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.

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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David Ermer
November 17, 2009 3:05 pm

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

November 17, 2009 3:14 pm

Might this have some application to the coronal heating problem?

November 17, 2009 3:30 pm

… 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.

Jim
November 17, 2009 3:32 pm

Cool! Errr … warm!!

Jack Wedel
November 17, 2009 3:41 pm

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
November 17, 2009 3:42 pm

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
November 17, 2009 3:43 pm

I see a giant capacitor.

Paul Linsay
November 17, 2009 3:46 pm

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
November 17, 2009 3:49 pm

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. http://www.ianschumacher.com/maximum_temperature.html

Robert Wood of Canada
November 17, 2009 3:53 pm

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.

Editor
November 17, 2009 3:59 pm

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.

morganovich
November 17, 2009 4:10 pm

ian-
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
November 17, 2009 4:12 pm

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?

kurt
November 17, 2009 4:12 pm

“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.

Paul
November 17, 2009 4:16 pm

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
November 17, 2009 4:20 pm

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.

Charlie
November 17, 2009 4:34 pm

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
November 17, 2009 4:36 pm

That 390 watts is for a perfect blackbody. According to
Hartwig Volz
http://www.klimanotizen.de/2006.06.17_Sea_Water_Emissivity_Volz.pdf
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:
http://www.acoustics.salford.ac.uk/feschools/waves/diffract3.htm
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
November 17, 2009 4:46 pm

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.

ShrNfr
November 17, 2009 4:47 pm

@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.

DaveF
November 17, 2009 4:48 pm

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?

michel
November 17, 2009 4:49 pm

“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?

JaneHM
November 17, 2009 4:58 pm

“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.”
Kurt,
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
November 17, 2009 5:11 pm

Paul,
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.

Steve
November 17, 2009 5:12 pm

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.

PaulH
November 17, 2009 5:13 pm

Steven Milloy over at JunkScience.com has an article about how the greenhouse effect actually works:
http://www.junkscience.com/Greenhouse/
It’s an oldie but a goodie.
Paul

George E. Smith
November 17, 2009 5:13 pm

“”” 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.

Steve
November 17, 2009 5:19 pm

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.

royfomr
November 17, 2009 5:26 pm

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!!

Bernie
November 17, 2009 5:36 pm

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
November 17, 2009 5:50 pm

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
November 17, 2009 5:59 pm

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
November 17, 2009 6:10 pm

” 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?

kurt
November 17, 2009 6:14 pm

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.

jt
November 17, 2009 6:15 pm

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
November 17, 2009 6:17 pm

“”” 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 !

Joseph
November 17, 2009 6:20 pm

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
November 17, 2009 6:40 pm

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?”

Merrick
November 17, 2009 6:47 pm

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.

Merrick
November 17, 2009 7:00 pm

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
November 17, 2009 7:00 pm

morganovich,
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?

AlexB
November 17, 2009 7:04 pm

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
November 17, 2009 7:13 pm

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 (http://pubs.giss.nasa.gov/docs/2007/2007_Mishchenko_etal_4.pdf) 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), http://www.remss.com/papers/wentz_science_2007_paper+som.pdf, 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.

E.M.Smith
Editor
November 17, 2009 7:14 pm

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

michel
November 17, 2009 7:35 pm

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.

michel
November 17, 2009 7:38 pm

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

AlexB
November 17, 2009 7:42 pm

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.

AStoner
November 17, 2009 7:51 pm

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.

steve
November 17, 2009 7:53 pm

“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
November 17, 2009 8:01 pm

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. 🙂

savethesharks
November 17, 2009 8:03 pm

Thank you for this very enlightening essay, Willis.
The fact that the swirling, eddying, jet-streaming, convectively-bubbled water-vapor-filled atmosphere of Earth, that behaves not unlike it were liquid….makes it almost impossible for those greenhouse-alarmists to pin down….because it ain’t steel, it ain’t glass, hell, it ain’t even a blanket.
Much more benign…and life-giving…than that.
Chris
Norfolk, VA, USA

Stephen Goldstein
November 17, 2009 8:05 pm

Great post and great post-post discussion 😉
I am trying to get used to using W/M^2 as equivalent to temperature — I agree it simplifies the explanation and, at the end of the day, any given level of equilibrium energy flux corresponds to an equilibrium temperature but still . . . .
I also want to add my 2 cents to two comments . . . .
morganovich (16:10:54) –
A compressible fluid under pressure is not warm because it is under pressure but because it was compressed, the increased temperature representing the work (energy) done on it by the compression process. Once compressed, the warm fluid will either cool to ambient temperature (consider a SCUBA diver’s air tank or a CO2 fire extinguisher) or, remain warm if there is an external source of energy (the Venusian atmosphere heated by the Sun).
jt (18:15:34)
I definitely do not understand your explanation “that an electron cannot emit a photon unless it can “find” another electron which is in a suitable configuration to receive said electron.”
It is night here and the sky is clear. If I understand you correctly, the reason that I can see Sirius, the Dog Star, is that 8.6 years ago, an electron on the Sirius photosphere “found” an electron in my retina and emitted a photon. Indeed, many electrons emitted many photons in quite a range of energies ’cause I see a very bright blue-white star.
Yes, I know you were thinking of long wave radiation on a terrestrial scale but why would wave length and distance change the fundamentals?

November 17, 2009 8:11 pm

I’m sorry but I thought the article was tremendous nonsense, and will cause immense damage to the reputation of this website.
If there are physicists watching this weblog then I’m sure they’ll fill you in on exactly why your “steel greenhouse” won’t work in the way Willis has described, and why adding “shells” to the planet will not raise the surface temperature one iota.
In a sense, what Willis has done is manage to make the same mistake as the people who think that the Earth is in “radiative balance” and try to work out a tedious budget of everything going in and out.
It’s wrong from first principles.

Editor
November 17, 2009 8:14 pm

Merrick (19:00:01) :
“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.”
Re: inside vs outside of the shell: your analysis is only true if the shell material is thermally saturated. If, instead, the materially is not thermally saturated there will be waves of energy passing through the material. That the atmosphere warms during the day and cools at night, cools under cloud and warms in broad daylight, cools in broad night time and warms under night clouds, says that the picture is far too complex to treat as a solid shell. That different levels of atmosphere travel at different speeds, are different temperatures, and have different gaseous concentrations tells me you need to treat it as a Matrioshka doll where each shell is its own lava lamp.
Re: inside of the shell vs Earths emissions: my point still stands, and its even more complex, see above. Terrain at different altitudes changes the number of atmosphere shells one is dealing with, and don’t get me started on ocean.
Now we can start getting into more real climate models, but ghu, you need a grid cell resolution of a few kilometers, not 250 or 100 km, and you need to analyse each cell as its own navier-stokes thermodynamic/fluid dynamic system, not just a basic heat in vs heat out childs toy.

Willis Eschenbach
November 17, 2009 8:15 pm

Tom in Florida (15:42:34) :

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

You are correct, this is a simplified model. However, including the emissivity changes the temperatures but not the radiation.

Willis Eschenbach
November 17, 2009 8:22 pm

Monckton of Brenchley (15:30:32) :

… 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.

Always a pleasure to hear from you. You are correct, the direct SB equation gives a sensitivity of 0.185 K/W/m2. However, this is for a 1 W/m2 change at the surface.
The UN IPCC, on the other hand, figures it for a 1 W/m2 change at the top of the atmosphere. This gives a larger number.
Next, they assume a large (and unverified) water vapor/cloud/whatever feedback on top of that, to give the 0.9 number. I think the UN IPCC feedbacks are far too large, my sense is that net feedback is negative or the temperature swings would be much larger. In addition, I think that there is a governor system which keeps the temperature within a narrow range. See my post here.
Many thanks, keep up the good work,
w.

Willis Eschenbach
November 17, 2009 8:25 pm

Bill Illis (16:20:51) :

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.

Good question, don’t think there’s an answer. IIRC, if you set it to 96.3% or so, it increases the downwelling radiation at the top of atmosphere (TOA) by about 3.7 W/m2. This is what is projected from a CO2 doubling … but this is a tinkertoy model with lots of assumptions.

Willis Eschenbach
November 17, 2009 8:32 pm

Bird Stewart Lightfoot (16:46:20) :

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.

The main radiation from the troposphere occurs at somewhere around 3000 -6000 ft. Let’s call it a kilometer and a half. The radius of the earth is about 6378 km, and the area is 511,185,933 sq. km. The area of the radiating surface at an elevation of 1.5 km is 511,666,935 sq. km. As this is an error of 0.1%, it is typically ignored in this type of analysis.
Energy flux in W/m2 is indeed conserved. If a planet is absorbing 235 W/m2, it must emit 235 W/m2 to stay in equilibrium.

Willis Eschenbach
November 17, 2009 8:34 pm

michel (16:49:44) :

“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?

In a perfect greenhouse, one shell gives a surface temperature of two times the incoming radiation, two shells give us a surface temperature of three times the incoming radiation, three shells gives four times, and so on.

Willis Eschenbach
November 17, 2009 8:36 pm

Steve (17:12:20) :

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.

Sorry for the lack of clarity. I meant that the total energy in the earths core gives a surface radiation of 235 W/m2.

Jeff L
November 17, 2009 8:41 pm

@ Monckton of Brenchley (15:30:32) :
“temperature change per unit change in radiative flux – is just 0.185 K/W/m2”
Amongst the many other key points this raises – such as much lower climate sensitivity than proposed by the IPCC as Lord Monckton points out – also consider how small this # is in the over all radiation budget. Assuming we have warmed about 1deg K (generous certainly) over the last 130 years, that would correspond to a net flux change + 0.185 W/m2 over the last 100 years. Right?
Look at the numbers in figures 3 & 4 for various components of the radiation budget. I have a hard time believing each of those components are known well enough that there isn’t at least a 0.185 W/m2 error bar on them. I havent researched this myself, so if that is incorrect, please give me a reference that shows the accuracy of each of the components of the budget.
So, the key implication from the last paragraph is that very minor changes ANY of the components of the budget could be just as easily responsible for the net temp changes observed as CO2. Why or how have these changes been ruled out?
Also, and maybe more significantly – what about the sun??? Here’s a question for both Lord Monckton & Leif (if you are reading). Hasnt the TSI over the last 130 years trended up by 3 W/m2? (1369 to 1372)? I remember some past post where Leif said these might be outdated #’s & the TSI is more constant (but I cant remember the details), so I am not sure, but based on the above calcs, don’t we only need an increase of 0.185 W/m2 to explain the observed temp changes? Seems pretty plausible / pretty easy to achieve.
Looking forward to your comments on this.

Willis Eschenbach
November 17, 2009 8:41 pm

michel (19:35:56) :

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.

Bear in mind thermal losses. Unless you have a vacuum and the various shells don’t touch each other anywhere, you will get thermal losses. These losses will increase rapidly with temperature, and will keep any physical system from getting too hot.
However, you will be able to get to “pretty warm” without much trouble.

Willis Eschenbach
November 17, 2009 8:45 pm

AlexB (19:42:17) :

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.

I include both convection and evaporation losses in the model I cited at the end of the article. One thing I learned from the model is how many assumptions you need to make to get anything that will come close to modeling the earth. For example, how fast does the evaporation increase as the temperature increases? While Clausius-Clapeyron gives a theoretical answer, in the real world we have spray and wind and changing albedo to complicate things. The GCMs claim to model this, but I has my doubts …

Willis Eschenbach
November 17, 2009 8:50 pm

Gary Hladik (20:01:30) :

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. 🙂

The UN IPCC gives climat sensitivity of ~ 3K per doubling of CO2. Doubling of CO2, in turn, gives a UN IPCC value of 3.7 W/m2 of forcing.
Dividing the first by the second gives a climate sensitivity of 0.8K/W/m2. Same number, different incarnation. Monckton is just using a different way of expressing the same thing.
Finally, the sensitivity is not constant, it is only valid at a certain temperature range. This is because the relationship between temperature and radiation is far from linear.

Willis Eschenbach
November 17, 2009 8:52 pm

John A (20:11:22) :

I’m sorry but I thought the article was tremendous nonsense, and will cause immense damage to the reputation of this website.
If there are physicists watching this weblog then I’m sure they’ll fill you in on exactly why your “steel greenhouse” won’t work in the way Willis has described, and why adding “shells” to the planet will not raise the surface temperature one iota.
In a sense, what Willis has done is manage to make the same mistake as the people who think that the Earth is in “radiative balance” and try to work out a tedious budget of everything going in and out.
It’s wrong from first principles.

I’m happy to be proven wrong, that’s how science advances. I await the information.
In the meantime, can you explain why my steel greenhouse won’t work?

Leo G
November 17, 2009 8:56 pm

wow, just want to say thanx for letting an average dude get to play with a climate model. It may be simple, but it really does help in understanding some more of the “science”.
I think i froze everyone to the ninth layer of hell…….
:)~

Alan D. McIntire
November 17, 2009 9:13 pm

In reply to michael- for shells, substitute ‘atmospheres’.
0 atmospheres will give you a wattage flux of 1 ( no greenhouse
effect)
1/3 atmosphere will give you a wattage flux of 1+ 1/3 at the surface.
3 atmospheres will give you a wattage flux of 1+ 3 = 4 at the surface,
n atmospheres will give you a wattage flux of 1 + n at the surface. The temperature will be proportional to the 4th root of 1 + n.
Of course greenhouse gases do not act as black bodies,
nor even as “gray’ bodies. They absorb only certain ranges of
radiation. Again take a planet
with no greenhouse atmosphere. Add 1 “standard greenhouse
atmosphere” of a gas that absorbs radiation
from 1/10 of the total outgoing spectrum, and is transparent in the
rest. What will the warming effect be?
You can compute this using the “harmonic mean”.
For the general formula,break the atmosphere into ranges, p0, p1,
p2…pn. p0+ p1 + …pn had better add up to 1. Take say 3 gases, with
effective ranges of atmospheric absorption = p1, p2, p3.
Throw in p0, the value for no absorption.
p1 has an effective “standard greenhouse atmosphere” of a1, p2 has a
“standard greenhouse atmosphere of a2, etc.
The total greenhouse effect will be
1
_______________________________________
p0/1 + p1/(a1+1) + p2/(a2+1) + p3/(a3+1)
For a concrete example, let p0 =0.3 and of course a0=0, let p1 =
0.6 and a1 = 1 ( maybe that’s water vapor)
let p2 = 0.1 and a2 = 10 (maybe that’s CO2).
Then, total watt flux will be
1
_______________________________ = 1/0.609 = 1.642
0.3/1 + 0.6/(1+1) + 0.1/(10+1)
The ground temperature will be (1.642)^1/4 = 1.132 times top of
atmosphere flux, with a top of
atmosphere flux of 255 K, surface temperature will be 255* 1.132 =
288.66 K.
Let’s double the amount of CO2. Assuming everything else stays
constant, the new temperature flux will be
1
___________________ = 1/0.605 = 1.653
0.3 + 0.3 + 0.1/(20+1)
The temperature at Earth’s surface will increase to (1.653)^1/4 * 255
= 289.14 K. Now
increase the level of CO2 by another factor of 10. Again, assuming
everything else stays constant,
the new temperature flux will be 1.665.
255 * (1.665)^0.25 = 289.66 so increasing CO2 by a factor of 10
only increased temperature by 0.52 K.
There is a limit here of 1.666….. No matter how much CO2 is
increased, forcing in watts can’t go past that level.
Caveats: The temperature assumes the earth is a black body. In
actuality, it approximates a gray body
with an average emmissivity of about 0.95. That implies that if the
effective temperature is 288 K,
watts radiated per square meter of surface will not be 390.7 but
0.95*390.7 =371.165 watts
The model is assuming classical theory rather than quantum theory.
Once quantum theory is
taken into consideration, temperature magnification can go higher.
Spectrum Broadening
Some on this site have stated that H20 can only absorb photons at
specific wavelengths. At some level this
may be true, but in practice it’s not, otherwise the argument about
saturation would be correct, and nobody would give a damn one
way or the other about further changes in greenhouse gases.
Suppose CO2 can absorb energy only at exactly 14 nanometers. In
practice, CO2 molecules are moving around with
speeds of rougly 400 meters/second. Since this is small relative to
light speed, non-relativistic physics can be applied.
Since 400 meters/second is rougly 1/750,000 the speed of light,
molecules moving towards a photon will see
wavelengths 1/750,000 shorter as actually exactly 14 nanometers in
length, and can interact with them. Likewise,
molecules moving directly away will see the wavelengths as
1/750,000
longer, and can interact with them, so
in practice there are a range of molecular speeds and wavelengths
interacting with the photons, which is why
you get a band rather than a single set of lines.
One result of quantum mechanical theory is Heisenberg’s uncertainty
princible
dE* dt> 1/2 h
where E is energy in joules, t is time in seconds, and h is Planck’s
constant, 6.626* 10^(-34) joule seconds.
Energy can’t be measured closer than 1/2(6.626) = 3.313*10^(-34)
joule
seconds.
Molecules are constanlty absorbing photons, then radiating the
photons
out in fractions of a second.
Say the average time between collisions is 1/1,000,000 second. Then
the energy absorbed can
vary by a factor of 3.313* 10^(-26) joules because of the uncertainty
principle. The collision rate
is a function of temperature and pressure. Increase the pressure and
you increase the frequency of collisions. Increase the temperature
and you increase the frequency of
collisions. Because of that uncertainty principle, the
gas can absorb energy at the standard energy + or – that 3.313 * 10^
(-26) joules.
The combinded broadening from a normal speed distribution of
molecules
and the broadening due to the uncertainty
principle is addressed in the Voigt temperature profile.
http://en.wikipedia.org/wiki/Voigt_distribution
Of course there’s not a neat solution for the above equation, you’ve
got to solve it numerically using something like:
http://en.wikipedia.org/wiki/Numerical_integration
That’s essentially what modtran does.
http://geosci.uchicago.edu/~archer/cgimodels/radiation.html
I picked up this information on the Voigt profile from a post by
Michael Hammer from
Jennifer Marohasy’s blog.
Specifically this letter:
“Comment from: michael hammer May 15th, 2009 at 11:22 pm
SJT: sorry but again i have to disagree with you. Line absorption
profiles follow a Voigt profile. I have integrated the area under
this
profile at various concentrations. When i also included the effect of
the analysis described in my first post on Jennifer’s site I then got
an almost perfect agreement to a logarithmic response. Hence the
analysis I did (which converts the loss from 10^-N to 1/N) needs to
be
incuded to get the logarithmic response.”

jt
November 17, 2009 9:14 pm

Stephen Goldstein (20:05:22) why would wave length and distance change the fundamentals?
As I read it it has nothing to do with wave length or distance. The argument is that among electrons energy transfer takes place along the light cone between two electrons and involves both the retarded and advanced Schrodinger waves. The two electrons enter into a common resonance state and while in that condition are not separated either in distance or in proper time because the time dilation and length contraction along the light cone reduce the relative distance along the photon trajectory between the interacting electrons to 0 and the past/future time interval along the photon trajectory to 0 as well. As a result the quantum of energy exchanged is never actually in a state of “free flight” between the two interacting electrons. The proposition is that every photon that is emitted is certain to be absorbed at some location in the universe and that an electron cannot emit except when in resonance with another which can absorb. In other words, electrons do not emit photons “on spec”. I refer you to Mr. Carver’s book for the details: Collective Electrodynamics – Quantum Foundations of Electromagnetism pub. The MIT Press ISBN 0-262-13378-4, 0-262-63260-8 (pb).

carlbrannen
November 17, 2009 9:17 pm

This post is wrong for a rather large number of reasons. It doesn’t take into account the radius and W/m^2 correctly, but most importantly, W/m^2 is not, nor ever has been, equivalent to temperature. It’s just a bad idea.
A better analogy to the greenhouse effect is “insulation”. You can get that by steel spheres and air gaps, but it’s better to analyze it as insulation. With different thickness of insulation, a fixed amount of heat flow results in a temperature difference proportional to the thickness.

yonason
November 17, 2009 9:37 pm

Willis Eschenbach (20:36:37) :
“In a perfect greenhouse, one shell gives a surface temperature of two times the incoming radiation, two shells give us a surface temperature of three times the incoming radiation, three shells gives four times, and so on.”
So, how many shells will it take to get the core to 2,000,000 degrees?

par5
November 17, 2009 10:09 pm

I think this is a great thought experiment- look at how many are getting involved, bringing in their expertise. Thanks Willis. Yes, the earth is capable of receiving more radiation than it emits- and visa versa. The first cartoon about the radiative budget (K/T) has always bothered me for this reason. Science at work on Anthonys’ blog…

Anton Eagle
November 17, 2009 10:12 pm

Logic. Science and logic should be inseperable.
This article is utter nonsense, and a simple logical analysis reveals this without even resorting to physics (and I am a physicist).
Using the author’s argument… the mythical planet radiates 235 W/m2. The steel shell absorbs this and re-radiates it all in both directions. The planet obligingly absorbs the re-radiated 235 coming off the shell, and re-radiates it, for a total radiation “budget” of 470 W/m2. The author conveniently stops at this point.
But wait… there’s more. If we are to follow the author’s logic, then the shell should absorb this 470 W/m2 and re-radiate it again… 470 out and 470 in. The planet should then absorb this 470… add it to the orignal 235 coming from the core, and now be radiating 705 W/m2… this would continue till everything anihilated at a gazillion degrees. Utter nonsense.
As requested by others above… please remove this article.

Thomas
November 17, 2009 10:14 pm

Wait what? You add a shell, it obtains thermal equilibrium with the earth such that it radiates the same magnitude flux. Right: so you’ve got:
F(earth) = Flux(shell) (1)
but then you say:
Flux(shell in) = Flux(earth) and also Flux(shell out) = Flux(earth)
This is obviously wrong because
Flux(shell) = Flux(shell in) + Flux(shell out)
which is simply in violation of (1)
So really you should have
Flux(shell in) + Flux(shell out) = Flux(shell) = Flux(earth)
If you want, you can assume that the shell radiates equally in and out and that would make each 1/2 of the earth’s flux.
….

November 17, 2009 10:29 pm

Stephen Goldstein (20:05:22) :
I am trying to get used to using W/M^2 as equivalent to temperature”

It isn’t. Firstly, radiation depends on a body’s resemblance to a black body. There is no reason a shell as postulated could not have different reflectance on the two sides (or even on parts of one side, as with the Earth and ocean/land/forest/bitumen etc.). Secondly, there is not a linear relation between WM^-2 and temperature, since radiation is proportional to T^4.
And I have my doubts that there is any reason to equate the reflectances of the inner and outer sides, since the rising temp on the inside will certainly give a different apparent reflectance.
As for the article, it starts out determined to be misunderstood by showing equal flux density at two different radii, which is just plain false. Sorry, Willis, but you will be misunderstood if you write fallacies and expect the reader to pick up on all the approximations. And there is, of course, no approximation with the Earth’s core and the surface, these radii are wildly different. If you want to promote this further, it should be rewritten in terms of total radiation and the approximation introduced only where appropriate.

Le Sage's theory
November 17, 2009 10:39 pm

The difficulty with Greenhouse theories comes from a basic misunderstanding of how atmospheric temperature is achieved. It’s mainly the result of cosmic presure on the Earths atmosphere.
“The theory posits that the force of gravity is the result of tiny particles (corpuscles) moving at high speed in all directions, throughout the universe. The intensity of the flux of particles is assumed to be the same in all directions, so an isolated object A is struck equally from all sides, resulting in only an inward-directed pressure but no net directional force (P1).
P2: Two bodies “attract” each otherWith a second object B present, however, a fraction of the particles that would otherwise have struck A from the direction of B is intercepted, so B works as a shield,…”
http://en.wikipedia.org/wiki/Le_Sage's_theory_of_gravitation
Temperature is proportional to pressure, therefor Earth atmospheric temperature is formed by background radiation pressure, and distance from the shielding effect of the Sun
http://www.chm.davidson.edu/vce/KineticMolecularTheory/PT.html
CO2 plays no part in this.

Ian Schumacher
November 17, 2009 10:54 pm

carlbrannen,
I’ve often thought that temperature itself is such bad concept. Its a state variable. It isn’t conserved. It depends on properties of the materials. Far better would be something like Energy density J/m^3. Energy density can be roughly translated into temperature when needed (for ideal gas, etc), we can integrate energy density to get energy which IS conserved. Various concepts come out of it naturally (energy flux energy density). Certain things come out of it easily, such as the concept of black body radiation and balance of energy flux with energy density in a cavity (for example).
Right away we could also see that the problem with Willis model is that he is starting off with an infinite internal energy source and therefore energy density inside the spheres is infinite to. How do we apply this internal energy source to the Earth with an external energy source. We can’t. Willis has put the Sun inside the Earth.
In a real model with external energy source, anything that we ‘put’ in place to restrict heat flux out will also restrict heat flux in. It is fairly obvious that putting a metal shell around the Earth would make it a whole lot colder, not warmer.
The key with global warming is conversion of wavelengths. Any global warming theory must address that. High energy short wavelengths come in, they heat the surface, lower energy longer wavelengths go out. This is what makes the Earth look more like a black body (absorbing more radiation then it otherwise would). Black bodies are often modeled as a small hole in hollow sphere. The greenhouse effect is a small hole in frequency spectrum, rather than space. The greenhouse effect makes the Earth more like a black body. But that is the limit, it can’t be MORE than a black body (hotter than a black body).

Willis Eschenbach
November 17, 2009 11:31 pm

Anton Eagle (22:12:14) :

Logic. Science and logic should be inseperable.
This article is utter nonsense, and a simple logical analysis reveals this without even resorting to physics (and I am a physicist).
Using the author’s argument… the mythical planet radiates 235 W/m2. The steel shell absorbs this and re-radiates it all in both directions. The planet obligingly absorbs the re-radiated 235 coming off the shell, and re-radiates it, for a total radiation “budget” of 470 W/m2. The author conveniently stops at this point.
But wait… there’s more. If we are to follow the author’s logic, then the shell should absorb this 470 W/m2 and re-radiate it again… 470 out and 470 in. The planet should then absorb this 470… add it to the orignal 235 coming from the core, and now be radiating 705 W/m2… this would continue till everything anihilated at a gazillion degrees. Utter nonsense.
As requested by others above… please remove this article.

The shell does absorb 470 W/m2. The point you seem to miss is that the surface area of the shell is twice the surface area of the planet.
Let’s take the surface area of the planet as X square metres. It is radiating a total of 470 W/m2 * X m2 = 470 X watts.
Now consider the shell. The surface area of the shell is 2X square metres. It is radiating 235 W/m2 * 2X = 470X watts … which is the same amount that it is absorbing. Energy in = energy out, and the system is at balance. That is what makes the system work, that a shell has two sides, so it has twice the surface area of the sphere.
So no, I see nothing in your claim that would require that I remove the article. Your “simple logical analysis” is simply incorrect.
Could we please hold off on requests to remove the article until we have at least discussed the objections to my article? This request, for action before consideration, is unseemly in a scientific discussion.

Willis Eschenbach
November 17, 2009 11:34 pm

yonason (21:37:01) :

Willis Eschenbach (20:36:37) :
“In a perfect greenhouse, one shell gives a surface temperature of two times the incoming radiation, two shells give us a surface temperature of three times the incoming radiation, three shells gives four times, and so on.”
So, how many shells will it take to get the core to 2,000,000 degrees?

Haven’t a clue, but the inner shells will melt long before that … not sure what your point is here.

Willis Eschenbach
November 17, 2009 11:37 pm

Thomas (22:14:36) :

Wait what? You add a shell, it obtains thermal equilibrium with the earth such that it radiates the same magnitude flux. Right: so you’ve got:
F(earth) = Flux(shell) (1)
but then you say:
Flux(shell in) = Flux(earth) and also Flux(shell out) = Flux(earth)
This is obviously wrong because
Flux(shell) = Flux(shell in) + Flux(shell out)
which is simply in violation of (1)
So really you should have
Flux(shell in) + Flux(shell out) = Flux(shell) = Flux(earth)
If you want, you can assume that the shell radiates equally in and out and that would make each 1/2 of the earth’s flux.

Thomas, it looks like you are stuck on the same point that Anton Eagle didn’t consider, which is that the surface area of the shell is twice that of the earth. As a result, it radiates half of the received energy inwards, and half outwards. Go back to Figure 1, which shows the relationships. The shell receives 470 W/m2. It radiates it over twice the area, which gives us 235 W/m2 in each direction.

Willis Eschenbach
November 17, 2009 11:45 pm

Ron House (22:29:46) :

Stephen Goldstein (20:05:22) :
I am trying to get used to using W/M^2 as equivalent to temperature”
It isn’t. Firstly, radiation depends on a body’s resemblance to a black body. There is no reason a shell as postulated could not have different reflectance on the two sides (or even on parts of one side, as with the Earth and ocean/land/forest/bitumen etc.). Secondly, there is not a linear relation between WM^-2 and temperature, since radiation is proportional to T^4.
And I have my doubts that there is any reason to equate the reflectances of the inner and outer sides, since the rising temp on the inside will certainly give a different apparent reflectance.
As for the article, it starts out determined to be misunderstood by showing equal flux density at two different radii, which is just plain false. Sorry, Willis, but you will be misunderstood if you write fallacies and expect the reader to pick up on all the approximations. And there is, of course, no approximation with the Earth’s core and the surface, these radii are wildly different. If you want to promote this further, it should be rewritten in terms of total radiation and the approximation introduced only where appropriate.

First, as I said, this is a thought experiment with a blackbody planet and a blackbody shell. So your objections about reflectance, which are correct and which apply to the real world, do not apply here.
Next, as I showed above (Willis Eschenbach (20:32:25), the difference in the radii is less that 0.1%. You are right, I should have stated that in the head post, but I remedied that in the response above, so I’m not sure why you are bringing this up again.
The same is true about the difference between the core and the surface radii. I addressed this above, at Willis Eschenbach (20:36:37).
Finally, I wish people would let up on the talk of “writing fallacies” and the like. I have written this as clearly as I can, and I am clarifying what is not clear as we go. I am not “writing fallacies”, I’m doing the best I can. If you have questions or objections, that’s great, that’s why I put this out … but there is no need to insult me in the process.

Willis Eschenbach
November 17, 2009 11:53 pm

Le Sage’s theory (22:39:15) :

The difficulty with Greenhouse theories comes from a basic misunderstanding of how atmospheric temperature is achieved. It’s mainly the result of cosmic presure on the Earths atmosphere.
“The theory posits that the force of gravity is the result of tiny particles (corpuscles) moving at high speed in all directions, throughout the universe. The intensity of the flux of particles is assumed to be the same in all directions, so an isolated object A is struck equally from all sides, resulting in only an inward-directed pressure but no net directional force (P1).
P2: Two bodies “attract” each otherWith a second object B present, however, a fraction of the particles that would otherwise have struck A from the direction of B is intercepted, so B works as a shield,…”
http://en.wikipedia.org/wiki/Le_Sage%27s_theory_of_gravitation
Temperature is proportional to pressure, therefor Earth atmospheric temperature is formed by background radiation pressure, and distance from the shielding effect of the Sun
http://www.chm.davidson.edu/vce/KineticMolecularTheory/PT.html
CO2 plays no part in this.

While it is good, albeit surprising, to get a message espousing the physics theories of the 17th Century, here we are discussing the Steel Greenhouse. I have fixed your Wikipedia link in what I quoted above so people can read your theory at their leisure, and I invite you to discuss it on another thread.
Thanks,
w.

Willis Eschenbach
November 18, 2009 12:05 am

carlbrannen (21:17:46) :

This post is wrong for a rather large number of reasons. It doesn’t take into account the radius and W/m^2 correctly, but most importantly, W/m^2 is not, nor ever has been, equivalent to temperature. It’s just a bad idea.

Please read up on the Stefan-Boltzmann equation that I list in the appendix. It allows us to calculation the amount of radiation which is emitted by a body at a certain temperature, or the temperature which a body needs to emit a certain radiation.

A better analogy to the greenhouse effect is “insulation”. You can get that by steel spheres and air gaps, but it’s better to analyze it as insulation. With different thickness of insulation, a fixed amount of heat flow results in a temperature difference proportional to the thickness.

Unfortunately, a “blanket” or “insulation” is a very poor model for the greenhouse effect. This is the difference between an insulated bottle and a Dewar flask (Thermos bottle). They operate on entirely different principles.
The Thermos bottle operates exactly like the steel greenhouse in my thought experiment, using a shell surrounding the inner chamber and separated from the chamber by a vacuum. This is why the Thermos is so much more efficient at keeping your coffee from losing heat than an insulated bottle. Half of the heat that is lost from the inner chamber to the shell is radiated back inward, keeping the contents warm and slowing heat loss.

Willis Eschenbach
November 18, 2009 12:08 am

Ian Schumacher (22:54:19) :


In a real model with external energy source, anything that we ‘put’ in place to restrict heat flux out will also restrict heat flux in.

Curiously, in the Earth this isn’t true. The atmosphere is basically transparent to incoming solar radiation, but it absorbs outgoing longwave radiation. Heat flux out is restricted, but heat flux in is not restricted.

Anton Eagle
November 18, 2009 12:15 am

Willis, your reply to my comment is even less logical than your original article. You state… “The shell does absorb 470 W/m2. The point you seem to miss is that the surface area of the shell is twice the surface area of the planet”… blah blah…
No, your shell is not necessarily “twice the surface area of the planet”. In fact, in your model, the distance of the shell from the planet has no bearing at all on the model…

and thus the area of the shell has no bearing on the model (the further the shell, the larger the area), and thus your attempt to show the radiation of the shell being allocated to some larger area falls apart, since the area can be any value that is larger than the area of the planet. Its not any specific quantity.
Your comments don’t even hold up consistently against your own model. Just give it up. Your model doesn’t make any sense… it’s non-physical.

Anton Eagle
November 18, 2009 12:30 am

Look. Lets simplify this argument.
You are treating this shell as if its a source of energy. But, its not a source of energy. The only source of energy in your model is the planet.
So, you have an energy source that is heating up a passive object. That passive object cannot then turn around and heat up the energy source. If it could, you would end up in some kind of infinite loop that violates the laws of thermodynamics. This simply cannot happen. Period. End of discussion.
Another way of looking at your model…
Pretend your shell is in contact with the surface. Its easy to see then, that the shell would simply be the same temp as the planet, and the planet would be the same temp as it was without the shell. Raising the shell up off the ground does not change this situation. Again, the shell does not magically heat up the planet… anymore than the surface material of the planet heats up the material below the surface.
You are violating laws of thermodynamics by essentially having energy flowing up hill (so to speak)… from the passive shell to the energy source. This just doesn’t happen.
If it could, then you could just as easily argue that two space heaters placed next to each other could heat each other up to millions of degrees with the same kind of infinite loop. Again… just doesn’t happen.

November 18, 2009 12:32 am

Exactly, Willis, very nice and analytic in Nature.
At the beginning, I was afraid that you would argue that the difference between 195 and 324 would permanently invalidate any such model.
The two figures are different simply because 195 refers to the radiation of the (colder) upper layers of the atmosphere (up), while 324 is the radiation of the (warmer) lower layers of the atmosphere. They don’t have to be equal. The whole greenhouse effect would disappear if there were no lapse rate – i.e. cooling of the air with the altitude which is close to the adiabatic one in the real system (as opposed to isothermal).
With the two shells, you may cure any contradictions of this sort because the two shells may have different temperatures and any discrepancy can be covered by non-radiative heat transfers in between the two shells. Of course, such additional terms in the budget are not arbitrary: heat convection is proportional to gradients, and so forth.
However, I am still not quite certain that you have fully accounted for the objects’ not being black bodies.
By the way, your figure 1) would suffer from an entertaining new huge problem – which is much smaller in reality. Your figure has the outer sphere which has approximately a 3x bigger radius than the Earth. Consequently, its surface area is 9x bigger, and “per squared meter” data should be recalculated to a much bigger area, changing them by a factor of nine which is not quite negligible. 😉
In fact, if the spheres were that different, it would still be a priori natural for them to have the same temperature at equilibrium. However, the inner sphere would be getting most of the energy from the outer sphere, and a negligible one from the Sun.
In reality, the atmosphere’t thickness is something like 0.3% of the Earth’s radius, making the external surfaces 0.6% bigger or so. Neglecting that the Earth is not round ;-), one could therefore get errors as big as 0.6% of 342 W/m^2 which is something like 2 W/m^2 only.
All the numbers are global averages. The actual flows depend heavily on the place, especially on the latitude, and there are additional heat transfers in between the different latitudes.

Michael D Smith
November 18, 2009 12:57 am

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.”
Nice to see you here, damn fine job on Glenn Beck. Next time maybe he’ll let you do the talking for the full hour.
The non-radiative effects (convection) and phase changes carry more heat into the upper atmosphere where there is a greater chance for energy to be radiated directly to space, less chance of radiative interaction with molecules on the way out. A small localized change in surface temperature can cause a convection burst (thunderstorm) and a large increase in convection height, improving both reflection of incoming solar radiation, and conveying sensible heat to a higher altitude where it can then escape to space via radiative processes with far less interference. Convection at lower altitudes will punch through the thicker lower layers allowing a much more direct path for radiative effects to become efficient and unhindered in delivering the heat to space.
Clouds and condensation are the balancing outgoing delivery mechanism of heat on this planet, and overwhelm the radiative effect with convection, and as a bonus also block incoming radiation, especially in the tropics, leading to a natural, self regulating thermostat effect. This results in the lower climate sensitivity we are seeing in many recent studies like Lindzen’s. The notion of an H2O positive feedback (which probably is present on a clear day) is squashed by this process.
While warmer air can hold exponentially more water vapor, presumably increasing greenhouse effects (an process the IPCC hangs its collective hat on), it is also this exact same property that vastly improves the chances of convective and phase change heat transport by thunderstorms. Once triggered, the radiative effects of H2O are completely overwhelmed by the storms, resulting in a very strong localized negative feedback. Cumulus clouds will have the same effect, but more in balance with the positive effects, resulting in less negative net feedback, but with the same result, much lower climate sensitivity than the IPCC would have you believe.
I realize that climate sensitivity is not usually discussed as a local phenomenon, but it should be, since it is the integral of all local phenomena.
This should be fairly obvious. Suppose some factor, like CO2, or weather, it doesn’t matter, produces a warmer day. More water will evaporate. And then it is more likely that a storm will develop. If one does, negative feedback by non-radiative effects. If one doesn’t, positive feedback by radiative effects. More positive feedback also evaporates even more water, further increasing the chances of a storm developing. If the storm develops, you can be guaranteed that the excess heat will, without a doubt, be taken to space in an orders of magnitude more direct way. The radiative changes will be fairly small no matter what, but the non-radiative effects will be most extreme where needed the most (tropics and mid-latitudes), and very mild where temperature differentials are smaller (toward the poles).
An important study would be to nail down these localized effects using measurements in very small increments of time and space. Once the local effects are understood, it should be a much simpler matter to integrate them over time and space using satellite cloud measurements to a more generalized climate sensitivity over various bands / latitudes of the earth. It seems most researchers are using data over far too much time and distance to derive the feedbacks properly, resulting in a very smeared picture of what’s going on. The real differences between radiative and non-radiative are very extreme but short-lived. Capturing this signal and integrating it seems to me to be the only way possible to better understand the broader effects and how sensitivity changes over time and space.
When the non-radiative effects are orders of magnitude more effective than the radiative ones, and we see this clearly on the surface of the sun (or in a boiling tea kettle), does this not scream net-negative feedback in the climate system?

Cyril
November 18, 2009 1:05 am

Thin foil metal (solid) are very efficient to protect you from very hot source, but as stated by doctor Pierre-Marie Robitaille (Ohio State University) in “An Analysis of Universality in Blackbody Radiation”
http://arxiv.org/ftp/physics/papers/0507/0507007.pdf :
“The application of the laws of Planck, Stefan and Wien to non-solids is without both
experimental and theoretical justification”
Neither the hearth (mainly ocean) nor the atmosphere are solid. This model will not work.

Ed Zuiderwijk
November 18, 2009 1:08 am

I’m sorry, but this is not even wrong …..

MartinGAtkins
November 18, 2009 1:25 am

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.
I haven’t had time to read other responses yet but.
I believe the model to be fundamentally wrong.
“To do this, the steel shell must warm until it is radiating at 235 watts per square metre.”
1. The entire surface of the outer shell would only have to radiate the total energy output of the core. Since the surface area of the outer shell has a greater area than the core then it would only radiate a fraction of the total energy at the outer surface area per unit that the expanded volume would dictate. Not 235 watts per square metre.
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.
2. The entire shell would radiate 235 watts in all directions so in a two dimensional world only 235/2 or 117.50 watts would be returned to the system.
If my obsevations are correct then post is misleading and of no help to those who wish to learn.

November 18, 2009 1:28 am

Re: Anton Eagle (22:12:14)
Logic. Science and logic should be inseperable.
This article is utter nonsense, and a simple logical analysis reveals this without even resorting to physics (and I am a physicist).
Using the author’s argument… the mythical planet radiates 235 W/m2. The steel shell absorbs this and re-radiates it all in both directions. The planet obligingly absorbs the re-radiated 235 coming off the shell, and re-radiates it, for a total radiation “budget” of 470 W/m2. The author conveniently stops at this point.
But wait… there’s more. If we are to follow the author’s logic, then the shell should absorb this 470 W/m2 and re-radiate it again… 470 out and 470 in. The planet should then absorb this 470… add it to the orignal 235 coming from the core, and now be radiating 705 W/m2… this would continue till everything anihilated at a gazillion degrees. Utter nonsense.

Thank you Anton. It wasn’t just me who spotted that the logic is simply busted.
If the shell were a perfect reflector, the temperature of the atmosphere would simply rise to the temperature of the input and stay there. Regardless of how many shells were outside, the temperature would never rise beyond the input.
Willis, next time contact a physicist before doing something like this.
This isn’t about the greenhouse effect, because you’ve misunderstood what the greenhouse effect is.
Case closed.

November 18, 2009 1:31 am

>>In a sense, what Willis has done is manage to
>>make the same mistake as the people who think
>>that the Earth is in “radiative balance” and try to work
>>out a tedious budget of everything going in and out.
Ummm. The Earth HAS to be in “radiative balance” over long time intervals, otherwise it would be warming up or cooling down.
.

November 18, 2009 1:36 am

Willis,
Let me demonstrate with a simple example:
Say you have a Dewar (vacuum flask) and you fill it with lukewarm soup. The energy is radiated back from the outer reflective skin (conduction and convection being minimal.
According to your recipe, the temperature of the soup would rise. And if you put that Dewar flask inside of another, the temperature would rise still further… and so on ad infinitum.
Except that that isn’t what happens at all. The soup remains lukewarm and slowly cools.
Its a basic mistake Willis, but you’re going to have to think about it first.

Jon-Anders Grannes
November 18, 2009 1:45 am

If i measure IR temp from clouds above I find that its much colder(density of the air) than on the ground, so this is not where the IR is re-emitted to ground.
The IR from the ground will have to “absorbed” by the air just above it and i really have a problem imagening that this air is re-emitting this energy as IR?
I imagine the climate on Earth mostly as consequence of density of the air an geographic position.
If we depleted the atmosphere over Earth from todays 1013 hPa to 500 hPa we would have an average temperature over Equator at the surface as it is today at 18.000 feet, that is below freezing!
So for me “greenhouseeffect” is just that energy(light and IR etc) is trasmitted trough the atmosphere and that resistance in the atmosphere, density of the air, leads to some warming of the media(atmosphere) this energy is going trough.
?

November 18, 2009 1:47 am

>>But wait… there’s more. If we are to follow the author’s
>>logic, then the shell should absorb this 470 W/m2 and
>>re-radiate it again… 470 out and 470 in. The planet
>>should then absorb this 470… add it to the orignal 235
>>coming from the core, and now be radiating 705 W/m2…
>>this would continue till everything anihilated at a gazillion
>>degrees. Utter nonsense.
The outer shell DOES re-radiate 470 W/m2 — 235 inwards and 235 outwards. That makes 470 to me and the outer shell is in equilibrium (as is the Earth).
And you are a physicist??
This is something I find with scientists in general. Quite brilliant in their specialist field, but all at sea in the real world. The trouble with climate, of course, is that it is a hugely multi-disciplinary field, and there are not that many people who have both the depth and breadth of knowledge to be able to deal with it logically – as this article has demonstrated.
.

SNRAtio
November 18, 2009 1:50 am

“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.”
OK, so now the whole tradition of which Kiehl/Trenberth is a part is proven wrong.

November 18, 2009 1:55 am

>>Ron House (22:29:46) :
>>As for the article, it starts out determined to be
>>misunderstood by showing equal flux density at two
>>different radii, which is just plain false.
Hardly. The shell is 2 km above the surface of a planet that has a radius of 6370 km. Hardly a difference in radii for a simple thought experiment.
.

lgl
November 18, 2009 2:01 am

Positive feedback from wiki: “Ai = (output voltage/input voltage) = A/ (1 − Aβ). Here A is the gain of the feed-forward active part of the amplifier without feedback, and β is the gain of the feedback element”
In your ‘Steelhouse’ A=1 and β=0.5 giving Ai=2 so that works fine.
Your problem is: “It has nothing to do with blankets, or mirrors, or greenhouse gases”. It has everything to do with greenhouse gases. O2 and N2 do not radiate (almost) so without the GHGs you would not get much downward LW.

Robert Wood of Canada
November 18, 2009 2:02 am

kurt (16:12:26) :
I stand corrected. I meant that the total radiative flux is proportional to Ts^4-Ta^4. Thanks for clarifying me, JaneHM.

Stephen Wilde
November 18, 2009 2:04 am

Willis,
You are on the right track but it’s horrendously difficult to create a description that avoids misinterpretation by others.
Your analogy is not perfect (in my humble opinion) but it helps to illustrate some important considerations that AGW theory ignores altogether.
You are dealing well with the naysayers who seem to lack specificity in proportion to the vigour of their opinions.
I consider the role of the stratosphere to be critical. It represents an area of transition between two separate energy transfer regimes (shells, if you will).
The troposphere works mostly via convection and the speed of the hydrological cycle. The upper atmosphere works mainly via direct radiative transfer.
The stratosphere is the buffer between the two and in a few days I will be releasing a fuller description of the significance of that (for climate changes) at climaterealists.com

michael hamnmer
November 18, 2009 3:15 am

This is an interesting post. Much of what it says I agree with but there are to me some glaring errors. To start, I totally agree with the steel shell model both the reasoning and the numerical conclusions. Where I start having problems is with the Kiehl and Trenberth model and the analysis flowing from that.
You see there is a second relationship which must also be met and that is the temperature of the radiating surface. Objects do not radiate because they receive radiation, they radiate because of their temperature. So one also has to check that the temperature data is correct.
Consider just the CO2 GHG component. The total absorbance of the CO2 in the atmospheric column at 280 ppm is 2000 absorbance. (For those who are not familiar with the term absorbance it is the amount of an absorber that absorbs 90% of the incident energy at the absorbing wavelength). Thus its not one shell or two shells but closer to 2000 shells. As the steel sphere correctly shows energy is only radiated from the outermost shell since energy radiated from innner shells is reabsorbed by shells above it. This means that energy is only radiated from the top of the atmospheric column containing the GHG.
For earth the most prominent green house gas is water vapour and the top of the water vapour column is the tropopause. For CO2 it could be somewhat higher. But herein lies a problem, the tropopause is too cold to radiate the 165 watts/sqM the Kiehl and Trenberth model say is radiated by the atmosphere. So, maybe the tropopause is is not the radiating layer after all Kiehl and Trenberth talk about energy radiated to space from within the troposphere and even talk about an equivalent radiation altitude? Well there is simple proof it has to be, you see the tropopause is colder both than the atmosphere below it and the atmosphere above it. How does the tropopause stay cold when it is completely surrounded by warmer atmosphere? There is only one way and that is that it is radiating energy to a heat sink and the only heat sink colder than it is outer space. So the tropopause has to be radiating directly to space. But could not the lower atmopshere also be radiating? Well no because the ability to radiate and the ability to absorb are both governed by the same factor – the emissivity. If the tropopasue is radiating strongly to space it will also be absorbing strongly any radiation received from below, so it would be blocking radiation emanating from lower in the troposphere. This means the tropopause has to be the effective top of the atmosphere from a GHG point of view.
Now to come back to the steel shell analogy – if there really are 2000 shells the energy radiated back to the surface would be astonomical or put another way, almost none of the energy from the surface would be escaping to space. But that would mean the surface of the earth would be more like venus and it clearly isn’t so there has to be a flaw. Indeed there is. CO2 is not like a steel shell because it only absorbs and radiates over a small range of wavelengths rather than at all wavelengths as a steel shell would. with 2000 odd shells, indeed very little energy escapes from earth at the wavelengthes at which CO2 absorbs. However energy can escape freely from wavelengths within the atmospheric window where there is little if any atmospheric absorption. Thus Kiehl and Trenberth underestimates the emission to space in the atmospheric window and overestimates the energy absorbed by GHG in the atmosphere. I did a numerical analysis of this in some depth in my first paper at jennifermarohasy.com/blog.
The suggestion of two shells, one being the troposphere and the other the stratosphere cannot be correct because the outermost shell has to be cooler than the next inner shell and in this case the tropopause is colder than the stratosphere.

Myrddin Wyn
November 18, 2009 4:09 am

Interesting article and debate.
I’m missing a step here though, the power source supplies 235watts/m2 for a surface area of ‘a’, if the shell has twice the area, 2a, shouldn’t it emit half the energy; 117.5 W/m2.
Per m2 I cant see how the shell emits twice the energy it receives from the power source, the shell would be a power multiplier, Eout = 2Ein, which is impossible – no ?

Marcus
November 18, 2009 4:20 am

Willis: I thought I would pipe up, since I had such vehement objections to your prior article on the Shindell paper, just to say that I thought this was a good post.
-Marcus

Vincent
November 18, 2009 4:22 am

Willis,
excellent article, but I there is one thing that doesn’t make sense to me. If you start with a ball radiating at 235 w/m2, then if you add the 235 w/m2 radiated back from the steel shell aren’t you creating energy out of nothing? Isn’t this an example of perpetuum mobile of the third kind? Sorry if I’m being a bit thick but if we start with 235 joules worth of energy at the surface, then the other 235 joules that were transmitted downwards came from the original 235 to begin with, so then we have created energy out of nothing.
HELP!

November 18, 2009 4:30 am

Quote:
John A (01:36:08) :
Let me demonstrate with a simple example:
Say you have a Dewar (vacuum flask) and you fill it with lukewarm soup. According to your recipe, the temperature of the soup would rise. And if you put that Dewar flask inside of another, the temperature would rise still further… and so on ad infinitum.
Its a basic mistake Willis, but you’re going to have to think about it first.

.
Will people please think before posting.
The soup analogy is wrong, because the soup does not have an internal energy source. The Earth system does (nuclear in this thought experiment, solar short-wave on the real Earth).
The soup analogy would only work if the soup had a small heating element inside it (nuclear powered?) giving of a constant 10w. Then it would heat up and up, depending on the number of vacuum layers.
.

PSU-EMS-Alum
November 18, 2009 5:01 am

You know what needs to be removed from this page?
Any comment that uses the term “re-radiate”.
This is a pet peeve of mine …. it demonstrates a complete lack of understanding of the concept of “radiation”.

Bill Illis
November 18, 2009 5:01 am

There is quite a difference between the Stefan-Boltzmann equations (the fundamental equations governing radiation physics and temperature) and the climate models.
The surface radiates at 390 watts/metre^2 at a temperature of 15.0C.
For the surface to go up 3.0C to 18.0C, another 16.5 watts/metre^2 would have to be added to the surface.
I don’t know how the climate models can reconcile their prediction of a 4.2 watt/metre^2 increase in the tropopause radiation results in a 3.0C increase in surface temperatures (and a new surface radiation level of 406.5 watts/metre^2).
Their mistake is by converting everything into “linear equations”. Hansen took a shortcut in the early 1980s using his early climate model results (for the last glacial maximum for example) where his climate model came up with -6.6 watts change. Temperatures declined by 5.0C so therefore, the sensitivity is 0.75C per watt.
http://europe.theoildrum.com/uploads/465/cv_hansen_fig3.png
Between this shortcut/mistake (which violates the Stephan-Boltzmann equations and was copied by all the following climate scientists) and through the climate model’s assumption of a constant linear lapse rate of 6C/kilometre when it is probably not constant), they have changed all the logarithmic radiation equations into linear ones. And they are not.
http://img524.imageshack.us/img524/6840/sbearthsurfacetemp.png
http://img43.imageshack.us/img43/2608/sbtempcperwatt.png
Shortcuts and mistakes that were not corrected and here we are.

Stephen Goldstein
November 18, 2009 5:02 am

Willis Eschenbach (23:34:12)
yonason (21:37:01)
So, how many shells will it take to get the core to 2,000,000 degrees?
Haven’t a clue, but the inner shells will melt long before that … not
sure what your point is here.
It was a jocular reference, I am sure, to Al Gore’s recent claim, in the context of geothermal energy, that the temperature of the earth’s core is “several million degrees.” The WUWT thread on Gore’s TV appearance and this claim immediately precedes this thread.

Steve Fitzpatrick
November 18, 2009 5:26 am

Alan D. McIntire (21:13:02) :
The wavelength for CO2 absorption is ~14 microns, not ~14 nm. 14 nm is close to the soft x-ray region.

Charlie
November 18, 2009 6:09 am

Re: Anton Eagle (00:15:58) :
Anton, you seem to be making the same mistakes that myself and, judging from the comments, many others made. There are a couple things that I missed when reading the article and the comments.
First this isn’t intended to be a climate model. It is in fact a model to show that our atmosphere does not act exactly like a greenhouse if I am understanding it correctly.
Second, let us look at the example of a ping-pong ball and a golf ball. We will assume that the golf ball has no dimples and is the same radius as the ping-pong ball, say 1cm radius. The surface area of the golf ball will be 4pi*r^2 or approximately 12.6 cm^2 as there is only an outside surface. The surface area of the ping-pong ball will be twice that, or about 25.2 cm^2 as there is an inside and outside surface.
Once I got my head around those items, the rest of the logic makes sense. Since the shell is radiating 235 W/m^2 from the side towards space (the net energy flow), it must also be radiating 235 W/m^2 from the other side towards the planet. Since the shell and planet must be at equilibrium, the energy radiated by the planet to the shell must raise to 470 W/m^2, where it will remain as the system is in “equilibrium” radiating a total of 235 W/m^2 towards space as it was before the steel greenhouse was added.
Also Re: Willis Eschenbach (23:34:12) :
I believe the comment by yonason was a very subtle dig at AlGore. If I understand that correctly he is referring to Mr. Gore’s comment that only a few kilometers below the crust the earth is a couple million degrees.
Charlie

Back2Bat
November 18, 2009 6:14 am

Thanks Anthony for what is apparently a very rich article. I’m gonna bookmark this jewel.

Tenuc
November 18, 2009 6:39 am

I like your “Tinkertoy” model of a world with a steel shell ‘greenhouse’. It reminds me of the hours of fun I had back in the ’80’s playing around with a Daisy World simulator I’d programmed in Basic on an old Dragon PC. Happy days. I’m going to have some fun playing with your toy :-))
I think much can be learned from messing around with these sorts of ‘toys’, although as pointed out by many of the earlier comments, models of our climate system have little or no correlation to what goes on in the real world. Earth’s climate system is an immensely complex dynamic chaotic system, with much turbulence and boundary effects involved in the physics, chemistry and biology which go into make the total system. Even the best of the non-linear models being used by the IPPC do not provide results which can be used to predict the direction of our climate, although as with simpler models they are useful learning tools.
I think all models fall down because the many of the initial assumptions are wrong and there is poor quantitative understanding of how each non-linear process works. There is too little accurate data and the granularity of that date is insufficient by several orders of magnitude.
An example of a possible wrong initial assumption of real world models is the fact that Stafan’s radiation law is only valid only for in vacuum black body radiation, where all wavelengths are present. It is particularly important to understand this if the effects of GHG’s on climate are to be understood. Couple of links to this issue here:-
http://www.colin-baxter.com/tips_and_ideas/bb_radiation_gh/index.html
http://www.worldscinet.com/ijmpb/23/2303/S021797920904984X.html
Thanks Willis for posting such an enjoyable and thought provoking thread – this is what keeps me coming back to WUWT!

TomVonk
November 18, 2009 6:54 am

Please scratch the above . I did a mistake . This one is the right version :
=========================================
Willis I think that you made a mistake when writing :
“R = sigma * epsilon * T^4 where R = radiation (W/m2)”
.
The units in the Stefan Boltzmann formula are W/m²/sr .
When the solid angle is omitted , it implies that the integration has already been done (generally along a plane surface) .
This is a frequent nevertheless particular case .
Concentrical spheres are not part of this particular case .
What would it mean to your example ?
.
The internal sphere of radius r emits some radiation P
(units W/m²/sr) .
The total power (units W) emitted is P.4.Pi.r².2.Pi
The solid angle here is 2.Pi .
OK .
Now this power (units W) arrives at the big sphere of radius R and will be absorbed and emitted . As both are equal we have :
Emitted power (units W) = P.4.Pi.r².2.Pi .
This power (units W) will be divided in 2 halves , one emitted by each face e.g P.4.Pi.r².Pi per face .
However as this big sphere has 2 faces it only sees a half space from each face . So the solid angle for each face is 2.Pi .
And the power per sterradian (units W/sr) for each face is :
(P.4.Pi.r².Pi) / 2.Pi = P.4.Pi.r²/2
As for the specific power (units W/m²/sr) it is : P.4.Pi.r²/2.4.Pi.R² = P/2 . r²/R² . This is for each face . For the whole outer sphere it is P.r²/R² .
.
What follows ?
1) The outer sphere has not the same temperature as the inner sphere in equilibrium because P.r²/R² is not P .
.
2) Specific powers (units W/m²/sr) do not conserve because of those m² and sr that stand in the way . Total powers do .
.
3) If r~R then both spheres have approximately the same temperatures and the specific power (units W/m²/sr) of the outer sphere is half of the inner sphere for each face (P/2) .
.
4) The absorbed part of the radiation returning to the inner sphere will decrease geometrically with reason r²/R² .
Thanks God for that because that will prevent the experiment of Michel going nuclear 🙂
.
I do not chime in with some others that the post should be removed because it is an interesting exercice of radiative interaction but it must be corrected .
Clearly it can’t be left standing that the Stefan Boltzmann law has W/m² for units when the right units are W/m²/sr .
Neither can be left standing that specific power (units W/m²/sr) conserves .

November 18, 2009 7:00 am

>>Vincent (04:22:19) :
>> If you start with a ball radiating at 235 w/m2, then if you
>>add the 235 w/m2 radiated back from the steel shell aren’t
>>you creating energy out of nothing?
No, its just that the diagram is poorly drawn, in my humble opinion.
a. There is 235 going from the Earth to the shell.
b. Then there is 235 radiating out from the shell into space (eventually).
But then there is another 235 bouncing around between shell and Earth and back again. This is not ‘created energy’, it is just bouncing around until it can finally escape as ‘b.’. So the radiation out from Earth is not instantaneous, as in the unprotected Earth, there is a time delay as the energy bounces around for a while before escaping. Draw the diagram like that, and see if it makes more sense.
.

Back2Bat
November 18, 2009 7:10 am

“If you start with a ball radiating at 235 w/m2, then if you add the 235 w/m2 radiated back from the steel shell aren’t you creating energy out of nothing?” Vincent
This bothered me a bit too till I thought of a heat source surrounded by a perfect insulator. Then the temperature of the heat source would rise indefinitely. The increase in temperature would (to my mind) simply be the accumulated energy from the heat source since it would be unable to dissipate any energy.
I hope this is correct and helpful; I am just a rusty engineer.

November 18, 2009 7:12 am

>>Myrddin Wyn (04:09:21) :
>>I’m missing a step here though, the power source
>>supplies 235watts/m2 for a surface area of ‘a’, if the
>>shell has twice the area, 2a, shouldn’t it emit half the
>>energy; 117.5 W/m2.
No, as per my post above.
The shell receives 470 units.
a. 235 from the Earth
b. 235 from the energy bouncing around in the Earth-shell void.
And the shell re-radiates 470 units.
c. 235 to space
d. 235 back to the Earth-shell void, where it bounces around for a while.
a + b = c + d, so the system is in equilibrium
And there is no ‘creation of energy’, as elements b. and d. are simply time-delayed manifestations of a.
There is only energy ‘a.’ (energy from the Earth). Energy a. (a 235-unit ‘package’) goes up to the shell, then down to Earth and back again a few times (bouncing around), and then finally escapes into space. But there is only one 235-unit energy ‘package’).
.

old construction worker
November 18, 2009 7:13 am

A couple of scientist wanted to know what the conman man thought was the greatest invention. So they went down to hills and pose this question to Bubba. He told them it was the thermos bottle. He said when I put something hot in it it stays hot and if I put something cold in it it stays cold. How do it know?

Brian
November 18, 2009 7:14 am

I agree withn John A. (20:11:22) this load of [snip] has no place on Wuwt.
Wuwt’s reputation has now been tarnished. This is not science but fantasy.
The thing is so sorely messed up that it doesn’t even qualify for print.
Has Wuwt fallen prey to a belief in pseudoscience? I am truly disappointed!
If I get started it will be 2 pages and space does not permit.
Warning: any serious consideration of this or any other AGW fantasies
will ruin your mind and lead to a belief in superstitions.

Stephen Goldstein
November 18, 2009 7:49 am

Back2Bat (07:10:21) :
“This bothered me a bit too till I thought of a heat source surrounded by a perfect insulator. Then the temperature of the heat source would rise indefinitely. The increase in temperature would (to my mind) simply be the accumulated energy from the heat source since it would be unable to dissipate any energy.
“I hope this is correct and helpful; I am just a rusty engineer.”
Exactly! Hence, in the context of nuclear power reactors, there is a need for post-shutdown core cooling to remove residual decay energy (think Three Mile Island and the consequence of a cooling system failure).

P Wilson
November 18, 2009 7:57 am

a simple diagnostic comparison that the SB constant is irrelevant to climatology:
100w/m2 is the optimum surface heat for floor based central heating. This creates 24C at floor level, cooling as hot air rises. 15C, or 59F is certainly cooler than 24C. It is improbable that the floor will re-absorb radiation it has already emitted.

Vincent
November 18, 2009 8:19 am

I think John A and Brian have made the same mistake that I made. I interpreted the other 235 w/m2 of back radiation as energy added to the original energy – ie energy out of nothing. Both Ralph and Back2bat have answered by question – thanks to both you guys.
I can see that if you take the radiation going from the steel shell into space, it is no more than the original value W, not 2W. Therefore, no extra energy is being radiated into space, but the inside gets hotter.
Obviously, if there was no energy input to begin with, the ball would continue to cool to absolute zero, since there would be a net loss of energy.
These sorts of problems are certainly tricky, and have fooled even some physicists. G&T come to mind.

TomVonk
November 18, 2009 8:37 am

Admin :
Please remove my first post (06:42:40) . I made a typo which inverted some figures and made it wrong . Thanks .
.
For Willis : the right post is the second
For those with a short attention span that doesn’t allow to read longer posts : 99% of those who posted here are confusing specific power (unit W/m²/sr) and power (unit W) .
The former does NOT conserve and the latter does .

Mark T
November 18, 2009 8:39 am

John A (01:36:08) :

Let me demonstrate with a simple example:

Your example is simple, but so simple you forgot one simple little point, which I will explain.

Say you have a Dewar (vacuum flask) and you fill it with lukewarm soup. The energy is radiated back from the outer reflective skin (conduction and convection being minimal.

Good so far.

According to your recipe, the temperature of the soup would rise. And if you put that Dewar flask inside of another, the temperature would rise still further… and so on ad infinitum.

Oops. You jumped to a conclusion that cannot be made based on his description – irrespective of the validity of his claim. Your system is not even close to the same as his system – his has a constant input power source, the sun, but I see none in your example.

Except that that isn’t what happens at all. The soup remains lukewarm and slowly cools.

Indeed, because there is no energy input to the system, the temperature will decay as a function of the reflected and radiated energy.

Its a basic mistake Willis, but you’re going to have to think about it first.

Perhaps you should take a dose of your own medicine.
Part of the problem I’m seeing here is that people don’t understand the basic mechanism of storage, which is a very simple feedback concept. While I understand that the concept of feedback is not part of the general curriculum of any typical education, people need to recognize this before making claims of psuedoscience or fantasy. To not do so is nothing more than arrogance.
Mark

Ian Schumacher
November 18, 2009 9:16 am

Willis,
If the energy source is external, your spheres must have a hole in them to let light in. You now have a spherical cavity with a hole in it, i.e. an approximation to a black body. No matter what you do inside this sphere you can not make it hotter inside then a black body as you model allows.
In the case of the earth the hole is in wavelength, not a physical hole. The result is the same though.

Hank Henry
November 18, 2009 9:17 am

I’m not sure I follow. We have a surface and a shell both at 235 w per meter but then the radiation from the shell, by radiating inward, increases the inner 235 w/m so the shell and the surface were really never both 235? It seems like we begin saying that ” the steel shell must warm until it is radiating at 235 watts per square metre” but then we continue and argue the shell “will warm the planetary surface until it reaches a temperature of 470 watts per square metre” So why wouldn’t we have to go back and say again that the steel of the shell must warm until it is radiating at 470 w/m? I think we are talking of temperature in a different way than I understand it.
To me the lesson of hot greenhouses, hot attics or hot metal sheds is that if air is confined and not allowed to convect upward to the sky it will heat up mostly because the cooling of convection has been stifled and less so because of any differential transmission of radiation by glass. In other words, sure it’s true greenhouses get hot, but once the roof of a shed or attic heats up and the air inside remains confined they get hotter than the outdoor temperature also.

November 18, 2009 9:28 am

For Ralph and other people who struggle with physics, believing in their intuition and think its physicists’ fault that they don’t understand “analogies”, let me ask a hypothetical question and see if anyone grasps the concept.
Suppose somehow that the glowing disc called the Sun is smeared out into a larger and larger disc. As it gets larger suppose the intensity of each little part gets less so that the total energy we get from the whole disc remains the same as its size grows. Now suppose the disc were smeared out all over the sky so that there is no distinction between night and day. We still receive the same amount of energy that we received before the smear out even though it would be received uniformly over all 24 hours instead of just during daylight hours. If this were to happen then the Earth’s average temperature would
a) increase
b) decrease
c) stay the same

November 18, 2009 9:40 am

It’s beginning to sound like man made a tragic error when he left the cave and decided to live on the fickle surface; does anyone know the internal temperature of our moon?

SteveBrooklineMA
November 18, 2009 9:48 am

I think Lubos Motl is too kind in his comment above. This analysis, which implies an error with K/T, errs in adding model assumptions to K/T that are not required in K/T’s approach. K/T is not a “single shell model” it is simply an energy transfer accounting. There is nothing that requires the K/T accounting to equate the downward radiated flux from the atmosphere to the ground with the radiated flux from the atmosphere to space. As Lubos points out, the atmosphere is hotter near the ground than in the high atmosphere. This is very different from a thin shell, which has only one temperature. You can argue with the values that K/T come up with, but this post does not somehow invalidate method of the K/T accounting.

Willis Eschenbach
November 18, 2009 9:58 am

Anton Eagle (00:15:58) :

Willis, your reply to my comment is even less logical than your original article. You state… “The shell does absorb 470 W/m2. The point you seem to miss is that the surface area of the shell is twice the surface area of the planet”… blah blah…
No, your shell is not necessarily “twice the surface area of the planet”. In fact, in your model, the distance of the shell from the planet has no bearing at all on the model…

and thus the area of the shell has no bearing on the model (the further the shell, the larger the area), and thus your attempt to show the radiation of the shell being allocated to some larger area falls apart, since the area can be any value that is larger than the area of the planet. Its not any specific quantity.
Your comments don’t even hold up consistently against your own model. Just give it up. Your model doesn’t make any sense… it’s non-physical.

Anton, please read all of the comments. The shell is only slightly larger than the planet. The difference between the outer area of the shell and the outer area of the planet is less than a tenth of a percent. So (as is typical for this type of discussion), I have ignored this trivial difference.
I stated this clearly above. So yes, the total area of the shell is twice the surface area of the planet, to within two tenths of a percent. I’m not sure why you continue to argue this point.

Charlie
November 18, 2009 10:09 am

Willis Eschenbach — the same misconceptions keep showing up in the comments.
You should update the head post with an addendum that addresses these common misconceptions.
1. Difference in size of shell vs earth.
Although you say in the text that the steel shell is ” few thousand meters” above the surface of the earth, the diagram exaggerates the difference in diameters. Your addendum or FAQ should explicitly state that “the steel shell is assumed to be close enough to the surface that for this simple model the outer surface of the shell and the surface of the earth are considered equal in area”.
2. The effect of albedo.
“For simplicity, this model treats both the earth and the shell as perfect blackbodies. Assuming an albedo other than zero will change the temperatures, but will not change the general conclusions on how the steel greenhouse works.”.
3. Confusion about blackbody radiation proportional strictly to the blackbody temperature, vs the NET radiation which is the difference between incoming and outgoing fluxes. The addendum/FAQ should emphasize that the inner surface of the shell emits downward at 235W/m^2 while receiving incoming radiation at 470W/m^2. There should also be a short discussion on the energy balance in the shell: 470W/m^2 received from the surface. 235W/m^2 radiated back down from the inner surface. This leaves and excess of 235 W/m^2 which in conducted through the very thin shell and radiated outward at 235W/m^2.
4. The shell has two sides. A blackbody at a given temperature radiates from all surfaces at a given flux. This ties back to point #3, above.
5. At least one post tried to point out a logical fallacy by saying that if this article was correct, then hot liquid in a thermos could be raised to an arbitrarily high temperature simply by adding more shells around it. That comment showed that they didn’t understand that the model had a constant energy source in the core, not a constant temperature mass. The real equivalent with a thermos bottle would be if one had an electric heater inside the thermos. With better insulation, the heater will get hotter, with no limit as the insulation approached perfection.
—————————————–
It’s an excellent article. Good enough that IMO it is worth responding to the common misperceptions by addressing them directly in the article.

Willis Eschenbach
November 18, 2009 10:16 am

Anton Eagle (00:30:43) :

Look. Lets simplify this argument.
You are treating this shell as if its a source of energy. But, its not a source of energy. The only source of energy in your model is the planet.
So, you have an energy source that is heating up a passive object. That passive object cannot then turn around and heat up the energy source. If it could, you would end up in some kind of infinite loop that violates the laws of thermodynamics. This simply cannot happen. Period. End of discussion.

While it is true that a passive object cannot heat an energy source infinitely, it can certainly heat it. Look at the energy flows in the K/T budget in Fig. 3 above. The earth heats the atmosphere, but the atmosphere also heats the earth.
In this case the shell heats the planet, but not forever, just until a new equilibrium is reached. If it were as you say, putting a shell around a planet would have no effect on the planet’s temperature, which flies in the face of experience.

Mark T
November 18, 2009 10:28 am

Good points, Charlie.
John A, I’m not exactly sure what point you are trying to make in your “smeared” sun example. It is immaterial to anything Willis’ has claimed. Certainly an assumption of an input averaged out over the whole surface makes computation simpler, but that is its only purpose, i.e., the assumption is made only to remove integration.
Are you the same John A that missed the external source point in the thermos example, as both Charlie and I have noted? If so, really, try a different approach, e.g., an approach that actually addresses Willis’ argument.
Mark

November 18, 2009 10:32 am

Robert Wood, and JaneHM have it correct.
The steel shell analogy fails. The S-B law uses the difference in temperature between the radiating body and the absorbing body, each raised to the fourth power.
If a steel shell were used as posited in the article, with equal heat transmission from the outer surface and the inner surface, then there would be a substantial temperature gradient across the steel shell. The inner surface would be much hotter than the outer.
There are real-world applications of this principle in thousands of applications, in particular double-walled cryogenic liquid storage tanks. The annular space between the inner and outer walls is evacuated, thus heat transferred is almost entirely due to thermal radiation.

MartinGAtkins
November 18, 2009 10:39 am

Fom the original post.
The planet is now being heated by 235 W/m2 of energy from the interior, and 235 W/m2 from the shell.
The universe is an open system. Even with a shell an entity will reach equilibrium with the input of the energy supplied.
The paper is rubbish.

Roger Knights
November 18, 2009 10:41 am

Typo–replace “it” with “is” in:
“Un-noticed by their programmers, however, it that…”

Willis Eschenbach
November 18, 2009 10:41 am

Luboš Motl (00:32:39) :

Exactly, Willis, very nice and analytic in Nature.
At the beginning, I was afraid that you would argue that the difference between 195 and 324 would permanently invalidate any such model.
The two figures are different simply because 195 refers to the radiation of the (colder) upper layers of the atmosphere (up), while 324 is the radiation of the (warmer) lower layers of the atmosphere. They don’t have to be equal. The whole greenhouse effect would disappear if there were no lapse rate – i.e. cooling of the air with the altitude which is close to the adiabatic one in the real system (as opposed to isothermal).
With the two shells, you may cure any contradictions of this sort because the two shells may have different temperatures and any discrepancy can be covered by non-radiative heat transfers in between the two shells. Of course, such additional terms in the budget are not arbitrary: heat convection is proportional to gradients, and so forth.
However, I am still not quite certain that you have fully accounted for the objects’ not being black bodies.

Thanks, Lubos, since you are a physicist your comments are much appreciated.
You are correct that the lapse rate explains the differences in the K/T budget upward and downward radiation. However, this doesn’t solve the underlying problem with the K/T budget — it doesn’t hold in enough heat. Unless there are two shells which are physically separated to prevent thermal and evaporative heat loss, the system will not heat the surface enough to explain the known surface temperature plus losses. There is a fixed physical limit on how much heat a single shell greenhouse system (with or without a lapse rate) can concentrate, and it’s not enough to explain the Earth’s temperature.
The objects are not black bodies, as you point out, which will affect the temperature of each object. Alternately, if the temperature is know, it will affect the amount of radiation.
However, as with the area of the shell and the area of the planet, this is not a large difference. For the shortwave (solar energy) part of the equation, it is explicitly accounted for by the surface albedo. For the longwave, most objects are very good at absorbing and emitting infrared radiation. Even snow and ice, which are blinding white at visual wavelengths, are almost black (excellent absorbers) in infrared. My well-thumbed copy of “Climate Near The Ground” gives the following emissivities for longwave radiation:
Water 0.98
Snow 0.986
Leaves 0.97
Grass 0.986
Coniferous Forest 0.97
Dry Peat 0.97
Dry sand 0.95
So if we assume 0.97 as the emissivity of the planet, it only makes a difference in the surface temperature of two degrees.
However, I must ask you to explain your statement that “The whole greenhouse effect would disappear if there were no lapse rate”. As my steel greenhouse shows, a greenhouse effect can exist without any atmosphere at all.

Roger Knights
November 18, 2009 10:42 am

PS–and delete the hyphen from “un-noticed”

Willis Eschenbach
November 18, 2009 10:45 am

Michael D Smith (00:57:13) :

Clouds and condensation are the balancing outgoing delivery mechanism of heat on this planet, and overwhelm the radiative effect with convection, and as a bonus also block incoming radiation, especially in the tropics, leading to a natural, self regulating thermostat effect. This results in the lower climate sensitivity we are seeing in many recent studies like Lindzen’s. The notion of an H2O positive feedback (which probably is present on a clear day) is squashed by this process.
While warmer air can hold exponentially more water vapor, presumably increasing greenhouse effects (an process the IPCC hangs its collective hat on), it is also this exact same property that vastly improves the chances of convective and phase change heat transport by thunderstorms. Once triggered, the radiative effects of H2O are completely overwhelmed by the storms, resulting in a very strong localized negative feedback. Cumulus clouds will have the same effect, but more in balance with the positive effects, resulting in less negative net feedback, but with the same result, much lower climate sensitivity than the IPCC would have you believe.
I realize that climate sensitivity is not usually discussed as a local phenomenon, but it should be, since it is the integral of all local phenomena.

I could not agree more. I discuss this at length on another thread here at WUWT. The greenhouse effect exists, but it does not control the temperature of the planet. That honor goes to clouds and thunderstorms.

November 18, 2009 10:45 am

>>For Ralph and other people who struggle with physics,
>>let me ask a hypothetical question and see if anyone
>>grasps the concept.
>>>Smeared out, 24-hour Sun.
The Earth would stay at the same temperature. But you have to understand that the climate models assume that the Sun has very little direct impact on the atmosphere (on the steel shell). Thus there is no Solar warming of the steel shell.
Thus the thought experiment remains the same. The internal energy source represents the Solar flux passing though a ‘transparent’ steel shell, and warming the Earth, while the outgoing energy from the Earth is faced with a suddenly opaque steel shell, and is forced to warm it up (change in wavelength).
.

Willis Eschenbach
November 18, 2009 10:50 am

Cyril (01:05:22) :

Thin foil metal (solid) are very efficient to protect you from very hot source, but as stated by doctor Pierre-Marie Robitaille (Ohio State University) in “An Analysis of Universality in Blackbody Radiation”
http://arxiv.org/ftp/physics/papers/0507/0507007.pdf :
“The application of the laws of Planck, Stefan and Wien to non-solids is without both
experimental and theoretical justification”
Neither the hearth (mainly ocean) nor the atmosphere are solid. This model will not work.

Relying on a single unpublished paper in arxiv which has no other support is generally not a good plan. The applicability of the Stefan-Boltzmann equation to liquids and solids has extensive support, both theoretical and experimental. It forms the basis of such things as infrared cameras, which show the temperature of water as easily as that of solids. It is also used in satellites to measure air temperature. I fear your basic thesis is flawed, but this is not the place to debate it.
Thanks,
w.

Willis Eschenbach
November 18, 2009 10:53 am

MartinGAtkins (01:25:06) :


I haven’t had time to read other responses yet but.

Dear friends, first rule of long threads. Read the other responses first, as many mysteries are solved therein. If you still have idea or questions or objections after reading the thread, then add them to the thread.
Thanks,
w.

November 18, 2009 10:59 am

>>>So, you have an energy source that is heating
>>>up a passive object. That passive object cannot
>>>then turn around and heat up the energy source.
Of course it can!!!!
Stand outside on a cloudy night, and you will invariably find the temperature quite balmy.
Then stand outside on a cloudless night (preferably the next day), and you will invariably find the temperature is quite cold in comparison.
The difference is that the clouds (your so-called passive objects) are heating up the surface of the Earth. (That term re-radiating, that someone does not like. Ok, absorption and radiation, so what.)
.
What this article has demonstrated, in an all-too-clear manner, is how easy it must be to dupe gullible politicians into believing the Global Warming hoax. Here we have a good cross-section of the educated public, and about half are totally stumped by a simple thought experiment involving a energy source and a surrounding barrier.
What we must have, therefore, is a degree course for politicians – a compulsory qualification that all budding politicians must acquire before standing for office.
.

Willis Eschenbach
November 18, 2009 11:03 am

Re: Anton Eagle (22:12:14)

Logic. Science and logic should be inseperable.
This article is utter nonsense, and a simple logical analysis reveals this without even resorting to physics (and I am a physicist).
Using the author’s argument… the mythical planet radiates 235 W/m2. The steel shell absorbs this and re-radiates it all in both directions. The planet obligingly absorbs the re-radiated 235 coming off the shell, and re-radiates it, for a total radiation “budget” of 470 W/m2. The author conveniently stops at this point.
But wait… there’s more. If we are to follow the author’s logic, then the shell should absorb this 470 W/m2 and re-radiate it again… 470 out and 470 in. The planet should then absorb this 470… add it to the orignal 235 coming from the core, and now be radiating 705 W/m2… this would continue till everything anihilated at a gazillion degrees. Utter nonsense.
Thank you Anton. It wasn’t just me who spotted that the logic is simply busted.
If the shell were a perfect reflector, the temperature of the atmosphere would simply rise to the temperature of the input and stay there. Regardless of how many shells were outside, the temperature would never rise beyond the input.
Willis, next time contact a physicist before doing something like this.
This isn’t about the greenhouse effect, because you’ve misunderstood what the greenhouse effect is.
Case closed.

Second rule of long threads. Any time you see “Period. or “Case closed.” or “Discussion over.” or the like … it isn’t.

For example, I said in the head post:

The planet is in interstellar space, with no atmosphere

You reply

If the shell were a perfect reflector, the temperature of the atmosphere would simply …

As you can see, with that kind of misunderstanding, the case can’t possibly be closed. You go on to say:

If the shell were a perfect reflector, the temperature of the atmosphere would simply rise to the temperature of the input and stay there.

Since energy is constantly being added from the decay of the radioactive elements in the planet’s core, the temperature inside a perfect reflector would rise until the reflector melted.
So no, the case is not closed. Nor is that kind of false certainty an asset to you. Yes, I think I’m right, but I might be wrong … I have been before. I suggest that if you want to have a discussion, you adopt a like attitude.

Gary
November 18, 2009 11:05 am

Willis,
Good analogy to radiation through an absorbing medium. However just on point. The comparison of the temperature with and without an atmosphere is theoretical only. If the earth did not have an atmosphere the solar insolation would not be 235 but around 430 W/m2. This is because there would be no reflection or absorption of incoming solar radiation. So the atmosphereic effect includes more than just greenhouse. The difference becomes much reduced.

Willis Eschenbach
November 18, 2009 11:05 am

Sorry, that last one should have been addressed to John A (01:28:22), who was quoting Anton Eagle.

Willis Eschenbach
November 18, 2009 11:08 am

SNRAtio (01:50:42) :

“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.”
OK, so now the whole tradition of which Kiehl/Trenberth is a part is proven wrong.

Slow down there, my friend. There is a tendency of AGW skeptics to say “If X is wrong then the whole edifice crumbles”. Generally, this is not true. Often if X is wrong it just means that X and perhaps a few other things are wrong.

Willis Eschenbach
November 18, 2009 11:11 am

lgl (02:01:11) :

Positive feedback from wiki: “Ai = (output voltage/input voltage) = A/ (1 − Aβ). Here A is the gain of the feed-forward active part of the amplifier without feedback, and β is the gain of the feedback element”
In your ‘Steelhouse’ A=1 and β=0.5 giving Ai=2 so that works fine.
Your problem is: “It has nothing to do with blankets, or mirrors, or greenhouse gases”. It has everything to do with greenhouse gases. O2 and N2 do not radiate (almost) so without the GHGs you would not get much downward LW.

Since I can build a steel greenhouse, clearly the greenhouse effect itself has nothing to do with greenhouse gases.
On Earth, as you point out, the particular greenhouse system found on the planet requires GHGs.

Willis Eschenbach
November 18, 2009 11:14 am

michael hamnmer (03:15:37) :

But herein lies a problem, the tropopause is too cold to radiate the 165 watts/sqM the Kiehl and Trenberth model say is radiated by the atmosphere.

As I note in the head post, in my two-shell model, the temperature of the tropopause is in agreement with measurements.

Jim
November 18, 2009 11:19 am

*********************
John A (20:11:22) :
I’m sorry but I thought the article was tremendous nonsense, and will cause immense damage to the reputation of this website.
**********************
I. Newton was wrong, but he didn’t take “immense damage.” Discussion and argument is good, that is a mainstay of science. Kick back and enjoy it 🙂

Willis Eschenbach
November 18, 2009 11:20 am

Myrddin Wyn (04:09:21) :

Interesting article and debate.
I’m missing a step here though, the power source supplies 235watts/m2 for a surface area of ‘a’, if the shell has twice the area, 2a, shouldn’t it emit half the energy; 117.5 W/m2.
Per m2 I cant see how the shell emits twice the energy it receives from the power source, the shell would be a power multiplier, Eout = 2Ein, which is impossible – no ?

Perhaps a refinement of the thought experiment would help. Imagine the planet without a shell. Now, in an instant, we add a cold shell. As you point out, it starts to warm. Because it has twice the surface area as the planet, as it warms it radiates energy both inwards and outwards.
The downwelling radiation from the shell, however, warms the planet. So rather than radiating 235 W/m2, it starts radiating more and more.
Equilibrium is finally reached when the situation is as shown in Fig. 1(B). At that point, the planet is receiving 470 W/m2. Of this, 235 W/m2 come from the core, and 235 W/m2 come from the shell. It radiates that same amount, 470 W/m2, so it is in thermal equilibrium.
The shell, as you point out, divides that in half, with 235 W/m2 radiated outwards, and 235 W/m2 radiated inwards. Thus the shell is in equilibrium as well.
Hope this clarifies it,
w.

George E. Smith
November 18, 2009 11:23 am

“”” PSU-EMS-Alum (05:01:09) :
You know what needs to be removed from this page?
Any comment that uses the term “re-radiate”.
This is a pet peeve of mine …. it demonstrates a complete lack of understanding of the concept of “radiation”. “””
How so ? Are you for banning any term YOU don’t understand ?
For a body to “re-radiate” is a verty common occurrence. For example, road tar surfaces receive radiation (solar spectrum) from incident “sunlight”; some of which is absorbed and some reflected, so the surface warms, and re-radiates in a completely different thermal spectrum that depends on the surface temperature and its spectral emissivity.
Yes I agree you could just say “radiates”. The point is that a body can be at a fixed stable temperature and be re-radiating all of the energy it is receiveing in the form of some other radiation; with the incoming and out-going spectra being quite different.
Where I see a lot of confusion, is when people talk about the atmosphere , with its GHGs “reflectign surface emitted LWIR radiation back to the surface.
Reflection is an optical process that returns the incident spectrum only modified by the spectral reflectance of the surface; and is quite independent of temperature. Well if you want to be pedantic, the spectral reflection coefficient might vary with temperature.
Some might argue that the term “re-radiate” should be reserved for cases where a molecule or atom absorbs a photon of a given energy, and later emits a photon of the same energy, as the excited state returns to normalcy.
since that is a rarity in earth’s lower troposphere, it is not of much interest in climatology.

Willis Eschenbach
November 18, 2009 11:34 am

Marcus (04:20:51) :

Willis: I thought I would pipe up, since I had such vehement objections to your prior article on the Shindell paper, just to say that I thought this was a good post.
-Marcus

Marcus, you are a gentleman and a scholar. Would that all people on both sides of the climate discussion were as gracious.
w.

Willis Eschenbach
November 18, 2009 11:36 am

Vincent (04:22:19) :

Willis,
excellent article, but I there is one thing that doesn’t make sense to me. If you start with a ball radiating at 235 w/m2, then if you add the 235 w/m2 radiated back from the steel shell aren’t you creating energy out of nothing? Isn’t this an example of perpetuum mobile of the third kind? Sorry if I’m being a bit thick but if we start with 235 joules worth of energy at the surface, then the other 235 joules that were transmitted downwards came from the original 235 to begin with, so then we have created energy out of nothing.
HELP!

No energy is created or destroyed. It is simply trapped inside the system, which raises the temperature. See my post at Willis Eschenbach (11:20:27), which may assist.

Willis Eschenbach
November 18, 2009 11:39 am

PSU-EMS-Alum (05:01:09) :

You know what needs to be removed from this page?
Any comment that uses the term “re-radiate”.
This is a pet peeve of mine …. it demonstrates a complete lack of understanding of the concept of “radiation”.

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”.
However, like my pet peeves of “baited breath” and “tow the line” and the constant misuse of “begs the question”, I fear it is ingrained in common usage, so I can’t help either of us.

Willis Eschenbach
November 18, 2009 11:43 am

Stephen Goldstein (05:02:24) :

Willis Eschenbach (23:34:12)
yonason (21:37:01)
So, how many shells will it take to get the core to 2,000,000 degrees?
Haven’t a clue, but the inner shells will melt long before that … not
sure what your point is here.
It was a jocular reference, I am sure, to Al Gore’s recent claim, in the context of geothermal energy, that the temperature of the earth’s core is “several million degrees.” The WUWT thread on Gore’s TV appearance and this claim immediately precedes this thread.

Aaaahhh, very funny, yonason, I missed that entirely. I should get out more …

Gary Hladik
November 18, 2009 11:49 am

Willis Eschenbach (20:50:08) :
1) Thanks for the explanation. Between that and your reply to Lord Monckton, I think I understand (though my head still hurts a little).
2) Thanks for the idealized “greenhouse” model. It really helps clarify the whole issue, plus it takes me back to my school days of weightless pulleys and frictionless surfaces (where the IPCC apparently still dwells).
3) Thanks for sticking around for the discussion. The apparent confusion over the model, in this relatively sophisticated readership, does not bode well for educating the general public. Definitely an uphill battle (hopefully not on a frictionless surface).
Charlie (10:09:01), great clarifications. Kudos.

Hank Henry
November 18, 2009 11:50 am

Here’s what I think. I think the greenhouse metaphor is broken. It never really was a simple model.
In 1906 American experimental physicist R. W. Wood published the
results of an experiment that demonstrated that a glass greenhouse was
not heated by trapped long-wave (infrared) radiation. In fact, the
glass windows excluded more infrared energy entering the greenhouse
than they trapped inside the greenhouse, which actually lowered the
temperature inside a glass greenhouse when compared with a comparable
quartz greenhouse.
Ok next question. How come you can fry an egg on the sidewalk? …. How come on a sunny day at the beach sand feels so hot on your feet?

lgl
November 18, 2009 11:51 am

Willis Eschenbach (11:11:01) :
The greenhouse effect itself is something in an atmosphere that radiates LW and in real atmospheres that something is GHGs and that fact makes your statement very misleading.

Alan S. Blue
November 18, 2009 11:59 am

The fluxes involved would seem amenable to recreation in an actual, physical model.
A sixty watt light bulb and a basketball come mighty close to the correct proportions.

Hank Henry
November 18, 2009 12:01 pm

R.W. Woods words on the greenhouse effect:
http://www.wmconnolley.org.uk/sci/wood_rw.1909.html

Willis Eschenbach
November 18, 2009 12:05 pm

TomVonk (06:54:18) :

Please scratch the above . I did a mistake . This one is the right version :
=========================================
Willis I think that you made a mistake when writing :
“R = sigma * epsilon * T^4 where R = radiation (W/m2)”
.
The units in the Stefan Boltzmann formula are W/m²/sr .
When the solid angle is omitted , it implies that the integration has already been done (generally along a plane surface) .
This is a frequent nevertheless particular case .
Concentrical spheres are not part of this particular case .
What would it mean to your example ?

Here is the definition of the Stefan-Boltzmann equation from Wikipedia, which is a good place to start:

The Stefan–Boltzmann law, also known as Stefan’s law, states that the total energy radiated per unit surface area of a black body in unit time (known variously as the black-body irradiance, energy flux density, radiant flux, or the emissive power), j*, is directly proportional to the fourth power of the black body’s thermodynamic temperature T (also called absolute temperature).

Note that what is being calculated is the energy radiated per unit surface area. You can measure this in square meters, or (for a sphere) in steradians. (A steradian is the solid version of a radian. A radian measures the angle of a circle based on pi. A steradian measures the area of a sphere based on pi.)
However, I know of no such measure as watts per square meter per steradian. The K/T budget given above does not use such a measurement.
The units of W/m2 is also derivable from the equation. The Stefan Boltzman constant is 5.67E-8 J s^-1 m^-2 K^-4. In english this is Joules per second per square metre per Kelvins to the fourth power. Multiplying this by Kelvins to the fourth leaves Joules per second per square metre.
But a Joule per second is a Watt, so we are left with Watts per square metre.

November 18, 2009 12:10 pm

I still do not buy it, assigning present surface temperature to “greenhouse effect”.
Mars has 15x more CO2 than Earth. This is effectively the same “greenhouse” as on Earth (Mars has no water vapor). But still, calculated and actual temperature on Mars is the same : 210K. Hint: Mars atmospheric pressure is just 600 Pa, Earth pressure is ~101,000 Pa.
“Greenhouse” effect is obviously non-existent without presence of bulk atmosphere, which can retain heat, absorbed from the surface. So it is the atmosphere itself, working as heating blanket. Question is, how much heat is removed from the surface by radiation, air convection and evaporative cooling.
If earth surface is “heated by back radiation”, then covering my face against night sky should be felt as lack of warming, as it does on sunny day. It does not happen, since my face is warmed by ambient air – nitrogen and oxygen, not some hypothetical arrow painted in scheme, made so to get the ins and outs into balance.
I believe clouds have some measurable effect, but for the rest call me a denier.

Seedload
November 18, 2009 12:19 pm

Doesn’t the steel shell have twice the surface area of the ball in the middle.
You know – two sides rather than one.

Willis Eschenbach
November 18, 2009 12:27 pm

P Wilson (07:57:56) :

a simple diagnostic comparison that the SB constant is irrelevant to climatology:
100w/m2 is the optimum surface heat for floor based central heating. This creates 24C at floor level, cooling as hot air rises. 15C, or 59F is certainly cooler than 24C. It is improbable that the floor will re-absorb radiation it has already emitted.

As the night-vision glasses used by armies show, everything emits infra-red rays. The amount and frequency of the waves is determined by their temperature. Air both absorbs and emits infrared rays. Why is it suddenly “improbable” that radiation from the air warms things around it?

Willis Eschenbach
November 18, 2009 12:33 pm

SteveBrooklineMA (09:48:33) :

I think Lubos Motl is too kind in his comment above. This analysis, which implies an error with K/T, errs in adding model assumptions to K/T that are not required in K/T’s approach. K/T is not a “single shell model” it is simply an energy transfer accounting. There is nothing that requires the K/T accounting to equate the downward radiated flux from the atmosphere to the ground with the radiated flux from the atmosphere to space. As Lubos points out, the atmosphere is hotter near the ground than in the high atmosphere. This is very different from a thin shell, which has only one temperature. You can argue with the values that K/T come up with, but this post does not somehow invalidate method of the K/T accounting.

My point is that unless the greenhouse system has more than one shell, with the two shells separated so that there is minimum thermal loss between them, it does not concentrate enough energy to reproduce the earth’s conditions.

Ian Schumacher
November 18, 2009 12:47 pm

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.
Your sphere must have a hole in them to let light in. We now have a spherical cavity with a hole in it. This is the typical example given to approximate a blackbody. It doesn’t matter how many additional shells are inside. It doesn’t matter what happens inside. The temperature inside can not exceed the temperature for a blackbody.
I assume you will claim the Earth is different because these are special steal spheres that are transparent to visible light, but opaque to IR, so you don’t need a hole for light to get in. But this doesn’t matter. This is just a different kind of hole. A hole in frequency spectrum.
A hole in frequency spectrum is not a way to ‘cheat’ thermodynamics and attain an internal temperature higher than a black body. The distribution of energy inside the sphere will continuously readjust to pour out the frequency hole, just as heat energy will continuously bounce around to pour out a physical hole.
If you disagree, then can you describe an experiment that would demonstrate this effect? Maybe a series of glass spheres inside each other with a black sphere in the center? I feel confident the temperature will not exceed that of a black body.

michael hamnmer
November 18, 2009 12:58 pm

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?
BY the way with my comment about the tropopause being too cold to radiate 165 watts/sqM as required by Trenberth if it were a black body which most certainly it is not, by Stafeans law it would have to be at a temperature of -40.7C yet it is actually at a temperature of about -57C so indeed it is significantly too cold. When you factor in that at the very least it cannot radiate between 8 and 14 microns (the atmospheric window) because if it did it would be opaque at these wavelengths it is very very much too cold to radiate 165 watts. The Trenberth data is extemely suspect.

carrot eater
November 18, 2009 1:00 pm

Willis, in the same spirit as Marcus, I will also chime in. Through Fig 2, I’m agreeable to your effort. Haven’t looked at the rest carefully yet. Along with Charlie above, I see the same misunderstandings repeatedly through the comments. So, some comments/suggestions:
-Doubly emphasize in the appendix that any surface with a temperature T will radiate at sigma*epsilon*T^4. This is true, no matter what the ambient or surrounding objects are. The surroundings only come into play if you want to consider the net heat transfer.
– People are missing that the two sides of the steel shell are at the same temperature. Perhaps it’d help them if I said the thermal conductivity of the steel shell is infinite; there is no temperature gradient across the shell.
– It is good that you drive home that a single-slab atmosphere isn’t representative. In reality, the radiation to space is from a colder temperature than the temperature at the bottom of the atmosphere.
-Emphasize that this is an equilibrium model, not a kinetic one. The temperatures and flows don’t change instantaneously to the final values if you add a steel shell.
– You might want to make it more clear in the first figures that the internally warmed planet is NOT the Earth. I was reading quickly the first time, and missed that you were making a fake planet there.

Willis Eschenbach
November 18, 2009 1:02 pm

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.

I fear I don’t understand this at all, which is why I didn’t reply. What do you mean by “heat something above the temperature of a blackbody”?
If you mean that no matter what you do, the sun can’t heat something above the temperature of a blackbody heated by the sun at the same distance, the Earth itself shows that this is not true. Blackbody temperature at 235 W/m2, the amount of incoming solar radiation entering our planetary system, is 255K, or -19 C. Thus the earth has “an internal temperature higher than a black body”, something which you claim is impossible under any conditions.
However, this is not “cheating thermodynamics” as you say, any more than using a Thermos bottle cheats thermodynamics to keep your coffee warm.

Jari
November 18, 2009 1:08 pm

Really sad to see this kind of rubbish published on this great site. The physics is completely wrong. Anthony, please do not let this kind stuff published on this site, it will take the value of the whole site down.

George E. Smith
November 18, 2009 1:11 pm

You know Willis’ model of a steel greenhouse is functionally the same, even if you postulate that the planet is a perfect sphere, and the steel shell is one micron thick, and is separated from the planet by a one micron gap.
So all of the discussion about the relative surface areas is somewhat off target; the analysis Willis gave doesn’t really depend on the gap. Now if the gap is in fact large so the shell OD is somewhat larger than the planet OD, the 235 W/m^2 number will change, inversely as the area of the shell. but his basic concept is unchanged by the shell spacing from the planet.
They system is a steady state, only because Willis asserted that the 235 W/m^2 and the original 254K surface temperature are maintained by internal heating as say from nuclear decay.
In reality, that energy source decays so the thing would eventually die out.
But it is not strictly correct to say the system is in equilibrium; the continual loss of energy to space would not occur if the system was in equilibrium.

Jim
November 18, 2009 1:13 pm

**********************
jt (21:14:58) :
“As a result the quantum of energy exchanged is never actually in a state of “free flight” between the two interacting electrons. ”
**********************
Wow, no wonder they had to get rid of the cathode ray tube type TV. If electrons are never in free flight, it would be impossible to to bend their path with a magnet, like most old TVs do.
BTW – Carver Mead’s work isn’t widely accepted. Does not mean it’s wrong.

Steve
November 18, 2009 1:14 pm

If you want to know what Earth would be like without a greenhouse effect, there’s no need to look as far away as Mars or Venus.
We have a moon right here, in the same orbit. Temps average over 100 Celsius in the day and drop below -150 Celsius at night. Since earth’s average daytime temp is lower, our atmosphere is obviously shielding us from solar energy during the day. And since earth’s average nighttime temp is higher, our atmosphere is obviously retaining some latent heat capacity through the night.
“Since I can build a steel greenhouse, clearly the greenhouse effect itself has nothing to do with greenhouse gases.”
It’s called the greenhouse effect, not the greenhouse gas effect. The greenhouse gasses are named after it, not the other way around.
And no, you have not built a steel greenhouse. As I pointed out earlier, your figure 2a clearly shows 2W inwards and 3W outwards, for a net of 1W outwards. This does not balance to zero, so you have a second source of heat W within your system.

michael hamnmer
November 18, 2009 1:15 pm

Willis;
I think much of your post is valid and is a worthwhile contribution to the debate about AGW but I see the the point about the cold tropoause as a massively important key to our understanding of the action of green house gases. Starting from the point of a cold tropopause one can readily show that the Kiehl and Trenberth data is almost certainly wrong and we need to remember that this data forms the basis of the models which are predicting dangerous warming. The K&T data exaggerates the impact of GHG because it claims atmospheric absorption which is much too high and also because it claims that atmospheric absorption and radiation significantly modulates the energy loss to space whereas in fact the concentration of GHG is so high in the atmosphere that they largely block radiation to space at certain wavelengths.
This is easily shown by looking at the energy spectrum as seen from space looking down towards the planet (Nimbus data). What you find is that in the region between 8 and 14 microns the equivalent radiation temperature is the surface temeprature of earth (which means the radiation comes from the surface – nothing else is warm enough) whereas at the GHG absorption wavelengths the equivalent radiation temperature is that of the tropopause. One can even see the comb effect where there are a number of absorbing lines close together (look below 8 microns) and the equivalent radiation temperature varies rapidly with wavelength between surface and tropopause temperature giving a very jagged plot until the lines get so close together that the interferometer cannot resolve them and one gets a very noisy average.

George E. Smith
November 18, 2009 1:16 pm

“”” 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.

Willis Eschenbach
November 18, 2009 1:22 pm

Juraj V. (12:10:02) :

I still do not buy it, assigning present surface temperature to “greenhouse effect”.
Mars has 15x more CO2 than Earth. This is effectively the same “greenhouse” as on Earth (Mars has no water vapor). But still, calculated and actual temperature on Mars is the same : 210K. Hint: Mars atmospheric pressure is just 600 Pa, Earth pressure is ~101,000 Pa.

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.

“Greenhouse” effect is obviously non-existent without presence of bulk atmosphere, which can retain heat, absorbed from the surface. So it is the atmosphere itself, working as heating blanket. Question is, how much heat is removed from the surface by radiation, air convection and evaporative cooling.

See the budgets above. The answer is 390, 24, and 78 W/ms respectively.

If earth surface is “heated by back radiation”, then covering my face against night sky should be felt as lack of warming, as it does on sunny day. It does not happen, since my face is warmed by ambient air – nitrogen and oxygen, not some hypothetical arrow painted in scheme, made so to get the ins and outs into balance.

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.

I believe clouds have some measurable effect, but for the rest call me a denier.

“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” …

A. Einstein Jr. Jr.
November 18, 2009 1:24 pm

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!
🙂

Ian Schumacher
November 18, 2009 1:25 pm

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.

George E. Smith
November 18, 2009 1:27 pm

“”” 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.

November 18, 2009 1:33 pm

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.

P Wilson
November 18, 2009 1:34 pm

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

Mark T
November 18, 2009 1:37 pm

Steve (13:14:42) :

And no, you have not built a steel greenhouse. As I pointed out earlier, your figure 2a clearly shows 2W inwards and 3W outwards, for a net of 1W outwards. This does not balance to zero, so you have a second source of heat W within your system

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

Willis Eschenbach
November 18, 2009 1:38 pm

Steve (13:14:42) :

… And no, you have not built a steel greenhouse. As I pointed out earlier, your figure 2a clearly shows 2W inwards and 3W outwards, for a net of 1W outwards. This does not balance to zero, so you have a second source of heat W within your system.

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.

P Wilson
November 18, 2009 1:42 pm

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.

Willis Eschenbach
November 18, 2009 1:48 pm

John A (13:33:44) :

Lots of people are not answering objections from physicists, let alone myself. Especially Willis.

I am attempting to answer all reasonable objections. If I have missed some, please let me know.

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.

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.

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.

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?

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 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.

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.

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.

michael hamnmer
November 18, 2009 1:55 pm

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.

Steve
November 18, 2009 2:04 pm

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?

Ian Schumacher
November 18, 2009 2:06 pm

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).

michael hamnmer
November 18, 2009 2:08 pm

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)

Willis Eschenbach
November 18, 2009 2:08 pm

P Wilson (13:34:31) :
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

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.

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

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.

Ian Schumacher
November 18, 2009 2:09 pm

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.

Anton Eagle
November 18, 2009 2:12 pm

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

Willis Eschenbach
November 18, 2009 2:14 pm

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”.

Joel Shore
November 18, 2009 2:17 pm

John A says:

If there are physicists watching this weblog then I’m sure they’ll fill you in on exactly why your “steel greenhouse” won’t work in the way Willis has described, and why adding “shells” to the planet will not raise the surface temperature one iota.

Willis, next time contact a physicist before doing something like this.

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

AlexB
November 18, 2009 2:19 pm

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!

Alan S. Blue
November 18, 2009 2:22 pm

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.

Willis Eschenbach
November 18, 2009 2:24 pm

Ian Schumacher (13:25:41) :

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.

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.

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.

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.

Richard Sharpe
November 18, 2009 2:33 pm

Willis says:

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.

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.

Joel Shore
November 18, 2009 2:34 pm

Bill Illis says:

There is quite a difference between the Stefan-Boltzmann equations (the fundamental equations governing radiation physics and temperature) and the climate models.
The surface radiates at 390 watts/metre^2 at a temperature of 15.0C.
For the surface to go up 3.0C to 18.0C, another 16.5 watts/metre^2 would have to be added to the surface.
I don’t know how the climate models can reconcile their prediction of a 4.2 watt/metre^2 increase in the tropopause radiation results in a 3.0C increase in surface temperatures (and a new surface radiation level of 406.5 watts/metre^2).
Their mistake is by converting everything into “linear equations”. Hansen took a shortcut in the early 1980s using his early climate model results (for the last glacial maximum for example) where his climate model came up with -6.6 watts change. Temperatures declined by 5.0C so therefore, the sensitivity is 0.75C per watt.

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.

Brian
November 18, 2009 2:39 pm

THANK You John A. Nice explanation! Couldn’t have said it better myself.

Joel Shore
November 18, 2009 2:51 pm

Gary says:

Good analogy to radiation through an absorbing medium. However just on point. The comparison of the temperature with and without an atmosphere is theoretical only. If the earth did not have an atmosphere the solar insolation would not be 235 but around 430 W/m2. This is because there would be no reflection or absorption of incoming solar radiation. So the atmosphereic effect includes more than just greenhouse. The difference becomes much reduced.

Actually, with no albedo (or absorption by the atmosphere), the amount of solar radiation absorbed by the earth would be ~342 W/m^2. And, in reality, the earth would still have an albedo without an atmosphere of ~0.09, so in fact without an atmosphere, the amount absorbed would be ~310 W/m^2. This would yield a surface temperature of ~272 K. When you compare this with the actual surface temperature of ~288 K and the temperature in absence of the greenhouse effect but no change in albedo of ~255 K, what we can say is the follows: The greenhouse effect due to all the greenhouse gases (water vapor, clouds, and the long-lived GHGs like CO2 and CH4) raises the temperature of the Earth by an amount of ~33 K (which is 288K – 255K); the albedo due to cloud reduces the temperature by ~17 K (which is 272 K – 255 K); the net effect of both the GHGs and the cloud albedo is ~16 K (which is 288K – 272K).

Willis Eschenbach
November 18, 2009 2:56 pm

Joel Shore (14:17:35) :

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.

Joel, always a pleasure to hear from you, thanks for the physicist’s vote of confidence.

(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.

You may be correct, but it is drawn as a shell model, where the atmosphere is considered as a single entity. My point remains — to get enough energy from a greenhouse system to model the earth, you need two physically separated shells. If there is thermal loss and air mixing between the shells, you won’t get the energy you need to explain the Earth’s temperature. The system, as my Tinkertoy model shows, is very sensitive to thermal loss between the shells. That’s the part that the K/T analysis glosses over.

(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.

You are generally correct. If the albedo goes from 30% to 31%, it has increased by 3.3%. And clouds are only part of the equation as you say. However, as my analysis shows, in the Pacific Ocean the change in clouds is responsible for a 60 W/m change between 10:30 and 11:30 each day …
And you are correct that when clouds come over, they warm as well as cool. However, the true change is masked by the use of averages. For example, in the tropics the mornings are generally clear, warming the earth when it is cool. After about 11:00, clouds form and cool the earth when it is warm. Since it is the tropics, the solar radiation is on the order of a kilowatt per square metre, and the afternoon change in net radiation is quite large. In addition, since over the tropics the humidity is quite high, adding clouds doesn’t change the downwelling radiation as much as it would in the desert. Finally, during the night the clouds disperse, allowing increased outgoing radiation.
The net result of the timing of these processes is critical. If we swapped day clouds and night clouds, the average cloudiness would be the same … but the result in terms of the change in net radiation would be huge.
As a result, the use of averages in these matters can be very misleading. That’s why it’s a Tinkertoy model.
Your comments are always welcome. They are clear, to the point, and backed by the science.
w.

LED
November 18, 2009 2:56 pm

The validity of this model is demonstrated in a practical
application. The inside of a Dewar flask (thermos bottle) used to store liquid gases contains not just a vacuum between the inner and outer surfaces, but many layers of plastic film that is metalized on both sides. This reflects the radiation as in the model, and greatly (order of magnitude) reduces the net energy flow to the stored gas.

November 18, 2009 2:59 pm

Willis:

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.

Nope. The entire system will be subject to Newton’s Law of Cooling. The planet’s surface will not be warmed by the shell above it.

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.
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?

The laws of thermodynamics forbid it. A cooler object warming a warmer one would imply that the cooler object can spontaneously lose entropy and that the total entropy of the total system (both planets) remains constant. That is impossible

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 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.

There’s no such thing as a “perfect Dewar” because the temperature of the Dewar must rise both inside and out. The Dewar radiates heat outwards into the space around it but because it has a lower entropy than the radioactive soup, it cannot heat the soup itself because that would imply it spontaneously losing entropy without doing any work.
All the Dewar does is reduce the rate of cooling. It does not provide heat back to the soup. The soup does not warm because it is contained and I defy you or anyone else to show any experimental evidence (or frankly, theoretical evidence) that it does.
Your thought experiment breaks the laws of thermodynamics in positing that a cool (lower entropy) entropy object can heat a higher (greater entropy) object. That cannot happen.
And I’m astonished that a few physicists are failing their thermodynamics course in supporting your deeply unphysical propositions.

Ian Schumacher
November 18, 2009 3:03 pm

Willis,
You site a solar heater as an example, but how does this prove your point? They say the solar heater can get up to 100C at mid-day, no clouds.
Ok, well the the sun in mid-day, no clouds can deliver 1366 w/m^2. Put that into Stephan-Boltzmann equation and you get a temperature of around 120C for an ideal blackbody. 120 is greater than 100. I’m missing how your example proves me wrong.

Mark T
November 18, 2009 3:07 pm

Anton Eagle (14:12:49) :

Okay, first off… I give up trying to get the author to see reason… apparently it just isn’t going to happen.

Um, based on what I’ve read from you, you’re the one in need of reason lessons.
For the record, the shell has twice the area because there is both and inside and an outside, both of which have approximately the same surface area as the planet itself. Well, there is about a 0.2% error, a point Willis has made clear on more than one occasion, yet you repeatedly don’t get the point.

So, again, can anyone show the results of a careful experiment that demonstrates that the so-called greenhouse effect is real?

Actually, you can do this at home with a dichroic mirror. Phil. often talks about doing just such an experiment. Once you wrap your head around the transient state in which , it is fairly easy to understand the equilibrium result.
Mark

November 18, 2009 3:14 pm

Can someone fix the tags in my previous comment. I think I didn’t close the first set of blockquotes correctly.
Thanks

Anton Eagle
November 18, 2009 3:36 pm

Mark T.
No… again… the shell does not necessarily have twice the area. It COULD have twice the area if its very close to the surface (give or take a percent or two like you said)… but the model he proposes does not in any way require the shell to be close to the surface… so it could just as easily be 10,000 miles above the surface without changing his model in anyway whatsoever. If the shell can be some arbitrary distance above the surface, then it can have any arbitrary area that is greater than twice the area of the planet. If you disagree, then show me where his model requires the shell to be at any particular distance from the planet.
There are many contradictions here.
Here’s another. Let’s accept for the moment his model. Then, lets add another shell just one inch further out than the first shell. According to the author, and his model, this second shell would cause a further increase in temp. Okay, so then add 8 more, each 1 inch further out from the previous. Now, we have a huge increase in temp (according to this flawed model).
However, how are 10 shells all 1 inch appart any different than 1 shell that is just 10 times as thick? And, since the thickness of the shell does not affect his (flawed) radiative balance, this results in the contractory situation where 10 shells 1 inch apart cause a much larger temp increase than a single shell exactly as thick as the sum of the ten shells. Huh? It makes no sense! He has created some kind of weird thermodynamic perpetual motion machine involving flawed radiative balances.
Stop focusing on the mathematics of the W/m2 accounting, and focus on the basic principles. If you do so, its easy to see that its nonsense.
-a

Frank
November 18, 2009 3:45 pm

Anthony, anybody,
Mercy KIll this silly thing. Its embarassing.

John Millett
November 18, 2009 3:57 pm

The thought experiment raises a question: How is a temperature gradient created in a vaccuum – 470 Wm-2 at the planetary surface to 235 Wm-2 at the inside surface of the shell?

P Wilson
November 18, 2009 3:59 pm

Willis Eschenbach (14:08:30) :
Its agreed that at 15C and 27C, a human is warmer than his surroundings and is there for unlikly to absorb heat from surrounding objects.
try on a google search (Cambridge journals) enter
“Description of a human direct calorimeter” and you should be able to read the measured radiation in w/m2 a huma produces.
for average human values in various situations on human heat exchange with the environment:
http://personal.cityu.edu.hk/~bsapplec/heat.htm
for various measurements of energy required to heat room:
http://www.energyinst.org.uk/content/files/4g.pdf
cites 70-100w/m2
i can’t readily find the value for the energy required to boil sunflower oil, as i’m not sure its on the internet, but it omes under the category of experiments in food engineering, so tomorrow i’ll find the source

Willis Eschenbach
November 18, 2009 4:01 pm

John A (14:59:54) :
Willis:

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.

Nope. The entire system will be subject to Newton’s Law of Cooling. The planet’s surface will not be warmed by the shell above it.
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.

You are making a simple mistake. You are confusing net warming with the energy interchange between two objects.
You are correct that net heat flows from warm to cold. But that says nothing about the individual flows.

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?

The laws of thermodynamics forbid it. A cooler object warming a warmer one would imply that the cooler object can spontaneously lose entropy and that the total entropy of the total system (both planets) remains constant. That is impossible

This is the mistake. You agree, I hope, that both planets radiate energy. The warmer planet radiates more than the cooler planet. As a result, an energy exchange occurs in both directions. And you are right that the net flow goes from the warmer to the cooler planet.
But that means nothing about the individual flows. If you agree that both planets radiate, what do you think is happening to the radiation from the cooler planet that hits the warmer planet? It has to warm the warmer planet … but not as much as the warmer planet is cooled by its own radiation.


And I’m astonished that a few physicists are failing their thermodynamics course in supporting your deeply unphysical propositions.

I’m not. There is nothing unphysical about radiation. All objects radiate. Their radiation adds energy to whatever object it hits. Do you seriously believe that when radiation hits an object, it first checks the object’s temperature to see whether to add energy to that object? It adds the energy regardless of the temperature of the object it strikes. This makes the object warmer than it would be without that radiation.

Anton Eagle
November 18, 2009 4:12 pm

Lets try to keep a clear accounting of what parameters affect this model, and what parameters do not affect this model.
First. This model is not dependent in any way on the distance of the shell from the planet. No matter what the distance, according to the model, the shell has to radiate 235 out to space, and then of course also must radiate 235 back to the planet. According to the model, this must be so if the shell is only 1 foot off the ground or 10,000 mile up. Again… according to this model.
The above fact inevitibly leads to the fact that the area of the shell is not a factor (that alone should raise some eyebrows). And thus, all attempts at carefully accounting for the radiative balance by stating that the shell is twice the area of the planet are nonsensical, because it doesn’t have to be twice the area… it could easily be much more without changing the basic premise of the model.
Second. This model does not in anyway depend on the thickness of the shell. Again, no matter how thick the shell… according to the model… the shell must radiate 235 out to space, and 235 back to the planet.
As I posted above, this leads to the non-sensical result that one thick shell results in a totally different radiative balance than 10 thin shells.
All you folks arguing that the model is valid are doing so by performing accounting on the radiative balance (235 out = 470 out – 235 in, etc.). But, by doing so you are accepting the basic premise of the model, and you can’t do that, because the basic premise is flawed because it inevitibly leads to contradictions.
Look, I’m not saying that there isn’t some way to make a model of the atmosphere that can be represented by a shell. I’m just saying that this model isn’t it. It violates basic priniciples of thermodynamics.
If you can’t resolve the above stated two contradictions, then you can’t defend the model. Period.

Willis Eschenbach
November 18, 2009 4:21 pm

Ian Schumacher (15:03:31) :

Willis,
You site a solar heater as an example, but how does this prove your point? They say the solar heater can get up to 100C at mid-day, no clouds.
Ok, well the the sun in mid-day, no clouds can deliver 1366 w/m^2. Put that into Stephan-Boltzmann equation and you get a temperature of around 120C for an ideal blackbody. 120 is greater than 100. I’m missing how your example proves me wrong.

Your numbers are wrong. You have given 1366 W/m2, which is the strength of the sun at the top of the atmosphere. The max sun on the earth’s surface (clear tropical noon) is about a kilowatt per square metre, because even on a clear day there is absorption in the atmosphere. See here for actual measurements. The maximum they found was 1044 W/mw, and rarely exceeded a kilowatt. So we can take a kilowatt as a practical maximum.
By Stefan-Bolzmann this is only 90C for a kilowatt, and 95C for the absolute maximum measured.
So if you think that you can boil water (100C) simply by sitting it in the noonday sun, you need more assistance than I can give you. Do you really think that people put the vacuum shell around the heater pipes for fun?

P Wilson
November 18, 2009 4:22 pm

If the two planets were at the same temperature they would be in equilibrium and wouldn’t radiate to each other. If terrestrial amtter is in equilibrium with the atmosphere then earth matter isn’t going to radiate much. Oceans might throw heat into the air or the sun heats oceans although at the optimum, a safe estimate would be to say that earth re-radiates around 5% of its heat, and not the 117% that is suggested that it has to get rid of in order to be in equilibrium

AlexB
November 18, 2009 4:29 pm

Physics 101 for black body radiation (epsilon =1):
Object A and B are in a vacuum and transfer heat to each other by black body radiation. Object A is at temperature T(a) and radiates a flux of R(a). Object B is at temperature T(b) and radiates a flux of R(b).
The gross radiative fluxes are:
R(a) = sigma*T(a)^4
R(b) = sigma*T(a)^4
The net radiative fluxes are:
For body A
=R(a) – R(b)
= sigma*T(a)^4 – sigma*T(b)^4
= sigma*(T(a)^4 – T(b)^4)
For body B
=R(b) – R(a)
= sigma*T(b)^4 – sigma*T(a)^4
= sigma*(T(b)^4 – T(a)^4)
All fair and above board. I can’t make it any simpler. So can we please stop arguing that this model contradicts basic physics because it doesn’t.

Willis Eschenbach
November 18, 2009 4:30 pm

P Wilson (15:59:21) :

Willis Eschenbach (14:08:30) :
Its agreed that at 15C and 27C, a human is warmer than his surroundings and is there for unlikly to absorb heat from surrounding objects.

Let me stop you there, because everything that follows flows from this misconception.
I say again, people are confusing net heat flow with individual heat flows. Conduction is different from radiation. When a warm object touches a cool object, there is only one measurable flow, from the warm object to the cool object.
When the energy is transferred by radiation, however, we can measure both individual flows as well as the net flow. The warm object cools, and the cool object warms (net flow). But there is a two way flow of radiation. This is an inescapable fact, because both objects are radiating. Radiation from the cool object strikes the warm object, and vice versa. Both radiation flows warm the object that they strike. The net flow goes from warm to cool, but there is an energy flow in both directions.

Willis Eschenbach
November 18, 2009 4:32 pm

John Millett (15:57:15) :

The thought experiment raises a question: How is a temperature gradient created in a vaccuum – 470 Wm-2 at the planetary surface to 235 Wm-2 at the inside surface of the shell?

There is no gradient. Temperature is a measure of the average of the speed of the molecules of what is being measured. But in a vacuum, there are no molecules …

tallbloke
November 18, 2009 4:33 pm

Willis, you seem to have a knack of posting ideas which engender interesting and informative discussion. I’m looking forward to reading this one thoroughly.
Thank you for your great insights.

P Wilson
November 18, 2009 4:36 pm

ok found something similar
Vijayan and Singh reported a convective heat transfer of 300-500 w/m2 to fry food in oil from frozen cited in “Advances in deep fat frying of food” 1997 CRC Press

Curt
November 18, 2009 4:45 pm

I am just boggled by the number of commenters here who cannot follow Willis’ very simple “steel greenhouse” thought experiment. It really is nothing more than a typical simple problem given to students in an introductory thermodynamics course. (If I teach such a course again, I will use it — OK, Willis?) It looks like a lot of commenters would flunk such a course.
I guess those of us who have studied thermodynamics formally are used to the practice of defining different “control volumes” in a system, then applying the laws of thermodynamics to each. (Any such control volume you define must obey the laws — the skill comes in defining ones that are both useful and reasonably easy to analyze.)
People who work for me recently inadvertently created such a steel greenhouse, quite literally. Part of what I do is to design power electronic systems. One of the challenges always is to get the internally generated heat out of the power transistors so they don’t fry themselves. To that end they are typically mounted onto a metal heat sink that conducts the heat out of the transistor, then gets rid of the heat through some combination of radiation and convection.
Anyway, in this design, we did not have room or budget for fins on the heat sink or a fan, so the heat elimination would be more radiative than convective — just a flat plate. For this reason, I specified that the sheet metal cover for the product should not go over the side with the heat sink.
However, when the first prototypes came back, I found that the sheet metal cover did include the side with the heat sink. It was only about a millimeter thick and a millimeter away from the heat sink. The initial testing showed that the transistors ran hotter than we wanted. So I had my engineers remove this “steel greenhouse” from the heat sink side. With the identical heat generation level in the power transistors, they ran 20C cooler. The younger engineers were amazed. Of course, we ship the product with this side “open”.
(Yes, there is some convection involved here. But the radiative barrier effect is very important.)
The thought experiments Willis provides can easily be verified in a lab. (How much relevance they have to real AGW issues is a separate question.)

stephan
November 18, 2009 4:53 pm

In physics one can only discuss measurable things. Radiation and its power flux can arbitrarily be decomposed into oppositely directed components, but this is not measurable. Any detector just above the earth will indicate that there is 235 W/m2 outward power flux in both situations of the first diagram. And so, provided the Earth and the steel behave as black bodies, the earth surface temperature will be exactly the same whether the steel is around or not.
Then this could be useful to explain that the greenhouse effect is not a simple blanket or mirror. One can only understand it by considering long- and shortwave radiation (which is perfectly measurable).

Charlie
November 18, 2009 4:59 pm

Anton Eagle (15:36:25) : “It COULD have twice the area if its very close to the surface (give or take a percent or two like you said)… but the model he proposes does not in any way require the shell to be close to the surface…… If you disagree, then show me where his model requires the shell to be at any particular distance from the planet.
Actually, he said it pretty clearly when he first describes the steel shell:
“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.”
——————————-
Note the “A FEW THOUSAND METRES ABOVE THE SURFACE”.
For a basic understanding of the greenhouse theory one can initially assume that the area of the earth’s surface and the area of ONE side of a steel shell just above the surface are equal.
The important point is that the total area of the shell is very close to TWICE that of the earth’s surface, with the inside surface of the shell radiating back towards the earth’s surface.

Mark T
November 18, 2009 5:02 pm

Anton,

Lets try to keep a clear accounting of what parameters affect this model, and what parameters do not affect this model.

Let’s do, and let’s also try to have a clear account of what the problem statement is.

This model is not dependent in any way on the distance of the shell from the planet. No matter what the distance, according to the model, the shell has to radiate 235 out to space, and then of course also must radiate 235 back to the planet.

Oops, you started out by failing the first question. If the outer shell is significantly larger than the planet, then the corresponding values at the planet surface will be significantly larger which will change all the equations. The only thing that will remain the same is the input/output relationship. Willis pointed out, several times now, that the reason we don’t have to adjust the values at the surface of the planet is precisely because the shell surface area is nearly (0.1%) the same as that of the planet and hence, the total surface is approximately 2:1 over the planet surface.

The above fact inevitibly leads to the fact that the area of the shell is not a factor (that alone should raise some eyebrows). And thus, all attempts at carefully accounting for the radiative balance by stating that the shell is twice the area of the planet are nonsensical, because it doesn’t have to be twice the area…

And thus, since you got the first point wrong, this one is wrong as well.

Second. This model does not in anyway depend on the thickness of the shell. Again, no matter how thick the shell… according to the model… the shell must radiate 235 out to space, and 235 back to the planet.

Correct, except that the values at the surface of the planet are different for a significantly larger shell, as well as the ratio of 2:1.

As I posted above, this leads to the non-sensical result that one thick shell results in a totally different radiative balance than 10 thin shells.

Given that the first assumption you made was incorrect, it follows that you cannot logically arrive at this conclusion, either.

All you folks arguing that the model is valid are doing so by performing accounting on the radiative balance (235 out = 470 out – 235 in, etc.). But, by doing so you are accepting the basic premise of the model, and you can’t do that, because the basic premise is flawed because it inevitibly leads to contradictions.

We understood the basic premise, you did not.

Look, I’m not saying that there isn’t some way to make a model of the atmosphere that can be represented by a shell. I’m just saying that this model isn’t it. It violates basic priniciples of thermodynamics.

No, actually, it doesn’t, at least not at the single shell level (the easiest to observe without some effort). John A is wrong, too, because he, like you, is missing one simple point (though different points).

If you can’t resolve the above stated two contradictions, then you can’t defend the model. Period

Given that there is no contradiction, then you can’t defend your conclusions regarding the model. Period.
If you would like, I can post some pretty pictures, maybe even an equation or two, that explains how the feedback in such a system works. Heck, I can even give you some Excel commands.
Mark

P Wilson
November 18, 2009 5:11 pm

Willis Eschenbach (16:30:19) :
If a chair is brough in from the cold into a room at 25C then placed in the centre, at least 5 2 metres from any other object, that chair will reach the equilibrium of the atmosphere, not the radiation from other objects, as those objects are already thermalised with theh atmosphere and aren’t giving off much radiation. Even air overpowers the effect of material radiation. According to the SB, different objects should be different temperatures at the same atmospheric temperature and should be radiating and absorbing from each other. In truth, they all thermalise with the atmospheric temperature so don’t give off or receive radiation from each other – air is a very poor conductor. I hope you’re not arguing that a human can radiate to the value of 36C to surrounding objects..
However, in more simple terms, if 100w/m2 is the average human radiate flow, then how can it be argued that at 15C, 235w/m2 is emitted from earth to the atmosphere? Take underflor central heating as a more common value and argue the same.
I’m not criticising your model at all, just maintaing that the SB figures have to be divided by 10 to give a realistic energy result. Heat is not the physical constant that physicist seem to think it is. Cooling can be autonomous. What for instance happens when you pull a re hot poker from a kiln? the air surrounding it heats, the poker cools until they are at atmospheric temperture. Heat disippates, yet we’re arguing that heat is a fixed constant, as though it has to migrate elsewhere

Charlie
November 18, 2009 5:15 pm

P Wilson (16:22:16) : “If the two planets were at the same temperature they would be in equilibrium and wouldn’t radiate to each other.”
This is wrong. There can be situations where there is no NET radiative transfer of energy between the two bodies, but they are always going to each be radiating per the Stefan-Bolzmann equation.
Even the assumption that two planets at the same temperature will have not NET radiative transfer is wrong. Even if they are the same size, just having different albedos would cause a net energy transfer.
Even if they are the same temperature and the same albedo, there will be net energy transfer if the temperature of the planets is different than the background and their sizes are different. You can envision this by imagining the limit case where a larger planet fills nearly one hemisphere of the view looking outward from a tiny speck of a planet. The larger planet sees incoming radiation from what is essentially all outer space (which we will assume is 4K for this thought experiment). Meanwhile, on the tiny speck of a planet, half of view looking out is 4K outer space, but the other half is the much warmer planet.

SteveBrooklineMA
November 18, 2009 5:15 pm

Willis, I have a question about your figure 4. Consider the 53 units of energy that are radiated from the ground and are not absorbed by the troposphere. Of these 52, 13 (13/52=24%) are absorbed by the stratosphere, the rest going out into space. However, out of the 321 units radiated outward from the troposphere, 271 (271/321=84%) are absorbed by the stratosphere, the rest going out into space. Shouldn’t these percentages be equal? Similarly, 339/392 of the radiation from the ground is absorbed by the troposphere, while 147/147 of the radiation down from the stratosphere is absorbed by the troposphere. Shouldn’t these portions be equal as well? Surely these layers can’t “tell” what the source of the radiation was and discriminate between them.

Ian Schumacher
November 18, 2009 5:17 pm

Willis,
I give you kudos for your model. Really there can be no harm from modeling and thinking about things. You have got a lot of flak I know, but really this area is hard for everyone. It is very hard to get an intuitive feel for how things work. I don’t claim to know or have an intuitive feel either. I’ve struggled with this for quite a while and finally I feel confident in claiming that without a focusing device of some kind that a black body is the limit of absorption. You can not have a body become hotter than a black body without a coherence (which the sun does provide directionally) and a focusing device to exploit this coherence. A magnifying lens and/or gravity.
But given that, here are some of the main issues I don’t believe you have addressed.
1.) The shells have double the surface area of the planet. Let’s assume there is some sort of constant flux outward from the surface. If there is 100 W/m^2 out, the flux in from the shell will only be 50 W/m^2 in and 50 w/m^2 out (not 100).
If you have infinite shells this mean you will asymptotically approach 200 W/m^2 flux, not ‘infinity’ flux. Two times flux means (2)^0.25 higher temperatures or a mere 20% higher, not even double.
2.) The model has a stated constant energy flux outward. This is not like a greenhouse in that a greenhouse needs a way for energy to enter (a hole). If you have a sphere with a hole in it, that is the standard approximation to a blackbody. The stated result for which is given by the Stephan-Boltzmann law. The maximum temperature that can be reached is equivalent to that of a black body. You are saying that it is possible to make a special hole that allows us to achieve temperatures higher than that of a blackbody. I don’t think so. All holes of all types will be equivalent.
As for the solar heater, you might be right about solar irradiance at surface, I couldn’t find a source. I’ll take your word for it. However I suspect that there is a focusing device at work here. i.e. A reflective back plane that adds to the effective ‘area’ of energy absorption. As you can see the result is close to black body and I feel confident that if focusing is taken into account the result is still consistent and we have not created a way to make a temperature higher than that of a blackbody with absorption alone.
cheers,
Ian

Willis Eschenbach
November 18, 2009 5:19 pm

Anton Eagle (16:12:49) :

Lets try to keep a clear accounting of what parameters affect this model, and what parameters do not affect this model.
First. This model is not dependent in any way on the distance of the shell from the planet. No matter what the distance, according to the model, the shell has to radiate 235 out to space, and then of course also must radiate 235 back to the planet. According to the model, this must be so if the shell is only 1 foot off the ground or 10,000 mile up. Again… according to this model.
The above fact inevitibly leads to the fact that the area of the shell is not a factor (that alone should raise some eyebrows). And thus, all attempts at carefully accounting for the radiative balance by stating that the shell is twice the area of the planet are nonsensical, because it doesn’t have to be twice the area… it could easily be much more without changing the basic premise of the model.

If as you say it makes no difference … then what difference does it make? How is this a “contradiction” as you claim?
However, you have forgotten the units and ignored the geometry. If the shell gets say ten times as large, the shell radiation in W/m2 will get very small. It will still emit a total of 235 X watts if X is the planet surface area, but the radiation per square metre will be smaller. This will reduce the warming of the planet. Instead of intercepting 235 W/m2, the planet will only intercept some small fraction of that. The shell will still have to emit the total radiation of the planet to space, but the planet will not be warmed as much by the radiation of the shell. Instead of being bathed in radiation of 235 W/m2, it will only intercept 23.5 W/m2 of radiation. Thus it will not warm up as much. Most of the shell radiation will simply be reabsorbed and maintain the shell temperature, rather than being intercepted by the planet.

Second. This model does not in anyway depend on the thickness of the shell. Again, no matter how thick the shell… according to the model… the shell must radiate 235 out to space, and 235 back to the planet.
As I posted above, this leads to the non-sensical result that one thick shell results in a totally different radiative balance than 10 thin shells.
All you folks arguing that the model is valid are doing so by performing accounting on the radiative balance (235 out = 470 out – 235 in, etc.). But, by doing so you are accepting the basic premise of the model, and you can’t do that, because the basic premise is flawed because it inevitibly leads to contradictions.
Look, I’m not saying that there isn’t some way to make a model of the atmosphere that can be represented by a shell. I’m just saying that this model isn’t it. It violates basic priniciples of thermodynamics.
If you can’t resolve the above stated two contradictions, then you can’t defend the model. Period.

Well, for starters, see the Second Rule at Willis Eschenbach (11:03:15). Putting “Period.” at the end of your post just indicates that you think it needs additional emphasis, but it adds nothing to the strength of your ideas.
Next, the shells are assumed to be vanishingly thin, as befits a thought experiment. This is because if there is a significant thickness to the shell, the inside will be warmer than the outside.
I fail to see any violations of thermodynamics. Two physicists whose opinions I greatly respect (Joel Shore and Luboš Motl) have said that they don’t see any violations. Your claims otherwise are not spelled out in enough detail to see exactly what laws are being violated, you just keep saying laws are violated, laws are violated. But what laws, and where?

Willis Eschenbach
November 18, 2009 5:23 pm

AlexB (16:29:05) and Curt (16:45:40), thank you for understanding the thought experiment. You are right, it is basic thermodynamics.

P Wilson
November 18, 2009 5:32 pm

Charlie (17:15:10) :
it pays to use smaller scale examples, so if a hot frying pan is left to cool, in a room the heat loss is from the pan – if objects are radiating energy they’re losing heat, until at such point the pan is the same temperaure as the air of the room. What happened to that heat? did it disappear? did it cool autonomously until it reached equilibrium to the 2nd law of thermodynamics? did the heat escape into the room and distribute itself around? Did it escape from the room and into the atmosphere? I don’t think there is an argument that heat has a permanent property that the last argument suggests – as it can thermalise to the atmosphere’s temperature when it reaches such a point that the flows between the two are barely measurable. I doubt that the heat of the pan will transfer itself to the next pan, back and forth.

Willis Eschenbach
November 18, 2009 5:33 pm

stephan (16:53:18) :

In physics one can only discuss measurable things. Radiation and its power flux can arbitrarily be decomposed into oppositely directed components, but this is not measurable. Any detector just above the earth will indicate that there is 235 W/m2 outward power flux in both situations of the first diagram. And so, provided the Earth and the steel behave as black bodies, the earth surface temperature will be exactly the same whether the steel is around or not.
Then this could be useful to explain that the greenhouse effect is not a simple blanket or mirror. One can only understand it by considering long- and shortwave radiation (which is perfectly measurable).

I’m not sure I understand you. Why could we not physically measure the radiative flux? This is done on the Earth all the time, and there are instruments specifically designed to measure it from both the ground and satellites … and I assure you that the upwelling and the downwelling flux are rarely equal. See e.g. here and here.

P Wilson
November 18, 2009 5:46 pm

Incidentally i’m not arguing that heat is “destroyed” in violation of the st law – but that the cooling of an object doesn’t depend on the heat it has received by another object, according to its entropy, as molecules and electrons become less excited during the cooling process.

P Wilson
November 18, 2009 5:50 pm

addendum oops. atoms, not molecules

Willis Eschenbach
November 18, 2009 5:56 pm

SteveBrooklineMA (17:15:39) :

Willis, I have a question about your figure 4. Consider the 53 units of energy that are radiated from the ground and are not absorbed by the troposphere. Of these 52, 13 (13/52=24%) are absorbed by the stratosphere, the rest going out into space. However, out of the 321 units radiated outward from the troposphere, 271 (271/321=84%) are absorbed by the stratosphere, the rest going out into space. Shouldn’t these percentages be equal? Similarly, 339/392 of the radiation from the ground is absorbed by the troposphere, while 147/147 of the radiation down from the stratosphere is absorbed by the troposphere. Shouldn’t these portions be equal as well? Surely these layers can’t “tell” what the source of the radiation was and discriminate between them.

What you say is certainly possible. However, the radiation is at a different frequency depending on the temperature at which it is radiating, so there is no reason to assume that the absorption will be equal. The set of numbers I chose were selected to replicate the main flows of the K/T budget. Other numbers will work as well. I encourage you and others to play with my model and see what other numbers work, or what other assumptions work.

Mark T
November 18, 2009 6:04 pm

Ian Schumacher (17:17:58) :

1.) The shells have double the surface area of the planet. Let’s assume there is some sort of constant flux outward from the surface. If there is 100 W/m^2 out, the flux in from the shell will only be 50 W/m^2 in and 50 w/m^2 out (not 100).

Um, no. If there is 100 W/m2 in from the sun, then there has to be 100 W/m2 out of the shell or else the planet will heat infinetely rather quickly – at whatever rate is generated by the imbalance. This is a pretty basic concept, so if you don’t understand it, you aren’t in any position to argue further.
This is also the reason, btw, that there is any storage, aka “gain,” in this system in the first place. The instant the shell is “applied” to the model a new transfer function is applied. Through superposition, this step can actually be applied to the input itself, which implies that the result can be obtained by modeling the step response of a simple 1st order feedback system.

If you have infinite shells this mean you will asymptotically approach 200 W/m^2 flux, not ‘infinity’ flux. Two times flux means (2)^0.25 higher temperatures or a mere 20% higher, not even double.

Nope. Since you got the first point wrong, which already doubles the energy, this is wrong by extension.

You are saying that it is possible to make a special hole that allows us to achieve temperatures higher than that of a blackbody. I don’t think so. All holes of all types will be equivalent.

Uh, no, he’s not saying that. The planet is radiating at its temperature, and the surface of the shell is radiating at its temperature. The difference, of course, is that the shell must radiate in all directions equally, but the output is constrained to be equal to the input.
There is a “hole,” btw, in that LWIR is passed without disturbance into the system, but 50% of the SWIR is radiated inward after being absorbed by the shell (in the feedback derivation, that would be a coefficient of 0.5, which results in a “gain” of 1/(1-0.5) = 2). This is an approximation, for a simple model of course, but the point is that most of the input resides in LWIR and most of the output resides in SWIR. Not perfect, but Willis makes that clear: it’s a simplification.
Willis, please note that when I put “Period.” at the end of my response to Anton, I should have also put /sarc. 🙂
Mark

Willis Eschenbach
November 18, 2009 6:15 pm

Ian Schumacher (17:17:58) :
Willis,

I give you kudos for your model. Really there can be no harm from modeling and thinking about things. You have got a lot of flak I know, but really this area is hard for everyone. It is very hard to get an intuitive feel for how things work. I don’t claim to know or have an intuitive feel either. I’ve struggled with this for quite a while and finally I feel confident in claiming that without a focusing device of some kind that a black body is the limit of absorption. You can not have a body become hotter than a black body without a coherence (which the sun does provide directionally) and a focusing device to exploit this coherence. A magnifying lens and/or gravity.

I show in Figure 1 how it can be done.

But given that, here are some of the main issues I don’t believe you have addressed.
1.) The shells have double the surface area of the planet. Let’s assume there is some sort of constant flux outward from the surface. If there is 100 W/m^2 out, the flux in from the shell will only be 50 W/m^2 in and 50 w/m^2 out (not 100).

Agreed. That is why in Fig. 1 the constant flux outward from the surface is 470 W/m2, and at the shell it is half of that, 235 W/mw in and 235 W/m2 out.

If you have infinite shells this mean you will asymptotically approach 200 W/m^2 flux, not ‘infinity’ flux. Two times flux means (2)^0.25 higher temperatures or a mere 20% higher, not even double.

I encourage you to draw up the two shell system which is the expansion of Fig. 1. At equilibrium, the outer shell is at 235 W/m2. The planetary surface is at 705 W/m2 (three times the source). The inner shell gets 235 W/m2 from the outer shell and 705 W/m2 from the surface for a total of 940 W/m2, so it radiates at 470 W/m2 (two times the source) up and down.

2.) The model has a stated constant energy flux outward. This is not like a greenhouse in that a greenhouse needs a way for energy to enter (a hole). If you have a sphere with a hole in it, that is the standard approximation to a blackbody. The stated result for which is given by the Stephan-Boltzmann law. The maximum temperature that can be reached is equivalent to that of a black body. You are saying that it is possible to make a special hole that allows us to achieve temperatures higher than that of a blackbody. I don’t think so. All holes of all types will be equivalent.

In a real planetary greenhouse, you have a “one-way” hole in the form of an atmosphere that allows solar radiation in and blocks longwave radiation from leaving. This is why we can achieve temperatures above blackbody on the Earth’s surface.

As for the solar heater, you might be right about solar irradiance at surface, I couldn’t find a source. I’ll take your word for it. However I suspect that there is a focusing device at work here. i.e. A reflective back plane that adds to the effective ‘area’ of energy absorption. As you can see the result is close to black body and I feel confident that if focusing is taken into account the result is still consistent and we have not created a way to make a temperature higher than that of a blackbody with absorption alone.
cheers,
Ian

While a focusing device would improve performance, it is not necessary.
w.

AlexB
November 18, 2009 6:22 pm

Re: Willis Eschenbach (17:23:33)
It’s a pleasure. Thank you for the excellent post, the most stimulating I have read in a while. I must say I admire your patients though. I would have torn my hair out long ago and the insides of my computer would now be engaging with radiative heat transfer with the troposphere from their scattered locations around my lawn.

Ian Schumacher
November 18, 2009 6:26 pm

Willis,
There is no such thing as a one-way hole. This is the key issue. If we state a theoretical one-way hole then yes you can achieve your result. But there is no such thing as a one-way hole in the real physical world. This would be Maxwell’s Demon.
Anyway we are at an impasse.
Cheers and goodnight.

Mark T
November 18, 2009 6:27 pm

Agreed. That is why in Fig. 1 the constant flux outward from the surface is 470 W/m2, and at the shell it is half of that, 235 W/mw in and 235 W/m2 out

Yes, I misinterpreted the way Ian was describing this, but your response puts it in the proper context, i.e., the surface of the planet is doubled. This means Ian’s subsequent statements regarding how the shells stack are still incorrect.
Mark

carrot eater
November 18, 2009 6:31 pm

AlexB (14:19:13) :
“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? ”
Exactly my thought as I’ve read through these comments. There seems to be no end of people who can’t tell the difference between the radiation emitted by a surface, and the net exchange between two surfaces. Anything with a temperature greater than 0 K (meaning, everything) will emit radiation, regardless of what is around it. Why is that so difficult?
Joel Shore: I was going to say the same thing about saturation. It’s a tinkertoy model, but from the direction it’s going, one can see where the saturation arguments go wrong. Yes, the original radiation from the earth at a given wavelength might be absorbed, but that isn’t the end of the story.
Really, I had no idea so many people around here actually disputed the basic mechanism itself, given that it’s based on pretty simple physics. I thought the main dispute was over the magnitude of the feedbacks. Instead, we’ve got people arguing about the magnitude of the SB constant? Oh well.

Ian Schumacher
November 18, 2009 6:34 pm

Mark T,
Really this whole argument is begging for an experimental isn’t it? I would like to, but admittedly, I’m too lazy.
Do you think that you can have a sphere with a hole in it to accept external radiation that can become hotter inside then a black body? Just wondering. That is the key. I’m trying to think what would such a thing look like. It really does require a one-way mirror, but no such device exists (see above). Ah well. Cheers to you also and good night (I’ll still check replies if you care to reply about ‘super’-black bodies, etc.

Stephen Goldstein
November 18, 2009 6:45 pm

Willis Eschenbach (16:30:19) :
. . . Both radiation flows warm the object that they strike.
Small point, I’m sure, but this explanation could be contributing to some of the confusion exhibited hereabouts . . . .
IMO, the verb “to warm” refers to something exhibiting an increase in temperature.
Your explanation clearly describes the two energy flows between the two and that the net energy flow is from the warmer body to the cooler. But this notion of the cooler body “warming” the warmer body also shows up here and elsewhere in the thread. (In at least on post you explain something like “warmer than it would be, otherwise.”)
Surely you agree that, in terms of temperature, until they reach an equilibrium temperature, since the net energy flow is from the warmer to the cooler, the cooler object only warms and the warmer object only cools.
Hope that helps.

Andy Beasley
November 18, 2009 6:47 pm

Very good article. I get the feeling from a lot of the posts that some people have a real problem with science or are being obtuse to confuse the issue on purpose (personally, I think the latter). This was intended to be a very simplified description of what is going on in the atmosphere with many assumptions to make it simple. We do this in physics and other disciplines all of the time. For example, a hockey puck slides across a frictionless surface…. Some of the posters here would then argue that there is no such thing as a frictionless surface. While that is true, for this problem, including friction does nothing to help us demonstrate an elastic collision. That is the point. The earth is not a perfect black body. The principle is the same and a black body is easier to work with. There may be some reflection from the steel sphere. OK, assume it has a perfectly non-reflective surface. And so on for the demonstration. Adding all of the variables back in results in a very complicated model and one that we don’t even really know if we have all of the factors included. It just brings all of the climate models into more question not less. Thank you Mr. Eschenback.

Willis Eschenbach
November 18, 2009 6:54 pm

Ian Schumacher (18:26:48) :

Willis,
There is no such thing as a one-way hole. This is the key issue. If we state a theoretical one-way hole then yes you can achieve your result. But there is no such thing as a one-way hole in the real physical world. This would be Maxwell’s Demon.
Anyway we are at an impasse.
Cheers and goodnight.

The atmosphere allows incoming solar radiation through. It blocks outgoing longwave radiation. That’s basic physics, I haven’t a clue how you can continue to argue with that.

Graeme W
November 18, 2009 6:59 pm

If I remember my high school physics correctly, there are three ways for energy to propogate: radiation, convection and conduction.
Anton Eagle, the difference between your thought experiment of ten thin shells vs one shell ten times thicker is that you are changing the energy transfer from radiation to conduction. The energy propogation in a thicker shell is via conduction which exhibits different characteristics to energy propogation via radiation. The thought experiment only considers energy radiation.
Thank you, Willis, for an interesting article. I’ll admit that I didn’t immediately understand how the two sides of the steel sphere affected the surface temperature, but I’m pleased that I worked it out before I read the comment about what would happen if the steel sphere was replaced with a perfect insulator.

Mark T