Guest post by Willis Eschenbach
There is a lot of misinformation floating around the web about how the greenhouse effect works. It is variously described as a “blanket” that keeps the Earth warm, a “mirror” that reflects part of the heat back to Earth, or “a pane of glass” that somehow keeps energy from escaping. It is none of these things.
A planetary “greenhouse” is a curiosity, a trick of nature. It works solely because although a sphere only has one side, a shell has two sides. The trick has nothing to do with greenhouse gases. It does not require an atmosphere. In fact, a planetary greenhouse can be built entirely of steel. A thought experiment shows how a steel greenhouse would work.
Before we start, however, a digression regarding temperature. The radiation emitted by a blackbody varies with the fourth power of the temperature. As a result, for a blackbody, we can measure the temperature in units of radiation, which are watts per square meter (W/m2). For objects with temperatures found on the Earth, this radiation is in the range called “longwave” or “infrared” radiation. See the Appendix for the formula that relates temperature to radiation.
This means that we can denote the temperature of a blackbody using W/m2 as well as the traditional measures (Fahrenheit, Celsius, Kelvin). The advantage is that while temperature (degrees) is not conserved, energy (W/m2) is conserved. So we can check to see if energy lost is equal to energy gained, since energy is neither being created nor destroyed by the climate.
For our thought experiment, imagine a planet the size of the Earth, a perfect blackbody, heated from the interior at 235 watts per square meter of surface area. How is it heated from the interior? Doesn’t matter, we’ll say “radioactive elements”, that sounds scientific.
The planet is in interstellar space, with no atmosphere and no nearby stars. The equilibrium surface temperature of this planet is, of course, 235 W/m2. To maintain the equilibrium temperature, it constantly radiates this amount of energy out to space. Coincidentally, this is the amount of solar radiation that makes it past the clouds to warm the Earth. If we convert 235 W/m2 to one of our normal temperature scales, it is -19 Celsius (C), or -3° Fahrenheit (F), or 254 Kelvins (K). It’s a possible temperature that the Earth might have if there were no greenhouse effect. That is to say … cold.
Now imagine that the planet gets completely surrounded by a thin black steel shell, located a few thousand meters above the surface, as shown in a cutaway view in the picture above, and in Figure 1 below. What happens to the surface temperature of the planet? (To simplify calculations, we can assume the shell has the same outer surface area as the surface. For an earth-sized planet with a shell two kilometers above the surface everywhere, the difference in area is only six-hundredths of one percent. This assumption makes no difference to the argument presented.)
In order to maintain its thermal equilibrium, including the new shell, the whole system must still radiate 235 W/m2 out to space. To do this, the steel shell must warm until it is radiating at 235 watts per square meter. Of course, since a shell has an inside and an outside, it will also radiate 235 watts inward to the planet. The planet is now being heated by 235 W/m2 of energy from the interior, and 235 W/m2 from the shell. This will warm the planetary surface until it reaches a temperature where it radiates at 470 watts per square meter. In vacuum conditions as described, this would be a perfect greenhouse, with no losses of any kind. Figure 1 shows how it works.
Figure 1. Building a steel greenhouse. (A) Planet without greenhouse. Surface temperature is 235 W/m2 heated from the interior. (B) Planet surrounded by steel greenhouse shell. Shell radiates the same amount to space as the planet without the shell, 235 W/m2. It radiates the same inward, which warms the planet to 470 W/m2 (29 C, 83°F, 302 K). [Clarification added] Note that the distance from the shell to the planet is greatly exaggerated. In reality, it is close enough to the planet that the difference in the areas can be neglected in practice.
The trick can be repeated again, by surrounding the planet and the shell with a second outer shell. In this two shell case, the planetary surface temperature (in W/m2) will be three times the initial surface temperature.
In nature, planets have atmospheric shells, of course, rather than steel shells. The trick works the same way, however. Rather than being warmed from the inside, the Earth receives 235 W/m2 from the sun. Solar radiation passes through the atmosphere. Outgoing longwave radiation is absorbed by the atmosphere, just as it is absorbed by the steel shell shown in Fig. 1.
So that’s the trick of the greenhouse. It has nothing to do with blankets, mirrors, or greenhouse gases. It works even when it is built out of steel. It depends on the fact that a shell radiates in two directions, inwards and outwards. This radiation keeps the planet warmer than it would be without the shell.
Now, it is tempting to think that we could model the Earth in this manner, as a sphere surrounded by a single shell. This is called a “two-layer” model, with the two layers being the surface and the atmospheric shell. In fact, a number of simplified climate models have been built in this way. Unnoticed by their programmers, however, is that is not physically possible to model the Earth as a two-layer system. Figure 2 shows why. Note that in all cases, the system has to remain in thermal equilibrium. This means that the surface must radiate as much as it absorbs, and the shell must also radiate as much as it absorbs. In addition, radiation from the shell to space (upwelling longwave radiation or ULR) must equal radiation from the shell to the Earth (downwelling longwave radiation, or DLR)
Figure 2. Single-shell (“two-layer”) greenhouse system, including various losses. S is the sun, E is the Earth, and G is the atmospheric greenhouse shell around the Earth. The height of the shell is greatly exaggerated; in reality the shell is so close to the Earth that they have about the same area, and thus the small difference in area can be neglected. Fig. 2(a) shows a perfect greenhouse. W is the total watts/m2 available to the greenhouse system after albedo. Fig. 2(b) is the same as Fig. 2(a) plus radiation losses Lr which pass through the atmosphere. Fig. 2(c) is the same as Fig. 2(b), plus the effect of absorption losses La. Fig. 2(d) is the same as Fig. 2(c), plus the effect of thermal losses Lt.
Figure 2(a) shows the same situation as Figure 1(B), which is a perfect planetary greenhouse. In this case, however, it is heated by an amount of energy “W”, which is coming from the sun. The planet receives solar radiation in the amount of “W” from the sun, and longwave radiation “W” from the atmospheric shell. The surface temperature is thus 2W. All energy flows are in Watts/square metre (W/m2).
Figure 2(b) adds two losses. The first is the reflection from the Earth’s albedo (Wo – W). This is energy which never enters the system and is reflected back into space. We are still getting the same energy entering the system (W). The second loss Lr is from radiation which goes from the surface to space through the “atmospheric window”. Because of the second loss, the surface of the Earth does not get as warm as in a perfect system. In a perfect system, the temperature of the surface is 2W. But including the radiation loss Lr, the surface temperature drops to 2W – 2Lr.
Figure 2(c) adds another loss La, which is the solar radiation which is absorbed by the shell. This cuts the radiation hitting the surface down to W – La. Including this loss, the surface temperature is 2W – 2Lr – La.
Figure 2(d) adds the final loss. This is Lt, the thermal loss. It is the sensible energy (energy you can feel) and latent energy (evaporation) which is transported from the surface to the shell by convection. Including all of these losses, the surface temperature of the Earth is 2W – 2Lr – La – Lt.
Now, why can’t the Earth be modeled in this manner? A look at the size of the various losses shows why. Here is the canonical global energy budget, called the “Kiehl/Trenberth” budget, or the K/T budget.
Figure 3. The Kiehl/Trenberth Global Energy Budget. This is a “two layer” representation, with the surface and the atmosphere being the two layers. Lr, the radiation loss, is the 40 W/m2 labeled “Atmospheric Window”. La, the absorption loss, is the 67 W/m2 labelled “Absorbed by Atmosphere”. Lt, the thermal loss, is 102 W/m2. This is the sum of the 24 W/m2 labelled “Thermals” and the 78 W/m2 labelled “Evapo-transpiration”. W, the energy from the sun, is the incoming solar radiation of 342 W/m2 less the 107 W/m2 that is reflected by the surface and the clouds. This means that W is 235 W/m2. SOURCE
What’s wrong with this picture? Note that the temperature of the Earth is 390 W/m2, labeled as “Surface Radiation”. This is 15 C, or 59°F. But from Fig. 2(d), we know that the surface temperature of a greenhouse system with a single shell, as shown in the drawing, is 2W (470) – 2Lr (80) – La (67) – Lt (102) = 221 W/m2. This is far below the known surface temperature of the Earth. In other words, a single shell greenhouse system simply isn’t efficient enough to give a surface temperature which is warm enough to allow for the known losses.
So where is the problem with the K/T budget diagram? The hidden fault is that the upward radiation from the atmospheric layer does not equal the downward radiation. There is 195 W/m2 going to space from the atmospheric shell, and 324 W/m2 going down to the surface.
In order to get enough energy to allow for the known losses, the simplest model requires two atmospheric shells. A perfect greenhouse with two shells would give a surface temperature of 3W, or 705 W/m2. This is enough to allow for the known losses, and still give a surface temperature which matches that of the Earth. Figure 4 shows one such possible system.
Figure 4. Simplest greenhouse global energy budget capable of representing the Earth. Note that all major flows in the K/T energy budget have been maintained. There are two shells, which must be physically separate. These are the lower troposphere and the lowest stratosphere. They are separated by the tropopause.
This budget fulfills all of the requirements for thermal equilibrium. The same amount is radiated upwards and downwards by each layer. The amount absorbed by each part of the system equals the amount radiated. Further verification of this energy budget is the 147 W/m2 emission from just above the tropopause. This converts to a Celsius temperature of about -50 C, which is a typical temperature for the lowest part of the stratosphere.
I have written a simplified Excel radiation/evaporation/convection model which encapsulates the above system. It is available here. Click on the “File” menu on the webpage and select “Download”.
I invite people to play with the model. It is what I term a “Tinkertoy” model, which is what I call the simplest possible model that can represent a particular reality. One of the interesting results from the model is that there can be very little thermal loss between the inner and the outer atmospheric layers. If there is more than a trivial amount of leakage, the surface cools significantly. Among the other insights yielded by the model is that a change equivalent to a doubling of CO2 (an increase of 3.7 W/m2 downwelling radiation at the top of the atmosphere) can be canceled by a 1% increase in the upper and lower cloud reflections.
Experiment with the values, it is an interesting insight into the energy flows in the simplest possible climate model that can represent the Earth’s greenhouse system.
APPENDIX
The formula that relates the temperature to radiation is called the “Stefan-Bolzmann” equation:
R = sigma * epsilon * T^4
where r = radiation (W/m2)
sigma = the Stefan-Boltzmann constant, 5.67 * 10^-8
epsilon = the emissivity of the body, which for a blackbody = 1
T^4 = absolute temperature in Kelvins (which is Celsius plus 273.15) raised to the fourth power
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“”” Ian Schumacher (12:47:38) :
Willis,
You haven’t replied to my objection above that you can’t heat something above the temperature of a blackbody using an external energy source. “””
Well Willis’ system has an internal energy source; not an external one; so your objection does not apply to his system.
Juraj V. (12:10:02) :
Not sure of your point. While there is a higher concentration (in parts per million of atmosphere) on Mars, the atmosphere is so thin that very little radiation is absorbed in the atmosphere.
See the budgets above. The answer is 390, 24, and 78 W/ms respectively.
This is a common misconception. People say “If there is radiation from the night sky, why don’t I feel a lack of warming when I go under my porch roof. After all, the porch roof should intercept the downwelling radiation.” The missing link is that not only is there infrared radiation from the night sky, there is even more infrared radiation coming from the porch roof.
It’s not like the sun, where the roof gives you shade because it is not radiating in the visible spectrum. In the infrared spectrum, the porch roof is radiating, and is radiating more strongly than the atmosphere.
A better guide is that during the winter, the clear nights are the coldest. This is because clouds absorb upwelling IR, and emit half of it back towards the earth. This warms the surface through downwelling radiation. The effect is quite perceptible when a single cloud passes over on a clear winter night, the warmth is immediately evident.
“Denier” is an ugly term, with overtones of the Holocaust deniers. In addition, what do you deny? Above, you deny that there is downwelling IR, despite it being physically measured by scientists around the world using instruments (radiometers) made specially for that purpose on a daily basis. This does not bode well for your career as a professional denialist.
I describe myself as a skeptic, as it is the duty of a scientist to be skeptical. As my grandma used to say, “You can believe half of what you see, a quarter of what you hear … and an eighth of what you say” …
Willis Eschenbach
“In theory you are correct. The energy which is absorbed is not “re-radiated”. Energy is absorbed. Energy is radiated. They are not the “same energy”.”
In fact they are. Just the other day I made a green dot on an infrared photon nearby, using an ordinary DVD marker pen, and had it absorbed by a CO2 molecule. About a millionth of a second later, a photon was emitted – and believe it or not, the green dot was still there, albeit a little smudged. If you don’t believe me – try it yourself!
🙂
Willis,
This higher temperature on earth’s surface is because of gravity potential energy field. The energy density will be approximately equal throughout the atmosphere, but molecules higher in the atmosphere have higher potential energy and lower kinetic energy and molecules lower in the atmosphere have lower potential energy and higher kinetic energy.
However, if we can, lets focus on an ‘on earth’ experiment that would demonstrate this effect so we can get rid of all the extra complexities of gravity, rotating spheres, etc.
So you think that black sphere inside a series of glass spheres (with vacuum between them) would result in a temperature several times higher than an ideal blackbody in the same conditions? I don’t think so.
George,
I pointed out the problem of internal heat earlier which was my original objection. And external energy source needs a hole and so we have a black body approximation.
“”” Willis;
There is no doubt your steel sphere analogy is correct. In fact one can buy home insulation based on it. The insulation consists of a stack of aluminium foils and is used where space is at a premium. However, with regard to your two shell model consider one simple issue. The tropopause is colder than both the rest of the troposphere and the stratosphere, how can you model explain that fact. How can the region between two shells be colder than either shell? For a region to be colder it must be losing energy to a heat sink colder than it is but the only thing colder than the tropopause at -57C is outer space at -269C. That means the tropopause has to be losing energy directly to space. How is that possible if there is an opaque shell above it? “””
Well that is because you are misunderstanding the second law of thermodynamics.
One version of that law says “heat” cannot flow unaided from a source at one temperature to a sink at a higher temperature.
That also is NOT happening in your Tropopause case.
But the second law says nothing at all about the radiation from a body above zero K temperature.
The earth emits a certain amount of LWIR radiation in a spectrum corresponding to a black body (roughly) at 255 or thereabouts Kelvins.
Some of that radiation falls on and is partly absorbed by the surface of the moon, which in the daylight sky could be much colder than the earth.
But exactly the same amoutn of LWIR energy in that 1/2 degree cone angle, also falls on the sun which is at 6000 K or so, and it passes through out layers of the sun, which are at million degree temperatures.
You see photons do not have any temperature associated with them, and they can go anywhere they please. “Heat”, on the other hand is the mechanical energy of motion or vibration etc of atoms or molecules, and Electromagnetic radiation does not need any molecular material to propagate.
So there is nothing to stop your colder tropopause from radiating energy right through any hotter layers of material, above or below.
Lots of people are not answering objections from physicists, let alone myself. Especially Willis.
On a planet with no atmosphere surrounded by a steel shell, where the planet is generating 235 W/m^2, then the radiation will warm the shell. But unless the shell is a perfect non-absorber, the heat of the shell will radiate out into space.
Thus the temperature of the shell will be less than the surface of the planet.
Since in thermodynamics cool objects do not heat warmer objects, the shell cannot heat the planet any more than the inside of a Dewar (or vacuum) flask heats the food contained therein.
If you put a radioactive element into lukewarm soup in a Dewar flask in order to maintain its temperature and put the flask into a refridgerator, then the outside skin of the Dewar would be lower than the inside skin which would be lower than the soup temperature.
In all of this, quite a lot of people have displayed astonishing ignorance of the “greenhouse effect”. Greenhouses do not warm up because of suppression of radiation but because of suppression of convection (blankets work the same way).
The atmospheric “greenhouse effect” works by suppression of radiation, not convection.
Jari (13:08:17) :
certainty the numbers are wrong. 235w/m2 has never been recorded as leaving any point on earth. Its more like 35w/m2 at 59F. Earth doesn’t give off much radiation. 500w/m2 is sufficient energy to bring sunflower oil to its boiling point
When was the SB constant first usadapted in climatology? It is important to know in order to correct it – as it exagerrates 10 times the radiative effect of gases and liquids, to which it isn’t supposed to be applied to in any case. Its like measuring the volume of a car by using boyles law. I susppect it is NASA
Steve (13:14:42) :
I would suggest you re-examine figure 2a. There is not a discrepancy. Of the 2 W radiated by the earth, exactly 1/2 of that is reflected back, which is made clear by the short 1 W arrow.
Mark
Steve (13:14:42) :
Consider each part of Fig. 2a separately. The Earth receives 2W, and radiates 2W. The shell receives 2W, and radiates 2W. The entire system receives W, and radiates W to space. Every part is in balance. There is no requirement that the flows total to zero, only that the amount absorbed by the entire system or any part of the system is equal to the amount radiated.
temperatures stabilise according to air temperatures than the value attributed from their “emissivity”. The emissivity of something is a function of its temperature, and if it is in equilibrium with the surrounding atmosphwere then its not radiating that much heat.
Try a standard 7-14 micron IR camera one of these nights and see just how little radiation leaves the earth.
John A (13:33:44) :
I am attempting to answer all reasonable objections. If I have missed some, please let me know.
As long as the planet is generating 235 W/m2 and the shell is radiating less than 235 W/m2, the system will continue to warm. It will not stop warming until the amount radiated to space equals the amount generated by the radioactive elements in the planet.
A classical example in physics is that of two planets, one warmer than the other. We know that both emit radiation according to the S/B equation. Now, some of the radiation emitted by the cooler planet strikes the warmer planet. Is it truly your argument that this does not add energy to (warm) the warmer planet?
In a perfect dewar, the inside skin of the dewar would be at the same temperature as the liquid. And yes, just as in my figures above, the shell is cooler than the radioactive liquid. Not sure what your point is here.
I’m not aware of a phenomenon called “suppression of radiation”. If you mean absorption of energy and subsequent radiation of energy, I’m not sure what your point is, as that is what I am saying.
And yes, greenhouses don’t work the way that people think.
George Smith;
You have misunderstood. Of course photons can travel from a cold object to a hotter object and it does not contravene the 2nd law of thermodynamics at all. The point is that the hotter objet would be radiating even more photons to the colder object. The cold object would receive more photns than it emits and therefore warm up. But the tropopause IS colder than the stratosphere. The only way this can come about is f as you say the tropopause radiates photons through the stratosphere to space. That bit is fine but there is a second requirement and that is that the strtatosphere does not radiate significant photons back to the tropopause despite the fact that it is warmer. Can that be – YES it most certainly can be the case. It means the stratosphere has very low emissivity so it is incapable of radiating at least it is incapable of radiating energy at wavelengths that the tropopause can absorb. But if it is not radiating it is also not absorbing (radiation and absorption of energy are controlled by the same factor – a consequence of Kirchoffs law) hence the statosphere cannot be an opaque (or “steel”) shell. It must be transparent at the greenhouse waelengths corresponding to CO2 and water. In fact the stratospeher is warmer because the UV light from the sun forms ozone which then absorbs UVC and some UVB enenrgy from incoming sunlight. This can be re-radiated to space at the ozone emission line at about 10 microns.
I suggest you all reexamine figure 2a. Heat source S isn’t providing any heat W to shell G. Miraculously, it’s going write through the steel shell, directly to sphere E. The vector line from S should stop at the surface of shell G. Shell G is the only source of heat W available to sphere E (we’re saying it’s opaque steel, right?)
Then you would see clearly that:
Sphere E receives 1W from G, but radiates 2W!
Shell G receives 1W from S, 2W from E, but only radiates 2W!
This thought experiment is all monkeyed up. To create a proper greenhouse your gap has to hold heat! This gap is “void”, so holds no heat. With a little visual jujitsu you are miraculously creating heat.
It’s not April Fools yet, but this is a joke thread, right?
Going back to internal energy source only with steel shell. As Seedload points out the shell has 2 sides and therefore twice as much surface area.
If earth radiates 100 W/m^2, then the shell will heat up and radiate 50 w/m^2 inward and 50 w/m^2 outward (it has twice as much area remember). As pointed out by John A , the shell will be cooler than the earth because it has one side exposed to warmth and one side exposed to cold (empty space). It will not be the same temperature as earth.
Now place a second shell. What happens? The outside shell will radiate 25 W/m^2 inward and outward. The inside shell will radiate 50 w/m^2 in and out. The Earth is no warmer than with a single shell.
And or course I’m just doing this to point out the flaws with this model, ignoring for the moment that this has no relationship to the greenhouse effect whatsoever. A real greenhouse needs to let energy in from outside. We need a hole. Holes are where energy can escape from also. A spherical cavity with a hole is an example and is ‘close’ to a blackbody, therefore at MOST we can expect the earth to have energy densities of a black body (yes gravity complicates the actual temperature on the surface).
John A; You make the same mistake that many many other mave made. You claim that the second law of thermodynamics states a cold object cannot warm a warmer object therefore the cold object cannot radiate heat back to the warmer object. Your argument is simply wrong and on this point I guarantee you I do know what I am talking about.
An object above absolute zero and emissivity above zero will radiate energy. When that energy is radiated the object does not know where the photons are headed, they simply are emitted. If it happens they are headed towards a warmer object they will strike that warmer object and transfer energy to it. What the second law states is that in such a circumstance the warmer object will be doing the same thing and since it is warmer (assuming both objects have the same emissivity) it will be emitting more photons thus on average the warmer object will send more photons to the colder object than it receives in return resulting in net heat flow from warmer to colder as required by second law.
In this case it might help to look at the situation a different way. If the atmosphere were not there the surface would be sending radiation to space (temperature -269K) so it would recieve almosty nothing back in return. Putting the atmosphere in place means the sruface is radiating not to very cold space but to the much warmer atmosphere (even though it is still colder than the surface). Thus it receives more energy back from the atmosphere than it would from space and hence it cools more slowly which is another way of saying it is warmed relative to if it were radiating to space.
You can easily try this for yourself. Go to a supermarket and stand in front of the vetical refrigerator section. Does it feel cold? Why? (because your body is radiating energy to a very cold sink and getting very llittle back in return). Now stand in front of a non refrigerated shelf. Does it feel warmer than when in front of the refrigerator? Why (because your body is now radiating to a warmer object and getting more heat energy back in return)
P Wilson (13:34:31) :
Jari (13:08:17) :
I’d need a citation for both of those claims. The earth is well known to give off radiation, and has been known to do so for years.
Again, please provide a citation that says that the SB law doesn’t apply to liquids or gases. The only paper I know of which makes that claim is the unpublished paper of Gerhard Gerlich and Ralf D. Tscheuschner. I don’t know of any serious physicists who believe their claims. For example, they say that the Stefan-Bolzmann constant is not a constant …
“Climate Near The Ground”, by Rudolph Geiger, is one of the canonical texts in the field. It was first published in 1927, long before NASA. You really should get a copy and read Chapter 1, “Earth’s Surface Energy Budget”. It will answer a number of your questions.
The second shell will warm up the first shell a little, so my math there is wrong, but the idea is the same in that the steel shell model has the steel spheres radiating 2 times to much energy.
Okay, first off… I give up trying to get the author to see reason… apparently it just isn’t going to happen.
However, I would like to venture into something that I don’t know much about, and could use some help.
Every article I have seen assumes that IR trapping from greenhouse gasses is a legitimate mechanism… but I have not found even one example of an experiment under laboratory conditions demontrating that re-radiation of trapped IR can warm anything. In fact, the only experiment I have heard about was run around 100 years ago, and involved to closed boxes, one covered with glass and one covered with a transparent salt compound. Glass is opaque to IR radiation, the salt compound was not.
If the “greenhouse” effect involved trapping IR in any way, then when placed in sunlight, there should be a temperature difference between the interiors of the two boxes (the glass covered box should be warmer) But, no temperature difference was observed. Now, I’m not saying that this 100 year old experiment is conclusive… but for the life of me I can’t find anything that refutes it! Can someone provide something?
In posts above, someone mentioned that the atmosphere acts like insulation on your house. This seems to me to be a good analogy. Your house insulation works by simply slowing the rate of heat loss of you house… but clearly there is no re-radiation occurring from your insulation back to the house. As also mentioned above, the atmosphere in sunlight mostly shields the earth from the sun (the moon in sunlight is much hotter), and then conversely slows the rate of heat loss on the night side. This essentially is a moderating effect, and does not seem to be dependent on re-radiation of IR trapping greenhouse gases in anyway, except to the extent that it slows the rate of heat loss at night (which may very well be a real – albiet a small – effect).
So, again, can anyone show the results of a careful experiment that demonstrates that the so-called greenhouse effect is real? From what I can see, re-reradiation plays no role at all, and convection etc. are the real players.
-Anton
George E. Smith (13:27:56) :
“”” Willis;
There is no doubt your steel sphere analogy is correct. In fact one can buy home insulation based on it. The insulation consists of a stack of aluminium foils and is used where space is at a premium. However, with regard to your two shell model consider one simple issue. The tropopause is colder than both the rest of the troposphere and the stratosphere, how can you model explain that fact. How can the region between two shells be colder than either shell? For a region to be colder it must be losing energy to a heat sink colder than it is but the only thing colder than the tropopause at -57C is outer space at -269C. That means the tropopause has to be losing energy directly to space. How is that possible if there is an opaque shell above it? “””
Depends on your definition of the tropopause. There are several. I am using it as the place where the temperature stops dropping with altitude. Typically, there is a region just above that where the temperature does not drop with altitude. Above that, the temperature rises with altitude. This area, where temperature does not change with altitude, is what I am calling the “lowest part of the stratosphere”. This is the part that is radiating to space, so there is no “opaque shell above it”.
John A says:
Well, both Lubos Motl and I are physicists who agree on little regarding AGW but both he and I agree that Willis’s model here is basically correct. Of course, it is as Willis says “a tinkertoy model” and there are various ways to make it somewhat more realistic (e.g., by having the shells be graybodies that don’t absorb all the IR radiation but let some of it through). However, the essential picture is correct.
I just have a few comments:
(1) I do, however, agree with SteveBrooklineMA (09:48:33) that there is nothing really wrong with the Kiehl / Trenberth diagram, which is indeed not meant to be a shell model at all but an accounting of the various energy flows that occur.
(2) I don’t think the statement, “Among the other insights yielded by the model is that a change equivalent to a doubling of CO2 (an increase of 3.7 W/m2 downwelling radiation at the top of the atmosphere) can be cancelled by a 1% increase in the upper and lower cloud reflections” is correct. What is presumably meant is that if you increase the albedo from, say, 0.30 to 0.31, then this would produce about the same magnitude forcing (opposite in sign) as doubling CO2. However, note that this is really an increase in albedo by something like 3.3% above its current value. Furthermore, since clouds only account for about 0.2 of that 0.3 albedo, it would take about a 5% increase in the current albedo due to clouds in order to cancel. And, even this number is in some sense deceiving because increasing clouds actually has two effects: a cooling effect due to the increase in albedo and a warming effect due to a decrease in the outgoing IR (‘longwave’) radiation. Which effect predominates depends on the type and location of the clouds, with low clouds tending to cool more than they warm and high clouds tending to warm more than they cool. The overall effect of the cloudiness on the earth is a cooling effect but there is enough warming due to the decrease in outgoing IR radiation that it offset a considerable fraction of the cooling effect due to the increase in albedo alone.
(3) There are some subtleties in going from this simplified picture to considering how the greenhouse effect is altered by the increasing the levels of greenhouse gases. In particular, some people here have worried about whether the effect of CO2 is already “saturated”, but it turns out that this is looking at things in the wrong way, since it is not so much whether the infrared radiation can make it out without getting absorbed at least once but rather consideration of the full radiative transfer problem where there can be multiple absorption and emission events in the atmosphere. And, what ends up mattering there is, essentially, what level the radiation that does escape into space (as opposed to being re-absorbed) is emitted from. The effect of increasing greenhouse gases is to push that effective emitting layer higher into the troposphere where it is colder (and this is where the lapse rate comes into the picture), which by the Steffan-Boltzmann Law means that less radiation is emitted back out into space. This puts the Earth’s climate system out of radiative balance and that balance is only restored by heating of the climate (although changes in clouds and thus albedo can also play a role in either reducing or increasing the amount of heating required to restore balance). See here for a historical discussion of this: http://www.aip.org/history/climate/simple.htm#L_0623
I am utterly astounded by the number of people claiming to be physicists who don’t understand fundamental physics. Which fundamental law does this model contradict? Considering space, the shell and the earth all in isolation they all emit the same energy as they absorb. I think what might be confusing some people is that the planet doesn’t reach this state instantly. Energy is accumulated in the earth and shell until the system reaches equilibrium. A larger amount of energy has to accumulate in the shell until the energy being received from space is equal to the energy being lost from space. No energy is being created, it’s just being accumulated in a transient state until all bodies (earth and shell(s)) are losing the same amount of energy as they are gaining.
This is as true of the Kiehl/Trenberth model as it is of the present model. If anyone can come up with a way that a planet with a shell around it can maintain equilibrium with space without the planet radiating more energy to the shell than it receives from space then I’d like to hear it. C’mon physicist, have at it!
The laws of thermo don’t say that no heat can proceed from a cooler object to a warmer object. Just that the net effect can not be so.
The gedankenexperiment would be to imagine the setup for a Maxwell’s Demon experiment. (Without a demon.) The requirements of thermo do not claim that the first molecule traveling across the line denoting the hot/cold split must be proceeding from the warmer side to the cooler side. Only that (a) it is statistically more likely, and (b) it is inevitable across the longer term.
That’s with convection, but the same thing applies to vibrating molecules in a solid transferring the energy via conduction, or to photons being emitted from anything.
A 200K object most certainly is emitting radiation – even when completely surrounded and solely exposed to a 2000K receptor. It would certainly be receiving a whole lot more energy from the surrounding 2000K object. But the claim that the 200K object isn’t emitting is rather odd.
Ian Schumacher (13:25:41) :
This leads to a temperature differential, as you point out. However, it does not lead to a higher temperature on earth. It leads to a lower temperature of the atmosphere. If gravity could warm things, we’d have perpetual motion.
Absolutely it would, no question. This principle has been used in some solar heaters. See here for an example. Google “solar heater vacuum” for a host of examples. ‘Fraid you’re wrong on this one.
Willis says:
Hmmm, what is the effect of difference in the diameters of the planet and the steel shell? It certainly changes the surface area, so that the energy being released by the planet is now spread over a larger surface area on the shell.
Bill Illis says:
This is nonsense. Everybody agrees that if there were no feedbacks in the climate system, then the resulting climate sensitivity, as dictated by the S-B Equation (using the effect radiating temperature of 255 K for the earth) is about 0.3 C per (W/m^2). The reason why Hansen gets a larger value is that there are feedbacks in the climate system and these feedbacks in essence modify (and, in most climate scientists’ opinions, increase) the number of W/m^2 increase that occur due to the change in CO2 levels alone.
For example, in a warmer climate, more water vapor is evaporated into the atmosphere and since water vapor is a greenhouse gas in the sense of absorbing IR radiation, this is a positive feedback, in essence increasing the W/m^2 from that due to CO2 alone. A negative feedback that occurs is that because the lapse rate in a warmer climate is expected to decrease(i.e., the upper troposphere is expected to warm more rapidly than the surface on average), it doesn’t take as large a surface warming to produce enough warming in the mid/upper-troposphere to restore radiative balance. (It turns out that both the water vapor feedbacks and the lapse rate feedbacks rely on a lot of the same physics of convection, so although there is variation from model-to-model in the strength of these feedbacks, those models that have a stronger positive water vapor feedback tend to have a stronger negative lapse rate feedback. So, the errors tend to cancel and the variation in the sum of these two feedbacks from model-to-model is less than the variations in each feedback individually.)
Melting of ice and the resulting drop in albedo is another positive feedback. Changes in cloudiness in a warmer climate can be either a negative or positive feedback and the uncertainty in this feedback is the major source of uncertainty in the IPCC’s estimate of climate sensitivity.