Green Plate Analyzed and Demolished

There’s this silly idea called a “greenplate effect” which is supposed to be an analogy to explain how the climate science radiative greenhouse effect works. The tldr on it is that a plate is heated on one side by some source and comes to some temperature, and then another plate is brought in behind the first plate and this new plate creates backradiation for the first plate after being heated by the first plate, and this backradiation causes the first plate to raise in temperature. The previous post reported on someone who performed the actual experiment, and found that the “greenplate effect”, i.e. the radiative greenhouse effect of climate science, does not manifest.

The rebuttal to the experiment was, get this, that the experiment wasn’t performed long enough! The experiment shows plots of the results, and the plots should have shown diverging curves if the RGHE existed, but it did not show this. And so the claim is basically that the RGHE may only be observed after the measurement to detect it, which thus explains why you didn’t detect it! How convenient!

This is the same claim other recent commentators on this blog have made claiming that the RGHE is not possible to measure by experimental apparatus: that all disparate and varied scenarios which should show the RGHE in operation to greater or lesser extents depending upon conditions, all coincidentally have the variable conditions such that the effect is perfectly cancelled out from measurement! How wonderful for them. Talk about a faith!

Once we realized that all that we need to do is to ask them for a single laboratory measurement under controlled conditions of the RGHE, then their approach to experimentalism became that no experimental demonstration is possible because 1) the RGHE can only be measured once you stop trying to measure it, or 2) in all scenarios where the RGHE should manifest to greater or lesser extent, there are perfectly-negating environmental conditions which make the RGHE unable to be detected.

Yes yes, I know…sophistry much? I made this comment regarding this new position of the climate scientists: 

[…] They COULD acknowledge the points I made, but they simply choose not to. Does the Sun heat the Earth on one side at high temperature? Yes, of course. But they simply ignore it. Does the adiabatic gradient establish that the near-surface air must be higher in temperature than any expected average? Yes, of course. But they can simply choose to ignore that.

Do we have bills of rights and freedoms, privacy laws, etc? Yes, of course. But they can just ignore those and get “health authorities” to rescind them.

There’s no such thing as law or reason. These people only care about power, and having power requires DEMONSTRATION of power, which they do by ignoring reason and law…because it requires power to do that.

Moving on:

In the original “greenplate” scenario created by some teacher named Eli Rabbet, they did not even discuss view factors of the source and first plate, which changes the outcome of the experiment. This just goes to show that these people have no idea what they’re doing. And so, we’ll analyze this “problem of the plates” correctly here for the first time. The most important concept in radiative transfer is something called view factors. The view factor is simply the VIEW of space, or the view of an object, that one object has relative to another. Basically, the view factor is how much solid angular area one object has as viewed from another object…how much “view of the sky” one object has as viewed from another.

There are two scenarios to start the problem off with in the first place: A) a distant point source, or B) a plane parallel source. These two options would the most common ways to simplify the problem and make it analyzable, but of course you could have some messy intermediate state between the conditions of A & B. But these A & B conditions make the problem quickly solvable.

A – Distant Point Source

Distant point source sending thermal EM to a single plate

With a point object as the source, then the view factor of the source relative to the plate is negligible, i.e., the point source occupies almost no or basically no angular view factor as viewed from the plate. Why is this important? It is important because it means that thermal radiation emitted from the plate on the side of the source can emit to open space, that there are no opposing vectors of input radiation from the source except for the infinitesimal vector pointed directly toward the source; being infinitesimal, it may therefore be ignored.

The result is that, for the equilibrium state of the plate, we must consider than it can emit freely to the entirety of both hemispheres of either side of the plate’s view. Thus, if we consider that the plate is very thin in terms of its thermal conductivity, so that both sides of the plate emit the same energy, then for a given input flux F from the source either side of the plate emits F/2, and then you may compute the resulting temperature via the usual way with the Stefan-Boltzmann Law: T = (F / 2 / σ)1/4

B – Plane Parallel Source

Plane parallel source sending thermal EM to a single plate

In this scenario we consider the plate and source as extending “infinitely”, or at least being very close together, such that the view factor of the source occupies the entire hemisphere of view from the first plate. How is this different? This results in every single vector of emission from the plate being met with an opposing vector of input from the source. That is, the plate cannot lose energy in the entire hemisphere of potential emission facing the source. Whereas with the point source the plate may emit to empty black space all around the infinitesimal point source, with the plane parallel source the plate now cannot emit to empty space because that space is now occupied by input from the source.

Thus, now to conserve energy the plate must emit all energy only on the side facing away from the source, since there is empty space on this hemisphere. And so, given an input F from the source, now the plate must emit F on the side facing away from the source, resulting in a higher temperature with no division by two of F of T = (F / σ)1/4.

Adding the Second Plate, the “greenplate”

Now that we have the two possible conditions established for the single plate (which the original author of this scenario completely left out of their discussion thus indicating that they have no idea what they’re doing), we may add a second plate “behind” the first plate.

A: Point source with second plate added

B: Plane parallel source with second plate added

And so how do we solve the two problems above? They have a lot of similar features. Of course, we must solve for thermal equilibrium by using the definition of thermal equilibrium, which is when temperature is constant. We utilize the First Law of Thermodynamics for this:

dU = Q = m Cp dT

The ‘d’ means “change” and we want the change to be zero, which means unchanging temperature dT = 0, which means thermal equilibrium. We already used this for the results above, without explaining it, as I know that most of you implicitly understand this equation and its universal presence. 

We consider again that the second plate is thin with respect to its thermal conductivity, and thus attains the same temperature on either side. The second plate in both scenarios experiences the view factor conditions of the first plate in scenario B: plane parallel view factors from its source, where the source for the second plate is the first plate.

A – Distant Point Source with Plate

This scenario is really simple to solve and again we don’t even need to formally refer to the First Law equation, although it is embedded in the solution. No energy is lost between the two plates, and the energy lost to space is emitted from the first plate on the hemisphere facing the point source, and from the second plate on the hemisphere facing away from the point source. Thus, you should be able to see in your mind’s eye that this scenario actually reduces to the single-plate case, where the gap between the plates is merely an infinitesimal slice of space where the second plate becomes the new outer surface of the first plate.

In other words, the two plates may only emit a total of F t0 space, which results in the same solution where F / 2 is emitted into the hemispheres from either open-facing plate surface. And so each plate now must have the same temperature of T = (F / 2 / σ)1/4.

Could introducing the second plate in this scenario result in a higher temperature for the first plate? If the first plate did achieve a higher temperature, then it would have to emit more than F / 2 to the hemisphere facing the source. However, because the plate is thin, then it also would emit this to the second plate. To conserve energy, then the second plate would have to emit less than F / 2 to the hemisphere facing away from the source…but how could the second plate possibly emit less than F / 2 when it in fact receives more than F / 2? Do you see how the logic breaks down when you invent backradiation heating? Very neat!

B – Plane Parallel Source with Plate

Just as in the original scenario, the second plate cannot lose energy in the hemisphere of the first plate since the view factor from the second plate to the first plate is completely occupied by the emission from the first plate to the second plate. And so, again, the only possible result is that the second plate emits the necessary energy to the hemisphere facing away from the first plate, thus attaining a temperature of T = (F / σ)1/4.

Could introducing the second plate in this scenario result in a higher temperature for the first plate? If the first plate did achieve a higher temperature, then it would have to emit more than F to the second plate. But since the second plate may only lose energy on the hemisphere facing away from the first plate, then the second plate would have to emit more than F, but F was all of the energy supplied in the first place. Again: Do you see how the logic breaks down when you invent backradiation heating? Very neat!


The argument of the “greenplate” effect is that if you take the single plate case, and simply cut the plate down the middle and create an internal gap, then you will increase the temperature of the plate remaining closest to the source. And thus, by extrapolation, if you simply keep on splitting the plates and creating more and more gaps, then you increase the temperature of the first plate as a function proportional to “n” the number of gaps! So, shine a candle at plate, but split the plate in half as many times as you want to make the plate as hot as you please!

Here is the original “Green Plate” post by this Eli Rabbet character:

If you analyze what he did when introducing the second plate, he treats the view factors as if the first plate is a point source relative to the second plate, and thus he splits the flux by two for the second plate. The view factor of the first plate relative to the second plate cannot possibly be the same as the view factor of the point source sun relative to the first plate, though. That’s his fundamental mistake, and again we will note that he doesn’t address view factors at all. He has no idea what he’s doing.

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145 Responses to Green Plate Analyzed and Demolished

  1. Joseph Postma says:

    The argument of the “greenplate” effect is that if you take the single plate case, and simply cut the plate down the middle and create an internal gap, then you will increase the temperature of the plate remaining closest to the source. And thus, by extrapolation, if you simply keep on splitting the plates and creating more and more gaps, then you increase the temperature of the first plate as a function proportional to “n” the number of gaps! So, shine a candle at plate, but split the plate in half as many times as you want to make the plate as hot as you please!

  2. Here is the original “Green Plate” post by this Eli Rabbet character:

    If you analyze what he did when introducing the second plate, he treats the view factors as if the first plate is a point source relative to the second plate, and thus he splits the flux by two for the second plate. The view factor of the first plate relative to the second plate cannot possibly be the same as the view factor of the point source sun relative to the first plate, though. That’s his fundamental mistake, and again we will note that he doesn’t address view factors at all. He has no idea what he’s doing.

  3. Max Polo says:

    Great post ! Thanks Joe

  4. Regis says:

    Interesting. Seems you need to be sure to use the right view factor when solving the problem.

  5. Regis says:

    Does the view factor from the sun enter into the solar constant?

  6. Yes it does, via the inverse square law. At the distance of the Earth, the sun occupies half a degree, and so is much closer to looking like a point source than an infinite wall.

  7. Weekly_rise says:

    Thus, now to conserve energy the plate must emit all energy only on the side facing away from the source, since there is empty space on this hemisphere. And so, given an input F from the source, now the plate must emit F on the side facing away from the source

    I am not sure that I agree with this assessment, but maybe I’m misunderstanding something. The plate should be emitting from all sides, the presence of the source does not prevent the side of the plate facing the source from emitting – there is no physical process by which this could happen. The sourceward side of the plate is, at equilibrium, emitting the same flux as the nighttime side of the plate (assuming the plate is a thin one). To be emitting no flux the sourceward side of the plate would have to be at a temperature of 0 Kelvin.

    What am I missing?

  8. Energy cannot be lost on the side facing the plane parallel source because energy is incoming from all possible escape vectors.

  9. Regis says:

    @weeklyrise , no object can emit thermal radiation if a hotter object is present in that direction. Second law of thermodynamics.

  10. Joseph E Postma says:

    The surface still emits thermal EM…it is just that this emission does not equate to loss of energy because there is incoming energy from all direction vectors of emission. The emission is “negated”, it is not that the emission does not exist at all.

  11. Regis says:

    Oh, so the energy change just the difference between emission and absorption?

  12. It is like the heat flow equation, where heat is the net difference between emissions:

    Q = s * (T1^4 – T2^4)

    Only the difference between emission is heat: this does not mean that the cooler body is not emitting, but that its emission is “negated” by the stronger emission from the warmer body, and only the greater portion of the warmer body’s energy is heat.

  13. Regis says:

    Huh, well now I’m trying to read it again and I don’t understand the plane parallel source section. If the plate can’t emit at all toward the source I get it. But if it can, shouldn’t the plate still emit F/2 to the source, just it gets F back so it’s not losing energy in that direction?

  14. The energy has to be equal and opposite for equilibrium…depending on how the view factors affect how equality may occur. With a filled view factor then equality is found with the emission from the plate being equal to the input from the source wall. If it worked like you say then only F/2 would be escaping to the other side…but there needs to be F going out from the other side.

  15. Regis says:

    Okay got it (I think). Is this right: the plate emits F on both sides. But the emission toward the source is cancels with the F absorbed from the source. Then the emission of F toward space balances the F absorbed from the source.

  16. Regis says:

    And then temperature is F/sigma^.25

  17. Weekly_rise says:

    “Okay got it (I think). Is this right: the plate emits F on both sides. But the emission toward the source is cancels with the F absorbed from the source. Then the emission of F toward space balances the F absorbed from the source.

    Doesn’t this explanation violate the conservation of energy? If the plate is emitting F on both sides it is emitting 2*F, so emitting a flux twice as large as the flux it is receiving. Am I reading that wrong?

  18. Regis says:

    Please ignore my previous post. I realized the plate would still have 2F emitted (F from each side) and only F absorbed. So the energy gain which is the difference would not be zero. I will have to think about this more and try to understand!

  19. @Weekly_rise – Regis

    The emission on the source-side is negated by input. So the total emission is only F on the other side from the source. And again this is only for the plane-parallel view factor case…for a point source then the view factor changes everything!

  20. CD Marshall says:

    This physicist is still at it like a dying man in quicksand who refuses to believe he is sinking.

    The same sunlight falling at 940 W/m^2 at the equatorial zenith is the same hitting the poles at 85 W/m^2.

    That is a distribution of energy due to geometry. The total amount of energy absorbed by the system stays the same.

    …But the temperature varies and that very crucial point creates the climate.

  21. Eli. Sounds Jewish

  22. Weekly_rise says:

    The emission on the source-side is negated by input. So the total emission is only F on the other side from the source. And again this is only for the plane-parallel view factor case…for a point source then the view factor changes everything!

    It’s possible that the plane/point source distinction is throwing me off, but this still doesn’t make sense to me. I agree that the sourceward side of the plate will “cancel out” with the incoming flux, but in this case we’d say that both the incoming and outgoing fluxes are being canceled – not just the outgoing. So if we assume F coming in and F going out on the sourceward side, then both Fs cancel and there is no net flux on the sourceward side. But then we have the problem of there being a flux of F on the nighttime side of the plate. Where does that energy come from?

    To use a stupid analogy, if a guy in front of me hands me a dollar and I hand him one dollar back, no money was exchanged. If we then say that I pass the guy behind me a dollar, we need to explain where that dollar came from.

    If instead we say that the guy in front hands me a dollar, and I hand him back fifty cents and hand the guy behind me fifty cents, then my net worth still hasn’t increased, but now we can account for where all the money is coming from. (It doesn’t matter the ratios – I could hand the guy in front seventy-five cents and the guy behind a quarter.)

    Hopefully that explains my thinking better.

  23. lurker says:


    “In possible climate breakthrough, Israel scientists engineer bacteria to eat CO₂
    Decade-long research at Weizmann Institute could pave way for low-emissions production of carbon for use in biofuels, food, and help remove excess global warming CO₂ from air”

    “In a remarkable breakthrough that could pave the way toward carbon-neutral fuels, researchers at the Weizmann Institute of Science have produced a genetically engineered bacteria that can live on carbon dioxide rather than sugar.”

    Some people say that they’re here to destroy the entire planet…to end life on Earth. Imagine: a bacteria engineered to consume all CO2 out of the atmosphere…turning the surface into a grey goo of bacteria, a barren rock with no life left. There is no safe scenario to engineer such a bacteria…once it is out, it is out, and will replicate exponentially until its food supply – CO2 – is depleted. This is an engineered TOTAL extinction level event.

    The whole argumentation strategy of climate alarmism seems particularly Jewish:

    “The more I argued with them, the better I came to know their dialectic. First they counted on the stupidity of their adversary, and then, when there was no other way out, they themselves simply played stupid. If all this didn’t help, they pretended not to understand, or, if challenged, they changed the subject in a hurry, quoted platitudes which, if you accepted them, they immediately related to entirely different matters, and then, if again attacked, gave ground and pretended not to know exactly what you were talking about. Whenever you tried to attack one of these apostles, your hand closed on a jelly-like slime which divided up and poured through your fingers, but in the next moment collected again. But if you really struck one of these fellows so telling a blow that, observed by the audience, he couldn’t help but agree, and if you believed that this had taken you at least one step forward, your amazement was great the next day. The Jew had not the slightest recollection of the day before, he rattled off his same old nonsense as though nothing at all had happened, and, if indignantly challenged, affected amazement; he couldn’t remember a thing, except that he had proved the correctness of his assertions the previous day.

    Sometimes I stood there thunderstruck.

    I didn’t know what to be more amazed at: the agility of their tongues or their virtuosity at lying.”

    from a man who tried to warn the world about the (((menace)))

  24. I’ve said this before but it bears repeating it seems: no energy is LOST on the source-facing side, because at equilibrium any vector of emission is met with a vector of input (for plane parallel view factor).

    Here’s the heat flow equation for plane parallel geometry:

    Q = sT1^4 – sT2^4

    At equilibrium Q = 0, and the emissions are thus equal: sT1^4 = sT2^4

    But the plate has a temperature, and so it emits sT2^4 on the other side, conserving the input of sT1^4.

  25. CD Marshall says:

    My brother is part Jewish traced his roots back to the lost tribe of something I forget, seems like it was Benjamin? He is ambidextrous, still a trait of that gene pool, apparently.

  26. @CD, WWF: Well…I’m not going to get in on all that…I like that quote and have used it before myself, but I replace “Jew” with “climate alarmist”…lol

  27. CD Marshall says:

    Question: What causes the parcel of air in the dry adiabatic lapse rat to descend once it has risen enough to be stably buoyant under its potential energy?

  28. @CD: I write an equation for this once…how did it go again…

    If you take a horizontal infinitesimal slice of space through the atmosphere, and consider the molecules passing through it…and also consider that this slice of space is in local thermodynamic equilibrium, then the molecules moving up through the slice have the vertical component of their velocity reduced by deceleration gravity, and the molecules moving down through the slice have their velocity increased by acceleration due gravity. Thus, the molecules above the slice have a lower average velocity and hence lower average temperature, and vice versa for the molecules below.

    So it occurs on the molecular level. It doesn’t actually require bulk convection, although the EXACT same principle applies when bulk convection occurs…it is the exact same gravitic concern applying to molecules either moving up or down…HOWEVER they are moving up or down, either through bulk convection as parcels of air, or as individual molecules but which is of course the exact same thing because a bulk parcel IS individual molecules.

    I wrote an equation for this once somewhere but hopefully that description makes sense. I think it was in a blog post actually! How could I find that…

    As for bulk convection, once a parcel reaches max altitude…it has maximally cooled, I guess it upward momentum then depleted and naturally falls back down. Plus there’s other parcel rising below it which are still slightly warmer, pushing the top parcels out of the way…but the only thing they can do is falls down as molecules since there’s enough space between molecules for them to do so, etc.

  29. I mean it’s convection…you have rising bubbles below bubbles which have reached max altitude…if the rising bubbles want to reach the same max altitude, then the air there will have to move out of the way. Where it can move to? Only one direction once it has reached max potential altitude, and is maximally cooled: back down. Of course these parcels move THROUGH each other…the cool parcel moving down through the warm parcel, because there’s enough space between the molecules for them to sieve through each other.

  30. “there’s enough space between the molecules for them to sieve through each other”

    Of course that creates turbulence because they don’t sieve through each other with perfectly zero interaction. Warm rising air…cold falling air…moisture…precipitation…cumulonimbus clouds…thunderstorms…tornadoes…hail…etc!

  31. “Warm rising air…moisture”

    Which is generated by HOT sunshine!!!

    Their approach to understanding the climate is to teach themselves that the Sun does not create the climate…! lol

  32. CD Marshall says:

    Guess I made the physicist mad he is so trying to not drown in his pseudoscience…

    ME “The energy is the same“ (roughly In and out of the system in a 24 hour period)

    No. The energy per time that the surface emits as radiation is roughly 1.65 times higher than the total energy per time absorbed by earth (surface + atmosphere) from the sun. Constantly. Measured fact. Still the earth (surface + atmosphere) emits to space energy per time of an amount very close to 1/1 to solar irradiation.

    No energy redistribution within the system can explain this delta in radiated energy per time from one part of the system to crossing the boundaries of the system. Doing so is thermodynamically forbidden by the 1st law. You are doing it by using evaporation, which is a „within the system only“ process.

    And you further can‘t explain it by any temperature variance, as you could easily calculate. You also can‘t explain it by some „full sphere“ vs. „hemisphere“ vs. „cross section“ or „angle“. That again is easily calculated.

    So, you have to deny the measurements, or acknowledge the measurements and find an explanation compatible with physics.

    You come around with numbers incompatible with evidence combined with an explanation incompatible with physics. In this very case minus by minus doesn‘t give a plus…

  33. “I wrote an equation for this once…”

    Found it:

    “With an atmosphere being sustained, however, it develops an important temperature gradient due to the local gravitational field. In mechanical vertical equilibrium, meaning that there is no bulk movement of the atmosphere either up or down, then for any given horizontal infinitesimal slice of atmosphere there must be conservation of mass for any gas passing through that slice. Why would any gas move through that slice? Simply because of thermal movement on the molecular scale. Some molecules will move up through the slice, and an equal number will move down through the slice when there is mechanical equilibrium. The conservation of mass equation is:

    ρ↑V↑ = ρ↓V↓

    where ρ is the density of the gas and V is its average velocity. The upward and downward arrows indicate upward movement and downward movement through the infinitesimal horizontal slice.

    But in an infinitesimal time period ‘dt’, and if the average velocities are initially ‘V0‘, how does the velocity change for the upward and downward moving molecules given that gravity is acting upon them? If the local gravitational field strength is ‘g’ (positive value), then the upward moving particles will have their velocity reduced by gdt, and the downward moving particles increased by gdt. And so:

    ρ↑(V0 – gdt) = ρ↓(V0 + gdt)

    and as a ratio:

    ρ↑/ρ↓ = (V0 + gdt)/(V0 – gdt)

    The right-hand-side is always larger than one, which then for the left-hand-side means that the density of the gas below the slice must be higher than the density above. And so density must decrease with altitude. At the same time, the particles crossing to below the slice will have a higher average velocity as compared to the ones crossing to above the slice, and so therefore temperature must decrease with altitude since this velocity corresponds to a component of the thermal molecular speed of the gas. An equal number of particles pass either up or down, but the particles moving above are slowed, whereas the particles moving below are hastened. Thus, lower density and lower temperature as altitude increases.”

  34. CD Marshall says:

    Of course the measurement is wrong for they divide the incoming soar irradiance by 4 and spread it as an average over the entire planet removing day and night which I told him time after time and he just ignores it.

  35. “The energy per time that the surface emits as radiation is roughly 1.65 times higher than the total energy per time absorbed by earth”

    The Earth emits over the globe precisely what comes in over the hemisphere. However, the input on the hemisphere is higher intensity than the average output. The surface emits LESS flux than the solar input flux.

    The lapse rate requires that the near-surface air temperature must be warmer than the expected average temperature. This is independent of any RGHE. If a RGHE existed, then the lapse rate would have to be different than what is calculated from a gravity-alone analysis. You would have to include RGHE effects along with gravity to get the lapse rate correct…the lapse rate would have to be much STEEPER than it is. As it is, the lapse rate is what is measured, equaling what is calculated without reference to a RGHE.

    We’re interested in understanding the climate. This requires treating sunshine as it actually exists…not as it doesn’t exist…hello!

  36. CD Marshall says:

    That is really good, take some time to digest it though. So the obvious point is once a parcel is stably buoyant it is still not static and each molecule then has to be addressed for it is still moving.
    The parcel or molecules would have to break LTE to descend, correct?

    I’m curious how a parcel pushes past the tropopause to move upwards. I guess thunderclouds would be the easiest route?

    I was going to re-read your paper, “A Discussion on the Absence of a measurable GHE”.

    You should rewrite that and try and get it published under another title with “Lapse Rate” in it.

  37. Weekly_rise: “To use a stupid analogy, if a guy in front of me hands me a dollar and I hand him one dollar back, no money was exchanged. If we then say that I pass the guy behind me a dollar, we need to explain where that dollar came from.”

    Let’s use the heat flow equation. You HAVE 1 dollar because you’ve been heated up already…as you are handed 1 dollar, you don’t give it all away initially…you keep some back until you’ve reached a holding of 1 dollar.

    Q = input – output

    Your output starts off at 0 because you start with nothing. The input is 1 and so Q is 1, and so you get 1. Now you’re holding 1.

    Now that you’re holding 1, you can output yourself. You can’t lose 1 dollar to the input because every time you try to give the source 1, it gives 1 back. However, you can give 1 to the guy behind you. So you do. Now youre down 1. Now you have nothing to give to anyone, you cant try to give to the source, and you can’t give to the guy behind you.

    But right at that moment, the source gives you 1 right away. You could try giving that 1 to the source, but it will just give it right back to you. So you give the 1 to the guy behind you.

    So in infinitesimal speed, you’re always actually holding 1, you’re trying to give 1 back to source but can’t, you do give 1 to the guy behind you and every moment that you do source gives you 1 replacement to keep you at a holding of 1.

  38. CD Marshall says:

    This guy is hilarious a denial reply for everything:
    “The surface emits LESS flux than the solar input flux.2
    No. That‘s just wrong. By evidence, because measured millionfold. That is the delta I speak of. It emits MORE.
    And it is wrong for one hemisphere as well for the spherical average.
    “The lapse rate requires that the near-surface air temperature must be warmer than the expected average temperature.“

    “Joseph has covered all of this in great detail.“_
    And in great errors, so maybe you should turn to textbooks and lectures than some YT science denier.
    If you can‘t answer this yourself you should maybe study the subject from several independent sources?”

  39. Regis says:

    “ But right at that moment, the source gives you 1 right away. You could try giving that 1 to the source, but it will just give it right back to you. So you give the 1 to the guy behind you.”

    Okay, now I understand. An object indeed can emit thermal radiation toward a hotter source, but the hotter source will just increase its output to cancel it (“it will just give it right back to you”)

    So to start, the source emits F toward the plate, and the plate emits F in both directions. This would obviously leave the plate out of balance. So to compensate, the source starts emitting 2F, and now everything works.

  40. No…the source does not emit 2F. Money is fermionic and energy is bosonic…maybe that’s the problem.

    You don’t lose ANY money to the source. The plate loses no energy towards the source.

  41. “An object indeed can emit thermal radiation toward a hotter source”

    This shouldn’t be a point of confusion. Please refer to the heat flow equation:

    Q = sT_hot^4 – sT_cool^4

    Energy is emitted by the cooler object, but not lost. The cooler object GAINS energy from the hotter source, even though it is emitting toward the hotter source. The energy emitted by the cooler object toward the hotter source does not increase the energy of the hotter source, because Q is from the hotter source toward the cooler object only.

    “the hotter source will just increase its output to cancel it (“it will just give it right back to you”)”

    To give it back to you, it does NOT have to increase its own output…it gives back what was given, and so this doesn’t requires it to give more of itself…it gives back what was given, not more of itself.

  42. Weekly_rise says:

    Now that you’re holding 1, you can output yourself. You can’t lose 1 dollar to the input because every time you try to give the source 1, it gives 1 back. However, you can give 1 to the guy behind you. So you do. Now youre down 1. Now you have nothing to give to anyone, you cant try to give to the source, and you can’t give to the guy behind you.

    But right at that moment, the source gives you 1 right away. You could try giving that 1 to the source, but it will just give it right back to you. So you give the 1 to the guy behind you.

    This doesn’t seem valid for the plate scenario though, because you have no choice but to give 1 to the source (objects above 0 Kelvin are all emitting). That’s where I’m getting hung up. You can’t give 1 behind you and none to the source, it has to be n to the source side and 1-n to the rear side.

  43. Q = sT_hot^4 – sT_cool^4

    Even though a cool object emits toward a hotter source…the sign of Q is negative for this orientation, meaning that Q goes the opposite way, from the hotter source to the cooler object. Since Q is what is needed to increase temperature, then the emission from the cooler body towards the hotter source does not increase the hotter source’s temperature.

    The emission from the cooler source is effectively “negated”.

    The cooler emits to the other side of the source.

  44. Well it’s likely not a good approach to use fermionic thinking to explain bosonic behavior. It’s actually a pretty fundamental mistake, which traps everyone who attempts it. We simply cannot use fermionic-world analogies for bosonic-world processes.

    So, luckily we have the math solved and so let’s refer to the heat flow equation for plane-parallel, and understand it in terms of its actual math, rather than in fundamentally incompatible analogies:

    Q = sT_hot^4 – sT_cool^4

    And the First Law:

    dU = Q = m Cp dT

    Equilibrium is dT = 0 therefore Q = 0 therefore T_hot = T_cool at equilibrium.

    The source supplies F, and that F shows up on the other side of the plate, therefore energy is conserved. Trivial, when you simply use the math, rather than fundamentally incorrect and incommensurate word-analogies.

    So there’s the result. You see the math. You see how energy is conserved. That’s it. Stop there. All the answers are there and there’s no more math to do.

    But then people invent a problem: let’s put this result, which is complete, final, consistent within itself, and finished…but into alternative contexts and word-analogies where we are fundamentally dealing with non-thermodynamic non-bosonic physics where the thermodynamic math would never have applied in the first place. And wow, oh look, now it’s confusing, therefore there must be something wrong with the self-consistent and complete solution arrived at in the original self-consistent context.

    You see how that works? Please stop doing that.

    Just please stop…it’s ridiculous. You have a self-consistent answer, using the math which applies to the problem. That should be it. Don’t do anything else. But then reinterpret the answer into terms to which the math doesn’t apply in the first place and which isn’t thermodynamics anymore? Of course you create problems doing that. The problem itself is invalid, because it’s a non-problem.

    Q = sT_hot^4 – sT_cool^4
    dU = Q = m Cp dT
    Equilibrium = dT = 0 = Q = 0 therefore T_hot = T_cool = T.
    Source = sT^4 = input
    plate = sT^4 = output to space

    A complete and total solution, with all requirements and equilibrium and conservation of energy satisfied given the actual definitions of those. There’s nothing else to do with it…nowhere else to go. Appreciate it. Look at it. Satisfy yourself of it within itself and its own terms. Acknowledge it!

    But then put the solution into terms in which the thermodynamic math would have never applied in the first place? STOP DOING THAT!

  45. @Regis…you need to do the math, and understand the math. Stop thinking about it the way you want to think about it, and think about it the way the math is telling you. The actual math…the thermodynamic math…not fermionic bank-account math.

  46. Kev-In-ZA says:

    Thank @Joe, Great explanation post on the green plate confusion. This should clear it up for all be the willfully obstinate or blind. But sadly many scientists are just to “clever” for their own good and think all S-B emission can simply be added up like an accounting ledger and converted into temperatures using the inversion of S-B law.

    Your explanation of the lapse rate concept using the molecules crossing a horizontal slice is very neat, and I have used a similar idea to try explain the idea to others but I focused on the trade-off of KE to PE up gravity gradient and reverse down gravity gradient. But like your horizontal slice description for explaining to others.

    Also good point you make that a strong “Back Radiation” RGHE should steepen the lapse rate above the theoretical dry lapse of -9.8K/km. But yet this is never observed. I further find it very perplexing that some people seem to think the atm vertical column would become isothermal if not for forced convection to “activate” the profile; and/or a RGHE to somehow “create” the lapse profile; while KE+PE=const is just such an obvious explanation to any engineer/scientist.

  47. Here’s a neat thing:

    Following on the point about thinking about the problem in terms of the math, rather than in terms of fermionic human experience…think of the problem from the perspective of light, from the mathematics of light.

    How does light experience the universe? How does light experience “the gap” it must “traverse” between origin and destination?

    Do the math!

    Apply the equations of relativity to light: for light, all spatial distance along the vector of travel is infinitely contracted to zero – at light speed, space contraction is INFINITE! Thus, light experiences no gap between source and destination. What about time? At light speed, time dilation is also INFINITE, thus, there is no time for light. For light, there is neither spatial depth nor temporal duration between origin and destination. Do humans have much experience thinking in these terms? No, not at all. But we can appreciate what the math says, and imagine that. Just like we have to do with thermodynamics.

    So we get this question all the time: “How do photons KNOW not to travel to the warmer source and heat it up?”

    Well…do the math. Literally, the math of photons literally says that photons are in contact with their destination at the moment of their creation. They know because they see their destination…they are touching their destination, resonating with whatever’s there, etc. Photons “know” because they’re there. You can’t put that in human terms, you can’t analogize that in human experience. But it’s what the math of relativity for photons says.

    We humans experience going from one place to another in a duration and via some effort. For photons, they effortlessly exist at their destination instantaneously. That’s amazing. You can understand it only if you understand and acknowledge the math…there’s no way, no way at all, to put that mathematical experience into human experience terms and analogies. WE think of photons travelling from one place to another…but that’s not what photons “think” – it’s not what they experience at all.

    Q: “How do photons KNOW not to travel to the warmer source and heat it up?”

    A: Because photons are already at their destination. The information about the destination is known for the photon.

  48. “some people seem to think the atm vertical column would become isothermal if not for forced convection to “activate” the profile; and/or a RGHE to somehow “create” the lapse profile; while KE+PE=const is just such an obvious explanation to any engineer/scientist”

    Exactly! The lapse rate develops at the molecular scale. You understand of course what I mean by “mechanical equilibrium”, but others may not: it means that the atmosphere itself is not moving. Mechanical movement is like a brick, moving, from one place to another.

    The atmosphere is in place…static as whole. However at the molecular scale there is movement, thermal motion of the molecules. If the atmosphere isn’t moving in bulk then there is conservation of mass for any given slice of atmosphere. If there wasn’t conservation of mass then the atmosphere as a whole would be moving through that slice. In general the atmosphere is fixed to the Earth and not moving away from or toward the surface in bulk. On average we do have conservation of mass for any given slice of atmosphere.

    So then the conservation of mass equation shows what happens to the molecules at the infinitesimal scale: they accelerate downward and decelerate upward, thus the average speed must increase downward and decrease upward. Thus: the adiabatic gradient.

    Yes it is a good point – the atmosphere cannot be isothermal in a gravitational field. Convection and what-not simply responds in bulk the exact same way that the molecules respond in infinitesimal…because of course the bulk is MADE UP OF what exists at and is occurring at the infinitesimal.

    You get convection obeying the lapse rate because the molecules of the gas are obeying the lapse rate!

  49. Rosco says:

    Anybody that equates a continually convecting atmosphere to a solid plate in an inane attempt to prove a point is simply an – well I won’t insult idiots !

  50. CD Marshall says:

    Physicist again:
    You are confusing flux and energy again. 910-980 W/m^2 is a flux, not the “full energy“. If you speak of input and output, you can use a flux, or at least a power, thus energy per time.

    1) The full energy per second (=power) absorbed by the system earth is ≈ 1.22410^17 J/s.
    2) The full energy per second emitted by the surface upwards is ≈2.020
    10^17 J/s.
    3) The full energy emitted by earth to space is ≈1.224*10^17 J/s.
    The first and third value are equal to very close approximation. The second value, power emitted by the surface upwards, is ≈1.65 times MORE than the balancing value with the total input of the sun that earth emits to space.

    No matter how you distribute the fluxes of IN and OUT and no matter how the water cycle redistributes energy within the system, this DELTA in the radiated power from surface to balancing IN/OUT over the boundaries of the system is there. Measurably there. It is an observation. Physics is made to explain observations.

    The question is, how physics explains this, for it would violate the 1st law with no further explanation.

  51. Regis says:

    Joe, after thinking more, I’m convinced you’re over complicating this. The heat transfer equation (Q=s Th^4 – s T”^4) is nothing more than the difference of Stefan-Boltzmann fluxes. No need to consider whether photons knowing where they’re going, or relativity, or fermionic vs bosonic statistics. All of the relevant physics is contained in the Stefan-Boltzmann law already. These topics you’re bringing up are just confusing the very simple physics at play.

    Objects always emit the same way — according to the SB law. They don’t change their emission depending on what other objects are around.

    So just add up the Stefan-Boltzmann fluxes. The plate is receiving F from the source. The Stefan-Boltzmann law says the plate emits the same flux G=s Tl^4 from each side. dU must be 0, so dU=0=F-G-G , so G=F/2 . It’s that easy.

  52. Joseph E Postma says:

    Just refer to this again CD:

    “The amount of energy emitted by the surface upwards as radiation exceeds the amount of energy emitted by earth to space. Every second. Always.
    Can you explain this simple fact of measurement with the physics you are showing here?”

    Yes. The gas is removing energy from the surface via convection, and the gas has low emissivity, and the lapse rate requires that the near-surface gas be higher in temperature than the expected average. So…not a problem!

    Now can he answer a question:

    Is there any validity whatsoever to thinking of the Earth as a flat plane with no day and night and with a solar input which cannot create the climate?

    To his question: we can address that in the details once we get things right to start with.

    To our question: the only answer is that it fatally undermines the entire current scientific approach to the question of the climate, thus undermining his own question as well.

    So who’s in a better position? We are, clearly!

    He’s not starting from a place where he can even be expected to formulate a correct answer! We are, however!

    His approach, his starting point to address this problem, requires that the colder atmosphere can heat a warmer surface. That’s an error. It’s an error because his approach is erroneous in the first place. Solar power is not -18C, and is not lower in flux than the output of Earth either at the surface or at the TOA.

    We DO NOT know that that energy is actually what is being emitted. This is an air temperature and the air has low emissivity…that power IS NOT measured anywhere…he is simply assuming it or inferring it based on UNIT = 1 emissivity for air, which air DOES NOT HAVE! Air at 15 degrees average actually has very low emissivity…in fact you can calculate based on his “inferred” values of emission and determine what the emissivity of air is, and it is a totally expected value. His total quantity of energy is NOT measured…he is only inferring it assuming unit emissivity for air, but air has much less than unit emissivity. Thus, it is easy to explain what he THINKS he’s seeing with very basic and simple physics. No problem!

    So add that in to the answers and position above: the air has low emissivity and he is wrong to assume it emits with unit emissivity.

  53. mtntim says:

    Here is a way to see Q is NOT necessarily 0 between any two objects in steady state. Take two blocks of copper and connect them with a wire. Supply a constant heating power S to block 1 (e.g. with a resistor), and remove a constant heat flow S from block 2 (e.g. with a thermoelectric cooler).

    Eventually each block will reach a steady state temperature. But is the heat flow between them zero? No! Obviously heat will flow between them through the wire — the amount of heat is exactly S.

  54. CD Marshall says:

    I’ve been dealing with this guy for a week. He is very specific. So he will query this, “Yes. The gas is removing energy from the surface via convection, and the gas has low emissivity, and the lapse rate requires that the near-surface gas be higher in temperature than the expected average.”

    He will argue energy is energy, so are you referring to thermal energy, or temperature? He is a physicist so he grills me over exact terms and specification.

    Which I find strange, since he is claiming E is coming out of nowhere and is more than the Earth is providing or the Sun which violates all physics. It is a strange thing, like aliens kidnapped them, reprogrammed them and they are not able to see the error.

    It’s like Dark City where they never question why it’s never daylight or why they neve leave that part of town or remember yesterday. Yet they are fine with it NOT making sense.

  55. boomie789 says:

    This guy explains how the “greenhouse effect” doesn’t actually exist and is derived from a misunderstanding of the important differences between energy and heat. He still claims that the earth has warmed by a degree though and this is due to depleting the ozone layer?

    “Major periods of rapid warming throughout Earth history are contemporaneous with major basaltic lava flows covering areas of hundreds to millions of square kilometers. The larger the flows, the greater the warming.”

    “Major explosive volcanic eruptions eject water vapor and sulfur dioxide into the lower stratosphere where they form aerosols or mists that reflect and scatter sunlight. Each major eruption causes cooling of approximately one-half degree Celsius, 0.9 degrees Fahrenheit, for two to four years. This short-term global cooling lowers ocean temperatures for up to a century. This is why several major explosive eruptions per century, continuing over tens of thousands of years, are observed to cool Earth incrementally down into ice-age conditions.”

    Ever heard of this guy?

  56. CD Marshall says:

    He is actually covering the heat part quite well.

  57. CD Marshall says:

    More from the that guy…
    “The flux is just a measure of energy per time averaged by math over the area. If you measure real fluxes, you‘ll find any values regarding on if you are in antarctica or sahara, to take the extremes.
    Since conservation of energy MUST be true, the total power emitted can be compared with the total power absorbed. Both need to be equal in a perfect steady state. And they approximately are for our system, which is the earth. No wonder, for this is claimed by the 2nd law.
    What we observe by measurements is, that the total power emitted by the surface of earth is ≈1.65 times the power earth „exports“ to space.

    This is completely regardless of how the fluxes are distributed. You are right that they could be distributed any way, point, hemisphere, whole sphere, something in between… So what?

    How can you explain this? THIS is the question.”

    At this point I’m not even sure what his question is anymore?

  58. Kev-In-ZA says:

    @CD, when talking quasi-steady state, Energy In must equal Energy Out on any control volume. That would apply firstly to the entire earth system at TOA (numerically around 240W/m2), but also to the surface (numerically around 165-170W/m2). If you focus on uni-directional Heat Flow it greatly simplifies thinking in many ways. Heat flow is uni-directional Hot to Cold only, and should balance with Net Energy Flow/Flux.

    So anyone claiming the Surface is lossing 1.65x what is lost by the earth system is implying the surface is undergoing rapid cooling. But saying the surface is emitting at 1.65x is nothing special, it just means the temperature is higher at the surface, as emission is not a Heat flux nor a Net Energy flux. I don’t have an accurate value, but in reality the surface only emits by radiative Heat-transfer around 50-70W/m2. Certainly nowhere near the full SB BB potential of an uninsulated surface at 288K of about 390W/m2.

    Btw, where is this happening..? Should I chip in…..

  59. CD Marshall says:

    Be my guest, he’s out out of my league but I did my best…

    Look under:

    CD Marshall
    3 weeks ago (edited)
    @Chris A.
    Found a reply I copied from Dr. Holmes when chaps on Joseph’s site had a few questions for him.

  60. @Regis (esttom) 2021/05/21 at 8:59 am

    “you’re over complicating this”
    “The heat transfer equation (Q=s Th^4 – s T”^4) is nothing more than the difference of Stefan-Boltzmann fluxes.”

    You don’t say, Sherlock?! lol

    “No need to consider whether photons knowing where they’re going, or relativity, or fermionic vs bosonic statistics.”

    Really? No need to discuss the physics of the situation and how the physics matters and changes and depends upon what is capable of occurring? Oh ok…that’s brilliant. Yes let’s have no discussion of the things that matter or are relevant…that’s a great way to do science. Hey! Just like how you start the understanding of the climate with flat Earth theory and cold sunshine because why not, right? Brilliant.

    “All of the relevant physics is contained in the Stefan-Boltzmann law already.”

    And as applied via the heat flow equation, which you now acknowledged, thank you!

    “These topics you’re bringing up are just confusing the very simple physics at play.”

    The topics relevant to the situation, to the physics? As interesting points of discussion since they’re relevant to the questions we’ve been asked before? No I think that in fact we should discuss those things, just like we discuss the relevance of the Earth being round and sunshine being which you prefer to also not talk about.

    “Objects always emit the same way — according to the SB law. They don’t change their emission depending on what other objects are around.”

    No one said that they did. But we do refer to the heat flow equation and the First Law of Thermodynamics. The heat flow equation contains the emission of the objects, and shows what part of the emission can act as heat.

    “So just add up the Stefan-Boltzmann fluxes.”

    You just acknowledged the heat flow equation in your second sentence: “The heat transfer equation […] is nothing more than the difference of Stefan-Boltzmann fluxes.” So why now are you ignoring that and asking to sum fluxes when you just acknowledged the difference of fluxes being heat? Your position is inconsistent. I just asked you to STOP CHANGING THE ANALYSIS to one where the heat flow equation isn’t used. STOP THAT!

    “The plate is receiving F from the source.”

    Yes, and it cannot lose energy on the hemisphere of the source, hence it loses F on the hemisphere opposite the source.

    “The Stefan-Boltzmann law says the plate emits the same flux G=s Tl^4 from each side.”

    You’re leaving out the heat flow equation again.

    “dU must be 0, so dU=0=F-G-G , so G=F/2 . It’s that easy.”

    I asked you to please stop using math which isn’t the correct math. You acknowledged the heat flow equation in your second sentence and now you ignore it. Stop being ridiculous. dU must be 0 THROUGH Q, through the heat flow equation.

    dU must be 0, so dU=0=Q=F-G, so F = G or T1 = T2. It’s that easy.

  61. boomie789 says:

    Get down to pre-industrial levels? We don’t want to get down to those levels! It’s bad for life on earth! Life on earth evolved with much higher concentration of CO2 in the atmosphere. We are actually quite low on the amount needed for life on earth to thrive.

    Down to those levels? O wait I forgot, these people don’t care about the truth.

  62. boomie…they want to end life on Earth.

  63. @mtntim 2021/05/21 at 9:14 am

    That’s not the same situation where only Q exists…the cooler device means that work is also being performed with power supplied from outside the system. So you have dU = Q + W = 0 where then it must be that Q = -W. We’re discussing the RGHE which exists with no work present, thus the equation is dU = Q = 0.

    Again: PLEASE STOP changing the conditions to entirely different scenarios. It’s ridiculous.

  64. @Regis, @mtntim, @esttom, @RT, @etc.

    Please try changing your IP address a little more carefully between accounts.

  65. Kev-In-ZA says:

    @CD, found it. But crikey, 255 sub-comments already. Will see how I get along reading all there… 🙂

  66. CD Marshall says:

    Yeah apparently I’ve been at it with this guy for 3 weeks.

  67. Regis says:

    @Joseph hmm maybe you’re right on your latest post. I will crack open my heat transfer textbook and check some more. Always an interesting subject.

  68. Rabbit is the featured meat in the Green Plate Special, served up nice and roasted like this:

  69. tom0mason says:

    boomie789 at 2021/05/21 at 2:47 PM
    Now your speaking my language! Regardless of the arguments of RGHE and CO2, the graphic you show spells out that when the global temperature is low life struggles as it does when atmospheric CO2 is low. When both are low, life heads towards an exit door.
    When the planet is warmer with lots of atmospheric CO2 (well above 210ppm and up to 2000ppm) all life benefits.
    Bottom line — A warmer planet with plenty of atmospheric CO2 equals proliferation of life.

  70. CD Marshall says:

    I worked at Burger King when they down sized the Whopper about the same time they downsized the Quarter Pounder. The trick with the Quarter pounder was 1/4 after it was cooked to a 1/4 before it was cooked, so still a 1/4 pound…technically.

    You can apply that to climate science easy enough.

    The exhaust is a quarter power of the total solar irradiance, but now it’s the quarter of the total solar irradiance.

  71. CD Marshall says:

    Or rather the total solar irradiance is now a quarter.

  72. CD Marshall says:

    Peter L. Ward
    1 year ago
    “This video applies to temperature in matter and to matter emitting and absorbing radiation. In this case, radiant energy (E) equals the Planck constant (h) times frequency (v). So thermal energy is simply frequency times a scaling constant. By Planck’s law, we can see that heat flows by simultaneous resonance at each and every frequency. What changes in resonance is the intensity or amplitude of oscillation. Resonance causes averaging, so the greater the difference in amplitude of oscillation, the greater the flow of each component of heat per second.
    This is why plots of cooling or warming of matter are asymptotic, which means they approach their final value as slower and slower rates.

    Temperature in a gas, on the other hand, is proportional to the average kinetic energy of translation of all the atoms and molecules making up the gas. Kinetic energy is proportional to velocity squared. Colder air contains less kinetic energy, lower mean velocity of translation. We observe that air cools with increasing altitude at the adiabatic lapse rate. The hotter, most energetic air rises into the cooler, less energetic air.”

  73. Kev-In-ZA says:

    @CD, wow, a long read but some interesting talking past each other, me thinks. You make some interesting points, but Chris A. seems to be mostly on the mark as I understand matters in the climate system.

    While I finds the foundational premise of Climate Alarmist to be really badly flawed and gibberish, as with most things, there are elements of truth mixed within that gibberish. And Joe is correct to call out some of that gibberish and pseudoscience for what it is.

    Chris A.’s valid logic questioning is insightful:
    …..{{Chris A. Quote: What physics say is
    a) that the amount of energy per time radiated to space by earth must match the amount of energy per the same time irradiated on earth by the sun.
    b) that the surface of earth has a temperature of +15°C on average
    c) that the energy per time radiated by such a surface at such a temperature exceeds the the solar energy per same time absorbed by earth by far
    d) that c) is calculated by the inequation average of (sum T^4) > average of (sum T)^4 which gives that the temperature at which according to S-B-law the surface of earth radiates the same energy per time than the sun radiates into the system per same time is exceeded by at least 33K.
    e) that the balance claimed by a) is maintained by the fact that not all of the surface radiation evades to space but partly the atmosphere radiates to space from altitudes being much colder than the surface}}
    So the problem is actually that the earth (per surface temperature and emission in the complete absence of IR-active gases and clouds in the atmosphere) emits too much theoretically.

    The above anomaly in surface temperature is largely true and necessitates explanation and detail. Climate Alarmist invoke what they call a GHE using backradiation to seemingly “heat the surface” to explain and quantify this. This particular explanation is pseudoscience, because the atmosphere cannot “Heat” the surface (hot to cold only). However, with a 2 body problem, the 2nd body can insulate the 1st body leading to a different equilibrium temperature. Hence I prefer to explain this anomaly due to Radiative Insulation together with a board Atmospheric Effect (hence avoiding the term GHE which is very misleading). The narrative explanation that Chris A. offers and quoted below is, IMHO, largely correct.
    …….{{Chris A. Quote: 1) A known and fixed quantity of energy enters the system, and it must all leave by radiation to outer space. The two must balance on average, to a very close approximation.
    2) The temperature of the radiating surface adjusts to emit precisely this amount of energy.
    3) Because of GHGs in the atmosphere, this radiating surface is [sic…not] the solid ground, but a fuzzy layer that averages 4-6 km up in the air.
    4) The layer 4-6 km up adjusts in temperature to emit precisely the required amount of radiation to space.
    5) The temperature of the rest of the atmosphere, both above and below, is locked in a fixed relationship to that of this middle layer by convective coupling. As air mixes vertically, the change in pressure with height compresses and expands the air, changing its temperature. The rate of change of temperature with altitude is a fixed constant (called the adiabatic lapse rate), that does not involve radiation in setting its value.
    6) The temperature of the surface depends only on the altitude at which the effective radiative temperature is achieved (i.e. 4-6 km) and the lapse rate between there and the surface. The surface temperature is T_eff + LR*h.
    7) As more GHGs are added to the atmosphere, the opacity of the atmosphere at LW increases and the radiating surface gets higher. As h increases, so does surface temperature. Global warming.}}

    The emission to space explanation is actually a lot more complicated than this narrative, because there is some Planck emission direct from the surface to space, plus there is some Planck emission direct from cloud-tops to space, and then there is also band emission from atmospheric H2O to space from various levels in the atmosphere depending on H2O presence, and also the other so-called GHG (mostly CO2) band emission from atmospheric components to space from the high troposphere varying according to CO2 concentration. And then on top of this you have feedback (some positive and mostly strong negative stabilizing factors) acting to adjust emission quantities according to local conditions to find a quasi-dynamic equilibrium state over time.

    Hence, imho so-called GHGs do significantly effect the earth surface temperature compared to: no atmosphere; as well as an atmosphere with no GHGs in it (eg only N2, O2 and Ar). The hairy gorilla is H2O in all its forms, while CO2 to a point is important. But by >100ppm, most of CO2’s heavy lifting has been done and the Radiative Insulation effect becomes largely saturated and of little cause for concern specifically that Catastrophic AGW is ahead.

    The so-called GHG’s are thus actually suppressing the emission to space (aka Radiative Insulation), but also providing the very mechanism for the atmospheric components to emit to space albeit at lower temperatures (following the lapse rate profile) at levels when the Radiative Insulation allows that emission. And this latter effect together with convention and latent heat and buffering and clouds participates in keeping surface temperatures a bit more stable and warmest at the bottom of the atmospheric column, aka an Atmospheric Effect.

    Hope this is of some use.

  74. CD Marshall says:

    Very helpful, thank you. I simply wasn’t getting some parts of his explanations.

  75. CD Marshall says:

    @Kev-In-ZA Did you mention any of this to him?

  76. Kev-In-ZA says:

    @CD, no, I chose not to engage. Nothing worth calling him out on so far that I could see. And I don’t know his views on CAGW, and that would be main difference of outlook superficially. I’m borderline “lukewarmer” at the moment.

  77. Ross Handsaker says:

    Point 7 above. “As more GHGs are added to the atmosphere, the opacity of the atmosphere at LW increases and the radiating surface gets higher. As h increases, so does surface temperature. Global warming.”
    If the radiating surface gets higher doesn’t this mean the atmosphere has increased in size, which would have a compensating cooling effect (the energy is warming a larger area).
    The radiative gases such as CO2 are good absorbers of energy and therefore also good emitters (to radiate means to emit). If CO2 was a good thermal insulator it would be used in window treatments rather than argon, or the much more expensive krypton or xenon.

  78. Kev-In-ZA says:

    @Ross H, my understanding (along with many other details) doesn’t extend to being able to answer that question without reference to others. My “other/s” for CO2 and Radiative Insulation at the moment is Howard Hayden who summarised the impact of this phenomenon also drawing on the recent work of Will Happer and Willian van Wijngaarden.
    This is a great explanation of the Radiative Insulation climate effect and impact (all else held constant)

    Click to access co2_clim.pdf

    (and some other backup)

  79. Kev-In-ZA says:

    @Ross H, sorry, seems I mis-read your questions in haste, and should be able to offer sensible answers.

    {{If the radiating surface gets higher doesn’t this mean the atmosphere has increased in size, ..}} The claim is not that the physical atmosphere changes in size, but rather that a different layer (higher up in the case of CO2 concentration increase because the IR opacity of the CO2 in those active bands has increased) will be doing the band emission to space for the CO2 active bands. Because of lapse rate, this high level will be colder and hence the relative emission will be less. So temperature of everything has to up-tick a little to get back into balance with energy In = energy out. But the total emission change due to this one part of total emission to space is actually quite small.

    {{The radiative gases such as CO2 are good absorbers of energy and therefore also good emitters (to radiate means to emit). If CO2 was a good thermal insulator it would be used in window treatments rather than argon, or the much more expensive krypton or xenon.}} The heat-transfer (HT) dynamics of double glazing are very different to the atmospheric emission scenario. I might have some detail incorrect, but double glazing is mainly about conduction-convection modification with minimal radiation HT effect of the gas at that scale. Best would be a vacuum gap, but that is a practical nightmare. The small air-gap provides a significant convection suppression and different gasses can be used to reduce conduction. All else equal, you then require a gas with low thermal conductivity and specific heat-capacity to further minimize conduction-convection. And Argon beats CO2 hands down and N2 in turn. So Ar about halves conduction-convection over air. Krypton/Xenon better still, but much more expensive.

  80. CD Marshall says:

    “A naked rock, the size of Earth and ~150 million kM from the Sun would have temperatures like the Earth’s Moon, ~95K-390K, avg 270K from NASA information…

    Kinda strange, considering all the hooraw about CO2 and AGW hypothesis, don’t you think?

    Not really, when you understand that the Moon has no appreciable atmosphere, with which to distribute the Sun’s warmth, and 95–390 is a range, with no basis for averaging the cold and hot conditions recorded.

    Moving on, there is no “cover” of carbon dioxide for Earth. CO2 is a component of the atmosphere here, not a blanket, and does little or nothing to retain the Sun’s energy.

    The reason Earth is more amenable to life is that we’re much larger and have enough gravity to hold on to our atmosphere, and the reason Earth is about 256K on average, is that we (ie, Earth) have water, and the energy of the Sun is necessary and sufficient to evaporate some each day and maintain a level of about 8000 ppm water vapor.

    It is not just the vapor that make the average so nice for us. Radiative transfer is only a small fraction of the distribution process and such transfer by carbon dioxide is only a fraction of that. Look at Mars, with only some 200 ppmv water and no addition to the “global warming” there, despite its 950,000 ppmv CO2 in its atmosphere.

    Despite the worry of the AGW proponents, the carbon dioxide level of Earth has little to do with the average temperature here. The energy of the Sun is received and distributed by the total composition of the atmosphere, by its reception by water and the conversion to vapor from 70+% of the surface during the cyclic process we know as the hydrologic water cycle.

    The Earth, with no atmosphere, would be like the Moon. With an atmosphere, perhaps some 4K warmer than the background…the Earth itself would heat the air above it by radiation and convection, even conduction.

    But we’re warmer than bb temperatures not only by radiation (a minor effect), or conduction (another lesser effect), but by convection the principal distribution and handling of the insolation from our nearby star (Sol), a consequence of both an atmosphere and the presence of liquid water, evaporable under the power of the Sun.

    No atmosphere: 256K, with exception based on Moon data

    Atmosphere: 260K, my estimate, or Moon + 4K

    Atmosphere over water: 288K, the reality

    Additional global warming from CO2: ~0.07K at 400 ppmv” -William Hoffman, Organic Chemistry. /Synthetic Organic Chemistry

  81. CD Marshall says:

    Although ghgs aren’t increasing surface temps, to do that you would need to increase emissivity of the surface which is not happening. At best non water vapor ghgs could do is set a limiter on temperature, much like a thermostat and I don’t see it ever contributing to warming other than molar mass.

    Bosonic energy does not add up like fermions.

  82. boomie789 says:

    I think this guy scored the highest on an IQ test
    Christopher Michael Langan (born c. 1952) is an American whose IQ was reported to be between 195 and 210,[1] although IQ tests are unreliable at such high levels.[2][3][4] He has been described as “the smartest man in America” as well as “the smartest man in the world” by the media.[5] Langan has developed a “theory of the relationship between mind and reality” which he calls the “Cognitive-Theoretic Model of the Universe” (CTMU).[6][7][8] His work is closer to the Spinozist God and Spinozism than most other systems. His work is analytical moral epistemology in the vein of Nicholas Rescher and Bertrand Russell.

  83. CD Marshall says:

    Anyone named Christopher is at least near genius level, just say’n.

  84. boomie789 says:

  85. CD Marshall says:

    Ghgs aren’t increasing surface temps,“

    I have simplified the case by just looking at the energy flows. The question still stands, for you have produced many words with no context to the question: HOW does the surface of earth radiates more energy per time constantly than earth in total radiates to space. That is an empirical observation. Physics is there to explain observations. How does your physic explain this observation?

    “480/2 in 240/4 out. That is correct.“

    480/2 = 240
    240/4 = 60
    You are telling earth takes in 240 W/m^2 constantly and gives out 60 W/m^2 constantly? Really?

    “Fine. Prove it.“

    Prove what? That the lapse rate is included in all models? No problem: Just look at Schwarzschild’s equation of radiative transfer which is base for all models. It includes dT/dh essentially. How to use and solve it you find at much places, but one benchmark is Ramanathan and Coackley 1987: Climate modeling through radiative convective models It is freely available.
    You numbers are of course not included, for they are simply wrong. Right numbers:
    a) absorbed by earth: 480 W/m^2 for one hemisphere
    b) emitted by earth: 240 W/m^2 for both hemispheres
    c) emitted by surface upwards 395 W/m^2 for both hemispheres
    My question is, how your physics explain the delta between b) and c). I’m still waiting.

    ME: I have no idea how to answer that. Anyone?

  86. boomie789 says:

  87. J Cuttance says:

    Boomie I literally L’edOL and got a please explain from my workmates

  88. boomie789 says:

    Throughout the pandemic, CT values in excess of 35 have been the norm, with labs around the world going into the 40s.

    Essentially labs were running as many cycles as necessary to achieve a positive result, despite experts warning that this was pointless (even Fauci himself said anything over 35 cycles is meaningless).

    But NOW, and only for fully vaccinated people, the CDC will only accept samples achieved from 28 cycles or fewer. That can only be a deliberate decision in order to decrease the number of “breakthrough infections” being officially recorded.

    Secondly, asymptomatic or mild infections will no longer be recorded as “covid cases”.

    That’s right. Even if a sample collected at the low CT value of 28 can be sequenced into the virus alleged to cause Covid19, the CDC will no longer be keeping records of breakthrough infections that don’t result in hospitalisation or death.

    From their website:

    As of May 1, 2021, CDC transitioned from monitoring all reported vaccine breakthrough cases to focus on identifying and investigating only hospitalized or fatal cases due to any cause. This shift will help maximize the quality of the data collected on cases of greatest clinical and public health importance. Previous case counts, which were last updated on April 26, 2021, are available for reference only and will not be updated moving forward.

    Just like that, being asymptomatic – or having only minor symptoms – will no longer count as a “Covid case” but only if you’ve been vaccinated.

  89. tom0mason says:

    The question that can not be asked too many times …

  90. justgivemeall says:

    Talk about no idea what they are doing beside spend money

  91. CD Marshall says:

    More from Chris A.
    “You can‘t answer the question not because of your level of radiation education. It is extremely basic physics of radiation needed here. There is a conceptual error that ypu have been infected with by reading J. Postma all the time.
    The question is posed in clarity: How can the surface of earth emit more energy per time than the earth in total emits to space?

    The answer is extremely simple, yet can never be found without considering that backradiation from the atmosphere to the surface adds to the internal energy of that surface. This consideration is forbidden by the logic you presented so far. But it is necessary in physics. That‘s all there is to explain this empirical observation.”

  92. J Cuttance says:

    CD your Chris A ticks all the boxes including condescension, incoherence and an unshakable faith in a cult. He’s a drainer.

  93. Kev-In-ZA says:

    @CD, to be honest, you’re a little in a “gotcha” situation with little room to manoeuvre. Chris A. is largely correct in his logic. (Note, I haven’t been back to the long YT thread, but relying on parroted quotes.) Some advice/comment.

    Perhaps the only logic I could call him out on given the quoted statement; is that “backradiation” no more adds to the “surface internal energy” than throttling the drain-plug in a filling bath of water increases the water level. The water to increase the level in the bath coming from the tap, not the throttling the drain-plug. But throttling the drain-plug causes the steady equilibrium level in the bath to increase. Insulation vs Heat-transfer. A subtle but important difference in the structure of the physics logic. As far as the Radiative Heat-Transfer equation Q=f.sigma(T1^4-T2^4) is concerned, the full T1^4 Internal Energy never left the Surface, only the T1^4-T2^4 part. Perhaps at a QM level, T1^4 did but instantaneously T2^4 was replaced at the speed of light.

    Perhap you can make something of that angle.

  94. Philip Mulholland says:
  95. J Cuttance says:

    Kev, hence the urgency to prove the greenplate theoty which turns out to be a greenplate fairy.

  96. justgivemeall says:

    When did global warming go from back radiation to adding insulation slowing cooling,they are two very different things. Which are they now trying to prove? I see more and more of the slowed cooling effect as if they have given up on the unprovable back radiation. If they have that’s one step forward but if they are now claiming something totally different they should state this is their argument and admit that back radiation doesn’t exist. I won’t hold my breath

  97. MP says:

    @ justgivemeall

    Their hypothetical claim that slower cooling by back-radiation has the exact same effect on the equilibrium temperature as hypothetical heating makes it even more bizarre

    They basically claim that you can stop the cooling of a hot lightbulb that is turned off, by shining the same w/m2 amount of much lower intensity light on it as the w/m2 heat loss from normal cooling

    If that was true there would be an experiment that proves it, and a nobel prize for the inventor of the experiment, since it would prove the latest and final back radiation theory..

  98. CD Marshall says:

    More from Chris A…
    “And all I ask is, how the radiation of the surface of earth can emit much more energy per time than a blackbody at 254K would (which is the temperature possible to reach by absorption of the flux the sun provides only).

    The answer is easy: Because it absorbs additional backradiation from the atmosphere, which is exclusively there due to the GHGs.

    The lapse rate cannot be the cause, because it is adiabatic and does not provide additional energy. Backradiation does. The lapse rate is a consequence of that: Energy in at the surface, energy out via atmospheric radiation to space from certain altitudes up. No GHGs, no overall lapse rate.
    The science of these very basics is settled. There are things unsettled with AGW, this is true. However, the basics of GHE are settled for over 100 years now.

    There is no other explanation up to date that doesn’t violate basic laws of physics.”

    So far he has remained unchallenged. Joseph stopped commenting to him after two or three posts. I don’t have anywhere near the physics education to challenge an actual physicist which he does appear to be one for his knowledge is far above that of an average person educated in science. My biggest concern on nailing this guy is he is teaching physics and thus teaching students this and probably using my posts as an example of wrong physics. As long as he remains unchallenged he is a hero in his student’s eyes and backradiation remains a physics fact.

  99. Joseph E Postma says:

    Backradiation of course does not provide new energy. If it can, then so can the adiabatic gradient.

    I already answered all this anyway last time I weighed in. The atmosphere has low emissivity. QED. The surface temperature is NOT measured, nor is its radiated output…these are only inferred from the air temperature…and air has low emissivity so can hold that temperature no problem without need of “extra energy”.

    Besides: is there a SINGLE laboratory experiment confirming that backradiation can have this effect? That should be simple…just a simple laboratory demonstration of backradiation heating, to confirm his interpretation where heat can flow from cold to hot.

    There is no such demonstration, and practical thermodynamic engineering, as in my new post, shows that temperatures DO NOT ADD, that backradiation DOES NOT ADD to a warmer surface.

    Thus, his interpretation is flawed. By the facts. The facts prove that his interpretation is flawed.

  100. Great capture, boomie.

    It is true that all that the elite are doing is dysgenics…people call them eugenicists, but they’re not – they’re dysgenicists. They don’t even seem to be eugenicists for themselves either…really. They seem to just be here as some alien force intent on destroying the planet, and that’s it…they’re just destroying everyone and everything and will be happy to see themselves destroyed at the end of it all too.

    The only thing I can hope in, now, is the realization that everyone who is zealous about mask-wearing and getting/enforcing the vax, are likewise those who believe in climate change, in the legitimacy of the current insane government and “democracy”, in the medical system, in the financial system, etc. How much better would this world be, getting all the retards to remove themselves? We’d have a paradise utopia in about 4 months. This would be a great way to halt the dysgenics program of those who are here to destroy the planet. Can we hope that anything sensible would ever happen on this planet?

  101. CD Marshall says:

    My last comment,
    “”Sunshine is not only a -18C heating potential, but is actually upwards of +120C! Sunshine is incredibly hot and powerful, able to generate the climate as we know it. Sunshine is powerful enough to heat through the latent heat phases of H2O and then this latent heat keeps the night-side of the planet much warmer than otherwise. The average surface temperature therefore could never be only -18C.

    And then add in the fact that the adiabatic gradient requires that the warmest air be found next to the surface, and any expected average must be at altitude, and then it becomes perfectly clear and justified that -18C could never be found at the surface.

    Why is the Earth’s surface not -18C?

    Because Earth is heated by sunshine to much higher than -18C.

    Because sunshine fills the latent heat sink at both the ice-water and water-vapor transitions, and this keeps the night side and the poles much warmer than otherwise as this heat comes back out when sufficient solar power is not present
    Because the adiabatic gradient requires that the expected average of -18C may only be found at the average of the atmosphere, not the extremity of the atmosphere closest to the surface.

    The answer to the above question is definitely not that the temperature of the colder atmosphere adds with the temperature of the warmer surface to create an even warmer temperature!” -JP

    Again as I said, thermodynamics equilibrium would forbid backradition from increasing the temperature. That’s the 0th Law of Thermodynamics as well.

    “If two systems are at the same time in thermal equilibrium with a third system, they are in thermal equilibrium with each other.”

  102. CD Marshall says:

    They are trying to claim backradiation is a third system or at least that is how it plays out in the math they are using.

  103. CD Marshall says:


    I am the author of the experiment and the article. My experiment disproves AGW idea because AGW is based absolutely in long wave radiation emitted by the surface being “trapped” by the atmosphere.
    In my experiment, as well as in Prof. Pratt’s experiment, I used a material which thermal and optical properties allows the solar short-wave and long-wave radiation, and surface long-wave radiation to go in and out the boxes. This condition permits us to verify if the effect of warming is due to any kind of retention of Long-wave radiation by the GHG in in the atmosphere, as AGW idea proposes. The experiment clearly shows that AGW idea is absolutely wrong; there is not any “trapping” of long-wave radiation in the atmosphere. In nature, the atmosphere and the oceans work more as distributors of thermal energy; therefore, they homogenize the temperature of the planet.

    I used also thick panels of silica glass, acrylic and cellulose acetate. The results recorded were the same. As soon as I took off the polyethylene film covering the boxes, the temperature decreased dramatically. Given that the polyethylene film permits the long-wave IR to go into and out the boxes, the increase of temperature before the elimination of the polyethylene film was not due to “long-wave IR trapped” by the inner atmosphere, but to another cause which was evident as soon as the polyethylene film is taken off the box, i.e. the blockage of convective heat transfer. Consequently, “trapped” radiation has nothing to do with the warming of the atmosphere.

    Regarding the thickness of the polyethylene film, the issue of coherence was taken into account and was calculated and measured for obtaining trustworthy results.
    Jerry, I hope my explanation has solved your question. You can send any questions to; I personally will answer them, Jerry.


  104. CD Marshall says:

    T= “the down-welling LW radiation appears to be globally a product of the air temperature rather than a driver of the surface warming. In other words, on a planetary scale, the so-called back radiation is a consequence of the atmospheric thermal effect rather than a cause for it.

    “What this means is that the thermal radiation emitted by the atmosphere is a consequence of the atmosphere having a temperature…NOT the cause of the atmosphere having a temperature. Think of the distinction here, the logic. It is CRUCIAL! What they are saying is of course consistent with physics – something which has a temperature will emit thermal radiation…that’s the Planck’s Law, etc. What the alarmists have done, as with everything else, is REVERSE things, so that they claim that the radiation emitted by the atmosphere is the cause of the atmosphere having a temperature, and thus can cause for itself higher temperature by emitting into itself. You see what they’ve done? Because they start with the first initial premise of reversing output for input as I’ve been discussing….then everything else that follows likewise becomes inverted and made backwards.

    “The thermal radiation emitted from the air is a consequence of it having a temperature ; it is not the cause of the atmosphere having a temperature . The cause of the atmosphere’s temperature came first, and then, the atmosphere can emit since it was given some temperature. So what’s the true cause? Of course the Sun and solar heat, and the lapse rate etc.” – Joseph Postma,

  105. justgivemeall says:

    It’s always the same with the agw crowd 240 outgoing so must be incoming,temp rises co2 goes up so co2 causes temp rise. And back radiation heating what’s already been warmed by the sun. It’s like they purposely make everything backwards to keep it confused. It’s no different than if ice melts that’s what made the temp rise or maybe the temp rose first.

  106. E. Nichols says:

    Hi Joe,

    I am an old school electrical engineer, you what climate change dogma reminds of is the asinine concepts we used to describe people who did not quite grasp thermodynamics. We use to use the concept of creating a electrical energy source by plugging a power strip into itself to create more voltage/current. Obviously this is asinine but it is analogous to climate change, the joking around was in fun! Clearly anyone dumb to accept such notion in my time would have deemed an abject idiot. I have investigated the matter thoroughly the scientists promoting this garbage either really stupid and cheated to obtain their credentials or hopelessly corrupt. In either case climate is is science fiction of the highest order. We need to expose this garbage for what it is, there is no way any true scientist or engineer would ever make such fundamental errors. Good work as always!

  107. Cheers E. Nichols. Nice to meet you.

  108. CD Marshall says:

    “Radiation is simply the amount of power per square metre. This power cannot be used and stored at the same time. Power cannot be raised without intensifying the source or adding another source of energy.”

    Which is the 1stLoT Which is simply the distinction of heat and work, a distinction that must exist to create Delta U=Q-W. Climate science uses work and heat as interchangeable summations of energy that always results in heat.

    In climate science , driven by politics, this logic does not exist and we get Rabbits who think they are physics experts.

    Funny that logic is the Kinetic theory of gases and potential energy. If W and Q were interchangeable then as gas rises it would increase in potential energy and heat, increasing both as it reaches the tropopause…

  109. CD Marshall says:

    “Authors who defend the “Greenhouse Effect” attempt to characterize it as a form of heat congestion (e.g. Archer, 2009). The problem with this defense is that no amount of heat congestion can result in an average power output exceeding the average power input.

    The defense is also subject to the limitations of “Kirchhoff’s Law”. “Kirchhoff’s Law” dictates that while emissivity and absorptivity are always equal for a given material or body, the equality of absorption (not absorptivity) and emission (not emissivity) of radiation defines thermal equilibrium between bodies that are not in thermal contact.

    Even the misconception that selective absorptivity makes it easier for radiation to get in than to escape, breaks down when both the atmosphere and the surface of the earth are treated as a whole body.

    Regardless of internal complexities, a whole body ultimately can only emit the exact amount of radiation it receives, or a lesser amount corresponding to a lower pre-equilibrium temperature if thermal equilibrium has not been reached. By increasing absorption, emission is increased – which was confirmed experimentally by Stewart (1858, 1860a, 1860b) and Kirchhoff (1859 & 1860).

    Moreover, this greater emission has a cooling effect on the atmosphere and Frankland (1864, p. 326) asserts that without this loss of heat by emission to space, atmospheric water vapor could not condense into clouds and precipitation. This cooling by radiative emission is further confirmed by Ellsaesser (1989) and Chillingar et al. (2008).

    Thus surface evaporation and subsequent condensation at altitude has a powerful cooling effect, which in addition to convection, offsets the high degree of heating that occurs at the surface.”

  110. Mark Munro ( says:

    Niblet No 6 – NO ‘greenhouse effect’ exists | Principia Scientific Intl. (

  111. Great work by Geraint!

    BTW Jerry Krause is a bot. It’s a Turing Test bot.

  112. CD Marshall says:

    Trolls always go with, “What do you think keeps the planet warm?”

    Misdirection don’t fall for it. Instead ask them what is keeping the planet cooler?

  113. CD Marshall says:

  114. CD Marshall says:

    I’ve watched videos on some of these bots, the bots that self learn as you talk to them are pretty creepy stuff. That’s why I try and test who I’m talking with by using looping algorithms sometimes, a bot won’t catch it as easy. Keep a conversation going like this…

    “Of course you can’t separate distinctions. CO2 doesn’t cause warming besides a minimal contribution to molar mass, the little IR interaction will never increase temperatures for it is not a heat source. But if you add butter to the popcorn it tastes better.

    Pollution is another topic entirely and walking your dog. Activists have to pool these points to cause distractions. That is why China, Northern Africa and Siberia are the largest contributors to air pollution. They do not follow clean air laws. Think with your head not your emotions. PR#6.

  115. That’s exactly what Jerry’s comments read like.

  116. CD Marshall says:

    Sometimes they might say “I like popcorn too” which is just weird. Instead of “WTF are you talking about!” Which is what a normal person would say.

  117. CD Marshall says:

    Sometimes people are just socially awkward or mentally awkward. My brother was telling me about this kid he met working on a house who was Rain Man incarnate, then he found out this guy was the aux co-pilot on commercial planes. Yeah, just think of that the next time you fly the friendly skies. The third guy in line to fly the plane in case of an emergency may be rocking back and forth in his seat reciting prime numbers.

  118. Max says:

    There is one thing I’m still unable to understand. What’s wrong with the approach of calculating the green plate temperature as follows (Tb=black plate temp; Tg=green plate temp) : let’s focus on the 1st Law applied to the green plate, i.e. dQ=0, ==> Qin = Qout, where Qin = s(Tb^4-Tg^4), Qout = sTg^4. This would give Tb^4=2*Tg^4 and not Tb=Tg.
    My understanding is that dQ=0 is to be applied individually to each plate, and this is what it would seem to come out when applying it to the green plate first…

  119. CD Marshall says:

    You have no idea how many climate scientists tell me heat is lost to space. I had a guy with a physics degree tell me how can you say it’s light, put your hand in front of a laser. I’m not sure what is going on with these guys anymore. I have to explain thermodynamics to people who claim to have degrees in these things.

  120. Max Polo says:

    Thanks, Nepal and CD Marshall, now I see it.
    Do things change if – instead of “space” – we have an actual physical body surrounding the black and green plates. Let’s assume for instance that we are inside a room, in vacuum condition, where the walls/roof/floor are large/heavy enough that they can be considered to be at constant temperature (cold sink). Isn’t “heat” the energy that comes out from the plates and goes to the walls, due to the differing temperatures?

  121. Heat is not something that “leaves” or “goes out” or “goes to”. Heat is an action, a verb, like work. Energy goes out, etc., but energy isn’t heat.

    You always have to follow the rules. To increase temperature you require heat. Hear doesn’t come from something colder. Surrounding yourself by an infinite universe of ice with you in a pocket in the middle won’t increase your temperature.

  122. Max Polo says:

    Thanks, guys. Joe, regarding your comment on the surrounding ice cubes. Let’s suppose I have my room with the walls at 20°C. With no irradiation on the black plate, I’ll have both black and green plates at 20°C as well. But when the sun shines on the black plate, it will warm the plate up until it reaches its equilibrium temperature. This temperature is not dependent on the presence or absence of the green plate as we have seen. But isn’t this equilibrium temperature also a function of the walls’ temperature? If my walls were at 0°C instead of 20°C, then the equilibrium temperature should be 20°C less than in the previous case…..or am I getting something wrong?

  123. CD Marshall says:

    Good concise answer.
    “Heat is, by definition, flow of energy that changes temperature.”

  124. Joseph E Postma says:

    They’re all passive objects. They can only be heated to the temperature supplied by sunlight to the first surface the light heats upon. After that, temperature cannot be augmented passively…because there’s no other heat.

    As Nepal said: heat is energy which changes (increases) temperature. Vice-versa: to increase temperature one requires heat.

  125. Max Polo says:

    Sorry guys, but I’m still a bit puzzled while trying to solve this textbook problem.
    We have the infamous two parallel plates in the vacuum, at a given distance from each other, such that the view factor F12 from plate 1 to plate 2 is a number less than 1. Plate 1 is energized and achieves a given temperature of T1. What’s the temperature of plate 2 when the equilibrium is reached?
    In this case, it’s clear (I think) that – if the plate’s distance is large – then plate 2 will reach (I believe) a temperature that will be closer to that of the surroundings at T0 than that of the plate 1 (T1).

    I wrote the energy balance to get this solution :
    T2^4 = [F12T1^4 + (2-F12)T0^4]/2

    If F12 –> 0 (infinitely distant plates) the fraction of energy from plate 1 to plate 2 will tend to zero and the temperature T2 will tend to that of the background (T0)
    If F12 –> 1 (extremely close plates) the temperature T2 will tend to T2 = [(T1^4+T0^4)/2]^(1/4) which will always be less than T1.

    Example :
    Extremely close plates : F12–>1
    T0 = 4K
    T1 = 1000K
    T2 –> 841K
    Extremely distant plates : F12–>0
    T0 = 4K
    T1 = 1000K
    T2 –> 4K

    How does that fit into your previous claim that T2 should become = T1?
    Not sure what I’m getting wrong (if anything).

  126. Max Polo says:

    Nepal, with all my goodwill, I find it hard to believe that a little plate (plate 2), located two thousand miles away from the energy source (plate 1), would become as warm as another little plate that is one millimeter from the source. In the first case, the view factor is 0,0000001, in the second case, it is 0,9999999. Do you mean that this only makes a difference in terms of heating rate (much higher in the second case) but not in the final equilibrium temperature? I do not think so. If you set Q=0 you always get T1 = T2 regardless of the view factor (= regardless of the distance of the plates). This does not sound right.
    I need to dig up my old heat transfer books, there may be some problems like this solved.

  127. Joseph E Postma says:

    The usual scenario is for plane parallel infinite plates. That is what Nepal would have been referring to. The equilibrium is always set by Q = 0 between two objects (not space as space is not an object), and yes of course distance has an effect WHEN the plates are not infinite. The distance factor is the view factor.

  128. Joseph E Postma says:

    @Max when you have non-unity or non-zero view factors then it all becomes a mess and you would need to use numerical integration techniques unless there’s a trick to simplifying the math to make it analytical. In any case the same rules always still apply: Q = 0 between two objects for equilibrium. This (Q = 0 between two objects) always guarantees conservation of energy because total emission (not heat, but emission) to space will be the same before and after.

  129. Max Polo says:

    @Joe, here is my calculation for two square parallel plates, each one 1×1 m, and 0.347 m apart, assuming blackbody behavior (view factor from: :

    A=1 m2
    view factor F12=F21=0.5
    T1 (given) = 500°C = 773.15K
    s=5.67E-8 W/m^2/K^4
    Q12 = sAF12*(T1^4-T2^4) = 0
    –> T2 = T1 = 773.15K=500°C

    So, unless I’m doing something wrong, the condition Q12=0 implies T2=T1 even for small-sized plates at a finite distance that makes the view factor reduce to 0.5.
    Let’s suppose to increase the distance of the plates to 2 m, so that the view factor falls to 0.039…yet the equation still requires T2=500°C which looks weird, to say the least.

  130. Joseph E Postma says:

    The view factor isn’t being factored in correctly…it can’t just be outside of the whole expression like that, but has to apply specifically to a single body, factored into one of the specific bodies. With the view factor outside of the whole expression it has no effect…the inverse square law for example, view factors in general, establish how much of the source’s energy the secondary body can intercept. And that’s not what the equation above is doing, simply factored into both bodies like that, on the outside of the brackets. You have to factor it into only one of the temperatures…not both temperatures.

  131. Max Polo says:

    Joe, now I see it…the heat balance becomes quite more complicated. I missed that point. Thanks for highlighting this. What I’ve tried to do is an experiment, that maybe I will be sharing here, to show once more how the “greenhouse theory” doesn’t work as advertised, well…doesn’t work at all. But I need to model my experiment arrangement, to compare results with theory. It’s a bit more complicated than I hoped, also because I have to account for convection…

  132. Max Polo says:

    Nepal, that’s the conclusion I had arrived at as well. But honestly, I find it hard to believe.

  133. Joseph E Postma says:

    At large distances from the plate, the flux from the plate starts to decrease as the inverse square law, because the plate starts to approximate a point source (at large distance from it).

    The view factor is the inverse square law when the source is a sphere or point. For a plate it is a bit more complex with distance from the plate. If the plate is infinite then the view factor is always unity.

    So no, the flux must diminish with distance from the plate when the plate is not infinite.

    The view factor between two spheres is also identical if the spheres are the same size. But the inverse square law applies in a given direction…it doesn’t cancel out.

    The way Max wrote above and Nepal is implying is that view factors cancel out. They don’t cancel out. This would be like saying the inverse square law cancels out.

  134. Max Polo says:

    Joseph, back to the plate problem: at large distances from the plate, the view factor F12 will be < 1. If A1=A2, then F21 = F12 (reciprocity theorem) so the heat flow equation will become Q=sA1F12*(T1^4-T2^4) which still gives T1=T2 if we set Q=0. But this does not sound possible for large distances: what am I getting wrong?

  135. Max Polo says:

    I will propose a different solution to the two-pane problem. It is based on experiments I did, although not in the vacuum, but correcting the results to account for the air convection effect. What turns out is a different solution than Joe’s, but still no “back-radiation heating” whatsoever is observed.
    If anyone is interested here is a brief description of my experiment :

    So here is the “theory” (simplified eliminating air convection effect and assuming background at 0K and 1 square meter pane area each side) to reproduce the empirical results :

    All that is needed is the Stefan-Boltzmann equation that states that the total radiant power [W] emitted by a blackbody is equal to the S-B constant times the body area times the absolute temperature elevated at the fourth power:
    W = sigmaAT^4

    1st CASE: single pane (pane 1) irradiated with flux S [W/m2] as a point source on one side

    Energy balance of (single) pane 1 :
    Irradiated Power-in: S
    Irradiated Power-out: 2sigmaT1^4
    Power Balance (steady state) :T1=(S/2/sigma)^0.25

    2nd CASE: double panes (pane 1 & pane 2), parallel at a distance such that their mutual view factor is f; pane 1 is irradiated with flux S as a point source on one side

    Energy balance of pane 2:
    Irradiated Power-in (the portion of radiant energy from pane 1 that gets absorbed from pane 2): fsigmaT1^4
    Irradiated Power-out: 2sigmaT2^4
    Power Balance (steady state) : T2 = T1(f/2)^0.25
    Note that the power balance equation can be also written as a heat flow equation (pane 2 heat-in from pane 1 equals pane 2 heat to the background): f
    sigma(T1^4-T2^4) = sigma(1-f)T2^4 + sigmaT2^4 that gives the same result solving for T2

    Energy balance of pane 1:
    Note that pane 1 is only energized by the external flux S on one side, and the portion of energy that it gets from pane 2 (= fsigmaT2^4) is not “new” energy, but is just the result of pane 1’s own irradiation: if there was no irradiation from pane 1, pane 2 temperature would be in equilibrium with the background at 0 K. Therefore, this energy from pane 2 cannot be accounted as “energy in” for pane 1. Pane 2 is just a passive “vector” of pane 1 energy to the background and does not introduce any new energy into the system. As such, the pane 1 power balance remains identical to the 1st case which means that its temperature won’t change [T1=(S/2/sigma)^0.25].

  136. CD Marshall says:

    Ken P. wrote this out how does it look? Apologies if it links out or something weird.

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