Guest Post by Willis Eschenbach
Pushed by a commenter on another thread, I thought I’d discuss the R. W. Wood experiment, done in 1909. Many people hold that this experiment shows that CO2 absorption and/or back-radiation doesn’t exist, or at least that the poorly named “greenhouse effect” is trivially small. I say it doesn’t show anything at all. Let me show you the manifold problems with the experiment.
To start with, let me give a curious example of the greenhouse effect, that of the Steel Greenhouse. Imagine a planet in the vacuum of space. A residue of nuclear material reacting in the core warms it to where it is radiating at say 235 watts per square metre (W/m2). Figure 1 shows the situation.
Figure 1. Planet in outer space, heated from the interior. Drawing show equilibrium situation
This planet is at equilibrium. The natural reactor in the core of the planet is generating power that at the planet’s surface amounts to 235 W/m2. It is radiating the same amount, so it is neither warming nor cooling.
Now, imagine that without changing anything else, we put a steel shell around the planet. Figure 2 shows that situation, with one side of the shell temporarily removed so we can look inside.
Figure 2. As in Figure 1, but with a solid steel shell surrounding the planet. Near side of the shell temporarily removed to view interior. Vertical distance of the shell from the surface is greatly exaggerated for clarity—in reality the shell and the shell have nearly the same surface area. (A shell 6 miles (10 km) above the Earth has an exterior area only 0.3% larger than the Earth’s surface area.)
[UPDATE: Misunderstandings revealed in the comments demonstrated that I lacked clarity. To expand, let me note that because the difference in exterior surface area of the shell and the surface is only 0.3%, I am making the simplifying assumption that they are equal. This clarifies the situation greatly. Yes, it introduces a whopping error of 0.3% in the calculations, which people have jumped all over in the comments as if it meant something … really, folks, 0.3%? If you like, you can do the calculations in total watts, which comes to the same answer. I am also making the simplifying assumption that both the planet and shell are “blackbodies”, meaning they absorb all of the infrared that hits them.]
Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2. At that point the system is back in equilibrium. The planet is receiving 235 W/m2 from the interior, plus 235 W/m2 from the shell, and it is radiating the total amount, 470 W/m2. The shell is receiving 470 W/m2 from the planet, and it is radiating the same amount, half inwards back to the planet and half outwards to outer space. Note also that despite the fact that the planetary surface ends up much warmer (radiating 470 W/m2), energy is conserved. The same 235 W/m2 of power is emitted to space as in Figure 1.
And that is all that there is to the poorly named greenhouse effect. It does not require CO2 or an atmosphere, it can be built out of steel. It depends entirely on the fact that a shell has two sides and a solid body only has one side.
Now, this magical system works because there is a vacuum between the planet and the shell. As a result, the planet and the shell can take up very different temperatures. If they could not do so, if for example the shell were held up by huge thick pillars that efficiently conducted the heat from the surface to the shell, then the two would always be at the same temperature, and that temperature would be such that the system radiated at 235 W/m2. There would be no differential heating of the surface, and there would be no greenhouse effect.
Another way to lower the efficiency of the system is to introduce an atmosphere. Each watt of power lost by atmospheric convection of heat from the surface to the shell reduces the radiation temperature of the surface by the same amount. If the atmosphere can conduct the surface temperature effectively enough to the shell, the surface ends up only slightly warmer than the shell.
Let me summarize. In order for the greenhouse effect to function, the shell has to be thermally isolated from the surface so that the temperatures of the two can differ substantially. If the atmosphere or other means efficiently transfers surface heat to the shell there will be very little difference in temperature between the two.
Now, remember that I started out to discuss the R. W. Wood experiment. Here is the report of that experiment, from the author. I have highlighted the experimental setup.
Note on the Theory of the Greenhouse
By Professor R. W. Wood (Communicated by the Author)
THERE appears to be a widespread belief that the comparatively high temperature produced within a closed space covered with glass, and exposed to solar radiation, results from a transformation of wave-length, that is, that the heat waves from the sun, which are able to penetrate the glass, fall upon the walls of the enclosure and raise its temperature: the heat energy is re-emitted by the walls in the form of much longer waves, which are unable to penetrate the glass, the greenhouse acting as a radiation trap.
I have always felt some doubt as to whether this action played any very large part in the elevation of temperature. It appeared much more probable that the part played by the glass was the prevention of the escape of the warm air heated by the ground within the enclosure. If we open the doors of a greenhouse on a cold and windy day, the trapping of radiation appears to lose much of its efficacy. As a matter of fact I am of the opinion that a greenhouse made of a glass transparent to waves of every possible length would show a temperature nearly, if not quite, as high as that observed in a glass house. The transparent screen allows the solar radiation to warm the ground, and the ground in turn warms the air, but only the limited amount within the enclosure. In the “open,” the ground is continually brought into contact with cold air by convection currents.
To test the matter I constructed two enclosures of dead black cardboard, one covered with a glass plate, the other with a plate of rock-salt of equal thickness. The bulb of a thermometer was inserted in each enclosure and the whole packed in cotton, with the exception of the transparent plates which were exposed. When exposed to sunlight the temperature rose gradually to 65 oC., the enclosure covered with the salt plate keeping a little ahead of the other, owing to the fact that it transmitted the longer waves from the sun, which were stopped by the glass. In order to eliminate this action the sunlight was first passed through a glass plate.
There was now scarcely a difference of one degree between the temperatures of the two enclosures. The maximum temperature reached was about 55 oC. From what we know about the distribution of energy in the spectrum of the radiation emitted by a body at 55 o, it is clear that the rock-salt plate is capable of transmitting practically all of it, while the glass plate stops it entirely. This shows us that the loss of temperature of the ground by radiation is very small in comparison to the loss by convection, in other words that we gain very little from the circumstance that the radiation is trapped.
Is it therefore necessary to pay attention to trapped radiation in deducing the temperature of a planet as affected by its atmosphere? The solar rays penetrate the atmosphere, warm the ground which in turn warms the atmosphere by contact and by convection currents. The heat received is thus stored up in the atmosphere, remaining there on account of the very low radiating power of a gas. It seems to me very doubtful if the atmosphere is warmed to any great extent by absorbing the radiation from the ground, even under the most favourable conditions.
I do not pretend to have gone very deeply into the matter, and publish this note merely to draw attention to the fact that trapped radiation appears to play but a very small part in the actual cases with which we are familiar.
Here would be my interpretation of his experimental setup:
Figure 3. Cross section of the R. W. Wood experiment. The two cardboard boxes are painted black. One is covered with glass, which absorbs and re-emits infrared. The other is covered with rock salt, which is transparent to infrared. They are packed in cotton wool. Thermometers not shown.
Bearing in mind the discussion of the steel greenhouse above, I leave it as an exercise for the interested reader to work out why this is not a valid test of infrared back-radiation on a planetary scale … please consider the presence of the air in the boxes, the efficiency of the convective heat transfer through that air from the box to the cover plates, the vertical temperature profile of that air, the transfer of power from the “surface” to the “shell” through the walls of the box, and the relative temperatures of the air, the box, and the transparent cover.
Seems to me like with a few small changes it could indeed be a valid test, however.
Best regards,
w.
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Greg House says (February 6, 2013 at 4:37 pm): “This is a wonderful idea real experimental testing, Mike, thank you. Has Roy Spencer done it yet?”
As I’ve pointed out in other threads, Dr. Spencer has no incentive to perform the experiment, as it would only confirm what he and just about everybody else already knows.
The real question is why his critics haven’t performed it. If they really believe it would overturn established physics, it would be worth a Nobel Prize at least, and would earn the undying gratitude ot a world saved from the imaginary threat of Thermageddon. That they haven’t performed such a definitive experiment and claimed their laurels speaks volumes.
Gino:
“Willis, HEAT is the transfer of energy. The transfer of energy is work (Joules), the rate at which energy is transferred is power (Watts). If no energy is transferred, no work is performed, If no work is performed, no power is used. If two bodies are at the same temperature the may “see each other” through radiation (see my standing wave post), but no work is performed transferring energy between the two bodies, therefore no energy moves between them, hence there is no power consumed.”
Thanks Gino, Willis has always had this problem with power and flux.
Gino says:
February 6, 2013 at 9:41 pm
“Higher temperatures emit more energy than lower temperatures so even though it “sees” the lower temperature object, the higher temperature object overpowers the lower on so to speak. That is the reason heat transfer calculations are based on temperature DIFFERENCE. Energy is transmitted at various wavelengths. Best explanation I’ve read involves wave theory. When two bodies at the same temp emit at the same wavelength, they form a standing wave. A standing wave does not transport energy. When they emit at different wavelengths, they ‘exchange’ energy as in absorb in one bandwidth and radiate in another with a net of zero to maintain a delta T of zero.”
I agree with Dino. It is my opinion that Dino’s way of thinking is the way one applies physics.
And the word “Backradiation” was never mentioned in radiation physics classes back then. Maybe it is “Newspeak”?
Anthony Zeeman says (February 6, 2013 at 5:04 pm): “Leaving a space, presumably containing a vacuum between the sphere and the shell would do nothing since this would have the same thermal resistance as from the sphere to space.”
You’re forgetting that the shell radiates both out and in. Very different from the one-way radiation from the surface only. Think of the shell as giving the sphere a “rebate” on its radiated energy. 🙂
“This would be the same as adding another heatsink on top of the existing heatsink on your cpu.”
No. The heat sink is in contact with the microprocessor. Willis’s shell is about the same diameter as the “planet”, is separated from it by a vacuum, and the whole shebang is surrounded by vacuum. For obvious reasons, my laptop is designed to work in air. It even has a fan to convectively cool the heat sink, which cools the cpu by conduction. As with Wood’s model greenhouses, radiative cooling is a minor factor. If I were Greg House, I’d say my laptop disproves the so-called “greenhouse” effect., but I’m not. 🙂
“Also, note that your soup doesn’t get hotter when you put it into a Thermos in spite of the reflected IR radiation from the silvered sides of the Dewar flask.”
To compare with Willis’s example, you need to add a constant heat source to the inside of the thermos. Now what happens?
So what happens if the steel inner ball and the outer shell is replaced by nitrogen gas. Does the nitrogen gas temperature approach infinity as they say it can’t radiate heat away?
wayne says:
February 6, 2013 at 11:06 pm
But, you and Gino are begging the question, by implicitly asserting that equilibrium has been achieved and that the equilibrium with the steel shell is the same as that without it. Allow me to interject some math.
Basically, without the shell, the planet is receiving heat at a rate which I will call P_core, the power from the nuclear core. It is radiating it at
P_planet_outgoing = sigma*T_planet^4*(4*pi*R_planet^2)
where sigma is the SB constant, T_planet is the temperature of the planet, and the last term in parentheses is the the surface area with R_planet being the radius of the planet. Thus, the temperature of the planet at equilibrium is
T_planet = (P_core/(sigma*(4*pi*R_planet^2)))^0.25
Now, the shell has an inner radius, which I will call R_inner, and an outer radius, which I will call R_outer. It has incoming and outgoing power at
P_shell_incoming = sigma*T_planet^4*(4*pi*R_planet^2)
P_shell_outgoing = sigma*T_shell^4*(4*pi*(R_inner^2+R_outer^2)
This is the nub of the issue: The shell has at least twice the surface area of the planet over which it radiates. At equilibrium, we have
T_shell = T_planet * (R_planet^2/(R_inner^2+R_outer^2))^0.25
[Tshell is at most 2^0.25 := 20% lower in temperature than T_planet] but, these being absolute temperatures, 20% can be quite a lot.
What is T_planet now, though? Is it higher than it was before? Yes. The planet power balance is now
P_planet_incoming = P_core + sigma*T_shell^4*(4*pi*R_planet^2)
P_planet_outgoing = sigma*T_planet^4*(4*pi*R_planet^2)
At equilibrium, these are equal, so
sigma*T_planet^4*(4*pi*R_planet^2) = P_core + sigma*T_shell^4*(4*pi*R_planet^2)
= P_core + sigma*T_planet^4*(R_planet^2/(R_inner^2+R_outer^2))*(4*pi*R_planet^2)
Leading to
sigma*T_planet^4*(4*pi*R_planet^2)*(1 – R_planet^2/(R_inner^2+R_outer^2) ) = P_core
so, with the shell
T_planet_with_shell = T_Planet_without_shell / (1 – R_planet^2/(R_inner^2+R_outer^2) )^0.25
Again, the maximum is about 20% higher.
Things are slightly different for the atmosphere, in that gaseous particles can radiate in any direction, and not all those directions will intersect with the Earth’s surface. But, over 45% of them will at the top of the troposphere, instead of the ~50% from a thin solid shell, so this is not such a big deal.
A bigger deal is that, the atmosphere is in contact with the surface, and can exchange heat through conduction and convection. On that score, the steel shell thought experiment appears to become a bit of an academic exercise which may not be relevant for a planet with a gaseous atmosphere.
Bart says:
February 6, 2013 at 11:54 pm
Erratum: “Tshell is at most 2^0.25 := 20% lower in temperature than T_planet…”
“To compare with Willis’s example, you need to add a constant heat source to the inside of the thermos. Now what happens?”
I’ll have them on the market in a blink! ☺
Don’t you just hate cold coffee. Think 5‹W/m²› would do it?
Bob Roberts says:
February 6, 2013 at 5:34 pm
I am fully aware of the physics of the way a greenhouse works, thanks, Bob – and I am thus also fully aware of the ‘misnomer’ status of the GHE. Unfortunately, until someone comes up with an alternative well accepted description for the ‘warming effect’ of the atmosphere, I think the consensus use of GHE is the only one we can realistically apply at the moment! (If someone has created a better term that can be widely used, perhaps hey can let us know?)
regards
Kev
I still haven’t heard an explanation, why 6,000 ppm of CO2 in Martian atmosphere (which is exact equivalent of 400 ppm CO2 and thousands ppm of water vapor here) has no visible effect on its surface temperature, e.g. its black body and average temperature is the same 210K.
http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html
Second, almost nobody considers one simple fact, that is, simple thermal insulation and heat retention of the bulk atmosphere – nitrogen and oxygen – does the job, which some are still trying to stretch on radiation diagram arrows. Moon night temperature is -150°C. Earth night is +10°C. Which kind of invisible, IR-based dragons breath warms our night by incredible 160°C? Photons bounced/radiated from tiny part of air molecules, really?
On my first day as a grammar school student (long, long ago but I recall it clearly) I became very upset when, after I gave my teacher the exact 50 cents required for the week’s “milk money”, she then gave “my” 50 cents to someone who had paid her a dollar. My mom eventually helped me to understand that the net result was correct and that it didn’t matter that my exact physical two quarters were no longer in the teacher’s possession; my milk was well and truly paid for.
It astonishes and dismays me that several rather didactic and even patronizing commentors are making similar conceptualization errors regarding basic radiative heat transfer as well as largely missing the point of Willis’ post.
A very wise man once wrote, “If any one thinks he is wise, let him become a fool, that he may become wise.” In other words, a humble, open mind is a prerequisite to learning.
Thank you for the clarifying and thought-provoking post, Willis. Well done.
Gino says:
February 6, 2013 at 5:59 pm
Not true. Thermodynamic entropy is defined in units of energy per degree of temperature. Once temperatures are matched (delta T = 0), then no energy is transferred between bodiesI
You didn’t read the post properly. Not the reference to NET transfer. If both bodies are at the same temperature there will be no net transfer of energy but both will continue to radiate to each other (balance therefore no change in energy state of either body). The vector defining the path of a IR photon will not stop because the surface it comes across is at the same temperature as its source. There will still be radiative transfer but NET transfer = 0. For your point to be true, that is no transfer, then the bodies would have to reach a state where they no longer emitted radiation.
Gino last point should have read:
Gino says:
February 6, 2013 at 5:59 pm
Not true. Thermodynamic entropy is defined in units of energy per degree of temperature. Once temperatures are matched (delta T = 0), then no energy is transferred between bodies.
You didn’t read the post properly. Note the reference to NET transfer. If both bodies are at the same temperature there will be no net transfer of energy, but both will continue to radiate to each other (balance: therefore no change in energy state of either body). The vector defining the path of a IR photon will not stop because the surface it comes across is at the same temperature as its source. There will still be radiative transfer but the net transfer = 0. For your point to be true, that is no transfer, then the bodies would have to reach a state where they no longer emitted radiation.
Mr. Eschenbach
thanks for attention
I remain convinced that whatever the conversion watt watt/m2, it does not change the fact that in Figure 2, you indicate that the core issue 235 w/m2 and, depending off the amount off energy in the core, and the surface (let suppose, it is an blackbody)and thus the surface of the sphere emits 235 w / m2 to stay équilibre.iff you forget a differences off the surface off the core, and the sphere.
Now you certainly indicate that, because You are aware that a diffrence in the surfaces matter.
and because you are in agreement that if the surface of the sphere is larger, the radiation for each squre m will decrease.
You then specify the sphere also emits 235 w/m2 inward.
But again, , you double the irradiation area!
By each square m off the sphere, You have 235 w émitted outwards, and also inwards, so the flux off energy is doubled?
Now, something else,
you indicate that the emission flux of 235 w/m2 of the core induce on the surface of the sphere equal flux irradiation outwards , but also inward.
I assume that if this time the core issues as indicated 470 w/m2, there is no reason that this time it would not be divided into two streams,
there should be 470 w/m2 inward as to the outside, if you follow your first pattern?
it is an circular patern?
–
Willis ….Lets reverse your set-up.
The steel hollow sphere is in deep space (-273C) with an initial internal vacuum.
Its heated by an external variable power supply carefully monitored to maintain the temperature at 50C.
The voltage and current are noted, say as V1 and I1, to get power P1
Now suddenly insert an object inside the sphere at -40C ( i can because its a thought experiment! ).
What happens next?
Does the real radiation from the colder object warm the steel sphere when absorbed?
I think not.
To restore the steel sphere to 50C the supplied power (P2) would have to increase for a time.
P2 > P1
Point being that in these apparent two object problems there is always a third (often ignored )temperature …..the surroundings.
In the example above the cold object at -40C is separated from the surrounding deep space by the steel sphere.
So no Virginia, colder objects don’t always ‘warm’ objects at a higher temperature.
Even If the steel shell was much larger than the planet surface, let’s say twice the surface area, it makes absolutely no difference to the argument. It this case the outer surface of the shell would have to radiate 117.5 W/sq.m to balance the 235 W/sq.m at the planet’s surface. This accords with the shell having a temperature of 213K. Because the shell is at 213K it also radiates from its interior surface 117.5 W/sq.m (Stefan’s Law). This energy falls on the planet which, because it has only half the surface area of the shell receives 235 W per sq.m of planetary surface.
So, the same calculation ensues. The planet is now receiving 235 W/sq.m from its interior and a further 235 W/sq.m ‘back-radiation’ from the shell, that is 470W/sq.m altogether and so the planet becomes much warmer because of the presence of the surrounding shell.
Did you notice the provocative ‘back-radiation’ I slipped in there?
Unfortunately, most commentators don’t seem to have the ability to grasp this, but it is so simple and fundamental it separates the sheep from the goats.
The shell game is a foolish game to play. The shell game is how the AGW pseudo scientist came up with their ludicrous 33 degrees warmer because of “GHGs” meme. Once again let’s review the critical “Do Nots” of atmospheric modelling-
1. Do not model the “earth” as a combined land/ocean/gas “thingy”
2. Do not model the atmosphere as a single body or layer
3. Do not model the sun as a ¼ power constant source without diurnal cycle
4. Do not model conductive flux to and from the surface and atmosphere based on surface Tav
5. Do not model a static atmosphere without moving gases
6. Do not model a moving atmosphere without Gravity
7. Do not model the surface as a combined land/ocean “thingy”*
As you can see the shell game commits the following scientific atrocities 2,3,5,6 & 7. It’s so bad at modelling the atmosphere it cannot even commit 4.
Does the surface emit IR? Yes. Does the atmosphere absorb some IR? Yes. Does the atmosphere radiate some IR back to the surface slowing it’s cooling rate? Yes. Is there a radiative GHE? Yes. Is the NET effect of radiative gases in our atmosphere warming? NO!
Radiative gases are critical for the continued vertical convective circulation observed below the tropopause. Without radiative energy loss at altitude, convective circulation would stall, and our atmosphere would heat and go isothermal. Increased radiative cooling of the surface under a non radiative atmosphere is no substitute for radiative cooling at altitude. Avoid the “Do Nots” and observe the science Willis avoids and you will find –
– Due to gravity, pressure gradient and resultant IR opacity gradient, an atmospheric column emits more IR to space than to the surface. 2 & 6
– Most energy emitted as IR to space from the atmosphere was obtained by the release of latent heat of evaporation above the level of max atmospheric IR opacity. 2 & 6
– Radiative gases emit more energy to space than the atmosphere ever intercepted from NET surface IR flux 2, 5 & 6
– Radiative gases return far, far less energy to the surface than emitted from the surface. 7*
– The conductive transfer of energy and the transfer of latent heat to the atmosphere is a function of surface Tmax not Tav. 3, 4 & 5
– Surface cooling by conduction + convection is more effective than gas conduction alone. 5
– Under a non-radiative atmosphere, increased radiative surface cooling will not result in significantly increased atmospheric cooling. 5 & 6
– Deep vertical convective circulation cannot continue under the tropopause without radiative gases. 5 & 6
And the summary? AGW is tripe. Radiative gases cool our atmosphere at all concentrations above 0.0ppm. Adding radiative gases to the atmosphere will not reduce the atmospheres radiative cooling ability.
* Willis has yet to concede that DWIR has no effect on the cooling rate of liquid water that is free to evaporatively cool. I base my claim on the several empirical experiments I have conducted myself. To this date I am unaware of any empirical work Willis has conducted in this area, or indeed any empirical experiment he has conducted.
Michael Moon says: February 6, 2013 at 9:37 pm
That shell is called ’insulation’ Mike; I think heating engineers know about it. But note, in Willis’s article he says nothing about doubling the heat input. That stays the same. Why not go back and read it very carefully and see if you can understand it this time.
.
There seems to be great confusion about fundamentals and the ‘imaginary’ 2nd Law of Thermodynamics.
Understanding electromagnetic radiation is a prerequisite to understanding the greenhouse effect. Radiation is a wave motion which transports energy through empty space at the speed of light by means of self-propagating electric and magnetic fields. It manifests itself in many forms, for example, as gamma rays, X-rays, light, infrared and radio waves. The only intrinsic difference between these various forms is that the waves have different wavelengths.
Radiation doesn’t ‘know’ if it is coming from a cold object or heading towards a hot object. How could it?
When radiation strikes the surface of a material one of three things may happen. It may be reflected, in which case it continues in a new direction with its energy conserved. It may pass through the material, like light through a window, radio waves through the walls of a house or like X-rays through your body. Or it may be absorbed by the material. Usually, a combination of two or three of these effects apply.
When radiation is absorbed the energy it conveys is also absorbed. Energy must be conserved and, most commonly, the absorption of radiation causes the absorbing material to heat up (the energy could, however, produce some alternative effects instead; photo-electric emission for example).
The Earth approximates to a black body in the long wave infrared region and so inevitably absorbs nearly all the down-welling radiation from the ‘cold’ atmosphere. And of course, this makes the surface warmer than it would otherwise be.
That shell is called ’insulation’ Mike
that something different no?
there is no matter off insulation here,
Willis
I have enjoyed your philosophy over the years, particularly your views on the safety valve action of thunderstorms, but your steel shell model here jars rather with my concept of the physics – bearing in mind that my degree is in archaeology, I might of course be missing a trick, but……
In my view your back radiation at 235 w.m-2 would only work if the steel shell is hotter than the core, which I don’t believe it is otherwise your model would be negated by the 2nd law of Thermodynamics. (Heat only flows from hot body to cold body, never in the other direction.) Otherwise the back radiation must just be deflected back to the steel shell, from whence it came and it wouldn’t heat that either, as it has just emanated from there and would eventually be re-radiated into space.
The same would apply, even if the core and shell were perfect black body radiators.
And as your analogy is, I assume, to ultimately represent in the end the earth and atmosphere, and particularly CO2 , the same principle applies – unless you assume that the CO2 molecules are hotter than the earth’s surface below the concept of ‘back radiation’ on which AGW seems to heavily revolve is not physically possible! Bit of a fatal flaw in the AGW concept.
Tx. Imarcus.
Your first model is OK for greenhouse gas theory, BUT IT IS COMPLETELY WRONG FOR PLANET EARTH!
Why is that? Well because the TOP os Earths’s atmosphere is actually HOTTER than the bottom. This is because UV radiation is absorbed by the top of the stratosphere to such an extent that the top of the atmosphere is the hottest part.
You might like to consider how this might change your simple metallic atmosphere model, Willis.
“And that is all that there is to the poorly named greenhouse effect. It does not require CO2 or an atmosphere, it can be built out of steel.”
This is untrue. What you see in the steel shell case is not “back-radiation” at all. It is merely reflection of the existing radiation from the planet – there is no change in wavelength. The steel shell will reflect the IR just as a mirrored surface will reflect visible light unchanged, and it will do so at the same angle of incidence too. CO2 actually re-radiates the IR but in all directions and at a different wavelength – quite different from your simplified model. Also, the CO2 only makes up 500ppm so it is nowhere near as dense as a steel shell at the photon level.
Furthermore, if the only heat source is actually outside the planet, how warm will the planet get then?
This steel shell model is typical of the over-simplified atmospheric models we get from Team-AGW to justify their claims. It has no business being considered here.
Konrad. says:
February 7, 2013 at 1:29 am
Konrad,
Your comment above contains many surprises, at least for me, but I find all of them plausible. You need to get published.
“Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards.”
Are you sure it warms up? If it is 100% reflective of IR it won’t absorb any energy – so it won’t warm up. What will be the temperature of the planets surface then? (Here’s a clue – it won’t just get hotter and hotter….)