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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lgl,
Oh, I see what you mean. Yes there is the question of coherence. If you have energy that is ordered in nature (same wavelength, same direction, that kind of thing) this is also know as coherence. If it has a predictable nature of some sort, then this can be exploited to concentrate it. If it is all in the same direction (for example) then your parabolic dish will work.
If energy is random (goes in random directions, random wavelengths following typical heat distribution) then it does not have coherence and is now just ‘heat’. Heat can not be focused.
This applies to short-wave or long-wave. Light from the sun is all coming from the same direction. This coherence can be used to easily focus it.
Ian Schumacher (14:04:30) :
Eh. What you’re getting at is already captured in the 235 W/m^2 irradiance mentioned above.
Perhaps you’ve already done this, but to drive the point home: while we have satellites that measure the solar irradiance, we can easily calculate an estimate.
In fact, anybody curious can try this:
Take the surface of the sun to be 5778 K, and treat the sun as a blackbody.
Look up the radius of the sun, and the distance between sun and earth. You will NOT need the radius of the earth, though you do assume it’s a perfect sphere.
Take the absorptivity of the earth to be 0.7.
From all that, I get 239 W/m^2 from the sun, absorbed at the earth’s surface. Pretty close to the 235 W/m^2 used by Willis.
SteveBrooklineMA,
You could also replace the two layers with a partially reflecting mirror. Set how much it reflects to whatever you want to make it equivalent to 1 layer, 2 layes, 10 layers .. even infinite layers.
Even simpler 😉
SteveBrooklineMA (14:30:01) :
Sweet as! Very nice, thanks. I’ll see if I can’t extend it.
Outstanding. Science moves forwards.
w.
Ian Schumacher (14:37:41) :
I’m not following you here. Collimators exist; they exist for infrared, as well. They’ll get all your rays nice and parallel. So?
I don’t follow the motivation here. Are you guys trying to use the IR given off by the earth’s surface (or your house, or body, or whatever else is handy) to do something useful, like work, or boil water? Is that the motivation?
The word ‘heat’ should be purely reserved for the transfer of thermal energy.
Ian Schumacher (14:37:41) :
See, thanks.
carrot eater (14:57:04) :
Yes, that was the motivation.
lgl (14:37:27) :
I just showed examples of practical devices that can focus IR using lenses. Others have given examples of practical devices that can focus IR using reflectors. Given that those devices exist, I’m not sure what your point is here. Are you saying that IR cameras don’t really exist, or that they can’t focus IR, or what?
carrot eater,
“Eh. What you’re getting at is already captured in the 235 W/m^2 irradiance mentioned above.”
Right! and how hot would a black body just outside our atmosphere have to be to generate this heat?
I want the value outside the atmosphere so that would be 341 or so.
(341/sigma)^.25 = 279 K
So a black body just outside our atmosphere that could be used as a source instead of the sun to produce the same energy on earth would be 279K.
Anyways I suspect I’m not convincing you and I appear incredible daft and I accept that 🙂 Think about this though. The earth can not use the sun to heat up to inifinity. The earth can not use the energy from the sun to become hotter than the sun. The earth is far away from the sun and the energy from the sun is spread out and the flux much much lower, so ….
What is the limit of the temperature of the earth?
carrot eater,
You can’t focus heat. if you could you could transfer heat from a cold source (just concentrate to make it hotter) to hotter source.
You can focus IR, yes, but only if it has coherence i.e. it is going a specific direction, and other forms of ‘order’.
Ian Schumacher (15:14:26) :
Again, please never use the word heat to mean anything but the transfer of thermal energy. I’m not even sure what ‘focusing heat’ would even mean, so let’s just never use that term. Seems like you agree with me on that.
lgl (15:05:33) :
Now, the second law is FINALLY going to bite. What is the temperature of the earth’s surface? What is the boiling point of water? I don’t care what you do, you’re not going to boil that water. The earth’s surface is simply too cold to be useful, unless you’re in Iceland or something and you’ve got a geothermally heated spot.
If you dig a hole in the ground and find some rocks at a high temperature, THEN you can do it. That’s the point of geothermal energy.
Willis Eschenbach (13:39:11) :
I’m not making assertions, I’m citing a definition.
“thermal equilibrium
The condition under which two substances in physical contact with each other exchange no heat energy. Two substances in thermal equilibrium are said to be at the same temperature.”
http://www.thefreedictionary.com/thermal+equilibrium
At thermal equilibrium, (T1^4-T2^4) = 0. There will be no net heat exchange, BY DEFINITION not assertion. Since radiant energy flux is directly proportional to that quantity, it will also = 0. By definition.
Also, energy in must equal energy out. If the flux per unit area coming out of a body of larger area is equal to the flux per unit area of a smaller body, there’s more energy coming out of the lager, in this case the shell. But since we started with a fixed energy flux, there can be no extra.
The shell cannot be as warm as the earth, and the earth cannot absorb energy from a shell which is at a lower temperature. However, since the shell is at a higher temp than background, though probably not by a lot, (background ~3 DegK, which is effectively = 0 for this problem), less heat would be transferred to the shell, and the earth would consequently retain more of it and so it would warm slightly.
If that doesn’t do it for you, then I don’t know what else to say.
regards
P.S., you’re the one who insisted the system be at thermal equilibrium, not be.
Ian Schumacher (15:09:47) :
You’re up to some odd mischief here. First, your heat source is colder than the earth’s current surface temp; that’s no good. Your heat source also seems to be one of Willis’s shells, wrapped all the way around the earth – not like the sun, at all. Maybe that’s what you intended, I don’t know. You’ve probably lost the reflected term, too. So whatever you’ve got with this weird new heat source, it’ll have a different result from the sun. Maybe that’s what you were getting at?
What is the limit of the temperature of the earth? That’s maybe interesting. Let’s leave the sun right where it is, and let’s leave it as a sphere. The minimum temperature is what Willis calculates above, in the case of no greenhouse. Assuming that an atmosphere-less earth would have the same albedo, at least.
The maximum temperature? That’s the more interesting question, and perhaps that’s where you’ve been steering me. Hmm. We can make a stronger and stronger greenhouse, but is there a thermodynamic limit? Clearly, we have to stay below the surface temperature of the sun, but is there a thermodynamic limit below that point? If there is, it isn’t obvious to me at the moment. That’s maybe the first interesting question in this whole thread.
Carrot eater,
Yes my numbers are wrong. Was just using the numbers you supplied for illustration there there is an equivalent temperature and it is lower.
Yes, for a point source use 1366W/m^ or whatever. The basic idea is the same and the idea I’ve been trying to get people to think about (but very poorly I agree). There is a maximum temperature, it isn’t infinitity. So what is it?
I’ve very glad that you are now on the case thinking about that. 🙂
Now I don’t feel like a lone lunatic (ok just less so).
yonason (15:52:45) :
Where ever you see Willis use the word ‘equilibrium’, go ahead and substitute in ‘steady state’.
Does that alleviate your difficulty?
Nobody is implying that the earth, the sun and the steel shell are all at the same temperature.
carrot eater,
It’s clear I think the maximum temperature is blackbody temperature of course. I think maybe I focused too much on that. Fine, if it isn’t blackbody, what is it?
Ian Schumacher (16:12:50) :
I see what you’re upto. Let’s agree to leave the sun where it is, as moving it or changing its shape is an unneeded diversion.
We can say without a doubt that no matter what we do, the earth must stay below 5788 K or whatever the temp of the solar surface is.
If we continue with the Willis cartoon, and continue piling on ‘steel’ shells, you could keep doing it until you approach the sun’s temperature, I guess. The ‘steel’ would of course melt and the whole thing is preposterous, but the cartoon lets you do it, doesn’t it?
Ian Schumacher (16:30:53) :
“It’s clear I think the maximum temperature is blackbody temperature of course.”
I don’t understand what you mean by that.
yonason (15:46:46) :
carrot eater has pointed to what I think may be the source of the confusion. I have used the “thermal equilibrium” to indicate a state where there is no further change of temperature with time. Or as Wikipedia says, “Thermal equilibrium is when a system’s macroscopic thermal observables have ceased to change with time.”
You are using it, on the other hand, to describe a state where all parts of the system have the same temperature. carrot eater is correct that “steady state” might be less subject to misinterpretation.
Does this help?
carrot eater,
I believe that distance from the sun (and the lower flux as a result) must also be a factor. So the maximum temperature will be way way below 5788K. Here is my reasoning (not really ‘science reasoning’, but a kind of gut feel ‘bounds’ check.
There is a maximum temperature a greenhouse effect can accomplish. Is it likely that this maximum is the same temperature for Mercury as it is for Pluto? No, I don’t think so. That just doesn’t seem possible. Mercury is closer, the flux from the sun is higher. I feel this must matter. If it doesn’t then theoretically we could warm ourselves from distance stars! Not too likely. Distance matter. Flux intensity matters. Being far away from the Sun means a maximum temperature much less than the sun.
Willis allows for infinity yes. Not a fault really as an approximation, but not useful for figuring out what the limit is. If this limit matters (is low) then Willis model might be of little use. Everyone but me is convinced the limit is high, so … no worries there yet 😉
I believe the maximum temperature is defined as the blackbody temperature of earth that would give a (earth surface area) flux to balance out the (earth cross section) received by the sun flux of 1366W/m^2 . Sorry if that part wasn’t clear.
Basically I think the maximum temperature of the Earth is the same as if the Earth was a super dark block of matter sitting there.
On a different note, (but somewhat important). Many point out “black body isn’t warm enough since the average temperature of the earth is already higher than this”. Maybe, but how is this average temperature determined? We don’t actually just take temperatures and average them do we? If we do, that clearly give one the wrong impression since temperature isn’t conserved. Averaging temperature is kind of meaningless. We would need to take the fourth root of the average of T^4 … or something at least to getting closer to a meaningful number than a dumb average. Anyways just a curiosity. The stated average is higher than blackbody .. sure, but does this stated average have real meaning? or is it just a bogus? At the moment, I have no idea, but of course I suspect bogus 😉
Ian Schumacher (17:04:20) :
Do you want to switch from the steel shells to an actual atmosphere now, held by the Earth’s gravity? At some point, the steel shells are not useful anymore; I could have put a billion of them up there, going up to the surface of the sun.
In either case, I’ve got work to do, so I might leave you to ponder it.
Instead of steel shells, I think it is easier to just have a mirror (a single shell with a mirrored inner surface). This will get the maximum possible. It’s equivalent to infinite shells. The mirror can be for a frequency range ie. (transparent to high frequency, reflect low frequency). That is the simplest and seems to me to be a model that would achieve the maximum greenhouse effect.
Yes for sure, this isn’t suppose to be our day job.
Once you start to look into it you see ‘ok’ I need to integrate the spectrum on incoming light and that has to balance the integration of the spectrum of outgoing light over the same region. That’s what I got to. That give you a pretty high temperature that is dependent on the cut-off frequency. BUT …. then I started to realize that the spectrum will not be static. It will continuously readjust itself and the high frequency transparent region will continously be repopulated. That was my thinking anways. Maybe you will conclude differently. I’m subscribed (rss) so if you post something a week from now I’ll take a look and comment.
Cheers.
Willis Eschenbach (16:54:34) :
“Thermal equilibrium is when a system’s macroscopic thermal observables have ceased to change with time.”
You are using it, on the other hand, to describe a state where all parts of the system have the same temperature.”
But that’s really just the same thing. (I know it sounds general enough to apply to “steady state” as well, but using it that way can be very confusing, and besides, that wasn’t the intended use.)
Anyway, so you meant “steady state,” and you invoked it in order that “…the whole system must still radiate 235 W/m2 out to space.
But there is still the problem of violating the First Law. Since the steel shell has a larger radius, it’s surface area is larger than that of the earth, and so for the TOTAL ENERGY emitted to remain constant, as required by the First Law, the shell must radiate less W/m2 than did the earth, so that when multiplied by the shell’s larger surface area, the total energy emitted still comes out the same.
I.e., the system still has to radiate the same amount of heat to space, but since it is now doing it through a greater surface area, the flux per unit area must decrease, otherwise energy would have been created, which violates the first law.
The earth will warm either until the total energy output per unit time = total energy produced per unit time, or until the heat source burns itself out. But that warming depends on how good an insulator the shell is, which depends in part on it’s thickness. It could be much hotter on the inside than on the out, because heat transfer through metals is via conduction, and so in this case the outside would be colder than inside, unless it was very thin, in which case it would pose a minimal barrier, and the earth wouldn’t have to warm very much at all. Also, if it’s a good reflector, heat will have to build up more inside than if it weren’t, but not because it’s radiating energy. Radiation and reflection are different processes.
Note – your initial example of the steel shell didn’t involve any energy source other than whatever “nuclear” reactions might be generating heat in the earth, so I’m not including the sun, yet.
Willis Eschenbach says:
No problem!
I think I will have to disagree with you here though (although I haven’t been able to follow the link because it seems really slow tonight). I don’t disagree with your statement that neither satellites nor climate models have the accuracy to directly detect such a small imbalance. However, as I understand Hansen’s paper, the imbalance is being detected instead by observing the amount of heat that is building up in the system due to the imbalance (and, nearly all of this heat buildup is in the oceans). So, even though we may not be able to measure the input and output fluxes with sufficient accuracy, we can infer their difference through the buildup of heat energy. (Not to say that there aren’t other potential issues, such as whether the heat content change was measured over a long enough period to give a result to that accuracy. But, I don’t think the fact that models or satellites can’t resolve the fluxes to this accuracy is not relevant.)