark plastic/leather interior. The hot steering wheel and seat are warmed by the visible light not any IR.

    Glass does not block IR. Visible light is REFLECTIVE. It bounces off things. IR is ABSORPTIVE in organic stuff, the car is absorbing IR Heat all over.

    You are correct of course that the surface is ALSO heated by the 7um and 15um IR emissions from the H2O and CO2 in the atmosphere that has been warmed by the IR emissions from the surface..

    Well you have to say that don’t you, because you’re following AGWScience which can make up any number of impossible things for you to believe before breakfast. The heat we feel from the Sun, DIRECTLY, is Thermal IR. Regardless of whatever else amount is coming to us, from the heated things around us for example, to exclude this direct Thermal IR from the Sun is just plain ludicrous. I’m finding it impossible to phrase it any other way.

    …………………………………………..

    Ira Glickstein, PhD says:
    March 1, 2011 at 6:11 am

    Re: Myrrh says: I’m really at a loss to understand any of this. How on earth does Visible light and near short wave heat the Earth?

    Myrrh, you really need to get outside more and sit in the Sunshine and feel the warmth! That is how visible and near-visible (“shortwave”) light warms the Earth.

    If you don’t or cannot get outside, turn on an old-fashioned incandescent light bulb and hold your hand near it (not too close you will get burned). Feel the heat? That is shortwave light because the filament is heated to temperatures similar to the Sun’s surface. You can tell it is shortwave because you can see the light.

    Well, again with the intellectual disdain from the so-calling themselves experts..

    Ira, please take note of my replies above. I am determined to get to the bottom of this because this is a KEY premise on which AGW energy is based and all other energy arguments depend, and it’s simply nonsense.

    For lightbulb information – http://www.newton.dep.anl.gov/askasci/eng99/eng99505.htm

    & especially:

    Physics 101: http://science.hq.nasa.gov/kids/imagers/ems/infrared.html

    Infrared light lies between the visible and microwave portions of the electromagentic spectrum. …The longer, far infrared wavelengths are about the size of a pin head and the shorter, near infrared ones are the size of cells, or are microscopic.

    Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared. The temperature-sensitive nerve endings in our skin can detect the difference between inside body temperature and outside skin temperature.

    Shorter, near infrared waves are not hot at all – in fact you canot even feel them. These shorter wavelengths are the ones used by your TV’s remote control.”

    So, WE CANNOT FEEL VISIBLE LIGHT – it is not hot. It does not carry HEAT. The Radiation we feel from the direct from the Sun which is HOT, is Thermal IR.

    You have excluded this in your Chart. We can all feel you’re wrong to do so..

    This AGW premise defies REAL Science, it is already FALSIFIED. Nothing that AGW calculates thereafter from this premise has any credibility whatsoever.

    Has it?

    Now, as I said, I’m finding it really difficult to find anything discussing this sensibly at all, most of what’s on line has been so thoroughly trashed by AGWScience propaganda in its utter mangling of basic science concepts. Here it is attributing to Visible light and near neighbours (Solar of AGW), that which is rightly only applicable to Thermal IR.

    But, I did find this just now: http://ishangobones.com/?p=509

    Which appears to be explaining the actual process I’ve been thinking of as Reflective v Absorptive.

    Thermal IR’s absorption method I’ve seen described as Conversion in those areas which deal with its absorptive energies, capacities, spectroscopy, medical and such, (as in a heat pad delivering Thermal IR to aching muscles, etc.), but I wasn’t concentrating on it, and I seem to recall a description similar to what’s being said here. Encyclopedia of planetary sciences says “Absorption is the conversion of electromagnetic energy to another form of energy through interaction with matter. The process can be viewed at several levels, but the end result is usually heat.”

    Anyway, Thermal IR is absorptive and Visible and Near IR are not, but are reflective in all general descriptions of the difference.

  186. wayne says:

    Al Tekhasski says:
    March 1, 2011 at 12:01 am

    I am sorry, but for the sake of accuracy, CO2 in air does not have any states that interact with 10-13um band. Therefore, hot air cannot possibly generate any photons in that range. Therefore the following statement is also wrong, nothing gets “transferred” into 10-um band from “heated air”:

    —-

    So you are saying that if you had a large clump of warm nitrogen in empty space held together by gravity that it could not radiate in a black body manner at all, zero, none, zip? Maybe check some good astronomy books on that matter that cover quantum aspects.

    Its radiation as a black body would be much lower (low emissivity at a frequency) than the few large emissivity emission lines in N2’s spectrum but the much smaller BB radiation would still be there per the temperature. And you cannot have a temperature without there being many, many molecules or atoms (LTE). All matter (not a single isolated molecule) radiates in a black body manner except at zero K. Now if you are talking of one isolated warm N2 molecule moving isolated in empty space you would be correct, it could not radiate except at it’s emission lines and once it reached it’s ground electron, vibration and rotational state it could not radiate at all, zero, zip, none. It is the collisions and interactions between multiple molecules (between the electrons really) that create the much smaller BB background radiation. That is how I see it from my years in studying physics.

    Do you see, Ira placed no numbers on exactly how much was split between those 7, 10, and 15 µm band portions. Just that those interactions do occur.

    The same is for CO2 molecules or any atoms or molecules when not in isolation (much like in the solar wind of particles or inter-galactic gases).

    If you have some good books or references countering this please leave them in a comment before leaving.

  187. Dave Springer says:

    George E. Smith says:
    March 1, 2011 at 2:08 pm

    Thanks for the further explanation between cold/hot and thick/thin gases. The Rutgers statement of Kirchoff’s First Law of Radiation was not clear about the distinction and I thought further clarification would probably be needed.

    Cold dense gases emit a continuous blackbody spectrum characteristic of their temperature. “Cold” means below the ionization energy threshold. “Thin” is a little more ambiguous but your description was good i.e. thin enough so that collisions are rare. In the earth’s atmosphere “thin” would begin beyond the mesopause at 90 kilometers where the temperature starts increasing with altitude and can rise to thousands of degrees F.

  188. Dave Springer says:

    CO2 IS INSULATION.

    IT DOES NOT HEAT ANYTHING. IT SLOWS DOWN HOW FAST THE SURFACE COOLS.

    I WONDER IF ALL CAPS WILL HELP. GOD KNOWS I’VE TRIED EVERYTHING ELSE TO GET THIS SIMPLE CONCEPT ACROSS.

  189. Dave Springer says:

    Tim Folkerts says:
    March 1, 2011 at 2:38 pm

    “For 400 ppm it clearly takes less than the height of the atmosphere, but how much less?”

    APPROXIMATELY 12,000 FEET OVER THE ARCTIC OCEAN. SEE THE GRAPH OF IR SPECTRUM FROM 20,000 FEET LOOKING DOWN THAT IRA POSTED IN THE COMMENTS TODAY.

    IN THE IR WINDOW YOU SEE A 265K BLACKBODY CURVE. IN THE 15UM CO2 ABSORPTION WINDOW YOU SEE A 225K BLACKBODY CURVE. THIS IS A TEMPERATURE DROP OF 40K. DRY ADIABATIC LAPSE RATE IS 1K PER 100 METERS SO THE POINT WHERE THE SPECTROMETER NO LONGER INDICATES ANY 15UM ABSORPTION IS 4000 METERS.

    THIS WILL VARY SOMEWHAT DEPENDING ON THE TEMPERATURE AND ALTITUDE OF THE SURFACE. AT HIGHER SURFACE ALTITUDES THERE IS LESS TOTAL CO2 IN THE COLUMN ABOVE IT AND IT WILL TAKE MORE THAN 12,000 FEET OF AIR TO GIT ‘ER DONE. THE TEMPERATURE OF THE SURFACE WILL SHIFT THE PEAK SURFACE EMISSION FREQUENCY AWAY OR TOWARDS 15UM WHICH MEANS LESS OR MORE, RESPECTIVELY, 15UM ENERGY TO ABSORB.

  190. davidmhoffer says:

    Dave Springer;
    ALL CAPS – LOL

    Richard Smith;
    You enjoy having your rants cut to shreds in one thread after another? Well at leat you’ve finally given in on the igloo… now if we can just get you past trying to define absorption and re-mission in a random direction as inventing perpetual motion…

  191. Dave Springer says:

    Al Tekhasski says:
    March 1, 2011 at 12:01 am

    “I am sorry, but for the sake of accuracy, CO2 in air does not have any states that interact with 10-13um band. Therefore, hot air cannot possibly generate any photons in that range. Therefore the following statement is also wrong, nothing gets “transferred” into 10-um band from “heated air”:”

    I AM SORRY BUT YOU ARE WRONG. COLD DENSE GASSES ALL EMIT A CONTINUOUS BLACKBODY SPECTRUM. SEE KIRCHOFF’S FIRST LAW OF RADIATION WHICH I POSTED EARLIER.

  192. robr says:

    Ira Glickstein, PhD says

    I have two statements/observations with respect to this radiation:

    1.) Forget thermo for a moment and let’s talk heat transfer. Two infinite planes separated by a distance in equilibrium (the earth being one and some height in the atmosphere being the other) – the radiosity of the one equals the emissivity of the other. The radiative heat transfer from the warmer to the colder is in proportion to the fourth power of the temperature difference. So the CO2 issue would necessarily raise the atmosphere temp, but where the temp difference between the earth and atmosphere is small, the difference between the atmosphere and space is large, so the fourth power factor of the temp would greatly lessen any lasting heating.

    2.) All this back scattering and stuff only works when the light is on. It is like shining a light in a mirror lined room, it bounces around good, but turn off the light and nothing. Once the sun goes down the only heat retained in the atmosphere is the latent heat of water vapor, all that radiation goes poof into space. Take the water vapor out of the air and I don’t care how much CO2 you have in the air, when night comes it is going to get real cold.

  193. Brian H says:
    March 1, 2011 at 11:15 am
    … BTW Ira, I’ve just gone thru the first part of your novel, and I wonder if you still “believe” the scenarios and causalities for climate behavior and fluctuation you posit and “explain” in Chapter 1. The verbiage circles round and round, but boils down to domination by CO2 changes, with some drastic natural and not-so-natural interventions and intervening events forcing it back down to “normal” levels.

    Which, IMO, is alla buncha hooya.

    Thanks Brian H for having a look at my free online novel. It was written over two years ago, prior to my learning about: 1) Climategate, 2) The official climate Team’s over-estimate of CO2 sensitivity, 3) Svensmark’s cosmic ray theory of cloud formation, and 4) How mistaken NASA’s (Hathaway) was in predictions regarding SC#24. Considering all that, I think I got an amazing amount correct, including the dominance of natural cycles and processes over GHG, when I wrote [Emphasis added]:

    Global warming turned out to be the result of a natural cycle of solar activity and the Earth’s orbit and tilt that increased solar radiation at the poles, augmented by human over-production of greenhouse gases. Average temperatures and mean sea levels increased by about a half degree Celsius and a half-meter between 2001 and 2031. By 2052, temperatures and ocean levels had gone down to the 2021 ranges. They were expected to slowly decrease towards normal values.

    This stabilization was due to: 1) Social effects of globalization that reduced reproductive rates below replacement levels, 2) Several genetic engineering disasters between 2020 and 2035 that killed tens of millions of people, 3) A nuclear exchange in the MidEast in 2021 that lead to a mini-“nuclear winter,” cutting average temperatures by five degrees for a year, and 4) A punitive tax on non-renewable carbon-based energy quadrupled costs and led to more efficient use of energy resources. Worldwide population in 2051 as compared to 2001 was halved and greenhouse gas production quartered.

    If I were to modify the above, I would add Svensmark’s GCR stuff and reduce the sea level rise by 50%. I still believe that CO2 levels are rising largely due to human activities and that they contribute to net mean warming, though not anywhere near the amount estimated by the official climate Team, and far less than natural processes not under human control. Sadly, I still think human-caused accidents are likely over the coming decades, and that the premise of the novel, Stephen Hawking’s prediction, will come true. He said: “I don’t think the human race will survive the next thousand years. Unless we spread into space. There are too many accidents that can befall life on a single planet. But I’m an optimist. We will reach out to the stars.” [Stephen Hawking, Physicist and Cosmologist, 2001]

    What do you predict?

  194. davidmhoffer says:

    There were a number of comments in the thread asking questions about, or proposing various lab experiments with cylinders of artificial atmosphere, or other approaches. They are invalid.

    They just don’t scale to reality and the reasons are well known. Even putting aside the existance of other kinds of molecules in the atmosphere, the cylinder in the lab with CO2 in it isn’t even close. For starters, it is at pretty much the same temperature from one end to the other. In the atmosphere, temperature declines with altitude. So the frequency range of photons that can be absorbed or emitted changes with altitude. And time of day. And latitude. And season.

    PLUS, the pressure in the lab cylinder is stable. In the atmosphere, pressure declines with altitude, so while the CO2 concentration may be stable in terms of parts per million, in terms of molecules per cubic meter, it declines.

    If that hasn’t completely discredited the lab cylinder experiment yet, let me continue by adding JUST water vapour to the picture. If we were to add H2O molecules to the equation, unlike CO2 which is evenly distributed throughout the atmosphere, the H2O molecules would be clustered at the bottom of the atmosphere near earth surface. In the real world, the holding capacity of the air declines with temperature, so the water vapour concentration may be as high as 40,000 ppm or less than 1,000 depending on temperature. And temperature varies with…altitude, time of day, latitude, and season. So down at earth surface 40,000 H20 vs 390 CO2, there’s not much point even doing the math to calculate the effect from CO2, it is meaningless. But as we move in altitude, latitude, season and so on to cooler temperatures and lower levels of water vapour, CO2 becomes more and more significant.

    Sorry, not done yet. What else goes on in the atmosphere? Convection! As air is heated at the equator, it rises, pulling in cooler air from the temperate zones. As the hot air rises, it cools. How? By on average emitting more photons than it is absorbing. And as it cools, the wavelength of photons that is released changes… the percentage chance that an emitted photon from a rising molecule of CO2 will escape increases as the molecule gains altitude because the path to space is shorter. And the cooler air being sucked down from the poles is doing the opposite.

    Then consider all of THOSE factors in terms of what wavelength of photons are being release and what wave length are being absorbed at any given point in time based on all of those factors and map that against the atmospheric window….

    Just like Ira’s physical model or the explanation of the atmopsheric window in this article, a lab experiment simplified down to a cylinder with air in it is fine for evaluating certain aspects of CO2 and IR, but for drawing any conclusions about the climate? Inadequate.

  195. Myrrh says:

    As it “re-radiates” in all directions, then it also must be said that it speeds up how fast the surface cools. Right?

  196. George E. Smith says:

    “”””” Bryan says:
    March 1, 2011 at 2:31 pm
    George E. Smith

    Did the classical EM radiation laws of Maxwell not run up against the ultraviolet catastrophe and hence the modification by Panck and Einstein introducing the quantised photon?
    I have always understood that continuous spectra are associated with radiating solids.
    For gases that radiate a line spectra is what I’d expect. “””””

    So I take it you are implying that the near Black Body radiation Spectrum from a radiating solid, is NOT a continuous spectrum; whiel also saying that it is.

    Was it not the introduction of quantization that led from the “ultraviolet catastrophe” that is quite unrelated to Maxwell’s equations for electromagnetic radiation to the highly successful BB radiation formula of Planck.

    If you reread your history, you will find, that what was quantized to turn the unsuccessful Raleigh Jeans formula into the successful Planck formula was not the quantization of radiant energy levels but of the average energy per degree of freedom of vibrating molecules due to their Temperature.

    The Jeans derivation assumed that the energy per degree of freedom (three for a translating only body) could have any continuous value. Planck’s breakthrough came in requiring the energy per degree of freedom to be quantized. The radiated frequencies are not quantized, and depend only on the Temperature.

    Remember that the theory of black body radiation is a purely hypothetical one since no such object actually exists; and as a consequence it is connected in no way to the actual physical properties of any material; just the Temperature; so it cannot possibly be related to any energy level structure, all of which are properties of real materials.

    And Planck’s derivation takes no account of the actual phase of the material doing the radiating; just its Temperature.

    (atomic )Gases emit both line spectra; that are a consequence of the electron structure of those atoms; but when ionised, they can also emit a continuum radiation that is in the Ultra-Violet beyond the infinity end of the series like Balmer series and others; and they are a continuum, since the electron that is captured by the ion to radiate the photon energy, can have any energy at all while it is a free electron, so the energy differnece of the transition from a free electron to a bound state can have any value. such atomic continuum spectra are observed in the spectra of many stars.

    The BB radiation spectrum arises not from discrete energy levels of any atoms or molecules, but from the acceleration of the electrons of the atoms or molecules as a result of Molecular collisions, ie Temperature.

    Einstein’s quantisation of photon energies, came about as a result of his work on the Photo-electric effect, which is what he got his Nobel Physics Prize for.

    Classical theory calculated the energy density of the EM field from Maxwell’s equations. From that energy density, and the crossection of a photoelectric effect material atom, such as Cesium for example, one could calculate how long it would take to intercept enough energy to release a bound electron from the material. So it was assumed that if one lowered the intensity of the “light”, it would take longer to collect enough energy from the EM field, so delaying the release of the Photo-electron.

    Instead, Einstein found that no matter how much the light was attenuated; the emission of the electron was alwqays instantaneous; there was no delay to accumulate enough energy (no delay in the technology of those days). But he also discovered by selecting individual wavelengths, that no matter how bright the illumination might get; some times there was no photo-electron emitted at all. Th3e longer wavelength illuminations failed to release photo-electrons, the shorter wavelength illumination no matter how dim, released electrons instantaneously, and the kinetic energy of those electrons increased for shorter wavelength light.

    He reasoned that the light must come in packets each containing a certain energy (E= h.nu), and all that happened when you attenuated the light, was the number of packets was less so you got fewer photo electrons emitted, but the KEs of those emitted were still the same no matter how dim the light. So far as I know to this day nobody has successfully explained the Photo-electric effect in terms of any classical theory of Physics.

    But all that did not send Maxwell’s equations to the dustbin of history; it is enshrined forever, in the velocity of light as c = 1/sqrt( mu-naught x Epsilon-naught). Those three fundamental constants are about the only fundamental Physical constants whose values are exact with zero error (by definition).

    So Maxwell still reigns supreme; just not over everything, and in particular Maxwell’s equations are not inconsistent with the Planck formula or the Stefan-Boltzmann formula (which is simply the integral of the former.)

    One can find excellent treatments of the various Radiation laws, in various Handbooks of Optics such as the one sponsored by the Optical Society of America and edited by Walter G. Driscoll. The section on BB-radiation is authored by Jay F Snell at Tektronix. One can also check “Applied Optics, and Optical Engineering” Vol IV by Rudolph Kingslake (Eastman Kodak).

    Snell incidently cites Kichoff’s law thusly:-

    “Kirchoff’s Law is a consequence of the necessary existence of an energy balance between emission qnd absorption for a body in an isothermal black enclosure, and in Temperature equilibrium with the enclosure.”

    Such conditions exist nowhere in nature; but we can actually build quite good approximations to such enclosures. Notice it says “black enclosure” meaning black in the black body sense, of total absorption of all radiation incident on the walls of that enclosure. Kirchoff’s law requires a wavelength specific balance. That is the absorption and emission of every single possible wavelength must match; but not it is a condition only of total equilibrium of a closed system.

    Earth’s atmosphere is NOT a closed system, and it is never in thermal equilibrium with anything. So Kirchoff’s law simply does not apply; no matter how hard people try to make it so.

  197. kuhnkat says:

    George E. Smith,

    “Did the classical EM radiation laws of Maxwell not run up against the ultraviolet catastrophe and hence the modification by Planck and Einstein introducing the quantised photon?”

    One of the misconceptions that even good scientists fall prey to is believing that mathematics drive science. Mathematics is a tool that can be used to DESCRIBE the real world and allow us to manipulate ideas about it and make projections if we do a good enough job of the description.

    I can draw a picture of something and easily transfer a number of data points to tohers through a picture with absolutely no understanding by wither of us of what we are seeing. Mathematics can do the same for the real world in physics.

    Sadly climate science has latched onto a few data points and algorithms and think it covers EVERYTHING while still struggling with how poorly their models do. At a much higher level quantum physicists do the same. It really is all RELATIVE!!

    HAHAHAHAHAHAHAHAHAHAHAHA

  198. Bill Illis says:

    Here is the outgoing longwave radiation map on Feb 20, 2011 from the CERES satellite.

    Can you see the continents? Can you pick out the oceans?

    Well, if the emission to space went right through the atmospheric windows, right through the spectra which are not intercepted by the GHGs, one should be able to see the hot parts of the continents, or the cold parts of the continents. Or something.

    What the map is really showing is that emission to spaces occurs at various heights in the atmosphere and cloudiness is a very important factor in this and that the atmospheric windows are not really apparent. Maybe this a bit esoteric, but it is telling you what is really happening in the atmosphere. It is not a AGW global warming map. It is reality. [This is always my default position, what really happens versus what the theory says should happen].

    http://img132.imageshack.us/img132/1866/renderdata.jpg

  199. wayne says:

    Ira, I mostly praise the way you presented this post, much clearer, but the only thing I will complain about of you post is in this statement:

    However, they differ in the mechanism. A real greenhouse primarily restricts heat escape by preventing convection while the “greenhouse effect” heats the Earth because “greenhouse gases”
    (GHG) absorb outgoing radiative energy and re-emit some of it back towards Earth.

    There you need to lose the statement “heats the Earth”. That is nonsense. You might say “keeps Earth warmer that without” or “slows the cooling of the Earth”, both implying the action of an insulator, but anything but what you said.

    For myself, I have now move on the scale from ‘warmist’ to ‘skeptic’ and now just a tad before ‘denier’ so I think both of those statements are also most likely incorrect, but still that’s a better way for you to state it. It’s just that “warms the Earth” breaks core thermodynamic principles and it unqualifiedly wrong. Best to admit it and restate.

    But I know you’ve already gotten an earful on that same matter.

  200. George E. Smith says:

    The room temperature (300K) band gap of silicon is 1.106 eV Einstein’s constant is 1.23980 electron Volt microns.

    So any photon of shorter wavelength than 1.121 microns, can be absorbed in bulk silicon, and release carriers, to be transported across a PN junction; to create a photocurrent flow. Well that assumes that the carrier lifetime is long enough for them to propagate to the junction and cross it, before they recombine.

    So silicon diodes can absorb sunlight shorter that 1.121 microns wavelength, down to at least 400 nm and create a photocurrent in a “solar cell”.

    But don’t try to get that same silicon junction to reverse its behavior when you pass a forward current through it, and emit any photons.

    Silicon is an indirect gap semiconductor, and the reversal of the absorption transitions are prohibited, so silicon does not form Light Emitting Diodes.

    So there’s a simple case of a solid that simply cannot radiate the same photon energies that it can easily absorb. Neither can Germanium which also makes good (low Voltage) solar cells.

    But GaAs with its 1.47 eV bandgap readily emits 844 nm photons, which is the bandgap photon energy, because it is a direct gap semiconductor, and that transition is allowed, without any momentum discrepancy, requiring a phonon interraction as well.

    So no Kirchoff’s law simply doesn’t apply to very many real world situations.

  201. George E. Smith says:

    “”””” kuhnkat says:
    March 1, 2011 at 5:04 pm
    George E. Smith,

    “Did the classical EM radiation laws of Maxwell not run up against the ultraviolet catastrophe and hence the modification by Planck and Einstein introducing the quantised photon?”

    One of the misconceptions that even good scientists fall prey to is believing that mathematics drive science. Mathematics is a tool that can be used to DESCRIBE the real world and allow us to manipulate ideas about it and make projections if we do a good enough job of the description. “””””

    I have no idea what your point is; do you have one you’d care to state ?

    And there is NO mathematics that can describe the real world or even attempts to. Mathematics can be used to describe the expected behavior of our theoretical modles of the real world; which are every bit as fictional as the mathematics is; we made all of it up, so none of it is real.

    We do try to construct our fictional models to try and mimic reality; but that is all they do; they do not explain reality.

  202. Al Tekhasski says:

    Wayne wrote:
    “So you are saying that if you had a large clump of warm nitrogen in empty space held together by gravity that it could not radiate in a black body manner at all, zero, none, zip? Maybe check some good astronomy books on that matter that cover quantum aspects.”

    and

    “Ira placed no numbers on exactly how much was split between those 7, 10, and 15 µm band portions. Just that those interactions do occur.” … That is how I see it from my years in studying physics.”

    Wayne, maybe you need to continue study physics for some more years and learn how to put numbers on magnitude of effects. I am sure you heard about “pressure broadening” and “collision-induced absorption” during your studies. Now consider that we are not living on a star. Please open your reference book and compare gas pressures and temperatures of stars with Earth atmosphere.

    Then try to comprehend the the difference between numbers. It is called “Physics”.
    Please also do understand that “greenhouse effect” does not occur because 1/1,000,000 fraction of 15um band energy gets re-distributed and re-emitted from 10-13um band.

    But of course you are formally correct. Instead of saying “hot air cannot possibly generate any photons in that range” I should have said “hot air cannot generate any remotely considerable amount of photons in that range”.

  203. Derek says:

    Dave Springer says:
    March 1, 2011 at 3:54 pm

    CO2 IS INSULATION.

    IT DOES NOT HEAT ANYTHING. IT SLOWS DOWN HOW FAST THE SURFACE COOLS.

    I WONDER IF ALL CAPS WILL HELP. GOD KNOWS I’VE TRIED EVERYTHING ELSE TO GET THIS SIMPLE CONCEPT ACROSS.

    Errr, NOPE, it don’t help, YOUR the one missing the point.
    (I hope the capitals help)

    Here is a more realistic way to “visualise” the “greenhouse effect” as commonly (and incorrectly) explained and viewed at present.
    from,
    http://www.globalwarmingskeptics.info/forums/thread-855.html

    Could someone please explain to me (understandably) why
    a thermal image of a greenhouse shows it radiating more than it’s surroundings. ?
    (I know the greenhouse radiates more because it is warmer than it’s surroundings, but
    it is supposed to trap radiation, yet (somewhat inconveniently) greenhouses, as far as I’m aware, radiate according to the Planck curve)
    I thought a greenhouse was supposed to “trap” radiation.
    Is Plancks law wrong, surely not.

    So, would the better question be. Why are the greenhouses surroundings so much cooler than the greenhouse, although
    the surroundings do not actually “trap” radiation, they do appear to be radiating far less than the greenhouse.. ?
    What then is cooling the surroundings. ?

    I am assuming the best explanation would be because the surroundings are cooled by something far more powerful than radiation looses,
    namely conduction and convection of sensible and latent heat.

    A greenhouse “works” because it reduces conduction and convection to it’s surroundings.
    It would appear reasonable to say observation of a thermal image of a greenhouse and it’s surroundings indicates that
    conduction and convection (of sensible and latent heat) is far more powerful than radiation looses, and
    is responsible for cooling the surroundings mostly.
    Is this a correct series of assumptions, or rather statements of the blitheringly obvious.

    I too am bored with the way “climate discussions” seem to be either circular, or descend into mud slinging.
    Maybe, just maybe, present discussions are discussing the wrong “things”..

    End of quote.
    The above has recently been misquoted on this blog by someone else.

    In the end, the REAL QUESTION, that is avoided, is simple.
    It is,
    Why do physicists using closed, isolated, far higher concentrations than are realistic, to suggest (and supposedly observe that) CO2 has a high specific heat capacity,
    YET,
    CHEMISTS USING THEIR MASS BALANCED EQUATIONS SHOW CO2 HAS A LOW SPECIFIC HEAT CAPACITY.
    (Hey, this caps thing is quite good isn’t it..)

    Maybe, just maybe, CO2 reacts differently in an open and mixed system, at far, far, far lower concentrations than a closed, isolated, at far, far, far, far, far, far, higher concentration of CO2 system.
    Dam, the EPA never thought of (or understood) that, AND niether did most “climate sceptics” either apparently.
    (brackets are quite good as well.)

    Increased CO2 in the upper atmosphere increases global cooling , obviously,
    BECAUSE,
    it increases (the planet’s ability to loose, or rather emit) radiation of energy to space.
    AND,
    CO2 helps move energy about FASTER in the lower atmosphere,
    BECAUSE,
    conduction and convection (of energy AND latent heat) is increased,
    BECAUSE,
    of the physical properties (which “we” really are not sure of) of CO2.

    But this is all irrelevant, CO2 is dwarfed into insignificance by H2O and
    it’s nefarious abilities, most undeniably including change of state,
    which at atmospheric temperatures, CO2 just can not touch.

    CO2 AIN’T INSULATION.
    It is the staff of life, this planet is almost entirely, carbon based life forms.
    (as James T Kirk noted, way back when, when Star Trek was young)

    Nowhere near enough people (and almost ALL politicians + bureaucrats ) understand how little “climate science” actually does know.

    Lines 2 and 3 of the infamous Rumsfeld quote describe the current state of the “settled science” of AGW and “climate science” best at present, in my opinion.
    ie,
    “Known, unknowns, and
    unknown, unknowns.”

  204. George E. Smith says:

    “”””” Bryan says:
    March 1, 2011 at 2:31 pm
    George E. Smith

    Did the classical EM radiation laws of Maxwell not run up against the ultraviolet catastrophe and hence the modification by Panck and Einstein introducing the quantised photon?
    I have always understood that continuous spectra are associated with radiating solids.
    For gases that radiate a line spectra is what I’d expect.

    You can flick between absorption and emission spectra for H2O
    http://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&Units=SI&Type=IR-SPEC&Index=1#IR-SPEC

    Same also for CO2
    http://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Units=SI&Type=IR-SPEC&Index=1#IR-SPEC “””””

    So that CO2 infrared spectrum for example, is the absorption spectrum for 390 ppm of CO2 in dry standard air, I take it; or am I presuming too much. You did check that the spectrum is for the system of primary interest; namely earth’s lower tropospheric atmosphere; didn’t you ??

    Your spectra look nothing at all, like the spectralcalc ones that Phil has pointed us to; can you explain that ?

  205. George E. Smith says:

    “”””” Myrrh says:
    March 1, 2011 at 5:26 pm
    http://wiki.answers.com/Q/What_percentage_of_the_electromagnetic_spectrum_is_visible_light

    some answers.. “””””

    So now we go from the sublime to the ridiculous. That wiki answer to the question it total BS; and you can use your own non wiki dictionary explanation of BS.

    The proposed standard Solar spectral irradiance (w/m^2/micron) contains 8.725% of the energy below 400 nm (UV) and 56.023% below 800 nm (visible, so that means 47.3% of the total solar spectrum is visible, and only 44% is in the IR from 0.8 micron to infinity. So there is more visible than IR

  206. davidmhoffer says:

    Was just thinking I’ve long since lost track of how many physicists, engineers, chemists, scientists of who knows how many stripes with credentials, real world experience, text books, articles and studies galore at their finger tips, having a raging argument about how photons interact with CO2 molecules. Such a tiny simple question about just that.

    But the climatologists at the IPCC with thousands of factors to consider, plus all the combinations of those factors… they’ve got a consensus. Wow, they must be really smart ;-)

  207. Al Tekhasski says:

    Dave Springer shouted: “I AM SORRY BUT YOU ARE WRONG. COLD DENSE GASSES ALL EMIT A CONTINUOUS BLACKBODY SPECTRUM. SEE KIRCHOFF’S FIRST LAW OF RADIATION WHICH I POSTED EARLIER.”

    You should be sorry, because you didn’t get the meaning of “dense gas” as it is used in astronomy and web articles about “continuous spectrum”. When astronomy talks about “dense gas”, they mean “optically non-transparent gas”, like star’s atmosphere or some other non-transparent object. Earth atmosphere is not optically dense, not in the 10-13um band at least. In climatology-speak it is called “IR transparency window”, for that very reason. As such, air does not absorb 10-13um EM waves, and does not emit any “remotely considerable amount of radiation” in that band, at least for any practical reasons.

  208. Phil. says:

    Domenic says:
    March 1, 2011 at 11:43 am
    to Phil and Cal

    Nope. The full characteristics of N2 and O2 in the long wavelength IR have never been actually tested.

    Indeed they have, as shown above they are exceptionally weak, it would require incredibly dry air to make the measurement since the spectrum of water is millions of times stronger. Probably very long path length FTIR spectroscopy.

    There is something very fishy in the spectral graphs bandied about from calculations regarding the emittance and reflectance of N2 and O2. They are not metals.

    Indeed they are not, perhaps you should realize that gases have very well known molecular structures and correspondingly well known spectra.

    Both N2 and O2 must be tested thoroughly in the long wavelengths and not guessed at.

    No one’s guessing at them, except perhaps you.

  209. Al Tekhasski says:

    Bill Illis submitted a CERES OLR image
    http://img132.imageshack.us/img132/1866/renderdata.jpg
    and stated:
    “one should be able to see the hot parts of the continents, or the cold parts of the continents. Or something.”

    Where did you get this picture? Is it possible that the signal was filtered to show absorption CO2 bands only, and not 10-13um? Could it be possible that ground emission in “atmospheric transparency window” is spotty and does not resemble land, so it is hard to recognize? Please keep in mind that 70% of surface are clouds that are blackbodies in IR, so it will not improve visibility either.

  210. Phil. says:

    Theo Goodwin says:
    February 28, 2011 at 7:51 pm
    bubbagyro writes:

    “CO2 is generated on the earth’s surface. It becomes more dilute as it diffuses upward. The models, similar to Ira’s, assume it is a narrow band. This is not true. The very dilute CO2 and water even in the stratosphere absorb radiation. The higher that occurs, the more is lost to space.”

    Please expand on this. My understanding is that the modelers assume that concentrations of CO2 are the same wherever they occur in the atmosphere – all the way up. In addition, they assume that the behavior of CO2 molecules regarding radiation is the same throughout the atmosphere. It has always seemed to me that these “uniformity” assumptions were just evidence of an unwillingness to do the necessary empirical research. In plain and simple terms, I would like someone to address the linked questions of “where are the CO2 concentrations and how does radiative behavior change depending on where they are?”

    Your assumption is not correct, concentration is not constant, mole fraction is to a good approximation. They do not assume that the radiative properties are the same through the atmosphere, line broadening changes with pressure and temperature.

  211. GaryP says:

    Very good article (and graphics). Most lay people have no idea that CO2 absorbs only a small fraction of the IR spectrum. Those of us who took undergraduate organic chemistry remember the use of IR absorption analysis and the fact that CO bonds absorb only certain wavelengths.

    Slightly OT, but VUK said:
    There is also possibility of a half way house:
    Solar input is constant,

    I don’t think that is assumption is a half-way house, I think that position is full on nut house. We have been studying solar output for such a short time, that making any statements about solar output being constant is, IMHO, ridiculous. Based on observations of many other stars, variable output of stars is certainly well known. Our star may be relatively constant in output, or may be in a relatively constant phase. But assuming that solar output is constant over thousands of years is a pretty big assumption that is based on nothing (once again, IMHO) other than wishful thinking.

    Certainly, with no convincing explanation for the major climate variations we know have occurred over 10K to multi-M year periods, I would be very hesitant to assume that changes in solar output could not be a contributor.

    We should study all aspects of climate, including the most important source of energy into our world climate system (the only other being radioactive decay heat, I believe) a little longer before we make simplifying assumptions that lead us to stupid answers (such as CO2 is the only thing that affects world climate–which is effectively what the IPCC and green lobby asserts today regardless of how much they weasel word it if pressed.)

  212. Phil. says:

    mkelly says:
    March 1, 2011 at 12:24 pm
    Phil. says:
    March 1, 2011 at 10:15 am

    “As stated above absolutely wrong, the only way that N2 and O2 molecules can shed energy is by collisional exchange with CO2, H2O etc. molecules which can then radiate.”

    Thanks for the information Phil. Your photo while nice does not answer the question of N2 and O2 at the same temperature. Since temperature drives the emittion frequency will they both be the same? I math (Weins Law) says yes, but I have found no info on the subject.

    No, why do you think that the N2 and O2 spectra shown are at different temperature?
    For two different gases the emission will depend on the transition concerned BB only provides the potential maximum, the line strength will have the dominant effect.

    By the way by your statement above you are in the camp that says without GHG we would be a very hot world indeed. Not at -18 C. Once a N2 or O2 molecule was heated via conduction with the ground it would stay heated for a very long time with no H2O or CO2 to reduce its energy. Correct?

    No, how do you propose that the atmosphere gets hotter than the surface?

  213. Oliver Ramsay says:

    Derek says:
    March 1, 2011 at 5:40 pm

    “Could someone please explain to me (understandably) why
    a thermal image of a greenhouse shows it radiating more than it’s surroundings. ?
    (I know the greenhouse radiates more because it is warmer than it’s surroundings, but
    it is supposed to trap radiation, yet (somewhat inconveniently) greenhouses, as far as I’m aware, radiate according to the Planck curve)
    I thought a greenhouse was supposed to “trap” radiation.
    Is Plancks law wrong, surely not.”
    ——————————–
    You didn’t show your thermal image, but, if it’s showing the greenhouse hotter than the tree adjacent, it’s because the greenhouse is hotter.
    If the greenhouse skin is polyethylene, the interior of the greenhoue will be what is seen for the greater part. If it’s glass, it will be the glass that is emitting.
    On a clear day, the greenhouse will not be much hotter than adjacent objects with similar exposure to the sunshine. It will be hotter than the air outside. The other objects would be as warm but some heat was convected away.
    On a clear night, it will be hotter than adjacent objects and hotter than the air. Once again, it’s about the heat loss on the outside, not heat gain on the inside.
    I have just taken readings with an IR thermometer from the outside of my greenhouse, looking in. It is nighttime There is a broken pane that is covered with polyE. The wall of an adjacent shed is 0C. The single-glazed pane of glass shows 7C, the poly shows 15C and through the open window, it shows 16.5C.
    Obviously, that only speaks to IR transmission of the materials, since the inside temperature arises from thermal mass or internal heat source.
    In practice, a poly greenhouse works about as well as a glass one, in spite of the difference in IR transmission. Conduction/convection cools much faster than radiation in these circumstances. And in many others!
    That’s my version!

  214. Jim D says:

    Bill Illis, that satellite picture looks like a water vapor channel. It is not total OLR, nor is it a window wavelength. It is very easy to see the surface diurnal cycle when you don’t look at the water vapor channel. The WV channel has other uses, but not to see the surface. Find other IR images. Typical ones are reversed so that clouds appear white (cold) and the day-time surface is black (hot), going grey at night.
    I reiterate, the atmosphere has no way to emit in the 10-micron window region. Its molecules are incapable of producing such wavelengths.

  215. wayne says:

    Al Tekhasski says:
    March 1, 2011 at 5:29 pm

    [snip the ad hominem]

    But of course you are formally correct. Instead of saying “hot air cannot possibly generate any photons in that range” I should have said “hot air cannot generate any remotely considerable amount of photons in that range”.
    —–
    Better. And no, I don’t have the numbers or even a rough estimate. Also I was not speaking of stars or star’s atmosphere but instead warm IR emitting gas clouds so you never see these in visual measurements or photographs. But, you can look at all of the spectrums of Earth’s atmosphere and see the imprint of the overall black, or better gray, body absorption and/or transmission low at the bottom.

    Do you have the numbers instead of pulling 1/1,000,000th from the top of your mind? Many here would like to have such knowledge, or do you prefer it to stay hidden?

  216. Richard Sharpe says:

    Alan McIntire says on March 1, 2011 at 6:40 am

    In reply to Katherine and to Phil’s dad. When the Earth’s temperature increases
    from 1 to 1 + p, the total radiation increases by a factor of (1 + p) ^4,
    Outgoing radiation increase at all wavelengths, but the PERCENTAGE increase at
    short wavelengths, which are not affected by CO2, increases at a much larger rate.

    Hmmm, so outgoing radiation at all wavelengths increases as temperature, but since the peak goes to lower wavelengths, it just does not increase as quickly at long wavelengths?

    So if CO2 results in 40 watts/500 total watts, with a 1% increase in the total to
    505 watts, CO2 will absorb less than 40.4 watts. There’s a negative feedback due to that 4th power increase in radiation with respect to temperature, as you discovered, there’s also a negative feedback with respect to CO2 due to a larger percentage of increased outgoing radiation in wavelengths not affected by CO2.

    Umm, if outgoing radiation increases at all wavelengths, how can that be a negative feedback? It seems that the amount of extra outgoing radiation decreases with each temp increase, so it is logarithmic or something.

    In any case, since the CO2 absorption bands are saturated, does it matter if CO2 increases. Also, does pressure broadening depend on partial pressure or total atmospheric pressure?

  217. Tim Folkerts says:

    Dave says
    “Enough already with the argument that certain gases that don’t absorb infrared can’t emit it…. The kinetic energy gained by the nitrogen molecules will be lost in a continuous blackbody spectrum characteristic of the temperature of the gas.”

    OK, enough with theory and arguments! Let’s study the data that you touted — the data looking up at downward IR. You postulate a continuous BB spectrum from the N2 & O2 in the atmosphere. Where is that BB radiation in the data?

    Looking up into the sky, it is clear that SOMETHING is emitting close to a BB radiation curve @ ~265 K (the atmospheric temperature near the surface), but only above ~14 um or below ~ 8 um (which happens to be where H2o & CO2 emit well). That radiation we can attribute to GHGs near the surface.

    Between 8-14 um, the energy is typically around 5% of the BB radiation levels that would be expected for ~265 K atmosphere. EVEN IF we attribute ALL that radiation between 8-14 um to N2 & O2, the much more common N2 & O2 molecules are emitting much less energy. At best, they are very poor radiators.

    No theory. No Kirchoff’s Laws. No models. Just data.

    I don’t know, but I suspect, that even the radiation between 8-14 um is not due primarily to N2 or O2, but rather to GHGs (possibly CO2 or H20, but potentially also O3, CH4 …). That would make the downward contribution from N2 & O2 molecules as even less. So that would make them very very poor radiators.

  218. Al Tekhasski says:

    Wayne remarks snidely: “Do you have the numbers instead of pulling 1/1,000,000th from the top of your mind? Many here would like to have such knowledge, or do you prefer it to stay hidden?”

    Glad to oblige. The number of six orders of magnitude can be seen from this picture, courtesy of “Phyl.:
    http://i302.photobucket.com/albums/nn107/Sprintstar400/CO2N2O2.png
    – intensity of lines in 10um area are about six orders of magnitude smaller than 15um band.

  219. Richard E Smith says:

    davidmhoffer said:

    “Richard Smith;
    You enjoy having your rants cut to shreds in one thread after another? Well at leat you’ve finally given in on the igloo… now if we can just get you past trying to define absorption and re-mission in a random direction as inventing perpetual motion…”

    David – you are referring to a different Richard Smith who was ‘ranting’ on WUWT a few days ago. It is a case of mistaken identity. I deprecate any ranting from either side. I have now re-registered as Richard E Smith to avoid confusion.

  220. Don V says:

    P. van der Meer, Al Tekhasski, Phil – you all get it! I completely concur with your clear thinking.

    I’m sorry Ira, but someone has convinced you to believe in a concept that violates the basic laws of thermodynamics. Unless you can convince me that this model does not violate both the first and second law of thermodynamics, I will have to conclude that you have drunk some of the kool-aid. I am not trying to insult you, I respect you, and would like to convince you of the truth, but to do that you will have to get back to the basics, and admit this is just plain wrong!

    There is so much wrong I would have to spend an inordinate amount of time writing, so I will focus on just one aspect of it. Your animated illustration at the far right shows the atmosphere absorbing energy radiated up from the earth at 15μ . Think about this carefully now. To do this which body has to be hotter? Earth right? OK so how can the atmosphere with only 15μ radiant energy warm up so hot that it exceeds the earths temperature and radiates back at two wavelengths 7μ and 10μ!? It can’t! The only way you can create a black body radiator at 7μ or at 10μ is to heat a body up to the hotter temp that corresponds to that wavelength. IF it radiates at any black body energy at all, to obey the first law of thermodynamics, it MUST radiate at energy equal to or longer than 15μ. So your illustration and explanation violate both the first and second law since 1) to be true it illustrates that energy in the atmosphere must have been created from nothing, and 2) it illustrates that a cooler body is capable heating up a warmer body.

    If I am wrong on this one point then please correct me, and teach me how this can be true. If not admit the mistake and correct the illustration. Then we can go on and correct the other things that are wrong with this “Trenberth” like illustration. While you are at it correct the mistake that shows 10μ radiation from a gas that is not even capable of producing this wavelength as well!

  221. mkelly asks:
    What difference is there between the spectrum of N2 and O2 both at the exact same temperature and radiating? I have not found anything on that.

    The HITRAN (high-resolution transmission molecular absorption) database contains the spectra of all the important gases that absorb in the infrared. Some of the N2 and O2 isotropologues (those with 2 different isotopes) do have measurable spectra in that range, but the 2 most common isotropologues don’t. This has been measured in the lab. Neither molecule has been observed to absorb or emit radiation at the frequencies associated with the temperatures of interest. However, if either molecule gets hot enough, it will emit visible light.

    Latter, mkelly continues:
    By the way by your statement above you are in the camp that says without GHG we would be a very hot world indeed.

    That is exactly correct – without greenhouse gases, the atmosphere would be above the boiling point of water both day and night, and the surface would freeze every night.

  222. Philip Peake (aka PJP) says:
    the effect is, indeed, logarithmic in nature (Log base e, not Log base 10).

    What? The only difference is a multiplicative constant! Except for a constant, ALL log relationships are identical. The base makes no difference at all.

  223. Dave Springer says:
    March 1, 2011 at 4:06 pm
    Tim Folkerts says:
    March 1, 2011 at 2:38 pm

    “For 400 ppm it clearly takes less than the height of the atmosphere, but how much less?”

    APPROXIMATELY 12,000 FEET OVER THE ARCTIC OCEAN. SEE THE GRAPH OF IR SPECTRUM FROM 20,000 FEET LOOKING DOWN THAT IRA POSTED IN THE COMMENTS TODAY.

    No, The spectrum from 20,000 feet (6 km) shows the energy emitted by CO2 from about 70 meters below the sensor.

    For your estimate, you should use the environmental lapse rate (6.5 K/km), not the dry adiabatic lapse rate (9.8 K/km). That produces a CO2 emission height of 20,177 feet (about 35 meters above the reported altitude).

    At the surface, 400 ppm Co2 absorbs 63% of the 15um band in less than 20 meters, 99% in 100 meters. Water vapor requires even less.

    In other post, you keep talking about “dense gases” without defining what “dense” is. As the pressure increases, the existing vacuum emission lines get broadened. At the surface of the Earth, these are merged into bands with significant gaps between them. At the surface of Venus (92 atm), the CO2 emission spectra is identical to the expected blackbody emission with no gaps.

    In truth, at the surface of the Earth, the spectra is continuous, but with ups and downs. As a result, an insignificant amount of energy is emitted (or absorbed) in the “gaps”.

  224. Bryan says:

    George E. Smith

    Thank you for your replies above.
    I agree with the general content but I have a different interpretation of the historical development of radiative physics.
    The classical Rayleigh-Jeans Law was developed in line with classical electromagnetic theory.
    Its departure from reality necessitated the quantum Planck radiation formula. Experiment confirms the Planck relationship.

    This lead gives I hope some hopefully uncontroversial background.
    Work through to the Planck formula and pick the wavelength option.
    There is a clear departure between the two formulas for wavelengths shorter than 3um.
    However for wavelengths longer than this the two formulas give the same result.

    http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

  225. Myrrh says:

    Ira – From the berkeley.edu link I posted above where I replied to you, http://wattsupwiththat.com/2011/02/28/visualizing-the-greenhouse-effect-atmospheric-windows/#comment-610932 is a short description of the difference between light waves.

    Under Thermal IR it says: “Very high intensity can heat up and kill living tissues”, compare with others eg Green “Normally harmless, can cause blindness or burn tissue at high intensity”

    The difference in effect is between heating up and burning re Thermal IR and Shortwave – that’s why UV, which we cannot feel either, burns the skin when it’s intense. If it was these shortwave energies heating the Earth, we’d bit a burnt crisp.

    These energies are cold, so the classic physics division into light and heat to describe them.

    This scam is a long time in the making, and as all cons it mixes up truth with the lies, that’s what makes a con seem plausible. It doesn’t matter how clever or not one is, if one takes on trust a vital piece of information for calculations which is wrong, it’s going to be gigo, it can’t be anything else.

    I see this a lot in scientists arguing for AGW, they take on trust something very basics in real physics from mangled versions in AGWScience, because it’s not in their own field. Because it’s repeated so often and by so many it’s assumed that such are actual real science facts. That’s why the con’s been so successful, imo.

    That NASA site I linked to is a prime example of the co-ordinated duplicity – the page I quoted from is from Real Science, before AGW it was known and understood and as you can see they’re still teaching it to children here, but if you look at some of their AGW slant pages you’ll not see it mentioned. One of their AGW pages shows a graphic with two earths, the first shows Solar, (Visible and near shortwave) entering and the second has Thermal IR going out. It isn’t actually a fib at that point, it’s just not the whole picture..

  226. Derek says:

    Oliver Ramsay says:
    March 1, 2011 at 8:02 pm

    ” That’s my version! ”

    It certainly is, and yet another attempt to misdirect in so many obvious ways,
    and at so many levels.

    ” You didn’t show your thermal image, ” – Imagine it please..

    ” but, if it’s showing the greenhouse hotter than the tree adjacent, it’s because the greenhouse is hotter. ”
    – Hotter than it’s surroundings dear chap, an obvious misdirect by you.

    ” If the greenhouse skin is polyethylene, the interior of the greenhoue will be what is seen for the greater part. If it’s glass, it will be the glass that is emitting.
    On a clear day, the greenhouse will not be much hotter than adjacent objects with similar exposure to the sunshine. It will be hotter than the air outside. The other objects would be as warm but some heat was convected away. ”

    I have not seen a shoal of red herrings before, well done that man.
    With misdirects as well.
    Congratulations Sir.

    ” In practice, a poly greenhouse works about as well as a glass one, in spite of the difference in IR transmission. Conduction/convection cools much faster than radiation in these circumstances. And in many others! ”

    Concluding by shooting your own foot off,
    was a bit of a let down however.
    Thanks for the laughs.

  227. davidmhoffer says:

    Richard E Smith;
    My apologies for the confusion. I was wondering why the sudden surrender on the igloo issue…

    As for the rest of your comment insisting that the laws of thermodynamics are being broken… I’m a hardcore, in your face, let’s debate any time anywhere anybody but only the direct science not the bunny rabbit population in northern Alberta has declined one percent so that’s why there is more snow in Europe this year SKEPTIC.

    And I’ve dug into this many times, my very first foray into the subject resulted in comments just like yours and Don V right after you. I was wrong. (50% of all the people who know me just sat up and said… huh what? he admitted to being wrong?)

    I think one of the best layman level explanations ever is Ira’s pervious article on this site http://wattsupwiththat.com/2011/02/20/visualizing-the-greenhouse-effect-a-physical-analogy/

    If you want something a little more technical, here’s my own explanation. Insulation isn’t quite the right word, and neither is delayed cooling. But it is what happens, the earth surface does warm if one considers ONLY radiative properties of CO2 and NO other effects, and the laws of thermodynamics are intact. Sorta ticked me off because I had to go looking deeper into the science…and the more I understood, the more of a skeptic I became. But skeptics need to get the basics and stop yelping about cold things being unable to warm hot things, and the laws of thermodynamics being broken. they aren’t.

    http://knowledgedrift.wordpress.com/2011/02/27/co2-exactly-how-does-it-warm-the-planet/

  228. Myrrh says:

    George E. Smith says,
    March 1, 2011 at 5:52 pm

    I posted it for the variations of explanations… You could add yours to it.

    But see my post above. How is the example of real light bulb which shows the proportion of Visible to Thermal IR at 5/95, and the real life physics from the steel industry which gives a billet of steel at temp X to have 100,000 times more IR than Visible not relevant to this? Is the Sun different?

    Objects heating up give off Thermal IR way before the object reaches a degree of heat capable of producing white light, or any visible light at all. One figure I found was that the Sun’s Thermal IR was at 80%. This fits in better with the lightbulb and steel examples, makes more sense to me.

    Anyway, the heat we feel on Earth coming from the Sun is Thermal IR, not Solar as Ira has it and as AGWScience teaches. Physics 101. See my post March 1, 2011 at 3:29 pm.

    davidmhoffer says: March 1, 2011 at 4:34 pm

    Re your artificial/real atmospheres and CO2 “well-distributed”, it isn’t any more than water vapour. CO2 is heavier than air so it would be, unless the lab cylinder is being agitated, at the bottom. CO2 will sink through the atmosphere displacing the air, because one and a half times heavier than air, unless there’s work being done to move it, wind. (Your lab cylinder example as problem with AGW, is also where AGW get their ‘Brownian’ motion from to produce the claim that CO2 is “well-mixed in the atmosphere” by “diffusion”; equal temperature requirement for Brownian doesn’t translate to real atmosphere, it’s convection which moves smells through a room, etc. Good explanation on http://en.wikipedia.org/wiki/Diffusion)

    CO2 doesn’t readily rise in air because it is heavier, so CO2 is really localised; what’s produced locally will tend to stay local (plants also breathe out CO2 as well as taking it in for photosynthesis). Something like a volcanic explosion which would force CO2 up into higher altitudes and from there carried in big wind systems, say the jet stream, would move it much further from local, but, say, a factory chimney would release it into whatever local weather system was happening. If a calm day it would simply sink to the ground, or on windy days spread around locally, or on windy wet days, it would come down with the rain. Which comes from the water vapour being lighter than air rising and condensing.

    I’ve never seen the raw data from the AIRS satellite survey, (I did see downloadable files but couldn’t get them to work), but their conclusion was that CO2 was not well-mixed in the atmosphere, it was lumpy, and they would have to re-think the data because the wind, local weather and other wind systems, had to be included in understanding CO2 distribution. The standard wind systems don’t cross hemispheres, there’s some mixing at the equator where they meet. http://www.earthfacts.net/weather/windmovement

  229. Myrrh says:

    My post to Ira before, before I posted to George and David, has not come up. Should I repost?

    [no]

  230. Richard111 says:

    If someone will provide a suitable subsidy I will conduct this simple test.
    Travel to one of the poles during the winter period and place a sheet of glass, say one square metre, with a hole in the middle, say 100 square centimeters, on any flattish ice surface.
    The glass is opaque to IR, the hole is not. Polar air is very low on water vapour and there is no sunshine. Thus backradiation from CO2 will pass through the hole in the glass and cause the ice below the hole to sublimate faster than the the covered ice. Knowing the time period and the mass of the missing ice we can calculate the energy provided by backradiation. Q.E.D.

  231. davidmhoffer says:

    George E Smith;
    I read the papers on CO2 being “lumpy” quite a while ago and only skimmed them. My recollection is that what you say is pretty much correct. That said, the distribution of CO2 is still FAR more even than water vapour in the atmosphere. Convection at the equator in particular drives CO2 upward as well as water vapour. As temps cool, the holding capacity of the air drops, driving water vapour out as droplets, snow, what ever the balance of conditions dictate. I’m guessing that some of the CO2 gets scrubbed out as well with the rain/snow, but the bulk of it continues upward. Point being that there is a band below which water vapour dominates the absorption spectrums where the two overlap, and a band above which CO2 becomes significant by comparison. Insignificant to climate, but still not representated well by a lab cylinder with artificial atmosphere or two plates of what ever with CO2 between them.

    I’d far rather spend time getting people to understand the fallacy of this thing called an “average” temperature, or the misleading way the IPCC calculates and presents sensitivity, but I don’t get a lot of traction on those issues. Everyone wants to fight about the laws of thermodynamics one photon at at time. Who needs warmists to debate with when there’s all these skeptics ready to take on skeptics?

  232. Bryan says:

    davidmhoffer

    …”and the more I understood, the more of a skeptic I became. But skeptics need to get the basics and stop yelping about cold things being unable to warm hot things, and the laws of thermodynamics being broken. they aren’t. “…….

    Well David, you can become even more of a sceptic because cold surfaces cannot heat warmer surfaces.

    Proof.

    Look at blackbody spectrum of a solid at 250K (say), then look at the blackbody spectrum of the body at 350K.

    Superimpose the spectra and compare.

    We notice two things.

    1. The higher temperature object produces short wavelengths that are absent from the lower temperature spectra.
    2. Now look at any wavelength shared by both.
    The higher temperature object produces much more radiation of that wavelength than the lower temperature object.

    To make it even more concrete lets say colder object radiates 1000 photons of wavelength 15um.
    The hotter object will radiate perhaps 1500 photons of 15um in the same time period.
    All the colder object does is insulates to some extent the warmer object.

    I will go further.
    Lets say the warmer surface was in thermodynamic equilibrium with its surroundings if then a colder object is brought near and it has a lower temperature than the surroundings then the colder object will increase the heat loss from the warmer object.

  233. Brian H says:

    Ricky111;
    Until the first snowfall.
    Oops!
    Bryan;
    Right, but if the cold object is warmer than the background, it will slow cooling of the warmest object. It’s all relative, as someone once observed.

  234. davidmhoffer says:

    Bryan,
    I’ve debunked that nonsense hals a dozen times in this thread and the previous thread on a physical model and I’m one of several who did so, not to mention that if you’d read the articles themselves and understand them in detail, you could ask a question about a specific part of the process you didnt understand or didn’t agree with.

    But instead you just skim through the very explanations of what the processes are, spout a thought experiment that is meaningless, and pronounce something impossible which you can falsify as simply as building an igloo and crawling in. So from now on i think I’ll respond to arm waving and pronouncements of conclusions with a strict time ought. Go sit in the igloo and cool off. Come out as soon as you are cooled off. Waiting for spring is an option.

  235. Phil. says:

    Myrrh says:
    March 2, 2011 at 2:44 am
    George E. Smith says,
    March 1, 2011 at 5:52 pm

    I posted it for the variations of explanations… You could add yours to it.

    But see my post above. How is the example of real light bulb which shows the proportion of Visible to Thermal IR at 5/95, and the real life physics from the steel industry which gives a billet of steel at temp X to have 100,000 times more IR than Visible not relevant to this? Is the Sun different?

    Yes it’s a damn sight hotter!
    If you’d take the trouble to look at the Stefan-Boltzmann equation you’d learn that cold surfaces like the earth emit in the IR around 10microns, your billet of steel around 2 microns, the sun around 0.5microns. The hotter something is the shorter the wavelength and the more energetic the light.

    Objects heating up give off Thermal IR way before the object reaches a degree of heat capable of producing white light, or any visible light at all. One figure I found was that the Sun’s Thermal IR was at 80%. This fits in better with the lightbulb and steel examples, makes more sense to me.

    You’re comparing objects at widely different temperature, and that figure for the sun is wrong.

    Anyway, the heat we feel on Earth coming from the Sun is Thermal IR, not Solar as Ira has it and as AGWScience teaches. Physics 101. See my post March 1, 2011 at 3:29 pm.

    The heat we ‘feel’ from the sun is about equal amounts of Vis and NIR, how much influence each has on a particular object depends on its absorbance at the particular wavelength, a black body will be heated by all wavelengths incident on it.

  236. Alan McIntire says:

    Check this out:

    http://www.geo.utexas.edu/courses/387H/Lectures/chap2.pdf

    See specifically figure 2.3 and 2.8, equation 2.16, figure 2.12 and related material.
    The example is for a graybody atmosphere. Note, as this article emphasizes, the atmosphere is not a graybody Say a gas is affected by a certain percentage of the spectrum, say 50%.

    From my link, a graybody atmosphere would have a flux of n + 1 = 1+1= 2^1/4
    or a flux of 1.4142 that of a planet with no atmosphere.

    This atmosphere blocks only half the radiation so the flux would be

    1/2 from ground to space atmosphere blocks 1/2, 1/4 from atm. to space,
    1/4 fromatm. to ground.

    ground 1 up , gets 3/4 from sun, 1/4 from atmosphere.

    So the total effect of a 1/2 atmosphere greenhouse is not half of a gray body, but
    1/3, the total additional flux is (4/3)^ 0.25, not (2)^.25 as in the example at the link.

    Now let the radion increase, say from the sun inreasing output. Now the gas, like CO2, which absorbs in a certain fraction of the infrared at the right tail of figure 2.3,
    is now absorbing only 1/3 of the flux. Again, balancing the above diagram,

    you’ll get a multiplicative effect of only (1 1/4)^0.25, not ( 1 1/3)^0.25 as in a 1/2 atmosphere model and not (2)^0.25 as in the article example.
    Although the TOTAL warming has gone up, thanks to the sun, the greenhouse
    multiplier has gone down- that’s what I meant by negative feedback.

  237. Vince Causey says:

    Richard Smith,

    You replied “No. Look at the Kiehl and Trenberth global energy flows diagram, for example. Earth receives 64wm2 (after deduction of thermals and evapotranspiration from insolation of 161 wm2) which somehow backradiation magnifies to an emission of 396wm2. This is more energy out than in.”

    Well, you may be right about kiehle and Trenberth, but all I’m saying is that if you take any body in general (not just the Earth), which has a source of heat, then if you have less energy leaving than entering, then it must warm up.

    The other points you have mentioned about being able to extract free energy if the GHG idea was true, I’m afraid I don’t understand. Perhaps you could explain in more detail why 1) GHG violates the laws of thermodynamics (if that is what you are saying) and 2) why this would lead to free energy.

  238. Domenic says:

    Are the data in HITRAN observed or calculated?

    The parameters in HITRAN are sometimes direct observations, but often calculated. These calculations are the result of various quantum-mechanical solutions. The goal of HITRAN is to have a theoretically self-consistent set of parameters, while at the same time attempting to maximize the accuracy. References for the source are included for the most important parameters on each line of the database.
    http://www.cfa.harvard.edu/hitran/

    The whole basis for the AGW argument rests on their misunderstanding of the true ‘greenhouse effect’. N2 and O2 are indeed members of ‘greenhouse gases’. ALL component gases of the atmosphere contribute to the ‘greenhouse effect’.

    It is because the AGW proponents ignore that fact, that the miniscule effects of a trace gas like CO2 is blown way out of proportion and they make absurd assumptions, claims, and predictions.

    For example:
    http://www.realclimate.org/index.php/archives/2005/04/water-vapour-feedback-or-forcing/

    Gavin Schmidt’s article here shows the complete fiction, fabrications of people completely deluded and ignorant of the basic sciences.

    Gavin noted his puzzlement over Lindzen’s observation here:

    “So where does the oft quoted “98%” number come from? This proves to be a little difficult to track down. Richard Lindzen quoted it from the IPCC (1990) report in a 1991 QJRMS review* as being the effect of water vapour and stratiform clouds alone, with CO2 being less than 2%. However, after some fruitless searching I cannot find anything in the report to justify that (anyone?). The calculations here (and from other investigators) do not support such a large number and I find it particularly odd that Lindzen’s estimate does not appear to allow for any overlap.”

    If Gavin had even a tiny background in heat transfer, and thermal properties, he would have recognized that the 98% H2O, 2% CO2 numbers come from their relative specific heat capacities times their respective amounts in the atmosphere.

    I actually think Lindzen was being generous regarding CO2 effects here. To me, the more accurate number would be
    CO2 0.05%
    ALL other ‘greenhouse gases 99.95%.

    CO2 literally can’t absorb enough heat to do much of anything even if it doubles or triples or quadruples….

  239. Domenic says:

    slight correction

    My last sentence should read

    “CO2 literally can’t absorb enough ENERGY to do much of anything even if it doubles or triples or quadruples….”

    (I am so used to thinking that heat and energy are interchangeable which I do automatically in my mind, which they indeed are, but that may not be so obvious to most people who wish to understand the science.)

  240. Dave Springer says:

    davidmhoffer says:
    March 2, 2011 at 2:25 am

    “Insulation isn’t quite the right word”

    It’s the best single word that most people can understand. Go a little deeper by saying it allows sunlight to pass straight through to warm the surface during the day but slows down the escape of that warmth at night and I think that’s pretty much the whole enchilada in terms the proverbial bartender can understand.

    I made up a rock & blanket analogy which I think is a good one. Take two black rocks and let them heat up in the sun during the day. At night put a blanket over one of them. In the morning remove the blanket and see which rock is the warmer one. That’s pretty much exactly how CO2 works – it allows the sun to warm things up during the day but prevents some of the warmth from escaping at night.

    Go even further and explain that the rock with the blanket over it won’t eventually melt from the retained heat. As the blanketed rock gets warmer and warmer with each passing day the blanket becomes less effective at retaining the heat and before long the blanketed rock will reach a stable maximum daytime temperature.

    Go even further by adding a second blanket after the stable temperature is reached. The rock will rise to an even higher stable maximum temperature but the temperature increase between no blanket and one blanket will be greater than the difference between one blanket and two blankets. Each additional blanket will have less warming effect than the previous blanket. That’s how CO2 works too. Nature gave us ten blankets to start with. Human activity (arguably) added four more blankets but our blankets aren’t nearly as effective as the blankets that were already there and each additional blanket we add in the future has less impact than the one added just previous to it.

    There’s nothing hard to understand about that and any objections that it’s not a 100% accurate analogy are pretty much pedantic.

    There’s no more a layman needs to know about how CO2 works to warm the planet.

    The rest of the debate gets far more complicated beginning with exactly how much surface warming there will be and what associated changes might come along with it.

    Addtional points that cannot be reasonably argued against are that atmospheric CO2 has been 10-20 times higher in the earth’s past, persisted that way for tens and hundreds of millions of years, and the earth was green from pole to pole during those times. There has never been a runaway greenhouse in the earth’s history. Green plants grow faster and use less water as atmospheric CO2 levels rise. Everything higher up the food chain benefits from more and larger green plants. The earth has experienced a number of runaway freezing events called ice ages. The earth has been in an ice age for the past 3 million years. Ice ages are bad for living things. The Holocene interglacial, the past 15,000 years, is a temporary respite from glaciers that bury New York City and all points north under a mile of ice. These interglacial respites last an average of 12,000 years while the glaciation side of the cycle last an average of 90,000 years. We’re overdue for the return of the ice age and should be justifiably concerned about doing things that will hasten its return. Worrying about doing things that will delay its returns is absurd. Worrying about global warming is absurd – we need all the global warming we can get.

  241. Bryan says:

    davidmhoffer

    The igloo insulates the person inside.
    It is not in itself a source of heat.
    It merely reduces the heat loss from the person.
    Its exactly the same situation as the insulating lagging of a hot water tank.

  242. Phil. says:

    Domenic says:
    March 2, 2011 at 7:22 am

    The whole basis for the AGW argument rests on their misunderstanding of the true ‘greenhouse effect’. N2 and O2 are indeed members of ‘greenhouse gases’. ALL component gases of the atmosphere contribute to the ‘greenhouse effect’.

    Absolute nonsense, take a look at the figure I cited above N2 and O2 contribute many orders of magnitude below CO2 and H20. It is you who have the ‘misunderstanding’ and in the face of the evidence to the contrary it seems more than that.
    Regarding the 98% number, the relative heat capacities have nothing to do with the GHE, that number is a fabrication by Lindzen.

  243. mkelly says:

    Phil. says:
    March 1, 2011 at 7:55 pm

    “No, how do you propose that the atmosphere gets hotter than the surface?”

    I don’t believe I ever said the atmosphere gets hotter than the surface, but if the pressure stayed the same then (PV=nRT) applies so the atmosphere would start at
    0 C. Then add heated molecules that cannot radiate and over time they would get to the temperature of the surface. No conduction exists between things of the same temperature.

  244. P. van der Meer says:

    >>>>>Ira Glickstein, PhD says:
    March 1, 2011 at 7:19 am
    I googled images for “outgoing infrared spectrum” and found this: http://www.skepticalscience.com/images/infrared_spectrum.jpg
    It should the spectrum looking up from somewhere in the arctic on a clear day and the view looking back down from 20 km. The incoming radiation would be the re-emitted IR from the atmosphere that you wanted. It clearly shows the 7 um, 10 um and 15 um bands Ira is talking about.
    THANKS Tim Folkerts for the link. I knew some WUWT reader would be alert enough to answer Fred Souder’s good question . GREAT LINK, let me repeat it: http://www.skepticalscience.com/images/infrared_spectrum.jpg
    Here it is for all to see:<<<<<

    The link shows a chart of upwelling radiation at 20km looking down and downwelling radiation at the surface looking up. As I see it, the upwelling radiation is only between 9.25μ and 10μ (1080 and 1000 cm^-1 wave number). But if the N2 and O2 molecules are radiating like a dense body without absorbing any radiation, as so many seem to be claiming, wouldn't one expect to see a BB radiation between ~8μ and ~14μ (1250 cm^-1 and 715 cm^-1 wave number). This would be the open window between the H2O and CO2 absorption windows.

    As that doesn't seem to be the case here, one can only conclude that the blip on the chart is due to something else unknown, but certainly not downwelling radiation from N2 and O2. The model as presented by Ira still does not reflect reality!

    My gut feeling is that if N2 and O2 were indeed radiating, it would have a disastrous effect on the cooling mechanism of this planet. But that would require more looking into!

  245. Phil. says:

    mkelly says:
    March 2, 2011 at 8:29 am
    Phil. says:
    March 1, 2011 at 7:55 pm

    “No, how do you propose that the atmosphere gets hotter than the surface?”

    I don’t believe I ever said the atmosphere gets hotter than the surface,

    No but you said this:
    “By the way by your statement above you are in the camp that says without GHG we would be a very hot world indeed. Not at -18 C.”
    So you certainly seem to think it for some reason, most likely a cold ice-covered surface with a troposphere like Triton’s.

    but if the pressure stayed the same then (PV=nRT) applies so the atmosphere would start at 0 C. </em.

    This is a bizarre statement with no basis in science, why do you think it should be at 0ºC, and why would PV=nRT lead to that?

    Then add heated molecules that cannot radiate and over time they would get to the temperature of the surface. No conduction exists between things of the same temperature.

    They’d cool with altitude just like the N2 atmosphere of Triton does, above that a stratosphere.

  246. Don V says:
    March 2, 2011 at 12:09 am
    … I’m sorry Ira, but someone has convinced you to believe in a concept that violates the basic laws of thermodynamics.

    Laws are like pie crusts, made to be broken (paraphrasing Lenin). I am sorry, but you have misinterpreted the laws of thermodynamics, at least in the case of radiation. No laws have been broken here.

    Unless you can convince me that this model does not violate both the first and second law of thermodynamics, I will have to conclude that you have drunk some of the kool-aid. I am not trying to insult you, I respect you, and would like to convince you of the truth, but to do that you will have to get back to the basics, and admit this is just plain wrong!

    A moral, sensible, well-bred person will not insult me, and none other can. (Paraphrasing William Cowper)

    There is so much wrong I would have to spend an inordinate amount of time writing, so I will focus on just one aspect of it. Your animated illustration at the far right shows the atmosphere absorbing energy radiated up from the earth at 15μ . Think about this carefully now. To do this which body has to be hotter? Earth right? OK so how can the atmosphere with only 15μ radiant energy warm up so hot that it exceeds the earths temperature and radiates back at two wavelengths 7μ and 10μ!? It can’t! The only way you can create a black body radiator at 7μ or at 10μ is to heat a body up to the hotter temp that corresponds to that wavelength. IF it radiates at any black body energy at all, to obey the first law of thermodynamics, it MUST radiate at energy equal to or longer than 15μ. So your illustration and explanation violate both the first and second law since 1) to be true it illustrates that energy in the atmosphere must have been created from nothing, and 2) it illustrates that a cooler body is capable heating up a warmer body. If I am wrong on this one point then please correct me, and teach me how this can be true. …

    Please have a look at the graphs (above in this comment thread) immediately after my comment with the following time stamp:

    Ira Glickstein, PhD says:
    March 1, 2011 at 7:19 am …

    Please look at the second graph and notice that it is labeled “(b) Surface looking up”. Notice that the wavelengths are at the top of the graph and there is considerable radiation coming DOWN to the Earth from the Atmosphere at ~7μ, ~10μ, and -especially- ~15μ. That is what I was trying to help visualize in my graphics.

    The second law of thermodynamics refers to the NET transfer of heat energy that must be from warmer to cooler bodies. But, as was noted earlier in this thread, if you are spraying me with a fire hose and I am simultaneously spraying you with a garden hose that has blue-tinted water, I will get a lot wetter, but you will get some blue water on your clothes.

    The first law of thermodynamics of course prevents the Atmosphere from creating energy out of nothing. But, it is not doing that in my graphic. It is simply returning some of the ~7μ, ~10μ, and ~15μ energy that was radiated UP from the Earth to the Atmosphere in a previous time slot of the animated graphic. The energy being returned is a portion of the energy radiated UP from the Earth, after it has been absorbed and re-emitted by the GHGs in the Atmosphere. If I give you $1000 in bills and change, and you give that money to your friends and they give some money back to you and you later return $400, possibly in different denominations, to me, you have not created that money, have you?

    I hope this clears things up a bit DonV.

  247. davidmhoffer says:

    Dave Springer;
    I used to use “insulation” and it is a pretty good description. Someone determined to “not believe” almost always comes up with the same objection. If you have a warm building, but with no heat source in it (just assume it is already warm) it will cool off slower with insulation than without insulation, but wrapping the insulation around it would never make the building temperature increase.

    So then we’re off on the merry go round, explaining that the insulation can let some energy in and some not out as fast, and the combination is a warmer building. And then…

    I still maintain that the various explanations are splitting hairs in the end. The IPCC hollers doubling CO2 = 3.7 w/m2 = +1 degree. But the quietly forget that they are talking about +1 at the “effective black body” temp of the earth, which is -19 as opposed to the average surface temperature which is +15. They don’t even make the pretense of converting via Stefan Boltzman to what that would mean at surface, which is more like 0.6, they even go so far as to say that there may be no change at all to surface temps, but climate will change anyway. Nor is there any discussion that there is not such thing as an “average” temp in the climate sense. If we accept +1 at effective black body and then extrapolate from SB to surface at +0.6, then we need to extrapolate even more…what happens in the tropics which are at +30? about +0.2 and at the poles at -40? about + 4 (guestimates for illustrative purposes). So, if the hot days in the tropics go from 30.0 to 30.2, will anyone notice? will the polar bears in their winter dens notice that its -36 instead of -40?

    My point being that there is a lot of science in IPCC AR4 that is accurate, but presented in such a way that it is completely misleading. Just getting through the effective black body discussion knocks half their argument out of existance, explains why water vapour doesn’t increase across the board as predicted, that most of the warming happens at nigh time lows, in winter, in arctic zones, and so pretty much doesn’t matter.

    Give me a white board and 30 minutes to go through that and I’ll swing far more warmists to the skeptic side than days and days and days of refining co2 models down to photon by photon accuracy.

  248. davidmhoffer says:

    Bryan,
    When you crawl into the igloo, exactly how does it reduce the heat loss from the person? Do the photons rising from the person’s skin get traffic reports telling them that there’s an igloo out there, slow down?

    No, the igloo doesn’t generate any heat. But it does absorb heat from you when you crawl in. Some of that heat it passes along to the outside, and some it radiates back at you. So you are in fact being warmed by heat from a colder surface radiating toward you.

  249. Bryan says:

    davidmhoffer

    The igloo insulates in exactly the same way as the jacket around a hot water tank.
    The material chosen for the insulator will reduce heat loss by conduction, convection and radiation
    The jacket is in fact a better radiator than the atmosphere.
    Yet no one would talk about the “greenhouse effect” of a water tank jacket.

  250. George E. Smith says:

    “”””” P. van der Meer says:
    March 2, 2011 at 8:36 am
    >>>>>Ira Glickstein, PhD says:
    March 1, 2011 at 7:19 am
    I googled images for “outgoing infrared spectrum” and found this: http://www.skepticalscience.com/images/infrared_spectrum.jpg
    It should the spectrum looking up from somewhere in the arctic on a clear day and the view looking back down from 20 km. The incoming radiation would be the re-emitted IR from the atmosphere that you wanted. It clearly shows the 7 um, 10 um and 15 um bands Ira is talking about.
    THANKS Tim Folkerts for the link. I knew some WUWT reader would be alert enough to answer Fred Souder’s good question . GREAT LINK, let me repeat it: http://www.skepticalscience.com/images/infrared_spectrum.jpg
    Here it is for all to see:<<<<<

    The link shows a chart of upwelling radiation at 20km looking down and downwelling radiation at the surface looking up. As I see it, the upwelling radiation is only between 9.25μ and 10μ (1080 and 1000 cm^-1 wave number). But if the N2 and O2 molecules are radiating like a dense body without absorbing any radiation, as so many seem to be claiming, wouldn't one expect to see a BB radiation between ~8μ and ~14μ (1250 cm^-1 and 715 cm^-1 wave number). This would be the open window between the H2O and CO2 absorption windows.

    As that doesn't seem to be the case here, one can only conclude that the blip on the chart is due to something else unknown, but certainly not downwelling radiation from N2 and O2. """""

    The mystery "blip" PvdM is the Ozone (O3) blip at around 9.6 microns.

    I presume these are calculated spectra; not actual measured data. I have no idea what all the multitude of BB dotted spectra at various Temperatures are all about. The 260 and 270 K plots are the only ones of any interest.
    In any case, you can'tr ead too much into these graphs; THEY DO NOT SUM TO 1.0.

    In other words, the bottom graph (b) of surface looking up, can reasonably be presumed to be downward radiation from the atmosphere, including atmospheric emissions, reflections, or scatterings. Personally, I would expect the atmosphere looking down to look pretty much the same as the atmosphere looking up. The upper graph (a) does NOT prove that it is not.

    Clearly, the bulk of that emission in (a) looking down is emitted from the earth surface, and most of it is simply passing through the atmosphere.

    The (a) diagram should be contrasted with a very similar graph published by Peter Humbug, in his January 2011 Physics Today article. In his fig 3 he plots data from a calculated model, and also actual measurements from the AIRS Instrument.

    The graph in PT much more closely tracks the 285 K BB curve, which is what I would expect, since the supposed mean global surface temperature is 288 K, and that is the principal source of the outgoing radiation.

    Where PH differs dramatically from the Skeptical Science graph you cite, is that the SS graph is a plot of Radiance; actually Spectral Radiance, in units of (m)Watts per m^2 per Steradian per cm^-1 and peaks around 100 .

    The Physics Today plot is a plot of FLUX measured in Watts per m^2 .steradian or W/m^2/sr if you want to put it that way.

    This is NOT a plot of Spectral Radiance as it has NO per cm^-1 units, and has a curve peak of around 0.11 in those FLUX units.

    The SS plots would have a similar peak value of 0.1, if the units were Watts per m^2/Sr/cm^-1, rather than milliWatts; but the units are entirely different for the Physics Today graph.

    In any case, the dissimilarity of the two graphs (a) and (b) merely demonstrates that most of the escaping radiation is emiited from the surface, and passes mostly through the atmosphere to space.

    One should note that if Kirchoff's law was applicable, and the emission spectrum and the absorption spectrum were identical by reason of that law, then by definition the radiation would be a complete black body spectrum at the equlibrium Temperature, and it wouldn't have any holes in it. The fact that the observed spectrum in the case of the AIRS instrument, does have holes in it, is prima facie evidence that Kirchoff's law is NOT applicable, and there is no equilibrium condition.

    The PT article contains some totally surprising claims; but then Peter Humbug is a Professor of Geophysical Sciences; and not just any Professor, but THE Louis Block Professor, at the University of Chicago, so I would take everything he says as gospel; even thoguh I don't undertand much of it; such as the earth heating to 800,000 K in a billion years, if it had no means of ridding itself of the energy it gets from the sun. If that were the case I would have expected the entire earth to have evaporated ages ago.

    The

  251. George E. Smith says:

    “”””” davidmhoffer says:
    March 2, 2011 at 3:24 am
    George E Smith;
    I read the papers on CO2 being “lumpy” quite a while ago and only skimmed them. My recollection is that what you say is pretty much correct. That said, the distribution of CO2 is still FAR more even than water vapour in the atmosphere. Convection at the equator in particular drives CO2 upward as well as water vapour. As temps cool, the holding capacity of the air drops, driving water vapour out as droplets, snow, what ever the balance of conditions dictate. I’m guessing that some of the CO2 gets scrubbed out as well with the rain/snow, but the bulk of it continues upward. Point being that there is a band below which water vapour dominates the absorption spectrums where the two overlap, and a band above which CO2 becomes significant by comparison. Insignificant to climate, but still not representated well by a lab cylinder with artificial atmosphere or two plates of what ever with CO2 between them.

    I’d far rather spend time getting people to understand the fallacy of this thing called an “average” temperature, or the misleading way the IPCC calculates and presents sensitivity, but I don’t get a lot of traction on those issues. “””””

    David, I don’t think we are in much disagreement about any of this. If you have gleaned an impression, that I considert he global CO2 “lumpiness” to be important; please let me disabuse you of that notion; I agree with your thought that H2O is far lumpier. For me personally, my ONLY interest in the CO2 lumpiness, is that it can lead to some understanding of what mechanisms are causing seasonal CO2 variations. And the relatively rapid variations in that lumpiness suggest to me that things like “residence Time” of CO2 is a way overblown issue. I don’t have any problems accepting the usual claimed amounts of CO2, although I don’t necessarily accept claims as to its origins; but none of that matters to me, because I don’t think CO2 is the controlling factor.

    As to your comment on the “Mean Global Temperature”; well I guess it has been some time since I last referred to that as being similar in importance to the average telephone number in the Manhattan phone directory. I’m in total agreement with you that the mean global Temperature is neither known nor of any real value if we did know it.

    I’m still mystified by repeated claims by the CO2 fans that H2O is not a GHG, but simply an amplification of the effect of CO2, and that absent CO2; the real green house gas, the earth would simply be a frozen ice ball.

    This claim has been most recently been made in the much publicised featured article in the Jan 2011 issue of Physics Today, by Peter Humbug; climate modeller supreme.

    Without the heating by CO2, he says there would be no water vapor in the atmosphere; well to be fair he didn’t say that but he did say it would all rain out. That to me DOES mean there would be NO CLOUDS.

    So let’s take his fozen ice ball, which presumably will be below 273 K

    Well the globally averaged solar input would be 340.5 W/m^2 based on the newly determined 1362 TSI number. That would put the radiation balanced equilibrium Temperature at 278.4 K Well that is before we take out the ozone hole at 9.6 microns.

    It’s not a very deep hole but maybe it can cut the incoming enough to squeeze under 273 K, and get to an ice ball Temperature.

    At this point the critice will say, that this frozen ice ball will reflect 80% of the sunlight back out into space, so the Temperature would go much lower. So it seems that PH is on solid ground; well solid ice.

    What I object to in this picture is two assumptions. We already have places on earth where the Temperature routinely gets below 250 K even with all the CO2 (well mixed), and those places still have plenty of clouds, made of ice crystals, so all the water has not rained out.

    My biggest objection is that the solar input to planet earth is NOT 340.5 W/m^2; that is the globally averaged value. The sun is actually delivering 1362 w/m^2 but only to a portion of the earth; half of it roughly; but non-uniformly.

    If you shine that 1362 W/m^2 blow torch on to the surface of a frozen ice ball at 250 K Temeprature, in the earth tropical zones, those places will NOT remain frozen. More importantly, if you remove ALL the CO2 and
    ALL the H2O, from the atmosphere; they can never freeze in the first place; but i’ll humor him and assume they are frozen; so the sun got shut off for a long time to allow that to happen.

    That energy is going to melt the surface of any ice ball, which will refreeze at night to become a solid ice rather than snow, and that ice will trnasmit a whole lot of solar energy, which will eventually get trapped by TIR in the ice “glass”.

    More importantly, H2O molecules will sublime off that frozen surface; and you WILL get H2O vapor in the atmosphere; even at 250 K Temperature.

    Well I won’t bore you with the gory details; but iceball earth is a crazy construct of Trenberth’s equally crazy cartoon of Earth’s energy balance. The real earth DOES NOT AVERAGE the sun’s energy input. Its thermal conductivity is not high enough to equilibrate to some isothermal Temperature of Trenberth’s model.
    It is heated by that 1362 W/m^2 attenuated by whatever absorptive gases there are INCLUDING BOTH CO2 AND H2O, which both absorb part of the solar spectrum; and that heat is stored mostly in the very slowly equlibrating oceans, so tomorrow morning when the sun rises again, the local heating from yesterday’s sunbathing, will have not dissipated, and so tomorrow it will continue to get hotter until we again reach today’s conditions.

    Iceball earth is a silly idea; not based on reality.

  252. George E. Smith says:

    “”””” Myrrh says:
    March 2, 2011 at 2:44 am
    George E. Smith says,
    March 1, 2011 at 5:52 pm

    I posted it for the variations of explanations… You could add yours to it. “””””

    Well Myrrh, I never said any such thing. I would appreciate it if you would not cite stuff as being originated by me; when I said no such thing. I have NO PROBLEM if you DO want to cut and paste things I HAVE said; I do that all the time; but pleae do NOT put anything under a label:-

    “”””” George E. Smith says,
    March 1, 2011 at 5:52 pm “””

    Unless it IS something I did say.

  253. mkelly says:

    Phil. says:
    March 2, 2011 at 9:19 am

    Phil, the -18 C comes from the standard model of radiated heat transfer given approx 240 w/m2 at surface. Thus having GHG’s take us up 33 degrees to 15 C.

    Without any GHG no H2o, CO2 etc the sun would essentially have free rein with no absorbtion, so surface would get 340 w/m2 thus raising up to 278 K or 5 C to start. I think the 340 number used is to low. It can easly be show via a pan of water and a thermometer that large protion of the surface get 700-1000 w/m2. Heck SOD has a graph on his web site showing the radiance of the sun on a plot of ground in Canada where measured solar radiation is about 700 w/m2.

    I think that ignoring the (PV=nRT) pressure of the atmosphere is a mistake. STP is 0 C. So not a bizarre statement. You are free to agree or disagree as you wish. So at most for me GHG’s will only add 15 degrees C so the effect of CO2 is less than thought. If any at all.

    Also, would N2 or O2 cool if they went up. If they cannot radiate just going up will do little. Granted some work will be done so energy will be lost but how much?

    As 99% of atmosphere is N2 and O2 we need to look at them. I am trying to find out. I have an idea they do radiate just like any matter at a temperature but I have not seen a temperature graph for each showing their respective pattens or amounts. If they do not I will change my idea.

  254. RJ says:

    No, the igloo doesn’t generate any heat. But it does absorb heat from you when you crawl in. Some of that heat it passes along to the outside, and some it radiates back at you. So you are in fact being warmed by heat from a colder surface radiating toward you.

    Is the above post serious or not?

  255. davidmhoffer says:

    George E Smith;
    EXACTLY! EXCELLENT!

    RJ says:
    March 2, 2011 at 11:44 am
    No, the igloo doesn’t generate any heat. But it does absorb heat from you when you crawl in. Some of that heat it passes along to the outside, and some it radiates back at you. So you are in fact being warmed by heat from a colder surface radiating toward you.>>>
    Is the above post serious or not?>>>>

    Oh for gosh sakes. Are you seriously asking if I’m serious? I quit, I’m not answering anymore. Go to the top, read the article, then read every comment afterward if you undersand it or not. When you get back down to the bottom, please advise if you think I am serious, and also if you think I am right. Get one of two wrong, and you have to go back to the top and do it again. Two of two wrong… I got no cure for that.

  256. George E. Smith says:

    You know Ilearned just enough Physics to be hazard to myself, and anyone who is foolish enough to believe what I write, without sanity checking it for themselves.

    Some times I rely on the fact that Phil is lurking out there somewhere, and if I totally screw up, that he will jump all over me.

    I never learned quite enough Wave or Quantum mechanics much beyond the Bohr Sommerfeld Atom; and I only recently realized that Maxwell’s equations didn’t go into the trash can as a result of a quite arbitrary, and unsupportable assumption by Bohr, that Maxwell be damned, an electron in orbit around a nucleus, did not have to radiate per Maxwell, so long as it stayed in one of those orbits. If I had studied another year beyond that, I would have found out that the Bohr Sommerfeld elliptical planetary orbit concept was what was wrong; despite its spectacular importance in explaining the mysteries of Atomic line spectra.
    If now the electron was just “somewhere there” (who said anything about moving), the continuous acceleration vanishes, and Maxwell is redeemed. Must have been since he is now enshrined in the exactly defined value of (c) and its Maxwellian components; epsilon-naught, and mu-naught.

    There was a time when I could actually solve the Schrodinger equation for some simple cases.

    So right now, in front of me on my desk, I have the definitive Extended Edition of the Feynman Lectures, as Well as Gerhard Herzberg’s Atomic Spectra, and Atomic structure, plus Feynman’s little QED book.

    So if I have to solve the Shrodinger equation for the entire atmosphere; sooner or later, I am going to figure out how the radiation of the atmosphere really works. And I’m not overjoyed with many of the explanations I have heard or read.

    But even that is just a curiosity of mine. I simply note that every single addition of another H2O molecule, or for that matter a CO2 molecule to the atmosphere anywhere on earth MUST reduce the total amount of direct solar spectrum energy that reaches the surface of the earth (solid or liquid); regardles of whatever else happens.

    And NOBODY ever observed it to get warmer (higher Temperature) in the shadow zone, when a cloud passes in front of the sun; it ALWAYS gets cooler (lower Temperature) And of course I also believe that it is the warmer more humid conditions at the surface, during the day, that creates the high level balmy cloud conditions at night; those clouds did NOT create the conditions on the ground.

  257. Myrrh says:

    George E. Smith – I was replying to what you said. Apologies if that wasn’t clear.

    If it had been a quote I would have used quotation marks or put it into italics.

  258. Al Tekhasski says:

    Bill Illis posted earlier an IR image of Earth, but did not submit optical conditions for the image. He was complaining that continents are not visible, therefore the infamous “IR transparency window” of real atmosphere seems to be in doubt. Here is what I found, an image in 11um band from GOES-10 satellite, 1998:
    http://goes.gsfc.nasa.gov/pub/goes/980221.goes10.ir4.gif
    Mexico, California, Florida, entire America is pretty much recognizable. Same goes for 12um (Ch5) band:
    http://goes.gsfc.nasa.gov/pub/goes/980221.goes10.ir5.gif
    Enjoy.

    I wonder, here are many posters interested in IR science. Could anyone return a favor and point to reasonably-resolved OLR spectra averaged for ALL_WEATHER CONDITIONS? Not like the one clear sky from arctics and another from tropics. Somehow I have difficulties to locate this kind of data. I know that Griggs and Harris went through some serious efforts to sort out 99% of all available IRIS (1970) spectra in order to select clear-sky for their particular purpose. I wonder where I can find the remaining 99% part of IRIS data. Or similar. Thanks in advance.

  259. Myrrh says:

    Re: Ira Glickstein, PhD says:
    March 2, 2011 at 9:22 am

    The second Law of thermodynamics refers to the NET transfer of heat energy that must be from warmer to cooler bodies. But, as was noted earlier in this thread, if you are spraying me with a fire hose and I am simultaneously spraying you with garden hose that has blue-tinted water, I will get a lot wetter, but you will get some blue water on your clothes.

    When, question to all here, did this idea of “NET” appear? I recall reading a discussion a few years ago when some applied scientist was arguing this point and saying that word wasn’t in the original statements of the law, he quoted from old physics text books he had learned from as well as current ones in use in his industry.

    Also, is this of any use in the Thermodynamics discussion going on here?

    Thermodynamics – Historical Background, Temperature, Expansion Coefficients, Thermostats, Water, Heat, The First Law of Thermodynamics – Conservation of energy

    http://science.jrank.org/pages/6816/Thermodynamics.html

    and, http://science.jrank.org/page/6814/Thermodynamics-second-law-thermodynamics.html

    From the second, bearing in mind the blue water in the exchange described by Ira above:

    “One of the most remarkable facts of nature is that certain processes take place in only one direction. For example, if a high temperature object is placed in contact with one of lower temperature, heat flows from the hotter to the cooler until the temperatures become equal. In this case (where there is no work done), the first law simply requires that the energy lost by one object should be equal to that gained by the other object (through the mechanism of heat flow), but does not prescribe the direction of the energy flow. Yet, in a situation like this, heat never flows from the cooler to the hotter object. Similarly, when a drop of ink is placed in a glass of water which is then stirred, the ink distributes itself throughout the water. Yet no amount of stirring will make the uniformly-distributed ink go back into a single drop. An open bottle of perfume … . These are all examples of the second law of thermodynamics, which is usually stated in two different ways. Although the two statements appear quite different, it can be shown that they are equivalent and that each one implies the other.”

    The first page goes through examples in various practical applications to explain these laws. As far as I can tell, (Don V et al?) none has any idea of there being some “NET” exchange concept.

    Re: Boltzmann -this is interesting, maybe, on http://science.jrank.org/pages/19739/Physics-SECOND-LAW-THERMODYNAMICS.html

    It say that “Bolzmann had less success answering a second set of objections regarding atomism. The British physicist William Thomson (1824-1907) and Scottish physicist Peter Tait (1831-1901) rejected atomism as a result of their adherence to the dynamical theory of matter, which rejected the existence of a void. Similarly, Ernst Mach put forward empiricist counterarguments, which rejected Boltzmann’s adherence to entities that could not be confirmed by direct observation.”

    Is this void aspect involved somehow in how the 1st and 2nd laws are understood differently by AGWScience and Real Science in the energy exchange argument? And, is this connected with the NASA rejection of Stefan-Bolzmann for their moon landing: “..NASA, proved there was no such thing as a greenhouse gas effect because the ‘blackbody’ numbers supporting the theory didn’t add up in a 3-dimensional universe:”

    http://sppiblog.org/tag/stefan-boltzmann-equations

  260. kuhnkat says:

    George E. Smith,

    “And NOBODY ever observed it to get warmer (higher Temperature) in the shadow zone, when a cloud passes in front of the sun; it ALWAYS gets cooler (lower Temperature) And of course I also believe that it is the warmer more humid conditions at the surface, during the day, that creates the high level balmy cloud conditions at night; those clouds did NOT create the conditions on the ground.”

    Careful, I think they are trying to pass laws so they can again burn heretics!! 8>)

  261. Myrrh says:

    Bryan says:
    March 2, 2011 at 10:31 am

    The igloo insulates in exactly the same way as the jacket around a hot water tank. The material chosen for the insulator will reduce heat loss by conduction, convection and radiation
    The jacket is in fact a better radiator than the atmosphere.
    Yet no one would talk about the “greenhouse effect” of a water tank jacket.

    In my attempt to work out just how big this “insulating blanket of CO2″ was of AGW claim, I calculated that to wrap around an area the size of the human body and if all the CO2 was gathered in one place it would amount to an insulating blanket around 7cm2 in a 2 square metre, 1 square inch in a couple of square yards blanket. (Roughly, taking skin area of human body and CO2 at 400 ppm). Some blanket! If this was “well-mixed” as AGW claims, where’s the CO2 blanket?

    Even if AGW use the unsubstantiated claim that CO2 stays in the atmosphere accumulating for hundreds and even thousands of years, take your pick, they make up the numbers to suit, and double current amounts actually stayed in the atmosphere, this would be nonsense as “insulating blanket” to not only stop heat loss globally but raise the global temperature of the Earth.

    So, no igloo, just a really really insignificant scattering of CO2 molecules around the frozen body in the snowy wastes.

  262. Ira, I have spent hours writing an essay in explanation of why we think differently and just now I have accidentally deleted it.- So for now, – Ok the Gods are telling me you are right.

    AGW is happening!

    But I will be back!

  263. Tim Folkerts says:

    Myrrh asks “When, question to all here, did this idea of “NET” appear? ”

    I think it has always been there. Thermodynamics has always involved statistics and averages. If Object A and Object B are in contact and at the same temperature, then there is no flow of energy either direction. But since the energies of particles in the two objects have a wide random variation in speeds, any given collision between an atom of A and an atom of B will transfer energy. There will be millions of transfers of energy from A to B. There will be millions of transfers of energy from B to A.

    There is no violation of the 2nd law. While A is transferring energy to B, B is transferring a similar amount of energy to A. Ie the NET transfer of energy is zero (at least when averaged over sufficiently long periods of time to average out short-term statistical fluctuations).

    Similarly, if A and B are interacting via photons, A will be transferring lots of energy to B, and B will be transferring lots of energy to A. The net flow is zero.

    Even if B is colder than A, there will be some collisions where B transfers energy to A. There will be some photons that transfer energy from B to A. There will just be MORE collisions and MORE photons transferring energy from A to B, so the NET transfer is from the warmer object to the cooler object.

  264. wayne says:

    I have to clarify myself on the aspect of oxygen and nitrogen absorption/emission in our atmosphere. After some detailed digging it appears there is a continuum absorption and emission but the exact nature and the mechanisms is where I was off a bit. Vibrational and rotational excitations are quantized and darn I already knew that, but for some reason that wasn’t registering, no sleep. ☺

    Here’s the gist gathered from some .mil and hitran and hartcode papers and sites: both oxygen and nitrogen do have bands in the infrared and both also have continuums about those lines. Oxygen’s is in the range of 1345-1820 cm-1(5.5-7.4 µm) and nitrogen is located in the range 2105-2620 cm-1(3.8-4.8 µm). I read HARTCODE is adding O2 & N2 continuum capabilities also. So how does the atmosphere as a whole handle a small portion of the energy within our atmosphere and help move it to space? Never found out how much.

    The long wave infrared radiation from the surface is as a near black body continuous spectrum and the tail of this distribution on the short wave side is where both are located. These photons in these ranges will be immediately be absorbed by the respective O2 or N2 molecules near the surface, that is a swag but a sure one. But is that it? No. Not to my best knowledge. That vibrational energy will mainly thermalized just as CO2 or H2O when excited and only sometimes re-emit a photon.

    But that raises another question. Can O2 or N2 get excited from within the atmosphere? Seems yes, but it can only occur where hit by a fast enough moving molecule that adds the right additional energy to put it into its respective continuum listed above or exactly at one of the strong spectrum lines, all by electron interaction. Then the same sequence happens as explained above, thermalized. I have yet to find a figure estimating the amount of energy that flow in this manner but would like to know.

    Now back to the 8 to 12 µm window. Does a continuum exist there? Yes. That is covered by the water vapors continuum that spans basically the entire IR spectrum to some degree or another and that is a very curious subject that everyone interested in this post should Google and read some about it. Many links. Search “H2O continuum in infrared”. Also that is why almost spectrums you ever see are taken in Antarctica, cold and little water vapor.

    I have yet to think through this, to see if this has any real influence, but much of this about continuums was all news to me.

    If anyone can fill in more info along these lines, please do so.

  265. Brian H says:

    Dave Springer says:
    March 2, 2011 at 7:51 am

    But but but — it might cause a tipping point tipping and it might tip any old which way because tipping points are so tippy, and all! Just a tip for you.
    ;)

    Warming = More Life. Cooling = More Death.
    Pick one!

  266. Brian H says:

    davidmhoffer says:
    March 2, 2011 at 9:38 am

    Bryan,
    When you crawl into the igloo, exactly how does it reduce the heat loss from the person? Do the photons rising from the person’s skin get traffic reports telling them that there’s an igloo out there, slow down?

    No, the igloo doesn’t generate any heat. But it does absorb heat from you when you crawl in. Some of that heat it passes along to the outside, and some it radiates back at you. So you are in fact being warmed by heat from a colder surface radiating toward you.

    Igloos and snow caves operate more or less like any enclosed space: they trap warm air. It helps that ice and snow reflect really well, of course. Re-radiation back atcha is trivial in an igloo.

  267. wayne says:

    Also on my [March 2, 2011 at 5:32 pm ] comment:

    I mentioned the Antarctica spectrums you see almost everywhere, this is why they are a distorted view of the Earth’s normal spectrum. The oxygen and nitrogen continuums are on the high frequency edge of the Planck gray body curve (assuming Earth’s emissivity to be about 0.96), so as it gets colder, at higher altitudes or high latitudes, the curve shifts to the lower frequencies where these two continuums are of little effect and not visible in spectrums.

    The same goes for the water vapor continuum, at locations where there is very little humidity, this continuum is not apparent. And where is that? Over deserts and the arctic or above the top of the atmosphere (TOA). And this might just be coincidence, but if you wanted to show spectrums that wrongly magnify CO2’s influence, where would you take them from? Right, high altitude, over deserts, or at one of the poles and that is exactly where all of the spectrums on the internet are located that I have come across anyway. Now that is curious.

  268. Domenic says:

    to Wayne

    You are looking in the right directions.

    Consider a single N2 molecule at the upper edge of the atmosphere, right on the border of outer space, if you will.

    Now, the boundary conditions of radiational heat transfer theory are two states (1) a blackbody perfect emitter/absorber and (2) a perfect reflector which is a non emitter/non absorber. Neither one exists in total in the real world. And all material objects (matter, atoms, molecules) exist in between.

    Now, our hypothetical N2 molecule at the edge of space, will, regardless of its absortivity, reflectivity, and transmissivity values given in ANY text books, models, calculation, whatever reference you will find… DESPITE ALL THAT….it will indeed radiate whatever internal energy it has into the 4 deg K thermal background radiation of outer space. And it can only give up that energy through radiation. Conduction and convection don’t work in outer space. Think about that.

    We know that because the only alternative is for N2 to be a ‘perfect reflector’ where it can somehow maintain an elevated internal energy despite its environment. And such things do not exist. At least there are none that I know of that have ever been found.

    This is a point I keep on bringing up. That ‘longwave’ emission/absorption data for atmospheric gases has never been properly tested under REAL LIFE CONDITIONS.

    They are calculated from various theories, opinions, assumptions, etc. The graphs, etc for absorption/transmission beyond 10 or 15 microns are at best just speculations, and at worst, just wild-assed guesses. Personally, I am of the opinion that they are more wild-assed guesses than anything.

  269. Myrrh says:

    Phil. says:
    March 2, 2011 at 5:59 am

    Re: Myrrh says to George E. Smith:
    “But see my post above. How is the example of real light bulb which shows the proportion of Visible to Thermal IR at 5/95, and the real life physics from the steel industry which gives a billet of steel at temp X to have 100,000 times more IR than Visible not relevant to this? Is the Sun different?”

    Yes it’s a damn sight hotter!
    If you’d take the trouble to look at the Stefan-Bolzmann equation you’d learn that cold surfaces like the earth emit in the IR around 10 microns, your billet of steel around 2 microns, the sun around 0.5 microns. The hotter something is the shorter the wavelength and the more energetic the light.

    Because it’s a damn sight hotter it begins emitting in the Visible Light spectrum, in other words, heat creates light. It doesn’t mean that light is hot.

    See the NASA page I posted from, Near IR is shortwave next to Visible Light and is COOL.

    Short wave energies are not Thermal Energies, they may well be more energetic, but that does not mean they have the ability to warm things.

    That’s why there’s a difference in terms used, Light energies and Heat energies.

    Thermal IR heats things, we feel it as heat from the Sun, we do not feel the rainbow of colours from the Sun as heat, they do not warm us up.

    If any of these rainbow of colours is intensified in some way, concentrated, it can burn surfaces and even burn through, but, luckily for us, Visible Light and short wave IR is Reflective, it bounces off things rather than penetrates. It is the longer wave Thermal IR that heats things, heats the Earth, heats organic matter, heats us, that we feel as penetrating heat from the Sun. We cannot feel the colours, they are Cool. See NASA page.

    Earth and we also emit around 10 microns, that’s Thermal IR. Our bodies naturally balance in and out heat to maintain constant temperature, our source of energy is food. The Earth just carries on cooling down from any heat it has obtained once the source of heat for it is gone, with no Sun at night in the desert for example, the temperature drops rapidly and the ground will give up its heat, or rather the colder atmosphere will take heat from it. Heat flows spontaneously from hotter body to colder, but I’ve also seen this explained as: if we grab a colder object, say snowball, it’s not the cold of the object we’re feeling, but the heat loss in our hand.

    So, it doesn’t matter that the Light is more energetic from a hotter object, it doesn’t have the ability to warm organic stuff, we and the Earth are organic. It can burn, UV we can’t feel but if particularly intense, as experienced by a body not used to it, it will burn the skin. It doesn’t penetrate, is not absorbed by the body to create warmth as Thermal IR is absorbed, it doesn’t even make it through even the first layer of skin, the epidermis. However ‘more energy’ it has is irrelevant, it can’t warm organic matter. Thermal energy, past the short wave Near IR, penetrates into matter and warms it.

    The AGWScience claim that Solar energies (Visible and the neighbour shortwave either side), is what warms the Earth, is total —- (insert favourite nonsense noun).

    My reference was to the information I gave on light bulbs and steel billet. Please, do find the links and read what they say. A light bulb gives off 95% of its energy as Heat, this is Thermal IR, a.k.a. Heat Energy. Light energy at 5% of the total energy given off is not giving us any heat, regardless that it is more energetic. It is giving off these two energies, light and heat, at the same time. It isn’t that one grows into the other… The other example of the steel billet at a certain temperature giving off 100,000 times more heat energy than visible light energy, shows amount, most of, the energy given off is in the form of Thermal IR..

    The ‘peak’ energy as Visible light is only a choice of reference. If you take the amount of energy that actually warms things up, the Thermal IR, then the peak energy in lightbulb and steel billet is Thermal IR, by a heck of a long way.

    Ira said to me (March 1, 2011 at 6:11 am): Myrrh, you really need to get outside more and sit in the Sunshine and feel the warmth! That is how visible and near-visible (“shortwave”) light warms the Earth.

    Well, I say different, I say it isn’t and I have gone to a great deal of effort to explain why I say it isn’t. I have given a NASA page which explains it exactly as I explain it, that the heat we feel from the sun is Thermal IR. That’s why it’s called Thermal IR or Heat Energy, because it is.

    On that same NASA page it explains that Near IR, which is included in the “Solar” Ira and AGW use because it is shortwave energy next to Visible, is COOL. They give an example of a remote control.

    Elsewhere I’ve also used the remote control to explain how Visible and short wave energies are Reflective (these are standard descriptions as Light/Heat energies), rather than Absorbtive as is Thermal IR. You can try it at home, change your TV channel by bouncing the light off the ceiling and walls rather than pointing directly at receiver.

    Thermal IR on the other hand, which you can also experiment with if you can borrow a Thermal IR heater, works by heat energy directly penetrating the objects in the room, the walls, the people, the furniture. It doesn’t warm the air, but organic matter. It doesn’t bounce off, it penetrates.

    So, Ira is wrong, it isn’t shortwave which warms the Earth. The graphic and AGW claims here are pure, unadulterated nonsense. Read that NASA page. Decide for yourself.

    Ira also said to me in that post: If you don’t or cannot get outside, turn on an old-fashioned incandescent light bulb and hold your hand near it (not too close you will get burned). Feel the heat? That is shortwave light because the filament is heated to temperatures similar to the Sun’s surface. You can tell it is shortwave because you can see the light.

    I’m sitting in a room at my computer which is directing light at me, just above the screen, it’s a low light, about 15″ from the top of my screen and a couple of feet away from me at an angle to me, is a 40 watt incandescent bulb. The top of my computer screen is hot. The lightbulb has warmed it. The bottom of the screen slightly tilted towards me is colder, colder than my hand. I can even feel the bulb’s heat on the top left of my forehead which is where it reaches me most from the angle I’m sitting. The light from the computer screen isn’t having any warming affect whatsoever..

    Ira’s “you can tell it is shortwave because you can see the light”, isn’t proof that that it’s the shortwave that is doing the warming. It only shows the temperature of the light bulb.

    It is ‘a proof’ of how hot the object is (but this not applicable to all materials, see Feet2thefire’s explanation elsewhere) in the light bulb, if we were so skilled; the steel worker can tell by the colour of the light how hot the steel is getting.

    All that shows is that the filament is hot enough to create high energy colours as well as producing Thermal IR I’m feeling as heat.

    If you switch on an electric ring on a stove and begin at a low setting, you will soon feel the heat, the Thermal IR, and when you turn this up to higher settings you will feel more and more heat, you will also notice a change in colour as it gets hotter and hotter and you’re feeling more and more heat from it. It’s not the colour giving off the heat. Is it? The colour is the effect produced by changes in the heated ring, just as the heat given off is an effect of the same changes.

    As the ring gets hotter so more heat is given off and the colour of the ring itself changes. They’re both effects. What we see in the Sun is the colour change of it as object because it’s hot enough to be that colour. The visible colours themselves aren’t hot.

    The peak wavelength of the Stefan-Bolzmann is only showing the peak of energies emitted by wavelength, this is in Visible light. If the hotter meant more short wave then the peak should be in far shortwave. http://sciencevault.net/ibphysics/astrophysics/stellarradiation.htm

    What is WRONG, in the descriptions of these and so all the confusion in AGW as Ira presents it here, is the wording used to describe what’s happening. “Hotter objects emit most of their radiation at shorter wavelengths: hence they will appear to be bluer while cooler objects emit most of their radiation at longer wavelengths.”

    It’s not true. As with the light bulb and steel, hotter objects emit most of their radiation in the longer wavelengths, heat. All that stuff under the curve to the right of Visible and Near IR. A small fraction only therefore is emitted as light, which happens to have shorter wavelengths; which get shorter in wavelength, so brighter, the hotter the object. The “peak” is only describing the wavelength peaking in luminosity, not the amount of energy generated.

    The peak comparison by wavelengths is not the same as peak comparison in amounts of energy.
    (Peak comparison in terms of amounts of energy would have to include other wavelengths, radio for example, but in comparing Solar to Heat as we’re doing here):

    So, if as Ira says “the filament of the incandescent light bulb is heated to temperatures similar to the Sun’s surface”, and we know such a lightbulb gives MOST of its energy in Heat and not Light, in Thermal IR, 95%, and only a small AMOUNT in Visible, only 5%, then what is really “the most” energy given off by the Sun?

  270. Phil. says:

    wayne says:
    March 2, 2011 at 5:32 pm
    I have to clarify myself on the aspect of oxygen and nitrogen absorption/emission in our atmosphere. After some detailed digging it appears there is a continuum absorption and emission but the exact nature and the mechanisms is where I was off a bit. Vibrational and rotational excitations are quantized and darn I already knew that, but for some reason that wasn’t registering, no sleep. ☺

    Here’s the gist gathered from some .mil and hitran and hartcode papers and sites: both oxygen and nitrogen do have bands in the infrared and both also have continuums about those lines. Oxygen’s is in the range of 1345-1820 cm-1(5.5-7.4 µm) and nitrogen is located in the range 2105-2620 cm-1(3.8-4.8 µm). I read HARTCODE is adding O2 & N2 continuum capabilities also. So how does the atmosphere as a whole handle a small portion of the energy within our atmosphere and help move it to space? Never found out how much.

    As shown on the figure I’ve shown here several times, more than a million times weaker than CO2 and H2O!
    http://i302.photobucket.com/albums/nn107/Sprintstar400/CO2N2O2.png

  271. Domenic says:

    to Phil

    Try to get your nose out of the fake world of assumptions, charlatans, naive scientists, fear mongers and dolts…. and into the real world.

    The data you keep pointing to are contrived.

    THEY HAVE NEVER BEEN ACTUALLY MEASURED.

  272. Myrrh says:

    Tim Folkerts @ March 2, 2011 at 4:45 pm

    Thank you, that’s helpful.

    I think there’s a word missing, or concept, (sorry, it’s late…). I was going to used “possible” in “There will [possibly] be millions of transfers of energy from A to B. etc.”, but that’s not quite right, because if the Law states categorically that heat never flows from cold to hot then it isn’t possible if these energies are carrying/creating heat.

    The millions is reduced to a discrete amount of possibilities, so, statistical net isn’t applicable as that only relates to possible range of movement of a molecule as a ‘base’, not the possible range of movement of a molecule restricted by the circumstances it finds itself in.

    Hope I’m making sense in that you can understand what I’m trying to say, even if you don’t agree… Will take another look at it tomorrow.

  273. Myrrh says:

    Aggh, I was trying to put Tim’s name in bold, and the rest normal, don’t know where the italics came from. I’m not quoting Tim, that’s all a reply from me to his post above; time for bed.

  274. RJ says:
    March 2, 2011 at 11:44 am
    No, the igloo doesn’t generate any heat. But it does absorb heat from you when you crawl in. Some of that heat it passes along to the outside, and some it radiates back at you. So you are in fact being warmed by heat from a colder surface radiating toward you.

    Is the above post serious or not?

    Not only serious, but seriously CORRECT.

    The problem you are having is most likely where is the energy radiated by the packed snow walls of the igloo coming from? Well, as davidmhoffer (who I believe is the author of the quote you reproduce) says in the quote itself, “…it [the igloo inner wall] does absorb heat from you [the living, breathing, 98.6ºF person in the igloo] and some of it is conducted through the wall to the outside of the igloo, but, BUT … some is radiated back to you, the person in the igloo.

    Any mass that is above absolute zero radiates thermal energy. The inner wall of the igloo, having absorbed some of the radiation from you, and because snow has air pockets and is not as good a conductor as an equal thickness of metal, it is cooler than you but warmer than a metal shell, so it does, in fact, radiate some of its warmth and make you warmer than you would be in a metal igloo.

    That said, the major reason an igloo helps keep you warm is the restriction of convection (which a thin metal or plastic shell would do as well) and reduction of conduction through the wall (which packed snow does better than metal of the same thickness).

    Of course you can help this process along by digging a wide hole at the bottom center, usually terraced with the people at the higher levels, so the cold air has a place to go. And, a firealso helps. See http://www.physlink.com/education/askexperts/ae579.cfm

  275. Phil. says:

    Myrrh says:
    March 2, 2011 at 6:42 pm
    So, if as Ira says “the filament of the incandescent light bulb is heated to temperatures similar to the Sun’s surface”, and we know such a lightbulb gives MOST of its energy in Heat and not Light, in Thermal IR, 95%, and only a small AMOUNT in Visible, only 5%, then what is really “the most” energy given off by the Sun?

    Nonsense which rambles on confusing the english meaning of ‘heat’ and ‘light’ deleted.

    Information: sunlight BB temp ~5800K, peaks at 0.5μm, ~5%UV, ~40%Vis, 55%IR
    an incandescent lightbulb on the other hand is limited by the mp of Tungsten so
    filament bb is less than 3600K so peaks at ~0.7μm, ~10%Vis, 90%IR (depending on temperature).

  276. Al Tekhasski says:

    Wayne wrote:
    “Now back to the 8 to 12 µm window. Does a continuum exist there? Yes. That is covered by the water vapors continuum that spans basically the entire IR spectrum to some degree or another”

    One (more) correction is in order here. Water vapor (as a gas) does not have much of continuum spectrum other than tails of peaks broadening of various kinds. However, a liquid water aerosols (aka “clouds”) do have continuous spectrum. Therefore, if we include clouds into the definition of “hot air” (from our preceding exchange), then my old formulation will be incorrect.

    However, my main objection to formulations of Dr. Glickstein still stands – the atmospheric greenhouse effect occurs not because of real cloudy air has certain non-Planck absorption-emission structure and has fluxes between different spectral bands via local with buffer gas.

  277. Phil. says:

    Domenic says:
    March 2, 2011 at 6:55 pm
    to Phil

    Try to get your nose out of the fake world of assumptions, charlatans, naive scientists, fear mongers and dolts…. and into the real world.

    The data you keep pointing to are contrived.

    THEY HAVE NEVER BEEN ACTUALLY MEASURED.

    So all those of us who ran their FTIR spectrometers containing dry air and observed no absorption are naive? Gerhard Herzberg who won a Nobel prize for Chemistry for his work concerned atomic and molecular spectroscopy, and used these techniques to determine the structures of diatomic and polyatomic molecules, was a ‘dolt and a charlatan’? Do us all a favor learn something about the subject before posting more of this dreck.

  278. Myrrh says:
    March 2, 2011 at 3:31 pm
    Re: Ira Glickstein, PhD says:
    March 2, 2011 at 9:22 am

    The second Law of thermodynamics refers to the NET transfer of heat energy that must be from warmer to cooler bodies. …

    When, question to all here, did this idea of “NET” appear? …

    “One of the most remarkable facts of nature is that certain processes take place in only one direction. For example, if a high temperature object is placed in contact with one of lower temperature, heat flows from the hotter to the cooler until the temperatures become equal. …”

    Please NOTE Myrrh (GRRRH!) that the quote you got is about CONDUCTION, not RADIATION. That is why it says “…if a high temperature object is placed in contact with one of lower temperature,…” If the objects are NOT in contact with each other (i.e., not connected by a good conductor, but RADIATING at each other with vacuum or air or some good insulator between) the hot to cold rule does not apply except to the NET transfer of energy. Got it?

    … Similarly, when a drop of ink is placed in a glass of water which is then stirred, the ink distributes itself throughout the water. Yet no amount of stirring will make the uniformly-distributed ink go back into a single drop. …”

    The above is certainly true, but only for statistical reasons. If the inky water is continually stirred, the molecules of ink will assume different positions with respect to the molecules of water. If, to take a very simple example, there were two molecules of ink and 14 molecules of water, in a 4 x 4 array, there are 16 x 15 / 2 = 120 different arrangements of the ink, only a small number of which have the two ink molecules adjacent. However, in this case, you will not have to stir much before you get a situation where the ink molecules are, in fact, adjacent.

    OK, do the math for a 10 x 10 array with five molecules of ink, or a 1000 x 1000 array with 50 molecules of ink, or a container of water that may have millions of molecules in each direction, with 5000 molecules of ink, which would give billions or trillions of possible arrangements, only a tiny percentage of which will have all the ink molecules adjacent to each other. However, if you have the time and gumption to wait, you will see all those ink molecules get back together. It may take the age of the Univese, but, it is bound to happen.

  279. Phil. says:

    Domenic says:
    March 2, 2011 at 6:41 pm
    They are calculated from various theories, opinions, assumptions, etc. The graphs, etc for absorption/transmission beyond 10 or 15 microns are at best just speculations, and at worst, just wild-assed guesses. Personally, I am of the opinion that they are more wild-assed guesses than anything.

    More rubbish, you do know that the UAH MSU measurements are made with the microwave spectra of O2, don’t you? According to you they’re based on wild-assed guesses of the spectra, based on the evidence your opinion is worth squat!

  280. Phil. says:

    Ira Glickstein, PhD says:
    March 2, 2011 at 7:58 pm
    However, if you have the time and gumption to wait, you will see all those ink molecules get back together. It may take the age of the Univese, but, it is bound to happen.

    And as you’d expect based on thermodynamics if you input energy into the system it can be reversed, for example pass it through a suitable absorbent and you can recover all the ink.

  281. Al Tekhasski says:
    March 2, 2011 at 7:40 pm
    … However, my main objection to formulations of Dr. Glickstein still stands – the atmospheric greenhouse effect occurs not because of real cloudy air has certain non-Planck absorption-emission structure and has fluxes between different spectral bands via local with buffer gas.

    Say what?

    I have no idea what the above paragraph means. Please restate your objections and I will try to answer, or I will change my mind if I agree with you.

  282. davidmhoffer says:

    Brian H; (and Ira, please take note)
    Igloos and snow caves operate more or less like any enclosed space: they trap warm air. It helps that ice and snow reflect really well, of course. Re-radiation back atcha is trivial in an igloo.>>>

    Practical experience trumps theory. I’m assuming you’ve never done any winter camping, or at the very least haven’t been taught how to do it properly. Any shelter constructed of snow, to be efficient and warm quickly enough to be usefull, has dimensions as small as possible while still allowing enought space for the number of people inside. In most cases this is accomplished by having each camper construct their own individual shelter, on of three basic designs that I will explain at the end of this response.

    Any camper who constructs a shelter such that it relies primarily on trapped warm air will be exposing themselves to potential carbon-monoxide poisoning. The shelter MUST be built in a fashion to drive several air exchanges per night via convection from the entrance at the bottom through a hole in the top of the shelter for that specific purpose. The ensures a continued supply of fresh air to the camper through the night and has the added benefit that the warmest air, the camper’s breath, is the primary driver of convection and so eliminates itself from the shelter, taking the exhaled carbon monoxide with it. The bulk of the warmth generated within these shelters is in fact heat radiated by the camper’s body, absorbed by the snow, and a portion of it re-radiated back. Oddly if one thinks about it, this is a case of the improperly termed greenhouse effect refering to absorption and re-emission being present, while the properly termed greenhouse effect of preventing convection is actually absent.

    There are three types of snow shelters, the most commonly known being the igloo. Oddly, from a camping perspective at least, it is the least common.

    Igloo – built from blocks of snow and/or ice cut to shape and used to construct a dome with a short tunnel as an entrance at the bottom, and a hole in the centre at the top of the dome to ensure air exchange. Too much work for a temporary shelter, so seldom used in camping. Of more value when a larger structure is required and/or a structure that will last weeks or months.

    Quinzee – very common as it can be constructed quickly even from only a thin layer of snow (igloos need thick packed snow or else ice). Loose snow is scooped by hand or shovel into a large pile as tall as possible. The snow is packed by rolling on it, or by hand, or flats of shovels. Allow to settle for an hour or two. On the SOUTH side of the pile dig a small hole at the bottom just big enough to allow one person in, and then hollow out the pile from the inside. The surface will have formed a stiff shell while the centre remains loose and easily scooped out. Poke a decent size hole in the top, and you are good to go. Do NOT dig the snow inside all the way down to the ground, leave a couple of inches on the bottom. Even at -20 C, with nothing to warm it but the afternoon sun shining through the small entrance at the bottom, exposed dirt will warm up enough to melt. Put a camper in there and the poor child will awake in the morning sleeping in mud.

    Less popular but very effective is to find a natural snow drift with one side as close to vertical as possible and at least three feet high. The campers hollow out a horizontal tunnel directly into the side of the snow drift just about the length of their body. They then sleep with their feet at the far end of the tunnel and their heads just poking out of the opening. You actually want the tunnel slightly downhill so that when the kids are sleeping they ate slightly uphill. The reverse is rather uncomfortable. The slight incline allows heat to escape via convection which is often desired because these tunnels heat up quickly and may get too warm for comfort. As the camper is breathing into the open air, not into the tunnel, there is no risk of carbon monoxide poisoning. I note also, this type of shelter is nearly 100% heated by snow absorbing heat from the camper and radiating some portion of it back into the tunnel.

  283. wayne says:

    Al Tekhasski says:
    March 2, 2011 at 7:40 pm

    Wayne wrote:

    “Now back to the 8 to 12 µm window. Does a continuum exist there? Yes. That is covered by the water vapors continuum that spans basically the entire IR spectrum to some degree or another”

    One (more) correction is in order here. Water vapor (as a gas) does not have much of continuum spectrum other than tails of peaks broadening of various kinds. However, a liquid water aerosols (aka “clouds”) do have continuous spectrum. Therefore, if we include clouds into the definition of “hot air” (from our preceding exchange), then my old formulation will be incorrect.

    However, my main objection to formulations of Dr. Glickstein still stands – the atmospheric greenhouse effect occurs not because of real cloudy air has certain non-Planck absorption-emission structure and has fluxes between different spectral bands via local with buffer gas.

    Apologize for the tangle above, it’s just that after following this area of science so long, at arm’s length, I probably don’t have that “many more years of study” that you called for.

    The H2O continuum I was speaking of was mentioned in one of the articles or papers I found, most just speaking of one range of frequencies or the other. It was questioning not if the continuum exists, multiple sites verified that, but exactly why a H2O molecule would absorb continuously across the IR spectrum. It brought up the suspect of the dimer tendency which adds so many more degrees of freedom. It didn’t go much deeper on that subject. Search for ”H2O dimmer continuum spectrum” for many links.

    Cite from one such paper’s abstract: http://jcp.aip.org/resource/1/jcpsa6/v128/i21/p214506_s1?isAuthorized=no

    … “Based on first-principle molecular dynamic simulations, we calculate the far-infrared spectra of small water clusters (H2O)n (n = 2,4,6) at frequencies below 1000 cm−1 and at 80 K and at atmospheric temperature (T>200 K). We find that cluster size and temperature affect the spectra significantly. The effect of the cluster size is similar to the one reported for confined water. Temperature changes not only the shape of the spectra but also the total strength of the absorption, a consequence of the complete anharmonic nature of the classical dynamics at high temperature.” …

    So to me the jury’s still out on that, and the scale. As physics usually is, if you get too simple of a view, you also have a high chance of being incomplete.

    On your second point I would just have you to keep an open mind with proper physics always applying, since each of these molecules (O2 N2 H2O CO2) absorb and then thermalize, that very fact guarantees a high chance of cross spectrum re-emission by a different molecule species, different frequencies. And in the ~10 µm case that would be via a water molecule, dimer, or even a multiple dimer.

    It sounds like this area is still evolving, so casual speculation is still rampant and rightfully so.

    I’ve been contemplating a cross language conversion of HARTCODE, thanks to Dr. Miskolczi, to a more simple common language for outsiders like myself without Fortran compilers and it is in this area of mysteries, as this post points out, that just might send me on that trek. That is if I can isolated enough time; that does fit squarely in my expertise.

  284. davidmhoffer says:

    O H Dahlsveen says:
    March 2, 2011 at 4:45 pm
    Ira, I have spent hours writing an essay in explanation of why we think differently and just now I have accidentally deleted it.- So for now, – Ok the Gods are telling me you are right.
    AGW is happening!>>>

    The Gods rarely speak plainly, leaving mere mortals to ponder their messages and derive meaning from them as best we can.

    The Gods may have suggested to you that Ira’s explanation of CO2 and LW radiance is right. I do not believe it follows that this means AGW is significant, or happening at all when the sum of all feedback is included.

    If there are any Gods following along, perhaps they would care to comment? Sorry Willlis, Anthony, Leif, Phil, Ira, Roy, George, and so many others I’ve learned so much from, but I believe you remain ranked at this time in the range of minor deities and demi-gods.

    I will allow however that Dr Ravetz, while certainly unable to demonstrate godly capabilities in terms of logic, reason, or fact, may have provided us with a glimpse of a godlike ego.

  285. Tim Folkerts says:

    Myrrh says: “I think there’s a word missing, or concept, (sorry, it’s late…). I was going to used “possible” in “There will [possibly] be millions of transfers of energy from A to B. etc.”, but that’s not quite right, because if the Law states categorically that heat never flows from cold to hot then it isn’t possible if these energies are carrying/creating heat. ”

    I didn’t mean “possibly”: I was thinking of two macroscopic chunks of material pressed together. This would put millions of atoms from the two pieces in contact, which will all be colliding many times per second with atoms from the other object. There WILL be many collisions here. There will be millions of microscopic transfers of energy every second. It is just the NET flow of energy for large collections of energy that is governed by the 2nd law, not the transfers during individual collisions.

  286. Domenic says:

    Phil

    UAH MSU measurements are made only at 55 gigahertz for the troposphere and at 57 gigahertz for the stratosphere. They are indicators at best.

    http://en.wikipedia.org/wiki/File:EM_spectrum.svg

    “Satellites do not measure temperature. They measure radiances in various wavelength bands, which must then be mathematically inverted to obtain indirect inferences of temperature.[1][2] The resulting temperature profiles depend on details of the methods that are used to obtain temperatures from radiances. As a result, different groups that have analyzed the satellite data have obtained different temperature trends. Among these groups are Remote Sensing Systems (RSS) and the University of Alabama in Huntsville (UAH). Furthermore the satellite series is not fully homogeneous – it is constructed from a series of satellites with similar but not identical instrumentation. The sensors deteriorate over time, and corrections are necessary for satellite drift in orbit. Particularly large differences between reconstructed temperature series occur at the few times when there is little temporal overlap between successive satellites, making intercalibration difficult.”
    http://en.wikipedia.org/wiki/Satellite_temperature_measurements

    Those measurements tell you absolutely nothing about the radiational heat transfer properties of EACH and ALL the greenhouse gases…which this discussion is about. Those measurements tell you nothing about the overall long wavelength emittance, absorption, or transmission of ANY the gases in the atmosphere.

  287. Domenic says:

    Phil wrote: So all those of us who ran their FTIR spectrometers containing dry air and observed no absorption are naive? Gerhard Herzberg who won a Nobel prize for Chemistry for his work concerned atomic and molecular spectroscopy, and used these techniques to determine the structures of diatomic and polyatomic molecules, was a ‘dolt and a charlatan’? Do us all a favor learn something about the subject before posting more of this dreck.”

    His work had nothing to do with radiational heat transfer.

    That is what we are discussing here.

    The ‘dolts’ are people like you.

    Dolts are those who try to make something they don’t understand appear to be something different from what it really is.

  288. Reed Coray says:

    Dr. Glickstein.

    I believe that if in the vacuum of space you place a blackbody object with (a) a constant (i.e., unchanging energy per unit time) internal thermal energy source, and (b) internal/surface thermal conduction properties such that independent of how energy enters the blackbody, the surface temperature of the blackbody is everywhere the same and you place that object in cold space (no background thermal radiation of any kind), eventually the object will come to a steady state condition–i.e., the object will eventually radiate energy to space at a rate equal to the rate of energy produced by the internal energy source. In the steady-state condition, the surface temperature (assumed uniform) of the object will be non-changing at a value such that according to Planck’s radiation law provides the radiation rate necessary to achieve steady state. If a second blackbody object (no internal thermal energy source but with thermal conduction properties such that independent of the direction of incident radiation on the second object, the second object’s surface temperature will be everywhere the same) is placed next to but NOT touching the original object, when the two-object system reaches steady state (i.e., for each object, the rate of energy leaving the object will equal the rate of energy entering the object), the surface temperature of the original object in the presence of the second object will be higher than it was in the absence of the second object.

    Some people attribute this increase in the temperature of the first object to “backradiation” from the second object. The argument goes that the radiation from the original object warms the second object, which then radiates energy from its surface–some of which is directed towards the original object. For lack of a better term, call the radiation from the second object to the first object “backradiation.” People then introduce the concept of “net” radiation–the difference between the radiation from object 1 to object 2 minus the radiation from object 2 to object one. I believe the second law of thermodynamics says that in the absence of any “work being done” the “net” energy exchange must be from the warmer object to the cooler object. However, I also believe it is possible to create two objects having the above properties (blackbody surfaces, uniform surface temperatures, only one object with an internal thermal energy source) such that as the second object (the object without an internal energy source) increases in surface area, the temperature of the first object (the object with an internal energy source) will at first increase but at some point will reach a maximum temperature; and as the size of the second object continues to grow, the temperature of the first object will start decreasing–never achieving a temperature that is lower than the first object’s temperature in isolation, but approaching that temperature. I also believe that if you allow conduction of energy between objects, depending on the rate of energy transfer via conduction, the temperature of the original object will be lower than the original object temperature in the absence of the second object. Note that the presence of conduction does not eliminate backradiation–it will still be present. If true, we have reached a state where (a) backradiation exists and (b) the temperature of the original object is lowered. As such, I find it difficult to buy into the claim that “backradiation results in a temperature increase to the original radiating surface.” To the best of my knowledge, you have never made this explicit claim. However, I infer from what you and others have written, that “greenhouse gas” backradiation increases the earth surface temperature. I believe backradiation is a contributor to final object steady-state temperatures, but not the controlling factor–in that if conduction is allowed, backradiation can simultaneously exist with lower surface temperatures.

    I have created a Power Point file with the mathematics of just such a two-object system. I am in the process of creating a shorter Word file that summarizes that file. If you are interested, I’ll figure out how to make a PDF file of the Power Point file and E-mail it (the PDF file) to you.

  289. Al Tekhasski says:

    Dr. Glickstein wrote: “I have no idea what the above paragraph means. Please restate your objections and I will try to answer, or I will change my mind if I agree with you.”

    My objections were briefly formulated here:
    http://wattsupwiththat.com/2011/02/28/visualizing-the-greenhouse-effect-atmospheric-windows/#comment-610460
    (please make sure to separate quotations from my actual remark, I forgot to properly close italics)

    See also http://climateclash.com/2010/11/28/g2-greenhouse-gas-effect/

  290. Oliver Ramsay says:

    The other neat thing about igloos is that you can plop a chunk of seal-meat down in the middle of it, then go out to hunt a polar bear. When you come back the meat is cooked to perfection. The radiation from the meat has been converted to heat and redirected at the said meat, thereby heating the dinner. Fortunately, the walls of the igloo are not melted in this process, so you can come back and exhale all the “carbon monoxide” you want.
    Okay, I’m sure the carbon monoxide was a simple mistake, but repeating that everything radiates all over the place, although uncontroversial, doesn’t make a good case for “heating”. Two objects in TE will radiate merrily. Neither will heat the other.
    Here’s another igloo trick. Take a thermometer out of your parka pocket and place it in the igloo. Observe the thermometer through the clear ice window you ingeniously included in the igloo wall. Phone us when the mercury goes up, because it’s probably Spring and the ice is breaking up.

  291. Phil. says:

    Domenic says:
    March 2, 2011 at 9:39 pm
    Phil wrote: So all those of us who ran their FTIR spectrometers containing dry air and observed no absorption are naive? Gerhard Herzberg who won a Nobel prize for Chemistry for his work concerned atomic and molecular spectroscopy, and used these techniques to determine the structures of diatomic and polyatomic molecules, was a ‘dolt and a charlatan’? Do us all a favor learn something about the subject before posting more of this dreck.”

    His work had nothing to do with radiational heat transfer.

    That is what we are discussing here.

    The ‘dolts’ are people like you.

    Dolts are those who try to make something they don’t understand appear to be something different from what it really is.

    That would be you when you produce such rubbish as this:
    “the graphs, etc for absorption/transmission beyond 10 or 15 microns are at best just speculations, and at worst, just wild-assed guesses. Personally, I am of the opinion that they are more wild-assed guesses than anything”, which is exactly what Herzberg worked on and wrote about.

  292. Phil. says:

    Oliver Ramsay says:
    March 2, 2011 at 10:07 pm
    The other neat thing about igloos is that you can plop a chunk of seal-meat down in the middle of it, then go out to hunt a polar bear. When you come back the meat is cooked to perfection. The radiation from the meat has been converted to heat and redirected at the said meat, thereby heating the dinner. Fortunately, the walls of the igloo are not melted in this process, so you can come back and exhale all the “carbon monoxide” you want.
    Okay, I’m sure the carbon monoxide was a simple mistake, but repeating that everything radiates all over the place, although uncontroversial, doesn’t make a good case for “heating”. Two objects in TE will radiate merrily. Neither will heat the other.
    Here’s another igloo trick. Take a thermometer out of your parka pocket and place it in the igloo. Observe the thermometer through the clear ice window you ingeniously included in the igloo wall. Phone us when the mercury goes up, because it’s probably Spring and the ice is breaking up.

    Check out the Weather Channel, Stephanie Abrams did exactly this within the last year, measured the temperature difference with time in both a snow cave and an igloo. Probably on line somewhere, certainly didn’t have to wait very long for the mercury to go up.

  293. Jim Masterson says:

    >>
    Phil. says:
    March 1, 2011 at 7:55 pm

    No, how do you propose that the atmosphere gets hotter than the surface?
    <<

    I can think of at least two cases: 1) radiation fog is caused by the surface radiating away energy and cooling enough to bring the temperature of air lying above it down to the dew point or below; 2) advection fog is caused by warmer (moist) air passing over a cooler surface. In both cases, the surface is cooler than the air above it.

    Jim

  294. Phil. says:

    Domenic says:
    March 2, 2011 at 9:31 pm
    Phil

    UAH MSU measurements are made only at 55 gigahertz for the troposphere and at 57 gigahertz for the stratosphere.

    No, the MSU/AMSU system uses more frequencies than that, they are measurements made using frequencies from the 50-60 GHz absorption band of O2.

    Those measurements tell you absolutely nothing about the radiational heat transfer properties of EACH and ALL the greenhouse gases…which this discussion is about. Those measurements tell you nothing about the overall long wavelength emittance, absorption, or transmission of ANY the gases in the atmosphere.

    On the contrary they explicitly measure the absorption by oxygen, just in that region of the spectrum where you said it was a wild-assed guess.
    Like this paper here:
    http://www.its.bldrdoc.gov/pub/journal_articles/liebe_journ_quant_spectrosc_radiat_trans_vol48-1992/liebe_journ_quant_spectrosc_radiat_trans_vol48-1992.pdf

  295. Al Tekhasski says:

    Domenic said:
    “Try to get your nose out of the fake world of assumptions, charlatans, naive scientists, fear mongers and dolts…. and into the real world.”

    It is my observation that, unfortunately, relative number of dolts (including all other eloquently described categories) is approximately equal in both, pro-AGW and anti-AGW groups. I would say it is about 98%. :-)

  296. izen says:

    Reed Coray says:
    March 2, 2011 at 9:40 pm
    “….I also believe that if you allow conduction of energy between objects, depending on the rate of energy transfer via conduction, the temperature of the original object will be lower than the original object temperature in the absence of the second object. Note that the presence of conduction does not eliminate backradiation–it will still be present. If true, we have reached a state where (a) backradiation exists and (b) the temperature of the original object is lowered….”

    No point in putting on a coat in cold weather then…

  297. Oliver Ramsay says:

    I said:

    “Here’s another igloo trick. Take a thermometer out of your parka pocket and place it in the igloo. Observe the thermometer through the clear ice window you ingeniously included in the igloo wall. Phone us when the mercury goes up, because it’s probably Spring and the ice is breaking up.”

    Phil. says:
    “Check out the Weather Channel, Stephanie Abrams did exactly this within the last year, measured the temperature difference with time in both a snow cave and an igloo. Probably on line somewhere, certainly didn’t have to wait very long for the mercury to go up.”
    ——————————
    I’m sorry I missed that!
    I found Stephanie online but not her igloo. She is pretty hot so she’d have to be extra careful that she wasn’t in the igloo with the thermometer.
    Do you happen to remember just how toasty it got in that igloo, in how much time?
    Do you think it would just keep getting hotter and hotter with time?
    Do you think it would work with a tarpaper shack, or does it have to be snow?
    Do you have other educational TV programs to recommend?

  298. davidmhoffer says:

    Oliver Ramsey;
    thanks for allowing that having been up for near 24 hours due to deadline with only swatting briefly at annoying gnats on WUWT as my occasional breaks, i may have messed up between di- and mono-

    I do not recall however, saying that an igloo would generate enough heat to cook meat. In fact, I don’t recall even saying it would warm up without an energy source at all. Sun shining in at the opening, or through that clever window made of ice. Or from little campers snug in their sleeping bags. Of course as the scout troop gets older, hits their teens, some of the rascalls sneak out to go visit the girls camp down the road. No harm done for the most part, although there was one case of mono… no… di… no… mono… sleep. need sleep.

    If the igloo or quinzee has a large enough vent to result in several air exchanges every night, the warmth inside couldn’t be from warm air being trapped, could it? So, its warmer inside trhe quinzee because….?

    C’mon Oliver. S’plain it. I’ll ridicule…. I mean read it in the morning.

  299. Brian H says:

    Ira Glickstein, PhD says:
    March 1, 2011 at 4:33 pm

    What do you predict?

    Well, I think the odds are getting better by the day/week/month that the mini-fusion project at LPPhysics.com will succeed. If it does, energy at 1/20 best current costs will soon (5 yrs. fwd) be plentiful (on-planet fuel is adequate for 10X current electric demand till about the Red Giant Sun stage). Space travel will become inexpensive and easy.

    Take it from there.

  300. RJ says:

    Thanks for the responses to my query re the Igloo above. And to all the other contributors.

    To be honest I have a huge problem with ice warming me up by returning my body heat back to me.

    It just does not seem correct. I can see that the air would warm up. So it will reduce my rate of cooling. But actually warm me up. This seems like fiction.

    I will keep an open mind on this though. But in the sky dragon book the statement in ch 15 ‘that a blanket can at best maintain your body temperature it can not give you a fever’ This makes sense. I can not see how there is any way to ever increase an objects temperature by reflecting back heat from that object. A small heat source in a perfect reflecting container will not and can not eventually produce a huge heat by reflecting the heat back.

    I guess my question is is this statement correct or not. Chapter 15 Hans Schreuder again (I am not looking for an answer and do not expect this to be resolved quickly)

    “The world has all too easily accepted greenhouse effect explanations that confuse the familiar reduction of CONVECTION heat loss with the production of radiative heat GAIN”.

    I will continue following discussions on this issue with interest.

  301. Bryan says:

    Ira Glickstein and davidmhoffer will need to revise their basic thermodynamics.

    Get hold of a physics textbook work through the section on the Carnot Cycle.

    They will find that there is no net heat only heat.
    It always moves from a higher temperature object to a lower temperature object never the reverse.(Clausius).

    Thermodynamic HEAT is unfortunately often confused with the vernacular meaning of heat.
    The idea of heat coming back from the colder object is almost always used with radiant transfer as backradiation.
    Coupled with this is the notion of “well if the colder object wasn’t there” and the thing being there instead is absolute zero (-273K).

    Well the same line of reasoning would give you “back conduction” and “back convection” .

    Object is at 350K, object B is at 270K and object C is at -273K.
    Arrange and substitute these objects with suitable steps to allow conduction and convection.
    This will give you “back conduction” and “back convection” .

  302. Myrrh says:

    Oliver Ramsay says:
    March 2, 2011 at 10:07 pm

    The other neat thing about igloos is that you can plop a chunk of seal-meat down in the middle of it, then go out hunt a polar bear. When you come back the meat is cooked to perfection. The radiation from the meat has been converted to heat and redirected at the said meat, thereby heating the dinner.

    So, the meat being dead meat is radiating energy which is bouncing back from the inside walls of the igloo and cooking it? Gosh, how long then does it take to cook a live child which davidmhoffer leaves lying around in igloos, who are generating even more heat than the chunk of dead seal-meat?

    Why then do Eskimos/Innuit build small fires in igloos to cook their food? Sometimes leaving meat to freeze and eating it raw?

    http://people.howstuffworks.com/igloo.htm/printable
    http://www.ehow.com/about_6513672_life-igloo.html

    Urban myth from ignorance of life in the snowy wastes or deliberate misinformation from AGW to promote the idea that back-radiation from a cold object can heat a warmer object? I’d go for the second. Or, third option perhaps, someone’s sark remark to an AGW pushing backradiation heat flow from colder to hotter object.., and being taken as entirely plausible by AGW crowd.

  303. Myrrh says:

    Ira Glickstein, PhD says:
    March 2, 2011 at 7:58 pm

    Please NOTE Myrrh (GRRRH!) that the quote you got is about CONDUCTION, not RADIATION. That is why it says “..if a high temperature object is placed in contact with one of lower temperature,..” If the object are NOT in contact with each other (i.e., not connected by a good conductor, but RADIATING at each other with vacuum or air or some good insulator between) the hot to cold rule does not apply except to the NET transfer of energy. Got it?

    Yes, yes, Ira, I’ve got it, but that was just an example. I’m trying to explore how this concept as you give it came into existence because it’s not anything to do with these laws as I can understand them, and more importantly, I’ve read countless times from people involved in heat transfer in the applied sciences who say the idea as you present it violates these laws. You can’t both be right. Shouting at me doesn’t make you right, but I’ll give it a go for emphasis too..

    It violates the 2nd Law of Thermodynamics which says that HEAT FLOW IS ALWAYS FROM HOTTER TO COLDER.

    THIS APPLIES TO ALL RADIATION STATES ACCORDING TO THE 2ND LAW.

    “THE SECOND LAW STATES THAT SPONTANEOUS NATURAL PROCESSES … THAT HEAT CAN SPONTANEOUSLY FLOW ONLY FROM A HIGHER-TEMPERATURE REGION TO A LOWER-TEMPERATURE REGION, BUT NOT THE OTHER WAY AROUND.”

    “The second law of thermodynamics may be summarized as follows:

    “When two isolated systems in separate but nearby regions of space, each in thermodynamic equilibrium in itself, but not in equilibrium with each other at first, are at some time allowed to interact, breaking the isolation that separates the two systems, and they exchange matter or energy, they will eventually reach a mutual thermodynamic equilibrium.”

    Radiation then, is such an exchange of energy. The 2nd law thus applies equally to radiation as it does to conduction.

    Spontaneously in nature. That means, that no extra work is being done to alter it. Otherwise, you end up with the ludicrous notion as described in the Trenberth graphic that back-radiation keeps bouncing between the ‘greenhouse gas’ and Earth, raising the global temperature of Earth. Or the igloo that cooks a chunk of raw meat.

    In the 1st Law – “This is a statement of conservation of energy. The net change in internal energy is the energy that flows in as heat minus the energy that flows out as the work that system performs on its environment.”

    This is the only mention of “net”, of net change, I don’t understand how that can apply to “net as transfer of radiation” outside of it relating to the flow of heat energy as above and by breaking the 2nd Law of heat flowing always from hot to cold and never the other way around, naturally spontaneous, that is, without work being done to change it.

    [I don't know how I'm doing here in the eyes of those arguing against AGW here by saying AGW violates these laws. I'm still working my way through it, do feel free to correct me but keep the explanations simple..]

    Ira says re the drop of ink in glass of water stirred – The above is certainly true, but only for statistical reasons. If the inky water is continually stirred, the molecules of ink will assume different positions with respect to the molecules of water. If, to take a very simply example, there were two molecules of ink and 14 molecules of water…

    This was given as an example of processes that are one way as heat transfer is one way. As the example of the perfume wafting into a room from a bottle it won’t spontaneously waft its way back into the bottle, or from the extreme therefore of the whole lot evaporating. Sure, given infinite time it’s possible.., or work can be done to separate out the smell from the room together with all the molecules of alcohol and water that made up the perfume and get it back into the bottle, but that would take an awful lot of work, of energy expended to achieve such a thing, same in re-constituting the ink in solution back into its original constituent parts, but, given that statistically that ain’t going to happen for all the spilt ink and evaporated perfume in the world unless you wan’t to wait for an infinitely long time for it to happen and then maybe it never will, (you are assuming it is bound to happen but it’s ‘statistically as likely not to happen as to happen’ has to be included, so there’s no “bound to” about it), or are willing to expend energy to do this for all the examples past present and future, then, for all practical natural processes purposes, the ink stays mixed and the perfume evaporated. One way process.

    So, radiation is included, not just conduction, which means that it is impossible for a cooler molecule to transfer heat to a hotter one because that would violate the 2nd Law.

    The explanations I give above in quotes come from wiki’s Laws of thermodynamics.

    As with the NASA page I showed you, you are contradicted here re radiation not constrained by these Laws as you are re claiming that it’s shortwave solar heating the Earth and not as normal Science has always stated, that it’s the Thermal energy we feel from the Sun and which heats the Earth.

    I have now worked out to my own satisfaction that Real Science is that it’s Thermal energy heating the Earth and not short wave Solar, by looking a the different aspects involved and understanding more about them.

    Prove that it’s short wave “Solar”heating the Earth which you claim in your diagrams above and is that we feel as warmth as you have continued to expand on the description to me, or have the courtesy to say you are wrong.

    To prove you are right you MUST contradict all the information given on the NASA page I posted.

    You’re not just arguing with me or my understanding. I am agreeing with what is written on that NASA page.

    Should I put that in CAPS, to get your attention? Unless you can prove NASA wrong, your energy premise is wrong and what follows from that premise will be out of context of the Real World energy balance which is by Thermal radiation from the Sun, these Thermal energies which are the Heat energies which are long wave IR.

    Here it is again:

    http://science.hq.nasa.gov/kid/imagers/ems/infrared.html

  304. Myrrh says:

    Phil. says
    March 2, 2011 at 7:32 pm

    Re: Myrrh says (March 2, 6:42 pm): “So, if as Ira says “the filament of the incandescent light bulb is heated to temperatures similar to the Sun’s surface”, and we know such a lightbulb gives most of its energy in Heat and not Light, in Thermal IR, 95%, and only a small amount in Visible, only 5%, then what is really “the most” energy given off by the Sun?”

    Nonsense which rambles on confusing the english meaning of ‘heat’ and ‘light’
    deleted.

    ? A lightbulb gives out 95% of its energy in Heat and only 5% of its energy in Visible Light. Heat is Thermal IR.

    It’s been confused by AGW.

    It may well “peak” in the Visible by energy states, but the Visible as % of the total energy emitted is relatively tiny compared with the total amount of energy emitted.

    The amount, therefore the real “most”, of the energy emitted from the lightbulb is Thermal IR.

    The Sun thus, if similar as Ira has it, should also be emitting the same kind of ratio of Visible to Thermal. [Assuming that the real difference between thermal and visible energies are understood, not as Ira has it that Solar is the Heat energy we feel from the Sun]. I have seen this figure put at 80% in Thermal for the Sun.

    How, your info on light bulbs and sun temps, the light bulb compares with the sun re what Ira said, I think, I could be wrong, he meant that the heat it takes to produce white visible light from the sun is the same as that it takes to produce visible light from the lightbulb. The sun even hotter will produce shorter wavelengths also.

  305. Phil. says:

    mkelly says:
    March 2, 2011 at 11:42 am
    Phil. says:
    March 2, 2011 at 9:19 am

    Phil, the -18 C comes from the standard model of radiated heat transfer given approx 240 w/m2 at surface. Thus having GHG’s take us up 33 degrees to 15 C.

    Correct.

    Without any GHG no H2o, CO2 etc the sun would essentially have free rein with no absorbtion, so surface would get 340 w/m2 thus raising up to 278 K or 5 C to start. I think the 340 number used is to low.
    Well you have forgotten to consider reflection (i.e. albedo)

    It can easly be show via a pan of water and a thermometer that large protion of the surface get 700-1000 w/m2. Heck SOD has a graph on his web site showing the radiance of the sun on a plot of ground in Canada where measured solar radiation is about 700 w/m2.

    The 340 value is an average so of course there are places and times where it is exceeded.

    em>I think that ignoring the (PV=nRT) pressure of the atmosphere is a mistake. STP is 0 C. So not a bizarre statement. You are free to agree or disagree as you wish.
    Thank you for your permission to disagree with junk science! STP is an arbitrary reference point and has no significance at all. All that PV=nRT tells us is that at the surface atmospheric density is a function of Temperature, n/V=P/RT where P is ~constant.

    Also, would N2 or O2 cool if they went up. If they cannot radiate just going up will do little. Granted some work will be done so energy will be lost but how much?

    Yes due to adiabatic expansion, just like the N2 atmosphere on Triton does.

    As 99% of atmosphere is N2 and O2 we need to look at them. I am trying to find out. I have an idea they do radiate just like any matter at a temperature but I have not seen a temperature graph for each showing their respective pattens or amounts. If they do not I will change my idea.

    You have been shown data here which shows exactly how weakly N2 and O2 interact with IR, but it doesn’t agree with you preconceived belief so you ignore it. Your ‘idea’ contradicts a century of science on the structure of molecules and their interaction with radiation as well as the experimental basis for it.

  306. Domenic says:

    to Al Tekhasski

    To me the dolts are heavily weighted towards the pro-AGW crowd.

    pro-AGW vs. anti-AGW is not a good way to put it. That seems to be a way that a pro-AGW person would look at it.

    I never consider someone who is trying to learn more about the science, its methods, and ramifications, to be a dolt. Most of the people here, by far, are in that category.

  307. Steve says:

    For those still confused about how O2 and N2 can absorb/emit energy, yet accomplish this with absorbing/emitting in the thermal IR spectrum, this article just up might help…
    http://wattsupwiththat.com/2011/03/03/feb-uah-global-temperature-anomaly-goes-slightly-negative/

    Note that the article specifies that the satellites detect the temperature of O2 by taking readings in the microwave spectrum. Thus, O2 absorbs/emits in the microwave spectrum (which is directly adjacent to the thermal IR spectrum).

    The physical chemists are right. A black body is an imaginary, ideal surface that can absorb/emit any wavelength. Individual atoms can only absorb/emit in specific quanta. The larger the variety of atoms in your object, the more it will approximate a black body. The more your object is composed of a single element, the less it will approximate a black body.

  308. Tim Folkerts says:

    Whether or not it is explicitly stated, “HEAT” ALWAYS MEANS “NET HEAT” = “NET TRANSFER OF ENERGY”.

    Look at the Maxwell-Boltzmann distribution, for example here: http://bouman.chem.georgetown.edu/S02/lect4/image8.gif. In any object at any temperature, there will be some fast moving, high energy atoms and some slow moving, low energy atoms.

    If you put a block of 100C metal into a cup of 0C water, we all agree there will be “heat” from the surface of the metal to the water. Now look microscopically at the individual atoms and individual collisions — some metal atoms will have lower energy
    than some water atoms. When a high energy water molecule collides with a low energy metal atom, the metal atom will gain energy. Let me repeat — THE WARM METAL GAINED SOME ENERGY FROM THE COOLER WATER.

    This is not a violation of the 2nd law, because there is, on average, a NET transfer of energy from the warm metal to the cooler water. Every second there are billions of instances where energy was transferred from the cooler water to the warmer metal; there are simply MORE transfers the other way! The same applies to transfer by radiation. There are many photons transferring energy from the cold object to the warm object; there are simply MORE photons transferring energy from the warm object to the cold object!

  309. Bryan says:

    Tim Folkerts

    Whether or not it is explicitly stated, “HEAT” ALWAYS MEANS “NET HEAT” = “NET TRANSFER OF ENERGY”.

    Tim what is your definition of Heat?
    Clausius was well aware of radiation when he postulated his famous Second Law.
    He was writing about the Macro World rather than the Micro World.
    The transfer of heat in the Macro World always moves spontaneously from the higher temperature object to the lower temperature object never the reverse.
    The thermodynamic meaning of heat implies the ability to do work in the given situation.
    This applies to the radiation moving from the hotter surface to the colder surface.
    It does not apply to the radiation moving from the colder surface to the warmer surface.
    So there is no dual heat transfer only heat transfer from higher to lower temperature objects

  310. Domenic says:

    to Phil

    You should learn to read more carefully.

    http://www.its.bldrdoc.gov/pub/journal_articles/liebe_journ_quant_spectrosc_radiat_trans_vol48-1992/liebe_journ_quant_spectrosc_radiat_trans_vol48-1992.pdf

    “Abstract-Over 5000 absolute absorption values for pressure-broadened O2 lines in dry air were measured at frequencies from 49 to 67 GHz in O.l-GHz-increments. The controlled laboratory studies were carried out at three temperatures (6, 30, and 54°C) for 11 pressure values ranging between 1.3 and 101 kPa.”

    1. They measured only in the 49 to 67 Ghz band. That is a very, very tiny fraction of the entire long wavelength band (10 microns and above, to meters of wavelength) of interest for radiational heat transfer in the atmosphere to outer space.

    2. They tested O2 only at 6, 30, and 54°C. That is not representative of O2 in the upper troposphere and stratosphere.

    One very little understood property of matter (atoms, molecules, etc) is that their real life emittance (thus absorption) changes as a function of their internal temperature and outside temperature (thermal radiation field). Emittance, thus absorption, changes as a function of its environment.

    The emittance and absorption found in a laboratory are only valid for THOSE SPECIFIC CONDITIONS. That is your mistake. You think the laboratory is the real world. It is not.

  311. Robert Stevenson says:

    The heat or energy balance at equilibrium should result in a sustainable biosphere, too cold and we all freeze to death to hot and we all boil to death. It has to be just right a ‘goldilocks scenario’ in fact. In the discussions of global warming a heat input to the equation which is rarely seen or mentioned is nuclear fission in rocks. I do recall some time ago, from a scientific paper, that nuclear fission was a very important heat contributor to the biosphere and that it contributed 50% of the heat necessary to sustain life. Without it would we be locked into a long ice age?

  312. Domenic says:
    March 3, 2011 at 7:00 am
    to Al Tekhasski

    To me the dolts are heavily weighted towards the pro-AGW crowd.

    pro-AGW vs. anti-AGW is not a good way to put it. That seems to be a way that a pro-AGW person would look at it.

    I never consider someone who is trying to learn more about the science, its methods, and ramifications, to be a dolt. Most of the people here, by far, are in that category.

    Domenic and Al Tekhasski: I agree with the highlighted statement above.

    IMHO, those in the climate science debate may be roughly grouped into:

    (1) Alarmists: Like “Chicken Little” who cried “the sky is falling” when he was hit by an acorn falling off a tree, the Alarmists think they are warning and saving the world from a “tipping point” that will cause “runaway warming” within a decade or so because of the coincidence of a warming trend and CO2 rise over the past century. This group uses non-scientific arguments (“coal trains are death trains”) and knowlingly bends data to the breaking point or stays silent while non-scientists like former VP Al Gore spout nonsense. IMHO, their belief system is more like a religious mission than true science. They may also harbor conspiratorial views of corporations, globalization, etc., yet, confoundingly, they trust the government, including nacent world government, to fix the problem.

    (2) Warmists: These are valid scientists who interpret the data in the most precautionary way possible, slanting it (but generally not breaking it) to emphasize what may happen due to human activities that are unprecedented in the history of the Earth. They generally favor action to curtail the human activities they see as dangerous.

    (3) Lukewarmers: These are valid scientists who interpret the data midway between Warmists and Skeptics. They favor moderate action to curtail some aspects of human activities, and tecnological adaptation to deal with whatever climate change may occur.

    (4) Skeptics: These are valid scientists who interpret the data in the most optimistic way possible, slanting it (but generally not breaking it) to emphasize the uncertainty about how much human activities may affect climate change and to point out the misinterpretation and misrepresentation of the data by the Warmists and Alarmists. They generally favor little or no action to curtail human activities others see as dangerous, believing (corrrectly IMHO) that increased CO2 is inevitable and that the supposed solutions would wreck world economies and, in the end, do little to curb warming, which is mostly due to natural cycles and processes that may well bring us cooling in coming decades. Besides, a moderate rise in CO2 and temperatures will be beneficial to much or most of the world population. Whatever happens, civilization will adapt. Skeptics usually distrust the ability of government, particularly nascent world government, to get much of anything right. (If the government was in charge of the Sahara Desert, within six months, there would be a shortage of sand :^)

    (5) Deniers: Mirror image of the Alarmists, they are like the ostrich who sticks his head in the sand to shut out the real events in the world. Deniers are sometimes motivated by religious faith that God will not allow humans to destroy ourselves, that humans are not capable of changing God’s plan, etc. This group uses non-scientific arguments (“the atmospheric ‘greenhouse effect’ in a scam) and ignorantly bends data to the breaking point or cheers while non-scientists like certain elected officials spout nonsense. IMHO, their belief system is more like a religious mission than true science. They may also harbor conspiratorial views of environmental groups and the government.

    So, where does AGW come in? Well, from my point of view (as a Lukewarmer-Skeptic) it is absolutely true, in a scientific sense, that human activities over the past century have changed climate to some extent, probably minor compared to natural cycles and processes. Thus, IMHO, the central three belief systems, from Warmist to Skeptic, all accept AGW as scientific fact. The main differences are “how much?” and “what should we do about it?” That is the basis for robust discusion.

    The outlier groups, Alarmists and Deniers, are non-scientific and should be ignored because they are not qualified for rational discussion.

  313. Robert Stevenson says:

    Forgot to tick the comments box.

  314. davidmhoffer says:

    Bryan;
    The thermodynamic meaning of heat implies the ability to do work in the given situation.
    This applies to the radiation moving from the hotter surface to the colder surface.
    It does not apply to the radiation moving from the colder surface to the warmer surface.>>>

    I’m very interested in how this works Bryan. How is it that radiation going in one direction is able to do work, but radiation going in the other direction is unable to do work? Do the photons carry with them documentation regarding their temperature of origin to determine if they have a work permit or not? Do they compare permits to other photons going the other way? Seems to me this would be required as the photon not knowing the temperature of the destination would need to periodicaly check the state of the union (these sound like unionized photons to me, but that is only an assumption on my part) to determine if they are permted to do work along the way. Or do the photons upon arrival, determine the temperature at destination, compare to documentation of temperature of origin, and only then make a determination if they are allowed to do work or not?

    This is very intriguing Bryan. Exciting. Who knew that a photon travelling at the speed of light, carrying a tiny bit of energy with it, and having charactersitics identical to other photons save for direction, would know if that packet of energy is allowed to do work or not? Does this not imply some sort of system capable of exchanging information between photons? Are the photons then indpendant decision makers arriving at their own conclusions, or do they await instructions to be communicated to them by a higher authority? Is this perhaps the very communication system that would confirm the existance of Gaia taking a firm hand in control of the universe?

  315. mkelly says:

    Phil. says:
    March 3, 2011 at 6:48

    “You have been shown data here which shows exactly how weakly N2 and O2 interact with IR, but it doesn’t agree with you preconceived belief so you ignore it. Your ‘idea’ contradicts a century of science on the structure of molecules and their interaction with radiation as well as the experimental basis for it.”

    I must have missed that. The graph I saw never said anything about temperature. That is my specific question. If it was intended to say that those areas are the only frequency they emit/aborb under all temperatures I misunderstood the graph.

    You seem so angry. I was not granting you permission to do anything I was trying to be polite/civil. If you chooe otherwise please let me know.

    Adiabatic lapse rate accepts the fact that pressure causes heat. Less pressure less heat. So PV=nRT is valid. I know very little of Triton.

    My heat transfer book shows insolation varys between 1063 w/m^2 at 90 deg to sun down to 41 w/m^2 at 5 deg. So if we are going to average then 55o w/m^2 should closer to what the surface receives. See what George said above.

    I did not forget reflection or albedo I just forgot to type in “etc.” after absorption. When at work hurrying to type something out it is possible to misspell or forget a word.

    I have very little preconcieved ideas about N2 and O2 radiation since I have seen little written about it and have found little. If they cannot as I said I can accept that.
    But as you have read here there are varied opinions of this.

  316. RJ says:

    Ira

    Have you read Slaying the Climate Dragon. Or followed the debate on this book

    http://judithcurry.com/2011/01/31/slaying-a-greenhouse-dragon/

    Referring to comment 5. No one said the greenhouse effect is a scam (CAGW is another matter). But science moves on. The greenhouse theory seems suspect as has been shown above. Name calling (deniers etc) will not change that.

    And a few extra parts of CO2 will not destroy the planet. Whether humans can (I doubt it though) is not relevant to this AGW debate.

  317. Phil. says:

    Myrrh says:
    March 3, 2011 at 6:32 am
    ? A lightbulb gives out 95% of its energy in Heat and only 5% of its energy in Visible Light. Heat is Thermal IR.

    It’s been confused by AGW.

    It hasn’t been confused at all except for those like you who don’t understand what they’re talking about. The light distribution emitted by and incandescent body is well described by the Stefan-Boltzmann law, visible light ranges from ~400nm to about ~770nm (some use slightly different limits).

    It may well “peak” in the Visible by energy states, but the Visible as % of the total energy emitted is relatively tiny compared with the total amount of energy emitted.

    The amount, therefore the real “most”, of the energy emitted from the lightbulb is Thermal IR.

    Not by any reasonable definition of ‘Thermal IR’, usually 3-15 microns (an application driven nomenclature associated with imaging, not any fundamental distinction). Most is probably in the NIR.

    The Sun thus, if similar as Ira has it, should also be emitting the same kind of ratio of Visible to Thermal.

    No because similar is in this case a factor of two as I showed above, particularly for a ‘cool’, longer life lamp. I’ve told you what the ratio is above, can’t you read?

    [Assuming that the real difference between thermal and visible energies are understood, not as Ira has it that Solar is the Heat energy we feel from the Sun]. I have seen this figure put at 80% in Thermal for the Sun.

    I’m sure you have, doesn’t make it right though, by the above definition ‘thermal IR’ from the sun is extremely low. All absorbed light from the sun, no matter what wavelength will heat up the earth, 1 W/m^2 absorbed will heat up an object equally, your fantasy notwithstanding.

  318. RJ says:

    Test

    My previous post has not appeared

    [Reply: That's because there were a lot of posts in the spam folder. Yours was one of them. Posted now. ~dbs, mod.]

  319. Phil. says:

    Bryan says:
    March 3, 2011 at 8:14 am
    The transfer of heat in the Macro World always moves spontaneously from the higher temperature object to the lower temperature object never the reverse.
    The thermodynamic meaning of heat implies the ability to do work in the given situation.
    This applies to the radiation moving from the hotter surface to the colder surface.
    It does not apply to the radiation moving from the colder surface to the warmer surface.
    So there is no dual heat transfer only heat transfer from higher to lower temperature objects

    There is energy transfer in both directions resulting in a net transfer of heat from the hotter to the cooler

  320. RJ says:

    Ira

    Have you read Slaying the Climate Dragon. Or followed the debate on this book. At judith Curry’s website for example

    Referring to comment 5. No one said the greenhouse effect is a scam (CAGW is another matter). But science moves on. The greenhouse theory seems suspect as has been shown above. Name calling (deniers etc) will not change that.

    And a few extra parts of CO2 will not destroy the planet. Whether humans can (I doubt it though) is not relevant to this AGW debate.

  321. Reed Coray says:

    izen says:
    March 3, 2011 at 12:08 am
    Reed Coray says:
    March 2, 2011 at 9:40 pm
    “….I also believe that if you allow conduction of energy between objects, depending on the rate of energy transfer via conduction, the temperature of the original object will be lower than the original object temperature in the absence of the second object. Note that the presence of conduction does not eliminate backradiation–it will still be present. If true, we have reached a state where (a) backradiation exists and (b) the temperature of the original object is lowered….”

    No point in putting on a coat in cold weather then…

    Izen, is it correct to infer that for a fixed internal energy source, you believe that the steady-state temperature of an object is always increased in the resence of backradiation?

    The key phrase in my statement was “depending on the amount of conduction”. Most coats are made of material that minimizes conduction. The simplist example I can think of to demonstrate that backradiation can simultaneously exist with lower temperatures is to consider a blackbody sphere with a fixed internal thermal energy source. To achieve steady state, the surface of the sphere must radiate energy away from the sphere at a rate equal to the internal energy source. The rate of energy radiated away from a blackbody sphere is proportional to the square of the sphere radius. Thus, for a fixed rate of internal energy generation, as the radius of the sphere increases the “steady-state” temperature of its surface decreases. Fix that radius at a finite value and determine the steady-state surface temperature. Now place a larger radius spherical shell around and centered on the sphere. Assume that the thermal conduction properties of this shell are such that its temperature is everywhere the same. Call this shell a “coat”. In the absence of conduction and convection, the presence of the shell (coat) will result in an increase in the sphere surface temperature. In the terminology of many, the increae in temperature is the result of backradiation from the surface of the shell. However, if thermally conducting rods are inserted that join the sphere and the shell such that the sphere surface and the shell surface can be brought to the same or very nearly the same temperature, then because the shell’s surface is larger than the sphere’s surface and thus can radiate the energy from the fixed internal energy source to space at a lower temperature, that temperature (both the shell and the sphere) will be lower than in the absence of the shell–notwithstanding backradiation from the shell’s inner surface to the sphere’s surface.

    On another topic. Several commentors have noted that for a situation where radiation is the only form of energy transfer, “heat” and “net” radiative energy transfer are the same thing. I agree. When discussing thermal energy transfer between objects by conduction, people seldom, if ever, discuss “backbumping”. For example, one model of conduction energy transfer between two objects made of the same material is that the “aggregate” speed of the random motion of the molecules of the object at the higher temperature is greater than the “aggregate” speed of the random motion of the molecules of the object at the lower temperature. When the two objects are brought into contact, some of the motion of the faster moving molecules of the higher temperature object is transfered to the motion of the slower moving molecules of the lower temoperature object; thereby increasing the “aggregate” speed of the colder object’s molecules and decreasing the “aggregate” speed of the warmer object’s molecules. However, since molecular speed is a distribution, not a single value, some of the molecules of the colder object will be moving faster than some of the molecules of the warmer object. When two such molecules collide, energy may be transfered from the colder object to the warmer object. Call this cold-to-warm object energy transfer “backbumping”. Most discussions of thermal energy transfer via conduction, don’t separately analyze “forwardbumping” and “backbumping”, they just discuss net energy transfer–i.e., heat.

  322. Oliver Ramsay says:

    davidmhoffer says:
    March 3, 2011 at 1:13 am

    “C’mon Oliver. S’plain it. I’ll ridicule…. I mean read it in the morning.”
    —————————
    I don’t doubt that you’ll ridicule…. nor that you’ll sort of read it. But understand it?
    Recall the sage counsel of your grandmother; she knew the score.
    Before sending you out to play in the snow, she gave you a big pair of woolen socks and said “Here, these will keep you warm!”
    You came in later and she handed you a steaming mug of cocoa, saying “This will warm you up”.
    She never got them reversed. How did you?
    Perhaps it’s this notion that if there’s any aperture in the dwelling, all the heated air will rush out and be replaced by frigid outside air in the blink of an eye. With a more nuanced approach you can size and locate your ventilation openings to compromise between convected heat loss and asphyxiation. If you get it wrong, the igloo walls will not save you with their back radiation. As an aside, most of the inhaled O2 of a breath is simply exhaled without finding its way into the bloodstream.
    It’s very bizarre to me that an igloo can heat a heat source but it can’t heat a non-heat source. Anything that is hotter than an adjacent thing is a potential heat source. However, the cooler thing is not a heat source to the hotter thing. Yes, they radiate back and forth, but they don’t radiate heat. Heating occurs when the average kinetic energy of the molecules in the thing rises. If it goes down, we say that it has lost heat, if it stays the same, we say it has stayed the same. In spite of all the to-ing and fro-ing of packets of energy.
    The vibrational excitement that an absorbed photon causes within a molecule does not necessarily get turned into translational motion of that molecule, which would be termed heating IF there were enough similar occurrences to elevate the AVERAGE kinetic energy of a whole bunch of molecules. Heat is not a property of individual particles.
    Photosynthesis is not a heat process.
    On the other hand….. If we built the igloos in the shape of a pyramid, we’d really be onto something.

  323. Tim Folkerts says:

    A few general comments and then I may slink back to the rest of life for a while …

    * Whatever Clausius or Carnot may have said 150 years ago, that is not the final word. Science moves forward. If you want to understand heat and entropy and the 2nd law at a fundamental level, you really should look at the more modern approach of Statistical Mechanics. (For example, a quick look at the online physics courses listed at MIT shows “Statistical Physics I & II”, and “Statistical Mechanics I &II”, but no “Thermodynamics” class!)

    * “Heat” is a very abused word – used to mean different things by different people in different settings.

    -> In thermodynamics, heat is akin to work — it is a process that transfers energy. You can be working on an object (applying force*distance), but you don’t “have work”. Similarly, you can be heating an object (perhaps holding a cold object in your hands) , but you don’t “have heat”. Delta(E) = δQ – δW where δQ is the heat that was applied TO the system, and δW is the work that was applied to the system.

    –> In other circumstance, people (including scientists and engineers) use “heat” to mean “thermal energy” or “internal energy”. In this meaning an object can “have heat”. This is, however, at odds with the formal thermodynamics meaning.

    –> Colloquially heat can mean “temperature” but this is never used in science as far as I know.

    * Thermodynamics deals with averages and net values (summations) (hence the term “Statistical Mechanics”. Pressure in a gas is the net force provide by trillions of collisions of individual molecules divided by the area where they hit. Heat is the net transfer of energy during trillions of interactions (photons, phonons, collisions, …).

    So energy does indeed flow from hot objects to cold object AND from cold objects to warm objects (via those photons, phonons, collisions …). It is only the NET flow (ie “heat”) will always be from hot to cold (for sufficiently large systems averaged over sufficiently large times).

    Specifically, energy does flow from the cooler atmosphere to the warmer surface via photons. But more energy flows the other way, so “heat” is doing what it is supposed to do. That flow of energy via photons from the atmosphere by itself can never raise the surface temperature above the temperature of the atmosphere (becasue more energy is going the other way). But that flow energy combined with the flow of energy from the sun can make the surface warmer than it would have been with no energy from the atmosphere.

    That’s the physics. You may go back to arguing the semantics of whether the atmosphere is “warming the ground” or simply “slowing the cooling” if you like.

  324. davidmhoffer says:

    Ira,
    Are you and I missing a business opportunity here? There was a suggestion that you and I need to get some new physics text books. I don’t know what you have in your personal library, but I dug into mine. I have some 15 yr old texts, some 35 year old texts, an encyclopaedia from the early ’70’s. The physics formulas in them all seem to be the same. The encyclopaedia references other text books going back to the 1800’s. I jumped in my car and drove to the bookstore, flipped through the stuff on the shelves, same, al the same.

    That must be where we’re going wrong! There must be new text books with new formulas these people are quoting from, but they aren’t in wide distribution yet. Seems to me some entrprneuarial spritit should allow for writing of now text books right away. I think what would make most sense is using the same formulas and documented experiments, no point doing all that research again. All we need do is print multiple versions, each version having its own set of conclusions.

    That way, people who have determined in advance that CO2 cannot absorb and emit LW without violating laws of thermodynamics could buy those versions, and people who have determined in advance cold things can’t warm up warm things could buy another version, and people who have decided in advance…we could even have custom texts printed (at additional cost of course) allowing for custom conclusions.

    we’ll make a killing! we can even present our selves as the authors of every version, making any opinion we want to argue correct since we’d be quoting from the newest texts available, which supersede all those centuries of science until now. We’d be able to say we wrote the book no matter what conclusion we were arguing, plus we’d have tons of customers happy as clams because they’ve got text books clearly concluding that their beliefs about physics are correct, and just as they haven’t had to read and understand the old text books, they won’t have to read and understand these either, but they’ll be so much easier to use because the conclusions are pre-selected and organized for easy reference.

    We could even publish versions either integrated with PNS, or not integrated.

    Ira, I really do think the field here for conclusion driven text books to be reverse engineered from the conclusions is about to blow wide open. We need to move fast, the IPCC already cornered the whole Climate Science market, and Ravetz is making his move on PNS, but the rest is open, wide open! Are you in?

  325. Reed Coray says:

    I believe Planck’s blackbody radiation law applies to “cavity” radiation. That is, it describes the amount of, directionality, and spectral distribution of electromagnetic energy radiated from a small “hole” in the surface of a cavity whose internal walls are at a uniform temperature. The amount of radiation is proportional to the area of the hole. This concept has been applied to the “surface” of a body. Specifically, for a differential area of the surface of a blackbody at a single temperature, the law gives the spectral distribution and the amount of energy radiated from that differential area into a differential solid angle. When discussing an atmosphereless earth, it’s reasonable to identify the transition between the earth and space as a surface–i.e., it exists in two not three dimensions. When considering a gaseous atmosphere, I’d appreciate it if someone could identify the “surface” to which Planck’s law can be applied. As I see it, Planck’s law does NOT apply to gases. Similarities may exist in the spectral nature of the radiation being emitted from a gas and the radiation being emitted from a differential area on a blackbody surface; but without a clearly defined surface, I don’t see how Planck’s blackbody radiation law can be applied. In particular, a part of Planck’s blackbody radiation law is the COSINE of the angle between (a) the direction of the solid angle into which the radiation is propagating and (b) the normal to the radiating surface. Would someone please tell define that “normal” for a gas?

  326. Phil. says:

    mkelly says:
    March 3, 2011 at 9:03 am
    Phil. says:
    March 3, 2011 at 6:48

    “You have been shown data here which shows exactly how weakly N2 and O2 interact with IR, but it doesn’t agree with your preconceived belief so you ignore it. Your ‘idea’ contradicts a century of science on the structure of molecules and their interaction with radiation as well as the experimental basis for it.”

    I must have missed that. The graph I saw never said anything about temperature. That is my specific question. If it was intended to say that those areas are the only frequency they emit/aborb under all temperatures I misunderstood the graph.

    That small weak band is the only one emitted between 2 and 20 microns, that’s the bulk of the BB emission due to the earth’s temperature.

    You seem so angry. I was not granting you permission to do anything I was trying to be polite/civil. If you chooe otherwise please let me know.

    Adiabatic lapse rate accepts the fact that pressure causes heat. Less pressure less heat. So PV=nRT is valid. I know very little of Triton.

    Pressure doesn’t cause heat, the adiabatic lapse rate is given by g for the planet divided by the specific heat at constant pressure of the atmosphere.

    My heat transfer book shows insolation varys between 1063 w/m^2 at 90 deg to sun down to 41 w/m^2 at 5 deg. So if we are going to average then 55o w/m^2 should closer to what the surface receives. See what George said above.

    You’ve forgotten about the other side of the planet!

    I did not forget reflection or albedo I just forgot to type in “etc.” after absorption. When at work hurrying to type something out it is possible to misspell or forget a word.

    OK but you get 340 as an average if you don’t allow for reflection (and you remember about the night-time).

    I have very little preconcieved ideas about N2 and O2 radiation since I have seen little written about it and have found little. If they cannot as I said I can accept that.
    But as you have read here there are varied opinions of this.

    It’s not a matter of opinion but of facts, the fact is that without a dipole N2 and O2 can’t absorb or emit in the IR band (the exceptional weak bands I showed are more than a million times weaker than CO2 and H2O bands).
    Gases do not emit as BB emitters under atmospheric conditions but at discrete frequencies, another fact. As someone once said you’re entitled to your own opinions but not your own facts, read any textbook on Molecular Spectroscopy.

  327. George E. Smith says:

    “”””” mkelly says:
    March 3, 2011 at 9:03 am
    Phil. says:
    March 3, 2011 at 6:48

    “You have been shown data here which shows exactly how weakly N2 and O2 interact with IR, but it doesn’t agree with you preconceived belief so you ignore it. Your ‘idea’ contradicts a century of science on the structure of molecules and their interaction with radiation as well as the experimental basis for it.”

    I must have missed that. The graph I saw never said anything about temperature. That is my specific question. If it was intended to say that those areas are the only frequency they emit/aborb under all temperatures I misunderstood the graph.

    You seem so angry. I was not granting you permission to do anything I was trying to be polite/civil. If you chooe otherwise please let me know.

    Adiabatic lapse rate accepts the fact that pressure causes heat. Less pressure less heat. So PV=nRT is valid. I know very little of Triton.

    My heat transfer book shows insolation varys between 1063 w/m^2 at 90 deg to sun down to 41 w/m^2 at 5 deg. So if we are going to average then 55o w/m^2 should closer to what the surface receives. See what George said above. “””””

    mk , only a casual reader here at WUWT would not be aware, that there is basically very little difference of opinion between what I understand about this radiative transfer, and what Phil says. I may put it a little differently; but I believe that Phil is much more conversant with this subject than I am; although I have no idea what he does.

    I’m basically relying on concepts I learned in school nearly 55 years ago, along with a bit of late night book cramming; whereas it seems Phil is into this stuff every day.

    I do a lot more hand waving, whereas Phil knows where to go get the goods.

    But I’m a firm believer that there is no future in trying to deny that the absorption of LWIR radiant energy by GHGs is a fairly well understood process, by means of which the atmosphere is warmed. There are of course other processes that heat the atmosphere including solar radiative heating; but what we call the “greenhouse” effect; at least to the point that it heats the atmosphere, is demonstrably real, and fighting that is a poor choice of causes to die for.

    I’m less sure of what subsequently happens, after that energy transfer heats the atmosphere, but it is obvious that as a consequencew some of the energy that would have escaped to space, is returned towards the surface, so the exit is at least delayed.

    The way I like to think about the consequences of that delay, is that during that delay, the sun continues to input energy to the earth at its standard rate, and it is that incremental solar input during the extended exit time of the outgoing LWIR radiation that is the source of the net heating; not the returned LWIR itself.

    And I have even offered an analagout situation of the water running into the bathtub, with the plug removed. The constant input rate stays fixed, but the outflow rate increases, as the water level rises in the tub, increasing the driving pressure.

    If you now partialy restrict the drain opening (GHG interception), you now need a higher water pressure to create the same outflow rate you originally had; but it is the faucet that supplies the extra water to raise the level back to the stady state level; not the obstruction.

    And what’s with the citation of the ideal gas equation of state; that only applies to a closed system; and the earth or any planetary atmosphere is anything but a closed system; so the ideal gas law, or any other equation of state, really adds no new understanding to the situation.

    I don’t know whether Phil and I agree on everything; he doesn’t post a lot of extensive writings; probably has better things to do (so do I); but I think we likely agree on a whole lot more than we might disagree; and in the past I have learned that I should take note of such disagreements, because it has proven to be me that is usually wrong.

    I’m far less interested in the role of CO2 than it may seem, because I’m fairly convinced that it is H2O that is in full control of the situation (specially clouds) so my interest in the details of CO2 is more curiosity than concern. I certainly do want to understand the inner Physics of the CO2 absorption bands; and more importantly the thermal radiation from gases; but I believe the answer to earth’s Temperature comfort range lies in the totality of the properties of the H2O molecule; not the CO2 molecule.

  328. Bryan says:

    Tim Folkerts

    I agree with the general outline of your post.
    I think you observed that I was using Heat in the correct orthodox thermodynamic sense.
    If we have a precise meaning for Heat then if we want a scientific discussion about the climate its better that we use it rather than each participant using their own.
    The hot and cold objects exchange photons.
    However Infra Red Electromagnetic Radiation is not Heat.
    Its correct to say that the net result is more photons go from the higher to the lower temperature.
    This means that the flow of Heat is from the higher to the lower temperature object.
    Quite a number of IPCC advocates have advanced the idea that the cold night atmosphere can HEAT the warmer Earth surface.
    Which as you well know is nonsense.

  329. Vince Causey says:

    Myrrh says:
    March 3, 2011 at 4:47 am

    You said “So, radiation is included, not just conduction, which means that it is impossible for a cooler molecule to transfer heat to a hotter one because that would violate the 2nd Law.”

    You based this on the following definition of the 2nd law: ““When two isolated systems in separate but nearby regions of space, each in thermodynamic equilibrium in itself, but not in equilibrium with each other at first, are at some time allowed to interact, breaking the isolation that separates the two systems, and they exchange matter or energy, they will eventually reach a mutual thermodynamic equilibrium.”

    Myrhh, you have read the 2nd law and have drawn the wrong conclusion. The 2nd law considers “two isolated systems in separate but nearby space, not in equilibrium with each other,” as you have pointed out. It then says that when they are allowed to interact they “exchange matter or energy. . .” This is the crucial point THEY ie both bodies – hotter and cooler – exchange energy. But the crucial conclusion from the 2nd law, is that as a result they “will eventually reach a mutual equilibrium.” This is because the hotter body is transmitting at a far greater flux density than the cooler body.

    The GHG hypothesis describess a cooler atmosphere radiating to a warmer surface, which some, including you, say is impossible because it violates the 2nd law. But we have seen that the 2nd law says both bodies will exchange energy, and they will eventually equilibriate – if they are isolated systems. If the Earth was not continually recharged with energy from the sun, it would cool down to the temperature of the atmosphere, and both of them would continual to cool as they equilibriate with space.

  330. jae says:

    Ira:

    Regarding the “denialists,” you say:

    “They may also harbor conspiratorial views of environmental groups and the government.”

    I would be interested in your definition of “consipracy theory.” Does it include the belief (which I possess) that environmental groups and government have the goal of amasssing more and more power so the state can dictate more and more about the details of peoples’ lives?

    Your use of the D-word disgusts me, BTW.

  331. Bryan says:

    davidmhoffer

    I said
    The thermodynamic meaning of heat implies the ability to do work in the given situation.
    This applies to the radiation moving from the hotter surface to the colder surface.
    It does not apply to the radiation moving from the colder surface to the warmer surface.
    You said
    I’m very interested in how this works Bryan. How is it that radiation going in one direction is able to do work, but radiation going in the other direction is unable to do work?
    I say
    Its really a very simple practical test.
    I can give you several examples of a higher to lower temperature situation doing work like a steam engine and so on.
    You cannot give me one example of low temperature to high temperature situation doing work.

  332. mkelly says:

    George E. Smith says:
    March 3, 2011 at 11:07 am

    “I’m far less interested in the role of CO2 than it may seem, because I’m fairly convinced that it is H2O that is in full control of the situation (specially clouds) so my interest in the details of CO2 is more curiosity than concern.”

    If it had been said years ago the H2O was causing a warmer atmosphere I would have nodded my head yes and moved on. But CO2 I think is a minor player if at all.

    Since the earth / atmosphere is a closed system I think the gas law applies. Pressure creates heat.

    I probably agree with you and Phil much more than he thinks, but asking questions never hurt. I have never seen a discussion of N2 and O2 anywhere and 99% of atmosphere better be able to be explained. As you seee there is a variety of opinions.

    I kept my heat transfer & thermo book from college and refer to it often to gain a larger more indepth understanding.

  333. Myrrh says:

    Tim Folkerts says:
    March 2, 2011 at 9:18 pm

    I didn’t mean “possibly”: I was thinking of two macroscopic chunks of material pressed together. This would put millions of atoms from the two pieces in contact, which will all be colliding many times per second with atoms from the other object. There WILL be many collisions here. There will be millions of microscopic transfers of energy ever second. It is just the NET flow of energy for large collections of energy that is governed by the 2nd Law, not the transfers during individual collisions.

    So you say. But at the meeting point of these two is the environment they are in, the 2nd Law says heat does not flow spontaneously from colder to hotter. I’m the one saying there should be a ‘possibly’ in there. That atoms in matter, waves in radiations, etc. might be able to do such things doesn’t mean that they can actually do such things. This “NET” flow of energy fitting 2nd Law, is imaginary. You may think you’ve calculated it all out “statistically” as an explanation, but that’s saying nothing at all, you’ve just added that to the Law. Average, your NET, is an artificial construct. It doesn’t acutally exist nor is it included in the Law as the Law stands.

    I think, this is of the same ilk as using ideal gas laws to describe real gas molecules. The ideal gas is imaginary, just as your Net is imaginary. The real gas is real. Ideal gas laws describe imaginary gases, they do not describe real gases. Just so, your Net flow of energy is an “ideal” assumption of what is happening, it does not actually describe what is happening or what is capable of happening.

    Just as an ideal gas has no volume etc. and so cannot actually describe what is happening to a real gas which has volume etc., so your Net transfer because ‘energy flows in every direction and collisions happening between them transfer energy doesn’t mean that the cooler are actually transfering that energy to a hotter.

    The transfer of energy at EACH collision MUST be from the hotter to the colder to be according to the Law. You’re fudging it by creating this idea of a Net Transfer when the Law says it applies to all energy too.

    I think this is part of the problem where AGWScience gets lost to the point where it thinks, as the energy graphic depicts here, that shortwave light energies can heat the earth and are now calling it Thermal Energy. Physical limitations in actual reality are jettisoned to this imaginary Net Transfer just as Carbon Dioxide is treated as an ideal gas without volume etc. capable of doing impossible things like defying gravity to spread in the atmosphere without work being done even though its heavier than air because it bounces off all the other molecules in the atmosphere ‘diffusing’ as if an imaginary ideal gas in a jar.

    If, heat does not always flow spontaneously from the hotter to the colder in all matter and energy as the Law states and you can prove it doesn’t as you claim it doesn’t, then you have falsified the Law.

    You cannot say that the Law doesn’t apply to your particular set of energies or matter just because you think it shouldn’t.

  334. George E. Smith says:

    “”””” Reed Coray says:
    March 3, 2011 at 10:54 am
    I believe Planck’s blackbody radiation law applies to “cavity” radiation. That is, it describes the amount of, directionality, and spectral distribution of electromagnetic energy radiated from a small “hole” in the surface of a cavity whose internal walls are at a uniform temperature. The amount of radiation is proportional to the area of the hole. This concept has been applied to the “surface” of a body. Specifically, for a differential area of the surface of a blackbody at a single temperature, the law gives the spectral distribution and the amount of energy radiated from that differential area into a differential solid angle. When discussing an atmosphereless earth, it’s reasonable to identify the transition between the earth and space as a surface–i.e., it exists in two not three dimensions. When considering a gaseous atmosphere, I’d appreciate it if someone could identify the “surface” to which Planck’s law can be applied. As I see it, Planck’s law does NOT apply to gases. Similarities may exist in the spectral nature of the radiation being emitted from a gas and the radiation being emitted from a differential area on a blackbody surface; but without a clearly defined surface, I don’t see how Planck’s blackbody radiation law can be applied. In particular, a part of Planck’s blackbody radiation law is the COSINE of the angle between (a) the direction of the solid angle into which the radiation is propagating and (b) the normal to the radiating surface. Would someone please tell define that “normal” for a gas? “””””

    Reed, in the “Cavity” case you cite, the radiation inside the cavity, that is presumed to have black walls (non thermally conducting), is of course isotropic. There is no preferred direction inside the cavity. But if you consider any planar “surface” element, inside the cavity, the total flux passing through that aperture vaies with angle in a Cosine fashion, so that the Radiance of the “aperture” is constant for any direction of observation. If that “aperure ” surface, happens to be a real planar aperture in the thermally impenetrable wall of the cavity, then radiation will be emitted from that aperture in the same cosice (Lambertian )pattern, and the total emitted energy, will simply be pi times the axial intensity (normal to the aperture). But the radiation inside the cavity remains isotropic. And in the closed case, the radiation is in equlibrium with the Temperature of the materials in the cavity (walls or contained gases or other materials) so Kirchoff’s law applies, and the radiation absorbed by any material must match fequency for frequency, and directionally, with what is absorbed by that material.

    In the atmosphere, you do not have a closed system, so such thermal equilibrium can hardly be present, given that all manner of things, are driving radiant energy into any volume of atmosphere, as well as conductive, and mass transport (convective) energy transfers are taking place. So I don’t see how Kirchoff’s law can apply to such an open system. It certainly doesn’t apply to a slab of pure silicon, which readily absorbs radiation , of higher frequency than the band gap energy; but it cannot then re-emit those same energies, because silicon is an indirect gap semiconductor, and the minimum energy gap does not occur at the same momentum state , so such transitions are generally prohibited.

    But the same slab of silicon can readily radiate a continuous thermal spectrum, that depends only on the Temperature of the material, and generally follows the Planck formula, with perhaps some spectral emissivity function. So clearly Kirchoff’s law does not apply to that clearly non equilibrium system.

    But for the earth atmosphere apparently based on AIRS satellite observations, as well as computer calculations, what is seen, looking down is a clearly blackbody like radiation spectrum, with holes in it corresponding to various atmospheric components like H2O, CO2, and O3 primarily. Arguably the existence of those holes is evidence that what is absorbing that radiation is clearly not then emitting it in the same directions, so Kirchoff’s law is not being followed.

    Now I can see that the surface should emita BB lke thermal continuum spectrum, based on the surface Temperature; and say CO2 will intercept some portion of that notably in the 666 cm^-1 frequency range (15 microns). Any re-emission would be isotropic, so the portion that continues upwards, must be attenuated; which is what is seen.

    But you still have the whole broad wavelength range of thermal radiation, so that is either directly from the surface; so should exhibit the surface Temperature characteristic (spectrum); or it is emitted from the atmosphere.

    We are asked to believe that a perfectly good solid or liquid surface is emitting BB like contiunous spectra, based only on the Temperature of the material; but the instant that material evaporates or sublimes, and becomes a vapor at the same tem[ertaure, the radiation must cease, because it is a gas, and gases do not emit black body like thermal radiation. Well that is what we are asked to believe.

    Perhaps it behoves us to enquire; what exactly is the Physical mechanism by means of which that solid or liquid was radiating a thermal spectrum in the first place. And why does that mechanism shut down only in the vapor phase (if it does) ?

    If a gas cannot emit a thermal continuum of EM radiation following the Temperature of the gas; but a sold, such as a gold brick can. How about a gold film that is say one Angstom thick or one nano metre thick. It is solid isn’t it ?
    How about a graphene layer, that is a single atom thick solid; can it emit a black body like (carbon is black) thermal spectrum.

    And if not; why not ?

  335. Tim Folkerts says:

    George E. Smith says: March 3, 2011 at 11:07 am
    “And I have even offered an analogous situation of the water running into the bathtub, with the plug removed. The constant input rate stays fixed, but the outflow rate increases, as the water level rises in the tub, increasing the driving pressure.”

    I tried to expand/improve on George’s analogy. What do you all think?

    * Water is spray downward from the shower head into the tub (like the sun’s photons “spraying” down onto the earth).

    * Pumps in the tub spray the water back up out of the tub (like the earth “spraying” IR photons outward).

    * The fuller the tub, the faster the pumps spray. There will be some equilibrium level of water for a given flow rate from shower head (like an equilibrium temperature of the earth). More inflow of water = higher level. [This is sort of like the driving force increasing as the water gets deeper in your analogy, but my version helps with the next stages of the analogy.]

    * Above the bathtub, there are some sponges that can absorb some of the water spraying up from the tub, but they are positioned so they don’t stop any of the water coming in from the shower head. (like the GHGs absorbing some of the outgoing IR but not blocking incoming solar photons).

    * Some of the water from the sponges will evaporate and leave the area, but some will drip back into the tub. This will add ADDITIONAL flow into the tub above and beyond the original flow from the shower head (like GHGs add ADDITIONAL photons to the earth, above and beyond what comes from the sun). And note that the sponges are not creating any water – they are just returning some of it (just like GHG’s do not create any energy).

    The existence of the sponges makes the tub fuller.
    The existence of GHGs makes the earth warmer. There IS a flow of energy from the GHGs to the surface, although the NET flow = “heat” is still upward.

    Now that I think about it, this is almost exactly like the animation Ira had way back in his blog entry – with streams of photons spraying around and getting absorbed and re-emitted!

    (There are still many shortcomings of the model. For example, you could envision two different tubs that get fill alternately by the shower head (like night and day). You could add “clouds” that reflect some of the water from the shower head away before it ever reaches the tub. But that is the nature of models like this — they are designed to illustrate key ideas, not to be 100% analogous.)

  336. davidmhoffer says:

    Bryan;
    You cannot give me one example of low temperature to high temperature situation doing work.>>>

    So as I understand it, being unable to answer the question I asked, having no response even to suggest that the question I asked didn’t represent the issue in some manner, not even a hint that the issue posed is not representative, you now retreat into that most devastating of classic debate challenges:

    Oh yeah? Give me an example.

    I guess you missed all that discussion about CO2 and water vapour in the earth’s atmosphere, igloos, quinzees, blankets, jackets, insulation, everything Phil said, everything George E Smith said so you want me to provide examples. OK.

    CO2 and water vapour in the earth’s atmosphere, igloos, quinzees, blankets, jackets, insulation, everything Phil said, everything George E Smith said.

    One more time Bryan. Photons carry energy. From the moment they are emitted to the moment they are absorbed. They don’t know or care what the temperature of the surface was that emitted them, they don’t know or care what the temperature of the surface was that absorbed them, and they don’t know how many other photons carrying how much energy there are nor what direction they are travelling in, nor what the net of all the photons in all directions is.

    The photon started Here and went There.
    H>>>>>>>>>>>>>T
    It carried with it and amount of Energy.
    As a consequence of which the two surfaces had a change in the amount of energy each had after the photon finished its journey.
    H-e T+e

    Was work done? Yes.
    And can you tell from the information provided which is warmer, Here, or There? Can you tell what the net energy flux between Here and There is?
    Can you tell in which direction the net energy flux is positive?
    Nope, the only question we can answer is that work was done, we can’t answer any of the other questions.

    Well, unless we would like to make up some new physics that fit our belief system. The religeon of I’ve made up my mind, stop confusing me with the facts.

  337. mkelly says:

    davidmhoffer says:
    March 3, 2011 at 1:18 pm
    ” They don’t know or care what the temperature of the surface was that emitted them,”

    They may care so they know what frequency to run around at. :)

    Was work done? Yes.
    Please tell what work was done. dQ= dU + dW Curious.

  338. kuhnkat says:

    DavidMHoffer,

    “The photon started Here and went There.
    H>>>>>>>>>>>>>T
    It carried with it and amount of Energy.”

    Don’t get carried away. Unless you were watching you don’t KNOW there was even a photon!!!

  339. George E. Smith says:

    “”””” davidmhoffer says:
    March 3, 2011 at 1:18 pm
    Bryan;
    You cannot give me one example of low temperature to high temperature situation doing work.>>>

    So as I understand it, being unable to answer the question I asked, having no response even to suggest that the question I asked didn’t represent the issue in some manner, not even a hint that the issue posed is not representative, you now retreat into that most devastating of classic debate challenges:

    Oh yeah? Give me an example. “””””

    David, it doesn’t take very much savvy, to grasp the notion that once a flock of Photons from any source are launched into the regions of space, they have no recollection of their origin, nor any knowledge about where they are going.

    It might be a billion years before they encounter anything that they can take any notice of. At that point they are too senile to recall that they came from a colder place than whatever this new thing is, so they are just going to land on it anyway.

    The point is that this is a one way open system.

    The second law in the form it was stated by Clausius, makes it clear that it applies to closed cyclic systems; where the means even exists for the energy to go in either direction. Then it becomes clear, that although each end can communicate with the other and even transport energy (in the form of photons shall we say) , the reverse is also possible, but the net result is that, in the absence of some other effect (as Clausius put it) namely the doing of work on the system, there can only be a NET transfer of energy in one of those directions; which Clausius asserted, is the direction from the hotter (higher Temperature) source to the colder (lower temperature) sink.

    Like you I am amazed that this very simple concept is unable to be grasped by so many. In an open non-cyclic system, the energy can go via radiation at least, any darn place it wants to.

    In the case of the downward (or back radiation if you like), it is no different from paying for your groceries at the cash register.

    You buy $7 worth of Negra Modello (specially after reading this); and you hand the teller a hundred dollar bill, and (s)he gives you back $93.

    You are still losing money; just not as fast as if the teller gave you no change back. You perhaps still rely on your employer to keep supplying you with more money in your pay check; he is the one making you not poor; not the teller who simply gives you back some change.

    Likewise it is the sun with its continuous pay check, that is keeping you warm, and the GHG “insulation” is simply slowing the rate at which you go broke.

  340. kuhnkat says:

    Tim Folkerts,

    the shower analogy stops at the Sun where it is Producing the energy. Everything is passively absorbing an radiating. There is a difference.

  341. davidmhoffer says:

    http://knowledgedrift.wordpress.com/2011/03/03/why-the-co2-greenhouse-gas-debate-doesnt-matter/

    I have provided the most comprehensive rebuttal possible to this debate in clear English, using the radiative properties of CO2 as calculated by the IPCC, extrapolated forward with the addition of twice as much CO2 as human’s have added to the atmosphere in our entire history so far added on top of what is already there, which at the highest consumption rates for fossil fuel we have ever achieved will require another 150 years to accomplish, and shown conclusively the catastrophe that awaits us. Having been spending far too much time in this thread and on this topic, I can with confidence suggest the following:

    1. a whole bunch of people will scratch their heads and ask….that’s it? that’s what we’re beating each other over the head with physics text books about?

    and

    2. An argument will immediately break out between Phil, George, Ira and who knows who else that I should have shown error bars. Followed by someone claiming I have the sign wrong, it should be cooling, not warming. And a troll carefully explaining in great detail that I didn’t do the log calcs correctly, he doesn’t know how to do them himself but his calculator does and mine are wrong. A slew of PNS groupies going see? We TOLD you nothing would happen if you didn’t listen to us, but I don’t really pay attention, I’m not really listeng to them, another troll carefully explaining that if we convert to coal to gas plus burn all the oil plus shut down all the nuclear reactors it will happen in 90 years…and there ought to be one comment from Khartoum complaining that my instructions on how to build a quinzie should have included the fact that it generates its own heat source from within and melts almost instantly, and is there any way to harness this commercially because it appears to be perpetual motion, free energy, maybe we can use it to heat the planet because there’s an ice age coming and it turns out the CO2 is pretty much useless.

  342. kuhnkat says:

    George E. Smith,

    ” they have no recollection of their origin, nor any knowledge about where they are going.”

    Please read the details of the 2 hole experiments as it relates to photons and electrons in quantum mechanics. Your statement is not correct.

  343. Domenic says:

    to George E. Smith
    You wrote: “Reed, in the “Cavity” case you cite, the radiation inside the cavity, that is presumed to have black walls (non thermally conducting), is of course isotropic. There is no preferred direction inside the cavity.”

    The walls do not have to be ‘black’.

    Cavity radiation is an interesting phenomenon. In a completely enclosed isothermal cavity, it makes no difference what material the interior walls are composed of. The internal radiation in the cavity will conform precisely to blackbody radiation at the temperature the material is at.

    Even if the walls were composed of N2, it will emit blackbody radiation. Thus N2 can indeed emit all wavelengths under certain conditions. And if you inject any photons into this hypothetical N2 constructed radiation cavity, of any wavelength, it will all be absorbed by the N2 molecules within. 100% absorption regardless of wavelength.

    How a gas, atom, molecule, etc responds to thermal radiation, or emits thermal radiation, is dependent on the conditions of the environment it is placed within.

  344. Oliver Ramsay says:

    @davidmhoffer says:
    March 3, 2011 at 2:44 pm

    ————
    David, perhaps you are putting more energy (or is that heat?) into articulating your derision and impatience than you are into understanding exactly what interlocutors are trying to say.
    That the air is heated by the earth is not in dispute.
    That the radiated energy of the GHGs makes the earth warmer than it otherwise WOULD have been is not disputed.
    That the radiated energy of cold ice contributes to slowing heat loss from a person is not disputed.
    That the radiated energy from cold ice can heat a warmer object to a higher temperature than it was before…. now that’s problematic.
    If you are convinced that that does occur I would be genuinely interested to see an explanation with some quantification. If the ice is -30C and the thermometer is 20C, how much warmer will it get? Why will it stop getting warmer? What happens once the temperature increase stops, does it start going down?

  345. cba says:

    Ira,

    nice job on the explanation. The important parts were correct and well stated.

    concerning those with problems on radiative matters,

    stefan’s law is often shown in an abbreviated fashion, P/a = epsilon * sigma * T^4. The full version is P/a = epsilon * sigma * (Tb ^4 – Ts^4), where Tb is the temperature of the body and Ts is the temperature of the surroundings (assuming epsilon – the emissivity – is the same for all).

    Except for the sun, the surroundings of Earth are pretty much the microwave background at 2.7K.

    If one has a BB at 300K and they place it inside a box at 300K, there will be radiation absorbed and emitted by the BB continually but the BB will remain at 300K. If the box were cooled to 2.7K, approximately zero so far as radiation rates are concerned, the the BB will cool down eventually to 2.7K and it will take a fairly short amount of time to cool to 290K. If the box were cooled only to 290K, the time it takes to cool the BB from 300K to 290K will be somewhat longer than the time it takes to cool down in the first case. In neither case is there a NET flow of energy from the colder box to the warmer BB.

    Once the BB reaches the temperature of the box,there will no longer be a net flow of energy in either direction yet there will be radiation being emitted and absorbed by box and BB.

  346. Tim Folkerts says:

    “Even if the walls were composed of N2, it will emit blackbody radiation. Thus N2 can indeed emit all wavelengths under certain conditions. And if you inject any photons into this hypothetical N2 constructed radiation cavity, of any wavelength, it will all be absorbed by the N2 molecules within. 100% absorption regardless of wavelength. ”

    True, but ONLY IF the “walls” are thick enough so that they are not significantly transparent. If you look into the top of an aluminum can (which is VERY reflective on the surfaces) with a small opening, the opening will look matte black
    http://www.instructables.com/image/FHFTT0XFFAIU486/First-find-a-can.jpg

    But if you look into the top of a transparent bottle, it will not look black. http://farm5.static.flickr.com/4106/5082190422_e140121b6d_z.jpg (If you made the walls 10 m thick, then it might start looking black and start acting like a BB cavity radiator.)

    For N2, even the distance thru the entire atmosphere is not enough to block most photons. So the “certain conditions” to make your scenario work include MUCH thicker “walls” than the entire atmosphere. The N2 walls will act like the “pop bottle radiator”, not the “pop can radiator”. Hence you would not expect anything close to BB radiation from earth’s N2. (This is why the sun DOES act like a BB. The atmosphere DOES go deep enough to eventually absorb all the photons — and it is a plasma not a gas.)

  347. George E. Smith says:

    The one area in which it seems that I differ from what Phil has said, is that regarding the emission of a (frequency) continuum spectrum of thermal (due to Temperature) radiation by neutral gases; say N2, or O2, or how about a mono-atomic gas like Argon.

    I understand (to some extent) how individual atoms can radiate specific frequency spectral lines, as a result of energy level transitions of an electron in that atom. (used to teach it in fact). I also understand to about the same extent, how those spectral lines can have a “fine structure” producing several lines instead of one; due to refinements first proposed by Arnold Sommerfeld and others. I even understand how some atoms have even a hyperfine structure to their atomic spectra; that is a result of the specific structure in the nucleus of that atom, rather than its electron configuration.

    I have a quite rudimentary understanding of how molecules that have atoms connected together, which are capable of undergoing internal molecular vibrations and other oscillatory motions, can also emit line spectra that generally fall into the Infra-Red frequency range, because of the masses of the oscillating components, and the strengths of the binding forces that hold them together. I’m less happy with my understanding of how it happens that those oscillations also have some “fine structure” that manifests itself in bands of closely spaced narrow lines; that have some intrinsic line width that is a result of Heisenberg’s Uncertainty Principle, and the lifetimes of the excited states that lead to those intrinsic line widths. The longer the lifetime of the excited state, the narrower is the frequency spread of the emitted photon. dE.dt > h/2pi or something like that.

    I don’t quite understand how those multiple lines appear, as that seems to be taught in chemistry, rather than in Physics classes; so I’m working on that.

    Now so far as I am aware, all of the preceding, applies to atoms or molecules essentially independently of the phase of the matter, except to the extent that that state alters the physical parameters of the interactions.

    The ability of molecules to interact with Electromagnetic Fields, evidently is highly dependent on the “antenna” characteristics of the particular molecule. The classical “radio” antenna of Hertzian waves is a structure where charges can move around due to electric currents in the antenna, and thereby set up intertwined electric and magnetic fields; which it turns out are pretty much always perpendicular to each other. Per Maxwell’s equations, if those electric currents are not stationary, so the electric charges have a constant velocity, then the antenna will radiate energy away from itself, in a manner that depends on the dimensions of the antenna, and the rates at which the currents are varied (the frequency). This is all taught in “RadioPhysics”; one of the things I actually studied. The most basic antenna is the electric dipole, consisting of separated electric charges of opposite sign. Well atoms and molecules contain separated electric charges; electrons and protons. In some atoms or molecules, these charges are symmetrically disposed, so that the net positive charge and the net negative charge, both have the same center of charge; and we tend to say the atom (or molecule) has no “dipole moment”; that being the charge at each end times the separation. So molecules like N2, or O2, are considered to have no net dipole moment, so they make really lousy antennas, either for transmitting or receiving. So we describe them as being “not Infra-red active”.

    Some molecules like H2O for example which is not at all symmetrical; there’s that 104 degree elbow, have a significant dipole moment; so it is very IR active.

    Well although I don’t understand much of the detail of this I do grasp the general concepts.

    So now what about “black body” or thermal radiation; EM radiation fields that have a continuous frequency spectrum, with a very defined spectral distribution that depends only on the absolute Temperature of the material, and doesn’t depend at all on the nature of the material. (remember that true black body radiation is a theoretical concept that doesn’t actually exist anywhere, but which is closely approximated by some real systems)

    There is nothing in the theory of BB radiation which contains the Physical properties such as bond strengths and the like of any actual material. Its origin is considered to be due to the acceleration of electric charges, as explained in Maxwell’s equations.

    Well any isolated neutral atom or molecule getting along in outer space is subject basically to nothing but the gravitational force; which is far and away the weakest of all the Physical forces of nature; so it doesn’t do a heck of a lot for any isolated neutral atom or molecule; which continues on in a straight line at a constant velocity according to Newton’s Laws (or maybe Einstein’s)

    So Maxwell says such a free roaming particle does not radiate EM energy; well if it did it would apparently evaporate itself, since there is nothing around to resupply it with energy.

    But now, if we put a whole lot of similar neutral particles close to each other, and bestow on them a Temperature which is not zero K , then we expect these particles to have on average an energy kT per degree of freedom of which there are at least three in ordinary three d space.
    In such an assemblage of particles, collisions occur between particles, so the amount of energy of any particle constantly changes in a completely unknown, and unknowable fashion; although it has statistical properties which can be prescribed from the concept of Temperature.

    Well now we have a system, which doesn’t have any defined structural state like a molecule or even an atom; but it does contain electric charges which must undergo acceleration during the interaction of two particles in a collision.

    According to Maxwell, during the time when that interaction is occurring, and acceleration of the charge is happening, the particles must radiate EM waves, which will result in lowering the average energy of the particles (if those photons escape), so the material will cool, unless some source of energy supply is present to resupply the particles with energy.

    I see no reason why two atoms (or molecules) in such a collision would not radiate according to Maxwell’s equations for electromagnetism.

    Now in the case of a gaseous material, the density of the particles is quite low. Avogadro’s number of atoms or molecules, in 22.4 litres of the gas at STP.

    Now because of the gaseous phase of the material the particle density is low compared to what it is in the liquid or solid phase of the same material.

    Consequently the number of particles radiating from some very thin layer (perhaps a mono-molecular layer) will be quite small, so the strength of the radiation will be likewise small. This presumably means the thin layer sample will have a very low value of emissivity; not unlike that of a single atomic layer in a gold film, would have a low emissivity.

    But I don’t see any reason why each of them won’t radiate a thermal continuum spectrum, that depends only on the Temperature. Of course establishing the Temperature of such a sparse sample would be rather difficult; well not to worry; I am sure Mother Gaia, knows exactly what the Temperature is, and she will see that the requisite amount of radiation is forthcoming.

    The above is to me a plausible explanation of how black body like radiation arises; regardless of what the nature of the source material is. I’ve never ever seen any formal physics denunciation of the very notion that gases simply refuse to radiate, because they are gaseous.

    At what point does an H2O molecule leaving the solid surface of a block of ice, or the slightly denser perhaps surface of a liquid sample, suddenly decide to stop radiating a thermal spectrum ? Simply doesn’t make any sense to me, but I can appreciate that the emissivity may very well reflect the lower particle density of the source material.

    So personally, I believe that ANY material that has a finite non-zero Temperature can and does radiate a continuum thermal spectrum dependent only on the Temperature (the spectrum that is).

    Yes the allowable average energies per degree of freedom, of the radiating particles apparently do have to be quantized; but nothing in the Planck derivation of the BB radiation law, is there any assumption of some energy level structure to the radiating system; such as occurs with either atomic or molecular line, and band spectra.

    Evidently in this belief, Phil and I apparently do not agree (I think). I’m still working on establishing a solid basis in my mind for that belief. I’m tending to believe that the aha explanation of the mystery lies in the fact that the negative electric charge is associated with a light particle of mass about 511 keV; while the positive charge is associated with the proton mass of about 938 MeV, a ratio of about 1836.
    While those “charge clouds”, maybe concentric in an isolated particle (between collisions), I suspect that they separate during collision so there is a definite electric dipole moment for the duration of the collision interraction.

    And I have to say, that my quantum mechanics, is not yet up to snuff, to explain the details of how this works; but I am increasingly confident that it does work, and that the collision dynamics is what makes it all possible.

    If I have this all totally screwed up, i suppose Phil or one of the many resident PhD Physicists who frequent WUWT, can straighten me out. I may have to throw myself on the mercy of Prof Will Happer at Princeton to get some steerage on this. Meanwhile the books will have to do; but If I figure it all out I will let you know.

    There is one aspect of this that is perhaps not so difficult to understand. The molecular band spectra (say of CO2) consist of very many pretty much equally spaced (in frequency) narrow lines. In practice, these lines are “temperature broadened” due to the Doppler effect, and the velocity distribution of the molecules, which of course has a Maxwell-Boltzmann distribution function. They also are “pressure broadened” as a consequence of the local gas pressure; which can be explained as a simple consequence of the mean time between collisions. An excited (CO2) molecule may have a lifetime (in isolation) that can be many milliseconds; but the collisions in the lower atmosphere occur many thousands of times faster than that, so the int5rinsic lifetime gets catastrophically shortened due to a collision which promptly terminates the excited state, and thermalizes the energy. As a result the line width increases per the Heisenberg dE.dt > h/2pi relationship, when the lifetime is cut short. The question is under what atmospheric conditions do thes broadened fine structured line spectra merge into a continuous blocking band. I don’t know (yet) but I’m sure it is known so I’m not worried.

    So what about my collision induced dipole moments for neutral atom radiations. Well the interaction time of two atoms or molecules in collision, is very much faster by many orders of magnitude, than the mean time between molecular collisions that result in pressure broadening of molecular absorption lines; and that means that the uncertainties in the emitted photon energies become extremely large.

    As a result the discrete quantized energies per degree of freedom, that were required to abolish the Ultra-Violet catastrophe, and yield the Planck function, result in extreme Heisenberg broadening of the intrinsic line width; which is why those merge completely into a continuous spectrum; even though the particle energy states are quantized.

    So I don’t have any problem at all, accepting a completely continuous thermal spectrum of emission resulting from a system of quantized equipartition energies of colliding molecules. That to me is as plain as daylight. The specific details of all of that are as yet not plain as daylight to me at all; but I am slowly getting the hang of it.

  348. George E. Smith says:

    “”””” Domenic says:
    March 3, 2011 at 3:41 pm
    to George E. Smith
    You wrote: “Reed, in the “Cavity” case you cite, the radiation inside the cavity, that is presumed to have black walls (non thermally conducting), is of course isotropic. There is no preferred direction inside the cavity.”

    The walls do not have to be ‘black’.

    Cavity radiation is an interesting phenomenon. In a completely enclosed isothermal cavity, it makes no difference what material the interior walls are composed of. The internal radiation in the cavity will conform precisely to blackbody radiation at the temperature the material is at.

    Even if the walls were composed of N2, it will emit blackbody radiation. Thus N2 can indeed emit all wavelengths under certain conditions. And if you inject any photons into this hypothetical N2 constructed radiation cavity, of any wavelength, it will all be absorbed by the N2 molecules within. 100% absorption regardless of wavelength.

    How a gas, atom, molecule, etc responds to thermal radiation, or emits thermal radiation, is dependent on the conditions of the environment it is placed within. ” ”

    Well if the walls consisted of N2, which ordinarily would not absorb much IR, it seems to me, that the cavity radiation would simply pass through the N2 “wall”, until it encounters some other sort of “wall”.

    If the wall, is not “black” in the totally absorbing sense, it would seem to me that it is in fact not “the wall”, and the radiation would propagate beyond it until it does find a “black” wall. I tend to agree that in the final equilibrium isothermal condition that any and all materials wihin the cavity “walls” will be at the same Temperature.

    But Hey, I am a novice at this stuff, and I am always eager to learn from the experts; sometimes it takes a little while.

    George

  349. davidmhoffer says:

    Oliver Ramsey;
    That the radiated energy from cold ice can heat a warmer object to a higher temperature than it was before…. now that’s problematic.
    If you are convinced that that does occur I would be genuinely interested to see an explanation with some quantification>>

    How about…read the original article and then work your way down the comments?

    Or go straight to why we’re splitting hairs and it doesn’t matter:

    http://knowledgedrift.wordpress.com/2011/03/03/why-the-co2-greenhouse-gas-debate-doesnt-matter/

  350. George E. Smith says:

    “”””” kuhnkat says:
    March 3, 2011 at 3:17 pm
    George E. Smith,

    ” they have no recollection of their origin, nor any knowledge about where they are going.”

    Please read the details of the 2 hole experiments as it relates to photons and electrons in quantum mechanics. Your statement is not correct. “””””

    Well off hand, I don’t quite see the connection; BUT, I’ll take your word for it; and see if I can understand what you are saying. I’m here to learn.

  351. 1DandyTroll says:

    Since the greenhouse, in general and for what ever purpose, is to create the optimal condition for what ever you wish to grow. One could say that with a greenhouse you want to retain the optimal summer conditions.

    High humidity, high water levels and through put, high proper nutrient levels, higher concentration of nitrogen and and much higher levels of CO2.

    Since most people sporting greenhouses aren’t growing pine trees heat actually have to be added sun and all. Depending on what greenery you grow heat is used to not place the plants in hibernation mode, some seeds wont even sprout if they’re not warm enough. Consider that most of land is in the northern most hemisphere and away from large bodies of warm water, so you need to add heat, and lots of it, by gas, coal, or nuclear, because god ol’ sodding Sol only does so much for as long as it’s rays reaches the ground, and H2O and CO2 what do they do but dissipate their retained “heat” the moment after they got it, until everything equalize with it’s surrounding, which happens before midnight if you don’t add heat.

    On the one hand you can only grow bananas north of Anchorage as it is, and only in the boiler room so to speak, if you add heat but who’s so dumb enough to say it would be bad to be able to actually grow pineapples, coconuts and rice even north of Anchorage as well, preferably without adding expensive wind powered derived heat.

    Point is, without the added heat the inside of a greenhouse gets to be the same temperature as the outside no matter what the levels and concentrations of humidity and CO2 you have inside. The only thing that makes a greenhouse hotter ‘an the outside in blazing sunlight is the enclosure, or rather the roof because if you open the roof but still keep your denser atmosphere heats go out to space anyhow.

  352. Jim Masterson says:

    >>
    kuhnkat says:
    March 3, 2011 at 3:17 pm

    George E. Smith,
    ” they have no recollection of their origin, nor any knowledge about where they are going.”

    Please read the details of the 2 hole experiments as it relates to photons and electrons in quantum mechanics. Your statement is not correct.
    <<

    If you mean the two-slit experiment where the particle-wave duality properties are tested, then you don’t understand what the experiment is doing. George is correct (as usual).

    Jim

  353. davidmhoffer says:
    March 3, 2011 at 10:27 am
    Ira,
    Are you and I missing a business opportunity here? There was a suggestion that you and I need to get some new physics text books. …

    Yup, there is a need for new physics textbooks that are customized to any bias. We could call our books New Clear Physics. We could have a textbook version with pre-canned exams that would ask the exact same questions each year, but change the answers.

    The lawyers tell about the town that had only one lawyer, and he was starving for business. Then another lawyer moved into town and they are both doing just fine with plenty of business for all. With different versions of our textbooks there would be conflict galore. We could each get a government grant to study the textbooks and conclude which wwer true and which false. Of course, our textbooks would be mostly true, but written with studied ambiguity. Each of our govenment-funded studies would come to different conclusions, so we’d get hired by industry and environmentalist think tanks to write reports to resolve the differences.

    The problem with the above paragraph is that it is a not too exaggerated account of how government-supported science is actually done now. (/sarc)

  354. Tim Folkerts says:
    March 3, 2011 at 12:45 pm
    … Water is spray downward from the shower head into the tub (like the sun’s photons “spraying” down onto the earth). …

    Now that I think about it, this is almost exactly like the animation Ira had way back in his blog entry – with streams of photons spraying around and getting absorbed and re-emitted! …

    Thanks Tim Folkerts for your similar physical analogy. For WUWT readers who might have missed my first entry in this series (with the bouncing balls) see here. More coming over the next couple weeks. Stay tuned.

  355. Domenic says:

    to Tim Folkerts

    Yes, of course, cavity radiation assumes no transmissivity through the walls.

    The N2 thickness required is not known.

    Now, with regards to N2 (and O2), the actual absorption properties must be tested at many, many wavelengths, and under the same conditions that N2 is found in the atmosphere. In the atmosphere, N2 is in temperature states from approx 300K to near 3K (from the surface of the earth to near space at night), and at the associated pressures depending on altitude. In addition, the source radiation needed to drive the signal through the sample must vary from the high temp solar equivalent to a very low temp 3K equivalent.

    Those conditions are not easy to duplicate in a laboratory. So, N2 has never really been tested through the full range of its enviromental conditions for all wavelengths of absorption. Most of the transmission/absorption data for the atmosphere comes from military research. But the military just wanted the ‘atmospheric windows’ to see clearly through, narrow wavebands are fine, and could care less about the rest of the spectrum and conditions.

    There was never any good reason to gather the actual real life N2 and O2 absorption data.

    Perhaps now there is. With all the whining of their ‘certainty’ by the AGW crowd, it may be necessary.

    I can almost guarantee you there will be surprises when, or if, the actual data is taken. I’ve seen it happen so many times in my professional experience.

    I spent a number of years overseeing design, manufacture, calibration, application and sales of IR sensors in the US, Europe and Asia.

  356. davidmhoffer says:

    Ira,
    I read your thoughts about the town with only one lawyer. In the town I grew up in there was a saying..when two farmers have a fight over a cow, two lawyers wind up with a side of beef each.

    It just hit me how similar that is to climate scientists. Of course there seems to be one subtle difference. They wind up with a side of beef each, but the farmers didn’t even know there was a fight.

  357. Reed Coray says:

    George E. Smith says:
    March 3, 2011 at 12:07 pm

    Thank you for your response. It may or may not be true, but I personally don’t question the statement that gases emit “blackbody-like” radiation–in part because “blackbody-like” is not a well defined term. What I have trouble with is applying Planck’s blackbody radiation law to matter that does not have a clearly defined surface area.

    My answer to your specific question:

    How about a graphene layer, that is a single atom thick solid; can it emit a black body like (carbon is black) thermal spectrum.

    And if not; why not ?

    I don’t know.

  358. Steve says:

    “Cavity radiation is an interesting phenomenon. In a completely enclosed isothermal cavity, it makes no difference what material the interior walls are composed of. The internal radiation in the cavity will conform precisely to blackbody radiation at the temperature the material is at….And if you inject any photons into this hypothetical N2 constructed radiation cavity, of any wavelength, it will all be absorbed by the N2 molecules within 100% absorption regardless of wavelength.”

    Someone has duped you regarding cavity radiation. If you make a box out of clear glass and shine a light into the box, those photons are still going to shine through the glass box just as if your were shining your light through a window. Glass doesn’t gain magical absorption properties because you shape the glass into a box.

  359. Domenic says:

    to Steve

    You didn’t read my posts correctly. Cavity radiation assumes no transmissivity.

    1. A thin layer of any of many substances on the outside of the glass can suffice, or

    2. The glass itself must be thick enough to stop transmission. Glass, like water, does eventually stop all transmission given enough thickness.

  360. Phil. says:

    Domenic says:
    March 3, 2011 at 7:15 pm
    to Tim Folkerts

    Yes, of course, cavity radiation assumes no transmissivity through the walls.

    The N2 thickness required is not known.

    Although at least thicker than the atmosphere?

    Now, with regards to N2 (and O2), the actual absorption properties must be tested at many, many wavelengths, and under the same conditions that N2 is found in the atmosphere. In the atmosphere, N2 is in temperature states from approx 300K to near 3K (from the surface of the earth to near space at night), and at the associated pressures depending on altitude.

    You may want to reconsider the 3K since the coldest part of the atmosphere is the mesopause at ~180K. Testing at many wavelengths are what spectrometers are made for, really easy.

    In addition, the source radiation needed to drive the signal through the sample must vary from the high temp solar equivalent to a very low temp 3K equivalent.

    No, this is irrelevant, the wavelength is the only parameter of interest, the temperature of the source isn’t ‘remembered’ by the photon!

    Those conditions are not easy to duplicate in a laboratory. So, N2 has never really been tested through the full range of its enviromental conditions for all wavelengths of absorption.

    Really, care to justify that assertion?

    Most of the transmission/absorption data for the atmosphere comes from military research. But the military just wanted the ‘atmospheric windows’ to see clearly through, narrow wavebands are fine, and could care less about the rest of the spectrum and conditions.

    There was never any good reason to gather the actual real life N2 and O2 absorption data.

    Oh you don’t think that it is the gas that usually fills the light paths in our spectrometers might be a factor? Or that N2 is frequently the diluent in the cell when measurements are made of other gases.

    The reason why we don’t “gather the actual real life N2 and O2 absorption data” is that there isn’t any to be gathered!

  361. Steve says:

    “You didn’t read my posts correctly. Cavity radiation assumes no transmissivity.”

    Considering your conclusion (that N2 can be forced to emit/absorb any wavelength if placed in a cavity), I’m thinking you didn’t read your post correctly.

    “Cavity radiation assumes no transmissivity. 1. A thin layer of any of many substances on the outside of the glass can suffice.”

    Ahhhh, so it isn’t the glass or N2 that is absorbing/emitting at new wavelengths, is it?

    “or 2. The glass itself must be thick enough to stop transmission. Glass, like water, does eventually stop all transmission given enough thickness.”

    Pure water? I hope you aren’t making this assumption because the ocean depths are dark. The ocean isn’t pure water – it is full of salts and organic debris. I’ll grant that given enough of anything, in the reality, you’ll run across enough impurities to stop all transmission (even an optic fiber). But then you really aren’t talking about the absorption/emission spectrum of your pure molecule, are you?

    The cavity doesn’t create quantum magic, it ensures that all reflected wavelengths are eventually absorbed. The molecules of the cavity walls don’t gain an ability to absorb wavelengths that they never could before. With each reflection, the photon loses a tiny amount of energy, becoming a slightly lower frequency, until it eventually does fall within a wavelength of the wall’s absorption spectrum. But the photon must begin as a wavelength that interacts with that wall’s molecules (reflection or absorption). Hence, no transmission is allowed.

    Saying,”If you can make a box of N2 that doesn’t transmit light that it normally does, you can force N2 to absorb light that it normally doesn’t.” is a tautology. Restating the case with different words doesn’t make it possible.

  362. Domenic says:

    to Steve

    Ahh…finally, a thinker here.

    Just as perfect blackbody does not exist, nor does a perfect reflector. That’s why I posted earlier that the opposite of a blackbody is a perfect reflector, as I saw a poster struggling with the idea of a blackbody.

    Thermal radiation comes into the atmosphere from the sun, is converted to different wavelengths, and leaves it again. N2 and O2 are indeed part of that process, shifting wavelengths through energy lost due to bouncing around. Kinetic energy for the molecules and wavelength shift for the photons.

    Because there are no perfect reflectors.

    There are no perfect reflectors.

    (Also, there no perfect transmission mediums other than a vacuum.)

    My whole point was that N2 and O2 are indeed part of the ‘greenhouse gases’.

    And there are vast quantities of N2 and O2 in the atmosphere. Especially compared to CO2.

    And yet, they are ignored.

    No one accounts for them.

    Phil doesn’t understand that. The AGW crowd doesn’t understand that.

    I am happy to see questioners and thinkers like you here.

  363. Take a quantity of pure N2 and place it into a sealed metal container. Heat the container to some high temperature. After a period of time, the N2 must assume a temperature approaching that of the heated container.

    Now, take a short hose and connect it between the container of hot N2 and and a much larger container of pure N2 that is at room temperature and pressure. As the hot N2 puffs into the room temperature container, take a video using an IR camera. Do you see the plume of hot N2? I do, which means the N2 is radiating IR energy. The container has nothing but N2 in it, so the radiation must be coming from N2.

    Based on previous comments and links in this thread, I think much of the radiation from the plume will be in the shorter IR bands, perhaps around 3-4μ, but I would not be surprised if some of it spreads out into the longer IR bands, up to 7μ to 10μ to 15μ or longer as the hot N2 mixes with the room temperature N2 .

    Anyone here expert enough to know for sure?

  364. Domenic says:

    Ira

    I need a complete description of the experiment.

    1. What kind of container was the second one?

    2. Was the hose evacuated of all gases except N2?

    3. Was the IR camera inside the second container or loooking through some sort of window?

    4. What was the spectral range of the camera?

  365. George E. Smith says:

    “”””” Phil. says:
    March 4, 2011 at 7:40 am
    Domenic says:
    March 3, 2011 at 7:15 pm
    to Tim Folkerts

    Yes, of course, cavity radiation assumes no transmissivity through the walls.

    The N2 thickness required is not known.

    Although at least thicker than the atmosphere?

    Most of the transmission/absorption data for the atmosphere comes from military research. But the military just wanted the ‘atmospheric windows’ to see clearly through, narrow wavebands are fine, and could care less about the rest of the spectrum and conditions.

    There was never any good reason to gather the actual real life N2 and O2 absorption data.

    Oh you don’t think that it is the gas that usually fills the light paths in our spectrometers might be a factor? Or that N2 is frequently the diluent in the cell when measurements are made of other gases.

    The reason why we don’t “gather the actual real life N2 and O2 absorption data” is that there isn’t any to be gathered! “””””

    Well I didn’t get any sleep all night thinking about this problem; maybe I’m having another CO2 Ice moment.

    I thought Phil and I were pretty much on the same page with this; but maybe not.

    I think we both agree that in the lower atmosphere (how about the two meter Owl box height) the generally 15 micron band radiation gets captured by CO2, but then is immediately thermalized through molecular collisions, long before the excited CO2 gets a chance to spontaneously decay, in a presumably allowed transition. So the captured energy gets shared with the regular gas molecules N2, O2, Ar. I’m comfortable with that and I think Phil and I agree on that.
    But at this point we apparently seem to diverge.
    I assumed that heated atmosphere being above absolute zero, then emits a “black body like” thermal continuum spectrum based on the Temperature. By the way, no mystery to “BB like” I use that term to mean a continuous spectrum that generally conforms to the Planck Radiation formula, and is bound by the Planck formula as a limiting envelope, but may have deviations here and there due to emissivity variations. But it is distinctly different from a line or band spectrum, which I take as characteristic of isolated atoms or molecules respectively. So then my atmosphere is emitting a spectrum that carries no signature of the species that grabbed the energy in the first place; or it even could have been solar radiation caused or simply conduction from the ground or somewhere else.

    One of Phil’s recent comments indicated to me, that he believes, that the N2/O2/Ar do NOT radiate, as per my concept; BUT those gases are in thermal equilibrium with the trace CO2/H2O/O3 /whatever other GHG molecules that are IR active, so they match in Temperature, and then the IR active GHG molecules radiate according to whatever vibrational/rotational etc states those molecules have; so that the “Atmospheric LWIR radiation” then does carry the signature of the various GHG species, rather than being “BB like”. I think Phil’s position is somewhat along those lines. I know we both agree that in the rarified stratosphere where mean free paths are long, the GHG molecules can spontaneously radiate at the same frequencies they absorbed.

    But back to the denser atmosphere where that isn’t possible.

    Now Tom Vonk, some time back talked about this same situation in several posts, and he basically said the same thing. That the ordinary atmosphere gases were in local Thermal Equilibrium with the GHG (CO2), so neither one was heating the other; or perhaps they all are heating each other; and then in Tom’s view (as far as I recall), then it was the CO2 etc that was radiating whatever LWIR was coming from the atmosphere, not the N2/O2/Ar.

    Now Peter Humbug says almost exactly the same thing in his “Physics Today” Featured Article. Now I assume, everyone knows that is my whimsical nickname for Raymond T. Pierrehumbert; the Louis Block Professor of Geophysical Sciences at the University of Chicago. And let me say here categorically, that is in no way derogatory; I’m not into name calling; it helps me remember his name, and it fits.
    Well if Foster Brooks or whoever likes to call himself Tamino, the sappy hero of Mozart’s Magic Flute, that’s ok with me; or Eli Rabbett for example. So Prof RTPH, is Peter Humbug, and with no malice.

    So he basically describes the same sort of scenario that Tom Vonk presented some time ago, and apparently is along the lines of what Phil has said here.

    Well among all the stacks of books, that crowd me out of my cube, there’s not a single one that describes the specific origins of the black body radiation; nor is there one that states categorically that it does not apply to gases. Those that mention the subject simply say EVERY body that is above absolute zero, emits black body (like) thermal radiation, that is characteristic of the Temperature, and may be modified by a spectral emissivity.

    Unfortunately most of those books are “handbooks”, not Physics Textbooks. I lost every single textbook I ever had or used in school 50 years ago, in a box that got shipped, and never delivered. I know we actually did the derivation of the Planck Formula and the Raleigh Jeans Formula, and there never was any mention of any “energy level structure”, just that the (average) energy assigned to each degree of freedom had to be quantized (in the Planck case). And It was my impression (at the time) that the radiation mechanism was simply the acceleration of electric charge; but I can’t say that at the time, I understood that to be as a consequence of the molecular or atomic interactions during collisions. That concept came to me only quite recently; although once again it was a post by somebody here at WUWT (can’t remember who) that led me to start thinking along those lines. And I think that is a very plausible mechanism; because I can see how during a collision, the electron cloud, and the nucleus, can get displaced from each other (CG wise) and create a dipole moment for the duration of the interaction. And my Quantum mechanics stops well short of being able to calculate that all out.

    In any case, now I’m not sure what to believe, but it seems that Phil, and RTPH, and Tom Vonk all have a similar view, that is not what I had thought. The forgotten poster here, had said something to the effect that a thermal continuum was possible (in gases) due to perturbations of the quantum states during collision; well as best as I can recall.

    Now I really don’t know anybody who is really handy that I know for sure would know the answer to this. Two names come to mind; but both are more solid state Physicists; both are recipients of the National (Presidential) Science Medal. One of them in my view should be a Nobel Laureate in Physics, but isn’t (yet). I’m going to try and contact him, if I can and ask him. The other one works (occasionally) in the building next to me, so I might be able to get to him; he’s a former student, of that other person.

    In any case the atmospheric look up/look down spectra should solve the puzzle; as Phil says here; there ain’t any N2/O2 LWIR spectral data to be had. Even Mother Gaia evidently doesn’t have it.

    So I may be having crow for lunch again today; but I do get so rejuvenated whenever that happens.

    I still believe that H2O is firmly in control of the climate; not CO2, but I sure am curious about that gaseous BB radiation question.

    And I think we are being pushed well off onto a back burner here; too much stuff up above.

    [thanks G , I really do appreciate it]

  366. pochas says:

    Ira Glickstein, PhD says:
    March 4, 2011 at 9:48 am

    “I do, which means the N2 is radiating IR energy. The container has nothing but N2 in it, so the radiation must be coming from N2.”

    ??? Ira, N2 is not a greenhouse gas. It is transparent to both incoming shortwave and outgoing longwave radiation. It does not radiate IR energy at any wavelength relevant to greenhouse theory. In a closed oven it will of course come to the equilibrium temperature inside the oven, but this involves conduction and convection and not radiation as long as no other gases are present. Here is a more complete graphic than the one you used. Transmission spectra for all of the greenhouse gasses are shown. Notice that Nitrogen is not present, and if nitrogen is transparent to IR it is also non-radiating in the IR bands.

    http://upload.wikimedia.org/wikipedia/commons/7/7c/Atmospheric_Transmission.png

  367. Tim Folkerts says:

    Ira Glickstein, PhD says: March 4, 2011 at 9:48 am
    Do you see the plume of hot N2? I do, which means the N2 is radiating IR energy.

    That is an insightful thought experiment. Have you ACTUALLY seen the radiated IR from that plume? I would love to see an IR video (or even a still picture) of such a thing.

    I suspect that it is not at all easy to see because I am pretty well convinced that N2 cannot radiate any appreciable amount of IR in the 5-50 um range associated with the applicable temperatures.

    There are a couple things you might see.
    1) If there was a small amount of H2O or CO2, then those molecules could radiate some IR. Using pure enough N2 or filtering for specific frequencies NOT related to CO2 & H2O should eliminate those possibilities
    2) The change in temperature will change the index of refraction, which could lead to a wavering effect (like you see over hot roads of hot roofs). IR light shining thru from the far wall could get distorted, which would create some sort of blurry features on the IR detector.

  368. Ken Coffman says:

    The Fire Hose Analogy

    The second law of thermodynamics refers to the NET transfer of heat energy that must be from warmer to cooler bodies.
    But, as was noted earlier in this thread, if you are spraying me with a fire hose and I am simultaneously spraying you with a garden hose that has blue-tinted water, I will get a lot wetter, but you will get some blue water on your clothes.
    – Dr. Ira Glickstein
    http://wattsupwiththat.com/2011/02/28/visualizing-the-greenhouse-effect-atmospheric-windows/

    I hate to rank on Dr. Glickstein; he seems like a smart and fair-mined man, but this kind of argument is used to justify the human-caused global warming theme. The idea is: outgoing infrared radiation stimulates atmospheric CO2 which heats up and “back radiates” to the earth…making us hotter. Honestly, at this point, does anyone believe this is physically possible? This “greenhouse effect” and analogy is complete nonsense.

    What can I do to help Dr. Glickstein? Let’s fix his analogy for him. There is no side channel or alternate path for the fire hose and the blue-water squirt gun. They must use the same path. So, I get the fire hose, you get the squirt gun. Open fire. How much of your blue water will travel into the nozzle of my fire hose? That’s right my friends. None!

    However, this is still not right because your finger working vigorously on the squirt gun’s trigger is an independent source of energy. This is not analogous to the human-caused global warming theory. CO2 does not have an independent source of energy. The only energy it has comes from other places…like collisions with the other atoms and molecules of the atmosphere, from infrared radiation and conduction from the earth’s surface. CO2 is not a battery, it is not being combusted and it is not a nuclear reactor.

    Let’s refine the analogy further. Instead of a squirt gun, you get a funnel with a blue die injector. A really teenie-tiny, small one to represent 390PPM of CO2 in the atmosphere. Bend the spout of the funnel back toward me…now you capture water from my fire hose and direct blue stuff back toward my nozzle.

    How much of your redirected water travels into my nozzle and up my fire hose?

    There’s a cardinal rule in the selling business…once the sale is made, shut the hell up.

    This is me shutting up now.

    Upgrade your paradigm. Buy the Dragon Slayer book. Thanks.

  369. Ken Coffman says:

    There is some excellent conversation going here.
    Would someone comment on this please? As I understand it, for N, N2, O, O2 and Ar…if there is mass and a temperature, this stuff will radiate, but the wavelength is really, really tiny, hard to measure, and we don’t know much about it because it is not really useful to us.
    Right? Not right?

  370. Don V says:

    Ira, I thank you for graciously for replying to my post. I know this is a late response, but I didn’t realize that you had replied.

    First, let me respond that I am not saying the “green house” effect, or delayed radiative loss, or whatever you want to call it, doesn’t occur to some extent. I agree. It most certainly does. Whenever the atmosphere above the surface, especially clouds, have accumulated energy during a hot sunny day, portions of that energy are prevented from direct and rapid release of it to the cold of outer space by back radiation. The atmoshpere does have a certain heat retaining “capacity” which is entirely dependent on radiative loss to outer space. However, by far the greatest contributor to this effect is water vapor, not CO2. Water has a MUCH higher energy retaining capacity than CO2 and serves as the largest energy “sponge” in the atmosphere.

    What I objected to was what appeared to be incorrect portrayal of this process in your animation. In the dark blue box all the way to the right you illustrate “15μ aborption” and you claim “7μ, 10μ, and 15μ” re-emission as a direct result of this “15μ absorption”. I agree that it is possible that water vapor might provide 7μ, and possibly 10μ, back radiation as a result of the numerous SHORTER wavelength NIR bands in which it receives energy. But that is not what the dark blue box is claiming. It strongly suggests 7μ re-emission that comes directly from 15μ illumination. How can that be? Your graphic is at the very least misleading. I know of only a few materials capable of down converting light in this manner, and they all have to be charged with much more energetic light to be able to “phosphoresce” in this manner (see: http://www.roithner-laser.at/cards.html ), and neither CO2, nor water vapor, nor CH4 are capable of this kind of behavior. Any energy that is taken up by these molecules at 15μ can ONLY re-emit at LONGER wavelengths – not shorter, higher energy wavelengths. If you are claiming that they are capable of re-emitting at a shorter wavelength after “excitation” at 15μ, I believe you are mistaken.

    There are only two substantial absorption bands that enable CO2 participates in this absorption/re-emission activity, and they are quite narrow when compared to all of the many NIR and IR bands that water vapor absorbs at. Every illustration I have seen of the combined absorption spectrum associates the shoulder at 15μ as being CO2’s main contribution. However the extremely low concentration of CO2 in the atmosphere almost guarantees that nearly all of the energy absorbed by CO2 at either of it’s absorbtion bands is almost immediately given up to one of the other atmospheric gases through thermal collision. The net effect is that very little, if any, IR radiation is “re-emitted” at 15μ. If this were not the case, and if your graphic was correct, then think about it, the atmosphere would appear to be somewhat “transparent” in the 15μ band, you would see energy in the 15μ appearing in the top graphic . . . . but your graphic doesn’t show any 15μ IR at all, coming “through” to outer space. Are you saying that the 15μ IR only radiates back towards the warmer surface? I don’t think so. Your analogy of a firehose and a garden hose doesn’t hold water either. The source of blue water coming from the garden hose must originate from the fire hose. Since ALL of the water leaving the fire hose is either consumed (ie. absorbed at 15μ) in the fire or goes right thru the house (doesn’t even see CO2 at all or leaves at 10μ), there is NO water left to dye blue and squirt back at a much lower pressure at me.

    Other’s have claimed that increasing CO2 has narrowed the 10μ window that allows the IR radiation caused by all of the shorter wavelength aborption bands of water and the one CO2 band to radiate to outer space. But this concept doesn’t hold validity either since the spectra from outer space showing the width of the IR band at this window has not shown any narrowing with increasing CO2 concentration! Every IR spectrum looking down looks the same. This suggests that the current concentration of CO2 in the atmosphere is high enough, thank you, to completely absorb all of the IR at 15μ, (and promptly give up that energy in thermal collisions) without re-emitting any of it, AND any change in concentration of CO2 will ONLY drop the altitude at which complete loss of IR radiation at 15μ occurs!

    Maybe the new post by Anthony reviewing the new paper by Hermann Harde titled “How much CO2 really contributes to global warming? Spectroscopic
    studies and modelling of the influence of H2O, CO2 and CH4 on our
    climate,” will clear up some of the confusion on this subject. It predicts the sensitivity of the climate to changes in CO2 concentration at 1/7 what the IPCC claims.

    What you think? I welcome any additional insights.

  371. Tim Folkerts says:

    Domenic says: March 4, 2011 at 9:16 am
    “My whole point was that N2 and O2 are indeed part of the ‘greenhouse gases’. ”

    The quantum theory of molecular vibrations says there should be no (significant) contributions from diatomic N2 in the IR range. The same theory says there should be contributions from triatomic H2O, CO2 and O3.
    The experimental data of downward IR presented earlier in the thread show clear contributions of H2O, CO2, and O3. In the regions where these three do not emit IR, there is practically no IR.

    Theory and experiment both agree that N2 (or other monatomic or diatomic gases) is not a significant GHG. What about the the theory or data do you disagree with? What theory or data can you present that suggest N2 IS a significant GHG?

    Now, N2 DOES play a role. The true GHGs will collide with the N2 and transfer energy to/from the N2 molecules. N2 acts as a “reservoir” of thermal energy simply by the N2 molecules gaining/losing speed when they collide with GHGs. In this sense they are “part” of the GHG since N2 surrounds and interacts with the GHGs.

    Unless you can refute clearly either the theory or the evidence presented so far, you have not convinced me that they are themselves GHGs that absorb/emit IR.

  372. Phil. says:

    George E. Smith says:
    March 4, 2011 at 10:19 am
    Those that mention the subject simply say EVERY body that is above absolute zero, emits black body (like) thermal radiation, that is characteristic of the Temperature, and may be modified by a spectral emissivity

    Hi George, I’ve emphasized what may be the key to your understanding, for N2 the spectral emissivity is zero for the wavelength range of interest. That’s why they showed the dotted lines of BB at different T on the downwelling plot, the BB sets the limit which is reached in spectral regions where emissivity is 1. So you can tell what temperature the emitting gas was at. Regarding the origins of BB radiation as far as I’m aware it requires a dense medium where the molecules (atoms) are in close proximity whereas in a gas they are acting as individuals. I’d tend to think of it as a very dense gas, in that case I’d expect extreme broadening so that all the lines would completely merge and ultimately you could end up with BB.
    Your point about collision induced effects is good, there are continua associated with such effects but they are extremely weak, there are also dimer effects (sticky collisions and such like). The proof of the pudding is in the eating and when we look at the measured atmospheric spectra there is nothing there from N2 or O2!

  373. Phil. says:

    Ken Coffman says:
    March 4, 2011 at 10:37 am
    There is some excellent conversation going here.
    Would someone comment on this please? As I understand it, for N, N2, O, O2 and Ar…if there is mass and a temperature, this stuff will radiate, but the wavelength is really, really tiny, hard to measure, and we don’t know much about it because it is not really useful to us.
    Right? Not right?

    Not right.

  374. wayne says:

    Don V says:
    “How can that be? Your graphic is at the very least misleading. I know of only a few materials capable of down converting light in this manner, and they all have to be charged with much more energetic light to be able to “phosphoresce” in this manner (see: http://www.roithner-laser.at/cards.html ), and neither CO2, nor water vapor, nor CH4 are capable of this kind of behavior. Any energy that is taken up by these molecules at 15μ can ONLY re-emit at LONGER wavelengths – not shorter, higher energy wavelengths. If you are claiming that they are capable of re-emitting at a shorter wavelength after “excitation” at 15μ, I believe you are mistaken.”
    — — —
    Don, if I may make a comment here. I like most of what you are saying, you seem to see it clearly but one area. I don’t interpret Ira ever implying electron level actions, only thermal ro-vibration interaction. Any upward movement to higher frequencies could only be via the tails of the Maxwell-Boltzmann distribution.

    IR active molecules mostly thermalize instead of emitting but there is always a certain fraction doing so. Once thermalized all of that energy just gets evenly distributed maintaining the equipartition principle between all of the degrees of freedom. But there are always some higher velocity molecules present also and this is where the re-excitation at higher frequencies occurs by the statistical chance and this all depends on the temperature. This gets deep in statistical mechanics and I could easily be wrong but that is the way I see it. But of course not phosphoresce, that is at a much higher temperature.
    ref: http://wattsupwiththat.files.wordpress.com/2010/08/vonk_kinetic_curve.jpg
    explained here: http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/

  375. Tim Folkerts says:

    @ Don V, March 4, 2011 at 10:38 am

    1) The heat capacities of H2O vs CO2 is not really an issue — the H2O and CO2 do not hold on to the energy from the IR absorption. Either of these GHGs will gain energy if they absorb IR from the ground (or from GHGs in the atmosphere), but then when they collide with other molecules (almost always N2 or O2) they will give away that extra energy until they are at the same average energy as the rest of the air. This would slowly warm the entire atmosphere.

    The GHG can also emit IR, which would remove energy from that molecule. These molecules would (on average again) take energy from the next molecule they hit (again most likely N2 or O2).

    The “heat retaining ability” of the atmosphere would be from the combined heat capacities of all the gases. Since N2 & O2 are by far the most common, they will contribute the bulk of the heat capacity of the atmosphere.

    The fact that there much more H2O than CO2 will allow the H2O to more effectively heat or cool the atmosphere as a whole. But this would be related primarily to their IR properties, not their heat capacity properties.

    2) I second what wayne said about the energy for the 7 um photons coming from the MB distribution of molecules in the atmosphere, not from some sort of 2 photon interaction of 15 um photons within a single CO2 molecule.

  376. George E. Smith says:

    “”””” Phil. says:
    March 4, 2011 at 10:57 am
    George E. Smith says:
    March 4, 2011 at 10:19 am
    Those that mention the subject simply say EVERY body that is above absolute zero, emits black body (like) thermal radiation, that is characteristic of the Temperature, and may be modified by a spectral emissivity “””””

    Thanks Phil. Yes you caught the essence of my thinking on BB emission from gases. And yes my assumption is that because a gas is rather rarified compared to solids, that any thin layer of gas, would necessarily have a spectral emissivity much less than one, reflecting the small number of atoms or molecules that are present in such a thin layer; which is why I asked the question about the graphene; does a thin enough solid layer also have a poor emissivity ?

    Now Ray PH in his article did say that this aspect of coninua emission from gases was not well understood (so I guess I’m in some really great company), but he talked about the colliding molecules “possibly” forming some, I suppose you could call it a super molecule that had its own energy levels for the duration of the collision interraction. Maybe so, but maybe not necessary, because it seems to me that any single neutral molecule or atom in collision, is going to undergo a distortion of the geometry of the nucleus, and the electron charges, and that is enough to create a radiating dipole moment; which is why I thought that was a plausible notion, and one inspired by someone else who had hinted at it. The very short collision interraction time, would certainly explain why a process with quantized energy values can be Heisenberg smeared into an effectively continuous spectrum; the grand daddy of line broadenings.

    It’s amazing how kmuch stuff we accept, thinking we understand it, and then find out we don’t understand much of anything.

  377. wayne says:

    Much of the misunderstanding I read here, dave you might perk up and find this interesting, has to do with the lack of understanding of what LTE (local thermodynamic equilibrium) is. It is NOT TE (thermodynamic equilibrium) you read so often in thermodynamics texts.

    The simplest, but true, visualization of LTE is that it only exists when there is hugely more collisions between atoms or molecules than there is radiation interactions. Above about 45+ km the atmosphere becomes radiation ruled and there it is in NLTE (non-LTE). To be in LTE the local parcel does not imply radiation equilibrium or even perfect TE.

    In the ionosphere at 1,000+K above 100km you would think you would be very warm but if floating in a spacesuit there you would freeze to death. Why? This is NLTE. Temperatures can only be defined and exist if that point being measured is in LTE. It would be better to just list the mean velocities of the particles when above some 45 km for there is really no temperature at all, it can not even be defined there.

    Wolfram’s definition of LTE is way wrong, totally. They are speaking only of radiation and in true LTE, radiation has a minor back seat. Wonder who wrote that.

    Don’t rely on Wikipedia to straighten you out either, they make it somewhat messy, but one statement made there is correct and simple:

    “It is important to note that this local equilibrium may apply only to a certain subset of particles in the system. For example, LTE is usually applied only to massive particles. In a radiating gas, the photons being emitted and absorbed by the gas need not be in thermodynamic equilibrium with each other or with the massive particles of the gas in order for LTE to exist.”

    and correctly on what creates the LTE in the first place,

    “… while local thermodynamic equilibrium (LTE) means that those intensive parameters are varying in space and time, but are varying so slowly that for any point, one can assume thermodynamic equilibrium in some neighborhood about that point.”

    and

    “The process that leads to a thermodynamic equilibrium…” (think local here in LTE’s case) ”…is called thermalization. … By interacting, they will share energy/momentum among themselves and reach a state where the global statistics are unchanging in time” albeit slowly.

    If radiation is flowing into, out of and across such of a parcel in LTE it *does* require that the flow to be smooth everywhere locally and constant or changing very slowly. You cannot have a laser pumping energy into such parcel and call it LTE even if it is dense for it is not smooth across all particles in that local volume. A small parcel up in the atmosphere being warmed by the Earth below or sun above is smooth and acceptably constant so LTE is maintained.

    That might help some reading here to keep their views correct in physic definitions when thermalization is brought into the conversation and radiation. If ever in doubt of my accuracy, just read.

  378. Steve says:

    Ira Glickstein, PhD,”Take a quantity of pure N2 and place it into a sealed metal container. Heat the container to some high temperature…As the hot N2 puffs into the room temperature container, take a video using an IR camera. Do you see the plume of hot N2? I do, which means the N2 is radiating IR energy…Anyone here expert enough to know for sure?”

    You don’t need an expert, Ira. Think about what you are proposing. What percentage N2 is the atmosphere between the aperture of your IR camera and the object you are taking a photo of (even if it is just a warmer patch of N2)?

    If N2 absorbed/emitted in the IR spectrum, every IR image taken on earth would look like a photograph taken in a thick fog. There’d be no way for the IR to get from object to aperture without hitting a significant number of N2 molecules. Since this doesn’t happen, we can take crisp IR photos.

  379. Domenic says:

    Well, I was waiting for Steve to return, but I’ll continue anyway.

    Regarding cavity radiation, Steve gave a good answer to what happens to an incoming photon.

    But that’s only part of it.

    He said nothing about outgoing radiation.

    Take our hypothetical N2 cavity again.

    Now put a tiny hole in it and put a detector there to look inside and measure the emittance.

    What will that detector see, what wavelengths, for a given temperature of the N2 inside that cavity?

    Will the N2 do what ‘Phil’ thinks it will do by reference to some textbook or some spectral source book, or other?

    (Hint: I’ve experimented with cavity radiation for over 25 years. So, if anyone wishes to bet on this stuff, put their money where there mouth is, I will be more than happy to do so.)

  380. Domenic says:
    March 4, 2011 at 10:11 am
    Ira

    I need a complete description of the experiment.

    1. What kind of container was the second one?

    Good questions Domenic. The second container is metal, at room temperature. If you want a different kind of second container, let just say so and it will be provided at no etra charge,

    2. Was the hose evacuated of all gases except N2?

    Yes, prior to connecting, the hose was flushed with pure N2 at room temperature.

    3. Was the IR camera inside the second container or loooking through some sort of window?

    The IR camera was at room temperature and is looking through a small hole in the second container. A slight positive pressure of pure N2 is maintained such that no outside air gets into the second container through that hole. Alternatively, there is rock salt window or some more modern material that is totally transparent to the entire IR range. your choice.

    4. What was the spectral range of the camera?

    The camera is sensitive to the entire range from UV (0.1μ) through far IR (70μ), and the video is “color” coded such that the viewer can determine which band or bands the plume occupies. (E.g., the UV range is shown in purples, the visual in blues, near-IR in greens, mid-IR in oranges, and far-IR in reds. If desired, any given band or bands may be expanded to the full color range for display since the video camera records the image digitally, characterizing each pixel according to its wavelength.)

  381. wayne says:

    Domenic:

    Yes! You do see my point. That is one proper view of reality, and they are sometimes so very hard to get, and I’m no exception. There are so many obscuring variables and you have to correctly remove all *known but unimportant* variables to finally get down to see what is really happening.

    One of my simple pet peeve on this line is viewing the Earth as a flat disk by climatologists when calculating energies, temperatures and radiation fluxes. You might as well wad it all up and throw it away.

    Five temperatures across the globe averaged (272+285+271+301+297)/5 = 282.2K so radiation to space ignoring emissivities is 5.67E-08•282.2²² = 359.59 Wm-2 right?
    WRONG!
    The mean radiation from these five regions is actually:
    5.67E-08•(272²²+285²²+271²²+301²²+297²²)/5 = 379.37 Wm-2.
    The radiation from varying temperatures will ALWAYS be more than that of a constant temperature at the mean.

    So why do we average temperatures then assume radiation flows? You see it every day. It is climatologists doing this and it’s a crying shame. :-(
    They may be lying by omission but probably just plain ignorance, bad teachers and texts.

    Same for “back radiation”… well, I won’t go into that right now.

  382. George E. Smith says:

    “”””” Domenic says:
    March 4, 2011 at 3:12 pm
    Well, I was waiting for Steve to return, but I’ll continue anyway.

    Regarding cavity radiation, Steve gave a good answer to what happens to an incoming photon. “””””

    Domenic, presumably you want this cavity to be isothermal, so that it has a definite Temperature.

    How then would one know at the peephole, whether the emerging radiation (BB) was emitted by the N2, or by whatever formed the enegy impermeable solid wall of the cavity. I’m inclined to believe that Kirchoff’s law does apply to this hypothetical situation. but it is not clear to me, that one can say that the N2 or whatever gas filling you might put in there must radiate (and absorb) the exact same spectrum of the cavity walls.
    The citation I gave earlier, from the “Handbook of Optics ” (Walter Driscoll), the author of the section on BB radiation specifically said the cavity walls must be “Black” in the radiation total absorption sense. I don’t think that is inconsistent with some other material being in the cavity, and in thermal equilibrium with it, so long as that body radiates and absorbs whatever spectrum IT has identically (the strange material that is.
    Something absorbing some part of the cavity radiation, but simultaneously radiating the exact same things is basically invisible to the cavity; which presumably can go on radiating its BB spectrum regardless of some included foreign material.

    I presume that freezing Platinum or Copper, Laboratory black body cavities, are not evacuated, but are open to the atmosphere; well maybe you do have to use them in a vaccuum chamber; because you certainly don’t want any physical window in them.

  383. pochas says:
    March 4, 2011 at 10:21 am
    Ira Glickstein, PhD says:
    March 4, 2011 at 9:48 am

    “I do, which means the N2 is radiating IR energy. The container has nothing but N2 in it, so the radiation must be coming from N2.”

    ??? Ira, N2 is not a greenhouse gas. It is transparent to both incoming shortwave and outgoing longwave radiation. It does not radiate IR energy at any wavelength relevant to greenhouse theory. In a closed oven it will of course come to the equilibrium temperature inside the oven, but this involves conduction and convection and not radiation as long as no other gases are present. …

    Tim Folkerts says:
    March 4, 2011 at 10:28 am
    … That is an insightful thought experiment. Have you ACTUALLY seen the radiated IR from that plume? I would love to see an IR video (or even a still picture) of such a thing.

    I suspect that it is not at all easy to see because I am pretty well convinced that N2 cannot radiate any appreciable amount of IR in the 5-50 um range associated with the applicable temperatures. …

    Pochas and Tim Folkerts: Thanks for your comments, I did not actually do the experiment. However, here is my thought process:

    1) The hot N2 has lots more heat energy in it that the room-temp N2.
    2) Once in the presence of the room-temp N2 it will exchange heat with the room-temp N2 molecules by conduction (when a hot N2 collides with a room-temp N2) and also by radiation. I believe the camera will record the radiation part, and, possibly some radiation that results from the collision part.
    3) After all the hot N2 is released into the second container, and some time has passed, all the N2 in the second container will become about the same temperature, and eventually go down to room temperature.

    The only part that is in question is my step 2, where, I suppose, you could conclude that all the heat tranfer is via conduction, with absolutely no radiation and therefore a blank video recording.

    OK, if that is a problem, repeat the experiment in the vacuum of Space. Take a container of pure N2 and heat it up, then open the valve and let the N2 escape into Space. Do you see a plume with the IR camera now? I do.

    The hot N2, we all know, will cool down. It cannot release its heat energy to other molecules or atoms because it is in empty Space. So the choices area: a) to stay hot (in other words, since heat is something like the average velocity of atoms, the N2 could just go off in all directions at high velocity, until they collide with something), or b) to radiate.

    I vote for “b) radiate”, but does any WUWT reader know for sure?

  384. Steve says:

    Domenic says: “What will that detector see, what wavelengths, for a given temperature of the N2 inside that cavity?”

    So IF you could build a 0% transmissive cavity out of nothing but pure N2? I think this falls in the “If my aunt had balls she’d be my uncle” category.

    In a 0% transmissive cavity you will detect a near continuous band of wavelengths in a range commensurate to the temperature of the cavity. That does not mean that the molecules in the walls of the cavity have gained the ability to absorb/emit in a continuous band of wavelengths – there is reflection.

    By definition there is no wavelength within our test bandwidths that will end up transmitted through the walls, so every photon is either absorbed or reflected. What happens if a photon just outside the absorption spectrum strikes the wall? It won’t be transmitted, right? (If it is, your walls must be made of something like N2) So the photon is reflected. A minuscule fraction of the photon’s energy is lost. The photon is now an ever-so-slightly longer wavelength. Compared to it’s previous wavelength, practically a continuous bandwidth. The photon continuous and strikes another wall. If it still isn’t within the absorption spectrum of the walls, the process continues until it is.

    So within your 0% transmissive cavity you can indeed detect a practically continuous bandwidth of photons. This has nothing to do with the molecules of the walls gaining an ability to absorb/emit new bandwidths.

  385. Phil. says:

    Ira Glickstein, PhD says:
    March 4, 2011 at 3:56 pm

    The only part that is in question is my step 2, where, I suppose, you could conclude that all the heat tranfer is via conduction, with absolutely no radiation and therefore a blank video recording.

    Yes.

    OK, if that is a problem, repeat the experiment in the vacuum of Space. Take a container of pure N2 and heat it up, then open the valve and let the N2 escape into Space. Do you see a plume with the IR camera now? I do.

    The hot N2, we all know, will cool down. It cannot release its heat energy to other molecules or atoms because it is in empty Space. So the choices area: a) to stay hot (in other words, since heat is something like the average velocity of atoms, the N2 could just goo off in all directions at high velocity, until they collide with something), or b) to radiate.

    Generally fly around until they hit something, take a look at the Earth’s thermosphere there’s plenty of hot N2 up there doing exactly that. Possibly get some emissions due to forbidden transitions occurring because of the very long lifetimes of the states (such as Lyman-Birge-Hopfield).

  386. Don V says:
    March 4, 2011 at 10:38 am

    Good to hear from you again DonV. Have a look at the graphs that appear after my comment time stamped: March 1, 2011 at 7:19 am

    Notice that, “(a) 20 Km looking down” (from near the top of the atmosphere, approximately Space, down towards the Earth), there is quite a bit of ~15μ energy (which in the description of my graphic includes all far-IR from about 12μ up) emitted by the Atmosphere. Note that there is a distinct depression between 14μ and 16μ, but still a fair amount of energy there, but not as much as at around 13μ and 17μ, which approximate the blackbody curve.

    Now, have a look at the same wavelength range in “(b) Surface looking up” (from Earth towards the Atmosphere and Space). There is a maximum amount of energy around ~15μ, but HOLY COW! while the range between 14μ and 16μ is right up there around the blackbody curve, the surrounding wavelengths are depressed.

    OK, my conclusion. The Earth radiates 14μ through 16μ at full force, but, when that radiation emerges from near the top of the Atmosphere, some of the 14μ and 16μ energy has been lost. Why? Well, CO2 and H2O molecules absorbed it and re-emitted some at their characteristic wavelengths, but, they also transferrred some of it, via collision (conduction) to other molecules in the Atmosphere, all in random directions.

    The energized molecules re-emit the radiation at a variety of wavelengths, including in the ~7μ and ~10μ portions, which is why, looking again at the entire spectrum of the graph, there is proportionately more energy in the ~7μ and -especially- ~10μ regions. Where did that energy come from? Well, the only place I can think of is from the ~15μ portion where it is missing.

    If little Johnny has blueberry juice all over his mouth and chin, and there is a piece of freshly baked blueberry pie missing, that is pretty good circumstantial evidence of where it came from!

    I hope that explains what is going on in the dark blue box in my graphic, where ~15μ goes in to the Atmosphere at the bottom, and a variety of radiation, including ~7μ, ~10μ, and ~15μ, comes out at both the top and bottom. The deeper we delve into this thing, the more it is starting to look like my original (bouncing balls) graphic, is it not?

  387. Steve says:

    “OK, if that is a problem, repeat the experiment in the vacuum of Space. Take a container of pure N2 and heat it up, then open the valve and let the N2 escape into Space. Do you see a plume with the IR camera now? I do.”

    You won’t detect any IR bandwidths that don’t show up in the absorption/emission spectrum using an IR spectrometer. Molecules can cool by emitting microwaves, radio waves…

    The temperature of the object determines which bandwidths within it’s emission spectrum will be emitted upon cooling. It doesn’t give the molecule a new emission spectrum (unless your temperature change causes a state change, which gives you a different molecular structure with a different emission spectrum).

  388. Domenic says:

    Hi Ira,

    The second container should have a matte, or black painted finish on the inside surface. If you left it shiny stainless steel, the thermal radiation would simply reflect throughout the vessel and mask any image.

    The entry hose on top needs to be as far from the opposite surface as possible. It also needs to be far from the IR camera lens to prevent heating of the lens.

    There should also be a second hose from the bottom of the second container to the first container to allow for fairly equal pressure to prevent cooling of the heated N2 gas as it enters the second container. Expanding gases cool themselves. Let it enter from slight natural convection, or low pressure rather than via high pressure. The hose should be insulated.

    If you do that, with a wide band .5 to 50 micron or 70 micron window on a high quality IR camera with high sensitivity, I am pretty confident that you will see the N2 entering. (I mention the .5 to 50 micron window because it is fairly easy to find. It uses a K Br coating on silicon lens. Going out to 70 microns might be tough to find.) The image should look like water or a vapor plume entering the chamber from the first hose, with the image most clear right after exit from the hose.

    I believe the highest sensitivity IR cameras are still those with liquid nitrogen cooled detectors. They have been around for years, and are extremely sensitive. I haven’t checked in a while, though.

    IR cameras are used to check for gas leaks in industrial facilities. They see the air plumes due to either heating or cooling of the surrounding air by the escaping gas. Gas under pressure, but at ambient temp will chill the air molecules as it escapes. Heated gases depend on the temperature and pressure of the piping they are escaping from and the outside ambient air temp. The escaping gas plumes are pretty easy to see with most IR cameras using a wide spectrum lens to make sure it is sees all possible wavelengths.

  389. Phil. says:

    Ira Glickstein, PhD says:
    March 4, 2011 at 4:32 pm
    Don V says:
    March 4, 2011 at 10:38 am

    Good to hear from you again DonV. Have a look at the graphs that appear after my comment time stamped: March 1, 2011 at 7:19 am

    Notice that, “(a) 20 Km looking down” (from near the top of the atmosphere, approximately Space, down towards the Earth), there is quite a bit of ~15μ energy (which in the description of my graphic includes all far-IR from about 12μ up) emitted by the Atmosphere. Note that there is a distinct depression between 14μ and 16μ, but still a fair amount of energy there, but not as much as at around 13μ and 17μ, which approximate the blackbody curve.

    OK, my conclusion. The Earth radiates 14μ through 16μ at full force, but, when that radiation emerges from near the top of the Atmosphere, some of the 14μ and 16μ energy has been lost. Why? Well, CO2 and H2O molecules absorbed it and re-emitted some at their characteristic wavelengths, but, they also transferrred some of it, via collision (conduction) to other molecules in the Atmosphere, all in random directions.

    Around 15μm the atmosphere is ~opaque until the temperature of ~225K above which the CO2 emissions are able to make it up to 20km. Between 13μm and 10μm and 9.5μm and 8μm the atmosphere is transparent and you can see BB from the surface at ~268K

    The energized molecules re-emit the radiation at a variety of wavelengths, including in the ~7μ and ~10μ portions, which is why, looking again at the entire spectrum of the graph, there is proportionately more energy in the ~7μ and -especially- ~10μ regions. Where did that energy come from? Well, the only place I can think of is from the ~15μ portion where it is missing.

    No from the surface.

  390. Steve says:
    March 4, 2011 at 2:50 pm
    …Anyone here expert enough to know for sure?”

    You don’t need an expert, Ira. Think about what you are proposing. What percentage N2 is the atmosphere between the aperture of your IR camera and the object you are taking a photo of (even if it is just a warmer patch of N2)?

    If N2 absorbed/emitted in the IR spectrum, every IR image taken on earth would look like a photograph taken in a thick fog. There’d be no way for the IR to get from object to aperture without hitting a significant number of N2 molecules. Since this doesn’t happen, we can take crisp IR photos.

    Good points, Steve, and you may be correct.

    However, IR photos are generally taken fairly close to 10μ (FLIR video, looking for objects whose temperatures range around the blackbody temperature of the surface of the Earth to, for example, expose people walking across a field) where there is a window in the H2O absorption spectrum, or sometimes around 4μ (detection of vehicles with hot engines or tires) where there is another window. If you try to take an IR image in other parts of the spectrum the water vapor will make the image look like fog or nothing will be clearly seen.

    The above is why I insisted that the second container be pure N2 with no H2O or CO2 to fog or block the image.

    As I said, you may be right that hot N2, even when it is injected into a room-temp container of N2 will not emit photons as it cools off. On the other hand, I think the hot N2, which has lots of energy, when it collides with room-temp N2, will release radiation somewhere in the IR spectrum, perhaps around 4μ, and possibly, given a fairly large mass of hot and room-temp N2 colliding and mixing, over a wider number of wavelengths. The reason I think so is that collisions may distort the molecules and change their exact geometry, which could affect the wavelength of the ejected photons. I am no expert in this area, so I do not know for sure.

  391. Domenic says:
    March 4, 2011 at 5:42 pm
    Hi Ira,

    The second container should have a matte, or black painted finish on the inside surface. … [and other specifics of the experiment] …

    If you do that, with a wide band .5 to 50 micron or 70 micron window on a high quality IR camera with high sensitivity, I am pretty confident that you will see the N2 entering. … The image should look like water or a vapor plume entering the chamber from the first hose, with the image most clear right after exit from the hose….

    IR cameras are used to check for gas leaks in industrial facilities. … The escaping gas plumes are pretty easy to see with most IR cameras using a wide spectrum lens to make sure it is sees all possible wavelengths.

    Thanks Domenic, you seem to confirm that plume I saw in my mind’s eye. And, not only that, but someone stole my idea for detecting escaping gas even before I came up with it :^). At what approximate wavelength(s) will the plume of hot N2 be observed?

  392. Domenic says:

    Hi Steve,

    Steve wrote:

    “In a 0% transmissive cavity you will detect a near continuous band of wavelengths in a range commensurate to the temperature of the cavity. That does not mean that the molecules in the walls of the cavity have gained the ability to absorb/emit in a continuous band of wavelengths – there is reflection…..and so on.”

    Yes, you used the reflections to account for decreasing energy states by increasing the wavelengths, to points where the photons would finally be absorbed by the N2 at its ‘supposed’ absorption spectra. Thus 100% absorption.

    But that is not what I asked in this part of the question.

    I am asking about what is being EMITTED only through the pinhole to the outside detector. From the inside to the outside. The inside is completely N2 cavity walls with no transmissive effects from the outside world.

    Are you trying to use DECREASING energy states via reflections to account for what you will ‘see’ as a continuous spectrum? A continuous spectrum that includes wavelengths of shorter length (thus HIGHER energy states) than those possible by ‘supposed’ published N2 absorption spectra at a given temperature? Thus, the assumed equivalent emission spectra at a given temperature?

    Because, if you isothermally heat that ‘all N2′ cavity up, above outside ambient temperature, the emitted radiation from the pinhole will follow near perfect BB radiation. It will not follow the ‘supposed’ N2 spectra for emission at all. And you can’t say that the emitted shorter wavelengths are from reflections due to entering photons from outside through the pinhole into the cavity, because the outside temperature is below the cavity temperature. Thus, there is NO source for the apparent shorter wavelengths except the N2.

    And yet, strangely enough, that is the way the real world works with cavity radiation. It doesn’t violate the laws of thermodynamics, but it does shift the emitted energy to BB like wavelengths regardless of the source molecules.

    I will leave you to ponder that mystery…

  393. Domenic says:

    Hi Ira,

    I don’t know the exact wavelengths, as I have never mearusred them. Nor have I seen anyone else measure them.

    As I’ve stated many times here, N2 and O2 have never been tested rigorously over the continuous spectrum. It’s tedious and difficult to do so, and there has never been a good reason to do so before. But I am pretty confident there will be detectable emissions in the .5 to 50 micron range from playing around with a .5 to 50 micron hand held detector many, many years ago, aiming it at the sky, the sun, etc and comparing the output to a detector using much narrower bands.

  394. Al Tekhasski says:

    I think it is a time to stop this discussion. Everything emits and absorbs in any frequency range. This is true but physically meaningless statement. The real question is how much. Good physics dictates to put numbers on magnitude of each effect, and reasonably sort things out in accord with their relative contribution to main effect.

    Yes, pairs of N2-N2 and O2-O2 do present in air at normal pressure and temperature when all molecules collide in gaseous mess. Yes, this is called “collision-induced absorption”, and it does exist, has been measured several times, and is fully theoretically accounted for:
    http://pubs.giss.nasa.gov/docs/2003/2003_Boissoles_etal.pdf
    At atmospheric conditions, the spectrum is continuous, centered at 100um, with measurable tails down to 30um. It is even considered important in certain cases of dry stratosphere when looking along a very slant path. Unfortunately I could not find a straight answer in their articles about relative contribution of N2-O2 collision-induced continuous background to vertical path of air filled by CO2, water vapor, and water droplets.

    I can only offer this simple estimate, that the entire relative energy of Earth IR emission is less than 0.02% above of 30um (see spectralcalc.com). Therefore, regardless of how strong N2-O2 absorb at its respective band, 30um-10000um, this effect cannot exceed 0.02%. Now you need to decide if this amount (about 0.05W/m2, or 50mW) is really important in explaining why the Earth surface is warmer than the planetary emission temperature is.

    Regarding the “hot N2 jet experiment” and depending on how hot it really is, it is possible that a good cooled IR camera will see some emission in 70um end; the modern IR cams are fairly sensitive.

  395. Phil. says:

    Domenic says:
    March 4, 2011 at 8:03 pm
    Hi Ira,

    I don’t know the exact wavelengths, as I have never mearusred them. Nor have I seen anyone else measure them.

    As I’ve stated many times here, N2 and O2 have never been tested rigorously over the continuous spectrum.

    Which is of course not true!

    It’s tedious and difficult to do so, and there has never been a good reason to do so before. But I am pretty confident there will be detectable emissions in the .5 to 50 micron range from playing around with a .5 to 50 micron hand held detector many, many years ago, aiming it at the sky, the sun, etc and comparing the output to a detector using much narrower bands.

    It’s easy to do, in fact every time I ran my FTIR spectrometer I did so, but of course there’s no signal. The only wavelength where anything could possibly show up would be the incredibly weak lines at 4.3μm, which is of no practical significance. Your confidence is misplaced.

  396. Tim Folkerts says:

    @Ken Coffman

    I hate to rank on you, because I do believe you also are fair-minded and trying to understand (and hopefully you will believe the same of me). However, the water hose analogy that has been used is just that — an analogy, and it has several shortcomings. Just to be clear — in the analogy, spraying water molecules are used as representing the IR photons.

    “There is no side channel or alternate path for the fire hose and the blue-water squirt gun. They must use the same path.”
    Unlike water molecules, photons can and do easily cross without colliding with each other. They can use the “same path” without any (significant) chance of blocking each other.

    “Instead of a squirt gun, you get a funnel with a blue die injector. A really teenie-tiny, small one to represent 390PPM of CO2 in the atmosphere. ”
    The original analogy was flawed here, and this further exaggerates the flaw. For IR photons there are similar numbers going each way, as i will explain now

    * The ground is a little warmer than the atmosphere, so that factor will mean some more photons going up than down (but since the back radiation is mostly from low layers, the atmosphere emitting the back radiation will not be that much cooler than the land so the effect from temperature will not be TOO great)
    * The ground is close to a black body for IR (emissivity = 1 for all IR frequencies), but the atmosphere has bands where it does not emit or absorb well (emissivity ~0) and other bands where it does emit or absorb well (emissivity ~1). So this factor by itself might mean ~ half as much going down as up (depending on just how big the “windows” are. (BTW, I am combining the emissions from H2O with the CO2 here. But the CO2 by itself would also return photons in significant bands of wavelength.)
    * Also, I believe the original graphic of the “windows” included water vapor (ie humidity) but not water droplets (ie clouds). Clouds cover much of the earth and would absorb and re-emit almost 100% of the IR photons.

    Combine these three factors , and there are nearly as as many photons going down as up. (In fact, if you believe Trenberth, the ratio is 324:390 ~ 83%. Even if you don’t believe the numbers exactly, the back radiation just from the clouds should be close to 70% since clouds cover about 70% of the earth.)

    The analogy should be more like a 4″ fire hose vs a 3″ fire hose if we are representing the relative numbers of IR photons.

    There are other aspects of the analogy worth discussing, but as you said “once the sale is made, shut the hell up.” :-)

  397. Al Tekhasski says:

    Before everybody start jumping on me, I’d like to correct myself. Energy fraction under Planck curve above 30um at 255K is somewhat larger than 0.02%. I pull this number out of my mind meaning that the N2-O2 spectrum corresponds to a blackbody with 100um peak, which would correspond to a body at about 40K with total emission of 45mW/m2. More, the entire region above 15um is already filled with CO2 and water over the top, see Ira’s spectrum above, so there is no way to see any N2 or O2 effect in practice.

  398. Tim Folkerts says:

    Ira,

    I would say this a little differently:
    OK, my conclusion. The Earth radiates 14μ through 16μ at full force, but, when that radiation emerges from near the top of the Atmosphere, some of the 14μ and 16μ energy has been lost. Why? Well, CO2 and H2O molecules absorbed it and re-emitted some at their characteristic wavelengths, but, they also transferrred some of it, via collision (conduction) to other molecules in the Atmosphere, all in random directions.

    The Earth radiates 14μ through 16μ at full force, but 100% of those photons are absorbed by the GHGs fairly near the surface. None of that radiation actually emerges from the TOA.

    GHGs near the TOA can and do radiate their own new 14μ through 16μ IR photons. However, since the atmosphere is much cooler there, they radiate much less energy in those bands — hence the much reduced intensity compared to the wavelengths that came all the way up though the “windows” where the atmosphere is nearly transparent.

  399. Don V says:

    Ira,
    I see something completely different in those two graphs than you do, and frankly what I see makes of alot more sense. At least to me it does.

    First things first. The description of these graphs say these are spectra of the atmosphere on a clear DAY in the arctic – get that? – DAY. Why the arctic? To minimize the effects that pesky water contributes to these spectra over most of the rest of the planet. So here is what is see. Not illustrated, but easy to visualize is the black body curve of what is creating both of these spectra – THE SUN – you just have to imagine another spectrum of the illumination source exactly following the top dashed line.
    Now what is the back drop for both of the spectra? Well for the ‘look down’ spectrum it is probably a snowy landscape right at 273K. What would you expect to see if there were no atmosphere? The spectrum of a BB radiator at 273K. For most of the graph where the sun isn’t exciting atmospheric gas molecules that is what you see. For the ‘look up’ scene the back drop is the the cold black of space. Were it not for the excited atmospheric molecules sitting between the detector and this back drop we would just see pretty much a flat line.

    So now throw in the sun, which is most likely illuminating whatever atmospheric gases are present from the side, (it is after all the Arctic), and you get the two spectra. From 20 Km we don’t see a large water signature radiating back out to space – which tells me there’s very little water vapor in the atmosphere in this scenario to be excited . And low and behold the lack of much water creates a window in which a large portion of the what would normally be the dominant absorber and radiator – water – is missing. In the visible spectrum you see a whole lot of electronic excitation states from all of the gases, and you clearly see one of CO2’s signature excitation bands – the 7μ mode being excited by THE SUN. You also see the unique line spectrum of ozone right in the middle of that 10 μ band.

    Starting right about 12 μ both graphs deteriorate into what looks like a much colder thermalization induced BB radiator for the whole of the atmospheric gas milieux. I’m sorry, but I do NOT see CO2’s signature absorption or emission peak at it’s most dominant, highest extinction coefficient band – 15 μ. Both graphs in this region show the same broad spectrum illumination characteristic of a weak BB radiator where every gas is in thermal equilibrium with the milieux, and none are particularly absorbing or emiting any characteristic spectra, even though one of them is a strong absorber.

    So what would convince me that CO2 actually plays the dominant role you claim? It is quite simple. I need to see a third and fourth graph of this same scenario WITHOUT solar illumination. It should be entirely possible to create a wide area BB radiator that ONLY radiates a black body spectrum at 273 K in the Arctic (ie no convection heating of the atmosphere over this cold light source). If the upward looking graph for this scenario shows 7μ, and 15 μ peaks radiating down to the surface, from this upward illuminating cold light source, and if the downward looking graph shows missing bands in those same regions when viewing the same light source then yes I will have to conclude that CO2 is contributing a significant ‘greenhouse’ effect. Why aren’t these most crucial graphs illustrated for this same scenario? I wouldn’t be surprised if they were taken, but since they showed NO CO2 signatures, they were sent to the circular file. They told the wrong story.

    Again, I believe that CO2 does play a minor role to keep our atmosphere warm, but I will reiterate, WATER DOMINATES when it comes to maintaining our atmosphere at a stable temperature. Water isn’t just a passive radiative player either. It is an active phase changing player storing and redistributing both heat energy and cooling all over the globe in very short time scales. Whereas CO2 is necessary in the photosynthesis cycle and therefore an increase can only help feed the world.

  400. wayne says:

    Not going to spend much time making these word pretty, probably too late for anyone to read it.

    I feel like I was partly to blame to start this very lengthy discussion on N2 & O2’s possibility to absorb and emit photons in out atmosphere but I do like the truth to be told if it is real no matter how tiny it’s effect. The fact that it is tiny can then be added as a caveat in any discussion.

    But let’s for a minute assume that they do. I’ll pick O2 at it’s 5.5-7.4 µm (1345-1820 cm-1) that HITRAN documentation claims. What is it’s effect on this whole discussion? I say very little and not just because that span of frequencies radiance is a small portion of the BB radiation from the ground at 288K, but instead that those photons within that range would not go but a few centimeters (swag) before being absorbed. There are just so many O2 molecules at this pressure of one atmosphere.

    And what if an O2 molecule a meter off of the ground did radiate? That photon once again would not travel far to find another O2 (matching absorption/emission frequencies) to absorb it and when absorbed it would have a high probability just to be thermalized as any of the other types of molecules in our atmosphere.

    If it was N2 this process would be even worse. N2 would not travel even one centimeter (swag) before recapture. So would this be “back radiation” from these photons if they have no chance to reach the ground unless they are emitted below your ankles? Not to me!

    Well what of H2O? Out of every 2500 molecules in the atmosphere there are 60 water molecules (per Wikipedia @ 2.5%). Nitrogen has 1950 of these 2500 and oxygen has 525 of them so water molecule emitted photons would travel even further before capture by another H2O molecule. Of every 2500 molecules there is one CO2 and I have read that CO2 emitted photons travel no more that 10 meters before capture near the ground so water might travel something like 50 centimeters to just be mainly thermalized. CO2 traveling the furthest.

    If all of the above is realistic then I don’t know if anyone else gets very curious feeling, but I do. Where is all of this tremendous “back radiation”? Really, from where? If all radiation at 100 meters of altitude high has no chance of hitting the ground, WUWT? The only place this process falls apart is near the TOA and above as even LTE fails and the photons have little chance of ever being captured.

    Well, now I have to be honest and admit that this has always been the way I have viewed the atmosphere, right… no back radiation of any meaningful amount. But these small jumps in radiation help to one, equalize, and two, to speed the movement of heat upward for the deck is stacked (probability) in that direction as all molecules get further and further apart so the capture distances get longer and longer.

    I have brought it up every few month’s here just to have no one ever acknowledge that they at least understood this view of the process whether they believed it was correct or not. Since everyone has been so intent on this very subject here, I just had to try one last time.

    Looking just a bit deeper you would find this whole thought process parallel’s that of
    Dr. Tom Vonk’s description of thermalization in this post: http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/

    (I feel like the fellow in the movie “I’m Legend”)… Anyone out there? Anyone? ☺

  401. Domenic says:

    to Al Tekhasski, and Wayne and any others

    My purpose with Phil and some of the others was simply to point out that when certain phenomena are measured in isolation, such as using his FTIR to measure absorption, the results can easily mislead the unwary when they try to apply the results to the ‘whole’, the real world.

    And it is the ‘whole’ that concerns us, not the bits and pieces.

    Personally, I don’t need any ‘model’ to see a very good picture of the ‘whole’.

    To me, the simple actual surface temperature records from Amundsen-Scott, Vostok, and other interior Antarctic stations provide the best possible data out there, even better than satellite data. And I don’t mean any ice core data, either. That’s just a load of speculation.

    I am talking about the real thermometer readings from 1957 to present. No speculation.

    The first reason the interior Antarctic data is important is because CO2 has also been measured there. And CO2 there has risen from 1957 onwards tracking the rise noted at Mauna Loa and other stations.

    The second reason the interior Antarctic data is important is because radiational heat transfer through the atmosphere is more dominant there than any other place on earth. Period.

    The third reason the interior Antarctic data is important is because there is NOTHING on earth that can drive those surface air temperatures down, except a decrease in the energy content of the atmosphere, OR direct radiational heat transfer from the surface to outer space. You can take your pick there, it doesn’t matter. It is the SUMMATION that is important. And that summation is exactly what the thermometer readings handily give us.

    There are lots of local effects, winds, moisture, UHI, etc that can drive the temperature up temporarily, but NONE of them can drive it DOWN.

    The fourth reason the interior Antarctic data is important is because you need only to look at the yearly low temperatures recorded. Each of those low temperature data points are the clearest signals (least amount of local noise) of any physical measurement that is possible on earth, or from a satellite in space. Period.

    AND THERE HAS BEEN NO HEATING trend whatsoever due to the increase in CO2 in the atmosphere above central Antarctica from 1957 to date. In fact, it appears to be cooling slightly.

    It’s that simple.

  402. Ken Coffman says:

    Domenic, my admiration for your intellect grows and grows. Thank you.

  403. Ken Coffman says:

    The concept of “back radiation” nonsense. It confuses cause and effect in a very fundamental way. Radiation is caused by delta-T, not the other way around. A passive material does not “back radiate”. Conduction, convection and radiation always work in the same direction (unless outside work is used)–they are mechanisms for transferring heat energy from hot to cold objects. Just because a formula can be used to describe a process does not mean mother nature will allow you to reverse the equation and use the effect to create the cause.

  404. Phil. says:

    Domenic says:
    March 5, 2011 at 5:24 am
    to Al Tekhasski, and Wayne and any others

    My purpose with Phil and some of the others was simply to point out that when certain phenomena are measured in isolation, such as using his FTIR to measure absorption, the results can easily mislead the unwary when they try to apply the results to the ‘whole’, the real world.

    And you achieve that purpose by making untrue statements about those properties? Seems a weird way to to it, I guess you’re admitting now that what you said about the absorption/emission of N2 and O2 in the IR was wrong?
    Nice way to change the subject though.

  405. Phil. says:

    Don V says:
    March 5, 2011 at 2:18 am
    In the visible spectrum you see a whole lot of electronic excitation states from all of the gases, and you clearly see one of CO2′s signature excitation bands – the 7μ mode being excited by THE SUN. You also see the unique line spectrum of ozone right in the middle of that 10 μ band.

    The 7μm band would have to be excited by IR not by the sun.

    Starting right about 12 μ both graphs deteriorate into what looks like a much colder thermalization induced BB radiator for the whole of the atmospheric gas milieux. I’m sorry, but I do NOT see CO2′s signature absorption or emission peak at it’s most dominant, highest extinction coefficient band – 15 μ. Both graphs in this region show the same broad spectrum illumination characteristic of a weak BB radiator where every gas is in thermal equilibrium with the milieux, and none are particularly absorbing or emiting any characteristic spectra, even though one of them is a strong absorber.

    What you’re looking at is the low resolution spectrum of CO2, see below for a calculated spectrum for the subarctic atmosphere with a surface temperature of 267K

    http://i302.photobucket.com/albums/nn107/Sprintstar400/rad05093716.gif

  406. Tim Folkerts says:

    Ken Coffman says: March 5, 2011 at 6:44 am

    The concept of “back radiation” nonsense. It confuses cause and effect in a very fundamental way. Radiation is caused by delta-T, not the other way around.

    I at least partially agree — the term “back”radiation can indeed a bit confusing and makes it seem “special” when there is nothing fundamentally special about it. The term “radiation” would suffice perfectly well.

    However, radiation is caused by Temperature, not by Delta-T.
    P= A (epsilon) (sigma) T^4.
    There is no need to mention other objects in the area or what caused the object to be warm.

    * The sun at ~ 5700 K radiates the same whether or not the earth was here to receive it. The temperature of the earth does not affect the black body radiation of the the sun.

    * a patch of ground radiates the way it radiates based on its temperature and emissivity. If the patch of ground is at 300 K, it will radiate as a 300 K black body (modified slightly by its emissivity which is slightly less than 1 for applicable). It doesn’t matter if the land was heated to 300 K by the sun or the atmosphere or an electric heater.

    * the atmosphere radiates the way it radiates based on its temperature and emissivity. The emissivity of N2 & O2 is practically zero across the applicable IR spectrum. The emissivity of CO2 and H2O and other GHGs is NOT always close to zero, but instead there are broad bands where the emissivity is not close to 1 (ie where these molecules can and do absorb and emit IR effectively given sufficient path length thru the molecules).

    The GHGs radiate in all directions. Some radiation heads up; some heads down; some heads east, some heads west ….

    “Back radiation” from the atmosphere is simply that part of the radiation that happens to be heading back toward the earth. It is still just regular old radiation as described by the S-B Law – the same as the radiation heading any other direction. The only special thing is the direction, not the fundamental nature of the radiation.

  407. Phil. says:

    Don V says:
    March 5, 2011 at 2:18 am
    So what would convince me that CO2 actually plays the dominant role you claim? It is quite simple. I need to see a third and fourth graph of this same scenario WITHOUT solar illumination. It should be entirely possible to create a wide area BB radiator that ONLY radiates a black body spectrum at 273 K in the Arctic (ie no convection heating of the atmosphere over this cold light source). If the upward looking graph for this scenario shows 7μ, and 15 μ peaks radiating down to the surface, from this upward illuminating cold light source, and if the downward looking graph shows missing bands in those same regions when viewing the same light source then yes I will have to conclude that CO2 is contributing a significant ‘greenhouse’ effect. Why aren’t these most crucial graphs illustrated for this same scenario? I wouldn’t be surprised if they were taken, but since they showed NO CO2 signatures, they were sent to the circular file. They told the wrong story.

    Well they would only have importance to you because you have a flawed view of how these gases are excited into the higher energy levels in order to emit. They are not excited by short wave solar radiation and trickle down the energy levels until they emit in the IR (your model). Since no one actually thinks that then the significance of the night-time data doesn’t occur. Your snide suggestion that because you haven’t seen them that the evil scientist trashed them is beneath contempt. Since seeing such a graph would convince you of the error of your ways I suggest you look at fig 8 below.

    http://www.osti.gov/bridge/purl.cover.jsp;jsessionid=6CB3171E19F77B57AEA27EB269BFEAC4?purl=/957049-3J26t7/

  408. Domenic says:

    Hi Ken,

    Thank you. That was very kind of you.

    Whatever intellect I have, and its not that great, I would gladly trade some of it for more intuitive abilities.

    It’s the intuitive abilities that let us go beyond ourselves and catch glimpses of the ‘whole’. The intellect just seems to tag along for the ride, telling the great stories.

    But I suspect you already know that. It’s probably what pushed you into becoming a publisher!

    Cheers,
    Domenic

  409. Domenic says:

    to Phil

    Nope.

  410. Reed Coray says:

    Ken Coffman says:
    March 5, 2011 at 6:44 am
    The concept of “back radiation” [is] nonsense. It confuses cause and effect in a very fundamental way. Radiation is caused by delta-T, not the other way around. A passive material does not “back radiate”. Conduction, convection and radiation always work in the same direction (unless outside work is used)–they are mechanisms for transferring heat energy from hot to cold objects. Just because a formula can be used to describe a process does not mean mother nature will allow you to reverse the equation and use the effect to create the cause.

    You make an interesting point. I agree that man’s theories (mathematical characterization) of how nature works have no effect on how mother nature actually works. As someone once said: “It’s just a matter of time before any theory is disproved.” However, this does not mean that man should abandon the use of theories. Man has benefited from many “tools” he developed using incorrect theories.
    In this light, I take exception to your statement: The concept of “back radiation” [is] nonsense. Maybe the word “back” is misleading, but Planck’s blackbody radiation law has proven useful to man’s understanding of the universe. Planck’s blackbody radiation law characterizes the properties (spectral distribution, directionality, magnitude) of electromagnetic radiation emitted from a planar differential area on the surface of a blackbody at a known temperature. The spectral distribution is characterized by specifying as a function of frequency the amount of radiated energy per unit frequency. The directionality is characterized by expressing the amount of radiation emitted into a differential solid angle and includes the factor of the COSINE of the angle between the normal to the plane of the differential area …and… the direction of the differential solid angle. The magnitude is characterized by some constants and the fourth power of the temperature. This formula seems to work experimentally, and as I understand it, can be derived from quantum mechanics principles. In fact, I read somewhere that one of the great accomplishments of quantum mechanics was to provide the theoretical basis for the radiation law, which up to then had been verified experimentally, but which seemed to contradict prevailing theory.

    Planck’s law does not differentiate between the surfaces of “active” and “passive” (I prefer the term “inert”) objects–it quantifies the properties of electromagnetic radiation emitted from a surface. As such, it can be applied to both an isolated object and multiple disjoint objects. For example, I believe the temperature of a passive blackbody object placed in the vicinity of an active blackbody object will affect the temperature of the active object. Specifically, consider an isolated, active (constant source of thermal energy), spherical, blackbody object whose surface temperature is everywhere the same. Once the rate that energy leaves the object via radiation equals the rate at which thermal internal energy is being generated, the temperature of the surface of the object will cease to change and settle at a fixed value. If a passive blackbody sphere is placed in the vicinity of the active sphere, the presence of the passive sphere will cause the steady-state temperature of the active sphere to change. I hope you agree; but I await your response. For the moment, however, let’s assume the temperature of the active sphere does change; and we want to compute the “new” temperature of the active sphere surface. If we assume a uniform surface temperature distribution for both spheres, Planck’s law can be used to relatively easily compute both the “new” temperature of the active sphere surface and the temperature of the passive sphere surface. I strongly believe that the surface temperature of the passive sphere will be less than the surface temperature of the active sphere; and energy will flow via radiation from the active to the passive sphere. Using Planck’s “model”, the rate of energy leaving the active sphere is the same IN ALL DIRECTIONS because the surface area of the active sphere is assumed to be uniform. Similarly, the rate of energy leaving the passive sphere is the same IN ALL DIRECTIONS. As I understand your thinking, you would argue that the rate energy leaves an object in the direction of another object is a function of the temperature difference between the objects, and as such the rate energy leaves the active sphere in the direction of cold space differs from the rate energy leaves the active sphere in the direction of the passive sphere. I have difficulty adopting this line of reasoning. Here’s why. Suppose you have a star, like our sun, whose internal energy source is for all practical purposes constant; and although other matter exists in the universe, that star is the only star. At some time, matter many light years from our star coalesces into a second star. Does the existence of the newly formed star affect the amount of radiation from the original star in the direction of the newly formed star? And if so, when? That is, is the amount of radiation affected at the time the star is created? Or is it affected only when the light from the newly formed star reaches the original star? Furthermore, if the newly formed star is at a higher temperature than the original star, does this mean that the original star ceases radiating in the direction of the newly formed star? This may in fact be the way nature works; but if so the process of analyzing the temperatures of a multi-object system becomes infinitely more complex than a model that quantifies the radiations from the individual objects and defines the “heat” between any two objects to be difference between the two radiation rates.

    One final point. You said A passive material does not “back radiate”. OK. Suppose in the passive object I place a small source of thermal energy–small in the sense that it produces a small increase in the temperature of the passive object. Has the passive object now become an active object? Surely the temperature increase of the “passive” object will have an effect on the temperatures of both the passive object and the originally active object. How do we quantify the rate, if any, that the energy from this small source of energy is radiated to the original active object? Is this effect (a) radiation, (b) backradiation, or (c) something else?

    Bottom line, I think we’re on the same side in that the statement that “backradiation warms the surface of the active object” is incorrect. ALL warming comes from the active object’s internal thermal energy source. However, I think the concept of individual object radiation, whether it’s backradiation, forward radiation, or just plain radiation, is a useful way to characterize radiative energy transfer.

  411. Myrrh says:

    Vince Causey says:
    March 3, 2011 at 11:33 am

    Re Myrrh says: “So, radiation is included, not just conduction, which means that it is impossible for a cooler molecule to transfer heat to a hotter one because that violate the 2nd Law.”

    You based this on the following definition of the 2nd law: “”When two isolated systems in separate but nearby regions of space, each in thermodynamic equilibrium in itslef, but not in equilibrium with each other t first, are at some time allowed to interact, breaking the isolation that separates the two systems, and they exchange matter or energy, they will eventually reach a mutual thermodynamic equilibrium.”
    Myrhh, you have read the 2nd law and have drawn the wrong conclusion. The 2nd law considers “two isolated systems in separate but nearby space, not in equilibrium with each other,” as you have pointed out. It then says that when they are allowed to interact they “exchange matter or energy..” This is the crucial point THEY ie both bodies – hotter and cooler – exchange energy. But the crucial conclusion from the 2nd law, is that as a result they “will eventually reach a mutual equilibrium.” This is because the hotter body is transmitting at a far greater flux density than the cooler body.

    The GHG hypothesis describes a cooler atmosphere radiating to a warmer surface, which some, including you, say is impossible because it violates the 2nd law. But we have seen that the 2nd law says both bodies will exchange energy, and they will eventually equilibriate – if they are isolated systems. If the Earth was not continually recharged with energy from the sun, it would cool down to the temperature of the atmosphere, and both of them would continue to cool as they equilibriate with space.

    First of all, apologies for not getting back to you earlier.

    You say I’m misunderstanding the 2nd Law because of not fully appreciating the phrase “THEY exchange matter or energy”. This is the crux of the matter, and why I say this is where AGW is confused, to the point where as above shortwave radiation is believed to ‘be thermal energy heating the Earth’, because this is how the ‘statistical’ has justified ‘its method ‘, by looking at energy exchange out of context of heat. It then has the temerity to say the 2nd Law doesn’t apply to it… Because of this it is assumed that the energy exchange from the cooler can happen from the hotter and in the exchange a ‘net’ result then conforms to the LAW, BUT, the Law says it can’t.

    There are two important things here. The Law as it stands CANNOT be broken. If a hypothesis, such as the cooler radiates to the hotter thereby heating it because etc. and so claiming the Law doesn’t apply in its particular instance, then THIS IS INCREDIBLY IMPORTANT TO SCIENTIFIC UNDERSTANDING – BECAUSE IF PROVEN IT FALSIFIES THE LAW!!

    Go collect your Nobel prizes for Science!

    I can’t see any recognition of the importance of this from those arguing such a thing can happen.. because there’s zilch understanding of what they’re saying by those saying it.

    AS THE LAW STANDS – this applies to all energy and matter states, on every level. It specifies energy and it is energy we are discussing, Electromagnetic.

    Now my understanding. I have not read anywhere, and it is nowhere stated in Real Science, that heat can flow from the cooler to the hotter. Investigating this, such claims as Ira has made here, that shortwave Solar is thermal and heats the Earth and is what we feel as heat, and elsewhere, that UV penetrates deeper than Visible that penetrates deeper than Thermal IR, etc., I have found where the error has occurred. How well I can explain this, well, obviously not very well since there’s a difficulty following my explanation, or you’re all just so sure of yourselves that you think you can dismiss what I say because I don’t come from a science background, whatever.

    But thank you for engaging with me, Ira still continues refusing to answer me even after I have shown him a NASA page showing his premise is utter nonsense, so I will have one more go at this.

    That LAW is not broken, it stands. It must apply to all energy states if it applies to any. That’s why it is a LAW, it applies in all cases.

    So, what is happening at the meeting place of the two previously isolated systems? Whatever it is it CANNOT be that AT ANY TIME in the exchange until equilibrium is reached that the Cooler is Heating the Warmer. Otherwise the 2nd Law is broken.

    I say that whatever energy is exchanged the energy of the cooler cannot be heating the warmer, and as this is Solar and Thermal radiation we’re talking about, whatever radiation is being transferred from the cooler to the hotter it cannot have the effect of adding heat to it.

    I have shown that not all IR is Thermal. Solar, Visible and the two shortwave either side, UV and Near IR, are not felt but if intense enough they can burn the cooler matter, but they cannot add heat to warmer matter. If, there is any exchange of energy from the cooler to the warmer, it will not energy capable of adding heat.

    If the cooler is hot then its ability to transfer any of that heat to a hotter is non-existant. Heat ALWAYS flows from Hotter to Colder. It may well be transferring cool Near IR as photons, but these will be INCAPABLE of adding heat to the hotter.

    That is the only way to read this LAW. If you have a hypothesis that doesn’t fit it exactly then the hypothesis is junk.

    As I tried to explain above, that the ‘statistical approach’ to science has come up with something that they think explains this imaginary ‘net transfer’, doesn’t mean that is has any actual basis in reality. Average ISN’T real, Ideal, as in gas, ISN’T real, these are mathematical constructs useful in some calculations, but, to use them and think of them as REAL is absurd.

    This idea of the statistical net exchange is what has confused AGW supporters here, they think it is real. They have no concept of the difference in Science between the terms REAL and IMAGINARY, as useful as imaginary can be in context for some calculations if one doesn’t understand that it isn’t referring to real world then all kinds of imaginary conclusions can be claimed, that have no basis in reality. And they can’t tell the difference and so can’t see what utter codswollop they’re saying.

    If there’s an average of 400 ppm of Carbon Dioxide in the atmosphere, does it mean that everywhere in every discrete portion of the atmosphere at every time? Of course AGW says, it is well mixed… If the rainfall in your home county is 10″ average a year, does it mean that it is 10″ every day and in every place? Average is imaginary. So, Ideal is imaginary in describing gases, there is no such thing as an Ideal gas in real life, all gases are Real. Look up the difference, see how AGW has used ideas from Ideal gas laws and applied them out of context to Real CO2 in the atmosphere.

    If AGW supporters understood that not all radiation is thermal and understood the difference between reflective and absorptive, or understood that the 2nd Law cannot be broken in any of its parts, or that ideal gas laws do not describe how CO2 behaves in the atmosphere, they would also see that the methodology behind this ‘net transfer obeys 2ndLaw’ is nonsense. Work out a better methodology sticking to the 2nd Law and sticking to actual Real properties of radiation and gases or continue living in AGW created cloud cuckoo land where Solar energies are thermal and CO2 acts like an Ideal gas with no volume and no weight and no etc.

    In other words, the explanation for ‘net transfer’ is a load of bull.

    And so many here think themselves scientists..

  412. wayne says:

    Domenic wrote:
    “My purpose with Phil and some of the others was simply to point out that when certain phenomena are measured in isolation, such as using his FTIR to measure absorption, the results can easily mislead the unwary when they try to apply the results to the ‘whole’, the real world.”

    I thoroughly agree. It’s frustrating that more research on the atmosphere as a whole cannot easily be located. I was searching for a word late last night to describe the radiation interaction in the lower troposphere and it finally came to me, “resonance”. That is the term that should replace “back radiation” in that sense at low altitudes. At high altitude it is next impossible for any radiation emitted from any molecule to go lower but moving upward is nearly unimpeded.

    Al Tekhasski:

    I really appreciate that paper on O2-O2, N2-O2 interactions, etc. You just intuitively feel that is real and must be occurring and it’s great to get some real empirical backup on that thought. It’s going to take a long time for this whole puzzle to be solved. Phil likes to marginalize such small effects because they are not large like H2O’s and CO2’s emissivities by also there is hugely more O2 and N2 there, so the overall effect on the total energy flows is not what it seems when looking at CO2 in a cylinder in a lab. We all seem to get that.

    Both:

    Wish someone would read my last comment and tell me where I might be wrong; I’m not very intuitively versed on the scales of radiation ‘path lengths’ and ‘optical path depths’. I feel I might have some scales incorrect and of course the emissivities must also always play into those interactions and lengths. See: March 5, 2011 at 3:10 am. It was quickly written and I missed a whole lot of needed commas.

  413. Ken Coffman says:

    Hello Reed. The way I see it, the radiation between a hot and a cold object already includes the net radiation intensity described by the delta-T. So, if we add in back radiation as the cold object warms, then this is “double accounting”. Regardless, the only thing the cold (passive or inert) object can do is modulate the cooling of the warm (heated) object. If there is perfect coupling between the objects, the most that can happen is both objects arrive at the same temperature…but never will the heated object be warmer than it was. Once they are the same temperature, then the net radiotion is zero and no energy is exchanged. This is the key contradiction of the global warming theory…given this analysis, how can we get higher and higher peak temperatures and/or higher and higher average temperatures? I thnketh not.

  414. Reed Coray says:

    Myrrh says:
    March 5, 2011 at 1:03 pm
    Vince Causey says:
    March 3, 2011 at 11:33 am

    In other words, the explanation for ‘net transfer’ is a load of bull.

    Myrrh, I think your characterization is a bit strong. For example, consider two stars (A and B) on opposite sides of a galaxy that are formed simultaneously (simultaneous in someone’s reference frame). Before the light (radiation) from star A has time to propagate to star B (and vice versa), isn’t it reasonable to say that star A is radiating a quantifiable amount of energy per unit time in the direction of star B, and vice versa? Does the arrival of the light at star A from star B alter/nullify this statement? I would say no. I believe “heat” is the exchange of thermal energy between objects. By this definition, for the time interval between the birth of the stars and arrival of the energy from the distant star, no heat is being exchanged. Energy still leaves each star; it just hasn’t become “heat” in the sense that “heat” is the exchange of thermal energy between objects. After the arrival of the energy from the distant star, “heat” exists. I believe the amount of heat is the difference between the energy rates in the directions of the opposing stars. The term “net” applies to the combined total (sum and difference) of all components. In this sense, “net radiation” is equal to heat; and the heat (net radiation) will be in the direction of the star at the higher temperature to the star at the lower temperature.

    When discussing radiative thermal energy exchange between two objects, it may very well be more appropriate to talk about the heat between objects and not mention the rate thermal energy leaves each object in the direction of the other object. However, such an approach precludes the use of Planck’s blackbody radiation law, which from a “rate” perspective describes only the rate that energy leaves a surface in a specified direction. In the absence of conduction and convection, I believe the heat exchanged between two blackbody objects can be determined by computing the difference (net) between (1) the rate radiative energy leaves object A in the direction of object B and (2) the rate radiative energy leaves object B in the direction of object A. If true, then a discussion of the “net rate of energy transfer” isn’t “bull”; it is just one way of determining the heat between objects; and hence can be used in part to determine the surface temperatures of objects in a multi-object system.

  415. Tim Folkerts says:

    Myrrh,

    I have one very simple request of you. Define precisely what you mean by “heat”. Until everyone agrees on the definition of this critical yet subtle idea, all the discussion in the world (any any side of the issues being discussed here) will be pointless.

  416. Phil. says:

    Myrrh says:
    March 5, 2011 at 1:03 pm
    You say I’m misunderstanding the 2nd Law because of not fully appreciating the phrase “THEY exchange matter or energy”. This is the crux of the matter, and why I say this is where AGW is confused,

    It’s nothing to do with AGW, it’s the fundamental basis of radiational heat transfer, read any book on the subject.

  417. Myrrh says:

    Reed – not all radiative energy is the same, different wavelengths have different properties, qualities, just as Real gases have different properties even though they are all gases. The ideal gas law CANNOT describe a real gas because the ideal gas is imaginary, it doesn’t exist, it’s a construct constrained by a description specific to itself alone; it describes no actual known gas. What one can imagine has to be from what is possible re real properties and what is actual environment.

    Someone said above, sorry not time to check who, that our atmosphere is a closed system – ? Why say that? It’s obvious to anyone giving it even a modicum of thought that this isn’t true, but people say it because the laws AGW use are from a mishmash of out of context concepts such as ideal gases in a closed system, so what has Planck got to do with it? If it isn’t relevant to actual real life conditions it can’t be used, because it doesn’t work.

    Somewhere up there I linked to the story why NASA junked Stefan-Boltzmann forty years ago for its real world calculations, because it wasn’t relevant to three dimensional reality; because it gave the wrong answers and they needed to be spot on. It seems deeply ironic that those anti this AGWScience are the ones called ‘flat earthers’…

    Re what you think heat is, and Tim, it doesn’t matter. Whichever way you think of it, it cannot disobey the Law. If you come up with a clever sounding explanation which breaks it and then you fudge it so that you obtain an imaginary ‘net’ effect which does obey the Law, then you have still broken the Law. As I said. I really don’t think those arguing for this have fully taken on board the extraordinary thing they are claiming here.

    Maybe it’s because of the overuse of ‘law’? Perhaps you (generic) think all laws can be played with at will. But somehow in this, laws specific to certain conditions only, from the ideal gas laws which are specific to imaginary gases only, to Planck and Bolzmann and the rest, are treated as if they are absolute laws applicable to all at every time and place while Absolute Laws, such as the 2nd, are dismissed as capable of being falsified, and moreover claimed falsified without any proof.

    Elementary physics has been thrown out here. If you (generic) can’t see how ludicrous it is to think that a real gas which has weight and volume behaves in the atmosphere as if an ideal gas which has neither, and think that because the ideal imaginary gas spreads to fill the container according to its given properties means that a real gas CO2 behaves this way in the atmosphere, you’ll be easily convinced that CO2 can diffuse and spreads as this imaginary gas and take it as perfectly logical then that because it is well mixed by this imaginary diffusion it stays that way and can stay in the atmosphere for hundreds and even thousands of years accumulating. And then, you’re shocked to be told it is heavier than air and sinks to the ground displacing air unless work is done to make it otherwise. Or hearing this you come up with the equally absurd solution, so sure there is someway to prove this view from AGWScience, that when you’re told the answer is that the wind is constantly mixing it so it stays mixed you don’t even stop to consider what that would mean in the real world… And even then, most of the time you imagine wind to be a wooden paddle stirring molecules in empty space..

    So, whatever your stars are doing or how they’re doing it, the colder is not heating the warmer at any time on any level of matter or energy.

    So, either prove that the 2nd Law is broken by your (generic) radiation methodology, or keep imaginated scenarios where it is broken out of the argument. These are irrelevant. It is truly insulting to Real Science to keep presenting these as if they’re not. We can all imagine several impossible things before breakfast..

    Context.

    Solar energy is not heating the Earth and is not what we feel as Thermal Energy. Start there.

    The rest should fall into place.

  418. Phil. says:

    wayne says:
    March 5, 2011 at 1:51 pm
    I really appreciate that paper on O2-O2, N2-O2 interactions, etc. You just intuitively feel that is real and must be occurring and it’s great to get some real empirical backup on that thought. It’s going to take a long time for this whole puzzle to be solved. Phil likes to marginalize such small effects because they are not large like H2O’s and CO2’s emissivities by also there is hugely more O2 and N2 there, so the overall effect on the total energy flows is not what it seems when looking at CO2 in a cylinder in a lab. We all seem to get that.

    Apparently you don’t, compared with the minuscule effect of the N2 and O2 there is not ‘hugely’ more of it. I do not ‘marginalize’ such small effects I tell it like it is and back it up with data. You on the other hand want to brag up N2 and O2 because you ‘intuitively feel’ that it must be significant to support your beliefs. Well science isn’t like that, your intuitive feeling doesn’t count, the facts are that N2 and O2 have negligible interaction with IR. Cut the handwaving and intuition show the data.

  419. Phil. says:

    Myrrh says:
    March 5, 2011 at 4:27 pm
    Solar energy is not heating the Earth and is not what we feel as Thermal Energy. Start there.

    OK, that’s nonsense.

    The rest should fall into place
    Yep, that’s nonsense too.

  420. Myrrh says:

    And Domenic – if you really think that you can get ‘background well-mixed CO2 levels’ sold as a ‘pristine site uncontaminated by local CO2 production’ from one of the most volcanic CO2 production regions on Earth then you’ve not thought it through..

    Do you know how they work out this mythical “background well-mixed CO2″ at Mauna Loa? Continuing in the tradition established by Keeling after less than 2 years of measurements on the world’s largest active volcano when he claimed he could measure man-made CO2 in all that , they decide what the numbers will be.

    Seriously, that’s exactly what they do. Look up the discussions on WUWT and read through the method. They decide what is “volcanic” and what is “man-made background” cut-off point. Keeling neither proved that he was measuring ‘background CO2′ or that there even was such a thing, nor that he was able to differentiate ‘man-made’ from volcanic.

    The AGW tradition was continued from him via his son and co-ordinated to include all stations kept in line to this base-line, and since has become even better organised.

    Keeling didn’t believe in real world measurements, he had an agenda and created his ‘science facts’ to prove it.

    And one last question to Ira re Near Infrared which is included in his Solar and which he says heats the Earth – why do you think it “is safe” in the following description?

    http://www.licor.com/translational/NIR-optical-imaging-overview.jsp

    Seriously, keep reading the NASA page I posted until this real world fact is thoroughly assimilated, Near IR is cool.

    This is Real Science.

    I am so fed up of this AGW junk science being taught to our children, they are incapable of understanding and will become incapable of creating real science knowledge as in this example.

    AGW doesn’t have to burn books and destroy the educated to produce an ignorant generation it can mold to its masters’ will, they have (whoever they are) already succeeded in creating such an ignorant generation by producing AGWScience to do its work.

    And knowingly or unknowingly, you are spreading it.

  421. wayne says:

    Phil, I have no “belief” but at least I now have the sites and papers given by others above that verifies nitrogen and oxygen’s role, however small for the emissivities at both sides of the IR spectrum are still hidden, to me at least. Thanks for not supplying them. And yes, there is massively more nitrogen and oxygen in our atmosphere. I noticed your objection to even that.

    Pull that “out of context” again as you did to both Myrrh and myself and I will plead with Anthony to apply his rules of conduct.

  422. Tim Folkerts says:

    @ Myrrh
    “Re what you think heat is, and Tim, it doesn’t matter. Whichever way you think of it, it cannot disobey the Law.”

    Do you realize how this sounds? You can’t or won’t define “heat”. The classical statements of the second law are all about heat, yet “it doesn’t matter” what heat is?

    If I think of heat as “temperature” then it must obey the 2nd law?
    If I think of heat as “net flow of energy between two objects” then it must obey the 2nd law?
    If I think of heat as “total thermal energy within an object” then it must obey the 2nd law?

    PS Are you familiar with the modern, statistical mechanics definition of the 2nd law? It doesn’t even use the concept of “heat”, going instead straight to probabilities and entropy. So we could go that way and then the definition of heat wouldn’t matter — but then you better be prepared to discuss canonical ensembles and microstates.

  423. Reed Coray says:

    Myrrh says:
    March 5, 2011 at 4:27 pm

    So, either prove that the 2nd Law is broken by your (generic) radiation methodology, or keep imaginated scenarios where it is broken out of the argument. These are irrelevant. It is truly insulting to Real Science to keep presenting these as if they’re not. We can all imagine several impossible things before breakfast.

    First, I disagree that it is “insulting to Real Science” to present and discuss “imagined/imaginary” situations. Such “thought experiments” (e.g., what would the universe look like if you traveled on the crest of a light wave?) are one of the traits that made Einstein one of the leading 20th century scientists. His thought experiments may not have solved any “real-world” problems, but they provided a framework in which real-world problems could be and were solved.

    Second, I don’t think I ever said the 2nd law is broken. In fact, I agree that it is not. What I did say was: In this sense, “net radiation” is equal to heat; and the (net radiation) will be in the direction of the star at the higher temperature to the star at the lower temperature.

    I also agree that applying ideal laws, formulas, rules or whatever you want to call them to real-world scenarios is fraught with risk. In fact, I think this is one of the weaknesses of the proponents of AGW. The earth and its atmosphere is oversimplified in that their treatment of conduction/convection is weak if not nonexistent. However, ideal laws, formulas, and rules are a starting point. Since almost any real-world scenario is too complex to be represented by a set of formulas, your objection to “applying ideal laws to real-world situations” can be made to the formation/application of any manmade set of rules to any real-world situation. In determining/predicting the temperature of real-world objects in real-world situations, where do you suggest we start? If we (generic) followed your line of reasoning, isn’t it useless to even start? Any model/formulation/theory won’t exactly represent any real-world situation unless it includes the interactions of all matter/energy in the universe, which is impossible.

    Just to make sure I don’t misrepresent you, I’d like to ask you three questions. Place a moon-like solid object (no or infinitesimal amounts of liquids and gases) with a small internal radioactive-decay source of thermal energy in space isolated from all other matter. First question: Will the temperature of the surface of this moon-like object be nonzero? Now place a second moon-like object lacking any form of internal energy generation in the vicinity of but not touching the original moon-like object. Second question: Will the original object’s surface temperature in the two-object case be the same or different than the original object’s surface temperature when in isolation? Third question: If different, will the temperature be higher or lower?

    Finally, I agree with Tim Folkerts when he wrote:

    March 5, 2011 at 3:01 pm
    Myrrh,

    I have one very simple request of you. Define precisely what you mean by “heat”. Until everyone agrees on the definition of this critical yet subtle idea, all the discussion in the world (any any side of the issues being discussed here) will be pointless.

    However, even though I agree with Tim, (a) I won’t step up to his challenge to you and provide my definition of heat, and therefore (b) I am guilty of continuing a discussion that may in fact be “pointless.”

  424. Myrrh says:

    Phil. says:

    Re: Myrrh “Solar energy is not heating the earth and is not what we feel as thermal energy.
    Start there.”

    Nonsense

    Prove it is nonsense! You are disagreeing with this page on Infrared produced by NASA for kids, http://science.hq.nasa.gov/kids/imagers/ems/infrared. It is the same Science as was well known and taught in schools pre AGWScience mangling. (I was taught in schools before AGWScience mangling, that’s why I understand that real gases have weight and volume, and what that means for CO2 in our atmosphere which is heavier than air..)

    What you are saying is what is nonsense here.

    [Infrared
    Infrared light has a range of wavelengths, just like visible light has wavelengths..
    "Near infrared" light is closest in wavelength to visible light and "far infrared" is closer to the microwave region of the electromagnetic spectrum. The longer, far infrared wavelengths are about the size of a pin head and the shorter, near infrared ones are the size of cells, or are microscopic.

    Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared.

    Infrared light is even used to heat food sometimes - special lamps that emit thermal infrared waves are often used in fast food restaurants!

    Shorter, near infrared waves are not hot at all - in fact you cannot even feel them. These shorter wavelengths are the ones used by your TV's remote control.]

    Solar energy as depicted in Ira’s graphic above and as indoctrinated by AGWScience, IS NOT CAPABLE OF HEATING THE EARTH, OR US.

    ORGANIC LIFE IS HEATED BY THERMAL IR WHICH IS NOT NEAR INFRARED, BUT FAR INFRARED.

    SOLAR, (VISIBLE, NEAR INFRARED AND UV) IS NOT THERMAL.
    WE CANNOT FEEL IT.
    IT CANNOT HEAT THE EARTH.

    Once you have fully re-adjusted to looking at the world as it really is, as real science has known and utilised practically pre AGW mangling, and as Real Science continues to know and utilise practically, the rest will fall into place. You’ll be able to untangle the wool that has been pulled over your eyes and knit yourself a real thermal blanket that’s bigger than the 1″sq in 2sq yds that CO2 can manage according to AGW.

    Go on, try it out for yourself. Get yourself a lightbulb which carries visible light but not heat and tell us how long it takes to cook your steak..

    http://www.ledsuperbright.com/led-hydroponic-grow-light-bulbs-96-watt-p-259

    Remember, Phil. and Ira, you’re arguing with that NASA page. Prove it wrong or change that graphic.

  425. Phil. says:

    wayne says:
    March 5, 2011 at 7:41 pm
    Phil, I have no “belief” but at least I now have the sites and papers given by others above that verifies nitrogen and oxygen’s role, however small for the emissivities at both sides of the IR spectrum are still hidden, to me at least.

    Yes it verifies what I said that their role is nonexistent!

    Thanks for not supplying them.

    I supplied you with both the N2 and O2 spectra, see above.

    And yes, there is massively more nitrogen and oxygen in our atmosphere. I noticed your objection to even that.

    Apparently not the caveat that I added comparing the weakness of the signal!
    N2 and O2 are order 1000x greater than CO2 whereas the line strengths are order 100,000,000x weaker.

    Pull that “out of context” again as you did to both Myrrh and myself and I will plead with Anthony to apply his rules of conduct.

    I haven’t a clue what you’re talking about, I’m talking about the science whereas you brought up your intuition.

  426. Phil. says:

    Myrrh says:
    March 5, 2011 at 9:01 pm
    Phil. says:

    Re: Myrrh “Solar energy is not heating the earth and is not what we feel as thermal energy.
    Start there.”

    Nonsense

    Prove it is nonsense! You are disagreeing with this page on Infrared produced by NASA for kids, http://science.hq.nasa.gov/kids/imagers/ems/infrared. It is the same Science as was well known and taught in schools pre AGWScience mangling. (I was taught in schools before AGWScience mangling, that’s why I understand that real gases have weight and volume, and what that means for CO2 in our atmosphere which is heavier than air..)

    Actually you don’t understand it at all and apparently think that the Gas Laws, Stefan-Boltzmann, Fick’s Laws of Diffusion and the Second Law of Thermodynamics don’t apply to the Earth’s atmosphere! (CO2 ‘sinking to the ground displacing air’ would defy both Fick’s Law and the 2nd Law).
    That page doesn’t exist apparently, and the statement that ““Solar energy is not heating the earth and is not what we feel as thermal energy”, is clearly nonsense. What is heating the Earth if not Solar energy?


    Infrared light has a range of wavelengths, just like visible light has wavelengths..
    “Near infrared” light is closest in wavelength to visible light and “far infrared” is closer to the microwave region of the electromagnetic spectrum. The longer, far infrared wavelengths are about the size of a pin head and the shorter, near infrared ones are the size of cells, or are microscopic.

    OK so far.

    Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat!

    No we don’t, far infrared extends from ~15μm to 1mm wavelength

  427. Oliver Ramsay says:

    Myrrh says:
    March 5, 2011 at 9:01 pm

    “Go on, try it out for yourself. Get yourself a lightbulb which carries visible light but not heat and tell us how long it takes to cook your steak.. ”

    ———————————-
    I will be in the stores first thing in the morning looking for a heatless lightbulb and a steak.
    In the meantime, here’s an interesting experiment you can try.
    Take the battery out of your flashlight and put one finger on one end and another finger on the other. Take note of any sensation that you might have.
    Then, open your electrical panel and locate a double pole breaker. Put one finger on the red wire contact and one on the black. Record any sensation in any manner you like.
    Clearly, if they don’t feel the same they must be different phenomena.

    On second thoughts, don’t do the experiment, since we would hate to lose the most entertaining commenter on WUWT.
    I actually suspect that you are pulling our legs, but I often misjudge humour!

  428. Al Tekhasski says:

    Myrhh wrote: “Remember, Phil. and Ira, you’re arguing with that NASA page.”

    This NASA page is absolute, unconditional GARBAGE. I would like to know who was that ignorant dolt who wrote that “Shorter, near infrared waves are not hot at all – in fact you cannot even feel them. These shorter wavelengths are the ones used by your TV’s remote control.”
    By this token, excimer lasers (like KrF, operating at very short, 0.248um UV light) cannot possibly heat targets to millions degrees in studies of confined fusion.
    Or cannot directly etch diamond surfaces,
    (try “Excimer‐laser etching of diamond and hard carbon films by direct writing”)
    Or evaporate landmines and other targets:
    https://www.llnl.gov/str/April02/Dane.html
    https://www.llnl.gov/str/October02/Dane.html
    https://www.llnl.gov/str/October04/Rotter.html
    (these BTW use nearly the same light-emitting diodes as in your TV remote).

    You’ve been kidded by NASA dupes. I feel very sorry for you.

  429. wayne says:

    Myrrh, NASA removed the page your comment above points to.

    But I can imagine, very incomplete and thereby misleading. But unfortunately your view of visble not being able to heat is not correct. All E/M carries energy by E = hv that can heat surfaces or cause chemical reactions. There is a big however here. First visible is a very narrow band so its portion of the solar spectrum is rather small. Second is that in order for visible to warm something the visble radiation must be absorbed.

    I am light colored and so I would not feel much warmth in shorts on a cold day at all from intense visible only radiation but let’s say my friend of much darker color would say “Hey, thanks for that warmth, I was frreezing MAO.” But a large portion of the visible is reflected back to space so in that respect you could say the Earth doesn’t receive much from the visible range on land but oceans absorb most of it.

    Most UV is reflected high above by O3 but what does get through can burn the heck out of you, no matter of your skin color. And that leaves us with the IR.

    Just because your TV control is in the near IR doesn’t mean a strong source in that range would not be indetectable, a dark red burner is peaking about there. It would once again depend on any reflection and the surfaces emissivity for the absorptivity is just 1-emissivity, it must be absorbed to feel it. Would think water vapor mainly with I now know a bit of N2 and O2 would absorb in this portion of the spectrum.

    But you are right, a great potion from the sun that does get absorbed by land would be in the IR and the darker oceans would absorb a larger portion of the visible along with nearly 100% IR.

    Do keep this discussion in proper physics, it’s not a story but correct. I for one always want this to not wander into non-reality with word of always and never which usually both are wrong.

  430. Myrrh says:

    So they removed the page! Ha ha. Maybe they read WUWT… Ah well, you’ve got the important bits here and I’m sure some of you have already seen it so know I haven’t fibbed here. How very, very sad. Reminds me of when the American Meteriological Association website had a page up in its educational section which scuppered the idea of CO2 being capable of global warming, after a few days of increased traffic they took it off, someone had managed to slip it in. This was around the time they gave Hansen their highest award – http://climaterealists.com/index.php?id=2734

    The saddest part of this, is that those promoting AGW, whether in ignorance or in knowledge of the junk science it is, no matter how high up the pecking order they think they are, are just useful dupes for those raking in the real money. The goreygavins are just cannon fodder like the rest of us.

    Visible energy is not the same as energy carrying heat into organic matter. The more energetic shortwave are reflective, not absorptive, visible light bounces off surfaces, in intensity they can burn but do not penetrate to give us what we know as heat from the Sun; what actually warms the Earth is Thermal IR.

    This was bog standard simple science fact before AGW began confusing it all. However erudite and complex mathematically expressed justifications for AGWScience, it will always be junk because it does not relate to the real world in the real atmosphere we have here on Earth.

    The real world understands real science – the reason Thermal Infrared is used in therapies is because IT WORKS. Because Heat is its property.

    You can argue and nit pick all you like about what heat is, but for those living in the real world feeling the real heat from the Sun and having been educated before AGWScience mangled everything or are in the real world science now using this real knowledge about Thermal IR, your arguments appear ridiculous.

    What is so difficult to understand about such examples – http://www.therichwaybiomt.com/infrared-rays.html

    And the irony here is that NASA has been one of the leading infrared lights in infrared research…

  431. Robert Stevenson says:

    CO2 in the atmosphere could be doubled to 700ppm without any further contribution to global warming.
    At 288 K the black-body radiation or total emissive power from Earth is 391W/m^2 (124 Btu/h-ft^2). CO2 in the atmosphere absorbs 79.8W/m^2 after 3600m (CO2 absorptivity 0.184 at 350ppm or PcL of 1.24 m.atm). H2O in the atmosphere absorbs 248W/m^2 after 120m (H2O absorptivity 0.573 for a PwL of 2.77 m.atm).

    Doubling CO2 to 700ppm would absorb the same 79.8W/m^2 after only 2000m ( CO2 absorptivity 0.195 at 700ppm 0r PcL of 1.4 m.atm).

  432. Domenic says:

    to Phil.

    you wrote: “I haven’t a clue what you’re talking about, I’m talking about the science whereas you brought up your intuition.”

    Of course you don’t have a clue.

    You’re not talking about science. You are simply more like a non-thinking copycat. That isn’t science.

    You understand neither science nor intuition. And thus, you cannot see beyond your own nose. That is your problem.

    If you ever wish ‘to boldly go where no man has ever gone before’, you had better learn to use your intuition.

    The alternative, you see, is to go ‘no where’.

  433. Phil. says:

    Domenic says:
    March 6, 2011 at 8:04 am
    to Phil.

    you wrote: “I haven’t a clue what you’re talking about, I’m talking about the science whereas you brought up your intuition.”

    Of course you don’t have a clue.
    Quite, that was in response to this statement by wayne, seems like you left out the context for that too! Here it is:
    wayne says:
    March 5, 2011 at 7:41 pm
    “Pull that “out of context” again as you did to both Myrrh and myself and I will plead with Anthony to apply his rules of conduct.”

    I talk about the science (including collision induced absorption), give data etc., all you do is handwave and say that measurements haven’t been made when they have and talk about fictitious ‘cavities’ made from solid N2 at 63K (which have no relevance to the emissions of N2 itself).

    You’re not talking about science. You are simply more like a non-thinking copycat.
    You understand neither science nor intuition. And thus, you cannot see beyond your own nose. That is your problem.
    If you ever wish ‘to boldly go where no man has ever gone before’, you had better learn to use your intuition.

    The alternative, you see, is to go ‘no where’.

    Well you’re not going anywhere with the nonsense you’re spouting!
    This appears to be your contribution to the science we’re discussing here:
    “But I am pretty confident there will be detectable emissions in the .5 to 50 micron range from playing around with a .5 to 50 micron hand held detector many, many years ago, aiming it at the sky, the sun, etc and comparing the output to a detector using much narrower bands.”
    You also seem to think that a photon of a certain wavelength changes absorption characteristics depending on the temperature of its source.

    As Al said above (my emphasis):
    Al Tekhasski says:
    March 4, 2011 at 9:08 pm
    Before everybody start jumping on me, I’d like to correct myself. Energy fraction under Planck curve above 30um at 255K is somewhat larger than 0.02%. I pull this number out of my mind meaning that the N2-O2 spectrum corresponds to a blackbody with 100um peak, which would correspond to a body at about 40K with total emission of 45mW/m2. More, the entire region above 15um is already filled with CO2 and water over the top, see Ira’s spectrum above, so there is no way to see any N2 or O2 effect in practice.

  434. Tim Folkerts says:

    Robert Stevenson says: March 6, 2011 at 4:31 am

    CO2 in the atmosphere could be doubled to 700ppm without any further contribution to global warming.
    At 288 K the black-body radiation or total emissive power from Earth is 391W/m^2 (124 Btu/h-ft^2). CO2 in the atmosphere absorbs 79.8W/m^2 after 3600m (CO2 absorptivity 0.184 at 350ppm or PcL of 1.24 m.atm). H2O in the atmosphere absorbs 248W/m^2 after 120m (H2O absorptivity 0.573 for a PwL of 2.77 m.atm).

    Doubling CO2 to 700ppm would absorb the same 79.8W/m^2 after only 2000m ( CO2 absorptivity 0.195 at 700ppm 0r PcL of 1.4 m.atm).”

    There are TWO effects that contribute to warming from increased CO2.
    1) The “edges” of the absorption bands do not absorb so well (dropping to zero over some finite range of wavelengths. By increasing the CO2 concentration, more energy will be absorbed at these edges of the bands.
    2) The energy emitted to space is what actually balances the incoming ~390 energy. This is related to the emission from the TOP layer of CO2, not the bottom layer of CO2. With more CO2, the effective “Top of Atmosphere” will be higher, which means cooler, which means less outgoing IR. To return to equilibrium, the top layer would have to warm up. This would in turn warm all the layers below.

  435. Tim Folkerts says:
    March 6, 2011 at 12:54 pm
    … There are TWO effects that contribute to warming from increased CO2.
    1) The “edges” of the absorption bands do not absorb so well (dropping to zero over some finite range of wavelengths. By increasing the CO2 concentration, more energy will be absorbed at these edges of the bands.
    2) The energy emitted to space is what actually balances the incoming ~390 energy. This is related to the emission from the TOP layer of CO2, not the bottom layer of CO2. With more CO2, the effective “Top of Atmosphere” will be higher, which means cooler, which means less outgoing IR. To return to equilibrium, the top layer would have to warm up. This would in turn warm all the layers below.

    Well-stated Tim Folkerts! I agree with you and you have real science on your side.

    Just because we reject the CAGW kool aide does not mean we have to consume the Disbeliever DAGW drink of choice :^)

  436. Myrrh says:

    Phil. says:
    March 5, 2011 at 10:34 pm

    Actually you don’t understand it at all and apparently think that the Gas Laws, Stefan-Bolzmann, Fick’s Laws of Diffusion and the Second Law of Thermodynamics don’t apply to the Earth’s atmosphere!

    You really haven’t been following what I’ve been saying, have you?

    Re-read what I’ve written. That statement is strawman garbled nonsense, I haven’t said anything like that. And specifically, you have reversed my very important point that the 2nd Law is very much is relevant to our atmosphere, the FULL LAW. Not with the fudging of this imaginary “net” which includes colder warming hotter.. I’m arguing that the 2nd Law applies to each and every state of matter and energy, just as it’s written.

    That page doesn’t exist apparently, and the statement that “”Solar energy is not heating the earth is not what we feel as thermal energy”, is clearly nonsense. What is heating the Earth if not Solar energy?

    Re-read what I’ve written if you want to continue to engage with me on this, I really don’t have the time to go through all the argument again if you’re going to present me as saying what I haven’t. I’m not going through it all again, but will answer relevant questions or elaborate on what I’ve written. Just this once more for your question here:

    SOLAR (according to and as used by AGW and as depicted in graphics such as Ira’s above), is THE VISIBLE and the SHORTER WAVELENGTHS EITHER SIDE, i.e. VISIBLE PLUS UV AND NEAR INFRARED.

    THESE ARE LIGHT, NOT HEAT ENERGIES. SEE BELOW FOR NASA EXPLANATION. THEY DO NOT HEAT THE EARTH. THEY ARE NOT HOT. THEY ARE REFLECTIVE, I.E. THEY BOUNCE OFF SURFACES. THERMAL INFRARED WHICH WE FEEL AS HEAT FROM THE SUN ARE ABSORPTIVE, THEY PENETRATE AND HEAT ORGANIC MATTER, THE EARTH AND US.

    SINCE AGW (AS IRA HAS DEPICTED ABOVE) CLAIMS THAT SOLAR ENERGIES HEAT THE EARTH, AGWSCIENCE SHOWS ITSELF COMPLETELY IGNORANT ABOUT LIGHT AND HEAT, ABOUT THE PROPERTIES OF THE DIFFERENT WAVELENGTHS.

    AGWSCIENCE SAYS THAT IT IS SOLAR ENERGY IN, (WHICH IS VISIBLE AND THE UV AND NEAR INFRARED) AND THERMAL IR OUT. UTTER UNADULTERATED NONSENSE. THESE DOWNWELLING SOLAR ENERGIES ARE NOT HOT.

    Hope that’s clear.

    See below for news on that page.

    And re what was written on the NASA page: Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat!

    No we don’t, far infrared extends from ~15mu to 1mm wavelength

    Shrug. The NASA page is back. It has been updated and moved to: http://missionscience.nasa.gov.ems/07_infraredwaves.html but will be available on the URL I posted until May 31, 2011, at which point you’ll automatically be re-directed.

    So, still up on http:/science.hq.nasa.gov/kids/imagers/ems/infrared.html

    And it still says the above which I’ve put in bold which it then goes on to describe as the heat that we feel from sunlight etc.:

    Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared. NASA

    So, go argue with NASA. Because:

    It still says: Shorter, near infrared waves are not hot at all – in fact you cannot even feel them. These shorter wavelengths are ones used by your TV’s remote control. NASA

  437. Myrrh says:

    Will come back to this tomorrow, night night.

  438. Tim Folkerts says:

    Myrrh

    Nd YAG lasers that operate in the near IR (1.06 um) are used to burn thru things – hot enough to vaporize the materials they are cutting. How can they do this if the near IR is not “hot” and can’t heat things? I guarantee you would get burned if you put your hand in front of such a high-powered near IR laser. (And for surgery, lower powered lasers are used to focus IR and/or visible light to heat and destroy various cells.)

  439. Phil. says:

    Myrrh says:
    March 6, 2011 at 5:03 pm
    Phil. says:
    March 5, 2011 at 10:34 pm

    “Actually you don’t understand it at all and apparently think that the Gas Laws, Stefan-Bolzmann, Fick’s Laws of Diffusion and the Second Law of Thermodynamics don’t apply to the Earth’s atmosphere!”

    You really haven’t been following what I’ve been saying, have you?

    Yes I have, the trouble is that you don’t understand what you’re saying!
    You’ve claimed that the ideal gas laws don’t apply in the atmosphere:
    “Just as an ideal gas has no volume etc. and so cannot actually describe what is happening to a real gas which has volume etc.”
    ……..
    claimed that Stefan-Boltzmann doesn’t work (based on some SPPI post),
    further claimed that Fick’s Law doesn’t work, separation of CO2 out from the atmosphere is contrary to the 2nd Law, apparently you don’t know that!
    “just as Carbon Dioxide is treated as an ideal gas without volume etc. capable of doing impossible things like defying gravity to spread in the atmosphere without work being done even though its heavier than air because it bounces off all the other molecules in the atmosphere ‘diffusing’ as if an imaginary ideal gas in a jar.

    Re-read what I’ve written. That statement is strawman garbled nonsense, I haven’t said anything like that.

    So yes you have!

    And specifically, you have reversed my very important point that the 2nd Law is very much is relevant to our atmosphere, the FULL LAW. Not with the fudging of this imaginary “net” which includes colder warming hotter. I’m arguing that the 2nd Law applies to each and every state of matter and energy, just as it’s written.

    Given your statement about CO2 separating by its mass it’s clear you don’t know what the 2nd Law means.

    As for that NASA page for schoolchildren there’s a good reason they’re in the process of replacing it, it’s nonsense!

  440. Phil. says:

    Tim Folkerts says:
    March 6, 2011 at 6:13 pm
    Myrrh

    Nd YAG lasers that operate in the near IR (1.06 um) are used to burn thru things – hot enough to vaporize the materials they are cutting. How can they do this if the near IR is not “hot” and can’t heat things? I guarantee you would get burned if you put your hand in front of such a high-powered near IR laser. (And for surgery, lower powered lasers are used to focus IR and/or visible light to heat and destroy various cells.)

    I used a Nd:YAG laser at 1.06 μm for Laser Induced Incandescence of soot, the temperature of the soot got up to ~4500K, not bad for ‘cold light’!

  441. Myrrh says:

    Al Tekhasski says:
    March 5, 2011 at 11:46

    This NASA page is absolute, unconditional GARBAGE. I would like to know who was that ignorant dolt who wrote that “Shorter, near infrared waves are not hot all – in fact you cannot even feel them. These shorter wavelengths are the ones used by your TV’s remote control.”

    And to all responding on this theme:

    This is so funny/sad, we get some NASA pages with real science and because you’re all so used to AGWScience mangling, which technique misattributes affects, properties, laws and so on, you can’t understand it. This is bog standard REAL WORLD SCIENCE. Utilised in countless applications where the difference in properties is understood.

    You’re all going to have to go through a rather strange paradigm shift if you’re going to understand what I’m saying here. Visible Light is NOT HOT. You have been conned into thinking that because a wavelength is in a higher energy state, shorter wavelength, it means it has the ability to penetrate and heat things up. But these wavelengths are REFLECTIVE, and not thermal. Which means they have the property of bouncing off matter, they do not penetrate into matter as do the longer wavelengths of IR.

    http://www.johnthawley.com/journal/2010/10/14/learn-to-see-the-light.html

    For real world science example. Same with the shorter IR wavelength Near, it is REFLECTIVE. A Near IR camera picks up on the shorter reflective wavelengths of IR bouncing off the subject. Near IR penetrates deeper than Visible, UV penetrates less than Visible.

    By this token, excimer laser (like KrF, operating at very short, 0.248um UV light) cannot possibly heat targets to millions of degrees in studies of confined fusion. Or cannot directly etch diamond surfaces,..

    UV light does NOT PENETRATE organic matter to any great extent, it doesn’t even get through the layer of skin on our bodies. It might just make it through the first layer, the epidermis. Why? BECAUSE IT IS REFLECTIVE LIGHT. It is its high energy state which can burn, but it doesn’t penetrate to heat the matter to any real extent unless it is intensified as in a laser.

    Do you see the difference? Longer wavelengths of IR become PENETRATING, they penetrate organic matter and heat it up, spreading.

    You’re confusing high energy states for high heat penetration which is the property of longer less energetic waves, as above in Ira’s diagram, this has been misappropriated by AGW to describe the wrong wavelengths.

    Because thermal IR waveslengths are less ‘energetic’ does not mean they are less potent in heating matter, but more potent. The shorter which are absorbed by the outside layer of matter will burn if in greater intensity, but they do not heat up matter.

    So, what we have is Visible light which is not hot going into UV at one end and into IR at the other. UV is even more energetic than Visible and that intensity of energy like a high powered drill can burn the top layer of matter. We cannot feel it, it is not hot, it is not producing penetrating heat in us. We do not notice we are being burned.

    All light can be intensified as in laser use, even Visible colours if intensified can burn. But note, burn, not heat. The longer ‘less energetic’ waves of IR when intensified produce more HEAT.

    The property of Visible light and Near IR and UV is reflective, they bounce off matter, they do not penetrate. So, in real science speak, these energies are simply called Light and ‘light’ energies which create heat are called Heat.

    This knowledge is well known and understood in all areas of the real world science which utilises it. Like growing plants hydroponically using Light to help photosynthesis and not adding extra Heat, by using a lightbulb that produces minimal Heat IR.

    By using intensified longer IR waves to penetrate into bodies in various healing therapies by adding greater heat internally.

    So first of all you’ve got to junk the mistaken idea from AGW that higher energy states produce Heat and are capable of Heating the Earth. They can’t. They can’t penetrate deep enough to heat up organic Earth, but can burn the surface by UV which is a higher energy state than Visible Light. Visible Light is benign. We cannot feel it. It does not burn us as the higher energy state UV. Sure, you can take any colour of the Visible spectrum and intensify it artificially and it can then be used as a laser to burn, but will not be as efficient as using a UV laser.

    Please, do try to get your heads around what I’m saying here, the world is full of applications which understand the real difference between the Light and Heat energies.

    Ira must change his graphic or he continues to promote junk AGWScience in this.

    Light energies, Solar of Visible, UV and Near IR, DO NOT HEAT THE EARTH.

    Neither plants nor rocks nor we are being heated by Solar.

    Visible will be taken in by organic surfaces to an extent, what isn’t bounced off, and has benefits not only for plants which use a couple of the colours for photosynthesis, but for us in producing vitamin D, for examples. They are Light energies not Heat energies, they are benign, we cannot feel them, they do not heat the Earth. We need both Light energies and Heat energies for Life.

    We are being misdirected by AGWScience by this reversal of properties, and by the misdirection of using ‘laws’ such as Boltzmann and Planck out of context into thinking that these ‘prove’ that the more energetic shortwave are Heat energies, and, that because the ‘peak’ is in Visible it means that there is a greater amount of this energy heating the Earth, by misappropriating the word “most”.

    Take an ordinary old fashioned light bulb, 5% of its emitted energies are in the Visible Light energies, 95% in the Infrared Heat energies. Which then is “most”?

    It takes an immense amount of heat to produce the shorter wavelengths of Visible Light, meanwhile it has already been producing great amounts of Heat (Long wave IR) and heating further continues to produce more and more of it. How is our Sun any different in this principle? The longer wavelengths or IR are able to pass through water vapour and such more easily, where visible light is stopped by being reflected away, think mist, fog.

    So, are you now clear about the difference?

    Remember, the shorter wavelengths of Visible and UV and IR intensified, in such as lasers, will burn because they are not hot but reflective lights, intensified to a greater degree they will burn deeper, but they are not spreading that heat, they are not warming the matter. These can be fine tuned to enable surgery even on the eye. If that intensity of high energy UV was matched by its ability to spread the heat to warm matter up the whole head would be cooked in an instant..

    High energy does not mean high penetrative power. The longer wavelengths intensified spread Heat deeper and deeper into matter, cooking it. Near Infrared penetrates deeper than Visible, not so deep as longer wave IR. It is used in a variety of ways in the medical sciences, I’ve given examples previously above.

    NASA has done an immense amount of research on Light and Heat via its Space interests, and has a very good understanding of how Light and Heat energies of IR can heal internally. Even used in cancer treatment.

    http://www.nasa.gov/centers/marshall/multimedia/photos/2003/photos03-199.html

    At the moment I have a sick dog, I’ve put on an infrared heater for him to speed up his recovery and keep him warm.

    NASA through its Space programme has been a real world leading light in research of all kinds, we’ve all got items around the house which have come from this research and its work has inspired further research from others. It is such a great pity that it has been hi-jacked by the junk AGWScience, so be aware of that when you go to NASA sites. If you get it straight, untangle, the AGWScience from RealScience, you’ll get a lot more sense out of the NASA sites. The RealScience is amazing.

    You’ll also be able to spot more easily how AGW takes one concept or law or property and misdirects you into believing it belongs to something it doesn’t.

    So Wayne, I’m the one here trying to keep the discussion to proper physics.. I hope I’ve made myself clearer now.

    Phil. there’s a technique for seeing both sides of the argument, you have to follow what the other person is saying first..

  442. Myrrh says:

    aggghhhh! Please Mods would you put a close italics after the first paragraph which is quote?

    [Like that? Robt]

  443. Myrrh says:

    Links to NASA research on http://www.lightforpain.com/research-media.html

    Just bear in mind that this research into the healing power of IR is grounded in seriously looking at its effects in difficult medical conditions and maintaining the health of astronauts, don’t let ‘fantasy face’ distract you from that fact.

    But do your own search, you’ll find lots of links to the work being done and findings in the spread of this knowledge: NASA atronauts healing infrared

  444. Myrrh says:

    Just like that! Many thanks, at least now it looks like it makes sense…

  445. Phil. says:

    Myrrh says:
    March 7, 2011 at 3:13 am
    So Wayne, I’m the one here trying to keep the discussion to proper physics.. I hope I’ve made myself clearer now.

    Phil. there’s a technique for seeing both sides of the argument, you have to follow what the other person is saying first..

    Amazing, a century of science completely ignored, there should be a warning attached to this nonsense.

  446. wayne says:

    Myrrh, I now see where you are coming from.

    Thank goodness there are enough people of science like you who have learned it best to ignore that century old Arrhenius type “science” in this case. I approach it from the astrophysics-solar side and watched weekly what the sun was doing in the prior decades and that is over for at least for a time. I am now trying to find out exactly why all GHGs have no (well, very tiny) effect and appears in gravity, no kidding, how the atmosphere stays supported. That is where science is heading. I thank Miskolczi on the IR optical thickness and Spencer on population density daily for their great insight, they are the real scientists.

  447. Tim Folkerts says:

    Myrrh,

    I think you (and perhaps the rest of us too) are too caught up in labels.
    * All photons carry and energy of E = hf [or h(nu) in some notations]. There are no “hot” or “cold” photons per se. We can certainly say 1 photon of visible light has much more energy than 1 photon of thermal IR.
    * All photons of any energy can reflect and can get absorbed to varying degrees by different materials. Shiny metals absorb <5% of visible and IR. Glass tends to absorb Near IR, but not visible or thermal IR. Every material is different.
    * 4.18 J of absorbed energy of any wavelength will provide 1 calorie of heating to the object that absorbs that energy.

    The question becomes "How much energy is available in the photons and how much actually gets absorbed in important real world situations?"

    IR photons tend to get absorbed well by soil and plants and concrete (the emissivity for IR is close to 1 for many IR wavelengths). So IR with an intensity of 1 W/m^2 hitting 1 m^2 of ground will provide close to 1 W of power to the ground (warming the ground, or at least slowing the cooling). That fact increases the importance of IR photons.

    Visible photons reflect much better (which we can see easily with our eyes since we see the reflected photons). So visible light with an intensity of 1 W/m^2 hitting 1 m^2 of ground will provide considerably less than 1 W of power to the ground.

    For 1000 W of sunlight, perhaps 40% = 400 W is visible. Of this, if 10% gets absorbed and 90% gets reflected (which is a quite white surface like typing paper!) , that would provide 40W of energy. For sunlight, about 3% = 30 W is longer wavelength than 3 um. Even if all that is absorbed, it will only be 30W — about the same as the power from the visible.

    WAIT! I just realized the perfect simple experiment! Set out two pieces of paper — one white and one black. Both will absorb IR well, so that is the same for both. (in fact, according to this table http://www.omega.com/temperature/z/pdf/z088-089.pdf, flat white paint will absorb IR slightly BETTER than flat black!) The only difference is how much visible is absorbed. If the black paper gets warmer in sunlight than the white paper, the ONLY explanation is that the absorption of the visible light from the sunlight made the difference.

  448. Domenic says:

    To Myrrh

    You are pointing towards some interesting areas, BUT your language terms, choice of words, and their traditional accepted meaning by scientists, is a bit muddled.

    For example, reflection is not a property of light, or any electromagnetic wave ALONE.

    It is a property of the INTERACTION of light, or any electomagnetic wave, with specific matter: atoms, molecules, cells, etc.

    It’s a dance between two. Not a solo act.

    So when you say something like you just did

    “The property of Visible light and Near IR and UV is reflective, they bounce off matter, they do not penetrate.”,

    that is quite a bit nonsensical.

    There are lots of examples of matter that can absorb them.

    It depends on the matter they interact with. The dance.

    Whether they are absorbed, or reflected, or transmit through, depends on the matter they interact with AND the specific wavelengths of interest. The dance between the two.

    But you are correct in that there are HUGE holes in science, LOTS of misunderstanding of how that interaction works.

    The variations are far, far greater than most scientists imagine. That is why I keep on insisting that the only valid tests are those that reproduce as closely as possible, the actual real life conditions of whatever phenomena is to be tested.

    The real world is loaded with surprises. Always will be.

    NASA’s tests on plant life that you pointed me to are of much more interest to me than those of the ‘healing effects’ on humans. The ‘placebo effect’ when working with humans and healing is very tricky to deal with in such tests. With plants, less so, of course. Plants are more indicative of overall natural processes.

    Now, theoretically, when a plant absorbs very certain wavelengths of energy, it need not produce any ‘heat’. Thus there is no increase in temperature whatsoever to detect, as a byproduct of the interaction. So, those wavelengths are not re-radiated back into the atmosphere contributing to any ‘greenhouse effect’. If those wavelengths are truly captured energy, they are simply converted into matter: new cells, new molecules, etc. with no waste heat. It’s simply the reverse of E = mc2. (I know, I know… the chemists are going to scream that it’s just chemical bonding energy, etc, etc…but that’s still a part of E = mc2.)

    So, I agree with you that there are many, many things that go on