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 8:34 pm
davidmhoffer says, February 6, 2013 at 8:12 pm: “2nd Law of thermo
Net energy flux between two bodies will flow from hotter to colder.”
=========================================================
No, the known historical statements do not contain the word “net” nor do they imply it, nor is there apparently any real scientific experiment confirming this notion.
I guess, that “net” thing is a trick of “climate scientists” to justify their “greenhouse effect”.
Nothing to do with ‘climate science’ it’s the basic physics of radiation heat transfer.
Arises from S-B law, light emitted by hot bb object = const.A.Th^4
light emitted by cool bb object = const.A.Tc^4
Net heat transferred between the two objects = const.A.(Th^4-Tc^4)
Since in our universe photons don’t carry with them a record of the temperature of the object which emitted them all photons of a given wavelength are absorbed equally regardless of the temperature of their source.
“mkelly says:
February 7, 2013 at 11:43 am
q = ε σ (Th4 – Tc4) Ac”
Emissivity is what changes when you add a shell to your greenhouse.
Cheers, 🙂
I live in the foothills of the mountains near the coast. Probably about 150 to 200m above sea level and about I km from the sea. From the back of my garden, the mountain slopes down to the sea and is covered in trees.
A couple of weeks ago, we had some windy weather. Probably force 5 gusting 6/7. The sun rises over the sea as seen from my back garden. Shortly after sunrise I was looking out and the low incidence of morning winter sun illuminated the spray being dragged off the top of the ocean. It was an impressive sight.
The skyline above the sea had a typical New York Skyscraper appearance. It is difficult to guage how high spray/spume was being sucked up but you could see a silhouette of what appeared to be mid to high level skyscrapers extending upwards perhaps 8 to 25 metres high. I am only guessing at the hieght in comparison to trees.
My point being that in these conditions force 5 gusting 6 to 7 which is not at all unusual for open ocean, and in these conditions there was clearly a layer of mist immediately above the ocean many metres high. That layer may often be transparent to the human eye, but in the backlit low incident sunlight conditions in which I was observing, it could clearly be seen. There is no reason to presume that a similar layer does not always exist in such windy conditions and to a lesser extent in lighter wind condiotions. Although this layer of fine mist may often be transparent to the human eye, it is not transparent to LWIR and would absorp DWLWIR as backradiated from upon high and would, because of the absorption characteristics of water, absorb that DWLWIR before it could reach the very top layer of the ocean itself.
Willis likes the real world but appears rarely to think about it when he comments upon DWLWIR (apart from his comments that clouds coming over head will quickly warm you up without actually dealing with convection)
Ryan says:
February 7, 2013 at 6:46 am
Lets consider what ACTUALLY happens with a CO2 molecule when a IR photon hits it. Remember that IR is not HEAT. It is an electromagnetic radiation with energy that can be converted to heat in certain circumstances. So a photon with energy in the IR spectrum reaches a CO2 molecule. What happens to it? Well if it is within the scope of the covalent bonds of the CO2 molecule it could be absorbed by the electrons in the bonds causing the electrons to be lifted to a higher energy level. But here is the crux of the matter – if the electron is already at that higher energy level then an incoming IR photon WILL NOT BE ABSORBED (at least not at the same frequency) – the photon will just travel straight through.
Again you are wrong, take a look at ro-vibrational spectra, Q-branch, just because a photon has been absorbed and a transition from v=0 to v=1 has occurred does not mean that a transition can not occur from v=1 to v=2.
But also bear in mind that lifting an electron to a higher energy level does not in itself cause heat to occur – the electron must drop back to its lower energy level and the assumption is that the electron then loses energy and that energy (probably) ends up at some point exhibiting itself as heat (just following the laws of entropy) but maybe not right away because the CO2 may just emit IR of a different wavelength (i.e. not heat as such) which may then be absorbed (or then again may not) by some other object which then gets a bit warmer.
At atmospheric pressure the most likely action is that collisional deactivation will occur due to collisions with surrounding molecules and those N2 and O2 molecules will heat up.
Michael Moon says:
February 6, 2013 at 9:37 pm
Willis,
You must consider something here. Energy, in the form of heating your house, or keeping your lights on with the electrical bill, has been the subject of, shall we say, INTENSE scrutiny, since
Edison in the 1870′s. When a large-gigantic-immense-overwhelming amount of money is spent on the subject of energy, the community of suppliers tend to research this subject, to learn how to satisfy their customers at minimal cost.
Check out double glazing!
If there were a way to heat anything by putting a shell around it which would magically double the “heat input,” otherwise known as Flux, then it would have been seen and exploited many many years ago.
Check out Dichroic Halogen Reflector (Lightbulb)!
mkelley asks: “Willis used the word “warms” which means the temperature increased. So there was a temperature change. Does this change your thinking?”
No. That doesn’t give me any reason to change my thinking 🙂
With no shell, the sphere will be 254 K (assuming it is a black body)
With a shell, the shell will be 254 K, and the sphere will be 302 K.
With two shells, the inner surface would be 334 K.
With three shells, the inner surface would be 359 K.
It all still works just fine. If you make it harder from IR radiation to leave, the system will warm up.
tjfolkerts:
“With no shell, the sphere will be 254 K (assuming it is a black body).
With two shells, the inner surface would be 334 K.”
This implies that the inner surface radiates at 3 times the flux density as with no shell. But surely it should be 4 times. This is my thinking –
The outer shell radiates at 1 times (say 1 unit) outwards and 1 unit inwards – 2 units in total. Therefore, the inner shell radiates 2 units outwards and hence 2 units downwards towards the sphere – 4 units in total. If it is radiating 4 units in total, then the sphere must be radiating 4 units upwards, not 3.
Where am I going wrong?
EForster says:
February 7, 2013 at 4:56 am
The missing link is that the shell radiates the same amount inwards as outwards. That radiation is absorbed by the planet, and adds to the energy of the planet. This leaves the surface warmer than it would be in the absence of the shell.
w.
Shawnhet says:
February 7, 2013 at 12:21 pm
“mkelly says:
February 7, 2013 at 11:43 am
q = ε σ (Th4 – Tc4) Ac”
Emissivity is what changes when you add a shell to your greenhouse.
Cheers, 🙂
The emissibity may or may not change, but you miss the point that when Th=Tc then the entire equation equals zero. No amount of change to the emissivity can fix that.
MikeB says:
February 7, 2013 at 11:37 am
As for being meaningless I’ll accepts Duke’s Physics Dept “Sloppy“, The reference to the molecule, was related to the 15u absorption band of Co2, as opposed to 4.5u to ~12u or 16.5 to 22u, where the photon is unlikely to be captured by Co2.
So, let me ask this, if a 15u wavelength photon is captured and thermalize by a molecule of Co2, what would it’s temperature be?
squid2112 says:
February 7, 2013 at 5:23 am
Wow, Willis, I have enjoyed so many of the things you have written here over the years, but this one, wow. Still clinging on to the “cold objects can make warmer objects warmer still” garbage?
I suppose I should write a post on the subject myself, but Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Still.
Thanks, squid. I love it that when you agree with me everything’s fine, I’m a man worth listening to, but when you don’t I’m suddenly some kind of ignorant fool “clinging” to “garbage” … a bit less abuse might be in order, my friend.
In any case, I suspect this is a sematics problem so let me restate it. Note that my restatement changes nothing about what I have said in the head post.
This is because rather than the warmer object receiving radiation from the slightly cooler object of say 400 W/m2, it is only receiving a paltry 315 W/m2 from the block of ice. As a result, it cools faster when it is next to the ice.
Can we agree on that, that a block of ice cools an object faster than a block of warm material?
Let’s take the next step.
This is because rather than the warmer object receiving no radiation at all (it being in outer space), it is receiving an extravagant 315 W/m2 from the toasty block of ice next door. As a result, it cools slower when it is next to the ice than when the ice is not there.
Now, this is the important part. Since the object cools slower with the ice next to it , it ends up warmer than it would be without the ice.
Now in this situation, in common parlance, we say that the ice is warming the warmer object. You may object to that choice of words, which is why I have provided the alternate statement above, that in outer space, a block of ice will slow the cooling of a nearby warmer object compared to the same situation with no ice.
Can we agree on that? If so, we have no disagreement, because that is the exact situation in the head post. The shell slows the cooling rate of the planet. As a result, the planet warms up until it is able to remove a NET 235 watts/m2 from the surface to restore the equilibrium. To do so, it must warm to 470 W/m2.
All the best,
w.
PS—To ward off the pickers of nits, yes, cosmic background radiation is not really zero but 3 W/m2 … so sue me.
mkelly says:
February 7, 2013 at 1:05 pm:”The emissibity may or may not change, but you miss the point that when Th=Tc then the entire equation equals zero. No amount of change to the emissivity can fix that.”
You’re right I don’t follow you, but changing the emissivity can allow the differential btw Th and Tc to grow. That was what I was referring to to wit(from your previous post):”A typical radiative heat transfer equation from Engineering Tool Box. Where in this formula can the radiation from the shell cause an increase in the temperture of the globe?”
Cheers, 🙂
KevinM says:
February 6, 2013 at 1:22 pm
….So Co2 AGW is real. The point to argue is that it is benign, and probably swamped by larger natural variability over short and long time scales.
>>>>>>>>>>>>>>>>>>>>>>>>>
The other point everyone neglects is the warmists only focus on one side of the whole picture. They focus only on the interaction of GHGs with earthshine. A look at the actual wavelengths for energy interaction (see Graph ) shows both CO2 and H2O interact with wavelengths in the solar as well as the earthshine bands.
The most important vibrational and rotational transitions for CO2 is
Center……Band interval
667…………..540-800
961
plus…………..850-1250
1063.8
2349………….2100-2400
Visible and near-IR absorption bands
2526………….2000-2400
3703………….3400-3850
5000………….4700-5200
6250………….6100-6450
7143………….6850-7000
Chart from http://irina.eas.gatech.edu/EAS8803_Fall2009/Lec6.pdf
Since earthshine is much lower in energy per wavelength band than sunshine, (See graph ) increasing GHGs could actually have a cooling and not a heating effect on the earth.
As far as I can see the only change that GHGs actually make is to make the weather milder by moderating the temperature swings. Day times are not as hot and night times not as cool ( see comment 1 and comment 2 and comment 3 ~ The comparison of the humid Brazilian rain forest and the dry N. African Desert show the day-night variation of ~ 10C with a high humidity vs a day-night variation of 35C without and an overall average temperature 8C higher for the desert.)
I do not have the in-depth physics knowledge to put an accurate number on it, but I really do not think a photon emitted by the earth hangs around for several hours before finally escaping to space even with the pinball machine effect of GHGs. Therefore you are just seeing the short lived effect of modifying the day/night temperatures. As I said the Warmists focus on the warming effect of the night temperatures only and ignore the cooling effect on the day time temperatures.
…………………
As far as the Wood experiment goes, It proves the ‘greenhouse effect’ in greenhouses is a property of glass. The Wood’s apparatus was too crude to actually capture the warmist’s greenhouse effect.
It would be interesting to do the experiment with very precise equipment and four boxes.
1. Pure Nitrogen
2. Dry Air
3. Air at 100% humidity
4. Pure Carbon Dioxide.
The experiment should be repeated several times with the boxes used for a different gas each time. (I still think the effect is too slight to be captured even with modern equipment)
Gail Combs
Thanks for your interesting post.
I will follow up your links.
Your post is well supported by traceable evidence.
Well done!
richard verney – Downward IR barely penetrates the oceans, but that strongly affects the “skin layer”, the top _millimeter_ of water. Increased air temperature and DLWR reduce the temperature gradient across that skin layer, reduce the energy flow out of the (warmer) near-surface ocean, and change the energy balance. That _reduction in cooling_ means visible light energy (penetrating much further) accumulates and warms the oceans until the thermal gradient is reestablished, until the heat leaving the oceans matches that arriving.
All of which is but one aspect of increased greenhouse gases _reducing_ energy loss to space, causing the climate to warm until outgoing = incoming energy.
See http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/ for a discussion of this, including both references and data demonstrating this relationship.
wayne;
This is where the I.P.C.C.’s Free Energy Oven fails.
>>>>>>>>>>>>>>>
I’ve read AR4 WG1 front to back and nowhere is such a claim made. If you understood the first thing about the physics then you might be able to understand the claim they are actually making and attack that instead of demonstrating that you haven’t a clue what you are talking about.
squid2112;
So you are suggesting then that the IR from the snow kept you warm?
>>>>>>>>>>>>>>>>>>
No. I am stating it as a fact. The house is warmer with snow packed around it than without, the snow is colder than the house. The IR from the snow can be directly measured.
Willis, I agree with your description. It got me thinking how we might tie this closer to our real situation. What I found is interesting so please feel free to correct me where I go off the tracks.
Let’s modify this experiment a bit and see where it takes us. For simplicity lets assume 240 w/m2 as our heat source. In addition,
1) A perfect conducting wire connects the heat source to the outer shell and transfers 80 w/m2 to the shell. This is done to simulate our atmosphere absorbing about 1/3 of the energy from the Sun.
2) Another wire connects the surface to the shell and transfers half the energy the surface receives to the shell. This is done to simulate latent heat and conduction that is convected away from the surface.
OK, at time T1 we should see the shell receiving 240 m2 (80 directly via 1 + 80 directly from 2 + 80 radiated from the surface.) It would radiate 120 to space and 120 back to the surface. If we continue over time the shell will eventually warm up to 480 w/m2 with 240 going to space creating an equilibrium condition. However, the surface will be radiating 200 w/m2 and conducting 200 w/m2 for a total 400 w/m2.
Essentially, the surface warmed from the initial 160 w/m2 to a new equilibrium of 400 w/m2. Note that the surface does not reach the 480 w/m2 in the base case Willis described. What this shows is the energy that never makes it to the surface cannot warm the surface through the GHE. If we consider our own atmosphere this means only the 160 w/m2 that reaches the surface can be considered in computing the GHE. In other words, the GHE is much stronger than sometimes indicated.
Also, it did not matter how the energy is transferred to the shell. If I remove 2) the final result is still the same. This appears to mean the GHE effect is independent of how the energy reaches our atmosphere. This would appear to raise questions for the concept of an ERL. There is no dependence on the height of the shell.
Note that you can’t heat the surface any more unless we add another shell.
We know the real atmosphere does not absorb all the radiation from the surface. If it did we would be at the maximum GHE. So, how much radiation goes directly to space? According to the KT cartoon that is 40 w/m2. And, this number must be reduced anytime we want to increase the GHE.
So, let’s take our situation above and reduce the surface radiation by 40 w/m2. This reduces the surface to 360 w/m2. But we know from measurement that our surface temperature is 288K which corresponds to 390 w/m2. In other words, our temperature already exceeds the value that we would expect if 40 w/m2 was passing through the the atmosphere without being absorbed. Interesting.
This should tell us that my assumptions above are a little bit off. More energy must be reaching the surface than assumed using the 160 w/m2 KT cartoon value or more energy is produced internally.
One other thing to consider is the maximum temperature of the surface in this example is 290K (corresponds to 400 w/m2). There is no way the GHE can warm our planet above this value given my assumptions. And, even if you assume we could trap the 40 w/m2 that is not currently absorbed the maximum only rises to 295K. Warm, but no boiling oceans and no Venus.
I should add that this experiment only uses a single shell. If our real atmosphere more closely appears as multiple shells then it would mean the surface could warm more. And yeah, I’m taking a lot of license here in comparing the real situation with this thought experiment. However, there does seem to be something to the idea that 160 w/m2 reaching the surface has a maximum GHE and we are darn close.
Vince, I had to think a little to make sure I was right about the radiation of the shells. (I actually worked out the gory details for arbitrary emissivities once — I should write that up some time).
Anyway, at the bottom layer (let P = 235 W/m^2)
+P from heater + 2P from above – 3 P radiated up = balanced energy
At next layer
+ 3 P from below + 1 P from above – 2P down -2P up = balanced energy
At top
+2P from below -P down – P up = balanced energy.
tjfolkerts says, February 7, 2013 at 9:59 am: “The Wood experiment is similar in some ways, but very different in others, which is the point hat Willis was making and which you apparently missed.
> The earth is surrounded by extreme cold (outer space), but Wood’s experiment is surrounded by materials at very similar temperatures to the experiment itself.
> The atmosphere of the earth is many KILOmeters in size, but Wood’s experiment is many CENTImeters in size.
These (and other rather fundamental differences) make Wood’s experiment a rather poor analogy to earth’s situation.”
=============================================================
Yes, there are 1,000,000 differences between the boxes in the Wood experiment and the Earth, but those differences are irrelevant, and the Wood experiment was not meant to be an “analogy to earth’s situation”. It deals only with the underlying mechanism of the alleged “greenhouse effect” as presented by the IPCC: effect of the trapped/back radiation on the temperature of the source. It does not even deal directly with the so called “greenhouse gases”, because it is not necessary.
Willis — I think I will have to sue you 🙂
The temperature of space is about 3 K, but the power is about 0.000003 W/m^2. (Of course, that makes your point even more emphatically.)
And one other nit-pick. I was a little confused when you said “In the vacuum of outer space, a warm object with a block of ice nearby will cool faster than a warm object with a slightly cooler object nearby.” The “sightly cooler object” is slightly cooler than the ORIGINAL object, but the first time I read it, I was thinking slightly cooler than the BLOCK OF ICE. Obviously that makes a huge difference in the interpretation. I’m not sure how to say that so there is less room for misinterpretation.
Gail;
(I still think the effect is too slight to be captured even with modern equipment)
>>>>>>>>>>>>>>>>>>
Actually it isn’t.
http://www.john-daly.com/artifact.htm
There’s a zip file at the top with criticisms of this experiment that I consider valid. At least it is a real experiment though, done with precision equipment and not some joke with cardboard boxes in thermal equilibrium with the salt and glass windows, neither of which are thermally isolated by a vacuum, and measured glass thermometers that are two orders of magnitude less accurate than what would be required to measure the radiative ghe in the first place. Did I say joke? It doesn’t ascend to joke status and the manner in which it is being misused to promote misunderstanding isn’t at all funny.
Phil. says, February 7, 2013 at 11:45 am : “Wood’s experiment is fatally flawed as a description of the atmospheric GHE for several reasons.”
========================================================
The Wood experiment was not meant to be a description of the “atmospheric GHE”, it was only meant to check whether the underlying mechanism of that alleged “atmospheric GHE” (effect of the trapped/back radiation on the temperature of the source) worked or not. It did not. Bad for the “atmospheric GHE” concept (as presented by the IPCC).
Phil. says, February 7, 2013 at 12:08 pm: “Nothing to do with ‘climate science’ it’s the basic physics of radiation heat transfer.
Arises from S-B law, light emitted by hot bb object = const.A.Th^4
light emitted by cool bb object = const.A.Tc^4
Net heat transferred between the two objects = const.A.(Th^4-Tc^4)”
==========================================================
This “net” thing does not “arise” from S-B law, and it has apparently never been proven experimentally. As long as it has not been proven experimentally, it remains a fiction. A sort of “radiation arithmetic” without any basis in science.
Gale Combs says
“The other point everyone neglects is the warmists only focus on one side of the whole picture. They focus only on the interaction of GHGs with earthshine.”
I agree, and it is their downfall.
For these wavelengths > 3um the classical Rayleigh – Jeans formula holds.
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html#c4
The classical model holds with no photons.
The Poynting Vector gives the direction of heat flow.
No reference whatsoever to backradiation.
Yet real heat transfer problems yield the correct answers!
Only with wavelengths < 3um(Solar) do we have to include quantum effects and the Rayleigh -Jeans model fails.
The Physics of Electromagnetic radiation in the vaccuum is a well founded robust theory. So much so, that the only exact fundamental constants of Physics are related to EM radiation theory.
The vaccuum of empty space has a Permittivity (Epsilon nought) that has an exact value. That constant gets involved in the capacitance relating to charge storage on conductors. Free space also has a magnetic Permeability (Mu nought) that has an exact value. That constant gets involved in the magnetic fields surrounding current carrying conductors.
Those two exact values also yield the velocity of electromagnetic radiation waves (c).
c = 1/sqrt (munought x epsilonnought) which thus also has an exact value.
Less well known or at least discussed is the remaining exact value fundamental physical constant which I will call Znouhgt, as I don’t recall right now what they really call it. It is the “characteristic impendance” of free space and is approximately 377 Ohms.
Well its exact value is 120 Pi Ohms. Paint your car with 377 Ohm paint and it will disappear. Well strictly speaking it will reflect no EM radiation of any knind, no matter what. It may not let you see through the car and out the other side .
I have no idea what if any effect either dark matter, or dark energy have on EM radiation if either exists.
So the point of this is that the vaccuum velocity of EM radiation, is not matched by the velocity of EM propagation in ANY real physical material medium (besides vaccuum). EM radiation always travels slower in any real physical material medium; well the phase veolicty is always lower.
The ratio of (c) to the velocity of EM radiation in a real medium is the refractive index of that medium.(N)
The characteristic impedance of the real medium is also different from 120 Pi Ohms (lower).
Whenever an EM wave travelling in a medium, encounters a different medium interface, the wave splits into two components; a transmitted wave which propagates into the denser medium, and a reflected wave which travels back into the original medium.
Going from the less dense medium, into the denser medium, also involves a phase change of pi radians, so the reflected wave is inverted. Going from the dense medium, into the less dense medium, there is no phase inversion. Most real media actually have a complex refractive index, due to surface phenomena at the interface, so there always can be a small phase shift, which manifests itself as an effective change in optical thickness of the medium.
The simplified reflection coefficient for EM radiation incident normally on the interface is given by:-
R = ((N1-N2) / (N1+N2))^2
N is 1.000…. for the vaccuum and something >1 for all normal real materials, so the reflection coefficient can never be zero except perhaps over some small frequency range (which might be of some use) But never over the whole DC -infinity sans ends range of EM radiation.
Maxwell’s Fisheye, and the Luneberg Lens are two hypotheticaloptical devices which have remarkable optical properties, but require that the material refractive index (graded index materials) be 1.00 at the surface of the device, and increase further from the surface. The Luneberg lens can be implemented approximately at microwave frequencies, with foam type materials. Otherwise they are hypothetical curiosities.
Ergo, no real physical body can absorb ALL em radiation that falls on it; so there can be no black bodies.
Luckily, we can emulate a BB over limited frequency ranges, so as to derive useful effects. And for such experimental devices, the bulk of the radiation closely approximates what the Planck and Stefan-Boltzmann, and Wien formulas for BB radiation predict. And that is why BB radiation theory is so important, even though it is quite fictional to the point that no physical observations we make of the real universe match it, except over small frequency or wavelength ranges.
Willis shell planet assumes that BB formulae are operating, yet his explanation of the model, with two completely different Spectra inside the shell, imply that the total radiant flux inside the shell is NOT a BB rtadiation spectrum at ANY isothermal Temperature.
All of which says nothing about the valididty of what we call the greenhouse effect, or whether real greenhouses or the Woods experiment, behave as is claimed.
Oddly there is a fifth Physical number that is exact.; but trivially so. That number is (g) the acceleration due to gravity. I’ll let you giggle the exact value for yourselves.
That just tells us when we say an acceleration is umpteen “g”s we meen umpteen times that number. OK so Ornithorynchus is a platypus. So we can just say umpteen platypi, rather than ornithorynchi; big deal. (g) is not exactly a fundamental physical quantity.