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.
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
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
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.
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.