Slaying a greenhouse dragon

by Judith Curry

On the Pierrehumbert thread, I stated:

So, if you have followed the Climate Etc. threads, the numerous threads on this topic at Scienceofdoom, and read Pierrehumbert’s article, is anyone still unconvinced about the Tyndall gas effect and its role in maintaining planetary temperatures?   I’ve read Slaying the Sky Dragon and originally intended a rubuttal, but it would be too overwhelming to attempt this and probably pointless.

I was hoping to put to rest any skeptical debate about the basic physics of gaseous infrared radiative transfer.  There are plenty of things to be skeptical about, but IMO this isn’t one of them.

Well, my statement has riled the authors of Slaying the Sky Dragon.   I have been involved in extensive email discussion with the authors plus an additional 10 or so other individuals (skeptics).  Several of these individuals  on John O’Sullivan’s email list actually agree with my assessment, even though they regard themselves as staunch AGW skeptics.

One of the authors, Claes Johnson, along with John O’Sullivan, expects a serious critique from the climate community.  Johnson says he intends to submit his papers to a peer reviewed journal.  I agreed to host a discussion on Johnson’s chapters at Climate Etc., provided that the publishers of Slaying the Sky Dragon would make Johnson’s chapters publicly available on their website (which they have).

Johnson’s first chapter is entitled “Climate Thermodynamics,” which presents an energy budget for the earth and its atmosphere that does not include infrared radiation.   The second chapter is entitled “Computational Black Body Radiation,”  which seeks to overturn the last 100 years of modern physics  and concludes that “back radiation is unphysical.”

For background info:

  • Claes Johnson’s website is here
  • Johnson’s blog is here, see specifically these posts ( here and here)
  • John O’Sullivan’s advert for the debate at Climate Etc. (note Monckton and Costella are in  my “corner” in criticizing the book and Johnson’s chapters).

I suspect that many undergrad physics or atmospheric science majors at Georgia Tech could effectively refute these chapters.  I’m opening up this discussion at Climate Etc. since

  • the Denizens seem to like threads on greenhouse physics
  • I’m hoping we can slay the greenhouse dragon that is trying to refute the Tyndall gas effect once and for all.

It will be interesting to see how this goes.  Claes Johnson has said that he will participate in the discussion.

Note: this is a technical thread, please keep your comments focused on Johnson’s arguments, or other aspects of Slaying the Sky Dragon.   General comments about the greenhouse effect should continue on the Pierrehumbert thread.

2,518 responses to “Slaying a greenhouse dragon

  1. It’s an interesting concept, that an atom cannot absorb (but only reflect) incoming EM at a cooler temp than its own blackbody emission temp at that instant. No idea if it’s true. My layman’s understanding of the thermodynamics constraint was just that it described net transfer, which must always be from hot to cold.

  2. I have to agree with omnologos on this. By inferring that all those skeptical of the man-made global warming meme (some, like us, skeptical of the greenhouse gas theory, itself) are supposed to be seeking a unified front as if we are a political or military force is, frankly, absurd.
    We prefer to leave ambitions to claim a consensus to the post-normal science green brigade; they appear to have abandoned the traditional tenets of the scientific method. Consensus is utterly meaningless- being proven right is the goal even when the so-called ‘consensus’ is adamant we are wrong.
    The statement, “I suspect that many undergrad physics or atmospheric science majors at Georgia Tech could effectively refute these chapters” is so funny coming from someone who is “too busy” to do what she infers is such a basic task, herself.

    • Well, I mainly found it interesting that a number of people on your self selected email list were highly critical of the book and Johnson’s chapters. Your email list does not begin to reflect the broader range of skeptical opinions.

      • Judy, I intentionally invited to participate those who I knew to have contrary views . This is the whole point of debate isn’t it? Let’s see some actual analysis please rather than insults and hand waving so far displayed by those made uncomfortable by what the book presents.

    • Thank you for mentioning me John as my comment has been snipped out. Perhaps I should be glad it deserved that much of an attention.

  3. Hi judith,

    I was positively surprised by the first chapter, which correspond to the mental model I have formed about GH effect, but I do not really see where it is in conflict with mainstream view nor why it is independent of infrared radiation. On the contrary, it explicitely agree with mainstream view, that is that TOA is variable in height and the higher the more GH gaz is present. The only thing it add is that lapse rate below TOA is related to thermodynamic and not radiation, and that lapse rate can vary with humidity and thus is a potential feedback (negative feedback). Up to here, I perfectly agree, appart that one should mention that it is an approximative model because all radiation does not happen a a precise TOA height, but that TOA is an average concept, the atmosphere is not perfectly IR opaque and then IR transparent, it is semi-transparent so radiation is a diffuse process and all radiation occuring at TOA is only a (usefull?) approximation.
    At this point, the model does not allow to predict the change of T_ground when CO2 is doubled, what would be needed is the change in TOA from CO2 doubling, and the various H2O feedbacks (on TOA itself, and on lapse rate). Still, this model seems to me much more useful and closer to reality than pure radiative model with an IR opaque shell-like atmosphere concentrated at TOA, and the (negative) feedback of H2O on lapse rate seems perfectly valid (and not mentioned explicitely on previous GH accounts I have read).

    This first chapter does not ring any physical alert bells though, so I guess reading the rest makes sense, and I am for now positively surprised by “Slaying…”…

    • oups, forgot to say: read the Pierrehumbert thread where I attempt to expose the mental model I built about GH effect. Done that only from the various GH threads here, at wuwt and rc, not from the “Slaying….” chapters….So u see why this first chapter was appealing to me :-)

      • ouch, started reading second chapter about blackbody…yikes, this one is definitely in crackpot territory, so “Slayer….” is a kind of mixed bag imho, if most chapter are long the first one, it is worthy, else (or if conclusion hold only if all chapters are true), then it will easily debunked…

      • Kai, the first chapter rests on the result of the 2nd chapter (they are both written by Claes Johnson), i.e. there is no back radiation and atmospheric infrared radiative transfer is not important in the earth’s energy balance. So if Ch 2 is crackpot, then Ch 1 is also.

      • Dr Curry,
        To help the readers understand, please:
        1. elaborate your definition of back radiation and your concept of it,
        2. explain whats wrong with ” atmospheric infrared radiative transfer is not important in the earth’s energy balance”.

        The Earth’s mass is so huge compared with atmospheric mass. The Earth’s IR energy emitted is so huge as compared with atmospheric absorption of IR energy. Will you care to do a comparison? Why is NASA’s radiation energy balance for K-12 incorrect?

      • Sam, I essentially agree with Pierrehumbert’s essay on this topic, see the previous pierrehumbert thread

      • Dr. Curry,
        I admire your tactics of diverting your GT students’ attentions for avoiding direct answers to direct questions soon I found the Figure 1 model there is not a true representation of the atmosphere radiation transfer, namely, lack of the cloud radiation transfer and lack of layers direct radiation transfer to the Earth surface.

      • kai,
        “… this one is definitely in crackpot territory”. This is very unrespectful to an an author who try to sort out radiation misconceptions, care to elaborate?

      • Is he really trying to sort out radiation misconceptions?

        Whether he tries it or not, the result leads to think the opposite. The book in, which the article appears is definitely trying to increase misconceptions.

      • Its easy to make a generalized comment. I find generalised comments do increase misconceptions. Will you be more specific, such as list them out item by item, concept by concept, misconception by misconception, page by page? Doing it this way helps the readers understand your points of view.

      • Sam
        I have done that kind of commenting in tens of messages. Repeating similar statements hundred or thousand times more, is not going to stop requests like yours.

        When you stop commenting, there will always be a new participant, who starts from the beginning again. That will go on as long as this site is active.

  4. Dr. Curry,
    I am sorry you felt obliged to give so much space to the ‘Dragon’ book.
    But if anything, it will unify skeptics by giving many something to agree on that fails as a skeptical case.
    I see this book as sort of a left hand paranthesis to Hansen’s Venus-ization of Earth as a right hand paranthesis, expressing clear markers where wishful thinking has taken over.

  5. Judy: I do not say that radiative transfer plays no role in climate. It would be
    helpful for the debate if you woul read what I write and not freely invent crackpot themes.

    • Well yes, you admit to solar radiation and black body radiation. But your treatment in the first chapter completely omits atmospheric gases (and cloud) infrared radiative transfer (and includes that ludicrously incorrect diagram from more than 10 years ago that somehow continues to exist on a NASA web site).

      • Having said that Johnson is wrong, I’d like to point out that his first chapter on climate thermodynamics – emphasizing heat transport by convection and evaporation, but not including radiation from the atmosphere, is no more wrong than Pierrehumbert’s article which does the opposite – making the incorrect claim that the surface temperature can be determined by calculation that only includes radiation, ignoring convection and evaporation.

        And yet of these two incorrect articles, Judith refers to one as “excellent” but says that the other could be refuted by undergraduates. I wonder if these same undergraduates could refute the Pierrehumbert article? I expect most could not, because the new generation of students are being brainwashed in the same way, for example by the GaTech course “EAS8803 – Atmospheric Radiative Transfer”. I note that the blurb for this course says that “Topics to be covered include the radiative balance at the surface”. I do hope that you have some students bright enough to realise that there is no radiative balance at the surface, and that one day this fact will dawn on those who design and teach the course.

      • Sorry, posted this in the wrong place in the thread.

      • Why NASA did not correct it and misled the general public for over 10 years with that incorrect diagram? Or NASA is incapable of understanding the subject of radiation? Or under the authority of James Hansen, no one in NASA dare to correct it?

      • This diagram apparently first appeared in a doc designed for K-12 education. The names Eric Barron (currently president of Florida State University) and John Theon were on the doc (back when theon was still employed at NASA and Barron was at Penn State, which places it in the mid 90’s). But I assume this diagram was drawn by a staff person, and Barron didn’t pay close attention. That is the only way I can explain this. Somehow John O’Sullivan spotted this (or at least publicized this). And it sits on a web site to the present day. In spite of my contacting several people about this. The bottom line is that there is too much form and not enough substance oversight on public communication documents (as opposed to satellite data quality issues, where there is a lot of oversight and checks and balance in place at NASA).

      • Over the 10 years, this diagram has misled the K-12 students, the teachers, the politicians and the world who visited the NASA site. This is a serious American educational flaw that NASA, Eric Barron and John Theon should be informed to correct the diagram or delete from the NASA website and owe the American Education and the world an apology.

        If you have not asked them to correct it, please do as an educator at the Georgia Tech.

  6. Ok fine then Judy: You don’t like the Kiehl-Trenberth diagram. So what is then wrong with it, as you see it? Maybe we share some insights?

    • I’m not clear which diagram you are discussing here. If it is Fig 5 of Chapter 2 of Johnson then it does closely resemble Fig 7 of Kiehl and Trenberth 1997. If the latter was ‘ludicrously incorrect’ then it was still given pride of place 10 years later (with added colour but no other changes apart from the caption) in IPCC AR4 WG1 Chapter 1, p 96 (2007). But I thought Dr Curry was referring to Fig 4 of Ch 1 of Johnson, which is also attributed to NASA, but which differs from the Ch 2 version in not showing any downward long-wave radiation. Is that also derived from K & T?

  7. Judy: You say that “I suspect that many undergrad physics or atmospheric science majors at Georgia Tech could effectively refute these chapters”.

    I suggest that you actually try this as a take home exam for your students.
    From your teaching they will understand that Kiehl-Trenberth is wrong
    but maybe they will find something they think is right. Go ahead!

  8. Apart from an over-indulgence in post-modern civility, the chapter on Climate Thermodynamics pursues the misconceptions underlying current AGW theory. A helpful touchstone for pdf files is a scan for the word equilibrium where used to describe what physical science calls steady states. I find three such instances in this chapter, all wrt the adiabatic lapse rate. Equilibrium states have no net fluxes of matter or energy entering or leaving. (Canonical ensembles allow fluctuations.) Equilibrium profiles are isothermal and the adiabat is not. Steady states require external fluxes to prevent them from relaxing to equilibria. The alert student should now be asking, how do I determine this flux needed to maintain an adiabatic profile?

    With CO2 doubling, one typically calculates a 2% flux reduction and then presumes a 2% increase in the thermodynamic potential difference (1/T) is needed to restore the flux level. Thus, given a 65K tropospheric differential, 1.3K. An alternative interpretation is that adding CO2 increases the resistivity of the troposphere, just as traces of phosphorus disproportionately increase the resistivity of a copper wire. Thermodynamics asks, what change in potential is required to restore the original rate of dissipation of free energy? In high school we learned the expression E^2/R, albeit in a different guise. Ergo, only a 1% potential change now compensates a 2% resistivity change to restore energy balance.

    When our student resolves the difference in these solutions, he should be able to answer his earlier question. Perhaps herein lies Sommerfeld’s dilemma – thermodynamics is not the intuitively obvious subject it may superficially appear. To paraphrase yet another quotation, ” …, and you’re no Arnold Sommerfeld.”

    • Is the presumption of a 2% change in (1/T) tied to a 2% change in flux found in textbooks, and generally accepted in the climate change literature?
      If so, then the generally accepted value for climate sensitivity is a factor of 4 too large. The reason is that the Stefan-Boltzmann law says j is proportional to T^4. Taking the derivative of both sides with respect to T, and then dividing both sides by the Stefan-Boltzmann law, and rearranging shows that the % change in T will be 1/4 times the % change in flux. Because 1/T contains T^1, the % change in (1/T) will also be 1/4 times the % change in flux. You and all other knowledgeable bloggers are asked to comment on and make any corrections to my calculations found at posted on Feb. 7 at 7:44 pm.

  9. Well, of course Johnson is wrong. It is perhaps instructive and useful to try to explain why. In the ‘blackbody’ chapter he seems to think that a warm body can warm a cooler one but not a warmer one. He says at one point (sadly no page numbers) that there is two-way propagation of waves, but only one-way propagation of energy. How does that work? Are there two types of EM wave, one transporting energy and one not?! We can also ask him this : an isolated backbody is radiating into a vacuum. Then a warmer body is brought in. How does the first body ‘know’ to stop radiating energy in that particular direction?

    Later on he tries to use equations – but his equation (4) is just wrong. Where does this equation come from? What is u supposed to represent? Why is radiation given by the third time derivative of u?

  10. The email debate of last week was the first geniune airing of the flawed Physics of AGW in all history. The fundamental flaws are explained in “OMG….Maximum CO2 Will Warm Will Warm Earth for 20 Milliseconds” posted at and at the website. Surprising that the truth was hidden in plain sight for so long. Since the show is now over, I felt it necessary to add one final comment “Climate Follies Encore” which explains the post 20 Millisecond exchanges. This has been the greatest education process, for the wisest among us, and we will now share.

    My chapter includes over 100 pages of footnotes and is supported by 60 articles in archive and Canada Free Press. We share a glorious future of truth. My thanks to Judy for enduring my repeated, well meaning barbs for over a year now. (co-author of SSD)

  11. To PaulM: You have not read and understood my argument: I present a differential equation modeling two-way wave propagation combined with
    radiation and with a dissipative effect making the energy transfer one-way,
    from higher to lower frequency. If you don’t like this equation, give me one you think is a better model. Just words is to diffuse to discuss.

    • Perhaps you should start by reading a standard text book on Radiation Heat Transfer and then move on to some papers (H C Hottel would be a good start) and learn about the subject rather than propose some wild theory? The fact of ‘backradiation’ has been well tested in many situations, furnaces, radiation shields for thermocouples etc., let’s see you apply your theory to such situations and see how it works?

      • Please define back radiation which confuses me even though I had written something about it. To me, back radiation is reflection from the back with wall or relective radiation. A thermocouple when placed at the center of a pipe gain heat from the flowing media as well as radiation directly from the wall concentrated at the thermocouple measured an erronous fluid temperature. With such a wall you get radiation concentration. Without a wall, the radiation is minmal.

        Similar analogy for the greenhouse situation, greenhouse has glass or sheets of clear plastics to trap most IR, without this layer of wall, no trapping of IR and hence no greenhouse. It is obvious.

        2 black bodies at different temperatures, they all emit IR with the resultant energy flow from the hotter to the colder in a free radiation condition. The colder can have an extremely small effect of slowing down cooling of the hotter unless a back wall from the colder reflect the radiation to other directions are reflected by the back wall.

        I have not read Mr. Claes Johnson’s article about the radiation. I will assume he is mostly correct in a free field radiation as in most climate situations. There is no back radiation. Furnace, thermocouples etc cases are not free field radiation cases which involved walls of reflecting radiations. Radiation involves walls of reflection has back radiation.

  12. curryja | January 31, 2011 at 8:31 am
    ” (and includes that ludicrously incorrect diagram from more than 10 years ago that somehow continues to exist on a NASA web site). ”

    As an “update” how about this diagram and text on Wikipedia,
    it appears a little more recent. The text does not appear “improved” either.
    So, no real changes to it appear warranted according to AGW.

    Maybe you could explain what makes the old one and the “new” one “ludicrously incorrect”,
    that might help in discussing what the “slayers” are showing, saying, suggesting, and raising for discusion.
    Heck, we might even get to a better understanding of where the science actually is at present.

  13. PS to PaulM: I start from the same equation as Planck did 100 years ago, but combine with finite precision computation instead of freely invented ad hoc
    statistics. Statistics is not physics, just imagination, and physical particles have little imagination.

    • “Statistics is not physics, just imagination, and physical particles have little imagination”.
      How dismiss 200 years of thermodynamics and physical statistics, with the only clue of a single metaphor: the “not-thinking” particle. Funny (what about: Einstein debunked, there is no light speed maximum: photons don’t care about cops and driving speed limits ?).

      Anything else more substantial, perhaps?

  14. Read the second chapter – it’s goofy, not physics. The initial clain to get rid of wave particle duality pretty much floored me, since this aspect has been very well experimentally shown. To accept this assertion means ignoring what you can see with your own eyes (and instrumentation) in a laboratory. A fatal flaw is confusing net energy flow with absolute energy flow – this is in the black body discussion. To say that a colder black body can’t radiate to a warmer black body (he calls this “back radiation”) is beyond ridiculous.

    Basically, he presents a circular argument without proving his ridiculous premise, throws a bunch of jibberish (maybe not jibberish, but I don’t call it physics) in the middle to make it all seem scientific, and then returns to his unproven assertion that a colder body doesn’t have black body radiation in the direction of a warmer body. Thus, besides claiming no one knows the nature of a photon (as part of an argument against the traditional treatment of blackbody radiation – yet single photon experiments have been run for decades) , he negates the Superposition Principle and relies on some mysterious instantaneous knowledge existing in one body about the temperature and direction of all other bodies in the universe. I think the spook guys would love to have this type of instantaneous directional communication device in their hands.

    Just to make it more clear, suppose you have two black bodies at different temperatures facing each other, with a shutter over each blocking all radiation. Remove the shutter in front of the colder body an very short time before removing the shutter in front of the hotter one. Then initially radiation would flow from the colder one toward the hotter one, and then reverse direction when the second shutter is opened.

    • The question is, can a cold body make a warm body hotter? Everything with mass and a temperature radiates. Who disputes that?

      Imagine a hot body (with an internal or external heat source) and a passive body floating in the vacuum of space. Can the passive body make the hot body hotter? Imagine the passive body gets closer to the hotter…it will absorb more radiation, right? It will get warmer. If there was such a thing as back-radiation heating, then the hot body gets more of it back. Then the bodies get so close together…that they touch. Now the radiation effect is greatly magnified (whatever radiation can do, conduction does much better).
      Does the hot body, at any time during this process, ever get hotter?
      Radiation from a passive source cannot make a hot body hotter.

      • Radiation from a passive source cannot make a hot body hotter.

        It certainly can, put a thermocouple in a flame and you’ll measure a certain temperature which is lower than the surrounding flame because of conductive losses down the wire and radiative losses to the surroundings. Surround the ThC with a silica tube and the temperature measured will increase due to radiation from the cooler tube.
        Check out ‘Suction Pyrometers’:

      • I can’t tell if you’re kidding, Phil.
        Transport your experiment into space so we can focus only on radiation effects. Then replace the flame heat source with a resistive one so it will work in a vacuum. Now, tell me how the passive thermocouple can increase the temperature of the heated body. The only thing it can do is cool the heated body…at various rates and with varying degrees of coupling, sure. But, under no condition can it make the heated body hotter. The passive body is never a source of heating for the source. Never.
        Now, what does that tell you about Trenberth and Keihl’s energy balance schematic? The earth’s surface is heated by back radiation from passive CO2 and water vapor?

      • I never kid, your complete failure at understanding the applicable physics, lack of reading comprehension and refusal to read the cited material makes responding to you a complete waste of time!

      • The topic is radiation, Phil, the supposed mechanism for global warming caused by increasing CO2 in our atmosphere. You love to talk about conduction and convection as if I don’t understand these concepts, but that is a hand-waving distraction. Focus, Phil. We’re talking about radiation…and how a passive body can heat a body with a heat source. I know how a passive body can cool a hot body…let us count the ways. Your GHG theory depends on passive materials heating hot materials.
        What are you going to do with radiation, Phil. Store it? Delay its transit time to space? You can reflect it, diffuse it, deflect it or focus it. You can’t store it or “back radiate” it to make a warm surface warmer.

      • You can’t store it or “back radiate” it to make a warm surface warmer

        It reradiates in all directions – the use of “back” is arbitrary and capricious, and assumes the location of another black body is somehow important. To be correct in what you say, it would have to stop radiating in a particular direction just because there is a black body in that direction – that’s ridiculous.

        You’re confusing net heat flow with absolute heat flow. A hot black body is in fact warmer if there is a cooler black body radiating toward it, simple because net heat flow is less.

        Dr Curry may confirm that I’m a definite skeptic, but I’m also a physicist and the linked chapter 2 and the posts here based on it are not even close to reasoned.

      • Reading comprehension still lacking I see!

        “radiative losses to the surroundings. Surround the ThC with a silica tube and the temperature measured will increase due to radiation from the cooler tube.”
        Missed this did you?
        And this, the first sentences in the cited reference:
        “When a bare thermocouple is introduced into a flame for the measurement of gas temperature, errors arise due to the radiative exchange between the thermocouple and its surroundings. In the standard suction pyrometers a platinum-rhodium thermocouple, protected from chemical attack by a sintered alumina sheath, is surrounded by two concentric radiation shields.”
        Yes Ken we are talking about radiation but unfortunately you don’t understand it.

      • If you are checking things out: try the 2nd la of Thermodynamics.


      • Phil,

        apparently you are confused by the slowing of a flux as you are not actually measuring the temp of the hot body, only the heated body, the thermocouple.

  15. The atmosphere is in thermodynamic equilibrium. There are slight variations which are caused by certain cyclical processes which the proponents of AGW mostly refuse to accept.

    CO2 concentration is not one of them. John Tyndall did not prove a damn thing about CO2 absorption. His equipment was far too primitive to distinguish between absorption, reflection, refraction, diffusion, scattering or anything else. He incorrectly concluded that all energy missing between the source and the pile in his half baked experiments had been absorbed by CO2. Above all he ignored Kirchhoff’s law.

    The conservation of energy falsifies the “greenhouse effect” because as per Kirchhoff’s law that which absorbs, equally emits. This fact is absent from Tyndall’s ramblings and exposes him for what he was.

    Nothing traps in heat, quote:

    “All matter–animate or inanimate, liquid, solid, or gas–constantly exchanges thermal energy in the form of electromagnetic radiation with its surroundings. If there is a temperature difference between the object in question and its surroundings, there will be a net energy transfer in the form of heat; a colder object will be warmed at the expense of its surroundings, a warmer object cooled. And if the object in question is at the same temperature as its surrounding, the net radiation energy exchange will be zero.”

    “In either case, the characteristic spectrum of the radiation depends on the object and its surroundings’ absolute temperatures. The topic of radiation thermometry for example, or more generally, non-contact temperature measurement, involves taking advantage of this radiation dependence on temperature to measure the temperature of objects and masses without the need for direct contact.”

    “The development of the mathematical relationships to describe radiation were a major step in the development of modern radiation thermometry theory. The ability to quantify radiant energy comes, appropriately enough, from Planck’s quantum theory.”

    According to Kirchhoff’s Law any substance which absorbs energy will equally emit that energy. CO2 has a lower specific heat capacity to O2 and N2. The atmosphere which is 99% N2 and O2 is in relative equilibrium. Therefore adding more CO2 at trace amounts to the atmosphere will simply force the CO2, with its lower specific heat capacity, into equilibrium with the rest of the atmosphere. The higher the concentration of CO2 the lower the overall atmospheric temperature will become.

    “A simple reproducible experiment”

    “Specific Heat Capacity of Gases”

    AGW theory requires that we suspend our knowledge of this obvious fact and accept that it is the 0.0385% CO2 which forces the other 99% of the atmosphere into equilibrium with itself.

    It is the same logic as claiming that by taking a pee in the ocean, you have warmed the ocean. When in fact your pee has been chilled by the ocean. It’s called semantics.

    It is interesting that Judith has played the appeal to authority card. It is also interesting that those who appeal to the authority of Tyndall and the RS (7GT/1Gt human v’s natural CO2!) fail to acknowledge that they are relying on primitive out of date 150 year old “science” which has not even been critically re-examined.

    Anyone who quotes John Tyndall as the man who proved the “physics” of the “greenhouse effect” displays nothing short of sheer ignorance. It is the ultimate in the bogus appeal to authority. John Tyndall was fool and a fraud. Above all he was an insider at the Royal Society. Tyndall’s experiments have as much value as Sir Paul Nurse’s implication in his recent Horizon “program” that natural processes account for 1 Gt CO2 while humans account for 7 Gt CO2, i.e. NONE.

    So can we quickly dispense with the pseudo science of John Tyndall and get back to reality? That appeal to authority was for yesterday’s people, those who had faith in the integrity of science, scientists and trust in the Royal Society.

    Those people are long gone. (Last seen heading south on highway 51 with Trust in the passenger seat and Faith at the wheel!)

    • You can see why I am personally not taking this on in any detail, it is just endless. You incorrectly state Kichoff’s Law. αλ = ελ, where Lambda should be a subscript. it says that at a particular wavelength, the fractional absorptivity equals the fractional emissivity, where the fractional part is relative to the intensity of black body radiation at that wavelength. So if an oxygen molecule at temperature 200K receives a bunch of solar radiation in the ultraviolet bands, it will also emit in the ultraviolet bands, but because the oxygen molecule is relatively cold, there is almost no actual energy emitted by an oxygen molecule with temperature of 200K. Your next sentence is a mistaken interpretation of very basic elements of the kinetic theory of gases. And on and on . . .

      My point in not rebutting all this personally is that I would need to spend an hour on each incorrect sentence to try to educate people that don’t already understand this. Roy Spencer and scienceofdoom have already tried. And there are hundreds of such sentences to rebuke.

      • I fully understand agree on your point on this issue.

        I can not understand your thoughts/position on “post normal science” and why “climate scientists” opinions should be given an preference in regards to governmental policy.

      • when did I EVER say that scientists should be given a preference in regards to governmental policy? I have been very actively fighting against that!

      • Then I have misunderstood your thoughts and the meaning applied to post normal science.

      • yes, that is my great frustration.

      • Post Normal Science, or Special Pleading?

      • I have to admit to misunderstanding it too, in that case…

      • Judy,

        The tail does not wag the dog.

        E in = E out.

        “My point in not rebutting all this personally is that I would need to spend an hour on each incorrect sentence to try to educate people that don’t already understand this. Roy Spencer and scienceofdoom have already tried. And there are hundreds of such sentences to rebuke.”

        DITO darling.

      • no, energy in does not equal energy out.

      • Yes it does!

      • No, energy in does not necessarily HAVE TO EQUAL, on virtually any time scale equal energy out.

        Does energy never get used?

        Does energy never get taken out of the system “permanently”?

        Of the energy taken out of the system, what determines when it is put back into the system, and how, as what?

      • Derek,

        Please consider the principal of the “conservation of energy”.

        “Does energy never get used?”

        I believe “converted” is the word you are looking for.

        “Does energy never get taken out of the system “permanently”?

        Of the energy taken out of the system, what determines when it is put back into the system, and how, as what?”

        Sorry, NO COMPRENDE ? ? ?

        Taken out by what Derek ?

      • Will,

        Sorry, NO COMPRENDE ? ? ?

        K&T “timescales”, do NOT compute Will.
        Hence “permanently”…, sedimentary rocks.

        Re “converted” – does that mean “some” will never be returned to escape from the “system” as heat (energy lost to space)?
        “Life”, and “work done”, being the obvious examples.

      • Derek I have made the point about the energy that does not leave the system in my paper here:

        I think you know what I mean when I say E in = E out.

        Apart from hair splitting, do you actually have a point?

      • E in = E out.

        That’s an equilibrium condition, Will. Given that we’ve been doubling the amount of CO2 we add to the atmosphere every three or four decades for the past century or more, we are nowhere near equilibrium. Nor will we be until (a) we hold constant the amount of CO2 we add each year and (b) nature catches up. Expect (b) to happen roughly three decades after (a). But don’t expect (a) to happen until we can no longer afford fossil fuel. And at that point (a) won’t happen anyway since our CO2 production will decline thereafter rather than holding steady.

        David Archer believes things will remain hot thereafter. I disagree: I believe that after we stop emitting CO2 the temperature will plummet even faster than it has been rising due to the way equilibrium works. Conceivably by 2150 we’ll be in a Younger Dryas type ice age, though at that point we’ll surely have figured out some way of preventing that.

      • Will, earths atmosphere is not a closed system. The sun adds energy to it, which it then rediates to open space. Hence E in = E stuck in system + E out.

      • That is probably the most enlightening statement in this thread!!!!

      • Ein – Eout = delta global energy storage (mostly as heat in the oceans and atmosphere)- only at top of atmosphere

      • Energy is also stored with chemical changes, plant growths, animal grows … and stored energy dissipated thru plant deaths, fossil fuel uses …

  16. To Harold: Yes physics is very goofy, in particular particle statistics physics.
    I start from the same wave equation as Planck and use a finite precision dissipative effect instead of jibberishy statistics. So what I do is less spooky
    than what you hint at. It is remarkable that in a discussion about the “greenhouse effect” physicists have nothing to say. To me it is a physical phenomenon that physicists should be able to grasp, but it seems they don’t.

    • I guess you didn’t get it. I’m a trained Physicist, now retired. I was pretty sure I just said something about the greenhouse effect, and particularly pointed how a simple thought experiment shows how wrong your theory is. I don’t intend to try to convince you, but my thought experiment should convince almost any reasonable idiot your theory is wrong.

      I don’t use different standards for either side of AGW. I have fairly rigorous standards,, which you have failed, and most ot the AGW papers also fail my standards. Sloppy work on the AGW’s crowd’s part doesn’t excuse sloppy work on the anti-AGW’s side. As for dragging Tyndall into the discussion and how the physics hasn’t been looked at, try reading some of Dr. Earl W. McDaniel’s and others’ books from decades ago on details of atmospheric excitation and radiation.

      Dr. Curry – FYI, Dr. McDaniel taught Physics at Georgia Tech, and had a great sense of humor.

  17. “Statistics is not physics, just imagination, and physical particles have little imagination.”

    If there’s a more rigorous, more unchallengable, more awe-inspiring development in physics than the development of statistical thermodynamics I am not aware of it.

  18. What are the main results of statistical thermodynamics with some form of informative content?

    • Claes
      This is a very interesting thread. If it wouldn’t be too much trouble, would you mind using the ‘reply’ button to respond to comments. It can be found next to the name and date/time of the person that you are replying to. It positions your response at the correct point in the blog and makes it easier for us lurkers to follow the argument. Many thanks.

  19. Having read just the first couple pages of the second chapter, I know this argument is going to ‘creative’.

    The first argument is rather interesting. Blackbody radiators absorb all frequencies of light (definition of ‘black’), but only emit radiation with a specific spectrum determined by the body’s temperature. I think that’s fairly standard physics canon.

    But the train gets off the tracks pretty quickly after that. The author makes the statement,

    ‘The net result is that warmer blackbody can heat a colder blackbody, but not the other way around.’

    which is obviously wrong.

    But why?

    Before this gem of a statement, the author goes through several analogies (which aren’t as informative as equations) proving to himself that only high frequency light that is not being emitted by the blackbody can increase the temperature of the body. This energy is absorbed, then Stokes shifted to lower energy by coupling to internal modes of whatever form of matter we’re discussing. In molecules, mostly vibrational and rotational degrees of freedom, along with intermolecular collision, play this role.

    The implied converse of this statement is that energy absorbed by the warmer blackbody from the colder blackbody is not outside of the frequency range of the warmer blackbody’s spectrum, therefore it doesn’t add heat and doesn’t increase the temperature!


    The question then becomes, if we are talking about blackbodies that absorb ALL frequencies of light, what happens to the energy in the lower frequencies absorbed by the warmer blackbody? Surely there is energy in those photons/waves. Because the warmer blackbody is, in fact, a blackbody, it MUST be absorbing those lower frequencies. What happens to that energy?

    I think most of us know that according to the conservation of energy, those lower frequencies absorbed by the warmer blackbody increase the temperature of that blackbody, even though the radiation was emitted by a colder blackbody. Kirchoff’s law is the mathematical manifestation of this fact. The emissivity of the blackbody in thermal equilibrium equals its absorptivity. Therefore, the thermal equilibrium of a blackbody can be shifted by changing its absorptivity in ANY SPECTRAL REGION, not just the high frequency region. This can be accomplished by increasing the inward flux of low frequency radiation due a nearby colder blackbody, as is the case with atmospheric greenhouse gases in the case of climate.

    So, I don’t doubt that the author’s math and equations are correct. Unfortunately for him, it’s the interpretation of those equations, along the lines of high pass filters and classrooms, that is flawed and ultimately leads to the incorrect conclusion that the greenhouse gas can’t exist. Not even that it doesn’t exist, but that it can’t. It’s brilliant in its simplicity, really.

    I would like to see this author handle the fact that we clearly observe a completely isotropic cosmic microwave background surrounding everything corresponding to a 4K collection of intra-solar and inter-stellar gases. I think that fact is fairly irrefutable proof that this guy is totally wrong.

    Moreover, why are we continuing to clamor to convince people like this that they are incorrect. Anyone who put enough time into convincing themselves of this type of theory after many, many attempts of others to ‘disprove’ them is not going to be swayed by observational evidence, proof of principle experiments or even reason. It’s better to not give their theories credence by taking the time to ‘debunk’ them.

    Comments welcome.

    • Comments welcome.

      You must be aware that the authors are partaking in these discussions on this thread. They’ve already posted a number of comments.
      So why are you referring to the authors in the third person?

    • “Surely there is energy in those photons/waves. Because the warmer blackbody is, in fact, a blackbody, it MUST be absorbing those lower frequencies. What happens to that energy?”

      Take a glass of water at 99 C and surround it with a dozen glasses of water, all also at 99 C. Does the water in the glass in the center get warmer? Does it cool off more slowly than it would if the other glasses were not present?

      • The net transfer of energy between the 99C glasses of water will be zero but all will be radiating energy at a rate appropriate for a single glass of 99C water. Again, all radiating but net transfer zero.

        As for cool down, yes the central glass will cool slower. I suppose you could say that the other glasses are insulating it. What is happening mechanically is that the outer glasses are exposed to a cooler room so the net energy flow between them results in heat loss to the room. The inner glass then is exchanging energy with an incrementally cooler glass of water so it experiences an incremental net loss of energy. Until final equilibrium is achieved, the inner glass will remain warmer than the outer glasses and the outer glasses will remain warmer than the room.

      • So at what point does the glass in the centre become warmer?

        Answer: At no point.

        Take note Roy Spencer and Science of doom, its called ENTROPY.

        Without a continuous energy input you have no net increase from so called “back radiation”.

      • But there is a continuous energy input into the Earth’s climate system; it is provided by the Sun.

      • “But there is a continuous energy input into the Earth’s climate system; it is provided by the Sun.”

        There is also continuous darkness.

        In reality, any given spot on the surface directly under the solar point receives at most 25% of the continuous energy input from the sun in any 24 hour period, but generally much less.

        You know that night/day warming/cooling thing?

      • Wow, this is really simple physics. I’m not a all sure how this can be so hard to follow.

        In the water glass case, there is no steady energy input to the center glass of water so it will loose energy to the environment. The surrounding glasses increase the time that it takes to cool by radiating heat ‘back’ to it as they also cool by radiating their energy.

        To increase the temperature of an warmer object by a cooler object, the warmer object must have an continuing energy input that must be dissipated. In that case, the presence of the cooler object near it radiating part of its energy toward the warmer object results in that energy being added to the total that the warmer object must dissipate. The temperature of the warmer object must increase again to radiate that somewhat larger total energy input.

        Of course, if you insist there is no such thing as a photon or that two streams of photons cannot pass each other traveling in opposite directions, we probably will never be on the same ‘wavelength’. I’ve spent my life around electronics, radio, and nuclear physics. Maxwell’s equations rock for many aspects of electronics and radio but they are just very handy tools. Because Maxewell’s math works a good share of the time does not mean it defines reality. It is not very useful for use when counting gammas to determine the activity level of a radioactive source. Calculations based upon photons and nuclear interactions are the tools that work there. You should use the tool that suits the job at hand. I believe the photon view is the correct one for radiated energy discussions.

        This is basic physics and basic engineering stuff. However, do not take this to mean that I believe doubling of the atmospheric CO2 is going to be a big problem. I don’t. I just prefer simple physics not be twisted to make a point.

      • jae

        Could you have picked a more complicated example, if you’re being rhetorical?

        Dr. Curry’s more creative students will no doubt be asking about partial gas pressure, room temperature, convection, conductivity, where the lights are in the room, and will you refill the glasses as evaporation causes volume change, to start with. ;)

        A glass ingot at 99 C, I could understand.

        Especially if you included conditions like, “in a closed system initially at STP,” and “surrounded in all directions with no significant gaps,” and “all ingots behave as uniform spherical black bodies,” etc.

        Does the water in the glass at the center get warmer? Unlikely, though Dr. Curry’s students could contrive extrinsic conditions to make it so, I am sure.

        Does it cool more slowly? Likely, for most sets of extrinsic conditions, I think Dr. Curry’s able students will find.

        More to the point, could you expand on your point, please, as it elludes me. (Though I’m sure Dr. Curry’s students would be able to explain it to me.)

      • I think we have to be very careful about how we set this problem up.

        We are using language of ‘external energy sources’ versus the system of interest, the surface of the earth in most of this discussion.

        In this case, the outer glasses ARE an energy source for the central glass. They just happen to be at the same temperature as the central glass, in stark contrast with the earth-like situation in which the sun has a dramatically different temperature from the earth.

        So let’s a assume the ‘other’ glasses are in a circle around the central glass and that the glasses can only emit energy out into the plane that contains all the glasses, for simplicity. Being all at 99 C, each glass will have some the same emissivity and emit the same spectrum. It is very likely that each glass can absorb most, if not all, of the energy emitted by the other glasses.

        Now, to me, there seems to be two parameters that matter the most. 1) the temperature of the surrounding air and 2) the distance of the 12 ‘other’ glasses from the central glass. The temperature difference between each glass of water and the surrounding air will determine the difference in the energy absorbed by the central glass and the energy that it gives off due to the air by conduction. The distance between the 12 ‘other’ glasses will determine what percentage of the emitted energy from those glasses can be absorbed by the central glass. The closer the ‘other’ glasses are to the central glass, the more of the emitted energy the central glass can absorb. The further away, the smaller the percentage of energy that can be absorbed.

        There *should* could be an air temp and distance from the ‘other’ glasses at which the central glass is taking in more energy from the surrounding air than it is giving off. In such a case, the central glass would increase in temperature as per the conservation of energy.

        It would be an interesting experiment to set up at least.

      • maxwell

        …..”It would be an interesting experiment to set up at least.”….

        Its been done.
        Sounds like the proof of the zeroth law of thermodynamics.

      • Bryan,

        ‘Sounds like the proof of the zeroth law of thermodynamics.’

        I don’t think this experiment would prove that there is no thermal motion at 0 K. I mean, that is the zeroth law of thermodynamics.

        Can you please explain a bit more what you are saying?

      • maxwell |

        There are a number of glasses at the same temperature
        “Being all at 99 C, each glass ”
        The zeroth law of thermodynamics may be stated as follows:
        If two thermodynamic systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.
        In the early days this was assumed but later questioned.
        It had to be experimentally determined and since we already had Laws 1,2 and 3 it was called the zeroth.
        Perhaps you are thinking about law 3 which is about absolute zoro

      • I have to agree with Zorro.. er, with Bryan, I mean.

        Zeroeth law well-established, a black body among bodies of equal temperature will not increase in temperature, though this says nothing of how quickly each will lose temperature or in what pattern.

        If the air were above the temperature of the glasses, or if there were certain complex salts that underwent a physical change in solution in the central glass, if it contained fissile materials in high enough concentrations, if exothermic chemical changes happened in the ‘water’, if the air pressure were suddenly increased compressing tiny soda bubbles (or at 99C, nearly boiling so water vapor bubbles), if the glass were in an atmosphere of pure reactive metal particulate suspension (potassium, say), if there were a series of lasers deflected toward the central glass, or electric currents, or sound waves.. there are all sorts of extrinsic conditions that might raise the temperature of the central glass.

        Which, as I said, a complex example.. and I still don’t follow the point of it originally being posted.

      • It looks like a jumped the gun under some confusion. Sorry for that.

    • How about debunking quickly this way – two black bodies at different temperatures separated by a perfect reflector. With the reflector in place, heat travels from each black body, bounces off the reflector, and returns to the originating black body. Under the theory that was proposed in chapter2, when the mirror is removed, the heat from the cooler black body must still return to the cooler black body – it has to act as if there is still a reflector in place, but not so for the hotter black body. A ridiculous result. Bad physics…

      • No reasoned rebuttal yet?

      • Harold,

        before quantum physics and the idea that a photon was an actual particle there was wave physics that was, and still is, experimentally proven. Those physics experiments described scattering, reflection, interference, and cancellation. Why would this section of physics suddenly become null just because you apparently have forgotten it?

        There are several possible explanations of why a cooler body would not heat a warmer body contained in this PROVEN area of physics and which are contained in the correct energy equations that give a NET energy flow.

      • When creating the waves and waves of orcs in the big battle of Gondor CGI, one of the directors apparently asked one of the programmers, “That one orc, why’s he running the wrong way?”

        The programmer answered, “Looks like he panicked.”

        Panic didn’t make the waves of orcs not waves of orcs, or invalidate the wave equations, or nullify the physics of Middle Earth.

        And while waves of panic can be observed in mobs, a one-orc wave isn’t really well-modeled by wave equations.

        Which are, after all, only proven mathematical models, not themselves mathematical axioms.

      • Bart R,

        we don’t need no stinkin’ wave EQUATIONS!!

        We have stinkin’ empirical data.

        That’s all I am asking of the cold object heating, or slowing the cooling of, the warm object by radiation. Empirical data.

        One of the thought experiments I really like is the one where the heated object is surrounded by a cooler sphere which is thermostatically controlled. My imagination tells me that the radiation from the sphere is cancelled by the radiation from the heater leaving the net to be drawn off by the cooling system of the sphere with nary an effect on the heater itself.

        I am told that this cool sphere will actually heat the heater. If the heater is made hotter by the cooler sphere, its radiation should be elevated measurably. If it isn’t measurable I really don’t care with respect to the climate disagreements.

      • Right, right. Data and thought experiments, very nice.

        But then why were you bringing up data and thought experiments in a discussion of wave equations, again?

        I’m not sure I follow the analogy built into your thought experiment. Which is the thermostat? Which the air conditioner?

        Help me with my own thought experiment:

        There is a mall full of inventory that is replenished through another channel, with customers entering and leaving all the time through multiple sets of doors. The doors are designed to allow customers in without hindrance, but to slow some customers on the way out by redirecting them randomly through the mall (the mall hopes to increase sales this way). There’s a ‘sellostat’ set by the mall manager designed to set how much the doors slow outbound customers, but the manager’s salary is set by sales figures, and every day he gets greedier.

        Can you see where my thought experiment is going, and maybe suggest improvements?

      • Uhh, Bart R, where did I suggest a thought experiment?

        Haven’t the vaguest on yours. I’m a simpleton remember?

        By the way, doors are NOT a way to allow people in without hindrance. A strip mall does that.

      • kk

        To your first question, that’d be:

        “One of the thought experiments I really like is the one where..”

        And one simpleton to another, remember what?

        I like your idea about using a strip mall rather than doors. Much clearer, and suggests better parallels.

        So, strip mall manager welcomes all buyers to his locale, and takes some steps (advertising posters facing people as they walk away from the strip mall, principally, but also shills who block the way out and chat up secret sales, and the smell of food from the strip mall’s food vendors) to hinder buyers as they leave.

        At first, the manager sets out only small hindrances, but he believes that they work because the theory of advertising tells him so, and his mall sells more and more as he puts more hindrances up, and he sees no reason to stop putting up hindrances since he’s rewarded by profit. He’s so rewarded by profit, he pays off the local officials to allow him to put up more posters, and muscles out any competing shills trying to get buyers to leave for their strip malls up the street.

        So, do you think the mall will be more crowded, the more the manager hinders buyers from leaving?

      • And I continued saying I would like it done as an experiment. I still would. I do not want to discuss it as a though experiment. It has been done to death.

      • kk

        Right, but all experiments start with design, with an analogy or meaning to the model built, with some hypothesis they test… And I’m still not sure what yours tests.

        Could you expand on that?

      • I wouldn’t say test so much as measure. It would seem that climate science, and maybe other fields, agree that there is backradiation and that either it slows cooling of the hotter body or heats it. I want at least one experiment, preferably several differing ones, that quantifies that relationship.

        If it slows cooling, by how much. If it heats the body, by how much. Does there appear to be conditions that increase or decrease the effect we need to research more…

        Why am I so adamant about real experiments? Because thought experiments are limited to the variables that we put into the experiment, you know, just like models. If we do not know about it or do not know it well enough to get a reasonable ball park figure, or cannot convert it to mathematics at a resolution that is useable, we have no way of knowing whether the results of our thought experiment is valid.

        The world has a reality that we need to include and we do not know all that reality. Look at the empirical experiments that detail an effect very well, yet, we do not see the result of the effect generally in reality due to offsetting effects.

        I will grant that there needs to be a lot of thought put into designing the experiments for these same reasons and there is where we find a good use for thought experiments. If we allow contamination the physical experiment will not be giving us results useful for the original purpose. Thought experiments can help us design the experiment to try and exclude contamination so we have a higher certainty of measuring what we think we are measuring.

      • kk

        I’m aware of multiple real measurements cited in this topic and elsewhere on this blog using advanced and proven equipment for decades.

        I’m aware of multiple real experiments cited in this topic and elsewhere on this blog and in countless other sources.

        On its face, the phenomenon of reflected radiation is so everyday commonplace that it would take extremely strong evidence, and with the addition of so much experimental proof, much better rebuttal than has yet been offered on these pages, to credit your words, “I want at least one experiment..” as anything but flat-out, and excuse me for being blunt, lie.

      • Bart R,

        when did Backradiation become REFLECTED radiation?? This is something I definitely missed along with 99.99% of papers and work I have never read that are obviously inexistence in spite of my ignorance.

        Please note, I am NOT trying to claim there isn’t literature, only that I am extremely limited in my exposure to the literature and reality.

      • kk

        You get the distinction between reflected radiation and back radiation?


        Means you have an advanced and subtle grasp of the topic, and should be able to handle the things talked about in the blog by people who do serious measurement, experimentation, analyses and interpretation of these things for a living. (Which would be not me.)

        You want quantified relationships, and that’s all well and good. Check out the slightly unpleasant quote referenced in or the very nicely put lower down this page for context and background about quantified measurements, and some of the problems with experimental interpretation present in the subject.

        I frankly don’t believe we are going to get to widely accepted experimental results along the lines of your suggestion any time soon, unless someone builds a pair of hermetically-sealed IR-transparent domes the size of Nebraska and experiments with changing their CO2 concentrations repeatedly under differing conditions of sunlight.

        Too much room for waffle, and too much brute force logic. And even then, questions of applicability will bedevil us.

        What we need is a guy with a teacup and some milk, and the ability to clearly explain so anyone can understand why the milk particulates suspended in it move.. erm, sorry. Wrong experiment. But you get my drift?

      • Bart R,

        I have just a thin veneer of knowledge and none of the math skills making it a very thin veneer.

      • Photon? Why did you switch frames? I said heat, nothing about photons – I’m using a classical EM frame. Maybe you thought I was talking about photons, since the waves would have to suddenly turn around and return to their source under the proposed theory, which doesn’t make sense to you. That’s my point, the proposed theory doesn’t make physical sense.

      • Harold,

        how does the HEAT travel without waves or particles??

      • Sorry Harold, I got lost.

        OK, back to waves. The classical wave experiments show that waves can interfere, cancel, and augment (sorry for the layman’s terminology). Interference partially cancels or deflects, cancellation negates and augmentation adds. What happens to the energy Harold??

        This has been shown to happen in experiments. Doesn’t it happen out there in the atmosphere? The thought experiment is that 2 bodies are radiating against each other. The colder body will be radiating at a lower energy peak, but the warmer body will be radiating at that wavelength also. Why won’t the waves at the same frequencies cancel or interfere?

        At the quantum level I am even less adept, but, I understand that the particles need to have a correct energy state to absorb energy. What happens if the bodies do not have the correct open energy state to absorb the wave/photon carrying the energy? Won’t it be deflected/reflected instead? Isn’t this a more reasonable explanation of what we see in the atmosphere between GHG’s and the surface and each other for that matter?

        Finally you suggested the wave would HAVE TO RETURN TO ITS SOURCE. . The wave would be deflected in another direction, although it would seem that it could be deflected back to the originating body.

        What amazes me is that there is all this partially understood and misunderstood knowledge being tossed around. Yeah, kinda like me.

      • kuhnkat, reference your comment about e/m waves from several sources interacting with each other, maybe you should take intensity and phase (and polarisation?) into consideration too. It may help to move your though experiment on if you gave some consideration to a practical experiment carried out around 1800 by English scientist Thomas Young described in “Instruction Manual and Experiment Guide .. ADVANCED OPTICS
        SYSTEM .. Experiment 4: The Wave Nature Of Light ( Alternatively you could set up youjr own experiment using a candle and some card board sheets with slits cut in them (

        If you’re interested, this interaction of e/m waves from a single source taking different routes to a common destination can present a surprising problem for Line-of-Site (LoS) radio communications links. Although the optical path from transmitting to receiving aerial may be unobstructed, the radio signal can be reduced (even to zero) as a result of cancellation of the signal travelling over several different paths (

        I’m sure that Joel, the thread’s resident expert in all things to do with theoretical physics, can explain it all in simple terms far better than I. He must do it all of the time when lecturing to his RIT students.

        Best regards, Pete Ridley

      • Pete, to be honest, I find kuhnkat’s posts quite painful to read. He understands just enough about the existence of interference to be led astray into a variety of completely wacky conclusions.

        First of all, interference occurs in only a very carefully prescribed set of circumstances. The light has to be of the same wavelength and to be “coherent”, which means that the waves are in lock-step with each other. Also, the geometry matters. Waves traveling in opposite directions (even assuming the coherence and all) don’t cancel each other out except at very specific locations separated by distances of half of wavelength, which means on the order of microns for what we are talking about. In between, they add together constructively. The result is a standing wave, such as is seen on a guitar string.

        So, I really don’t see anything useful coming out of kuhnkat’s ramblings. They are just an attempt to turn the nonsense that we know Claes and the other Slayers are spewing into something intelligible. But, you can’t produce sense from nonsense by adding more nonsense.

      • Hi Joel,. I felt confident that a top lecturer like you woujld be able to present a compexy subject like e/m wave interactions in a simple manner. Well done, but can you please try to avoid unusual words like “coherent” and concepts like “standing waves” which might confuse us simple lay people.

        Best regards, Pete Ridley.

      • Pete,

        (1) The adjective “top” as in “top lecturer” is yours, not mine.

        (2) I gave brief descriptions of what “coherent” and “standing waves” mean. However, I also wanted to use the correct terminology so that people can easily look on the web to find more detailed description. For example, here is the Wikipedia page on coherence: and here is their discussion of standing waves:

      • The problem was, Joel, that Ridley turned to his library to look it up. But he couldn’t find either “coherent” and “standing waves” in his dog-eared copies of The Elders of Zion and Mein Kampf. So it’s good that you’ve given him links to look the terms up.

        The other problem is, he won’t. Like Kuhnkat, Ridley uses ignorance as a war club to bludgeon his enemies.

        Pete also prefers using his time and bandwidth to search for Holocaust deniers, Neo-Nazis, and Jihadists like Daniel E. Michael to quote. (Did you READ the Michael letter Ridley quoted yesterday that ends with “Death to America!!!”)

      • Joel,

        As the earth emits a relatively continuous band it emits the same wavelengths that are emitted by the GHG’s. While I readily agree that the amount of interaction is probably quite small, if we toss out enough minimal influences we make other amounts larger. (one of many issues with models)

        How about an actual experiment to measure the backradiation effect. Something like a tube with earth at one end and a short wave source at the other. Use at least two runs, one with atmospheric gasses with no GHGs and one with GHGs computed to give the actual backradiation of a column in the open atmosphere. Measuring how fast the earth is warmed with and without GHGs should give a rough idea of how much the backradiation effect is. Or, has this been done and can you point me to the paper?? Simply shining IR through a tube of co2 tells us little about the effects of the radiation emitted by that co2 or h2o or ch4… on the ground. For the truly anal we could use differing types of material such as granite, dirt, wet dirt, loam, sand… to see how the effect is modified if there are measureable differences.

        Actually a third run with close to a vacuum would be good to show that there is no difference between non-GHGs and a vacuum insofar as the rate of warming of the surface. That is, there is negligible backradiation from non-GHGs matching their negligible absorption.

        This is the type of straightforward experiment that MIGHT convince some sceptics and deniers that there really is a measurable, significant in relation to the earth system, increase in warming speed. It should be able to clarify which of the ideas of no effect, slows radiation from the earth, or warms the earth is correct. I would note that some significant warmists apparently believe there is a real warming. An actual series of experiments should be able to sort this mess out!!

        It is really silly to have all these conflicting discussions over the number of angels that can dance on the head of a pin when we should be counting them with electron microscopes or other detectors. (well I guess there is the issue of finding the pin they are dancing on or luring them to our dance)

      • Kuhnkat: I don’t even understand your experiment…and I don’t really see why scientists should waste their time running it.

        For one thing, the basic physics of the radiative transfer in the atmosphere and specifically radiative forcing of CO2 is well-accepted and well-tested science by everyone who has even a small modicum of respect within the scientific community (e.g., Roy Spencer and Richard Lindzen accept it). So, the issue comes down to feedbacks and that is not something that can be settled by such a simplistic experiment.

        For another, I am under no illusions that we can ever convince “skeptics” who doubt such basic tenets of science to become AGW believers. Such people are like Young Earth Creationists: they don’t disbelieve AGW because they doubt the science; rather they believe any bogus nonsense attacks on the science because they are ideologically opposed to the actions that follow from addressing AGW. If you guys can’t even comprehend and accept basic science about which there is no serious controversy whatsoever and instead believe nonsense, how am I ever to convince you on the issue of feedbacks and climate sensitivity, which actually require weighing the balance of the evidence? It is like telling me that if I can only get a Young Earth creationist to abandon the belief that the earth is only 6000 years old, he will actually fully accept evolutionary theory…Ain’t gonna happen!

      • Hi Joel, I agree with your comment (yesterday at 9:09 pm) about the heat retaining effect of water vapour and some trace atmospheric gases preventing some of the IR energy that is emitted by the earth from radiating back out unobstructed (AKA the Greenhouse Effect) and that humans adding a tiny amount of CO2 could result in a small (beneficial?) rise in temperature. Ias you say there are not many respected or knowledgeable scientists who consider otherwise.

        On the other hand I’ll be very very surprised if you can provide a sound analysis of your own that convinces true sceptics that the balance of evidence indicates that a global climate catastrophe looms as a result of our continuing use of fossil fuels.

        Rest assured that the use of fossil fuels will continue for many many decades yet and all of the scare-mongering by the power hungry, the UN, the politicians and the environmental activists will not change that.

        I still haven’t seen your refutation of the analysis carried out by Roger Taguchi showing that the feedback effect is negligible. Was it too hard for you? OK, here a simple question. If positive feedback due to increased water vapour arising from a slight increase in global temperature due to our use of fossil fuels is able to cause a global climate catastrophe in the next 90 years why didn’t such a disaster happen during the Roman warming or during the MWP? I’m sure that you can explain that in simple enough terms for lay people like me to understand, but please don’t try to argue that the rate of warming now is far greater than ever experienced during the past 300M years or that Mann was correct and there was no such thing as the MWP.

        BTW, have you started reading “The Hockey Stick Illusion” yet – no, I thought not.

        Best regards, Pete Ridley

      • Joel,

        the experiment is to see how fast the material warms with and without ghg’s in the atmosphere giving an empirical figure for the effect of backradiation in a carefully controlled experiment.

        Why is this important? Because deniers like me say there is none. Luke-Warmers and warmers believe in varying amounts of slowing of the surface cooling, and some alarmists say the backradiation actually raises the temperature of the material above the level that the short wave can make it. Even if everyone suddenly went sane and decided there was only a reduction in the rate of cooling (faster warming also) it would be good to actually quantify by empirical experiment exactly what the magnitude of the effect is.

        You say:

        “the basic physics of the radiative transfer in the atmosphere and specifically radiative forcing of CO2 is well-accepted and well-tested science by everyone who has even a small modicum of respect within the scientific community”

        Yet, that statement says NOTHING about the magnitude of the effect on the earth itself. I am sure you agree that different materials would react differently even if your theory is correct. Being able to put constraints on the effect in the models would be a real contribution outside of just making some people happy that their position was proven.

        The Climate Science community appears to me to be adverse to the drudge work of detailed science. It is time they stopped talking about saving the earth and started doing the real work necessary to prove the hypotheses and giving us more information on what may need to be done.

      • Why is this important? Because deniers like me say there is none. Luke-Warmers and warmers believe in varying amounts of slowing of the surface cooling, and some alarmists say the backradiation actually raises the temperature of the material above the level that the short wave can make it. Even if everyone suddenly went sane and decided there was only a reduction in the rate of cooling (faster warming also) it would be good to actually quantify by empirical experiment exactly what the magnitude of the effect is.

        This one paragraph shows how hopelessly confused you are about something that is just basic physics! You make this distinction between “rais[ing] the temperature of the material above the level that the short wave can make it” and “a reduction in the rate of cooling”. There is no such contradiction between those two pictures: CO2 slows the rate of cooling and, in doing so, it causes the temperature of the earth to be warmer than it would be in its absence because the earth is heated by the sun and its steady-state temperature is determined by the balance between the rate at which it receives energy from the sun and the rate at which “cools itself” by sending energy back into space.

        The fact that you have been unable to comprehend this shows how you are unwilling to allow yourself to comprehend the most basic of scientific principles.

        Yet, that statement says NOTHING about the magnitude of the effect on the earth itself.

        The magnitude of the radiative effect of CO2 is not under debate in any serious quarters. Roy Spencer and Richard Lindzen and the rest of the scientific community all agree it is 3.8 W/m^2 (+/- 5%, or at most 10%). The magnitude of the resulting temperature change is still under debate, but this involves the question of feedbacks, which alas can’t be settled by any experiment smaller than the entire scale of the earth. (Which is not to say we can’t learn a lot about feedbacks from empirical data. In fact, we can and have. See, for example, here: )

      • Joel,

        “CO2 slows the rate of cooling and, in doing so, it causes the temperature of the earth to be warmer than it would be in its absence because the earth is heated by the sun and its steady-state temperature is determined by the balance between the rate at which it receives energy from the sun and the rate at which “cools itself” by sending energy back into space. ”

        You have very bad radiation, the Earth cooling and the energy content concept here. 0.04% CO2 in the atmosphere has absolutely minimal energy content in it when comparing the energy content of the atmosphere (orders of magnitude more than CO2) not to mention the LW radiation energy from the Earth surface (orders of magnitude larger than atmosphere). I guess you know the mathematical differention of infinitely small -> 0, thats CO2 capable of warming the air -> 0, warming the Earth -> 0 and CO2 capable of slow cooling -> 0. CO2 cooling warms the Earth is absolutely absurd if you have any energy concept at all. Warming and cooling are mainly due to huge amount of water presents on the Earth. The movement of water causes most weather changes. I would advise you to appreciate the energy contents in them and study the physical properties of water, CO2 and the energy they are involved or you will never learn and keep on misinforming the general public wasting your life unless you have an agenda in order to stay on the gravy train.

        The Earth receives the Sun energy, stores (chemically and physically) some of it, reflects some of it, refracts some of it, conducts some of it, convects some of it, radiates (naturally including decays, volcano eruptions, human consumptions of food and fossil fuels) some of it. The Sun itself also in an ever changing state of emitting energy. There is no steady state temeperature, only instantaneous temperature.

        The fact that you have been unable to comprehend this shows how you are unwilling to allow yourself to comprehend the most basic of scientific principles of energy, cooling, heating and radiation.

      • “Warming and cooling are mainly due to huge amount of water presents on the Earth” should be amended as “Warming rate and cooling rate are mainly due to huge amount of water presents on the Earth”

      • Sam NC: It would have been more precise of me to talk about the rate at which energy is emitted or absorbed by the earth. Yes, the conversion of this into a rate at which temperature changes involves the heat capacity which, as you note, is largely due to thelarge amount of water. However, this doesn’t change the end result, i.e., the final steady-state temperature, but just how long it takes to get there. [Of course, this ignores water vapor or cloud feedbacks, which can affect the end result.]

        The fact that you have been unable to comprehend this shows how you are unwilling to allow yourself to comprehend the most basic of scientific principles of energy, cooling, heating and radiation.

        Well, if I fail to comprehend this, I am in good company with basically all of the scientific community. Why do you think you understand these things better than the National Academy of Sciences, the authors of the major physics textbooks which discuss global warming, etc., etc.? You are just fooling yourself…It is the Dunning Kruger effect ( ).

        Look, if you want to believe nonsense, I can’t stop you. Go play with your fellow travellers who believe the Earth is only 6000 years old and all the rest of the folks who would rather believe pseudoscience than science that conflicts with their ideology. Ignorance can only be cured if someone wants to learn. You want to remain ignorant and so you will.

  20. To Judy: It is clear that you miss the points I want to make. Of course there are endless little things you can focus on and question, but in the spirit of Leibniz I ask you to try get the main message. I am not saying that my model is perfect. I try to make a point about radiative heat transfer based on a mathematical analysis of the same equation Planck tried to use but gave up with. If you focus on this equation, do see something of interest in my analysis? What is your model for radiation? Does it contain “backradiation”?
    Is it a stable phenomenon in your model?

    Next, you said you did not like Kiehl-Trenberth, and I asked you why? I do it again.

    And have you given your students my chapters for homework? It could be an educational experience, and students need assignments, right?

    • “I am not saying my model is perfect”

      Your model and main message are fundamentally flawed, as was easily shown.

  21. To Maxwell: A warm body also absorbs low frequency waves but re-emit them
    and thus avoid getting heated by low-frequency stuff. Like an educated person
    simply does not get heated up by silly remarks from uneducated, only by remarks from more educated. Right?

    • Claes, to me, that seemed to be a weak response to a very clear post by Maxwell. You wanted Judith to give you the opportunity to debate the science contained in your book, so debate it properly rather than handwaving away difficult objections.

    • Mr. Johnson,

      Is the irony lost on you?

      ‘A warm body also absorbs low frequency waves but re-emit them
      and thus avoid getting heated by low-frequency stuff.’

      Without warming the warm body with low frequency light from the colder body, there is lack of energy conservation. In order to emit more low frequency light (ie the low frequencies already being emitted and the absorbed low frequencies from the colder body) the thermal equilibrium must change, coming to a higher temperature according to the Stefan-Boltzmann law. Raising the temperature costs energy.

      So there are two options 1) your theory violates the conservation of energy because the emission of low frequency light by warmer blackbody doesn’t change in response to increase flux of low frequency light from a colder blackbody or 2) conservation of energy is preserved and your thesis (cold blackbody can’t heat warm blackbody) is wrong.

      I’ll let you pick which options you want.

      With respect to your poorly thought out classroom analogy, I am constantly learning from people who have less education than me. On an almost daily basis in fact. So not only is your analogy not informative in the context of energy transfer via radiation, it’s as fundamentally incorrect as your physical theory.

      Any other thoughts?

  22. One thing that may be overlooked, in these discussions on whether or not a cold object can heat a warm body though exchange of radiation is that, a photon doesn’t know where it came from, the only thing it “knows” is its frequency. All the properties, momentum, wavelength and energy are directly related to its frequenncy and vice-versa. Measure one of the four and you know all of them.

  23. The trouble with photon particles carrying energy back and forth is that it is
    an unstable phenomenon, or do really think there is a highway with left
    and right lanes connecting two bodies? Why would photon respect such
    traffic laws? Which equation is describing the physics you are hinting at?

    • Why does there have to be left and right lanes, or what happens when two photons traveling in opposite directions reach the same point in space?

      Do they collide, or interact in anyway?

      Or do you have anything other than handwaving to support this statement from your book?

      “We argue that such two-way propagation is unstable because it requires
      cancellation, and cancellation in massive two-way flow of heat energy is
      unstable to small perturbations and thus is unphysical.”

      Why does it require cancellation and why is it unstable?

    • Why would it be unstable? In second chapter you obtain an equation witch is the same as Boltzmann law for 2 bodies and infinitely small T difference. So conclusion about stability should be the same. By the way, your equation is not symmetrical, meaning that cold and hot temperature have not the same influence. So how do you generalise to a N>2 body problem? Looks trickier than classic Boltzmann to me.

      But more important, you throw out quanta interpretation. Sure it is not intuitive, but since Boltzmann it has been used to derive a huge amount of physical equations, and explain a lot of experimental results. Throwing out quanta to obtain radiative transfer equations you like better is only the begining of the story, because now you will have to reinterpret THE major part (more important imho than relativity) of modern physics ( post WWI physics). This is not out of question, but it is a huge task, and a task far far far too big to start from just radiative heat transfer…even if historically it was the start up of quantum mechanics. To make such a body of inference collapse, a single new fact may be sufficient, but the new fact will usually not be the same as the one at the origin of the old theory, and the new theory should be as powerful as the one it replace. Not bearing well for your new interpretation, so yes, even if I like the first chapter a lot, the second one is definitely in crackpot territory…

    • I notice you have avoided the challenge of applying your theory to a real world problem such as heat loss from a pipe, the concept of a cooler body transferring heat to a warmer has great success in these situations and has been tested many times. Cut to the chase, try some of the problems on page 582 of Mills, ‘Heat and Mass Transfer’.
      Here’s a link in case you don’t have a copy.

    • BTW, your interpretation of radiative heat transfer is very easily testable experimentally: consider heat exchange between a hot body at T_h and a cold body at T_c=T_h/2.

      classic equation (eq 20 in chapter 2) gives R =sigma (T_h^4-T_c^4)=15/16*sigma T_h^4
      your new equation (eq 21) gives R =4*sigma *T_h^3*(T_h-T_c)=2*sigma T_h^4, i.e. almost 2 times the heat transfer predicted by S-B. Quite easy to test using simple calorimetric experiment, no?

  24. Believers in the greenhouse effect will not honestly take in information counter to their belief, no matter how it is couched. As long as everyone pretends that there is a legitimate scientific debate being engaged here, it is obvious that situation will continue unchanged. Meanwhile, the truth lies elsewhere than the mass of climate scientists, and the hapless public, supposes. What follows is a comment I started to post on Claes Johnson’s site a few days ago, but didn’t because I realized no one was listening. I’ll put it here just because I exist, and the facts exist, and it has to be said, and eventually admitted by everyone:

    You need to establish first how the atmosphere is basically warmed: By atmospheric absorption of direct solar infrared irradiation, or by surface absorption of visible radiation followed by surface emission of infrared. Climate scientists, and their defenders, who tout the greenhouse effect, believe the latter [which leads to the infamous backradiation], and ignore the former.

    But as I have tried to communicate, to other scientists and to the public (see my blog article, “Venus: No Greenhouse Effect”), comparison of the atmospheric temperatures of Venus and Earth at corresponding pressures, over the range of Earth atmospheric pressures (from 1 atm. down to 0.2 atm.), shows the ONLY DIFFERENCE between the two is an essentially constant 1.176 multiplicative factor (T_venus/T_earth) which is just due to the relative distances of the two planets from the Sun. Nothing more. It has nothing to do with planetary albedo, or with the concentration of carbon dioxide or other “greenhouse gases”. The only (small) deviation from this general condition is in the strictly limited altitude range of the clouds on Venus (pressures between about 0.6 and 0.3 atm. only), where the Venus temperature is LOWER (not higher, despite the carbon dioxide atmosphere) by just a few degrees than the strict 1.176 x T_earth relationship, due no doubt to the cooling effect of water (dilute sulfuric acid) in those clouds.

    The only way this overwhelming and definitive experimental finding (T_venus/T_earth = essentially constant = 1.17 very closely, encompassing the data of two detailed planetary atmospheres) can be explained is that the atmospheres of both planets are heated by the SAME PORTION of the solar radiation, attenuated only by the distance from the Sun to each planet. This means absorption of visible radiation at Earth’s surface, followed by surface emission of infrared (heat) radiation into the Earth atmosphere, cannot have anything to do with the basic warming of the atmosphere, because Venus is largely opaque to the visible solar radiation, and it cannot reach Venus’s surface (and is thus not part of that common portion warming both atmospheres).

    So the first unarguable fact is: Earth and Venus are both warmed by direct atmospheric absorption of the same infrared portion of the solar radiation. There is no speculation, no theory in this statement: It is an amazing (because so many scientists believe otherwise) statement of experimental fact, based on the actual detailed temperature and pressure profiles measured for the two planets (which have been available to climate scientists promoting the greenhouse effect for nearly 20 years now, which means they are incompetent). And it completely invalidates ANY “greenhouse effect” of additional warming by adding carbon dioxide to the atmosphere: Venus has 96.5% carbon dioxide (compared to Earth’s 0.04%), yet its atmospheric temperatures relative to Earth’s atmosphere have nothing to do with that huge concentration, but only and precisely to the fact that Venus is closer to the Sun than is the Earth. Venus’s surface temperature is far higher than Earth’s, because Venus’s atmosphere is far deeper than Earth’s. To tell the public — and to teach students — otherwise is to recklessly spread an obvious falsehood and steal hard-earned knowledge from the world, thereby misusing and ultimately defaming the authority of science in the world.

    Stop playing around with theoretical put-downs, and talking past each other, and admit that the Venus/Earth data completely and unambiguously invalidates the greenhouse effect.

  25. Claes starting point is not concerned with the climate change issue as such.
    His contribution is to question if Plank and Einstein were correct to abandon classical wave theory in favour of the quantisation of electromagnetic radiation.
    To be sure Plank and Einstein were deeply unhappy with the situation and regarded the concept of the photon as a “fix” or even a “trick” which would give way to some fuller explanation of phenomena like the photoelectric effect and so on.
    IMHO the photon explanation is the best we have at the moment but I’m glad that imaginative people like Claes are ready to reexamine the fundamentals from time to time.
    I’m sure if a real problem about heat transfer required a solution Claes would produce a solution that competent Physicists would agree with.
    He would probably use the Poynting vector to give the direction and magnitude of heat flow.
    Which of course as Clausius pointed out is always from higher to lower temperature bodies.
    On the climate change issue he would say I’m sure that the colder atmosphere cannot increase the temperature of the warmer Earth Surface.

    • His contribution is to question if Plank and Einstein were correct to abandon classical wave theory in favour of the quantisation of electromagnetic radiation.

      And he, in turn, throws out the superposition principle ( the two black bodies’ radiation patterns can be solved for indepandantly, and then added together), which holds for classical wave theory. I don’t see switching to a classical EM frame, and then having to destroy a central tenet of the classical EM theory an advance. You can’t have it both ways – classical EM holds and classical EM doesn’t hold. The very frame it’s put itno says his theory is flat 100% wrong.

  26. Maxwell posts:
    “But the train gets off the tracks pretty quickly after that. The author makes the statement,
    ‘The net result is that warmer blackbody can heat a colder blackbody, but not the other way around.’
    which is obviously wrong.
    But why?”

    One warm body in dark space radiates energy in all directions except back at itself (ignoring internal self-balancing). Two warm bodies in dark space do that but also each warms the other which reduces the rate at which they cool. This is true regardless of relative temperatures. The cooler body radiates the warmer body (can’t be helped – it doesn’t know it is the cooler body and science doesn’t care) and that unavoidably slows the rate of cooling of that warmer body.

    Unlike electricity passing through a straight taut wire (and a tapered wire will demonstrate distributed radiated energy), no part of the wire radiates any other part of the wire. It is like that solitary radiating body in space. The thermal distribution is a consequence of the local resistance and thermal conductance.

    Not the case with radiated energy. Each object paints any other visible object and that object is compelled to react to that energy.

    • dp,

      it was a rhetorical question, but I appreciate your answering it in the context you used. It’s more practical than my own and hopefully will get through to more readers.


    • OK, I’m late to this and have what may be a very dumb question. But, dp, in the scenario you posit isn’t it possible, depending on the temperature, size, and proximity of the two bodies, that the cooler may actually increase in temperature, at least for a period of time, while it is never possible that the warmer object would increase in temperature? And isn’t that the point some are making, i.e. the colder body can not warm the hotter body?

      • The warmer body does indeed warm the colder body, but at the same time the warmer body gets also warmer than it would be without the colder body. It would still radiate as much as without the colder body and this radiation would disappear to the empty space. What the colder body does is that it is also radiating (although less) and some of this radiation is going to hit the warmer body and bring some heat to it. Some additional heat is heating the body whatever its source is.

      • Pekka,

        I have seen this explained before.

        If the colder body causes the warmer body to heat then the radiation of the warmer body will increase and it should be measurable. If we cannot measure it the effect is so small as to be ignored in the context of the climate debate (much larger effects are ignored by the Models). Can you point us to papers showing the experimental data on this?

        No one else has bothered to beat us over the head with the actual empirical data, that I have seen, and my head is really hard so takes a lot to penetrate it.

  27. Claes, I could recommend some books on statistical thermodynamics if you’re interested. It seems to me that if one is going to dismiss it as “jibberishy” one ought to know something about it, if one is not to be be considered a crank.

  28. To David: I have tried to learn from books on statistical thermodynamics but
    I belong to the large group of mathematicians who cannot understand what
    this theory tells you about reality.

    As Harry DH says: A constructive debate requires constructive minds. To argue
    with a three year old who has decided to not do something, requires something
    else than good old logic.

    Yes; it is a good idea to go back and understand that Planck and Einstein and Schrodinger were not happy at all with particle statistics. Maybe they had
    some good reasons not to be which are still valid.

  29. Right, they are challenging Planck and Einstein so we should prove it.

    From the chapter on Blackbody radiation:

    “7.13 Stefan-Boltzmann’s Law for Two Blackbodies
    The classical Stefan-Boltzmann’s Law R = T 4 gives the energy radiated
    from a blackbody of temperature T into an exterior at absolute zero temperature
    (0K). For the case of an exterior temperature Text above zero,
    standard literature presents the following modification:
    R = T 4 − T 4
    ext, (20)
    where the term T 4
    ext conventionally represents ”backradiation” from the
    exterior to the blackbody. It is important to understand that this is a convention
    which by itself does not prove that there is a two-way flow of
    energy with T 4 going out and T 4
    ext coming in.
    In our analysis, there is no such two-way flow of heat energy, only a
    flow of net energy as expressed writing (20) in the following differentiated
    R  4T 3(T − Text) (21)
    with just one term and not the difference of two terms. The mere naming
    of something does not bring it into physical existence.”

    If you have two bodies, or one body radiating to an exterior, which can be considered as two bodies. They both are radiating, and how do they know of the existence of the other, which would be required to determine the magnitude of the net flow of energy.

    Pretty much requiring inanimate objects having knowledge of other inamimate objects is what your analysis requires.

    We can detect the cosmic background radiation, and those photons, when they enter a detector, must add that energy to the detector in order to satisfy consevation of energy, which warms the detector, slightly. That cosmic background radiation is just blackbody radiation extremly red-shifted.

    I guess we are getting somewhere, as those who are trying to disprove the greenhouse gas effect, realize that in order to do that, they must attack Einstein, Planck and the Photon, and you wonder why they are labeled crack-pots.

  30. I often find challenges to my existing perspectives to be enlightening, because in responding, I’m forced to review my own understanding, and on ocassions, revise it. In this case, however, the claim that a cooler body can’t cause a warmer body to become warmer still (if that is indeed claimed) is so nonsensical that it would be hard to learn anything from refuting it. Instead, I will simply suggest a simple experiment.

    I assume most of us are located in what is now a relatively cold time of year. Here is what I suggest. When the temperature outside is 2 deg C and your body skin temperature is, say 35 C (measured by a thermometer taped to your skin and insulated to shield it from the outside), go outside dressed only in a short-sleeve shirt and shorts, wait for about an hour, and then take your temperature. It will be lower – record the value. It might be around 32-33 C.

    Now go back in the house, and put on heavy clothes and an overcoat, taken from the closet at 20 C (obviously colder than your body skin temperature). Again, take your temperature after an hour.

    Did the 20 C clothes cause your 32 C temperature to go up or down?

    The mechanism of warming by the clothes is primarily convective, while the warming of the surface from the atmosphere is primarily radiative, but the principle is the same – a cooler body can cause the temperature of a warmer body to rise. For this to happen, of course, the cooler body must itself be exposed to heat that originated in an even warmer source than the current temperature of the warmer body. In the absence of such a source, a cool object can’t raise the temperature of a warmer object (although it can cause it to cool less than if the warm object were simply radiating to space). For your skin, that heat is generated by metabolism sufficient to maintain your internal temperature above the 32 C skin temperature, and the clothes retard its escape. For the atmosphere, the heat comes from the sun, and is transmitted to the atmosphere by absorption of solar radiation and IR radiation from the surface. Ultimately, of course, the net heat flow is from warm to cold – from the sun, via various routes, to the Earth, and then to space. In the meantime, the greenhouse effect operating on the atmosphere makes the Earth’s temperature habitable.

    • Fred, with all due respect, everyone here understands conduction and its little brother convection (and convection’s little brother advection) just fine. The nonsensical greenhouse gas theory is based on radiation and radiation balance causing heating. Bringing conduction, convection and insulation into the conversation is off-topic and a distraction…and certainly seems intentional to me…like a magician trying to distract the audience from the things going on in his left hand.

    • Fred Moolten |

      ……”In this case, however, the claim that a cooler body can’t cause a warmer body to become warmer still (if that is indeed claimed) is so nonsensical that it would be hard to learn anything from refuting it. Instead, I will simply suggest a simple experiment.”……..

      During a school lesson the Physics teacher might say the force of gravity causes “bodies” to accelerate towards the Earth at 9.81m/s2.
      A pupil might ask “is the body alive”.

      Fred we are not talking here about heat sources that have a means of regulating their power output such as an animal.

      Will putting clothes on a bronze statue at a temperature of say 350K cause its temperature to rise above 350K if the ambient temperature is say 275K?
      Of course not!
      All the clothes can do is to insulate the body i.e. to reduce the rate of heat loss from the object.

      • I believe most readers will understand the point I made.

      • Fred Moolten

        Most readers will conclude you don’t know much about heat transfer!

      • I’ll take my chances, Bryan.

      • Fred, I do hope you are joking here as the reason I will be warmer after putting on the 20c clothes is the energy I’m burning and turning into heat (you know, calories) will not be lost as quickly allowing my body to warm.

      • Ken – Your explanation is correct, but there was no joke intended. The point is simple – as long as a heat source is available for the cooler object to operate on, that object can raise the temperature of a warmer object. In this example, the heat source is body metabolism. For the greenhouse effect, the heat source is the sun. The inability of a cooler object to raise the temperature of a warmer object applies when there is no source of heat for the cooler object to divert back toward the warmer object, but that is not the case with our atmosphere.

      • Bryan,

        if a reader came to the conclusion that Fred was discussing convection or conduction it would point more to reader’s inability to decipher the most important aspect of the his example rather than a real lacking on the part of Fred. Yet, here you are.

      • Fred seemed to have interlinked lines of confusion .
        Power sources that regulate their output.
        Insulation does not imply that the insulator transmits heat by any method to the source of heat.

      • Actually I don’t understand Fred.
        Let me see, if I put a brick out in the sun and it warms to X degrees, and if I then split the brick in 2 and seperate them a couple of centimetres, they will “become warmer still”?

      • Possibly yes, because of increased surface exposure to sunlight, but it depends on air temperature, the absorptivity of the bricks for solar and IR wavelengths, their IR emissivity, conductivity and temperature at the surface they are resting on, and other variables.

      • I don’t know why you would introduce all those variables. It’s the one brick under the one sun sitting on the one surface. All I do is tap it with my trovel and split it in half (like a good brickie would) the properties of the 2 halves are identical. If it’s T rises due to the greater surface area, what has that got to do with the discussion about a cool body increasing the T of a warm body via radiation?

        OK we’ll void the extra surface area by placing a brick under the sun until it reaches X degrees. We now get a 2nd brick from the shed and place it next to the first one.
        Will the T of the first one now rise above X degrees because a 2nd brick was placed next to it?

      • Yes, under many circumstances (see Frank Davis’s link below to Spencer’s blog).

      • that means we could….for instance…..increase the surface temperature of the Moon from 107DegC to a somewhat higher T by placing an atmosphere around it?

      • The mean (day/night average) lunar temperature could be increased by an atmosphere containing greenhouse gases.

      • Why are you mentioning the mean T? The moon has a T of 107DegC during the day. If we introduce a cool body next to it (an atmosphere) will it increase the moons T? It’s a simple question expanding on our discussion so far. You didn’t introduce ‘mean’ or day night into the brick example.

        So what will the new daytime T be?

      • BH – It will increase it more at night, but it would also increase the daytime temperature as long as the cool body did not shield the moon from the sun.

      • Splitting a brick is a very good example. If we split the moon in two halves, will they warm each other, so each alves become warmer? I dont thnk so.

      • At equilibrium, the temperatures won’t change as long as the new surfaces have the same physical properties (emissivity/absorptivity) as the original surface. That is because the moon’s temperature is determine by the level at which radiative loss to space equals radiative gain from sunlight. Since splitting the moon won’t change the incoming solar energy in W/m^2, the outgoing flux and therefore the surface temperature won’t change.

      • actually if you split the moon in half, the two halfs would both be cooler as the combined surface area would be greater

      • Rob – the extra surface area would both absorb and radiate more heat. The temperatures would remain unchanged, because they are dictated by solar absorption on a W/m^2 basis. Surface area is therefore irrelevant.

      • Fred- If you slice a moon or planet in half wouldn’t it expose the warmer core of each half, which would result in greater heat loss

      • just kidding

      • Baa Humbug

        Indeed, what a number of the IPCC adherents miss out is that a colder object can make a warmer object colder than it would be in the absence of the colder object.
        Why do they have this blindspot?

      • I don’t understand your (Moolten’s) point either. Clothing does not heat up a human body via “back-radiation” or “back-conduction.” There is no heat transfer from cold to hot without work input (Clausius), and clothing (and likewise the atmosphere) cannot add work input.


        “The total surface area of an adult is about 2 m^2, and the mid- and far-infrared emissivity of skin and most clothing is near unity, as it is for most nonmetallic surfaces. Skin temperature is about 33 deg C, but clothing reduces the surface temperature to about 28 deg C when the ambient temperature is 20 deg C. Hence, the net radiative heat loss is about Pnet = 100 W.”

        Clothing “reduces the skin surface temperature” because the human body has to supply heat energy to the colder clothing to increase the temperature of the clothing. Clothing does limit convection, as do glass panes in a greenhouse, but CO2 has no such ability. Thus, the analogy fails.

        Please point me to a textbook of physics which contains the terms “back-radiation” or “back-conduction.”

      • Without the clothing, the temperature would decline even further. If you are skeptical, try the experiment I proposed.

  31. “I belong to the large group of mathematicians who cannot understand what
    this theory tells you about reality.”

    It is impossible to take anything you say seriously when you make statements like this. When classical thermodynamics fails to explain the specific heat of your atomic crystalline solid, to where do you turn? Maybe you can guess what technique Einstein used to model the solid.

  32. No I am serious, as serious as Einstein when he distanced himself from statistics as a way of understanding physics.

    • As in the statistical emission properties of a theoretical S/B, solid two-dimensional black-body disc.

      As aposed to the physical emission properties of a real, fluid, three-dimensional grey-body gas.

      The Stefan/Boltzmann BBD argument is a infra-red herring.

      It leads nowhere because it is an apple and oranges comparison.

      We can easily compare a body of CO2 to a body of air and clear up the argument in seconds.

      “An easily reproducible experiment”

      This simple experiment demonstrates that CO2 in the atmosphere is forced in to equilibrium by and with, the O2 and N2.

      Not as AGW has it, the other way around.

    • Interesting. You choose a classical frame for your work, and then use a statement about the interpretation of QM wave functions to boslter your argument .. but fail to note that Einstein didn’t distance himself from statistical mechanics, etc.

      I see no clarity in your thoughts or arguments, merely throwing in red herrings instead to answering the obvious inconsistencies which result from your theory .

  33. Judy says that something is wrong with the KT energy budget, but refuses to
    tell what is wrong. What kind of debate is this? Is it some kind guess play?
    So Judy, please tell me now what it is you find is wrong with KT?

    • Exactly when and where have I said something is wrong with the KT energy budget? KT’s numbers are almost certainly inexact. Attempting to do some sort of globally averaged energy balance may not be the best way to go about it. But that does not mean that atmospheric infrared back radiation does not exist.

  34. To Fred Molten: Can you give me the equations you are using showing that
    heat by itself (without external input of energy) can move from cold to warm?
    Of course putting on clothes makes it possible to keep a higher body surface temperature but the heat comes from the catabolism of your body, not from your clothes, at least if you live in Sweden.

  35. The present physical theories are perfectly able to describe all basic processes that need to be considered in analyzing atmosphere and they have been tested extensively in very many different setups. There are no reasons to replace any of this knowledge by some conflicting physical laws. Most physicists are, however, unaware about, how much of the physical understanding can be described in several different ways. Handling of electromagnetic radiation is one good example.

    One of my former colleagues did theoretical research on laser physics. Most descriptions of lasers start immediately with quantum field theory, but his approach was based on classical electromagnetic field theory and it was very successful. It was not in contradiction with quantum mechanics, but the mathematical approach was very different.

    I can see in Claes Johnson’s texts superficial similarities with that approach. The way quantization is brought into the calculations can be chosen from several alternatives. In some approaches it can make sense to state that there are not forward radiation and back radiation, but only the net radiation. If the final results differ from the conventional approach they are certainly wrong as the conventional approach has been validated so well, but the alternative approach may also be correct as long as it leads to the same results.

    I do not believe that the alternatives would often be easier to understand or of any particular value, but I would be careful before declaring some non-conventional approach automatically wrong. The case of analyzing lasers that I mentioned at the beginning is proof of the fact that sometimes one may indeed find advantage from postponing the quantization and using classical formulation as far as possible.

    Using obscure alternative formulation and vague argumentation as evidence on weaknesses in the conventional understanding of physics is another matter. When it is done in parts of physics, which have been applied widely for years without any conflict with observation, I would not give any weight on such claims.

    • Thank you Pekka

      • Steven Mosher


        I’ll suggest a cage match.

        Johnson versus maxwell.
        no other commenters allowed. People can then see that Johnson will not be able to maintain his position. we will them ask him to admit his honest error and ask the publishers to correct the book.

      • As publisher of the North American and Oceania version…I accept this challenge. I’m happy to publish errata and a new edition if and when the errors reach a critical mass.
        I’m not sure how to prove anything when the topic gets this esoteric…I prefer lab experiments where the data verifies or falsifies a claim. No models. No dueling weblinks or appeals to authority in any form. It makes things tough when you need a vacuum to isolate the experiment from conduction and convection effects. We’ll see how it goes, I suppose.
        Good idea, Steve.

      • Steven Mosher

        Thank Ken. I suggested the same thing for the IPCC. We need to make room for the admission and correction of honest error. The IPCC could not do it. I do not trust them as a consequence and thus am forced to look at primary research on my own to come to a considered judgement.

      • Now that we have aired some stuff, I agree that the discussion is best left to those with a degree in physics (maxwell, pekka, and there are others among the denizens of climate etc that have not shown up).

      • I’d be down for this ‘cage match’ if I thought it would do any good. Alas, we’ve seen that even when faced with the idea that his theory violates the conservation of energy (the 1st law of thermodynamics, the very theory he claims supports him), he is unwilling to concede or even engage.

        It’s my opinion, based on this fact and the lack of transparent discussion perpetuated by some other commenters, that science is not of interest to these people. Maybe it is an ‘honest’ mistake that Johnson has gotten to this place, but I see his poorly thought out analogies beginning Chap. 2 as a way for him to rationalizing away the physical meaning of some of the most well-known and thoroughly tested laws physics has given us thus far. In such a case I have to wonder how much honesty is involved…

    • Pekkka,

      a good historical example of what you are saying is the Drude model. It posits that electrons are classical in enough numbers when confined in a solid. There is a basic kinematic equation describing the force acting on each electron that, when solved for the appropriate situation, gives an answer that fits ‘reasonably well’ to observations. You may be familiar with this model if your friend works on lasers.

      But the Drude model, and other so-called ’empirical models’, is flawed physically. Just as your friend’s laser theory is flawed. That is to say, it is practical for a well-trained experts to use such a theory because he/she understands its flaws and faults. It works for back of the envelope calculations which are quite important in the lab.

      What happens when we are trying to determine a ‘physical understanding’, however?

      In such cases, it’s my opinion that we must do our damnedest to get to the meat of a problem. Even if that means dispelling a computationally practical and useful formalism like the Drude model. Because the Drude model doesn’t give us transistors or quantum wells or superconduction…or lasers for that matter. Having relied on the Drude model takes away from our understanding of reality.

      In the same way, while Mr. Johnson’s attempts might seem like an interesting facet of science, they fundamentally take away from a broader understanding of reality. There is no basis in it’s being real other than the words on a pdf. It is especially problematic since so many here are willing to simply regurgitate his memes without any skepticism at all.

      I think the most important aspect of doing science, as Mr. Johnson claims he is doing, is determining whether or not you can handle being wrong. If you cannot handle such an outcome, as Mr. Johnson’s reaction to the criticism he has faced here makes me think, you are not interested in science. I don’t think Mr. Johnson is interested in science.

      I’d be interested in your take on that.


  36. Steven Mosher

    You and John Sullivan utterly mis understand the concern about the “united front”
    If, for example, the AGU were to offer some session time to discuss skeptical
    issues, the first question is WHICH skeptical positions should be given time?
    If, for example , a research center were to open its laboratory time to test skeptical ideas on GCMs WHICH skeptical positions should be given time.
    It was a PRAGMATIC discussion about a PRACTICAL problem.

    Now then warmists could pick the WORST skeptical ideas and only discuss those. this is what realclimate does.

    • Give an some specifics for a really good skeptical idea that Real Climate has ignored.

      Paul Middents

      • As far as I remember RC has covered just about every paper which has been promoted by the skeptics in recent years.
        In the end there aren’t good skeptical ideas and bad skeptical ideas, there are just good and bad ideas, and good ideas will generally get proper consideration.
        Maybe there are some exceptions – if someone can provide evidence that there are good, credible ideas out there which are not being considered then fine, until then I remain, well, skeptical.

      • Andrew Adams,

        I will be glad to give you an example of bad ideas that RC still supports. Hockey Sticks. Have they admitted yet that Mann’s and associated work are all severely flawed and should be withdrawn? That they do NOT support the claims they make?

      • Hi kuhnkat, I suspect that none of the “Hockey Team” ground staff at RealCLimate have been allowed to read respected investigative science journalist Andrew Montford’s excellent exposé “The Hockey Stick Illusion” ( This was declared by another respected investigative science journalist Matt Ridley ( as being “…a rattling good detective story and a detailed and brilliant piece of science writing .. ”.

        Ref. your comment yesterday at 11:48 pm, for Andrew to “ .. Dig harder man!!! .. ”, he had the opportunity on 1st July and because he refused to remove his blinkers he threw it away. Investigative science journalist? – pull the other one.

        Best regards, Pete Ridley.

      • Montford is a “respected investigative science journalist” by what standards exactly? Has he won awards of his colleagues? Has his work appeared in prestigious publications?

        I haven’t read Montford’s book but my experience is that those who have make lots of charges regarding Mann that they can’t actually defend, most likely because they are false. (At least if they are true, noone has provided evidence to rebut my evidence that they are false.) I am not sure whether they got this info from Montford but that has been my impression.

      • I strongly recommend reading Montford’s book. It is very well written and extremely well documented.

      • Joel, instead of waffling from a position of ignorance try reading the book and following the references, do your own assessment then go over to the blogs of Steve McIntyre’s blog ( and Andrew Montford ( and try to convince them that you know better than they do. Let me know how you get on. They may let you co-author a paper with them on the subject.

        Best regards, Pete Ridley

      • Pete and Dr. Curry:

        Well, next time I find it in a bookstore, I will look through it and see what it has to say about the “censored” directory and about the Tiljander proxies. If it just repeats the same unsubstantiated nonsense that I see from people like “Smokey” on WUWT, I will be very unimpressed. If not, then maybe it is more worthwhile.

      • Joel, rather than reading the books, try going to their websites and reading the archives. Especially Climate Audit, Steve McIntyre’s site, as he was central to debunking the Hokeystick. You might even ask him directly about the “censored” directory with r2 information that was not published as he wrote about it first I believe.

        Of course, even if that was an inflated anecdote by some unknown person, the fact is that the r2 statistics for the Hokeystick FAILS! The difference is whether Mann knowingly misled people or is just sloppy and ignorant about the statistical methods he uses.

        Here is a start at CA:

        Be sure to ask Steve directly about how he knows the “censored” directory really came from Mann’s FTP server.

  37. To Pekka: You seem to agree that macroscopic physics cannot be modeled by
    quantum mechanics, and so macroscopic equations are needed for atmospheric radiation. Now macroscopic radiation seems to be well described by Maxwell’s equations, modulo the difficulty of the ultraviolet catastrophy, which destroys everything. What I suggest is a rational way to avoid the catastrophy and keep the great advantage of Maxwell’s equations as compared to primitive particle statistics. Isn’t that something to think of a bit, in the spirit of Planck and Einstein, rather than dismissing without reflection?

    And the radiative transfer equations are much cruder than Maxwell, right?

    • Claes,
      I do not agree that quantum mechanics cannot be used in those parts of atmospheric physics, where it has been used. What I was saying that in some situations the agreement with quantum theory can be obtained in surprisingly many different ways. Even for effects where the quantum theory differs from traditional classical physics the correct results may sometimes be obtained in ways where the quantum effects are somehow hidden.

      Hamiltonian formulation of mechanics allows for presenting some quantum effects in less common fashion etc.

      Einstein was not happy with quantum theory. I think that the main reason is related to conceptual difficulties in joining it with general relativity described in the elegant ways that he had developed. His dissatisfaction came out also in his statement about God not playing dice or in his paper with Podolski and Rosen, which has now been proven to be in conflict with experiments after Bell had formulated his inequality along the lines of that paper. In this case Einstein erred and quantum mechanics prevails.

      The problems in interpreting quantum mechanics are also related to some of the possibilities of doing the calculations differently. The quantum mechanics is, however, extremely successful in giving correct predictions with high accuracy. Thus it is a very good and valid physical theory in pragmatic sense. Most physicists do not worry about the philosophical problems and do their work successfully. Whether the philosophical difficulties turn out to have some relationship to the next paradigm, which would solve the problems of Einstein and unify gravity and quantum mechanics in a elegant way, remains to be seen. Perhaps not by our generation, but our children or grandchildren.

      I am still not telling the name of my former colleague, but I can tell that he has been later professor at KTH. When we were working at the same institute, we had some very interesting discussions on the foundations of quantum mechanics.

      • I add that sometimes it has also turned out that results generally thought to depend on quantum mechanics turn out to be true in more general settings. This is not very common and I cannot give examples, but I have certainly heard about such cases.

    • Claes,
      Concerning back radiation I certainly believe that it is a useful concept and that the radiative energy transfer can be handled most easily by including it in the calculation. I cannot figure out, how all correct results could be obtained without considering it explicitly. On macroscopic level avoiding it may be possible, but on the more detailed microscopic level it seems almost impossible, but only almost.

  38. To Judy: OK so now you say the KT is basically correct and that backradiation is a real physical phenomenon. Very good because we now have something concrete to discuss.

    May I then ask you about the equations describing your effect of backradiation? Without equations anything is possible.

  39. Mr.Johnson:
    In your description of a IR camera you admit that the instrument , directed appropiately, show radiation. At the same time you negate that this radiation reflect some reality. Could you please explain this ? I am extremely confused.

    • I think that what Claes is saying, is that the radiation you measure is a result of the temperature. Not the other way around. Sounds good to me.

  40. To Judy: Do you claim that radiative transfer equations model backradiation?

  41. I hope that this is relevant to the discussion:

    Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Still
    July 23rd, 2010 by Roy W. Spencer, Ph. D.

    • Frank Davis

      Google the famous “Pictet Experiment”

      Its of great historical importance and quite relevant to this discussion.

      IPCC adherents miss out is that a colder object can make a warmer object colder than it would be in the absence of the colder object.
      Why do they have this blindspot?

    • Thanks, Dr. Curry, for hosting this debate.

      @ Frank Davis:

      Dr. Spencer says in his article: “So, once again, we see that the presence of a colder object can cause a warmer object to become warmer still.”

      However, the process he refers to is not heat transfer by radiation from a colder system to another system, but a kind of isolation, like in a thermos. Yet, the colder system is not providing “more” energy to the warmer system, but just would be avoiding that the warmer system emits heat towards the colder system. This argument is not true because the thermal energy is transferred to the colder system, invariably, unless the colder system is a perfect reflecting material or the colder system has a very low heat capacity. I would make Dr. Spencer to recall that the Earth is not a thermos; his argument could be possible if the highest layer of the Earth’s system, i.e. the thermosphere, had a mass density higher than that of the surface. It’s not the case for the real Earth.

      On the other hand, if you wish to consider QM on this thread, you must include also the well-described by Einstein induced emission, which has been corroborated experimentally and in the construction of some devices, and the well-know and demonstrated radiation pressure. These two real physic phenomena debunk any idea of a “backradiation” from the atmosphere warming the surface.

      • Nasif,

        I think you are confused. Roy’s posts on this topic are very clear and definitely show that there is in fact backradiation toward the surface from the atmosphere. He ultimate experiment used an IR thermometer to measure the actual temperature of the air several hundred feet above via the IR light it emits.

        Even more confusing are your statements concerning stimulated emission and radiation pressure. Can you please explain specifically how those physical processes play a role in radiative transfer or the lack there of in the atmosphere?

      • @ maxwell… No more than you are.
        Roy’s “experiment” only demonstrates that there is energy flow by radiation, whatever his conclusions could be.
        Both processses, induced emision (or induced radiation) and radiation pressure, influence radiation. If you know what those terms mean, you won’t be so confused, as you are, on “backradiation” issues.

      • Nasif,

        you’re damn right I’m confused. You still haven’t provided a physical mechanism for how those interesting terms have to with the most important aspects of radiative transfer.

        On the point of Roy’s experiment, what type of energy transfer is he measuring? If the IR thermometer is conducting energy from the surrounding air, his thermometer would have measure about 300 K. Instead, his thermometer read around 200 K. Where does that difference come from in terms of energy transfer? Are you saying that the thermometer can conduct energy from the upper reaches of the lower atmosphere without conducting through all the layers?

        That would be a monumental theory!

        Also, if you’re going to charge that a particular person doesn’t understand some terms you use, you should make sure you know what you’re talking about. I have extensive experience in classical optics, quantum optics, atomic and molecular spectroscopy and nonlinear optics (I built an optical parametric amplifier over the past week in fact) so I KNOW those terms you’re using have absolutely nothing to do with this discussion, the greenhouse effect or ‘backradiation’. It’s a purely quantum mechanical, spontaneous effect.

        It was an interesting go at it though.

      • Dear Maxwell,
        Please, visit again Roy’s experiment and see what the box floor is and on what kind of surface it was placed on. You’ll get the answer.
        Regarding induced emission, you should not forget the natural photon streams, so upwards, during nighttime, as downwards, during daytime. On the other issue, if you make the proper calculations on radiation pressure, you’ll find that the downwelling radiation heating up the surface is not possible in the real world.
        If Dr. Curry, the owner of this blog, grants me permission to go out of topic, I will proceed to answer properly your questions.

      • Nasif,

        the hole just keeps getting deeper.

        The experiment to which I was referring had Roy traveling around in his convertible sedan pointing an IR thermometer into the air. Not his make-shift holhraum.

        In that case, where he is clearly measuring the temperature of the atmosphere directly several hundred feet above him, how does energy interact with the thermometer to produce a reading of 200 K?

        I’ll give you a hint, it has nothing to do with radiation pressure.

        It’s becoming more and more clear to me that you are using words that have one meaning to you, but a totally different meaning to actual optical scientists. You ought to look into the ways in which these terms are used in scientific circles so that you can more easily communicate your points in a scientific debate.

      • Dear Maxwell,
        Don’t go further on this or you’ll get disappointed on your own limitations about those concepts. Take your book on Radiative Heat Transfer and you’ll see I’m absolutely correct. I don’t want to go further on discussing those concepts because they are out of topic and I respect the admonitions of Dr. Curry on the purpose of this blog thread.
        Well, what the ground on which Roy placed his box and what the floor of the box was? Could you be so kind as to tell us what was it, specifically?
        Third, when he was “meassuring the temperature travelling around on his convertible sedan”… maxwell, tell me honestly, don’t you know how thermometers work and what thing makes them work?

      • Nasif,

        I’m trying to determine if your lack of comprehension of my comments is due to the possibility you are not a native English speaker or just plain stupidity. I’m willing to give you the benefit of the doubt and assume the first option, but not for too much longer.

        Again I’m discussing Roy’s use of a IR thermometer, not his makeshift holhraum. Please make an important mental note of that fact and stop your persistent confusion over this fact. It’s making you look dumb.

        An IR thermometer measures IR light (heat) emanating from a body or gas. Therefore, if Roy is pointing this thermometer at the sky, the thermometer reads the temperature of the sky via its IR emission. Therefore, the atmosphere is emitting IR radiation toward the ground that began its journey in ‘life’ at the surface, making it ‘backradation’. QED.

        As for radiative transfer, I’ve extensively studied ‘Introduction to Three Dimensional Modeling’ by Washington and Parkinson. From this text I am able to recover what both quantum mechanics and thermodynamics imply should be a downwelling IR emission from the atmosphere.

        Do you have other certified texts that you feel are better than Washington and Parkinson?

        Furthermore, you continue to lack any sort of meaningful physical description of what you are talking about. Based on the plethora of these facts so far, I must say I don’t think very highly of your opinion on this matter.

        It’s been real though.

      • Maxwell, I have an analogy I think is quite good.

        If you have a 6Volt potensial, and a “current sink” at 1 Volt, you will have a current from 6 Volt to 1 Volt. Increase the “sink” to 4 Volts.

        The 6 Volt source will drain slower. But the current is still seen as going from 6 Volts to 4 Volts. But we dont talk about “back-current”. That would be confusing.

      • Maxwell,

        darn, you probably will never see this to answer my question, but, just on the offchance that someone does and can:

        I am under the impression that the atmosphere absorbs virtually all of the IR radiation (except the window) within about 15ft of the surface. if this is so, exactly what was Dr. Spencer measuring from the ground?? Wouldn’t downward IR also be absorbes so that all he would be able to measure would be about 20 feet over his head and not an average of several hundred feet????

  42. Again Judy: Which equations do you claim model backradiation? As I said I want equations and I want the equations to be motivated or derived mathematically. Which equations are you referring to?

    • The equation that models backradiation is the Planck law for the intensity of light emitted by a blackbody at a specific temperature. This equation is carried out for every layer in the atmosphere, which has a stratified temperature profile. The absorption of light, all frequencies, is modeled by the Beer-Lambert law which is easily derivable from Maxwell’s equations via the electromagnetic wave equation.

      If we wanted to get down and dirty with the most fundamental equation governing the behavior of absorbing material to first order in the perturbation due to the interaction with light, we would have to use the quantum master equation with a phenomenological coupling to the vacuum field. We can go to second order in the perturbation to get to scattering processes if we liked as well.

      Ever wonder why the sky is blue?

      This process would allow us to see absorption and spontaneous emission (the dominant form of emission in the atmosphere) on a per atom/molecule level. The Beer-Lambert and Planck laws get the overall average effect of the quantum master equation in this context.

      So from first principles, we can easily calculate (grad school quantum problems) the rate of absorption and emission of a particular molecules when the light in question is on resonance with a particular allowed quantum transition, the linewidth of that transition based on different broadening processes and the necessary equipment to test the predictions of any such calculation. From there, we can sum over all the molecules in our volume and get an answer to compare to the observational laws used in climate models.

      You can see whether the agreement between these methods is good. Let me know how it goes.

      • Maxwell,

        doesn’t CO2 absorb and emit based on its molecular bond configuration as opposed to planck energy?? Maybe someone can jump in and explicate what the difference is if any??

      • For example, CO2 emits at 15 microns at an intensity according to the number of molecules and their temperature using the Planck function for temperature that wavelength (which actually peaks not far from 15 microns for normal atmospheric temperatures). This emission is seen at the ground as part of the back-radiation, together with all the other CO2 and H2O bands in clear sky that make up all the back-radiation.

      • kuhnkat,
        Not one opposed to the other but both combined.

        The molecular properties determine which wavelengths have strong emission and absorption, i.e. they determine the emissivity, which is equal to absorptivity. Planck’s law tells how strong the emission is at those wavelength as the strength is a product of the emissivity and Planck’s law for black body at that wavelength.

        When the emissivity of a gas is strong for a particular wavelength, it means that its is significant already for a thin layer and very close to one for a thick layer in accordance with the Beer-Lambert law. Then the the strength of emission at that wavelength is the same as for a black body of the temperature of the gas. This is true for those wavelengths, but at other wavelengths the gas does not emit at all or very little.

      • Then Planck’s laws are applicable to emission whether it is from level changes in single atoms or bond interactions.

      • As Pekka pointed out above it sets the upper limit at any wavelength, if there is no allowed transition at a particular wavelength then the emission will be zero no matter what the Planck value is. The Co2 band at 15μm will emit strongly up to the Planck limit. O2 can emit in the UV at around 220nm but in the atmosphere the Planck limit will be generally so low that this emission will be very weak (Judith made this point earlier).

  43. To Lucia: If you had read the equations I refer to as Navier-Stokes you would have seen that they express conservation of mass, momentum and total energy
    and are the basic equations of thermodynamics describing transformation
    between kinetic and heat energy through work. Are you familiar with thermodynamics?

    • Claes,

      The basic equations of thermodynamics are called “The first law of Thermodynamics” and “The 2nd law of Thermodynamics”. One of the clues is that these equations contain the word “thermodynamics”. In contrast, conservation of momentum is “mechanics” and the navier-stokes equations are basic equations for fluid mechanics.

      Conservation of mass is used in analyses, but that doesn’t transform the equation into “a basic equation of thermodynamics”.

      Are you familiar with thermodynamics?
      I’m laughing myself to tears here. I am familiar enough to know that you are making errors. :)

      • This is lucia.

      • Kim— You are correct. I don’t know why wordpress auto-filled the name incorrectly. I should have seen that.

      • The mistake is to think that it is possible to consider thermodynamics and fluid mechanics as separate in a gas or liquid.

        They are inseparable in the context of free atmospheric thermalisation.

        To imply otherwise is erroneous, perhaps even fallacious.

  44. To Lucia: The convective adjustment that you think is science, is just an ad hoc fix up without any mathematical basis. If you are allowed to adjust what your
    equations tell you, then you can get anything you want.

    • Still waiting for you to apply your model to a real world example such as those which I showed above. Most applications of standard radiation heat transfer have a substantial overlap between the incoming spectrum and the emitting spectrum by the way.

    • Claes–
      Since your paper suggests you think the first law of thermodynamics is the 2nd law, and the navier stokes equations is the basic equation of thermodynamics, I am not surprise that you think the convective adjustment is just an adhoc fix up. To understand the physical motivation, will need to apply thermodynamics. At my blog I gave you a tip on how to distinguish the first law from the second:

      The second law should contain an inequality symbol ≤, a symbol that represents entropy (S is often used), and a symbol to represent temperature (T is a popular choice, but rebels sometimes use θ). Also, if I recall correctly, it generally contains no work term (i.e. W would not appear.)

      As for this:

      If you are allowed to adjust what your equations tell you, then you can get anything you want.

      Yes. I agree.

      • Also lucia.

      • The above is me– Lucia.

      • Lucia,
        I did not check carefully, but I think the equations that Claes is presenting do present correctly the second law. The inequality is hidden in the requirement that D ≥ 0.

        The formulation is not the one we all have seen most often, but I think it is correct.

        The same statement that Claes presents correct formulas in a less conventional way seems to apply to the other chapter as well, but there I have doubts on, whether all equations are correct or only some of them. I did not read in this text at all carefully or study the equations more than superficially as I do not think that his approach is useful even when it is correct. Many of the claims in the text are strange if not outright wrong.

      • Pekka–
        Specifying 0≤D where D is dissipation is a consequence of the 2nd law of thermodynamics. However, it does not turn those equations into the 2nd law.

        That equation may be a correct representation of something but it is not the 2nd law of thermo. This is not a matter of notation. Other puzzling things about that equation may have something to do with non-conventional representations — for example, it’s not clear to me that it’s even a correct formulation for the first law. But in order to pinpoint the problems, I need to know whether that’s supposed to be a control volume formulation or an analysis on a fixed volume, and possibly where the boundaries are etc. My impression is it’s supposed to be a control volume with the top at the top of the troposphere– but if so quite a few terms may be missing. (Or not. It depends on whether we have a control volume whose shape is permitted to change– in which case…. well…)

      • Lucia,
        My purpose is not to defend the book or conclusions presented by Claes Johnson in the book. I certainly disagree on very many things. I am only noting that texts that are obviously wrong, when they lead to definitely wrong conclusions may not be wrong in all of their details. Most people seem to agree that this chapter is actually correct in what it describes. Its content may be used in reaching wrong conclusions outside its range of validity, but that is another matter. It is also possible that the unconventional way the equations are presented contributes to wrong conclusions, but even so the equations may be correct.

        Claes Johnson presents two inequalities in eq. (2). They are equivalent when combined with the first law /eq. (4). Of course this is not the most general presentation of the first and second law, but for the problem considered they appear to be equivalent with the general formulation. It is clear that using these laws as more basic than the general formulation may lead to errors. Perhaps such an error is really done, when considering radiative processes in the other chapter. I am not really interested enough to even check.

        Also in this chapter the formulas (5) and the related discussion are obscure. If not for other reasons then at least in the total neglect of considering units properly. The equations can only be valid in units where temperature is dimensionless (i.e. 1 K = 1) and the unit of acceleration is inverse of the unit of length. Furthermore it is stated that specific heat capacity cp = 1. Whether all that is possible at all is certainly not obvious. But then again all that is more or less forgotten when the next formulas are standard knowledge.

        The whole paper is confusing and may well be misused, but even so it is good avoid erroneous claims about its content.

      • Pekka–
        I have never suggested things that are wrong in their results must be wrong in all their details. I am pointin
        I am saying is that those equations are not “The second law of thermodynamics”. The reason I am saying they aren’t is that they aren’t.

        In undergraduate fluid mechanics problems, students solving pipe flow and other simple problems, often use an equation referred to as “the mechanical energy equation” or sometimes “the energy equation”. It is derived from conservation of mass and momentum, sort kinda-sorta like the first law of thermo and includes a dissipation term. The 2nd law requires that dissipation term to be positive.

        So, using that equation lets students impose the requirements of the 2nd law on their analysis. However, you don’t get to call that equation “the second law of thermodynamics” merely because it permits students to correctly incorporate the effects of dissipation on pressure drop in pipeflow.

        Likewise, what Claes writes down is not the 2nd law of thermodynamics.

        Moreover, I find your clain that

        Of course this is not the most general presentation of the first and second law, but for the problem considered they appear to be equivalent with the general formulation.

        To be rather dubious.

        In fact, based on the text, I’m not convinced it is possible to pin down what “the problem considered” really is.

      • Lucia,
        At least I agree on your last point. Reading the text of CJ it is often very difficult to pin down what he is writing about or where he is aiming to.

  45. Dear Friends,
    I come late to the interesting discussion, so I did not read through all. Therefore I do have a remark.
    A flat hot body with two sides, unit heat capacity and with time dependent temperature Th(t), starting at Th(0)without an internal or external energy source cools from both sides with the rate dq/dt = sigma Th(t)^4 per unit area.
    Now you put a cold body with Tc(t) adjacent, facing exactly one side without touching, the hot body cools from this side with the rate dq/dt = sigma*(Th(t)^4 – Tc(t)^4) per unit area and with dq/dt = sigma Th(t)^4 per unit area from the other side. Therefore the hot body in both cases is cooling all the time, since Th(t) is always greater or equal to Tc(t).
    However, the hot body Th(t) stays in the second case warmer all the time than in the first case. But this is different from saying it gets warmer than initial Th(0). If Tc(0) is smaller than or equal to Th(0), then Th(t) is always smaller than Th(0).
    Of course as Roy Spencer showed a hot body with an internal or an external energy source can get warmer than Th(0), if you put a cold body adjacent to it.
    Best regards

    • Dear Günter…
      Anyway, the colder system IS NOT heating up the warmer system, but cooling it, continuously, if we wish, but only up to the point of equilibrium, i.e. when both systems reach the same energy density. And even so, the internal or external source of heat would continue heating up the warmer system.
      Take off the internal or external operator, for example, and you’ll see the colder system cannot heat up to the warmer system but quite the opposite.
      It is the internal or external PRIMARY heat source what heats up the system, not the colder system. The latter is Dr. Spencer’s argument.

      • Dear Nasif,
        that’s what I wrote if you reread my paragraph..
        Therefore I said: “Of course as Roy Spencer showed a hot body with an internal or an external energy source can get warmer than Th(0), if you put a cold body adjacent to it.”
        Of course it is the energy source that heats the body up.
        I think it is important not to confuse “getting warmer” or “keeping warmer” with a energy source that heats a body up.

      • Dear Günter,
        Yes, you’re right. I misinterpreted the last paragraph of your post. Sorry… You’re also right on not confounding “getting it warmer” and “keeping it warming”.
        All the best,

    • In his discussion of a hot plate next to a cold plate, Dr Roy Spencer says:

      The 2nd Law of Thermodynamics: Can Energy “Flow Uphill”?
      In the case of radiation, the answer to that question is, “yes”. While heat conduction by an object always flows from hotter to colder, in the case of thermal radiation a cooler object does not check what the temperature of its surroundings is before sending out infrared energy. It sends it out anyway, no matter whether its surroundings are cooler or hotter.

      Yes, thermal conduction involves energy flow in only one direction. But radiation flow involves energy flow in both directions.

      Of course, in the context of the 2nd Law of Thermodynamics, both radiation and conduction processes are the same in the sense at the NET flow of energy is always “downhill”, from warmer temperatures to cooler temperatures.

      But, if ANY flow of energy “uphill” is totally repulsive to you, maybe you can just think of the flow of IR energy being in only one direction, but with it’s magnitude being related to the relative temperature difference between the two objects.

      Clearly Spencer thinks that radiative heat transfer is completely different from conductive heat transfer, and can go ‘uphill’. He writes:

      The only way I know of to explain this is that it isn’t just the heated plate that is emitting IR energy, but also the second plate….as well as the cold walls of the vacuum chamber.

      Does that mean that while radiative heat transfers don’t ‘check’ to see which way to go, conductive heat transfers actually do ‘check’?

      • Frank,

        Does that mean that while radiative heat transfers don’t ‘check’ to see which way to go, conductive heat transfers actually do ‘check’?

        The separation is not that clear. On molecular level even conduction may “go uphill”, but this is not visible and can be ignored. In conduction as in radiation energy goes in both directions at micro level. In conduction this is related to the motion of energetic atoms or molecules or to vibrations (phonons) in solids. The distances are usually very short. Therefore only the collective conduction is observable and described by an equation that describes only the net flow.

        In radiation it is often possible to set measuring equipment to detect separately radiation in each direction. One photon may go over a large distance etc. The back radiation is thus observable and it may also be that the easiest way of calculating the net energy transfer represents separately the two directions. In some cases it may be easier to consider directly the net flow, but as I said above, this is not always true.

      • On molecular level even conduction may “go uphill”, but this is not visible and can be ignored. In conduction as in radiation energy goes in both directions at micro level.

        Ah! So there is ‘back-conduction’.

      • Does that mean that while radiative heat transfers don’t ‘check’ to see which way to go, conductive heat transfers actually do ‘check’?

        I would express it differently. Conduction describes the *process* by which heat flows along an existing temperature gradient. Radiation is a something that a body *does* based on its temperature and emissivity. The former process directly involves both/all bodies that define the local temperature gradient; the latter by definition only depends on the characteristics of the radiating body itself.

        At least that’s the way I look at it.

      • The approach used in describing conduction can easily be extended to part of radiative heat transfer, to those wavelengths with strong absorption. Heat is transferred in accordance of essentially the same diffusion type differential equation in atmosphere by radiation near the center of the 15 um IR band.

        For wavelengths with weak absorption this approach does not work well, because such radiation does not proceed with small steps in diffusive fashion but by long leaps to a point where the temperature may be significantly different or even escape through the whole atmosphere. Most backscattering occurs in the region where the diffusion-like process describes the heat transfer rather well. On this basis one could describe all this with the diffusion equation and remove most of the back scattering from being considered explicitly.

        The way the calculations are done does of course not affect what really happens, but it affects often the way this is described.

      • Frank,

        Radiative heat transfer consist of two radiative energy flux, one from hot to cold and one from cold to hot.
        Radiative heat or net radiative energy flows from hot to cold, radiative energy in both directions.
        It is a little bit confusing, since “energy” and “heat” are sometimes used interchangeably, which is strictly speaking a bit wrong. However, scientists are doing that occasionally and the reader needs to bring it into context. Bad style, though.
        The second law as stated by Clausius reads: “There is no change of state that only results in transferring heat from cold to hot.”
        Note, it is not energy in general.
        Heat in this context should not be interchanged with energy.

        Best regards

      • Frank Davis

        Look at the blackbody spectrum of an object at say 300K.
        Superimpose the BB spectrum of the identical object at 400K
        Now using the spectra predict what would happen if these two objects were brought closer together so that they radiate to each other.

        We notice that;
        1. The hotter object has at the short wavelength end, frequencies absent from the lower temperature object.

        2. Pick any wavelength that both objects have in common.
        You will notice that the hotter object is emitting more radiation than the colder once.
        Now examine the hot surface;
        It is emitting more radiation of every wavelength than it is receiving.

        You can now hopefully appreciate that a colder object can never increase the temperature of a hotter object.

      • I understand your point, Bryan. Perhaps you agree with Guenter Hess’s comment just before yours, in which he wrote:

        Radiative heat transfer consist of two radiative energy flux, one from hot to cold and one from cold to hot.

        If that’s how it is, then if it were possible to block or divert the radiative flux going from the hotter object to the colder object, while continuing to allow the radiative flux from the colder object to the hotter object (a sort of diode), then the colder object would heat the hotter object.

      • Frank Davis

        For the hotter object to radiate to the colder it must “see” the colder object.
        Since light rays must be able to travel backwards (rectilinear propagation) no such diode effect is possible.
        We therefore are forced to agree with Clausius that even for radiative transfer Heat only travels from the hotter object to the colder object.
        Yes I agree with Guenter Hess’s comments.

  46. I would like to take this section of Chapter 1 as a point of departure for my comments. It says:

    “We have formulated a basic model of the atmosphere acting as an air
    conditioner/refrigerator by transporting heat energy from the Earth surface to the top of the atmosphere in a thermodynamic cyclic process with radiation/gravitation forcing, consisting of ascending/expanding/cooling air heated by low altitude/latitude radiative
    • descending/compressing/warmingair cooled by high altitude/latitude
    outgoing radiation,
    combined with low altitude evaporation and high altitude condensation.
    The model is compatible with observation and suggests that the lapse
    rate/surface temperature is mainly determined by thermodynamics and not by radiation.”

    Yes, of course they’d like to formulate a simple “model” that works this way, as some of their other conclusions might nicely fall in line, and in so doing, to re-write some laws of physics in the process, but unfortunately, their simple thermodynamic model is simply not the way the real atmosphere of the planet works, nor in fact the way the laws of physics work.

    It takes hardly anything more than a few basic real world observations to provide proof that radiational balance is a far more potent regulator of atmospheric temperatrue then the authors of this book would like in their “simple” model. But then, isn’t that the point they are trying to refute? For observational proof, take the role of water vapor as an GH gas, using the predicted GCM forecasts that the planet will see higher night time temperatures due to the increase in water vapor keeping more LW radiation near the surface. Witness to this is the fact that 37 U.S. cities and hundreds of other cities across the globe set night time high temperature readings in 2010, a year in which saw a record in precipitation. Based on a their simple thermodynamic cyclic process, this result would not be expected as that additional night time heat at the surface would surely have been carried away via convective thermal processes and added to the TOA output. This increase in global water vapor, measured over the past few decades is exactly as predicted by every GCM when using well established and quantified GH physics with the addtional radiative forcing caused by the additional accumulation of CO2 and water vapor in the atmosphere. Warmer night time temps are exactly what one would expect when considering the real world (i.e. measured) absorbtion and retransmission of LW radiation by increasing amounts of GH gases in the troposphere.

    Futhermore, one only needs to step outside on a calm cloud-less winter night and then step outside on a similar night when is has a nice overcast sky to feel the radiative GH effects of the water vapor in those clouds.

    I would ask the authors this: how would their model explain the warmer night time ground temperatures as measured throughout the world if not for the LW radiative effects of additonal GH gases?

    • Futhermore, one only needs to step outside on a calm cloud-less winter night and then step outside on a similar night when is has a nice overcast sky to feel the radiative effects of a smaller delta-T between the earth’s surface and the water vapor in those clouds.

    • Some data/citations, please?

    • Futhermore, one only needs to step outside on a calm cloud-less winter night and then step outside on a similar night when is has a nice overcast sky to feel the radiative GH effects of the water vapor in those clouds.

      A word of caution – a clear night can feel much colder than an overcast one, even if the air temperature, as measured by thermometer, is the same.
      That’s because your perspiration evaporates more readily in drier air, so the perception of temperature can be largely subjective.

  47. R. Gates:

    “I would ask the authors this: how would their model explain the warmer night time ground temperatures as measured throughout the world if not for the LW radiative effects of additonal GH gases?”

    So, how do you explain the Medieval Warm Period?

    • Steven Mosher

      1. our knowledge of the extent and amplitude of the MWP is VERY uncertain.
      2. The presence of large amplitude warmings is evidence FOR long term natural oscilations, it is NOT evidence against the physics of radiation.
      3. The final temp is the result of many forcings, not merely C02.

      Basically, your comment is OT to the discussion of the physics of the tyndall gas effect

      • Thank you Steven, I couldn’t have said it better myself, though I would welcome a discussion of the MWP on some other thread, perhaps in the context of Dansgaard-Oeschger and their likely Holocene cousins, the Bond events, a subject which fasninates me to no end…

      • not to hijack the thread but mostly for my own clarification, can we also agree to the converse: that the existence of the physics of radiation are not evidence against long term natural oscillation?
        Discussions such as this one may frustrate some, but I do feel they go a long way to clarifying what aspects of the science are clear and where and why there is uncertainty and/or a lack of clarity.
        Moreover, can we acknowledge basic processes but still differ as to their relative impact, rate and magnitude of change, and, of course, our ability to adapt to the changes they invoke?

  48. To R Gates: Yes the model is simple but the point is that it is more complete (with thermodynamics) than a model with radiation only, which is the basic model of CO2 climate alarmism based on a “greenhouse effect” from radiation alone.

    • I would agree that both forms, thermodynamic and radiative, need to be included in any full understanding of the climate dynamics, but specifically, when speaking to the well-established science behind the GH properties of atmospheric gases, I believe the simple thermodynamic model falls far short, and can simply not explain or predict real world effects of GH gas increases as well as a GCM’s can when considering their full LW absorption/retransmission radiative effects.

      • Dear Mr. Gates,

        it is the other way around.

        The main physical reason for the effect of GH gases is not „back radiation“ , but rather the effect on the TOA balance, which is a decreasing outgoing longwave radiation (OLR), before reaching a new stationary state.
        “Back radiation” is only an internal energy flux that does not alter the energy content of the earth system. Changing OLR, however changes the energy content.

        The concept of emission height or “cooling to space” together with thermodynamics/lapse rate is enough to explain the greenhouse effect. Heat transfer by radiation, latent heat or sensible heat is enough. “Back radiation” is a parameter included in heat transfer by radiation ,though.

        Absorption/Reemission or “back radiation” alone cannot explain the greenhouse effect: I know that there are texts out there that try that, but they stay incomplete.

      • I agree that back radiation shouldn’t be invoked as the “cause” of surface and atmospheric warming. A TOA flux imbalance is required for the temperatures to change, but the mechanism by which the imbalance is transmitted by the atmosphere to the surface involves back radiation. If downward radiation to the surface didn’t increase as a result of greenhouse gas forcing and the consequent TOA imbalance, the surface wouldn’t warm.

      • “…but the mechanism by which the imbalance is transmitted by the atmosphere to the surface involves back radiation…”


      • In the context of the greenhouse effect surface and troposphere warm simultaneously because of the TOA imbalance, we have a radiative – convective equilibrium. The sun warms the surface. The net effect of longwave radiation is cooling to space, integrated across the globe.
        Back radiation increases with temperature, not the other way round. Back radiation is a parameter in the energy balance of the surface, even though you can measure downwelling radiation. Downwelling longwave radiation can heat a patch of surface, if the air is warmer on top of it. However, globally integrated downwelling longwave radiation is more than balanced by sensible heat, latent heat and radiative energy from the surface. Otherwise we would not have an decreasing temperature gradient with height on average.

      • Back radiation increases with air temperature, and in turn increases the temperature of the surface. That is how atmospheric heating from an energy imbalance is transferred to the surface. If the lapse rate is linear, the temperature changes equally at all laltitudes. In reality, lapse rates may not always be perfectly linear, but the approximation is a reasonably good fit with observations.

        It is not correct to imply that downwelling radiation only heats the surface if the air is warmer on top of it. It heats the surface even when the air is cooler, as is typically the case.

      • To avoid confusion about terminology, my point is that back radiation from an atmosphere cooler than the surface makes the surface warmer than it would be otherwise. The net IR flow is from the surface upward.

      • Steven Mosher

        Thanks Guenter.

        You will note however that now the conversation has shifted from Johnson defending his mistakes to you explaining how things really work. They are of course related.

      • Are we discussing CO2 greenhouse effect, or general atmospheric warming?

        It is important to note that there is no way to determine if “downwelling” IR has been emitted from CO2 or any other atmospheric molecule.

        All molecules and therefore all gas molecules emit IR.

        So “downwelling” IR should be expected. But that does prove a net increase in energy, or “greenhouse effect”.

        If you cannot show with real world experiment that more CO2 = higher temperature, you fail.

        “More CO2 = less temperature”

        Why? Because of specific heat capacity.

      • But that NOT does prove a net increase in energy, or “greenhouse effect”.

        I should have said!

      • “It is important to note that there is no way to determine if “downwelling” IR has been emitted from CO2 or any other atmospheric molecule.”

        The origin of downwelling IR can be identified by its spectral signaature. Almost all will be from CO2 and water.

      • Incorrect.

        The spectral signature is not determined by the substance that emits IR but by the temperature of that substance compared to the surrounding ambient temperature when the IR was emitted.

      • Claes Johnson, do you agree with Will since he seems to be on your “side” of the debate?

      • I am on no ones side Judith.

        I just happen to know that adding CO2 to the atmosphere does not cause warming. In fact it causes cooling.

        I have demonstrated it with experiment. I have given an explanation with supporting references with regard to specific heat capacity.

        Further evidence :

      • Will,

        If the experiment you reference is the one in the link given in an earlier comment: More CO2=Less Temperature”, then you should know that this experiment, even if conducted with utmost care and precision (which I doubt), proves quite the opposite of what you’re stating. For the container with pure CO2 SHOULD BE, by the very processes you claim don’t occur, be cooler than the one with “ordinary air”, as that “ordinary air”, would, I presume, contain ordinary water vapor, and as such, with a much greater percentage of “ordinary water vapor” and would naturally show a greater GH effect (assuming of course that all the other varibles are the same).

        In addition the experiment is flawed for many other reasons, for the title states “more CO2 = less temperature,” and in such an experiment one would expect to have a control container that is kept under the same conditions as all the others, and then one would expect that the only varible to change would be the amount of CO2 in a serios of other containers. One could then produce a series of data points that would show how the temperature of the container varied with the only variable being the change in the amount of CO2.

        All this aside, I highly suspect that the container with “pure CO2” is indeed that, as one can see condensation on the inside, and since CO2 (under these pressure and temperature conditions) is a non-condensing gas, then that condensation is most likely water vapor, so the entire experiment is invalid as the container is certainly not “pure CO2”.

      • Will

        I just happen to know that adding CO2 to the atmosphere does not cause warming.

        I agree.

        Another proof:

        In a scientific argument, the judge is the observation, not the theory!

      • Visit the HITRAN database. Each IR emitter has a spectral signature. The temperature of the emitter vis-a-vis its surroundings is irrelevant, and in fact, the temperatures are for practical purposes identical – i.e., they exist in local thermodynamic equilibrium (LTE). The temperature of the emitter does influence the quantitative balance in the intensity of one spectral line vs another from that emitter, but the wavelength of the CO2 and H2O lines is almost completely unaltered by temperature – at least within the atmospheric range of temperatures.

      • “the wavelength of the CO2 and H2O lines is almost completely unaltered by temperature – at least within the atmospheric range of temperatures.”


        All you need to know:

      • Will – Using emphatic language (“Nonsense”) doesn’t strengthen a case that can’t be made. To the extent the site you link to is informative, it confirms my statement. It refers to positions, intensities, and line widths of CO2 and H2O, but with no suggestion that the wavelengths oft these molecules are shifted by temperature. Any such change under atmospheric conditions would be miniscule. If you have data to the contrary, link to it specifically rather than citing a long list of article titles.

      • Wrong again Will, the spectrum is determined by the identity of the emitter, however it can not emit more at any wavelength than that defined by S-B.

      • Which is determined by its absolute temperature. Which in turn is determined by its surrounding ambient temperature as per its altitude.

        Say above 5km @ -80º C .

        As for your comment below :

        “Not true, O2, N2 and Ar notably in our atmosphere do not!” (emit IR)


        So that would mean that 99% of the atmosphere cannot cool to space via radiation at TOA right?

        Come on!

      • Judith Curry wrote in her posting:

        “I was hoping to put to rest any skeptical debate about the basic physics of gaseous infrared radiative transfer.”

        We see again that there is little sign of that becoming true.

      • Downwelling IR is not “Backradiation”.

        Downwelling IR does not add energy to the system because it is energy which is already present. It does not cause net E increase.

        Let us leave the subject of downwelling radiation there.

        The so called “Backradiation” is the energy we expect to find from the claimed “greenhouse effect”.

        The ability of a substance to absorb/emit, or radiatively transfer IR does not say anything about its ability to store that energy.

        Increasing CO2, increases the radiative transfer properties of the atmosphere in the far infra-red region. How is that even remotely like a “greenhouse effect”?

        How does a decrease in overall resistance of a poor conductor such as air, produce an increase in temperature? It is unphysical.

        It is the opposite of reality.

        “The physics of deep convection have been formulated since 1958 and are based on sound thermodynamics and measurements on location. The trends of the temperature in the high atmosphere in the last half century are very negative, starting on this height where the convection reaches. That means that more CO2 has a cooling effect rather than a warming effect.”

        See here:

      • So that would mean that 99% of the atmosphere cannot cool to space via radiation at TOA right?

        Correct, now you’re getting it, which is precisely why the change in CO2 concentration is so important (999645 ppm of the atmosphere does not absorb or emit IR).

      • Phil you are silly.

        ALL substances above 0K emit IR. That is not controversial physics.

        Your misleading statement has been repeated many times by the warmist’s but repetition cannot make it true.

        Why are you here making such false statements and clouding the issue?

      • Steven Mosher

        “It is important to note that there is no way to determine if “downwelling” IR has been emitted from CO2 or any other atmospheric molecule.”

        Wrong. Please tell me you have nothing to do with the design of aircraft, sensor systems, or other devices meant to protect our country.

        Start with this design guideline.

      • LOL – Very good Mosh. As a former weapons instructor I appreciate why you attached this citation. Those rocket scientists certainly knew a thing or two about missle guidance, CO2 and the IR spectrum.

      • Another non-sequitur Steven?

      • All molecules and therefore all gas molecules emit IR.

        Not true, O2, N2 and Ar notably in our atmosphere do not!

      • Imho Johnson first chapter is quite good, and is consistent with the explanation Guenter gives. Chapter 2, on the other hand, is…hum, well, it is clearly inferior to classical black body radiation, which is the most polite thing I can say ;-)

        Problem is that below the troposphere, heat is exchanged by both radiation and convection (with latent heat release ), only conduction can mostly be ignored. So no simple model, either purely convective or purely radiative, is complete. However, all flux analyses I have seen show clearly that more heat is transported by convection (and a lot more when latent heat release is present) than by conduction. It follows that, if a simple model including only one heat transfer mechanism has to be chosen, better to use a convective one.

        Moreover, convective lapse rate is a stability condition, so I see it (and, from what I get, classic climatology “above the atmosphere=rigid shell level” see it the same) as a limit for temperature gradient that can not be exceeded, due to stability reason. It thus makes sense that one can derive a max ground temperature from TOA temperature using this lapse rate, without knowing exactly how much the heat flux. Above TOA, we have radiative transfer, so we know TOA temperature from S-B law. Heat flux is then determined by conservation of energy, convective heat flux is just what is missing to ensure equilibrium.

        The only error I see with this model is that it is too simple: 1D, and it does not take into account the fact that radiation is diffuse, so all radiation to space does not occur at a precise TOA level, it is only an average notion.

        But still, compared to simple shell-like purely radiative atmosphere (1D also, all those shells and the earth are considered perfectly conductive in the horizontal directions), the model with the lapse rate is head and shoulder above: at least it does not neglect the largest heat transfer to keep only the smaller radiative one because it is tractable.

        This is one of the biggest error in climatology vulgarisation: the CO2 blanket is completely wrong, but it may be enough for those allergic to science/mathematics. The radiative shell (or multishell) models are mathematically complex enough do deter those one, and thus is presented as a simple but usefull model. It is not, it is almost as wrong as the CO2-reflective blanket, and frankly, it paint a very poor image of climatology for those scientifically-minded enough to understand it, but who start to evaluate it compared to an earth-like atmosphere …

      • The first chapter has a major error in assigning the 10 C/km lapse rate to radiation while also referring to it as the dry adiabatic lapse rate. Radiation has nothing to do with the 10 C/km dry adiabatic lapse rate. A radiative equilibrium is isothermal, not isentropic. This mess confuses the whole later argument about lapse rates.

      • Jim D
        Bad news and good news

        First the Bad news
        …….”a major error in assigning the 10 C/km lapse rate to radiation while also referring to it as the dry adiabatic lapse rate. “……

        The dry adiabatic lapse rate is given by dT/dh = -g/Cp
        Where g = Gravitational Field Strength
        Cp = Heat Capacity.
        In other word the temperature acquired by air molecules after contact with the surface drops by almost 10K per Km of ascent.

        Now in the case of the dry adiabatic troposphere although water vapour may be absent, CO2 being well mixed should be there as usual.

        However it seems to play no part that I can see.
        Even more alarming, in this Nasa description of the atmosphere with various conditions specified there is no mention of greenhouse gases!
        Surely the radiative effects of CO2 must get at least a tiny mention, shouldn’t they?

        Now the good news
        The greenhouse theory has been banished to the TOA.
        The radiative gases radiate long wavelength EM radiation to space to attempt an overall radiative balance for the Earth.
        It acts like the drain hole at the bottom of the bath.
        The Sun acting like the water flowing from the bath taps.
        If the drain hole is too narrow, water level rises(temperature); if too wide temperature falls.
        Now back to a dry atmosphere; the temperature lapse rate will still fall at 9.8K/km in the troposphere.
        The net effect then of changing CO2 and H2O vapour is to move the tropopause up and down.
        Now this truncated version of the Greenhouse Theory is one that I think is very plausible.

      • In some way we can agree that the tropospheric lapse rate is fixed by the dry and moist adiabatic lapse rates, and therefore its whole temperature profile is linked to the surface temperature, which is in turn affected by a radiative balance. CO2 can’t change the lapse rate, which is based on physical constants, such as g, cp, latent heat constant, gas constants, etc., but can only affect the surface temperature to raise the effective radiating level of GHGs. The troposphere’s only degree of freedom is the surface temperature in this simplified model that represents CO2 effects in one atmospheric column.

      • The lapse rate is determined by thermodynamics of moist air as long as there is a sufficient heat flow from the surface to the upper atmosphere to keep the real lapse rate at the adiabatic limit. That requires that the surface is warm enough to release the required amount of energy excluding that part that escapes through the atmosphere without being absorbed. The heat flow is a combination of radiative transfer, convection and advection of latent heat. Convection is the part that guarantees automatically that the temperature gradient cannot exceed the adiabatic lapse rate. Therefore the strength of the radiative transfer does not influence the result as long as the surface is warmed so strongly that the adiabatic lapse rate would be exceeded without convection.

        Adding CO2 influences the situation in at least two ways. The first is due to the reduction in the amount of energy that escapes without being absorbed. Due to this effect less energy is leaving directly from the surface. The same applies also to the low clouds. In equilibrium all this reduction must be compensated by increased radiation from the upper atmosphere and increased heat flow from the surface to the upper atmosphere.

        The second effect occurs around tropopause. The increased CO2 concentration moves the effective radiating altitude of CO2 higher up.

        Combining both effects we notice that the radiation that escapes from the upper atmosphere must be both stronger and originate higher up. Both requirements lead to an increase in the temperature of the atmosphere at a fixed altitude if upper troposphere near tropopause. The two effects are separate. The first comes from the increase of CO2 at lower altitudes, the second from its increase at tropopause. My understanding is that the first effect is stronger than the second, but I have not done any calculations to support this conjecture.

      • Pekka
        Look at a description of the broad outlines of the atmospheres structure with particular emphasis on the troposphere.
        There is no mention of the greenhouse effect.
        The effect of water vapour is explained through the mechanism of latent heat.
        Of course CO2 and H2O radiate in the IR.
        It just doesnt seem to be that important.

      • Bryan
        That is a description of certain issues. That something else is not mentioned there is not an argument against that. I didn’t notice anything there that would in some way contradict what I wrote here or in numerous other messages on this site.

        It is also dishonest to pick one sub-chapter from the tutorial stating that it does not discuss greenhouse effect when the previous sub-chapter does discuss it. I think you might try to avoid being dishonest.

      • Pekka
        …”I think you might try to avoid being dishonest.”..

        I try to avoid using language like that.

        I have no way of knowing how honest you are but I give you the benefit of the doubt.

        I was genuinely surprised when I came across the NASA document.
        Beforehand I would have thought that the radiative effect of CO2 would have to be accounted for even in a dry adiabatic Earth atmosphere.
        In fact it would be a good experimental method of isolating the CO2 effect from the H2O effect in the limit.
        There seems to be a growing body of opinion that the radiative effects of CO2 are either minor or self cancelling.
        A number of IPCC advocates are now promoting this and say the real and significant greenhouse effect is to be found at TOA.

      • To put it simply, CO2 affects the absolute temperature, not the lapse rate in a dry atmosphere. This is why it is important. It displaces the whole temperature profile according to its radiative effect.

      • Bryan,
        I have become less polite to you after your baseless insulting comments towards me some times ago.

        I told you that the previous sub-chapter of the same tutorial tells that the CO2 is important. Why do you neglect that and choose to concentrate on the next, which discusses other things.

        If you find the chapters contradictory, the fault may be in your understanding of the content and its significance. For that the only help comes from studying the basics. Trying to make guesses from more advanced texts (even when they are tutorials like in this case) leads often to such misunderstandings that are visible on this site all the time.

      • Jim D
        I think we are in close agreement about the broad outlines.
        On the dry adiabatic atmosphere I used to be a bottom up advocate.
        Surface temperature determined by Sun/Earth interaction.
        Gravity giving rise to lapse rate of 9.8K/km.
        This very simple structure then modified by convection, latent heat and radiative effects till the convective impetus petered out at the tropopause.
        Above the tropopause the radiative effects adjusted to keep the Earth energy in/out in balance.
        However recently I find the top down approach quite compelling.
        The TOA conditions acting like a gate.
        The consequences of the gate being too narrow being passed back down by the same dry adiabatic lapse rate to determine the surface temperature.

      • Pekka
        I’m sorry if I addressed you in a way that you found disrespectful.
        I think I used the word IPCC apologist rather than my usual term IPCC advocate so I must have been loosing my cool.
        I think that one undisputed plus for Judith’s site has been to tone down the insult level.
        However if you are a sceptic you have to develop a much thicker skin.
        For a laugh go onto a site like Deltoid and pretend to be Nasif Nahle.
        You wont get out alive!

      • Bryan,
        The net discussions are often difficult. Short messages cannot always transmit the tone correctly. Some of the participants are provocative by purpose, and some others write claims that they know to be false, even deliberate lies.

        In climate science and in particular in the physics behind the climate science there is very much that I have full confidence in based on my schooling and understanding based on that. There are many other things I have much less confidence in and also conjectures that I consider more likely to be false than true.

        In these discussions I comment most often on issues I am certain about. Trying to do that as well as I can and getting answers that show no evidence on willingness to learn, is often frustrating and leads to doubts about the goals and even honesty of other participants. All concrete hints to the same direction strengthen these suspicions. At the same time I know perfectly well that many points are difficult and cannot be verified personally without specialized education.

        I try to stay polite, but sometimes it leads to a point, where I start to think that I am played with and that I am making fool of myself unless I react strongly. I know that this is going to happen also in the future, if I continue to comment on climate sites.

      • Bryan, I think the dry adiabatic atmosphere can be thought of from both perspectives, top and bottom, which both lead to a requirement that the whole temperature profile is displaced in the warmer direction when CO2 is added.
        My view is that more CO2 initially reduces outgoing IR but also causes the surface to warm, which in turn convectively forces the atmosphere to warm, increasing the outgoing IR till it balances again.

      • I just came across this discussion, and since it was a discussion rather than an argument, I thought I would offer my perspective. In general, a TOA radiative imbalance due to impeded loss of IR to space is translated into more energy at each layer, ultimately impacting the surface temperature. In turn, this further warms the atmosphere over time as the surface temperature rises. The immediate result of atmospheric warming is an increase in lapse rate beyond the adiabat due to greater warming at low than at high altitudes. This results in static instability that triggers a convective adjustment restoring an adiabatic profile (which in most regions eventually proves closer to a moist than dry adiabat due to latent heat transfer with release at higher altitudes).

        The radiative changes are very rapid. The dry convective adjustment (according to Andy Lacis) is slower, and the full change including the latent heat effects occurs over many days or longer.

        The “super-adiabat” would tend to enhance surface warming because of the higher lapse rate. On the other end, the moist adjustment creates a negative lapse rate feedback that reduces the warming effect. This, however, is accompanied by a positive water vapor feedback, and the combined water vapor/lapse rate feedbacks are generally computed to show a net positive effect.

      • Fred,

        are you finally going to tell us about the hot spot? I believe you need a hot spot for there to be any appreciable top down warming don’t you?

      • kuhnkat,
        Is this really so difficult?

        Nobody claims that there would be warming in the sense you imply – nobody at least of people supporting main stream climatology. Therefore there is absolutely no need for such a hot spot.

        This is not in contradiction with the fact that atmosphere radiates to surface and contributes to a temperature increase. If you do not understand the point after all these discussions and hundreds of messages where it has been explained in different words, then I propose looking in the mirror.

      • Pekka,

        you are a very reasonable, intelligent, respectful person. I respect you for your knowledge and comportment. Unfortunately I am often none of the above.

        Frank started discussing heating at elevation which is caused by bottleneck in IR emissions. He did not give a mechanism for the purported bottleneck. He also talked about heating from the top down.

        With emissions bottlenecks, heating from top down, backradiation, and eventual heating of the surface, exactly what am I supposed to assume he is talking about??

        I have actually read explanations of this effect and have always been confused about how the bottleneck comes about. The statements seem to say that the heating will raise the effective emission altitude as the heated atmosphere expands. As the new higher altitude is supposed to be cooler than the old average altitude less IR can be emitted.

        Hopefully you can clear this up for me. If the atmosphere expands from warming, doesn’t that say the higher altitude will be about the same temperature as the old altitude? That is, the altitude will average higher but the temp will be about the same because everything is warmer.

        If we are saying that this warming will not happen it would seem to me that the temperature is more controlled by the lapse rate and convection, in which case there will be no significant warming in the first place without major perturbation.

        Thank you for any clarification you can give on this “hot spot” issue.

      • kuhnkat,
        Nobody of us is capable of always finding clear expressions for his messages. While many issues are not really complicated, they involve anyway numerous details and attempts to explain the issues in limited space and simpler language requires leaving something out. All too often happens that just those things left out are for some reason in the mind of the other party of discussion.

        Another problem is that the concepts are not defined precisely. What means “warming a body”? In these discussions some participants expect that the effect that warms must be the final source of heat or energy that rises the temperature to its final value. A colder body can never do that for a warmer one.

        Many others mean by the sentence “body A warms body B” that taking the A away would lead to a colder B. This is very often possible even when A is colder than B, if B is heated also by some other source. I have still difficulties to understand why this second way of interpreting “A warms B” is not understood by everybody.

        I commented to the most recent post of Judith that many people can much better form general views on issues than present scientific type arguments in their support. It is very common, that the role of detailed arguments is overvalued. They are overvalued often both by those who are competent in presenting them and by others for whom a more general intuition works much better.

        This is also a source of dispute and confusion, when people are sure that they are right in the main issue, but cannot justify it by a detailed arguments. There is too much belief that detailed arguments are the way of winning argumentation, even when that does not work at all. In climate issues this fact comes up all the time. Even for experts a more general and intuitive approach may give more reliable results than trying to prove by detailed arguments when not enough is known about those details.

      • You certainly nailed it! Very good.

      • You will note however that now the conversation has shifted from Johnson defending his mistakes to you explaining how things really work.

        This may be because it’s now past midnight in Sweden.

  49. Judith,
    I want to comment that I am increasingly an admirer of your approach, especially on this technical thread.
    By letting others take a turn at being the authority, people seem to come to more openly examine their own ideas and knowledge – including errors. By just minding the store, wrong assumptions and weak knowledge claims are brought to the surface by others, instead of driven underground by your authority. It’s a better learning process than confrontation.


  50. To complement the many comments made above indicating that the radiative transfer principles contributing to the greenhouse effect, including the role of back radiation (downwelling longwave radiation) are consistent with the laws of physics, it’s worth pointing out that the back radiation predicted from these equations has been confirmed by measurement. For a general overview, readers should revisit the Radiative Transfer Models post to review the links Judith Curry has cited, with particular reference to the Atmospheric Radiation Measurement (ARM) program – the post is at Radiative Transfer

    For a particularly informative description of the ARM program, see –
    ARM Prrogram

  51. Claes,

    You write:
    “Let us now sum up the experience from our analysis. We have seen
    that the atmosphere acts as a thermodynamic air conditioner transporting
    heat energy from the Earth surface to a TOA under radiative heat forcing.
    We start from an isentropic stable equilibrium state with lapse rate
    9.8C/km with zero heat forcing and discover the following scenario for
    the response of the air conditioner under increasing heat forcing:
    1. increased heat forcing of the Ocean surface at low latitudes is balanced
    by increased vaporization,
    2. increased vaporization increases the heat capacity which decreases
    the moist adiabatic lapse rate, if the actual lapse rate is bigger than the actual moist adiabatic rate,
    then unstable convective overturning is triggered,
    4. unstable overturning causes turbulent convection with increased heat
    The atmospheric air conditioner thus may respond to increased heat forcing
    by (i) increased vaporization decreasing the moist adiabatic lapse rate
    combined with (ii) increased turbulent convection if the actual lapse rate
    is bigger than the moist adiabatic lapse rate. This is how a boiling pot of
    water reacts to increased heating.:”

    I think your model is incomplete, since the “heat forcing” as you name it is external and you describe only energy flux that is internal. “Heat forcing” increases the energy content of the earth system and therefore leads to increased temperature on the long run to decrease your so-called “heat forcing” by increasing outgoing longwave radiation (OLR). Your model leads necessarily also to increased temperature. You describe radiative-convective equilibrium as well. So what is different in your model compared to the classical model
    Best regards

  52. “If they are wrong, prove it”

    Done already

  53. One thing that always puzzles me when IR and the GHE are discussed is why on a nice clear summer day in Atlanta I don’t melt. I mean, we supposedly have an AVERAGE downwelling radiation of 324 wm-2. I would imagine that the downwelling radiation at noon on a humid day in Atlanta would be higher than the average due to all the water vapor in the air. Let’s make it 25% higher, or 405 wm-2. Now, let’s add the sunshine, which is certainly greater than 900 wm-2 at noon. So we now have 1305 wm-2 on my greybody. Using the SB equation, with emissivity of 1, that translates to 116 C. Something doesn’t add up.

    • Hi Jae… Excelent observation! You could calculate the energy the human body would absorb, from those 405 W/m^2, by knowing that it has an average absorptivity of 0.7. Imagine the hard work the body would perform for getting rid of that excess of energy!

      • A black body radiates 400 W/m^2 at a temperature 0f 17 C. It’s the sunlight that would cause a problem for an object unable to shed heat via perspiration, reflection, conduction, or respiratory heat loss. With 900 W/m^2 absorbed, its temperature would equilibrate at 82 C.

      • At an ambient temperature of 40 °C, a normal human body absorbs 43.4 W. That figure represents an intensity of 160.71 W/m^2. However, Jae mentions c.a. 1305 W/m^2 the energy emitted by the atmosphere, if the stuff of backradiation were true. Fortunately, as Jae points out in his post, it’s not true because, if it were true, the human body would absorb the dizzying amount of 913.5 W, which would represent an intensity of 3,383 W/m^2.
        On the other hand, if you are considering an idealized blackbody emitter, emitting 400 W/m^2, then the human body would absorb 280 W, which corresponds to an intensity of absorption of 1,037 W/m^2.
        Now, let’s consider a blackbody-ambient at 17 °C; the human body would be losing, not gaining, 23.17 W (-23.17 J/s), which corresponds to -85.82 W/m^2.

      • Your figures aren’t well explained. The average human has a surface area of about 1.7 m^2, so I’m not sure what you mean when you imply that 160 W/m^2 corresponds to 43.4 W absorption.

        More importantly, an ambient temperature of 40 C is very hot (and represents much higher than average back radiation). It is equal to a Fahrenheit temperature of 104 F, which is very difficult for humans to tolerate on a sustained basis, although they can adapt temporarily through sweating and panting.

        It is incorrect to state that 1305 W/m^2 is emitted by the atmosphere. Most of that figure comes from the assumed value of 900 for sunlight, which would be an immense problem for an individual who could not adapt, and would be unsustainable for any extended period. Back radiation has little to do with it.

        Finally, in the example I gave, which you cite, of ambient temperature at 17C, this is easily tolerable, because human metabolism generates enough heat to compensate for the heat loss. In fact, tolerable climates for humans require some degree of heat loss to the environment, because we can’t shut down our metabolism, and so if we couldn’t lose heat, we would quickly die.

        In essence, the values I gave in my earlier comment are correct, and the most significant problem in the cited example is the sunlight.

      • Fred:
        “It is incorrect to state that 1305 W/m^2 is emitted by the atmosphere. Most of that figure comes from the assumed value of 900 for sunlight, which would be an immense problem for an individual who could not adapt, and would be unsustainable for any extended period. Back radiation has little to do with it.”

        I originally thought you digged the conversation, bro., but it appears that you don’t have a clue!

      • Dear Fred,
        0.27 m^2 exposed to radiation, unless it is naked.
        40 °C is a usual temperature, here, during summer.
        The average absorptiviy of the skin, in a normal human being, is 0.7.
        I never said you’re wrong. I only made the calculations for the conditions you specified in your post. At 17 °C the human body would lose 23.17 W of energy, which would be transferred to the environment. It would be a problem if we were endothermic organisms. Fortunately, we are self-regulating thermodynamic systems; otherwise, we should spend many hours under the sunbeams, as lizzards, for example.
        Now, if you say that a blackbody at 17 °C is emitting 400 W of thermal energy, how much Watts it would emit in my location when the temperature can reach, easily, 40 °C in summer?

      • Nasif – A true black body at 40 C (313 K) would radiate about 544 W/m^2 in accordance with the SB equation.

        Humans can’t afford to sustain a body temperature of 40 C for very long. At 37 C body temperature, they lose heat by all the mechanisms I mentioned above, not just radiation. I’m sure humans can tolerate an ambient temperature of 40 C for intervals, but I doubt they can tolerate it for a very long sustained period, day and night, without some exogenous cooling source, such as drinking cold water.

      • Dear Fred,
        Exactly! An idealized blackbody at 40 °C would emit 544 W/m^2, which is not the case if we consider the real system atmosphere-lithosphere. The external operator, for the case of my location, where we undergo up to 40 or higher degrees Celsius during the summer daytime and 30 or more degrees Celsius through the nighttime (and, believe me, we have survived it through many days), cannot be other but the Sun, and you will agree on this because the atmosphere cannot “store” such load of heat. Primarily, because the absorptivity of the whole atmosphere, including a 4% of water vapor, is quite low (by the order of 0.01 when considering the mean free path length of photons and the time they spend to leave the Earth’s atmosphere). That’s why, I sustain that the current models on TAO (or TOA) are absolutely flawed.

      • I’m not sure what your point is. The emissivity of the atmosphere in the IR range of greenhouse gas emssion and absorption is certainly less than unity, but although the emissivity of any small atmospheric layer, even near the surface, is small due to the low concentration of greenhouse gases, the total downwelling longwave radiation comes from multiple layers and is substantial.

        Radiative transfer codes derived from the Schwartzhcild radiative transfer equations, in conjunction with observed values of CO2, H2O, and surface temperature, yield values for both OLR and downwelling radiation that match observations very well, confirming the validity of the principles on which they are based.

      • Fred… I’m referring to the time that a photon takes to abandon the atmosphere, as wide as it is, and to the distance that a photon can travel without touching a molecule air, those molecules that can absorb it or scatter it. From the databases of both parameters, we find that the air, as dense as it is, we find that the emissivity of the air, 4% of water vapor included, is 0.01; no more. The atmosphere is not a blackbody.
        Perhaps those observers on the downwelling radiation are observing other things, except any downwelling radiation?

      • Here is the reductio ad absurdum. A human body is like a black body at 37 C which emits about 525 W/m2. Now according to this theory proposed above, nothing can emit towards a human body that isn’t as warm as it, so when you go out at night you are losing heat at 525 W/m2. Wouldn’t you cool down really fast even on a balmy night with a 20 C ground temperature?
        The fact is, everything emits towards everything else regardless of relative temperature. We do have incoming radiation to us at night even from the cooler ground. Go out and try it. Explain how this is different from the atmosphere radiating towards the warmer ground.

      • Wow! Jim! You have got rid of S-B Law! Please, tell me, are you related in some way to the Hockey Stick producers? Besides, you made us, humans, real blackbodies!
        Jim, a human body has a temperature of c.a. 37 °C. If it (the human body) is exposed to an environment at 17 °C, it would lose 23.17 W, i.e. the energy transferred by radiation from the human body to that environment at 17 °C, according with the S-B Law derived formulas. No more. The formula is quite easy:
        Q = e (A) (σ) (Te^4 – Thb^4)
        Where e is the emissivity of the system (human body in this case), A is the area exposed area of the human body, σ is Stefan-Boltzmann constant, Te is the ambient temperature in K, and Thb is the average temperature of a normal human body.
        Go on, make your calculations.

      • See, you now have the ambient air radiating towards the body when the slaying book says it can’t because it is colder.

      • Jim…
        I’m not having the air radiating towards the body, but quite the opposite. The body is losing energy, not gaining it from the environment. Under those conditions, the body is pushed to generate more thermal energy, from metabolism, to maintain his energy state in a quasi-stable state.
        In summer, only when the environmental temperature is higher than the body’s temperature, the body gains energy from the environment; however, the thermoregulating system starts working to get rid of the excess of thermal energy absorbed.

        If you applied the S-B formula correctly, you had to obtain a negative result, which means that the body is losing energy, not gaining it; the body must generate more energy through the cellular respiratory process and other mechanisms for not cooling off, in this case.

      • The Te term in your equation comes from back-radiation is all I am saying. If you believe your equation, you implicitly agree with back-radiation. I am not saying your equation is wrong, I am saying it proves back-radiation exists.

      • The Te is the temperature of the environment and it comes from the energy it has absorbed from the surface.

      • Q = e (A) (σ) (Te^4 – Thb^4)
        Where e is the emissivity of the system (human body in this case), A is the area exposed area of the human body, σ is Stefan-Boltzmann constant, Te is the ambient temperature in K, and Thb is the average temperature of a normal human body.

        e (A) (σ) Te^4 Heat received from surroundings
        e (A) (σ) Tb^4 Heat emitted by body

      • Jim and Phil…

        You’re much confused.

        It’s the energy from the human body to the environment… Have you noticed that the energy flows ALWAYS from the warmer system to the colder system? Backradiation doesn’t apply because it is the human body what is radiating, not the environment.

        Again, for this case, the human body is LOSSING energy, NOT gaining it.

      • No, Thb is from the body to the environment, Te is from the environment to the body, which is why they have opposite signs. Since the environment is colder than the body, this is the term the slaying book says should be zero. We clearly agree the book is wrong on this matter. The environment is preventing the body from losing heat at an unrealistic rate of 525 W/m2 in the same way as the atmosphere prevents the ground from losing heat at an unrealistic rate (where a similar formula applies with Thb being from ground temperature, Te from the atmosphere).

      • LOL…

        Thb is temperature of the body in K, and Te is temperature of the environment in K. :)
        If you read well my posts, I’m always referring to an “idealized” blackbody. Got it or start again?

      • Nasif, you have already contradicted slaying the dragon by having the Te term, but you haven’t realized it yet. I suggest you argue with those authors about that term. I am not arguing about it.

      • I’m afraid it’s you who’s confused Nasif, the environment radiates according to its temperature and the body absorbs it, the body also radiates according to its temperature. The net effect is that when the body is warmer than its surroundings the body loses heat (when the environment is hotter than 37ºC the body gains heat).
        The environment doesn’t stop radiating because the warmer body is present, ‘back radiation’ is always present. that’s what the term, e (A) (σ) Te^4, represents.

      • Nope, confusion is on your side. I’m afraid you think the environment is never colder than your body. The formula is the S-B equation, and you’re blatantly misinterpreting and twisting it, as usual in AGW idea.

      • Good Grief, why dont you give it a rest. Nahle is correct.

      • Phil, you’re absolutely wrong. If you eliminate the term Tb^4 from the formula, you would be referring to the energy of the atmosphere. It has nothing to do with “energy received from surroundings”. You have only one term, the temperature of the environment, and it is the result of the FLOW of energy IN the environment.

      • To Phil…
        Anser this question for me: what the value of “e” could be in the formula that you say it is “energy received from surroundings”? If you are referring only to the temperature of the air, then you have to introduce the value of “e”, and in the case of the human body, you have to introduce the value of “e” for the emissivity of the human body.

        It’s very simple. You’ve dissected the formula and you’re referring to two different things.

      • To Phil…
        Anser this question for me: what the value of “e” could be in the formula that you say it is “energy received from surroundings”?

        I suggest you look up Kirchoff’s Law, as the heat is being absorbed it should be ‘a’ not ‘e’, however a=e.

      • @Phil

        My question:
        “To Phil…
        Anser this question for me: what the value of “e” could be in the formula that you say it is “energy received from surroundings”? ”

        Phil’s answer:

        “I suggest you look up Kirchoff’s Law, as the heat is being absorbed it should be ‘a’ not ‘e’, however a=e.”

        You didn’t answer my question… Yours is blah, blah blah.

        I repeat, it is S-B equation. Check your books.
        If your environment has e = 1 and a = 1, you’d be scorched… Mmm…

      • Fred??
        Everyone keeps telling me that we ADD all incident radiation, no matter where it is from, to determine what the temperature should be. What are YOU saying here??

      • What I’m saying is the Noel Coward song line – “Mad dogs and Englishmen go out in the midday sun.”

      • Is THAT your scientific basis for all of your comments?

    • jae, absolutely correct. Another simple example of how the purely radiative calculation as proposed by Pierrehumbert and other climate scientists gives completely the wrong answer. The correct answer of course, is that you need to take into account other mechanisms of heat transfer such as convection and evaporation. This point has been made dozens of times on all these threads.

      • Bruce Cunningham

        There is convection etc, within the Earth’s atmosphere, but the only way that heat (energy) is released from the Earth to outer space is through radiation. Reflection of incoming solar radiation by clouds is the big question. Until someone can accurately define how this changes the amount of incoming energy, predictions of future temperatures cannot be accurately calculated.

    • No sweat?


    1) Water vapor in the air changes from 1% to 4%.

    2) CO2 in the air is about 0.038%.

    3) Since the industrial revolution, the proportion of CO2 in air has increased by 0.01 % (from 280 to 380ppm)

    3) Both water vapor and CO2 are greenhouse gasses.

    4) A natural change in 3% of water vapor in air does not cause global warming.

    5) How can a change in 0.01% of CO2 (3/100th of the natural change in water vapor) due to human use of fossil fuel cause global warming?

    • Your question is somewhat off-topic, but relative humidity has not increased over the past century, while CO2 has risen almost 40 percent over its pre-industrial concentration. Water vapor, in fact, has such a short atmospheric lifetime that its absolute humidity value cannot remain elevated in the absence of some other factor that causes the atmosphere to warm and thereby retain more water. That is why it operates as a feedback mechanism amplifying warming mediated by CO2, solar increases, or other forcings rather than acting as a forcing in its own right. If the average relative humidity had in fact increased by 300 percent, the warming would have been immense.

      I believe this has been discussed in the threads on feedback and on climate sensitivity. You might want to review the previous discussions before proceeding further, so as not to repeat material already covered.

      • Water vapor, in fact, has such a short atmospheric lifetime that its absolute humidity value cannot remain elevated in the absence of some other factor that causes the atmosphere to warm and thereby retain more water.

        I don’t agree with “short atmospheric lifetime” argument regarding water vapor.

        Does not every half a day, the temperature drops at night?

        Is the “atmospheric lifetime” of water vapour less than half a day?

        Do not tell me what I discuss here. This is not your blog!

      • Girma

        Do not tell me what I discuss here. This is not your blog!

        Interesting point.

        In light of what Dr. Curry has said elsewhere:

        Note: this is a technical thread, please keep your comments focused on Johnson’s arguments, or other aspects of Slaying the Sky Dragon. General comments about the greenhouse effect should continue on the Pierrehumbert thread.

        Now that we have aired some stuff, I agree that the discussion is best left to those with a degree in physics (maxwell, pekka, and there are others among the denizens of climate etc that have not shown up).

        ..could you help me out by relating what you say directly to the topic at hand, and how the two connect?

        Much obliged.

      • Thanks, Bart. I tried to say it very tactfully, but your direct approach is better.

      • Please take discussion of water vapor to the Pierrehumbert thread.

    • Derry MCCarthy

      you might find an answer of sorts here

      by Robert H. Essenhigh, Department of Mechanical Engineering, The Ohio State University, Columbus, USA. In press in the journal ‘Energy and Fuels’, but now available at ACS website

      • Derry… The residence time of carbon dioxide in the atmosphere could be as long as you wish… The important thing here is that the lapse time for the thermal energy to stay in the atmosphere is quite low: 0.0097 milliseconds! The mean free path lenght of one photon of thermal energy is 21 m. Besides, from experiments realized by many physicists, at its current concentration in the atmosphere and under the current physical conditions of the atmosphere, the carbon dioxide cannot absorb-emit more than 0.002 of thermal energy.

      • “carbon dioxide cannot absorb-emit more than 0.002 of thermal energy.”

        Hi Nasif.
        Is this figure a percentage? If so, what is it a percentage of? Surface emission?

    • As you rightly imply, it cannot.

  55. The bottom line in all this is that there is absolutely no proof–or even a reasonable demonstration–of an “atmospheric greenhouse effect.” All planets/moons with an atmosphere have a surface temperature that is much higher than the SB equations–based on the IR from those bodies–at about 100 mbar–suggest. It is high time that the “climate science community” HONESTLY faces the questions that are posed by the skeptics (and stop with the dishonest, unconvincing, meaningless, disgusting, and typically liberal insult of “denialists). The “community” has already lost the public and only has politicians and rent-seekers on its side. The smart ones are already publishing papers refuting the stupid, ever-present “catastrophe” of our times (aka, Chicken Little). Grow up!

  56. Fred Molten,
    one can make an interesting thought experiment about “back radiation”.
    Let’s assume we have the earth system as a stationary state with 280 ppm CO2, well mixed. Normal lapse rate.
    In the first case, we bring in a thin layer of CO2 that contains a similar amount of CO2 compared to the whole atmosphere in a thin layer next to the surface.
    In the second case, we bring in a thin layer of CO2 that contains a similar amount of CO2 compared to the whole atmosphere in a thin layer next to the top of the atmosphere.
    Both layers are equilibrated with respect to temperature.
    “Back radiation” is highest in the first case, but surface temperature is lowest. It is the emission height that counts.
    As I said, it is the cooling to space that rules. That is, why I don’t think “back radiation” is a necessity to explain the greenhouse effect.
    Best regards

    • I don’t believe there is any way to warm the surface without back radiation. In its absence, radiative imbalances in the atmosphere would change atmospheric temperature but not surface temperature (except for the minimal effects of conduction).

      Regarding your thought experiment, my assessment is the following, at least at first consideration. If we ignore water vapor as well as non-radiative phenomena, I believe that the same number of CO2 molecules will absorb the same number of photons, regardless of altitude. At equilibrium, they will emit as much energy as they absorb, and the temperature of that layer will therefore rise until it suffices for that emission to occur. For the high altitude case, this would cause a temperature inversion such that temperature is much higher at the height of the absorbing layer than it is below. This is clearly an unphysical situation, but something vaguely similar occurs in the stratosphere, where ozone absorbs solar UV, resulting in a temperature inversion.

      There may be other factors that I’m ignoring in addressing your thought experiment, but my first paragraph rather than the second is what I would emphasize – the surface can’t warm unless it receives the radiation needed to warm it.

      • There is only one way the physics really works, but there are many ways of putting this into words and more than one way of formulating the equations used to calculate the correct results.

        There are no limits on the number of ways the physics can be misrepresented and we have already seen pretty many in comments on this site. Countering these erroneous claims is made more difficult by the fact their details may well be in agreement with some of the correct descriptions while the errors are in putting these pieces together. Some of the erroneous theories are pure nonsense from start to end, but not all of them.

        There is a continuing argumentation on whether one mechanism can heat an object which is actually receiving heating through many processes or from many sources. Then one may claim that any single process cannot heat it, if the processes are individually weaker than cooling of the object. Such arguments are presented as if all heat sources would not add up whatever their mechanism is and as if each of the heat sources would not have its share in the total heating. How can this kind of argumentation be supported by so many?

  57. Claes Johnson,

    Your statement that “back radiation” is fictional, a figment of the imagination for any length of time longer than a fraction of a second, I totally agree. I will read your paper (book) as I get time and I might not totally agree with the methods you use to describe this. Maybe so.

    I have always viewed “back radiation” as a null operator:

    — 2 units or energy leaves a surface cooling the surface by that 2 units.
    — That 2 units are absorbed by molecules (GHGs) warming the gases locally.
    — 1 unit of energy is radiated to space and lost to the system and also cooling the gases by 1 unit.
    — 1 unit is radiated back to the surface to be reabsorbed warming the surface by 1 unit and also cooling the gases by 1 unit.

    — NET EFFECT: In the end the surface has cooled by 1 unit and 1 unit is lost to space, all in a few milliseconds.

    All other effects have totally cancelled. One way to view this is a reduction of effective emissivity of the surface by at factor near one half.

    That seems very close to your initial statements I was reading and I agree, there is no real warming. After reading onward I may not agree with the exact methods you use to place this effect into a physics framework but I will read it, that takes time.

    • Yes, Wayne, but if the surface was at a temperature that demanded it radiate 2 units and it only radiated a net 1 unit, then it is not in equilibrium anymore and its temperature must go up. (I think I got that right, I normally lurk on the technical threads and keep my head well down!) Regards, Rob

      • Hi Rob,

        I started to write you a detailed explanation, but after reading many of your comments, I’m afraid it would be pointless if you are not able to take my example above and limit it the exact case I gave. The two units must be radiated upward, those two are not all radiated upward (your injection of temperature), and those two units must be absorbed and not transported directly to space without absorption (window).

        If you can not grasp even that simple example there probably is no hope of you understanding Dr. Miskolczi’s methodology he used in his latest papers and which is very close to my example above.

        Kind regards. I like to lay low too. Open your mind, the AQUA AMSU temperature just hit the same temperature that was read thirty years ago, how can that be? If I were you I would get real curious right now. I have already found my answers.

  58. Michael Larkin

    Whatever else can be said of this thread, I am enormously grateful for the earlier link to Roy Spencer’s explanation of the GHG effect which is worth repeating:

    This is the clearest explanation for non-specialists like me that I have ever come across. I’ve saved it to my hard drive.

    I have a question. Spencer says that with no atmosphere, the earth’s surface would be around 0 deg. F (-18 deg C or 255 deg K). Suppose all GHGs (but nothing else) were removed from the earth’s atmosphere (I’m assuming there would be no water on the planet). Would the temperature be greater than 0 deg. F?

    I’m hoping that’s on topic, because I’m trying to establish in my own mind whether just the presence of an atmosphere pretty much as dense as the one we have now, but sans GHGs, would in some way produce warming. I hope it makes sense to ask the question.

    • afaik, yes: what I think would happen is that all radiation would occur at the surface, because the atmosphere would be perfectly transparent for all wavelength, at by K., would also emit no EM radiation (In reality, it would not be like that, but I guess it is the idealised situation you have in mind).
      So, the surface T at equilibrium would be computed the same as in the no-atmosphere case.
      But I am not so sure about the T profile in this transparent atmosphere. Quite fast, we should reach the lapse rate for this gravity field and adiabatic fluid, by convection. I think, after some time, conduction should produce uniform T, which seems to be the no -heat flow limit regime (well, assuming 1D problem)… but I am not sure uniform T is the equilibrium in a gravity well, some equirepartition principle may mean that T goes down the higher you go (some interpretation of virian theorem would say so too, which makes sense: monoatomic gases modeled as elastic spheres, should have a lower velocity at top of atmosphere…else they would reach escape velocity) which would falsify simple conductive transfer, except if “total” temperature incorporate somehow potential energy. Interesting question, I would be interested about what gaz kinetic theory specialists would have to say about that, all in all my hinch would be for non-constant T and conduction process acting with a “total” T incorporating potential energy….

      • yes, definitely a non-constant T at equilibrium due to gravity: after all, simple heat transfer linearly proportional to T gradient is not a fundamental law, it is derived from kinetic gas theory, one of the hypothesis being, iirc, no volume forces. Gravity is a volume force, so I am almost sure the Fourier law for conduction is not strictly valid in this case (it is a first order phenomenological law, nothing fundamental there), but that heat transfer must incorporate gravitational potential energy…..

      • Yes the convective equilibrium lapse rate is g/cp, about 10 K/km, so I would expect something like that. It is complicated by variations in surface heating with latitude and the diurnal cycle, so it is not clear what temperature this would equilibriate to over the surface, but since the non-GHG atmosphere has no other cooling mechanism than contact with a colder surface, the surface temperature would somehow control its eventual equilibrium temperature profile.

    • “I have a question. Spencer says that with no atmosphere, the earth’s surface would be around 0 deg. F (-18 deg C or 255 deg K). Suppose all GHGs (but nothing else) were removed from the earth’s atmosphere (I’m assuming there would be no water on the planet). Would the temperature be greater than 0 deg. F?”

      Michael – Removing only GHGs would have slightly greater cooling effects than removing the entire atmosphere. This is because atmospheric molecules (O2, N2, CO2, etc.) scatter some sunlight back to space, and in their absence, all solar radiation would reach the Earth’s surface.

      The 255 K figure assumes no other changes. In fact, in the absence of water, there would be no ice, snow, or clouds, and the Earth’s albedo (percent of sunlight scattered or reflected back to space) would decline significantly. As mentioned above, some scattering would still occur from air molecules, and some from light-reflective surfaces such as sand, but it would be far less than the current 30 percent figure. As a result, the Earth would absorb more heat, and warm well above 255 K. I don’t know what the exact temperature would be. It would be colder than today, but probably by only a modest amount.

      • A small correction – Above, I should have omitted CO2 from my example of light-scattering molecules, because you were asking what would happen if it were removed. Of course, N2, O2, argon, etc., would remain, and their contributions would be little diminshed by the removal of a minor constitutent by volume such as CO2.

      • Michael Larkin

        Thank you for your clear and not-too-technical response, Fred. Might have seemed a peculiar question, but it elicited useful extra information for me.

  59. To summarize my position:
    1. Radiative heat transfer is carried by electromagnetic waves described by Maxwell’s equations. The starting point of a scientific discussion of radiation
    should better start with Maxwell’s equations than with some simplistic ad hoc model like the ones typically referred to in climate science with ad hoc invented “back radiation” of heat energy. If there is anything like “backradiation” it must be able to find it in Maxwell’s wave equations. In my analysis I use a version of Maxwell’s wave equations and show that there is no backradiation, because that would correspond to an unstable phenomenon and unstable physics does not persist over time.

    2. Climate results from thermodynamics with radiative forcing, and radiation alone cannot tell anything of real significance, such as the effect of changing the atmospheric radiative properties a little: It is not clear if more clouds or
    water vapour will cause global cooling or warming, or the effect of a small change of CO2. Climate CO2 alarmism is based on a postulate of a climate sensitivity of + 1 C which is a formality without known real significance.

    I welcome specific comments on these two points.

  60. 2. Agreed. And I am not too comfortable with the model hierarchy used in Climatology: pure radiative models are imho correct,but they do represent the main heat transfer in earth system well…so are useless for earth. TOA+lapse rate is better, but I think they are not so solid mathematically, I do not really like the treatment of it. Should be consolidated, and then it is 1D, so predictive value is not clear, but at least this model could have heat transfer similar enough to actual heat transfer on earth to be somewhat useful.
    Finaly, there are GCM….but they are huge, use numerical methods I do not like (FD for something with complex continental shapes – yuck), and introduce a lot of approximation (solving NS equation on earth lenghtscale is ridiculous…so it is not NS that is solved, but some kind of approximation of it. Never really have seen the PDO that are solved in fact, which is in itself very worrying. Lot of blackboxes modelling different process connected to each other (radiative module – ocean module – salinity module – biological C cylce module), so it is more an ad-hoc model that something starting from first principles or even a solid set of PDO. OK, not easy to do better, but the validation is pitifull for this kind of model, which live and die by extensive validation.

    1. Not agree: Maxwell’s equation are ok, but you need quanta (or a replacing full theory) to deal with radiative heat transfer: It is not even needed to accept black body treatment by Plank to know Maxwell alone will not be up to the task: those EM are radiated by molecules, that can not be modeled by
    maxwell: remeber the paradox for Bohr atom model of orbiting electrons? Why does the electrons not fall down in the nucleus, when all its kinetic energy should be dissipated by bremstrhallung/synchrotron radiation? This was a fundamental problem (before or at the same time as BB radiation) that was solved by quantization. You may not like Quantum mechanics (I myself have trouble with it, it seems like an unfinished and overly complex theory), but it is extremely successfull, maybe the biggest success of physics. Going against it is a huge task, there is a reason it was accepted between the wars although it is quite often counter-intuitive: it explains and predict a lot, much more than simple BB radiation.

    By the way, you continue to mention that backradiation would be unstable. A few posts (some of mine too) challenged this. You still not have explained why you believe it would be unstable, just that it is and is a flaw of S-B model. This is not a tenable position, you need to show how S-B is unstable and how your theory is not. Good luck!

    • So if you don’t accept Maxwell’s equations for radiation, which are then your equations and what do they tell you?

      • Maxwell equation are valid for propagation. For emission/absorption, you need to take into account the quantitized nature of emmiters/absorbers when those emitters are molecules or atoms. Which is the case for the IR wavelength of interest. If you want to use continuous Maxwell down to atomic lenghtscale and energies, you predict unstable atoms. Everything should go back to neutronium, which will be a problem for predicting EM radiation with Maxwell equations ;-)

      • Actually I believe that it is possible to describe the situation without the need of standard way of introducing the quantization. The quantum field theory of electromagnetism (QED) is used in practical calculations as perturbation theory in the form of Feyman diagrams, but this is not necessary in principle. Similarly the quantum transitions of molecular states are introduced in the spirit of Copenhagen interpretation of quantum mechanics. This is again not necessary while very useful in practice.

        Both choices are valuable practical tools in quantitative physical analysis, but they are not really required. In principle one can formulate the whole problem by writing the full equations to describe all molecules in the atmosphere and all radiation by Schrödinger equation and Maxwell’s equations and possibly introducing modifications related to QED. There is no basic reason to assume that these equations cannot be used in another way, which does not involve the traditional way of quantization at micro level but aiming directly to answering some macroscopic questions. It may even be possible that this approach gives many results more easily and directly than the standard procedure.

        What I have seen in the text of Claes Johnson is certainly not a complete and valid presentation in this line of thought, but it may be partially correct and it might be possible to continue in this direction and reach correct results. I have full trust that the final results would agree with the results of the standard approach, but it is likely that the same results would indeed be reached in a way that does not include back radiation. This would be an extension of the idea of wave-particle dualism. The description in terms of waves does not include back radiation, but it would still give the same quantitative results.

        Agreeing with accepted physics does not require dogmatic adherence to the standard way of describing the details.

      • agreed, that’s what is a little bit disturbing about QD imho: not easy to draw where quantum description start, and where classical physics end. For example, a lot of classical QD imply wave/particles in external potential….but those potential are themselves caused by phyical objects, so by W/P assemblies. Why are they represented by perfectly know and unchanging potential fields them, like some kind of ghost of classical Newtonian entity? I guess QD has progressed since (I only have some training about early stage QD, probably from the Plank/Einstein era, and still it is vulgarisation).

        But I have 2 problems with C. J. approach. one is that is is hopeless imho to try to use Maxwell equations only, you have to introduce some quantization, or an equivalent effect, to avoid molecules to radiate even at 0K just by electron orbiting. Or you can say that bohr atom model is not correct, but this is just another way to make QD come back through the backdoor…As you said, QD can be introduced in many ways (which I find slightly disturbing, but I also agree with you that QD is one of the most (if not the most) succesful physical theory), but Maxwell has to be complemented somehow. C.J approach seems to be “add a phenomenological structural damping for elementary resonators”. I am fine with that, even if I think it explain less than quanta as introduced by planks and so is a poorer approach.
        The problem number 2, unfortunately, can not be bystepped just by saying that choosing the method is a matter of personal preference: the radiative exchange presented is not equivalent to S-B law, we have R = 4 s T³ (T-T_cold) versus R = s (T⁴-T_cold⁴).
        Not the same, and I prefer S-B for symmetry reason (the fact that each body radiates without having to know his surrounding is a huge plus in S-B), but here personal preference has no play: the difference is so high as to be easily tested by simple calorimetric experiment.

        Maybe I have misunderstood C.J., and his derivation is in fact strictly equivalent to S-B. But then, why the fuss? it is only a re-interpreation of the same formula, and by definition, should have exactly the same effect, being used for computing calorimeter calibration, heat exchange in a turbine, or GH effect…

      • kai,
        I am not for CJ, I am only noticing that much that has been used against it is not valid argumentation but presents lack of knowledge about the variety of the ways the same basic physics can be approached in practice.

        The full dynamic equations are very complex and cannot be solved directly. Therefore some ways have been developed for solving then stepwise. The standard approach goes through the micro physics. The method is based on perturbation theory which is equivalent to introducing photons. The method implies also discussing the emission and absorption of the photons by transitions between the ground state and vibrational state of individual molecules. Each photon is a separate entity having a random phase of EM fields in relation to other photons. This is in accordance with a state collapse in the Copenhagen interpretation of quantum mechanics. Thus we describe the wide macroscopic phenomena as a combination of a huge number of independent microphysical phenomena. This leads to good results because the higher order terms of the perturbative analysis are very small and the coherence between micro-processes very weak.

        While the above approach has provided very good results, it is not the only possible approach of making the original insolvably difficult problem solvable. Another approach would be to look at the macroscopic problem and use some clever averaging and smoothing to make the field equations solvable. I am not at all sure that this can be done in practice, but it is not excluded. If the approach works in is likely to involve solving Maxwell’s equations with some clever way of describing the interaction of electromagnetic fields with molecules. This interaction must conform with quantum mechanical description of molecules, i.e. with the Schrödinger equation, but this may be done without the use of the state collapse of Copenhagen interpretation. Like the Schöringer’s cat the molecules will remain both alive and dead, i.e. it is not known whether they are in the exciter on in the ground state.

        What I have written is highly speculative and would have its right surrounding in a site, where different interpretations of QM are discussed, such as Copenhagen interpretation, many worlds, hidden variables etc. What I have written is in line of my own longstanding thoughts on these issues and I do not know, how many others would agree on them.

      • One may ask, how is the above second approach consistent with the fact that we can measure backradiation with a measuring device. There is no problem in that. In that approach electromagnetic field is present everywhere in space and the measuring device is interacting with this field.

        In this approach the field is no more forward or back radiation, it is just EM field in a state consistent with all matter that interacts with it. Gas that is conventionally described as radiating back-radiation is influencing this field reducing the energy carried by the field upwards, but there is not specific back-radiation.

  61. Backradiation is unstable because it would correspond to a negative dissipative effect, which is unstable just like the backward heat equation with negative diffusion. This is well known and supported by solid math. You cannot unsmooth a diffused image by negative diffusion. If you don’t believe try it
    in photoshop.

    • It is not: “back radiation” (I put it in quotes, because there is only radiation, not main or back) is always lower than the other one. You can not analyse stability of “back radiation” only, you have to include all radiative exchange in your stability analysis. Including all radiative exchange, S-B always predict net heat exchange from hot to cold, never the opposite. In a many-body case, it is thus a diffusive equation, not a negative diffusive equation, and it is thus stable. If you do not agree with this statement, you just have to provide an example using S-B relations that would lead to unstable situation, where entropy would decrease (hot will get hotter, cold colder). Best would be to start with isothermal and find a perturbation that grow, but even in non-isothermal situation, if you can find an example where entropy is decreased by S-B, it would be enough ;-)

    • Claes,

      for clarity can we also stipulate that “Backradiation” in this context refers to the net energy increase caused by the so called “greenhouse effect”.

      As opposed to downwelling radiation which is the result of general emission based cooling of air at 30km alt and above.

    • There’s no ‘negative diffusion’ and no instability. The cooler body transfers heat to the warmer one, but the warmer one transfers more heat to the cooler one, so the net heat flux is always from the warmer to the cooler.

  62. Sometimes experiments serve better than words at resolving differences of description.

    Consider a 1-D system of two black body plates set at 100K and 400K. The Stefan-Boltzmann flux is 1446 W/m^2. Next, insert two intermediate plates, positions otherwise irrelevant. The steady-state temperatures become (100K, 304.53K, 361.62K, 400K) and the flux drops to 482W/m^2. Next insert a central fifth plate. The temperatures are now (100K, 283.67K, 336.69K, 372.36K, 400K) and the flux 361 W/m^2. Adding the fifth plate has lowered the temperature of one plate and raised that of another.

    Does Johnson’s physics yield these numbers? (I have no idea!) If not, there’s a simple experiment to do. If so, where’s the beef?

    • “Sometimes experiments serve better than words at resolving differences of description.”

      Sorry Quondam,

      I did not see an experiment, I just saw words. Please go and perform your experiment, preferably recorded to video, see how that goes.

      I doubt you will achieve the same results as your “thought experiment”.

      • I have no doubt that he would since that is standard Radiational Heat Transfer Engineering which is applied in such situations every day! I invited Claes to apply his method to such problems several times but he ignores it. I guess the mathematician likes to derive his new equation but can’t test it against real world situations.

  63. Perhaps on this point we could also ask:
    – How much time does the energy represented by a photon from the Sun spend in the Earth system before it is lost to space?

    – How many individual molecules does that energy represented by a photon from the Sun spend time in before it is lost to space?

    – Why does the surface only warm by 0.017 joules/m2/second during the height of the day when the sunshine is beating down at 960.000 joules/m2/second.

    – Why is there no “time” component in any of the greenhouse radiation physics equations.

    The vacuity of the greenhouse gas hypothesis to answer these questions, to my mind, is its undoing especially since we are now finding so much empricial evidence telling us CO2 causes no warming.

    • Are these your questions, John? Bill Illis asked the same questions at lucia’s last night.

    • John,

      I think those are very good questions to ask in my opinion. I have always thought that the best way we can understand the effect of manmade CO2 was to calculate the extra time energy spends in the earth climate system in response to an increasing greenhouse effect. I think that way of thinking about the problem gives us the best handle on how big of an issue it really is.

      I think the essential problem, however, is that the transient nature of climate is neglected in most of these treatments. As a molecular physicist who studies time-dependent transient behavior of absorbing molecules, it seems to me that this is the area in which climate science needs the most work.

      I heard a talk by Ricky Rood in which he told an audience member that the typical atmospheric transient is gone in a few days, yet La Nina and El Nino events, which represent the coupling of the atmosphere and oceans, have very long time transients. These could be represented in the amplitude fluctuations of the El Nino/La Nina events, their phase or their damping and range from a few weeks to a few years, maybe even decades. We don’t even know yet. So his answer struck me as quite odd.

      The steady state solution in most important cases in a limiting case. Since we (the community of scholars and interested public) are convinced this case is pretty well understood, it’s time to move on to transient scenarios that better model the real world each person sees on a year to year basis. I think from there we might be able to answer the questions you pose. I think they deserve an answer.

    • John – let me address your various points one at a time.

      1. Time is an important component of computations involving radiative warming of the Earth and atmosphere as a function of the concentration of CO2 and other greenhouse gases. However, this is not because of the time needed for radiative energy transfer within the atmosphere, which is almost instantaneous. Rather, it is because heating of the surface is a time-related function of specific heat capacity, combined with elements of thermal conductivity, and in the oceans, turbulence and convective mixing. More below.

      I don’t know where your figure of 0.017 W/m2 for solar heat uptake come from – can you provide a reference to the relevant data? However, I’m not sure the figure is very meaningful. Land warms (and cools) much faster than water, but 70 percent of the Earth’s surface is ocean, and most of the heat from the sun and from back radiation originating in the atmosphere is stored in the ocean. Because ocean heat capacity is so enormous, diurnal changes in radiation entering from above exert appreciable temperature effects only near the surface. Mixing of the upper layers quickly averages out these effects, so that temperature changes in the entire mixed layer are very unresponsive to short term variation in radiation. For this layer, one tends to think in terms of months and years, and for the entire ocean, centuries to millennia – not hours. In essence, most of the W/m^2 radiated into the ocean is absorbed, the remainder being reflected as a function of albedo, which in the case of water is relatively small. Of course, increased absorbed radiation is met with an increase in emitted radiation, along with an increase in latent heat transfer via evaporation and convection. I suspect the figure you cited, if accurate, may refer to very superficial layers of the ocean, but in any case, one must specify what “surface” is involved when citing such statistics. Ultimately, the warming from the exposure you describe will be greater than the figure you cite.

      2. The number of molecules among which a photon’s energy is diffused is astronomical because of thermalization. The vast majority of excited CO2 molecules, for example, are de-excited by collision with neighboring gas molecules, thereby raising the average kinetic energy (i.e., the temperature) of their surroundings. Since the energy of each collision is immediately distributed widely via further collisions, one would have to calculate a mean number based on the Boltzmann distribution. I’m sure it could be done, but I’m not sure how informative it would be for our purposes.

      3. By similar reasoning, I’m not sure how informative we would find an analysis of the mean time a photon’s energy spends in the climate system, although the calculation could probably be done. Perhaps it would provide a clue as to the warming potential of greenhouse gases, but if so, it would be a very indirect means to that end. In explaining the greenhouse effect to non-scientists, CO2, water, and other GHGs are sometimes described as “delaying” the escape of radiation to space, but the description is misleading. It is true that energy radiated from the surface, and absorbed and reradiated many times before escaping is delayed in a temporal sense, but the time delay, which is extremely small by our mundane concepts of time, is not the mechanism underlying the warming. Rather, warming occurs because of a temporary imbalance between the incoming solar radiation and the longwave radiation escaping to space due to the fact that the GHGs intercept upwelling radiation and cause it to be reradiated in all directions including downward. This imbalance is translated into increased radiative energy absorbed within each layer of atmosphere down to the surface, and a balance can be restored only when each of these entities warms sufficiently so that outgoing longwave radiation, which depends on temperature, returns to its former level. Because escape is impeded by higher GHG levels at any given altitude, energy must reach a higher altitude for adequate escape, and since higher altitudes are colder, they must be warmed from below to mediate IR emission sufficient for a full restoration of balance. In essence, the greenhouse effect can be quantified not by asking “how long?” but rather by “how high, and how cold?”, and computing the results over a spectrum of wavelengths. These theoretical calculations are now well confirmed by observational data.

      4. I’m surprised by your claim that empirical evidence refutes a warming role for CO2. I’m familiar with the climate science literature, including data from recent and current measurements, as well as data extending back more than 400 million years – all converging from multiple sources to demonstrate a very substantial role for CO2. It would be illegitimate in science to insist that any phenomenon, including a warming role for CO2, can be demonstrated with 100 percent certainty, but in this case, the level of certainty is high enough to approach 100 percent. I’m unaware of any evidence at all that suggests the absence of CO2-mediated warming, and so I believe your statement is simply wrong. However, I would be interested in appropriate data references that have led you to make your claim. In truth, though, the realistic element of uncertainty is not whether CO2 warms the climate appreciably, but to what extent. This quantitation has been the subject of numerous discussions here and elsewhere.

      • Fred,

        that was a thorough answer and quite informative.

        I did take notice of one particular statement.

        ‘…although the calculation could probably be done. Perhaps it would provide a clue as to the warming potential of greenhouse gases, but if so, it would be a very indirect means to that end.’

        I think this statement is only meaningful if we assume the climate system is a strictly steady state system. Obviously, for a steady state system time dynamics are not interesting because we’ve assumed that they have dissipated, whatever they were. That is the definition of steady state.

        The climate system, however, is inherently dynamical and its the transient in the climate system that cause the up ticks/down ticks in snowstorms, hurricanes, floods (when people aren’t causing them) and the other ‘wonderful’ events we witness in this world.

        I also think that it is incomplete to think that the radiative transfer happens instantaneously. I agree that when we focus on the gases in the atmosphere thermalization occurs very quickly and when a lone CO2 or water is excited and along enough, radiative decay happens faster than we can perceive. That said, it may be possible for transients in the atmosphere to manifest themselves in other aspects of the climate system and get propagated for much longer times.

        Couplings to the oceans, cryosphere and biosphere are very poorly understood at this point in time, especially because they are heterogeneous. I can imagine reasonable cases in which transients of greenhouse effect could cause plant growth that impacts an ecosystem for many years or cause the overturning of a current in a different way or melts/freezes portions of glacier, in all cases causing changes that last much longer than ‘instantaneous’.

        To the zeroth order, I think the steady state picture provides a useful tool. I just wonder if we’ve used most to all of its utility.

      • Bill Illis has answered these question’s at lucia’s Blackboard. Hey, Judy, how about a main post for this gem. It’s shiny.

      • Maxwell – I agree completely that in our current non-steady state, time constants are an important element in determining climate dynamics. I tried to make that point when I mentioned the very long times involved in ocean heat storage. I can’t agree with John that these elements are neglected, and in fact, both models and observational studies are often aimed at quantifying the time relationships. My more limited point was that radiative changes in the atmosphere in response to a change in radiative balance at the top of the atmosphere occur extremely rapidly. It is the non-radiative elements of climate dynamics, including convection in the atmosphere and energy transport and storage in land and oceans that consume more time.

  64. I’ve been looking for easier ways to understand what’s happening to global temperature and why. The concepts of back radiation and the second law of thermodynamics both seem to me to make the reasoning very complicated, with the result that one can use these concepts to prove anything you want, including that the planet is cooling, or that it is warming. We’ve seen endless examples of this sort of reasoning not just on Climate Etc. but all over the web.

    So I asked myself, is there one single phenomenon to which all such questions can be reduced, which doesn’t allow the outcome to be argued either way according to what one believes?

    I think there is. It is how many photons are leaving Earth. Or how much radiation if you don’t like thinking about photons.

    There seems to be no serious debate as to how much radiation is arriving. The intensity of sunlight at 1 Astronomical Unit (AU) from the Sun, which is where we are, is around 1370 W/m2. The area of the Earth capturing this as a disk is around 127 .5 million sq. km (precisely one quarter of the area of the surface of the Earth as a sphere). And Earth’s albedo is around 0.3, meaning only 70% of the intercepted insolation is heating Earth.

    Multiplying these together gives 1.37 * 127 * 0.7 = 121.8 watts, with the decimal place 3+12 = 15 places to the right. This comes to 122 petawatts, a phrase that’s easily googled if you want to check the math.

    For equilibrium, that is, in order to maintain a steady temperature, Earth must radiate 122 petawatts to outer space. Each photon of that radiation can come from only two places: the Earth’s surface, or a molecule of one of the greenhouse gases in the atmosphere.

    These two sources of radiation behave very differently. Earth’s radiation is sufficiently broadband as to be reasonably modeled as radiation from a “black body” at around 288 K. In sharp contrast the greenhouse gases radiate at certain wavelengths called emission lines. These lines coincide in wavelength, if not always exactly in strength, with absorption lines.

    The radiation leaving Earth can therefore be classified into two kinds: the black body radiation leaving the surface of the Earth, and the emission lines leaving the atmosphere.

    The last line of these tables shows that 80% of the blackbody radiation leaving Earth’s surface is between 7.62 and 32.6 microns in wavelength. Some of these wavelengths are open to the escaping radiation while some are blocked by the absorption lines of the atmosphere’s many greenhouse gases.

    The two dominant greenhouse gases are H2O or water vapor and CO2 or carbon dioxide, having respective molecular weights of 18 and 44. (There are variants of these with an extra neutron or two in each atom but those are in a distinct minority and hence can be ignored here.)

    Human population has been growing exponentially for many thousands of years, doubling around every 90 years or so in the past couple of centuries. The per capita fuel consumption has also been growing exponentially over this period, with the result that we are doubling our contribution of CO2 to the atmosphere every three or four decades.

    The late David Hofmann, shortly after his retirement as director of NOAA ESRL Boulder, claimed a more precise doubling period of 32.5 years, along with 1790 as the approximate date when the residue remaining in the atmosphere from our additions was 1 part per million by volume (ppmv) of CO2. He assumed this residue to be added to a natural base of 280 ppmv during the previous few centuries.

    Barring any strenuous objections to these numbers I’m happy to go along with them. The upshot is that we can estimate CO2 over the past few centuries as 280 + 2^((y − 1790)/32.5) where y is the year. For example if y = 2010 then this formula give 389 ppmv which is in excellent agreement with the CO2 level measured at Mauna Loa.

    All this arithmetic is mainly to make the point that we are increasing the CO2 in the atmosphere, while adding a little corroborative detail.

    Of the photons escaping from Earth’s surface, some are at wavelengths blocked more or less strongly by CO2. Call a wavelength closed when the probability that a photon leaving Earth’s surface will be absorbed by a CO2 molecule before reaching outer space is less than 1/2, and open otherwise. (Sometimes 1/e instead of 1/2 is used, in conjunction with the terminology of unit optical thickness, but it doesn’t make much difference to the outcome and 1/2 is easier to relate to.)

    The HITRAN08 database of CO2 absorption lines lists 27995 lines in the above-mentioned range from 7.62 microns to 32.6 microns. Currently 605 of those lines are closed. According to Hofmann’s formula CO2 will double by 2080, which will close a further 120 lines. This will leave 27,270 absorption lines of CO2 still open, of which only a further 2502 lines will close when and if the CO2 level rises to 40% of the atmosphere by volume, a more than lethal level for all mammals.

    Now the closed lines aren’t truly closed because they can emit as well as absorb. These account for the photons radiated to space from the atmosphere, as opposed to from the surface of the Earth.

    It is tempting to argue that increasing CO2 will increase the radiation from these closed lines. To see why this is wrong, picture the CO2 molecules in the atmosphere as grains of white sand on a black sheet of cardboard. When there are very few grains the cardboard looks black, but as the grains fill up it gradually turns white. Furthermore the more grains there are, the higher above the cardboard are the visible grains.

    The same effect is happening with CO2 molecules that both absorb and emit. For any given wavelength, with very little CO2 an observer in outer space looking at just that wavelength sees the surface of the Earth. As the CO2 level increases the observer starts to see CO2 molecules covering the Earth’s surface. And as the level continues to increase, the visible CO2 molecules are found higher and higher, just as with the grains of sand. But the higher they are, the colder, at least up to the tropopause (the boundary between the troposphere and the stratosphere). So radiation from CO2 molecules decreases with increasing level of CO2 in the atmosphere. This is not true of the CO2 molecules in the stratosphere, but there are too few of them to make a significant difference.

    This is a complete analysis of the impact of increasing CO2 on how much radiation leaves the Earth at each wavelength. It describes what’s going on both simply and precisely, unlike accounts based on back radiation and other phenomena which are far harder to analyze accurately.

    This analysis ignores the impact of feedbacks, most notably the increase in water vapor in the atmosphere expected from the temperature increase induced by the increasing CO2. That increase could work either way: more water vapor could block heat at other absorption lines since water vapor is a greenhouse gas. But water vapor also conducts heat from the surface to the clouds, a cooling effect. Hence the net effect of such feedbacks needs to be analyzed carefully.

    However the feedback cannot result in an overall cooling, since the feedback depends on CO2 raising the temperature in order to evaporate more water. The question is only whether the feedback reduces the warming effect of CO2 by some factor between 0 and 1, a negative feedback, or enhances it by a factor greater than 1, a positive feedback. It cannot reduce the warming effect to zero since then there could be no feedback.

    This pretty much covers the whole thing.

    • Vaughan: “This pretty much covers the whole thing. ”
      Nope, you missed out entirely geothermal energy loss from Earth’s core. Where’s that 5000 C degrees of heat going? IN = OUT or BOOM!
      Your equation means ‘BOOM!’

      • John, you raise an excellent question, one that was asked in the 19th century. Based on the thermal insulating qualities of the Earth’s mantle and crust, Lord Kelvin calculated that the heat at the core must be leaking out at a rate that would prove that Earth could not have formed more than 50 million years ago.

        However the geologists were unable to reconcile Kelvin’s figure with what they were observing in the geological record, which suggested the Earth was billions of years old. This huge discrepancy was a great puzzle for a while, until it occurred to physicist Ernest Rutherford to calculate the heat that could be generated by a small quantity of radioactive material (uranium etc.) in Earth’s crust. He found that it would not take much to exactly balance the amount of heat leaking out through the crust. If this were not so, in the four billion years of Earth’s life the core would long ago have cooled down to something closer to the surface temperature. In effect the small amount of radioactivity in the crust is acting like a stove to keep Earth’s core at a steady temperature over billions of years.

        Global warming has only kicked in strongly over the past half century. Compared to the billions of years in which the core could have cooled down but didn’t, half a century is nothing timewise.

    • @ Vaughan Pratt…

      You say:
      This analysis ignores the impact of feedbacks, most notably the increase in water vapor in the atmosphere expected from the temperature increase induced by the increasing CO2. That increase could work either way: more water vapor could block heat at other absorption lines since water vapor is a greenhouse gas. But water vapor also conducts heat from the surface to the clouds, a cooling effect. Hence the net effect of such feedbacks needs to be analyzed carefully.
      I have led myself to make the calculations, from the observational and experimental derived formulas, and have found the results corresponding to Photons Mean Free path and to Photons Lapse Time before the absorbent molecules of the atmosphere hit or diffuse them. I have done it for each component of the atmosphere and for the whole atmosphere. Most relevant results are as follows:

      Crossing time-whole column of mixed air (r = 14 Km, wv = 0.04) = 0.0097 s
      Crossing time dry atmosphere (r = 14 Km) = 0.0095 s
      Lapse time rate-whole mixed air (r = 14 Km, wv = 0.04) = 20.78 m
      Absorptivity-whole mixed air = (r = 14 Km) = 20.79 m
      Crossing time-water vapor at 0.04 (r = 14 Km) = 0.0245 s
      Lapse time rate-water vapor at 0.04 (r = 14 Km) = 8.05 m
      Crossing time-whole column of carbon dioxide (r = 14 Km) = 0.0042 s (4 milliseconds)
      Absorptivity-whole column of carbon dioxide = (r = 14 Km) = 46.8 m
      Total aborptivity of the whole mixture of air (r = 14 Km, wv = 0.04) = 0.01
      Total emissivity of the whole mixture of air (r = 14 Km, wv = 0.04) = 0.0096
      Total absorptivity of dry air (r = 14 Km, wv = 0.04) = 0.01 (rounded up from 0.0099)
      Total emissivity of dry air (r = 14 Km, wv = 0.04) = 0.0094
      Total absorptivity of water vapor at 0.04 = 0.024
      Total emissivity of water vapor at 0.04 = 0.0237
      Total absorptivity of carbon dioxide at 0.0004, whole column = 0.0039
      Total emissivity of carbon dioxide at 0.0004, whole column = 0.0039
      Overlap water vapor/carbon dioxide, absorptivity = 0.024
      Overlap water vapor/carbon dioxide, emissivity = 0.0235

      Those are well reviewed results, supported by observation and experimentation.

      Now tell me, do you think the “downwelling” radiation heats up the surface? Why to talk about a “downwelling” radiation when we perfectly know that the possibility for the energy to be emitted is, equally, at every trajectory?
      Besides, there is a photon stream, stronger than any photon stream coming from the atmosphere that nullifies any backradiation” or “downwelling radiation from the atmosphere. The term “backradiation” is absolutely invented and incorrect; why? Because the air is not a mirror.
      Why dismissing convection, when we perfectly know that it is the prevailing way of heat transfer in the atmosphere?

      • WordPress have mixed up all the lines corresponding to the data. Please, go to the following table:

      • The term “backradiation” is absolutely invented and incorrect; why? Because the air is not a mirror.

        Since both you and I have rejected the concept of “back radiation” as not helpful (if not for exactly the same reasons—in particular I don’t consider it incorrect, just harder to work with) it sounds like we’re both more or less on the same page regarding that aspect.

        Why dismissing convection, when we perfectly know that it is the prevailing way of heat transfer in the atmosphere?

        I agree that convection can make a difference in the thermal insulating qualities of the atmosphere, for example by transporting heat from the surface upwards, e.g. via thermals. What I was focusing on however was the heat leaving Earth for outer space, which cannot be accomplished by convection because there is no significant flow of matter from Earth to outer space. Radiation is the only way available to Earth to shed the 122 petawatts of heat that the Earth is constantly absorbing from the Sun. Hence to understand how an increase in CO2 could heat up the Earth it suffices to consider how increasing CO2 blocks some of the departing radiation.

        One point I neglected to make is that the heating resulting from blocking radiation raises the temperature of the Earth until it is once again shedding 122 petawatts, the amount of heat it is absorbing from the Sun. The additional lines closed by increasing CO2 make for a smaller atmospheric window through which to push those 122 petawatts. In order to get the same amount of heat through this smaller window, Earth’s temperature has to increase. This is analogous to having to raise the voltage across an increasing resistance if you want to maintain a constant current.

  65. Judith,

    The misconception of science is that it is suppose to be a balanced system.
    It is far from it.
    Our concept is so far out of balance with what is actually happening due to the past down theories that apply to ALL of the planet at the same time.
    Hmmm. Round Planet rotating.

    • Claes, the issue is this. For the past many decades climate researchers and physicists have put their equations, data and analyses out there. The story of IR emission by gases hangs together very well in terms of observations, theory, and radiative transfer modeling. The challenge is in your court to demonstrate that any of this is incorrect, and to put forward a coherent case that convinces people that are knowledgeable of the observations, theory, and modelling. IMO you have failed to do this. This isn’t about exchanging equations. The body of physics and chemistry that underlies the calculations of gaseous absorption and emission made by line-by-line radiative transfer models is well understood, apart from some issues related to the water vapor continuum absorption under very high humidity conditions (this is understood in terms of the observations, but not theoretically, and hence is parameterized empirically in the models).

      • “The story of IR emission by gases hangs together very well in terms of observations, theory, and radiative transfer modeling.”

        And yet still all the counter arguments which have been presented here supported by hard evidence and real-world observation are far more compelling.

        No thought experiment or computer model can change reality. As Richard Feynman said:

        ” It doesn’t matter how beautiful your theory is, it doesn’t matter how smart you are. If it doesn’t agree with experiment, it’s wrong.”

      • Judy: You just repeat a mantra without mathematical basis. I prove that the “backradiation” of the KT energy budget which you say you believe describes real physics, is not to be found in Maxwell’s equations, which have shown to model almost all of macroscopic electromagnetics. You say nothing about this proof. You are still convinced and probably teach your students that in some mysterious way a cold body sends out some mysterious particles which in some mysterious way heats a warmer body. It is a mystery in every step from scientific point of view, but mystery is not science. I have demonstrated that “backradiation” is fiction, and it
        is now up to you show that my proof or assumption is incorrect, or accept it as correct. Can we agree on this? So what is wrong with my argument?
        Have you read it?

      • Your argument is incapable of explaining radiational heat transfer which is used in practical situations everyday where theoretical predictions are confirmed by measurement. You have dodged the challenge to apply your ‘theory’ to a practical situation, until you do you’re just hand waving. Show your working for us to follow your calculations, until you do it’s just ‘hot air’, I’ll await your calculations.

      • Sure, they are on the way. There are many things you can compute from Maxwell’s equations.

      • You talk too much… Demonstrate that Claes is wrong with your own numbers. I’ll be waiting here… … … … …

      • Do you dispute that if you put an infrared radiometer on the surface of the earth and point it upwards, that it will measure an IR radiance or irradiance (depending on how the instrument is configured)? Go to for decades worth of such measurements. And that this infrared radiation comes from IR emission by gases such as CO2 and H2O and also clouds? If you say yes, well this is what people are calling back radiation (a term that I don’t use myself). If you say no, then I will call you a crank – all your manipulations of Maxwell’s equation will not make this downwelling IR flux from the atmosphere go away.

      • Dr. Curry,
        The IR irradiance from the lower temperature/frequency/entropy atmosphere cannot heat the higher temperature/frequency/entropy Earth, as explained by another author of “Slaying” here:

        even though “back-radiation” can be measured by a thermocouple or thermister that has been cooled by liquid nitrogen to temps lower than the atmosphere in order to measure said “back-radiation.” [alternatively, less expensive units can measure “back-radiation” at ground temperature by e.g. a thermister increasing or decreasing resistance (depending on the type) due to the thermister losing heat to the atmosphere and a mathematical correction is applied to measure temps lower than the sensor]

      • How about a little thought experiment, or actually a quiz, anyone?

        Imagine two blackbodies, one has emitted a 9um photon, which will interact with the other blackbody, the other has emitted a 10um photon which will interact with the first blackbody.

        Now both blackbodies will be warmed by the photon it interacts with.

        The question is:

        Which blackbody is warmer, the first or the second?

        I will listen only to those who can answer the question.

      • Warmer blackbodies emit more energetic radiation.

        Photons are very small.

        I have nothing to add.

      • Blackbodies make me warm.

      • I have nothing to add.

      • Given the amount of information you have provided for your ‘quiz’, it is not possible to tell which body is warmer. Two bodies at different temperatures can both emit photons at both 9 and 10 um. The distribution of frequencies emitted is very large.

        If you are more precise and specific with the question, I should be able to answer it.

      • The question is vague because the real question is whether or not there is a two-way flow of energy between the two blackbodies or not.

        If Cleas Johnson is right, then Planck, Einstein, and the Standard Model is wrong, and there should be some exchange of Nobel Prizes.

        And maybe a photon can carry more than one peice of information.
        Like it needs to know where it has been and where it is going.

      • bob,

        I don’t know why the question is vague, but it is.

        More to your point though. If one blackbody gives off energy, and another gives off energy, why wouldn’t they flow energy to each other?

        Johnson is basically saying that one blackbody (the warmer one) KNOWS that the other blackbody is colder. And he is saying with by only referring to this fact via the source-less Maxwell’s equations, according to his comments here.

        Did I miss something?

      • Now that I realize it was your question originally I take back my previous comment.

        You won’t understand it.

      • No, you didn’t miss a thing.

        That’s what I have been trying to say, that Cleas Johnson requires that the photons know where they are going and where they have been.

      • If Einstein is right, then Claes is right.

      • Nasif,

        you do realize that Einstein was the first to propose the existence of the photon, don’t you?

      • @ maxwell…

        Don’t you? I know also Einstein deduced induced emission many years before it was confirmed by observation/experimentation.

      • you do realize that Einstein was the first to propose the existence of the photon, don’t you?

        In 1678 Huygens proposed that light was a wave, contradicted in 1704 by Newton who claimed light consisted of particles. Newton’s particle theory was generally accepted over Huygens’ wave theory until 1801 when Young’s two-slit experiment showed that Huygens was right. The wave account then survived for a century until Einstein showed that Newton was right too.

        However Huygens had no idea what the wavelength was, while Newton had no idea how big the particles were or how a mirror could reflect them. So neither of them had as much claim to their respective theories of light as Young and Einstein, who were the first to actually observe respectively the wave and particle forms of light.

        Newton called the particles “corpuscles” while Einstein called them “light quanta.” The snappy term “photon” was introduced later.

      • The IR irradiance from the lower temperature/frequency/entropy atmosphere cannot heat the higher emperature/frequency/entropy Earth,

        Yes, but what it can do however is reduce the loss of heat. When (for each square meter of surface) you have U watts of heat going up and D watts going down, with D < U, the net loss of heat from the surface is U − D.

        If U is 396 W and D is 0 W then the net loss of heat from the surface is 396 W. If however D is 333 W then the net loss of heat is only 63 W. This does not contradict the 2nd law of thermodynamics because the net flow of heat is still from the hotter to the colder entity, there just isn't as much flow between two entities that are at relatively similar temperatures. Although 63 W might seem like a lot of heat, in terms of temperature the difference is only 289 − 277 = 12 degrees. (100*sqrt(sqrt(396/5.67)) = 289 K.)

        There is incidentally a fundamental error in Dr. Anderson’s website. He says “Each time a greenhouse gas molecule absorbs ground radiation energy, it sends half of it back to the surface.” While it’s true that half the energy goes up (not necessarily straight up) and half goes down, the latter need not reach the surface because it may be intercepted by another GHG molecule first. That possibility is one of the things that makes it extremely hard to calculate just how much heat GHGs intercept.

        This is why I recommend my much simpler way of calculating it, namely solely in terms of the number of photons escaping to outer space. Those are the only ones capable of cooling Earth: if none escape the temperature will rise enormously. Those photons reradiated from the atmosphere bounce around the atmosphere, sometimes hitting the ground and sometimes escaping to space, and are much harder to reason about. Rather than even try to reason about them, just ignore them altogether on the ground that only those photons that escape to space make any difference to global temperature.

      • Vaughan,

        a couple of nit picks. Some of the IR goes sideways and depending on height some of the generally downward doesn’t even go to earth as the earth is round and not infinite, so, less than half goes in the direction of the ground.
        Increasingly less with altitude.

        As far as what is moving between earth and GHG’s, part of the argument is whether observed IR really transfers a quantum of energy that translates to heat.

        Various arguments include the fact that a photon is a wave front until it actually transfers its energy to something, which means in quantum mechanics it simply may not do it where we think it should.

        As there really do appear to be teleconnections between “particles” the photon may KNOW not to transfer its energy to the higher temperature bit just like in conduction where the material KNOWS not to move energy from cold to hot.

        Finally, fitting in with the idea of a slower cooling of the surface, the warmer surface may simply reradiate the energy from the incoming IR without it affecting the temperature.

        Then there is the older solid science of wave interference. Long before quantum theory was relatively solid it was known that waves interfered and cancelled each other. Why that is not considered as a possibility for colder not heating warmer or not slowing the warming I simply don’t understand. The energy equations show a NET energy flow and the interference, scattering, and cancellation could be components of creating this NET flow. In the case of a NET flow it should be noted that there would be NO slowing of the rate of radiation from the hotter surface unless the scenario where the photon coming from the colder source is absorbed and reradiated is correct. I am unsure why the lower energy photon would be able to cause a quantum increase in the warmer material though. Again, where are the quantum mechanics to explain this stuff!!

        My problem with the reradiation of the colder sourced IR is that there is an additive effect that would seem to cause more warming or at least extending the cooling time. This should be measurable. If it isn’t the effect probably isn’t large enough to worry about. The problem with the current numbers is that they do not appear to break out the effects of conduction from depth in the surface. This is a small effect, but, so is the amount of CO2 heating that is alledged to cause feedback with water vapor.

        So many choices and so few people with the skills to guide us to the correct conceptualization of what is actually happening.

      • Kuhnkat,
        The quantum electrodynamics (QED) developed by Feynman and others is an extremely successful theory in describing how photons are created and how they interact. It has been tested empirically to a better accuracy than perhaps any other physical theory. From QED we know how photons interact with material. We know that the photons do indeed release their energy in well understood ways. The is no change that the alternatives that you propose might be true.

      • Yes. And I am almost sure you can check that low frequency photons can heat “high” temperatures bodies, if, like most westerners, you have a microwave.

        The food you put in is usually between 270 and 300 K. It is very rapidly heated to 370+K by photons at 2.4 GHz, about 0.1 m wavelength. I do not have blackbody emission curve at hand, but this should be the typical max emission wavelength for a bb opf a few K, maybe 10 K max, no? Much lower than the food temp, for sure. So why is it heated? Because magnetron is a coherent source? I doubt it, nowhere in the heating proces is coherence required afaik…

      • Kai,

        why do you forget the concept of heat pumps? Is it just convenient to ignore the actual physics.

      • Kunhkat,
        sorry, but here, I completely fail to see any relationship between heat pump and microwave heating.

        Appart from the fact that a fridge and microwave oven are often quite close to each other in a typical kitchen or in the mall ;-)

        Seriously, you will have to elaborate a lot more before I consider my failure to see any connection something to be corrected…

      • Kai,

        how much energy does you microwave “consume” to heat your food and how efficient is it??

      • It is very efficient. Do you think microwave would have been introduced for industrial food heating if they were not? (the first one were much more powerful than the current one – they were scary ;-) ).

        Efficiency for a heating apparatus is something extremely easy to achieve though, depending on how you measure it.
        typically almost 100% of input energy is converted to heat…because everythin is ultimately converted to heat!
        So I guess you refer to heat energy IN FOOD/input energy instead total heat energy/ energy input (which is usually near 100% -possibly escaping sound and EM wave are absorbed too far to be counted).

        For the “food” efficiency, microwave ovens are very efficient. Magnetron are nice efficient device (I am quite fascinated with tube-age power electronics, klystron, fusor, all that stuff. Nice that everybody have his own magnetron nowadays) , and not much heat is lost outside the oven nor transmitted to recipient (well, it depend in what you put your food). Or maybe you refer to efficiency compared to a perfect carnot cycle (hence you heat pump reference). Sorry, I do not know of any food warming technology using heat pump. Maybe there is, but I never saw any. If there is, I guess for large amount of food it can me more efficient than microwave…
        Why, you are just about bringing to market an ultra-efficient combined fridge/oven based on heat pump? Congrat to you, but what does it have to do with C.J. theory about radiative heat transfer???

      • Try this Kai,

        it actually explains in detail how ir excites h2o molecules. Notice that if a molecule does not have the correct configuration it will NOT be excited by this method.

        So, what molecules exactly are preferentially excited by 15 micron IR from CO2??

      • Pekka,

        it is interesting you couch your closing sentence to me with the phrase “there is no chance” when theoretical physicists tell us that the universe may just be one of those chances that you blithely suggest doesn’t exist.

        Quantum mechanics, as I am sure you are much more aware than I am, is based on statistics. Statistics allow for many stranger things than my simpleton maunderings.

        But, I am a hardheaded simpleton. Can you refer me to experiments showing the increased radiation from a heated object caused by moving a cooler object close to it??

      • You could probably do an experiment yourself at home that would test the ability of a cooler object to raise the temperature of a warmer object. It wouldn’t be perfect, but should give a reasonable approximation.

        Start with a cool room, and in the center, a 100 Watt light bulb. Turn on the bulb, and let the room temperature equilibrate. Also place a thermometer against the light bulb (shielded from outside influences) and record the temperature at the surface of the bulb. At this point, the bulb is radiating 100 W and the room is losing 100 W through walls, windows, etc.

        Now surround the bulb at a distance of about 1 meter with wire mesh at room temperature. The purpose of using mesh is to provide space for air currents to escape so as not to interfere with convection. We can also leave the mesh open at the top so that rising heated air will not affect it. Also, because the conductivity of air is very low, we can reasonably assume that most heat transfer will occur by radiation – admittedly, it would be better to perform the experiment in a vacuum, but that wouldn’t be practical.

        Place a thermometer on the mesh (again shielded, so that it records only mesh temperature). Allow equilibration. The room temperature will not change, because 100 W are still flowing into the room – the amount from the warmed mesh compensating for the reduction due to heat absorption by the mesh.

        Here are my questions:
        1. Do you agree that the mesh will warm due to radiation absorbed from the light bulb?
        2. Do you agree that the mesh will remain cooler than the light bulb surface, because not all the 100 W are absorbed by the mesh?
        3. Do you agree that the warmed mesh will radiate some of the wattage it receives back to the light bulb?
        4. Do you agree that the surface of the light bulb will also continue to receive 100 W from its internal heating element?
        5. Do you agree that the internally generated 100 W plus the W from the mesh will exceed the wattage the light bulb surface was receiving prior to being surrounded by the mesh?
        6. Do you agree that at equilibrium, the light bulb surface will now be radiating the W described in 5?
        7. What do you think will happen to the temperature of the light bulb surface? Why?

      • Fred,

        if what you are suggesting starts to happen the filament increases its resistance changing the energy flux.

      • Assume a filament that emits a constant 100 W. How would you answer the questions?

      • Fred,

        I am happy to read about real experiments and discuss them to the extent my garbled knowledge allows. Are you planning on doing this one with appropriate instrumentation?

      • Kuhnkat – I’m not planning to do the experiment, because I don’t feel a need to prove anything. However, I would still welcome your thoughts about how it would come out on the basis of the questions I asked. I also wrote those with the thought that other interested readers besides yourself might appreciate the reasoning that has been expressed by many of us regarding the ability of a cooler object to raise the temperature of a warmer one, as long as the cooler object didn’t depend on its own energy but could gain energy that originated from an external source.

        If you would like to answer the questions simply from the perspective of a thought experiment, I hope you’ll go ahead.

      • Fred,

        How can my misconceptions contribute to the advancment of the discourse??

      • For distinction between Downwelling IR and “Backradiation”, as already discussed and apparently ignored.

        And for an important understanding of why the “Backradiation/greenhouse effect” in unphysical pseudo science :

        Also apparently being ignored.

  66. As some of my writing in this chain may appear obscure and even support Claes Johnson’s texts, I want make clear that I do not see anything wrong with the standard description involving photons, back radiation and transitions between ground state and the vibrational state of CO2 molecules. I wanted only to tell that the same physics with the same conclusions may perhaps be formulated totally differently. This alternative formulation would be closer to, what Claes Johnson has presented, but would definitively not change the results of the standard approach, which rest on solid experimental and theoretical knowledge of physics. Thus I disagree totally with all his statements that would modify the final conclusions.

    • Well Pekka, either there is backradiation or there isn’t. It can’t be just a play
      with words unless physics is a swamp where something can mean anything.

      • Claes,
        The physics is the same, but it can be described in different ways. The only way that si well developed and known to work includes back radiation. It may be possible to drop the particles and stick to fields without (second) quantization, but nobody has developed theory on that basis.

        The wave-particle duality is reality when ways are searched for describing quantum physics in classical terms. People cannot discuss directly in quantum physics. Therefore such different classical type descriptions are used although there is just one real quantum physics behind.

        Back radiation is a part of the particle type description. It would not be part of the wave type description if that would really exist. The physics would still be the same. Using Maxwell’s equations is a small step in this direction, but it has not been made complete (by you or anybody else as far as I know).

  67. Tomas Milanovic

    Claes Johnson

    To your claim 1) aka non existence of “back radiation”.
    As you prefer equations , so just a few very simple ones.
    Let us consider 3 interacting systems.
    S1 is the void
    S2 is the atmosphere
    S3 is the Earth
    We will consider that we know some things about the Earth and the void but the atmosphere is complicated . There are clouds , moving gases , many mysterious and complex processes.
    So we will consider S2 as a black box where the only knowable parameters are the energy fluxes at the interfaces.

    The only assumption we will take is that S2 (atmosphere) and S3 (Earth) are in a steady state . They may transport and transform energy internally as they want but they neither store it nor release it.
    For S1 (Void) we will assume that it is in an approximate radiative equilibrium with S2+S3.
    If we call the energy fluxes F (W/m²) then we have the following equations :
    At the interface S1-S2 we have F1->2 = F2->1
    At the interface S2-S3 we have F2->3 = F3->2
    There is no contact and no interface between S1 and S3.

    That is 2 equation , 4 unknowns.
    However we can measure F1->2 and F2->1 and find that they are 340 W/m² and indeed approximately equal.

    Remark : Of course the conservation of energy would require that I write the equation for the whole system and use energy (units J) for a certain time scale .
    However once I have the TOTAL in and out energy , without loss of generality I can always divide the result by the surface of the interface and by the time to get back to fluxes (W/m²) which are more familiar . This of course doesn’t mean that it is assumed that the real fluxes are 340 W/m² everywhere . They aren’t . This “average” value is just what represents the energy conservation.

    Back to S3 (Earth) . It is behaving like a grey body with an excellent approximation and emits according to F = ε.σ.T⁴.
    When we integrate that over the whole surface and divide by the surface to get homogeneous units for all fluxes , we get a value of about 390 W/m² .
    But the radiation is not the only component of the F3->2 flux .
    We have also convection , conduction and latent heat transfers .
    These 3 components can be computed and estimated to about 100 W/m².

    Now only 1 unknown is left , the energy flux from the atmosphere to the Earth , and it is necessarily 390 + 100 = 490 W/m² .
    What can that be ?
    Even if the radiation from S1 (Sun/Void) goes completely through the atmosphere and we know it doesn’t, it is only 340 W/m².
    There would be still 150 W/m² missing.
    Convection and conduction towards the Earth is very weak because it is generally warmer than the atmosphere . Part of the latent heat may possibly return.
    But whatever part of the 100 W/m² come back to Earth , it is still not enough .
    As what is missing is neither convection/conduction nor latent heat it can only be radiation.

    Conclusion : the atmosphere radiates “back” on the Earth (hence “backradiation”) at minimum 50 W/m² but actually probably significantly more because not all incoming 340 W/m² get through and not all 100 W/m² of convection/latent return to the Earth .

    Thus it appears clearly that one doesn’t need any quantum mechanics , second thermodynamics principles or complex radiative transfers to conclude that the “back radiation” is a necessary consequence of the dynamics of the interacting systems S1,S2,S3 , as long as they conserve energy and are in a steady state at least approximately in a temporally averaged sense what we indeed observe.

    Of course one can then become much more specific and explain how the “backradiation” can be deduced from the first principles too . But I won’t repeat what has already been written 100 times above , I wanted merely to prove its existence which can be of course confirmed either directly or by measuring the fluxes I defined above .

    To your 2)
    I largely agree with this opinion . I have exposed on other threads the arguments why I believe that. It has mostly to do with the fact that the system is governed by non linear dynamics which lead to spatio-temporal chaotic solutions.
    Analytical or statistical considerations of spatial averages alone destroy all spatial correlations and have no possibility to recover the right dynamics.
    As for the computer models, their resolution doesn’t allow and will never allow to really solve the dynamical equations.
    What the computers produce are plausible states (e.g states respecting more or less the conservation laws) of the system but they are unable to discriminate between dynamically allowed and forbidden states.
    This inability to discriminate between allowed and forbidden states becomes of course worse when the time scales get bigger .

    • But you are dismissing something very important, the total emittancy of the carbon dioxide, which, from experimentation and observation, is quite low. Well applied, the algoritms give 0.02 for CO2 and 0.01 for the whole mixture of the air, including water vapor. I must say that the algorithms derived from experiments give a ridiculous total emittancy for CO2, which is 0.002, at its current partial pressure in the atmosphere. Those are important parameters that are not taken into account by the current models. Carbon dioxide is not a blackbody, according to the most elemental definition of blackbody, but a graybody. The ignorance on this physics issuesintentonally or not, has taken to many people believe in backradiations heating up, or keeping heat, the surface.

      • Sorry, it should have said:

        “Well applied, the algoritms give 0.004 for CO2 and 0.01 for the whole mixture of the air…”

      • Tomas Milanovic

        Nasif I make no assumption about the blackbox atmosphere , what it contains and what it does .
        I just observe and measure the fluxes at interfaces and apply energy conservation for systems in a steady state .
        From there follows necessarily the existence of a radiation flux from the atmosphere to the Earth .
        I do not attempt to say how much or by what mechanism because others have developped it ad nauseum .
        I demonstrate that observation tells us that the number is strictly positive what is enough to establish its existence .

    • “At the interface S1-S2 we have F1->2 = F2->1
      At the interface S2-S3 we have F2->3 = F3->2
      There is no contact and no interface between S1 and S3.

      That is 2 equation , 4 unknowns.
      However we can measure F1->2 and F2->1 and find that they are 340 W/m² and indeed approximately equal.”

      ?? Isn’t that 2 equations and TWO unknowns? If you know F1->2, you already know F2->1, if they are equal.

  68. One area that Claes approach may give a new way of looking at a problem that has often been discussed on SoDs site.

    That is what is the fate of the radiation from the colder object when it arrives at the hotter object.

    To keep things simple lets say both objects are blackbodies.

    Three tenable approaches are generally given.

    1. No radiation from the colder object arrives.
    2. The radiation arrives but is simply subtracted from the greater amount of radiation of every wavelength leaving the hotter object.
    3. The radiation arrives and is completely absorbed.

    Lets see how the 3 approaches deal with a simplified problem.
    Let the colder body be at 290K

    Lets consider an area of 1m2 some way from the colder object.
    With the hotter object absent;
    this area has a flux of 100W/m2 passing through it. (This means 100joules per second pass through the area)
    If examined the spectrum of the radiation would be BB centred around 15um.

    Now bring the hotter (1000K) object to this area.

    Approach 1 says the radiation from the colder object no longer arrives at this area.
    I consider this to be unphysical and will now drop this as it seems unreasonable.

    Approach 2 says the subtraction of the radiation will still leave more radiation of every wavelength leaving the hotter object.
    This satisfies the Stephan Boltzmann equation and also means that the colder radiation has no effect on the temperature of the hotter object.

    Approach 3 says the 100Joules per second is totally absorbed and add to energy of hotter object.
    The temperature of the hotter object is increased even if only slightly.
    Effectively this means that 100J/s centred around 15um is transformed into 100J/s centred around 4.3um.

    I would say this improvement in the “quality” of the radiative energy is forbidden by the second law of thermodynamics.

    Further although approach 3 seems to satisfy the Stephan Boltzmann equation there may be a conflict there if the temperature of the hotter object increases significantly.

    For these reasons approach 2 seems to be the only correct solution.

    • As you said: this point has been discussed dozens of times on SoDs site.
      And your error is always the same: approach 3 is not “forbidden by the 2d law”. Approach 2 is impossible: it would suppose that the hotter object magically “knows” that the radiation comes from a colder object.
      The approach 3 is the correct one.

      • Ort
        Perhaps you could expand your reasoning as to why approach 2 is wrong.

      • The approach 3 is the correct one.

        Quite, it is a blackbody therefore it absorbs all incident radiation.

        Effectively this means that 100J/s centred around 15um is transformed into 100J/s centred around 4.3um.
        I would say this improvement in the “quality” of the radiative energy is forbidden by the second law of thermodynamics.

        Only if the same number of photons of higher energy were emitted, however this does not happen fewer photons would be emitted to balance the additional incoming flux.
        An example in my lab I used a Nd:YAG laser which emitted at 1066nm which I then passed through a crystal which doubled the frequency to give me 533nm output. Two 1066 quanta are combined by the crystal lattice which then emitted a single photon at 533, no thermodynamic laws broken.

      • Phil. Felton
        So you are saying that 100J of radiative energy at say 15um is thermodynamically equivalent to 100J of radiative energy at 4.7um?
        See Hockey Schtick post above.

      • Yes 100J is 100J, just fewer photons in the 4.7μm band.

      • Phil. Felton
        So you are saying that 100J of radiative energy at say 15um is thermodynamically equivalent to 100J of radiative energy at 4.7um?

        ….”Yes 100J is 100J, just fewer photons in the 4.7μm band.”…..

        Now you must feel that this is on shaky ground.

        With other physical equivalents of the “crystal” you could input low quality radiation say from seawater290K(radiative equivalent 15um) and by suitable “crystals” transform it in stages into 4.7um radiation equivalent to 1000K with no losses when absorbed.
        With such a device ships would have no need of fuel simply extract it from seawater.
        I think this is a clear violation of the second law
        I think method 2 is correct

      • Well what you think doesn’t matter. Frequency doubling (and tripling) crystals exist and don’t violate the second law as does two photon excitation microscopy.

      • @Phil…

        Oops! You’ve touched entropy. Does the entropy of a crystal diminishes or it increases? Does the entropy of that crystal surrounding increases or it decreases? Does the entropy of other crystals behave homogenously?
        Would that crystal preserve its structure as long as the universe exists? You’ve got a biiig problem, and you did it alone.

      • Phil. Felton
        The point you bring up is very interesting.
        If a crystal can double the frequency of radiation with no energy loss then I will have to revise my understanding of the second law.
        I have been to several websites to get more background information.
        I have so far been unsuccessful.
        The more relevant ones seem to be behind pay walls.
        If you could provide a link to the thermodynamics of frequency doubling crystals it would be a great help

      • I am ignorant in this area so please let me ask a couple questions.

        Are these crystals in a passive system similar to a crystal used to display a spectrum? My thought is that if it is in a powered system there may be a pump effect whereas a passive system would have much less possibility for a pump effect to be happening.

        Why would anyone consider that combining two photons of one frequency into one photon of another frequency with no change in net energy be a plus or minus to either side of the argument? It would apparently conserve mass and energy and the frequency change is proportional?

      • kuhnkat

        As far as I know these crystals are only used in lasers.
        The total power output will be less than the input.
        So it might be using work to achieve what would not happen spontaneously a bit like a refrigerator.
        However if someone can prove that a crystal can without any loses double the frequency of radiation then I will need a rethink on the second law.

      • Beats me! They’re a passive system, the crystal lattice absorbs two photons and is excited to emit a single more energetic photon (double frequency). This gives a good account:

      • Phil. Felton
        It seems the radiation has to be of very high intensity like a laser.
        Later on they talk about increasing the efficiency.
        They dont specify whether this is energy efficiency however.
        I will need to keep looking.

      • Thanks gentlemen.

      • I would assume a trade-off between frequency and amplitude.

      • BryaN,

        You are safe. The picture with the article shows a residual wave in addition to the desired second harmonic. It looks like only part of the beam is doubled and they filter out the residual for the microscopy.

      • kuhnkat
        Yes it looks like the a fraction of the fundamental went through and the desired output the second harmonic is then utilised.
        Its strange that to make sense of this phenomena we have to use the language of wave physics.
        Why should a particle phenomena like the photon have harmonics?
        It lends support to Claes ideas.
        I would like to see a further analysis of the thermodynamics of this system.

      • Bryan,

        actually I do not see it as strange at all. If there weren’t serious issues with the whole wave versus particle bit it would not have taken so many great minds so long to come up with the current compromises. The fact they settled on quantum theory as the explanation in no way invalidates the experimental data on what appeared to be waves at work.

        I think there is an issue with people thinking of a physical particle when quantum theory doesn’t really say there are physical particles. My limited reading seemed to indicate that electrons moved closer to waves than waves moved to electrons. They are both just convenient ways for us to think about a sets properties and how they interact.

      • One of the ways Climate Scientists obfuscate the physics is confusing energy and temperature. They are separate at the level that is being discussed in climate science.

        Getting a particular frequency out of a CO2 molecule does not mean it is the temperature as assumed by planck radiation. The frequency is determind by the molecular bond and not black body emission. The temperature would be indicated by the number of photons emitted by the CO2 molecule at atmospheric temperatures. Apparently CO2 has do be at combustion chamber temps for planck radiation to become significant.

      • In the cases I am aware of, the crystal which will take two photons and add them to create one photon of twice the energy, are very carefully selected or designed materials for having that effect on a particular wavelength photon. They do not do this for other frequencies of radiation incident upon them. So, the case you describe is a very special case and, yes, no violation of physics occurs in that case. But water, rock, and dirt do not generally have this property.

      • Charles
        The radiation from the laser does seem to have some unusual properties.
        For instance it does not obey the inverse square law.

      • huh? It sure does obey inverse square law, if not energy conservation would be violated. Simply, it has very high directivity, the divergence angle of a typical laser beam is very low (often almost as low as his frequency allow). But within this very small solid angle, it sure obey inverse square law: in a perfectly transparent medium, the intensity of the laser will be much lower (and the surface illuminated much larger) 1 light year away, an even a few km away, the broadening is already noticeable…

      • Dr. Anderson,

        ‘But water, rock, and dirt do not generally have this property.’

        The important distinction with respect to nonlinear optics is that the nonlinear optical process necessitates coherence in space and time between the mixing beams. From there one can get sum and difference frequency and harmonic generation. That’s why lasers are used in such situations.

        But those are not the only kinds of nonlinear effects possible in a material. There are many more incoherent nonlinear processes in which the different photons acting upon a material are not coherent. Excited state absorption and spontaneous light scattering are two such situations which have fairly high cross-sections.

        So while many rocks do not have the property of being crystals with specific bi-refringent properties, there are still many nonlinear optical processes that can occur, all which you neglect in the piece that has been featured in the comments here.

      • Ort,

        How does one electron magically know the state of another electron in quantum mechanics???

        Magic is apparently how our world works. It does what it does and we must figure out the rules and make up explanations that are palatable to our limited minds.

    • Thanks for that Bryan,
      Physicist Charles Anderson also just posted an explanation on why “back-radiation” from colder objects does not heat warmer objects:


      “…If the Earth’s surface is at a certain temperature, then it too will have a black body-like emission spectrum. Now suppose that CO2 absorbs a particular wavelength of infrared radiation out of that spectrum and then re-emits that energy at that wavelength back to the Earth’s surface. Can that photon absorbed by the surface raise the temperature of the surface? No. The reason it cannot raise the temperature of the surface is because to do so, the radiative spectrum has to move to the left in the diagram above. The shorter wavelengths on the left correspond to higher frequencies and to higher energies. For the surface to become warmer due to the absorption of the photon from a greenhouse gas, higher energy vibrational states must become occupied in the Earth’s surface materials. A photon from a lower temperature emitter cannot warm the surface to a higher temperature because that lower energy photon cannot excite the necessary higher energy vibrational modes. That photon can slow down the cooling of surface at night, since its emission at night will cool the surface and the returned photon will be at a higher energy than the surface is by the time the photon returns. This is the equivalent of the process when we put hot coffee in a thermos, thereby slowing down its cooling rate. But, the returning photons from the reflective wall in the thermos never heat the coffee to a higher temperature than it was at when it was poured into the thermos.

      Whether a photon is absorbed by a material or not is dependent upon the electronic and vibrational states in the material which can be excited and the energy of the photon. The fact that a photon is incident upon a material does not mean it will be absorbed. The greenhouse gas theorists recognize this when the material is nitrogen or oxygen molecules, but they assume the Earth’s surface can absorb whatever strikes it, at least if it is a low energy or longwave infrared photon. But, just as visible light passes through window glass without absorption, this is not necessarily the case. The light photon is not absorbed in glass because glass has a wide energy band gap in which there are no occupied or unoccupied electronic states. Still higher energy ultraviolet energy may excite available unoccupied electronic states, which in turn will de-excite in time. Until they do, they can warm the glass. But visible light just passes through. The same is the case with some of the low energy, longwave infrared radiation returned from greenhouse gas molecule de-excitations. The Earth’s surface will not accept them since the excitable vibrational states are already excited and vibrating assuming that its temperature has not dropped since the returned photon was emitted by the ground. There simply is no available energy state able to accept it.”

      • Hockey Schtick

        Thanks for the link, he knows what he is talking about being a specialist in materials physics

      • Yeah but he’s made a few mistakes. Following his argument, by emitting photons the surface necessarily cooled the instant those photons left leaving energy states ready to absorb any returning photons.

      • In the equilibrium case of solar radiation flux upon the surface, the Earth’s surface temperature is constant and the emission of a photon does not cool the surface. Of course at night, with no incident solar radiation, the surface is constantly cooling as infrared photons are emitted. In that case, a photon absorbed by a water molecule or CO2 may result in emission of a photon from that molecule and the photon may be absorbed by the cooling Earth’s surface, thereby retarding the cooling. Where I said the emission of the photon from the Earth’s surface cooled it, I was talking specifically about the phenomena of cooling at night. I wanted to make sure the reader knew that I was not denying that the presence of infra-red absorbing molecules in our atmosphere can contribute to a retardation of surface cooling at night and to make it clear how it did this, when it could not do it in the case of the surface at a constant or increasing temperature.

      • Your support is clumsy: don’t you care that this theory is in contradiction with what Bryan said?
        His theory and Bryans “approach 2” cannot be correct at the same time. Choose one side.
        (anyway, they are both wrong)

      • Ort: no, it is in agreement with Bryan’s approach 2. And tell us exactly why you know “both are wrong”

        Phil Felton: doesn’t matter – obviously the process continues to cycle, with no heating of the hotter object

      • Of course, here are the conceptual differences:
        Brian: no backradiation (supposedly because the 2d law) with a theorical case of two blackbodies (so, total absorptivity).
        Charles Anderson: backradiation, but absorptivity of the Earth surface = 0 for the longwave radiations (all the confuse 2d paragraph). That’s obviously false: you can check any textbooks for the absorptivity vs. wavelength for all the different type of opaque materials.

      • Ort: no, Bryan’s approach 1 is “no backradiation,” which he dismisses. Approach 2 is that there is “backradiation,” but the colder objects “backradiation” cannot heat the hotter object. This is exactly what materials physicist Charles Anderson explains in detail, and you fail to understand why the absorptivity is effectively 0 by a hotter temperature/frequency/entropy body from a colder body – did you even bother to go to his blog post instead of just reading the small excerpt?

      • I did and he’s wrong!

      • The explanation is false. A black body goes never to a state where it would not absorb all radiation reaching it. The absorption leads unavoidably to the transfer of the energy of the absorbed radiation to the heat content of the body. The body is also radiating at the same wavelength, but the rate of radiation is not changed by the absorption. Thus the incoming radiation influences the heat balance of the body.

        The same applies very closely also to most surfaces of earth for infrared radiation, because they are almost black in the infrared region.

      • Pekka
        ……”but the rate of radiation is not changed by the absorption. Thus the incoming radiation influences the heat balance of the body.”…..
        This seems to be self contradictory

      • Bryan,
        I was not fully precise. There will be an effect through increased temperature of the body. I meant that there is no immediate effect related to the absorption. For a real surface even this is not quite true, but only a very good approximation, but for a black body it is true.

      • Pekka
        With reference to the option2 and 3 in my post above.
        To all realistic intents and purposes there is little practical difference between them.
        The heat by calculated by SB equation goes from hotter to colder body.
        Option 3 has the unfortunate implication of upgrading the quality of the radiation from the colder object which conflicts with the second law.
        Also the possibility of an increase in temperature is a signature of Heat transfer from colder to hotter which Clausius said was forbidden.

      • Bryan,
        I answered to your other message on this point. Your argument is in error.

      • There is a curious definitional issue here. A black body is often defined as a body that will absorb all wavelengths of radiation incident upon it. This is a case however in which we need badly to talk about real materials, such as those in the surface of the Earth.

        I extensively use a technique called FTIR spectroscopy to identify and characterize materials in my laboratory. The technique commonly uses infrared radiation covering the range from 2.5 microns to 25 microns in wavelength. A material placed on a IR transparent window, such as diamond, is irradiated as the IR wavelength is varied and any absorption results in a scattering of the IR radiation so that much less is reflected back to the IR detector. If real materials absorbed all IR in this broad range of wavelengths, the technique would be pretty useless. The range of IR radiation wavelengths covers most of the spectrum of radiation from a material emitting IR at a temperature of 288K. Near IR spectroscopy covers the longer, low energy tail of the 288 K emitter and while absorption here tends to be greater, it is still much less than 100%. That makes near IR spectroscopy a useful technique also for studying many materials. Most of the Earth’s surface is covered with water and the biggest window for water in the range of IR radiation near that of a 288 K emitter is pretty well aligned with the peak of the emitter spectrum. So water does not absorb all incident IR. Plants certainly do not either. Indeed, we often perform FTIR on plant materials and food products extracted from them. Near IR spectroscopy is also used on plants and food products extensively. FTIR is used less frequently on minerals because they commonly are not very good absorbers.

      • Sorry. I do not actually do near IR spectroscopy. I should have remembered that it applies to the IR wavelengths in the tail of the solar spectrum, not in the tail of the spectrum of an emitter at 288K. Near IR is therefore irrelevant in this discussion.

      • Your ‘theory’ of non absorption by a surface of thermal radiation from colder emitters (you don’t say what happens to the incoming radiation), is clearly invalidated by the fact that microwave ovens work, (see Kai’s posts elsewhere). The usual frequency used is 2.45 GHz (wavelength 122mm), your surface at 300K doesn’t emit much radiation at that wavelength! So why does that get absorbed in an oven?
        While you’re on here why don’t you explain that when you use your FTIR spectrometer you don’t have to do it in a vacuum because O2 and N2 don’t absorb IR, some of the ‘sceptics’ on here don’t believe that. Perhaps your practical experience will convince them?

      • The specific frequencies that 99% O2 and N2 absorb emit at are filtered out.

        Such a device would be worse than useless if that was not the case.

      • PF;
        Same answer as to most AGW silliness: it’s the H2O, st**id.

      • In that case, please explain me the approach 2 , and don’t forget Brian was talking about two black bodies.

        Now, about Anderson: “why the absorptivity is effectively 0 by a hotter temperature/frequency/entropy body from a colder body “.
        You fail to understand that the absorptivity of a surface, which is the proportion of radiation absorbed vs reflected, at a given wavelength, is a constant property of the material. No matter where 15um photons come from (from a cold body, a hot body, a distant body, a shaking body, an “active” body, a “passive” body), the ratio of absorbed 15 um photons is the same.

        Another time, you can check easily textbooks for the absorptivity vs. wavelength for all the different type of opaque materials : the position of Anderson is untenable.

      • Ort |
        So you quite happy that 100J of radiative energy at say 15um is upconverted to 100J of radiative energy at 4.7um, without any work being done?

      • Without an explanation, this question does not make sense for me. Details, please.

        (sorry to have mispelled your name in my last comment)

      • Ort
        If you look at one consequence of option 3 it means that
        100J of radiative energy at centred at 15um is up converted to 100J of radiative energy centred at 4.7um, without any work being done?
        This is contrary to the second law.
        This is why option 2 is correct.
        It satisfies the Stephan Boltzmann equation without violating the second law

      • It has nothing to do with the second law.

        For the black body the wavelength of the incoming radiation makes no difference, when the amount of energy is the same. 100J heats by 100J. After the absorption it is in the heat of the body and for that the type of the incoming energy makes no difference, only its quantity in energy units.

        As stated by really many writers the black body absorbs also any wavelength whatever its own temperature.

        The second law has nothing to say about this. It tells that more radiation goes from the hotter body to the cooler than wise versa. It does not say anything about what happens when radiation hits a body.

      • “If you look at one consequence of option 3 it means that
        100J of radiative energy at centred at 15um is up converted to 100J of radiative energy centred at 4.7um, without any work being done?”

        You seem not to understand the Stefan Bolzmann law (radiation of the body occurs, even in a vacuum), and the laws of thermodynamics neither. In fact, your assertion itself is totally confused and erroneous, linking two independent phenomenons with apparently an implicit energy equality (is that what you call “2d law”?) which can not be applied to your body, which is not a closed system! You already had long, clear, detailed, repeated and explanations of this on SoDs site by different contributors, more patient than I am; so it seems I am losing my time.

      • Ort
        “If you look at one consequence of option 3 it means that
        100J of radiative energy at centred at 15um is up converted to 100J of radiative energy centred at 4.7um, without any work being done?”

        Back to the previous question you included the quote but did not answer the implication.
        Instead you ignored it and went into a irrelevant rant.
        If the increase in quality of the radiation does not happen then options two and three are the same.
        If it does happen the second law is violated.

      • There is no such thing like your imaginary direct process of “upconversion” by “work”, so, I repeat, your rhetorical question as formulated does not make sense.

        Emission of thermal radiation is a function of temperature of the body (and if not a black body, a function of emissivity, a material property) and that’s that. Period.

        If ever there is some incoming radiation, whatever its wavelength, it will be absorbed (black body). But no matter if there is or not some incoming radiation from other bodies, and what could be its spectrum, the status of the outside world has no effect on emission of thermal radiation.

        Now, in all the possible configurations, if doing the sum of all the energy exchanges (including the radiating ones) between the black body and its environment, you find: E_in > E_out, then the temperature will increase (in function of the mass and of the heat capacity). If E_in<E_out, it will decrease; if E_in = E_out, no change. There is nothing "thermodynamically wrong", here.

        With your choice of same energy values (the last case), you tried at the same time to imply an imaginary direct causal link between emission and absorption, by the means of "upconversion", your word. You are supposing that the emission of thermal radiation is due to photoexcitation: you have invented some physics. You are now free to repeat ad nauseam "2d law, 2d law!", but don’t expect another response from me.

      • Ort
        If you read the original post it was the consequence of to the radiation of having the hotter object there as opposed to its absence as it passed through the defined area.
        Absent ; 100J at BB spectrum centred around 15um.
        Option 2 no effect on the temperature of the hot body other than to reduce the heat loss from the hot body.
        Option 3.
        To increase the temperature of the hot body.
        The 100J Joules is upgraded to be centred around 4.3um.
        This violates the second law as stated by Clausius.
        Heat flows from a hot object to a cold object never the reverse.
        The increase in the “quality” of the radiation reduces entropy =>against 2nd Law.

        If the problem was solved using vectors, there would be a single vector pointing from hot to ciold

      • The case you cite is different. The incident radiation comes from a hotter source, not a colder source or one of an equal temperature.

        When I measure the absorptivity of radiation in my lab, I use a light source with a filament or emitter which is hot compared to the material I am reflecting and absorbing radiation upon. LEDs are pretty cool compared to a tungsten filament, but they are still warmer than the room temperature object being examined for its absorption of light.

        Note also that IR detectors image objects warming than themselves, not objects cooler than themselves.

      • Great explanation, IMHO!

      • This guy Anderson is a total crank when it comes to the greenhouse effect at least.

        First, the whole idea that the ground cannot absorb low frequency light because of vibrational states is complete garbage. The density of states of low frequency motion in a solid (like the surface of the earth) is much higher than the density of high frequency motions. Moreover, the vibrational energies of water and CO2 are several hundred to thousands of wavenumbers. That corresponds to temperatures over 1000 K. You’re telling me that the ground can’t absorb photons corresponding to a temperature of 1000 K? Really?

        On top of that fact, what percentage of the earth is hotter than 1000 K? Like the 0.00000001% that exposes lava lakes? So even if the structure of materials like the surface of the earth were such that there was a low density of low frequency motions, the energy from GHG emission could still be absorbed virtually everywhere on the face of the planet!

        So this guy can even come up with a meaning, fake physical theory! And you take him at face value?!

        Moreover, all this talk about how a colder body can’t transfer heat to a warmer body is only meaningful in the macroscopic limit. Unfortunately, that means when we’re talking about the collisions between molecules, which Anderson does at length, we are in the MICROSCOPIC limit and energy can transfer to a molecule upon a collision, even if that molecule is in a lower energy state. Temperature and heat are not defined for a single pair of molecules, therefore we are not violating the second law of thermodynamics as well! Make an important mental note of this fact.

        There are also sooo many collisions happening, that it is very likely energy gets transferred from a kinetically excited O2 or N2 into a state of CO2 or water from which that molecule can decay radiatively. The chances of this processes happening increase as one increases in altitude because collisions become less frequent (density decreases) while the radiative decay rate stays relatively the same. In the stratosphere, the rate can even increase due to increases in temperature.

        So all in all, one needs to be slightly more skeptical of these types of claims. To just believe it because it says what you want to hear is not very scientific. In fact, it may be the exact opposite from scientific. Then again, I’m beginning to expect that from some of you.

      • Does your rant actually refer to Anderson’s paper or to something else? What is all this stuff about 1000 K, e.g.? Where does he say that the “ground cannot absorb low frequency light because of vibrational states?” Maybe I missed something?