The Steel Greenhouse

Guest post by Willis Eschenbach

There is a lot of misinformation floating around the web about the greenhouse effect works. It is variously described as a “blanket” that keeps the Earth warm, or 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 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 metre (W/m2). For objects with a temperatures found on the the Earth, this radiation is in the range called “longwave” or “infrared” radiation. See the Appendix for the formula which 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 metre 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 the temperature that the Earth would 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 metres 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? (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.)

In order to maintain its thermal equilibrium, 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 metre. 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 of 470 watts per square metre. 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, or 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. And in fact, a number of simplified climate models have been built in this way. Un-noticed 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, labelled 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 fulfils 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.

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 cancelled 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.


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


newest oldest most voted
Notify of
David Ermer

Figure 2?
Nice article, Thanks!
REPLY: WordPress didn’t like the Macintosh generated graphics – fixed now -A

Might this have some application to the coronal heating problem?

… but if, as K&T and you both agree, the surface radiation is ~390 W/m2, equivalent to 288 K (today’s mean global surface temperature) by the SB equation assuming emissivity = 1 for longwave radiation, then the first differential – i.e. temperature change per unit change in radiative flux – is just 0.185 K/W/m2, while the UN gives 0.313 K/W/m2, and also assumes temperature feedbacks on top, giving a final climate sensitivity parameter close to 0.9 K/W/m2, around five times too big. Discuss.


Cool! Errr … warm!!

Jack Wedel

I know this is OT, but I just couldn’t resist this release from NASA.
The image is from November 17, 2009, days after a large snow storm swept over China, covering much of the country in white. This image shows part of the North China Plain near the city of Shijiazhuang. On the plain, the snow is white where it fell on fields or natural landscapes. Artificial surfaces in cities, towns, and roads are gray. While the larger cities and towns would be visible on a day without snow or clouds, many of the smaller towns and roads would be difficult to distinguish from the surrounding landscape. In the snow, however, small towns and the roads that connect them stand out clearly.
The storm that brought the snow came unusually early in the winter. The snowfall, the heaviest in decades, killed at least 32 people, destroyed 300,000 hectares of crops, and caused more than 15,000 buildings to collapse, reported Xinhua, the Chinese news agency. The storm closed roads and curtailed train and air travel. Airports in many large cities, including Shijiazhuang, closed.

Tom in Florida

You stipulate that the Earth acts like a perfect blackbody. Didn’t I read here on this blog that it doesn’t?

Steve in SC

I see a giant capacitor.

Paul Linsay

The more I look at the K/T diagram, the more I think it’s wrong. Without convection and the multiple phases of water vapor, the earth’s surface would be close to or above the boiling point of water. Radiation is not what keeps the temperature at roughly 20 C. A correct understanding of the atmosphere would derive the K/T diagram, not use it as a starting point.

Ian Schumacher

One possible problem with your thought model of a greenhouse is that you have the energy coming from the Earth. This a completely unlike a greenhouse where energy first comes from outside in that anything you put to absorb and re-radiate heat inward will also stop heat from getting to the surface in the first place. This is the problem with the greenhouse theory applied to Venus (for example). The atmosphere of Venus is so thick that the light can’t even reach the surface to begin with. Venus hot temperature must be internally generated. If this wasn’t the case then our oceans should be boiling (infinitely thick absorber of IR and below, so thick light never reaches the bottom, just like Venus!)
Imagine your thought scenario with infinite shells (solid steel miles thick as an approximation). The temperatures on the surface would be infinite. This must be wrong, but why? It’s because of the idea of energy coming from inside at constant rate no matter what is not physical.
Maybe my thinking is muddled somewhere, but that’s how I see it at the moment. For an alternate though experiment how about this.

Robert Wood of Canada

Whoa there! I haven’t finished this article but I must make a couple of points:
1. The steel plate will not radiate equally in both directions. The radiation is in proportion to Ts^4 – Ta^4. (Ts = temperature of radiating surface; Ta = temperature of ambient surroundings). Now, the planet’s surface is around 270K while the outside (space) is around 4K.
Clearly. most energy is radiated to space.
2. The Earth’s surface temperature is not a function of greenhouse gas warming, but of thermodynamics. It is largely a function of three things: mass of planet; mass of atmosphere and distance from the Sun.

Ok, mathematical quibble here:
The area of the surface of the earth, having a given radius, is going to always be less than the inner surface of the atmospheric shell, which is also going to be less than the outer surface of the atmospheric shell.
For this reason, the shell will receive less W/sqm from Earth radiation than is emitted by the surface, because the radiation will spread out by the inverse square law just as sunlight gets less intense the further from the sun you get, and the shell will radiate more to space than back to Earth.


i’m just sort of taking a stab here, but it occurs to me that one should be careful comparing water to gasses. our oceans behave nothing like the venusian atmosphere. venus has 90 atmospheres of pressure. that’s enough to make a gas or liquid very, very hot.
the exception is water. waters is weird, weird stuff. the fact that is in non-compressible is what allows deep oceans to be cold despite the pressure. the earth itself, by contrast, gets very hot as you go deeper. even stone and iron are more compressible than water.
further, the state changes of water require massive energy and are a big part of our climate cycle. raising one CC of water one degree C takes one calorie. but to change that same amount of ice to water takes about 80 calories. the state change to gas requires even more energy. this allows the ocean’s evaporative process to shed massive amounts of surface heat in a way that the venusian atmosphere could never do.

C Colenaty

Would it be reasonable to speculate that the overall framework of this model might provide an explanation of ice ages as resulting from some sort of periodic fluctuation in the atmosphere?


“Robert Wood of Canada (15:53:58) :
Whoa there! I haven’t finished this article but I must make a couple of points:
1. The steel plate will not radiate equally in both directions. The radiation is in proportion to Ts^4 – Ta^4. (Ts = temperature of radiating surface; Ta = temperature of ambient surroundings).”
This is incorrect. Emitted radiation is proportional only to the temperature of the emitting body. Convection and conduction are the modes of heat transport that are proportional to the temperature differential across a boundary.


Something is puzzling me about the “greenhouse effect”
CO2 molecules are supposed to re-radiate IR Radiation Energy. Surely, if you double CO2 in the atmosphere, the amount of IR energy being re-radiated will be halved per CO2 molecule? From my understanding of Physics, energy can’t be created from nothing, so why would the CO2 theory mean increased re-radiation of IR energy?

Bill Illis

Thanks very much Willis,
In your model, the orange cell (long-wave absorption by GHGs) is set to 95%, what levels do we set it at for changing GHG levels.
I am interested in simulating the last glacial maximum mainly right now – so CO2 would be around 185 ppm. Just changing the Albedo percentage to a number I believe it was (0.333) and then changing the orange cell to 89% simulates the surface temperature properly but the troposphere temperature drops to -11.9C, which is not the way I understand the climate models would simulate it. So how does the orange GHG cell change.


This is the clearest explanation of the greenhouse effect that I have seen anywhere.
Thank you!!!!
An oversimplified qualitative explanation works well for some things, but in this case, actually attaching some numbers to the various models really put the explanation in focus for me.

Alan D. McIntire

That 390 watts is for a perfect blackbody. According to
Hartwig Volz
the emissivity of sea water is somewhere between 0.92 and 0.965.
Splitting the difference, you get 0.9425.
You only need a surface flux of 390 * 0.9425 = 368 watts to
get 15 C.
In response to Mike Lorrey, the scale height of the atmosphere
is about 8 km. The radius of the earth is about 6400 KM,
so the difference in area between the two surfaces is
(6408/6400)^2 =1.0025. As Mike said, it’s a quibble.
I remember in high school we performed an experiment with water waves. We had a lever repeatedly strike the water at a specific point, and what we got was a series of concentric circles spreading out.
Stick a divider with a slit in the water, and the water passing
through the slit formed a new series of concentric circles, sort of
like this:
The only reasonable explanation is that at each point, a new
concentric series of waves is generated, but most waves interfere
with each other, cancelling out. The same thing happens in
our atmosphere. At each point, a series of concentric waves is
generated, but most ot those cancel each other out. The net result is two virtually parallel patterns between the earth and atmosphere. Half the radiation from the shell will go to space, half will go back to earth into that insignificantly smaller shell,

Bird Stewart Lightfoot

This article should be removed. The author obviously knows nothing about radiative heat transfer.
All of the numbers based on constant heat flux are meaningless because the surface area of the emitting surface is smaller than the surface area of the receiving shell. The actual radiative heat transfer is, ideally, proportional to the fourth power of the temperature, and the analysis must be based on this.
While total energy is conserved, energy flux is not.


@Robert Wood of Canada on your point 1. If the earth were in deep space with no energy input, what you say is true. However, there is 342 W/m^2 being pumped into the surface. This energy is partially absorbed and partially reflected. Depending on how “black” the steel is, and I think the article regards it as an ideal black body with zero albedo on the interior and exterior the fact that open space is 4 degrees K is irrelevant. Its the 342 W/M^2 input that has to be dispensed with.


I know that this is a little off-topic, but I thought I’d just mention that today’s “Independent” (UK) reports that Prof. Corrine Le Quere of the University of East Anglia (where else?) thinks the world is heading for a 6 deg C rise in temperature by the end of the century. Apparently this will mean the mass extinction of almost all life and that Copenhagen is the last chance to stop it. Head for the hills, folks, if this is true, or do I smell the whiff of desperation in the air?


“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.”
OK, I’ll bite. How much would 10 shells raise the surface temperature by? Or 100 shells?


“kurt (16:12:26) :
“Robert Wood of Canada (15:53:58) :
Whoa there! I haven’t finished this article but I must make a couple of points:
1. The steel plate will not radiate equally in both directions. The radiation is in proportion to Ts^4 – Ta^4. (Ts = temperature of radiating surface; Ta = temperature of ambient surroundings).”
This is incorrect.”
no Robert is actually correct (although he didn’t express it precisely). The net ‘radiation’ is the emission (Ts^4) minus the absorption (Ta^4). Every blackbody above 0 K is both emitting and absorbing thermal radiation
JaneHM (University of Cambridge PhD in Theoretical Physics)

Clayton Hollowell

As you increase the quantity of CO2 molecules in the atmosphere, the fraction of IR radiation absorbed increases. The relationship is far from linear, however. Roughly speaking, the probability that any photon in the absorption band of CO2 passes the barrier is an exponential relation, but the absorption band changes with the CO2 pressure.


You can’t say that the earth radiates 235 watts per square meter from the center and that this amount still ends up being 235 watts per square meter at the surface. As you go out in radius, the square meters of surface increases. For the total amount of energy radiated to be the same, the watts per square meter have to go down.
For example, if you measure the watts per square meter radiated by the sun at a distance of 1 AU from the sun’s surface, you aren’t going to get the same measurement of watts per square meter if you back away to 5 AU.
I haven’t done my geometry in a while, but I recall something about the surface area increase being proportional to the square of the increase in radius.


Steven Milloy over at has an article about how the greenhouse effect actually works:
It’s an oldie but a goodie.

George E. Smith

“”” Robert Wood of Canada (15:53:58) :
Whoa there! I haven’t finished this article but I must make a couple of points:
1. The steel plate will not radiate equally in both directions. The radiation is in proportion to Ts^4 – Ta^4. (Ts = temperature of radiating surface; Ta = temperature of ambient surroundings). Now, the planet’s surface is around 270K while the outside (space) is around 4K.
Clearly. most energy is radiated to space. “””
Well why would the energy radiated by a black body be dependent on the ambient temperature ? Now I agree that the ambient temperature will then be radiating according to that temperature and that will be absorbed by our black body, so certainly the net energy radiated to space would be according to 25464 -4^4 if those are the assumed temperatures.
But I don’t see what would stop the shell inner surface from radiating the same 235W/m^2 as the outer surface does, and the BB planet would have to absorb all of that. So now there is no net energy flow between the shell and the planet, as both are at the same temperature.
So now the planet has to start warming due to its internal nuclear decay source, until its temperature gets up to where it radiates 470W/m^2, so the temperature has to increase by 4th root of 2 (1.1892) which makes it 302 K surface temperature for the planet, and 254 K for the shell.
Also due to the Wien Displacement law, the BB spectrum emitted by the planet, is peak shifted by that same 1.1892 factor to a shorter wavelength (well 1/1.1892); so the shell and the planet are now emitting a totally different spectrum.
The system is not in equilibrium (since there is a net flow of energy outwards of 235 W/m^2), and that state is maintained only by the internal nuclear decay, supplying the energy that is being lost to space.
And of course in the case of a real planet with a real atmosphere, and primarily an outside source of energy (the sun); a whole host of other mechanisms would come into play.


And is there really that much of a metaphorical difference between a blanket, a mirror and a steel shell? Call it whatever you want – it is reflecting waves of radiation back down to the surface.


Now I’m really confused.
I’d always thought that the T in the steigan-BriffaMann eekwayshun stood for Tree!
This Stefan bloke and his pal Boltzmann have to come clean about who Kelvin is!!


I think the model is an excellent aid to understanding the earth’s heat balance. I think the earth’s temperature is also stongly controlled by water and clouds as only small changes in the surface temperature produce big changes in the clouds. It would be possible to add a linkage from the surface temperature to the cloud reflectivity to see how this could stabilise the surface temperature.
The radiation situation from the outer shell is complex because the sun only heats one half at a time and the flux is not uniform. However the average flux can represent the steady state average condition.

Gary Hladik

Bird Stewart Lightfoot (16:46:20) and Steve (17:12:20), all energy fluxes are apparently relative to the surface area of the “planet” in this model. As Alan D. McIntire (16:36:32) points out, the surface area of a shell at 8 km altitude is pretty much the same as the surface area of the earth (or rather, an earth-sized sphere).

David L. Hagen

Stockwell at Niche Modeling discussed on the steel greenhouse model
Steve McIntyre reviewed: Sir John Houghton on the Enhanced Greenhouse Effect January 8th, 2008
Miskilczi found errors in the original infinite thickness climate model assumptions that caused an error of a temperature step jump at the Earth’s surface. He formed a semi-infinite climate model correcting that error. See
The new climate theory of Dr. Ferenc Miskolczi
PS the steel greenhouse model should balance the total flux of the surface integral at the earth’s surface and at the steel shell, not the flux per unit area.

David Ermer

” Bird Stewart Lightfoot (16:46:20) :
This article should be removed. The author obviously knows nothing about radiative heat transfer.”
Where can we find this problem worked out in the correct way?


JaneHM (16:58:14) :
“no Robert is actually correct (although he didn’t express it precisely). The net ‘radiation’ is the emission (Ts^4) minus the absorption (Ta^4). Every blackbody above 0 K is both emitting and absorbing thermal radiation.”
I don’t think that Robert was talking about “net” radiation. The model shown in the first figure of the article posits that the shell receives 470 W/m2 which inpinges on the interior surface of the shell. What the post was questioning was the assumption that the radiation emitted by the shell is distributed in equal portions inwards and outwards.


Re Kurt, Robert Wood of Canada. In his book “Collective Electrodynamics” Carver A. Mead argues (convincingly in my view) that there are no “free” photons, indeed that photons are a superfluous concept inasmuch as the spacetime interval between an emitting electron and a receiving electron is 0 and both the spatial separation along the light cone and the temporal duration along the light cone are 0. He models radiative energy transfer in free space and radiation transfer between atoms from pages 73 to 113. It follows from what he writes that an electron cannot emit a photon unless it can “find” another electron which is in a suitable configuration to receive said electron. That implies that an excited electron in a greenhouse gas molecule in the atmosphere cannot radiate toward the ground unless it can “find” another electron on the surface in a ground state which is capable of absorbing the photon which is to be radiated. There have been repeated assertions in blog posts that a GHG molecule cannot “choose” which way to emit its photon, but Mead’s book suggests that, in effect, it can, inasmuch as it is more likely to find a receptive partner in a colder region than in a hotter one. That in turn suggests that Robert’s objection may have some validity.

George E. Smith

“”” Steve (17:19:24) :
And is there really that much of a metaphorical difference between a blanket, a mirror and a steel shell? Call it whatever you want – it is reflecting waves of radiation back down to the surface. “””
Well actually there is a vast difference. In the case of the steel shell, there is a partial reflection and a partial absorption; and the spectra of those two components are different because of the temperature and emissivity differences.
In the case of the real atmosphere, there is very little reflection of LWIR by the atmosphere and it GHGs; but a respectable amount of absorption in the right spectral bands. That leads to heating of the atmosphere, which then leads to radiation of a different spectrum to that absorbed. Whereas the LWIR absorption may have been by GHG species such as CO2, H2O of CH4 etc, these are just minor components of the atmosphere, and they transfer the energy to the main atmosphere in multiple molecular collision processes.
So it is the ordinary atmosphere that is radiating the atmospheric LWIR radiation; except at very high prehaps ionoospheric levels where the mean free path is such that the GHG species can spontaneously decay to the ground state, before a collision occurs.
Now the water (H2O) in the form of a liquid or ice crystals, can and does scatter and reflect a good bit of the radiation that impinges on it, but in the case of the LWIR, I would expect it still to be mostly absorbed by the water rather than refected.
The incoming soalr spectrum is something else and our eyesa re witness to the fact that reflection does occur; but that is energy at solar spectrum wavelengths, and when it reaches the surface it is treated quite differently from the LWIR, since most of it lands in the oceans (an oxymoron), and propagates deeply until it is finally absorbed. The LWIR on the other hand doesn’t get much past the first 10 micorns of water surface layer, which then leads to prompt evaporation rather than long term storage of the energy in the ocean.
Trying to morph the steel greenhouse into the earth, and its atmosphere, is about like pushing a grand piano out of a fifth story window, and expecting to hear a Beethoven Sonata when it crashes onto the pavement below.
Ain’t gonna happen !


Willis, you said:
“In order to maintain its thermal equilibrium, 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 metre. Of course, since a shell has an inside and an outside, it will also radiate 235 watts inward to the planet.”
Willis, this cannot be true. The “W/m^2” is a measure of a quantity of energy received during a time across an area. The steel shell in your model cannot emit this quantity of energy both “up” and “down”, as that is unphysical. You are effectively doubling the quantity of the energy when you claim that.
Your model requires revision.

John F. Hultquist

Figure 3 is labeled as “the Kiehl/Trenberth ….”
This seems to be an update of a very common sort of cartoon-like sketch of Earth’s energy budget that has existed in much this form for many years. Introductory texts in Earth science classes have had them for years. I have a mostly black & white printing of a ‘Modern Physical Geography’ text by Strahler and Strahler (2nd Ed., 1983) that uses data found in ‘Physical Climatology’ by Sellers (1965). So 1965 is as far back as my personal sources go but the point is “Why do K & T get the credit for this sort of sketch?”


JaneHM – your statement regarding *net* radiation is correct. However, the radiation at issue in the statement is not the *net* radiation but the actual radiation being emitted by the surface which clearly is proportional to T^4. Of course the surface is receiving radiation as well, which is not at all dependent upon the surface temperature but the incoming radiation.
Now, the ambient temperature *is* important in systems where conductive and convective heat transfer are present, but this simple model doesn’t include them explicitly.
Robert Wood – *if* the shell had real thickness *and* we stipulated that the shell radiated *exactly* the same *total* amount of energy outward as it did inward then you are right that the power per unit area (W/m^2) would be slightly less from the outward surface than from the inward surface. But that is not the model and the modeler couldn’t have been clearer. The model says that the metal shell is the same temperature on the two surfaces. That means that the radiated power per unit area is identical outward and inward. As you point out, the outer surface has slightly larger area so, in the model, there is in fact *more* radiation per unit area moving outward as compared to inward. But for the reasons I stated here and not what you said where you argue that radiation from the inward direction is 270K while the outward direction is 4K. Even that argument is wrong because from the outward direction HALF of the shell “sees” 4K from space but from the other half of the outward direction sees the temperature equivalent of 1360 W/m^2 because of the sun. OK, it’s really 680 W/m^2 on average on the solar side because of Earth’s curvature – for an average of 340 W/m^2 over the whole surface if we assume that 4K ~= 0 W/m^2. But it’s certainly not 4K.


Robert Wood – my apologies. I read through all comments and had yours and Mike Lorrey’s combined in my head when I wrote that last comment. I still think my argument regarding convection/conduction and that the outward direction isn’t really at 4K are correct, however.
Mike Lorrey – I believe my comments regarding temperature and surface radiance are correct with respect to your statement. W/m^2 up and @/m^2 down are the same so (slightly) more W up than down in a system with a shell of real thickness.

Ian Schumacher

Well why doesn’t the ocean act like the proposed greenhouse of venus? Go down into ocedan deep enough and you can reach 90 atmosphere also. So pressure is not the answer (I assume 90 atmosphere can be found in the ocean, not sure how deep one has to go).
Water absorbs IR. It is water vapor in the atmosphere that provides most of the ‘greenhouse’ effect. Why not water in the ocean at 100% humidity 😉
So what is the difference between the ocean and atmosphere of Venus from a ‘pure’ density/pressure, IR absorption point of view?


Re: jt(18:15:34)
“It follows from what he writes that an electron cannot emit a photon unless it can “find” another electron which is in a suitable configuration to receive said electron.”
If that is true then how does a laser work then? To me it contradicts stimulated emission inside a laser resonator which is electrons emitting photons which stimulate other electrons to emit other photons which resonate with no determinate target.

Steve Fitzpatrick

Interesting post Willis.
I note that the magnitudes of the various energy fluxes are quite large compared to man-made GHG forcing, even if amplification by tropospheric moisture is assumed to be correct. It would be easy to not recognize a change in one of the fluxes that off-sets (or partially off-sets) man-made GHG forcing.
Recent work by Mishchenko et al ( suggest a substantial reduction in the optical density of the atmosphere between 1994 and 2005, indicating an increase of 2% – 3% in the intensity of solar radiation at the surface under clear sky conditions. This aerosol driven increase in energy reaching the surface is much larger than any increase caused by GHGs during that same 11 year period: a potential aerosol driven forcing of 3.4 to 5 watts/M^2. (GHG increases would add only a small fraction of a watt/M^2 over the same period.) Yet since about 2000 (or 2001) the average surface temperature has been essentially flat, and since 2003 the ocean heat content (according to Argo) has been flat to slightly falling…. in spite of falling aerosols and increasing GHG concentrations. So there must be some off-setting factor(s) that is(are) ‘canceling’ expected warming.
Wentz et al, Science 317, 233 (2007),, show that satellite measurements indicate increases in surface temperature have increased the net rate of ocean evaporation and global rainfall by close to 7% per degree K (in line with the Clausius-Clapeyron equation). By comparison, climate models consistently predict a much lower increase (1%-3% /K) in evaporation and rainfall. Since latent heat transport (and surface cooling of the ocean) must increase in proportion to the rate of evaporation, perhaps Wentz et al have identified a reason why the models appear to overstate climate sensitivity: the actual latent cooling increases by about 4 watts per square meter more than the models predict for each degree rise in surface temperature.
In addition, it only rains where there are clouds, so an increase in evaporation/precipitation can reasonably be expected to increase tropospheric cloud cover and increase net albedo, further off-setting radiative forcing. So maybe Richard LIndzen was right all along, and there is there an important ‘Iris’ effect that helps regulate the surface temperature.

Well, I *like* the article.
Just the notion of a “steel greenhouse” is kind of cool, er, so to speak 😉


I think its an amazing invention, the steel greenhouse. If I understand it properly.
Right now it is looking like I can produce any temperature I want, its just a matter of how many shells my greenhouse has. I am enormously excited about this, and am going to do a small experiment tomorrow. Unfortunately I only have five or six cans which will fit inside each other, but I will buy a few more tomorrow, and put them all together. Then I will put a battery powered torch in the inside one, and wait for the explosion.
It is a sad commentary on human nature that the first application that comes to mind for this brilliant invention is as usual a bomb. But there you go. It depends how many boxes I can buy. if I can find someone to sell me a few hundred cheap, well, you might even hear the explosion, accompanied by a mushroom cloud, where you live. Looking forward to it. I have always wanted to go out with a bang, and this should do it.


[snip – we don’t need discussion of potential terrorist acts on this thread]


I know what your talking about Willis.
A very good model. If only some people would realise that working with GROSS values of radiation is acceptable despite what thier high school physics teacher might have told them. Also that the difference in areas is negligable.
There is no provision in your model however for the effect of thermal losses between the troposphere and stratosphere such as is caused by verticle wind shear.
Verticle wind shear correlates well with temperature so as temperature increases you can expect more thermal losses across the tropopause which will reduce the increase in radiation reflected back to the surface which should reduce the expected surface temperature increase.


Everyone claiming to have found a problem based on his statements of 235 w/m2 are missing his calculation starting point. The outer surface of the black body sphere is where all the values of watts are calculated from. Thus, it does not matter what the actual amount of watts per square meter is at the center of the sphere, or at the interior of the shell, because the radiation is calculated to be measured at the surface of the sphere, not of the center of the sphere or the interior surface of the shell.
Thus, if the base black body radiates 235 watts per square meter and sized to be 1 square meter and the shell is sized to be 2 square meters on the interior surface, then the shell would be radiating 117.5 watts per meter squared from it’s surface, but that gets concentrated to 235 watts when it finally reaches the surface of the base black body sphere. Thus all calculations are based on what the effect is at the surface of the initial black body sphere regardless of sizes and distances.


“Well actually there is a vast difference. In the case of the steel shell, there is a partial reflection and a partial absorption…”
Same with a blanket, so zero difference. Make your blanket out of steel wool, if you must.
In the second diagram, Figure 2, it shows the sun transmitting energy W through the outer sphere, and the outer sphere G transmitting it’s own W to the inner planet, which creates net two W to the earth. Add up your vectors of two W in and 3 W out and you get a net of negative one W. If the sun is the only source of heat, net W flow should be zero (for every unit of heat the sun pumps in, eventually one unit should escape). What is shown is a system with two bodies emitting heat of W. It’s either both S and G, or both S and E.
Since Figure 1 shows the steel shell alone (no sun) emitting energy inwards as well as outwards, I’ll assume the steel shell is the second heat source.

Gary Hladik

Monckton of Brenchley (15:30:32) : “…giving a final climate sensitivity parameter close to 0.9 K/W/m2, around five times too big. Discuss.”
This one really has me confused. Up to now I’ve only seen “climate sensitivity” defined as the temperature increase expected from a doubling of atmospheric CO2. I assumed all the temperature rise was attributed to “radiative forcing” directly and indirectly increased by CO2. I thought the magnitude of the radiative increase was in dispute, but the effect of a known change in radiation wasn’t. Now it is? WUWT?
BTW, according to the Stefan-Bolzmann Equation, a sensitivity of 0.9 K/W/m2 implies the average temperature of the earth is about 170 Kelvins, or -103 degrees Celsius.
I’m putting another blanket on the bed tonight. 🙂