TRMM Satellite Coming Home Next Month

May 22nd, 2015 by Roy W. Spencer, Ph. D.

Japan's Hayabusa satellite renters the atmosphere in June, 2011.

Japan’s Hayabusa satellite renters the atmosphere in June, 2011.

NASA’s Tropical Rain Measuring Mission, the first satellite to carry a rain radar, has been on-orbit since late 1997, but last year it finally ran out of the fuel required to keep it maintained at its relatively low altitude, 400 km.

So, TRMM is “coming home” after a very successful mission measuring tropical rain systems for over 17 years.

Back when the TRMM concept was being pitched by NASA-Goddard scientists at HQ, I was pitching the competing mission representing NASA-Marshall. In retrospect, John Theon (the Program Manager at the time) made the right decision and gave the go ahead to develop TRMM.

I helped campaign for the design of the TRMM Microwave Imager (TMI), but by the early 1990s our global temperature monitoring work was taking up most of my time and to everyone’s surprise (since my original expertise was rainfall measurement from satellites), I chose not to be part of the TRMM Team.

TRMM also carried one of the CERES Earth radiation budget instrument packages, which allowed researchers to document the diurnal cycle in cloud effects on reflected sunlight since TRMM was placed in a non-sun-synchronous orbit, as well as the Lightning Imaging Sensor (LIS) which was developed here in Huntsville, and a Visible and InfraRed Scanner (VIRS).

I’ve been tracking the fall of the TRMM satellite, and as can be seen it is now descending rather rapidly:

TRMM-altitude-since-Jan-2014

If we zoom in, we get a better idea of it’s trajectory in the last couple months, and Space-Track.org is now calculating a reentry date of June 19:

TRMM-altitude-since-April-2015

Once the satellite reaches about 90 km altitude, it reenters very quickly. Because the rate of descent is nonlinear, and depends upon the satellite orientation, which might be tumbling and causing variable amounts of atmospheric drag, it is almost impossible to determine where the satellite will fall…it could be anywhere between 35N and 35S latitude, as suggested by this single day of TRMM radar coverage:

TRMM-orbits

The June 19 date could also change substantially…it might be many days off. For example, in just one day, the reentry date was moved up by a day by the Space-Track folks.

I’d like to congratulate all of the many engineers and scientists here in the U.S., in Japan (which provided the radar), and throughout the world, who made the TRMM mission such a great success.


385 Responses to “TRMM Satellite Coming Home Next Month”

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  1. Stephen Richards says:

    My big fear is what happens when AMSU sats return to earth. IT will leave the door open to the conmen worldwide.

  2. dave says:

    There are many years of life in them.

  3. jimc says:

    All good things must come to an end? What goes up must come down? The 2nd law?
    No, job well done.

    • dave says:

      “What goes up must come down.”

      It was quite disappointing. And no, she didn’t accept it was because of the 2nd law.

    • Gordon Robertson says:

      @jimc…”The 2nd law?”

      f = ma gets it up and f = mg brings it down.

      If the trajectory takes it into a doable orbit, it’s momentum will keep it going in a tangential direction while mg provides the normal force. The orbit is the resultant between mv and mg, until such time as it loses momentum (mv).

      If the trajectory is more directly into space it carries on forever and ever once it escapes the clutches of mg. Unless it comes under the influence of mg on another body.

      If mg > ma, it wont get into orbit and will crash back to Earth.

      That’s all I remember from my engineering physics courses and my formulae are woefully inadequate.

      Let’s just hope that the moon doesn’t lose mv. I’m not sure with a mass as large as the moon if it would accelerate sufficiently to smash into us or whether it would slowly nudge into us as it spiraled inwardly.

      Need to drag out my slide rule that I used early on in my engineering studies before calculators appeared to save the day. The professors were in a quandary trying to decide whether to allow calculators on exams but they finally relented.

      • jimc says:

        Gordon, as I recall, the height of the moon’s orbit is actually increasing as its angular velocity decreases. (Something about angular momentum must be conserved as the earth’s tide burns up lunar kinetic energy.)
        The 2nd law in the sense that drag converts TRMM’s KE to heat and not vice versa.
        I could probably still use a slide rule too.

  4. KevinK says:

    Actually, its “home” was in orbit, it was only built and tested here on the surface (most likely in a vacuum chamber to simulate “home” conditions) temporarily.

    It will still be home, just burnt up into a lot of gas and a few solid specks. A very efficient recycling job, started as basic chemical elements and returned to same.

    Kudo’s to all involved.

    Cheers, KevinK.

    • Curt says:

      Yes, NASA tests these satellites in vacuum chambers before launch. These chambers have super-cooled walls (using liquid nitrogen) to eliminate “back radiation”.

      Cheers to you!

      • KevinK says:

        “These chambers have super-cooled walls (using liquid nitrogen) to eliminate “back radiation”.”

        Actually Curt, it depends on the environment that the satellite will be operating in. Indeed some satellites are tested with “cooled” walls inside a vacuum chamber using nitrogen shrouds. And the Apollo Crew Capsule was tested in front of a whole bank of light bulbs that simulated the SUN (that would be “forward radiation”).

        Other satellites (like the future JWST) are tested in a (freaking huge) vacuum chamber specially modified to use liquid helium shrouds(inside another liquid nitrogen shroud) to simulate the environment “their home” where they will be operating.

        And none of these tests eliminate “back-radiation”, these tests simply emulate (i.e. match) the environment that the satellite will come to call “home” after it is launched.

        Just to clarify, I do not dispute the existence of “back-radiation”, or the existence of “back pressure” in old fashioned steam powered railroad locomotives. Yes, it exists, but it has no effect on the average temperature of the Earth (Oceans, Landmass, or Sky).

        Cheers, KevinK.

        • Curt says:

          Hmm – My reply ended up below. And the second sentence should read:

          “The JWST just underwent 3+ months of testing inside the helium shrouds.”

          • KevinK says:

            Actually Curt, SOME of the JWST support equipment just went through some testing inside the He shrouds. The JWST telescope has not been assembled yet, so it cannot be tested in any environment.

            I recently got an update from “folks on the ground” about all the JWST related testing at JSC chamber “A”. I never heard the term “back radiation” mentioned once…

            Cheers, KevinK.

          • Curt says:

            Kevin:

            Why do you keep bringing up irrelevant details?

            Any power-dissipating device, like all electronics, must in the steady state have a higher temperature than the ambient it is in. I’ve been working in electronics for 35 years now, and have never seen otherwise. It is also a trivial conclusion from the equations of thermodynamics and heat transfer.

            When the electronics is in a vacuum, the only way for it to dissipate heat is by radiative means, so it must be hotter than the objects it is in radiative exchange with. If the walls of the vacuum chamber are at earth surface ambient temperatures (~300K), the electronics will be at a higher temperature than this.

            But if the walls are super-cooled to liquid nitrogen temperatures or below, the electronics will be in radiative exchange with these walls that emit virtually no thermal radiation, and the electronics will be at temperatures much lower than earth ambient.

            The whole point of using the He shrouds is to virtually eliminate the thermal radiation that is present at earth ambient temperatures, so to mimic the almost total absence of ambient radiation it will find in deep space.

  5. Thanks, Dr. Spencer.
    TRMM did good work.

  6. Curt says:

    Kevin:

    I’m well aware of all those permutations of the vacuum chamber testing – I was trying to keep it brief. The JWST – 3+ months of testing inside the helium shrouds.

    But the whole point of going through the expense and trouble of using these super-cooled shrouds, the ONLY reason, is to dramatically reduce the ambient (“back”) radiation for the spacecraft. There is no other point to it.

    They simulate the environment the craft will see in space precisely BY virtually eliminating the ambient radiation.

    If the chamber walls were left at typical earth surface ambient temperatures, they would be radiating about 400 W/m2 toward the craft. That 400 W/m2 most certainly does have an effect on the temperature of things that absorb it. (To assert otherwise requires a “disappearance” of that energy, which is a blatant 1st Law violation.)

    Looked at from the satellite’s viewpoint, to be in steady state, the satellite must reject as much power as it gets from its power source. In a vacuum, the only method to output power is radiative exchange with the environment. It must have a higher temperature than its radiative surroundings.

    So without cooled walls, it must have a higher temperature than earth ambient. This would not be an adequate test of the thermal conditions of its space environment, when it would mostly be in radiative exchange with the 3K effective blackbody temperature of deep space.

    Cheers!

  7. geran says:

    ‘Yes, it exists, but it has no effect on the average temperature of the Earth (Oceans, Landmass, or Sky).”

    ++++++

    Gosh KevinK, it seems you got it right once again!

  8. Curt says:

    So…what happens to the energy in this radiation that you admit exists when it hits an object that has a higher temperature than the emitting object?

    What happens to the energy in this radiation when it hits an object that has a lower temperature than the emitting object?

    If the two answers are different, what is the physical mechanism that creates this difference?

    • KevinK says:

      Ok, Curt, this back radiation continues to travel through the system at the speed of light (generally considered quite quick). It then bounces off the surface (actually gets absorbed and remitted) and makes another trip through the atmosphere at the speed of light (still considered quite quick).

      The “Radiative Greenhouse Effect (aka the rGHE) merely delays the flow of energy though the system. A delay is not the same thing as a reduction in velocity, they look kind of the same, but they are different..

      A thermal insulator (aka a blanket) works by reducing the velocity of energy flow (heat). An optical delay line (like that produced inside an optical integrating sphere) merely delays the flow of energy (all of the energy entering the system leaves the system after a delay, ie shortly afterwards). None of the energy is trapped.

      Cheers, KevinK

      • Curt says:

        Ahh, so you admit that “back radiation” is absorbed by the warmer body, which means that it adds to the internal energy of the warmer body. That’s progress!

        But you have not yet answered the question as to what the difference between radiation from a warmer body and from a colder body is. Let’s look at radiation of the same wavelength. You seem to think that what happens on the receiving end is different in the two cases.

        Is it? And if so, how does the receiving body distinguish between these two cases?

        You keep talking about heat transfer in terms of physical velocity (e.g. meters/sec) instead of power (Joules/sec). I have never seen this type of analysis in any thermodynamics or heat transfer text. Can you point me to one that shows this type of analysis?

        Oh, and do you really believe that insulating a power-dissipating object with a blanket does not result in a higher thermal energy level of the object (i.e. “traps heat”)? Seriously?

    • dave says:

      “…what happens to the energy…?”

      If radiation is absorbed, its energy goes into modifying the movement of chance-encountered, light [low mass], atomic particles – usually electrons.

      “If the two answers are different,…”

      They are never different.

      See:

      “Electrons and Ether Waves”, The Robert Boyle Lecture at Oxford University, 1921, given by Sir William Bragg F.R.S Nobel Laureate in Physics, reprinted in The Scientific Monthly
      Volume XIV, 1922.

      His conclusion is essentially:

      “We know what happens; it is simple…”

      He goes on:

      …but completely mystifying.”

      The lectures of another Nobel Laureate in Physics, Richard Feynman, sixty years later, talked about the same “road-block”.

      “You may be feeling baffled by now. Well, I am baffled too. We have ways of calculating certain effects whose results are completely supported by experiment. But, in any satisfying sense, nobody understands anything about quantum interaction. The truth is, we have been completely stuck since the very beginning.”

      Still, as they both said, we do know WHAT happens.

    • Gordon Robertson says:

      @Curt…”….what happens to the energy in this radiation that you admit exists when it hits an object that has a higher temperature than the emitting object?”

      That’s a question for which I have been trying to find an answer.

      Obviously, you are referring to infrared energy, which is not heat. The heat is represented by the average kinetic energy of the atoms in the cooler and hotter objects. The IR is only an enabler for the transfer of kinetic energy.

      If the IR is from a hotter body, and it reaches a cooler body, it has a higher intensity and frequency range. It has the capacity to raise the energy level in atoms in the cooler body, which raises the KE, making the body warmer.

      What happens when IR from a cooler body strikes a hotter body? According to the second law it cannot raise the temperature of the hotter body therefore it cannot affect the average KE of the hotter body.

      The reason it cannot raise the KE in the hotter body is that IR from a cooler body has neither the intensity nor the frequency range to affect kinetic energy in the hotter body (my theory).

      It makes no sense that it simply disappears. Maybe it passes on through, not affecting anything in the hotter body.

      Afterall, IR is broadcast in a spheroid naturally from the cooler body and because part of its radiation path is intercepted by the hotter body does not mean the hotter body has to absorb the energy.

      Anybody??

  9. geran says:

    Curt! I think I remember you.

    Yeah, you are the one with the “stained blue dress” syndrome.

    When confronted with the actual science, you tried to redefine words. I don’t remember all the details, but it was hilarious.

    Were you able to get the stain out?

    • geran says:

      Okay, I am starting to remember more. You claimed to be an engineer, which I so easily disproved. Then, you claimed to be a college professor. So, humor us. What is your degreed specialty?

    • Curt says:

      Oh, I remember you too. You were the guy who tried to get me to agree to meaningless statements like “radiation has a temperature”. When I pointed out how ridiculous this was, you accused me of evading the question.

      You got the meaning of Wien’s Displacement Law totally backwards. It allows you to compute the peak wavelength/frequency of blackbody radiation for a given temperature. But you have completely twisted it around to compute a temperature value from a wavelength/frequency of radiation, which simply shows that you have absolutely no comprehension of the underlying concepts.

      • dave says:

        “…meaningless statements like “radiation has a temperature”…”

        Enclosed radiation can have “a temperature.” The statement, therefore,is imprecise rather than meaningless.

        • nigel says:

          “Enclosed radiation can have ‘a temperature.'”

          Of course:

          (1) Make a cavity black body with silvered, inner walls and introduce a large black body with a temperature of say 1,000 Deg K. The cavity will be filled with steady radiation.

          (2) Allow some of the radiation to escape into another cavity, with silvered inner walls, which has a small cold black body in it. That body will reach a temperature of 1,000 Deg K. – i.e. exactly as if the RADIATION ENCLOSED in the cavity were a material body at 1,000 K touching the cold body.

          (4) Open the second cavity. The FREED radiation no longer has ‘a temperature’. After all, it is no longer ‘mixing with itself’, it has lost any identity or unity of action.

          • Massimo PORZIO says:

            Hi Nigel,
            just to be fussy, I only partially agree with you.

            (2) IMHO the large hot BB should reduce a very little its temperature depending on how much heat had been transferred to the cold BB because of the net radiation exchange. I.e. the final temperature should be a little lower. So, by this point of view, radiation has not a temperature by itself indeed, even if enclosed. This because the total radiation increased (the cold BB wasn’t 0K) but the average temperature of the bodies decreased a little.

            Do you agree?

            Have a great day.

            Massimo

          • Massimo PORZIO says:

            Uhmmm…
            Thinking a little more about what I wrote, the averaged temperature of the 2 bodies probably doesn’t change, it’s the temperature of both the bodies which changes.
            The averaged temperature should become the bodies temperature.

            Massimo

          • Curt says:

            The thing is that geran believes that radiation of a given wavelength “has a temperature” that you can calculate by inverting Wien’s Displacement Law. Further, that this radiation cannot be absorbed by any object with a higher temperature.

            So he computes that 15 um radiation “has a temperature” of 193K (-80C) and cannot be absorbed by anything of a higher temperature.

          • geran says:

            No Curt, your “interpretation” of what I wrote is incorrect. But, nice try.

        • Gordon Robertson says:

          @dave “Enclosed radiation can have “a temperature.”

          A ‘colour’ temperature, which is the ‘equivalent’ of a temperature given off by a hot body but does not contain heat per se.

          If you heat a piece of metal until it begins to glow different colours, it will begin getting red then progress through different shades until it gets white in appearance. Those colours are representative of difference frequencies of infrared energy given of by the atoms in the metal.

          You could interpolate that radiation intensity down through the non-visible radiation spectrum and claim something as cool as ice has a radiation temperature.

          The point is that electromagnetic energy, which is radiation, has no heat content, therefore no real temperature. Although IR is given off by a red hot metal object, the actual heat is in the atoms of the metal, in their average kinetic energy.

          The radiation from the hot metal can intercept another substance and cause the atoms in that substance to increase their kinetic energy, making them hotter. However, if the IR does not intercept anything with mass, it will carry on through space with no heat and no temperature.

    • Curt says:

      If you really want to go there..

      I studied mechanical engineering, undergraduate and graduate, at MIT and Stanford, with significant coursework in thermodynamics and heat transfer. (None of my textbooks remotely agree with your conception of these topics.)

      I have worked as an engineer for over 30 years. As I mentioned to Kevin, controllers that I designed were used to make the mirrors (with nanometer-level accuracy) for the James Webb Space Telescope that is under development and test now.

      Periodically, UCLA asks me to teach engineering courses for undergraduate and graduate students.

      Your turn…

      • geran says:

        Okay, that explains your confusion.

        Your institutionalized “training” does not equip you to recognize the errors in the IPCC equation (“radiative forcing”):

        Delta F = 5.35 * ln {C/C0} W/m-2

        Which then leads to your additional confusion about AGW/CO2/GHE.

        • Gordon Robertson says:

          @geran “Your institutionalized “training” does not equip you to recognize the errors in the IPCC equation (“radiative forcing”)”

          No insult intended, geran, but I think you have that backwards. The IPCC is run by climate modelers who have taken terrible liberties with physics, especially in the field of thermodynamics.

          When it comes to the training of an engineer, especially from MIT, I am going with the engineer over any pseusdo-science coming out of the IPCC and its climate modelers.

          • geran says:

            But Gordon, what if that MIT engineer clings to the IPCC pseudoscience?

            (And, no offense taken.)

          • Curt says:

            Geran:

            I took my engineering thermodynamics and heat transfer courses long before there was an IPCC, and before there was much organized climate science at all. There is nothing in the least controversial in the engineering world about my arguments. I have not changed my views on these topics since I studied them, any more than I have changed my views on Newtonian mechanics.

            Any thermodynamics or heat transfer textbook starts its explanation of radiative heat transfer with the concept of “radiative exchange”. That’s “exchange”, as in BACK and FORTH. (You really ought to try reading a textbook on the subject some time. You might actually learn something!)

            Here’s how one of my heat transfer textbooks, written in the 1960s, introduces the topic:

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

            Consider two black surfaces A1 and A2, as shown in Fig. 8-8. We wish to obtain a general expression for the energy exchange between these surfaces essentially one of determining the amount of energy which leaves one surface and reaches the other. To solve this problem the radiation shape factors are defined as

            F12 = fraction of energy leaving surface 1 which reaches surface 2
            F21 = fraction of energy leaving surface 2 which reaches surface 1
            …

            The energy leaving surface 1 and arriving at surface 2 is

            Eb1 A1 F12

            and that energy leaving surface 2 and arriving at surface 1 is

            Eb2 A2 F21

            [where Ebn is the total radiative energy leaving the surface n as a function of its temperature and emissivity, as derived in the previous section]

            Since the surfaces are black, all the incident radiation will be absorbed, and the net energy exchange is

            Eb1 A1 F12 – Eb2 A2 F21 = Q1-2

            If both surfaces are at the same temperature, there can be no heat exchange, that is, Q1-2 = 0. Also

            Eb1 = Eb2

            so that

            A1 F12 = A2 F21

            The net heat exchange is therefore

            Q1-2 = A1 F12 (Eb1 – Eb2) = A2 F21 (Eb1 – Eb2)

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

            Note the concept of “exchange” and how all the radiative energy leaving each body and reaching the other is absorbed, even when one has a lower temperature than the other. So it states that the radiative energy from the colder body IS absorbed by the warmer body. This is not in the least controversial in the engineering world!

            Note also that the “heat transfer” is the “net” resulting from the difference in the two radiative energy flows.

            Note also that there is no specific specification of which body has a higher temperature than the other. The only thing that distinguishes the higher-temperature body in this analysis is that it has a higher numerical value. (In this simple example of blackbodies, emissivity is 1, so the radiative power density is just sigma * T^4.)

          • Gordon Robertson says:

            @Carl “Note also that the “heat transfer” is the “net” resulting from the difference in the two radiative energy flows”.

            Carl, that makes no sense because heat is the average kinetic energy in atoms and the net energy to which you refer is electromagnetic energy. I am fully aware that the kinetic energy can be transferred between separate bodies using EM but it is a mistake to suggest that heat and that EM are one and the same.

            The reason heat can be transferred from a hotter body to a cooler body only (without compensation) is because heat can only transfer from a region of higher KE to a region of lower KE. That applies whether we are talking radiative transfer, conductive transfer, or convective transfer.

            Clausius, who wrote the 2nd law, stated clearly that radiative energy can flow both ways but that heat can be transferred in one direction only, from hot to cold, without compensation. That implies heat and the net energy, which is IR, are not the same. It also implies that IR transferred from cold to hot has no effect.

            I have no argument that IR is the transfer agent but when you start confusing IR with heat, it leads to the alarmists notion that a positive net EM energy flow ‘is’ heat transfer. That confuses the type of energy that is heat with the type of energy that is EM. They are entirely different types of energy.

            The net energy flow of IR is described by the 1st law and Clausius wrote the 2nd law because the 1st law would allow perpetual motion under certain circumstances. The positive feedback in AGW-based climate models is perpetual motion and contravenes the 2nd law. Those models are based on what you describe, the net flow of IR.

            According to Gerlich and Tscheuschner, who both have expertise in thermodynamics, it is not permissible to talk about the net flow of infrared energy separately from the transfer of heat. IR is EM, not heat. The 2nd law is about heat, not IR.

            There are two processes taking place, the flow of EM and the transfer of heat. It is incorrect, according to G&T, to separate the two processes, inferring that IR is heat.

            Climate alarmists like Stefan Rahmstorf have used that argument about net energy flow to claim that transfer of heat from the cooler atmosphere to the warmer surface does not violate the 2nd law. It certainly does. For one, the 2nd law refers to heat only, not IR. Clausius did not discuss IR in relation to heat other than radiative transfer of heat.

            For another, net IR flow does not describe heat transfer in a system. It enables heat flow by transferring kinetic energy via IR but it does not determine the direction of heat transfer. The direction is determined by the average kinetic energy, either within a body, between bodies in contact, or between bodies separated by distance.

            Furthermore, Rahmstorf claims that back-radiation (IR) from the cooler atmosphere can be added to solar energy to warm the surface to a temperature beyond which it is heated by solar energy. That is the positive feedback built into climate models.

            When you start playing with math, vis-a-vis net IR energy flows, and you lose the context, which is heat transfer and kinetic energy, you can create or infer energy which is not there.

            If your statement about net energy flow was correct then the heat transfer should refer to conduction and convection, and it doesn’t.

            If you read Clausius, and how he developed the mechanical theory of heat, he goes on and on about heat transfer and it’s work equivalent. Heat is related to work and is directly interchangeable with it, even internally at an atomic level. EM is not related to work.

            When a metal heats, it expands, because the atoms, restricted to general position, expand their mean free path of vibration. Heat the metal enough and the bonds will break causing melting.

            That increase in length in their mean free paths represents work. Overall, that increase in mean free path causes expansion.

            Thermodynamics is about heat, not IR. The properties of heat, in general, cannot be described by EM theory. It’s more of a convenience that the amount of IR in radiative transfer corresponds to the amount of heat transferred. It is a mistake, however, to presume much more than that.

            Heat and EM have distinctly different properties.

          • Gordon Robertson says:

            @geran “But Gordon, what if that MIT engineer clings to the IPCC pseudoscience?”

            I did not mean to imply that engineers were always right. I implied that if an engineer disagrees with the IPCC pseudo-science I am going to back the engineer. šŸ™‚

            I gather that Carl agrees with the IPCC position and I have left him a lengthy rebuttal.

            There is another engineer, Pierre Latour, who has laid out an excellent argument against the IPCC pseudo-science. He has also debated Roy on the 2nd law.

            I don’t pretend to understand half of what Pierre and Roy are talking about.

            http://www.principia-scientific.org/skeptical-arguments-that-don-t-hold-water-pierre-latour-s-rebuttal.html

          • geran says:

            Curt: So, do I take things out-of-context from my old text books? No, that is silly. What you need to do is UNDERSTAND the textbooks, then put it in your own words, SUCCINCTLY. Since the IPCC violates the laws of science, you need to explain those violations, as you choose to believe them.

          • Curt says:

            Gordon:

            I’m afraid you are tying yourself in knots over terminology, especially with regard to “heat”. Here’s a great quote you should ponder:

            From Atkins, Peter, “Four Laws That Drive the Universe”, Oxford University Press, 2007:

            “In everyday language, heat is both a noun and a verb. Heat flows; we heat. In thermodynamics heat is not an entity or even a form of energy: heat is a mode of transfer of energy. It is not a form of energy, or a fluid of some kind, or anything of any kind. Heat is the transfer of energy by virtue of a temperature difference. Heat is the name of a process, not the name of an entity.

            Everyday discourse would be stultified if we were to insist on the precise use of the word heat, for it is enormously convenient to speak of heat flowing from here to there, and to speak of heating an object. The first of these everyday usages was motivated by the view that heat is an actual fluid that flows between objects at different temperatures, and this powerful imagery is embedded indelibly in our language. Indeed, there are many aspects of the migration of energy down temperature gradients that are fruitfully treated mathematically by regarding heat as the flow of a massless (“imponderable”) fluid. But that is essentially a coincidence, it is not an indicator that heat is actually a fluid any more than the spread of consumer choice in a population, which can also be treated by similar equations, is a tangible fluid.”

            No, heat is NOT the average kinetic energy in atoms. Temperature IS a measure of the average kinetic energy, but that is very different.

            When I stated that (radiative) heat transfer is the “net” resulting from the difference in the two radiative energy flows, I simply stated the utterly uncontroversial core concept on the subject that has been taught to generations of engineering and physics undergraduates (nothing to do with climate science).

            You also need to brush up on your history of science. When Clausius formulated his statement of what we now call the 2nd Law in 1850, this was well before Maxwell’s equations of electromagnetism, before atomic theory was well accepted, and way before probabilistic statistical mechanics was well developed. So Clausius had nothing to say about IR exchanges, as the underlying mechanism was not at all understood then.

            Climate science has many problems, but “inventing” radiative exchange for heat transfer is not one of them.

          • Curt says:

            geran:

            If you cannot even understand the direct applicability of the textbook excerpt I quoted above (I have given up on getting you to agree…), there is no hope.

            We have been discussing here the concept of radiative exchange, with the embedded idea that radiative energy from a colder object can be absorbed by a warmer object (although more is always transferred the other way).

            A lot of people on these blogs think the IPCC cabal made up this concept. So I show a quote from an old textbook that pre-dates the IPCC that explains exactly this concept, and you don’t understand how it fits in to the discussion. Amazing!

            On the other hand, you bring in to the discussion the very different (and irrelevant to what we were debating) sensitivity equation. I’m now sure you did it as a distraction from the core argument you were losing.

            By the way, I’m still waiting for your explanation of the difference in effect on a 288K target object of 15um radiation from a 278K (colder) and a 298K (warmer) object. If there is a difference, what is the physical mechanism that creates this difference? So far, crickets from you…

          • geran says:

            No Curt, you get it wrong again.

            I don’t complain about the concept of IR radiation going in all directions. I don’t even complain about the possibility of it striking a warmer object. No violations of the laws of physics yet.

            But, when Warmists and Lukewarmers then claim that the radiation from a colder surface can warm (raise the temperature of) a hotter surface, then there is a problem.

            I don’t think you really want to address that problem, hence your “spectrum vs. wavelength” nonsense. Or, take your analogy of the cashier receiving a $20 bill for a $5 dollar item. The cashier returns $15 change to the customer. That does not violate any math or science. That is where your analogy ends, and you try to subsequently claim “victory”. But, your analogy is incomplete. If the cashier returns $25 to the customer, then there is a problem. A problem you wish to avoid.

            And, as to your “cherry-picked” question about 15 µm radiation, you should know that all wavelengths would be handled with the same physics. Your problem is to show how radiation from a cold surface can raise the temperature of a hotter surface. And, you can’t do that!

            And, the IPCC equation, which you also seek to avoid (for good reason), is something that you must address if you are going to believe in the IPCC “science”.

          • Curt says:

            geran:

            My reply is at the bottom of the thread.

          • Gordon Robertson says:

            Curt…re heat….from Gerlich and Tscheuschner’s Falsification Of The Atmospheric CO2 Greenhouse Effects Within The Frame Of Physics

            http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf

            Page 16 of 115″

            “Heat is the kinetic energy of molecules and atoms and will be transferred by contact or radiation”.

            Both are physicists who work and teach thermodynamics.

            Why the heck are you as an engineer reading abstract theories on heat? What the heck did you think it was?

            You are in electronics. Why do you think a resistor ‘heats up’? Because electrons are colliding with atoms in the resistor material, causing their average kinetic energy to rise.

            When you touch your finger on a 2 watt resistor running a healthy current what is it that makes you withdraw your finger as your finger begins to cook? The resistor is damned hot. That is not temperature, that is not imaginary, that is heat.

            Microscopically, when you touch the resistor, your are contacting rapidly vibrating atoms with a relatively high kinetic energy. That energy is transferred to your finger, causing it to burn.

            If someone wants to claim the transfer itself is heat they are seriously misguided. Or that heat is only a concept. I have read that rubbish before.

            In electronics, some people insist that holes in semiconductors, left behind when an electron jumps to another hole have mass. When Shockley. who developed the theory, was asked about it, he claimed the holes are imaginary, that they were developed to help visualize the movement of electrons, especially through a P-type material.

            Temperature is a measure of heat. Temperature measures the average kinetic energy of atoms by proxy.

            In the work of Clausius he shows you how he derived the concept of entropy. He took infinitesimal amounts of heat at a specific temperature T and summed them. The infinitesimal heat amount is dq and the temperature at which it is taken is T. So the sum of dq/T in a process is the entropy.

            If that is applied to heating a metal bar, the quantity dq signifies a change of energy at a specific temperature T, as the metal warms. The quantity q, is energy, it is heat energy.

            Heat is energy, it’s not some abstract notion created by a theoretical physicist. It is also the kinetic energy of atoms in a substance to differentiate it from the kinetic energy in an electrical process, or a nuclear process, or a chemical process, etc.

          • Gordon Robertson says:

            @Curt…re heat…some more.

            About temperature vs heat:

            If you are in a room and someone asks you how hot it is in the room, what do you do? You might check a thermometer on the wall or the thermometer in a thermostat.

            The wall thermometer is likely a glass tube filled with mercury. The mercury expands and contracts in a narrow glass vial indicating the relative temperature of the room. The thermostat uses a bimetallic strip that expands and contracts with temperature.

            The thing to see clearly is that temperature does not exist as a physical entity, it is a human invention used to measure the relative degree of something else by some sort of proxy. There is no energy in a room called temperature.

            What is the thermometer measuring? If not placed correctly it could be measuring the wall temperature but more likely it is measuring the air temperature as molecules of air collide with the glass vial in which the mercury is kept.

            Those molecules of air are continually moving rapidly and colliding with each other, the walls, the ceiling, the doors, the windows, and the floor. Because they are moving, they have kinetic energy but you can’t measure the KE of a single molecule. Therefore experiments were carried out to determine the pressure created on container walls to get an average air pressure, which can be related to kinetic energy.

            It was noted that the pressure increased when the container was heated and decreased when the container was cooled. The conclusion was that the average kinetic energy was increasing and decreasing as the container was heated and cooled.

            Back to the room. You look at the thermometer and it indicates 20 C, which is classified as room temperature. C refers to Celsius where 0C is ‘defined’ as the freezing point of water. Please note that temperature is defined and that something else is causing the defined temperatures to rise and fall.

            That something is the average kinetic energy of the air molecules in the room, given that the wall or something else is not contributing to the measured temperature. So we know the room has an average temperature of 20 C as indicated by the expansion of mercury. We also know the room has energy in the air but what kind of energy is it?

            Is it nuclear energy? Nope. Is it chemical energy? Nope. Is it electrical energy? Nope. Is it electromagnetic energy? Maybe partially but EM has no temperature associated with it. So, it has to be some other kind of energy.

            Some people would imply that it is thermal energy and they try to separate thermal energy from heat. Engineers do not. To an engineer, heat and thermal energy are one and the same.

            Curt, old boy, if you are an engineer, you need to get practical and get on board. Only theoretical physicists, and those susceptible to thought experiments, split hairs with regard to heat and thermal energy. Thermal implies heat, and vice-versa.

            What we feel in a room is heat energy associated with the kinetic energy of air molecules. We may be influenced by direct radiation from a space heater, or a tiny effect from EM radiating from walls and the carpet, but in a general room without the convective heat from a hot air furnace blower, we are feeling heat energy.

          • Curt says:

            Gordon:

            I first studied thermodynamics in the Chemistry department at MIT, a very rigorous theoretical course. And from what I learned there (and later), my jaw drops at how bad the G&T paper is. An excellent takedown can be found starting here:

            http://scienceofdoom.com/2010/04/05/on-the-miseducation-of-the-uninformed-by-gerlich-and-scheuschner-2009/

            Re-read the quote I cited from Peter Atkins’ book “Four Laws That Drive the Universe” here:

            http://www.drroyspencer.com/2015/05/trmm-satellite-coming-home-next-month/#comment-192290

            Atkins is a professor of chemistry at Oxford. The book is excellent; I highly recommend you read the whole thing. And in this quote, he accurately and succinctly describes the confusion resulting from imprecise terminology.

          • Gordon Robertson says:

            @Curt…”I first studied thermodynamics in the Chemistry department at MIT, a very rigorous theoretical course. And from what I learned there (and later), my jaw drops at how bad the G&T paper is”.

            You have not explained what is wrong with the G&T paper. They are both specialists in thermodynamics and you are referencing your studies in chemistry.

            In the SoD rebuttal, it is painfully obvious that the author is engaged in ad homs, cherry picking, and talking around subjects he obviously does not understand. Nowhere in the paper does he address the key issues in the G&T paper. In one part, he thoroughly confuses his understanding of the G&T paper by making a complete mess of the 2nd law.

            SoD describes the 2nd law as the ‘imaginary second law’ when it is applied to the greenhouse effect. So, he went looking for the ‘real 2nd law’ and claimed to have found it on Wikipedia. Since the wiki explained the 2nd law with reference to entropy, he claimed he never did like entropy. He could not understand what he was reading about entropy.

            He goes on to quote Clausius but it is not a quote from Clausius it is his interpretation of the quote. Clausius did not say that heat cannot ‘generally’ flow from a cooler body to a warmer body, he said that heat can only flow from a warmer body to a cooler body without compensation.

            Here is SoD’s murder of the 2nd law:

            http://scienceofdoom.com/2010/03/15/the-imaginary-second-law-of-thermodynamics/

            It clearly reveals this guy is a layman who does not have the slightest clue about thermodynamics.

            The fact that you have to reference SoD rather than use your MIT degree in engineering to address the G&T claims, that even I can understand from studying physics as part of my engineering studies, suggests to me that the only time you have been to MIT is on a bus tour. I no longer accept that you are an engineer.

            Pierre Latour is a real engineer and he explains the physics related to climate science. Anything I have read from you references questionable sources.

          • Gordon Robertson says:

            @geran….”But Gordon, what if that MIT engineer clings to the IPCC pseudoscience?”

            geran…my apologies. After Curt’s last reply, in which he cited a totally biased article about Gerlich and Tscheuschner on scienceofdoom, I no longer accept that he is an engineer. Maybe he is, but having studied engineering I no longer think he is.

            I have read articles on SoD before and the guy who runs the site makes ridiculous inferences like ‘the imaginary 2nd law’.

            http://scienceofdoom.com/2010/03/15/the-imaginary-second-law-of-thermodynamics/

            In this paper he makes such a mess of the explanation of the 2nd law it is painfully obvious that he has no idea what it is.

            To begin with, he has lost his books on thermodynamics and consults Wikipedia to refresh his memory. On the wiki, the 2nd law is explained using references to entropy, and since SoD ‘never liked entropy’ he looks further afield, on Wikipedia. He finally finds a quote he attributes to Clausius, which is in none of the works I have read written by Clausius.

            SoD quotes Wikipedia, “Heat generally cannot flow spontaneously from a material at lower temperature to a material at higher temperature….”

            Clausius said nothing of the kind. He stated that heat can only flow from a warmer body to a cooler body without compensation.

            Gerlich and Tscheuschner have laid out a masterful expose on the greenhouse effect which has yet to be debunked. They use pertinent thermodynamics explanations for everything they claim, yet the major rebuttals to their paper barely address most of the points on thermodynamics they raise.

  10. dave says:

    Ah, we are wandering into the territory where people mix up the concepts of “black body” and “black hole.” The glowing heart of a steel furnace is a “black body.” If you stand next to it, it will absorb all of the heat power your warm flesh sends out towards it. Absorbing well through a surface is the definition of a black body, not BEING black.

    One analogy to a black body would be your bank account. It will “absorb” cash whether it is a rich man or a poor man who pushes it across the counter to the teller. Being liberal with your money, you will naturally disburse it, in ways dependent on the amount showing at any time as available. Your recipients will “absorb” all such spending whether you are disbursing it to a beggar or Mr Money Bags.

    The analogy to a black hole would be where you are a miser – take and not give. If you only took from men richer than yourself, that would be analogous to the idea that a body cannot absorb a photon from a source hotter than itself. Just not the way the physical world works, however.

    It is called a black body because one can make a good absorbing surface with soot from a candle – “lamp black.” A quirk of the jargon of Victorian experimenters. “Blacked over” as an adjective slid into “black” as an adjective which slid into “black body” as a substantive.

    The whole thing is a bit like the nonsense of saying imaginary numbers are by definition non-existent. When it is only a case of a bad name which has stuck. Geologists still call certain types of rocks “acid” because of a completely exploded idea of their composition. These abuses of notation and nomenclature are everywhere in science. It does not say much for scientists’ profundity that they are as prone to fall into the resulting verbal traps as any tyro.

    • dave says:

      “..cannot absorb a photon from a source hotter than itself.”

      Meant, of course:

      “cannot absorb a photon EXCEPT from a source hotter than itself.”

      • Rich says:

        “cannot absorb a photon EXCEPT from a source hotter than itself.”
        Another way to look at this is to think of the fact that the temperature is actually the wavelength (even though it is a photon due to the wave particle duality as described by AE.) This temperature is even used in computer/TV displays to express the color of the display. Temperature is wavelength is color. The color of an object is a process of your eyes detecting the wavelength of the photons reflected and absorbed (thus not seen – invisible to your eye) and your eye informs your brain that it is red, or blue, or green, etc. If you have a radio tuned to a certain frequency it will not recognize frequencies (wavelengths) of a different frequency. The hotter an object is the faster the molecules are vibrating. The colder an object is the slower an object is. The photons emitted by an object will be at the temperature of the object and this temperature is in their wavelength and color.

        In nuclear physics, the probability of a molecule of a material absorbing a neutron from fission to create another fission is measured in barns to provide a number describing its extremely small probability. A barn is defined as 10-28 m2 (100 fm2) and is approximately the cross-sectional area of a uranium nucleus. This area is “Area” is affected by the temperature of the atom it strikes/passes through.

        Cold objects can absorb photons emitted from warmer objects and the colder the body is the higher the probability. However, if the photon is colder than the body it is striking, it no longer has enough energy to dislodge electrons in the elements of that material and thus can not change its temperature. There is no way, as the warmists claim, that a cold photon can strike a hot body and increase the number of electrons released, or cause electrons to move to different energy levels thus causing the release of photons of energy of a higher wavelength from that body. Nor will it slow down the release of photons causing the body to increase in temperature. The physics is just not there. Think about it.
        We are receiving photons of energy generated essentially at the time of the Big Bang. and they are still (13 billion years later) the same color, same wavelength, same temperature. Disbelievers of the fact that cold objects heating up put that in your brain and contemplate it. How did they stay at the same color, frequency, wavelength if that stupid theory is true.

        • Curt says:

          No, temperature is NOT wavelength. Most substances at a given temperature emit radiation with a broad spectrum of wavelengths. The radiation at any wavelengths carries NO information as to the temperature of the emitting substance, or even whether the radiation was from thermal emission at all.

          Many substances at both 278K (5C) and 298K (25C) will emit significant radiation of 15 um wavelength. When this radiation hits a substance at 288K (15C) that absorbs radiation of this wavelength (which most solid and liquid substances on earth will do), it does not matter which of these two sources, one hotter and one colder, was the source. The radiation will be treated the same, as there is NO physical mechanism for the target to tell the difference.

          • Rich says:

            If temperature has no wavelength, then how does an IR temperature detector work. I have been under the impression that it is measuring the wavelength of the emitted energy. I just can not see how it could be measuring temperature from a distance otherwise.

          • Rich says:

            A quickly found document on theory of IR Temperature detectors.
            http://support.fluke.com/raytek-sales/Download/Asset/IR_THEORY_55514_ENG_REVB_LR.PDF

          • geran says:

            A perfect example of how Curt has confused himself:

            Curt says: “No, temperature is NOT wavelength. Most substances at a given temperature emit radiation with a broad spectrum of wavelengths. The radiation at any wavelengths carries NO information as to the temperature of the emitting substance, or even whether the radiation was from thermal emission at all.”
            ++++++++
            First Curt says it is not “wavelength”, it is “spectrum”.

            Curt says: “Many substances at both 278K (5C) and 298K (25C) will emit significant radiation of 15 um wavelength. When this radiation hits a substance at 288K (15C) that absorbs radiation of this wavelength (which most solid and liquid substances on earth will do), it does not matter which of these two sources, one hotter and one colder, was the source. The radiation will be treated the same, as there is NO physical mechanism for the target to tell the difference.
            ++++++++
            Then, he says it is not “spectrum”, it is “wavelength”!!!

          • Curt says:

            geran:

            Your reading comprehension is as bad as your scientific comprehension. My points were simple and clear:

            1. Thermal emission from a substance is not at a single wavelength. It covers a spectrum, often a broad spectrum.

            2. From this it follows that substances of many different temperatures will emit radiation of a given wavelength. (Was that so hard?) I used 15 um wavelength simply as an example.

            3. Many folks claim that radiation from a colder body cannot contribute to the internal energy of a warmer body, whereas radiation from a hotter body can.

            4. Using 15 um radiation as an example, and a receiving object at 288K, I pointed out that 15 um radiation could come from a hotter object (e.g. 298K) or a colder object (e.g. 278K).

            5. Radiation carries NO information as to its source. So when 15 um radiation strikes the 288K object, there is NO physical mechanism by which the object can discern whether that radiation came from a hotter body or a colder body, absorbing it in one case, and rejecting it in the other.

          • Rich says:

            RE @Curt May 26, 2015 at 9:30 AM
            That does not seem to agree with the explanation in the Fluke explanation – Link Above.

            I seem to get the impression from that text that a spectrum (blackbody shaped, sort of lopsided bell curve) will be radiated. The peak of this spectrum will be different for deferent temperature and can thus provide a means of relating to a temperature. That is very similar to the AM radio frequency transmission and reception of audio frequencies. Most other radio transmissions are also similar but will have a different shaped spectrum. All of which will transmit frequencies hundreds of hertz, thousands of hertz, hundreds of megahertz, and thousands of megahertz from the principle frequency at a decreasing level the further you are from the main frequency.

            So according to you, even radios don’t work.

          • geran says:

            Curt, you just keep “spinning” your own confusion. You say things that everyone knows, but then you mix in things you are confused about, resulting in nonsense.

            Keep it simple.

            For example, what is the wavelength, or spectrum if you prefer, for a molecule of CO2, at 270K, in the IPCC equation (below)?

            Delta F = 5.35 * ln {C/C0} W/m2

            Just a simple question….

          • Curt says:

            Rich:

            The Fluke app note you link to does not get down to the lowest level of the principle of operation.

            Inexpensive IR thermometers have two sensors whose electrical resistance is a function of temperature. One of these sensors is well shielded from the external world, and its resistance is a function of the ambient temperature. This is the reference sensor.

            The other sensor is at the focal point of the optics, in maximum radiative exchange with the object pointed at. If the object being pointed at is hotter than the ambient measured by the reference sensor, this sensor will absorb more radiative energy from that object that it emits to it, so it will have a higher temperature, and higher electrical resistance, than the reference sensor.

            If the object being pointed at is colder than the ambient measured by the reference sensor, this sensor will emit more radiative energy to the object than it absorbs from it, so it will have a lower temperature, and lower electrical resistance, than the reference sensor.

            In both cases, the bigger the difference in temperature between the target object, the bigger the difference in electrical resistance between the two sensors.

            The only wavelength sensitivity of the sensor comes from the wavelength band that the lens will transmit. As the app note points out, you want different lens materials for measuring different temperature ranges.

            So the measurement of electrical resistances of the two sensors gives the difference in temperatures of the sensors, from which can be derived the net radiative power imbalance of the under-lens sensor with the target object (over the wavelengths that the lens can transmit), from which the temperature of the temperature of the object can be derived. The inexpensive sensors assume an emissivity value (~0.95) in computing the temperatures.

            These sensors are NOT looking for a peak frequency. They are effectively looking at the magnitude of the net radiative power flow (the area under the curve).

          • Curt says:

            geran asks, “what is the wavelength, or spectrum if you prefer, for a molecule of CO2, at 270K, in the IPCC equation (below)?

            Delta F = 5.35 * ln {C/C0} W/m2”

            Leave it to geran to ask an ill-posed, incomplete, and largely irrelevant question, as usual.

            Figure 3 on this page:

            http://home.pcisys.net/~bestwork.1/

            has a good answer to the question I think geran was trying to ask. It shows the absorption spectrum (remember that absorptivity equals emissivity at any and all wavelengths) for CO2, comparing it to the 270K Planck curve.

          • geran says:

            Curt, from “your” Figure 3, my calibrated eyeballs come up with about 0.08 W/m2 (net energy), if I’m generous.

            The IPCC equation yields almost 2 W/m2, for 400/280 CO2 concentrations.

            Are you saying that the IPCC equation is WRONG???

            I think that is called “denial”.

            But, your comment (desperately grasping at straws) is funny, as usual.

            The fact is, Curt, this equation is the crux of the IPCC/AGW/CO2/GHE nonsense. The equation has NO basis in fact. It has no mathematical derivation (PROOF), nor does it have any empirical evidence. In fact, the last 20 years of temp data is empirical evidence that the equation is BOGUS.

            Isn’t it interesting how Lukewarmers will pervert and corrupt well-establshed science just to “bitterly cling” to this flawed, non-scientific equation?

          • Curt says:

            Geran:

            I should have realized that that graph was far too advanced for your limited understanding. The red curve shows virtually total absorption by CO2 in the wavenumber range of 600 to 800 cm^-1.

            The blue curve shows the Planck (blackbody) curve for 270K. In this wavenumber range, it is above 0.3 W/m^2*cm.

            Since CO2 in over the height of the atmosphere is virtually a perfect absorber in this wavenumber range, it is also almost a perfect emitter. So at the 270K level you asked about, over this wavenumber range, you get over 0.3*(800-600) = 60 W/m^2 downwelling radiation.

            So your calibrated eyeball was only off by almost three orders of magnitude — but what’s a factor of a thousand between friends, anyway?

            This ~60 W/m^2 is a significant fraction of the total measured downwelling longwave infrared radiation of 300+ W/m^2. (Most of the rest comes from water vapor.) It is NOT a delta (change) value from recent increased concentrations.

          • geran says:

            Hilarious Curt!

            You trapped yourself again. I didn’t do it to you, you did it to yourself. I just played along with your distraction, and you fell for it.

            You act as if you have never seen the IPCC equation before. Earlier, you had tried to obfuscate the issue by confusing spectrum with wavelength in seemingly some futile effort to justify “cold” warming “hot”, a violation of 2LoT. So, when I threw out the equation, and asked you to explain the spectrum/wavelength thingy, you went off to Never-Never land. So, I just gave you some more rope….

            So, now that we’ve had our fun, what is the spectrum/wavelength implied by the IPCC equation:

            Delta F = 5.35 * ln {C/C0} W/m2

            (This is the second time I have asked you this. You probably just missed it the first time….)

  11. RIch says:

    PLEASE EXPLAIN THIS
    I am having trouble comprehending that the 30 some degrees that the Earth is warmer than a chunk of rock at the same distance from the Sun is all from CO2, as the AGW believers do. Can someone please provide me with a calculation of the “Global” temperature of the Earth assuming just that the atmosphere had NO, zero, CO2. That is the existing atmospheric gasses less CO2. I have searched the internet for several years trying to find any calculation similar to this to no avail. This number and the fact that they contribute that much solely to CO2 just defies scientific logic.
    Last week I saw one of those stupid scientific shows on tv where they do dangerous and stupid things. On this show I saw a clip of where they dropped a red hot ball into water. It stayed hot and sort of made weird noises for a while. And that is where I saw exactly what I was looking for. As I work in the power industry, I know full well if a boiler pipe does not have a good flow of water going past it quickly enough to prevent steam bubbles from forming around the pipe it will destroy it self faster than I can type this comment. This clip clearly showed this.
    They dropped a red hot smooth ball into a glass of water. It quickly enveloped itself with steam which slowed down the transfer of energy to the water (sounds like s GHG to me) eventually the bubble collapsed and it sizzled like you would expect from shoving a hot poker into water or any other none mirror smoothed object.
    The Earth is surrounded by a bubble of atmosphere also. That bubble, even with absolutely ZERO CO2 is going to slow down the transfer of energy, insulate, act as a GHG, whatever you want to call it and by all laws of physics and thermodynamics MUST make the EARTH warmer than that theoretical rock at the same distance from the SUN. Note: The ball must be smooth if rough it will not hold the bubble. And the Earth is many times smoother that that ball relative to size.
    Where is the calculation? One for no water, one for no CO2, one for neither and there is your smoking gun that CO2 AGW is BS.
    Clip discussed above – https://www.youtube.com/watch?v=l3LOhYRY59Y

      • Rich says:

        Does this site have a description on the affects of JUST the insulating capacity of the atmosphere? The atmosphere will have a specific rate of heat transfer. Approximate values are shown here, http://www.physicsclassroom.com/class/thermalP/Lesson-1/Rates-of-Heat-Transfer
        Any object that has a low rate of heat transfer acts like an insulator. the lower it is the better that insulator is. Air, oxygen and Nitrogen are all about 0.024 making them fairly good insulators. The insulation of the atmosphere provided by the several miles of air must have some effect on the surface temperature. My question is What is that amount. If you don’t think air makes a good insulator find out more about it. 50 years ago I bought a house. The first winter I lived in it, it seemed like the furnace ran continually. Upon crawling into the attic, I discovered there was NO insulation in the attic. As I was just an E-4 in the NAVY, I had just enough money to pay the mortgage and buy food. However, in my garage was a big pile of empty cardboard boxes. I took them, flattened them out, and spread them out in the attic on top of the rafters. Putting in a few roofing nails to hold them in place. After I did that instead of filling up my oil-tank once a week it only needed filled once a month for the rest of the winter. AIR is an insulator. (even without CO2)

        • daveI says:

          “…insulation…must have some effect on the surface temperature…”

          It would – except for the fact that the lower atmosphere is convective, and this process, together with evaporation of sea-water and condensation into clouds, completely dominates (gross) conduction effects.

          The lower atmosphere is therefore called the “troposphere” – as this is Greek for “place of overturning.”

          Shortly, somebody will be here to write a million words about another theory.

          A war “trophy” originally was a marker on a battle-field where an invader had literally turned his back and fled.

        • dave says:

          Rich,

          Because of my original typo you have misunderstood me.

          A substance CAN absorb a photon emitted from a cooler substance; and a bulb smeared with lamp black, or a patch of dark earth, or a constructed cavity, or a plasma such as the sun, WILL absorb all rays of radiation which fall on it. That is the definition of a black body and it is a simple experimental fact that black body behavior, or approximation to it, is common. When you stand next to a camp-fire, you are heating the burning wood with your radiated body heat as truly as the glowing wood is heating you. The fact that the two processes differ enormously in amount, is neither here nor there.

          It is a matter of FACT that some substances are reflective, some are selectively absorbent, some are generally absorbent, some are transparent; and that it also depends on the wavelength involved and the thickness of the material*.

          Your implicit argument – using terminology such as “ejecting” electrons – is about ONE way in which photons interact with electrons, namely by causing jumps between atomic orbitals, or out of atoms completely. This is called the (narrow) photo-electric effect. There are OTHER ways and the sum of all the ways is called the (general) photo-electric effect. For example, the absorption of infra-red by water molecules is an example of radiation affecting electronic MOLECULAR orbitals, which are much closer together, in energy terms, than atomic orbitals.

          Once again, the muddle in scientists’ heads has led an enquirer astray. Possibly, you are thinking of something called an “absorption spectrum.” This has nothing to do with “the spectrum absorbed” by a black body!

          If you cared to read the reference I gave, much would be explained – without any maths – by Sir William Bragg F.R.S.

          *a photon might “flirt” with trillions of trillions of electrons before it “gets engaged to one” and then “marries him.”

  12. AndyG55 says:

    Hi Roy,

    A bit off topic… sorry.

    Do you have an update of this graph, using UAH 6.0?

    Thanks,

    http://kaltesonne.de/wp-content/uploads/2015/05/modell1-1024×921.gif

  13. Roy W. Spencer says:

    Curt knows what he’s talking about. Listen and learn.

    • cunningham says:

      Doctor Spencer, in your opinion, when people say that “cooler” co2 cannot make a “warmer” surface warmer, are they confusing the green house effect with heat transfers through molecular collisions? I’m just trying to get some clarity here… thanx

      • Curt says:

        cunningham:

        There are several sources of confusion among people who say that “cooler substances cannot make a warmer substance still warmer”.

        One is mistaking gross energy transfers for net heat transfer. The early 18th/19th century scientists working on thermodynamics, theorizing by analogy, conceived of heat transfer as like a fluid flow, calling this hypothesized fluid the “caloric”.

        With the development in the late 19th and early 20th centuries of atomic theory, statistical mechanics, and quantum mechanics, no serious scientists have favored the caloric theory of heat for a century now. But much of the old terminology was widespread enough that it has stayed in usage, and in some regards is still useful, as long as its limitations are understood.

        The term “heat flow” is now realized to be a metaphor, as there is no physical substance flowing like a fluid. (It amuses me that many people who object vigorously to the imperfect metaphor of the “greenhouse effect” blindly accept the even poorer metaphor of “heat flow”.) But at a high level, it can still be useful to think of heat transfer this way, as a uni-directional fluid flow.

        However, if you want to understand the underlying mechanisms, this is no longer sufficient. The analogy I like to use is making change for a purchase. If I’m a cashier and you give me a $20 bill to purchase a $5 item, you would expect me to give you $15 in change.

        But by the logic of these people, I would be making a horrible mistake. If you are buying something from me, I shouldn’t be giving you money! This is the key mistake of confusing gross flows and net flows.

        In radiative heat transfer, we now understand that the heat transfer is a result of energy exchanges between the objects. Objects radiate energy based on their temperature and emissivity, without any regard to what they are radiating toward. (How would they know, anyway?) The hotter object radiates more energy to the colder object than the colder does to the warmer, but the colder object does radiate a non-zero amount of energy to the warmer. The difference between these two is what we call the (net) “radiative heat transfer” from the hotter to the colder object.

        The next source of confusion is not considering the larger context of such an exchange. If we have these two objects exchanging radiative energy, but isolated from the rest of the universe, the net radiative heat transfer from the hotter to the colder object simply lowers the temperature of the hotter, and raises the temperature of the colder, until they reach the same temperature.

        However, when these two bodies are not isolated from the rest of the universe, it’s not so simple. Let’s say the warmer body has a separate and constant source of power. This body’s temperature is determined by the balance between the input power from the source, and the temperature-dependent radiative exchange with the cooler body.

        The higher the temperature of the cooler body, the higher the temperature the warmer body must have to have a large enough net heat transfer to match the amount of the power source input, and so reach steady state conditions. So if you replace a very low temperature cooler object with one of a higher temperature (but still cooler than the other), the result will be an increase of temperature of the warmer body. This is true even though the net heat transfer is always from the warmer body to the cooler body.

        So did the cooler body “warm” the warmer body? Now we are dealing with yet another common source of confusion, the imprecision of everyday language. The heat flow is always from the power source to the warmer object to the cooler object (and then to the cold rest of the universe), steady state and changing. So in that sense, it did not.

        It would be more precise to say that replacing a lower temperature cool object with a higher temperature cool object (but still of lower temperature than the warmer object) led to a higher steady state temperature of the (separately powered) warmer object. But it is difficult to speak that pedantically all the time.

        I’ve just scratched the surface, but that’s enough for now.

    • geran says:

      Curt knows his stuff pretty well. There are a couple of people that are even funnier, but Curt does a good job.

      Of course, everyone has their climate comedy favorites, like “CO2 produces warming”, “Earth warms the Sun”. The competition is fierce.

    • Gordon Robertson says:

      @Roy W. Spencer “Curt knows what he’s talking about. Listen and learn”.

      Curt referred me to a link at scienceofdoom in which the owner could not explain the 2nd law, because he had lost his textbooks. So he went looking for an explanation…on Wikipedia. Because the explanation was given in terms of entropy, and he ‘never did like entropy’, he went looking further afield in Wikipedia.

      The point is obvious. This guy at SoD is slamming a paper written by two experts in thermodynamics yet he cannot even explain the 2nd law.

      The link to which Curt referred me was a supposed debunk of the Gerlich and Tscheschner paper on the GHE. It should have been apparent to him that the paper was largely about the SoD author smirking at G&T and using red-herring arguments to discredit them. Not once did he address the pertinent points in the G&T paper.

      Curt scoffed at the paper, suggesting it was rubbish, based on a course he had taken in chemistry. G&T are both physicists who work in the field of thermodynamics as teachers and researcher respectively.

      Scientists who know what they are talking about do not scoff at other scientists as a response to their work. They analyze what is claimed and offer a scientific rebuttal. Curt obviously cannot do that.

      He (SoD) finally posted a comment, implying it to be from Clausius, in which he claimed heat could not ‘generally’ FLOW from a cooler body to a warmer body’ (heat does not flow…it is transferred atom to atom, like electricity) Clausius said no such thing, he made it crystal clear that heat cannot be transferred from a cooler body to a warmer body WITHOUT COMPENSATION.

      If Curt knows what he’s talking about, why does he not explain it using his understanding of physics. All he is doing is regurgitating thought experiments like those in which you indulge.

      Don’t get me wrong, I like you and I really appreciate the work you have done at UAH bringing truth to the AGW debate. However, I am appalled at your lack of understanding of basic thermodynamics.

      I recall Dick Lindzen warning you to be careful about expounding on disciplines with which you have no expertise. You admitted that he had said that. Yet you have the temerity to debate with Pierre Latour, an engineer with a lot of experience using thermodynamics. You were using thought experiments, not scientific fact.

      • Curt says:

        Gordon:

        You seem awfully disappointed that I didn’t have an instant rebuttal to a rambling incoherent 115-page paper that I hadn’t looked at in years.

        Well, I’ve waded into that sewer again, and put a few comments of my own at the bottom of the thread.

        • Gordon Robertson says:

          @Curt…”Well, I’ve waded into that sewer again…”

          Curt…it’s comments like that against two physicists that make me question your background. Gerlich teaches advanced math in the field of thermodynamics and Tscheuschner is a researcher in thermodynamics. They both have Ph.Ds in the field.

          You already referred me to the SoD article about G&T which was laden with egregious errors. His paper lead to another in which he referred to the ‘imaginary 2nd law”.

          He ran off about the 2nd law, claiming he had lost his textbooks on it so he looked it up on Wikipedia. The first wiki article explained the 2nd law in term of entropy which he claimed ‘he never did like’. So he went off again to find something by Clausius.

          He misquoted Clausius, claiming that Clausius inferred heat can ‘generally’ not FLOW from a cooler body to a warmer body. Those were SoDs words yet he made them appear as if they came from Clausius.

          Clausius was very precise in his wording. He was very intelligent and had a penchant for detail. He stated clearly that heat can only be transferred from a warmer body to a cooler body without compensation.

          I don’t understand how someone like you, claiming to be an MIT trained engineer, could point me to such an ad hom-based attack on G&T.

          Even you are at it, calling their paper material from a sewer. That’s completely unprofessional and not how an engineer would talk.

          • Curt says:

            Gordon:

            Let’s look at what Clausius actually said (8th memoir), shall we?

            “The principle may be more briefly expressed thus: Heat cannot by itself pass from a colder to a warmer body; the words “by itself”, however, here requires explanation. Their meaning will, it is true, be rendered sufficiently clear by the exposition contained in the present memoir, nevertheless it appears desirable to add a few word here in order to leave no doubt as to the signification and comprehensiveness of the principle.

            In the first place, the principle implies that in the immediate interchange of heat between two bodies by conduction and radiation, the warmer body never receives more heat from the colder one than it imparts to it. The principle holds, however, not only for process of this kind, but for all others by which a transmission of heat can be brought about between two bodies of different temperature, amongst which process must be particularly noticed those wherein the interchange of heat is produced by means of one or more bodies which, on changing their condition, either receive heat from a body, or impart heat to other bodies.”

            He EXPLICITLY endorses the concept of radiative exchange, with some transferred from the colder body to the warmer body (but less than in the other direction). I agree with him, as does SoD. Do you?

            I work with PhDs on a daily basis. Most of them are no brighter than those with lesser degrees. And I generally find them, outside of their immediate and tiny specialty, to be no more knowledgeable (and often less knowledgeable) than those with lesser degrees. That’s what I see with G&T — climate science is so multi-disciplinary, and outside their little sub-specialties, they are lost.

            Oh, and about civility. If G&T had been civil in their paper, I would be more civil in talking about them. Their paper reminds me of a joke my father, a physical chemist with a PhD from MIT who had to deal with a lot of German PhD chemists, used to use: “‘arrogant German Scientist’ is a redundancy!”

  14. Rick Adkison says:

    May she come home in a blaze of glory!

  15. dave says:

    Rich asked for a calculation of the effective temperature, as defined by astronomers, for the Earth, on the assumption that there is no water and, generally, no atmosphere.

    THE ANSWER IS:

    268.5 K

    The calculation is as follows. Let x be the required number.

    Handbooks give the effective temperature of (the ball of rock) Mars as 217 K and its average distance from the sun as 1.5303 A.U.

    The intensity of sunlight at 1 A.U. (where the earth orbits) is therefore greater than where Mars is, by 1.5303^2 = 2.3418 times.

    Then, assuming the “Stefan-Boltzmann Law” applies, and the Earth’s heat budget is in long-term equilibrium,

    (x / 217) ^ 4 = 2.3418

    which gives x = 268.5 K.

    Both the Earth and Mars have eccentric orbits and the above calculation is an approximation to a varying situation.

    The solar intensity reaching the Earth varies by 7% over a year, and that will seasonally add or subtract 2 K from the number.

    Also, the ACTUAL temperature an inch above the surface will be undefined, since, by assumption, there will be vacuum; the ACTUAL temperature an inch below the surface will everywhere differ greatly from the effective temperature, as a result of latitude considerations; and there will be large diurnal variation of ACTUAL temperatures, down to a depth of a meter or so.

    If there were an atmosphere of oxygen and nitrogen, the usual assumption is that, over time, it would uniformly take on the temperature of 268.5 K. Somebody will be along shortly to say that is not true because gravity produces a lapse rate.

  16. Curt says:

    geran:

    You still are missing the most basic points.

    On the financial analogy (and yes it’s an analogy but it’s a good one): we are agreed that the cashier giving change to the customer is valid, as long as the amount is less than the amount given to the cashier in the first place. But your argument on radiative transfer is effectively that giving any change is wrong.

    You claim that my argument is tantamount to saying that the cashier should give me more in change that I originally gave him. Not at all! I have never argued that the colder body transmits more energy to the hotter body than the hotter body transmits to the colder body (and neither has the IPCC). But I do argue that the colder body transmits an amount greater than zero to the warmer body, and that this energy is absorbed by the warmer body.

    So if the hotter body is transmitting more to the colder body than vice versa, how can the radiation from the cold surface raise the temperature of the hotter surface? I explained this clearly in my response to cunningham, which you read but clearly did not understand.

    If it is just the two bodies isolated from the rest of the universe, this will not happen. But if the warmer body has a separate power source, the presence of a cooler body (that is warmer than the ambient or a previous body) WILL serve to increase the temperature of the warmer body.

    How? Consider the spending analogy again, but assume there is a steady income. Let’s say you earn $240 per week, and each week you make 12 $20 purchases. Week to week, your bank account balance is the same. But this week, you start getting $15 change on each of these 20 purchases (less than you hand the cashier). At the end of the week, your bank account will be $180 greater than at the start.

    So even though the change was always less than what you handed the cashier, the result was that your bank balance increased. Did the change increase your wealth? As I said to cunningham, it is a matter of semantics. Your bank balance would not have increased unless you also had your paycheck deposited each week — if not, you would just decrease the rate at which your bank balance decreased. But the fact is, that your bank balance increases with the change provided compared to no change provided, even though the change is always less than what you gave to the cashier.

    This is directly analogous to the heat transfer problem. The paycheck deposited is equivalent to the power input — from the power supply for the satellite, or from the sun for the earth. The “no change” spending is equivalent to the surface radiating to the virtual absolute zero of deep space (no “back radiation”). The spending with change is equivalent to radiating to a temperature colder than the surface, but still warmer than space. This leads to a higher energy level and temperature of the surface. This is true even though the “back radiation” from the colder substance to the warmer substance is ALWAYS less than the “forward radiation” from warmer to colder.

    As to the (fundamentally irrelevant to this discussion) IPCC equation you keep trying to drag in, the argument is that increased concentrations of CO2 lead to increased absorption at the outside boundaries of the absorption bands, mainly the 15um band. So the relevant wavelengths this equation deals with are approximately 14um and 16um.

    That this increased absorption occurs to some extent can be demonstrated in controlled laboratory experiments. However, prominent physicists like Freeman Dyson and Will Happer think that the IPCC has overestimated the rate of increase of absorption by about 40%. And I think the IPCC has done a horrible job in handling the knock-on (“feedback”) effects from this increased absorption, and probably has significantly exaggerated them.

    So I have answered your question of doubtful relevance. So now it is time for you to stop dodging my very relevant question: Radiation of a given wavelength (15um was just a convenient example) can come from a body colder than the target or warmer than the target. Is the result of this radiation from the two sources any different? If so, what is the underlying physical principle that permits them to be handled differently?

    • Massimo PORZIO says:

      Hi Curt,
      shouldn’t the work done to keep up the atmospheric molecules be accounted the same way of the back-radiation for computing the temperature offset?
      I mean, when the energy come from a separate power source which essentially starts warming the planet surface, using the very same argument of GHGs, shouldn’t the bi-directional heat transfer at the surface/air interface rise the surface temperature?
      There is a bunch of energy used to throw up the molecules in the sky which returns to the ground because of gravity.
      The only difference I see is that in the GHGs case the returning energy flux is limited by the number of GHGs molecules, while the energy exchanged for keeping the molecule up is directly proportional to the temperature.

      I would like to know your opinion about that, am I missing something?

      Have a great day.

      Massimo

      • Curt says:

        Massimo:

        In physics, mechanical work (the kind you are referring to) is force acting over distance. Technically, it is the integral of the force over the physical distance of motion of that force.

        A static force — the earth “holding up” the atmosphere, or conversely, the atmosphere “pressing down” on the earth’s surface — does not cover movement over a distance, so does no work. There is no net energy transferred by such a static force.

        Hydroelectric power is electric work created by the gravitational force acting over the distance the water falls. But the water has to end up lower than it started to create this electric power. But the atmosphere has already “fallen”.

        • Massimo PORZIO says:

          Hi Curt,
          first of all thank you for consider my issue, I’m just an electronic engineer and thermodynamics is not my field.

          Anyways, I’m not arguing that gravity itself does any work at all, my point is that without a ground temperature there is no atmosphere because there is no KE in its gases. The only reason there is KE in the atmosphere is the continuous exchange of energy at the ground boundary (which should zero net only at steady ground temperature).
          there should be a continuous work done by the ground to keep the molecules up, which has the effect of returning back the very same energy to the ground.
          So, I don’t see any difference with the work done by the ground radiation on the GHGs molecules which are bended and returns (part of) the radiation back as consequence of the their return at rest.
          Could you explain me where the two works differ?

          Have a great day.

          Massimo

    • geran says:

      Curt says: “But your argument on radiative transfer is effectively that giving any change is wrong.”
      __________

      Curt, my argument is NOT as you state. In your analogy, I clearly stated that giving the correct change was not a problem. I stated: “Or, take your analogy of the cashier receiving a $20 bill for a $5 dollar item. The cashier returns $15 change to the customer. That does not violate any math or science.”

      Why do you choose to be so disingenuous? Trying to misrepresent what I stated is a certain sign you know that you have lost. Don’t make yourself such a loser. Get on the right side of science and you won’t have to twist, alter, skew, manipulate, misrepresent, or “spin”.

      • Curt says:

        Oh, so your argument is just that getting the $15 change will not increase the wealth of the buyer (compared to not getting any change). That’s so much better!

        You completely misunderstand the 1st Law, whereas I thought you completely misunderstood the 2nd Law. My apologies!

        And yes, this is exactly what you are arguing when you state that radiation from a cooler body can never increase the temperature of a warmer body (even when the warmer body has a separate power source).

        • geran says:

          Curt, I guess that is the closest thing to a sincere apology that I will get, so, I accept your apology.

          And, I already answered your question about 15 µm photons (upthread):

          “And, as to your “cherry-picked” question about 15 µm radiation, you should know that all wavelengths would be handled with the same physics. Your problem is to show how radiation from a cold surface can raise the temperature of a hotter surface. And, you can’t do that!”

          • Curt says:

            geran:

            If the radiation is absorbed by a body, the energy in that radiation increases the internal energy of that body. That is what the 1st Law demands. And that increase in internal energy will increase the temperature of that body (or possibly go into phase change or some such, but we are talking simple systems here).

            And you have seemingly agreed that the radiation will be absorbed by the body the same regardless of whether it came from a body of higher or lower temperature than the absorbing body. (Actually, you keep answering a different question from the one I asked.)

            So we can conclude that the temperature of the absorbing body will be higher by virtue of absorbing this radiation than if it did not receive this radiation, even if this radiation came from a colder body. QED. (If this radiation came from a colder body, that body will receive more radiative energy from the warmer body than it transmits to the warmer body, as the 2nd Law demands, but this lesser amount of radiation still leads to a higher temperature of the warmer body than if it did not transmit any radiation.)

            If the warmer body has a separate constant power source, this extra radiation from the colder body will lead to a higher steady-state temperature of the warmer body than if the radiation were not there.

            Let’s consider an example using the vacuum chambers KevinK brought up — what started this discussion. We have a spherical satellite with 1 m^2 of surface area (think Sputnik) with an emissivity of 1.0 (blackbody) and a 240 watt internal radioactive power supply. It is placed in this vacuum chamber with the walls super-cooled to near absolute zero so that the walls emit virtually no thermal radiation.

            For this satellite in thermal steady state, it must emit 240 watts from its surface, which it can only do by radiation. With no significant ambient radiation from the super-cooled walls, the surface will be at a temperature that emits 240 W/m^2. At emissivity 1.0, a temperature of 255K is necessary to do this, so that will be the steady-state surface temperature.

            But now we warm the walls up to 227K so they emit (also emissivity 1.0) 150 W/m^2. So now the energy balance for the satellite is 240 W in from the radioactive power source, 240 W out from thermal radiation, and 150 W in from the walls’ thermal radiation. Despite the net heat transfer of 90W from the satellite to the walls (as the 2nd Law demands), the satellite is receiving more energy that it is outputting, so its internal energy and so temperature increase (as the 1st Law demands).

            The temperature will continue to increase until the satellite reaches energy balance again. With 240W from the internal supply, and 150W from the walls, it needs to be hot enough to emit 390W from its 1 m^2 surface. This requires a temperature of 288K.

            So even though the walls were always colder than the satellite, the radiation from the walls at 227K led to an increase in the surface temperature of the satellite compared to the case where the walls were near absolute zero and emitting virtually no radiation.

            This type of problem is given early in an introductory undergraduate thermodynamics course, with this type of analysis expected of any student who is going to pass the course.

          • geran says:

            Curt, why do you continue to do this to yourself? There are rich people that enjoy petty crimes, like shop-lifting. They get a thrill out of violating the law. Do you get a thrill out of violating laws? Do you get a thrill out of debate tricks, trying to see if you can get away with them, or not?

            Once again, as if you get a thrill out of it, you have misrepresented my words. You say:

            “And you have seemingly agreed that the radiation will be absorbed by the body the same regardless of whether it came from a body of higher or lower temperature than the absorbing body.”

            I never used the word “absorbed”, and I never said “regardless of whether it came from a body of higher or lower temperature”.

            As I stated earlier: Don’t make yourself such a loser. Get on the right side of science and you won’t have to twist, alter, skew, manipulate, misrepresent, or “spin”.

          • Curt says:

            geran:

            You can’t even keep up a logical discussion. I asked you multiple times what would happen to radiation of the same wavelength hitting a target body when this radiation came from warmer body and a colder body. I asked it with specific temperatures and wavelengths and again without.

            When I repeated the question yet again, you objected, saying you had already answered the question (even though technically you answered a different question). So when I went with that, you objected to that.

            So I will repeat the question for a final time to see if you actually have the capability to do simple analysis:

            There is a body (let’s call it a blackbody for simplicity) of temperature T0 (we can use 288K as an example, but it doesn’t matter).

            There is a second body of temperature T1, where T1 > T0 (we can use 298K as an example, but again, it doesn’t matter as long as it is greater), emitting radiation toward the body that is at lower T0.

            There is a third body of temperatuare T2, where T2 < T1 (we can use 278K as an example, but once again, it doesn't matter as long as it is less), also emitting radiation toward the body that is at higher T0.

            Question 1: What happens to the radiation of a certain wavelength (we can use 15 um as an example, but it doesn't matter) from the body at T1 when it hits the blackbody at lower T0?

            Question 2: What happens to the radiation of this same wavelength from the body at T2 when it hits the blackbody at higher T0?

            Question 3: If the answer to questions 1 and 2 is different, what physical mechanism explains the difference?

            As I look at your previous answers, you talked about all wavelengths being treated the same. That was NOT the question!

          • geran says:

            (Curt, I just now saw your questions.)

            A black body emits a spectrum of wavelengths. Assuming all three bodies are ideal black bodies, then:

            Question 1: Most of the IR radiation that strikes the lower temp body will be absorbed.

            Question 2: Most of the IR radiation that strikes the higher temp body will be reflected.

            Question 3: Absorption of IR radiation is dependent on wavelength of the incoming radiation and the temperature of the target (neglecting shape of the absorber).

          • Curt says:

            geran:

            Finally, you’ve taken a stab at the questions! But you still haven’t actually answered the questions I asked.

            I very carefully asked each time about a specific wavelength and whether there would be any difference in what happens at the target when this wavelength of EM radiation comes from a source colder than the target versus hotter than the target. I’d still like to see your answers to those specific questions.

            But already I’m shaking my head. A blackbody by definition absorbs all radiation incident upon it, and yet you claim most of it would be reflected in some cases. Can you source that claim? I can’t find it in any of my texts.

          • geran says:

            Sorry Curt, I was trying to make the scenario more “real world”. I did not want to confuse you any further than you already are.

            If you stick with the strict definition then all radiation is absorbed.

            Hope that helps.

          • Curt says:

            geran:

            What is the “strict” definition as opposed to the “geran” definition?

            But you still haven’t actually answered my three questions. Why do you keep dodging them?

          • geran says:

            Curt, I have answered your questions, but you have not answered mine.

          • Curt says:

            geran:

            I answered your question in great detail here:

            http://www.drroyspencer.com/2015/05/trmm-satellite-coming-home-next-month/#comment-192300

            You still have not directly answered mine.

            I think you finally admitted that radiation from a colder body would be absorbed by a warmer body the same as radiation of the same wavelength from an even warmer body — which is progress — but since you refused to state it explicitly, I can’t be sure.

  17. Norman says:

    geran,

    I have been reading your interaction with Curt and I do believe he is on the correct side of science. Not sure where you are coming from, maybe reading some blogs and calling it science. Curt is using established textbook science in his posts. It is possible these texts are wrong (though a lot of science has been established by experiments in the past like the gas laws). You state “Get on the right side of science”

    If you want to be on the right side of science the best approach would not be to endless argument with the textbooks (thought to be established science by many in the field and used often in design issues, actually making things that work based upon the concepts found in such textbooks). You should do experiments to prove Curt wrong. Set up an experiment, describe how you did it, what you expect and then state the results you obtained. This is real science, not blog reading.

    Experiments to try (which I may soon do). Take a heat lamp and put a high temperature thermometer on its surface. let the temperature reach equilibrium. Now move a heat lamp of same watts (producing the same amount and spectrum of IR) near the first and see if it is able to rise the temperature of the thermometer. Or even better would be get a concave mirror and have it reflect back (backradiation) the energy emitted by the heat lamp onto the surface to see if it can actually raise the surface temp. If it can’t you have a point. If the mirror can do such then you may have to reconsider and try to understand correctly what Curt has been saying to you.

    • geran says:

      Norman–I don’t argue with the established science (textbooks). I argue against pseudoscience. Maybe you missed my example of the IPCC equation. That equation is “pseudoscience”. It has NO scientific basis. It is a dream of Warmists. They want to “believe” it is valid. It is NOT.

      If you plan on the experiment you mentioned, “pseudoscience” would be to use a concave mirror. A concave mirror “focuses” E/M energy, just as a solar oven, or a magnifying glass (which can burn paper). So, as you “believe” Carl is on the “correct side of science”, and you “believe” heating something with a concave mirror “proves” the GHE, then your “belief systems” trump your devotion to the scientific method.

      Why not do a simple experiment. Put a thermometer in your freezer. Leave it overnight. If “back-radiation” can warm a hotter surface, the ice in your freezer should be able to warm the thermometer.

      Careful you don’t burn your fingers reading the thermometer….

      • Massimo PORZIO says:

        Hi geran,
        I don’t have any proof that the “back-radiation” warms our planet ground, but (if I get it right) Curt doesn’t say what you wrote.
        He say that if there is an external source of energy that enter the system through the warmer body, this warmer body become warmer if its radiation was back-radiated by a second body, this should be valid independently by the temperature of this second body, no matter if it was colder or warmer than the first body, what it matter is that the second body has a temperature above the surroundings universe.
        About your freezer/thermometer analogy, IMHO it has nothing to do with Curt argument.

        Have a great day.

        Massimo

        • geran says:

          “Curt doesn’t say what you wrote.”

          ???? Could you give me the exact quote? I’m not sure what you are referring to.

          Massimo, your English is much better than my Italian, but I’m still having trouble with this:

          “He say that if there is an external source of energy that enter the system through the warmer body, this warmer body become warmer if its radiation was back-radiated by a second body, this should be valid independently by the temperature of this second body, no matter if it was colder or warmer than the first body, what it matter is that the second body has a temperature above the surroundings universe.”

          You may be referring to some “thought experiment” that Curt came up with. So, I’m not sure of all the conditions. If you put energy into a system, and that energy can not leave the system, then yes, you would expect to see a temperature rise. I’m not sure how that applies to Earth’s atmosphere. But, again, your English is much better than my Italian. (I do understand “pizza”, however.)

          “About your freezer/thermometer analogy, IMHO it has nothing to do with Curt argument.”

          Sorry Massimo, we have to disagree here. The freezer/thermometer analogy is ALL about “cold” warming “hot”. So, it is relevant to what Curt is trying to peddle. If the Warmists and Lukewarmers could warm a hot surface using only a cold surface, they would be showing us that experiment. I haven’t seen it, have you?

          • Massimo PORZIO says:

            Hi geran,
            sorry, maybe I was not clear with my “I don’t have any proof that the “back-radiation” warms our planet ground…”.

            In fact, I’m still looking for a proof of that, in fact some posts above I’m asking for a difference between the work done to fire up in the sky the gases molecules (all the atmospheric gases molecules), work which is fully returned to the ground surface, and the (theorized ) GHGs effect.

            The only case I can report for a source which rise the temperature of it’s target above it’s own (lower) color temperature is the CO2 laser. Where the colour temp is 9.6 to 10.4um(max colour temp 302K/29.7°C), but locally the temperature on the target can rise up to levels that melts some rocks.
            But it’s a very different issue, it has nothing to do with back-radiation.
            I’m the first who asks about a true measurement of the effective incoming and outgoing radiation at the TOA, which in my opinion is not still achieved.

            “Sorry Massimo, we have to disagree here. The freezer/thermometer analogy is ALL about “cold” warming “hot”.”
            Maybe I get it wrong, but I understood that Curt’s warmer body was continuously supplied with an external source of energy which I don’t see in the freezer/thermometer analogy.

            Have a great day.

            Massimo

            By the way, despite I live in the north (better pizza makers come from Naples indeed) I very like pizza and in the backyard of my house I’ve a pizza’s oven šŸ™‚

          • Massimo PORZIO says:

            Hi geran,

            (sorry, I posted this message above, as if I was referring to your previous post, me silly; so I repost it again here below)

            I don’t have any proof that the “back-radiation” warms our planet ground, but (if I get it right) Curt doesn’t say what you wrote.
            He say that if there is an external source of energy that enter the system through the warmer body, this warmer body become warmer if its radiation was back-radiated by a second body, this should be valid independently by the temperature of this second body, no matter if it was colder or warmer than the first body, what it matter is that the second body has a temperature above the surroundings universe.
            About your freezer/thermometer analogy, IMHO it has nothing to do with Curt argument.

            Have a great day.

            Massimo

          • Massimo PORZIO says:

            Aaarrg!
            I switched to Mozilla Firefox and it doesn’t update automatically the page after the posting of a message, so I
            believed to have reply you in the wrong place…

            It’s half past midnight here, much better I go to sleep šŸ™

            Have a great day.

            Massimo

  18. Norman says:

    geran,

    I do think you have some form of scientific misconnection somewhere in your thought process. I have read many of your exchanges and Curt (not Carl) is not talking about the IPCC equation on climate sensitivity. I would would agree that this equation is not science. It is not based upon any established physics and it is pseudoscience. It is a guess at best or speculation on possibilites.

    You are completely wrong in your understanding of the concave mirror. The mirror is not doing any magic it is only returning energy giving off by the heat lamp. It has no other source of energy to return. The reason I suggest a concave mirror is to retrun more of the energy given off by the lamp. Because of inverse square law you will lose a lot of the emitted radaiation. The point would be can returning energy add energy back to the source. That is all the experiment is designed to do. Thinking this is psuedoscience because it concentrates the energy is bizarre thought process outside of scientific logical thinking and I not sure what logical process you are basing this upon.

    If you ever read Curt’s posts with the desire to understand them you would see how totally irrelevant your freezer example of “back-raditaion” is to what he has been describing. If a source has its own power (like a heat lamp) then if you return energy to its surface it will be warmer than if you did not return any energy. It is a relative state compared to another. One with returning energy and one where no energy returns. The one with returning energy will be warmer than the one without.

    If you looked at earlier threads of Roy Spencer. The GHE does not warm the surface, it just slows the cooling rate. Why is this concept beyond your ability to understand. The Earth still cools with GHE, just not as fast.

  19. geran says:

    Norman, okay, I see what you were trying to say. You do not want to focus any additional energy with the concave mirror, just the energy from the source. I agree, that would not invalidate your experiment. Thanks for clearing that up.

    And yes, I meant “Curt”, not “Carl”.

    Your next point: “If a source has its own power (like a heat lamp) then if you return energy to its surface it will be warmer than if you did not return any energy.” Well, that depends on how you set up the experiment. If you set it up wrong, as Anthony Watts did in his light bulb “experiment”, then you could see some minor warming due to restricting conduction and convection. However, if you do the experiment properly, any “back-radiation” energy would not raise the bulb filament above its equilibrium temperature.

    Finally, you wrote: “The GHE does not warm the surface, it just slows the cooling rate. Why is this concept beyond your ability to understand. The Earth still cools with GHE, just not as fast.” Sorry Norman, but that is not what the IPCC says. They state “warming”, as in “global warming”. They believe the GHE will raise the average temperature of the Earth. I’m not sure why this concept is beyond your ability to understand.

  20. Curt says:

    geran:

    Take a standard incandescent light bulb. First wrap it with transparent but poorly conductive film (e.g. “Saran Wrap”). Take voltage and current readings.

    Next, wrap it with reflective and highly conductive aluminum foil. Take voltage and current readings.

    The electrical resistance of the tungsten filament increases with temperature in a well-known relationship.

    When I did this experiment with 40W bulbs, I got a consistent 10K increase in filament temperature, even though:

    (a) the foil provided less conductive insulation than the film;

    and

    (b) the power in the filament is actually reduced due to the increased electrical resistance.

    It takes professional-grade precision instrumentation to see the difference (home multimeters aren’t sensitive enough), but it is very real and repeatable.

    By the way, if you think Anthony’s placement of a mirror several inches away from the side-facing flood bulb had any detectable effect on the conductive/convective losses from the bulb, then you have no concept of how natural convection works.

    (Hint: It’s vertical, not horizontal. The air heated by the bulb rises, drawing more air in from below. He created absolutely no restrictions to this flow.)

    • JohnKl says:

      Hi Curt,

      Copper proves excellent at reflecting mid and long wavelength IR (98 percent) but proves an excellent conductor like copper allowing it to reach high temperatures and thus frequently used (like copper) in pots and pans to allow faster cooking at higher temperatures than most other materials. Your 10km has a lot to do with it’s conductivity.

      Have a great day!

      • JohnKl says:

        That should have read 10k (as in Kelvin). Sorry I’m typing on a Samsung cell phone and it’s late evening.

        HAVE A GREAT DAY/NIGHT!

        • Curt says:

          JohnKl:

          The metal foil (aluminum not copper, but still highly thermally conductive) provided very little resistance to thermal conduction from the hot bulb to room ambient temperature, but still resulted in the filament getting 10K hotter.

          It was the radiative reflection back to the filament that resulted in the higher filament temperature.

          • JohnKl says:

            Hi Curt,

            Again Aluminum like Copper conducts very well. In fact it’s used as a superconductor and it will heat faster and much hotter than most material (again like copper). After aluminum and copper cooking pots and pans heat faster and more intensely than other material. Being contiguous with the filament there appears no rational reason why aluminum foil cannot convey thermal energy back to the filament and certainly you do not appear to have done so! It shouldn’t surprise you that it would since chefs and many common folk use wrap food in aluminum foil to INCREASE TEMPERATURE and cook the food faster. It’s what effective high conducting metals do!

            If I have missed some point of yours please clarify.

            Have a great day!

  21. Norman says:

    geran,

    geran your quote: “Sorry Norman, but that is not what the IPCC says. They state “warming”, as in “global warming”. They believe the GHE will raise the average temperature of the Earth. I’m not sure why this concept is beyond your ability to understand.”

    The term “global warming” is a relative state and does not imply that the GHE will actually warm the Earth in violation of the 2nd Law of Thermodynamics. The Earth system is not a closed system. It is receiving a constant input of new energy from the Sun (not all locations some are in the dark at night) but the Earth is always receiving new energy. Global Warming means that the Earth is warmer than it would be without GHG and not that the GHG are sending more energy down to the surface than what is available.

    You do not like Anthony Watts experiment. I always refer people the the reality of a thermos bottle. A real good experiment to allow you to comprehend what is actually being claimed by Curt and others (and not a false interpretation of the situation from the likes of Doug Cotton) would be to have two different types of thermos bottle. Both thermos would have vacuum to prevent most conduction. Difference between the two would be one has a mirror to redirect radiation back into the fluid and the other one does not. Pour equal amounts of liquid in each. Have a timed heater add energy to each in equal amounts (cylce of heating and no heat added) and see what happens after equilibrium conditions are reached. I would predict the one with the mirror would end up at a higher temperature.

    This does not mean the mirror is heating the fluid (which it is not), the mirror is stopping radiation loss and redirecting the energy that the fluid was giving off back into the fluid. It will not warm the fluid above its emission temperature but it will keep it from losing energy.

    The rate of energy loss is what is important and it is what Curt has been saying and Roy Spencer spent a whole thread on this issue (couple threads back). The backradiation does not warm the surface above its emission temperature and it actually does not even stop it from cooling. It slows the rate of cooling and when new energy is added the following day the surface will be warmer than if no backradiation was present.

    In the thermos the backradiation from the mirrored surface keeps the fluid warmer longer than it would be in a situation with no mirrors.

    I think you can understand it but will need to look at it differently than how you are currently understanding what is going on.

    Thanks for you time. Hope this will be a beneficial post.

    • geran says:

      Norman, sorry, but the issue is global warming. They have switched on you because the planet is not warming. Go back and study the IPCC literature—-

      Earth is warming due to CO2 (GHE).
      Earth will continue to warm–see numerous model projections.
      We will soon hit a “tipping point”.
      Yada, yada, yada.

      And, in your fascination with thermos bottles, don’t forget the difference between “reflection” and “back-radiation”. They are very different things, with differing effects.

      Hope this will be of benefit.

      • fonzarelli says:

        Norman says: “… The back radiation does not warm the surface above its emission temperature and it actually does not even stop it from cooling. It slows the rate of cooling and when the new energy is added the following day, the surface will be warmer than if no backradiation was present”

        Geran, at the risk of adding to the confusion, i’ve decided to interject here. I figure it would be better for me personally than continuing to bang my head against the wall which makes my head ache (and doesn’t do much for the wall either…).
        Forget the ipcc for a moment… Do you agree or disagree with the above quote of norman’s? Is there anything that you would add to it (or take away from it)? Anything to add clarity to what norman said? Anything thing to add clarity as to why norman would be incorrect?

        • geran says:

          Hi Fonz. What the “Lukewarmers” are now trying to do is walk away from the GHE. They are catching up with the established science. They want to cling to the GHE, but yet they know it is wrong. Go figure.

          Earth’s atmosphere, considered as a “control volume” can NOT warm the Earth. It can only cool the Earth. Everyone wants to use the word “thermodynamics”. Many can only spell the word thanks to spellcheck. But thermodynamics is a WELL established science. Go figure.

          Earth gets its energy from the Sun. (Geothermal is insignificant compared to solar. Okay, geothermal is magnitudes below solar. Okay, geothermal is multi-magnitudes below solar. I hope I made my point.)

          So, the atmosphere can only cool, not warm. It can handle anything the Sun puts out. CO2 is NOT a “heat source”. IPCC “science” is wrong, invalid, false, ludicrous, inane, un-scientific, and incorrect. Did I say, “WRONG”?

          Okay, now that I have finished my rant, I will try to answer your intelligent question.

          When the Lukers now state that “the atmosphere slows the cooling”, they are retreating from “the atmosphere warms”. That appears to be a huge concession, but it still involves the “greenhouse gas” issue. They want to cling to their belief that GHE “slows the cooling”.

          Heat transfer, like thermodynamics, is a well established science. Heat transfer is seldom instantaneous. Heat transfer takes time. People have sued McDonalds because the coffee burned them. Duh.

          Earth’s atmosphere does not “slow the rate of cooling”. The atmosphere “adjusts” the rate of cooling. The rate of cooling is established by the laws of physics. To slow the rate of cooling, you would need to add elements/compounds to the atmosphere that cannot conduct, nor radiate. Ask Lukewarmers if they know of any such compounds….

          • Norman says:

            geran,

            I would need a good working explanation of what you consider back-radiation and how it works differently than energy redirected by a mirror.

            Here is my current understanding (which could be wrong if good evidence is presented to demonstrate this understanding to be flawed). You claim a difference with reflection and back-radiation. I consider them similar in property that both will redirect energy.

            The Earth at given temperature will emit so many photons of IR at 15 microns per unit of time. These photons move in straight lines away from the source (Earth’s surface). If nothing is in the way to impede their flow they move away from Earth’s surface (source) at the speed of light and the energy is gone, the surface cools at a given rate based upon the total energy loss.

            If you have a mirror above Earth the 15 micron photon is now redirected back to the surface, absorbed and re-emitted and will once again be sent back to the surface. If the mirror is really good the process can continue quite some time (like in fiber optics) and no energy is lost. The surface does not warm above its current temperature but at the same time it does not cool.

            Now with CO2 a 15 micron photon is emitted from Earth’s surface. The Carbon Dioxide molecule absorbs this photon, vibrates and then reemits the 15 micron photon in any random direction. It has a 50% chance of returning to the surface or going into space to be lost forever. If no CO2 were present the photon would automatically leave and have zero chance of returning to Earth’s surface. Both mirror and CO2 redirect the 15 micron photon. Why are these two processes different on a basic level of what is taking place?

          • geran says:

            Norman, when a photon strikes a “reflector”, it is “bounced” away. It is the same photon-same energy, same wavelength. When a photon is “absorbed”, it is destroyed. It no longer exists. The energy it contained is absorbed by the atom it strikes. Later, if the atom emits a NEW photon, it typically has less energy (longer wavelength).

          • Norman says:

            geran,

            You make this claim: “Norman, when a photon strikes a “reflector”, it is “bounced” away. It is the same photon-same energy, same wavelength. When a photon is “absorbed”, it is destroyed. It no longer exists. The energy it contained is absorbed by the atom it strikes. Later, if the atom emits a NEW photon, it typically has less energy (longer wavelength).”

            This may be the case with a typical surface that has so many modes of vibration it will emit a spectrum of radiation when energy is added to it. This is not the case for gaseous molecules like CO2 or H2O. Carbon dioxide will only emit energy at discrete wavelengths and only absorb energy at these same wavelengths. If it absorbs a 15 micron IR photon it will also only emit at certain wavelengths (this is quantum physics). You may need to update your understanding of this.

            This may help to clarify my post.
            http://dl.clackamas.edu/ch104/lesson5spectra.html

          • geran says:

            Norman, you have learned well the debate tricks used by Curt. You were asking about surfaces, such as your thermos bottle. In fact, you even mentioned a “mirror” in your question. When I answered you, you tried to imply that my answer was wrong because I was not addressing CO2 molecules.

            Hint: When you have to resort to such tricks, it only indicates you have lost the argument.

      • Norman says:

        geran,

        For what it is worth I do agree with you concerning the fear mongering of the IPCC. I do not believe Curt is a representative of the IPCC. I think is more a political body than a scientific one and one with a biased agenda. But that does not mean I agree with you back radiation is unable to slow cooling rate or is significantly different in operation than mirrored walls within a thermos that redirect radiation emitted from a hot liquid back into the liquid maintaining a hot temperature much longer than would exist is the mirrored surface was not present.

  22. geran says:

    Curt, you should be really proud of this one. It is hilarious! One of your best.

    1) “It takes professional-grade precision instrumentation to see the difference (home multimeters aren’t sensitive enough), but it is very real and repeatable.”

    ******This after taking multiple paragraphs to explain how to do the nonsensical “home” experiment! Hilarious. It must be something to be the only one that is capable of seeing such results. I’m sure we can just “trust” your conclusions. How about this: You could not get any statistically meaningful results that can be verified, so you imagined what you wanted. Much like the IPCC equation. Just make stuff up and hope no one notices.

    2) “By the way, if you think Anthony’s placement of a mirror several inches away from the side-facing flood bulb had any detectable effect on the conductive/convective losses from the bulb, then you have no concept of how natural convection works.

    (Hint: It’s vertical, not horizontal. The air heated by the bulb rises, drawing more air in from below. He created absolutely no restrictions to this flow.)”

    ******Hey Curt! Anthony did the experiment on a TABLE! You know what a table is right? A table DOES affect “drawing more air in from below”. Hilarious! And, “situation 1” had NOTHING in front of the light bulb. “Situation 2” had the mirror restricting convection. It was a “magic trick”, and you swallowed it like you swallowed the GHE.

    Hilarious.

    • Massimo PORZIO says:

      Here I fully agree geran,
      some years ago I tried myself some experiments like Anthony’s one, and I learnt that it’s not so easy to take account of such environmental factors (any damned thing around it matters).

      I also noted the big difference between reflection and back-radiation you highlighted to Norman.
      The first surely warms up the source, while about the second I never get any proof of that.
      But my experiments where really bad done here in my home workshop so I still hope that someone reports of a real experiment which demonstrate that whether it exists or not here one day.

      Have a great day.

      Massimo

      • fonzarelli says:

        Massimo, recall my grandfather was from san marco not too far from naples. He made a great pizza which was only surpassed (in greatness) by his spaghetti. If there ever was a place where heaven and earth came together, it would have to be grandpop’s kitchen there in allentown, pennsylvania. He had his garden right outside his kitchen where he would grow some of the ingredients that he’d use. And he always insisted on using fresh grated parmesan… I don’t know the equivalent in italian so i’ll just settle for the french “OO LA LA” !!!

        • Massimo PORZIO says:

          Hi Fonzie šŸ™‚ ,
          Naples and the neighboring coastal cities are very esteemed for the way they cook foods, especially seafoods.
          If you had ever eat the “scialatielli alla pescatora” then you knew what I mean (awfully too good, the scialatielli are an hand made pasta looking as spaghetti but no longer than 10 cm and thicker about 4-5 mm, slurp!!!).

          Have a great day.

          Massimo

    • Curt says:

      geran:

      Wow! You are as utterly clueless about experimental science as you are about theory!

      Since you are supposedly all knowing, you tell me what the current change to be expected of the filament of a 40W incandescent bulb is when its temperature increases by 10K (at 120VAC). Then tell me what resolution and repeatability your measurement device would require to determine this reliably.

      These calculations are not difficult for someone who knows his stuff. Let’s see if you can do them!

      And you don’t need precision measurements when you are looking at the bulb surface measurements. You can use a simple thermocouple, or even a kitchen infrared thermometer.

      The steady-state bulb surface temperature (remember that glass is highly absorptive/emissive in the longwave infrared) is tens of degrees higher with an absorptive shell (black anodized or black painted metal) than with a transparent (clear glass – highly transparent to visible and shortwave infrared) shell of the same size. The only difference is the “back radiation” from the absorbent shell!

      And (this is for Massimo), if you replace the absorbent black shell with a reflective shell, you get steady-state bulb surface temperatures tens of degrees even higher.

      Oh, and you can still see these results if you spray paint the surface of the bulb black so none of the radiation from the filament escapes the bulb directly.

      Your comments on Anthony’s experiments were even funnier! You obviously have never worked seriously with convective heat transfer.

      First, you made the absurd comment about the table below the bulb, which is in the same place both with and without the mirror on the side. Talk about contradicting yourself!

      And you still think that a mirror many inches away horizontally could have any noticeable effect on the convection from the bulb! You just have no idea how difficult it is to limit the free convection of an open device like that bulb.

      When Anthony did that experiment, I duplicated his results. I also used as my control a non-reflective surface in the same location as the mirror. That surface made no difference to the temperature, whereas the mirror did.

      Out of curiosity, I also tried to limit the convection by moving the non-reflective surface closer horizontally to the bulb, putting it below and above the bulb (above should at least theoretically have had an effect), and I could not create a difference!

      I actually test my theories. It doesn’t look like you can say the same…

      • Massimo PORZIO says:

        Hi Curt,
        “And (this is for Massimo), if you replace the absorbent black shell with a reflective shell, you get steady-state bulb surface temperatures tens of degrees even higher.”

        I agree because it is reflective.

        About the black anodized or black painted metal compared to the glass shell: did you do it, or it’s a thought experiment?
        Because, if you measured that, IMHO it could be a good proof that the warming back-radiation works that I’m looking for.
        Anyways, if you really did it, what was the inner surface black shell temperature compared to the bulb glass outer temperature?

        Have a great day.

        Massimo

        • Massimo PORZIO says:

          I would explain my last question, I asked it because the outer black shell converted to heat higher color temperature photons, if that inner surface of the shell was at a little higher temperature than the bulb surface it could be just due to convection.
          Again, have a great day.

          Massimo

        • Massimo PORZIO says:

          Uhmmmm…
          I’ve to correct my enthusiastic “IMHO it could be a good proof that the warming back-radiation works”, because thinking about it a little more during the lunch time, it doesn’t proof that a LWIR back radiation can warm it’s source, since in the lamp bulb case the source (the filament) emits photons at a very high color temperature, say 2700 K, and the outer bulb temperature is much lower say 370 K.
          IMHO that experiment demonstrated what anybody already should know for sure, that is that a black body absorb any visible photon and convert it to heat.
          There is no proof that the inner filament, that is the photons source, really increased in temperature because of the added LWIR back radiation.
          And even when you stated that you saw an increase of the filament temperature using an aluminium foil, IMHO there is a good possibility that the increase wasn’t due to the LWIR back-radiation, but that it was due to the Vis-SWIR reflection.
          Do you agree?

          Have a great day.

          Massimo

          • Gordon Robertson says:

            @Massimo…please don’t allow people to confuse basic physics using thought experiments. From the WUWT link by Curt, he is once again confusing infrared energy with heat.

            In the WUWT article is this statement by Curt:

            “This is part of a more general claim by Slayers that radiation from a colder body cannot transfer any energy to a warm body and lead to a higher temperature of the warm body than would be the case without the presence of the colder body”.

            I think the Slayers have been misquoted but the basic premise is correct, that radiation from a cooler body cannot raise the temperature of a warmer body which is in proximity. That’s especially true if the warmer body warmed the cooler body. To suggest a reciprocal warming in such a context is plain bad physics, and it represents a perpetual motion machine.

            Curt is confusing heat and infrared radiation. Clausius stated clearly in his treatise on heat that EM energy can flow both ways but that heat can only be transferred from the warmer body to the cooler body without compensation. By that he meant if you take heat from a cooler body to warm a warmer body, you MUST replace the heat in the cooler body.

            That will not happen naturally, it requires external compensation.

            That is done in a refrigerator using electric power to drive a motor, which drives a compressor, that compresses a refrigerant.

            HEAT CANNOT BE TRANSFERRED FROM A COOLER BODY TO A WARMER BODY WITHOUT COMPENSATION LIKE THAT.

            Please don’t let anyone con you into thinking otherwise.

            Alarmists like Curt are totally wrong in their assertions that a net positive flow of EM energy can satisfy the 2nd law. The reason is simple. The 2nd law is about HEAT and EM is NOT HEAT. Balancing an EM energy flow does not satisfy the 2nd law.

            It gets completely ridiculous when a physicist like Stefan Rahmstorf claims that back-radiation from the atmosphere can be added to solar energy to raise the net energy input to the surface. They should take away his Ph.D for making a foolish claim like that.

            When IR leaves the surface, it is represented by a massive, continuous flux. Although people like to think in terms of a one-to-one relationship between a photon of surface IR and a molecule of GHG, that is absurd. Einstein stated clearly that no one knows if such a flux is a wave or a mass of photons.

            The IR flux radiated by the surface is massive and the number of ACO2 molecules that can intercept that flux are pityfully small in comparison. When the flux is radiated per unit time, that represents a loss at the surface. There is a further loss going through the atmosphere. ACO2 back-radiation must make up that loss since its heat came from the surface.

            Please note that the heat in GHGs is not the IR that came from the surface. The GHG molecules have bonds that absorb the IR and that raises the kinetic energy in the bond, causing it to vibrate harder. That raised KE and increased vibration IS heat.

            Not only that, ACO2 back-radiation, if it exists, radiates in all directions. It is also of a lower intensity than the surface IR that warmed it and in a restricted frequency band. So you have a tiny fraction of the surface IR received in ACO2 molecules, then re-radiated at a lower intensity and in a restricted frequency band. That so-called back-radiation must make up all surface flux losses before it can even begin to warm the surface.

            However, it can’t. The 2nd law does not allow a cooler body to transfer HEAT to the warmer surface THAT WARMED IT.

            Curt is wrong. He is talking about infrared energy, not heat. Proving his point, however goes well beyond thought experiments and into statistical mechanics and Feynman diagrams.

            A dreadful error has been perpetuated by alarmist climate scientists who insist that a net positive energy flow can satisfy the 2nd law.

            I am embarrassed to say that the error reaches to distinguished skeptics like Fred Singer, who has completely misunderstood the argument about the 2nd law with regard to the GHE and AGW. The truth is that thermodynamics is well beyond the understanding of many scientists.

            Clausius was a brilliant scientists who methodically worked out his arguments in thermodynamics. Many modern scientists have the temerity to claim he was wrong because he did not know about atom structure and infrared energy. That is bs, he talks about both in his work.

            What bugs me most is people redefining entropy as they see it without understanding how Clausius derived it. He used math to explain entropy but he also explained it in words, which most scientists I read today cannot do. To them it is an abstract notion.

            Clausius said of entropy, it is the sum of infinitesimal changes in heat at the temperature T at which the differential change takes place. If you read explanations of the 2nd law based on that sum of dq/T, many of them don’t get it at all.

        • Curt says:

          Massimo:

          I did perform the experiments, and I published the first round of them a couple of years ago:

          http://wattsupwiththat.com/2013/05/28/slaying-the-slayers-with-watts-part-2/

          I did several follow-up rounds, with more refinements and variations, that I have not published.

          The most important comparison is between the glass shell and the absorptive black shell. And yes, the black shell has a lower temperature than the bulb surface.

          As I mentioned before, I later repeated the experiment with a bulb that I had spray-painted black so none of the radiation from the filament could escape without being absorbed. I saw the same elevated bulb surface temperature with the black shell compared to the clear shell.

          • Massimo PORZIO says:

            Hi Curt,
            I feel ashamed because I give a look to Anthony’s website almost with a daily basis, but I missed your very well done article.
            I don’t really understand how I could miss it, that because I very esteem Anthony and his blog. The only reason I post very few times there is that he has already an incredible number of posters and I have great problems reading the English language, so I should spend almost all my time reading all those comments before write something that maybe someone already wrote.

            Probably you missed my reply to your post about gravity:
            http://www.drroyspencer.com/2015/05/trmm-satellite-coming-home-next-month/#comment-192314

            I would be glad if you could explain me where stand the difference between the two works. That is, why should not accounted the energy needed to keep the air gases up in the sky?

            Have a great day.

            Massimo

      • geran says:

        Curt, the filament of an incandescent light bulb is designed to reach a certain temperature, based on the designed voltage and current (wattage). Manufacturers use tungsten alloys that are really well designed. The filament may not actually reach that designed temperature (i.e., reach equilibrium) due to heat energy loss. The filament radiates to the bulb. High energy photons pass through the bulb, providing visible light. Much of the energy supplied by the filament ends up as “heat”, not visible light. The bulb gets very hot. In a sense, it becomes a small room heater. That is why incandescent bulbs are considered to be “inefficient”. (In a house in winter, the “waste” heat energy warms the room, so it is not really “wasted”.)

        So, understanding how a light bulb functions, it is ridiculous to claim a mirror reflecting the emitted photons back to the filament will raise its temperature above the designed equilibrium temperature. There are, of course, “magic tricks” that can be performed, involving conduction and convection, that make it appear the bulb is actually being “warmed” by the reflected photons. But, the science does not agree.

        People that try to ‘prove” such nonsense apparently have fun playing “scientist”.

        • Curt says:

          geran:

          The steady-state temperature of a tungsten filament, like any other substance, is determined by the real-word energy inputs and outputs, some of these inputs and outputs being temperature related. This is very basic introductory thermodynamics.

          Designers will tweak their design so that under expected conditions, it will operate at a certain temperature — or more properly, within a certain temperature range. Many devices come with all sorts of instructions about what these expected conditions are, and with all sorts of warnings about the dire consequences — like overheating — that come from operating outside these expected conditions.

          Do you really believe that because the bulb designers optimized their design to have a certain filament temperature in unobstructed conditions, it is thermodynamically impossible for the filament to run hotter?

          • Massimo PORZIO says:

            The “trick” to keep the temperature of the filament into a “safe range” is to use material such tungsten which assert a resistive positive temperature coefficient. Dimensioning the right volume/surface of the filament the designer of the bulb establish the working temperature. This is the reason that Curt experienced a slightly reduction of the current when the filament warmed up because of his experiment’s setup.

            Curt, by the way, here in Italy all filament bulbs are filled with inert gases, they are not “vacuum filled”, so the filament/bulb dissipation path is not only radiative but also convective, as far you know are the US bulb made the same way?

            Have a great day.

            Massimo

          • geran says:

            Curt, why do you continue to do this to yourself?

            I never said anything such as you imply here: “Do you really believe that because the bulb designers optimized their design to have a certain filament temperature in unobstructed conditions, it is thermodynamically impossible for the filament to run hotter?”

            I specifically stated that reflected photons could not raise the filament (emitter) above its equilibrium temperature. You are now trying to go in a new direction.

            Either you are seriously confused, or you are just trying another one of your tricks.

            Why do you continue to do this to yourself?

          • Curt says:

            geran:

            You quibble about semantics to cover your complete ignorance of basic thermodynamics! When photons are reflected back to an absorptive filament that has a constant power source, this extra power input from the photons absorbed by the filament is of course going to increase the internal energy, and therefore the temperature of the filament.

            This is trivial energy balance stuff that you would learn in the first week of an introductory thermodynamics course.

            I have been able to demonstrate this trivially in a “high school physics” experiment as I have already explained to you.

            If you don’t believe me, you might want to research “infrared reflective coatings” on halogen incandescent bulbs. The whole point of these coatings, applied to the surface of the glass, is to reflect the infrared component of the radiation back to the filament to raise its temperature above what it would be with just clear glass around it, thus causing the filament to radiate with higher intesity and at shorter wavelengths, so emitting a larger percentage of visible light.

            Better yet, you might want to buy one — you get a lot of lumens per watt!

            Oh, and if you think these bulbs are bunk, you should file a complaint with the Federal Trade Commission for advertising fraud, and with the Patent Office for patent fraud!

          • geran says:

            Curt, once again you take off on a tangent.

            I stated: “It is ridiculous to claim a mirror reflecting the emitted photons back to the filament will raise its temperature above the designed equilibrium temperature.”

            Then, you came up with “infrared reflective coatings”. Reflective coatings help the filament get closer to its equilibrium temperature. A halogen bulb does not violate the laws of physics, just because you believe it does. Once again, your tricks just make you look silly. Stick with the subject so you do not get even more confused.

            But, of course, you will not do that. So, here’s an idea for you. Completely cover all of the walls, ceiling, and floor, in your room with mirrors. Then, turn on a 40 Watt light bulb. As you know, from your deep understanding of science, the mirrors will cause the light bulb to get even hotter. But, the mirrors on opposing walls will cause the other mirrors to get hotter. Then other mirrors will get even hotter as the light continues to be re-reflected. In a very few minutes, the room will get so hot it will start to melt. But, before that happens, you can convert all of the heat energy to electrical energy and sell it back to the utility company. You can literally make millions, right in your home! A dream come true.

            Try it….

            (Are you and Norman brothers?)

          • geran says:

            Massimo, I’m happy to explain what I mean by “equilibrium temperature”. I’ll try an analogy.

            Consider a pot of water on a stove. The water is at the temperature of the room (its environment). Now, you turn on the burner beneath the pot. The water starts to warm. It is warming (temperature rising) nbbecause it is receiving more energy from the burner than it is losing, by heat transfer, to the environment. After a while, the water begins to boil. At that point, the temperature no longer rises because the pot/water system is losing the same energy as it is gaining. It is said to be at its “equilibrium temperature”.

            (Oh, and don’t worry about me getting over-heated by Curt. I am both intrigued, and entertained, by his nonsense).

          • Curt says:

            geran:

            Now you are going off into cloud cuckoo land. You have been stating over and over again that reflected radiation back to a hot source like a light-bulb filament cannot increase the steady-state temperature of that source. So when I claimed that I could do this simply by wrapping aluminum foil around a standard bulb, you said I could not possibly be correct.

            But when I showed you the case of light bulb designers exploiting this exact same physical principle intentionally to increase the temperature of the filament above what it would be without the reflected radiation, you are okay with that because that was “designed”.

            I’m sorry, but in all of my thermodynamics and heat transfer classes, I never saw anything in the laws of thermodynamics that distinguished between cases of designed intent and others! Is that the 7th Law? Can you point me to a single reference that shows this distinction?

            (The simple fact is that, all other things being equal, the filament of a halogen incandescent lamp with an infrared reflective coating has a higher steady-state temperature than one without. You are pathetically trying to weasel out of this incovenient fact by bringing in design intent!)

            And then you go on to claim that if (in the non-design case, I guess) that if the reflected radiation could increase the temperature of the filament, it would do so without end. Haven’t you ever heard about a converging infinite series? The difference between converging and diverging infinite series is covered in high-school math, but it is obviously completely beyond your comprehension!

            By the way, I already told you in this thread that I did the experiment you are demanding I try. I made the surroundings of a 40W bulb almost completely reflective. It raised the steady state temperature of the filament by 10K. Note that I say “steady state”. It did not run away. It raised the bulb surface temperature by about 200K. (For extra credit, explain the reason for the difference.)

            P.S. Can you tell me where to go to register the designed equilibrium temperature of my designs? You see, customers have been heating up some of the electronics I have designed past what I intended as their equilibrium temperature (by using them in ways I did not intend). According to you, this should not be possible. But then I realized, I had never notified the authorities of the Laws of Thermodynamics what that “design equilibrium temperature” is, so they could not know…

          • geran says:

            Curt, see why I told you to keep it simple? You throw so much stuff out there you eventually get confused.

            I doubt it, but let me see if I can help get you out of your own confusion.

            When the light bulb is warming (increasing temperature) to its equilibrium temperature, energy in exceeds energy out.

            A photon reflected from a mirror cannot raise the temperature of the filament above the temperature the photon was emitted.

          • Curt says:

            geran:

            Your pot of boiling water is a horrible example of equilibrium temperature in a system, because it will stay at that phase-change temperature with significant changes in power input and output.

            Just another minor example of how you can’t get anything right in this field.

            Now, it would be better if you had a lower power input resulting in a steady-state temperature well below boiling, say at 60C.

            But there is nothing magical about this temperature. It all depends on exactly what the overall conditions are, even for this constant power input from the burner.

            There are many things that you could do to raise the steady state temperature, and many other things you could do to lower the steady state temperature.

          • geran says:

            Curt, with your record of falsifying what I state, you have the credibility of an agenda-driven, non-scientist, which is how you earned the “blue dress” award.

          • Curt says:

            geran:

            Why have you steadfastly refused to answer my very basic and straightforward three questions? I have repeatedly asked you, and you always find some reason to answer different questions or otherwise avoid them.

            You are playing a politician’s game of never directly answering a question to provide the maximum chance of weaseling out of a difficult situation.

        • Massimo PORZIO says:

          Hi geran & Curt.

          I feel embarrassed because I don’t believe you both are young boys.
          Your skilfulness in the field let me think that you are old about like me or maybe older.
          So take the below for what it is, that is just a suggestion to turn your temperature down about this contention.
          Because despite the filament temperature is still unknown, it seems to me that both you became hotter šŸ™‚
          Take it easy, it’s just a discussion.

          @geran
          could you explain me what do you mean with “equilibrium temperature”?
          I’m not joking, I just haven’t get your point of view about it. Just to explain better my issue with it, I tell you my concept of “equilibrium temperature”, because maybe I get it wrong. IMHO the “equilibrium temperature” happens when the incoming energy to the body equates the outgoing one (no matter the kind of energy: electric or thermal).
          As I said many times, I am an electronic engineer and for me reading the current reduction at the electric “port” of the filament means that the filament itself had increased its temperature respect the original filament “equilibrium temperature”, this because it’s a positive temperature coefficient element. So, the negative feedback, introduced by the increased filament resistance, reduced the input electric energy. But the filament has surely increased its temperature. What could have increased the filament temperature other than the returning photons and the returning KE of the inner inert gas?
          I added the gas KE just because (even if rarefied) it transfer heat between the filament and the bulb.

          @Curt,
          In the past, 3 or so years ago, I conducted an experiment with a dual filament car stop/parking light bulb.
          I did probably a silly thing, because it was obvious that turning on the high power filament (the 25W stop light) and after turning on the low power filament (the 5W parking light), the current in the high power reduced because the filament is a PTC. What left me dubious is that when I analysed the spectrum of the two filaments alone and together, the colour temperature apparently didn’t change at all.
          My optical spectrum analyser has a resolution of a tenth of nanometre and reducing the power until the filament was practically emitting only in the NIR band, the BB emission looked as it just reduced in power not in colour temperature. Any ideas about it?

          Have both a great day.

          Massimo

          Have a great day.

          Massimo

          • geran says:

            Massimo, my response ended up above your comment! I must not have had enough coffee yet….

        • Curt says:

          geran:

          You say, “A photon reflected from a mirror cannot raise the temperature of the filament above the temperature the photon was emitted.”

          But I have done exactly that! As have the makers of the bulbs with infrared reflective coatings.

          The filament would be radiating at a certain temperature without the coating. Introducing the coating would raise the temperature.

          Trivial, trivial stuff, but beyond your capabilities

          • geran says:

            Curt, before you actually buy all those mirrors to “heat” your house next winter, you might want to study the 1LoT, and something called “standing wave”.

            Ah, heck, go ahead and buy the mirrors to heat your house. You know more than the established science anyway.

          • Curt says:

            geran:

            You can buy in-wall insulation with a reflective foil surface for your house. Why do they sell this? (Before you object, the main insulating mechanism here is still reducing conduction.)

            It is very common practice to wrap hypothermic individuals with reflective “space blankets” to raise their body temperatures.

            Why the extra expense of the reflective surface in both these cases?

            Again, if you think these are fraudulent, have you reported these companies to the Federal Trade Commission?

  23. Norman says:

    geran

    It seems Curt and Massimo have actually done some real experiments. You claim Curt is wrong in what he is stating or twisting things. Have you yet performed and test or experiment to prove his statements false or do you rely upon ideas put out by others not based upon established physics and stating things that have not been stated?

    You claim the IPCC is saying that CO2 is warming the Earth’s surface. Where do they say this? All radiation balance graphs I have ever seen always show the Earth’s surface (and these are just hypothetical averages not actual physical realities) emitting more radiation than is being returned by atmosphere. I have not see one which shows the atmosphere sending more radiant energy to the surface than the surface is emitting. The current average calculation I have seen is for the Earth’s surface to be emitting at 390 watts/meter^2 with backradiation at 340 watts/meter^2. This would not warm the surface as more energy is leaving the system than is returning. It is slowing the rate of energy leaving the system. The effect is because the Earth is constantly receiving solar energy that the surface will be warmer with backradiation present than if there was none (only a relative state between two conditions). But the backradiation is not warming the surface. There is no violation of the 2nd law of thermodynamics. There is only a lack of understanding of the process. If you work to understand what is going on it will become very clear to you.

    • geran says:

      Norman, I take it you just felt like rambling along without making any sense.

      Have fun!

      • Norman says:

        geran,

        Maybe the flaw is with your thinking ability. Have you thought to self reflect. I did make a few typos but the content is quite sensible.

        It might help if you would be a little more specific on what parts you are having trouble comprehending.

        • geran says:

          Norman, okay, you want specifics of your rambling nonsense:

          First you claim the Earth is NOT warming–
          “This would not warm the surface as more energy is leaving the system than is returning.”

          Then you claim the Earth IS warming’–
          “The effect is because the Earth is constantly receiving solar energy that the surface will be warmer with backradiation present than if there was none (only a relative state between two conditions.

          Then you claim the Earth is NOT warming–
          “But the backradiation is not warming the surface.”

          And all in the same paragraph!

          You have your own debate going with yourself!!

          Hilarious.

    • Gordon Robertson says:

      @Norman “Have you yet performed and test or experiment to prove his [Curt’s] statements false or do you rely upon ideas put out by others not based upon established physics…”

      Norman, the odd part is that Curt is the one questioning the early work of Clausius on the 2nd law and thermodynamics in general. I have no interest in Curt’s thought experiment’s or any conclusions he has reached concerning thermodynamics. If his experiments stand up to the scientific method, then let him publish his work challenging the 2nd law.

      The laws of thermodynamics have been established for more than 150 years now and one of the basic laws states clearly that heat cannot be transferred from a cooler body to a warmer body without compensation. Another way of stating that is that heat cannot be transferred naturally from a colder body to a warmer body. It requires human intervention.

      There is a link provided by Massimo to a WUWT article by Curt which challenges the Slayer’s group. That group is comprised of very intelligent professionals who have experience in thermodynamics.

      One of them, mathematician Claes Johnson, has produced math that challenges Planck’s quantum theory. He has done it by representing EM frequencies as tiny oscillators, and using Newtonian physics and harmonic oscillators, he as emulated what can only be done with quantum theory.

      Here’s an article by him that reveals the absurdity of some claims of Slayer detractors.

      http://claesjohnson.blogspot.ca/2011/07/sky-dragon-strikes-back.html

      One detractor claims to have falsified Slayer claims by heating a needs in a microwave. I don’t know where these clowns come from.

      Steel placed in microwaves, or any high frequency field, heats because the HF produces eddy currents in the steel. That is not the same thing as a cooler body warming a warmer body through IR.

      A microwave is not a cool source. Most microwaves generate 1500 watts at a microwave frequency.

      • Gordon Robertson says:

        sorry…that should read, One detractor claims to have falsified Slayer claims by heating a needle in a microwave.

    • Gordon Robertson says:

      @Norman…”You claim the IPCC is saying that CO2 is warming the Earth’s surface. Where do they say this?”

      It’s built into model theory. If you read the book on atmospheric radiation by Craig Bohren, he describes two models for what you describe. The first model is the heat trapping model which he concluded is a metaphor at best, and at worst, plain silly.

      The atmosphere does not work like the simple model of a photon of radiation from the surface being trapped by a molecule of anthropogenic CO2. No one knows how it works, according to Bohren, and describing it would require highly complex physics.

      Besides, no one knows if a photon exists. A photon is ‘defined’ in physics as a particle of electromagnetic energy that has momentum and no mass. The photon is modeled on a particle of air (an atom or molecule) in sound theory, where atoms of air are compressed and rarified into wave fronts. Whereas wave fronts work in sound theory it was desired to create particles of EM to simplify analysis in EM theory.

      That was especially true after Planck introduced the concept of quanta. Planck’s quantum constant is integral to the definition of a photon.

      When asked about it late in his career, Einstein suggested that some scientists think they know whether EM is a particle or a wave but that they are wrong. He claimed that after a lifetime of studying the problem, he still did not have the answer to that problem.

      So here we are in climate science suggesting that GHG molecules can trap photons of IR. It is just not as simple as that. Just as the GHE is a dumb metaphor so is the heat trapping theory.

      Bohren proposed another model that better fits the climate model theory and the one you are asking about. That model is two surfaces radiating against each other, like the atmosphere and the surface. That’s what climate models propose. The CM theory goes that the surface radiates IR to GHGs in the atmosphere which absorb it and radiate it back.

      One climatologist, Stephan Rahmstorf, has suggested that such back-radiated IR can be added to solar energy to raise the surface temp beyond what it is heated by solar energy alone. That surface heating allegedly evapourates more water vapour to the atmosphere, warming it.

      That is a positive feedback built into climate models. The 2nd law of thermodynamics does not allow such a positive feedback.

      There is a seriously fundamental flaw in the theory of two surface radiating against each other. At scienceofdoom, they have used a system like that where the Sun is radiating against the Earth and another star is brought into the vicinity.

      The second star is an independent heat source. That is not the case in our Sun-surface model. In that model, the Sun warms the planet’s surface with short wavelength EM causing the surface to radiate long wave EM (IR). That long wave IR heats GHGs molecules which allegedly back-radiate even longer wave IR to the surface.

      What’s wrong with that theory? Is it not apparent? The GHGs are DEPENDENT on the surface for their radiative energy. You CANNOT take IR radiated from the surface and radiate it back to the surface so as to increase its temperature.

      That is perpetual motion, yet it is programmed into climate models.

      Besides all that, as if that is not bad enough, a huge mistake has been made in the presumption that the radiated energy, which is long-wave EM, is heat. Heat has distinct properties as does EM and the properties are not the same.

      Work and heat are equivalent but the same equations cannot be used to describe both since they use different units. You can claim that IR has an equivalence to heat with one HUGE exception. Heat can only flow from a warmer body to a cooler body without compensation.

      For that reason alone it is not permissible to claim that a positive net IR energy exchange between a warmer and a cooler body satisfies the 2nd law. The 2nd law is about heat only and stipulates that it can be transferred in one direction only. That rules out the theory that heat can be transferred from GHGs in a cooler atmosphere to a warmer surface that warmed it.

      Why people are arguing that point using thought experiments is not clear. I’ll tell you one thing, most of the arguments are based on an equivalence between heat and IR that is not defined in thermodynamics. It’s a bad assumption that is wrong.

      • mpainter says:

        Plus, water vapor, the only GHG that counts, is actually a coolent, every molecule representing the removal of energy from the earth’s surface to be radiated aloft into space. And in this process simultaneously increasing albedo, reducing insolation via cloud formation.
        AGW hypothesizes that water vapor warms, instead of cooling. That added CO2 increases water vapor which does not cool, but warms. And that water vapor raised aloft and reduced to ice means warming for the planet, instead of cooling.
        All this based on physics fundamentally wrong because it is so imperfectly understood: photons, etc., hence the apparent contradictions.

        • Gordon Robertson says:

          @mpainter…”AGW hypothesizes that water vapor warms, instead of cooling”.

          Wouldn’t that be something, if the increasing CO2 caused cooling?

          I think the AGW reasoning is that increasing water vapour will increase back-radiation, hence the mythical positive feedback.

          If we had positive feedback in our atmosphere we’d have been doomed long ago. My experience with PF in electronics is that it multiplies almost instantaneously, and that’s with only a fraction of the output signal being fed back in phase to the input of an amplifier.

          Over 100s or 1000s of years we’d have been fried by now.

        • mpainter says:

          The “black body” view of the thermodynamics of the earth’s atmosphere (the AGW view) divorces water from its context in the natural process of evaporation–> convection–>condensation. This makes for a contrived, flawed representation of atmospheric processes. It is a theoretical house of cards that collapses when it’s poked.

  24. tonyM says:

    Curt:

    I do really appreciate your experimental efforts to in this field which is a most difficult endeavour to get it right. If it was simple then climatologists would have conducted many such experiments to demonstrate clearly the GHG effect under controlled conditions.

    Just goes to show the greatness of past scientists like Maxwell etc who seemed to conceptualise with ease (but of course it could not have been so easy).

    I will side with Massimo and Geran. Anything which restricts heat dissipation is going to increase the tungsten T which is the source of the heat in an enclosed container.

    I also note that no one has said that GHG will not restrict heat dissipation via radiation. Even Doug Cotton says so via his pseudo scattering which is just another way of saying suppressing radiation. Virtually all say that this will not increase the T of the passive surface directly. I think that is what you say too.

    I refer you to an experiment which seems pertinent:

    http://www.tufts.edu/~rtobin/Wagoner%20AJP%202010.pdf

    The authors should have taken it much further by, for example, having boxes with far taller sides, stopping the heat source at say a given time, fully balancing the molecular weights (as an air mix say.

    Although they have the usual caveats that this does not disprove GHG effect, I personally think it does or rather could show that it does not “enhance” the T. I don’t accept that with 100% CO2 vs Argon one would not be able to measure a difference (that’s why I suggest much taller sides to the box which would also increase the experimental time and see what happens to T without the heat source).

    Yeah, I used the word “enhance” undefined but I think self explanatory when taken in the context of the experiment.

    • Gordon Robertson says:

      @tonyM….”I also note that no one has said that GHG will not restrict heat dissipation via radiation”.

      There’s not nearly enough of it, Tony, by a long shot.

      Meteorlogist/physicist, Craig Bohren, referring to that heat trapping theory, claimed it is a metaphor at best, and at worst, plain silly. Bohren has written a book on atmospheric radiation.

      Visualize IR being radiated from the Earth’ surface from each atom and molecule on the surface. Could you every count them?

      Now visualize molecules of anthropogenic CO2. According to the IPCC, based on 390 ppmv of CO2, all CO2 makes up 0.04% of atmospheric gases. ACO2 makes up about 0.001%.

      Remember that all ACO2 claimed to have been emitted since the pre Industrial era is included in that 0.001%.

      You could count those molecules using Avogadro’s number and the estimated weight of ACO2 in the atmosphere.

      Now imagine that incredibly dense flux of IR from each and every atom/molecule in the Earth’s surface (and oceans) and try to imagine ACO2 molecules trapping a significant amount.

      No one has ever measured it.

      Even if they could, the presumption is a one-to-one relationship between an emitted photon and a receptor ACO2 molecule. Problem is, no one knows how that process works. It cannot be described on a one-to-one basis, it requires advanced physics and Feynman diagrams.

  25. tonyM says:

    Gordon Robertson:
    Thanks for your reference but I don’t wish to do a lot of work if it does not seem so productive for me. I am inclined to accept the pressure broadening and the chuncks taken out of the IR spectrum re CO2 and accept that it does absorb and emit irrespective of the low concentration.

    If we are talking the same Craig Bohren then he must be good mates with Dr M Mann as both relate to the same institution (sarc). He is reported as saying to try 380ppm arsenic in your tea and see if it has any effect as an analogy.

    I note that Coca Cola specify less than 1ppm acetaldehyde for their PET bottle. PVC resin used for beverage containers was required to contain less than 1ppm vinyl chloride monomer. Cans are double coated to avoid even the slightest iron contamination for beer. Pot permanganate 0.001 molar (18 ppm) is still very purple. Clearly small amounts can have an effect.

    I don’t buy your assertion that ACO2 represents only 0.001% in the 0.04% even with outgassing if that is what you mean.

    Not sure of your direction when you say we do not know how the process works. I don’t think we really know how anything works yet we accept them because our concepts and formulae do yield the correct empirical outcomes. I have never seen an electron or electron probability distribution shell for example. Even Feynman was somewhat bemused:

    “ ..We always have had … a great deal of difficulty in understanding the world view that quantum mechanics represents. At least I do, because I’m an old enough man that I haven’t got to the point that this stuff is obvious to me. Okay, I still get nervous with it. …. It has not yet become obvious to me that there’s no real problem. I cannot define the real problem, therefore I suspect there’s no real problem, but I’m not sure there’s no real problem.”

    “I think I can safely say that nobody understands quantum mechanics.”
    — Richard P. Feynman

    • mpainter says:

      Tony, you are talking about toxins,of which CO2 is not, except in the minds of the alarmists.
      You say:
      “Clearly a small amount can have an effect”

      Very true. For example, if CO2 were to fall much under 200 ppm, life could hardly be sustained at such a paucity of this all-important gas. Add 200 ppm to that and life thrives.
      Here is the truth: atmospheric CO2 is entirely beneficial, the more the better.
      Pay no attention to the alarmists. They are wrong in every assertion.

      • tonyM says:

        mpainter:

        Must be April 1st.

        First I have Gordon repeating his story and taking both my comment and Craig Bohren’s out of context or embelleshing them.

        Now I have you doing the same but asserting your pet beliefs without reference and exhorting me to look on you as my guiding light. This is accentuated by your belief that I refer to acetaldehyde and iron restrictions because of toxicity issues. Suggest you go look up their toxicity as this is daft.

        I will definitely call you if I ever need your special skills!

        I actually do thank Gordon as Bohren’s book is a good reference and anyone wanting to digest it can look up:
        ftp://ftp.sron.nl/pub/antonion/Bohren_Clothiaux%20-%20Fundamentals%20of%20Atmospheric%20Radiation_An%20Introduction%20with%20400%20Problems.pdf

        Gordon’s idea that I believe the CO2 traps photons is just nonsense. Photons are extinguished forever on absorption. In the process of trying to equilibrate, energy can be shared with other molecules via collisions and new photons can always be created by emission. But this interaction is always ongoing and does not rely on absorbing any photons from the surface or elsewhere. In fact most of the net energy transferred to the atmosphere from earth’s surface is via convection/conduction/latent heat and not radiation.

        Bohren says “trapping” which is one way of saying “closing the atmospheric window” is nonsense and leans more strongly to the whole atmosphere radiating and sending energy to the earth surface of a simplified radiation only model. He was talking about the interpretation or visualisation of the same formula.

        Gordon has taken it totally out of context as Bohren believes very much in back radiation. Here is the actual text and interpretation comment:

        “We prefer the increased emission interpretation for a few reasons. According to this interpretation we are warmed at the surface of Earth by two sources of radiation: the sun and the atmosphere. With this interpretation the atmosphere is actually doing something (emitting) whereas according to the other interpretation it only prevents something from happening. Moreover, the notion that the atmosphere traps radiation is at best a bad metaphor, at worst downright silly.”

        • mpainter says:

          Tonym, it is May 31.
          I suggest that you do as you advise me and look up acetaldehyde, which you seem to regard as non-toxic. It is a proven carcinogen. How toxic do you like your soda water?
          You sneer at my “pet beliefs given without reference” is noted. Were you referring to any particular comment? and which, pray tell. I note that you make no attempt to refute except in your incorrect assertion that acetaldehyde is not toxic. You need to avail yourself of somebody’s special skills.
          Otherwise,
          Think Happy Thoughts
          Such as the benefits of atmospheric CO2.

          • tonyM says:

            You continue to compel me to believe it is April 1st.

            Did you actually read what I wrote. My comment was:
            “This is accentuated by your belief that I refer to acetaldehyde and iron restrictions because of toxicity issues. “

            My reference to acetaldehyde and iron is NOT related to toxicity at all (particularly at the levels expected in packaging materials). Had you done some research you might grasp that. Similarly for iron.

            Heavens a PET bottle left in the sun will generate more than 1ppm acetaldehyde. I guess Coca Cola must have their warning message written in invisible ink.

            If I wanted to talk about toxicity I would have pointed out that all substances are toxic, even water in excess! Try 5% CO2 in your bedroom and see how well you feel given your truth slogan is: “Here is the truth: atmospheric CO2 is entirely beneficial, the more the better. “

            Have a nice day. I promise to call you if I need your special expertise

          • mpainter says:

            tonyM,
            I refer to your statement above: “Suggest that you go look up their toxicity as this is daft”.
            Did you not mean what this statement implies? I inferred from this statement that your view was that acetaldehydetake is non-toxic. If you did not mean that, I must say that you have a strange way of asserting that acetaldehyde is a toxin. April fools?

            Your 5% CO2 is far beyond any level that could be ordinarily encountered in nature, even through burning all the fuel on the planet. So share your understanding, what does 5% CO2 do to one? Or is your vague claim simply more alarmism? Or more April fools?

            I will stand on my statement concerning the benefits of atmospheric CO2 .

          • tonyM says:

            MPainter:

            I am having a good laugh; you seem capable of turning every day into April 1st. Please after midday your tricks should cease as is the usual custom.

            Ignoring a whole qualifying clause attached to the suggestion to look up its toxicity will hardly enhance your comprehension.

            In any case, asking you look up its toxicity infers that it has a toxicity. This is the norm with any substance! Just its toxicity level is well beyond what we were talking about for PET bottle production. The same goes for any iron contamination via pinholes in the beer can coating. In excess, iron too is toxic.

            Had you looked it up I had hoped you would have also come across its properties and might have realised that it is quite widespread in foods like coffee, bread, ripe fruits and fermentation processes. I guess you avoid all these things and plenty more. Alcohol can be turned into acetaldehyde in the body. This is life or have you decided to stop living now that you are aware of its widespread existence.

            Mentioning iron in the same context would surely have given you a good clue since we shop for foods with a relatively high content. Some women do take iron tablets. This is common knowledge; you are an exception.

            Now go see if you can find out why these two are restricted to such low levels. It is not due to toxicity OK? Should I repeat it for you?

            Strange how you want to weasel out of your absolute statement that more CO2 is always beneficial. If you perchance feel ok with 5% CO2 night perhaps you can try 10% and just step up in increments. Do report back. Unless you are superman I doubt whether you will be alive to furnish more than two reports. But as you feel it is beneficial just let it hang around your house at the 5% level continuously.

            Don’t you have the capability of doing a quick search for yourself?
            “A value of 40,000 ppm is considered immediately dangerous
            to life and health based on the fact that a 30-minute exposure to 50,000 ppm produces intoxication, and concentrations greater than that (7-10%) produce
            unconsciousness (NIOSH 1996; Tox. Review 2005) “

            Does that sound like alarmism or being reasonably factual.

          • mpainter says:

            Tony, thanks for your reply, I think.

            It sounds like alarmism since such levels could never be achieved by natural means nor by the use of fossil fuels, though you burn every drop of oil and lump of coal. You raise the bugaboo of 5% when you know such levels can only be contrived artificially. So no, I am not concerned about increasing atmospheric CO2, which is entirely beneficial at any level, the more the better. Are you?
            One additional benefit of increased atmospheric CO2 is milder winters in the higher latitudes, according to some scientists. Unfortunately, this particular benefit has yet to materialize.

          • tonyM says:

            mpainter:

            I guess I was a bit brusque but put it in the context that I felt Gordon had misrepresented Craig Bohren, who in reality comes across as a measured, very intelligent scientist, and then you followed up with very absolutist statements about CO2. So I let fly a bit. Sorry.

            The problem with your statements were that they had no basis and are simply the same as alarmists but in reverse. Alarmists do carry on about CO2 as a pollutant (officially defined as such in the US) that will lead to Thermaggedon. At least they posit some mechanism; your absolutist position did not.

            Let me give you some points to consider:

            Localised effect:
            I am limited in my examples but asphyxiated miners and people building near retired coal sites etc would hardly be comforted by your absolute claim.

            World Effect:
            There is far, far more CO2/GHG sequestered in the oceans and permafrost than in the air. Warming will result in out-gassing.

            Now couple this with the fact that historically we are still in an Ice Age. Measured even in the last half million years the T could naturally rise much further. Bear in mind that past peaks are smoothed and do not reflect the more extreme higher T’s which would have occurred.

            Now increase the T and CO2 naturally and then add additional ACO2. Can you really say you are in a position to claim adamantly that it can have no effect? How?

            Food:
            Major crops synthesise via C3 and C4 chemical routes. The efficiency will be variable. Some will do worse under hotter conditions. Some will do worse under very high CO2 or other variations. Can you really claim that the out-gassing + ACO2 + T levels will still be ok for a much increased population?

            I am not a climatologist and so the issues they will see will be far broader and deeper than my limited knowledge. I would take my cue from Bohren:

            “Carbon dioxide is an infrared-active gas (I hate the term “greenhouse gas”), and hence all else being equal (an important qualification) we expect more downward infra-red radiation (and a heating effect) from the atmosphere with an increase in carbon dioxide. The detailed consequences of this, however, are unknown and possibly unknowable. By consequences I mean length of growing season, distribution and amount of rain, distribution and amount of sunshine, etc. And the economic and social consequences are even more uncertain. However the climate changes, it is likely that some regions of the planet will gain, others will lose.”

            As he says elsewhere:
            “Fortunately [for me], I’ll be dead before the consequences of global warming become dire, if indeed they do. But I would like to stick around long enough to see this drama played out.”

            Guess that applies to me too.

            Have a nice day.

          • mpainter says:

            Tony,
            Relax, think happy thoughts and ignore the alarmists. The world will prosper under the benefits of increasing atmospheric CO2. Already, the world has greened from the Sahel to the tundra. Crop yield has increased, particularly when soil moisture is at critical levels, as during droughts.
            All life forms will benefit from increased CO2, the fundamental substance of life.

            Some interesting climate facts:
            During the mid Eocene, about 50 million years ago, the earth was about 6-10 °C warmer than today. London was a tropical forest with laurel species similar to those found in the humid climes of Indonesia. This attested by fossils from the London Clay. Anchorage, at 60 °N latitude was subtropical, with palms. At 80° N lat., on Ellesmere Island, there were deciduous forests. Life flourished like it never has since. Do not let the alarmists frighten you with their invented bugaboos, but keep your wits and laugh at them.

    • Gordon Robertson says:

      @Tony “If we are talking the same Craig Bohren then he must be good mates with Dr M Mann …”

      Based on the arsenic analogy it is the same Bohren, but Bohren is a skeptic. I doubt that he’d be impressed by the alarmist antics of Mann.

      I got the 0.04% figure for all CO2 from the IPCC. The same figures once appeared in the Department of Energy website which they took from the same IPCC graph. The alarmist EPA have likely removed the page.

      The IPCC posted a graph showing all CO2 emitted to the atmosphere between 1990 and 1999 at a CO2 concentration of 390 ppmv. Based on that concentration they claimed in words that anthropogenic CO2 was a small percent of the natural CO2 0.04% in the atmosphere.

      If you add up all the sources of CO2 on the graph and calculate the % of ACO2, it comes to a bit under 4% of all CO2. That was confirmed on the DOA site before the page was removed.

      That means the 4% ACO2 of the 0.04% of all CO2 in the atmosphere comes to about 0.04 x 0.04 = 0.0016% of atmospheric gases.

      As I said, the amount of ACO2 since pre Industrial era is built into the 390 ppmv.

      • tonyM says:

        Gordon Robertson:
        I would agree as I doubt that many serious scholars would wish to be that close to M Mann (hence my sarc tag) even though he is a great money spinner no doubt. Agree that Bohren is a skeptic but as any good scientist should be. He is not skeptical that CO2 has a warming effect.

        I could be wrong, but I suspect you were looking at total annual emissions to atmosphere (cumulative) of which ACO2 would be the 4%. That does not mean that man’s net contribution is only 4% of the 400ppm which remains in the air. Isotope studies suggest it is due to man.

        If CO2 started at 280ppm say in 1950 and man did not increase its output then barring major changes the CO2 level should stay in reasonable stasis around 280ppm.

        ACO2 annual output has increased. Out-gassing from T increases can’t account for all the increase to 400ppm even ignoring the extra sink through additional greening. There are no other plausible routes for CO2 increases in the air that anyone suggests. The mass balance points to the cumulative extra output by man beyond the original “stasis” plus outgassing. Alternatively extra CO2 just appears out of thin air!

  26. Massimo PORZIO says:

    This is for Gordon Robertson.

    Hi Gordon,
    I write this cumulative message for you just because I believe it will be easier for us to continue the discussion later on a single message, once you responded to me.

    Curt’s experiment, IMHO reported an indisputable increment of filament temperature because he measured a reduction of the current flow keeping the voltage at the filaments terminals stable. Which is what one expect to see when a PTC resistor (as the lamps filament it is) is warmed up.
    Now we have to establish what it was responsible of that warming that evidently occurred despite the reduction of the incoming energy to the filament itself.
    My question about the temperatures of the outer bulb and the inner black anodized aluminium shell it was all about to establish if the warming of the bulb could be due to convection between the two surfaces.
    The fact that he reported an higher temperature at the outer bulb surface, let me think that it has received energy back from the shell in a way different that convection.

    Otherwise how do you explain that?

    Please note that what we call “temperature equilibrium” is an equivalent equilibrium from the power flow incoming from the electric outlet and the power flow exiting the external surface of the shield to the ambient which is considered so much capable of get that power flux without altering its own temperature at all. We are not talking about the thermal equilibrium between the filament, the bulb glass and the external shield, which never happens because of the power flow that is constantly passing through the system.

    That should be what it happens in nature at the TOA, the Vis/SWIR incoming flux is the equivalent of the electric supply, while the outgoing LWIR flux is the equivalent of shield outgoing heat.

    You wrote:
    “I think the Slayers have been misquoted but the basic premise is correct, that radiation from a cooler body cannot raise the temperature of a warmer body which is in proximity. That’s especially true if the warmer body warmed the cooler body. To suggest a reciprocal warming in such a context is plain bad physics, and it represents a perpetual motion machine.”

    I don’t believe it is perpetual machine. It was that, if and only if, there wasn’t any power flux flowing through the whole system of bodies.
    The whole system is composed by the power flux source (the Sun), the warmer body (the Earth surface), the colder bodies (the GHGs molecules, but I suspect that also all the other molecules in the air are involved in the process because of gravity) and finally the power sink (the universe background).

    In a message @mpainter you wrote:
    “If we had positive feedback in our atmosphere we’d have been doomed long ago. My experience with PF in electronics is that it multiplies almost instantaneously, and that’s with only a fraction of the output signal being fed back in phase to the input of an amplifier.”

    Not really so, or better is not always so. You are doing the very same mistake that someone highlighted to me that I was doing some years ago on this very same blog.
    That is, you (like me, I’m an EE too) forget the Barkhausen’s criterion, in effect if the product of the direct gain by the feedback gain (or loss) is <1 a positive feedback can exist without any runaway.

    Anyway, as always said I could be wrong of course.

    Have a great day.

    Massimo

    • Gordon Robertson says:

      #Massimo “That is, you (like me, I’m an EE too) forget the Barkhausen’s criterion, in effect if the product of the direct gain by the feedback gain (or loss) is <1 a positive feedback can exist without any runaway".

      I'll only reply to this point for now.

      I am not arguing that PF cannot exist without runaway, I am only saying that when losses are involved, and no external amplifier is supplied, the only feedback available can be negative feedback.

      I don't think Barkhausen applies here since it is primarily related to whether a circuit will oscillate or not. Oscillation is generally a controlled positive feedback since one is looking for a stable oscillation at a fairly constant amplitude.

      Oscillators are generally designed with tank circuits in which the oscillation takes place between a capacitor and inductor. Or you can have RC combos to shift the phase from output to input. It's done in a controlled environment so runaway wont happen.

      In an audio amplifier, negative feedback is used by sampling the output signal and sending it back to the input 'out of phase' through a passive device. An audio amplifier is designed to have a frequency response from about 20hz to 20,000 hz, or more.

      As that range of frequencies is fed back to the input, there are phase shifts and some of the frequencies will shift in phase so as to become in phase with the input signal, causing a positive feedback and oscillation.

      The frequencies at which that happens are called pole frequencies. The feedback is used to emphasize certain frequencies and de-emphasize others to broaden and flatten the frequency response of the amplifier.

      That has nothing to do with what were are talking about wrt the atmosphere. In fact, nothing applicable to normal electronics circuits would compare because we don't want raw positive feedback in an amplified circuit.

      Where you see it naturally is when a PA system is set up with the speakers too close to the microphones. Audio from the speakers is picked up by the mic and fed back to the input. Such feedback is frequency specific, depending on the natural resonant frequency of the room, and other factors. However, if you put a mic right in front of a speaker, it will likely cause a runaway effect resulting in the familiar squeal.

      To cut the squeal instantly, turn of the amp. An amplifier is required for positive feedback except in a very few natural situations like a resonant system. There are no resonant systems in the atmosphere.

      I don't think Barkhausen applies to the atmosphere at all, only to a circuit with gain and feedback. Amplifier gain is inherent in the Barkhausen effect and there are no amplifiers in the atmosphere.

      • Massimo PORZIO says:

        Hi Gordon,
        Barkhausen applies to any system with a feedback. In your PA example it starts the Larsen effect at a certain frequency because, as you correctly said, the feedback path has some resonances which change phase and amplitude along the PA response frequency, but the point is that the Larsen effect is started only when the AB product of the system is >1, when that product is below that limit it doesn’t happen.
        In Larsen effect the runaway could never happen because even if the PA was DC coupled, both the microphone and the loudspeaker are unable to handle DC signal at their electrical port.

        You wrote: “I am not arguing that PF cannot exist without runaway, I am only saying that when losses are involved, and no external amplifier is supplied, the only feedback available can be negative feedback.”

        Yes, this is another mistake I did too at the beginning. But you must remember that in our “electronic way of think” losses turns to heat, which is exactly the handled signal in the atmospheric feedback system. So there are no losses in this context, because the losses are the signal at the same time.

        If you read my posts in last years here, you should know that I’m not an AGW proponent, if I have to define me about this issue, I would place myself in the lukewarmers field.
        Until I was unaware of Curt experiment (which I still ask myself how I could incredibly lost it on WUWT), I didn’t have any proof of warming by back-radiation, now I’m almost convinced that he demonstrated it, because the setup of his experiment under my EE perspective lead me to say “yes it’s back-radiation… or at least back-reflection”.

        Anyway, I still have doubts about what WL photons did the work.
        Curt said “I had spray-painted black so none of the radiation from the filament could escape without being absorbed. I saw the same elevated bulb surface temperature with the black shell compared to the clear shell”.

        Well, thinking a little more, I believe that Curt didn’t excluded any visible back-reflection, and I would like to see a comment from Curt about what I’m writing here below.

        Curt, if you take a laser pointer and aim a black (not glossy) cardboard with a little angle respect the normal to the cardboard plane, you can see almost no visible light reflection from the cardboard, but if you put the same black cardboard below a glass (also thin) panel you clearly see the laser light back-reflected. This happened in your black painted bulb experiment too. So maybe that the filament temperature has increased not because of the LWIR, but because of that back-reflected visible photons.
        Maybe that the LWIR photons back-radiated did nothing to the filament and exited the system after few rebound on the bulb glass.

        Sorry, it’s me I’m an irreducible sceptic šŸ™‚

        Have a great Sunday.

        Massimo

        • Gordon Robertson says:

          @Massimo “Yes, this is another mistake I did too at the beginning. But you must remember that in our “electronic way of think” losses turns to heat…”

          I don’t think losses turn to heat in the atmosphere. In electric circuits, losses are heat in many cases because electrical energy is lost as electrons collide with atoms.

          There is such a thing as I^2 x R heat loss in an electric circuit but that is a property of a pure resistance when an electrical current runs through it.

          In the atmosphere it is different. Solar energy heats the surface. The surface is made of atoms and molecules and the short wavelength radiation causes their kinetic energy to rise. At the same time, the atoms and molecules radiate long wave radiation which is dependent on their temperature.

          That long-wave IR represents a loss at the surface (a reduction in surface KE) and if solar energy did not replace it, the Earth would cool. So you have an equilibrium state at the surface between incoming short-wave solar energy and outgoing long-wave IR.

          It is the long-wave surface IR that is absorbed by GHGs, warming them. However, the very energy that warmed them represents a loss at the surface. Therefore, any exchange between the GHGs and the surface in a reverse direction must make up that lost energy from the surface (that warmed the GHGs) BEFORE it can raise surface temperature.

          Solar energy is just there. It has no interest in making up for losses at the surface due to LWR losses. Therefore, to produce positive feedback, those losses must be made up by back-radiation from GHGs, that are of a lower intensity and representing only a fraction of the lost surface energy that warmed them.

          SWR from the Sun and LWR emitted from the surface have to be in equilibrium at the surface, otherwise the surface would warm or cool.

          An exchange between GHGs and the surface represents a negative feedback until such a time as back-radiated IR can overcome the losses that warmed them and produce a warming that represents a positive feedback.

          Here’s where AGW alarmists are going wrong. Some, like Stefan Rahmstorf, claim that the back-radiated energy from GHGs can be added to solar energy. However, there is a time factor missing in his assertion. The solar energy that caused the surface warming leading to the long-wave radiation has already happened. You cannot come back later and add back-radiation from past solar energy to new solar energy and claim it is additive.

          That back-radiated energy is directly related to the surface LWR that warmed the GHGs. The warming of the GHGs caused a loss at the surface, and before the BR can add to surface heat, it first has to make up for the losses.

          That’s why the 2nd law is in effect. Clausius was prompted to create the 2nd law because Carnot had claimed there were no losses in a heat engine. The cycle involving surface LWR warming GHGs and the GHGs back-radiating LWR is a form of heat-engine and it has losses.

          That’s why positive feedback is not available. PF requires not only a situation with no losses, it requires amplification, and there is no amplifier.

          As I said, the LWR emitted due to solar energy at t1 heats GHGs in the atmosphere. There is a loss at the surface due to the LWR at t1 that has to be made up before back-radiated IR can be added to solar energy.

          If you could maintain the surface temperature at t1, that may be possible, but the surface temperature is an equilibrium state between solar energy and LWR at t1.

          There is no doubt a temperature inertia between incoming SWR and outgoing LWR, since SWR from the Sun goes deeper into the surface, especially the ocean, and tends to be stored before being emitted as LWR. However, it is not correct to claim that back-radiation in general is additive when combined with solar energy.

          For one the spectrum of SWR, plus the intensity, are quite different than the LWR back-radiated from the atmospheric GHGs. Any back-radiated LWR from ACO2 will have an intensity and bandwidth less than the initial LWR at t1 that warmed the ACO2.

          In summary, there is no way back-radiation from GHGs, especially, ACO2, can produce a positive feedback effect causing the surface to warm beyond the temperature it is warmed by SWR from the Sun.

          • Massimo PORZIO says:

            Hi Gordon,
            “I don’t think losses turn to heat in the atmosphere. In electric circuits, losses are heat in many cases because electrical energy is lost as electrons collide with atoms.”

            At the time of my scholarship, a physic professor told me that any work done finally decay in the “less noble state of energy” (he literally said), that is heat.

            Maybe I was not clear in my posts before, I suppose that in an energy flow system, as the atmosphere should be, only everything that is escaping to the outer space can be considered a loss indeed. In this perspective anything is returning a part of that escaping energy is a positive feedback applied to an “amplifier” with a gain less than 1.
            In a feedback system it is A*Beta that must be lesser than 1 not A only.
            So (and here I could be not clear before) I accept the positive feedback by back-radiation, but I never intended that A*Beta could be greater than one, so no runaway effect at all is possible in my opinion.
            The forward unity amplifier is the surface of the planet without any atmosphere, while the positive feedback is the atmosphere (the whole atmosphere, comprising all gases).
            Note that only when all the emitted surface radiation/work is back-radiated/returned the system reaches the instability point.

            You wrote: “That back-radiated energy is directly related to the surface LWR that warmed the GHGs. The warming of the GHGs caused a loss at the surface, and before the BR can add to surface heat, it first has to make up for the losses.”
            I agree, but when the photon is re-emitted back there is an another photon at that WL which is ready to exit to the outer space that can’t exit because of the returning one (or exits and it is immediately substituted by the returning one). That photon is not coming from the reduced temperature surface, because the surface received other energy from the Sun, while in the meantime the previous photon which momentarily reduced the surface temperature had been absorbed and re-emitted back by the CO2 molecule. The photon did a work binding the CO2 molecule, and after doing that work it returned to EM energy as the photon to the ground (if it has not been thermalized to the surrounding molecules of course).

            Otherwise, how do you explain the increase of temperature of Curt’s experiment filament?

            Have a great day.

            Massimo

        • Gordon Robertson says:

          @Massimo…from wiki…”The Barkhausen effect is a name given to the noise in the magnetic output of a ferromagnet when the magnetizing force applied to it is changed”.

          Also…”In electronics, the Barkhausen stability criterion is a mathematical condition to determine when a linear electronic circuit will oscillate…”

          Please note, in statement 2, an amplifier is required.

          I don’t see what Barkhausen has to do with positive feedback in general.

          The Larsen Effect describes acoustic feedback and that kind of positive feedback cannot happen without an electronic amplifier. As I said, turn off the power to the amp and it stops instantaneously.

          There are no amplifiers in the atmosphere and there is nothing related to magnetic noise. I don’t see what Barkhausen or Larsen has to do with alleged PF in the atmosphere.

          • Massimo PORZIO says:

            Hi Gordon,
            “I don’t see what Barkhausen has to do with positive feedback in general.”
            As I wrote above in my previous reply to you, a forward amplifier can have a gain less then 1. So until the feedback is lesser than 1/A in gain can be positive without oscillation or runaway (if the two blocks don’t have one or more zeroes in the origin of the axes of their own Bode plots).
            Of course as explained above, that product can’t be reached in atmosphere because a positive feedback greater than 1 is impossible (1/A for A1).
            But a positive feedback in atmosphere exists.
            Nothing to take care anyway.

            “The Larsen Effect describes acoustic feedback and that kind of positive feedback cannot happen without an electronic amplifier. As I said, turn off the power to the amp and it stops instantaneously.”
            I agree, it is because there is an evident loss in the feedback path. I cited it just because you wrote about PAs and that’s the right name to give to the oscillation induced in a PA system when the microphones ear the loudspeakers with a total loop gain of one or more at certain frequencies.

            Have a great day.

            Massimo

    • Gordon Robertson says:

      @Massimo…”Now we have to establish what it was responsible of that warming that evidently occurred despite the reduction of the incoming energy to the filament itself”.

      Massimo, we don’t have to establish anything, the science has already been established, more than 150 years ago. When you dicker with the ambient temperature around a tungsten filament in a bulb, it does not prove that back-radiation is heating the filament hence reducing the power demand on the filament. All it tells you is that the ambient temperature in which the bulb is operating has increased.

      I think you are right in that the change in current is related to the PTC of the tungsten. If you raise the ambient temperature surrounding the tungsten, it’s resistance will increase and the current through it will drop.

      There is no way in such an experiment to prove that back-radiation has caused the temperature of the tungsten to rise.

      • Massimo PORZIO says:

        Hi Gordon,
        “There is no way in such an experiment to prove that back-radiation has caused the temperature of the tungsten to rise.”

        What could it be otherwise?
        The system where well defined by Curt’s experiment.
        The bulb was at a temperature higher than the external shield, so no convection should has been involved.

        And yes, to shut up Curt’s conclusions, I believe we should find an alternative to back-radiation, because with that temperature values between the bulb and the shield, the environment shouldn’t matter at all.

        Have a great day.

        Massimo

  27. Curt says:

    Gordon:

    You object that I didn’t have an instant rebuttal to the 115-page G&T paper prepared, and that I referred you to SoD’s scornful take on it.

    I have now had a chance to look over the paper, and it deserves every bit of scorn SoD heaps on it. It is very largely an exercise in setting up strawmen and knocking them down. They think by pointing out how simplified explanations don’t capture the full complexity of the situation, they can disprove the entire theory.

    Firstly, the whole premise of the paper, “Falsification of the Atmospheric CO2 Greenhouse Effects…” is confused. There is no “Atmospheric CO2 Greenhouse Effect”. There is an “Atmospheric Greenhouse Effect”, and CO2 is a part of it. Water vapor, by all accounts the most important of the “greenhouse gases” is barely even mentioned.

    They spend an inordinate amount of the paper explaining that “real” greenhouses don’t work primarily by radiative absorption. Classic strawman! The term “atmospheric greenhouse effect” is a metaphor, and like all metaphors, it is not perfect. They simply waste a large part of the paper with this nonsense.

    They start their analysis with the conductivity of the atmosphere, when anyone who has really done atmospheric heat transfer calculations knows that conduction is by far the least important mechanism of heat transfer. They claim that mainstream analysis ignores conduction totally, which it does not. (You cannot explain nighttime temperature inversions without conduction from the warmer low atmosphere to the colder surface that is radiating to the upper atmosphere.)

    They make a big point of the fact that increasing CO2 concentrations barely changes the conductivity or capacitance of the atmosphere, when no one claims that it does. On page 12 they claim that “If such an extreme effect [33K warming] existed, it would show up even in a laboratory experiment involving concentrated CO2 as a thermal conductivity anomaly.”

    This just shows that they do not even understand the fundamental arguments here. No one is arguing that the atmospheric greenhouse effect is due to conductive effects, but instead to radiative effects. The fact that they do not understand this means that they are not even in the game so to speak.

    They repeatedly claim that we have no measurements to confirm the atmospheric greenhouse effect. Baloney! (I’d use a stronger term, but I respect the sensibilities of our blog host.) We have lots of good measurements of the full spectrum and magnitude of shortwave and longwave radiation, upwelling and downwelling, at various latitudes and altitudes, including above the atmosphere.

    In section 2.3, they completely botch the comparison of incoming solar radiation to surface thermal radiation. They conclude that “the radiation of the ground is about four times weaker than the incoming solar radiation.” But we have very reliable measurements that the radiation of the ground (~400W/m2) is more than 50% larger than the incoming solar radiation absorbed by the earth (~240W/m2). This imbalance must be explained. The (poorly named) “atmospheric greenhouse effect” explains it.

    In section 3.7, they do a better job of it, for example accounting for the fact that the earth’s surface area is four times that of the “disk” of intercepted solar flux. But there are two key problems with this section. First, they do not realize or acknowledge that this section completely contradicts their analysis in section 2.3.

    Second, they point out (correctly) that averaging the solar flux over time and area, then computing temperature, is not accurate when radiant flux is a non-linear function of temperature. They claim that this invalidates the greenhouse effect arguments, when in fact it makes them stronger, as the surface temperatures that would exist without radiatively absorptive gases would be even more than 33K below what they really are.

    There are appalling problems on virtually every page of the paper. I could go on and on and on, but there is no point.

    • Gordon Robertson says:

      @Curt…appreciate you taking the time to comment on the G&T paper but I think you are reading to much into it and over-stating your case. Like many other people, you seem to be angry that G&T would question the GHE and AGW.

      You seem to be inferring that the GHE is a metaphor, that everyone knows that, and that it doesn’t matter. They are disagreeing with you, as experts in thermodynamics, and everyone else who infers the GHE is part of physics theory. It’s not. And since AGW is an extended GHE, it is equally fictitious and not based on physics.

      You object right off to their paper’s title, “Falsification of the Atmospheric CO2 Greenhouse Effects”. I think you have to give them a bit of license here because their original paper is printed in German and they both live in Germany. Since they say greenhouse effect(s) rather than greenhouse effect, I think they are lumping the GHE and AGW together. AGW is definitely about anthropogenic CO2.

      Perhaps it’s wrong to do so in English, but I am reading between the lines. I did not have to read between the lines in most of the paper.

      They state their purpose in the Abstract:

      1)that a fictitious mechanism is described in which a planetary atmosphere acts as a heat pump driven by an environment that is radiatively interacting with but radiatively equilibrated to the atmospheric system.

      Note: from wiki…”A heat pump is a device that provides heat energy from a source of heat to a destination called a “heat sink”. Heat pumps are designed to move thermal energy opposite to the direction of spontaneous heat flow by absorbing heat from a cold space and releasing it to a warmer one. A heat pump uses some amount of external power to accomplish the work of transferring energy from the heat source to the heat sink”.

      I have the same problem with the GHE and the AGW which is an extended GHE. No one has ever satisfactorily explained how this system works. There are vague notions of radiation from GHGs that are supposed to act in some way to trap heat or to back-radiate it to the surface.

      G&T are obviously claiming that the fictitious heat pump, which somehow transfers heat from a cooler atmosphere to a warmer surface has been calibrated between atmospheric radiation and surface heat. That’s what they are objecting to later in the paper.

      You have argued that. You have claimed their is a direct relationship between net radiated IR between the atmosphere and surface and transferred heat. Your claim is simply not true in thermodynamics. That’s what G&T are claiming.

      2)according to G&T:

      According to the second law of thermodynamics such a planetary machine can never exist. Nevertheless, in almost all texts of global climatology and in a widespread secondary literature it is taken for granted that such mechanism is real and stands on a firm scientic foundation. In this paper the popular conjecture is analyzed and the underlying physical principles are clarified.

      By showing that:

      (a) there are no common physical laws between the warming phenomenon in glass houses and the fictitious atmospheric greenhouse effects,

      (b) there are no calculations to determine an average surface temperature of a planet,

      (c) the frequently mentioned difference of 33C is a meaningless number calculated wrongly,

      (d) the formulas of cavity radiation are used inappropriately,

      (e) the assumption of a radiative balance is unphysical,

      (f) thermal conductivity and friction must not be set to zero, the atmospheric greenhouse conjecture is falsified

      • Gordon Robertson says:

        @Curt…with the abstract out of the way, you can ignore 1.1 and 1.2. All they are claiming is that a doubling of CO2 will have negligible effect and that the IPCC lives in a world of consensus and not scientific fact.

        Near the end of 1.2 they claim:

        “…in many calculations climatologists perform calculations where idealized black surfaces e.g. representing a CO2 layer and the ground, respectively, radiate against each other. In reality, we must consider a bulk problem, in which at concentrations of 300 ppmv, at normal state, 8 x 10^6 CO2 molecules are distributed within a cube V with edge length 10 microns, a typical wavelength of the relevant infrared radiation.

        In this context an application of the formulas of cavity [blackbody] radiation is sheer nonsense”.

        Note: This is an excellent point, trying to apply blackbody radiation formulae to CO2 molecules which have dimensions of a typical IR wavelength.

        Then the zinger:

        “Global climatologists claim that the Earth’s natural greenhouse effect keeps the Earth 33C warmer than it would be without the trace gases in the atmosphere. About 80 percent of this warming is attributed to water vapor and 20 percent to the 0.03 volume percent CO2. If such an extreme effect existed, it would show up even in a laboratory experiment involving concentrated CO2 as a thermal conductivity anomaly. It would manifest itself as a new kind of `superinsulation’ violating the conventional heat conduction equation. However, for CO2 such anomalous heat transport properties never have been observed.

        Note: claiming such a trivial amount of ACO2 can have a heating effect of up to 25% is just plain ridiculous. In all atmospheric GHGs, which are about 1% of the atmosphere, water vapour accounts for 96%. ACO2, based on a 390 ppmv density makes up about 0.001%. How does that translate into a heating effect of 25%?

        I have seen a lower estimate of 9% but even that is ridiculous.

    • Gordon Robertson says:

      @Curt “No one is arguing that the atmospheric greenhouse effect is due to conductive effects, but instead to radiative effects. The fact that they do not understand this means that they are not even in the game so to speak”.

      There is not one thing in the paper…not one…to indicate what you are saying is true. Both these guys have Ph.Ds in physics and work in the field of thermodynamics. To claim they do not understand what you claim is ludicrous.

      “They repeatedly claim that we have no measurements to confirm the atmospheric greenhouse effect. Baloney! …. We have lots of good measurements of the full spectrum and magnitude of shortwave and longwave radiation, upwelling and downwelling, at various latitudes and altitudes, including above the atmosphere”.

      What you are missing entirely is that no one can prove, using that radiation, that a greenhouse effect or an AGW effect exists. That’s all they are saying. There is no scientific proof that relates atmospheric radiation to a greenhouse effect or to anthropogenic warming.

      People presume that because CO2 is in the atmosphere, and because it has the ability to absorb IR, that it is contributing to catastrophic global warming and climate change.

      No one has proved that. The IPCC, after reading scads of peer reviewed papers cannot claim that, they can only state it is likely. Sorry, but likely in physics holds no water. It might for politicians.

      To show you how messed up the IPCC are with their likelies, after the 2013 review, they admitted there has been a warming hiatus since 1998. Then they followed that up by raising the likelihood that humans are causing global warming.

      Same thing with radiation. There is no demonstrable effect relating any radiation to so-called greenhouse warming.

      Throughout the course of this paper, G&T supply complex math and evidence to back their claims. You have not provided one iota of evidence that a GHE or AGW exists.

      It is your opinion that the GHE exists and that is not good enough.

      It is far more plausible that the oceans have contributed to warming the planet.

      “But we have very reliable measurements that the radiation of the ground (~400W/m2) is more than 50% larger than the incoming solar radiation absorbed by the earth (~240W/m2). This imbalance must be explained. The (poorly named) “atmospheric greenhouse effect” explains it”.

      There is no proof for what you claim, Trenberth-Kiehle admitted that in their radiation budget calculations. Not one of the implied radiation budget figures is measured, it is all implied.

      Since G&T have supplied tons of math to back their claims and you have supplied nothing but the waving of your arms and copious ad homs, I have to conclude they are right and you are wrong.

      Besides, they have tons of experience in thermodynamics and Gerlich teaches advanced math in thermodynamics. Claiming he is full of it when he teaches that to university students is highly illogical.

      • Curt says:

        Gordon:

        You stated, “There is not one thing in the paper…not one…to indicate what you are saying is true” in response to my claim that G&T were trying to refute the non-existent argument that the atmospheric greenhouse effect was a conductivity phenomenon.

        Look on p12, near the bottom. G&T say:

        “Global climatologists claim that the Earth’s natural greenhouse effect keeps the Earth 33C warmer than it would be without the trace gases in the atmosphere…If such an extreme effect existed, it would show up even in a laboratory experiment involving concentrated CO2 as a thermal conductivity anomaly.” (My emphasis.)

        What further evidence do you need that they think this effect is being argued as a conductivity phenomenon? There is also the fact that they spend 5 pages (pp 6-10) on thermal conductivity properties of the atmosphere. These are 5 pages “to indicate what [I am] saying is true.”

        You spend most of the rest of your response completely conflating the two very separate ideas of whether there is a radiative greenhouse effect or not, and whether recent increases in CO2 concentration will significantly increase that effect. These are very distinct issues! G&T go back and forth between the two without acknowledging it, one of the things I find infuriating about their paper.

        You claim of the Kiehl and Trenberth papers that “Not one of the implied radiation budget figures is measured, it is all implied.” Hogwash! They are based on thousands and thousands of measurements, which you would know if you had actually read their papers. You may quibble about how they integrate all of the data, and you could get answers that are a few W/m2 different, but nothing comes remotely close to closing the huge power gap that I mentioned.

        As I pointed out to mpainter below, we actually have better measurements for radiative transfer than for other modes of heat transfer, because it is easy to stick sensors in the way. Getting from there to resulting thermal effects only requires applying very basic thermodynamic analysis to get approximate figures at least.

        • Gordon Robertson says:

          “If such an extreme effect existed, it would show up even in a laboratory experiment involving concentrated CO2 as a thermal conductivity anomaly.” (My emphasis.)”

          I don’t understand why you are having a problem with this.

          G&T stated very clearly that they are talking about a laboratory experiment, not the atmosphere, with reference to conductivity.

          They are claiming that IF an extreme effect such as a rare amount of atmospheric CO2 is claimed to have, could cause as much warming as claimed, it would show up in a laboratory experiment involving CONCENTRATED CO2 as a thermal conductivity anomaly.

          In other words, if CO2 had the thermal effect claimed by alarmists, using a more concentrated amount in a lab would produce thermal effects that no one has ever seen.

          Another way of stating that exposes the CO2 warming thing for what it is. If CO2 can warm the atmosphere using a trace amount then we could fill transparent containers with it, that pass infrared energy, to a much higher concentration than is found in the atmosphere, and leave them sitting around a room. If the theory is correct, the bottled CO2 should absorb infrared from a heat lamp or equivalent, and heat the room on its own.

          We could sit transparent bottles of CO2 on the roof, or the ground, with water pipes running through them, and absorb infrared from the Sun or the ground. Free energy, wonder why no one has thought of it?

          Alternately, you seem to be claiming that if you remove the tiny percent of CO2 that is in air, from a greenhouse, the greenhouse would not warm. The air in a greenhouse is 99% nitrogen and oxygen, yet it warms just fine.

          You don’t think N2 and O2 can do the same in our atmosphere???

          • Curt says:

            Person A: We think that putting small amounts of substance X into the air will reduce radiative heat transfer through the air. This can result in a net warming of a body surrounded by cold ambient.

            Person B: Even large amounts of substance X in the air will not reduce conductive heat transfer through the air.

            Gordon, if you cannot see the problem in Person B’s logic, there is no point in continuing this discussion. (Especially when there can be no conductive heat transfer to space, but there can be radiative heat transfer to space.)

            A common laboratory experiment for kids is to take an infrared imaging camera that images things well in standard air even through a transparent container, then put an identical container with concentrated CO2 in between the camera and the target. The CO2 very obviously blocks the thermal infrared from the target.

            G&T, whom you agree with, state that the (supposed) atmospheric greenhouse effect is very different from how real greenhouses work. I agree! Real greenhouses work by their glass suppressing conductive/convective losses. Why would changing the radiative properties of the gas inside make a difference?

            You say,”You don’t think N2 and O2 can do the same in our atmosphere???”

            No I don’t! And I have a century of spectroscopic data to back me up, all well understood before there was any “climate science”. N2 and O2 are virtually completely transparent to longwave infrared. H2O and CO2 are significantly opaque to longwave infrared.

            We even understand very well why this is. H2O and CO2, as tri-atomic molecules, have bending and stretching vibratory modes with natural frequencies corresponding to longwave infrared. N2 and O2, as diatomic molecules, do not.

        • Gordon Robertson says:

          “You claim of the Kiehl and Trenberth papers that “Not one of the implied radiation budget figures is measured, it is all implied.” Hogwash! They are based on thousands and thousands of measurements, which you would know if you had actually read their papers”.

          I have read the Trenberth-Kiehl radiation budget, in fact, I have it before me.

          http://climateknowledge.org/figures/Rood_Climate_Change_AOSS480_Documents/Kiehl_Trenberth_Radiative_Balance_BAMS_1997.pdf

          At the beginning of the Abstract:

          “The purpose of this paper is to put forward a new estimate, in the context of previous assessments, of the annual global mean energy budget…”

          Note the word ‘estimate’. If you read through the paper they draw on a liberal number of estimates and guesses from climate models.

          Shortly after the abstract:

          “Despite these important improvements in our understanding, a number of key terms in the energy budget remain uncertain, in particular, the net absorbed shortwave and longwave surface fluxes.

          With regard to the radiative energy component, there is also a wide range in the estimates for the contribution of the individual gaseous absorbers to the radiative forcing of the climate system. Using detailed radiation models for the shortwave and longwave spectral regions, we show what role the various absorbers play in determining the radiative balance of the earth’s system and their dependence on the presence or absence of clouds…”

          If you forward to page 10 you will see the famous Trenberth-Kiehl radiation budget diagram. Note the 324 watts of back-radiation. Complete science fiction.

          Now look at the current version (NASA) of the same diagram:

          https://tallbloke.wordpress.com/2010/06/14/apocalypse-recalculated-whatever-happened-to-back-radiation/

          Oops, the back-radiation is gone.

          Curt, how did you ever get sucked into this pseudo-science circus?

          • Curt says:

            I repeat that the K&T figures are based on may thousands of measurements. Now, do they have measurements for every point on earth at every moment in time? Of course not!

            So they must perform calculations to interpolate between the measured points, both in space and time. They do not claim that these calculations are perfect, so they properly call them “estimates”.

            Now the question is, how good are these overall estimates derived from real measurements. Personally, I don’t think they are good enough to resolve any possible changes over time from increased CO2, where we are talking about fractions of a single W/m2.

            But they are easily good enough to resolve the fundamental difference between the incoming solar radiation (~240 W/m2) and the power losses from the surface (~500 W/m2). Even if the overall estimates are off by 20 W/m2, this doesn’t come close to closing the gap.

            You call back radiation “complete science fiction”. Yet your hero Clausius specifically acknowledged it 150 years ago. I have already quoted a relevant passage to you, but I will quote it again:

            “The principle may be more briefly expressed thus: Heat cannot by itself pass from a colder to a warmer body; the words “by itself”, however, here requires explanation. Their meaning will, it is true, be rendered sufficiently clear by the exposition contained in the present memoir, nevertheless it appears desirable to add a few word here in order to leave no doubt as to the signification and comprehensiveness of the principle.

            In the first place, the principle implies that in the immediate interchange of heat between two bodies by conduction and radiation, the warmer body never receives more heat from the colder one than it imparts to it. The principle holds, however, not only for process of this kind, but for all others by which a transmission of heat can be brought about between two bodies of different temperature, amongst which process must be particularly noticed those wherein the interchange of heat is produced by means of one or more bodies which, on changing their condition, either receive heat from a body, or impart heat to other bodies.”

            He specifically and explicitly describes radiative heat transfer as a two-way heat exchange. I agree with Clausius. Your arguments imply that you do not!

            As to the two energy budget diagrams you cite. Can you really not understand the difference between gross flows and the resulting net flow?

            If the purchase of a $5 item involves the purchaser giving the cashier a $20 bill and the cashier giving BACK to the customer $15 in change, if the store simply records the transaction as receiving $5 from the customer, is that inconsistent?

          • mpainter says:

            Gordon, thanks for the link.
            It is the science such as the Kiehl, Trenberth diagram that fosters my skepticism:
            Note the 168 W of insolation vs 324 W of back radiation, which figure is absent from the NASA diagram. Note also the ” absorbed at the surface” claim on their diagram regarding the 324 W of backradiation.
            Thus more energy is absorbed by the surface from back radiation than from insolation, according to their science.
            This aspect of AGW has always pegged my skepticism meter.
            I, for one, do not understand the claims that such figures given for back radiation are by measurement.
            Your link says it all, IMHO, as to the reliability of the AGW science, at least as put forth by the likes of Trenberth and Kiehl.

    • mpainter says:

      Curt,
      I also question your assertion that “we have very reliable measurements” with respect to the radiative emission from the surface. In fact, as Gordon points out, this is all assumed, not measured.

      • Curt says:

        mpainter:

        We do have many good measurements of the radiative emission from the surface, both as to magnitude and spectrum. They agree well with theory. Most surface substances, including the water that covers 70% of the earth’s surface, behave like very high-emissivity graybodies (e~=0.95) so their emissions are very close to what an ideal blackbody would emit at those temperatures.

        Thanks to satellites, we also have many good measurements of this thermal radiation looking down from space. We see significant “bites” taken out of this Planck spectrum at the wavelengths where H2O and CO2 are known to absorb and emit.

        A typical one of these measurements shows the spectrum close to the Planck curve for the surface temperature of ~290K in the “atmospheric window” of the 8-14um band where neither H2O nor CO2 absorbs.

        Below 8um wavelengths, the spectrum is close to the Planck curve for the ~240K temperature above which there is virtually no water vapor. In these wavelengths, H2O absorbs and emits very well, but CO2 not so well.

        In the 14-16um wavelength range, where CO2 is highly absorptive and emissive,, the spectrum is close to the Planck curve for the ~220K temperature that exists high in the atmosphere where it becomes thin enough that most radiation in these bands escapes to space.

        (Yes, it’s more complicated than this, but this is a very good approximation.)

        The area under this curve must integrate to about 240 W/m2 to be in at least approximate balance with the incoming solar radiation power. (Not even the most ardent alarmist thinks the radiative imbalance is greater than about 1.0 W/m2. The interesting arguments are about tenths of a W/m2.)

        A regular Planck curve with no “bites” out of it from atmospheric radiative absorption would have this area under the curve for a temperature of ~255K (-18C). The “three-piece” spectrum actually measured by the satellites comes from a significantly higher temperature (~290K) in the atmospheric window, but lower temperatures in the absorption/emission bands of H2O and CO2.

        The area under these two curves is basically the same, so the earth is close to being in radiative balance with space. But the surface temperature is significantly higher in the case with the absorptive gases. Simply put, this difference is the “radiative greenhouse effect”.

        Oh, and the measurements of downwelling (“back”) radiation show that this comes overwhelmingly in the absorption/emission bands of H2O and CO2, with virtually nothing in the atmospheric window band of 8-14um wavelengths.

        The fact that radiative transfers can go directly through gases and even vacuums actually makes it much easier to get direct measurements than with other modes of transfer, because it is easy to stick sensors right in the path. So the idea promulgated by G&T, and repeated by Gordon, that we don’t have good measurements in this case, is the exact opposite of the truth!

        • mpainter says:

          Curt, thanks for your reply.
          I have seen the curves that you refer to.
          I believe that satellite imagery looking down detects most IR comes from cloud tops. Hence, you are seeing latent energy radiated at height, not surface IR emission. I do not see any way to measure the magnitude of surface emission by satellite. You give W/sq. m derived theoretically. I feel that you have talked around the issue. Specifically, what actual _measurements_ of surface emission have been reliably measured?
          An IR astronomer declared that atmospheric back radiation is about 97% water and 3% CO2. Here is a scientist who is most familiar, in a practical, daily way with the the matter. He said ” I have been to Antarctica 4 times, where precipitable H2O is only .02 mm, and water is still the predominant source of back radiation. No astronomer puts any heed in claims about CO2.”

          • Curt says:

            mpainter:

            Thanks for your questions. The spectrum of radiation measured from the satellite I was referring to was “clear sky”. I should have clarified that at the start.

            It is not the satellites that measure the longwave radiation that leaves the surface, but it is the same instrumentation just a meter or so above the ground.

            The question I have been dealing with is whether there is a “radiative greenhouse effect” at all and whether the “back radiation” from these radiatively absorptive gases can lead to higher surface temperatures that would exist with a transparent atmosphere. Several commenters here vociferously disagree with these propositions. So be it.

            As to the question of how much of this “back radiation” comes from CO2 and how much from H2O, generally it is mostly from CO2. If you look at the spectrum shown in Figure 2 of this paper:

            http://www.aeolus.wsu.edu/vonw/pubs/TownEtAl_2005.pdf

            showing measurements at the South Pole, where there is very little water vapor, the contributions from H2O and CO2 look pretty similar to me.

          • mpainter says:

            Curt, again, thanks for your response and the link. The abstract of your linked study states ” two-thirds of the radiative flux was due to water vapor and one-third to CO2″
            This confirms the IR astronomer whom I quoted, who claimed that back radiation from water predominated over CO2 even in Antarctica, the driest place on earth.

            It seems as though he know the issue very well and that the contribution of CO2 to the radiative flux is very slight, indeed.
            It can be shown in other ways that the role of CO2 is greatly exaggerated, respecting that gas’s contribution to the GHE.

            It bears repeating that an IR astronomer deals in a very practical way with the atmosphere and the IR flux therein. In my estimation, no scientist can be more familiar with the issue.

          • Curt says:

            mpainter:

            I agree that H2O is the dominant greenhouse gas. I have never said otherwise. Gordon was taken aback the other day when I criticized the title of the G&T paper he likes: “Falsification of the CO2 Greenhouse Effect..” because the title (and most of the paper) completely ignore the larger effect of H2O.

            I have emphasized above that the question as to how much added CO2 would create further warming is very different from the question of whether there is a radiative atmospheric “greenhouse effect” at all.

          • mpainter says:

            Curt, thanks for your response.
            At June 1, 12:22am, you wrote:
            “Generally it is mostly from CO2”, referring to back radiation.
            I assume that you misspoke and meant “..it is mostly from H2O” instead.

            Thanks for your input.

          • Curt says:

            Yes, I meant “it is mostly from H2O”. Thanks for pointing that out. (It was late…)

        • Gordon Robertson says:

          “We see significant “bites” taken out of this Planck spectrum at the wavelengths where H2O and CO2 are known to absorb and emit”.

          That’s interesting, considering that the H20 and CO2 spectrums overlap. And how do you extract the ACO2 spectrum?

          Roy has already done an article on that. They were using isotopes of carbon to distinguish CO2 from ACO2 but according to Roy’s article the isotope argument holds no water.

          I am afraid our understanding of the IR emissions from the surface are still in the dark ages. Trenberth-Kiehle admit as much.

          • Curt says:

            There are significant wavelength bands where the H2O and CO2 absorption spectra do not overlap. Also, the concentration of H2O in the atmosphere varies widely, and we can see the differences in the spectrum observed by the satellites looking down.

            So we can do a good (but not perfect) job of separating out the effects.

            I never claimed to be able to extract the ACO2 effect. The fact that fossil fuel CO2 emissions are very slightly lower in 13C and devoid of 14C is not relevant here.

            These spectral measurements are not yet good enough to discern the effect of added CO2 in the atmosphere. They are easily good enough to demonstrate conclusively that there is an “atmospheric greenhouse effect” (and yes, that is an imperfect metaphor).

  28. MikeB says:

    Curt,

    It’s nice to hear from someone on this blog who actually understands something. Well done. Of course, you are still outnumbered by the relentless ignoratii, but still refreshing.

    • Gordon Robertson says:

      @Mike “Of course, you are still outnumbered by the relentless ignoratii, but still refreshing”.

      That would be Mr. Ignoratii to you, Mike. šŸ™‚

    • mpainter says:

      Such a nice fellow you are. I found this exchange of Curt, Gordon, Norman, tonyM, German, Massimo, and others as intensely interesting. But you characterize this discussion as one by “ignorati”.
      But perhaps you now regret using such an expression.

      • mpainter says:

        My comment intended for MikeB.

      • mpainter says:

        Correction:
        German=Geran
        !#*#!!* spell checker

      • Massimo PORZIO says:

        If I have to respond to MikeB, I would ask him what he means with ignoratii.

        I’m Italian and even if the Latin is no longer in use by years here, my language directly derives by it.
        So, “ignoratii” is the plural of “things or people which/who are ignored”.

        If MikeB instead intended to say that I am (amongst others) an ignorant in this matter, I fully agree with him.
        Yes only one who is aware of his ignorance and is not an
        apathetic by nature, asks questions in the field and is sceptic.

        Anyway, ignorant in Latin is “bardus”

        Have a great day.

        Massimo

        • Gordon Robertson says:

          @Massimo “Anyway, ignorant in Latin is “bardus””

          Is the plural bardii? Since a poet in English is a bard, that could be a bunch of poets. šŸ™‚

          • Massimo PORZIO says:

            Hi Gordon,
            please don’t believe the I know the Latin language. I just know a veeeery little, just because my sister studied it at the time of college and I remember her when she had to learn all the forms of verbs, nouns, articles and adjectives. Latin was a very complex and redundant language for me, to be honest to much redundant for me indeed.

            Anyway looking to the Latin dictionary on the shelf, the plural of “bardus” is “bardi”.
            But asking my sister it seems that in the case used by MikeB it is more correct to use “imperitus”, which plural is “imperiti”. My sister said that “bardus” is most like “dumb” not ignorant.
            About the English “bard” etymology, AFIK it come from the ancient Celtic language, not from Latin. So it has a complete different meaning (almost the contrary indeed šŸ™‚ )

            Have a great day.

            Massimo

          • Massimo PORZIO says:

            By the way Gordon,
            if you wonder about a word which comes from two different languages with almost the opposite meaning, then you must know that one of the most known Latin proverb today could have exactly the reverse meaning that it had at the times of Roman Empire!
            I’m arguing about “verba volant scripta manent”.
            If you ask to a latinist, he probable tell you that the meaning is: “spoken words fly away, written words remain”, even Wikipedia reports it.
            Well, it’s true that the proverb come from a speech of Caius Titus of the Roman Senate, what some researchers still are questioning is what Caius Titus wanted really say with it.
            In ancient time the people who was alphabetized were very few. So, today always more Latinists support the point of view that Caius Titus was intending to say “if you write it but you don’t speak it to the people, what you write probably remains unknown to the most of people”.

            That’s fascinating to me. The same phrase from the same language, not only could have changed its meaning along time, but now it has a meaning which is exactly the contrary of the one it had in ancient times!!!
            Wow!!!

            Have a nice day.

            Massimo

  29. Norman says:

    Gordon Robertson

    http://ocw.mit.edu/courses/aeronautics-and-astronautics/16-050-thermal-energy-fall-2002/lecture-notes/10_part3.pdf

    On page HT-63 They describe radiative heat transfer between two objects, one warmer than the other.

    Heat transfer is described as a net transfer. Bot objects radiate energy toward each other (which is energy regardless if you believe in wave or particle description, quantum physics says light and electrons are both wave or particle and it depends upon how the experiment is set up). The heat transfer is the net radiation flow, the heat will flow from the hot object to the cold object because more radiation is flowing from it to the cold object than from the cold object to the hot one.

    You may also realize if you have two equal temperature emitters in space near each other, even though a lot of energy is moving between them, no heat is being transferred.

    I also am not sure where you get the idea that water vapor or carbon dioxide that absorb IR given off by Earth’s surface will radiate at a lower energy level. I have not seen anything in physics or chemistry to verify this statement by you. Gaseous molecules emit radiation only in discrete quantum of radiation regardless of their temperature (unless you get real hot, like plasma temperature then the molecule no longer exists). Carbon dioxide does not absorb a 15 micron IR photon and then emit a longer wavelength. It re-emits the same photon as it absorbed. It does not act like a surface.

    I also do not know why you are so into the amount of carbon dioxide present. Satellite IR spectrum do indicate their is enough CO2 present to block a large amount of IR given off from the Earth’s surface from leaving to outer space (where the satellite is reading from).

    • mpainter says:

      Norman, my understanding is that CO2 does not emit the absorbed photon but transmits the absorbed energy kinetically to other air molecules. Do you hold that this is incorrect? Your views on this, I pray.

      • Norman says:

        mpainter

        I am not sure what the actual mechanisms are at the molecular level. I do not know if the vibrational energy of the excited Carbon Dioxide bond will transfer energy to an oxygen or nitrogen molecule or re-emit it as an IR photon. I think the physics there is very complex and I am not sure how to verify with an experiment.

        The convincing point for me is the IR spectrum taken from outer space. The Earth is emitting as close to a blackbody at the surface temperature. You can generate the suspected curve of energy. When you superimpose the satellite actual data there is missing radiation from the bands of carbon dioxide and water vapor.

        • mpainter says:

          Norman, thanks for your reply. My understanding is that this kinetic energy transfer from radiative gas–> air molecule (N2,O2) is what determines ambient air temperature and that it occurs too rapidly for the CO2 molecule to be able to emit a photon.
          I suspect that this is another of those indeterminables of climate science.

        • fonzarelli says:

          Norman and mpainter, the late ernst beck also held the view that the absorbed photons are turned into kinetic energy (and not re-emitted). If this is the case would not the satellite data look the same with the same bands turning up missing? Is there any evidence that energy at the 15 band is being re-emitted toward the surface (as opposed to the ir detected by the satellites in outer space)? I’d really appreciate some input here so thank you (both) in advance…

          I must say that this has been a very enlightening thread with valued input from everyone. Norman, i especially appreciate your clarity. I also think that gordon raises some good challenges to supporters of the greenhouse effect which should be addressed…

          mpainter, i just turn my (‘scuse my frenchin’) spellcheck OFF and correct the misspellings manually. The last thing that i ever care to do is apologize to appell for ANYTHING!!! (not even for calling him “apple”…)

          OH & MASSIMO, remember these are mostly americans here at this blog and as such don’t know how to ‘take it easy’ !!! I may be the anomaly here because i had grandpop and grandmom to teach me. (I think americans who don’t have close relatives from the old country are really missing out on something special…)

          • Norman says:

            fonzarelli

            I could be wrong here, it is just a thought. If the IR energy absorbed by GHG (water vapor and carbon dioxide) were mostly converted to heat by exchanging energy with oxygen and nitrogen molecules it would seem the lapse rate should be even less than observed. Lapse rate is stated to be caused by a parcel of air expanding as it rises and cooling as internal energy is used up in the expansion against the surrounding air. With water vapor present in the parcel the lapse rate is much less than dry rate since the water vapor condenses as it cools releasing energy and making the lapse rate less steep. If IR was converted to heat directly in the atmosphere by process you described it would seem this energy would make and even less lapse rate.

            I am not certain what actually happens to the absorbed photons. Tim Folkerts (on another thread) explained that even if CO2 exchanged its energy with non-radiative molecules then these same non-radiative molecules would return this energy when they collided with the CO2 negating any direction in such interactions and leaving the effect the same. Missing IR as seen from outer space.

            On Roy Spencer’s earlier thread about how GHG keep the Earth surface from freezing at night, he provides a model to play with. If there was no back-radiation it would seem the surface would cool very rapidly unless you could provide a good source of energy return to slow the cooling.

          • mpainter says:

            Fonzarelli, good to hear from you. There seems to be no definite answer, only theory.
            CO2 will re-emit the photon if the atmosphere is rarefied and rate of collusions much reduced, apparently. But how do you get 15 band IR to that height under saturation condition? And so forth puzzle, puzzle.
            My question: how is the shifted energy (from absorbent bands to kinetic energy) eventually radiated? What are the thermodynamics of this?

          • mpainter says:

            Fonzarelli,
            You say : “Is there any evidence that energy at the 15 band is being re-emitted towards the surface?”

            A very good point. If this band is not re-emitted in the troposphere, wherefore the back radiation from CO2? Yet this is seen spectroscopically, …?

    • Gordon Robertson says:

      @norman

      “The heat transfer is the net radiation flow, the heat will flow from the hot object to the cold object because more radiation is flowing from it to the cold object than from the cold object to the hot one.

      Norman…they are talking about blackbody radiators. Blackbody radiators are theoretical and they can only be partially represented by very hot sources.

      Furthermore, blackbody radiation is about electromagnetic energy, not heat. On the page to which you referred me, they are talking about EM transfer.

      Another thing you need to understand is that the bodies to which they refer are independent heat sources which are extremely hot. In the surface-atmosphere model that is not true. The surface is the independent heat sources which supplies heat via radiative transfer to the cooler, dependent GHGs in the atmosphere.

      The paper is written for engineers, and although my background is in engineering, no engineer studies thermodynamics that deeply. There’s no need for a deep study, engineers are concerned mainly with external applications of heat that can be studied without getting into atomic level theory.

      I got the basics but I have read deeply on what Clausius had to say. He describes the process precisely as to how he developed his theories on heat and how he developed the 2nd law and the theory of entropy.

      The MIT paper has skimmed over the topic of radiative transfer. There is not enough information there for a professional engineer to apply it to a radiative process.

      Norman, no insult intended, but I encourage you to sit down and visualize the difference between electromagnetic energy and thermal energy. When you talk about net transfer of energy, are you referring to EM transfer or thermal transfer? There are two processes taking place in a model like the surface – atmosphere.

      Heat in a body like the surface is represented by the kinetic energy in it’s atoms/molecules. If you heat the body more the KE rises and if you cool it, the KE is lowered.

      That KE IS NOT EM or IR.

      The KE is a property of the moving/vibrating atoms/molecules. The EM comes from the electrons in the atoms or the bonds in the molecules.

      Heat can only be transferred between bodies by EM, however. Hotter bodies emit higher intensity photons from their orbital shells and from their molecular bonds (in some molecules).

      That energy is NOT heat!!!

      If EM, or IR, travels through space it has no heat related to it. EM does NOT have heat. Neither does it have colour. It is a broad-spectrum radiation with no properties related to heat. It can do no work, but heat can.

      If EM comes in contact with matter, it can cause the atoms or molecules in the matter to produce heat by absorption into an atoms electrons or a molecular bond. Absorption into an electron causes the electron to rise to a higher energy level, increasing the KE in the atom. Or, the EM can be absorbed into a molecule’s covalent bond, raising the KE.

      Heat does not travel through the space between separated bodies. All the heating and cooling is done locally. The hotter body can only get cooler (if it’s heat loss in not replenished) and the cooler body can only get warmer (till it reaches thermal equilibrium with the heating body).

      There is no NET heat transfer. Heat is transferred in one direction only. There is net EM transfer, but if an electron in an atom is residing at a higher energy orbital in a hotter body, there is no reason why IR from a cooler body should cause it to rise to an even higher orbital. Therefore, the hotter body cannot increase it KE or it’s heat.

      That’s my story, and I’m sticking to it.

  30. Norman says:

    geran

    Your post from above: “Norman, okay, you want specifics of your rambling nonsense:

    First you claim the Earth is NOT warming–
    “This would not warm the surface as more energy is leaving the system than is returning.”

    Then you claim the Earth IS warming’–
    “The effect is because the Earth is constantly receiving solar energy that the surface will be warmer with backradiation present than if there was none (only a relative state between two conditions.

    Then you claim the Earth is NOT warming–
    “But the backradiation is not warming the surface.”

    And all in the same paragraph!

    You have your own debate going with yourself!!

    Hilarious.”

    You really are having problems linking thoughts together. It may seem confusing to you but it is not in my lack of explanation, it is your inability to understand this, maybe that is why it is “hilarious” to you.

    Again. The back-radiation is not warming the surface. The energy from back-radiation comes from the surface and is returning, (some not all). If you had 100% return of energy to its source it would not warm the source, it would keep the temperature the same. If the source has energy added to it from somewhere and you return 100% of the energy it will be warmer than it was before energy was added.

    I think you have a problem understanding the Earth-Sun system. The Earth’s system is not a closed or isolated surface. If the Earth surface had no input of energy (Sun is gone) even with back-radiation it would cool and continue to cool. Does this make sense to you or is it rambling in your mind. I really do not know how to simplify it so you might have a slight chance of comprehending what is written (but one can only understand things if they want to).

    The solar energy is what is warming the surface. If you slow the rate of cooling for any reason but keep adding energy the surface will be warmer than it would be without this addition. The item slowing the cooling is not warming the surface, but with this in place the surface will become warmer than if not put in place if the surface has an input of energy. It really is simple, not confusing and most can understand the point. You do not because you do not want to.

    • geran says:

      Norman:

      An emitter emitting peak energy at 15µm wavelength would represent a temperature of about 193 K. (That’s -80 ŗC, -112 ŗF)

      And you believe 193 K is going to “keep the temperature the same”?

      (No wonder Lukers believe they can heat their homes with ice and mirrors!)

      • Curt says:

        geran:

        Can’t you get anything right?

        Wien’s displacement law allows you to calculate the peak wavelength for blackbody radiation given the temperature. CO2 does not emit remotely like a blackbody. (Ironically there are those in this thread who complain that mainstream climate science incorrectly treats CO2 like a blackbody.)

        I have asked you 5 times already in this thread what happens to 15um radiation from a colder body when it reaches a blackbody that is at a higher temperature.

        The next question you have been evading is what happens to 15um radiation from a body warmer than the blackbody when it reaches the same blackbody.

        And finally, if your answers are different to the two questions, what physical mechanism explains the difference?

        You have been dodging the questions for a week now…

        • geran says:

          Curt, Curt, Curt. You are so obsessed. I should not respond because I know you are not a scientist and you have no credibility. But, I can’t help myself because you provide so much hilarity.

          Curt: Can’t you get anything right?
          >>>>>Do you mean after I perceive your obsession?

          Curt: Wien’s displacement law allows you to calculate the peak wavelength for blackbody radiation given the temperature. CO2 does not emit remotely like a blackbody.
          >>>>>So, what does CO2 emit remotely like? (LMAO)

          Curt: I have asked you 5 times already in this thread what happens to 15um radiation from a colder body when it reaches a blackbody that is at a higher temperature.
          >>>>>No, you have not asked that 5 times. You have been ingenuous numerous times. You have misrepresented what I said numerous times. Yet, you try to claim the “moral high ground”. That’s why you won the “blue dress” award.

          Curt: The next question you have been evading is what happens to 15um radiation from a body warmer than the blackbody when it reaches the same blackbody.
          >>>>>> No, I have not been evading this question. You have deceived yourself again.

          Curt: And finally, if your answers are different to the two questions, what physical mechanism explains the difference?

          You have been dodging the questions for a week now…

          >>>>>>What are the “two questions”?

          Curt, not to imply that you have been evading MY questions, but is “Norman” your brother?

          (See. I only have ONE question for you. It should be an easy “yes” or “no”.)

          • Norman says:

            geran,

            No Curt is not my brother. I have interacted with him in reading his posts. He does seem to possess a much better physics education than you seem to possess. Your attitude seems to be one of attacking things that you choose not to understand and pat yourself on the back for some perceived clever behavior that seems to only amuse yourself. Not a real productive form of discussion. I would much rather be proven wrong with some good link based upon some established physics or some experiment you performed to show the error in my thinking. This type of behavior is much more beneficial than thinking so highly of your own opinion that you can not hear anything but your own mind.

          • geran says:

            “Yes” or “No” was the answer…from Curt, but YOU answered.

            You chose a paragraph of denigration.

            It’s kinda informative, huh?

            (Are you two sisters then?)

          • Curt says:

            geran:

            A whole lot of bluster to try to obscure the fact that you have NOT, once again, answered my questions.

            I think it’s because you cannot! Prove me wronhg!

          • Norman says:

            geran,

            As the saying goes “You are a legend in your own mind”. You are about as productive to useful conversation as the Bot Doug Cotton.

            I think you post to be clever and amuse yourself. Sad you waste your intellect on this High School level interaction.

            “Sisters” how long did that one make you laugh. A true stroke of genius I think. You ought to join Saturday Night Live and show your just how clever you are and make a good living at it as well. I think you would enjoy that more than trying to determine the truth of Climate Change science.

          • geran says:

            Curt says: “Prove me wronhg [sic]!”
            +++++

            When I think Curt can not get any funnier, he just continues to amaze.

            And, I see Curt has not been able to find an answer to my question.

            Curt: Wien’s displacement law allows you to calculate the peak wavelength for blackbody radiation given the temperature. CO2 does not emit remotely like a blackbody.
            >>>>>So, what does “CO2 emit remotely like”?

          • Curt says:

            Sigh … once again to educate geran in the basics.

            “So, what does ‘CO2 emit remotely like”?

            If you have to ask this, you should not be involved in this discussion. But I will humor you.

            You can find plenty of plots of the absorptivity profile of CO2 that look like the one in Figure 4 here:

            http://www.bom.gov.au/info/climate/change/gallery/4.shtml

            This was well understood long before any climate change frenzy.

            A blackbody would be solid at 100% all along the spectrum.

            Remembering that emissivity equals absorptivity at any and all wavelengths (as per Kirchhoff) this is also a plot of the emissivity profile. To get the emission profile of CO2 at a given temperature, this profile is multiplied by the Planck (blackbody) curve for that temperature.

            While Wien’s Displacement Law gives the wavelength of peak emission of the Planck curve for any given temperature, it does not give the wavelength of peak emission of a substance like CO2 that “does not remotely emit like a blackbody”.

            CO2 has high emissivity in the 14-16 um band, but virtually zero emissivity in the 5-14 um band, and above 16 um. So, the peak emissions from CO2 will be at about 15 um at temperatures from over 300K (when blackbody peak is less than 10um) all the way down to absolute zero.

            So for you to say that because CO2 is radiating with a peak at 15 um, it “would represent a temperature of about 193 K” is completely wrong, and just goes to show that you don’t understand the very basic concepts of radiative heat transfer.

          • geran says:

            Curt, once AGAIN, you misrepresent my words! (See why you are a “pseudoscientist”? A scientist does NOT have to misrepresent what someone else said.)

            Here’s what you said: “So for you to say that because CO2 is radiating with a peak at 15 um, it “would represent a temperature of about 193 K” is completely wrong, and just goes to show that you don’t understand the very basic concepts of radiative heat transfer.”

            Here’s what I ACTUALLY said: “An emitter emitting peak energy at 15µm wavelength would represent a temperature of about 193 K. (That’s -80 ŗC, -112 ŗF)”

            I was not referring to a CO2 molecule. I was referring to an emitter that was close enough to a black body that Wien’s Law would apply. You just did not get it.

            But, if I had been talking about a CO2 molecule, it would be even funnier. And you do not know enough to figure it out. (Hint: What is the energy of a 15 µm photon emitted by a CO2 molecule?)

          • Curt says:

            geran:

            Weasel words! The whole conversation has been about the 15 um radiation from CO2.

            To answer your question, the energy of a 15 um photon emitted by CO2 (at any temperature), or by any other substance, is 1.324 x 10^-20 Joules.

            Now, I’ve answered all of your remotely relevant questions, but you have evaded my questions for a week now:

            ******************************************
            There is a body (let’s call it a blackbody for simplicity) of temperature T0 (we can use 288K as an example, but it doesn’t matter).

            There is a second body of temperature T1, where T1 > T0 (we can use 298K as an example, but again, it doesn’t matter as long as it is greater), emitting radiation toward the body that is at lower T0.

            There is a third body of temperature T2, where T2 < T1 (we can use 278K as an example, but once again, it doesn't matter as long as it is less), also emitting radiation toward the body that is at higher T0.

            Question 1: What happens to the radiation of a certain wavelength (we can use 15 um as an example, but it doesn't matter) from the body at T1 when it hits the blackbody at lower T0?

            Question 2: What happens to the radiation of this same wavelength from the body at T2 when it hits the blackbody at higher T0?

            Question 3: If the answer to questions 1 and 2 is different, what physical mechanism explains the difference?
            **************************************

          • geran says:

            Norman says: “Both sides seem extreme and have lost the ability to discuss the science and be humble enough to realize you do not know it all and use this blog as a source of learning rather than projecting some superior attitude.”
            +++++++

            Norman, maybe projecting your superior attitude, judging others, and lacking humility cause you to lose your ability to discuss science.

          • geran says:

            To answer your question, the energy of a 15 um photon emitted by CO2 (at any temperature), or by any other substance, is 1.324 x 10^-20 Joules.
            ++++++++

            Bravo Curt, you got it right! You found the right equation, and did the calculation correctly.

            The point is that a photon is identified by three values–frequency, wavelength, and energy. If any ONE of the three values is known, the other two can be calculated. This is true for EVERY photon out there.

            And, guess what, those values determine whether a photon will be absorbed or reflected by a surface. Just because a 15 µm photon strikes a surface, it does not mean it will be absorbed and raise the temperature of the surface.

            Science is NOT about things being the way you want them to be.

            Also Curt, I already answered your questions. You just did not get the answers “the way you want them to be”.

          • Curt says:

            geran:

            No, you have NOT answered these questions. (Show me where you have answered them!) Like an evasive politician, you have answered questions different from what was asked, to try to get out of an awkward corner. That is not scientific discussion.

        • Gordon Robertson says:

          @Curt…”Ironically there are those in this thread who complain that mainstream climate science incorrectly treats CO2 like a blackbody”

          The heat-related laws of Planck, Kircheoff, and Boltzmann are all for blackbody calculations.

          Applying Kircheoff’s emission/absorption equations to general heat calculations is just plain wrong. They apply only to radiation between idealized blackbodies running at very high temperatures.

          • Curt says:

            Gordon:

            I’m sorry, but you have absolutely no idea what you are talking about, and so you get this exactly backwards.

            The whole point of Kirchhoff’s emission/absorption equations is to be able to deal with non-blackbodies.

            For example, if a body has an absorptivity of 0.33 at a certain wavelength (and angle) it will absorb 1/3 of the radiation at that wavelength (and angle), whereas a blackbody would absorb all of it.

            Kirchoff’s Law says that the emissivity must equal the absorptivity and any and all wavelengths (and angles), so this same body would emit 1/3 of the radiation at that wavelength and angle that a blackbody would at the same temperature.

            Abd this is for all temperatures, not just “very high temperatures”.

            The fact that you get this basic topic, which is introduced very early in any EM physics or engineering heat transfer course, completely wrong just shows that you do not begin to have the technical background to discuss these issues properly.

    • Slipstick says:

      “If you had 100% return of energy to its source it would not warm the source, it would keep the temperature the same.”
      That statement is 100% incorrect.

      • Norman says:

        Slipstick,

        The statement is made with the assumption the source is a truly closed and isolated system. No energy in no energy out. How would the temperature change in this situation?

        • Slipstick says:

          If there is no energy in or out, what is causing the source to emit the energy that is then returned to the source? Consider the following: If you were to place a filament and a huge supply battery connected to the filament in an evacuated box with a perfectly reflecting, and therefore perfectly insulating, interior surface (your conditions rendered physically), what do you think would happen?

  31. Massimo PORZIO says:

    Hi there,
    in a post above I wrote “My optical spectrum analyser has a resolution of a tenth of nanometre and reducing the power until the filament was practically emitting only in the NIR band, the BB emission looked as it just reduced in power not in colour temperature. Any ideas about it?”
    This night I did it again and instead of what I wrote there, I well found a relation between the maximum radiation and the power flux reported by the spectrum analyser which matches Wien displacement law. I probably reminded the behaviour of LEDs which is a complete different emitter.

    Very sorry, next time that I try to remember things of some years ago I should check it before write silly things like that.

    Have a nice day.

    Massimo

    • geran says:

      Massimo, confirming Wien’s displacement law is valid experimental science. Congratulations!

      Now, if you could only teach the scientific method to the “Lukewarmers”.

      Se avete mai avuto la possibilitą di comprarmi un bicchiere di vino , io ti devo un bicchiere di vino!

      • Massimo PORZIO says:

        Hi geran,
        my God, here is 1AM and I’m still here.

        It was a long time that I didn’t use the optical spectrum analyser, so after the above measurements I decided to check its integrity and the time ran away faster than I thought!
        Luckily tomorrow here in Italy is an holiday, so I don’t have to go to office early in the morning.

        Yes, Wien displacement law holds of course, it was the very reason I was dubious of the stupid things I wrote in my former post.

        Anyway as I wrote to Gordon, if I have to define myself about the CO2 back-radiation, I would call myself a lukewarmer at this point.

        That is I don’t know what really happen to an heat emitting body when a photon of the very same WL returns to its surface, that is I don’t know if it is absorbed by the body structure or it just don’t allow other same WL photons to be emitted by the same structure.
        For what I read, I suppose that it could increase the temperature of the body, if and only if the body receives a continuous flux of power from an external source.
        Sorry, I know that you are probably sure that it can’t happen, and I’m sure that you are much more entitled than me to discuss this field, but being an EE at this point I don’t have any other explanation other than the back-radiation (or maybe back-reflection), to justify the warming of the filament in Curt’s experiment. Since he read a reduction in supply current, my question is: what warmed the filament if not the returning photons?

        Even if the bulb was filled with an inert gas (probably argon), since its pressure didn’t changed other than for the tiny change due to the change in temperature as established by the gas law (not vice versa), what could have increased its temperature if the electrical input power reduced?

        Because the fact that the filament reduced its resistance, it is an unequivocal sign of that for me.

        Waiting for your opinion, now I go to sleep, it’s 1:40AM here.

        Have a nice evening and night.

        Massimo

        • geran says:

          “Anyway as I wrote to Gordon, if I have to define myself about the CO2 back-radiation, I would call myself a lukewarmer at this point.”
          >>>>>>>>
          Massimo, you don’t want to go there. Get out the physics books. “Back-radiation” is a failed “science”. Even the IPCC has moved on to “radiative forcing”, which is the new “failed science”. You are a EE, so you will figure it out….

          “That is I don’t know what really happen to an heat emitting body when a photon of the very same WL returns to its surface, that is I don’t know if it is absorbed by the body structure or it just don’t allow other same WL photons to be emitted by the same structure.”
          >>>>>>>>>
          As you suspect, each photon gets to respond to a surface as it occurs. But, at the “macro” level, arriving photons are reflected if their wavelength is too long. (It’s a problem EEs face daily.)

          Even if the bulb was filled with an inert gas (probably argon), since its pressure didn’t changed other than for the tiny change due to the change in temperature as established by the gas law (not vice versa), what could have increased its temperature if the electrical input power reduced?

          Because the fact that the filament reduced its resistance, it is an unequivocal sign of that for me.
          >>>>>>>>>
          No, you always follow the energy. The resistance of an incandescent filament does not really change. The filament, once heated sufficiently, begins emitting E/M energy. As the temp increases, the current drops. At equilibrium temp, the current stabilizes. The resistance of the filament appears to change, but that is caused by E/M emissions, not resistance.

          • Curt says:

            Another spectacular own goal from geran!

            “The filament, once heated sufficiently, begins emitting E/M energy.”

            The filament is always emitting E/M energy, even at room temperature. Basic, basic stuff! And you get it completely wrong!

            The resistance is a function of the temperature.

            The temperature is a function of the overall energy balance, including the temperature-related emissions.

            The kind of stuff you learn in an introductory thermo class — I recommend you take one some day.

          • geran says:

            Well, if you want to be excessively pedantic Curt, then what I said was more correct than what you said. (But, that’s how it always is, huh?)

            A filament at ZERO K would NOT be emitting, until it was heated.

            Funny.

          • Curt says:

            Except you can’t actually reach absolute zero (another unforced error!)…

          • geran says:

            Curt, with all your pseudoscience, you reached ZERO a long time ago.

          • Massimo PORZIO says:

            Hi geran (and Curt),

            I don’t know what IPCC states, because for what I read about it, there is little science and more political agenda in their work. So having big trouble in reading the English language I should spent many time trying to understand things that (at this point) seems just a complete waste of time for me.

            You wrote “As you suspect, each photon gets to respond to a surface as it occurs. But, at the “macro” level, arriving photons are reflected if their wavelength is too long. (It’s a problem EEs face daily.)”

            I know that when and object face one other object at almost the same temperature the photons couldn’t enter the structure of the other object. Somewhere in the Internet there is a nice photo of a guy leaning against a wall of plaster looked through a FLIR camera. In that photo it is clearly visible his specular thermal image reflected on the wall.
            Which it means for me that his 36/37°C IR emission never entered the colder wall, but that’s not because of the difference in temperature between the emitting body and the receiving body, it is because of the roughness of the surface of the receiving body, which at that WL works as a mirror does at the shorter WL of the visible radiations.
            I never found a proof that a photon of the very same WL of the one emitted by the surface, really can’t avoid the exit of other photons from that surface.

            “No, you always follow the energy.”
            The only known balance we should follow is the energy flow at the TOA, don’t you agree?

            “The resistance of an incandescent filament does not really change. The filament, once heated sufficiently, begins emitting E/M energy. As the temp increases, the current drops. At equilibrium temp, the current stabilizes.”
            Except for the initial radiation due to the environment temperature I agree with you here.
            I write the caveat because being also an Ham-radio (IK1IZA), I well know that for example the noise produced by an attenuator at the front-end of a radio receiver depends exactly on Boltzmann’s constant and the resistors temperature in that case.

            “The resistance of the filament appears to change, but that is caused by E/M emissions, not resistance.”
            Here I disagree instead, because anyway the current fallen down in Curt’s experiment and for that the input power to the system fallen down too, and the only reason I see for that it is that the filament received a flux of energy from
            somewhere else or it is that the outgoing flux reduced.
            I both cases I currently don’t have any explanation other than the back-radiation for that.

            Of course, as always said, I could be stupidly wrong.

            Have a great day.

            Massimo

          • geran says:

            geran said: “The resistance of the filament appears to change, but that is caused by E/M emissions, not resistance.”

            Massimo said: “Here I disagree…”

            Massimo, we may both be right, depending on the wording. Here’s a more detailed explanation of what I was referring to–A 60 Watt incandescent bulb has a “cold” resistance of about 18 Ohms. Within about 5 seconds of turning the bulb on, the current stabilizes at about 0.5 Amps. (120 Volts X 0.5 Amps = 60 Watts) So, the “calculated” resistance is then 120/0.5 = 240 Ohms.

            It “appears” as if the resistance has increased from 18 to 240 Ohms. But, in reality, the power distribution is more like: (0.5)^2 * 18 = 4.5 Watts, and 55.5 Watts going to visible light and high energy IR. The IR being “loss” causes the resulting efficiency is very low, maybe as low as 5-10%.

            So, see, depending on the phraseology, we are both correct!

          • Curt says:

            Massimo:

            It is indeed a real change in electrical resistance as a function of temperature. You would get the same voltage-to-current ratio — the definition of electrical resistance — if the high temperature of the filament was due to it being put into an oven while unpowered (then measured with a very small voltage applied by an ohmmeter) as when the high temperature was due to large amounts of electrical current through the filament.

            At the moment you turn on the “cold” light bulb, you have a very high power dissipation for a short period. In this example using North American voltages, the V^2/R electrical power that is thermalized at the instant power is applied is 120 * 120 / 18 = 800 watts. (Yes, this is real. Engineers often use light bulbs as “dummy loads” to see if a power circuit can properly supply “inrush current”.)

            With a very high power input, and low power output (both radiative and conductive), the internal energy, and so the temperature, of the filament increase rapidly. As the temperature increases, the resistance also increases. Also, both the radiative and conductive transfers increase with higher temperature.

            The combination of reducing power input and increasing power output lead to very quick steady state temperature conditions. For common incandescent bulbs in typical operating systems, the operating electrical resistance of the filament is 12 – 15 times the room temperature resistance.

            So in the common steady state conditions, we have about a 240-ohm resistance, and so about a 0.5 amp current and about 60 watts of electrical power input to the filament. For it to be steady-state, the filament must output 60 watts thermally. At these standard conditions, about 5/6 of the output is radiative, so about 50 watts of radiative power. Then about 10 watts is passed conductively to the argon gas, then on to the glass bulb (not much actual convection inside the bulb), and then conductively/convectively to ambient.

            Of the radiative output, 90 – 95% has too long a wavelength for our eyes to detect, so the “luminous efficacy”, as it is formally called, is in the 5 – 10% range. But the limitations of our eyes have nothing to do with the thermodynamic processes in the bulb.

          • Massimo PORZIO says:

            Hi geran,

            “It “appears” as if the resistance has increased from 18 to 240 Ohms. But, in reality, the power distribution is more like: (0.5)^2 * 18 = 4.5 Watts, and 55.5 Watts going to visible light and high energy IR. The IR being “loss” causes the resulting efficiency is very low, maybe as low as 5-10%.”

            Ok, on this I fully agree with you, but is an aspect which I didn’t care in my previous argumentation about Curt’s experiment.
            What have convinced me that a back-radiated photon can warm the surface which previously emitted it, is the fact measured that avoiding lots of the outgoing radiation from the bulb, its filament unequivocally warmed, that because at the same voltage applied it absorbed less current as its resistance was higher than it should be for that input power.

            If Curt’ had supplied a constant current instead of a constant voltage, then the filament probably stabilized at an even more higher temperature.

            Do you get my point?

            Anyway, in my opinion the work done to keep the molecules in the sky is the real culprit of the 33 K difference between the theorized and the measured temperature at ground.
            My question is: what is the phenomenon which does the work of keeping the molecule high in the sky?
            At sea level in a sunny day the pressure is 1kg/m^2. Where come from the energy which throw away from the surface all those “heavy” molecules?
            And since since any work done against gravity finally it is returned as heat to the ground, we have an increase in temperature which is proportional to the original temperature without the atmosphere.
            Because it is true that when any single molecule is fired up, the temperature of the ground decrease a little, but it is also true that since there is still a continuous flux of SWIR coming from the Sun, when that molecule come back releasing its KE to the ground it warm a little more the surface which was already returned at the equilibrium temperature because of that continuous flux.

            Of course this is all fantasy of mine, I could be plain wrong, but I would someone explain me the difference between the back-radiation warming and the process to keep the molecule high in the sky against the gravity.
            If that work really doesn’t exist I see a perpetual motion in the atmosphere.

            Thank you for the patience of reading my posts.

            Have a great day.

            Massimo

          • Massimo PORZIO says:

            Hi Curt,

            Yes, I fully agree to your last message in this thread.
            I couldn’t say it better.

            Thank you very much.

            Have a nice day.

            Massimo

  32. Norman says:

    An actual experiment that can be performed to show what a GHE works like. It does not heat the Earth’s surface but it will slow down the cooling rate relative to a state with less amounts of GHG.

    Here is the experiment. Take a can and have a thermometer. Surround the center can with other cans.

    Do two tests. Pour boiling water in the center can and put ice water in all the other cans surrounding the center can and monitor the rate this boiling water cools. Heat will only flow one way, from the hot can to the colder cans. But monitor the rate.

    Now start again again add boiling water to the center can but now instead of ice water have hot water in the surrounding cans. Heat will still flow from the center can to the surrounding ones but will the rate of cooling of the center can slow down and will it be warmer at the end of time periods equal to the first test? Will the warmer cans surrounding the hot center one slow the rate of cooling?

    Now if you want to mimic the Earth/Sun system you would have a heater in the center can and turn on the heater for a period of time (but still put ice water in the outside cans in one test and hot water in the other).

    Will the center can be warmer in the second test? If you do this experiment it should shed light on what is going on with the GHE.

    Also convection, conduction and evaporation will only effect regional temperatures on Earth. They will not alter the global temperature, only radiative energy gain and loss effect the global temperature. Energy enters the Earth system by radiation and that is the only way it can leave the system.

    • geran says:

      “…mimic the Earth/Sun system…”

      So you think this is a model of the Earth/Sun system?

      Hilarious.

    • Slipstick says:

      Your experiment is a valid test of the laws of thermodynamics (which, by the way, are macro effects and have no bearing on the interactions of individual photons) but it is in no way representative of the Earth/Sun system. As an aside, you might want to consider that your experiment deals in differences of nearly 100 deg. C, while just a few degrees C is the difference between the Earth having significant polar ice sheets and us having to build massive sea walls to protect coastal cities.

      • Norman says:

        geran and Slipstick,

        I do not think either of you are following the point of this experiment. Many who completely deny Carbon Dioxide can end in a higher surface temperature than without any of such a gas will claim the Second Law of Thermodynamics forbids it. They claim a colder body cannot make a warmer body warmer. This statement would be true in a closed system that is only losing energy, if the system has a constant input of energy it is not.

        geran has a terrific difficult time understanding it (but at least the effort does amuse him so it has some benefit). A surface that has a constant flow of energy to it (via conduction, radiation, electrical, chemical) will warm to a given temperature based upon how much energy is being put into it and how much energy it is losing. If it is gaining energy at a greater rate than it is losing energy it will warm.

        The warming of the Earth’s surface is a relative comparison between different conditions. In one condition you have no restriction of energy loss, the Earth’s surface would be colder than if something (whatever it might be) were preventing some of the energy from leaving the system.

        The Second Law states that a colder object cannot transfer its heat to a warmer body making it cooler and the warm body warmer. It does not make claims at what rate a body will cool based upon what environment it happens to be in. The experiment purpose is to show that warmer objects around a hotter central object will slow the rate of cooling and that the center object will be warmer relative to a situation with cooler objects surrounding the center one.

        • Slipstick says:

          As I mentioned in a previous post, the Second Law is a macro effect; it has no bearing on the interactions at the quantum level, i.e., a single photon interacting with matter. A photon does not have a “temperature”; it is not “heat”. Heat is a cumulative effect of energy interacting with matter and one form of energy that has this effect is the electromagnetic, which is transferred in quanta we call photons. If a photon’s energy is imparted to matter, the energy state of the matter, which we measure as temperature, will rise, regardless of the initial temperature of the matter. Of course, matter at a higher energy state is more likely to emit photons, and photons at a higher energy, than matter at a lower energy; this is what makes the Second Law true. But, again, this is a cumulative macro effect and does not preclude a photon emitted by matter at a given temperature being absorbed by, and increasing the energy state of, matter at a higher temperature.

          • geran says:

            “But, again, this is a cumulative macro effect and does not preclude a photon emitted by matter at a given temperature being absorbed by, and increasing the energy state of, matter at a higher temperature.”
            +++++++++

            A low energy photon can NOT raise the temperature of a hotter (higher energy) surface.

          • Norman says:

            geran,

            “A low energy photon can NOT raise the temperature of a hotter (higher energy) surface”

            One photon would not do much even if higher energy.

            A large amount of low energy (which means what exactly?) photons can warm a hotter surface if energy is prevented by some method of leaving that hotter surface.

            A surface temperature will raise if more energy is entering it than leaving it.

          • Curt says:

            geran:

            You say, “A low energy photon can NOT raise the temperature of a hotter (higher energy) surface.”

            So you do claim that radiation has a temperature! You angrily denied it when I pointed this out upthread…

          • David A says:

            geran says:
            “A low energy photon can NOT raise the temperature of a hotter (higher energy) surface.”

            Yes, it does. (Just what do you think happens to that photon’s energy when it is absorbed by an object?)

          • geran says:

            Curt says: “So you do claim that radiation has a temperature!”
            +++++++
            Nope. (Curt, at some point your constant misrepresentations just become lies.)

            David A. says: “Yes, it does. (Just what do you think happens to that photon’s energy when it is absorbed by an object?)”
            +++++++
            Nope. If cold warms hot, you could boil water with ice cubes. The planet would be in thermal runaway. Use your brain!

          • Curt says:

            David:

            It is now completely obvious that geran’s intellect is too limited to consider anything more than an isolated two-body problem. Add a separate power source, even a constant one, to the problem, and his synapses overload.

            So in geran’s world, if he, with his ~100 watt metabolic power generation, started to get hypothermia in a 0C environment, he would turn down the opportunity to move to a 20C environment (other things being equal) to get his body temperature back up, as it wouldn’t help because “cold can’t heat hot”.

            A potential candidate for a Darwin Award, if you ask me!

          • Curt says:

            geran:

            You said: “A low energy photon [radiation!] can NOT raise the temperature of a hotter [higher temperature] (higher energy) surface.”

            The plain meaning of “hotter” is higher temperature. (The molecules have higher average kinetic energy.) You seem not to realize it in your fundamental confusion, but you are clearly stating that the surface has a higher temperature than the photon of radiation, which is a meaningless statement unless you think that the photon of radiation has a temperature.

          • geran says:

            Curt, at some point your constant misrepresentations just become lies.

          • Curt says:

            geran:

            I quoted you directly, word for word. I can’t help it if you don’t have the technical sophistication to understand the implications of your own statements.

            And why do you continue to misrepresent that fact that you haven’t actually answered my 3 questions for a week now? Answering your own substitute questions doesn’t count!

        • Gordon Robertson says:

          @Norman “They claim a colder body cannot make a warmer body warmer. This statement would be true in a closed system…”

          It’s true in all systems.

          I have been reading Planck all night on radiation and heat then got onto Bohr on atomic structure. Something occurred to me based on something both talked about.

          When an atom absorbs energy it warms. However, the absorption is due to the electron(s) changing energy levels. The electron will not absorb the energy required to change energy levels unless the frequency and intensity of the absorbed energy is specific. It has to be a fraction of the energy required to keep the electron in orbit.

          That’s what quanta is about. The energy to be absorbed has to be in a specific ratio to the energy of the electron.

          That would explain why energy from a cooler object is not absorbed by a hotter body.

          Several people are using Kircheoff’s laws of emission and absorption to suggest IR is absorbed by both bodies when a hotter object radiates against a cooler object. That is only true in blackbodies and it is a purely theoretical application.

          In lower temperature systems like the atmosphere-surface model, the cooler radiation from the atmosphere does not have the correct intensity and frequency to affect the electrons orbiting atoms in the hotter body.

          As Planck claimed, if the energy cannot be absorbed it dissipates.

          You are now claiming the GHGs in the atmosphere act like a blanket. Bohren referred to that theory as a metaphor at best, and at worst, plain silly. If you could see the calculations and theory that goes into such a claim you’d soon drop it.

          As Bohren put it, GHGs in the atmosphere are not like a truant officer nabbing school children.

          • Norman says:

            Gordon Robertson,

            Only visible light (and maybe the Infrared just below red light) and higher frequencies involve the motion of an electron up or down the positive gradient of the nucleus. The greater the electron jump, the more energy given off.

            With Infrared and microwaves the source of energy is not electron jumps but the variations in the electromagentic fields caused by bond streching and bending. Electons do not move from higher to lower orbitals in these types of radiation.

            The quantum is smeared when you have a whole lot of potential energy states of connected molecules (a normal surface). Some materials are nearly blackbodies where they will absorb any radiant energy that strikes them (having multiple states to occupy).

            You seem convinced that radiant energy cannot be absorbed by a hotter object unless that radiant energy happens to be coming from a hotter source. How does this reasoning make sense? I think Curt has already asked it of geran. If a 15 mircon photon was coming from the sun (which is a very hot source) or a 15 micron photon was coming from a cold asteroid how would the Earth’s surface know to reject the one from the cold asteroid but accept the one from the hot sun? It is really illogical and absurd physics. I think they peddle that stuff at PSI but it is not spreading into the larger world of physics since it has no logical continuity. Thanks for you thoughtful posts.

            Gordon, a link to information you may find of value in understanding how radiant energy of different wavelengths is being generated and by what forces.

            http://hyperphysics.phy-astr.gsu.edu/hbase/mod3.html

          • David A says:

            Norman wrote:
            “Only visible light (and maybe the Infrared just below red light) and higher frequencies involve the motion of an electron up or down the positive gradient of the nucleus. The greater the electron jump, the more energy given off.”

            No. The energy levels in an atom are quantized. Electrons makes quantum jumps among these energy levels, giving off or absorbing radiation. The “gradient of the nucleus” doesn’t matter — that’s what Bohr’s model (and quantum mechanics) was all about.

            “With Infrared and microwaves the source of energy is not electron jumps but the variations in the electromagentic fields caused by bond streching and bending. Electons do not move from higher to lower orbitals in these types of radiation.”

            No. Electrons certainly move from/to higher/lower orbitals in these types of radiation. The energy levels of a molecule being bent or stretched are also quantizied. Electrons transition between these quantized levels, and give off radiation in quantized amounts.

          • Norman says:

            David A

            I do think you are not correct in your understanding of causes of radiation emission of various frequencies (or wavelengths). In IR radiation the vibration of the atoms in the molecule are what result in the photon emission or absorption. The electrons in the atoms are not jumping in orbitals. The energy of electrons jumping from a higher orbital state to a lower one is considerably higher (visible light and above). In IR it is the weaker electric fields of atoms in a molecule moving closer or further away. The electric gradient is much less and that is why the radiation emitted is not as energetic.

            Here is a link to demonstrate the point. Maybe update your science files on this issue. What I have stated was from my Chemistry learning and if you have good sources which state different I will be open to them. Your own personal declarative statements are not sufficient for me, sorry if this insults you.

            Read and enjoy.
            http://www.wag.caltech.edu/home/jang/genchem/infrared.htm

          • David A says:

            “In IR radiation the vibration of the atoms in the molecule are what result in the photon emission or absorption. The electrons in the atoms are not jumping in orbitals.”

            Yes, but those molecular vibrations (and rotations) are quantized.

          • Norman says:

            David A

            Your statement “Yes, but those molecular vibrations (and rotations) are quantized.” I do agree with this. That is why carbon dioxide only emits at a few wavelengths of IR regardless of the temperature of the gas. If the bond energy is at an excited (higher energy) state in a CO2 molecule it will go to a lower energy state and emit an IR photon of discrete energy based upon this quantized bond energy. Thanks.

          • David A says:

            “As Planck claimed, if the energy cannot be absorbed it dissipates.”

            For blackbodies — objects that absorb all radiation incident on them — incident energy will warm the body. It doesn’t matter where that energy came from — the object has no way to distinguish where the energy is coming from, it’s just energy.

            Radiation carries energy. If it’s incident on a blackbody, that energy will warm the blackbody.

    • tonyM says:

      Norman:
      Problem I see with the experiment is that you are not isolating it to the radiation only so that the T of medium surrounding the centre can will be changed and hence will slow its cooling when hotter cans surround it.

      It may be a valid experiment in a vacuum but even then what we are testing is radiation to varied T differences. But my question would then be so what as no one has suggested that the centre can will not be affected by its environment T with which is in radiative contact.

      The real experiment would be if we have an environment of high CO2+air mix compared to a non GHG gas mix (keep molecular weight similar). Now test that.

      Such an exsperiment has been done and I posted earlier:
      tonyM says:
      May 29, 2015 at 8:34 PM

      I don’t think CO2 will achieve anything.

      • tonyM says:

        I must be developing a lissssp; experiment vs exsperiment!

      • Norman says:

        tonyM,

        Thanks. The point of my experiment was not designed to prove GHE. There are many that say GHE violates the Second Law of Thermodynamics. The point of this test is to show that the rate of cooling can be determined by the surrounding environment.

        Roy Spencer has his model a few threads back to show how GHE slows cooling. It does not heat the surface. The back-radiation of the GHE is from the Earth’s surface and not all is returning to the surface. GHG redirect the flow of IR. Some leaves the system and some is returned to the surface slowing the cooling rate.

        Earth with or without GHG would still cool without solar flux but with GHG it would cool slower and I do not see how that is a violation of the Second Law of Thermodynamics.

        • tonyM says:

          Hi Norman:

          If the only issue is, in the simplest form with no other interaction, whether GHG would return radiation and slow heat loss then I would agree with you that back radiation is there, slows the loss of energy and does not violate the 2nd Law.

          I think most would agree with you.

          Problem is that it is more complex for the same 2nd Law will also say it won’t wait around and not respond if there are alternatives available for heat dissipation. There will be a response via extra evaporation, conduction, convection interconnected with a whole host of components like heat sinks and circulation, gravity, clouds, aerosols, bulk of atmosphere which is not GHG capable and many more that I can’t think of immediately.

          External particles/rays which can interact with extra moisture. There is also ample opportunity for any “extra” energy to dissipate at night. The cooling rate slows down sometime well before sunrise; perhaps this could shift a little as well and compensate.

          This now is far more complex; it can be considered as a highly buffered system. My gut feel suggests that the theoretical 4wm-2 for CO2 doubling would be largely offset. But, then, that is what the whole argument is about. Warmers say it will be enhanced by water.

          The answers will be unlikely to come from theoretical calculations or paramatised models. Ultimately in science the only valid evidence is the empirical evidence. That evidence, to my way of thinking, just does not support the CO2 forcing and positive feedbacks whether we calculate over the last 17 years, the last 70 years or the last 140 years. It just is not there to support the alarmist proposition.

          Further much of the avg T increase is due to the Arctic. Who is to say that the avg T captures heat change well. An avg T metric is really questionable. The colder Arctic will simply increase its T disproportionately for the same amount of energy as say in the tropics (simply due 4th power of T) and hence distort that average.

          • Norman says:

            tonyM

            I strongly argee with your statement about the positve feedbacks. I am not arguing in favor of CAGW (which has all these positive feedbacks programmed into models that do no seem to be supported by empiriacl reality). My purpose in some of my posts is just to try and get the science valid and accurate. I think many get their science from blogs and think they know much more than those who have taken the time to go through difficult textbook physics at the University levels. I sent a link for Gordon Robertson from a thermodynmaics textbook showing that it is standard physics to take into account back radiation on how heat will flow in a system. There is the claim that a photon emitted by a colder source cannot be absrobed by a warmer one. I do not know why this is considered a valid or good claim. Doug Cotton points out it comes from a mathematician. I did not see any experimental evidence to support his claims in his paper. Lots of math which does not mean in any way he is correct.

            The processes you describe can effect the Earth’s surface temperature but would not let you know if the overall system was warming or cooling. The only valid method to determine global warming (the totality and not just the surface) is to measure radiant energy in and radiant energy out. If they are not equal it will mean the Earth system is either warming or cooling at the time of the measurement. The complexity of energy flow in the system is vastly complex as you can have a heat wave in one region (say India) and then another location is experiencing very cold conditons. The Climate models are supposed to be able to mimic these flows and arrive at some point that is in tune with the real world. Problem with models is that a biased person can make them perform as they choose. Really good science needs to be free of bias by either side.

            I see both sides and really biased. geran and Doug Cotton’s views and David A and Slipstick. Both sides seem extreme and have lost the ability to discuss the science and be humble enough to realize you do not know it all and use this blog as a source of learning rather than projecting some superior attitude.

          • David A says:

            Norman: Positive feedbacks are not “programmed into models.” Feedbacks, positive and negative, arise from the laws of physics.

            It’s the laws of physics that are programmed into models. Feedbacks are among the output.

          • David A says:

            “There is the claim that a photon emitted by a colder source cannot be absrobed by a warmer one. I do not know why this is considered a valid or good claim. Doug Cotton points out it comes from a mathematician.”

            It’s simply a consequence of Planck’s Law: a photon of frequency f has energy E=hf. The temperature of the source (or the destination) doesn’t enter into it, or any other factors. When the photon is absorbed by an object, the object absorbs this energy.

          • fonzarelli says:

            Norman, i think your last paragraph here says it all. If people can win the game without the science, then that’s what they’re going to do. (If kerry calling skeptics “flat earthers” will stifle and thus win the debate, then that’s what he will do…) Tactical “alinski” type debating skills will rule the day unless very strong evidence to the contrary (in the case of agw, cooling…) disproves the theory.

          • mpainter says:

            David A
            Cheer yourself.
            The thought police know about this blog and they are keeling files on all the suspects. Norman’s day is coming and he will find himself hauled in to give account before the Lord High Grand Climate Inquisitor.

          • tonyM says:

            Hi Norman:
            I think we are in agreement on most things and the complexity of this field. A couple of points are worth exploring.

            Can a photon from a lower T source be absorbed by a higher T target? Leaving aside that we are not talking in a macro Thermodynamic sense, if this holds then two such photons can be absorbed in the same way and so on to a large number of photons.

            Your experiment with multiple cans versus a single centered can at higher T could represent this case. A bit more clearly, place a smaller ring inside a much bigger ring.

            There are more photons coming from the cooler outer ring towards the centre one than it is emitting. One can design the experiment to be so. At the limit even if both were the same T this would hold.

            Now if multiple photons are absorbed they are extinguished and this results in a higher atomic/molecular vibration and hence higher T.

            This would lead to the conclusion that the centre ring should increase its T which would violate the 2nd Law.

            This is why I think Doug Cotton is on the right track with pseudo scattering and absorbing only above the threshold. This also has the effect of suppressing some of the heat loss. This is considering it at single atomic/molecular level but in macro terms the end result is the same and in keeping with the Thermodynamic Laws which is how it must be for it to be valid.

          • Norman says:

            David A,

            Have you read Roy Spencer’s take on Climate Models from way back. It is interesting.

            http://www.drroyspencer.com/2009/07/how-do-climate-models-work/

            I am doing research on climate models to try and get better understanding of how they actually work. I thank you for you reply as it will require I do not just believe what others have said but find out for myself on this issue.

        • Norman says:

          tonyM

          I am thinking about your ring example. I do not have an answer at this time. I will try to figure out something that makes sense to you.

          • MikeB says:

            Tony, see ….
            http://scienceofdoom.com/2010/07/26/do-trenberth-and-kiehl-understand-the-first-law-of-thermodynamics/

            In particular, see how a constant heat source of 30,000 watts can generate 1,824,900 watts, just by adding insulation (creating energy from nothing?)

            In fact, you can learn everything you want from that site; no need to guess.

          • Curt says:

            MikeB:

            The 1st LoT is conservation of energy, not conservation of power. Power is rate of energy transfer — and it is not constrained by the law of conservation of energy.

            In your own personal financial accounting, where the “law of conservation of money” applies (since you are not a government…), the rate at which you transfer money between your own accounts is not constrained.

            If you earned $30,000 a year and spent $30,000 a year, you could still transfer $1,824,900 back and forth between savings and checking during that year. That is the point SoD was making.

        • Gordon Robertson says:

          @Norman… “I do not see how that is a violation of the Second Law of Thermodynamics”.

          Anything that violates the transfer of heat from GHGs in a cooler atmosphere to a warmer surface – that warmed the GHGs – is a violation of the 2nd law.

          People trying to get around the 2nd law are using net EM energy flow as an argument while failing to understand that EM is not heat and that the 2nd law applies only to heat.

          You are trying to talk around it using abstractions. Roy has done the same in several thought experiments to get around the 2nd law.

          Why don’t you work from the premise that the 2nd law is correct and that heat cannot be transferred from a cooler atmosphere to a warmer surface? Why try to find loopholes rather than questioning the GHE and the AGW?

          The same thing has happened with entropy. Clausius defined entropy in relation to heat as part of his treatise on heat. He defined entropy in words as the sum of infinitesimal changes in heat (dq) at a specific temperature (T) at which the change takes place, over a process.

          If the process is reversible, the sum (or integral) of dq/T of a process is zero. However, most processes are not reversible in the universe therefore the entropy is positive.

          Entropy becomes the sum of dq/T and applies to heat. Later on, scientists began using it in a manner entirely different from what it was defined. They associated entropy with a general disorder in the universe unrelated to the heat in processes.

          Botlzmann, and later Planck, tried to relate entropy to probability. As far as I am concerned, that’s where science took a wrong turn. That’s when physics got lost in math and that situation was complicated by the fact that the math was fudged to the point that no one can visualize what is going on. Feynman said as much.

          Planck admitted that, you cannot visualize what he is talking about.

          Today, in climate science, we have people applying those fudged equations of Planck and Boltzmann, and claiming they override the 2nd law of thermodynamics, which is based on Newtonian mechanics.

          I don’t think that is right. The 2nd law is based on sound scientific reasoning and proof. The fact is, that without external compensation, it is not possible for HEAT to be transferred from a cooler body to a warmer body.

          Of course, in the world of Boltzmann and Planck, the math applies to blackbodies, which are highly theoretical very hot bodies. The equation also involves electromagnetic radiation and the heat transfer is highly idealized between blackbodies.

          In the real world, we need to come back to Clausius and his very practical approach to real bodies. With Planck and Boltzmann, they have converted the 2nd law to entropy applications and in doing so have related entropy to probability. That may work with extremely hot sources like stars, where it is not practical to measure heat transfer directly.

          In the real world, we can measure heat transfer by observing the work it does. From doing such experiments the 2nd law has been validated along with the principle that heat can only flow from a warmer body to a cooler body without compensation.

          I really don’t understand why you guys are debating this and trying to find a way around it. It’s like trying to build a perpetual motion machine because the 2nd law has claimed it wont work.

          • Curt says:

            Gordon:

            You continually say, “EM is not heat and that the 2nd law applies only to heat.”

            Why do you keep disagreeing with your idol Clausius, who explicitly stated otherwise?

            Here’s the key quote:

            “the principle implies that in the immediate interchange of heat between two bodies by conduction and radiation, the warmer body never receives more heat from the colder one than it imparts to it.”

          • Norman says:

            Gordon Robertson,

            I do not know what source you use to suggest anyone is stating a violation of the 2nd law when it comes to the GHE. On a previous post (I think to geran) I explained to him that the energy from the Earth’s surface is 390 watts/meter^2 (as an average calculation) and the returning energy is 324 watts/meter^2. The returning energy is not self generated, it is being redirected from the surface back to the surface. It is not a violation of the 2nd law when the energy from the cooler source is less than the energy from the hotter source. This is what the 2nd law states. A violation of the Law would be if the Earth emitted 324 watts/meter^2 and the atmophere was returning 390 watts/meter^2.

            The atmosphere is not warming the surface. The atmosphere is redirecting the flow of EM back to the surface keeping it warmer than it would be if no redirection was taking place. If the sun was gone the Earth would cool with or witout GHG in atmosphere. With GHG the cooling rate would be slower. That is all I have ever seen with GHE theory. I do not comprehend how this is being incorrectly interpreted to mean the atmosphere is actually warming the surface. Where does this idea come from in your understanding. Do you have links to support this claim from Climate science or any paper written in the field? Or is this a PSI claim based upon misunderstanding?

          • Gordon Robertson says:

            @Curt “Here’s the key quote:

            “the principle implies that in the immediate interchange of heat between two bodies by conduction and radiation, the warmer body never receives more heat from the colder one than it imparts to it.”

            Note the word ‘implies’. You cherry-picked his words out of context. Furthermore, you have failed to supply the context in which the heat transfer is taking place. There is a difference between two bodies interacting that are independent sources of heat and two bodies in which one is dependent on the other for heat.

            Here’s the whole quote:

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

            The principle may be more briefly expressed thus: Heat cannot by itself pass from a colder to a warmer body; the words “by itself”, however, here requires explanation. Their meaning will, it is true, be rendered sufficiently clear by the exposition contained in the present memoir, nevertheless it appears desirable to add a few word here in order to leave no doubt as to the signification and comprehensiveness of the principle.

            In the first place, the principle implies that in the immediate interchange of heat between two bodies by conduction and radiation, the warmer body never receives more heat from the colder one than it imparts to it. The principle holds, however, not only for process of this kind, but for all others by which a transmission of heat can be brought about between two bodies of different temperature, amongst which process must be particularly noticed those wherein the interchange of heat is produced by means of one or more bodies which, on changing their condition, either receive heat from a body, or impart heat to other bodies.

            On considering the results of such processes more closely, we find that in one and the same process heat may be carried from a colder to warmer body and another quantity of heat transferred from a warmer to a colder body without any other permanent change occurring. In this case we have not a simple transmission of heat from a colder to a warmer body, or an ascending transmission of heat, as it may be called, but two connected transmission of opposite characters, one ascending and the other descending, which compensate each other. It may, moreover, happen that instead of a descending transmission of heat accompanying, in the one and the same process, the ascending transmission, another permanent change may occur which has the peculiarity of not being reversible without either becoming replaced by a new permanent change of a similar kind, or producing a descending transmission of heat. In this case the ascending transmission of heat may be said to be accompanied, not immediately, but mediately, by a descending one, and the permanent change which replaces the latter may be regarded as a compensation for the ascending transmission.

            Now it is to these compensations that our principle refers; and with the aid of this conception the principle may be also expressed thus: an uncompensated transmission of heat from a colder to a warmer body can never occur. The term “uncompensated” here expresses the same idea that was intended to be conveyed by the words “by itself” in the previous enunciation of the principle, and by he expression “without some other change connected therewith, occurring at the same time” in the original text.

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

          • Gordon Robertson says:

            Curt…more from the Clausius paper:

            Clausius has written in the Mechanical Theory of Heat:

            “Heat consists in a motion of the ultimate particles of bodies and of ether, and that the quantity of heat is a measure of the Vis Viva of this motion”.

            Vis Viva is an old term for kinetic energy. He also proposed the word ‘ergal’ for potential energy, likely because the German word was about a page long.

            Ether was a word used in the old days for a substance believed to exist in empty space. Einstein claimed it does not exist but Miller disagreed. To this date, there is disagreement. Recently Wal Thornhill suggested it could be comprised of neutrinos.

            Clausius states clearly that heat is the motion of atomic particles in bodies and that the quantity of heat is the kinetic energy of the particles.

            Later he states: “…we must also take into consideration another quantity which shews by how much the heat actually existing in one unit-weight of a substance (i.e. the vis viva [kinetic energy] of the motion of its ultimate particles) is increased when the substance is heated through one degree of temperature. This quantity we will name the body’s true heat-capacity.

            It’s plain that Clausius knew about atoms, kinetic energy, and potential energy, and that he referred to the amount of heat in a body as the kinetic energy of the atoms.

            He goes on to distinguish heat capacity from specific heat:

            “It would be advantageous to confine this term ‘heat-capacity’, (even if the word ‘true’ be not prefixed) strictly to the heat actually existing in the body; whereas for the total heat which must he imparted for the purpose of heating it under any given circumstances, and of which work-heat forms a part, the expression ‘specific heat’ might be always employed”

            In the chapter in which he develops the second law, he does a brilliant, intricate, analysis of heat producing work and proposes “Heat cannot, of itself, pass from a colder to a hotter body.”

            It is extremely important to understand that Clausius developed this statement from direct experimentation and proof by demonstrating it in action in a heat-work reversible process.

            Keep in mind with the following statement that he is still developing the 2nd law and that he follows the statement with a disclaimer:

            “It is true that by such a process (as we have seen by going through the original cycle in the reverse direction) heat may be carried over from a colder into a hotter body: our principle however declares that simultaneously with this passage of heat from a colder to a hotter body there must either take place an opposite passage of heat from a hotter to a colder body, or else some change or other which has the special property that it is not reversible…”

            Disclaimer: “This simultaneous passage of heat in the opposite direction, or this special change entailing an opposite passage of heat, is then to be treated as a compensation for the passage of heat from the colder to the warmer body; and if we apply this conception we may replace the words” of itself” by “without compensation,” and then enunciate the principle as follows: ” A passage of heat from a colder to a hotter body cannot take place without compensation.”

            He has made it clear that transfer of heat from a colder body to a warmer body is NOT reversible and that it must be accompanied by a simultaneous transfer of heat from the hotter body to the colder body.

            In a refrigerator or a heat pump, the compensating heat is supplied via electric motors, compressors, expanders and special refrigerants. Heat cannot flow from cold to hot without such compensation.

            Then he develops the entropy version of the 2nd law:

            “This Principle may be expressed in words as follows: If in a reversible cyclical process every element of heat taken in (positive or negative) be divided by the absolute temperature at which it is taken in, and the differential formed be integrated for the whole course of the process, the integral obtained is equal to zero.

            If the integral dQ/T corresponding to any given succession of variations of a body, be always equal to zero provided the body returns finally to its original condition, whatever the intervening conditions may be, then it follows that the expression under the integral sign, viz. dQ/T must be the perfect differential of a quantity, which depends only on the present condition of the body, and is altogether independent of the way in which it has been brought into that condition.

            If we denote this quantity by S, we may put

            dQ/T = dS or dQ = TdS

            an equation which forms another expression, very convenient in the case of certain investigations, for the second main principle of the Mechanical Theory of Heat.

            He explains later that S is entropy.

          • Curt says:

            Gordon:

            Nothing in the extended Clausius quote you cite contradicts my claims. He continues to explain that in all cases heat transfer, both conductive and radiative, is comprised of a two-way exchange of heat.

            Without what he calls “compensation”, the heat flow from warmer to colder body is always greater than the heat flow from colder to warmer body (but there is heat flow from colder to warmer).

            I agree with him on this. But you have said repeatedly that there is NO heat flow from colder to warmer in the case of radiative heat transfer. Do you still believe this?

          • Kristian says:

            Curt, you do of course realise that when Clausius speaks of “heat”, that would be the modern equivalent to “energy” (conductive or radiative). In modern terminology, only the ‘net’ of the two opposing “energy” flows constitutes the HEAT transferred. This is simply to avoid confusion, the very confusion you appear to be fighting here on this thread.

            In other words, there is never any ‘heat’ flowing both ways in a thermal exchange, because ‘heat’ is – by (modern) definition – only the net flow of energy, the actual thermal transfer of energy between bodies or a body and its surroundings as a direct result of a temperature difference.

            Only the ‘net’ flow of energy, the actual thermal transfer, the HEAT, Q, is able to do anything in terms of temperature. It is the only thermodynamic flux. As Clausius stated: “… an uncompensated transmission of heat [energy] from a colder to a warmer body can never occur.” Which is to say that you are not allowed to pick one of two opposing potential ‘fluxes’ of energy in a thermal exchange and ignore the other, thus letting yourself pretend that this individual energy ‘flux’ can and does act and work as if it were itself an independent heat flux to the receiving object.

            Which is what the “back radiation” explanation of the rGHE really boils down to:
            https://okulaer.files.wordpress.com/2014/10/drivhuseffekten.png
            (Derived from Stephens et al. 2012.)

          • mpainter says:

            Kristian,
            In other words, the AGW bunch have their thermodynamics in a tangle?
            You say “It is the only thermodynamic flux”.
            Then the atmospheric radiative flux is not a thermodynamic flux except in the sense that net flux is to space. Correct or not?
            If correct, then ” down welling” IR is nonsensical because it is not the net of the flux, if I understand correctly.

          • mpainter says:

            Kristian,
            In other words, the AGW bunch have their thermodynamics in a tangle?
            You say “It is the only thermodynamic flux”.
            Then the atmospheric radiative flux is not a thermodynamic flux except in the sense that net flux is to space. Correct or not?
            If correct, then ” down welling” IR is nonsensical because it is not the net of the flux, if I understand correctly.

          • Kristian says:

            mpainter says, June 6, 2015 at 7:24 AM:

            “In other words, the AGW bunch have their thermodynamics in a tangle?”

            It’s hard to say if it’s intentional or out of sloppyness, but they do insist on treating the potential DWLWIR ‘flux’ from the cool atmopshere to the warm surface as a separate heat flux (in the sense that they expect it to give thermodynamic results as if it were). The “back radiation” explanation of the rGHE (and of surface temperatures) is thus clearly in violation of the 2nd Law of Thermodynamics and therefore flawed.

            This doesn’t mean, however, that the described effect violates any laws of nature. The question is, to what extent is this described effect of theirs a ‘temperature effect’, and to what extent is it merely a ‘temperature result‘?

            “You say “It is the only thermodynamic flux”.
            Then the atmospheric radiative flux is not a thermodynamic flux except in the sense that net flux is to space. Correct or not?”

            Correct.

            “If correct, then ” down welling” IR is nonsensical because it is not the net of the flux, if I understand correctly.”

            It is not ‘nonsensical’ per se. Conceptually, it can be useful, and in radiative physics it is useful as a mathematical construct. But in the postmodern world of ‘Consensus Climate Science’ it is explained and promoted in such a way that people end up utterly confused by it. Which I guess is precisely how and where the Establishment wants them …

            “It’s extra heat to the ground. Only it’s not. But it is. And it’s not. It works like heat. But it isn’t heat. Not really. Just a little. It is heat. Just not ‘net heat’.”

            And so on and so forth …

          • Ball4 says:

            Bryan: “Zemansky also recommended dropping the term ‘heat energy’ and substituting ‘internal energy’ in most applications to remove misunderstandings.”

            Thermal energy is generally regarded as a simplification of therm[odynamic intern]al energy. The late Mark Zemansky actually wrote in 1970*: “the concept of thermal energy is by all odds the most ambiguous term employed by writers of elementary physics and by chemists”. With his words ringing in your ears, become more specific, toss “thermal energy” simplification onto the thermo. scrap heap along with Zemansky rec. for confusing and contradictory “use and misuse of the word heat” as in your “heat flow”. Once science realized that heat doesn’t exist separate from energy in any single object, heat cannot then flow from one object to another.

            *The Physics Teacher, 1970, Vol. 8, pp. 295-300.

            ——

            Kristian and mpainter: Back to basics.

            Instantaneous total thermodynamic internal energy U = KE + PE enclosed by the border of an arbitrary volume of interacting point masses w/no gravitational field is, in joules:

            U = KE + p*V (again no gravity field)

            By 1st law U is conserved, infinitesimal changes to U over time become when accounting across the border of the point masses for g=0:

            dU/dt = d(KE)/dt + d(p*V)/dt = 0

            Add a gravitational field PE as in the atm. and rate of change internal thermodynamic energy changes to:

            dU/dt = d(PE+ KE)/dt + d(p*V)/dt = 0

            Commonly & generally in a g field since internal energy U of point masses is not necessarily constant, can increase & decrease this last eqn. is denoted:

            dU/dt = Q + W (plus sign if work is done by the system like an engine doing work on surroundings)

            Kristian incorrectly writes “the HEAT, Q,” when discussing an arbitrary volume of planetary atmospheres since 1) Q is a heating rate (rate of change internal energy as a consequence of temperature differences between the bounded system and surroundings), 2) with gravity, the Q term also has rate of change of gravitational PE which is not “heat” and 3) W is a working rate

            Kristian should correctly write “the HEATING rate, Q,”.

            Note: dU/dt = Q + W is deceptively simple, it is almost useless as it cannot be solved to obtain functional relationships among thermo. variables.

            ——-

            In more detail, since we know about molecules, for a 0 g bounded system of point masses, the thermodynamic internal energy U within the boundary is the sum of point mass KE about the center of mass and the PE arising from forces between point masses (p*V) thus U=KE+p*V.

            Molecules however are not point masses and the story gets more interesting from there by adding a gravity field where dU/dt = Q + W as shown.

            Macro U for bounded molecular air in a planetary atm. can change by: 1) bounded system interacts with surroundings at different temperatures (heating or cooling), 2) exert a force on the system boundary over a distance (working). For an isolated system, Q=W=0 thus dU/dt=0 and U is constant.

            More interestingly, in addition molecules also both emit radiant energy and absorb radiant energy from their surroundings, processes that are vital to the workings of the atm. So Q also has a relevant component that results from 3) interactions of molecules with radiation.

            dU/dt = d(PE+KE+RE)/dt + d(p*V)/dt

            Radiant energy (RE) (wave or particle (photon)) can interact with matter but does not interact with other radiant energy (other waves or particles (photons)). Conclude: atm. DWIR photons do not interact with L&O UWIR photons.

            These text book basics have withstood countless experiments, both lab & in-situ, & get so misused on blogs that going back to them from time to time becomes necessary even for me.

          • mpainter says:

            Ball4, thanks for your comment.
            Referring to Wikipedia, it gives that Q is the convention for heat transferred and that Q (bold italics with dot) symbolizes rate of heat transfer.
            Is this incorrect, in your view?

          • Ball4 says:

            mpainter – In context, Kristian 11:38pm was discussing a flow which implies a change in the amount of something per unit time i.e. a time rate of change. Wikipedia.org is a fair entry point to a subject but is not a text book so the references given are vital. Here is Wiki ‘Heat’ Ref. 32 W.B. Jensen 2010 they use as a history of symbol Q:

            http://www.che.uc.edu/jensen/W.%20B.%20Jensen/Reprints/182.%20q%20and%20Q.pdf

            Commonly in text books the dot above a quantity means the time rate of change of that quantity so Qdot is commonly d(Q)/dt. Each text book might define symbols differently; it is incumbent on the writer to define symbols used and the reader to understand them when used. M. Zemansky pointed out the need to reduce thermo. terminology confusion in 1970 and even gives earlier ref.s to the effort. Following that advice will reduce blog communication errors. Which today exist by the quantity: tons.

        • Gordon Robertson says:

          @David “It’s simply a consequence of Planck’s Law: a photon of frequency f has energy E=hf. The temperature of the source (or the destination) doesn’t enter into it, or any other factors. When the photon is absorbed by an object, the object absorbs this energy”.

          You are not quoting Planck accurately. He claimed that some emissions are simply not absorbed, that they dissipate.

          I was reading both Planck and Bohr last night. Bohr is more specific. He claims that an electron in an atomic orbit will only absorb energy under specific conditions. The energy of the emission has to be a specific fraction of the electron’s orbital energy.

          Hydrogen only emits and absorbs in specific frequency bands, as do most elements. A hydrogen atom will not absorb any energy outside those bands.

          It’s been a while since I studied this stuff in an astronomy course I took. Here’s a wiki explanation:

          http://en.wikipedia.org/wiki/Hydrogen_spectral_series

          • Curt says:

            Gordon:

            You say, “He claimed that some emissions are simply not absorbed, that they dissipate.”

            What happens to the energy in the radiative emissions when “they dissipate”?

          • Gordon Robertson says:

            @Curt…”What happens to the energy in the radiative emissions when “they dissipate”?”

            I am guessing that it converted to a different form of energy.

            When solar energy is absorbed by the surface, it dissipates. It no longer exists as EM and converts to thermal energy. The resultant thermal energy dissipates into IR as it is emitted.

            Conservation of energy only implies that energy as a whole is conserved. It allows for conversion from one form of energy to another.

          • Curt says:

            Gordon:

            In your earlier comment you said that EM emissions that are not absorbed are “dissipated”. Now you say that EM emissions that are absorbed are dissipated.

            Radiation can be absorbed, reflected, or transmitted (possibly with scattering). If a photon of radiation is absorbed, it disappears, and its energy goes to increase the energy of the absorbing body, often (but not always) as thermal energy.

            If the photon is reflected or transmitted, it continues on without affecting the energy of the absorbing body. Its own energy is unchanged.

            “Dissipation” of energy is not a helpful concept in this context.

          • Gordon Robertson says:

            @Curt ““Dissipation” of energy is not a helpful concept in this context”.

            I got it from Max Planck but unfortunately I was not able to ask him to explain. šŸ™‚

            I think it is fair to use the word dissipate when we are talking about energy that dissipates as a particular form of energy, in fact, disappears as that kind of energy, then appears in another form of energy.

            If we talk about electrical current in a circuit, passing through a pure resistor, the electron has charge which I take to be electrical energy. That electrical energy can be dissipated in a resistor as an i^2 x r loss and converted to thermal energy.

            The mechanics of how it is converted is not clear to me, at an atomic level. I’d sure like to know more about it but unfortunately, in quantum theory, the atomic relationships have been converted to probability functions hence difficult to visualize.

            I don’t have expertise in statistical mechanics as applied to heat and I am trying to understand what Planck and Bohr are talking about. Amazingly, the math I learned in engineering keeps me in the ball park while reading their stuff but it is so dense that I need time to absorb it, even at a superficial level.

            It would have been nice had Planck or Bohr talked about the direct effect of a lower intensity IR on an atom operating at a higher kinetic energy. However, Planck was more into broad-spectrum EM as applied to black-bodies.

            If a photon of lower intensity IR from a cooler body impacts an atom with it’s electrons in higher energy orbitals, will it be absorbed? According to my understanding of Bohr, the energy level of the photon must be a specific ratio of the electron’s orbital energy or it will not be observed.

            I am aware of your point. If it is not absorbed, and it does not pass through, what becomes of it?

            Good question.

          • Bryan says:

            Gordon
            I have come lately to this thread but would like to say I agree with almost all your comments.
            However I think you should cut Curt a little slack.
            I have noticed very similar comments to Curt’s from other chemists.
            I’m sure they are all very intelligent but they share some common mistakes.
            For example Josh Halpern, Frank,DeWitt Payne and Curt over at SoD’s site (and the book Curt quotes from is written by another chemist), all are uncomfortable with the concept of heat transfer.
            I put it down to oversimplifying or truncating the second law.
            Clausius and Carnot were practical men who were motivated by trying to extract useful or high quality energy from diffuse low quality thermal energy.
            The second law of thermodynamics is a direct result of this endevour.
            If E = energy extracted and Th and TC the thermal reservoirs then the maximum useful energy extracted

            = E(Th – Tc)/Th

            Now how many chemists go through the derivation of the second law from the Carnot Cycle?
            Would they ever need to know that?
            I’m sure some do but most of them seem to seem to be left with the second law is about ‘more energy in one direction(hot to colder) than the other’
            And that’s about it.
            So in a crowded curriculum its easy to see why some parts are dropped.
            Some better quality heat transfer books might not cover the Carnot Cycle either.
            So the electromagnetic flows between two objects are often described as radiation.
            A cruder oversimplification is to call them ‘heat flows’ and for most purposes no harm is done.
            Physics students and others from departments that think thermodynamics should have robust fundamental orthodox basics will follow a more lengthy introduction.
            I used Zemanskys ‘Heat and Thermodynamics’ textbook(1934) which incidentally was recommended by Feynman.
            Zemansky updated thermodynamics to include two way radiative flow facilitated by photons.
            Zemansky also recommended dropping the term ‘heat energy’ and substituting ‘internal energy’ in most applications to remove misunderstandings.
            Zemanskys interpretations are reflected in all modern physics textbooks.

          • Curt says:

            Gordon:

            Fundamentally, EM radiation is either transmitted, reflected, or absorbed. We understand this down to the photon level.

            In the first two cases, the photon just “keeps going”. Our most powerful telescopes have detected photons that have been traveling for over 10 billion years, so there is really no known limit.

            In the third case of absorption, the photon is “extinguished” and its energy incorporated by the absorbing body in one form or another. Often this increase is manifested as thermal energy (“heat”).

            We may be just quibbling about semantics, especially with the possibility of inexact translations thrown in, but when I think of thermodynamic “dissipation”, I think about the the entropy-increasing thermalization of energy.

            But possibly another way of looking at it is that EM radiation that is not absorbed locally “escapes” the system of interest, and so is lost to the local system, and therefore “dissipated”.

          • Curt says:

            bryan:

            My degrees are in mechanical engineering, where we learn a lot about Carnot cycles, and even real-world thermodynamic cycles!

            Because we learn so much about them, I can definitely say that someone who keeps talking about Carnot efficiency in the context of pure heat transfer (no work generation) doesn’t have any real understanding of the basics of thermodynamics.

            You say, “most of them seem to seem to be left with the second law is about ‘more energy in one direction(hot to colder) than the other’”.

            Let’s see, where could “they” have gotten that. Perhaps from Clausius (8th memoir):

            “The principle may be more briefly expressed thus: Heat cannot by itself pass from a colder to a warmer body; the words “by itself”, however, here requires explanation. Their meaning will, it is true, be rendered sufficiently clear by the exposition contained in the present memoir, nevertheless it appears desirable to add a few word here in order to leave no doubt as to the signification and comprehensiveness of the principle.

            In the first place, the principle implies that in the immediate interchange of heat between two bodies by conduction and radiation, the warmer body never receives more heat from the colder one than it imparts to it.”

            That Clausius, just a dumb chemist who never really understood thermodynamics like Bryan does…

          • Bryan says:

            Curt
            You are obviously an ideologue with no real attempt to understand the science.

            Fighting a rearguard operation as you are; a few slogans is all you have to offer, how pathetic.

            Your ‘analysis’ of the G&T paper above is a classic example.
            Hot air and venom with not one error detected.

            My point was the development of the second law emerged from studies involving changing low quality thermal energy into higher quality electrical or mechanical energy.
            This then can be used as a test to determine if something is a heat transfer in a given situation or not.
            Anyone with a physics background knows this!
            But the fact that you don’t, speaks volumes!

            Your posts here can be summed up as trying to show insulation is a heat source.
            If you really think that then hand back your mechanical engineering degree as it seem to be a worthless piece of paper.
            Worse still is, if you realize that insulation is not a heat source and yet you try convince others that it is, is beneath contempt

          • Bryan says:

            Curt

            If it was possible to split the spontaneous two radiation streams I would be delighted.
            However unfortunately this is impossible I think.

            However if it could happen it would be a great teaching aid for dealing with reluctant learners such as yourself.

            Intercepting the hot to cold stream with a transducer it would be possible to obtain higher quality energy(or work) within the limitations of the second law.
            The Carnot cycle predicts the maximum energy possible as no machine can be more efficient that a Carnot machine.
            Intercepting the cold to hot stream with a transducer it would not be possible to obtain higher quality energy(or work).
            So the cold to hot stream cannot be a heat transfer.

            The Carnot cycle formula is relevant to all thermal to electrical energy transformations.
            The solar or photocells powering the satellite are constrained to work within the Carnot efficiency limit.
            Hit Wikipedia and solar cell to find thousands of references to Carnot and the second law

          • Curt says:

            bryan:

            The Carnot efficiency expresses the maximum theoretical amount of thermodynamic work (as a fraction) that can be extracted from a “heat flow” from a higher temperature reservoir to a lower temperature reservoir, as a function of the absolute temperatures of those reservoirs.

            Here we are talking about pure heat transfers with no (zero) work extracted, so we are not anywhere close to the limits of the Carnot efficiency. This is true whether we consider it a one-way heat flow using the fluid-flow analogy, or whether, like Clausius (and Zemansky), we consider it as the net result of two opposing radiative power flows.

            The Clausius statement of the 2nd Law has nothing to do with possible work generation.

            I have never claimed that you could separate out the lesser cold-to-hot radiative power flow from the greater hot-to-cold radiative power flow. They are inextricably tied together by a common geometry.

            But saying you cannot separate them is completely different from saying the cold-to-hot flow does not exist, or that it cannot have any effect on the hotter body.

            You say that “Zemansky updated thermodynamics to include two way radiative flow facilitated by photons.” Well, it wasn’t his “update” as it had been well understood before his textbook was written, but I agree that he was correct to include it, as I agree about the “two way radiative flow”. You seem not to. Why do you disagree with Clausius and Zemansky?

            You talk about “intercepting the hot to cold stream” of radiation and “intercepting the cold to hot stream” of radiation, and the difference between these. What ridiculous and meaningless statements! This implies that the radiation carries information about the temperature of its target before it reaches that target! Completely absurd, and just evidence that you are confused about the most basic concepts!

          • Bryan says:

            Curt says

            “I agree about the “two way radiative flow”. You seem not to.”

            Where did you get such an absurd idea?

            Do you have problems with your eyesight?
            Can you not read plain text?

            Indicate where I have written that I don’t believe in a two way radiative interaction between hotter and colder objects.
            Failure to do so would indicate to me that your reading comprehension sucks.
            This would also explain your rambling pointless posts.

            Why some mechanical engineers cannot understand the technical language of thermodynamics must indicate a failure of their degree course syllabus.

            So here it is for the nth time

            A hot and a colder object have a two way energy flow interaction.
            However there is only a one way spontaneous heat transfer and it is always from the higher to lower temperature object.
            You will find this definition in every physics textbook dealing with thermodynamics.
            All physics departments on the planet teach this plain statement.
            Go do some reading.

          • Curt says:

            Bryan:

            For someone who claims to believe in the two-way exchange of radiative energy, you reach some pretty strange conclusions!

            You put down chemistry students by saying that “most of them seem to seem to be left with the second law is about ‘more energy in one direction(hot to colder) than the other’ “.

            How is this different from what you now say: “A hot and a colder object have a two way energy flow interaction. However there is only a one way spontaneous heat transfer and it is always from the higher to lower temperature object.”

            Let’s see if you can follow through on the implications of this “two way energy flow interaction” by solving a simple problem of the type given in an introductory undergraduate course. We’ll use the satellite-testing vacuum chamber that KevinK brought up that started this whole thread.

            You have a small spherical satellite with a 1.0 m^2 surface area. It has an internal continuous 100-watt power supply from a radioactive source. You are testing it in a vacuum chamber under two different conditions.

            In the first case, the walls of the chamber are held at a very low 77K by liquid nitrogen circulating through the walls.

            In the second case, the walls of the chamber are allowed to come to earth ambient temperature of 288K.

            What is the steady-state surface temperature of the satellite in each case? You can use an emissivity for the both the satellite and the walls of 0.95, or if you want it a little simpler, assume perfect blackbody emissivity of 1.0.

            If your answers for the two cases are different, what explains the difference?

  33. barry says:

    Nice tribute, Dr Spencer.

  34. Slipstick says:

    Norman,
    I’m sorry you feel I’m projecting a superior attitude or trying to belittle. That’s not my intent, with the exception of a couple of, quite frankly, cranks whose “scientistic” hypotheses, flawed analysis techniques (calculating a trend using two points being a favorite), and absurdly simplistic models are truly worthy of derision. I am simply trying to stem the tide of physics misinformation and misinterpretation that floods this forum and, I guess, my frustration is affecting my responses. Going forward, I will try to be more cogent of to whom I am replying and more considerate of the tone. Be well.

  35. Norman says:

    Slipstick,

    Thank you for the thoughtful reply. I may have falsely assumed you were on the opposite spectrum from geran and Doug Cotton from my readings of your posts. I am not against any position or view if it can be backed up with strong science or experimental evidence. I like this blog as a learning space with many views and light moderation. Very little banning of thoughts going on. Skeptical Science banned me a few years ago.

    I think if your goal is to correct the flawed science that is terrific.

    Sorry for any incorrect assumptions I may have had about your goals or character. Also you be well.

    • geran says:

      Hey Norman, thanks for the “mention”. It just shows that I’m living in your brain. (Maybe I can undo some of the damage there….)

      • Norman says:

        geran,

        Even though I think you have some very flawed science I do think you have a wit and some humor and add some entertainment to this blog. I think you get most your information of science from various blogs but will not crack a textbook and view such as a vial untouchable. Real science takes lots of effort and time. Reading a PSI article is easy and takes a few minutes of effort. You would do much better in undoing damage with some solid evidence to support the claims you make. Some good solid links. You make many statements and believe them to be unquestionable truth but you provide very little to back them up.

        • geran says:

          Hey Norman, project much?

          http://en.wikipedia.org/wiki/Psychological_projection

          (Remember, I am inside your head.)

        • fonzarelli says:

          Norman, it looks like this thread will go the way of all threads before the week is over when Dr. Spencer posts his monthly update (and everyone will forget about this one). There has been much brilliant (and i truly mean brilliant…) talk here, but it seems that everybody has missed the “crux” of where the debate should be. Curt proposed it when he asked what the mechanism is that a photon would be received from a warmer but not a cooler source. I think (correct me if i’m wrong here geran) that geran’s answer to this was as follows:

          Geran june 3 4:11am The point is that a photon is identified by three values- frequency, wavelength, and energy. If any ONE of the three values is known, the other two can be calculated. This is true for every photon out there. And guess what, those values determine whether a photon will be absorbed or reflected by a surface. Just because a 15 um photon strikes a surface, it does not mean it will be absorbed and raise the temperature of the surface.

          (geran if there is anything that you would like to add or take away here feel free to do so…)

          Gordon has also touched upon this, but i don’t believe that i’ve seen much in the way of a rebuttal to this from either you or curt. AND it would be a shame (a darn shame!) for y’all to have come this far on the subject without having done so. This is the most in depth discussion that i’ve seen on this subject. Some how the topic always seems to get bogged down with the image of a turkey being cooked in a freezer full of ice and it gets no further than that. It’s been nice to see everyone air there differences. HOWEVER, it would be really nice to see the discussion focused on the question posited by curt. Again, it would be a shame for y’all to come this far and not quite cross the finish line…

          • Curt says:

            fonzarelli:

            geran’s question and answer was just a poor attempt at an diversion from mine, and irrelevant to the discussion at hand. It does NOT answer the question at all!

            I have asked him repeatedly for a week whether there is any difference in behavior at the target whether the 15 um photon comes from a source hotter than the target or colder than the target. He has completely dodged the question. Don’t hold your breath that he will actually answer it. His refusal is what is keeping us from “cross[ing] the finish line”!

            What do you think? Do you think there is any difference in what will happen to the 15 um photon at the target when the source is hotter than the target compared to when it is colder?

          • geran says:

            Curt, I answered the question. You will have to search the thread to find it. That will not be hard for you, as you are obsessed with it.

            You’ve already won the “blue dress” award and proved you’re willing to misrepresent what others have said, so satisfy your obsession and go searching. And, when you’re finished you can prove “cold” can warm “hot” by baking a pizza in your freezer.

            (Fonz–You quoted me correctly. Maybe you could teach Curt to do the same??)

          • Curt says:

            No geran, like a dishonest politician, you answered the question you wanted, rather than the question that was actually asked of you, so you could keep plausible deniability about the issue.

            So any time someone tries to infer what your answer really is, you can accuse them of dishonesty! It gets really old!

          • geran says:

            Curt, notice that your entire comment (8:18 pm) was your opinion, or “belief system”. No facts, just opinion. When I accused you of misrepresentation, I presented both what you said and what I said–exact quotes.

            I can sense your frustration. (I take it the “baking pizza in the freezer” experiment isn’t going so well.) Trying to support IPCC/GHE/CO2/AGW nonsense would make me frustrated also.

            Get on the right side of the issue–it’s much more fun.

          • Curt says:

            geran:

            In my comment here:

            http://www.drroyspencer.com/2015/05/trmm-satellite-coming-home-next-month/#comment-192595

            I quoted you directly, and analyzed your statement word for word. You accused me of misrepresenting you to the point of lying. No facts, just opinion. No quotes.

            You claim that you have directly answered my questions on 15 um radiation (or radiation of any given wavelength). All you have to do to prove it is to give me a link as I have just done.

          • geran says:

            Nope, you didn’t “quote” me directly. You “inserted” your own interpretation (misrepresentation). Sorry, it doesn’t work that way.

            You have already found my answers. You just didn’t like the way I worded them, as it is too hard for you to “misinterpret”. (I’m learning about you!)

            When will the pizza be ready?

          • Curt says:

            geran:

            I followed standard academic/professional style in quoting you directly, with my annotations in square brackets to show that they were mine, not yours. You act like you have not seen that before, even though you did the exact same thing when you quoted me.

            As to your answers, if you think these answers:

            “A black body emits a spectrum of wavelengths. Assuming all three bodies are ideal black bodies, then:

            Question 1: Most of the IR radiation that strikes the lower temp body will be absorbed.

            Question 2: Most of the IR radiation that strikes the higher temp body will be reflected.

            Question 3: Absorption of IR radiation is dependent on wavelength of the incoming radiation and the temperature of the target (neglecting shape of the absorber).”

            remotely answer my questions about a specific wavelength, then you have no understanding of the underlying issues whatsoever.

            I am very amused however, that you think “ideal black bodies” will reflect radiation! Did you sleep through the first day of class?

          • Curt says:

            Ahh — italic tags got screwed up!

          • geran says:

            Curt, if I confused you, I will be happy to explain. When I used the term “ideal” black body, I was emphasizing no less than perfect emittance. I assumed you wanted to understand absorption, so I wanted to specify that the emitter was “perfect”. My direction was to make the answer easy to understand, but I should have said “ideal emitter”.

            But, that was before I had learned about you. I will be much more “legalistic” in my answers from now on.

            (Has the pizza in the freezer even started to thaw yet? Maybe you should use “dry ice”. When the dry ice sublimates, you will get even more heating from the “atmospheric CO2”. Be careful that you don’t burn it though.)

          • Curt says:

            geran:

            Why do you keep digging yourself in deeper?

            One of the most basic tenets of radiative physics is that emissivity equals absorptivity at any and all wavelengths. So a perfect emitter is also a perfect absorber — an “idealized blackbody” absorbs all radiation of all wavelengths incident upon it.

            Hint: Why do you think they call it “black”?

            Did you sleep through the second day of class as well as the first?

          • Norman says:

            geran,

            It seems you are missing a very important element in your understanding of radiation. If you have a black body or even near to one that surface will absorb nearly all the radiant energy that strikes it and it will become randomized energy (heat in terms of kinetic energy) regardless of the source (hotter or colder makes no difference). But at the same time (this is the part you are not getting and why you come up with the goofy and irrelevant pizza humor…cooking pizza with ice) the surface is constantly radiating away energy. That is what you lack in your view and why Curt cannot explain elementary thermodynamics to you.

            Here is a way for you to understand it (you will twist this most simple statement it is your nature). The hotter surface is emitting two photons for every one a colder surface is sending it and that is why ice cannot heat pizza. I do not think it can be made any easier to understand. But, and this is the big one, the colder surface will slow down the rate the hotter one cools!!! If the cold surface was not present the hot surface would lose two photons with no compensation. With the other surface present it will lose two and absorb one so the overall loss now is one photon instead of two. Why make the easiest concept impossible to grasp???

          • geran says:

            LMAO!

            The climate comedy tag team has shown up!

            Curt is now confused about what he was not confused about before, when I tried to un-confuse him. Now, he is confused about what he said was confusing him.

            And, Norman has jumped in to explain that my “baking a pizza in the freezer” attempted farce of IPCC/GHE/CO2/AGW “science” is wrong! Norman, so do you no longer bitterly cling to the IPCC nonsense then? Do you now realize that atmospheric CO2 can NOT warm the planet?

            (This is climate science, folks! Enjoy the show.)

          • geran says:

            Curt, this is too funny. I can’t even reply right now.

            I just wanted to see if you agree with Norman, before my response.

            Do you agree fully with:

            “The hotter surface is emitting two photons for every one a colder surface is sending it and that is why ice cannot heat pizza. I do not think it can be made any easier to understand. But, and this is the big one, the colder surface will slow down the rate the hotter one cools!!! If the cold surface was not present the hot surface would lose two photons with no compensation. With the other surface present it will lose two and absorb one so the overall loss now is one photon instead of two.”

            (I’m not sure it can get any funnier than this.)

          • Norman says:

            geran,

            Who has claimed that CO2 will warm the surface? No climate scientist I have read has made this claim. I do not see Curt making this claim.

            GHG slows the cooling rate. You may fail to understand there is a constant input of new energy into the system from the Sun. The Sun is adding more photons to the mix. If you slow the rate of energy loss with a constant input (the Sun is fairly constant) the surface will warm above the temperature it would be if you did NOT SLOW the rate of cooling (which is what GHG’s do). They absorb IR and redirect it. Where the photon emitted by the surface would leave to space without GHG present some will be redirected to the surface. The emissivity of the Earth is above 0.9 so almost all radiant energy directed to the surface will be absorbed.

            Listen to reason man. The climate scientists claim is that the surface radiates at an average (overall every meter of Earth, some places more some less) of 390 watts/meter^2. This can be translated to actual energy of 390 joules of energy/second are leaving one square meter of surface. Joules will determine the ultimate temperature of a surface based upon its composition. So you have 390 joules/sec-meter leaving the Earth surface and the atmosphere is redirecting 324 joules/sec-meter back to the surface. That means the Earth’s surface is losing 66 joules/sec-meter. This will cool the surface not warm it so I need you to explain why you claim climate scientists (Roy Spencer included) are saying that a cold atmosphere is warming the surface. Where does this faulty thinking come from. You and many others seem to have this ingrained in your core thought process but what logical reason do you still believe this to be a true thought. Clean up the cobwebs and sweep the dust from your thought process.

            If you claim Curt or my understanding to be flawed at least understand it. Curt really is very correct in all his thermodynamics posts. I have not seen a fault or flaw in one yet and it is all textbook. I linked to a textbook on thermodynamics you did not see it. Let me link to in here and please look you will see how very wrong you are in your current thought process and maybe give Curt a humble apology and thank him for the free education.

            http://www.kostic.niu.edu/352/_352-posted/Heat_4e_Chap13-Radiation_HT_lecture-PDF.pdf

          • geran says:

            Norman–Wrong, confused, and rambling with just the right touch of bloviation.

            All in all, one of your funnier comments.

          • Norman says:

            geran,

            It is quite obvious that you are the true meaning of the word Internet Troll. You have nothing real to say. You really do not know anything at all about science or thermodynamics. You believe you are super clever and awesome and maybe you are but you are not here to understand, learn or inform. All your posts are very empty of thought and valid science. You cobble some words together and maybe influence a couple people now and then.

            Dense and boring you really are as you get sounding the same after awhile.

            A real waste of potential you are. Trolling to amuse yourself for some unknown reason. Enjoy yourself, you are probably the only one who does.

            Anyway did you at least click on the thermodynamics link? Probably not since you are a Troll and not interested in anything but being annoying and trying to provoke an emotional response.

            One question to ask you. How old are you? I hope under 18 as your tactics appear as someone who has not attained an adult thinking level yet. If you are older than what a sad state!

          • Curt says:

            geran:

            You have refused to answer my three questions for a week now. (My thermo professors would never have let me get away with answering different questions from what I asked…) Why should I answer yours?

            I tell you what, if you answer those three questions directly, without any evasion, plus the one directly below, I’ll answer your latest.

            Here’s the new question.

            You’ve been trapped outside for a long period on cold night, lets call it 0C, in light clothing. You are getting hypothermia, and your body temperature has dropped from its normal 37C to 34C. You are in serious trouble.

            I see your predicament, and offer to let you come into my house, which is 25C inside. Nothing in the house is warmer than this?

            Do you turn down my offer, because this is still colder than your body temperature, and “cold cannot make warm warmer!” (as you have repeatedly stated)?

          • geran says:

            Curt: The answer, of course, is a warm house would be more comfortable than outside. But, you talk about senseless, irrelevant questions? Your body is a “heat source”, duh. Body heat actually warms a cooler enclosure. No violation of physics.

            Norman: Sorry buddy, but your nonsense is not worth trying to answer. Your “two photons” vs. “one photon” rambling was hilarious, but there is no way you can get back to science after going so far down such a road.

            For everyone else, here is how the pseudoscientists corrupt science: They state their “belief”. Then, they write about a lot of actual physics. They, they restate their belief as is they have proved something. The idea is to throw out so much “smoke” that their false science is covered up.

            An example of their technique: The state their belief “Oranges are blue”. Then, they ramble along for 5-10 paragraphs explaining color spectrum, how fruits are different, how the eyes recognize colors, how photosynthesis works, etc., etc., etc. Then, they state “See, oranges are blue.” The science they cite is usually correct, it is just that it is not relevant to their point.

            If you try to argue with them, they just say that you do not understand the science. That’s why, when they try to claim the atmosphere can violate the laws of thermodynamics, I throw out things like “baking a turkey (or pizza) in a freezer”. The purpose is to force them to face reality. They don’t like that….

          • Norman says:

            geran,

            Objectively it is you who use psuedo-science and think it is real.

            Pal I linked you to an actual thermodynamic textbook. Did you look at it? You will find your “science” whatever you want to call it, is wrong, has no basis in reality. Try to undertand these equations in thermodyanmic textbooks have proven true in the real world. Engineers use them to design heat exhchangers, power plants, and many other products that generate and involve heat. If they were not correct engineers using them would know very soon. Things they designed for a given function would not work as expected which could be very bad.

            You state “If you try to argue with them, they just say that you do not understand the science. That’s why, when they try to claim the atmosphere can violate the laws of thermodynamics”

            You really do not understand the science and NO ONE is making the claim the atmosphere is violating the laws of thermodynamics. Why do you keep this STRAWMAN attack going?

            I think you should read from an actual thermodynamics text book. Learn the real science and come back and post rather than constantly making an idiot of yourself. So how old are you?

          • Curt says:

            geran, you say: “Your body is a “heat source”, duh. Body heat actually warms a cooler enclosure.”

            The earth has a “heat source”, too. It’s called the sun, duh!

            This makes the two situations effectively equivalent, thermodynamically.

            Situation 1A: Metabolism heats body. Body heats very cold ambient (273K, 0C). A certain body temperature could be maintained steady state, unfortunately not high enough. We call this hypothermia.

            Situation 1B: Metabolism heats body. Body heats cooler (298K, 25C) enclosure, enclosure heats very cold ambient. A higher body temperature can be maintained. No hypothermia.

            Situation 2A: Sun heats earth. Earth heats very cold ambient of deep space (3K). A certain surface temperature can be maintained steady state, but lower than we consider comfortable.

            Situation 2B: Sun heats earth. Earth heats cooler (~255K, -18C) enclosure of radiatively opaque gases, enclosure heats very cold ambient of deep space. A higher surface temperature can be maintained.

          • geran says:

            Situation 2A and 2B are not real world. They exist only in Curt’s head.

            Curt has been baking his pizza in the freezer all day. He is amazed it is still frozen!

            (Curt, did you try “dry ice”? How about the ultimate heatng devices–mirrors?)

          • Curt says:

            Situation 2A is the moon.

            Situation 2B is the earth.

            These are not hard concepts, but they are completely beyond you.

          • geran says:

            First Curt says: ‘Situation 2A: Sun heats earth.”

            Then Curt says: “Situation 2A is the moon.”

            Oh, I get it. You think the Earth is the Moon!

            (Isn’t the word “looney” derived from “lunar”?)

          • Curt says:

            You can’t actually be that pathetically stupid, can you?

            Yes, I guess you can!

          • Kristian says:

            Hi, Curt.

            I see you finally got around to the clincher: It is in fact the Sun doing the actual heating; the atmosphere only reduces the cooling rate of the surface. It acts as an insulating layer around the solar-heated object, not as a second heater.

            A ‘three-body problem’.

            Why didn’t you simply incude the Sun right away? Yes, I know you take it as a given, that it should be obvious to everyone. But it’s not obvious to these people.

            OF COURSE a cooler atmosphere cannot make a warmer surface even warmer (that is, directly raise its temperature in absolute terms). It can only ever make it cool more slowly than what space can, by itself being a less cold ‘cold reservoir’ to the surface than space. But bring in the Sun and you get surface warming (temperature rise), because the Q_in of Earth’s surface remains (almost) the same, but its Q_out is vastly reduced, until its temp has risen sufficiently to balance the two.

            The eternal confusion of the Postma acolytes is just this: They simply don’t get the Q_in vs. Q_out part. They don’t think the Q_out matters (or even exists as a separate entity), only the Q_in. And so, when you argue that the cooler object forces the warmer one to become even warmer, then they think you mean that Q_in is increased, when what you do mean is that Q_in is held constant, but Q_out is reduced. IOW, they think you mean that the cooler object actually heats the warmer one directly, when in fact it is only itself heated at a slower rate by the warmer object than what, say, space would be. It is the Sun – the third body and the actual ‘heat source’ – that does the real heating.

            I wouldn’t want to perpetuate such confusion. I would rather try to dispel it (at an early stage).

          • Kristian says:

            On second thought, reading through geran’s imbecile responses here, I fear it wouldn’t be possible anyway, no matter how hard you tried …

          • geran says:

            Hi Kris.

            I’m glad to see you types are admitting “It’s the Sun”. That’s progress, now if we can just get you to drop the “magic gas” routine.

            Oh, and when all you got is ad hominem, that means you lost the debate.

            I’ll give you the last word here, and see you next time.

            Keep learning and you’ll eventually get it right!

          • Norman says:

            geran,

            Don’t let the door hit you on your way out!

  36. Mike M says:

    Is there a link to the rainfall data in the form of “total global rainfall per month/week/day”?

    • Mike M says:

      I should have added what it is I wanted to do – simply plot global rainfall, and/or its derivative, against or in comparison to, global temperature. (I went to the TRMM web site but the descriptions are too esoteric for me to figure out what it what.)

  37. Norman says:

    fonzarelli and tonyM

    I believe this webpage may answer many if not all your answers you have asked. It may take some time to work through it but it should be rewarding.

    http://www.kostic.niu.edu/352/_352-posted/Heat_4e_Chap13-Radiation_HT_lecture-PDF.pdf

    tonyM page 7 of this link should answer your ring question quite well. This is even better than rings as all radiation is enclosed with two spheres. The inner sphere will transfer all the radiation it can emit to the larger outer sphere but the larger sphere will not transfer all its radiation to the smaller sphere. Some radiation will be aimed from one wall of the larger sphere to the opposite side without ever hitting the inner sphere. It would be the same with the rings. All the energy from the inner ring will hit the outer ring but not all the energy of the outer ring will fall upon the inner ring. Some will be directed to the opposite side without ever hitting the inner ring. (Is that the answer you requested?)

    fonzarelli, I also think this weblink will answer your questions about photon absorption. Maybe geran could take a peek and learn some real physics (we can only hope!!).
    Page 21 should explain the photon situation.

    Hope this helps. If not I like to research the topic as it does help me learn and better my understanding.

  38. Massimo PORZIO says:

    Hi Curt,
    I read the SoD thought experiment linked above by Mike B (the one about the PVC spherical chamber) and I agree with that.

    Just a question for you.
    In that experiment the source temperature never changed, while in your bulb real experiment it does.
    Is it due to the different emissivity of the filament at the LWIR (lower) respect to its emissivity at the visible light (higher)?

    I fact SoD held the emissivity to the same value, so the filament temperature didn’t change.

    Have a great day.

    Massimo

  39. Curt says:

    Massimo:

    Even in my initial experiment that I published, any change in power generated due to the changed resistance (as a function of temperature) of the filament was absolutely trivial compared to the changes in the bulb surface temperature from the different shells.

    I showed the current readings to demonstrate that the increase in temperature could not have come from increased power from the filament.

    • Massimo PORZIO says:

      Hi Curt,
      excuse me, I was not discussing the validity of your result, I would just know your opinion on what does the difference between your real experiment and the SoD thought experiment.
      Could it be, as I supposed, that it is because the returning photons aren’t on the very same WL where the filament has been designed to get maximum emission/absorption?

      I’m not thinking that one of two results are wrong, I’m just intrigued by this discrepancy.

      Have a great day.

      Massimo

  40. Curt says:

    Massimo:

    SoD never specifies the temperature of the source in his thought experiment — he simply specifies it as a constant power source. He also doesn’t specify the method of heat generation or the material.

    These kinds of simplifications are very common in college thermodynamics problems — it’s so students can focus on the real issues of the problem and not “get lost in the weeds”. (Do you know that expression?)

    • Massimo PORZIO says:

      Hi Curt,
      “get lost in the weeds”, uhmmm…
      No, I don’t know indeed, but I can imagine the meaning.

      Anyway, the simple fact that the filant increased the temperature, while the input energy reduced, shouldn’t mean that the filament changed its emissivity?

      Excuse me, but when I wrote to MikeB that I’m ignorant in this field I wasn’t joking. I attended the thermodynamic class so long ago, and in last 30 years I used it very little, and never in these details.

      So maybe I’m telling silly things here.

      Have a great day.

      Massimo

      • fonzarelli says:

        Massimo, curt’s not talking about marijuana here (wink!). Think of a doug cotton comment. THAT would be “lost in the weeds”…

        • Massimo PORZIO says:

          Hi Fonzie,
          yes I get it.
          Many posts above you argued about the Americans who don’t know the way to “take it easy”. Well, even here in North Italy most of us forget it.

          These aren’t the days of “The Eagles”, no more:

          https://www.youtube.com/watch?v=FhH3mRkKDX8

          I suggest to listen to The Eagles’s “Hotel California” too, a “must” for relief your soul. šŸ™‚ šŸ™‚ šŸ™‚

          Have a great weekend.

          Massimo

  41. Massimo PORZIO says:

    HI geran.
    Some posts above you wrote “Your body is a “heat source”, duh. Body heat actually warms a cooler enclosure. No violation of physics.”

    I think that I started to get your point.

    Have a nice day.

    Massimo

  42. Norman says:

    geran,

    You have a reputation going I see. Even stevengoddard called you out.

    “stevengoddard says:

    October 18, 2014 at 2:34 am

    geran,

    You are making very stupid comments”

    You made many of the same mindless comments on his blog that you do here.

    If you want to contribute your intellect and reason please read some REAL thermodynamics material (where backraditation is part of the equation for energy flow) instead of using your own limited opinion or what you somehow have concluded is absolute truth.

    If geran makes a statement it is the absloulte truth. He has been endowed by the Lord God to have total wisdom and truth residing within. Thinking and reasoning are not part of the gift just making statements. You questions if I and Curt were brothers. I would wonder if you are Doug Cotton’s son. The only difference between you and him is he can’t post three times without adding Venus and Uranus in a post.

  43. geran says:

    “If geran makes a statement it is the absolute truth.”

    Gosh Norman, and I didn’t think you ever got anything right….

  44. tonyM says:

    Hi Norman;
    Thank you for your efforts; good find.

    I had envisaged this problem and thought that at there would be a maxima which would exceed the loss even if one also had to resort to similar T. But it does not happen.

    I was thinking of two flat plates of different size or a cube facing two identical cubes.

    These also fail but for a different reason with the larger area emitting more energy to the inert medium than would the smaller face which now interacts with an active larger area (we would only be testing the sole cube to see whether it increases its T, which won’t happen).

    There does not seem a way to construct an experiment to obviate this.

    The only thing I can think of is something like a one way mirror with photons being sent from a lower T source to the object already facing that mirror without reciprocal interaction. We would be trying to establish whether the T of the object actually increases once these extra photons hit.

    Perhaps there is refracting/reflecting material or surface which have this this “one way” capability. I have no idea. Perhaps some engineers can help. Heeelllllllllp!

    I know you feel page 21 vindicates your idea of lower T source photons being able to be absorbed but that is not what those formulae prove. The T^4 difference shows the flow is always in the direction of the lower T (if lower T photons are actually absorbed they need a simultaneous compensating back-flow).

    Even this does not disprove Doug Cotton’s pseudo scattering. All he is saying is that the lower T photons don’t get absorbed and thermalised at all as those lower energy quantum states are already occupied. Those lower energy states are equally prevented from emitting by the incoming photons which are thereby scattered.

    This would parallel any volume of air at thermodynamic equilibrium instantly ejecting an equivalent photon to one coming in. Normally the life of a photon thermalised by CO2 would take milliseconds to respond and is not instant so there is time to pass energy on to other molecules via KE loss.

    This is how I have understood the interaction.

  45. Norman says:

    tonyM

    The equation shows that the temperature of the colder surface (surface 2) is having an effect on the heat transfer from the hotter surface 1. Use the equation and put in different values for surface 2 and see how it effects the heat transfer. As surface 2 temperature goes up surface 1 is radiating away less energy because of the photons from surface 2 being absorbed by it and cooling at a slower rate (losing energy at a slower rate).

    If Doug’s conjecture had merit the temperature of the surface 2 would have no effect on the energy emitted by surface 1 (unless surface 2 was warmer) since none of this energy can be absorbed. The equation shows that surface 2 temperature has an effect on surface 1. These equations are valid and are used in manufacture and design of objects used where heat exchange is needed.

    As Curt has asked geran, how does a surface know the temperature a photon came from? That means the surface will selectively absorb any 15 micron photons from the Sun (if they can make to through the atmosphere) but somehow know to reject them if they are from a colder source?

    Did you look at Curt’s experiments on WUWT that he linked to? Those may answer some of your questions on photon absorption. Massimo and Curt has discussion over these experiments.

    • tonyM says:

      Hi Norman,
      You say: “…As surface 2 temperature goes up surface 1 is radiating away less energy because of the photons from surface 2 being absorbed by it and cooling at a slower rate (losing energy at a slower rate). “

      In this second scenario (limited to snap shot) we have S2 at an increased temp. It is radiating more photons as per the T4 formula. Photons from S2 are being absorbed by S1 (ok go along with this for now).

      From your comment S1 is now also ‘radiating away less energy’ and ‘losing energy at a slower rate’ than before.

      Can’t agree here as a conclusion I can form from your argument is that the T of S1 has increased as it has had a net increase in photon transfer to it compared to where it was in scenario 1.

      Closer to reality is S1 has no change in photon emission if it has not changed its T irrespective of the T of S2. This is how the formula on P21 is derived from the individual emission rate S-B calcs.

      As we are focusing on photon transfer we have used photons as some constant equivalent energy unit. S-B would be in watts. That’s fine as long as we understand each other. Fully aware that photons can be different as the energy is given by hv. If has not come out clearly then kindly point it out.

      You say:
      “If Doug’s conjecture had merit the temperature of the surface 2 would have no effect on the energy emitted by surface 1 (unless surface 2 was warmer) since none of this energy can be absorbed. “

      You are not right here. The increase in photons from S2 simply lifts the cut off point of absorption by S1 (on the Plank curve). This means S1 loses fewer photons due to the pseudo scattering below that cut off level. He in effect actually takes the formula on P21 to the extreme i.e. at face value even though he does not say it that way.

  46. Norman says:

    tonyM,

    Thanks for the correction. I did speak incorrectly when I made the claim ” As surface 2 temperature goes up surface 1 is radiating away less energy because of the photons from surface 2 being absorbed by it”

    Surface 1 would not be radiating away less energy, it would radiate based upon the surface temperature and its emissitivity. I do stand by the statement it would cool slower.

    I am thinking on the pseudo scattering effect or absorption and re-emission. Would not the process have the same effect regardless of what is actually taking place at the surface of 1? If pseudo scattering or absorption both slow down the cooling rate (which is all any climate scientist I have read has claimed and all the energy budgets I have found show this effect) what does it matter how you describe the interaction? It would only matter for the sake of the beauty of Truth concept of what science is trying to find.

    I have a potential experiment that may prove one or the other effect. You have a heated surface held to constant temperature to maintain a uniform radiation flux, At an angle you use a source of filtered light that is 15 microns, at the opposite side of the angle you have a detector (diffraction grating in front to filter out all but 15 micron photons) with a polarized filter in front of the detector. You establish the conditions first by just measuring the energy of 15 micron photons that reach the detector from the heated surface to establish a baseline. Now you add a source of polarized 15 micron photons to strike the surface. The polarized filter in front of the detector is rotated so that none of the generated 15 micron photons can get through. One thing I do not know about this experiment is if a polarized photon will change its polarization upon hitting a surface but not being absorbed.

    My take on the experiment is that if the 15 micron polarized photons are not changed by interaction with the surface (pseudo scattering, bounce off with no net change in photon properties). That would mean you would have no increase in photon detection with the polarized light source on since none of the photons would make it through the polarized filter. If the photons were absorbed by the surface they are gone, thermalized with no polarized identity. If the surface now has more energy at the bends and stretches that generate 15 micron photons and emits more of these photons they would not have a unique polarization that the original photons had and now you will have a chance for any new photons getting to the detector and giving a higher detection than the baseline.

    Another question about the pseudo scattering. Why would a surface have a preference of choosing internal energy to stimulate surface bonds as a more likely choice than a photon hitting the surface? Both are energy sources to replace energy that is being continuously radiated away. Why would a photon hitting the surface scatter as opposed to a kinetic energy vibration from layers of molecules below the surface?

    • tonyM says:

      Yes a higher T for S2 does slow the heat loss from the S1.

      Yes agree that the net result of the equation is basically the same in a macro sense, mostly.

      Claes Johnson suggests it is an issue in atmospheric physics. You may have dismissed him prematurely. He has a lot there on his site so I don’t pretend to understand it fully and only dabbled in a few sections.

      Claes actually concludes that a two way stream is unphysical, certainly for an atmosphere, and shows that the two-stream  gross-flow  Swartzchild model leads to :

      “… conclude that the two models are not identical, and so the models give different temperature distributions. 

      The lesson is that radiative heat transfer should better be modeled using the physical one-stream net-flow correct form of Stefan-Boltzmann’s law (5). Using the unphysical two-stream gross-flow form (1-2) can lead to unphysical results. “

      Here he is contrasting the two way flow to the equation on P21, the net one stream. But read it for yourself as it quite detailed. He also points out the T^4 one way formula is for transfer to a zero T.
      http://claesjohnson.blogspot.se/2015/04/unphysical-schwarzschild-vs-physical.html

      Doug seems to have embellished this into pseudo scatter but it leaves me confused as he, in effect, accepts a two way flow (but with scattering without absorption). Claes seems to suggest one stream (although he is talking for modelled purposes). But they should be the same. Dunno!

      In trying to respond to your pseudo scattering question, already occupied quantum bond states are not available much like a Pauli exclusion principle for example. Doug is actually saying there is no absorption and no emission of/from those occupied quantum states and only energy above the cut off overlapping Plank curves is then available for “net” emission. I think I have his concept right with this net emission filling the “quantum” states of surface B in preferential order at that T, I imagine.

      It is not possible to understand clearly the quantum world and how it interacts. For example how does an electron know its partner’s quantum spin when they are separated and other entanglement at a distance issues? “Spooky” said Einstein. An electron is certainly a bit more physical than a photon which has no mass at all! Interesting and I guess will always leave one scratching his head.

      The only practical difference that I can see is in Claes’ comments.

      Let me think on your experiment before I comment and thank you for your efforts.

      • JohnKl says:

        Hi TonyM,

        Just a brief point. From what I’ve read of Claes Johnson in the past, he doesn’t believe in the existence of photons! He accepts quantum emissions of energy, but not the usual definition of photons as commonly referenced. Later I’ll have more to state.

        Have a great day!

        • Norman says:

          JohnKl,

          I was under the impression (from my studies) that a photon exhibited both properties depending upon the way an experiment is set up.

          We have the photoelectric effect where EM energy can actually knock an electron free from a metal surface (behavior of a particle).

          There is also the double and single slit experiments performed with both photons and electrons. If only one slit is available for light or electrons to travel through, the pattern on the opposite side is a round circle (like a stream of particles). If two slits are available the electron or light interfere with itself and you get a band of regions which resembles a wave. It is a duality. Light can act like a wave or particle.

          • JohnKl says:

            Hi Norman,

            My post referred to Claes Johnson’s views and not necessarily my own. However I do share his skepticism in this case.

            In the example you provided of the photoelectric effect I see no reason why a wave phenomena cannot produce the same effect of knocking electrons free. Waves have generally been understood as disturbances (usually periodic) in observed phenomena.

            The double slit experiment using two holes and/or slits resembles Young’s experiment resulting in several bands in what appears to be a wave-like phenomena. This result proves highly suggestive of wave action and not particle action. The presence of a circle resulting from a single slit doesn’t disprove wave action. Thank you for the reply and…

            Have a great day!

          • Norman says:

            Hi JohnKl,

            This explanation of the wave/particle duality of light (and electrons) may help in your quest for the truth on this issue.

            http://hyperphysics.phy-astr.gsu.edu/hbase/mod1.html

            The link goes into details of the photoelectric effect and why it can only be the product of particle action with respect to light. They give details to the reasons for coming to these conclusions.

            Hope this helps eliminate confusion. I am not sure Claes is an expert in this field. He may be a phenomenal mathematician but it may not translate to physics so well. If I want to learn advanced math he could be a great teacher but I think I would look elsewhere when it comes to physics.

            Have a really great day!

          • JohnKl says:

            Hi Norman,

            Thank you for a thoughtful post and link. in my opinion your link proves the wave nature of electrons and radiation rather than the particle nature or even a dual nature. Your link stated “analysis of data from the photo electric experiment showed that the energy of the ejected electrons was proportional to the frequency of the illuminating light.” Frequency and/or wavelengths is a wave function.

            Keep in mind Louis de Broglie proved the wave nature of radiation. The particle nature of radiation is projected from the discontinuous nature of energy emission but that doesn’t mean that radiation takes the form of a discrete point in space or time or packet. For example Louise de Broglie Believed the electrons did not exist one single spot in its orbit it had a wave has a wave nature and exist at all places in the allowed orbit.

            Albert Einstein apparently had a similar view and stated “since the theory of general relativity implies the representation physical reality by a continuous field the concept of particles or material points cannot play a fundamental part nor can the concept of motion.”

            “A careful analysis of the process of observation in atomic physics has shown that the sub atomic particles have no meaning as isolated entities.
            Quantum theory thus reveals a basic oneness of the universe it shows that we cannot decompose the world into independently existing smallest units.”
            Fritjof Capra, The Tao of Physics, on Quantum Theory

            http://www.spacemotion.com/physics-quantum-theory-mechanics.htm

            Norman please note I visiting my folks in Phoenix and do not have access to a computer so my responses will be infrequent. It’s not because I’m being rude but have difficulty spending time typing into my small samsung phone.

            Have a great day!

        • tonyM says:

          JohnKl

          It does not surprise me that Claes has that view. He is in good company.

          I stated earlier that even Feynman was somewhat bemused:
          “ ..We always have had … a great deal of difficulty in understanding the world view that quantum mechanics represents. At least I do, because I’m an old enough man that I haven’t got to the point that this stuff is obvious to me. Okay, I still get nervous with it. …. It has not yet become obvious to me that there’s no real problem. I cannot define the real problem, therefore I suspect there’s no real problem, but I’m not sure there’s no real problem.”

          “I think I can safely say that nobody understands quantum mechanics.”
          Richard P. Feynman

          Add Einstein who was the one to suggest the photoelectric effect was due to light being a stream of packets of energy (photons) rather than waves (won Nobel Prize for it). Towards the end of his life he said:
          “All these fifty years of conscious brooding have brought me no nearer to the answer to the question, ‘What are light quanta?’ Nowadays every Tom, Dick and Harry thinks he knows it, but he is mistaken. (Albert Einstein, 1954) “

          You have a nice day too.

        • JohnKl says:

          Hi Norman,

          Albert Einstein quote should have Read:

          “Since the theory of general relativity implies the representation of physical reality by a continuous field the concept of particles or material points cannot play a fundamental part, nor can the concept of motion.”

          Have a great day!

      • JohnKl says:

        Hi TonyM,

        One other point. Claes Johnson claims the two way stream is un-physical because thermal energy can flow from colder to warmer (up the temperature gradient).

        Have a great day!

  47. Norman says:

    tonyM

    “Using the unphysical two-stream gross-flow form (1-2) can lead to unphysical results.”

    The equations that engineers use are based upon the gross-flow and would have known long ago if these led to unphysical results I would think as things designed would not be working like they are supposed to. You may see massive equipment failures as heat exchangers are not removing or adding heat as required by a subsystem that leads to failure for the overall system.

    • tonyM says:

      Hi Norman:
      I may not have been very clear, but I thought I did qualify Claes statement to the atmosphere. Claes is specifically analysing the Schwarzschild radiative heat transfer in a horizontal slab atmosphere

      It would not be doing justice to Claes work to simply compare it to what engineers have been doing in other fields.

      Otherwise we can use the argument that because engineers mostly use σ Stefan-Boltzmann’s constant at a given value in the T4 formula then that value must always apply. It is not a universal constant of physics.

      Additionally quoting Gerlich et al “the T4-law will no longer hold if one integrates only over a filtered spectrum, appropriate to real world situations” even though engineers successfully use it under other conditions.

      Have a nice day.

  48. Norman says:

    tonyM

    I am trying to read through the Claes article you linked to in a post above. He states (Claes): “Schwarzschild’s two-stream model is unphysical in the sense that the gross downward flux F − is directed against the temperature gradient and thus violates the 2nd law of thermodynamics.”

    I do not understand why this is a true or correct statement. If the downward flux is less than the outgoing upward flux how can it warm the surface? A violation of the 2nd law (in my reading of it, maybe there are other interpretations of it) would be if a cold surface got colder via transfer of its energy to a warmer surface making that surface warmer. This does not violate the first law as the total energy of the system is conserved but such events as this are never observed. I do not know why they claim that energy from a cooler surface moving to a warmer surface violates the second law. Maybe you or John can help me see why they believe this. A cold surface is still radiating energy away from it and any surface in its path will be hit by this energy and if it is not reflected or transmitted it will be absrobed. That a hotter surface is receiving energy from a cooler surface does not imply in any way that the colder surface is warming the hotter surface. The hotter surface is radiating away energy at a faster rate than the incoming energy can warm it so it still cools, just at a slower rate as given in the thermodyanics equations.

    I do not know how Claes forms his conclusions. If you read the comments to his article most do not agree with him.

    • tonyM says:

      Hi Norman:
      Claes says the T4 formula is based on emitting to a zero T. It is supposed to be an independent emission. Yet its use dictates that something cooler cannot send more energy to something warmer which implies a dependent relationship i.e. the s(Ta4-Tb4) which works on the difference.

      Then he shows that the results of the two formulae are different for the modelled atmosphere other than for an extreme opaque or clear atmosphere.

      But this on its own does not prove anything as the atmosphere model may be, and surely is, wrong anyway which could be the cause of the difference. (but I don’t know why).

      I don’t see anything wrong with your or Doug’s approach. I prefer Doug’s as it seems more in keeping with the quantum world interaction. But then if we are never going to see it and if the predictions give the correct outcome it matters little unless some fundamental situation shows otherwise.

      Have a nice day.

  49. Bryan says:

    Curt says

    “You put down chemistry students by saying that “most of them seem to seem to be left with the second law is about ‘more energy in one direction(hot to colder) than the other’ “.

    I don’t put down chemistry students here’s what I actually said

    “I have noticed very similar comments to Curt’s from other chemists.
    I’m sure they are all very intelligent but they share some common mistakes.”

    I go on to speculate that some of these “very intelligent” students perhaps suffered from an oversimplified thermodynamics curriculum.
    Otherwise I cannot account for the fact that it is almost entirely non physics graduates like yourself that find difficulty with words like ‘heat’
    For example I have often had discussions with physicists like Joel Shore and Tim Folkerts and we disagree on many things.
    However ALL PHYSICISTS like Joel,Tim and myself agree that heat transfer (by any means including radiation) only occurs spontaneously from a higher to a lower temperature.
    Now other chemistry students who follow a more rigorous thermodynamics curriculum will achieve the same level of understanding as the physics students.

    This is not unusual and many chemists have made a big contribution to thermodynamics.

    Finally I note here that this is just another example of your poor reading comprehension.

    Why else would you misconstrue my posts?
    It certainly would not be deliberate now,.. would it?

  50. Curt says:

    Bryan:

    I will turn it right around on you. Where on earth do you get the idea that I believe in anything other than that spontaneous heat transfer is ever other than from higher temperature to lower temperature?

    I have been quoting Clausius repeatedly and approvingly in this thread, including directly to you, to that effect.

    So I defy you to show me anywhere that I have stated otherwise.

    And I notice that you, who claims to understand this subject so well, will not answer a trivial problem in the subject, of the kind that an undergraduate early in an introductory course could answer in a couple of minutes.

    You dodged similar questions from SoD for weeks. Why? The only reason I can think of is that you don’t know how to answer them!

    I repeat the question for you here:

    You have a small spherical satellite with a 1.0 m^2 surface area. It has an internal continuous 100-watt power supply from a radioactive source. You are testing it in a vacuum chamber under two different conditions.

    In the first case, the walls of the chamber are held at a very low 77K by liquid nitrogen circulating through the walls.

    In the second case, the walls of the chamber are allowed to come to earth ambient temperature of 288K.

    What is the steady-state surface temperature of the satellite in each case? You can use an emissivity for the both the satellite and the walls of 0.95, or if you want it a little simpler, assume perfect blackbody emissivity of 1.0.

    If your answers for the two cases are different, what explains the difference?

  51. Bryan says:

    Curt

    Who TF are you or SoD to set problems?

    In the case of SoD the problem was near identical to Willis Steel Greenhouse.
    The ‘solution’ has been rehashed and published several times and yours is no different.
    I said that given all the assumptions I agreed with the Willis solution for what its worth.
    I went further and gave the general dynamic equation.
    What is more interesting was that SoD was shown on his previous thread to have made at least three unsubstantiated and false claims about G&T.
    He was boxed in and starting a new thread was his only way out.
    If I thought you were worth it I would give you the links but I really cant be bothered
    Its pathetic that climate alarmists have so little science background that all they seem to dredge up is variations on the Steel Greenhouse.
    Another poor attempt here is being made to confuse insulation with a heat source.

  52. Curt says:

    Bryan:

    You continue to insist in your infinite wisdom that generations of science textbooks are wrong and that you know better, but when asked to demonstrate the most basic competence yourself, you dodge it completely.

    Now, in your latest response you agree that radiative insulation between a powered object and cold ambient leads to a higher temperature for that object than would be the case without the insulation.

    That is the fundamental statement of the radiative greenhouse effect.

    You have also agreed that there is radiative energy transfer from a colder object to a warmer object (but always less than from warmer to colder). So what is your objection to the “back radiation” meme (as long as it is shown that “forward radiation” is greater)?