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Saturation of IR Wavelengths

May 9, 2022

By Paul Homewood

 

I though I would throw this one open for comments.

I may summarise them into a post later:

 

 

 

 image

228 Comments
  1. May 9, 2022 6:02 pm

    The jig is up on the U.N. IPCC’s greenhouse gas warming hoax.

    It’s not that IR wavelengths are saturated, it’s that atmospheric gas molecules aren’t Planck black body radiators, hence they can’t stop Earth’s surface from cooling via radiation, period.

    http://www.historyscoper.com/whyaregreenhousegastheoriesdeadwrong.html

    https://www.quora.com/Who-was-the-first-scientist-to-suggest-that-carbon-dioxide-from-industrial-processes-could-create-a-greenhouse-effect/answer/TL-Winslow

    What are you doing to spread the news and get the IPCC laughed out of existence?

    • Adam Gallon permalink
      May 10, 2022 12:40 pm

      Your hypothesis drops to bits.
      Now, who should I believe?
      T.L. Winslow, https://medium.com/@historyscoper_33955/about who appears to have a very high opinion of himself.
      Or William Happer, Emeritus Professor? https://www.scienceunderattack.com/blog/2021/4/5/how-near-saturation-of-co2-limits-future-global-warming-74

      • May 11, 2022 8:50 am

        Happer every time

      • May 15, 2022 9:31 pm

        @Gallon
        My hypothesis drops to bits? Who should you believe, moi or some academic?

        When it comes to science, you don’t believe, you know and understand.

        Drops to bits? I missed that long refutation. So why can’t you quote my killer disproof of the IPCC greenhouse gas warming hoax point by point and furnish a scientific refutation, but only flame me with ad hominem? How much does the IPCC pay you?

        Happer spent his career sucking up to academia. I spent my life outside academia studying instead of doing the dogwork he had to in order to stay employed, and always thought for myself, which he doesn’t and couldn’t to work up in the system that’s been taken over by the IPCC octopus. Since he can’t refute my paper even to discredit me and vindicate his rep it’s QED.

        I not only refuted the IPCC and all its apologists like Happer, I refounded thermal physics as it applies to Earth’s atmosphere, causing the need for all IPCC papers to be junked and the real work to build on mine when it’s out of the way.

        Everybody here has their big chance to get on the bandwagon before it’s pulled out of the station. Don’t make me say I told you so.

        I wish you could also enjoy my new paper exposing the IPCC for a global Marxist lie machine that’s poised to destroy Western civilization by controlling CO2 emissions.

        http://www.historyscoper.com/whatisthedifference.html

  2. May 9, 2022 6:05 pm

    All I can say is the subject of IR absorption in the atmosphere is hideously complex. Does anyone actually understand that Wijngaarden and Happer paper ?!

    https://arxiv.org/abs/2006.03098

    There does seem to be an issue, in that when people attempt to calculate the greenhouse gas (GHG) forcing from physics, e.g that paper, they come up with smaller values than the IPCC use.

    However most climate models don’t attempt to calculate the forcing from first principles, because it is so complex with so many unknowns. The GHG forcing numbers in models come from subtracting known sources of warming from the measured warming. This is my understanding anyhow.

    • It doesn't add up... permalink
      May 9, 2022 7:44 pm

      I actually do have a good understanding of the paper, as I studied the underlying science: I read it almost as if I were doing a peer review. It’s an excellent piece of work, because it shows that calculating what the absorption spectra should look like at the ToA from the detailed data produces results that are in very good agreement with actual satellite measurements over a range of atmospheric conditions and latitiudes, which implies that not much if anything of significance is being left out. That gives a high degree of confidence in extrapolating the calculations to hypothetical atmospheres with higher concentrations of GHGs will also produce results that agree with what would be measured were they to occur.

      Of course, the climate depends on weather rather than radiation balances, and that is not specifically modelled. However, the overall consequences on the energy budget of the earth are clear: IPCC models run too hot.

    • Phoenix44 permalink
      May 10, 2022 8:54 am

      Climate scientists claim that the forcing is an “emergent property” of the models, which is complete BS. I believe there’s a fudged disconnect between the modelers and the scientists who don’t understand models that suits both parties, in the same way there was for Covid. The modellers know they are winging it, the scientists suspect that’s true but don’t want to ask, everyone else gets taken it by it. The simple fact is that if we knew what was what, you wouldn’t need a vast Black Box model to demonstrate how temperatures will increase with CO2. The big models would simply be adding detail and granularity.

      • dave permalink
        May 11, 2022 6:32 pm

        Running a simulation anjd finding an unanticipated result may well be ‘a finding’ but it is certainly not an ’emergent property’ ; which, in the philosophy of science, is a phenomenon such as consciousness.

        We know consciousness is caused by a network of quadrillions of neural synapses in an active brain but have not got a clue how it is caused or even what it is that is caused. It just ’emerges.’

        If some climate scientists are indeed talking about ’emergent properties’ not knowing that the term is already co-opted in Science this proves once again that they are hopelessly uneducated.

  3. John Culhane permalink
    May 9, 2022 6:05 pm

    In case anyone is looking for the link. There are a number of presentation by Wil Happer on the subject that I have watched over the years.

    The greenhouse effect, summary of the Happer and Van Wijngaarden paper
    https://clintel.org/the-greenhouse-effect-summary-of-the-happer-and-van-wijngaarden-paper/

  4. The Informed Consumer permalink
    May 9, 2022 6:21 pm

    When we can count precisely how many volcanoes, vents and fissures on the all the Ocean beds then we are completely pi55ing in the dark about atmospheric CO2, saturated or otherwise.

    In 2018, 91 volcanoes were discovered under the west Antarctic ice sheet. We have only been exploring there for 100 years or so.

  5. Neil Mahony permalink
    May 9, 2022 6:29 pm

    I asked Dr. Happer directly.

  6. May 9, 2022 6:33 pm

    See Wijngaarden & Happer 2020

  7. Jordan permalink
    May 9, 2022 6:37 pm

    My suggested method is “count the photons” as a way to question the suggested Enhanced Greenhouse Effect (EGHE). This assumes all photons carry the same energy, which is approximately true within a band known as “IR”.
    Let’s say the EGHE has incrementally warmed the surface layer to the tune of 2 additional photons heading skyward. Lets say these 2 photons are absorbed by GHGs, so we have maximum chance of something being returned to the surface to maintained the EGHE warming.
    The atmosphere has just added 2 photons worth of energy, and assume these are immediately radiated. However the atmosphere has no sense of direction, and radiates equally upward and downward. Therefore 1 photon is returned to the surface.
    On that single round-trip of the EGHE, it has lost half of the energy it started with. So there is no source of energy to provide another 2 photons attributable to the EGHE.
    On that reasoning, an EGHE cannot raise the surface temperature and sustain the temperature rise beyond the time it takes for any starting energy to be lost each time photons attempt to complete the round trip from the surface to the atmosphere and back to surface.

    • It doesn't add up... permalink
      May 9, 2022 8:09 pm

      This is basically exactly what the Wijngaarden and Happer paper does in a rather more sophisticated fashion because it looks at photons of all energies and their probabilities of absorption and emission given the details of atmospheric column composition including varying molecular composition with altitude, and varying underlying pressure and temperature, all of which affect the calculations, calculating layer by layer and energy by energy.

      • Jordan permalink
        May 9, 2022 10:50 pm

        Thanks for the comment IDAU … maybe time for me to take a look at that paper.
        The GCMs are said to conserve energy, which immediately leads to questions about how they might account for the above “round trip loss” challenge.
        As I understand things from descriptions offered by others, the GCMs carry out an energy balance at the top of the atmosphere. This allows their developers to claim conservation of energy.
        Balance at the top of the atmosphere could easily hide a multitude of sins where the balance is not so checked.
        My guess is that the models contain a perpetual motion process below the top of the atmosphere, with assumptions that energy (IR photons) can leave the surface and return to the surface without loss, and can sustain this in perpetuity. Which gets to the unphysical energy accumulation and exaggerations of warming at the surface. It will be amazing what equations can appear to prove, so long as the checking and testing is inadequate.

    • Richard C (NZ) permalink
      May 10, 2022 12:23 am

      Jordan >”This assumes all photons carry the same energy, which is approximately true within a band known as “IR” ”

      Certainly is not true. IR-A/B in the solar spectrum, IR-C in the terrestrial spectrum. Lookup Wikipedia: Infrared, for infrared categories.

      IR-A/B energy-per-photon is measured in electron Volts (eV), IR-C in milli electron Volts (meV), 3 simple orders of magnitude less. Lookup Wikipedia: Electromagnetic Spectrum, for energy-per-photon table.

      This is the root of the IPCC’s scientifically fraudulent attribution (actually speculation) of ocean heat rise in ZetaJoules (ZJ, 21 zeros) which is 93% of the planet’s observed energy increase (atmosphere a negligible 1%, land rivers lakes and ice 6%). They assume that IR-C radiated down to the surface from GHGs heats the ocean beyond that of solar heating – impossible.

      Solar IR-A/B penetrates and heats the ocean at a most effective depth of 1m in the tropics 20N+ – 20S+. A north-south cross-section of Pacific Ocean sub-surface temperature shows this clearly. Terrestrial IR-C penetrates at maximum about the thickness of a human hair (100 microns) but is most effective at 10 microns and therefore, given weak energy-per-photon, is not the surface heating agent – solar IR-A/B is. It’s real power vs apparent power.

      The radiative heating debate should be the attribution of the most significant 93% ocean heat rise – not the negligible 1% atmospheric heat rise. The ocean is the planet’s largest heat sink by far but the atmosphere can hardly be considered a heat sink, more of a surface-to-space energy transfer medium.

      • Richard C (NZ) permalink
        May 10, 2022 12:47 am

        >”They [IPCC] assume that IR-C radiated down to the surface from GHGs heats the ocean beyond that of solar heating – impossible”

        Of that downwelling IR-C, CO2 is a minor component. Wang & Liang (2009):

        “[29] The dominant emitters of longwave radiation in the atmosphere are water vapor, and to a lesser extent, carbon dioxide. The water vapor effect is parameterized in this study, while the CO2 effect on Ld is not. The effect of CO2 can be accurately calculated with an atmosphere radiative transfer model given the concentration of atmospheric CO2. Prata [2008] showed that under the 1976 U.S. standard atmosphere, current atmospheric CO2 contributes about 6 W m−2 to Ld, and if atmospheric CO2 concentration increases at the current rate of ∼1.9 ppm yr−1 [Intergovernmental Panel on Climate Change, 2007], this will contribute to an increase of Ld by ∼0.3 W m−2 per decade. Therefore, the total variation rate in Ld is 2.2 W m−2 per decade.”

        To put this in perspective, total Ld (DLR) fluctuates around 400 W.m2 day and night in the tropics e.g. Darwin (CO2 still 6 or 7 W.m2 note). No-one in their right mind attempts to harness this energy (like solar collectors do) because it does no work – it is only apparent power, not real power. And no point trying to heat water with DLR because it is not a water heating agent – it lacks the requisite radiation-matter “tuning”.

      • Richard C (NZ) permalink
        May 10, 2022 3:28 am

        >”IR-A/B in the solar spectrum, IR-C in the terrestrial spectrum”

        Spectrum graph at OpenLearn – Climate Change, Section 1.2.1
        https://www.open.edu/openlearn/nature-environment/climate-change/content-section-1.2.1

        Figure 5 Wavelength spectrum of solar radiation (red) and terrestrial radiation (blue).

        “The plots are schematic, in the sense that the vertical scale is not defined, but each shows how the radiative power is apportioned among the range of wavelengths emitted”

        Except solar radiative power is real, terrestrial radiative power is apparent.

        To the right of the solar Visible spectrum is IR-A/B (red). Further right is the Terrestrial spectrum IR-C (blue).

      • Jordan permalink
        May 10, 2022 5:31 pm

        Richard C (NZ) > “Certainly is not true.”
        Thanks for the clarification, and for putting me right on that.
        The thought experiment of counting photons is appealing, as it argues a net loss on the round trip to the atmosphere and back. I’d like to be able to hang onto the idea as it is a way to question whether the EGHE contradicts conservation of energy.
        The assumption of carrying the same energy was to keep it simple. If I drop the assumption, and instead move onto a thought experiment of counting photons in bands (each band containing photons with approximately the same energy), The W&H paper (mentioned above) draws emissions spectra, so could run the same photon counting (and photon losses on the “round trip” of the EGHE) if the spectrum is divided into bands?
        Does this allow me to cling onto my thought experiment in a somwhat more refined form? Or am I completely busted on this one.

      • Richard C (NZ) permalink
        May 11, 2022 1:08 am

        Jordan >”Does this allow me to cling onto my thought experiment in a somwhat more refined form? Or am I completely busted on this one.”

        Perhaps refresh the basics of energy vs power which is electron Volts (eV or meV) and Watts per metre squared (W.m2) in this case

        1 eV in Joules is equivalent to the numerical value of the charge of an electron in coulombs i.e. energy.

        Watts (Joules/second) is a measure of the rate at which energy flows i.e. power.

        If you are going to count photons (your choice) you have to know the respective energization of each photon stream and how it interacts with the matter in question (radiation-matter “tuning”). Again, solar IR-A/B photon energization is in eV units but terrestrial IR-C in meV – big difference.

        Solar IR-A/B is “tuned” to water – terrestrial IR-C isn’t, which is why solar IR is a water heating agent but terrestrial IR isn’t. The IPCC climate scientists are completely ignorant of this.

        A good radiation-matter “tuning” example is UV-A, UV-B, IR-A/B radiative effect on the human body (UV-C no effect).

        UV-B is tuned to the epidermis.
        UV-A is tuned to the dermis.
        IR-A/B penetrates both epidermis and dermis and is tuned to underlying tissue/muscle and given 57-60% water composition on average and IR-A/B-water tuning is therefore the major heating agent for the human body.

        In short, you have to decide what effect your photon stream will have on the matter under consideration – is it real effective power or just apparent ineffective power?

      • Richard C (NZ) permalink
        May 12, 2022 2:59 am

        Jordan >”Does this allow me to cling onto my thought experiment in a somwhat more refined form? Or am I completely busted on this one.”

        Forgot to say keep on that track. If nothing else you will learn a heap of what is really the base issue here and maybe one day put others on the right track – me included.

        Don’t know how much knowledge you already have baked in but I recommend looking up Photon Energy and Photon Energy Transfer. Wikipedia has some fundamental but worthwhile stuff e.g. the Photon Energy page has 3 examples including photon counting in respect to photosynthesis:

        “During photosynthesis, specific chlorophyll molecules absorb red-light photons at a wavelength of 700 nm in the photosystem I, corresponding to an energy of each photon of ≈ 2 eV ≈ 3 x 10−19 J ≈ 75 kBT, where kBT denotes the thermal energy. A minimum of 48 photons is needed for the synthesis of a single glucose molecule from CO2 and water (chemical potential difference 5 x 10−18 J) with a maximal energy conversion efficiency of 35%”

        Also,

        How do you calculate the number of photons? [and similar on web]
        https://socratic.org/questions/how-do-you-calculate-the-number-of-photons

        And,

        Basic Health Physics – 03 – Interaction of Photons with Matter

        Click to access ML11229A667.pdf

        This last is high energy ionizing radiation (solar/terrestrial is low energy non-ionizing) but very much radiation-matter tuning. Scroll down to:
        page 34 Photoelectric Interaction (shielding) – lead, tungsten, uranium
        page 84 Probability of Interactions – water and lead.

  8. Harry Passfield permalink
    May 9, 2022 6:37 pm

    Isn’t this the point that CO2 effects are logarithmic?
    And, most certainly the reason why Methane is not as much a problem as the CC-disciples would want us to believe: Its spectrum is saturated.

    • Graeme No.3 permalink
      May 9, 2022 10:47 pm

      The Beer-Lambert law perhaps?

      And has anyone thought that CO2 might absorb incoming IR (27% of sunlight is in the solar spectrum) so increased CO2 at the top of the atmosphere would be radiating more back into space?

  9. Martin Brumby permalink
    May 9, 2022 6:39 pm

    I try not to comment on the ultra-scientific stuff – blackbody emissions and absortion / the finer points of Thermodynamics / Milankovitch cycles and so forth. (I had to do thermodynamics at University but it wasn’t a favourite subject. I mugged up enough to pass in my Finals, but never needed it in 50 years as a Chartered Engineer.)

    No point in getting closely involved if you aren’t really up to speed.

    But although it doesn’t greatly assist Mike Gilding, I must say that there is a workaround.

    If the majority of people / organisations give you the benefit of their wisdom in annointing some hypothesis as “The Settled Scientist”, but refuse to discuss both this and alternative hypotheses by intelligent commenters / scientists / experts in related fields, you can be 97% sure the “The Settled Science” is baloney. If it is based on modelling and the initial conditions are unsure, the actual physics, chemistry, mechanics etc is shakey and the are a host of fudge factors built in, you can be absolutely certain it is complete tripe.

    So far as Moana Loa (and perhaps the amazing fact that CO2 was always 275 ppm before the Industrial Revolution), some scientists that seem trustworthy, have accepted both, So I try to hide my cynicism.

    • Phoenix44 permalink
      May 10, 2022 8:56 am

      Very true. I use an even simpler method – if the predictions are Apocalyptic and demand we change our economic system, they are 99.9% likely you be wrong.

  10. May 9, 2022 6:39 pm

    Doh ! didn’t notice W & H had been cited in the post.

  11. dodgy geezer permalink
    May 9, 2022 6:44 pm

    As far as I can see, the actual science is completely irrelevant. All that is required is for some complicated model to claim that we will all fry in 10 years, any disagreement with this to be refused publication, and a few thousand activists to walk down Main Street with placards.

    Hey Presto – $10tn for the woke entrepreneurs, and taxes go up 20%.

    This situation certainly needs to be addressed, but no one will rectify it by academic study and field experiments.

    • Gamecock permalink
      May 9, 2022 10:03 pm

      True. The Left claims the moral high ground. Their arguments are moralistic. Facts have no power against them.

    • Phoenix44 permalink
      May 10, 2022 8:57 am

      Climate change is the problem socialists have sought for 30 years to justify applying their “solution”.

  12. Jack Broughton permalink
    May 9, 2022 6:50 pm

    The paper by David Coe, which was covered in detail on this site a few months back, is probably the only attempt to actually calculate the effect of CO2 (as well as CH4) on the earth’s atmosphere using a well proven radiative absorption model called HITRANS. It showed that both so-called “Greenhouse gases” have very little “greenhouse effect” indeed.

    As KB comments the IPCC and many modellers (and of course MET office staff) have used a simplified, almost “Carpenterian” model to obtain ECS and Forcing factors. It is frightening how the scientific community have accepted what are poor and unfounded opinions for fact.

    The IPCC reports claim the balance all of the radiative factors to obtain their net warming value, however it is well known that the effects of dust are totally underestimated and the atmospheric dust load that was used is out by a large factor. The average values and “confidence-level” bands that they associate with the values of emissivities, which are the basis for all models, are at best poor science, at worst lies.

    • It doesn't add up... permalink
      May 9, 2022 8:14 pm

      Coe’s paper takes broadly the same approach as Wijngaarden and Happer, but their paper is slightly more sophisticated than his.

      • May 10, 2022 1:46 am

        The Coe paper was not published in a proper, peer reviewed, science journal. Coe himself said he doesn’t understand W & H.

  13. Richard North permalink
    May 9, 2022 6:50 pm

    I have to admit I have also wondered – why Mona Lua? Also I seem to remember Ian Plimer writing that many readings from there are discarded which he suspects is another temperature homogenisation-like scam.

    • Graeme No.3 permalink
      May 9, 2022 10:57 pm

      As a volvano Moana Lua emits lots of CO2 and gusts of wind carry it to the sensors, so sudden changes are rejected. As I remember the rejected figures are around 83% of the total.
      In any case there are other sites elsewhere which give the same CO2 level (allowing for seasonal changes). Indeed CO2 seems to be well mixed in the atmosphere, somewhat surprising in that 90% of supposed human emissions occur in the Northern Hemisphere but the warming is claimed to be world wide.

      • Broadlands permalink
        May 10, 2022 1:44 am

        Those other sites…Atmospheric CO2 is not just measured at Mauna Loa. It is also measured at several other sites in the Pacific…

        https://cdiac.ess-dive.lbl.gov/trends/co2/sio-keel.html

      • Mike Gilding permalink
        May 13, 2022 4:11 pm

        Thanks Graeme No 3. I had not realised that many of the readings are rejected and agree that CO2 is fairly even over the surface.
        Even so, it seems a poor choice of location.

  14. May 9, 2022 6:56 pm

    The concept of CO2 “Saturation” measures what fraction of 15 micron (u) IR is absorbed when passing through a volume of CO2. In that sense 15u IR from the surface is fully absorbed in the atmosphere, because its mean path length is measured in meters. However, atmospheric CO2 also emits 15u IR, so the net 15u flux within the atmosphere does not diminish.
    Greenhouse warming by CO2 is not related to such “saturation”. Rather, warming occurs because CO2 emits 15u IR to space from high and cold altitudes, where the 15u emission rate is much lower (see Stefan-Boltzman relation). Because the reduction in the flux of escaping 15u IR is exponentially related to temperature, the effect of increasing atmospheric CO2 does have a decreasing influence on temperature (approximately logarithmic). That’s why greenhouse temperature increases are often expressed in terms of doubling of the greenhouse gas concentration. Each doubling of atmospheric CO2 reduces the escaping 15u IR by approximately three watts per meter squared and is equivalent to slightly more than 1 deg-C temperature increase.

  15. May 9, 2022 6:56 pm

    The concept of CO2 “Saturation” measures what fraction of 15 micron (u) IR is absorbed when passing through a volume of CO2. In that sense 15u IR from the surface is fully absorbed in the atmosphere, because its mean path length is measured in meters. However, atmospheric CO2 also emits 15u IR, so the net 15u flux within the atmosphere does not diminish.
    Greenhouse warming by CO2 is not related to such “saturation”. Rather, warming occurs because CO2 emits 15u IR to space from high and cold altitudes, where the 15u emission rate is much lower (see Stefan-Boltzman relation). Because the reduction in the flux of escaping 15u IR is exponentially related to temperature, the effect of increasing atmospheric CO2 does have a decreasing influence on temperature (approximately logarithmic). That’s why greenhouse temperature increases are often expressed in terms of doubling of the greenhouse gas concentration. Each doubling of atmospheric CO2 reduces the escaping 15u IR by approximately three watts per meter squared and is equivalent to slightly more than 1 deg-C temperature increase.

  16. Mike Jackson permalink
    May 9, 2022 6:58 pm

    I thought the saturation of the relevant wavelengths (mainly by water vapour) had been established years ago. The climastrologists simply stuck their fingers in their ears and shouted a bit louder (as is their wont!).
    Until politicians get the message that “climate science” has b***er all to do with climate — and there are hundreds of quotes from their own mouths from Strong’s “Isn’t the only hope for the planet that the industrialized civilizations collapse? Isn’t it our responsibility to bring that about?” to Edenhofer’s “One has to free oneself from the illusion that international climate policy is environmental policy. This has almost nothing to do with environmental policy anymore.” — we can argue CO2 measurements until we’re blue in the face.
    Net-zero is unachievable in the 21st century real world and the attempts to achieve it can only result in increasing poverty worldwide. Nothing we can do will prevent the plucked-from-the-air (by Schellnhuber by his own admission) 2°C threshold from being exceeded if that is the course that nature is currently on.
    And as Informed Consumer implies, as long as subsea volcanoes continue to erupt the oceans will outgas increasing amounts of CO2 making our contribution look increasingly and pathetically irrelevant!

    • dave permalink
      May 10, 2022 9:13 am

      It is BECAUSE the CO2 process IS ‘saturated’ that the effect of an increase in CO2 is LOGARITHMIC rather than linear. The papers referred to rework this rather obvious conclusion by calculation.That is all there is to say about them..

      When a photon is released by a CO2 molecule in the atmosphere it can go in any direction, not simply up and down. The effect is that the ‘same’ photon (if one may be permitted to speak so loosely) is absorbed and reradiated about a million times in the atmosphere before working its way out to space. (However, It is still only in flight for a total time of a second since each individual flight is so short. FAR more time is spent waiting for the CO2 molecule to ‘fire’.) More CO2 means shorter, and therefore more of them, flights to achieve escape and this is where the conclusion is obvious that each doubling of concentration is equivalent in the (temporary) trapping effect. Throughout the process the CO2 is at the temperature of its surrounding air, of course. The way in which this happens is called ‘relaxation.’

      As far as I am aware ‘serious’ members of the Climate Science cult gave up long ago on pretending the increase would have a linear effect and have resorted to ad hoc fixes involving feedbacks of one sort or another. to achieve the required doomsday forecasts.
      They have also had to throw away the Stefan-Boltzmann law because it absolutely prohibits ‘run away’ warming.

      There is a different red herring, to do with the WIDENING of the absorption bands. This does not happen at the present amounts of CO2 and so whether this would be similar to a ‘desaturation’ need not be considered.

  17. will davis permalink
    May 9, 2022 7:42 pm

    I do not believe that the sighting of CO2 recording is imortant. Is the increase in CO2, whever recorded, having an measurable effect world temperatures is the question? I have read H & W papers, as much as I can understand, their conclusions appear to be perfictly logical, from an engineering point of view. Over 40 years of satilite atmospheric recording the average temperature has remained with one Deg C but the increase of CO2 % is?

    The other burning question is how much of any temperature change is natural (W. Soon and others)?

  18. M Fraser permalink
    May 9, 2022 8:11 pm

    Just at base level;
    a) Is it warmer during daylight hours because there’s a daytime increase in CO2?
    b) Here, north of the equator in the months between the vernal and autumnal equinox’s is the climate warmer due to increased CO2?
    OR
    Perhaps it may correspond with the Earth’s proximity and inclination to the huge gaseous fireball it orbits!
    Putin may be just a small problem when compared to the eco religion we’re being subjected to!!

    • dave permalink
      May 15, 2022 9:48 am

      “…North of the Equator…Earth’s proximity…”

      I know you are being ironic, and mean ‘the Sun is always there’. But, it is an interesting fact that the Earth is farther away from the Sun during the Northern Hemispere’s summer than during its winter. That is to say the aphelion (at present*) occurs in July. The solar flux received by the entire Earth is 7% lower at aphelion than at perihelion.

      Incidentally, a famous video interviewing newly minted Arts graduates of Harvard University revealed that most of them thought winter happened when the Earth was closer to the Sun. They did not seem aware that the Southern Hemisphere does not have its seasons at the same time as the Northern.

      * During the “Climatic Optimum,” ten thousand years ,ago the situation was reversed and therefore the contrast between the Northern Hemisphere’s summer and winter was more pronounced.

  19. Pablo permalink
    May 9, 2022 8:32 pm

    “It appears certain that on our Earth surface temperatures lower than -73ºC., or 200ºK. can not occur, possibly because of the almost total absorption
    by the atmosphere of all radiations beyond 13 microns.

    …the position of the normal maximum in the energy-curve for the lowest arctic temperature very nearly coincides with the great absorption band of carbon dioxide.”

    From “Atmospheric Radiation” by Frank W. Very 1900

  20. May 9, 2022 8:36 pm

    I believe that the conventional argument is that adding more greenhouse gases always causes warming, assuming everything else stays the same. The key notion is where the radiation to space comes from. If you could see in infrared, and looked at the Earth from space, for most wavelengths the radiation would be coming from a certain height in the atmosphere. The outgoing IR photons in the atmosphere get absorbed until a certain height, the height at which the greenhouse gases are thin enough to give a high chance of no further absorption. That height continues to rise with rising greenhouse gas levels, hence the temperature of the effective radiating surface continues to go down, hence there is less outgoing radiation, hence the Earth must warm to restore balance.

    Gravity plays a key role, as it causes the air temperature to fall with increasing height. Clouds and the stratosphere are complications.

  21. David Wojick permalink
    May 9, 2022 8:38 pm

    Perhaps the most interesting aspect of the W & H paper is that the IPCC gets the same results. In fact AR4 is one of the paper’s references. The IPCC just does not draw the obvious conclusion.

    Note too that the paper also shows H2O to be as “extremely saturated” (their language) as CO2, so the purported water vapor positive feedback that supposedly causes the dangerous warming cannot occur.

    • Mike Gilding permalink
      May 13, 2022 5:21 pm

      The IPCC was created to report on Anthipogenic Warming. Nothing else (see their TORs). So it is unsurprising that they do not agree with W&H. If they did, they would simply become an unessesary expense. Members of the IPCC would cease ‘saving the planet’, collecting a generous paypacket and living the life of Reilly in Switzerland.

  22. Pablo permalink
    May 9, 2022 8:45 pm

    “The absorbent action of carbon dioxide and the permanent gases is almost invariable; but the absorption bands of aqueous vapour are much stronger in the summer than in winter, and the selective scattering of short waves also increases in summer. One result of this variation is that the direct rays of the midday sun, received upon a a normal surface, are more powerful in winter than in summer, in spite of the greater distance traversed by the sunbeam through the air in winter.”

    “The direct effect of the sun’s rays upon a normal surface is less in the tropics than in temperate regions, and less at sea level than upon a mountain top, owing to the difference in the aqueous component of the air; and the ability of the solar radiation to maintain a high temperature in the torrid zone or at sea level is due to the accumulation of the thermal energy imparted to the Earth’s surface by reason of the retention of the escaping radiation from that surface by a moist and highly absorbent atmosphere rather than to the direct power of the sunbeam.”

    “As long as the mantle of water vapour remains unbroken, thermal fluctuations are kept within narrow limits. Storms may make inroads upon the continuity of this aqueous atmospheric envelope,
    but evaporation of moisture restores the rents.”

    “The gradual accumulation of moisture in higher and higher atmospheric layers during the summer, clothes the temperate regions with so deep a protective covering of moist air, that summer conditions are prolonged in the autumn to a time which is astronomically the correlative of late winter. The absence of this deep protective layer, whose formation can only be effected gradually, permits late frosts in spring, long after the sun has resumed his ascendency.”

    “…in addition to the large specific heat and mobility of water, conducing to the slowness of oceanic temperature changes; but more important as a retainer of oceanic heat is the extension of the
    great band (6,3 microns) to greater wave-lengths in the absorption of the layer of air nearly saturated with moisture, which always hangs over the water.”

    From “Atmospheric Radiation” by Frank W. Very 1900

  23. May 9, 2022 8:59 pm

    A most accessible presentation by H & W is this one available as a transcript:

    Climate Change and CO2 Not a Problem

    The most pertinent image is this one:

    The important point here is the red line. This is what Earth would radiate to space if you were to double the CO2 concentration from today’s value. Right in the middle of these curves, you can see a gap in spectrum. The gap is caused by CO2 absorbing radiation that would otherwise cool the Earth. If you double the amount of CO2, you don’t double the size of that gap. You just go from the black curve to the red curve, and you can barely see the difference. The gap hardly changes.

    The message I want you to understand, which practically no one really understands, is that doubling CO2 makes almost no difference.

    • David Wojick permalink
      May 9, 2022 10:32 pm

      Well said! Yes the tiny difference between those two lines if the increased forcing from doubling. Compared to the big low-concentration dip it is as nothing.

      Mind you this is all without clouds. W & H are working on clouds now. Happer tells me they actually decrease the forcing change even more.

      • May 9, 2022 10:44 pm

        Thanks David, though I hasten to affirm those are Happer’s words, not mine.

  24. May 9, 2022 10:29 pm

    What would the Earth’s mean surface temperature be if the atmospheric pressure there was 92 bar, like Venus? Assume any level of so called greenhouse gas you want.

  25. May 9, 2022 10:32 pm

    Why does no one read my killer disproof of the IPCC greenhouse gas warming hoax and learn how sweet pure thermal physics really works with Earth’s climate?

    [[My suggested method is “count the photons” as a way to question the suggested Enhanced Greenhouse Effect (EGHE). This assumes all photons carry the same energy, which is approximately true within a band known as “IR”.]]

    Zonk! Each photon carries an energy inversely proportional to wavelength. The surfaces of the Sun and Earth are Planck black body radiators, which attempt to radiate all their photons at a wavelength inversely proportional to temperature with quantum precision. The power-wavelength curve isn’t a single sharp line only because of the Second Law of Thermodynamics AKA ENTROPy. All the photons off the Wien Displacement Law wavelength are WASTE HEAT. Planck black body radiation is Nature’s way of transmitting HEAT across space. Heat is measured by temperature. The Wien wavelength is precisely determined by temperature. When one black body radiates onto another, the second absorbs its photons and raises its internal temperature based on its heat capacity, moving from one black body radiation curve to another until the temperatures and radiation curves equalize. A colder black body cannot raise the temperature of a hotter black body AT ALL. The source photons are just dissipated by entropy into the Heat Death of the Universe.

    All attempts to create a greenhouse gas effect by counting photons and balancing energies are MOOSE HOCKEY, because they totally ignore Nature’s ironclad Second Law of Thermodynamics, and the fundamental process of entropy dispersal.
    .
    https://en.wikipedia.org/wiki/Entropy_(energy_dispersal)

    From day one atmospheric physicists starting with Fourier thought that atmospheric gases were black body radiators, and didn’t even know about quanta, making all their greenhouse gas warming theories into garbage. It took until 1900 for Planck to announce the quantum and the Planck Black Body Radiation Law, after which it took longer to realize that atmospheric gases weren’t black body radiators, but only captured and reemitted photons at certain narrow wavelength ranges. This makes all attempts at justifying greenhouse gas warming theories futile, because the only black bodies involved, the surfaces of the Sun and Earth, heat and cool nicely on their own, with atmospheric gases unable to stop it. Even worse, the only wavelengths that CO2, H2O, and CH4 are active at are outside the normal Earth surface temperature range of -50C to +50C, meaning that all the massive torrents of surface radiation photons just pass those molecules by, allowing them to reach the top of atmosphere untouched, as seen in NASA Goddard’s own satellite data. Once the real heat photons have left the surface, it’s COOLED. And that’s FINAL. There’s no need to “count photons” hoping to find a pony in the river of manure. There is no pony in the manure. CO2 doesn’t have any power to reraise the surface temperature to the same level that the Sun did, much less add more heat and raise it. It’s not even junk science. It’s LIES.

    Not just portions, but the entire U.N. IPCC greenhouse gas warming hoax is a bloomin’ onion of lies. There’s no truth in it. They don’t even treat the Earth as a spinning ball being basted by solar radiation, heating the Earth’s surface and giving it time to cool each day. Their sick inversion of climate science has turned the Earth into a stationary flat disk that’s irradiated by a yearly average of solar radiation, from which they pretend to calculate “global average temperature”, (GAT) a totally nonphysical fairy land statistic that bears no resemblance to reality, then pretend to find tiny increases correlated with Mauna Loa CO2 readings.

    The game isn’t to recalculate GAT, but to throw it out, and just recognize that greenhouse gases can’t do diddly to reverse the Second Law of Thermodynamics. Thus the entire IPCC edifice designed by leftist anti-fossil fuel activists can be dumped into the trash, and the whole organization laughed off. The next step is to get it disbanded and demand the money be refunded, plus damages asses for all the evil they’ve caused. Only then can real research in climate science be funded by govts., with all-new scientists who understand thermal physics, like moi.

    Please take time to read my killer disproof articles cited in the first reply to this article, and be prepared to learn true thermal physics, not the hijacked upside-down version of the IPCC.

    • May 10, 2022 1:58 am

      “A colder black body cannot raise the temperature of a hotter black body AT ALL.”

      I see a lot of this viewpoint. It must be a misrepresentation of what is going on, because otherwise insulation wouldn’t work !

      • It doesn't add up... permalink
        May 10, 2022 3:58 am

        But that isn’t how insulation works. It works by slowing the passage of heat. It doesn’t prevent it, or provide heat that isn’t provided to it by a hotter source first. It remains true that a cold body cannot of itself heat a hotter body (‘cos the cold in the cooler would get hotter as a ruler and that’s a physical law!).

      • May 10, 2022 6:40 pm

        IDAU: yes insulation works by slowing the passage of heat. That is exactly what CO2 in the atmosphere does.
        Since the Earth’s surface is heated by sunlight, it’s temperature increases when the insulation is increased. The insulation (CO2) is not heating the surface, it is slowing down the radiation of heat.

      • It doesn't add up... permalink
        May 11, 2022 1:26 am

        The process is slightly more complicated for a GHG. Heat of the air is determined by the local kinetic energy of the molecules, summarised by the gas equations PV=nRT=(mnc^2)/3 pressure x volume = number of molecules x temperature x gas constant = 1/3 number of molecules x molecular mass x mean square molecular velocity, equals the kinetic energy of the gas, with the actual speeds of the individual gas molecules being spread between low and high according to the Maxwell-Boltzmann distribution. But photons of radiation are passing through the atmosphere from the sun and from the warm earth and ocean.

        When a photon gets into very close proximity to a molecule, if its energy matches the energy required to change the energy of the molecule initially between permitted quantum states of the electron orbitals it gets absorbed, and the electrons are excited into higher states. This changes the electrical configuration of the molecule, and allows the nuclei more freedom of movement within the expanded electron shells in ways that depend on the orbital changes which are associated with particular vibrational modes (bending, (a)symmetric stretching, rotations etc). In this way the energy can be changed to vibrational energy of the molecule.

        However, the molecule may emit a photon with an energy corresponding to a permitted quantum change, and the time between absorption and emission contributes to the delay in the transmission of the radiated heat through the atmosphere.

        There is another effect: a molecule that absorbs a photon converts radiative energy into its own kinetic energy, so of itself it heats up. It may give up some of its kinetic energy in collisions with other molecules before it gets the chance to re-radiate a photon (the vibrations can have the effect of giving a punch to another molecule, rather than just a billiard ball type collision). There are two consequences of this: it can no longer emit as energetic a photon as before, since it has lost energy transferred to another molecule, and the molecule that has acquired kinetic energy from the collision may not have modes of photon emission available (e.g. non GHGs). Either way, the energy stays trapped in kinetic form (i.e. heat), or can only be released as a lower energy photon, unless later collisions result in a molecule acquiring sufficient kinetic energy from its encounter to be able to emit a more energetic photon. Obviously this process heats the atmosphere and delays the transfer of energy.

        The overall effect depends on the detailed statistics of gas kinetics and quantum chemistry as well as spectra of the originating radiations. There is also an overlay from the bulk properties of the atmosphere: its temperature and pressure dependence on altitude and gravity which will affect the relative importance of kinetics (collisions) and quantum effects for instance, and likewise changes in composition.

      • May 11, 2022 12:02 pm

        IDAU thanks for the physics. It seems you are agreeing that GHG’s do indeed delay the transfer of heat energy through the atmosphere.

        Imagine an object suspended in a vacuum chamber. The object has a resistive heater built in so it can be heated by an electric current, and it carries a thermometer which enables its temperature to be read from outside the chamber.

        There are two such setups, identical in every respect, except one object is surrounded by reflective aluminium foil (inside the vacuum chamber).

        When equal currents are switched on, will one object reach a higher equilibrium temperature than the other? Yes, the one surrounded by the aluminium foil reflector will be hotter.

        Why is this? The aluminium foil at no point is hotter than the object itself. It is cooler during the heat up stage, and it will rise to a temperature at which the energy radiated from its outside surface is the same as the object with no surrounding foil. It never reaches a temperature exceeding the object itself.

        However the people with this mindset would interpret this as a cooler object (the foil) warming the object. It’s an incorrect way of viewing the situation.

    • Richard C (NZ) permalink
      May 10, 2022 3:41 am

      >”Each photon carries an energy inversely proportional to wavelength”

      Notes on this and Solar IR-A/B vs Terrestrial IR-C wavelength spectrum graph upthread in my replies to Jordan. The IPCC adopts IR-A/B properties for Terrestrial IR-C radiation – but they are dead wrong.

    • Ed Bo permalink
      May 10, 2022 9:45 pm

      TL, you ask: “Why does no one read my killer disproof of the IPCC greenhouse gas warming hoax?”

      Because you have no freaking idea how any of this works! You would be in deep trouble early in any introductory thermodynamics or heat transfer course.

      You have no clue how to do a basic 1st Law energy balance analysis. You do not begin to understand how radiative heat transfer works.

      Just one example: CO2 is highly absorbing of 14-16 um radiation, REGARDLESS of the temperature of the body emitting that radiation, whether high or lower than the absorbing CO2.

      You would do well to spend a couple of years learning the basics of the subjects. I encourage you to study these subjects in an engineering context, where they must get these things right. Then you would have the possibility of providing a meaningful contribution to these discussions, rather than just making a fool of yourself.

  26. May 10, 2022 1:54 am

    I trust people have noticed, as David Wojick hints at above, that W&H obtained a climate sensitivity of 2.2-2.3 degrees for a doubling of CO2, which is actually within the range quoted by the IPCC, although towards the lower range of it.

    So it isn’t like W&H have completely overturned the IPCC on this. Perhaps they will do that when they include clouds in their models, but we have not seen that paper yet.

  27. Peter Qualey permalink
    May 10, 2022 2:24 am

    One can talk ‘proper’ science till one is blue in the face; of course W & H are right, GCM’s forget the sun is shining (cf C Monckton of Brenchley over at WUWT), satellite data is not fudged (sorry, homogenised, pasteurised till the right answer appears), journals just use ‘pal’ review, various vested interests have politicians by the unmentionables, and the public remain fed a diet of climate offal, etc…, but temperatures still rise – slowly, very slowly.
    Sanity will only return if I) temperatures (now flat for 7yrs 7mths) begin to fall and/or ii) cost of living increases can be shown to be a consequence of government meddling in the energy market, ie the ‘unreliables’ are an insane idea for a developed economy.
    W & H is, however, a breath of fresh air.

  28. cookers52 permalink
    May 10, 2022 5:35 am

    The science is settled, but there appears to be some uncertainty on what that means.

    The belief that human activities affect the weather is as old as humanity itself, doesn’t really matter what the science says.

  29. Dudley Marks permalink
    May 10, 2022 6:16 am

    Many years ago I acquired a graph showing the saturation you mentioned. 
    Cant remember Y axis but it was some measure of heat v’s X axis in ppm CO2.
    High slope to 100ppm CO2 after which slope was low but non zero
    Dudley Marks

  30. May 10, 2022 9:35 am

    At the Earth’s South Pole you’re looking at a temperature difference of 144 Kelvin compared to the Moon’s.

  31. Peter permalink
    May 10, 2022 10:24 am

    The UN, together with its agencies, IPCC, WCRP (controls the models) and WMO are pursuing a political agenda and use politically manipulated science to drive the project. The resulting pseudoscience is full of things that are just plain wrong. They also ignore proper science that would harm their cause, such as the excellent work of W&H.

    What are our government funded climate scientists doing about this? We can all see the answer to that.

  32. May 10, 2022 3:42 pm

    Regarding Mauna Loa, it’s also directly downwind of CO2-spewing China, so not exactly representative of worldwide average levels or rates of increase.

    • Ed Bo permalink
      May 10, 2022 10:08 pm

      The overwhelmingly dominant winds on the island of Hawaii, where Mauna Loa is, are trade winds from the northeast, not remotely from China. These winds are predominantly the return path of the Hadley cell, from air that rose from the equatorial area of the Pacific. It’s very well mixed.

      • It doesn't add up... permalink
        May 10, 2022 11:45 pm

        It’s also at about 10,000ft, not far from the summit on the North side. When the wind blows in the wrong direction they do get vapours from the volcano cone. But mostly the prevailing wind blows them away. Lots of tropical Pacific Ocean emitting CO2 upwind seasonally.

  33. Athelstan. permalink
    May 10, 2022 6:17 pm

    I think that we all forget something.

    NB, the puny bit mankind puts into the Atmosphere, estimated to be circa 20 ppm.

    Listen up, at 20 ppm whatever is the significance – it’s just undetectable noise and IR values, so blooming what? Hot air and lies, the majority of its emissivity ex the UN. The great global warming scandal is just that and something Bernie Madoff would have been proud to call it his own.
    Mauna Loa, and CO2 measurements – they’re pulling our pi88ers.

  34. Peter permalink
    May 10, 2022 7:14 pm

    The amount of CO2 in the atmosphere is mainly controlled by Henry’s law. A huge amount is in the oceans and an unknown amount is in the biosphere.

  35. Chris Speke permalink
    May 10, 2022 9:08 pm

    International Journal of Atmospheric and Oceanic Sciences
    2021; 5(2): 29-40 http://www.sciencepublishinggroup.com/j/ijaos
    doi: 10.11648/j.ijaos.20210502.12
    ISSN: 2640-1142 (Print); ISSN: 2640-1150 (Online)

    The above piece by David Coe appeared last year and seems to broadly agree with the W and H paper.

    • May 10, 2022 10:02 pm

      Science Publishing Group credibility is less than zero. There is no real peer review. Anything “published” there can be dismissed without thought by the other side. They laugh at it.
      The Coe paper does not broadly agree with W&H. Coe finds a climate sensitivity of 0.7 degrees compared to the W&H value of 2.2-2.3 degrees.
      Both authors can clearly use HITRAN; the problem comes after that stage. Coe takes a big shortcut there and you have got to question if he has the expertise to do that.

  36. LOL@Klimate Katastrophe Kooks permalink
    May 11, 2022 10:32 am

    Jordan wrote:
    “The atmosphere has just added 2 photons worth of energy, and assume these are immediately radiated. However the atmosphere has no sense of direction, and radiates equally upward and downward. Therefore 1 photon is returned to the surface.”

    Not exactly… there is a mean free path length / air density / altitude relation… as altitude increases, air density decreases and mean free path length increases exponentially. Vice versa as altitude decreases.

    Thus, the atmosphere effectively ‘pumps’ IR photons upward.

    Note: All the numbers below are just for illustrative purposes.

    Consider a photon emitted at 1 km altitude… it has a mean free path length in a downwelling direction of 20 m, and in an upwelling direction of 30 m.

    Let’s assume it’s emitted downwelling, travels 20 m, is absorbed, and re-emitted upwelling. It now can go 30 m before being absorbed. So it’s now 10 m higher altitude than when it started.

    Now, the numbers are just PFMA, and in reality the mean free path length changes exponentially with changing altitude, but you get the picture.

    Near the surface, at altitudes higher than ~20 m, effectively no photons reach back to the surface, given that the extinction depth near-surface is so short (currently ~9.7 m, decreasing with increasing CO2 atmospheric concentration).

  37. LOL@Klimate Katastrophe Kooks permalink
    May 11, 2022 10:45 am

    No, sorry, that should be an extinction depth of ~10.4 m. It’d be ~9.7 m if atmospheric CO2 concentration doubled.

    From my prior writings:

    If ‘backradiation’ from CO2 atmospheric emission causes CAGW, where is it coming from?

    Near-surface extinction depth is ~10.4 m at current CO2 concentration, and a doubling of CO2 concentration would reduce that to ~9.7 m. The troposphere is essentially opaque to 13.98352 µm to 15.98352 µm (to account for the absorption shoulders of CO2) radiation. In fact, it’s opaque to that radiation right up to ~15 – 20 km (TOA for that wavelength of radiation). That’s where the effective emission height of CO2 is.

    CO2’s absorption of IR in the troposphere only has the effect of thermalizing that radiation and thus increasing CAPE (Convective Available Potential Energy), which increases convection of air to the upper atmosphere (carrying with it the latent and specific heat of polyatomic molecules… more polyatomic molecules will carry more energy and will more readily emit that energy in the upper atmosphere), which is a cooling process.

    Mean free path length for radiation decreases exponentially with decreasing altitude and vice versa due to air density changing inversely exponentially with altitude, therefore the net vector for radiation in the 13.98352 µm to 15.98352 µm band is upward, so the majority of ‘backradiation’ which could possibly reach the surface would be from that very thin layer of atmosphere which is within ~10.4 m of the surface, and the great majority of that energy is being thermalized and convected. So where’s this ‘backradiation’ energy coming from that’s going to cause catastrophic anthropogenic global warming, especially considering that the maximum able to be absorbed by CO2 is 8.1688523 W/sr-m^2, and the maximum able to be absorbed by anthropogenic CO2 is 0.29652933849 W/sr-m^2?

    At 287.64 K (the latest stated average temperature of Earth) and an emissivity of 0.93643 (calculated from NASA’s ISCCP program from data collected 1983-2004), at a photon wavelength of 14.98352 µm (the primary spectral absorption wavelength of CO2), the spectral radiance is only 5.43523 W / m^2 / sr / µm (integrated radiance from 13.98352 µm – 15.98352 µm of 10.8773 W/sr-m^2 to fully take into account the absorption shoulders of CO2).

    Thus the maximum that CO2 could absorb in the troposphere would be 10.8773 W/sr-m^2, if all CO2 were in the CO2{v20(0)} vibrational mode quantum state.

    While the Boltzmann Factor calculates that 10.816% of CO2 are excited in one of its {v2} vibrational mode quantum states at 288 K, the Maxwell-Boltzmann Speed Distribution Function shows that ~24.9% are excited. This is higher than the Boltzmann Factor calculated for CO2 because faster molecules collide more often, weighting the reaction cross-section more toward the higher end.

    Thus that drops to 8.1688523 W/sr-m^2 able to be absorbed. Remember, molecules which are already vibrationally excited can not absorb radiation with energy equivalent to the vibrational mode quantum state energy at which they are already excited. That radiation passes the vibrationally excited molecule by.

    That’s for all CO2, natural and anthropogenic… anthropogenic CO2 accounts for ~3.63% (per IPCC AR4) of total CO2 flux, thus anthropogenic CO2 can only absorb 0.29652933849 W/sr-m^2.

    CO2 absorbs ~50% within 1 meter, thus anthropogenic CO2 will absorb 0.148264669245 W/m^2 in the first meter, and the remainder 0.148264669245 W/m^2 within the next ~9 meters.

    CO2 absorbs this radiation regardless of any increase in atmospheric concentration… extinction depth is ~10.4 m at 14.98352 µm wavelength. A doubling of CO2 atmospheric concentration would reduce that to ~9.7 m. Thus any tropospheric thermalization which would occur at a higher CO2 atmospheric concentration is already taking place at the current CO2 atmospheric concentration. Thus the net effect of CO2 thermalization is an increase in CAPE (Convective Available Potential Energy), which increases convective transport to the upper atmosphere, which is a cooling process.

    Tropospheric thermalization is saturated. A doubling of CO2 doesn’t appreciably reduce extinction depth at the band centered around 14.98352 µm. But upper-atmospheric radiative shedding of energy to space is not saturated… and more CO2 molecules will cause more upper-atmospheric cooling, increasing buoyancy of lower-atmosphere air and thus increasing convection. IOW, polyatomic molecules (such as CO2) increase thermodynamic coupling between heat source (in this case, the surface) and heat sink (in this case, space) due to the fact that they have higher specific heat capacity than the monoatomics (Ar) and homonuclear diatomics (N2, O2).

    An increased CO2 atmospheric concentration will emit more radiation in the upper atmosphere (simply because there are more molecules absorbing energy in the lower atmosphere, more molecules convectively transporting energy to the upper atmosphere and advectively transporting energy poleward, and more molecules capable of emitting radiation in the upper atmosphere), thus more radiation will be emitted to space, and that represents a loss of energy to the system known as ‘Earth’, which is a cooling process.

    This illustrates what I’m stating:
    http://imgur.com/Zxq4KlB.png
    That’s a MODTRAN plot at 287.64 K for 415 ppm vs. 830 ppm CO2 for 13.98352 µm to 15.98352 µm radiation (to fully account for the absorption shoulders of CO2). It assumes no water vapor, no CH4, no O3 present. Note that the troposphere plots aren’t appreciably different, whereas the 100 km plots (ie: at the edge of space) are appreciably different. IOW, a doubling of CO2 atmospheric concentration doesn’t appreciably change the upward or downward radiative flux in the troposphere (because the extinction depth for those wavelengths at 415 and 830 ppm is low enough that it’s thermalizing nearly all of that radiation, the net effect being an increase in CAPE (Convective Available Potential Energy), which increases convection, which is a cooling process), but it does appreciably change how much energy is exiting the system known as ‘Earth’, and that represents a cooling process. One can clearly see the effect of CO2 upon energy emission to space, as delineated by the shoulders of the emission spectrum of CO2 in the 100 km plots. That cools the upper atmosphere, and since the lapse rate is ‘anchored’ at TOA and since the heat transfer equation must (eventually) balance, that means the lower atmosphere must cool toward the temperature of the upper atmosphere (because a higher concentration of polyatomic molecules shifts the lapse rate vertically, and radiatively cools the upper atmosphere faster than the lower atmosphere can convectively warm it), and thus the surface must cool with an increasing CO2 atmospheric concentration. This is what is taking place, we’re just working through the humongous thermal capacity of the planet, which warmed due to a now-ended long series of stronger-than-usual solar cycles (the Modern Grand Maximum), but it is cooling (in fact, it’s projected that we’re slipping into a Solar Grand Minimum which will rival the Dalton Minimum, and may rival the Maunder Minimum).

    https://i0.wp.com/wattsupwiththat.com/wp-content/uploads/2019/02/Figure-3.png
    Zoomed in…
    https://i0.wp.com/wattsupwiththat.com/wp-content/uploads/2019/02/Figure-4.png
    Note the extreme right-hand edge of that chart… negative and decreasing at an accelerating rate.

    Why do you climate kooktards deny that global cooling is occurring? Climate deniers. LOL

    Spectral Cooling Rates For the Mid-Latitude Summer Atmosphere Including Water Vapor, Carbon Dioxide and Ozone
    https://web.archive.org/web/20190331141324if_/https://co2islife.files.wordpress.com/2017/01/spectralcoolingrates_zps27867ef4.png

    Note the CO2-induced spectral cooling rate (positive numbers in the scale at right) extends right down to the surface of the planet, whereas CO2 shows just a slight bit of warming (negative numbers in the scale at right) only at the tropopause (ie: just above the clouds, where it absorbs a greater percentage of cloud-reflected solar insolation and radiation from cloud condensation).

    Polyatomic molecules shift the lapse rate vertically, more of them shifts the lapse rate more vertically (which attempts to decrease temperature differential between different altitudes by transiting more energy from surface to upper atmosphere), while also radiatively cooling the upper atmosphere faster than the lower atmosphere can convectively warm it… ie: they are coolants.

    This is borne out empirically by the long-term and dramatic upper-atmosphere cooling and by the fact that OLR increased by ~7 W/m^2 over ~72 years even as surface temperature showed no statistically significant trend for more than two decades (said increased OLR partly caused by the increasing CO2 concentration making available more molecules capable of efficiently convectively transporting energy to the upper atmosphere, and advectively transporting energy poleward, then radiatively emitting it).

    https://i.imgur.com/FLhcY0B.png

  38. LOL@Klimate Katastrophe Kooks permalink
    May 11, 2022 11:11 am

    Ed Bo wrote:
    “Just one example: CO2 is highly absorbing of 14-16 um radiation, REGARDLESS of the temperature of the body emitting that radiation, whether high or lower than the absorbing CO2.”

    He says, denying 2LoT, denying that temperature is a measure of energy density, denying that energy density determines radiant exitance, denying that warmer objects will have higher energy density at all wavelengths than cooler objects.

    Temperature is equal to the fourth root of energy density divided by Stefan’s Constant (ie: the radiation constant)

    e = T^4 a
    a = 4σ/c
    e = T^4 4σ/c
    T^4 = e/(4σ/c)
    T = ^4√(e/(4σ/c))
    T = ^4√(e/a)

    Since we’re using the Kelvin temperature scale, which has its base at 0 K, we can calculate temperature (and thus energy density) as above. Energy density at 0
    K is zero, thus temperature at zero energy density is, of course, 0 K.

    q = ε σ (T_h^4 – T_c^4)

    ∴ q = ε σ ((eh / (4σ / c)) – (ec / (4σ / c))) Ah

    Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).

    W m-2 = W m-2 K-4 * (Δ(J m-3 / (W m-2 K-4 / m sec-1)))

    ∴ q = (ε c (eh – ec)) / 4

    Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).

    W m-2 = (m sec-1 (ΔJ m-3)) / 4

    For graybody objects, it is the energy density differential between warmer object and cooler object which determines warmer object radiant exitance. The climate alarmists misinterpret the S-B radiant exitance equation for graybody objects. Warmer objects don’t absorb radiation from cooler objects (a violation of 2LoT in the Clausius Statement sense and Stefan’s Law); the lower energy density gradient between warmer and cooler objects (as compared to between warmer object and 0 K) lowers radiant exitance of the warmer object (as compared to its radiant exitance if it were emitting to 0 K). The energy density differential between objects manifests an energy density gradient, each surface’s energy density manifesting a proportional radiation pressure.

    The S-B equation for graybody objects isn’t meant to be used to subtract a fictive ‘cooler to warmer’ energy flow from the incorrectly-calculated and thus too high ‘warmer to cooler’ energy flow, it’s meant to be used to subtract cooler object energy density (temperature is a measure of energy density, the fourth root of energy density divided by Stefan’s constant) from warmer object energy density. Radiant exitance of the warmer object is predicated upon the energy density gradient.

    Their problem, however, is that their take on radiative energetic exchange necessitates that at thermodynamic equilibrium, objects are furiously emitting and absorbing radiation (this is brought about because they claim that objects emit only according to their temperature (rather than according to the energy density gradient), thus for objects at the same temperature in an environment at the same temperature, all would be furiously emitting and absorbing radiation), and they’ve forgotten about entropy… if the objects (and the environment) are furiously emitting and absorbing radiation at thermodynamic equilibrium as their incorrect take on reality must claim, why does entropy not change?

    The second law states that there exists a state variable called entropy S. The change in entropy (ΔS) is equal to the energy transferred (ΔQ) divided by the temperature (T).

    ΔS = ΔQ / T

    Only for reversible processes does entropy remain constant. Reversible processes are idealizations. All real-world processes are irreversible.

    The climastrologists claim that energy can flow from cooler to warmer because they cling to the long-debunked Prevost Principle, which states that an object’s radiant exitance is dependent only upon that object’s internal state, and thus they treat real-world graybody objects as though they’re idealized blackbody objects via:

    q = σ T4

    … thus the climate alarmists claim that all objects emit radiation if they are above 0 K. In reality, idealized blackbody objects emit radiation if they are above 0 K, whereas graybody objects emit radiation if their temperature is greater than 0 K above the ambient.

    But their claim means that in an environment at thermodynamic equilibrium, all objects (and the ambient) would be furiously emitting and absorbing radiation, but since entropy doesn’t change at thermodynamic equilibrium, the climastrologists must claim that radiative energy transfer is a reversible process.

    Except radiative energy transfer is an irreversible process, which destroys their claim. In reality, at thermodynamic equilibrium, no energy flows, which is why entropy doesn’t change. A standing wave is set up between two objects at thermodynamic equilibrium, with the standing wave nodes at the surface of the objects (this is standard cavity theory… deny it, and you deny wide swaths of science), thus no energy can be transferred to or from the objects.

    Should one object change temperature, the standing wave becomes a traveling wave, with the group velocity proportional to the energy density differential, and in the direction toward the cooler object.

    All idealized blackbody objects above absolute zero emit radiation. Idealized blackbody objects do not emit (nor absorb) according to the energy density gradient. Idealized blackbody objects don’t actually exist, they’re idealizations.

    Real-world graybody objects with a temperature greater than zero degrees above their ambient emit radiation. Graybody objects emit (and absorb) according to the energy density gradient.

    That interaction through radiation pressure determines radiant exitance of each object. So while the climate alarmists claim that there’s no way a photon could possibly ‘know’ the temperature of the object upon which that photon supposedly incides, it absolutely does ‘know’ because that photon must pass through the EM field (the photon being nothing but a quantum of EM energy; per QFT, a persistent perturbation of the EM field above the average field energy density) between objects, and thus the energy density gradient between objects… and if the EM field energy density (ie: the energy density gradient between the objects) is such that the chemical potential of the EM field due to that energy density gradient becomes higher than the chemical potential of the photon from a cooler object, that photon won’t even reach the warmer object… it will be subsumed into the background EM field (there is no law of conservation for photons).

    • May 11, 2022 3:17 pm

      Genuine question.
      Imagine we have two identical vacuum chambers. In chamber A we have a cube of material which can be heated electrically to a known temperature.

      In chamber B we have two such cubes, close to each other but not touching. All three cubes are identical.

      We heat the cube in chamber A to (say) 800K.

      In chamber B, we heat cube #1 to 800K and cube #2 to 400K.

      We then turn off the current to all cubes simultaneously and plot the temperature change with time of all three cubes.

      So what would we see? If what you say is correct, the cooling curve of the chamber A cube should be identical to that of cube #1 in chamber B. The presence of 400K cube #2 in chamber B should make no difference to the cooling curve of cube #1.

      Is this what you are saying?

      • LOL@Klimate Katastrophe Kooks permalink
        May 11, 2022 7:27 pm

        Assuming all three cubes are the same size and composition, you’re already starting with more internal energy in the two-cube chamber B, so I’m not sure the point you’re attempting to make.

        Why “if what I say is correct”, would the cooling curve of the Chamber A cube be identical to that of cube #1 in Chamber B?

        Are you attempting to conflate a cube (which doesn’t convect) with an atmosphere (which convectively, advectively and via specific heat capacity and latent heat capacity removes ~76.2% of all surface energy)? Sure you are.

        That leaves only 23.8% for radiant exitance from the surface. Which is the bigger effect… 76.2% or 23.8%? Now consider what happens to that bigger effect should one change the overall composition of the atmosphere and thereby increase the specific heat capacity and/or latent heat capacity of the atmosphere.

        Question: Why do A/C units use high-DOF polyatomic molecules as a refrigerant, whereas the fill gas in double-pane windows is generally a low-DOF simple monoatomic gas?

        Hint: Because the first maximizes energy transfer, whereas the second minimizes it.

        Now compare CO2 (a polyatomic) or H2O (a polyatomic) to Ar (a monoatomic) or N2 (a homonuclear diatomic).

      • May 12, 2022 10:06 am

        The cube experiment was about seeing if a cool body could warm a hotter body. If what you say about radiant energy is true, the 400K cube should not transfer any energy to the 800K cube. This means the cooling curve of the 800K cubes in both chambers should be identical.
        They are in vacuum chambers precisely to isolate the effect of radiant energy and so convection does not confuse the result.

  39. LOL@Klimate Katastrophe Kooks permalink
    May 11, 2022 2:51 pm

    “But… but!… but!!!… it’s the positive feedback between CO2 and water vapor that causes the CAGW!”, you may blurt.

    Yeah, no.

    Water vapor isn’t a ‘global warming’ gas… it acts as a literal refrigerant (in the strict ‘refrigeration cycle’ sense) below the tropopause.

    ———-

    You know, the refrigeration cycle (Earth) [A/C system]:

    A liquid evaporates at the heat source (the surface) [in the evaporator], it is transported (convected) [via an A/C compressor], it emits radiation to the heat sink and undergoes phase change (emits radiation in the upper atmosphere, the majority of which is upwelling owing to the mean free path length / altitude / air density relation) [in the condenser], it is transported (falls as rain or snow) [via that A/C compressor], and the cycle repeats.

    That’s kind of why, after all, the humid adiabatic lapse rate (~3.5 to ~6.5 K km-1) is lower than the dry adiabatic lapse rate (~9.81 K km-1).

    The effective emission height is ~5.105 km. That also happens to be where atmospheric pressure is 1/2 that at sea level, which isn’t a coincidence.

    7 – 13 µm: >280 K (near-surface)
    >13 – 17 µm: ~260 – ~240 K (~5km in the troposphere)

    The emission profile is equivalent to a BB with a temperature of 255 K, and thus an effective emission height of 5.105 km.

    The lapse rate is said to average ~6.5 K km-1. 6.5 K km-1 * 5.105 km = 33.1825 K. That is not the ‘greenhouse effect’, it’s the tropospheric lapse rate. The climate alarmists have conflated the two.

    Polyatomic molecules (CO2, H2O) reduce the adiabatic lapse rate (ALR), not increase it (dry ALR: ~9.81 K km-1; humid ALR: ~3.5 to ~6.5 K km-1) by dint of their higher specific heat capacity and/or latent heat capacity convectively transiting more energy (as compared to the monoatomics and homonuclear diatomics), thus attempting to reduce temperature differential with altitude, while at the same time radiatively cooling the upper atmosphere faster than they can convectively warm it… they increase thermodynamic coupling between heat source and sink… they are coolants.

    9.81 K km-1 * 5.105 km = 50.08005 K (dry adiabatic lapse rate, due primarily to homonuclear diatomics and monoatomics), which would give a surface temperature of 255 + 50.08005 = 305.08005 K. Sans CO2, that number would be even higher.

    Water vapor (primarily) reduces that to 272.8675 K – 288.1825 K, depending upon humidity. Other polyatomics (CO2) contribute to cooling, to a lesser extent. The higher the concentration of polyatomics, the more vertical the lapse rate, the cooler the surface.

    Also remember: the atmosphere is stable as long as actual lapse rate is less than ALR… and a greater concentration of polyatomic molecules reduces ALR… thus convection increases.

    That’s kind of why, after all, neither CO2 nor H2O are used as a filler gas in double-pane windows… if they were such terrific ‘heat trapping’ gases, they’d be used as such. They’re not. Low DOF (Degree of Freedom), low specific heat capacity monoatomics generally are.

    So one could say that in regards to water, we live inside the equivalent of the evaporator of a gigantic world-sized A/C unit, with water as the refrigerant.

    It is the monoatomics and homonuclear diatomics which are the actual ‘greenhouse’ gases… remember that an actual greenhouse works by hindering convection.

    Monoatomics (Ar) have no vibrational mode quantum states, and thus cannot emit (nor absorb) IR. Homonuclear diatomics (O2, N2) have no net magnetic dipole and thus cannot emit (nor absorb) IR unless that net-zero magnetic dipole is perturbed via collision.

    In an atmosphere consisting of solely monoatomics and diatomics, the atoms / molecules could pick up energy via conduction by contacting the surface, just as the polyatomics do; they could convect just as the polyatomics do… but once in the upper atmosphere, they could not as effectively radiatively emit that energy, the upper atmosphere would warm, lending less buoyancy to convecting air, thus hindering convection… and that’s how an actual greenhouse works, by hindering convection.

    Thus, common sense dictates that the thermal energy of the ~95.94 – 99.74% (depending upon humidity) of the atmosphere which cannot effectively radiatively emit (N2, O2, Ar) must be transferred to the so-called ‘greenhouse gases’ (CO2 being a lesser contributor below the tropopause and the largest contributor above the tropopause, water vapor being the main contributor below the tropopause) which can radiatively emit and thus shed that energy to space.

  40. LOL@Klimate Katastrophe Kooks permalink
    May 11, 2022 7:47 pm

    Perhaps this will explain better the mathematical fraudery the climastrologists have been committing:

    https://i.imgur.com/QErszYW.gif

    ———-
    The climate alarmists misuse the S-B equation, using the form meant for idealized blackbody objects upon graybody objects:
    q = σ T^4
    … and slapping ε onto that (sometimes) …
    q = ε σ T^4

    Their misuse of the S-B equation inflates radiant exitance far above what it actually is for all graybody objects, necessitating that they carry that error forward through their calculations and cancel it on the back end, essentially subtracting a wholly-fictive ‘cooler to warmer’ energy flow from the real (but calculated incorrectly and thus far too high) ‘warmer to cooler’ energy flow… which leads especially scientifically-illiterate climate alarmists to conclude that energy actually can flow ‘cooler to warmer’ (a violation of 2LoT and Stefan’s Law).

    https://i.imgur.com/2Bd25TG.png

    The S-B equation for graybody objects isn’t meant to be used to subtract a fictive ‘cooler to warmerenergy flow from the incorrectly-calculated and thus too high ‘warmer to coolerenergy flow, it’s meant to be used to subtract cooler object energy density (temperature is a measure of energy density, the fourth root of energy density divided by Stefan’s constant) from warmer object energy density. Radiant exitance of the warmer object is predicated upon the energy density gradient.

    If anyone wants an empirical example, I can provide my tear-down of the Kiehl-Trenberth graphic.

    I can show how the climate alarmist take on radiative energetic exchange leads one to believe that a cooler but higher emissivity object can warm a warmer but lower emissivity object… in this case I use an ice cube (at 273.15 K) and aluminum foil (at 600K), using the very calculations the climate alarmists use. Then I do it correctly, wherein I arrive at the correct answer to a precision of 3.8 parts per 100 trillion.

    I can also analogize thermodynamics to electrical theory, complete with a circuit simulator demonstration, since people can think in terms of amps, watts and volts more easily than radiant exitance, radiant flux and energy density. They can think more easily in terms of two batteries of different voltages connected ‘+ to +’ and ‘- to -‘ than they can think of two objects at different temperatures (and thus energy densities).

    • Ed Bo permalink
      May 12, 2022 3:42 am

      LoL@KKK:

      Interesting ideas – sent me scurrying back to my textbooks, where I could find absolutely no support for them.

      I did come across this old classic, though. I guess Clausius was a “climate alarmist”…

      Clausius DIE MECHANISCHE WÄRMETHEORIE 1887

      ” What further regards heat radiation as happening in the usual manner, it is known that not only the warm body radiates heat to the cold one but that the cold body radiates to the warm one as well, however the total result of this simultaneous double heat exchange is, as can be viewed as evidence based on experience, that the cold body always experiences an increase in heat at the expense of the warmer one. ”

      I just did the calculation for your cold ice cube and hot aluminum foil example, using the “climate alarmist” method you criticize. To keep it simple, I used two 1 m^2 parallel plates close together (so edge effects can be ignored) well isolated from the rest of the universe. I used 0.95 emissivity for the ice, and 0.05 for the foil (close to real-world values).

      I’m curious as to the results YOU got with the “climate alarmist” method, and with your method. Also, how did you verify the correctness to 3.8ppt?

      BTW, the result I got with the “climate alarmist” method using Clausius’ radiative exchange method (just as I learned in my engineering classes) was 350 watts transfer from the hot foil to the cold ice.

      • May 12, 2022 10:21 am

        It’s my belief that LOL has made one or more fundamental errors. You point out the same one that I have thought of, i.e. that cool bodies do indeed radiate IR which can be absorbed by a hotter body nearby. I lack the physics and maths to prove what is going on however !

      • AC99 permalink
        May 13, 2022 8:44 pm

        Let’s make your problem even simpler so we don’t have to deal with reflection or transmission. Let’s make all of the plates blackbodies just to make the math simpler and the problem cleaner. Let’s also say that your plates are emitting radiation to 0K surroundings. Then what are the governing equations and their solution?

        For a single plate with electrical energy in at 100 Watts, the plate will emit 50 Watts out from its right side and 50 W out from its left side. Thus it emits 50 W/m^2, and to do so to 0K surroundings the Stefan-Boltzmann equation, q = sigma T^4, implies that its steady state temperature must be 172K.

        Now, with the second plate in place and only the first plate with 100W being supplied, the heat transferred from the heated plate to the second plate must be, q12 = sigma (T1^4 – T2^4). The heat transferred from the heated plate to the 0K surroundings must be q10 = sigma T1^4. The energy balance for the heated plate is then q12 + q10 = 100, or 2 T1^4 – T2^4 = 100/sigma. Next, the second plate transfers heat to the 0K surroundings in the amount q20 = sigma T2^4. Then, the energy balance for the second plate is q12 – q20 = 0, or T1^4 – 2 T2^4 = 0. Finally, if we solve these two simultaneous energy balance equations we get 3/2 T1^4 = 100/sigma -> T1=185K, and 3 T2^4 = 100/sigma -> T2=156K.

        Do you agree with my calculations?

      • AC99 permalink
        May 13, 2022 8:44 pm

        Let’s make your problem even simpler so we don’t have to deal with reflection or transmission. Let’s make all of the plates blackbodies just to make the math simpler and the problem cleaner. Let’s also say that your plates are emitting radiation to 0K surroundings. Then what are the governing equations and their solution?

        For a single plate with electrical energy in at 100 Watts, the plate will emit 50 Watts out from its right side and 50 W out from its left side. Thus it emits 50 W/m^2, and to do so to 0K surroundings the Stefan-Boltzmann equation, q = sigma T^4, implies that its steady state temperature must be 172K.

        Now, with the second plate in place and only the first plate with 100W being supplied, the heat transferred from the heated plate to the second plate must be, q12 = sigma (T1^4 – T2^4). The heat transferred from the heated plate to the 0K surroundings must be q10 = sigma T1^4. The energy balance for the heated plate is then q12 + q10 = 100, or 2 T1^4 – T2^4 = 100/sigma. Next, the second plate transfers heat to the 0K surroundings in the amount q20 = sigma T2^4. Then, the energy balance for the second plate is q12 – q20 = 0, or T1^4 – 2 T2^4 = 0. Finally, if we solve these two simultaneous energy balance equations we get 3/2 T1^4 = 100/sigma -> T1=185K, and 3 T2^4 = 100/sigma -> T2=156K.

        Do you agree with my calculations?

  41. LOL@Klimate Katastrophe Kooks permalink
    May 12, 2022 10:15 pm

    Ed Bo is attempting to appeal to authority from the guy who wrote the Clausius Statement sense of 2LoT, as his justification for the climastrologists misusing the S-B equation and thus their entire CAGW facade violating 2LoT. LOL

    KB is just so very sure that I’ve made a fundamental error, but he can’t point it out. LOL

    Both of them could stand to learn a bit of actual science, rather than desperately clinging to the warmist propaganda.

    The climate alarmists misuse the S-B equation, using the form meant for idealized blackbody objects upon graybody objects:
    q = σ T^4
    … and slapping ε onto that (sometimes) …
    q = ε σ T^4

    Their misuse of the S-B equation inflates radiant exitance far above what it actually is for all graybody objects, necessitating that they carry that error forward through their calculations and cancel it on the back end, essentially subtracting a wholly-fictive ‘cooler to warmer’ energy flow from the real (but calculated incorrectly and thus far too high) ‘warmer to cooler’ energy flow… which leads especially scientifically-illiterate climate alarmists to conclude that energy actually can flow ‘cooler to warmer’ (a violation of 2LoT and Stefan’s Law).

    https://i.imgur.com/2Bd25TG.png

    The S-B equation for graybody objects isn’t meant to be used to subtract a fictive ‘cooler to warmerenergy flow from the incorrectly-calculated and thus too high ‘warmer to coolerenergy flow, it’s meant to be used to subtract cooler object energy density (temperature is a measure of energy density, the fourth root of energy density divided by Stefan’s constant) from warmer object energy density. Radiant exitance of the warmer object is predicated upon the energy density gradient.

    Their problem, however, is that their take on radiative energetic exchange necessitates that at thermodynamic equilibrium, objects are furiously emitting and absorbing radiation (this is brought about because they claim that objects emit only according to their temperature (rather than according to the energy density gradient), thus for objects at the same temperature in an environment at the same temperature, all would be furiously emitting and absorbing radiation), and they’ve forgotten about entropy… if the objects (and the environment) are furiously emitting and absorbing radiation at thermodynamic equilibrium as their insane take on reality must claim, why does entropy not change?

    The second law states that there exists a state variable called entropy S. The change in entropy (ΔS) is equal to the energy transferred (ΔQ) divided by the temperature (T).

    ΔS = ΔQ / T

    Only for reversible processes does entropy remain constant. Reversible processes are idealizations. All real-world processes are irreversible.

    The climastrologists claim that energy can flow from cooler to warmer because they cling to the long-debunked Prevost Principle, which states that an object’s radiant exitance is dependent only upon that object’s internal state, and thus they treat real-world graybody objects as though they’re idealized blackbody objects via:
    q = σ T^4
    … thus the climate alarmists claim that all objects emit radiation if they are above 0 K. In reality, idealized blackbody objects emit radiation if they are above 0 K, whereas graybody objects emit radiation if their temperature is greater than 0 K above the ambient.

    But their blather means that in an environment at thermodynamic equilibrium, all objects (and the ambient) would be furiously emitting and absorbing radiation, but since entropy doesn’t change at thermodynamic equilibrium, the climastrologists must claim that radiative energy transfer is a reversible process.

    Except radiative energy transfer is an irreversible process, which destroys their blather. In reality, at thermodynamic equilibrium, no energy flows, which is why entropy doesn’t change.

    All idealized blackbody objects above absolute zero emit radiation. Idealized blackbody objects do not emit (nor absorb) according to the energy density gradient. Idealized blackbody objects don’t actually exist, they’re idealizations.

    Real-world graybody objects with a temperature greater than zero degrees above their ambient emit radiation. Graybody objects emit (and absorb) according to the energy density gradient.

    It’s right there in the S-B equation, which the climate alarmists fundamentally misunderstand:
    https://i.imgur.com/pFBcBl0.png

    They cite Clausius out of context… Clausius was discussing a cyclical process by which external energy did work to return the system to its original state (for irreversible processes), or which returned to its original state because it is an idealized reversible process… except idealized reversible processes don’t exist. They’re idealizations. All real-world processes are irreversible processes, including radiative energy transfer, because radiative energy transfer is an entropic temporal process.

    If your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation. – Arthur Eddington: The Nature of the Physical World. (1929)

    Their mathematical fraudery is what led to their ‘energy can flow willy-nilly without regard to energy density gradient‘ narrative (in their keeping with the long-debunked Prevost Principle), which led to their ‘backradiation‘ narrative, which led to their ‘CAGW‘ narrative, all of it definitively, mathematically, scientifically proven to be fallacious.

    One can use electrical theory to arrive at the same conclusions…

    [1] Images of circuit simulator:
    https://i.imgur.com/6vvuwHb.gif
    https://i.imgur.com/sLHKnn1.gif

    [2] Circuit simulator:
    https://tinyurl.com/yzo8hak9

    The bottom circuit in the circuit simulator is how reality works… objects interact. That interaction through radiation pressure determines radiant exitance of each object. So while the climate alarmists claim that there’s no way a photon could possibly ‘know’ the temperature of the object upon which that photon supposedly incides, it absolutely does ‘know’ because that photon must pass through the EM field (the photon being nothing but a quantum of EM energy) between objects, and thus the energy density gradient between objects… and if the EM field energy density (ie: the energy density gradient between the objects) is such that the chemical potential of the EM field is higher than the chemical potential of the photon, a photon from a cooler object won’t even reach the warmer object… it will be subsumed into the background EM field (there is no law of conservation for photons).

    You’ll note that the top two circuits in the circuit simulator are how the climate alarmists calculate radiant exitance… (ie: they put each object into its own separate system that doesn’t interact with other objects, and they essentially claim all objects emit to 0 K, equivalent in electrical terms to discharging to ground). That’s not how reality works, and their doing so leads them to conclude that a lower-temperature but higher-emissivity object can radiatively warm a higher-temperature but lower-emissivity object… in fact, their conclusions lead one to believe that wrapping an ice cube in aluminum foil (with the foil at 600 K!) will radiatively warm the foil and cool the ice, which is ludicrous.

    Ice has an emissivity of 0.96 – 0.99. We’ll use 0.96, with the ice temperature at 273.15 K.

    Aluminum foil has an emissivity of 0.04.

    We’ll put the temperature of the aluminum foil at 600 K.

    Now, we’ll do the calculations as the climate alarmists do them, using their assumption that the objects are emitting to 0 K. That treats the objects as though they’re idealized blackbody objects (ie: emission to 0 K), just with ε0K emit, and that their radiant exitance is not affected by ambient energy density.

    Aluminum foil:
    q = ε σ Th4 = 0.04 * 5.670374419e−8 W m−2 K−4 * 129,600,000,000 K4 = 293.95220988096 W m-2

    Ice:
    q’ = ε σ Tc4 = 0.96 * 5.670374419e−8 W m−2 K−4 * 5,566,789,756.30100625 = 303.03150940877251107800754 W m-2

    q – q’ = -9.07929952781251107800754 W m-2

    Yes, folks, the climate alarmists claim that ice at 273.15 K (32 F, 0 C) can warm aluminum foil at 600 K (620.33 F, 326.85 C). In effect, they claim that if you wrap an ice cube in aluminum foil, it’ll warm the aluminum foil as the ice cube radiatively cools to 0 K.

    Now, let’s do it correctly:

    T = 4√e/(4σ/c)

    e = T4 4σ/c
    e = T4 a

    c = 299792458 m s-1

    Th = 600 K
    εh = 0.04
    eh = 9.8051902920439712996382317262965e-5 J m-3

    Tc = 273.15 K
    εc = 0.96
    ec = 4.2116846355194792664142496540057e-6 J m-3

    σ = 5.670374419e−8 W m−2 K−4
    a = 7.5657e-16 J m-3 K-4 = 4σ/c = 7.5657332500339284719430800357226e-16 J m-3 K-4

    σ / a = 74948114.5 W J-1 m
    σ / a * Δe = W m-2

    9.8051902920439712996382317262965e-5 J m-3 – 4.2116846355194792664142496540057e-6 = 9.3840218284920233729968067608959e-5 J m-3 energy density differential

    9.3840218284920233729968067608959e-5 J m-3 energy density differential * 74948114.5 W J-1 m * (εh = 0.04) = 281.3258969889278120384163525 W m-2

    https://i.imgur.com/SgvQvLb.gif

    Again, the 0.0036929889278120384163525 W m-2 differential is due to rounding on the Hyperphysics webpage.

    In fact:
    (281.3222048179386 / 5.6703e-8 * 5.670374419e-8) – 281.3258969889278120384163525 = 0.000000000000038296252606814321993545315062695 W m-2

    That’s a precision of 3.8 parts per 100 trillion, just by accounting for the rounding of σ on the Hyperphysics page. And the remaining differential is still due to rounding (of the final result) on the Hyperphysics page. LOL

    In fact, the Kiehl-Trenberth diagram…
    https://i.imgur.com/RJMT26o.png
    … does exactly as I stated… it treats a real-world (graybody) surface as if it were an idealized blackbody object, with emission to 0 K ambient and ε = 1. That’s the only way that diagram can get to 390 W m-2 surface radiant exitance.
    https://i.imgur.com/nJh4Z7U.png
    That’s proof-positive that they’ve misused the S-B equation to fit their narrative. Had they used the actual emissivity, they couldn’t have arrived at 390 W m-2 (see below), and had they used the proper form of the S-B equation for graybody objects, they’d not have even gotten close to 390 W m-2 (see below).

    Their use of the wrong formula increases radiant exitance of graybody objects far above what it actually is:
    https://i.imgur.com/4ulMbq6.png
    … which necessitates that they carry those incorrect values through their calculation and subtract a fictional ‘cooler to warmer‘ energy flow from the real (but calculated incorrectly and thus too high) ‘warmer to cooler‘ energy flow.

    Thus, some of the alarmists come to believe that energy actually can flow ‘cooler to warmer‘ (the basis of their ‘backradiation‘ blather). This violates 2LoT in the Clausius Statement sense… energy cannot flow from lower energy density to higher energy density without external energy doing work upon the system to push that energy against the energy gradient. Do remember that a warmer object will have higher energy density at all wavelengths than a cooler object.

    The equation for the radiation energy density is Stefan’s Law and a is Stefan’s constant.
    e = aT^4

    ∴ T = 4^√(e/a)

    In other words, temperature is equal to the fourth root of energy density divided by Stefan’s constant. It is a measure of energy density.

    Keep in mind that Stefan’s constant above equals 4σ/c (which is sometimes known as the radiation constant).

    Which is why: U = T^4 4σ/c
    The above formula is the Stefan-Boltzmann relation between energy density and temperature.

    If ΔU = 0, then (ΔU * c/4σ) = 0, thus no energy can flow.

    U has the same physical units as pressure (J m-3) and U ∝ T. That is radiation pressure, which sets up the energy density gradient.

    This agrees with Planck’s Law: ρ(T) = aT^4 = T^4 4σ/c.

    The S-B equation integrates Planck’s Radiation Formula (which calculates the energy density for a given wavelength) over all wavelengths.

    F = U – TS
    Where:
    F = Helmholtz Free Energy (J)
    U = internal energy (J)
    T = absolute temp (K)
    S = system final entropy (J K-1)
    TS = energy the object can receive from the environment

    If U > TS, F > 0… energy must flow from object to environment.
    If U = TS, F = 0… no energy can flow to or from the object.
    If U < TS, F < 0… energy must flow from environment to object.

    Free energy is defined as the capacity to do work. If U = TS, p_photon = u/3 = p_object, energy cannot flow because no work can be done. Helmholtz Free Energy is zero. Photon chemical potential is zero.

    So in the real world, the energy density gradient determines radiant exitance, energy does not flow willy-nilly without regard to energy density gradient and 2LoT applies always and everywhere.
    https://i.imgur.com/IgxATSg.png

    • LOL@Klimate Katastrophe Kooks permalink
      May 12, 2022 10:53 pm

      “just with ε0K emit, and that their radiant exitance is not affected by ambient energy density.”

      A glitch, somehow… should read:
      “just with ε<1"

      Not sure how that got scrambled, it's not scrambled in my text files.

      • LOL@Klimate Katastrophe Kooks permalink
        May 12, 2022 10:55 pm

        Testing…

        Now, we’ll do the calculations as the climate alarmists do them, using their assumption that the objects are emitting to 0 K. That treats the objects as though they’re idealized blackbody objects (ie: emission to 0 K), just with ε0K emit, and that their radiant exitance is not affected by ambient energy density.

      • LOL@Klimate Katastrophe Kooks permalink
        May 12, 2022 10:57 pm

        Weird… it corrupts the text each time.

        I’ll just leave the last part of that paragraph here to see if it gets corrupted, and I’ll replace the greater than sign with “greater than”:

        Remember, the climate alarmists claim all objects greater than 0K emit, and that their radiant exitance is not affected by ambient energy density.

    • LOL@Klimate Katastrophe Kooks permalink
      May 13, 2022 12:09 am

      Analogizing thermodynamics to electrical theory, the climastrologists essentially claim:

      For a 1.5 V battery and a 12 V battery connected ‘+ to +’ and ‘- to -‘, electricity flows both ways, just more electricity flows from the 12 V battery to the 1.5 V battery than flows from the 1.5 V battery to the 12 V battery.

      For a 1.5 V battery and another 1.5 V battery connected ‘+ to +’ and ‘- to -‘, electricity flows both ways between the batteries, but it cancels out so there is no net flow.

      And that is so ludicrous as to be laughable. But thanks for proving to the world that you anti-science loons deserve to be the world’s laughingstocks. LOL

      I know, I know… now you’ll get all mad, stomp your feet, spittle-fleck vitriol, make desperate appeals to authority by taking quotes from the greats of science out of context… I’ve seen it all far too many times to keep count of, and I’ve already countered countless of your ilk. They all ran away, even the warmist physicists and climastrologists. Mikey “Mouse” Mann won’t even say my name. LOL

      Best you just slink off to the safety of the shrubbery to nurse your fragile psyches, said psyches mortally wounded by the realization that you were gullible enough to be taken in by a poorly-told climate fairy tale. Don’t be like all the others… don’t make this intellectual giant grasp your head between thumb and forefinger, squash your soft skulls, then toss you aside like a bug. LOL

  42. LOL@Klimate Katastrophe Kooks permalink
    May 12, 2022 10:18 pm

    KB wrote:
    “The cube experiment was about seeing if a cool body could warm a hotter body. If what you say about radiant energy is true, the 400K cube should not transfer any energy to the 800K cube. This means the cooling curve of the 800K cubes in both chambers should be identical.”

    Not so, but not because of what you claim (that the cooler object warms the warmer object radiatively). Can you figure it out?

    Think in terms of energy density gradient.

    • May 12, 2022 10:28 pm

      No I can’t figure it out.

      WHY cannot IR radiation emitted by the 400K cube be absorbed by the 800K cube?

      If the two cubes were radioactive sources, they would absorb radiation from each other, despite both of them being radioactive. What’s the difference?|

      • LOL@Klimate Katastrophe Kooks permalink
        May 12, 2022 10:40 pm

        Another hint: Specifically think of the difference in the energy density gradient on each face of the two warmest cubes. Isn’t the energy density gradient lower for the one with another cube facing it, as compared to the lone cube with nothing facing any of its faces? Sure it is.

        In fact, now that you know that, you can calculate the difference using the S-B equation on each face of each cube.

        I suggest converting temperature to energy density and calculating that way, as I show how to do in another post. That way you’ll start to understand exactly how the climastrologists have been fooling you all along.

  43. LOL@Klimate Katastrophe Kooks permalink
    May 12, 2022 10:28 pm

    Ed Bo wrote:
    “Clausius DIE MECHANISCHE WÄRMETHEORIE 1887

    ” What further regards heat radiation as happening in the usual manner, it is known that not only the warm body radiates heat to the cold one but that the cold body radiates to the warm one as well, however the total result of this simultaneous double heat exchange is, as can be viewed as evidence based on experience, that the cold body always experiences an increase in heat at the expense of the warmer one. ””

    Clausius was stating the current knowledge at the time. He went on in the next page (Page 296) to write:
    “If this assumption be correct, the principle enunciated above must be false, and the proof, deduced by means of that principle, of the second fundamental principle of the Mechanical Theory Of Heat would thus be overthrown.”

    Of course, 2LoT wasn’t overthrown, it’s a fundamental physical law. What was overthrown was the blather that Ed Bo is spewing. LOL

    Clausius wrote in his paper entitled “Entropy” (and reiterated in his “The Mechanical Theory of Heat” book):

    Carnot assumed, as has already been mentioned, that the equivalent of the work done by heat is found in the mere transfer of heat from a hotter to a colder body, while the quantity of heat remains undiminished.

    The latter part of this assumption – namely, that the quantity of heat remains undiminished – contradicts our former principle, and must therefore be rejected if we are to retain that principle.

    If we now suppose that there are two substances of which the one can produce more work than the other by the transfer of a given amount of heat, or, what comes to the same thing, needs to transfer less heat from A to B to produce a given quantity of work, we may use these two substances alternately by producing work with one of them in the above process. At the end of the operations both bodies are in their original condition; further, the work produced will have exactly counterbalanced the work done, and therefore, by our former principle, the quantity of heat can have neither increased nor diminished. The only change will occur in the distribution of the heat, since more heat will be transferred from B to A than from A to B, and so on the whole heat will be transferred from B to A. By repeating these two processes alternately it would be possible, without any expenditure of force or any other change, to transfer as much heat as we please from a cold to a hot body, and this is not in accord with the other relations of heat, since it always shows a tendency to equalize temperature differences and therefore to pass from hotter to colder bodies.

    It seems, therefore, to be theoretically admissible to retain the first and the really essential part of Carnot’s assumptions, and to apply it as a second principle in conjunction with the first; and the correctness of this method is, as we shall soon see, established already in many cases by its consequences.

    So Clausius wasn’t claiming that heat would flow from a colder to a warmer body, he was hypothesizing as means of refuting Carnot’s claim that the quantity of heat remains undiminished.

    If taking quotes out of context is the best you’ve got, you’re far out of your depth, Ed. LOL

    • Ed Bo permalink
      May 13, 2022 3:30 am

      LOL@KKK –

      You just keep getting more ridiculous! I quoted Clausius in the proper context – discussing how radiative heat transfer works. You, on the other hand, quoted him blatantly out of context.

      Your quote says: “If this assumption be correct, the principle enunciated above must be false, and the proof, deduced by means of that principle, of the second fundamental principle of the Mechanical Theory Of Heat would thus be overthrown.”

      You imply that “this assumption” refers to the section I had just quoted. But it does not! The sentence immediately preceding your quote concerns “the assumption that rays of heat can be concentrated by reflection in such a way, that at the focus thus produced a body may be rasied to a higher temperature than is possessed by the bodies which emit the rays.”

      That is a very different issue, and in no way invalidates the statement that I quoted. I try not to attribute to malice when confusion could explain an error, but it’s very hard for me to consider this an honest mistake on your part.

      Your second selection, where Clausius takes on Carnot’s assumptions, is also on a very different issue. Here Clausius takes on Carnot’s analysis of heat engines, which used the “caloric” theory of heat. Clausius is saying here that a heat engine producing work from a high-temperature source does not reject as much heat to the low-temperature sink as it takes from the high-temp source.

      Again, a very different issue from the idea of radiative exchange. Confusion once more on your part?

      Next, let’s look at your example of what you call “climate alarmist” analysis with the cold ice cube and hot foil problem. You imply that because the high-emissivity but cold ice cube would emit more than the low-emissivity but hot foil (303 vs 293 W/m2) that this analysis predicts that heat transfer is from the cold ice to the hot foil.

      WRONG!!! You made a very basic and fatal mistake there! You forgot that absorptivity equals emissivity (Kirchhoff’s law of thermal radiation), and so are using the wrong equation. So the foil only absorbs 4% of the 303 emitted toward it (~12) and reflects the other 96% (~291) back to the cube, which absorbs 96% (~279) of that reflecting ~12 back to the foil, and so on.

      Meanwhile, the ice absorbs 96% of the 293 the foil radiates toward it (~281), only reflecting 4% back (~12).

      If you follow these until convergence, you see that there is a strong resulting heat transfer from the hot foil to the cold cube. When I run the numbers here, I get a resulting 281 W/m2 from foil to cube, just like you do! So by your arguments, this actual analysis gets the right answer!

      This “absorptivity equals emissivity” concept is one of the most basic tenets of radiative heat transfer. Yet you seem utterly ignorant of it! Talk about being out of your depth…

      You also claim that objects stop radiating in an isothermal environment. Consider the case of conductive heat transfer. By your logic, the molecules of an isothermal solid would not vibrate, because that would cause irreversible heat transfer (even if net zero) across any plane in the solid. Your argument is equally ridiculous.

      Let’s do the math quickly. You use the equation dS = dQ/T for the entropy change from heat transfer. So radiation output of dQ results in a reduction in entropy of dQ/T. But if it receives the same dQ from the outside from a body at the same T, this results in an increase of the same dQ/T, for a net dS of 0. The same holds true for the other body.

      This IS a reversible process. It is only irreversible, with increasing entropy, if the temperatures are different.

      Your analyses are all built on a foundation of quicksand. You need to spend a lot of time going back to the basics to build a solid foundation, and having a little humility along the way would really help.

      • LOL@Klimate Katastrophe Kooks permalink
        May 13, 2022 3:49 am

        Ed Bo, the first climate loon to stomp his feet and spit vitriol, wrote:
        “I quoted Clausius in the proper context”

        No, you didn’t. You neglected to mention that on the very next page, Clausius debunked the then-current knowledge that radiative energetic flow didn’t depend upon energy density gradient. I did so, and now you’re all butthurt and denying reality. LOL

        Ed Bo wrote:
        “You also claim that objects stop radiating in an isothermal environment. Consider the case of conductive heat transfer. By your logic, the molecules of an isothermal solid would not vibrate, because that would cause irreversible heat transfer (even if net zero) across any plane in the solid. Your argument is equally ridiculous.”

        You’re conflating two different concepts because you’re desperate to defend your CAGW religious belief.

        I suggest you look up the term “quiescent state”, then come back here to apologize to everyone for spewing misinformation. LOL

      • LOL@Klimate Katastrophe Kooks permalink
        May 13, 2022 3:55 am

        Ed Bo wrote:
        “This IS a reversible process. It is only irreversible, with increasing entropy, if the temperatures are different.”

        There’s the climate loon, claiming radiative energetic transfer is a reversible process… not even warmist physicists are bold enough to attempt that, but not you! No, sir, not you! You’ll spew whatever you need to spew to defend your religious belief in the poorly-told climate fairy-tale. LOL

        And in so doing, you deny every single fundamental physical law.

        All real-world processes are irreversible. Reversible processes are idealizations.

        You’ll be getting right on providing a link to even one real-world process which is an idealized reversible process… we’ll wait right here. LOL

      • LOL@Klimate Katastrophe Kooks permalink
        May 13, 2022 4:07 am

        Ed Bo wrote:
        “You imply that “this assumption” refers to the section I had just quoted. But it does not! The sentence immediately preceding your quote concerns “the assumption that rays of heat can be concentrated by reflection in such a way, that at the focus thus produced a body may be rasied to a higher temperature than is possessed by the bodies which emit the rays.””

        And Ed Bo remains completely incognizant of how that completely destroys his argument and corroborates mine. LOL

        Hint: Check the fundamental physical laws. LOL

      • LOL@Klimate Katastrophe Kooks permalink
        May 13, 2022 4:45 am

        Ed Bo wrote:
        “You made a very basic and fatal mistake there! You forgot that absorptivity equals emissivity (Kirchhoff’s law of thermal radiation), and so are using the wrong equation.”

        Yeah, there’s a huge problem with Kirchhoff’s Law, and it’s been invalidated for all but certain laboratory conditions.

        Further, its statement that absorptivity equals emissive power (not emissivity) is only true at thermodynamic equilibrium of an object with its ambient, and only because both are zero.

        So you don’t know the difference between emissivity and emissive power, which is why you confused the two in regards to Kirchhoff’s Law. Don’t worry, even Kirchhoff himself confused absorptive power and absorptivity in his first iteration of Kirchhoff’s Law. LOL

        Think about it… if absorptivity and emissive power were always equal, an object could never change temperature. But that fact never occurred to you, did it, Ed? LOL

        Nor, apparently, were you cognizant of the fact that Kirchhoff’s Law is a ratio, not an equality.

        In Kirchhoff’s original parlance:
        E/A = e
        Not E=A.

        In modern parlance:
        Eν/αν= f(T, ν)

        In other words, the ratio between emissive power and absorptivity is equal to specific intensity.

        See, Ed, there’s so much that you think you know, but what you know is so wrong. LOL

        For instance… how does an object absorb radiation it just emitted (your blather about the ice cube and aluminum foil)… in order for energy to flow, work must be able to be done (free energy)… no free energy, no work able to be done, no energy flow.

        So now you’re essentially claiming that an object can emit a photon, that photon can be reflected, and that photon can then do work upon the object while the object is at the same temperature as it was when that photon was emitted (unless you’re claiming that ice changes temperature as it melts, which means you have a far deeper fundamental misunderstanding of phase change and shouldn’t even attempt discussing anything more complicated. LOL).

      • LOL@Klimate Katastrophe Kooks permalink
        May 13, 2022 5:21 am

        Here you go, Ed:

        https://web.ics.purdue.edu/~gonza226/ME612/Lecture-10.pdf#page=5

        Thermodynamic equilibrium
        After an external perturbation (of finite duration in time), all systems tend to evolve to a quiescent and spatially homogeneous (at the macroscopic length scale) terminal state where the system’s macroscopic observables have constant limiting values.

        https://www.sciencedirect.com/science/article/pii/S0003491615003504

        The quiescent state of an isolated system in which all properties remain constant on the t_TD time scale is called the state of thermodynamic equilibrium.

        https://muycapaz.wordpress.com/2008/11/01/sciam-does-nature-break-the-second-law-of-thermodynamics/
        The Achilles’ heel of thermodynamics is that, strictly speaking, it applies only when the system under study is in a quiescent state called equilibrium.

        I mean, it’s only in the definition of “thermodynamic equilibrium”, Ed, so it’s understandable that you wouldn’t have picked up on that. LOL

        The world expects an apology from you for spewing disinformation, Ed. LOL

      • LOL@Klimate Katastrophe Kooks permalink
        May 13, 2022 6:37 am

        Ed Bo wrote:
        “WRONG!!! You made a very basic and fatal mistake there! You forgot that absorptivity equals emissivity (Kirchhoff’s law of thermal radiation), and so are using the wrong equation.”

        Kirchhoff’s law I address in another post… it’s been invalidated except for certain laboratory conditions, you don’t know the difference between “emissivity” and “emissive power”, you’re apparently not aware that it’s a ratio not an equality, nor are you aware that absorptivity and emissive power are only equal at thermodynamic equilibrium because both are zero.

        If I used the “wrong equation”, then you must admit that Kiehl and Trenberth did as well, in their KT diagram, given that I was using the equation the climastrologists use. You must now admit that every single climastrologist using:
        q = σ T^4
        … and slapping ε onto that (sometimes) …
        q = ε σ T^4
        … for real-world graybody objects is doing it wrong, which invalidates large swaths of climate “science”. LOL

        Nicely foot-bulleted, Ed. LOL

        Especially given that you then go on to perform a complicated iterative calculation (that the climastrologists don’t do, as evidence by the KT diagram, as I outline in another post), then you state:

        “When I run the numbers here, I get a resulting 281 W/m2 from foil to cube, just like you do! So by your arguments, this actual analysis gets the right answer!”

        Which means you’re attempting to state: “You’re absolutely right! But you’re wrong.”

        Can’t have it both ways, Ed. You know I’m right, you get the same correct answer as I did when I calculated upon energy density (then corroborated that with the S-B equation to a precision of 3.8 parts per 100 trillion)… but you didn’t use the equation the climastrologists use, as I showed in the post about the KT diagram.

        I try not to attribute to malice when confusion could explain an error, but it’s very hard for me to consider this an honest mistake on your part. LOL

        Feeling a bit punch-drunk, Ed? LOL

      • Ed Bo permalink
        May 13, 2022 9:18 pm

        LOL@KKK –

        I have been looking for a source to corroborate your claim that Kirchhoff was fundamentally wrong in his assertion that absorptivity equals emissivity. I have looked both on-line and in dead-tree sources and found nothing. Every source I find backs up “absorptivity equals emissivity”.

        And this relationship has been used very successfully for a century now to solve radiative heat transfer problems. How is that possible?

        Do you really still contend that metal foil absorbs all radiation that strikes it, as you used in your above calculations? Haven’t you ever noticed that metal foil reflects a lot of light (radiation)? Seriously?

  44. LOL@Klimate Katastrophe Kooks permalink
    May 13, 2022 3:15 am

    Waiting for the climate loons to drag out their out-of-context quotes from Planck, Feynman, Maxwell, Lorentz, etc… do you think they’ll be surprised that their attempt has been attempted many times in the past, and each and every out-of-context quote put into context thus proving the climate loons wrong? Do you think they’ll be surprised that I’ve got all the prior writing on just this topic archived for just such an occurrence? Likely, they will be. They always are. They are always so convinced that they are original thinkers, and they never are. They are merely propaganda parrots. Not a critical thought between the lot of them. LOL

  45. Ed Bo permalink
    May 13, 2022 5:14 am

    LOL@KKK –

    I’m supposed to take my science advice from a person who thinks that shiny metal foil absorbs all of the radiation incident on it, rather than reflecting ~96% of it back? Who makes a factor of 25 error in his equations and won’t admit it? Who doesn’t understand Kirchhoff’s radiation law, which has been unchallenged for well over a century?

    I’m supposed to take my science advice from a person who thinks that a process that does not increase entropy (using his own equations) is not reversible? When the definition of an irreversible process is one that increases entropy?

    I’m supposed to take my science advice from a person who can’t see the exact parallel between isothermal radiation exchange and conductive exchange? Who can’t see that vibrating molecules on both sides of a plane in an isothermal solid transfer energy back and forth across that plane, a process that he asserts is irreversible – against all evidence?

    I’m supposed to take my science advice from a person who falsely accuses others of quoting scientific authorities out of context while blatantly doing that himself? (Clausius asserting that you cannot concentrate radiation to yield a higher temperature than that of the radiation source in no way debunks HIS OWN claim that there IS radiative exchange between bodies. I notice you do not even TRY to explain why it does.)

    • LOL@Klimate Katastrophe Kooks permalink
      May 13, 2022 5:26 am

      Unchallenged?! Kirchhoff’s law has been invalidated except for certain laboratory conditions… but you go right on denying reality, Ed. The world will sit back, watch, point and laugh. LOL

      Is the world supposed to take their advice from someone who denies the definition of thermodynamic equilibrium, and who claims there is such a thing as “isothermal radiation exchange” because he denies the fundamental physical laws? LOL

      Clausius stating that one cannot attain a higher temperature than the emitter no matter how focused the beam absolutely invalidates your claim that energy can flow without regard to energy density gradient, Ed. You just can’t figure out how that might be because you’re scientifically illiterate. And now you’re throwing a tantrum. LOL

  46. Ed Bo permalink
    May 13, 2022 5:24 am

    LOL@KKK –

    Here is a link to the heat transfer textbook that MIT uses to teach its thermal engineering students.

    Click to access AHTTv510.pdf

    It explicitly uses the same paradigm of “radiative exchange” that I have been using here. You can find it easily in both the introductory overview chapter, and at the very beginning of the chapter on radiation heat exchange.

    This fundamentally has not changed since I studied thermal sciences at MIT in the 1970s, before the present climate hysteria.

    Is it your assertion that MIT has for many decades been teaching its thermal engineering students, who go on to design some of our most sophisticated thermal systems, fundamentally wrong science???

    • LOL@Klimate Katastrophe Kooks permalink
      May 13, 2022 5:33 am

      Yes, they have. Study what I wrote and you’ll see that I’m correct. This is why you, who attended MIT and who learned that material, think there is such a thing as “isothermal radiative exchange”. LOL

      You are a product of that teaching, and here you are spewing incorrectitudes. LOL

      The way MIT teaches it is a handy shortcut way of accounting for energy flows, but it’s not the proper way of doing it because it leads the especially scientifically illiterate to conclude that energy can flow willy-nilly without regard to energy density gradient, that energy can flow against an energy density gradient without external energy pushing that system energy up the energy gradient… leading those who are especially, especially scientifically illiterate to claim there is “isothermal radiative exchange”. LOL

      I suggest you study up on what happens in a cavity at thermodynamic equilibrium. Is there not a standing wave set up in the cavity space? Are not the standing wave nodes at the wall surface (and hence no energy can be transferred either in or out of the walls)? Should one wall change temperature, does not that standing wave become a traveling wave with the group velocity proportional to the energy density gradient and in the direction of the cooler wall?

      Oh, but that’d invalidate all of your blather… you wouldn’t dare invalidate your own blather. Your poor little head would explode. LOL

    • LOL@Klimate Katastrophe Kooks permalink
      May 13, 2022 6:14 am

      Click to access AHTTv510.pdf

      Consider a wall connecting two thermal reservoirs as shown in Fig. 1.2. As long as T_1 > T_2, heat will flow spontaneously and irreversibly from 1 to 2. In accordance with our understanding of the Second Law of Thermodynamics, we expect the entropy of the universe to increase as a consequence of this process. If T_2 → T_2, the process will approach being quasistatic and reversible. But the rate of heat transfer will also approach zero if there is no temperature difference to drive it. Thus all real heat transfer processes generate entropy.

      And entropy production means it’s an irreversible process.

      Do remember that temperature is a measure of energy density, equal to the fourth root of energy density divided by Stefan’s constant. It is the temperature difference and hence the energy density gradient which is the impetus for energy to flow, whether by conduction or via radiation.

      But then, that’s on page 20 of that college textbook you learned at MIT… I guess you didn’t get that far into the book. LOL

      • Ed Bo permalink
        May 13, 2022 9:03 pm

        LOL@KKK –

        You continue to amaze me with the depth of your confusions. In an attempt to refute my claim that radiative exchange continues to occur in an isothermal environment (and after saying that the conductive case is not analogous) because any transfers must be irreversible, you quote from a textbook section on conductive transfer entitled “Reversible heat transfer as the temperature gradient vanishes”.

        This section expresses exactly what I have been arguing — that if there is a temperature difference, the resulting heat transfer is irreversible, but if there is no temperature difference, the process is “quasistatic and reversible”.

        Note that they do not claim that the underlying processes have stopped in an isothermal evironment. In fact, they use the term “quasistatic” to emphasize that this has not happened!

        You still have not shown me any errors in my demonstration using your equations (dS=dQ/T) that isothermal exchange does not increase entropy and therefore is reversible.

  47. avro607 permalink
    May 13, 2022 5:06 pm

    Thanks for your efforts LOL, the second law is rescued from those who wish to pervert the truth.

    • Ed Bo permalink
      May 13, 2022 9:19 pm

      You are seriously admiring a guy who thinks that metal foil absorbs all radiation that hits it?

      • LOL@Klimate Katastrophe Kooks permalink
        May 14, 2022 3:02 am

        You’re seriously still harping about my using the incorrect method used by the climastrologists, as a means of proving that they’re committing mathematical fraudery, Ed?

        You poor butthurt brainwashed little idiot. LOL

      • Ed Bo permalink
        May 15, 2022 1:25 am

        You are most certainly NOT using that method – you have simply demonstrated that you cannot understand it, even when it is explained carefully to you. The method does NOT assume that a body with emissivity of 0.04 absorbs all of the radiation incident upon it.

  48. Ed Bo permalink
    May 13, 2022 10:35 pm

    LOL@KKK –

    You continue to disparage the “radiative exchange” method of analyzing radiation heat transfer, claiming it leads to ridiculous results, such as predicting a cold ice cube would transfer heat to hot foil.

    But it is clear that you don’t understand this method properly at all. Properly done, it gives the same answers as your “energy gradient” method. Equation 10.24 (book page 565, file page 577) in the Leinhard text I linked covers your problem exactly. Just plug in your temperatures and emissivities to the equation and you get the same 281 W/m2 transfer from hot to cold as in your energy gradient method.

    (I first solved the problem iteratively to understand the individual steps involved in non-blackbody transfer. And no, I am not asserting that the radiation reflected back to the ice cube does thermodynamic work on the cube. Absorption of impinging radiation does NOT mean work is done.)

    At best, you have a philosophical objection to this method, arguing that it does not reflect (no pun intended) the underlying reality. But I would disagree even on that point. The radiative exchange method is truer to the underlying reality, not requiring distant objects to directly affect local sub-atomic actions like thermal photon emission, as your method does.

    And furthermore, the radiative exchange method is a lot more straightforward to apply to complex situations. There is a reason that every engineering textbook I have ever seen uses radiative exchange.

    • SamH permalink
      May 13, 2022 11:32 pm

      It is easy to disprove LOL@KKK’s statement that radiation cannot pass from a region of low energy density to a region of high energy density. Consider a hot and cold plate exchanging heat radiatively. Now shine a bright flashlight in the space between the plates. This increases the energy density between the plates, higher than the energy density above or below the light beam. If LOL@KKK were correct, radiation would not be able to flow through this region from either direction, and heat flow would stop.

      Obviously, it does not.

      • LOL@Klimate Katastrophe Kooks permalink
        May 14, 2022 8:29 pm

        I guess SamH isn’t aware that photons following the path of least time (not a straight line), that the energy density due to photons is directional.

        For instance, say we have Sam’s two plates, one top and one bottom. They are exchanging photons from warmer (top) to cooler (bottom).

        Now to the left and right, we put two more plates. The left plate is near 0 K (because being pedants, if I said ‘at 0 K’, the libtards would mewl “Nothing can be 0 K, so you’re wrong!!!”), the other is above the temperature of either of the two horizontal plates.

        The radiation going top-to-bottom isn’t affected by the radiation going right-to-left. The energy densities in the two directions are different.

        { We are, of course, neglecting the fact that some photons emitted by the vertical plates will incide upon the horizontal plates and vice versa… we make the assumption of straight-line plate-to-plate photon travel. If SamH wishes to do the overwhelmingly complicated calculations for every conceivable angle of emission, he may do so… but they’ll only prove him wrong. LOL }

        Why? Because the EM field energy density is created by the EM field, the EM field is comprised of photons, photons follow that nearly-straight-line (depending upon conditions… around a large stellar object, gravitational lensing would occur) path of least time, and photons do not interact below the pair-production energy.

        Nice try, Ed, but huge fail which only serves to further display your lack of knowledge. Try harder. LOL

      • SamH permalink
        May 14, 2022 8:41 pm

        LOL@KKK , of course you’re right that the radiation going from up-to-down would not be blocked by the flash light from the side. The reason is that your entire theory of energy density blocking radiation is crap.

        Instead, you are trying to salvage your theory by inventing “directional energy density.” But there is no such thing: energy density is a scalar quantity, and by definition does not have a direction. “Directional energy density” is an entirely made-up, ad hoc fabrication to save your pseudoscientific theory.

      • Ed Bo permalink
        May 15, 2022 1:28 am

        Sam – Take it easy on LOL. He doesn’t have the sophistication to understand that he just torpedoed his own arguments.

    • LOL@Klimate Katastrophe Kooks permalink
      May 14, 2022 3:00 am

      Ed Bo wrote:
      “You continue to disparage the “radiative exchange” method of analyzing radiation heat transfer, claiming it leads to ridiculous results”

      I don’t just “claim” it, I mathematically prove it. How else do you explain the K-T diagram?

      Yes, subtracting a wholly-fictive ‘cooler to warmer’ energy flow from a real but incorrectly calculated and thus far too high ‘warmer to cooler’ energy flow arrives at the same result as doing it properly… except it inflates radiant exitance for all graybody objects and leads especially scientifically-illiterate climate loons to conclude that energy can flow from cooler to warmer… and hence we have you here embarrassing yourself by denying the fundamental physical laws and wide swaths of long-established science. LOL

      Why can’t you just admit that what occurs in a cavity at thermodynamic equilibrium also occurs outside a cavity at thermodynamic equilibrium? That a standing wave is set up, that the wave nodes are at the wall surfaces (and hence no energy can flow into or out of the walls), that should one wall change temperature that standing wave becomes a traveling wave with the group velocity proportional to the energy density gradient and in the direction of the cooler wall.

      Oh, wait… I’ll answer that… because then you’d have to admit you bought into a poorly-told climate fairy tale which has led you to deny the fundamental physical laws. LOL

      But thanks for admitting that my method is the correct method and the method you were inculcated with is the simplified, dumbed down simpletonian method taught to those who cannot grasp the correct method. LOL

      Perhaps you should contact Kiehl and Trenberth and all the other climastrologists doing it incorrectly and apprise them of the Leinhard method… or would you rather attempt to explain how it is that Kiehl and Trenberth arrive at 390 W m-2 surface radiant exitance, which they could only have arrived at if they treated a real-world graybody surface as if it were an idealized blackbody object (emission to 0 K and emissivity = 1). LOL

      • Ed Bo permalink
        May 15, 2022 3:25 am

        LOL@KKK –

        You say, “I don’t claim it. I mathematically prove it.”

        Baloney! I’ve pointed out that you made an incredibly basic mistake, and you cannot acknowledge it. I pointed you to the exact equation in the MIT textbook explaining the “radiative exchange” method that produces what you consider to be the “right” answer, and somehow you still can’t get it.

        Then you say, “thanks for admitting that my method is the correct method and the method you were inculcated with is the simplified, dumbed down simpletonian method taught to those who cannot grasp the correct method.”

        Your reading comprehension is as poor as your scientific comprehension. I admitted no such thing. I demonstrated that the methods generate the same answers in these cases. It takes an incredible reality disconnect to call generations of professors and students at top universities like MIT “simpletons”.

        Next you say, “Perhaps you should contact Kiehl and Trenberth and all the other climastrologists doing it incorrectly and apprise them of the Leinhard method.” They ARE using the “Leinhard method”. In a simplified conceptual diagram, they round up typical surface emissivities of 0.97 to 1.00. I’d prefer it if they were a little more precise, but that’s not the big issue with that diagram.

  49. LOL@Klimate Katastrophe Kooks permalink
    May 14, 2022 3:06 am

    Ed Bo wrote:
    “You continue to amaze me with the depth of your confusions. In an attempt to refute my claim that radiative exchange continues to occur in an isothermal environment (and after saying that the conductive case is not analogous) because any transfers must be irreversible, you quote from a textbook section on conductive transfer entitled “Reversible heat transfer as the temperature gradient vanishes”.”

    “Approaches” is not “is”, Ed… you libtards seem to have trouble grasping those simple nuances. That text I quoted stated that as the temperature (and hence energy density) gradient tended to zero, so too did the heat transfer. At zero temperature differential, there would be zero heat transfer… unless you’re now going to deny the mathematics of the S-B equation. LOL

    Go on, do it, Ed. You know you want to. You’ve denied pretty much everything else (fundamental physical laws, that thermodynamic equilibrium is defined as a quiescent state, that no energy can flow if there is no energy density gradient and hence tacitly that temperature is a measure of energy density). LOL

  50. LOL@Klimate Katastrophe Kooks permalink
    May 14, 2022 3:11 am

    Ed Bo wrote:
    “You still have not shown me any errors in my demonstration using your equations (dS=dQ/T) that isothermal exchange does not increase entropy and therefore is reversible.”

    It is you making the claim in direct contradiction to known science that “isothermal radiation exchange” (your words) takes place… it is incumbent upon you to prove it.

    It is you making the claim in direct contradiction to known science that radiative energetic transfer is an idealized reversible process… it is incumbent upon you to prove it.

    Because long-established science says you are wrong.

    So prove your claims or STFU, Ed. Do so mathematically, if you’re able, which you’re not… because you’re wrong. LOL

    You see, Ed… you’re not fighting me, you’re fighting all of established science… and you are, as is usual for brainwashed libtards, diametrically opposite to reality. LOL

    • Ed Bo permalink
      May 15, 2022 3:42 am

      LOL@KKK – I HAVE proved my claim mathematically. I know it is a difficult claim, that two equal and opposite dQ/T values cancel out, resulting in zero entropy change, which yields a reversible thermodynamic process.

      It is YOU who has to back up a ridiculous claim, that the presence of a remote object at the same temperature somehow inhibits the molecular processes that create photon emission. And to cover the general case, explain how this occurs faster than the speed of light.

      And seriously, you need to understand when reversible “microscopic” (molecular) processes such as photon emission and absorption lead to irreversible macroscopic changes and when they do not. At present you are totally confused on this.

      • LOL@Klimate Katastrophe Kooks permalink
        May 15, 2022 4:05 am

        You’ve proven nothing of the sort, Ed. All you’ve demonstrated is that you don’t understand what you purport to understand, which is why we see you now spewing utter hogwash such as radiative energetic exchange being an idealized reversible process, in direct contradiction to long-known scientific reality.

        We’re still waiting for you to provide a link to even one real-world process which is an idealized reversible process. Why haven’t you done that, Ed? Should be a simple task, right, Ed? Just show us any scientific text which states that radiative energy transfer is an idealized reversible process… why can’t you do that, Ed? You should do that right away, Ed… otherwise people might start thinking that you’re pulling ‘facts’ from your sun-don’t-shine again. LOL

  51. LOL@Klimate Katastrophe Kooks permalink
    May 14, 2022 3:19 am

    Ed Bo wrote:
    “I have been looking for a source to corroborate your claim that Kirchhoff was fundamentally wrong in his assertion that absorptivity equals emissivity. ”

    Still unable to grasp the fundamentals, Ed? And apparently you can’t even figure out how to work a search engine. LOL

    It’s not “absorptivity equals emissivity”, it’s “absorptivity equals emissive power only at thermodynamic equilibrium, and only because both are zero“.

    As you’ve already been told in a prior post, Kirchhoff’s law has largely been invalidated for all but certain laboratory conditions, it is not an equality, it’s a ratio, and you apparently still can’t fathom the difference between ’emissivity’ and ’emissive power’.

    Stop embarrassing yourself, Ed. LOL

    • Ed Bo permalink
      May 15, 2022 3:48 am

      As I pointed out in another response, a source YOU linked says:

      “Kirchhoff’s law states, in simpler language: For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.”

      Who’s embarrassing himself?

      • LOL@Klimate Katastrophe Kooks permalink
        May 15, 2022 4:08 am

        And as I point out in response to your inanity:

        Now you’re laughing at the left-leaning Wikipedia because they misconstrue Kirchhoff’s words, because they, like you, don’t understand the difference between ’emissivity’ and ’emissive power’.

        Perhaps you should inform them of their mistake. Looks unprofessional to be so loose with their language. LOL

        Here, since you seem to be so ineducable that you cannot even work a browser to search the terms and definitions yourself:

        Emissive Power: the total amount of thermal energy emitted per unit time, per unit area of the body for all possible wavelengths.

        Emissivity: the ratio of the total emissive power of the body to the total emissive power of a perfect black body at that temperature

        See the difference? NO?! LOL

        How many foot-bullets are you going to do today, Ed? LOL

  52. LOL@Klimate Katastrophe Kooks permalink
    May 14, 2022 3:27 am

    Ed Bo wrote:
    “Note that they do not claim that the underlying processes have stopped in an isothermal evironment. In fact, they use the term “quasistatic” to emphasize that this has not happened!”

    And now you demonstrate that you don’t know the meaning of the term “quasistatic state”, just as you don’t know the meaning of the term “quiescent state”, just as you cannot fathom the difference between “emissivity” and “emissive power”, just as you cannot fathom that Kirchhoff’s Law is a ratio, not an equality, just as you cannot fathom cavity theory applied outside a cavity, just as you cannot fathom EM interaction between a photon (a quantum of EM energy and the EM field predicating whether that photon will incide upon an object, be totally reflected or be subsumed into a higher chemical potential EM field.

    No… you were taught a simpletonian method of thermodynamics and radiative theory predicated upon the long-debunked Prevost Principle, and upgrading your knowledge would necessitate upgrading your poor underpowered brain. And we all know that isn’t going to happen. LOL

  53. SamH permalink
    May 14, 2022 7:29 pm

    LOL@KKK:

    Amazing how you accuse Ed Bo of “denying… wide swaths of long-established science,” while YOU deny the long-established science taught in MIT thermodynamics textbooks as “not the proper way of doing it.” YOU are the one denying established science in favor of kooky theories.

    In any case, you reveal your ignorance at every turn, and have been wrong on just about every point in the last few posts.

    “And now you demonstrate that you don’t know the meaning of the term “quasistatic state””

    There is no such thing as a “quasistatic state,” only quasistatic transformations between states.

    “Kirchhoff’s Law is a ratio, not an equality”

    This is nonsense. A physical law must contain a statement, and a ratio alone cannot constitute a statement. Kirchhoff’s law is an equality, which can involve a ratio if you choose to write it in that manner. Namely, for a body at equilibrium, at each wavelength α=I(T)/E(T), where α is the absorptivity, I is the emitted power, and E is the universal black body function. Equivalently, α=ε, where ε is emissivity.

    “It’s not “absorptivity equals emissivity”, it’s “absorptivity equals emissive power”

    Wrong, it is α=ε. Your units aren’t even right. Absorptivity is a unitless quantity, so it can’t be equal to a power.

    “only because both are zero”

    Wrong. Give me one source for this. Textbook or peer-reviewed, please.

    “Kirchhoff’s law has largely been invalidated for all but certain laboratory conditions”

    Wrong. Kirchhoff’s law is a simple consequence of the second law of thermodynamics for reciprocal emitters, and can also be derived from Maxwell’s equations. Only in the past few years has any experiment demonstrated a violation of Kirchhoff’s law, and only in highly engineered magneto-optic materials. From https://energyzhao.github.io/pdf/Zhao%20et%20al.%202019%20Nonreciprocal%20Thermal%20Emission.pdf : “The vast majority of thermal emitters obey Kirchhoff’s law… Kirchhoff’s law is a consequence of the Lorentz reciprocity theorem of Maxwell’s equations.” To deny Kirchhoff’s laws, you have to deny both Maxwell’s equations and the second law of thermodynamics. Do you?

    • LOL@Klimate Katastrophe Kooks permalink
      May 14, 2022 8:09 pm

      Sam H wrote:
      ““Kirchhoff’s Law is a ratio, not an equality”

      This is nonsense.”

      I’m sorry that you can’t do simple math: And you’re another of those who can’t fathom the difference between “emissivity” and “emissive power”. LOL

      Oh, look what Gustav Kirchhoff himself wrote:
      https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation
      “For a body of any arbitrary material emitting and absorbing thermal electromagnetic radiation at every wavelength in thermodynamic equilibrium, the ratio of its emissive power to its dimensionless coefficient of absorption is equal to a universal function only of radiative wavelength and temperature. That universal function describes the perfect black-body emissive power.”

      So it doesn’t apply to graybody objects, and it only applies at thermodynamic equilibrium, and it has failed validation tests and is thus invalidated for all but certain laboratory conditions (see below).

      I know, I know… you’re embarrassed that you’ve yet again displayed your reading comprehension problems… and now you’ll vigorously defend your incorrectitude because you’ve become emotionally invested in being wrong. LOL

      In Kirchhoff’s original parlance:
      E/A = e
      Not E=A.

      In modern parlance:
      Eν/αν= f(T, ν)

      In other words, the ratio between emissive power and absorptivity is equal to specific intensity.

      Any more of reality you’d like to deny? Oh… wait… you deny all of it because you’ve been brainwashed into believing that the shortcut method of accounting for energy flow as taught my most institutions of higher learning actually represents reality, complete with energy flowing willy-nilly without regard to the energy density gradient, thus you deny the fundamental physical laws even in the face of the mathematical equations proving you wrong upon even a cursory examination of said equations. LOL

      It’s sad that you climate tardlings are forced to cling to ancient and outdated information such as the Prevost Principle from 1791 in order to make your CAGW fairy-tale work, but when that’s all you’ve got, that’s the best you can do. We all know there’s no way you’re capable of updating your knowledge base. LOL

      https://www.researchgate.net/publication/3165633_On_the_Validity_of_Kirchhoff's_Law_of_Thermal_Emission

      Click to access 1708.0053v1.pdf

      Kirchhoff’s Law of Thermal Emission:
      What happens when a law of physics fails an experimental test?

      https://books.google.com/books?id=a434DAAAQBAJ&lpg=PA11&ots=JTkEDql9ks&dq=Kirchhoff's%20Law%20invalidated&pg=PA11#v=onepage&q=Kirchhoff's%20Law%20invalidated&f=false
      “The discovery that under suitable conditions even gases can emit continuous spectra later invalidated Kirchhoff’s model.”

      • LOL@Klimate Katastrophe Kooks permalink
        May 14, 2022 10:56 pm

        From the paper above:

        Kirchhoff’s law, in fact, remains valid only for the perfect absorber.

        And a “perfect absorber” (ie: an idealized blackbody object) does not (cannot) exist… it is an idealization and a provable contradiction. An idealization that you climate loons use to commit your mathematical fraudery in furtherance of your CAGW scam, using the form of the S-B equation meant for idealized blackbody objects:
        q = σ T^4
        … and slapping ε onto that (sometimes) …
        q = ε σ T^4
        … for graybody objects.

        https://i.imgur.com/QErszYW.gif

        You treat real-world (graybody) objects as though they are idealized blackbody objects (emission to 0 K and (sometimes) ε = 1)… doing so is effectively subtracting a fictional ‘cooler to warmerenergy flow from the real (but calculated incorrectly and thus far too high) ‘warmer to coolerenergy flow. That’s a handy shortcut way of accounting for energy flows, but it gives the wrong radiant exitance values for all graybody objects, which is why you have to carry those incorrect values through your calculations and cancel them on the back-end.

        The S-B equation for graybody objects isn’t meant to be used to subtract a fictive ‘cooler to warmerenergy flow from the incorrectly-calculated and thus too high ‘warmer to coolerenergy flow, it’s meant to be used to subtract cooler object energy density (temperature is a measure of energy density, the fourth root of energy density divided by Stefan’s constant) from warmer object energy density. Radiant exitance of the warmer object is predicated upon the energy density gradient.

        Even worse is that the climastrologists then further misuse the S-B equation by converting it into what they call a “forcing formula” (as used in IPCC AR6):
        4 ε σ T3
        … which builds-in a warming trend:
        ———-
        For 288 K, a 1 K negative temperature change, ε=0.93643 (ref: NASA ISCCP program):
        4 ε σ T3 = 5.07369679087621 W m-2

        For 288 K, a 1.00525093764635 K negative temperature change, ε=0.93643 (ref: NASA ISCCP program):
        q = ε σ T4 = 5.07369679087621 W m-2
        ———-
        For 288 K, a 1 K positive temperature change, ε=0.93643 (ref: NASA ISCCP program):
        4 ε σ T3 = 5.07369679087621 W m-2

        For 288 K, a 0.99474906235365 K positive temperature change, ε=0.93643 (ref: NASA ISCCP program):
        q = ε σ T4 = 5.07369679087621 W m-2
        ———-
        So their bastardized equation gives the result for a warming of 0.99474906 K while claiming it’s a warming of 1 K, and gives the result for a cooling of 1.00525093 K while claiming it’s a cooling of 1 K.

        And that’s likely why they bastardized the S-B equation… to hide the fact that they continue to treat real-world graybody objects as idealized blackbody objects, and to build-in whatever warming trend they possibly could in order to help sustain their alarmist narrative.

        You’ve been taken in by a complex mathematical scam… and the end result of that mathematical scam is to dis-empower and impoverish you while empowering and enriching the elitists pushing the scam via the implementation of world-wide Communism… I’ve got quotes from IPCC and UN officials admitting this, if you’d like to see them.

        Stop being their “useful idiots”… it never ends well for the “useful idiots”. Start thinking for yourself.

      • Ed Bo permalink
        May 15, 2022 3:35 am

        Your own cited source says this:

        “Kirchhoff’s law states, in simpler language: For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.”

        ROFLMAO!

  54. LOL@Klimate Katastrophe Kooks permalink
    May 14, 2022 8:44 pm

    SamH wrote:
    “LOL@KKK , of course you’re right that the radiation going from up-to-down would not be blocked by the flash light from the side. The reason is that your entire theory of energy density blocking radiation is crap.

    Instead, you are trying to salvage your theory by inventing “directional energy density.””

    Oh, sure… I mean, those lasers just scatter their energy density all over the place… that’s why they’re able to cut extremely small kerfs… oh… wait. They’re able to cut extremely small kerfs because energy density is directional. LOL

    Any more of reality you’d like to deny today, Sam? LOL

    • LOL@Klimate Katastrophe Kooks permalink
      May 14, 2022 8:52 pm

      E = n * h * f, where n = number of photons, h = Planck’s constant, f = frequency

      That’s how a laser is able to cut, right, Sam? By increasing energy density via coherent light, right?

      Are you now claiming that photons have an effect outside of their radius? Because they do have a radius…

      A sinusoid is a circular function.

      You’ll note the peak amplitude of the sinusoid is analogous to the radius of the circle, the peak-to-peak amplitude is analogous to the diameter of the circle, and the frequency of the sinusoid is analogous to the rotational rate of the circle. You’ll further note the circumference of the circle is equal to 2 pi radians, and the wavelength of a sinusoid is equal to 2 pi radians, so the wavelength of the sinusoid is analogous to the circumference of the circle.

      Thus the magnetic field and electric field (oscillating in quadrature) of a photon is a circle geometrically transformed into a spiral by the photon’s movement through space-time. This is why all singular photons are circularly polarized either parallel or antiparallel to their direction of motion. This is a feature of their being massless and hence having no rest frame (if a photon had a rest frame, no rest mass and no momentum equals nothing, so massless particles must remain in motion), which precludes their exhibiting the third state expected of a spin-1 particle (for a spin-1 particle at rest, it has three spin eigenstates: +1, -1, 0, along the z axis… no rest frame means no 0-spin eigenstate). A macroscopic electromagnetic wave is the tensor product of many singular photons, and thus may be linearly or elliptically polarized if all singular photons comprising the macroscopic electromagnetic wave are not circularly polarized in the same direction.

      • LOL@Klimate Katastrophe Kooks permalink
        May 14, 2022 9:04 pm

        This app uses the Schrodinger equation to compute energetic transfer from free space to an object:
        https://physics.weber.edu/schroeder/software/BarrierScattering.html
        – Set it to whatever energy level you want, as long as ‘Wavepacket Energy’ and ‘Barrier Energy’ are equal.

        – Set the ‘Width’ slider to maximum. This will be akin to free space to the left, and an object to the right of the display.

        – Set ‘Ramp’ to some arbitrary value (the higher the energy density of the barrier, the wider the energy gradient (ie: the ‘ramp’).

        – Select the ‘Density / Phase’ option-box.

        If you’re doing it right, you’ll see the free space wavemode nodes are always at the ‘ramp’, and you’ll see the wavemodes set up a standing wave (which will vary in space because ‘Wavepacket Energy’ has an uncertainty added to it). Note that even if ‘Wavepacket Energy’ is equal to ‘Barrier Energy’, no energy is imparted to the barrier (ie: the object). If the code glitches, start with ‘WavePacket Energy’ lower than ‘Barrier Energy’, then ramp it up after it’s running.

        This is akin to stating that a particle (atom or molecule) in an excited quantum state cannot absorb energy which is equal to the energy which excited that quantum state, unless the particle has another quantum state available to be excited which is equal to that energy.

        The electric field of a non-resonant photon, ~100,000,000 times smaller than the Coulomb field seen by the bound electron(s), slightly changes the phase of the bound electron(s), just as it changes the phase of the incident photon… no energy is exchanged if the photon is not absorbed, only the phase is shifted, altering the vector of the photon. This is how reflection from a step potential works. This is dependent upon the differential between energy density of the photon and atom / molecule. In the case of absorption, the photon constructively interferes with the available resonant quantum states of the atom or molecule. No phase shift.

  55. LOL@Klimate Katastrophe Kooks permalink
    May 15, 2022 1:49 am

    Ed Bo wrote:
    “Sam – Take it easy on LOL. He doesn’t have the sophistication to understand that he just torpedoed his own arguments.”

    Says the climate kook tacitly denying the existence of the Poynting vector… and therefore photon relativistic momentum… and therefore radiation pressure… and therefore energy density… and therefore energy density gradient… and therefore the underlying mechanism which regulates radiant exitance. LOL

    Perhaps you should just stop while you’re behind, Ed. You’re not up to the task. LOL

  56. LOL@Klimate Katastrophe Kooks permalink
    May 15, 2022 1:56 am

    Ed Bo wrote:
    “You are most certainly NOT using that method – you have simply demonstrated that you cannot understand it, even when it is explained carefully to you. The method does NOT assume that a body with emissivity of 0.04 absorbs all of the radiation incident upon it.”

    If Kiehl and Trenberth get 390 W m-2 surface radiant exitance in their KT diagram, they do so by disregarding energy density gradient and emissivity… they can only arrive at 390 W m-2 by assuming emission to 0 K and emissivity = 1, which means they assume that for the entirety of the planet, surface, atmosphere, everything.

    And lo and behold, that’s exactly what the climastrologists do:
    —–
    The IPCC assumes in its equations emissivity = 1 and thus absorptivity = 1, which is clearly unphysical. CO2 is not a hypothetical perfect blackbody.

    The correct form of the S-B formula: h = (ε (A) (σ)(Ts ^4 – Ta ^4)) / (A * ΔT)
    The incorrect IPCC formula, applied to gray bodies: h = ((σ)(T^4)) / (A * ΔT)
    Note the lack of ε in the IPCC formula… they assume emissivity = 1 and thus absorptivity = 1 at thermal equilibrium, a hypothetical perfect blackbody. Also note that they implicitly deny that radiation pressure of the atmosphere reduces emissive power of the surface… they violate 2LoT by claiming that a graybody surface emits as though it’s in a 0 K ambient, and absorbs all radiation incident upon it.
    —–

    You just can’t stop shooting yourself (and your climate loon compatriots) in the foot, can you, Ed? LOL

    • LOL@Klimate Katastrophe Kooks permalink
      May 15, 2022 2:25 am

      Even worse is that they then further misuse the S-B equation by converting it into what they call a “forcing formula”, as used in IPCC AR6:
      4 ε σ T3
      … which builds-in a warming trend:
      ———-
      For 288 K, a 1 K negative temperature change, ε=0.93643 (ref: NASA ISCCP program):
      4 ε σ T3 = 5.07369679087621 W m-2

      For 288 K, a 1.00525093764635 K negative temperature change, ε=0.93643 (ref: NASA ISCCP program):
      q = ε σ T4 = 5.07369679087621 W m-2
      ———-
      For 288 K, a 1 K positive temperature change, ε=0.93643 (ref: NASA ISCCP program):
      4 ε σ T3 = 5.07369679087621 W m-2

      For 288 K, a 0.99474906235365 K positive temperature change, ε=0.93643 (ref: NASA ISCCP program):
      q = ε σ T4 = 5.07369679087621 W m-2
      ———-
      So their FUBAR equation gives the result for a warming of 0.99474906 K while claiming it’s a warming of 1 K, and gives the result for a cooling of 1.00525093 K while claiming it’s a cooling of 1 K.

      And that’s likely why they FUBAR’d the S-B equation… to hide the fact that they continue to treat real-world graybody objects as idealized blackbody objects, and to build-in whatever warming trend they possibly could in order to help sustain their alarmist narrative.

  57. LOL@Klimate Katastrophe Kooks permalink
    May 15, 2022 3:34 am

    Ed Bo wrote:
    “Baloney! I’ve pointed out that you made an incredibly basic mistake, and you cannot acknowledge it.”

    Given that I was using the equations that the climastrologists use, the “incredibly basic mistake” is on the part of the climastrologists.

    And given that the climastrologists misuse the S-B equation by treating real-world graybody objects as though they’re idealized blackbody objects (emission to 0 K) just with emissivity less than 1 (sometimes… they often treat graybody objects exactly as if they were idealized blackbody objects (emission to 0 K and emissivity = 1), as I show via the IPCC’s use of the S-B equation and their use of their FUBAR “forcing formula”, you’ve just condemned whom you wish to defend. LOL

    You just can’t stop shooting yourself in the foot, can you, Ed? LOL

    • Ed Bo permalink
      May 15, 2022 7:47 pm

      LOL@KKK –

      You say, “I was using the equations that the climastrologists use”.

      No you weren’t! You assumed that obtaining graybody exchange from individual graybody emission equations was as simple as obtaining blackbody exchange from individual blackbody emission equations.

      It is NOT, and that is YOUR error, which you refuse to own up to. I have cited the proper equation, (10.24 in the MIT text), and somehow you cannot acknowledge it.

      • LOL@Klimate Katastrophe Kooks permalink
        May 15, 2022 9:38 pm

        Oh, sure, Ed… in fact, that’s why I wrote:

        Now, we’ll do the calculations as the climate alarmists do them, using their assumption that the objects are emitting to 0 K. That treats the objects as though they’re idealized blackbody objects (ie: emission to 0 K), just with ε less than 1. Remember, the climate alarmists claim all objects greater than 0K emit, and that their radiant exitance is not affected by ambient energy density.

      • LOL@Klimate Katastrophe Kooks permalink
        May 15, 2022 9:42 pm

        Ed Bo wrote:
        “You assumed that obtaining graybody exchange from individual graybody emission equations was as simple as obtaining blackbody exchange from individual blackbody emission equations.”

        Are you one of those who thinks only idealized blackbody objects emit blackbody radiation, and graybody objects emit “graybody radiation”, Ed? LOL

        That seems to be a common trait amongst you climate loons. LOL

        Would you like to learn about blackbody radiation… what causes it, what can and cannot emit BB radiation, etc., Ed? If you’re willing to learn, I can teach you… oh, wait… you’re a climate loon, you’ve already demonstrated that you’re unwilling to learn if it contradicts your CAGW fairy-tale. LOL

  58. LOL@Klimate Katastrophe Kooks permalink
    May 15, 2022 3:51 am

    Ed Bo wrote:
    “Your own cited source says this:

    “Kirchhoff’s law states, in simpler language: For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.”

    ROFLMAO!”

    Now you’re laughing at the left-leaning Wikipedia because they misconstrue Kirchhoff’s words, because they, like you, don’t understand the difference between ’emissivity’ and ’emissive power’.

    Perhaps you should inform them of their mistake. Looks unprofessional to be so loose with their language. LOL

    Here, since you seem to be so ineducable that you cannot even work a browser to search the terms and definitions yourself:

    Emissive Power: the total amount of thermal energy emitted per unit time, per unit area of the body for all possible wavelengths.

    Emissivity: the ratio of the total emissive power of the body to the total emissive power of a perfect black body at that temperature

    See the difference? NO?! LOL

    How many foot-bullets are you going to do today, Ed? LOL

    • Ed Bo permalink
      May 15, 2022 7:41 pm

      LOL@KKK –

      Are you seriously so dense that you can’t see that the two statements are simply describing the same thing in different wording?

      The statement you quoted says that

      EmPower/alpha = BBPower.

      Now to use a very sophisticated mathematical manipulation (watch carefully!), we get

      alpha = EmPower/BBPower

      But EMPower/BBPower is the DEFINITION of emissivity! So the two statements are equivalent!

      (Sometimes I wonder why I am spending my time arguing with someone who has repeated trouble with middle school algebra…)

      • LOL@Klimate Katastrophe Kooks permalink
        May 15, 2022 9:35 pm

        Another kook who claims that the unitless value of emissivity which can range from 0 to 1 is the same as the object’s total emissive power. LOL

        So I take it you’re done denying that it’s a ratio, then. LOL

        In Kirchhoff’s original parlance:
        E/A = e
        Not E=A.

        In modern parlance:
        Eν/αν= f(T, ν)
        Where:
        Eν = emissive power
        αν = absorptivity
        f(T, ν) = specific intensity

        IOW, you’ve conflated specific intensity with absorptivity with emissivity… likely because you’re incapable of understanding the difference, just as you’re unable to understand the difference between emissivity and emissive power, just as you’re unable to understand that temperature is a measure of energy density, that energy density is what regulates radiant exitance (it’s right there in the S-B equation, FFS), that “isothermal radiation exchange” doesn’t happen and violates the fundamental physical laws… there’s so much you don’t understand. LOL

        e = T^4 a
        a = 4σ/c
        e = T^4 4σ/c
        T = 4^√e/(4σ/c)
        T^4 = e/(4σ/c)

        q = ε σ (T^4_h – T^4_c)

        ∴ q = ε σ ((e_h / (4σ / c)) – (e_c / (4σ / c))) A_h

        Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
        W m-2 = W m-2 K-4 * (Δ(J m-3 / (W m-2 K-4 / m sec-1)))

        ∴ q = (ε c (e_h – e_c)) / 4

        Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
        W m-2 = (m sec-1 (ΔJ m-3)) / 4

        For graybody objects, it is the energy density differential between warmer object and cooler object which determines warmer object radiant exitance. The climate alarmists misinterpret the S-B radiant exitance equation for graybody objects. Warmer objects don’t absorb radiation from cooler objects (a violation of 2LoT in the Clausius Statement sense and Stefan’s Law); the lower energy density gradient between warmer and cooler objects (as compared to between warmer object and 0 K) lowers radiant exitance of the warmer object (as compared to its radiant exitance if it were emitting to 0 K). The energy density differential between objects manifests an energy density gradient, each surface’s energy density manifesting a proportional radiation pressure.

  59. AC99 permalink
    May 15, 2022 11:46 am

    @Ed Bo
    @LOL@Klimate Katastrophe Kooks
    @SamH

    I attempted to make Ed’s problem above even simpler, but I think it got lost in the discussion. I’ll try again here.

    Let’s make Ed’s problem even simpler so we don’t have to deal with reflection or transmission. Let’s make all of the 1m^2 plates blackbodies just to make the math simpler and the problem cleaner. Let’s also say that your plates are emitting radiation to 0K surroundings. Then what are the governing equations and their solution?

    For a single plate with electrical energy in at 100 Watts, the plate will emit 50 Watts out from its right side and 50 W out from its left side. Thus it emits 50 W/m^2, and to do so to 0K surroundings the Stefan-Boltzmann equation, q = sigma T^4, implies that its steady state temperature must be 172K.

    Now, with the second plate in place and only the first plate with 100W being supplied, the heat transferred from the heated plate to the second plate must be, q12 = sigma (T1^4 – T2^4). The heat transferred from the heated plate to the 0K surroundings must be q10 = sigma T1^4. The energy balance for the heated plate is then q12 + q10 = 100, or 2 T1^4 – T2^4 = 100/sigma. Next, the second plate transfers heat to the 0K surroundings in the amount q20 = sigma T2^4. Then, the energy balance for the second plate is q12 – q20 = 0, or T1^4 – 2 T2^4 = 0. Finally, if we solve these two simultaneous energy balance equations we get 3/2 T1^4 = 100/sigma -> T1=185K, and 3 T2^4 = 100/sigma -> T2=156K.

    Do you agree with my calculations? If not, what are your answers for the temperatures of the plates and what equations are you using to arrive at your answers?

    • LOL@Klimate Katastrophe Kooks permalink
      May 15, 2022 9:18 pm

      From physicist Dr. Charles R. Anderson, PhD:
      https://objectivistindividualist.blogspot.com/2017/11/solving-parallel-plane-black-body.html
      “I have long argued that the reason that radiant energy only flows from the warmer to the cooler body is because the flow is controlled by an electromagnetic field and an energy gradient in that field. I will offer that proof in this paper.”

      Physicist Dr. Charles R. Anderson, PhD presents the consensus solution to the parallel plane black body radiator problem and demonstrates that it is wrong (this is where he demonstrates that the consensus solution violates Stefan’s Law in that using the consensus take on radiative energetic transfer results in double the energy density in a cavity space at thermodynamic equilibrium than we would calculate from Stefan’s Law). His solution is not a ‘non-mainstream’ view, the mathematics and concepts are taken from the book Thermal Physics, Second Edition by Philip M. Morse, Professor of Physics at MIT, co-founding editor of Annals of Physics, co-founder of MIT Acoustics Laboratory, first Director of Brookhaven National Laboratory, founder of MIT Computation Center.

      In short, physics and thermodynamics have been corrupted by the climastrologists to the point that we have tardlings like SamH and Ed Bo arguing against themselves… they claim their incorrect take on radiative energetic exchange doesn’t violate the fundamental physical laws, but in order to make their CAGW fairy-tale work, they must treat real-world graybody objects as idealized blackbody objects, they must claim radiative energetic transfer is an idealized reversible process and they must claim that energy can flow without regard to the energy density gradient and hence that “isothermal radiation exchange” happens… all of which violate the fundamental physical laws.

      • AC99 permalink
        May 15, 2022 10:08 pm

        So what are Charles Anderson’s equations for this problem? What temperatures are obtained for the plates when solving his equations?

  60. SamH permalink
    May 15, 2022 5:20 pm

    LOL@KKK:

    Your posts are full of inconsistencies. You quote equations, then immediately contradict them with your own writing. I will provide some examples, then ask you to address two simple points, which you should be able to do in a few sentences. Examples:

    You: “It’s not “absorptivity equals emissivity”, it’s “absorptivity equals emissive power”

    Also you: “Eν/αν= f(T, ν)”. Here Eν is emissive power, αν is absorptivity, and f(T, ν) is the universal blackbody function.

    This equation directly contradicts your claim that “absorptivity equals emissive power.” You have attempted to state Kirchhoff’s law in 2 different ways, one of which is wrong.

    Please note that this equation can be rearranged as Eν/f(T, ν) = αν . Then note that emissivity, by definition, is the ratio of emissive power (Ev) to the blackbody function, f(T, v). So finally your equation reduces to εν = αν . That is, **absorptivity equals emissivity**.

    ——

    You: “Kirchhoff’s law… **is not an equality**”

    But again, you said Kirchhoff’s law is “Eν/αν= f(T, ν).” **Do you know what we call it when you have two things with an equals sign between them? That’s right, an equality.**

    ——

    1: Please pick **one** of your many definitions of Kirchhoff’s law. It’s impossible to have a productive discussion when you are contradicting yourself.

    2: And please, **provide a source for your claim that absorptivity and emissivity go to zero at equilibrium.** A direct quote saying that would do.

    • LOL@Klimate Katastrophe Kooks permalink
      May 15, 2022 8:51 pm

      SamH wrote:
      “Please note that this equation can be rearranged as Eν/f(T, ν) = αν . Then note that emissivity, by definition, is the ratio of emissive power (Ev) to the blackbody function, f(T, v). So finally your equation reduces to εν = αν . That is, **absorptivity equals emissivity**.”

      Again, it’s sad that you can’t do simple math, Sam. So you’re saying that a unitless value that ranges from 0 to 1 is equal to the total emissive power of an object… because if you’re saying that “absorptivity equals emissivity”, you must also be saying “emissivity equals emissive power”, because Kirchhoff’s Law states that absorptivity equals emissive power at thermodynamic equilibrium and only for idealized blackbody objects… does that make sense to you? LOL

      As to “provide a source for your claim that absorptivity and emissive power go to zero at equilibrium”… check the fundamental physical laws.

      It’s you climate loons claiming that energy can flow willy-nilly without regard to energy density gradient, and in so doing, you directly contradict the fundamental physical laws.

      If you disagree, you’ll get right on explaining why temperature is literally a measure of energy density (the fourth root of energy density divided by Stefan’s constant), and the S-B equation subtracts the energy density of the cooler object from the energy density of the warmer object, wherein at zero temperature (ie: zero energy density), there is zero radiant exitance… so there certainly cannot be any radiant exitance against the energy density gradient… and that is the whole point of my posts… the climastrologists have been misusing the S-B equation (and FUBAR’ing it into their kooky “forcing formula”) by using the form meant for idealized blackbody objects upon real-world graybody objects.

      What determines radiant exitance in your kooky world? Object temperature alone? Because that belief went out with the long-debunked Prevost Principle. LOL

      • LOL@Klimate Katastrophe Kooks permalink
        May 15, 2022 10:17 pm

        “wherein at zero temperature (ie: zero energy density)”

        Should be:

        “wherein at zero temperature (ie: zero energy density) differential”

  61. SamH permalink
    May 16, 2022 2:20 am

    What the heck are you not understanding here? You keep saying:

    “Kirchhoff’s Law states that absorptivity equals emissive power”

    That would correspond to an equation:

    αν=Eν

    The αν is “absorptivity,” the equals sign is “equals,” and the Eν is “emissive power.”

    However, that is not correct. The actual form of Kirchhoff’s law (an equation which you yourself have copy pasted) is

    Eν/αν= f(T, ν)

    Hopefully this will clear up your confusion…

    • SamH permalink
      May 16, 2022 2:38 am

      Also: you accuse me of the following:

      “So you’re saying that a unitless value that ranges from 0 to 1 is equal to the total emissive power of an object…”

      Are you kidding me? No, I am not saying that, but it is exactly what YOU are saying when you claim “absorptivity equals emissive power.” Absorptivity is a unitless value that ranges from 0 to 1.

      Are you just trolling or what?

      • LOL@Klimate Katastrophe Kooks permalink
        May 16, 2022 2:45 am

        I state exactly what Kirchhoff states, that specific intensity is equal to the ratio of emissive power and absorptivity only for idealized blackbody objects and only at thermodynamic equilibrium.

        Kirchhoff’s exact words:

        For a body of any arbitrary material emitting and absorbing thermal electromagnetic radiation at every wavelength in thermodynamic equilibrium, the ratio of its emissive power to its dimensionless coefficient of absorption is equal to a universal function only of radiative wavelength and temperature.

        You are the one attempting to claim that the unitless value known as ’emissivity’ which ranges from 0 to 1 is the same as emissive power and hence that “absorptivity equals emissivity”.

        If you have a problem with the wording of Kirchhoff’s Law of Thermal Radiation, I suggest you take it up with Kirchhoff… you climate loons believe in all sorts of unscientific droolery, so speaking to the long-dead should be right up your alley. LOL

    • LOL@Klimate Katastrophe Kooks permalink
      May 16, 2022 2:51 am

      What the heck are you not understanding here, Sam?

      I specifically stated exactly as Kirchhoff himself stated:

      For a body of any arbitrary material emitting and absorbing thermal electromagnetic radiation at every wavelength in thermodynamic equilibrium, the ratio of its emissive power to its dimensionless coefficient of absorption is equal to a universal function only of radiative wavelength and temperature. That universal function describes the perfect black-body emissive power.

      The ratio of emissive power and absorptivity is only equal for idealized blackbody objects at thermodynamic equilibrium.

      If they were always equal, an object could never change temperature. Think about it, Sam. Use that head of yours for more than just a hat rack for once in your life. LOL

      Thus far, you’ve denied the wording of the Kirchhoff Law of Thermal Radiation on the points of it being a ratio and it being a ratio of emissive power and absorptivity.

      And now you’re desperately bleating out-of-context quotes of mine (I specifically said they’re only equal at thermodynamic equilibrium) because you’ve displayed your stupidity and you’re butthurt because of it. LOL

      What’s it like, Sam? Being wrong, knowing you’re wrong, getting caught in incorrectitude after incorrectitude, but not being intelligent nor honest enough to admit your mistakes and learn from them? LOL

    • SamH permalink
      May 16, 2022 2:53 am

      “I state exactly what Kirchhoff states, that specific intensity is equal to the ratio of emissive power and absorptivity”

      That’s all well and good, except that you also said:

      “Kirchhoff’s Law states that absorptivity equals emissive power”

      So why don’t you admit that you were incorrect?

      • SamH permalink
        May 16, 2022 2:54 am

        You continually made two different, contradictory statements of Kirchhoff’s law. You need to accept that one of them is false.

      • LOL@Klimate Katastrophe Kooks permalink
        May 16, 2022 2:56 am

        Awww, are you snipping my quote to take it out of context, Sam, you disingenuous little butthurt kook?

        My exact quote:

        Kirchhoff’s Law states that absorptivity equals emissive power at thermodynamic equilibrium and only for idealized blackbody objects… does that make sense to you? LOL

        Why’d you leave out the “at thermodynamic equilibrium” bit, Sam? Why were you attempting to make it seems as if I had stated that they were always equal, Sam? Why are you such a lying butthurt kook, Sam? LOL

  62. LOL@Klimate Katastrophe Kooks permalink
    May 16, 2022 3:00 am

    SamH dribbled:
    “You continually made two different, contradictory statements of Kirchhoff’s law. You need to accept that one of them is false.”

    I’ve done nothing but state exactly what Kirchhoff himself has stated… it is you who is taking my quotes out of context (leaving out the “in thermodynamic equilibrium” bit) in a desperate attempt at claiming that I’ve made “two different, contradictory statements about Kirchhoff’s law”.

    For a body of any arbitrary material emitting and absorbing thermal electromagnetic radiation at every wavelength in thermodynamic equilibrium, the ratio of its emissive power to its dimensionless coefficient of absorption is equal to a universal function only of radiative wavelength and temperature. That universal function describes the perfect black-body emissive power.

    You’ve destroyed the few remaining shreds of credibility you had, Sam. Best you just slink away with tail tucked. You’ve proven yourself to be a lying kook who can never admit when he’s wrong, and who is nearly always wrong. LOL

    • SamH permalink
      May 16, 2022 3:11 am

      Oh okay, let’s use the full context. You said:

      “Kirchhoff’s Law states that absorptivity equals emissive power at thermodynamic equilibrium and only for idealized blackbody objects”

      Still wrong. Absorptivity is not equal to emissive power, ever, in thermodynamic equilibrium or out of equilibrium.

      Now can you please just admit that you were wrong? It’s like pulling teeth, Christ.

      • LOL@Klimate Katastrophe Kooks permalink
        May 16, 2022 3:18 am

        Bwahahaha! Now you’re denying Kirchhoff’s own words:

        For a body of any arbitrary material emitting and absorbing thermal electromagnetic radiation at every wavelength in thermodynamic equilibrium, the ratio of its emissive power to its dimensionless coefficient of absorption is equal to a universal function only of radiative wavelength and temperature. That universal function describes the perfect black-body emissive power.

        How butthurt must you be to deny the words of the man who came up with Kirchhoff’s Law of Thermal Radiation? VERY, VERY BUTTHURT. LOL

        And as you’re proven wrong on topic after topic, your butthurt grows, and the voices screaming in your head compel you to humiliate yourself further and further. What next, Sam? Do continue to entertain the sane folk. LOL

      • SamH permalink
        May 16, 2022 3:24 am

        Kirchhoff’s words aren’t wrong, yours are.

        If you refuse to have a rational discussion and admit your mistake, go pound sand.

      • LOL@Klimate Katastrophe Kooks permalink
        May 16, 2022 3:26 am

        SamH wrote:
        “Oh okay, let’s use the full context.”

        You admit, then, that you intentionally left the blurb “in thermodynamic equilibrium” out of my quote because you wanted to make it appear as if I’d stated that absorptivity and emissive power were always equal, so you could go off on a straw-man tangent of claiming that I’d presented two different and contradictory statements of Kirchhoff’s Law of Thermal Radiation… and now you’re denying the words as written by Kirchhoff himself.

        Oh, things aren’t looking good for you, Sam… you’ve just exposed yourself as a lying kook and so very, very butthurt that you’re denying reality. A meltdown from you shouldn’t be long in coming. LOL

  63. LOL@Klimate Katastrophe Kooks permalink
    May 16, 2022 3:29 am

    SamH wrote:
    “Kirchhoff’s words aren’t wrong, yours are.”

    SamH wrote:
    “Still wrong. Absorptivity is not equal to emissive power, ever, in thermodynamic equilibrium or out of equilibrium.”

    Gustav Kirchhoff wrote:
    “For a body of any arbitrary material emitting and absorbing thermal electromagnetic radiation at every wavelength in thermodynamic equilibrium, the ratio of its emissive power to its dimensionless coefficient of absorption is equal to a universal function only of radiative wavelength and temperature. That universal function describes the perfect black-body emissive power.”

    You poor, sad, sorry sack of stupid, do you think anyone’s buying your backpedaling bafflegab, Sam? LOL

    But thanks for demonstrating that you’re pulling your ‘facts’ from your coal-chute, and that you’re so butthurt that you’re now denying that you’ve denied Kirchhoff’s own words. LOL

    • SamH permalink
      May 16, 2022 4:46 am

      1) If you honest-to-god can’t see the difference between your statement and Kirchhoff’s, something is seriously wrong with you. The only question is whether the problem is in your eyes or your brain.

      2) All of your insults sound like Donald Duck if he got into QAnon

      • LOL@Klimate Katastrophe Kooks permalink
        May 16, 2022 6:27 am

        Says the kook who literally claimed Kirchhoff was wrong, then denied doing so, then denied denying having done so.

        Your butthurt has gone recursive, Sam. LOL

  64. LOL@Klimate Katastrophe Kooks permalink
    May 16, 2022 10:55 am

    White paint: Absorptivity: 0.16 Emissivity: 0.93

    Copper (mechanically polished, 250 K): Absorptivity: 0.88 Emissivity: 0.6

    0.5 μm thick Magnetic Weyl Semimetal Film, 8.3 μm wavelength: Absorptivity: 0.96 Emissivity: 0.06

    Magneto-optical photonic crystal atop Al mirror: Absorptivity: 0.05 Emissivity: 0.99

    In fact, a static electric field applied to an object can increase emissivity, whereas a static magnetic field can reduce emissivity. The dominant source of blackbody radiation is transient oscillating dipoles induced by inter-molecular thermal vibrations within a material, is it not? Well, a static electric field increases the activity of those transient oscillating dipoles, whereas a static magnetic field damps it.

    All of the above violate Kirchhoff’s Law of Thermal Radiation… because Kirchhoff’s Law should never have been a law, as it only really pertains to idealized blackbody objects, which do not and cannot exist.

    Kirchhoff, in attempting to extend the idealized blackbody case to other materials (notably perfectly reflecting materials) cheated (as did Planck) by placing a small piece of graphite or carbon in the cavity, which acted as a thermalizer of the radiation in the cavity space. In a perfectly-reflecting cavity, radiation cannot equilibrate to a blackbody spectrum except for that thermalizer being present.

    So Kirchhoff’s Law of Thermal Radiation doesn’t extend to most materials, and has thus been invalidated for all but idealized blackbody objects (the closest approximations of which are only available in laboratory settings).

    This has knock-on effects upon the concepts of Planck time, Planck temperature, Planck length, and Planck mass, all derived from Kirchhoff’s Law of Thermal Radiation. They are no longer universal and are thus devoid of fundamental meaning in physics.

    Planck’s attempt at validating Kirchhoff’s Law in The Theory of Heat Radiation is filled with errors… he had to redefine blackbodies, he ignored absorptivity at the interface of the blackbody, he used polarized light (when thermal radiation is never polarized), and he, like Kirchhoff, cheated a bit by using a small chunk of carbon or graphite as a thermalizer (what he called a ‘catalyst’) in a perfectly-reflecting cavity (which cannot otherwise exhibit a blackbody spectrum)).

    Planck further erred in clinging to a long-debunked radiative model (Prevost Theory of Exchanges and Prevost’s Principle), and his follow-on assumptions stemming from that led to his treating real-world (graybody) objects as though they radiatively emit willy-nilly without regard to the energy density gradient.

    Planck correctly stated:

    Conduction of heat depends on the temperature of the medium in which it takes place, or more strictly speaking, on the non-uniform distribution of the temperature in space, as measured by the temperature gradient.

    Do remember that temperature is equal to the fourth root of energy density divided by Stefan’s Constant… temperature is a measure of energy density. In other words, Planck correctly stated that energy can only flow (the definition of ‘heat’) via conduction if there is a temperature (and therefore an energy density) gradient.

    Where Planck erred is in his clinging to the Prevost Theory Of Exchanges (and its core tenet, the Prevost Principle) in regard to radiative energy, which led him to eschew scientific reality (that energy only flows if there is an energy density gradient), to wit:

    But the empirical law that the emission of any volume-element depends entirely on what takes place inside of this element holds true in all cases (Prevost’s principle).

    Energy is energy no matter the form, and it follows the same rules no matter the form.

    The long-debunked Prevost Theory of Exchanges (first replaced by the Kinetic Theory of Heat, then by Quantum Thermodynamics) assumed that energy flowed without regard to energy density gradient, because only an object’s internal state determined radiant exitance. This led Planck to make the further incorrect assumption in keeping with the Prevost Theory of Exchanges:

    We shall now introduce the further simplifying assumption that the physical and chemical condition of the emitting substance depends on but a single variable, namely, on its absolute temperature T.

    He correctly stated that energy transfer via conduction was predicated upon there being an energy density gradient, but for radiative energetic exchange, he clung to the Prevost Principle (core tenet of the Prevost Theory of Exchanges, a long-debunked hypothesis from 1791 which was predicated upon the long-debunked Caloric Theory and which postulated that radiant exitance of an object was solely determined by that object’s internal state, thus that energy could flow willy-nilly without regard to energy density gradient).

    Except the Prevost Principle would only work for an idealized blackbody object, and they don’t actually exist… they’re idealizations. And the object would have to be in an isolated system, and they don’t actually exist… they’re idealizations.

    A graybody object’s radiant exitance isn’t solely determined by that object’s internal state, as the S-B equation plainly shows:

    Idealized Blackbody Object (assumes emission to 0 K and ε = 1):
    q_bb = ε σ (T_h^4 – T_c^4) Ah
    = 1 σ (T_h^4 – 0 K) 1 m^2
    = σ T^4

    Graybody Object (assumes emission to greater than 0 K and ε less than 1):
    q_gb = ε σ (T_h^4 – T_c^4) Ah

    The ‘Ah’ term is merely a multiplier, used if one is calculating for an area larger than unity [for instance: greater than 1 m^2], which converts the result from radiant exitance (W m-2, radiant flux per unit area) to radiant flux (W).

    Temperature is equal to the fourth root of energy density divided by Stefan’s Constant (ie: the radiation constant).
    e = T^4 a
    a = 4σ/c
    e = T^4 4σ/c
    T^4 = e/(4σ/c)
    T = 4^√(e/(4σ/c))
    T = 4^√(e/a)

    Since we’re using the Kelvin temperature scale, which has its base at 0 K, we can calculate temperature (and thus energy density) as above. Energy density at 0 K is zero, thus temperature at zero energy density is, of course, 0 K.

    q = ε σ (T_h^4 – T_c^4)

    ∴ q = ε σ ((eh / (4σ / c)) – (ec / (4σ / c))) Ah

    Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
    W m-2 = W m-2 K-4 * (Δ(J m-3 / (W m-2 K-4 / m sec-1)))

    ∴ q = (ε c (eh – ec)) / 4

    Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
    W m-2 = (m sec-1 (ΔJ m-3)) / 4

    For graybody objects, it is the energy density differential between warmer object and cooler object which determines warmer object radiant exitance. Warmer objects don’t absorb radiation from cooler objects (a violation of 2LoT in the Clausius Statement sense and Stefan’s Law); the lower energy density gradient between warmer and cooler objects (as compared to between warmer object and 0 K) lowers radiant exitance of the warmer object (as compared to its radiant exitance if it were emitting to 0 K). The energy density differential between objects manifests an energy density gradient, each surface’s energy density manifesting a proportional radiation pressure.

    Thus, the climastrologists cling to the long-debunked Prevost Principle (whether they know it or not… and if they don’t know it, then they really have no business being anywhere near anything related to science) when they claim that energy can radiatively flow without regard to energy density gradient… and that leads to all manner of unscientific drivelry… ‘backradiation’, ‘Global Warming Potential’, ‘carbon footprint’, the incorrect usage of the S-B equation in the K-T (Kiehl-Trenberth) diagram and even in instruments such as pyrgeometers and FTIR spectrometers, and ultimately in their core narrative: CAGW.

    And that’s how an incorrect radiative hypothesis from 1791 and an improperly-extended law of thermal radiation from 1859 still affects the weak-minded today… those weak-minded individuals now claim, based upon those incorrectitudes, that radiative energetic exchange is an idealized reversible process and hence that even at thermodynamic equilibrium objects are furiously emitting and absorbing radiation.

    Of course, that’s twaddle… all real-world processes are irreversible processes including radiative energetic exchange (which is temporal and thus entropic), and a proper understanding of the S-B equation (see animated graphic above) shows that at thermodynamic equilibrium for graybody objects, the system reaches a quiescent state, no energy flows. Photons being nothing but quanta of EM energy, if no energy flows, no photons are emitted nor absorbed at thermodynamic equilibrium… the photons in the intervening space between the objects at thermodynamic equilibrium set up a standing wave, the wavemode nodes are at the object surfaces (and hence, given that nodes are the zero crossing point, no energy can be transferred either in or out of the surface) and should one object change temperature, that standing wave becomes a traveling wave with the group velocity proportional to the temperature (and hence the energy density) gradient and in the direction of the cooler object. This is standard cavity theory applied outside a cavity… deny it and you deny wide swaths of long-established science.

    But deny it the climate loons most certainly do… by treating real-world graybody objects as through they’re idealized blackbody objects (emission to 0 K), just with emissivity less than 1 (sometimes… other times they treat graybody objects exactly as though they’re idealized blackbody objects (emission to 0 K, emissivity = 1); by claiming that radiation is an idealized reversible process; by claiming that “isothermal radiation exchange” is a reality… none of which is physical, none of which is scientific, none of which represents reality.

    2LoT (in the Clausius Statement sense… “No process is possible whose sole result is the transfer of heat from a cooler to a hotter body”) states that energy cannot flow from a lower to a higher-energy region without external work being done upon the system… not via conduction, not via radiative means, not macroscopically, not at the quantum scale, not ever. Do keep in mind the definition of heat: “an energy flux”. Thus: “No process is possible whose sole result is an energy flux from a cooler to a hotter body” without external energy doing work upon the system.

    http://www.solvayinstitutes.be/pdf/Proceedings_Physics/1948.pdf#page253

    This means that, at every point of the region of space that is in (pointwise) radiative equilibrium, the total, for all frequencies of radiation, interconversion of energy between thermal radiation and energy content in matter is nil (zero).

    https://en.wikipedia.org/wiki/Chemical_potential#Sub-nuclear_particles

    Therefore, at thermodynamic equilibrium, the chemical potential of photons is always and everywhere zero. The reason is, if the chemical potential somewhere was higher than zero, photons would spontaneously disappear from that area until the chemical potential went back to zero; likewise, if the chemical potential somewhere was less than zero, photons would spontaneously appear until the chemical potential went back to zero.

    https://objectivistindividualist.blogspot.com/2015/05/the-greenhouse-gas-hypothesis-and.html

    In this case, not only do no thermally-emitted photons from the cooler plane become absorbed by the warmer plane surface, but none are even incident upon that surface.

    • AC99 permalink
      May 16, 2022 2:37 pm

      “White paint: Absorptivity: 0.16 Emissivity: 0.93”

      The numbers being reported here are averages of the absorptivity weighted towards the solar radiation spectrum, and the emissivity weighted towards temperatures around room temperature. This is why the reported numbers are different even though at each wavelength the absorptivity is equal to the emissivity. This is explained in the following Wiki article.

      https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation#Theory

      • LOL@Klimate Katastrophe Kooks permalink
        May 16, 2022 8:12 pm

        Fair point. That doesn’t invalidate, however, the fact that Kirchhoff’s Law of Thermal Radiation can be violated, as several studies now show.

        Fundamental physical laws aren’t violated, if they were, they wouldn’t be laws.

        The words “fact”, “theory”, “hypothesis” and “law” have very specific definitions in science:
        ———-
        Hypothesis: A tentative explanation of an empirical observation that can be tested. It is merely an educated guess.
        ———-
        Working hypothesis: A conjecture which has little empirical validation. A working hypothesis is used to guide investigation of the phenomenon being investigated.

        Scientific hypothesis: In order for a working hypothesis to be a scientific hypothesis, it must be testable, falsifiable and it must be able to definitively assign cause to observed effects.

        Null hypothesis: Also known as nullus resultarum. In the case of climate science, the null hypothesis should be that CO2 does not cause global warming.

        A Type I error occurs when the null hypothesis is rejected erroneously when it is in fact true.

        A Type II error occurs if the null hypothesis is not rejected when it is in fact false.
        ———-

        Fact: An empirical observation that has been confirmed so many times that scientists can accept it as true without having to retest its validity each time they experience whatever phenomenon they’ve empirically observed.

        Law: A mathematically rigorous description of how some aspect of the natural world behaves.

        Theory: An explanation of an empirical observation which is rigorously substantiated by tested hypotheses, facts and laws.

        Laws describe how things behave in the natural world, whereas theories explain why they behave the way they do.

        For instance, we have the law of gravity which describes how an object will behave in a gravitational field, but we’re still looking for a gravitational theory which fits into quantum mechanics and the Standard Model and explains why objects behave the way they do in a gravitational field.
        ——————–

        Hence, Kirchhoff’s Law of Thermal Radiation should actually rightfully be deemed a hypothesis… it cannot be a law, a law is a mathematically rigorous description of how some aspect of the natural world behaves… and given that Kirchhoff’s Law of Thermal Radiation only really pertains to idealized blackbody objects, and given that idealized blackbody objects are idealizations which do not exist and are in fact provable contradictions, Kirchhoff’s Law of Thermal Radiation at best describes something which does not and cannot exist.

  65. AC99 permalink
    May 16, 2022 11:52 am

    @LOL@Klimate Katastrophe Kooks

    I looked at your link, and for blackbody objects Anderson’s formula is the same as what I used to solve Ed Bo’s modified problem. Do you agree with the equations that I wrote down and their solution for the plate temperatures?

    • LOL@Klimate Katastrophe Kooks permalink
      May 16, 2022 8:18 pm

      You seem anxious to get me to agree with your solution… that’s a game played by several past kooks who lied, said their equations agreed with some or other physicist’s equations (when their equations absolutely didn’t agree and were in fact twisted to prove some point or other), then crowed about some purported ‘victolly!’ based upon their lies.

      Without doing the calculations myself, if your equations are the identical solution to physicist Dr. Charles R. Anderson, PhD’s and if you arrive at the same conclusions as he does, then why do you need me to agree with your solution? You’ve got what you wanted from the equations.

      • AC99 permalink
        May 16, 2022 10:30 pm

        “You seem anxious to get me to agree with your solution”

        I’m not sure if you agree with my solution or not. I am just trying to find out if you think that there is a different solution and how you arrive at it.

        My equations are the same ones used in textbooks, and your comments suggest that you do not agree with the textbooks. So I am wondering if you have a different solution and how you arrive at it.

        “if your equations are the identical solution to physicist Dr. Charles R. Anderson, PhD’s”

        He did not provide a solution to this problem. His formula for the heat transfer between two blackbody plates is the same as mine.

  66. AC99 permalink
    May 16, 2022 10:26 pm

    “That doesn’t invalidate, however, the fact that Kirchhoff’s Law of Thermal Radiation can be violated, as several studies now show.”

    That’s true. Kirchhoff’s “law” is not really a law in the same sense as the first law of thermodynamics. There are some esoteric materials that violate it, but most materials, like white paint and copper do abide by it.

    • SamH permalink
      May 17, 2022 5:19 am

      You are underselling Kirchhoff’s law here. The vast majority of materials obey it _exactly_: any reciprocal material has emissivity exactly equal to absorptivity, for each wavelength and direction, at thermal equilibrium. Very few materials have been observed to violate this law. For example, the paper by Robitaille that LOL@KKK linked, is utterly flawed.

      Moreover, a slightly modified Kirchhoff’s law is true for _all_ thermal emitters, and cannot be violated. From https://www.pnas.org/doi/epdf/10.1073/pnas.1701606114 , this law guarantees

      “the overall equality of the sums of the emissivities and the absorptivities for any object, including nonreciprocal ones”

      Thus, the emissivity and absorptivity, summed over all directions, are equal for each wavelength, at equilibrium.

      • SamH permalink
        May 17, 2022 5:29 am

        Put another way, Kirchhoff’s law is a law in the same sense as the laws of thermodynamics: because it is a direct result of the laws of thermodynamics, restricted to reciprocal thermal emitters. And the modified form I posted above is exact for _all_ thermal emitters, guaranteed by the second law of thermo.

        Finally, the claim by LOL that Kirchhoff’s law “pertains only to idealized blackbody objects” is incorrect. It pertains to any reciprocal emitter.

      • LOL@Klimate Katastrophe Kooks permalink
        May 17, 2022 6:49 am

        SamH wrote:
        “For example, the paper by Robitaille that LOL@KKK linked, is utterly flawed.”

        He claims, while finding himself utterly unable to point out these purported flaws. You shiteflinging monkey, who do you think believes your drivelry? LOL

        Point out these purported flaws now, or STFU, Sam:
        https://sci-hub.se/10.1109/TPS.2003.820958

        SamH wrote:
        “The vast majority of materials obey it _exactly_: any reciprocal material has emissivity exactly equal to absorptivity, for each wavelength and direction, at thermal equilibrium. ”

        Nicely qualified and weasel-worded, Sam… only Helmholtz reciprocal materials, only at thermal equilibrium… except it can be argued that selective radiators do not obey Kirchhoff’s Law “exactly”, nor can they. That’s just about every single polyatomic gas in existence on this planet. Nor do semi-transparent films made of anisotropic material obey Kirchhoff’s Law “exactly”. Nor does any material undergoing stimulated emission obey Kirchhoff’s Law “exactly”. Nor does water obey Kirchhoff’s Law “exactly” due to phase change. Nor does anything influenced by a static magnetic field obey Kirchhoff’s Law “exactly”. Nor does anything influenced by a static electric field obey Kirchhoff’s Law “exactly”… and do remember there’s an atmospheric electric gradient of ~120 V / m. Of course, fissionable materials don’t obey Kirchhoff’s Law “exactly”, nor do materials undergoing nuclear fusion. And due to red-shift / blue-shift, objects in motion don’t obey Kirchhoff’s Law “exactly” (especially as they reach significant fractions of c). And diffraction is neglected under Kirchhoff’s Law so objects smaller than the wavelength of incident light don’t obey Kirchhoff’s Law “exactly”… the problem growing worse for longer wavelengths.

        Seems your “vast majority” is whittled down to non-fissionable stationary materials not influenced by static magnetic or electric fields, which are unable to undergo phase change, which are at thermodynamic equilibrium and which are known Helmholtz reciprocal materials.

        Which aren’t that many materials after all. Certainly not the “vast majority”. LOL

        And you’ll likely still deny that at thermodynamic equilibrium, emissivity and absorptivity are only equal because both fall to zero in the zero-energy-density-differential quiescent state known as thermodynamic equilibrium, wherein no energy can flow, no photon production can take place, no photon absorption can take place, as physicist Dr. Charles R. Anderson, PhD shows on his website, to which I’ve linked previously.

        Why, I bet you’re not even aware that you’ve yet again shot yourself in the foot… the very paper you linked to destroys your own argument. In fact, Kirchhoff’s Law should be invalidated if for no other reason than that he cheated in attempting to extend the blackbody case to all objects by putting a small piece of carbon or graphite in the cavity to act as a thermalizer… in effect, he was working with the blackbody case all along! So it shouldn’t be the “Kirchhoff Law of Thermal Radiation”, it should be the “Miller-Zhu-Fan Law of Thermal Radiation”, because that paper fully invalidates Kirchhoff’s Law. LOL

        Old outdated knowledge is swept away, replaced with more precise knowledge, right? Except for you libtards, who always, always, always double-down on stupid. LOL

        Wait a minute…
        “Law 4 generalizes the directional Kirchhoff radiation law for reciprocal optical objects”

        Directional?! Weren’t you the kook who claimed that radiative energy density wasn’t directional, and in so doing, you denied the existence of the Poynting vector? LOL

      • AC99 permalink
        May 17, 2022 10:49 am

        No argument here.

        Any chance you can confirm my results for the plate calculation that I provided?

      • SamH permalink
        May 17, 2022 1:21 pm

        AC99, I can confirm your numbers are right for the plate problem.

        LOL, I will wait to respond so you have a chance to provide your answer to the temperatures of AC99’s plate problem.

  67. LOL@Klimate Katastrophe Kooks permalink
    May 17, 2022 12:56 am

    AC99 wrote:
    “My equations are the same ones used in textbooks, and your comments suggest that you do not agree with the textbooks.”

    I don’t deny that the shortcut method of calculating energy flows as taught by most institutions of higher learning arrives at the correct results, I disagree with the way they do so and the side effect of doing it that way leading people to believe that energy can flow willy-nilly without regard to energy density gradient.

    So essentially, they take each graybody object in isolation, effectively putting each object into its own isolated system and calculating as though each object were emitting to 0 K. They then carry these incorrect radiant exitance values through their calculations and cancel energy flows on the back end, arriving at the correct net energy flow.

    Except that’s not how the S-B equation is intended to be used. It’s meant to be used to subtract cooler object energy density from warmer object energy density… temperature being a measure of energy density (the fourth root of energy density divided by Stefan’s Constant).

    Objects interact via the EM field and that field’s energy density gradient. Photons being a quanta of energy (per QFT, a persistent perturbation of the EM field above the field average energy density), those photons must traverse the EM field between the objects and thus the energy density gradient between the objects.

    So while the climastrologists claim that there’s no way a photon could ‘know’ apriori what the temperature (and hence the energy density) of an object is before it incides upon that object, the photon most certainly does ‘know’ because of that EM field.

    Take a look at my analogization of thermodynamics to electrical theory… the top two circuits are what the climate alarmists do. Each object emitting to 0 K (analogous to discharging to ground) and not interacting with the other object.

    The bottom circuit is how reality works. Objects interact.

    Always break the problem down to the energy density of each object.

    • AC99 permalink
      May 17, 2022 10:52 am

      “I don’t deny that the shortcut method of calculating energy flows as taught by most institutions of higher learning arrives at the correct results”

      So does that mean that you agree with the results of my calculation?

      Is there some simple problem where you use different equations from those presented in textbooks and thus different solutions?

      • LOL@Klimate Katastrophe Kooks permalink
        May 17, 2022 5:56 pm

        Again, without having examined your claim that your equation is identical to that used by physicist Dr. Charles R. Anderson, PhD, I’ll not make any judgement as to your calculations. If your solution agrees with the solution Anderson’s equations would get and arrives at the same conclusions as Anderson’s equations lead to, then you’re all good, you don’t need my agreement… why you insist upon attempting to acquire my agreement, while continuing to fail to prove that your equation is identical to that used by physicist Dr. Charles R. Anderson, PhD and arrives at the same conclusions as Anderson’s equations makes one suspicious that you’re pulling that old kook ‘bait and switch’ trick of claiming your equation is identical to Anderson’s, while it’s not.

        And again, I use the same calculations as the S-B equation uses, just stripped down to energy density… as the S-B equation is meant to be used. My calculations arrive at the same solutions as the S-B equation, without leading to the fallacious conclusion that energy can flow willy-nilly without regard to the energy density gradient and thus that “isothermal radiation exchange” takes place in violation of 2LoT and Stefan’s Law.

        If one is going to do the calculations, one might as well do them correctly throughout, no?

  68. AC99 permalink
    May 18, 2022 12:43 am

    @LOL@Klimate Katastrophe Kooks

    “why you insist upon attempting to acquire my agreement”

    Generally, that’s how discussions progress. SamH has confirmed that he agrees with my solution. If we can all agree on this one then maybe we can find one where we disagree and then go through the mathematical steps. That is why I am asking you how you would solve it. If your solution is the same as the textbooks, then I don’t see the issue. It is only if your solution differs from textbooks that there would be an issue.

    “My calculations arrive at the same solutions as the S-B equation”

    So then your solution gives that the single heated plate achieves 172K at steady state, and that heated plate will increase in temperature to a new steady state of 185K when a second plate is introduced, which will get to 156K.

    If we all agree on that, then let’s try to find a problem where we do not agree,

    • LOL@Klimate Katastrophe Kooks permalink
      May 18, 2022 5:33 am

      That’s just the thing… we agree on the end result (but only if one is calculating for net energy flow… if one is calculating for radiant exitance of each object, the climate alarmist method gets the wrong result), the net energy flow, as both methods arrive at the correct net energy flow.

      Where I demur is in the method by which that end result of net energy flow is achieved, and the conclusions led to by the shortcut method commonly taught in most institutions of higher learning, and the fact that those institutions of higher learning no longer teach their students that it is merely a shortcut method for calculating net energy flow and does not actually represent reality. They used to. They no longer do. That knowledge has apparently rotted out of academia as academia rots from the inside out.

      As I said, the climate alarmist way of doing it is to treat each object in isolation and assume each object emits to 0 K and doesn’t interact with any other object. That inflates radiant exitance of each graybody object, necessitating that the climate alarmists carry those incorrect values through their calculations and cancel them on the back end… essentially subtracting a wholly-fictive ‘cooler to warmer’ energy flow from the real (but calculated incorrectly and thus far too high) ‘warmer to cooler’ energy flow. They only invoke the interaction between objects on the back-end to get their calculation to balance via hand-wavium, given that they treat the objects in isolation on the front end. This is exactly what the K-T diagram does, which is the only way they can arrive at 390 W m-2 surface radiant exitance… and they do so because that’s the only way they can sustain their ‘backradiation CAGW’ narrative.

      That’s not how the S-B equation is meant to be used, and leads the weak-minded climate alarmists to conclude that energy actually can flow ‘cooler to warmer’, a violation of 2LoT and Stefan’s Law. From that incorrectitude, the climate alarmists have extrapolated a whole universe of further failures of logic… global warming potential, carbon footprint, backradiation, their FUBAR ‘forcing formula’ used in IPCC AR6, the incorrect usage of the S-B equation in the Kiehl & Trenberth diagram and even in equipment such as pyrgeometers and FTIR spectrometers, and ultimately in their core narrative: CAGW.

      The S-B equation is meant to be used by subtracting cooler object energy density from warmer object energy density, temperature being a measure of energy density, equal to the fourth root of energy density divided by Stefan’s Constant.

      It is the energy density gradient which determines warmer graybody object radiant exitance. That’s why q ↓ as ΔT ↓, why q → 0 as ΔT→.0 for graybody objects, because as ΔT→.0 Δe→.0.

    • LOL@Klimate Katastrophe Kooks permalink
      May 18, 2022 6:14 am

      AC99 wrote:
      “SamH has confirmed that he agrees with my solution.”

      That’s not something to brag about… remember, SamH denies that energy density is directional so he tacitly denies the Poynting vector, he tacitly denies the fundamental physical laws, he agrees with Ed Bo that energy can flow without regard to the energy density gradient in violation of 2LoT and Stefan’s Law, he denies that it is energy density gradient which is the underlying mechanism which regulates radiant exitance, etc., etc., etc.

      In fact, SamH agreeing with anything you state makes any sane person more suspicious of your solution.

      Prove your calculations arrive at the same solution and conclusions as that which physicist Dr. Charles R. Anderson, PhD’s calculations would arrive at. I suspect you’re about to claim that because you arrive at the same net energy flow result, that somehow “proves” that energy flows from cooler to warmer… another old kook tactic that always fails when one considers entropy and the fact that radiative energetic exchange is a temporal entropic irreversible process, not an idealized reversible process.

      But it’s not the end result that’s in contention… it’s the method of arriving at that net energy flow. You’ll have to show that not only do you arrive at the correct net energy flow, but also the correct radiant exitance of each object while considering the mutual interaction of the objects via the EM field energy density gradient.

      The shortcut method taught in most institutions of higher learning does arrive at the correct end result (but only when calculating for the net energy flow)… but it does so by incorrectly calculating radiant exitance of each object, then canceling that error on the back end. It does so by assuming each object to be isolated from other objects and emitting to 0 K (ie: no EM field energy density gradient, hence no object interaction). In effect, it treats real-world graybody objects as if they are idealized blackbody objects (emission to 0 K) just with emissivity less than 1 (sometimes… other times the climate alarmists treat real-world graybody objects exactly as though they’re idealized blackbody objects (emission to 0 K and emissivity = 1), as the Kiehl and Trenberth diagram does.

      • AC99 permalink
        May 18, 2022 1:24 pm

        “SamH denies that energy density is directional”

        Given that energy density is a scalar quantity, I do not understand why you think that it can have a direction.

        Do you mean the gradient of energy density? That would be a vector and would have a direction.

  69. Ed Bo permalink
    May 18, 2022 6:48 am

    LOL@KKK –

    So I’m trying to apply your analysis method of energy density gradients…

    Let’s take a red giant sun with a surface temperature of about 3000K, 100 light years from earth. It’s emitting toward the ~3K of deep space and is very close to a blackbody, so it has no trouble emitting photons.

    A set of these photons travel unimpeded down the energy density gradient for 99+ years. So far so good. But then the earth’s sun, with surface temperature of 5770K crosses their path, reversing the energy density gradient.

    So do we have a time machine that goes back 100 years and suppresses the emission of these photons because the gradient reversed?

    But wait! The earth, with surface temperature of ~300K, sweeps across the path of these photons, reversing the energy density gradient again, and after going back 100 years again to cause re-emission, these photons are absorbed by objects on the earth’s surface, such as telescope sensors.

    The radiative exchange method has no trouble handling this case, but it would take ridiculous contortions to try to make the energy density gradient method work…

    • LOL@Klimate Katastrophe Kooks permalink
      May 18, 2022 10:16 am

      That you somehow arrive at the conclusion that time reversal occurs is another commonality between you climate kooks… I’ve argued with one of you climate kooks who claimed that reversal of particle direction resulted in time reversal, because he was incapable of understanding CPT symmetry. You’re sounding more and more like that particular kook, whom I drop-kicked across the width and breadth of CFACT on a daily basis for more than 3 years. He was wrong about so much that it took multiple posts just to list all his fundamental misconceptions… in bullet-point format. LOL

      In the case of the photons, they’re already emitted. Should they pass into an EM field energy density gradient with a chemical potential which exceeds the chemical potential of the photons, those photons will be subsumed into the background EM field. Their chemical potential is then zero, their Helmholtz Free energy is then zero, they can do no work.

      The energy density gradient and hence the energy flow will then be in a different direction… unless you’re going to yet again claim that photons can flow up an energy density gradient without external energy doing work upon that system energy to push it up the gradient. Isn’t that what 2LoT states, Ed? That energy cannot flow from lower temperature (lower energy density) to higher temperature (higher energy density) and hence up an energy density gradient (temperature being a measure of energy density, the fourth root of energy density divided by Stefan’s Constant), without external energy doing work upon that system energy to push it up the gradient? Isn’t that how your refrigerator works, Ed? Or does it spontaneously cool by expelling energy to a warmer ambient with no external energy required in your kooky world? Are you going to deny 2LoT again, Ed? LOL

      Should an object (say, a planet) of lower chemical potential then cross the path of the sun and temporarily lower the energy density, photons (not necessarily the same photons subsumed into the background EM field, but you’d have no way of knowing whether they were or not… nothing distinguishes one photon from another for photons of the same wavelength and polarization) would spontaneously manifest from the background EM field to equilibrate the energy density differential.

      This would take place until the chemical potential gradient, the energy density gradient was zero, whereupon the system would reach a quiescent state wherein no photons can be generated, no photons can be absorbed, no energy flows. A standing wave is set up between the objects, the wavemode nodes at the object surfaces (and since nodes are a zero crossing point, no energy can flow either in or out). Should one object change temperature, that standing wave becomes a traveling wave with the group velocity proportional to the energy density gradient and in the direction of the cooler object.

      This is standard cavity theory, applied outside a cavity.

      Applied universally, TE is otherwise known as “heat death of the universe”. Look it up and marvel at how wrong you are about everything, Ed. LOL

      Ed Bo wrote:
      “The radiative exchange method has no trouble handling this case, but it would take ridiculous contortions to try to make the energy density gradient method work…”

      He says, neglecting to do those calculations to prove his contention, and getting so much wrong as he attempts to wrap his poor underpowered brain around reality and fails to do so, coming up with some tardbabble about time reversal if photons reverse their vector. LOL

      The “energy density gradient method” (your words) is the same method as is used by the S-B equation when the S-B equation is properly used, you loon… temperature is a measure of energy density. Are you going to deny that, too, Ed? LOL

      Temperature is equal to the fourth root of energy density divided by Stefan’s Constant (ie: the radiation constant).

      e = T^4 a
      a = 4σ/c
      e = T^4 4σ/c
      T^4 = e/(4σ/c)
      T = 4^√e/(4σ/c)
      T = 4^√e/a

      q = ε σ (T^4_h – T^4_c)

      ∴ q = ε σ ((e_h / (4σ / c)) – (e_c / (4σ / c))) A_h

      Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
      W m-2 = W m-2 K-4 * (Δ(J m-3 / (W m-2 K-4 / m sec-1)))

      q = (ε c (e_h – e_c)) / 4

      Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
      W m-2 = (m sec-1 (ΔJ m-3)) / 4

      For graybody objects, it is the energy density differential between warmer object and cooler object which determines warmer object radiant exitance. The climate alarmists misinterpret the S-B radiant exitance equation for graybody objects. Warmer objects don’t absorb radiation from cooler objects (a violation of 2LoT in the Clausius Statement sense and Stefan’s Law); the lower energy density gradient between warmer and cooler objects (as compared to between warmer object and 0 K) lowers radiant exitance of the warmer object (as compared to its radiant exitance if it were emitting to 0 K). The energy density differential between objects manifests an energy density gradient, each surface’s energy density manifesting a proportional radiation pressure.

  70. SamH permalink
    May 18, 2022 1:30 pm

    @LOL:

    1. Just to be sure, are you agreeing that the single heated plate reaches 172 K, and introducing a second plate causes the first to reach 185 K? If everyone has finally agreed on the numbers, that seems like a starting point to begin a discussion of a similar effect caused by CO2. If at some point the textbook method of radiative exchange produces the wrong result, you can provide the correct numbers and prove your case.

    2. Since you’re specifically calling this out again, I must correct you:

    “remember, SamH denies that energy density is directional so he tacitly denies the Poynting vector”

    The Poynting vector does not correspond to energy density. It corresponds to energy flux (W/m^2), which is directional, i.e. a vector. Energy density (W/m^3) is not directional, it is a scalar quantity.

    • AC99 permalink
      May 18, 2022 5:35 pm

      One minor correction. Energy density is J/m^3. But it certainly is a scalar quantity and does not have a direction.

      • SamH permalink
        May 18, 2022 6:05 pm

        Thanks for the correction, that’s what happens when I write at 7 am…

      • LOL@Klimate Katastrophe Kooks permalink
        May 18, 2022 11:35 pm

        Energy is a scalar because energy is defined as the dot product of force and displacement.

        E = F · S
        Where: F = force vector, S = displacement vector

        Since the dot product of two vectors is a scalar, energy is a scalar.

        Energy density is energy per unit volume. A scalar divided by a scalar is a scalar, so energy density is defined as a scalar.

        The gradient of a scalar field is what?

        Energy density gradient is the gradient of the scalar energy density, is it not? And are not all gradients vectors?

        Are not vectors directional? They have a magnitude and a direction? So the gradient of the energy density is most certainly a vector.

        And it is the gradient of the energy density which determines radiant exitance of the warmer object, which we calculate by subtracting the energy density of the cooler object from the energy density of the warmer object to arrive at the energy density gradient.

        Think deeper. LOL

    • SamH permalink
      May 18, 2022 6:03 pm

      Thanks for the correction, that’s what happens when I write at 7am…

    • SamH permalink
      May 18, 2022 6:08 pm

      Thanks for the correction, that’s what happens when I write at 7 am…

  71. AC99 permalink
    May 18, 2022 1:30 pm

    “That’s just the thing… we agree on the end result (but only if one is calculating for net energy flow… if one is calculating for radiant exitance of each object, the climate alarmist method gets the wrong result)”

    What experiment has proven that the methods taught in countless texbooks is wrong? It can only be experiments that discern right or wrong.

    It looks like you are admitting that the temperatures obtained by “alarmist” methods are always correct. So that would mean that no measurements of temperature could ever prove that the “alarmist” methods are wrong.

    • LOL@Klimate Katastrophe Kooks permalink
      May 18, 2022 10:55 pm

      The fact that at thermodynamic equilibrium, if the shortcut method of calculating net energy flow taught in most institutions of higher learning actually represented reality, it would require:

      1) the redefinition of thermodynamic equilibrium as a quiescent state to something that isn’t a quiescent state.

      2) the redefinition of radiative energetic transfer as an idealize reversible process rather than what it is… a temporal, entropic irreversible process.

      3) the redefinition of 2LoT to allow energy to flow up an energy density gradient without external energy being required to do work to push that system energy up the energy density gradient… in effect claiming that energy can flow willy-nilly without regard to the energy density gradient, and hence that energy can flow when no work can be done, and hence that work can be done without energy flow.

      In short, an overturning of everything we’ve learned about thermodynamics to date.

  72. AC99 permalink
    May 18, 2022 1:35 pm

    @LOL@Klimate Katastrophe Kooks

    Can you explain how to calculate your formula, q = (ε c (e_h – e_c)) / 4 ?

    What is “ε”? Is it the emissivity of the hot object, the cold object, or something else? Am I correct in interpreting that e_h and e_c are calculated without any consideration of their emissivities?

    • LOL@Klimate Katastrophe Kooks permalink
      May 18, 2022 10:48 pm

      Emissivity only applies to objects which are emitting, yes?

      Energy only flows down an energy density gradient unless external energy does work to push that system energy up the energy density gradient, yes?

      That’s 2LoT in the Clausius Statement sense in a nutshell, yes?

      That’s merely the S-B equation, stripped down to energy density, as it’s meant to be used.

      • AC99 permalink
        May 18, 2022 11:15 pm

        “Emissivity only applies to objects which are emitting, yes?”

        It’s my understanding that all materials have an emissivity. But your response along with your history of comments suggests that you mean the emissivity of the hot object.

        Does that mean you think the hot object will transfer the same amount of energy to the colder object no matter what the absorptivity of the colder object is?

        What if the colder object had an absorptivity of 0.01? Will the amount of heat transfer be the same as to an object with absorptivity of 0.9?

        If you thin so, than I believe we have found a problem where you disagree with the textbooks.

  73. LOL@Klimate Katastrophe Kooks permalink
    May 19, 2022 12:36 am

    AC99 wrote:
    “Does that mean you think the hot object will transfer the same amount of energy to the colder object no matter what the absorptivity of the colder object is?”

    Absolute not.

    Emissivity only applies to an object which is emitting, and between two objects, one warmer and one cooler, the cooler object cannot even emit into the higher energy density of the intervening space due to the energy density gradient.

    Objects don’t emit ‘uphill’. And emissivity and absorptivity cannot be equal per Kirchhoff’s Law if energy is being transferred because there is no thermodynamic equilibrium.

    { Remember, if the object were forced to always emit the same amount it absorbed, it could never change temperature… the reason an idealized blackbody object is a provable contradiction… it assumes the object emits to 0 K (maximum radiant exitance without regard to the actual energy density gradient), it assumes it absorbs all radiation incident upon it (and hence must be at 0 K… otherwise energy would flow up an energy density gradient in violation of 2LoT), it assumes it emits all radiation it absorbs… how does an object at 0 K emit to 0 K (energy flow without energy density gradient), how does an object at 0 K emit at all (zero energy density at 0 K), and given that it must emit all it absorbs, it can never rise above 0 K. That’s why idealized blackbody objects do not (cannot) exist… they’re idealizations and contradictions. }

    What is the emissivity of an object which is not emitting, per the definition of emissivity (the ratio of the total emissive power of a body to the total emissive power of a perfectly black body at that temperature)? Zero, is it not?

    Did you think absorptivity and emissivity were fixed? LOL

    For absorptivity, there is no equivalent “resistance” (to put it in electrical theory terms) akin to emissivity. Absorptivity is only affected by a surface’s reflectivity and transmissivity.

    α = absorbed / incident
    ρ = reflected / incident
    τ = transited / incident

    α + ρ + τ = incident = 1

    For opaque surfaces τ = 0 ∴ α + ρ = incident = 1

    Stop using idealizations to describe real-world processes. It’s tripping you up and leading you to false conclusions.

    • LOL@Klimate Katastrophe Kooks permalink
      May 19, 2022 1:16 am

      Now, let us assume that two objects are at thermodynamic equilibrium. Hence there is zero energy density gradient. Photon production is zero. Existing photon chemical potential is zero. Existing photon Helmholtz Free Energy is zero. Now work can be done by existing photons. No energy can flow. There are no quantum states which are unexcited and hence available to absorb radiation, there is no production of radiation.

      α = absorbed / incident
      ρ = reflected / incident
      τ = transited / incident

      α + ρ + τ = incident = 1

      For opaque surfaces τ = 0 ∴ α + ρ = incident = 1

      At thermodynamic equilibrium, ε = α = 0 per Kirchhoff’s Law of Thermal Radiation.

      If α + ρ = incident = 1

      0 + ρ = incident = 1

      All photons are reflected by the two objects, a standing wave is set up in the intervening space. Due to boundary considerations, the standing wavemode nodes must be at the object surfaces, and given that wavemode nodes are a zero-crossing point, no energy can be transferred either in or out of the objects.

      Should one object change temperature, that standing wave becomes a traveling wave with the group velocity proportional to the energy density differential and in the direction of the cooler object.

      This is standard cavity theory, applied outside a cavity. Deny it and you deny wide swaths of long-known and empirically-corroborated scientific reality.

      • LOL@Klimate Katastrophe Kooks permalink
        May 19, 2022 1:17 am

        “Now work can be done”

        – should be –

        “No work can be done”

  74. AC99 permalink
    May 19, 2022 1:33 am

    @LOL

    “Absolute not.”

    OK. Then, how much heat is transferred from a plate with emissivity of 0.6, reflectivity of 0.4, and temperature of 300K, to a plate with absorptivity of 0.4, reflectivity of 0.6, and temperature of 200K?

    This is not a cavity problem, but rather an infinite parallel plate problem.

    • SamH permalink
      May 19, 2022 1:45 am

      Maybe to put a fine point on it, a second question should be asked: how does the heat transfer change if the cold plate changes to absorptivity 0.1, reflectivity 0.9, still 200 K?

      • AC99 permalink
        May 19, 2022 1:48 am

        That’s definitely the next question. I’m just trying to understand how he gets answers to simple problems. I still don’t understand what equations he uses and what properties.

    • LOL@Klimate Katastrophe Kooks permalink
      May 19, 2022 5:05 am

      Warmer object:
      ε = 0.6
      ρ = 0.4
      T = 300 K

      Colder object:
      α = 0.4
      ρ = 0.6
      T = 200 K

      Graybody: q = ε σ (T^4_h – T^4_c)
      Blackbody: q = σ T^4

      σ = 5.670374419e−8 W m−2 K−4
      a = 7.5657e-16 J m-3 K-4
      e = T^4 a

      σ / a = 74948114.5 W J-1 m
      σ / a * Δe = W m-2

      Given that emissivity is defined as the ratio of the total emissive power of a body to the total emissive power of a perfectly black body at that temperature, we’ll calculate that BB total emissive power first:

      e_h = 300^4 K^4 * 7.5657e-16 J m-3 K-4 = 8100000000 K^4 * 7.5657e-16 J m-3 K-4 = 0.000006128217 J m-3

      q_bb = 74948114.5 W J-1 m * 0.000006128217 J m-3 = 459.2983093968465 W m-2

      The 0.0040093968465 W m-2 differential between my solution and the S-B equation is due to rounding on the Hyperphysics page.

      (459.2943 / 5.6703e-8 * 5.670374419e-8) – 459.2983093968465 = 0.0020185421535 W m-2.

      That drops to 0.0020185421535 W m-2 differential just by accounting for the rounding of σ on the Hyperphysics page, and the differential is still due to rounding (of the end result) on the Hyperphysics page.

      Ok, so we’ve got our idealized blackbody radiant exitance, and given that the warmer object has an emissivity of 0.6, and going by the definition of emissivity, we’ve got our warmer graybody object maximum radiant exitance:
      q_gb = 459.2983093968465 W m-2 * 0.6 = 275.5789856381079 W m-2

      The 0.00240563810786 W m-2 differential between my solution and the S-B equation is due to rounding on the Hyperphysics page.

      (275.57658000000004 / 5.6703e-8 * 5.670374419e-8) – 275.5789856381079 = 0.00121112529214000052497398726699 W m-2

      That drops to 0.00121112529214000052497398726699 W m-2 differential just by accounting for the rounding of σ on the Hyperphysics page, and the differential is still due to rounding (of the end result) on the Hyperphysics page.

      So we’ve got the maximum radiant exitance of the graybody object: 275.5789856381079 W m-2

      Now to account for the 200 K of the cooler graybody object:

      e_h = 300^4 K^4 * 7.5657e-16 J m-3 K-4 = 8100000000 K^4 * 7.5657e-16 J m-3 K-4 = 0.000006128217 J m-3
      e_c = 200^4 K^4 * 7.5657e-16 J m-3 K-4 = 1600000000 K^4 * 7.5657e-16 J m-3 K-4 = 0.000001210512 J m-3

      σ / a = 74948114.5 W J-1 m
      σ / a * Δe = W m-2

      74948114.5 W J-1 m * (0.000006128217 J m-3 – 0.000001210512 J m-3) =
      74948114.5 * 0.000004917705 = 368.5727174172225 W m-2 maximum * 0.6 emissivity = 221.1436304503335 W m-2


      (221.14170000000001 / 5.6703e-8 * 5.670374419e-8) – 221.1436304503335 = 0.00097189066651000013124349681674691 W m-2 differential from the S-B equation by accounting for the rounding of σ on the Hyperphysics page, and the differential is still due to rounding (of the end result) on the Hyperphysics page.

      And now we account for the absorptivity of the cooler object, per the definition of absorptivity:
      221.1436304503335 W m-2 * 0.4 = 88.4574521801334 W m-2

      So in reality, the warmer object will not emit 459.2983093968465 W m-2 (emission to 0 K, ε_h = 1), nor will it emit 368.5727174172225 W m-2 (emission to 200 K, ε_h = 1), nor will it emit 275.5789856381079 W m-2 (emission to 0 K, ε_h = 0.6) , nor will it emit 221.1436304503335 W m-2 (emission to 200 K, ε_h = 0.6, α_c = 1), it will emit 88.4574521801334 W m-2 (emission to 200 K, ε_h = 0.6, α_c = 0.4). That’s all the energy density gradient can sustain and that’s all the cooler object can absorb.

      The initial photons reflected from the cooler surface won’t re-incide upon the warmer object (they’d have to climb the energy density gradient to do so, and by the time they did, their chemical potential would be zero, they could do no work upon the warmer object… look up ‘reflection from a potential step’ to see that their phase is changed, hence their vector is changed, they are reflected), they will contribute to the intervening space energy density and hence slow radiant exitance of the warmer object until its radiant exitance is equal to the irradiance able to be absorbed by the cooler object.

      In effect, the warmer object ‘extends’ its energy density into the intervening space, that lowers the energy density gradient seen at the warmer object’s surface, which lowers warmer object surface radiant exitance.

      Do remember that the two objects interact via the EM field and its energy density gradient. It is that gradient which determines radiant exitance of the warmer object. Also remember that nature is ‘lazy’… as Maupertuis wrote, “Nature is thrifty in all its actions.”… in our two-object system, it’s not going to have the warmer object spewing out a bunch of photons that will only be reflected (because there’s nowhere for them to go… they cannot be reabsorbed by the warmer object, they’d have to climb the energy density gradient and by the time they did, they’d have a chemical potential of zero, they could do no work; they cannot be absorbed by the cooler object due to its reflectivity), it’ll damp down warmer object radiant exitance until the photons emitted can be absorbed by the cooler object.

      Of course, that assumes a View Factor of 1, and it assumes a system of only two stationary objects. Things get far more complicated with multiple objects, all at different temperatures, and even moreso when things start moving around.

      Think of it in terms of water… the climate alarmists essentially claim that for two lakes at exactly the same level and connected via a canal, water flows from each lake to the other at some rate (calculated by analogizing gravity to energy density and hence temperature… in this case, they would have to calculate the flow rate from each lake for a gravitational gradient of some unknown large amount equivalent to emission to 0 K), but since both flow rates are equal, the net flow is zero… that’s insane. In reality, for two lakes at the same level and connected via a canal, no water flows. It’s the same for photons and energy density… for two objects at the same temperature (the same energy density), no photons flow.

      Think of it in terms of electricity… the climate alarmists essentially claim that for two batteries of the same voltage electrically connected ‘+ to +’ and ‘- to -‘, electrons flow from each battery to the other at some rate (calculated by analogizing voltage to energy density and hence temperature… in this case, they would have to calculate the current flow rate from each battery for a voltage gradient from battery voltage to ground), but since both current flow rates are equal, the net flow is zero… that’s insane. In reality, for two batteries of the same voltage electrically connected ‘+ to +’ and ‘- to -‘, no current flows. It’s the same for photons and energy density… for two objects at the same temperature (the same energy density), no photons flow.

      Anyone else remember these old sliderules?

  75. AC99 permalink
    May 19, 2022 1:58 pm

    @LOL@Klimate Katastrophe Kooks

    Your result of 88.4574521801334 W m-2 differs from the value given in textbooks on heat transfer for this problem. The result from heat transfer textbooks is 116.4 W/m^2.

    So to condense your method, you do the following calculations:

    Calculation 1: ε_h σ T_h^4 = 275.6 W/m^2
    Calculation 2: ε_h (σ T_h^4 – σ T_c^4) = 221.1 W/m^2
    Calculation 3: multiply calculation 2 by α_c to get the final formula of:

    q = α_c ε_h (σ T_h^4 – σ T_c^4) = 88.5 W/m^2

    This formula is at odds with those in textbooks, and comparing to what you wrote before, q = ε σ (T^4_h – T^4_c), it must mean that the ε in your prior formula is ε = α_c ε_h.

    Your result is different from textbook results by over 20%.

    • AC99 permalink
      May 19, 2022 2:14 pm

      Your final formula is also different from the one that you gave a link to previously. Quoting from that link:

      “For our two parallel plates above, if both are gray bodies, then between the plates

      Δe = eH – eC = εH a TH4 – εC a TC4

      PHI = (σ/a) Δe = εH σ TH4 – εC σ TC4

      PCI = 0.

      Here we see that the emissivity which determines the electromagnetic field energy density at the surface is also playing the role of the absorptivity at the absorbing colder surface. So of course, Kirchhoff’s Law of thermal radiation that the emissivity equals the absorptivity of a material in a steady state process applies. There is really nothing at all to prove if one starts with the primary fact and boundary condition that the energy density is the fundamental driver of the thermal radiation of materials.”

      So that formula is both different from yours and different from the one in textbooks.

      • LOL@Klimate Katastrophe Kooks permalink
        May 19, 2022 4:11 pm

        No, it’s not. It’s the S-B equation in energy density format, nothing more, as physicist Dr. Charles R. Anderson, PhD’s equations are the S-B equation, using energy density gradient in a more roundabout way.

    • LOL@Klimate Katastrophe Kooks permalink
      May 19, 2022 4:17 pm

      The textbook assumes that the radiant exitance is affected only by the energy density gradient, not the absorptivity of the cooler object, so the warmer object just keeps pumping out photons regardless of whether they’re absorbed by the cooler object. That’s a violation of the Principle of Least Action.

      Now you’ll have to explain where that extra energy goes, in a system with only two objects, given that those photons, were they to ascend the energy density gradient toward the warmer object again, would have zero chemical potential, zero Helmholtz Free Energy, could do no work upon the warmer object, and thus could not be absorbed by the warmer object. And they’re reflected by the cooler object. And if they go to contributing to the energy density of the intervening space, they reduce the radiant exitance of the warmer object.

      Let’s see these textbook calculations.

  76. AC99 permalink
    May 19, 2022 4:13 pm

    @LOL@Klimate Katastrophe Kooks

    To be clear, you are claiming that the hot plate only emits 88.5 W/m^2 and then the cold plate absorbs all of that radiation that is emitted even though the cold plate has nonzero reflectivity.

    Is that right? My understanding is that a surface with reflectivity of 0.6 will reflect 60% of the radiation that is incident upon it.

    I am then left to wonder, how much radiation you claim is emitted by the hot plate if the cold plate has an absorptivity of 0.4 and a transmissivity of 0.6 (instead of a reflectivity of 0.6).

    How would those conditions alter your solution procedure?

    • LOL@Klimate Katastrophe Kooks permalink
      May 19, 2022 4:21 pm

      Absorptivity, emissivity and reflectivity are variable. For instance, at thermodynamic equilibrium, emissivity is 0, absorptivity is 0 and reflectivity is 1 for all objects at thermodynamic equilibrium.

      Objects are coupled through the EM field energy density gradient. They interact. You’re still attempting to take each object in isolation and do the calculations without regard to that energy density gradient.

      Which is the whole point of my posts.

    • LOL@Klimate Katastrophe Kooks permalink
      May 19, 2022 4:39 pm

      AC99 wrote:
      “I am then left to wonder, how much radiation you claim is emitted by the hot plate if the cold plate has an absorptivity of 0.4 and a transmissivity of 0.6 (instead of a reflectivity of 0.6).”

      That just means that the object has the capability of transmitting 60% of that which it absorbs… that doesn’t mean it will, just that it has that capability. Boundary conditions always determine behavior.

      Of course, you have to make some assumptions here.

      I’m assuming two objects in an isolated system, so even if it can transmit 60% of that which it absorbs, how’s it going to emit? The boundary of the system is at the boundary of the objects except for the faces of each object facing the other object. So the system is comprised of warmer object, intervening space between warmer and cooler object, and cooler object.

      Next, if you assume that the system boundary is bigger and the cooler object can emit to the system, do you hold system energy density constant, or do you allow it to rise as the objects emit to the intervening space? Holding it constant is unphysical. Allowing it to rise means you have to quantify system volume.

      Next, you’ll have to assume an energy density of the intervening space.

      None of which you’ve done.

      I took the energy density of the intervening space between the two objects into account in my calculations, which is why the warmer object can only emit what the cooler object can absorb.

      Of course, properly setting up the initial conditions of a problem seems to be a common problem with you climate kooks… the kook I drop-kicked across the width and breadth of CFACT on a daily basis for more than 3 years stepped on rake after rake in that regard, getting each problem wrong because he’d delineated initial conditions improperly. LOL

  77. AC99 permalink
    May 19, 2022 4:47 pm

    “No, it’s not. It’s the S-B equation in energy density format, nothing more, as physicist Dr. Charles R. Anderson, PhD’s equations are the S-B equation, using energy density gradient in a more roundabout way.”

    Can you show how these are the same?

    Your procedure leads to: q = α_c ε_h (σ T_h^4 – σ T_c^4)

    Anderson’s equations are: PHI = (σ/a) Δe = εH σ TH4 – εC σ TC4, PCI = 0, and he says that absorptivity is equal to emissivity at steady state.

    I do not see how your formula is the same as his? What are the mathematical steps you are using to prove these are the same?

    • LOL@Klimate Katastrophe Kooks permalink
      May 19, 2022 5:07 pm

      Already done, several times. Did you not pick up on it, or did you not understand it? LOL

      Check my bolds and hang your head in shame at the fact that you’ve engaged mouth before brain. LOL

      AC99 wrote:
      “Anderson’s equations are: PHI = (σ/a) Δe

      e = T^4 a
      a = 4σ/c
      e = T^4 4σ/c
      T^4 = e/(4σ/c)
      T = 4^√e/(4σ/c)
      T = 4^√e/a

      σ = 5.670374419e−8 W m−2 K−4
      a = 7.5657e-16 J m-3 K-4 = 4σ/c = 7.5657332500339284719430800357226e-16 J m-3 K-4

      σ / a = 74948114.5 W J-1 m
      σ / a * Δe = W m-2

      q = ε σ (T^4_h – T^4_c)

      ∴ q = ε σ ((e_h / (4σ / c)) – (e_c / (4σ / c))) A_h

      Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
      W m-2 = W m-2 K-4 * (Δ(J m-3 / (W m-2 K-4 / m sec-1)))

      ∴ q = (ε c (e_h – e_c)) / 4

      Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
      W m-2 = (m sec-1 (ΔJ m-3)) / 4

  78. AC99 permalink
    May 19, 2022 4:50 pm

    “Let’s see these textbook calculations.”

    Just about any textbook on radiative heat transfer has the formula for the heat transferred between two opaque graybody plates. Do you own any textbooks like that?

    • LOL@Klimate Katastrophe Kooks permalink
      May 19, 2022 4:55 pm

      Sure I do. Let’s see your calculations. You do know, of course, that for a system comprised of only warmer object, intervening space between warmer and cooler object, and cooler object, with no other volume in the system, that the photons have nowhere else to go but to the warmer object, to the intervening space, or to the cooler object.

      They cannot be reflected and re-absorbed by the warmer object… the photons would ascend the energy density gradient and would have zero chemical potential, zero Helmholtz Free Energy, zero ability to do work. They cannot flow to the cooler object, that just reflected them.

      They flow to and contribute to the energy density of the intervening space, the only other place they can go. And that higher energy density reduces radiant exitance of the warmer object.

      That’s why the warmer object can only emit what the cooler object can absorb.

      Your equations assume the warmer object just keeps pumping out photons regardless of how many photons are absorbed by the cooler object, and those photons go God-knows-where. You’ll get right on explaining where they go. Or are you going to claim that they just disappear, a violation of 1LoT? LOL

  79. AC99 permalink
    May 19, 2022 4:52 pm

    “Absorptivity, emissivity and reflectivity are variable. For instance, at thermodynamic equilibrium, emissivity is 0, absorptivity is 0 and reflectivity is 1 for all objects at thermodynamic equilibrium.”

    Do you have any textbook that states that “emissivity is 0, absorptivity is 0 and reflectivity is 1 for all objects at thermodynamic equilibrium”?

    I have never read such a claim before I encountered you.

    • LOL@Klimate Katastrophe Kooks permalink
      May 19, 2022 4:57 pm

      It’s only in the definitions of emissivity and absorptivity. Go look them up, FFS. LOL

      Then go study cavity theory and marvel at how much of reality you’ve gotten wrong. LOL

  80. AC99 permalink
    May 19, 2022 4:57 pm

    “That just means that the object has the capability of transmitting 60% of that which it absorbs”

    That is not my understanding of transmissivity. A material with transmissivity of 1 allows all radiation to pass through it without absorbing any of the radiation that is incident on it.

    Do you have any textbooks that state your definition of transmissivity?

    • LOL@Klimate Katastrophe Kooks permalink
      May 19, 2022 5:15 pm

      Mheh… yeah, you’re right. I engaged mouth before engaging brain. LOL

      Doesn’t change the fact that in a two-object system with the system boundaries at the object boundaries except for the mutually-facing object faces, that cooler object can’t emit nor transmit… there’s nowhere for photons to go except the intervening space between objects, and the energy density is too high for the cooler object to ever emit there. If we assume the photons transmitted are reflected at the system boundary back through the cooler object and into the intervening space, they’ll only contribute to the energy density of the intervening space, reducing warmer object radiant exitance to that which the cooler object can absorb.

      This goes on until thermodynamic equilibrium is attained between warmer object, intervening space, and cooler object, whereupon a state of quiescence is attained, no photon emission, no photon absorption, no energy flow, no energy density gradient, no impetus for photon production.

  81. AC99 permalink
    May 19, 2022 5:26 pm

    “Already done, several times. Did you not pick up on it, or did you not understand it?”

    Perhaps both. I still don’t understand how you can claim that your formula:

    q = α_c ε_h (σ T_h^4 – σ T_c^4)

    is the same as Anderson’s: PHI = (σ/a) Δe = εH σ TH4 – εC σ TC4

    It is mathematically impossible to get from your equation to Anderson’s. In fact, if you put in the numbers for the 300K and 200K plates, Anderson’s formula gives PHI = 239.3 W/m^2, which is different from your result of 88.5 W/m^2.

    • LOL@Klimate Katastrophe Kooks permalink
      May 19, 2022 5:37 pm

      That’s not my formula, that your interpretation of my formula for a very specific problem which you apparently aren’t capable of understanding and apparently didn’t properly set up the initial conditions for. LOL

      All of the equations are identical. Some break it down to energy density, some use energy density gradient, some use temperature (a measure of energy density).

  82. AC99 permalink
    May 19, 2022 5:30 pm

    “Sure I do. Let’s see your calculations.”

    Then you should be able to find that the formula for the heat transfer between two opaque graybody plates is:

    q = ε_h ε_c σ (T_h^4 – T_c^4) / (ε_h + ε_c – ε_h ε_c)

    This formula is different from yours. My calculations are based on this formula.

    • LOL@Klimate Katastrophe Kooks permalink
      May 19, 2022 5:33 pm

      That’s using the effective emissivity kludge. That only applies in certain situations.

      Let’s see your calculations, not your equations. Show your work. LOL

  83. AC99 permalink
    May 19, 2022 5:40 pm

    “It’s only in the definitions of emissivity and absorptivity. Go look them up, FFS. LOL”

    OK, I did. Absorptivity is the fraction of incident radiation that is absorbed by an object. I have found no source that claims that absorptivity is variable (unless you mean that absorptivity is dependent on the wavelength of the radiation).

  84. AC99 permalink
    May 19, 2022 5:43 pm

    “That’s using the effective emissivity kludge. That only applies in certain situations.”

    I don’t see any effective emissivity. The formula applies to two infinite parallel plates.

    My calculations will be the same as those presented in textbooks. There are multiple ways to derive this equation. But the point is that you are claiming that the textbooks are wrong.

  85. AC99 permalink
    May 19, 2022 5:45 pm

    “That’s not my formula, that your interpretation of my formula”

    So how is your formula different from what I wrote? You provided a description of your calculation and the formula that I wrote down follows that exactly.

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