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, it 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 (heat) 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: http://atoc.colorado.edu/~dcn/ATOC7500/members/Reading/KiehlTrenberth.pdf

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 at http://homepage.mac.com/williseschenbach/.Public/global_energy_budget.xls

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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482 Responses to The Steel Greenhouse

  1. David Ermer says:

    Figure 2?

    Nice article, Thanks!

    REPLY: WordPress didn’t like the Macintosh generated graphics – fixed now -A

  2. Matt says:

    Might this have some application to the coronal heating problem?

  3. … 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.

  4. Jim says:

    Cool! Errr … warm!!

  5. Jack Wedel says:

    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.

  6. Tom in Florida says:

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

  7. Steve in SC says:

    I see a giant capacitor.

  8. Paul Linsay says:

    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.

  9. Ian Schumacher says:

    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

  10. Robert Wood of Canada says:

    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.

  11. Mike Lorrey says:

    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.

  12. morganovich says:

    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.

  13. C Colenaty says:

    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?

  14. kurt says:

    “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.

  15. Paul says:

    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?

  16. Bill Illis says:

    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.

  17. Charlie says:

    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.

  18. Alan D. McIntire says:

    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,

  19. Bird Stewart Lightfoot says:

    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.

  20. ShrNfr says:

    @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.

  21. DaveF says:

    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?

  22. michel says:

    “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?

  23. JaneHM says:

    “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)

  24. Clayton Hollowell says:

    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.

  25. Steve says:

    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.

  26. PaulH says:

    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

  27. George E. Smith says:

    “”” 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.

  28. Steve says:

    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.

  29. royfomr says:

    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!!

  30. Bernie says:

    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.

  31. Gary Hladik says:

    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).

  32. David L. Hagen says:

    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.

  33. David Ermer says:

    ” 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?

  34. kurt says:

    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.

  35. jt says:

    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.

  36. George E. Smith says:

    “”” 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 !

  37. Joseph says:

    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.

  38. John F. Hultquist says:

    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?”

  39. Merrick says:

    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.

  40. Merrick says:

    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.

  41. Ian Schumacher says:

    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?

  42. AlexB says:

    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.

  43. Steve Fitzpatrick says:

    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.

  44. E.M.Smith says:

    Well, I *like* the article.

    Just the notion of a “steel greenhouse” is kind of cool, er, so to speak ;-)

  45. michel says:

    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.

  46. michel says:

    [snip - we don't need discussion of potential terrorist acts on this thread]

  47. AlexB says:

    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.

  48. AStoner says:

    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.

  49. steve says:

    “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.

  50. Gary Hladik says:

    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. :-)

  51. savethesharks says:

    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

  52. Stephen Goldstein says:

    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?

  53. John A says:

    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.

  54. Mike Lorrey says:

    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.

  55. Willis Eschenbach says:

    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.

  56. Willis Eschenbach says:

    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.

  57. Willis Eschenbach says:

    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.

  58. Willis Eschenbach says:

    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.

  59. Willis Eschenbach says:

    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.

  60. Willis Eschenbach says:

    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.

  61. Jeff L says:

    @ 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.

  62. Willis Eschenbach says:

    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.

  63. Willis Eschenbach says:

    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 …

  64. Willis Eschenbach says:

    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.

  65. Willis Eschenbach says:

    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?

  66. Leo G says:

    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…….

    :)~

  67. Alan D. McIntire says:

    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.”

  68. jt says:

    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).

  69. carlbrannen says:

    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.

  70. yonason says:

    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?

  71. par5 says:

    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…

  72. Anton Eagle says:

    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.

  73. Thomas says:

    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.

    ….

  74. Ron House says:

    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.

  75. Le Sage's theory says:

    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.

  76. 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).

  77. Willis Eschenbach says:

    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.

  78. Willis Eschenbach says:

    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.

  79. Willis Eschenbach says:

    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.

  80. Willis Eschenbach says:

    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.

  81. Willis Eschenbach says:

    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.

  82. Willis Eschenbach says:

    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.

  83. Willis Eschenbach says:

    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.

  84. Anton Eagle says:

    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.

  85. Anton Eagle says:

    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.

  86. Luboš Motl says:

    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.

  87. Michael D Smith says:

    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?

  88. Cyril says:

    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.

  89. Ed Zuiderwijk says:

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

  90. MartinGAtkins says:

    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.

  91. John A says:

    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.

  92. ralph says:

    >>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.

    .

  93. John A says:

    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.

  94. Jon-Anders Grannes says:

    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.

    ?

  95. ralph says:

    >>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.

    .

  96. SNRAtio says:

    “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.

  97. ralph says:

    >>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.

    .

  98. lgl says:

    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.

  99. Robert Wood of Canada says:

    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.

  100. Stephen Wilde says:

    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

  101. michael hamnmer says:

    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.

  102. Myrddin Wyn says:

    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 ?

  103. Marcus says:

    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

  104. Vincent says:

    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!

  105. ralph says:

    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.

    .

  106. PSU-EMS-Alum says:

    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”.

  107. 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.

    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.

  108. Stephen Goldstein says:

    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.

  109. Steve Fitzpatrick says:

    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.

  110. Charlie says:

    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

  111. Back2Bat says:

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

  112. Tenuc says:

    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!

  113. TomVonk says:

    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 .

  114. ralph says:

    >>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.

    .

  115. Back2Bat says:

    “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.

  116. ralph says:

    >>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’).

    .

  117. old construction worker says:

    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?

  118. Brian says:

    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.

  119. Stephen Goldstein says:

    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).

  120. P Wilson says:

    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.

  121. Vincent says:

    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.

  122. TomVonk says:

    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 .

  123. Mark T says:

    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

  124. 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.

  125. Hank Henry says:

    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.

  126. John A says:

    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

  127. Ken Roberts says:

    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?

  128. SteveBrooklineMA says:

    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.

  129. Willis Eschenbach says:

    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.

  130. Charlie says:

    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.

  131. Willis Eschenbach says:

    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.

  132. Mark T says:

    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

  133. Roger Sowell says:

    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.

  134. MartinGAtkins says:

    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.

  135. Roger Knights says:

    Typo–replace “it” with “is” in:

    “Un-noticed by their programmers, however, it that…”

  136. Willis Eschenbach says:

    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.

  137. Roger Knights says:

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

  138. Willis Eschenbach says:

    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.

  139. ralph says:

    >>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).

    .

  140. Willis Eschenbach says:

    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.

  141. Willis Eschenbach says:

    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.

  142. ralph says:

    >>>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.

    .

  143. Willis Eschenbach says:

    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.

  144. Gary says:

    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.

  145. Willis Eschenbach says:

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

  146. Willis Eschenbach says:

    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.

  147. Willis Eschenbach says:

    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.

  148. Willis Eschenbach says:

    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.

  149. Jim says:

    *********************
    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 :)

  150. Willis Eschenbach says:

    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.

  151. George E. Smith says:

    “”” 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.

  152. Willis Eschenbach says:

    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.

  153. Willis Eschenbach says:

    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.

  154. Willis Eschenbach says:

    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.

  155. Willis Eschenbach says:

    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 …

  156. Gary Hladik says:

    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.

  157. Hank Henry says:

    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?

  158. lgl says:

    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.

  159. Alan S. Blue says:

    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.

  160. Hank Henry says:

    R.W. Woods words on the greenhouse effect:

    http://www.wmconnolley.org.uk/sci/wood_rw.1909.html

  161. Willis Eschenbach says:

    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.

  162. Juraj V. says:

    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.

  163. Seedload says:

    Doesn’t the steel shell have twice the surface area of the ball in the middle.

    You know – two sides rather than one.

  164. Willis Eschenbach says:

    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?

  165. Willis Eschenbach says:

    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.

  166. 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.

  167. michael hamnmer says:

    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.

  168. carrot eater says:

    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.

  169. Willis Eschenbach says:

    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.

  170. Jari says:

    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.

  171. George E. Smith says:

    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.

  172. Jim says:

    **********************
    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.

  173. Steve says:

    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.

  174. michael hamnmer says:

    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.

  175. George E. Smith says:

    “”” 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.

  176. Willis Eschenbach says:

    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” …

  177. A. Einstein Jr. Jr. says:

    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!

    :-)

  178. 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.

  179. George E. Smith says:

    “”” 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.

  180. John A says:

    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.

  181. P Wilson says:

    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

  182. Mark T says:

    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

  183. Willis Eschenbach says:

    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.

  184. P Wilson says:

    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.

  185. Willis Eschenbach says:

    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.

  186. michael hamnmer says:

    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.

  187. Steve says:

    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?

  188. 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).

  189. michael hamnmer says:

    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)

  190. Willis Eschenbach says:

    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.

  191. 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.

  192. Anton Eagle says:

    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

  193. Willis Eschenbach says:

    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”.

  194. Joel Shore says:

    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

  195. AlexB says:

    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!

  196. Alan S. Blue says:

    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.

  197. Willis Eschenbach says:

    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.

  198. Richard Sharpe says:

    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.

  199. Joel Shore says:

    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.

  200. Brian says:

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

  201. Joel Shore says:

    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).

  202. Willis Eschenbach says:

    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.

  203. LED says:

    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.

  204. John A says:

    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.

  205. 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.

  206. Mark T says:

    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

  207. John A says:

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

    Thanks

  208. Anton Eagle says:

    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

  209. Frank says:

    Anthony, anybody,

    Mercy KIll this silly thing. Its embarassing.

  210. John Millett says:

    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?

  211. P Wilson says:

    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

  212. Willis Eschenbach says:

    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.

  213. Anton Eagle says:

    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.

  214. Willis Eschenbach says:

    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?

  215. P Wilson says:

    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

  216. AlexB says:

    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.

  217. Willis Eschenbach says:

    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.

  218. Willis Eschenbach says:

    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 …

  219. tallbloke says:

    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.

  220. P Wilson says:

    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

  221. Curt says:

    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.)

  222. stephan says:

    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).

  223. Charlie says:

    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.

  224. Mark T says:

    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

  225. P Wilson says:

    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

  226. Charlie says:

    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.

  227. SteveBrooklineMA says:

    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.

  228. 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

  229. Willis Eschenbach says:

    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?

  230. Willis Eschenbach says:

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

  231. P Wilson says:

    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.

  232. Willis Eschenbach says:

    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.

  233. P Wilson says:

    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.

  234. P Wilson says:

    addendum oops. atoms, not molecules

  235. Willis Eschenbach says:

    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.

  236. Mark T says:

    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

  237. Willis Eschenbach says:

    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.

  238. AlexB says:

    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.

  239. 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.

  240. Mark T says:

    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

  241. carrot eater says:

    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.

  242. 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.

  243. Stephen Goldstein says:

    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.

  244. Andy Beasley says:

    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.

  245. Willis Eschenbach says:

    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.

  246. Graeme W says:

    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.

  247. Mark T says:

    Ian Schumacher (18:34:13) :

    Really this whole argument is begging for an experimental isn’t it? I would like to, but admittedly, I’m too lazy.

    Maybe. I already mentioned one. I have another, but it is conservation of mass, not energy, and the mechanism for actually implementing the mass idea would be rather complex (though doable).

    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?

    This is the point you keep making that doesn’t make sense. There is no set temperature of a black body. If there is sufficient energy storage, the temperature will necessarily be higher.

    Keep in mind, Willis is making a few ideal assumptions that cannot absolutely hold in the real world, which necessarily implies his results are upper limits.

    It really does require a one-way mirror, but no such device exists (see above).

    A dichroic mirror, actually, may get you close. It lets different wavelengths pass in different directions. For Willis’ example, the shell is transparent to long wave IR, but reflects 50% of short wave IR. Since the earth absorbs LWIR, but then radiates SWIR, you have feedback, allowing energy to increase within the system, though only during the transient state. Once equilibrium is reached, there is no more warming and things continue as Willis has shown (well, in the ideal).

    Mark

  248. Joel Shore says:

    Anton Eagle says:

    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.

    There is nothing wrong with this. The assumption in the model is that each shell is a blackbody, meaning it absorbs all the radiation impinging on it and emits according to the Stefan-Boltzmann Law with epsilon = 1. So, indeed, having 10 blackbody shells leads to a higher temperature than having one blackbody shell.

    If you wanted to have something that converged to a specific result in the limit that you take the number of shells to infinity, you would want to scale epsilon for the shells so that the atmosphere has a constant absorption as you increase the number of shells.

    John A says:

    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

    As Willis explains, these are statements about net flows. It is not as if a cooler object detects a warmer object approaching it and magically stops emitting radiation to it. They both emit radiation to each other but the warmer body emits more to the cooler body than the cooler does to the warmer.

    The Second Law of Thermodynamics is not magic. It has its basis in statistical physics. Consider another example with molecular flow instead of heat flow: Say, you have two chambers with a wall between them, one at a pressure of 1 atmosphere and another evacuated to 0.01 atmosphere. Now, let’s say you punch a hole in the wall between them. What will happen is that there will be a net flow of molecules from the high pressure to the low pressure chamber. However, this does not mean that magically no molecules from the low pressure chamber move into the high pressure chamber. In fact, some will. It is just that many more will go the other way. And, the reason, from a statistical point of view, that the net flow is from the high pressure chamber to the low pressure chamber is simply that there are many more molecules impinging on the hole from the high pressure side than from the low pressure side. And, once you are dealing with situations involving any sort of macroscopic amount of material, the statistics become overwhelming, so that the chance of the pressure in the high pressure chamber increasing while the pressure in the low pressure chamber decreases becomes so astronomically small that it ain’t going to happen even if you carry out the experiment once a second for the entire lifetime of the universe. (If you did the same experiment with, say, only 5 molecules in one chamber and 1 in the other, then while most of the times [on average 5/6] that you tried the experiment you would see the number of molecules in the high pressure chamber decrease and the number in the other increase, if you ran the experiment enough times you would in fact see a case where the 1 molecule in the low pressure chamber went into the high-pressure chamber so that the flow went the “wrong way”. The Laws of Thermodynamics are an emergent property of macroscopic systems that involve enough molecules that the probabilities of net flows going the wrong way become astronomically small.)

  249. P Wilson says:

    although the cooler planet emits radiation toward the warmer planet, supposing that one were nominally 100K and the other 150K, at some point, supposing they were fixed to receive one another’s thermal influuence they would thermalise, if there were not the presence of another radiating body, although radiation goes in all directions, not just towards other hypothetical planets, and may lead to Kelvin’s heat death hypothesis where there was no thermal energy left. Kelvin extrapolated that heat loss is endemic to nature, based on this 2nd law of thermodynamics, but for the purposes of our climate, we still have the sun.

    carrot Eater wrote that some are apt to forget that anything above absolute zero emits radiation – well yes it does, because its losing heat, but depends on the temperature surrouding it. If an area of space is 10K and an object suddenly enters it at 4K, eventually the 4k object will reach 10K, supposing the absence of other radiative sources. Its improbable that the 10K region of space will lose heat to a 4K object

  250. P Wilson says:

    constants are constants. If this planet emits 235w/m2, and eeven if totally blocked to outgoing radiation, its still 235 w/m2 ad infinitum, so it won’t accumulate to 235*some multiplier or else 235 outgoing+235 incoming. Its emitting as much as it blocks, which is efectively the same energy

  251. carlbrannen says:

    The basic problem with this post is that it gives a complicated analogy to a complicated system. It just doesn’t simplify things much. Furthermore, it reverses the heat source unnecessarily. And Figure 1 puts people who know how surface scales with radius, off their feed.

    Re: “Please read up on the Stefan-Boltzmann equation …” This is a mean spirited statement not intended to illuminate. Like one of the other commenters, I’ve designed heat sinks for commercial production. To convert temperature to W/m^2 requires assuming an emissivity which depends to an incredible degree on the choice of finish. In addition, the emissivity of surfaces typically changes with temperature.

    My point is that “temperature” is a simple concept that everyone understands. “W/m^2″ is not. And the relationship between them is a mess. An analogy with an insulated bottle is simpler:

    Light from the sun gets into a (transparent) bottle and heats it up. The amount it heats it up is proportional to the thickness of the insulation. As an insulator, CO2 provides a thickness that is proportional to the logarithm of the amount of CO2 which is why global warming is not much of a threat.

  252. P Wilson says:

    put another way, 60F will not surpass 60F with the shell number increase, just that that 60F will cover a wider area distribution than 1 shell

  253. Willis & Mark T,

    Ok I am dense, I fully admit that, but I am open to learning ;-) And for some reason this problem fascinates me. So I’m thinking through the shells again. Ok step by step least see if I can get this.

    Iteration:
    1.) 100 W/m2 out … hit shell
    2.) shell give 50 back and 50 out to space.
    3.) 50 from shell hits earth, warms it up, goes back out as 50 watts, hits shell.
    4.) 25W/m2 back, 25W/m^2 out to space.

    We can see were this is going, so:

    From our special source we have 100W/m2 out .. as stated .. always.
    hitting the surface we have 50 + 25 + 12.5 … = 100W/m2
    going out from surface we have same … 100W/m2
    From shell in we have 50 +25 + 12.5 …. = 100 W/m2 in.
    From shell out we have .. same = 100 W out.

    We have a source of 100 W/m^2 and we have 100 W/m^2 leaving … great. Everything seems in balance.

    Outer shell will always have 100W/m^2 leaving from it.

    Ok so next shell 100W/m^2 leaving, means 200W/m^2 from first shell and 200 W/m^2 from surface (plus out source of 100W/m^2).

    Ok, you are right. I should have thought through it more carefully the beginning. My mistake. Basically this is like extra layers of insulation I guess.

    Now, point #2.

    In you thought experiment we have this constant source. This source is the ‘sun’ through the visible window. Ok so to apply this to the steel shell model the sunlight has to come in to hit the surface. We can do that by punching a hole in the steel shell that lets in 100W/m^2 equivalent amount of sunlight energy. Or we can say the shell has a special property. It is a high-pass filter that lets in most of the sunlight, but rejects (acts like a solid shell) for low frequencies. Let’s pick the second one.

    This high frequency light comes in, hit the cold surface, warms it up a bit and emits a black body spectrum of frequencies outward, most of these being below our cutoff frequency of our shell.

    The radiation that is above the cutoff will easily exit the shell. The remaining radiation will (after bouncing around back and forth a bit such) generate new radiation above the cutoff frequency.

    Let’s make the model even simpler and assume that the outside shell doesn’t radiate any energy at all outward. Its a perfect reflector. A perfect reflector is equal to your shells inside shells if you have infinite of them. To show this, imagine your internal energy source again. You have a perfect reflector shell. The flux will keep increasing forever given infinite time to build up. This is the same as with infinite shells, but easier to manage mentally.

    So we have a sphere that is a perfect reflector. All energy comes through our frequency hole. All energy leaves through this same hole.

    100 W/m^2 comes in through the hole. reflects around. How much comes out? 100W/m^2.

    What is the temperature inside?

    The distribution of radiation frequency is give by Plank’s law. At first glance it seems that we need to have a temperature large enough that when we integrate the radiation above our cutoff frequency we get 100w/m^2 total.

    But this is wrong for the same reason that thinking that the radiation in a spherical cavity with a physical hole in it will need to achieve a temperature so that radiation aimed directly at the hole goes out (integrate over all radiation at the right magic angles to get out). Its wrong because it ignores the constant reflections and ‘repopulating’ of these angles (for the hole example) or repopulating the radiation above the cutoff frequency (for the case of a frequency hole).

    Just like the physical hole, the frequency hole is the same.

    A hole in frequency is the same as a hole in space. A cavity with a hole in it will achieve temperatures near a black body. Doesn’t matter what kind of hole this is.

    At least that’s my theory ;-)

  254. idlex says:

    Le Sage’s Theory of Gravitation.

    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.

    The wikipedia link doesn’t work. Or rather, there is no content. This one looks rather more promising:

  255. John A says:

    Joel Shore:

    The Second Law of Thermodynamics is not magic. It has its basis in statistical physics. Consider another example with molecular flow instead of heat flow: Say, you have two chambers with a wall between them, one at a pressure of 1 atmosphere and another evacuated to 0.01 atmosphere. Now, let’s say you punch a hole in the wall between them. What will happen is that there will be a net flow of molecules from the high pressure to the low pressure chamber. However, this does not mean that magically no molecules from the low pressure chamber move into the high pressure chamber. In fact, some will. It is just that many more will go the other way. And, the reason, from a statistical point of view, that the net flow is from the high pressure chamber to the low pressure chamber is simply that there are many more molecules impinging on the hole from the high pressure side than from the low pressure side.

    Unfortunately you are wrong.

    When the hole is opened, the total entropy of the system increases which takes energy out of the system (because entropy is a form of energy in the form of disorder) even if it is isolated from the rest of the Universe.

    If you look up “Maxwell’s Demon” and its resolution you will see why cold does not flow to hot even in a statistical sense.

    In Willis’ case he’s arguing that because everything radiates (everything above absolute zero) then the cooler shell radiates back to the planet’s surface. No it doesn’t.

    If you took a bowl of hot soup and surrounded it on all sides with ice, then the radiative flux from the ice does not heat the soup at all. Instead the ice melts, taking energy away from the soup and converting that energy into a change of state from ice to water, and increasing entropy of the total system.

    People here think its all about radiative fluxes and bouncing backwards and forwards when the reality is that the laws of thermodynamics are violated.

  256. John A says:

    Here’s the link to the story of how Maxwell’s demon works in statistical thermodynamics: http://www.auburn.edu/~smith01/notes/maxdem.htm

  257. Mark T,

    Yes i keep mentioning this hole in a sphere thing. Well its not me really. That the example given in text books as an approximation to a blackbody. A sphere with a hole in it is ‘close’ to an ideal blackbody. They say (the texts, Stephan-Boltzmann, etc) that a spherical cavity with a hole in it will achieve temperatures of a blackbody.

    So that’s why i keep going back to it. Blackbody temperature for a given external flux is the maximum temperature.

    I don’t think it matters what kind of hole. You need a one-way device to achieve a higher temperature. Such a device would be equivalent to Maxwell’s demon and appears (to my demonstratively weak mind) to violate the second law of thermodynamics since it would allow heat from a cold side to move to a hotter place.

    Basically many people are saying we ‘can’ create a special hole that allows us to have a temperature hotter than a black body in the same environment. I’m doubtful, but listening, but I admit that for this particular argument we are probably at an end, without an actual experiment to show it quite simply one way or the other.

  258. Willis Eschenbach says:

    P Wilson (19:47:09) :

    constants are constants. If this planet emits 235w/m2, and eeven if totally blocked to outgoing radiation, its still 235 w/m2 ad infinitum, so it won’t accumulate to 235*some multiplier or else 235 outgoing+235 incoming. Its emitting as much as it blocks, which is efectively the same energy

    Clearly you misunderstand the thought experiment. The planet emits 235 w/m2 at equilibrium. In other words, at that point it is emitting as much as it absorbs.

    Now, suppose we wrap it in a perfectly insulating blanket. The radioactive elements are still producing heat, but it has nowhere to escape. As a result, the planet’s temperature starts to rise.

    So while constants are constants, the temperature of the planet in the thought experiment is not constant. That’s why e.g. ovens have insulation. The energy coming from the gas flame is constant … but the oven without insulation is not the same temperature as the the temperature of the same oven with insulation.

  259. Hank Henry says:

    I think of conductive heat transfer as being the rather direct transfer of the motion of one atom or molecule to another. Would it be wrong to think that the radiative mode of heat transfer is also going on within a solid or liquid or gas?

  260. par5 says:

    Willis- with this reverse greenhouse (heat from within) will there still be changes in the atmosphere ie cool and warm phases? Will the steel green house provide a constant temp worldwide? Will there be polar ice caps?

  261. DeWitt Payne says:

    Willis,

    You have the patience of a Saint.

  262. Willis Eschenbach says:

    carlbrannen (19:57:41) :

    The basic problem with this post is that it gives a complicated analogy to a complicated system. It just doesn’t simplify things much. Furthermore, it reverses the heat source unnecessarily. And Figure 1 puts people who know how surface scales with radius, off their feed.

    How is a planet with a steel shell around it a “complicated analogy”?

    Re: “Please read up on the Stefan-Boltzmann equation …” This is a mean spirited statement not intended to illuminate. Like one of the other commenters, I’ve designed heat sinks for commercial production. To convert temperature to W/m^2 requires assuming an emissivity which depends to an incredible degree on the choice of finish. In addition, the emissivity of surfaces typically changes with temperature.

    I am sorry you felt it was mean spirited. It was not meant as such. I said it because you claimed that

    W/m^2 is not, nor ever has been, equivalent to temperature

    I assumed you said that because you didn’t understand that for blackbodies we can measure temperature equally well either in W/m2 or in Kelvins. My bad.

    Regarding emissivity, in my thought experiment I specified both the planet and the shell as being black bodies. How is this “complicated”? I really don’t know what to say.

    My point is that “temperature” is a simple concept that everyone understands. “W/m^2″ is not. And the relationship between them is a mess. An analogy with an insulated bottle is simpler:

    As I mentioned before, this cannot be done in units of temperature, because unlike W/m2, Kelvins are not conserved.

    Light from the sun gets into a (transparent) bottle and heats it up. The amount it heats it up is proportional to the thickness of the insulation. As an insulator, CO2 provides a thickness that is proportional to the logarithm of the amount of CO2 which is why global warming is not much of a threat.

    Yours is a very poor way to explain the greenhouse effect. Why? Because the real greenhouse effect in the atmosphere does not work like that. I am trying to explain how a greenhouse

    actually works

    , not give an inaccurate analogy to insulation. I am explaining it by positing a thought experiment of a steel atmospheric greenhouse, which works by exactly the same principles of the real atmospheric greenhouse. It is not the same as insulation, as you claim, which is why a Thermos bottle works better than an insulated bottle.

  263. anna v says:

    Willis Eschenbach (20:31:54) Re:the thought experiment

    Reminds me of the igloo and the Eskimoes (?spelling) being naked inside with a small whale oil lamp for heating in the middle of subzero arctic temperatures.
    A real experiment.

  264. Willis Eschenbach says:

    Hank Henry (20:42:03) :

    I think of conductive heat transfer as being the rather direct transfer of the motion of one atom or molecule to another. Would it be wrong to think that the radiative mode of heat transfer is also going on within a solid or liquid or gas?

    Radiation goes on within a gas, but not, as far as I know, within a solid

  265. Willis Eschenbach says:

    par5 (21:22:15) :

    Willis- with this reverse greenhouse (heat from within) will there still be changes in the atmosphere ie cool and warm phases? Will the steel green house provide a constant temp worldwide? Will there be polar ice caps?

    Since my thought experiment has no atmosphere, none of them are possible.

  266. Curt says:

    John A: You are completely and utterly wrong. You would get flunked out of any undergraduate engineering thermodynamics or heat transfer course. If I were your instructor in such a course (and I have taught courses with this subject matter), and you kept maintaining these views, not only would I flunk you, but I would be having a conversation with the dean about how the hell you got into the program in the first place.

    Bodies radiate based on their emissivity and their temperature alone, without regard to what they are radiating towards. They have no way of “knowing” what they are radiating towards. This is easily demonstrable in a lab. Put two plates of different temperatures parallel to each other, and you can still measure the radiative energy from the colder plate with a radiometer.

    A radiometer that is warmer than the device radiating towards it can still measure the radiation from that object — according to your arguments, that would be impossible.

    In any introductory engineering heat transfer text, you will see that the net radiative heat transfer between two objects (1 and 2) is given as:

    K * (T1^4 – T2^4)

    where K is a system constant incorporating the emissivities and geometric properties of the two object. This expression automatically incorporates the fact that the cooler body is radiating towards the warmer body, albeit less than the warmer body is radiating towards the cooler body.

    By your argument, this equation is completely wrong. Your claim is that the temperature of the cooler body is irrelevant, and that the net heat transfer from the warmer body is the same whether the cooler body is 0.001C colder or at absolute zero.

    If you really believe your argument, you should not be spending your time on blog wars like these. You should be doing whatever it takes to shut down any system that was designed using the expression I gave above (which is ever power plant, every power electronic system, among many others), because according to you, they were designed under completely fallacious assumptions, and therefore could be totally unsafe.

  267. Willis Eschenbach says:

    DeWitt Payne (21:24:17) :

    Willis,

    You have the patience of a Saint.

    I believe that the only stupid questions are the ones that you don’t ask. And I think that if I put up a controversial post, it is my responsibility to at least try to answer the questions and objections.

    However, I have been known to lose it …

  268. yonason says:

    anna v (22:20:27)

    Snow is a very efficient insulator, and is what keeps the ground from getting so cold in the winter that many plants and organisms wintering over underground would otherwise be killed. It has nothing to do with radiation. The body heat and whatever other heat sources they have warm the air which is trapped, much like in a greenhouse.

  269. Rabe says:

    @jt:
    I had that same idea of non-flying photons some years ago. The consequences wouldn’t be obvious here on earth because there are always plenty of ‘receivers’ around. But thinking about a test of this hypothesis leads to very nasty unsymmetries on the space time relationship of the two objects regarding the direction of a transfer. Ignoring those I came up with the following: When, say in the ISS, the sky is scanned with a laser (or other light source) there would be measurable differences in the energy consumption depending on the nuber of receivers available in the respective direction. Showing that amount of consumed energy on a screen would show the number of receivers in the scanned area. Taking as target our sun what should we expect to see at time ‘now (here)’ at the place where the sun was 8 min’s before now, it is now, will be in 8 min’s? When the transfer takes no time from the viewpoint of light the ‘negotiation’ will. And this is the reason why we will not have a look into the future and seeing now if our sun exploded within the next 16 minutes.

  270. Anders L. says:

    I am layman when it comes to physics (I am a computer programmer by profession), so I am not going to comment on the physics itself. It is quite interesting, though, to see how the concept of thermodynamics gets tossed around and how the commenters accuse each other of not understanding the first thing about it. It is almost like hearing programmers debate the pros and cons of Java.

    However, I am a little at loss trying to figure out the purpose of this entire thought experiment. As far as I can understand it simply says that anything that blocks the outgoing radiation will make the planet warmer. A layer of steel will do it. Probably a layer of cotton as well, or a layer of old newspapers. Or – as in the case of the Earth – an atmosphere. And – all things equal – as long as nothing changes, everything will remain the same, once the system has reached thermal equilibrium.

    But the interesting thing, really, is what happens when things do change. What happens when the system is NOT in thermal equilibrium, as is the case with our planet right now? According to Hansen et al 2004, the Earth is receiving 0.85 +- .15 W/m2 more than it is emitting into space. How fast will the surface temperature of the planet rise or fall as a response to shifting properties of the “shell” (for example, when the mix of greenhouse gases in the atmosphere is changed)?

  271. julien says:

    Interesting model.
    Is there a way to re-design this shell system using an external radiation flow ?

    The analogy may be accurate, but it does not feel so at first glance, and the number of comments confirm this.

  272. Stephen Wilde says:

    Well all I can say is that if the comments here are a reflection of the confusion currently existing amongst professional scientists then there is no surprise about the media panics arising from virtually all new observations of natural real world changes, not just climate shifts.

    I see that many contributors here are far more scientifically qualified and experienced than I am yet seem unable to see the wood for the trees where multiple factors need to be considered simultaneously and their respective significances unravelled and weighted appropriately.

    Willis has given us a limited but useful scenario that brings to the fore certain aspects of AGW theory that are quite frankly untenable.

    In doing so he has had to ignore certain aspects which his simplified model cannot deal with but they are matters it does not need to deal with to make the point.

    Hell, that’s the whole reason for thought experiments and simplified modelling. They are not designed for complete and total accuracy. They are designed to emphasise and more clearly illustrate limited but important issues.

    In fact all those ‘wonderful’ climate models that excite us so much are no more accurate in their description of reality than is Willis’s model but at least he is using his for a specific and limited illustrative purpose and not pretending that his model is adequate to justify political policy decisions which have the potential to redraw human societies worldwide and forever.

  273. Allan M says:

    I hate it when people give me the anal ogies ( or any other sort of ogies!). I also hate it when people choose to move indescriminately between the statistical world of quantum theory and the classical model. As far as I know, no-one has yet got(ten) the Nobel Physics Prize for a unified theory.

  274. Nick Stokes says:

    I too admire the patience of Willis and Joel. This is a simple radiation example which does illustrate the greenhouse effect well. Textbook stuff. It is described in, for example, this set of Oxford University undergrad lecture notes (top page 9).

  275. yonason says:

    “In order to maintain its thermal equilibrium, the whole system must still radiate 235 W/m2 out to space…” – from paragraph just above fig.1

    In other words, by definition, the steel shell is arbitrarily assigned the same temperature as the earth? If so, then, from thermodynamics, there will be no heat transfer between them, because what drives radiant heat exchange (or any other, for that matter) is the difference in temperature (here it’s proportional to [T1^4 - T2^4]). So, all things being equal, the shell’s output per unit area should really be lower, because the rate at which the earth generates heat remains a constant. And, because the shell has a larger surface area, and the Net Power (see below) is constant, the per unit area output can’t be the same as it was for earth alone.

    Now, if you insist that the per unit area output remain the same as it was for the smaller area of the earth, your problem would appear to be as follows:

    The equation for a blackbody is P = esA(T1^4 – T2^4), where P (Net Power) has units of Watts, not Watts/squared meter. And because of the larger area of the shell (it’s radius is greater than that of the earth), for the radiation emitted by the shell to be maintained at the same Watts/square meter, the “Net Power” of the shell would then be greater than what the earth’s was alone. That will only happen if the earth increases it’s untill-now-fixed output.

    Of course, even without the increase in earth’s output, it’s temp would rise slightly because it was no longer radiating to empty space at T~0, but now to the Shell at T2′. Nevertheless, the slight increase in earth’s temp., due to encasing the earth in the shell wouldn’t be sufficient to drive the shell’s output at the same level per unit area as the earth’s alone, the earth would have to increase it’s net power output for this scheme to work. But wasn’t it fixed?

    “Bridge to engine room, Scotty . . . ?!”

    NOTE – The addition of another shell would again necessitate another call to the engine room, otherwise the requirement that the new shell be in thermal equilibrium with the rest of the system is not going to happen.

    What would happen without the unrealistic escalating demands on the earth’s ability to generate heat would be that each successive shell gets colder and colder, while the earth would warm slightly each time, with each successive warming less than that of the previous one. Compare that result with what’s expected for CO2, where for each fixed increment of CO2 added, the temp goes up less than it did for the previous one.
    http://www.junkscience.com/Greenhouse/co2greenhouse-X4.png.

  276. carrot eater says:

    Ian Schumacher: The reason you are struggling so much with the blackbody idea from the old physics textbook is that you are misinterpreting what it means. The rule is simply: at a given temperature, nothing can emit more radiation than a blackbody. Perhaps, stated mathematically: the emissivity of a surface cannot be greater than one. Willis is setting the emissivity of his earth to one, so there is no problem. That doesn’t keep it from changing temperature, given changing conditions around it.

    The cavity with the pinhole illustration is only given because it approaches the perfect blackbody: something with perfect absorption, as well as maximum and diffuse emission.

    *Actually, this brings a slight clarification: Willis’s earth emits like a perfect blackbody, but it does not absorb like one. Which is perfectly fine.

  277. Vincent says:

    John A

    I have followed your arguments about the second law of thermodynamics, and after reading your reply to Joel Shore’s excellent explanation, I conclude that you are making the same mistake over and over and over.

    The example you give of the ice surrounding the soup shows that you completely misunderstand the problem. The mistake you make is that you are not seeing the whole picture. You evisage ice surrounding soup and your experience tells you that the ice does not make the soup warmer, but quite the reverse. But in your thought experiment, your bowel of soup is in a nice warm kitchen, and then you surrounded it with cold ice. The result obviously is then the soup cools faster and the ice melts. But your mistake is to have imagined the soup to be in the warm kitchen in the first place. Instead, lets imagine the soup (in a pressure vessel of course) is placed in space, out of the sun, near absolute zero. The soup will cool down pretty damn quick on its own. But now surround it with the ice, and the rate of cooling is less.

    Your reference to Maxwells demon, explains that the entropy of the system increases, since the Demon has to do work to make the warm partition hotter. But in the example with two radiating bodies at different temperatures that was given by Willis, the warmer body does indeed loose heat to the cooler body, in accordance to the second law of thermodynamics. However, some photons of energy from the cooler body will still strike the warmer body. Why does that violate the second law? Please try and think of a physical mechanism that would stop this from happening. And do not just say “it violates the second law” because that is a tautoligious argument. Why, cannot this photon from the cooler body strike the warmer?

    The consequence of a photon from the cooler striking the warmer, is that the warmer will not cool down so quickly as it would have done if it was completely surrounded by empty space at absolute zero. So therefore, yes, the cooler body, will have reduced the rate of cooling of the warmer body.

  278. carrot eater says:

    Willis Eschenbach (23:46:55) :
    “Radiation goes on within a gas, but not, as far as I know, within a solid”

    To clarify: all matter radiates, no matter where it is. However, radiation emitted by a molecule within a solid will generally be absorbed by the immediately neighboring molecules: it doesn’t get very far. It may or may not be intuitive then, that within a given solid, we can neglect heat transfer by radiation.

  279. P Wilson says:

    Willis Eschenbach (20:31:04)
    i’m trying to establish whether the optimum energy that this hypothetical planet emits is 235w/m2, of its own radiation in the first instance

  280. P Wilson says:

    looking at another analogy, the energy emitted by a 100 watt incandescent lightbulb, that emits heat and light across a wide range of frequencies, but lets just use heat and say that three feet from the bulb, it is IR, and we were to put a globe of aluminium foil around it to prevent convection, and in another simultaneous experiment we were to line the foil at the same distance with black paper or another blackbody material. After two days of constant emission, is the lightbulb hotter than if it were not surrounded by either of these two insulators, or is the interior of each globe at roughly a slighly lower temperature than the recorded energy leaving the surface of the lightbulb?

  281. P Wilson says:

    Willis Eschenbach (00:05:34)

    “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.”

    ok lets take this example as i’m still struggling here with the thermodynamic process. If there were a heat source in this flask emitting energy at a temperature that measured 25C over a period of time, how exactly does the flask increase this energy? If the flask were measured with a thermometer would the entire contents be stable at 25C, or would it increase overtime. Given that the optimum possible source was measured 25C, how can reflected energy increase it? I see how it is done in theory, but nothow it is borne out in practice?

    It would be interesting to hear if you’ve done a heat experiment of this nature

  282. Charlie K says:

    To all of those arguing that a cooler object will not radiate towards a warmer object I have an example.

    Consider two light bulbs, a 40W and a 100W bulb. If we place them in fixtures that are 1M apart and turn them on, they will both radiate. Even though the 100W bulb is “hotter” than the 40W bulb, the 40W bulb will still radiate on the side that is facing the 100W bulb.

    If we consider the same situation using perfect IR bulbs in a vacuum, we can see that it is reasonable that the 40W IR source will still radiate on the side that is facing the 100W IR source. However, if we measure the energy flow through an imaginary plane located between the two sources, we will see that the NET flow between the sources will be towards the 40W “cooler” IR source.

    – Charlie K

  283. Charlie K says:

    @ Anton Eagle (15:36:25) :
    “However, how are 10 shells all 1 inch appart any different than 1 shell that is just 10 times as thick?”

    Graeme W (18:59:18) touched on this point in his post. In the case of 1 shell that is 10 times as thick, you have changed the heat transfer from radiation to convection. The effectiveness of a heat shield is proportional to the number of layers, and not their thickness assuming that the layers are isolated from each other. See the Wikipedia article on Multi-Layer Insulation for a better description of how MLI works than I can give here.

    http://en.wikipedia.org/wiki/Multi-layer_insulation

    Sorry, I don’t know how to make the links in comments, so you’ll probably have to copy & paste the url.

    – Charlie K

  284. Allan M says:

    As a source of heat, I would not like to spend a cold winter’s night under a steel blanket!

  285. TomVonk says:

    Willis

    You wrote that you answered all critics but you didn’t mine .
    You stopped only at the question of units .
    .
    1)
    The question of units .
    The SB law is derived from the Planck law by integrating for all frequencies and OVER A HALF SPHERE
    [I (nu,T) dnu domega] where I units are W/m²/sr .
    As the sr (solid angle unit) disappeared in the integration , follows that SB law is only valid for convex bodies . The SB constant is also only valid for this case .
    Whenever one has to calculate a radiation flux in a specific geometry one has to keep in mind that SB is only valid for half spheres . If I substituted a non convex shell in your model it would give wrong results from A to Z . I will show later why the solid angles are important .
    .
    2)
    Irradiance (units W/m²) “conservation”
    This one simply contradicts the laws of physics . As I have shown the irradiance of the shell is P.r²/R² where P (units W/m²) is the irradiance of the inner sphere , r and R corresponding radiuses of the sphere and the shell . Obviously P.r²/R² < P so irradiance (units W/m²) is NOT conserved . In other words the shell has a lower temperature and irradiance than the sphere .
    You may say that when r~R then P.r²/R² ~ P but you may not say that irradiance is "conserved" .
    Of course when you increase R , like John A is saying , this postulated "conservation" is violated much more . In the limit at R infinite the temperature of the shell is 0 K and it's irradiance 0 !
    It doesn't mean that your model is "wrong" but it means that you are forbidden to talk about irradiance "conservation" .
    This must be corrected .

    3)
    The solution of the apparent paradox that the shell receives 240 W/m² but SEEMS to emit 2 times "more" e.g 480 W/m² .
    Here one has to recall what I said in 1) . The SB law is only valid for a half sphere . But the problem is that a 2 face object doesn't see only a half sphere , it sees the full sphere . So it will emit in a solid angle of 4Pi instead of only 2 Pi for which the SB law is valid .
    So emitting 480 W/m² per 4Pi sr (in the whole space) is perfectly equivalent to emitting 240 W/m² per 2 Pi sr (in a half space) and the SB law is safe . What's more important is that Planck's law
    (I units are W/m²/sr) is safe too .
    Here I don't contradict your model , I give only a rigorous explanation of the "paradox" that perturbated most of the posters .
    4)
    Relevance to the atmosphere .
    Here it is for me 0 and again for a solid angle problem .
    Atmospheric radiation can be assimilated to ponctual sources .
    But ponctual sources see the whole space . Moreover they form no convex surfaces . SB law doesn't apply . That should be enough .
    Please note that this remark doesn't relate directly to the steel "model" itself . While it has a qualitative feature that LOOKS like one specific atmospheric process , it has no chance to show the dynamics that the atmosphere does .
    The atmosphere can't clearly be reduced to a BB equilibrium radiation problem .
    But it's true , there is a vague partial qualitative analogy .
    .
    5)
    Synthetically as a physicist I range myself with the opinion of L.Motl .
    You did no glaring errors (the irradiance "conservation" is wrong but doesn't impact much because you considered a case where r~R) , there is a qualitative similarity to the GHE but the model will not simulate even partly the atmospheric dynamics .

  286. carrot eater,

    You might be right, but here is my mental problem illustrated (very poorly I might add) http://www.ianschumacher.com/img/diagram.png

    Imagine instead of photons, or whatever your favorite energy metaphor, we are dealing with energetic particles. Imagine there is a flow of these particles going in a very specific direction (like a water hose kind of thing) and these particles are being directed through a hole in a sphere. My problem is that the density of the energy of the particles (proportional to the pressure, assuming particles are all equal mass) is the same inside as outside.

    In order for them to not be the same, we need a special hole. one that lets particles in more easily then it lets them out. I don’t believe such a thing exists in the physical world and as I and others have mentioned, this would be equivalent to Maxwell’s demon. A thought-experiment mischief maker that allows us to violate the second law of thermodynamics.

    It is my belief that there is no special filter that can be made for energetic particles of any kind.

    Basically you can not get the ‘pressure’ inside higher than the pressure flow of the particles through the hole.

  287. carrot eater,

    Just a follow up, you said

    “*Actually, this brings a slight clarification: Willis’s earth emits like a perfect blackbody, but it does not absorb like one. Which is perfectly fine.”

    I don’t think that is true. I believe at equilibrium emissivity and absorption must be equal. Kirchoff’s Law I think.

  288. P Wilson says:

    Vincent (04:14:30) :

    these demonstrations also incur the 2nd law, but at different magnitudes. The soup will lose its heat and the ice will reach the temperature in this lost in space region you refer to, although the change in rates would be related to the pressure. They will both thermalise to near absolute zero. Remove the pressure chamber altogether and throw the soup into space.

    Are you arguing that space will reach soup temperature, and soup will reach this modified space temperature? They both thermalise to around the hypothetical absolute zero in a short period of time.

  289. P Wilson says:

    Vincent

    “However, some photons of energy from the cooler body will still strike the warmer body.”

    I’ll put my eggs in the fridge, and see if they are cooked tomorrow morning.

  290. lgl says:

    So many who do not understand positive feedback? Try this:
    You put a shell around the planet. Using 256 because that gives nicer numbers.
    The shell receives 256 W/m2 and emits 128 from the outside and 128 from the inside.
    1st iteration: the surface receives 256+128=384 W and must emit 384 W
    The shell now receives 384W and will emit 192 out and 192 in.
    2nd iteration: surface receives 256+192=448 W and emits the same, shell emits 224 inward.
    3rd iteration: surface receives 256+224=480 W
    4th iteration: surface receives 256+240=496 W

    So the runs add 128 W, then 64, then 32, 16, 8, 4, 2 …
    and after an indefinite number of times this adds up to 256 W i.e the double of the initial.

  291. carrot eater says:

    Willis Eschenbach (00:05:34) :

    You skipped over the main advantage of having a vacuum layer in a thermos bottle: the elimination of conduction and convection as possible modes of heat transfer. If all you have is radiation (or conduction leakage out the top; the vacuum can’t go all the way around), the heat transfer is slower.

  292. beng says:

    Willis, thanks for your apparently infinite patience. Some people have really exposed their misconceptions on this post — the strawmen have piled up even after you stated that it’s a simplified “toy” model right at the start (and I can’t see any basic conceptual problems w/it, as it is). Kudos for extending it to a better representation than the original Kiehl/Trenberth budget.

    Michael Mann is supposedly a physicist too. :)

  293. Hank Henry says:

    In reply to: Willis Eschenbach (23:46:55).

    On the subject of radiative transfer and transmission within solids…

    I thought about this more as I went to sleep last night. Thinking of the example of glass or clear water, I decided that it must happen, because a clear substance is surely capable of letting radiation through. So I am guessing that in an opaque substance the radiative transmission happens but it also gets stopped very quickly – in the same way that it is stopped at the surface of the opaque substance. Now the question is whether my ideas about heat transfer as vibrational are backward. I want to say that that way of thinking is still valid because I was shown Brownian motion in freshman biology. I also think I remember that Einstein studied Brownian motion and made some mathematical calculations concerning them.

  294. Joel Shore says:

    John A: Curt and Vincent have answered you very well so I will just make a few more comments here on what you have said.

    If you look up “Maxwell’s Demon” and its resolution you will see why cold does not flow to hot even in a statistical sense.

    I don’t see what Maxwell’s Demon has to do with what I said; in fact, if anything it tends to re-enforce what I said as Maxwell’s Demon essentially serves to counter the statistics by making an “intelligent decision” as to which molecules to let through and which to block. Furthermore, I said pretty much the opposite of what you seem to think I said. What I said was that the statement that cold does not flow to hot is just a statistical one, albeit one that becomes completely statistically overwhelming for any sort of macroscopic objects. And, my point is that it applies to the NET heat flow because for any sort of macroscopic objects it becomes ridiculously improbable to have NET heat flow from cold to hot even though there are exchanges of heat going in both directions.

    In Willis’ case he’s arguing that because everything radiates (everything above absolute zero) then the cooler shell radiates back to the planet’s surface. No it doesn’t.

    Sure it does. However, even more heat is radiated by the planet to the shell, so the net flow is from the warmer planet to the cooler shell. [What some people get hung up on is how this can then result in the planet being warmer...i.e., if the net flow is from the planet to the shell, why doesn't the shell make the planet cooler? However, the answer to this is carefully considering the case you are comparing to which is the case where there is no shell at all. In that case, all of the heat that the planet radiates would simply escape out into space. By contrast, when you have the shell there, then some of the heat that the planet radiates (actually, in this simple example where the shell is a perfect blackbody, all of that heat) is absorbed by the shell which subsequently radiates part of it back to the planet. This means the planet will end up warmer than it does in the case where none of the heat that it radiates finds it way back.]

    If you took a bowl of hot soup and surrounded it on all sides with ice, then the radiative flux from the ice does not heat the soup at all. Instead the ice melts, taking energy away from the soup and converting that energy into a change of state from ice to water, and increasing entropy of the total system.

    Sure, you are correct that the NET flow of heat is from the soup to the ice. However, that does not mean that there is no heat flow from the ice…just that there is more in the other direction.

    People here think its all about radiative fluxes and bouncing backwards and forwards when the reality is that the laws of thermodynamics are violated.

    There is nothing in Willis’s description that violates the laws of thermodynamics. In particular, all net flows of heat are from warmer objects to colder objects. The only part of your complaint here that is true is that it is not about the fluxes “bouncing backwards”. What happens is that the shell absorbs the radiation from the earth and then because it has a nonzero temperature (determined by the balance of the energy it receives to what it radiates), it itself also radiates. In this regard, here is the webpage of a retired atmospheric science professor who is sort of stickler for describing the greenhouse effect correctly: http://www.ems.psu.edu/~fraser/Bad/BadGreenhouse.html He is a bit militant in what the correct pedagogy is for my taste, but I think he does have a valid point that thinking about it in the wrong way can lead people astray.

  295. Joel Shore says:

    Anders L. says:

    But the interesting thing, really, is what happens when things do change. What happens when the system is NOT in thermal equilibrium, as is the case with our planet right now? According to Hansen et al 2004, the Earth is receiving 0.85 +- .15 W/m2 more than it is emitting into space. How fast will the surface temperature of the planet rise or fall as a response to shifting properties of the “shell” (for example, when the mix of greenhouse gases in the atmosphere is changed)?

    The answer to that question is that it depends on how many Watts of power it takes to provide enough energy to raise temperatures back up to the point where radiative balance is restored. And, this means that one has to calculate what is called the “thermal inertia” or “thermal capacity” of the system. The atmosphere itself does not have that large a thermal inertia, but the oceans do…In fact, just the top ~2.5m of the oceans have as much thermal inertia as the entire atmosphere. (There is also some thermal inertia in the land but not as much as in the oceans.) So, it is the oceans that really slow down the re-establishment of radiative balance.

    The oceans are complicated and can be thought of as having a mixed layer on top, which is generally about 100m in depth (but can vary quite a bit) that is in pretty good contact with the atmosphere. The deep ocean below that tends to have only slow exchange of heat with the mixed layer. How quickly most of the temperature rise occurs is pretty sensitive to assumptions regarding the deep ocean and its communication with the mixed layer. If the deep ocean were completely isolated so you just had to heat the mixed layer, you’d have an approximately exponential response with a the time scale for the of ~5-10 years. The effect of the deep ocean is to lengthen the time for re-adjustment and also make the approach to the new radiative balance more non-exponential.

  296. P Wilson says:

    Joel since you’re explaining it in absence of Willis, its still confounding, given that the source of emission – a black body planet surrounded by a black steel enclosure – if the optimum radiation is 235w/m2, then surely the optimum exchange will be no more than 235w/m2, since that is the optimum energy in the system. This could surely be verified by experimental procedure, since it seems to be universally agreed by consensus that 235w/m2 produces a temperature of -19C.

    Thats not an impossible temperature to reproduce in a closed experiment.

  297. carrot eater says:

    Ian Schumacher (06:47:03) :
    “I don’t think that is true. I believe at equilibrium emissivity and absorption must be equal. Kirchoff’s Law I think.”

    Be careful. We have been to some extent been glossing over this, but emissivity and absorptivity are functions of wavelength, and even direction. Because such spectral dependence complicates matters, undergraduate classes generally don’t concentrate on them.

    With some generality, you can say that at a given wavelength, the emissivity and absorptivity are equal. But remember that emissivity and absorptivity themselves can vary with wavelength. For the earth, we are looking at very different bands of wavelength: the earth is absorbing or reflecting visible light, while it is emitting infrared.

  298. P Wilson says:

    lgl (06:56:12)

    its assumed that the emission is a constant process, so that at some stage the shell will always be receiving and emitting 256w/m2. What happens in reality? They go into a stable state of thermal equilibrium where there is no change of energy flow and both remain constant at the temperature that accords with 256w/m2.

  299. carrot eater says:

    Ian Schumacher:

    I think my last comment might be a bit confusing, so let me put it thus:

    Say the earth has an absorptivity around 0.7 in the visible wavelengths. That does not tell you anything about the absorptivity in the infrared wavelengths. Therefore, it also does not tell you anything about the emissivity in the infrared wavelengths. Thus, there is no problem in using an emissivity of 1 for the relevant band of infrared. If the actual emissivity in the appropriate IR band is measured to be 0.97 to 0.99 ( Willis Eschenbach (10:41:34) : ), well, for the purposes of a tinkertoy model, using 1 is good enough.

  300. carrot eater says:

    Joel Shore (07:21:07) :

    That source really is too militant about semantics. For example,

    “Does the atmosphere trap radiation? No, the atmosphere absorbs radiation emitted by the Earth.”

    Is this really so necessary? “Trap” has no definitive physical meaning, but if a layman used ‘trap’ as a synonym for ‘absorb’, I would not get bent out of shape. Would you?

    Willis: I should commend you on one point: that your picture goes beyond an isothermal slab atmosphere. Many websites don’t really mention that the IR emission that makes it out to space is from colder parts of the atmosphere.

  301. lgl says:

    julien (01:27:30) :

    “Interesting model.
    Is there a way to re-design this shell system using an external radiation flow ?”

    Yes, replacing the shell with greenhouse gases.

  302. Roger Clague says:

    I welcome this post. This is a physical model of the whole system. It not based on the chemistry of molecules such as CO2.

    It is a model commonly used in climatology and is similar to that proposed by Professor Lindzen of MIT, as below.

    http://www-eaps.mit.edu/faculty/lindzen/198_greenhouse.pdf

    As many have said, it has problems.

    1. Its basic assumptions are not close to reality.
    2. Its logic does not follow it own fundamental assumption.
    3. It gives a poor result and is then modified.

    A thin metal shell is very different from a 30km layer of gas.
    Energy is created from no-where.
    85degreesF is way off.

    A better shell model is described here.

    http://freenet-homepage.de/klima/atmoseffect.htm

    The increased temperature is due to the pressure of the air. This is called Charles Law, discovered over 200 years ago. The fact that more matter alone can increase temperature is not something we experience. So we look for other explanations.

  303. P Wilson says:

    Thats not a one way process either, as if you increase air pressure in a chamber, for eg, the temperature increases above ambient according to the pressure. THEN, when the optimum temperature is reached, temperature declines even if pressure increases

  304. chris y says:

    Those posting demands for removal of this posting need to spend some time in the library, or in a lab with an operating vacuum furnace or <10K cryostat. There is considerable peerreviewedlitchurcher (copyright ClimateAudit) to back up the veracity of Willis' analysis. For example, this free article popped up after 12 seconds of effort with google-

    http://jpsj.ipap.jp/link?JPSJ/11/1184/

    Y. Ishikawa, S. Sawada, "Effective use of radiation and thermal shields in a vacuum furnace," J. Phys. Soc. Japan, 11 (November 1956), pp 1184-1190.

    They actually have experimental measurements of the efficacy of a stainless STEEL radiation shield in a vacuum furnace. Its nice to find measurements that verify theory.

    References therein go back even further into the annals of physics history.

    A. W. Lawson, R. Fano, Rev. Sci. Inst. 18, (1947), pp 727.
    A.E. De Barr, Rev. Sci. Inst. 19 (1948) pp 569.
    J. B. Garrison, A. W. Lawson, Rev. Sci. Inst. 19 (1948) pp 574.
    A. Braun und G. Busch, Helv. Phys. Act. 9 (1946), pp 33.

    They are known as radiation shields, and are commonly used in high temperature furnaces and cryogenic containers. In both cases, radiation losses are very important.

    Nice work, Willis.

  305. Fred Lightfoot says:

    Want to start this all over again ? ICE !!!!!!

  306. lgl says:

    P Wilson (08:02:49) :

    The system reaches equilibrium when the shell emits 256 W/m2 to space, and then it also emits 256 to the surface, 512 in total, which must all come from the surface, the only source. What I tried to describe was the transition phase, aften placing the shell.

  307. Charlie says:

    Ian Schumacher (06:40:44) : says

    In order for them to not be the same, we need a special hole. one that lets particles in more easily then it lets them out. I don’t believe such a thing exists in the physical world and as I and others have mentioned, this would be equivalent to Maxwell’s demon. A thought-experiment mischief maker that allows us to violate the second law of thermodynamics.

    It is my belief that there is no special filter that can be made for energetic particles of any kind.

    You may believe that there is no special filter that can be made for energetic particles of any kind, but your belief is wrong. As for a filter that lets in visible light but reflects infrared, these are quite common and ordinary. As several other posters have mentioned, they are called dichroic filters. They are also called dielectric mirrors, and sometime hot mirrors and cold mirrors. You will find them are some low voltage halogen bulbs. You may have noticed that most dentists have a lamp that is very bright, but without much heat. That’s because the lamp has filters built in that pass the visible light on to the patient work area, but allow the IR heat to escape out the backside of the lamp assembly.

    The atmosphere works in a similar manner. It allows in visible light with relatively low visible absorption, but absorbs most infrared. For a thought experiment such as Willis’s article, it is very reasonable to postulate a shell that allows the sun to deliver an average of 235 W/m^2 on the earth’s surface, but which totally absorbs (and then re-radiates) the infrared emissions from the heated surface of the earth.

    —————————————-

    It is clear that you have seen diagrams that illustrate blackbodies as hollow spheres with a small hole in the shell. Real blackbodies used for scientific experiments and measurement are often built in this manner. The reason is actually pretty simple. A blackbody is defined as having an albedo (reflection) of zero. No real life surfaces have a truly zero reflection. But if you have a hollow sphere with just a tiny entrance hole, then any light (or infrared) that enters through the hole will bounce around many, many times before managing to find its way back out of that tiny hole. This makes the effective albedo looking into the hole much lower than the albedo of any one point on the surface of the sphere.

    Since the albedo looking into the small hole is very close to zero, the radiation coming out of that hole will be very close to the theoretical predicted for a blackbody whose temperature is that of the inside of the hollow sphere.

    It appears that you are somehow conflating this method of emulating a blackbody with Willis’s though experiment where sunlight passes through his steel sphere without being absorbed.

    My recommendation to you it to, at least in the beginning, ignore the way that the sun gets to the surface. Just think about the even more simplified model where there is a isotope decay heat source at the center of the earth that generates sufficient energy to have a net outward radiative flux of 235 W/m^2 at the Earth’s surface.

  308. P Wilson says:

    I agree that its a good model: However, that piece of research Y. Ishikawa et al is about maintaining the furnace at 1300C, whereas the consensus here is that it will go way beyond 1300 because of these shields

  309. P Wilson says:

    lgl (08:34:09)

    Ah thanks but… its transmitting 256 in two directions, though perhaps not 512 in total. If that surface is 10C it re-radiates 10c worth to the source, and 10C away. Not quite the same as 20C total, as the temp is fixed at 10C.

  310. Vincent says:

    P Wilson wrote:
    “Vincent

    “However, some photons of energy from the cooler body will still strike the warmer body.”

    I’ll put my eggs in the fridge, and see if they are cooked tomorrow morning””

    Ask this question. In which scenario do the eggs cool faster – a) In a fridge or b) in space at absolute zero? That would be an analogy to Willis’s steel sphere. The presence of the sphere allows the planet to cool less quickly than if it was exposed directly to space at absolute zero.

    You dodged the question I posed at the end of my previous post. If a cooler body emits a photon in the direction of a warmer body, can you explain why that photon cannot then strike that body and impart its energy? If you cannot, then your reply about eggs in the fridge is just an argument from ignorance. If you do manage to give a credible explanation that I can verify, then I will accept that I was wrong and you were right. Deal or no deal?

  311. chris y says:

    Here is another practical implementation of a radiation shield.

    GE released an improved incandescent light bulb in the late 1990’s (I knew the inorganic chemist that developed the coating process at GE R&D) that contained a small cylindrical bulb that enshrouded the filament. This assembly was then contained within a regular glass envelope. The small glass cylinder was coated with a multilayer dielectric stack (up to 30 layers!) to reflect near-IR radiation back onto the filament while allowing visible light through, resulting in additional heating of the filament and a concomitant reduction in electrical power (up to 30%) required to maintain the same filament temperature. It won a greenie award from the State of New York back when Cuomo was governor. I don’t know if they are still being sold.

  312. Carrot eater,

    I think that Kirchoff’s law says emissivity and absorptivity are equal at all wavelengths and angles at equilibrium. That is my interpretation.

    Charlie,

    I do believe there are filters allow through different wavelengths. I don’t believe there are filters that can let photons through in one direction and ‘trap’ them so that energy builds up to higher density then the incoming photon stream. The photons will interact with the environment and some of them will become high energy and escape and at equilibrium the energy density of the incoming stream and the ‘trapped’ photons will be the same.

    Imagine I have a box with energetic particles of some kind bouncing around. If a one-way energy filter were to exist, then I put that filter in the middle of the box to split the box into two parts. Now our filter goes to work, letting high energy go one direction only. The box spontaneiously becomes warm on one-side and cold on the other.

    We KNOW that particular filter doesn’t exist, but somehow we think we can get ‘around’ this problem if we are dealing with photons and not molecules (for instance).

    What if we directed particles in through a small hole and had them bounce around. Wouldn’t that trap them? Only particles heading in very special directions could get out. Well yes, it does trap them, to a limit, but once the pressure builds up inside is equal to the pressure outside, we have reached equilibrium.

    With photons it is the same. Yes we might use a filter that rejects low energy light, but this is no different then our small hole before. Just as particles in our previous example will keep bouncing around to find their way out, photon energy will redistribute to get out the frequency hole as well. So yes our device would trap photons, but to a limit. Once photon energy density is equal to incoming energy density then the limit has been reached. This is the blackbody limit.

    Anyways I’m going to think about it some more. It really is a fascinating subject.

  313. carrot eater says:

    Ian Schumacher (09:16:30) :

    “I think that Kirchoff’s law says emissivity and absorptivity are equal at all wavelengths and angles at equilibrium. That is my interpretation. ”

    You are forgetting that emissivity and absorptivity themselves are not constant over all wavelengths. Please refer to a text that goes through the derivation, and explores the restrictions.

    The emissivity at a given wavelength is generally equal to the absorptivity at that same wavelength. (I’m assuming for now that angle doesn’t matter). So, if Willis wants to use an emissivity of 1 at the long infrared, then he would have to use an absorptivity of 1 at the long infrared as well. And in fact, he does. The earth absorbs ALL of the infrared emitted by the steel shell: there is no reflection term there.

    Willis does have less than perfect absorption by the Earth for visible light. This would imply an Earth emissivity of less than one in the visible range, but that doesn’t come up because the earth isn’t hot enough to visibly glow.

  314. P Wilson says:

    Vincent (09:04:39)

    Yes a photon from a cool object can hit a warm object – just like a raindrop could be thrown on a glacier, although the tendency is for thermal equilibrium. With Longwave radiation the tendency is for the cooler object not to increase the temperature of the warm object, and if they are at the same air pressure, and ambient temperature -take 2 identical apples, one at 8C the other at 15C an inch apart, in a room at 20C – the likely outcome is that each apple reaches equilibrium with the ambient temperature of its own entropy, as if they were not in each other’s presence. the limiting factor is ambient temperature.

    Eggs cool faster in space at absolute zero than in the fridge, but that doesn’t mean that cool fridges (eg at 7C) transfer heat that were put there at room temperature

  315. Carrot eater,

    My understanding was that e(f) = a(f) So yes e(f) is a function of frequency (or wavelength, whatever your preferred view), but Kirchoff’s law says at equilibrium then are equal at each frequency.

    So while, at first there is a difference (i.e. when the system is not at equilibrium), but once equilibrium is reached then they are equal at all frequencies. Again that is my interpretation from what I have read.

    To add to my energy density comment above. Energy density is a way to represent temperature. I like it better because temperature is very messy. Energy is conserved, etc, makes life a lot easier.

    If we were to take a snapshot of a volume of space capturing the stream of photons from the sun hitting earth. We could calculate what the energy density is in that region of space. This will have an associated equivalent temperature (for an ideal gas, etc). Inside the earth, there is also an energy density (with an associated temperature).

    The energy density in space from the sun can not be lower than the energy density in the earth atmosphere, otherwise energy is being transferred from a low energy density to a higher one. Or in other words, heat is being transferred from a low temperature region to a higher one. This can’t be done, no matter what kind of filters we put in place.

    Can it? Can energy density heating an object be less then the energy density of the object itself? At this time, I don’t believe so.

  316. Anton Eagle says:

    Okay, I can admit when I’m wrong. And I am.

    The radiative process described in the first model work as stated… with one caveat… W/m2 is not temperature, but is rather a consequence of temperature.

    I know its probably just a slip of the tounge, but posts have repeatedly stated that the model results in a doubling of the temperature (from 235 to 470). That is not the case, it results in a doubling of the energy flux. Temperature is the 4th root of this and so this results in a temperature increase of only about 18.9%. I’m sure this is well understood by the author (as he did correctly state this at one point).

    Also, although additional shells would increase the energy flux by 235 per shot (in this model), they increase the temperature by every decreasing amounts (as compared to the energy flux increase). For example, a second shell would increase the temp by an additional 10.7% (the energy flux increases by 50%), a third shell by a further 7.5% etc.

    So, looking at this from the point of view of actual temperature, it starts to look a lot more reasonable. The so-called “doubling” of the temp was throwing me off.

    Of course, the increases stated only work for an idealized blackbody, which doesn’t exist, and other losses would further mute the effects, but you’re right… the basic pricipal is sound.

    Sorry to have been so harsh in my (incorrect) rebutal. Please accept my apologies.

  317. Vincent says:

    P Wilson

    “Eggs cool faster in space at absolute zero than in the fridge, but that doesn’t mean that cool fridges (eg at 7C) transfer heat that were put there at room temperature”

    I think we agree then. I thought for a moment you were saying that Willis’ model was impossible because it violated the second law of thermodynamics.

  318. Hank Henry says:

    Greenhouse effect = picture-book science.

  319. DeWitt Payne says:

    Tom Vonk,

    Consider two parallel planes of infinite extent with different temperatures, Ta and Tb. Are you saying that the energy transfer per unit area between the planes cannot be described by the Stefan-Boltzmann equation: E=sigma*(Ta**4-Tb**4) because they are not hemispheres? Each point on the surface of either plane emits over 2pi steradians and also sees radiation coming from 2pi steradians. For a steel shell close to sphere with the radius of the Earth, a parallel plane approximation is pretty good.

  320. chris y says:

    P Wilson- “However, that piece of research Y. Ishikawa et al is about maintaining the furnace at 1300C, whereas the consensus here is that it will go way beyond 1300 because of these shields”

    That’s irrelevant when discussing the physics of the situation. The paper derives the general solution for n layers, then explores the solutions for the cases of 1 layer and 2 layers, including effects due to variations in spacing between layers and due to variations in emissivity. I suspect the general solution in the paper will also predict much higher source temperatures as n increases.

  321. lgl says:

    P Wilson (08:57:43) :

    Yes, 256+256=512, always. “re-radiates 10c worth” has no meaning, stick to energy, and see Anton Eagle (10:12:57) :
    B t w, thanks to Anton. We are all making errors all the time, it’s the cost of creativity.

  322. carrot eater says:

    Ian Schumacher (10:12:35) :

    “My understanding was that e(f) = a(f)”

    Yes (assuming angle doesn’t matter), and that’s all you need. Willis obeys this throughout. For the earth, e(IR range) = a(IR range) = 1. a(visible) < 1 so e(visible) would also be < 1, but it never comes up.

    I'm not sure where you're going with energy density. You're best off going back to the temperature of the two bodies. The sun is much hotter than the earth; thus NET radiation exchange will be from sun to earth, by a huge margin. No need to complicate things beyond that.

  323. yonason says:

    P Wilson (04:36:11) :

    Keep going with the analogy of making aluminum shields for lightbulbs.

    Observe that to come as close to possible to being isothermal, you would have to wrap the bulb with foil, (unfortunately you would then have to deal with substantial conduction, but of course we can ignore such rigorous treatment for now. I doubt anyone will notice).

    Now for the second half of the experiment, make that shield a foot radius from the bulb. (Note that the top will rapidly get very hot, which will be from convection, but again don’t worry that anyone will notice such a minor detail. If need be, we can draw a vacuum down on the room to prevent that.)

    While waiting for this new shield to come to thermal equilibrium with the bulb you might want to call out for a pizza, this could take a while, or, if you are really hungry, check the fridge to see if your eggs are done.

    When finished with the experiment, you can make party hats for everyone else from the excess foil. It seems they already have sufficient Kool-Aid.

    Cheers.

  324. Juraj V. says:

    Hmm looks like there is no scientific consensus even in the basic mechanism of “global warming” :-o

  325. Willis Eschenbach says:

    Anders L. (01:26:20) :

    But the interesting thing, really, is what happens when things do change. What happens when the system is NOT in thermal equilibrium, as is the case with our planet right now? According to Hansen et al 2004, the Earth is receiving 0.85 +- .15 W/m2 more than it is emitting into space. How fast will the surface temperature of the planet rise or fall as a response to shifting properties of the “shell” (for example, when the mix of greenhouse gases in the atmosphere is changed)?

    Hansen’s paper uses a cherry picked time frame and the usual Playstation™ models to get that number. I discussed this at the time, hang on … OK, thanks for waiting, it’s here.

    In short, there is no way to determine whether the earth is in or out of balance, except that at any given instant we can pretty much guarantee that it is out of balance.

  326. Willis Eschenbach says:

    julien (01:27:30) :

    Interesting model.
    Is there a way to re-design this shell system using an external radiation flow ?

    The analogy may be accurate, but it does not feel so at first glance, and the number of comments confirm this.

    See Fig. 2(a), which is the same as Fig. 1(b) except with external radiation flow.

  327. Willis Eschenbach says:

    yonason (03:43:14) :

    “In order to maintain its thermal equilibrium, the whole system must still radiate 235 W/m2 out to space…” – from paragraph just above fig.1

    In other words, by definition, the steel shell is arbitrarily assigned the same temperature as the earth? If so, then, …

    yonason, I can’t make any sense of this. Please boil it down to the essentials and resubmit it. Brevity is clarity.

  328. Gary Hladik says:

    Joel Shore (07:21:07), thanks for the “Bad Greenhouse” link. I like the way it compares the Earth’s atmosphere to the sun’s photosphere. In a sense, I suppose one can think of the sun as the “steel greenhouse” model on steroids.

    It also makes life here on our cold earth seem a bit more “sunny”. :-)

  329. Willis Eschenbach says:

    TomVonk (06:30:56) :

    Willis

    You wrote that you answered all critics but you didn’t mine .
    You stopped only at the question of units . …

    My apologies, Tom, but I find your writing impenetrable. Please clarify and simplify what your objections are, and I’ll be happy to answer.

    w.

  330. Willis Eschenbach says:

    Joel Shore, thanks for your assistance with the tsunami … much appreciated.

  331. Willis Eschenbach says:

    Anton Eagle (10:12:57) :

    Okay, I can admit when I’m wrong. And I am.

    The radiative process described in the first model work as stated… with one caveat… W/m2 is not temperature, but is rather a consequence of temperature.

    I know its probably just a slip of the tounge, but posts have repeatedly stated that the model results in a doubling of the temperature (from 235 to 470). That is not the case, it results in a doubling of the energy flux. Temperature is the 4th root of this and so this results in a temperature increase of only about 18.9%. I’m sure this is well understood by the author (as he did correctly state this at one point).

    Also, although additional shells would increase the energy flux by 235 per shot (in this model), they increase the temperature by every decreasing amounts (as compared to the energy flux increase). For example, a second shell would increase the temp by an additional 10.7% (the energy flux increases by 50%), a third shell by a further 7.5% etc.

    So, looking at this from the point of view of actual temperature, it starts to look a lot more reasonable. The so-called “doubling” of the temp was throwing me off.

    Of course, the increases stated only work for an idealized blackbody, which doesn’t exist, and other losses would further mute the effects, but you’re right… the basic pricipal is sound.

    Sorry to have been so harsh in my (incorrect) rebutal. Please accept my apologies.

    Sir, you are a gentleman. This is how science works. We throw our ideas out and let people try to find flaws with them. If they find flaws, we nod our heads, say “Dang … wrong again. Not the first time,” and move on. I can’t count how many times I’ve been wrong, that’s how my own scientific understanding moves forwards.

    One caveat. You say:

    W/m2 is not temperature, but is rather a consequence of temperature.

    One is not a consequence of the other. Both are just different ways of measuring the same thing, which is the average speed of the molecules involved. We could also measure it in meters/second. None of these are “a consequence” of the other. They are different ways of expressing the same thing.

    My thanks to you for your honesty and style. Well done.

  332. SAM says:

    Willis – Nice article!

    I’ll admit that as I read the article, I had similar reservations regarding a seeming confusion between energy balance vs. energy flux balance and the radii impact. I think you sufficiently explained your reasoning in the followup comments, but I do wonder if it is prudent not to generalize on assumed small radii differences.

    Most readings I have come across use W/m2 to illustrate earth’s energy budget analysis, similar to your Figure 4. But a radii difference within atmospheric altitudes can give relevant W/m2 deviations.

    As an example (and please doublecheck calculations for yourself) I refer to a statement by Anders L. (01:26:20) that quotes a max 1 W/m2 imbalance from Hanson et al. While I believe your explanation is very reasonable, I would like to note that 1W/m2 out of 492W/m2 surface flux (Fig4) can be offset by an altitude of approximately 20K ft. So, when I see energy balance charts such as fig4 given in units of W/m2 and attempting to tally in/out numbers from the surface, clouds, and space, I cannot help but question how anyone can assert an imbalance of 1 W/m2.

  333. Willis Eschenbach says:

    Juraj V. (11:16:41) :

    Hmm looks like there is no scientific consensus even in the basic mechanism of “global warming” :-o

    Most scientists have a pretty good handle on it. I published the “Steel Greenhouse” to clarify the mechanism, and to point out that the usual “single shell” model used in innumerable places is inadequate to model the earth.

  334. lgl says:

    I have always wondered why we can’t concentrate LW the way we do with solar SW. Is it because the surface can’t be made smooth enough or what? I thought the main thing with a Thermos was just that, reflecting the (LW) radiation back inside.

  335. Willis Eschenbach says:

    SAM (12:56:40) :

    Willis – Nice article!

    I’ll admit that as I read the article, I had similar reservations regarding a seeming confusion between energy balance vs. energy flux balance and the radii impact. I think you sufficiently explained your reasoning in the followup comments, but I do wonder if it is prudent not to generalize on assumed small radii differences.

    Most readings I have come across use W/m2 to illustrate earth’s energy budget analysis, similar to your Figure 4. But a radii difference within atmospheric altitudes can give relevant W/m2 deviations.

    As an example (and please doublecheck calculations for yourself) I refer to a statement by Anders L. (01:26:20) that quotes a max 1 W/m2 imbalance from Hanson et al. While I believe your explanation is very reasonable, I would like to note that 1W/m2 out of 492W/m2 surface flux (Fig4) can be offset by an altitude of approximately 20K ft. So, when I see energy balance charts such as fig4 given in units of W/m2 and attempting to tally in/out numbers from the surface, clouds, and space, I cannot help but question how anyone can assert an imbalance of 1 W/m2.

    The paper by Hansen asserting a 0.85 W/m2 imbalance is a joke, in my opinion. Even with satellites, we cannot determine either incoming or outgoing radiation to anywhere near that accuracy. Don’t know if you took a look at my citation above, but it discusses this question as regards Hansen’s paper.

    Climate models are of no help in this regard. Their errors are legion, and large. Here’s an example, from the people who wrote the model:

    2.4 Principal Model Deficiencies

    Model shortcomings include ~25% regional
    deficiency of summer stratus cloud cover off the west
    coast of the continents with resulting excessive absorption of
    solar radiation by as much as 50 W/m2, deficiency in absorbed
    solar radiation and net radiation over other tropical regions by
    typically 20 W/m2, sea level pressure too high by 4-8 hPa in the
    winter in the Arctic and 2-4 hPa too low in all seasons in the tropics,
    ~20% deficiency of rainfall over the Amazon basin, ~25%
    deficiency in summer cloud cover in the western United States
    and central Asia with a corresponding ~5°C excessive summer
    warmth in these regions. In addition to the inaccuracies in the
    simulated climatology, another shortcoming of the atmospheric
    model for climate change studies is the absence of a gravity
    wave representation, as noted above, which may affect the nature
    of interactions between the troposphere and stratosphere.

    The stratospheric variability is less than observed, as shown by
    analysis of the present 20-layer 4°×5° atmospheric model by
    J. Perlwitz (personal communication). In a 50-year control run
    Perlwitz finds that the interannual variability of seasonal mean
    temperature in the stratosphere maximizes in the region of the
    subpolar jet streams at realistic values, but the model produces
    only six sudden stratospheric warmings (SSWs) in 50 years,
    compared with about one every two years in the real world.

    Hansen et. al, Climate Simulations 1880-2003

    Given Hansen’s own admission of 50 W/m2 errors, the idea that his model can detect a 0.85 W/m2 imbalance is … well … I’ll call it “extremely improbable” and leave it at that.

  336. FOIA says:

    [A lot is happening behind the scenes. It is not being ignored. Much is being coordinated among major players and the media. Thank you very much. You will notice the beginnings of activity on other sites now. Here soon to follow. ~ ctm]

  337. yonason says:

    Willis Eschenbach (11:24:58) :

    It’s clear enough. At “thermal equilibrium” the temperatures are all the same (see link I provided to definition).

    The temperature of the shell is a “boundary condition” that you are choosing to solve your equation. That and the term “isothermal” are arbitrary because, while necessary to get the result you want, they are not consisstent with the initial conditions you’ve already given, or the final state that will evolve from those initial conditions after the addition of the shell. You already have enough information to solve for the temp of the shell, which will not come to the same as the earth’s initial temp., unless more heat is added. In fact, even then they will not both be the same, so it will never be isothermal.

    The earth has, like P Wilson’s analogy of the lightbulb, a fixed output as postulated by your initial conditions. It cannot heat the iron shell to the same temp., without adding even more heat, but that violates your initial condition of constant output. It will also drastically raise the temp., of earth, thereby violating your condition that the system be isothermal.

    While you are correct that the rate of heat emitted per unit area is the same for all perfect blackbodies at the same temperature, a larger surface area will radiate more total energy. It has to come from somewhere (conservation of energy). That’s why I looked at the Net Power, which is an absolute measure of available energy. Keep it constant, and the shell never warms to the same as the initial temp., of the earth, and the earth warms only slightly because the presence of the shell slows down the efflux of heat (it’s an insulator).

    You are looking at emissivity, which is not a measure of total output, which is why your accounting doesn’t balance. Naturally the earth’s temp., will go up drastically if you add as much energy as needed to bring the shell to the same initial temp., of the earth. But it doesn’t come from the earth, which may be why John A made his comment about Maxwell’s demon?

    Reconsider the shell and do the following thought experiment. Keep increasing it’s distance from the earth, and ask yourself what will need to happen in order to maintain it at a constant temp., and then ask where that energy comes from.

    Summing up: Your drastic increases in earth’s temp., appear to be an artifact of improper boundary conditions and incorrect assumptions.

  338. carrot eater says:

    lgl (13:03:55) : “I have always wondered why we can’t concentrate LW the way we do with solar SW.”

    “Concentrate” meaning what?

    Willis Eschenbach (12:38:02) :

    “None of these are “a consequence” of the other. They are different ways of expressing the same thing.”

    Well, if it’s blackbody emission, yes. But if you don’t know the emissivity, then you can’t go from the total emitted flux to temperature.

    TomVonk (06:54:18) :

    You’re overly worried about the solid angle. Yes, the flux found from S-B is integrated over all directions. And this is perfectly fine: all of the energy emitted by the earth ends up at the first shell. All of the energy emitted by the inner surface of his first shell ends up at the earth. No need to worry about angles or directions or solid angles; the view factor is 1 in both directions.

  339. John A says:

    Curt:

    ohn A: You are completely and utterly wrong. You would get flunked out of any undergraduate engineering thermodynamics or heat transfer course. If I were your instructor in such a course (and I have taught courses with this subject matter), and you kept maintaining these views, not only would I flunk you, but I would be having a conversation with the dean about how the hell you got into the program in the first place.

    Very touching. Unfortunately for you I took the course and passed it. I would be asking the Dean why someone like you was arguing against results known and reproduced since the 19th Century, and why such a crackpot is on the faculty.

    Bodies radiate based on their emissivity and their temperature alone, without regard to what they are radiating towards. They have no way of “knowing” what they are radiating towards. This is easily demonstrable in a lab. Put two plates of different temperatures parallel to each other, and you can still measure the radiative energy from the colder plate with a radiometer.

    Duh. I did not make the statement that the cooler objects cannot be detected by warmer objects. I made it clear that all objects above absolute zero radiate.

    A radiometer that is warmer than the device radiating towards it can still measure the radiation from that object — according to your arguments, that would be impossible.

    Duh. See above.

    In any introductory engineering heat transfer text, you will see that the net radiative heat transfer between two objects (1 and 2) is given as:

    K * (T1^4 – T2^4)

    where K is a system constant incorporating the emissivities and geometric properties of the two object. This expression automatically incorporates the fact that the cooler body is radiating towards the warmer body, albeit less than the warmer body is radiating towards the cooler body.

    Duh. Except that the laws of mathematics would have me point out that T1 must be greater than T2 otherwise the net radiative heat transfer is negative.

    The heat flow is one way from hotter to colder, which is exactly the point I’ve continued to make, consistent with the laws of thermodynamics.

    By your argument, this equation is completely wrong. Your claim is that the temperature of the cooler body is irrelevant, and that the net heat transfer from the warmer body is the same whether the cooler body is 0.001C colder or at absolute zero.

    I make no such claim. I have made no such claim. I pointed out that cooler objects do not warm warmer objects because the net radiative heat transfer would be negative and because it would imply that the cooler object spontaneously loses entropy without doing any work.

    If you really believe your argument, you should not be spending your time on blog wars like these. You should be doing whatever it takes to shut down any system that was designed using the expression I gave above (which is ever power plant, every power electronic system, among many others), because according to you, they were designed under completely fallacious assumptions, and therefore could be totally unsafe.

    Unfortunately for you, all power plants, power electronic systems blah blah blah obey the laws of thermodynamics.

    SImple experiment: put a sample of a radioactive element (which has a measured temperature) inside of a Dewar flask and measure if the temperature of the radioactive element rises as a result of radiative reflection from the inner walls of the flask.

    A Nobel Prize awaits if it does.

    But I’m not holding my breath.

    Hence or otherwise, grow up. Crack a book on introductory thermodynamics and read. You may be pleasantly surprised at how much you can learn from a book.

  340. Willis Eschenbach says:

    yonason (13:18:36) :

    Willis Eschenbach (11:24:58) :

    It’s clear enough. At “thermal equilibrium” the temperatures are all the same (see link I provided to definition).

    The temperature of the shell is a “boundary condition” that you are choosing to solve your equation. That and the term “isothermal” are arbitrary because, while necessary to get the result you want, they are not consisstent with the initial conditions you’ve already given, or the final state that will evolve from those initial conditions after the addition of the shell….

    I’m sorry, but this is still not “clear enough” at all. You have asserted, not shown but asserted, that at equilibrium all the temperatures are identical.

    You keep saying no, it doesn’t work out like that … but simply saying so doesn’t advance the discussion. For example, you say “That and the term ‘isothermal’ are arbitrary …”, but since I haven’t used the term “isothermal” in this entire thread, I haven’t a clue what you are talking about.

    Try to boil your assertions down to a single paragraph where, instead of assertions, you pose questions that point to what you think I’ve done wrong. That might help.

    Best regards,

    w.

  341. Willis Eschenbach says:

    John A (13:35:21) :

    Curt:

    In any introductory engineering heat transfer text, you will see that the net radiative heat transfer between two objects (1 and 2) is given as:

    K * (T1^4 – T2^4)

    where K is a system constant incorporating the emissivities and geometric properties of the two object. This expression automatically incorporates the fact that the cooler body is radiating towards the warmer body, albeit less than the warmer body is radiating towards the cooler body.

    Duh. Except that the laws of mathematics would have me point out that T1 must be greater than T2 otherwise the net radiative heat transfer is negative.

    The heat flow is one way from hotter to colder, which is exactly the point I’ve continued to make, consistent with the laws of thermodynamics.

    You are still ignoring the difference between heat flow and net heat flow. The net heat flow is one way, as you correctly point out. But as the equation clearly shows, it is the net of two individual heat flows in two different directions. See Fig. 1 for an example. The net heat flow is from the planet to the shell. But that is made up of two individual heat flows, one from the planet to the shell, and one from the shell to the planet.

    SImple experiment: put a sample of a radioactive element (which has a measured temperature) inside of a Dewar flask and measure if the temperature of the radioactive element rises as a result of radiative reflection from the inner walls of the flask.

    A Nobel Prize awaits if it does.

    Of course it will rise. The temperature of anything will rise until the system loses heat at the same rate that heat is added. Since the radioactive element is constantly producing heat at a steady rate, the temperature inside the Dewar will increase until the loss from the outside of the Dewar is equal to the heat produced inside the Dewar. But since the Dewar slows the heat loss, the contents of the Dewar must be much warmer for the outside of the Dewar to radiate the heat produced.

    Until then, the heat lost by the entire system will be less than the heat produced, so the temperature inside the Dewar must rise.

    Hence or otherwise, grow up. Crack a book on introductory thermodynamics and read. You may be pleasantly surprised at how much you can learn from a book.

    We are here to attack scientific ideas, not individuals. If you want to convince us that you are right, attacking the person you are discussing an idea with is bad tactics, as well as bad manners.

    w.

  342. lgl says:

    carrot eater (13:23:57) :

    “Concentrate” meaning what?

    Hm.. perhaps focus works better in english, using a parabola for instance.

  343. Carrot eater,

    “I’m not sure where you’re going with energy density. You’re best off going back to the temperature of the two bodies. The sun is much hotter than the earth; thus NET radiation exchange will be from sun to earth, by a huge margin. No need to complicate things beyond that.”

    The sun is way hotter yes, but it is also very far away. The energy flux from it has decreased significantly by the time it reaches us. Imagine there was an equivalent energy source just outside our atmosphere that generated heat identical to the heat from the Sun when it reaches our atmosphere. What would be the equivalent temperature of this source? That is the hottest we can get then, not the temperature of the sun 8 light minutes away.

    That’s why the focus on energy density. Energy density just outside the atmosphere, just before the light enters the atmosphere is the upper limit of the average energy density on earth.

  344. lgl,

    Long wave radiation can be concentrated also. A microwave dish for receiving microwaves from satellites is an example.

  345. carrot eater says:

    John A: Pick out a single instance in the Figures above where net heat flux is going from a cold object to a warmer one. If you are unable to do so, then I suggest toning down the attitude.

  346. Willis Eschenbach says:

    lgl (13:03:55) :

    I have always wondered why we can’t concentrate LW the way we do with solar SW. Is it because the surface can’t be made smooth enough or what? I thought the main thing with a Thermos was just that, reflecting the (LW) radiation back inside.

    Lenses can concentrate longwave, applications are common.

  347. lgl says:

    Ian Schumacher (14:05:39) :

    But I can’t point a disk to the ground and focus the 400 W/m2 (avg. yes) emitted to boil water.

  348. SteveBrooklineMA says:

    Willis-

    I worked out steady-state formulas that allow for making calculations in your two-layer model without iteration. In case you are interested, they are in here:

    http://home.comcast.net/~stevehaker/GlobalWarmingCalculator/Willis1.xls

    I was thinking of using these formulas and your two-layer model to extend this single-layer “Global Warming Calculator” I made a couple of years ago:

    http://home.comcast.net/~stevehaker/GlobalWarmingCalculator

    In any case, thanks for your post. I am impressed at how many comments you have replied to!

  349. lgl says:

    Willis Eschenbach (14:16:08) :

    So I can point a 3 m2 lense to the 15 C warm ground and get 1 kW in the focal point. Don’t think so.

  350. lgl,

    Oh, I see what you mean. Yes there is the question of coherence. If you have energy that is ordered in nature (same wavelength, same direction, that kind of thing) this is also know as coherence. If it has a predictable nature of some sort, then this can be exploited to concentrate it. If it is all in the same direction (for example) then your parabolic dish will work.

    If energy is random (goes in random directions, random wavelengths following typical heat distribution) then it does not have coherence and is now just ‘heat’. Heat can not be focused.

    This applies to short-wave or long-wave. Light from the sun is all coming from the same direction. This coherence can be used to easily focus it.

  351. carrot eater says:

    Ian Schumacher (14:04:30) :

    Eh. What you’re getting at is already captured in the 235 W/m^2 irradiance mentioned above.

    Perhaps you’ve already done this, but to drive the point home: while we have satellites that measure the solar irradiance, we can easily calculate an estimate.

    In fact, anybody curious can try this:
    Take the surface of the sun to be 5778 K, and treat the sun as a blackbody.
    Look up the radius of the sun, and the distance between sun and earth. You will NOT need the radius of the earth, though you do assume it’s a perfect sphere.
    Take the absorptivity of the earth to be 0.7.
    From all that, I get 239 W/m^2 from the sun, absorbed at the earth’s surface. Pretty close to the 235 W/m^2 used by Willis.

  352. SteveBrooklineMA,

    You could also replace the two layers with a partially reflecting mirror. Set how much it reflects to whatever you want to make it equivalent to 1 layer, 2 layes, 10 layers .. even infinite layers.

    Even simpler ;-)

  353. Willis Eschenbach says:

    SteveBrooklineMA (14:30:01) :

    Willis-

    I worked out steady-state formulas that allow for making calculations in your two-layer model without iteration. In case you are interested, they are in here:

    http://home.comcast.net/~stevehaker/GlobalWarmingCalculator/Willis1.xls

    I was thinking of using these formulas and your two-layer model to extend this single-layer “Global Warming Calculator” I made a couple of years ago:

    http://home.comcast.net/~stevehaker/GlobalWarmingCalculator

    In any case, thanks for your post. I am impressed at how many comments you have replied to!

    Sweet as! Very nice, thanks. I’ll see if I can’t extend it.

    Outstanding. Science moves forwards.

    w.

  354. carrot eater says:

    Ian Schumacher (14:37:41) :

    I’m not following you here. Collimators exist; they exist for infrared, as well. They’ll get all your rays nice and parallel. So?

    I don’t follow the motivation here. Are you guys trying to use the IR given off by the earth’s surface (or your house, or body, or whatever else is handy) to do something useful, like work, or boil water? Is that the motivation?

    The word ‘heat’ should be purely reserved for the transfer of thermal energy.

  355. lgl says:

    Ian Schumacher (14:37:41) :

    See, thanks.

    carrot eater (14:57:04) :

    Yes, that was the motivation.

  356. Willis Eschenbach says:

    lgl (14:37:27) :

    Willis Eschenbach (14:16:08) :

    So I can point a 3 m2 lense to the 15 C warm ground and get 1 kW in the focal point. Don’t think so.

    I just showed examples of practical devices that can focus IR using lenses. Others have given examples of practical devices that can focus IR using reflectors. Given that those devices exist, I’m not sure what your point is here. Are you saying that IR cameras don’t really exist, or that they can’t focus IR, or what?

  357. carrot eater,

    “Eh. What you’re getting at is already captured in the 235 W/m^2 irradiance mentioned above.”

    Right! and how hot would a black body just outside our atmosphere have to be to generate this heat?

    I want the value outside the atmosphere so that would be 341 or so.

    (341/sigma)^.25 = 279 K

    So a black body just outside our atmosphere that could be used as a source instead of the sun to produce the same energy on earth would be 279K.

    Anyways I suspect I’m not convincing you and I appear incredible daft and I accept that :-) Think about this though. The earth can not use the sun to heat up to inifinity. The earth can not use the energy from the sun to become hotter than the sun. The earth is far away from the sun and the energy from the sun is spread out and the flux much much lower, so ….

    What is the limit of the temperature of the earth?

  358. carrot eater,

    You can’t focus heat. if you could you could transfer heat from a cold source (just concentrate to make it hotter) to hotter source.

    You can focus IR, yes, but only if it has coherence i.e. it is going a specific direction, and other forms of ‘order’.

  359. carrot eater says:

    Ian Schumacher (15:14:26) :

    Again, please never use the word heat to mean anything but the transfer of thermal energy. I’m not even sure what ‘focusing heat’ would even mean, so let’s just never use that term. Seems like you agree with me on that.

    lgl (15:05:33) :

    Now, the second law is FINALLY going to bite. What is the temperature of the earth’s surface? What is the boiling point of water? I don’t care what you do, you’re not going to boil that water. The earth’s surface is simply too cold to be useful, unless you’re in Iceland or something and you’ve got a geothermally heated spot.

    If you dig a hole in the ground and find some rocks at a high temperature, THEN you can do it. That’s the point of geothermal energy.

  360. yonason says:

    Willis Eschenbach (13:39:11) :

    I’m not making assertions, I’m citing a definition.

    thermal equilibrium
    The condition under which two substances in physical contact with each other exchange no heat energy. Two substances in thermal equilibrium are said to be at the same temperature.”

    http://www.thefreedictionary.com/thermal+equilibrium

    At thermal equilibrium, (T1^4-T2^4) = 0. There will be no net heat exchange, BY DEFINITION not assertion. Since radiant energy flux is directly proportional to that quantity, it will also = 0. By definition.

    Also, energy in must equal energy out. If the flux per unit area coming out of a body of larger area is equal to the flux per unit area of a smaller body, there’s more energy coming out of the lager, in this case the shell. But since we started with a fixed energy flux, there can be no extra.

    The shell cannot be as warm as the earth, and the earth cannot absorb energy from a shell which is at a lower temperature. However, since the shell is at a higher temp than background, though probably not by a lot, (background ~3 DegK, which is effectively = 0 for this problem), less heat would be transferred to the shell, and the earth would consequently retain more of it and so it would warm slightly.

    If that doesn’t do it for you, then I don’t know what else to say.

    regards

  361. yonason says:

    P.S., you’re the one who insisted the system be at thermal equilibrium, not be.

  362. carrot eater says:

    Ian Schumacher (15:09:47) :

    You’re up to some odd mischief here. First, your heat source is colder than the earth’s current surface temp; that’s no good. Your heat source also seems to be one of Willis’s shells, wrapped all the way around the earth – not like the sun, at all. Maybe that’s what you intended, I don’t know. You’ve probably lost the reflected term, too. So whatever you’ve got with this weird new heat source, it’ll have a different result from the sun. Maybe that’s what you were getting at?

    What is the limit of the temperature of the earth? That’s maybe interesting. Let’s leave the sun right where it is, and let’s leave it as a sphere. The minimum temperature is what Willis calculates above, in the case of no greenhouse. Assuming that an atmosphere-less earth would have the same albedo, at least.

    The maximum temperature? That’s the more interesting question, and perhaps that’s where you’ve been steering me. Hmm. We can make a stronger and stronger greenhouse, but is there a thermodynamic limit? Clearly, we have to stay below the surface temperature of the sun, but is there a thermodynamic limit below that point? If there is, it isn’t obvious to me at the moment. That’s maybe the first interesting question in this whole thread.

  363. Carrot eater,

    Yes my numbers are wrong. Was just using the numbers you supplied for illustration there there is an equivalent temperature and it is lower.

    Yes, for a point source use 1366W/m^ or whatever. The basic idea is the same and the idea I’ve been trying to get people to think about (but very poorly I agree). There is a maximum temperature, it isn’t infinitity. So what is it?

    I’ve very glad that you are now on the case thinking about that. :-)

    Now I don’t feel like a lone lunatic (ok just less so).

  364. carrot eater says:

    yonason (15:52:45) :

    Where ever you see Willis use the word ‘equilibrium’, go ahead and substitute in ‘steady state’.

    Does that alleviate your difficulty?

    Nobody is implying that the earth, the sun and the steel shell are all at the same temperature.

  365. carrot eater,

    It’s clear I think the maximum temperature is blackbody temperature of course. I think maybe I focused too much on that. Fine, if it isn’t blackbody, what is it?

  366. carrot eater says:

    Ian Schumacher (16:12:50) :

    I see what you’re upto. Let’s agree to leave the sun where it is, as moving it or changing its shape is an unneeded diversion.

    We can say without a doubt that no matter what we do, the earth must stay below 5788 K or whatever the temp of the solar surface is.

    If we continue with the Willis cartoon, and continue piling on ‘steel’ shells, you could keep doing it until you approach the sun’s temperature, I guess. The ‘steel’ would of course melt and the whole thing is preposterous, but the cartoon lets you do it, doesn’t it?

  367. carrot eater says:

    Ian Schumacher (16:30:53) :

    “It’s clear I think the maximum temperature is blackbody temperature of course.”

    I don’t understand what you mean by that.

  368. Willis Eschenbach says:

    yonason (15:46:46) :

    Willis Eschenbach (13:39:11) :

    I’m not making assertions, I’m citing a definition.

    “thermal equilibrium
    The condition under which two substances in physical contact with each other exchange no heat energy. Two substances in thermal equilibrium are said to be at the same temperature.”
    http://www.thefreedictionary.com/thermal+equilibrium

    At thermal equilibrium, (T1^4-T2^4) = 0. There will be no net heat exchange, BY DEFINITION not assertion. Since radiant energy flux is directly proportional to that quantity, it will also = 0. By definition.

    Also, energy in must equal energy out. If the flux per unit area coming out of a body of larger area is equal to the flux per unit area of a smaller body, there’s more energy coming out of the lager, in this case the shell. But since we started with a fixed energy flux, there can be no extra.

    The shell cannot be as warm as the earth, and the earth cannot absorb energy from a shell which is at a lower temperature. However, since the shell is at a higher temp than background, though probably not by a lot, (background ~3 DegK, which is effectively = 0 for this problem), less heat would be transferred to the shell, and the earth would consequently retain more of it and so it would warm slightly.

    If that doesn’t do it for you, then I don’t know what else to say.

    regards

    carrot eater has pointed to what I think may be the source of the confusion. I have used the “thermal equilibrium” to indicate a state where there is no further change of temperature with time. Or as Wikipedia says, “Thermal equilibrium is when a system’s macroscopic thermal observables have ceased to change with time.”

    You are using it, on the other hand, to describe a state where all parts of the system have the same temperature. carrot eater is correct that “steady state” might be less subject to misinterpretation.

    Does this help?

  369. carrot eater,

    I believe that distance from the sun (and the lower flux as a result) must also be a factor. So the maximum temperature will be way way below 5788K. Here is my reasoning (not really ‘science reasoning’, but a kind of gut feel ‘bounds’ check.

    There is a maximum temperature a greenhouse effect can accomplish. Is it likely that this maximum is the same temperature for Mercury as it is for Pluto? No, I don’t think so. That just doesn’t seem possible. Mercury is closer, the flux from the sun is higher. I feel this must matter. If it doesn’t then theoretically we could warm ourselves from distance stars! Not too likely. Distance matter. Flux intensity matters. Being far away from the Sun means a maximum temperature much less than the sun.

    Willis allows for infinity yes. Not a fault really as an approximation, but not useful for figuring out what the limit is. If this limit matters (is low) then Willis model might be of little use. Everyone but me is convinced the limit is high, so … no worries there yet ;-)

  370. I believe the maximum temperature is defined as the blackbody temperature of earth that would give a (earth surface area) flux to balance out the (earth cross section) received by the sun flux of 1366W/m^2 . Sorry if that part wasn’t clear.

    Basically I think the maximum temperature of the Earth is the same as if the Earth was a super dark block of matter sitting there.

    On a different note, (but somewhat important). Many point out “black body isn’t warm enough since the average temperature of the earth is already higher than this”. Maybe, but how is this average temperature determined? We don’t actually just take temperatures and average them do we? If we do, that clearly give one the wrong impression since temperature isn’t conserved. Averaging temperature is kind of meaningless. We would need to take the fourth root of the average of T^4 … or something at least to getting closer to a meaningful number than a dumb average. Anyways just a curiosity. The stated average is higher than blackbody .. sure, but does this stated average have real meaning? or is it just a bogus? At the moment, I have no idea, but of course I suspect bogus ;-)

  371. carrot eater says:

    Ian Schumacher (17:04:20) :

    Do you want to switch from the steel shells to an actual atmosphere now, held by the Earth’s gravity? At some point, the steel shells are not useful anymore; I could have put a billion of them up there, going up to the surface of the sun.

    In either case, I’ve got work to do, so I might leave you to ponder it.

  372. Ian Schumacher says:

    Instead of steel shells, I think it is easier to just have a mirror (a single shell with a mirrored inner surface). This will get the maximum possible. It’s equivalent to infinite shells. The mirror can be for a frequency range ie. (transparent to high frequency, reflect low frequency). That is the simplest and seems to me to be a model that would achieve the maximum greenhouse effect.

    Yes for sure, this isn’t suppose to be our day job.

    Once you start to look into it you see ‘ok’ I need to integrate the spectrum on incoming light and that has to balance the integration of the spectrum of outgoing light over the same region. That’s what I got to. That give you a pretty high temperature that is dependent on the cut-off frequency. BUT …. then I started to realize that the spectrum will not be static. It will continuously readjust itself and the high frequency transparent region will continously be repopulated. That was my thinking anways. Maybe you will conclude differently. I’m subscribed (rss) so if you post something a week from now I’ll take a look and comment.

    Cheers.

  373. yonason says:

    Willis Eschenbach (16:54:34) :

    “Thermal equilibrium is when a system’s macroscopic thermal observables have ceased to change with time.”

    You are using it, on the other hand, to describe a state where all parts of the system have the same temperature.”

    But that’s really just the same thing. (I know it sounds general enough to apply to “steady state” as well, but using it that way can be very confusing, and besides, that wasn’t the intended use.)

    “It is observed that a higher temperature object which is in contact with a lower temperature object will transfer heat to the lower temperature object. The objects will approach the same temperature, and in the absence of loss to other objects, they will then maintain a constant temperature. They are then said to be in thermal equilibrium. Thermal equilibrium is the subject of the Zeroth Law of Thermodynamics.
    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

    Anyway, so you meant “steady state,” and you invoked it in order that “…the whole system must still radiate 235 W/m2 out to space.

    But there is still the problem of violating the First Law. Since the steel shell has a larger radius, it’s surface area is larger than that of the earth, and so for the TOTAL ENERGY emitted to remain constant, as required by the First Law, the shell must radiate less W/m2 than did the earth, so that when multiplied by the shell’s larger surface area, the total energy emitted still comes out the same.

    I.e., the system still has to radiate the same amount of heat to space, but since it is now doing it through a greater surface area, the flux per unit area must decrease, otherwise energy would have been created, which violates the first law.

    “* Energy can neither be created nor destroyed. It can only change forms.

    * In any process in an isolated system, the total energy remains the same.

    * For a thermodynamic cycle the net heat supplied to the system equals the net work done by the system.

    The First Law states that energy cannot be created or destroyed; rather, the amount of energy lost in a steady state process cannot be greater than the amount of energy gained.”
    http://www.answers.com/topic/laws-of-thermodynamics

    The earth will warm either until the total energy output per unit time = total energy produced per unit time, or until the heat source burns itself out. But that warming depends on how good an insulator the shell is, which depends in part on it’s thickness. It could be much hotter on the inside than on the out, because heat transfer through metals is via conduction, and so in this case the outside would be colder than inside, unless it was very thin, in which case it would pose a minimal barrier, and the earth wouldn’t have to warm very much at all. Also, if it’s a good reflector, heat will have to build up more inside than if it weren’t, but not because it’s radiating energy. Radiation and reflection are different processes.

    Note – your initial example of the steel shell didn’t involve any energy source other than whatever “nuclear” reactions might be generating heat in the earth, so I’m not including the sun, yet.

  374. Joel Shore says:

    Willis Eschenbach says:

    Joel Shore, thanks for your assistance with the tsunami … much appreciated.

    No problem!

    The paper by Hansen asserting a 0.85 W/m2 imbalance is a joke, in my opinion. Even with satellites, we cannot determine either incoming or outgoing radiation to anywhere near that accuracy. Don’t know if you took a look at my citation above, but it discusses this question as regards Hansen’s paper.

    Climate models are of no help in this regard. Their errors are legion, and large. Here’s an example, from the people who wrote the model:

    I think I will have to disagree with you here though (although I haven’t been able to follow the link because it seems really slow tonight). I don’t disagree with your statement that neither satellites nor climate models have the accuracy to directly detect such a small imbalance. However, as I understand Hansen’s paper, the imbalance is being detected instead by observing the amount of heat that is building up in the system due to the imbalance (and, nearly all of this heat buildup is in the oceans). So, even though we may not be able to measure the input and output fluxes with sufficient accuracy, we can infer their difference through the buildup of heat energy. (Not to say that there aren’t other potential issues, such as whether the heat content change was measured over a long enough period to give a result to that accuracy. But, I don’t think the fact that models or satellites can’t resolve the fluxes to this accuracy is not relevant.)

  375. carrot eater says:

    yonason (18:55:21) :

    You aren’t the first person to bring this up; it’s been addressed above. Willis was lazy and neglected the difference in surface area, just so he could keep using fluxes. Not a big deal. Assuming the shell is close to the earth, the error is small. If you really want, redo the math with a different radius. Keep in mind that this cartoon works no matter how close the shell is to the earth. So put the shell 1 centimeter above the earth’s surface. The difference in area is trivial.

  376. DeWitt Payne says:

    Joel Shore,

    All recent measurements of ocean heat content show a rate of increase of about half of Hansen’s 0.85 W/m2 at worst and probably more like 0.25 W/m2 over the last two decades. Since 2003, OHC has been increasing even more slowly, possibly not at all. Maybe there’s some loss to the deep ocean, but the usual estimate of the proportion of heat transfer I’ve seen is ~90% to the upper ocean. Hansen’s estimate has not been upheld by measurement.

  377. Denny says:

    Mr. Eschenbach,

    I have a question but I think it “might” be related in the explaination of “different energies”. I state different in that there’s “sensible temperature” and “radiant temperature”. I’m going to use my Wood/Coal addon in my House as an example.

    I burn wood mainly because I don’t have to pay for it. I love wood heat because it’s a “Bone Warming” heat and I was raised with it on the Farm. I hope you’ve heard that phrase before. Now should I decide to go buy some coal and “mix” it or use it by itself, the feel of the “Bone Warming” is different, it actually changes in the “feel” of the Heat in my living space and of course around my Wood/Coal Stove. My question is “Why does it feel different? Is is because of different IR emitting against the Heat Exchanger? I think this might be the case because of
    two “different” types of fuel.

    I’ve been a HVAC Service Technician for over 15 years and when I walk into a Home in winter, I will know with most certain accuracy what kind of Heat is being produced.

    You input would be most kind if you would. I think it would also explain the differences in “energy” being used in the Complex Climate Change System that we all endure.

    Regards,
    Denny

  378. Willis Eschenbach says:

    Willis Eschenbach says:

    Joel Shore, thanks for your assistance with the tsunami … much appreciated.

    No problem!

    The paper by Hansen asserting a 0.85 W/m2 imbalance is a joke, in my opinion. Even with satellites, we cannot determine either incoming or outgoing radiation to anywhere near that accuracy. Don’t know if you took a look at my citation above, but it discusses this question as regards Hansen’s paper.

    Climate models are of no help in this regard. Their errors are legion, and large. Here’s an example, from the people who wrote the model: …

    I think I will have to disagree with you here though (although I haven’t been able to follow the link because it seems really slow tonight). I don’t disagree with your statement that neither satellites nor climate models have the accuracy to directly detect such a small imbalance. However, as I understand Hansen’s paper, the imbalance is being detected instead by observing the amount of heat that is building up in the system due to the imbalance (and, nearly all of this heat buildup is in the oceans). So, even though we may not be able to measure the input and output fluxes with sufficient accuracy, we can infer their difference through the buildup of heat energy. (Not to say that there aren’t other potential issues, such as whether the heat content change was measured over a long enough period to give a result to that accuracy. But, I don’t think the fact that models or satellites can’t resolve the fluxes to this accuracy is not relevant.

    Take a look at the link when things get back, I can’t access it either at the moment. From memory, Hansen et. al used a ten year period during which the model was a close fit to the ocean temperature observations to make his claims. However, I took a look at the earlier ocean temperature observations from the same source he used, and the model Hansen used does a very poor job of replicating those. It is out on the order of ±6 W/m2. Given this size of error, both the size and the precision of his claimed imbalance (0.85 W/m2) are simply not credible.

    Always good to hear from you,

    w.

  379. Willis Eschenbach says:

    Denny (19:50:35) :

    Mr. Eschenbach,

    I have a question but I think it “might” be related in the explaination of “different energies”. I state different in that there’s “sensible temperature” and “radiant temperature”. I’m going to use my Wood/Coal addon in my House as an example.

    I burn wood mainly because I don’t have to pay for it. I love wood heat because it’s a “Bone Warming” heat and I was raised with it on the Farm. I hope you’ve heard that phrase before. Now should I decide to go buy some coal and “mix” it or use it by itself, the feel of the “Bone Warming” is different, it actually changes in the “feel” of the Heat in my living space and of course around my Wood/Coal Stove. My question is “Why does it feel different? Is is because of different IR emitting against the Heat Exchanger? I think this might be the case because of
    two “different” types of fuel.

    I’ve been a HVAC Service Technician for over 15 years and when I walk into a Home in winter, I will know with most certain accuracy what kind of Heat is being produced.

    You input would be most kind if you would. I think it would also explain the differences in “energy” being used in the Complex Climate Change System that we all endure.

    Regards,
    Denny

    Radiation is emitted by everything. The only difference in the radiation is the frequency/wavelength, which changes with the temperature. Hot steel glows red, but if your hand was that hot it would glow the same color red as well …

    So there is no difference in the type of radiation. But there may be a difference in how much energy is emitted as radiation versus hot gas. This is because as wood burns, typically there are a lot of flames. Also, the wood ash tends to flake off and expose new surfaces, that burst into flame. Coal, on the other hand, tends to end up with a shell of ash and clinker surrounding it, and with much less visible flame. So that’s a possible explanation.

    My guess would be, though, that your sense of smell is what tips you off. Houses heated by coal have a distinctive smell to me, which I can smell even if the stove is not lit.

    That’s my guesses, for what they’re worth.

  380. Willis Eschenbach says:

    yonason (18:55:21) :

    Willis Eschenbach (16:54:34) :

    “Thermal equilibrium is when a system’s macroscopic thermal observables have ceased to change with time.”

    You are using it, on the other hand, to describe a state where all parts of the system have the same temperature.”

    But that’s really just the same thing. (I know it sounds general enough to apply to “steady state” as well, but using it that way can be very confusing …

    Indeed it is confusing. yonason, again I can’t get to whatever is the nature of your complaint, or the source of your misunderstanding. Let me suggest again that you look at Fig. 1 and ask some pointed, short questions that show exactly what you find wrong about that Figure.

  381. par5 says:

    Willis Eschenbach (23:47:52) :

    par5 (21:22:15) :

    Willis- with this reverse greenhouse (heat from within) will there still be changes in the atmosphere ie cool and warm phases? Will the steel green house provide a constant temp worldwide? Will there be polar ice caps?

    Since my thought experiment has no atmosphere, none of them are possible.

    Willis- If you raise the temp of a black body in space, with a core temp as you decribe, surrounded by a shell as you descibe, then the surface temp will rise high enough to melt and evaporate water. Even if no water is on your planet, there will still be outgassing. You will have an atmosphere- like it or not. The temp of your planet can not be determined without the gas tables, and knowing the mass and pressure. I concede your physics because I see your POV, and agree. Will you concede that your experiment must attain an atmosphere?

  382. anna v says:

    Denny 19:50:35

    On the difference of “feel” of heat coming from coal or from wood. Coal burns at a much higher temperature than wood, therefore its radiation curve is different, with higher frequencies. This will be felt next to the stove burning the coal/wood.

    Now for the house, it is not clear.
    If it is heated air circulating then the same holds as above, the convected air has the spectrum of the air next to the stove.
    If it is water radiators it must be a mind thing: because one has felt the difference next to the stove the mind assumes it carries out in the house. Else it is the smell difference between coal and wood, as Willis has suggested.

  383. ralph says:

    “In order to maintain its thermal equilibrium, the
    whole system must still radiate 235 W/m2 out to space…”
    – from paragraph just above fig.1

    In other words, by definition, the steel shell is arbitrarily
    assigned the same temperature as the earth?

    No. The Earth rises to 470 W/m2, due to the ‘insulating’ (re-radiating) properties of having the shell around it – just as a naked flam will turn into an oven if you place a shall around it.

    That is the whole point of the experiment. And, as I see it, the ultimate point of this exercise, is to demonstrate that a single layer system would not warm the real Earth sufficiently, and that we need two layers in the atmosphere.

    The experiment is a VERY simple model to demonstrate this hypothesis by Willis. Whether he is right in his TWO LAYER system, I do not know, but the thought experiment to demonstrate this IS DESIGNED TO BE VERY SIMPLE AND CLEAR TO UNDERSTAND.

    I fail to see why so many people are having problems with it. But it does demonstrate the problems we have in explaining scientific points to politicians, who have to decide where to place the nation’s money. It must be as easy to bamboozle politicians as it is to bamboozle the contributors to this blog. Hence the problem we have with AGW taking over the political world.

    .

  384. ralph says:

    >>>At thermal equilibrium, (T1^4-T2^4) = 0.
    >>There will be no net heat exchange, BY DEFINITION
    >>not assertion.

    Yes, NET heat exchange – but that does not mean that there is no heat exchange. Place two white-hot suns next to each other and there will be LOTS of heat exchange going on – but no net exchange.

    .

  385. Willis Eschenbach says:

    par5 (21:55:26) :

    Willis Eschenbach (23:47:52) :

    par5 (21:22:15) :

    Willis- with this reverse greenhouse (heat from within) will there still be changes in the atmosphere ie cool and warm phases? Will the steel green house provide a constant temp worldwide? Will there be polar ice caps?

    Since my thought experiment has no atmosphere, none of them are possible.

    Willis- If you raise the temp of a black body in space, with a core temp as you decribe, surrounded by a shell as you descibe, then the surface temp will rise high enough to melt and evaporate water. Even if no water is on your planet, there will still be outgassing. You will have an atmosphere- like it or not. The temp of your planet can not be determined without the gas tables, and knowing the mass and pressure. I concede your physics because I see your POV, and agree. Will you concede that your experiment must attain an atmosphere?

    A “thought experiment” is an experiment wherein we can specify conditions in order to simplify and understand a system. I specified that there is no atmosphere. Why? I don’t know. Perhaps the planet is made out of solid Cantobtainium with a radioactive core. In any case, 470 W/m2 (the eventual steady-state temperature of the planet with a shell) is 28C, which is far too cool to melt much of anything.

    But in any case, no, there is no atmosphere.

  386. par5 says:

    Cantobtainium? Are you no longer patient? Thought experiments have no ramifications?

    “The planet is in interstellar space, with no atmosphere and no nearby stars.”

    I’m pretty sure this was before you built your shell, yes? Perhaps it doesn’t matter?

  387. TomVonk says:

    Willis you said :

    “My apologies, Tom, but I find your writing impenetrable. Please clarify and simplify what your objections are, and I’ll be happy to answer.”
    .
    Fair enough even if I think that I have been too long only because I tried to be as simple as possible and to demonstrate every step of the argument mathematically .
    My basic assumption is that I assume that you are familiar with the derivation of the SB law . If not , then you will benefit of looking it up .
    So that means that the SB law is just a consequence of some MORE basic law and this more basic law is the Planck’s law .
    OK , you’re with me ?
    1)
    Now the Planck’s law gives the spectral radiance I (units W/m²/sr/Hz) .
    If you ask what power will be emitted by some surface into some solid angle you will integrate the Planck’s law for all frequencies and within some solid angle and obtain the SB law .
    The particular case used here is the case where the Planck’s law is integrated for solid angles in a “half sphere” e.g the plane surface sees only a “half space” .
    That means concretely that the SB constant is not an universal constant but is only valid for plane surfaces emitting in a half space . For instance a ponctual source or a double faced shell emit in the whole space and the SB law for them has an SB constant which is DOUBLE of the one you used .
    I went farther by saying that the constant you are using is only valid for convex bodies . No problem for your model because you have indeed chosen convex bodies but it would be wrong to think that the SB constant you use is valid for any body . Especially bodies with cavities are tricky .
    2)
    W/m² (or irradiances) generally do NOT conserve .
    For spheres they definitely don’t (they do for INFINITE parallel plans) .
    The irradiance of your shell is P.r²/R² where P is the irradiance of the inner sphere . As P.r²/R² < P follows that :
    a) the shell has a lower temperature than the inner sphere
    b) the shell and the inner sphere are not in thermal equilibrium (e.g there is a net flux going from the inner sphere to the shell)
    c) The difference of temperature between the shell and the inner sphere (so the net flux) increases as R increases .
    .
    I could have written only very shortly the following synthesis but then you could have rightly asked me where is the proof .
    So consider that if you find what follows not clear , then what is above is the proof .
    .
    ========================================
    1) SB constant is not universal and depends on the solid angle over which the Planck's law is integrated .
    .
    2) The apparent paradox between the shell emitting "two times as much" as the inner sphere is due to the fact that the double faced shell has an SB constant which is 2 times the one of the inner sphere .
    .
    3) The shell and the inner sphere are not in a thermal equilibrium and have different temperatures .
    .
    4) W/m² (irradiances) do not conserve . The W/m² of the shell are r²/R² times smaller than the W/m² of the inner sphere .
    ========================================
    .

  388. par5 says:

    I find your remark condescending (cantobtainium) because of my chemistry background. I conceded your point on the physics, yet you refuse to see the ramifications of your own thought experiment. I’m calling Shenanigans.

  389. P Wilson says:

    lgl (10:37:03)

    it still makes little sense. if 256 re-radiate back to the source and 256 exits the barrier, its still 256 one way and 256 the other that can’t be added together. (Addidion is the unified magnitude of several units) . translating it to temperatures. if an object radiates to an energy measured at 20C to a shell for a sufficent time to heat it to that temperature, then the shell isn’t going to be 40C because radiation leaves it bi-directionally, 20C each way. Temperatures don’t add up mathematically like that. If they did you’d have a litre of water at 10C and a litre at 30C combined to give 2 litres at 40C

    blow up a balloon

  390. carrot eater says:

    TomVonk (04:22:04) :

    1. What is conserved

    Several people have become upset that he has conserved fluxes (W/m^2), not conserved flows (W). It’s been addressed a few times. If the shell is thin and near the planet, the error is trivial. If you don’t like it, redo it using actual areas and flows. You’ll see nothing much changes, so long as the shell is near the planet.

    2. Stefan-Boltzmann

    Your point about Stefan-Boltzmann is an absolute red herring.

    Yes, the derivation starts with Planck’s law. You assume the surface is a diffuse emitter (intensity has no directional dependence), so integrating over the hemisphere of vision is trivial (it just adds a factor of Pi), then you integrate over wavelength and you get the Stefan-Boltzmann law. So what?

    a. Curvature of the emitting surface

    Stefan-Boltzmann gives the total power emitted by a surface. You think it should only be used for plane surfaces? Fine, then break up the earth or the shell into tiny differential areas (each of equal area); each differential area dA is essentially planar. Apply Stefan-Boltzmann to each of those areas dA to get the power emitted at each, dP. Now, add it all up. the differential areas dA all add up to the surface area of the sphere or shell, 4*pi*r^2. Adding up all the dP will just get you…exactly what you’d get by applying Stefan-Boltzmann to the entire curved body, because dP is the same for every area dA. So the fact that the body is slightly curved is completely inconsequential, so long as you know its surface area.

    b. Solid Angle

    Are you then worried about the solid angle, then? Why? Stefan-boltzmann gives the total power emitted by the surface, in all directions. And that is exactly what is needed here. Every last bit of power emitted by the planet will reach the inner surface of the shell. Every last bit of power emitted by the inner surface of the shell will reach the planet.

  391. SteveBrooklineMA says:

    Willis-

    Here is a web-based calculator for your two-layer model:

    http://home.comcast.net/~stevehaker/GlobalWarmingCalculator-II/

    I put your figure in there for reference. I hope you don’t mind. I will take it out if you prefer.

    I had fun making it. Playing around with the numbers is interesting.

  392. P Wilson says:

    The SB may give total power emitted by surface, although other calibation equipment gives better measurements of power emitted form a surface. at 100w/m2 uniformly over a flat surface, is enough energy to produce 24C. 300-500w/m2 is enough energy to cook food from frozen in oil (see citiations above)

    Why is this equation used to describe anything but incandescent/hot or superheated metals? If a black box were the absorbing heat at maximum potential, then it theoretically emits as much as it receives, but only if it is out of equilibrium. If it is in equilibrium then the rate for this theortical plane is 235w/m2 no more, no less between the two spheres. (if you thing the SB constant is a valid measure of energy).

  393. P Wilson says:

    this is why AGW theory is all fraud. If earth receives this amount of solar energy, there is no reason why it should emit as much, or even more. if it is in a fair equilibrium with the incoming solar that hits the surface then earth gives off very little radiation. Repeat: When two objects are at the same temperature in the same vicinity and are in thermal equilibrium, energy is not being radiated from these objects, as objects only give off radiation when they are getting colder relative to their surroundings. Bring the temperature down then the said objects cool and give off radiation until they reach the new equilibrium

  394. carrot eater says:

    P Wilson: You aren’t just arguing against AGW, you’re arguing against physics that has been successfully used by a lot of people for a lot of things for a long, long time. When that is the case, you must consider the strong likelihood that the physics is just fine, but you misunderstand it.

    “The SB may give total power emitted by surface,”

    Stefan-Boltzmann gives the maximum possible power – that emitted by a blackbody.

    “although other calibation equipment gives better measurements of power emitted form a surface. ”

    To the extent that something is not actually acting as a perfect blackbody, yes, some measurements are needed.

    “at 100w/m2 uniformly over a flat surface, is enough energy to produce 24C.”
    “300-500w/m2 is enough energy to cook food from frozen in oil”

    By itself, you can’t say any of these things. You need to describe the entire system.

    “Why is this equation used to describe anything but incandescent/hot or superheated metals?”

    Because it works. For some reason, you are having a difficulty with infrared radiation. I am not sure why.

    “If earth receives this amount of solar energy, there is no reason why it should emit as much, or even more.”

    If the temperature of a body is stable, that means that energy is coming to that body at the same rate it is leaving.

    “as objects only give off radiation when they are getting colder relative to their surroundings.”

    Absolutely incorrect. All matter gives off radiation, proportional to T^4. Any body is always radiating, and receiving the radiation from surrounding bodies. The NET exchange between the bodies will be from the higher temperature body to the lower temperature body.

    P Wilson, all of these things have been described before, above. All of them are easily learned in a physics text. All of them are used in the design of useful things. All of them can be observed. If you’re clever enough, it can even be derived from first principles. One really should pause before declaring that some fundamental principle of physics is invalid.

  395. Curt says:

    Denny (19:50:35):

    Excellent question! I think you are noticing the phenomenon that different wavelengths of IR radiation penetrate different depths into the human body (as you instinctively suspect when you talk about “bone-warming heat”). This was something my heat transfer prof loved to demonstrate in the lab, because he knew it would really “get” us. I still remember the demonstration 30 years later!

    So if the coal burns significantly hotter than the wood, as Anna V claims, then yes, you could “feel” the difference if you are attuned to it.

    This lets me segue into another issue on this thread — radiative power from cooler bodies to warmer bodies. (Yes, it does occur, even though John A will never concede the point, but of course, it is less than the radiative power from the warmer body to the cooler body.)

    One underemphasized benefit of having good insulation inside your walls is that it makes the room surface of the wall warmer so it radiates more heat back to the people in the room. This lets the people have an equivalent comfort level with the air temperature of the room several degrees below what they would want with cold walls.

    If you are attuned to it, you can “feel” this difference too when you walk into a room.

    Also, there is a significant difference in “feel” between radiative heat sources, where the surfaces are heating the air, and forced-air heat sources, where the air is heating the surfaces.

  396. P Wilson says:

    carrot eater (10:03:18)

    The SB works well for IR radiation such as the sun, and heated metals in engines or but not normal temperature, or LW surfaces in equilibrium. If a black box in your room was next to a white box, according to SB they should be different temperatures and the black should be emitting more radiation than the white box, meaning they ought to have different temperatures. However, if they are both in thermal equilibrium at that room temperature then they’re the same temperature. Normal temperature matter just doesn’t don’t give off so much radiation that the constant gives it. (if its in equilibrium)

  397. BillG says:

    Willis,

    Sorry to be late to the party but it has been difficult to find the time to read the 400 comments to get to this point. Obviously you have generated great interest.

    Your model appears correct for a solely radiative heat transfer process occurring in a vacuum. Unfortunately radiative heat transfer plays a relatively small role in the heat transfer processes occurring between a sphere and a relatively dense, gaseous atmosphere under the influence of a gravitational field.

    Heat transfer flux for all three modes of heat transfer (conduction, convection and radiation) is basically a function of the temperature differential between the source and receptor and the heat transfer coefficient. In your model, the sphere heats up due to the reduced net heat transfer output flux from the sphere that results when the ∆T between the sphere and shell decreases (the input flux to the sphere (Sun) remains constant). In a planet with an atmosphere all three modes of heat transfer will operate in the same manner. If the atmosphere is colder than the surface, heat transfer will occur from the surface to the atmosphere until and if the surface and atmosphere reach the same temperature. Since the total heat transfer is the sum of the three heat transfer modes, the heat transfer coefficients for each process determine which modes will predominate.

    Since the heat transfer coefficient for conduction in gases is very small, conduction plays only a minor role (primarily to initiate convection from the surface).

    Radiative heat transfer will behave quite differently in a gaseous system in contrast to your model where all bodies are solids. First, radiative heat transfer will only occur with the approximately 2% of the gaseous molecules that are IR active. Second, in the lower and mid troposphere, the IR active molecules will experience approximately 3000 – 10,000 collisions with other molecules during the relaxation time between absorbance and emission. Thus the transferred heat will be thermalized among the non IR active molecules and the temperature of the total gas mixture will rise. Very little emissivity will occur. In other words, in the lower and mid troposphere, the heat transfer coefficient for radiation heat transfer in a gas is very low (especially at or near ambient conditions). Heat flux absorbed by the gas from radiation will be converted to convective heat transfer. (In the upper troposphere, where molecular density and the chance for molecular collisions is much lower, emissivity will play a much larger role and is a factor in the total OLR).

    Convection and the gravitational field are the primary players in tropospheric thermodynamics. The heat transfer coefficient for convection is large (and becomes larger as turbulence increases). You illustrated part of the convection role very well in your previous article on the thermostat hypothesis (this was a very good article, by the way). But you did not explain the other part of convection which is subsidence. Meteorologists frequently use the term “convective overturning” which illustrates the convection/subsidence cycle that is true for a convection process. “What goes up must come down”.

    The gravitational field is important in that as mass is convected to a higher altitude via expansion and work on its surroundings, thermal energy is traded off for potential energy. As a result, the temperature of the mass parcel decreases due to adiabatic expansion and the dry adiabatic lapse rate is created. But the total energy content remains constant. At higher altitudes radiation heat transfer becomes more significant and OLR is emitted and the parcel cools. This is one reason why the greybody temperature of the earth is lower than the surface temperature. Subsidence then occurs and the mass parcel descends and trades off potential energy for thermal energy and the parcel increases in temperature due to adiabatic compression. This is why the surface temperature is higher than the greybody temperature. This subsidence conversion of potential energy to thermal energy represents the bulk of the 324 W/m2 down arrow in the K&T cartoon. Very little is due to “back radiation”.

    I tried to keep this as brief as possible and a lot of detail is missing. But if you or anyone else would like to pursue this approach, I would be glad to elaborate more fully. But I am late to the table and I’m not sure if many are still plugged into this thread. But either way, thanks for listening.

  398. Willis Eschenbach says:

    par5 (04:19:34) :

    Cantobtainium? Are you no longer patient? Thought experiments have no ramifications?

    “The planet is in interstellar space, with no atmosphere and no nearby stars.”

    I’m pretty sure this was before you built your shell, yes? Perhaps it doesn’t matter?

    “Cantobtainium” is a humorous name for an element which we can’t obtain, which I sometimes use in thought experiments to indicate a substance with certain properties which real substances might not have. I could just as well have said “Steel”. Neither steel nor cantobtainium vaporize at 470W/m2.

    Remember, this is a thought experiment, designed to simplify the issues so that the main principles become clear. The substances used are immaterial.

    And no, I haven’t lost my patience, although some posters tempt me to temporarily mislay it …

  399. P Wilson says:

    If you take the human basal metabolic rate as a standard of comparison – it is the calories required against calories burned, converted into w/m2, which for most people is 58w/m2, increasing on average to 100w/m2. In the following experiment radiation is measured fromleaving individuals.

    in google enter:

    Description of a human direct calorimeter

    and it takes straight to the cambridge journals which conducted the experiment.

    It stands to reason that if a human optimum emites 100w/m2 of radiation through a duration, that earth is emitting much less per square metre.

  400. Willis Eschenbach says:

    par5 (04:19:34) :

    Cantobtainium? Are you no longer patient? Thought experiments have no ramifications?

    “The planet is in interstellar space, with no atmosphere and no nearby stars.”

    I’m pretty sure this was before you built your shell, yes? Perhaps it doesn’t matter?

    A “thought experiment” often posits conditions which are not physically possible, in order to clarify some thorny question. Einstein used them frequently:

    Here, and throughout his lifetime, he made up a thought experiment (Gedanken experiment). Gedanken expermiments cannot actually be performed, but nevertheless are a useful tool for testing a theory. Einstein imagined a person travelling at the speed of light and asked how that person would perceive the light wave.

    Now, you could object to Einstein’s thought experiment as you have objected to mine, for a host of valid physical reasons … but both are useful nonetheless.

    “Cantobtainium” is my humorous name for a mythical element which, as the name suggests, we can’t obtain. It is very useful in thought experiments, as it has no known properties. As Foghorn Leghorn used to say, “That’s a joke… I say, that’s a joke, son.” However, I could have just said “steel” instead, neither one vaporizes at 470 W/m2.

  401. P Wilson says:

    Its a thought experiment. Well fine. A lot of physics contains thought experiments not yet proven when applied to areas they don’t belong. Would you use Boyles law to measure the area of a room, or would you use a standard measuring tool calibrated in centimetres?

    The SB is pulled into climatology to justift some quite preposterous figures, applied to gases and liquids equally, irrespective of their molecular properties.

  402. Willis Eschenbach says:

    TomVonk (04:22:04) :

    The particular case used here is the case where the Planck’s law is integrated for solid angles in a “half sphere” e.g the plane surface sees only a “half space” .
    That means concretely that the SB constant is not an universal constant but is only valid for plane surfaces emitting in a half space . For instance a ponctual source or a double faced shell emit in the whole space and the SB law for them has an SB constant which is DOUBLE of the one you used .
    I went farther by saying that the constant you are using is only valid for convex bodies . No problem for your model because you have indeed chosen convex bodies but it would be wrong to think that the SB constant you use is valid for any body . Especially bodies with cavities are tricky .
    2)
    W/m² (or irradiances) generally do NOT conserve .
    For spheres they definitely don’t (they do for INFINITE parallel plans) .
    The irradiance of your shell is P.r²/R² where P is the irradiance of the inner sphere . As P.r²/R² < P follows that :
    a) the shell has a lower temperature than the inner sphere
    b) the shell and the inner sphere are not in thermal equilibrium (e.g there is a net flux going from the inner sphere to the shell)
    c) The difference of temperature between the shell and the inner sphere (so the net flux) increases as R increases .

    First let me say that you are correct. Second let me say that the differences between what you say and the results I show above are trivial, on the order of 0.2%. This is because for all practical purposes, there is no difference between the situation I describe above and two infinite parallel planes.

    This is because the radii of the sphere and the shell are equal to within less than a tenth of a percent. As is common in analysing this particular situation, this tiny error is ignored, and they are treated as though they were infinite parallel planes. See e.g. the Kiehl/Trenberth diagram I show in Fig. 3.

    I think that answers your questions, if not, let me know.

  403. Willis Eschenbach says:

    P Wilson (05:29:08) :

    lgl (10:37:03)

    it still makes little sense. if 256 re-radiate back to the source and 256 exits the barrier, its still 256 one way and 256 the other that can’t be added together. (Addidion is the unified magnitude of several units) . translating it to temperatures. if an object radiates to an energy measured at 20C to a shell for a sufficent time to heat it to that temperature, then the shell isn’t going to be 40C because radiation leaves it bi-directionally, 20C each way. Temperatures don’t add up mathematically like that. If they did you’d have a litre of water at 10C and a litre at 30C combined to give 2 litres at 40C

    blow up a balloon

    I’ll repeat an explanation I gave upstream. Assume the surface area of the planet in Fig. 1(b) is “X” square metres. Total radiation from the planet is 470 W/m2 times X square metres, which is 470X watts.

    The surface area of the shell is twice that of the planet, or 2X. Total radiation from the shell is 270 W/m2 times 2X square metres, which is 470 watts.

    Finally, the reason for using W/m2 is that you can add them together, but you can’t add temperatures together.

  404. Willis Eschenbach says:

    SteveBrooklineMA (07:53:29) :

    Willis-

    Here is a web-based calculator for your two-layer model:

    http://home.comcast.net/~stevehaker/GlobalWarmingCalculator-II/

    I put your figure in there for reference. I hope you don’t mind. I will take it out if you prefer.

    I had fun making it. Playing around with the numbers is interesting.

    Steve, that’s awesome. many thanks. Give me some time to play with it.

    w.

  405. Willis Eschenbach says:

    P Wilson (08:09:46) :

    The SB may give total power emitted by surface, although other calibation equipment gives better measurements of power emitted form a surface. at 100w/m2 uniformly over a flat surface, is enough energy to produce 24C. 300-500w/m2 is enough energy to cook food from frozen in oil (see citiations above)

    Why is this equation used to describe anything but incandescent/hot or superheated metals? If a black box were the absorbing heat at maximum potential, then it theoretically emits as much as it receives, but only if it is out of equilibrium. If it is in equilibrium then the rate for this theortical plane is 235w/m2 no more, no less between the two spheres. (if you thing the SB constant is a valid measure of energy).

    You seem to be operating under a misconception. Maximum tropical sunlight on a flat surface at midday is on the order of a kilowatt per square metre. Call it two thirds of that in the US. But if you think that you can cook dinner by just laying it out on a wooden plank at noon, you are very mistaken, tropics or not.

    I did a quick search for your “citation above” on google but found nothing. Please supply a URL.

  406. Willis Eschenbach says:

    P Wilson (13:07:07) :

    carrot eater (10:03:18)

    The SB works well for IR radiation such as the sun, and heated metals in engines or but not normal temperature, or LW surfaces in equilibrium. If a black box in your room was next to a white box, according to SB they should be different temperatures and the black should be emitting more radiation than the white box, meaning they ought to have different temperatures. However, if they are both in thermal equilibrium at that room temperature then they’re the same temperature. Normal temperature matter just doesn’t don’t give off so much radiation that the constant gives it. (if its in equilibrium)

    I fear I’m going to have to stop responding to you, P Wilson. The SB equation applies from the temperature of the interstellar microwave background to the temperature of the sun, and all temperatures in between. Please read up on it, as this is not the thread for disputing well known basic physics.

  407. Willis Eschenbach says:

    BillG (13:11:21) :

    The gravitational field is important in that as mass is convected to a higher altitude via expansion and work on its surroundings, thermal energy is traded off for potential energy. As a result, the temperature of the mass parcel decreases due to adiabatic expansion and the dry adiabatic lapse rate is created. But the total energy content remains constant. At higher altitudes radiation heat transfer becomes more significant and OLR is emitted and the parcel cools. This is one reason why the greybody temperature of the earth is lower than the surface temperature. Subsidence then occurs and the mass parcel descends and trades off potential energy for thermal energy and the parcel increases in temperature due to adiabatic compression. This is why the surface temperature is higher than the greybody temperature. This subsidence conversion of potential energy to thermal energy represents the bulk of the 324 W/m2 down arrow in the K&T cartoon. Very little is due to “back radiation”.

    I was with you until I got to this paragraph. The downwelling longwave radiation from the atmosphere is not some theory that you can wave off with your own theory. It is a physical phenomenon which is measured routinely by scientists, and has been measured for decades all over the world. Do a google search on ‘surface downwelling longwave measurements’ to see thousands of examples.

  408. Willis Eschenbach says:

    P Wilson (13:26:41) :

    If you take the human basal metabolic rate as a standard of comparison – it is the calories required against calories burned, converted into w/m2, which for most people is 58w/m2, increasing on average to 100w/m2. In the following experiment radiation is measured fromleaving individuals.

    in google enter:

    Description of a human direct calorimeter

    and it takes straight to the cambridge journals which conducted the experiment.

    It stands to reason that if a human optimum emites 100w/m2 of radiation through a duration, that earth is emitting much less per square metre.

    Sorry, missed this one. You have misread the reference. The values that it is giving are in watts, not watts per square metre.

  409. Dyspeptic Curmudgeon says:

    Interesting thought experiment, but I think it is totally and completely wrong.

    You have improperly deleted all reference to the energy inputed to the system from the sun and discussed only the energy produced by the interior of the earth. In reality, together those are in equilibrium. Put a steel greenhouse around the earth, and the surface will be heated by the radiation coming from the inside of the shell (instead of directly by the sun). To which the energy from the earth’s core will continue to be added.
    Your experiment is fatally flawed in real terms. Moreover you have conflated cause and effect. The surface will not double in temperature because of the effect of the radiant energy of the shell. The surface and the shell will stabilize (at the same temperature since this is a ‘perfect’ experiment.) In your version, the energy source is the earth’s core giving a 235 W/m2 level. In reality, the OUTSIDE of the shell will be heated, and will radiate about 234 W/m2 inward (I find it hard to believe the the earth’s core produces much more than 1 W/m2 at the surface, although I am almost surely wrong about the exact number….it is probably whatever equates to 3 degrees C…which is the average deep oceanic water temperature.)

    To claim that since the shell ‘must’ radiate both inwards and outwards, that the surface necessarily will be heated to

    Interestingly I was recently pointed to a paper which discusses the Greenhouse effect and falsifies it *on a physical basis*. CO2cannot do the job.

    http://arxiv.org/pdf/0707.1161v4 (link is to a pdf file at Cornell U).

  410. Dyspeptic Curmudgeon says:

    Opps, missed something in the editing. Should be:

    ‘To claim that since the shell ‘must’ radiate inwards and outwards, that the surface necessarily will be heated to twice the 235 W level, since it is already at that temperature, and will also receive that amount of radiated energy is mindboggling. Yes you *can* add numbers representing similar units but that does not necessarily mean that you *can* add meaningfully add those items of those units and have the result mean something.

    Short thought experiment: 2 small steel plates suspended closely in parallel in your kitchen at room temperature: say 20C. Remove one plate. Does the temperature (black body radiation) of the other change? NO. The temperature is the result of external forces. Replace the plate. Heat the back of one, with a heat lamp, until its temp is stable over the entire front surface.No direct light impinges on the other, and the room stays the same. Say the front of the plate is now 40C. What is the temperature of the front face of the other plate? In your thought experiment that plate would be 20C (in W/m2) plus what is radiated from the other 40C (in W/m2)…. Nah….Does not compute…and it does not happen like that. Agreed, the cool plate might climb a little but not much.

    Even allowing for perfect conduction of energy between the plates, what you hypothesize DOES NOT HAPPEN.

    An Inconvenient Truth perhaps, but like the goreacle you are wrong.

  411. P Wilson says:

    Willis Eschenbach (14:08:45) :

    whoops you’re right. A human can be anything from 0.75-2m2

  412. P Wilson says:

    Its in a textbook:
    Vijayan and Singh

    “Advances in deep fat frying of food”
    1997 CRC Press

    Not sure if it can be found online.

    However, we’re all agreed black tarmac surface will absorb more heat and transmit more heat than the equivalent area of grey or white matter. However, once equilibrium is as close as it can be then theoretically, re-radiation should occur at night, or at the point when the temperature goes down, and presumably this is where the greenhouse effect should be felt the most – yet matter emits heat very quickly – and quickly thermalise to new temperatures, so as not to give off that much radiation. That can be seen by thermal imaging equipment at night that traces 7-14 microns. they’re specifically used to trace humans

  413. Gary Hladik says:

    Willis, I see you’re still here answering questions. That’s the mark of a true educator.

    This all reminds me of the earlier discussion on whether CO2 could freeze out at earth’s south pole. As here, a number of misconceptions had to be addressed, and as here, I had to hit the references pretty hard to sort the whole thing out.

    It’s a good thing you presented a simple model, otherwise you’d be here until some hacker exposed the whole CAGW fraud. :-)

    Oh, wait…

  414. carrot eater says:

    P Wilson (13:07:07) :

    P Wilson, you are continually confusing yourself into thinking things which aren’t true, and then blaming the Stefan-Boltzmann law for it.

    If this matter so much to you, please go to the hardware store and buy an infrared thermometer. It’ll cost you about $50. Then aim it at different things around the room. You’ll find that the thermometer works. And it works because all those things around the room are emitting radiation, just as the SB law says they should.

    It’ll be as direct a confirmation of the principles as I can offer you.

  415. P Wilson says:

    of course they’re emitting some radiation. Just not as much energy as is inferred from an equation

  416. carrot eater says:

    P Wilson: Buy an infrared thermometer. It uses Stefan-Boltzmann. It works. The only reason you think it doesn’t work is because you’ve confused yourself in applying it, and you’re saying things like “300-500w/m2 is enough energy to cook food from frozen in oil”, which has zero meaning.

    It’ll only not work if you point it at some shiny metallic surface with a low emissivity.

  417. Steel shells, mirrors, blankets, they are all fine metaphors but I have always felt they failed at getting to the heart of how the greenhouse effect works i.e. letting light in at one frequency and rejecting it at others.

    They don’t give us any idea of limits, if there are any, what they are and if they are important.

    I had written up a document some time ago about ‘limits’ being those of a blackbody, but the document wasn’t that well put together and I don’t think convinced many. The resistance of most to even considering this concept seems strange to me. To accept, without thought, that we can create something that absorbs more energy than a blackbody … I still don’t understand that. Ca la vie.

    I have taken yet another attempt at writing up a description of why there must be limits to the greenhouse effect and why this limit is almost certainly that of a blackbody. I hope this document is much clearer then the previous one. The document can be found here:

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

    I accept again that most will reject it. That’s ok. The unquestioned wisdom that greenhouse effect can build up to infinity is strong. I accept that I’m probably wrong even. I make a lot of mistakes … But I would ask then that those who quickly dismiss this idea consider then, ‘what is the limit?’ Is it inifinity? Unlikely. Is it the temperature of the Sun. Maybe, but if so, then with the the right atmosphere, Pluto could become as warm as Mercury. Distance from the Sun wouldn’t matter. Somehow I doubt it. So in that case, what is the limit? Think about it …

    Cheers.

  418. Willis Eschenbach says:

    Dyspeptic Curmudgeon (14:16:36) :

    Dyspeptic, please read the entire thread. Some things, such as your link to the arxiv paper, have already been discussed.

    Interesting thought experiment, but I think it is totally and completely wrong.

    You have improperly deleted all reference to the energy inputed to the system from the sun and discussed only the energy produced by the interior of the earth. In reality, together those are in equilibrium. Put a steel greenhouse around the earth, and the surface will be heated by the radiation coming from the inside of the shell (instead of directly by the sun). To which the energy from the earth’s core will continue to be added.
    Your experiment is fatally flawed in real terms.

    Also, please re-read what I wrote in the opening post. For example, I said “The planet is in interstellar space, with no atmosphere and no nearby stars.” No nearby stars means that there is no radiation from the sun.

    Moreover you have conflated cause and effect. The surface will not double in temperature because of the effect of the radiant energy of the shell. The surface and the shell will stabilize (at the same temperature since this is a ‘perfect’ experiment.) In your version, the energy source is the earth’s core giving a 235 W/m2 level. In reality, the OUTSIDE of the shell will be heated, and will radiate about 234 W/m2 inward (I find it hard to believe the the earth’s core produces much more than 1 W/m2 at the surface, although I am almost surely wrong about the exact number….it is probably whatever equates to 3 degrees C…which is the average deep oceanic water temperature.)

    Again, what is shown in Fig. 1 is a thought experiment, not a description of the earth..

    To claim that since the shell ‘must’ radiate both inwards and outwards, that the surface necessarily will be heated to

    Interestingly I was recently pointed to a paper which discusses the Greenhouse effect and falsifies it *on a physical basis*. CO2cannot do the job.

    http://arxiv.org/pdf/0707.1161v4 (link is to a pdf file at Cornell U).

  419. Willis Eschenbach says:

    Dyspeptic Curmudgeon (14:16:36) :

    Interesting thought experiment, but I think it is totally and completely wrong. …

    Dyspeptic, please re-read first the original post, then the entire thread. You keep talking about the sun heating the steel shell, when I said “The planet is in interstellar space, with no atmosphere and no nearby stars.” So there is no solar heating, because no star is nearby.

    Your other objections have been dealt with in the thread, including your reference to the arxiv paper.

  420. par5 says:

    Willis, I think that I have solved your conundrum for you. I did not realise that the planet Cantobtainia was completely sealed with a thick layer of cantobtainium (which prevents outgassing). At first, I thought that my questions were valid, on topic and had merit. I was wrong and I do apologise. Anyway, I glanced at my atomic chart shower curtain today and realised that cantobtainium has to be some sort of radioactive isotope or trans-uranium element. Giggle. So you see, it is your encasement of cantobtainium that is producing your 235W/m^2.

    It is a win/win situation for both of us. You get to keep your planetary greenhouse without an atmosphere, and I have the pleasure of knowing that your planet isn’t going to last much longer due to its high rate of decay.

    Can we now shake hands, thank each other for the polite conversation and walk home happy and content?

  421. Denny says:

    anna v, 22:10:40

    Thank you for your input! I thought about that also, Wood doesn’t burn as hot and I liked what Willis stated about “more flame” produced by Wood than Coal. Coal being denser, it makes sense.

    Anna, a wood/addon is a stove that has a box around it and a Phlenum on top. The heated air is sent around the Fire box by a Squirel Cage Blower, out the Phlenum and into the Duct system that carries the heated air to the Rooms. When I use “only” Wood for heat, it creates the specified sense of warmth I mentioned in my first post. The warm air surrounding the Living Space has this feeling. It’s false only in the way that the RH is very dry. Without Humidifing, my RH goes down to below 30%. Not good for your health I might add.

    This feeling I’m talking about in the heated conditioned air going into the Living area, changes with a noticable difference in “confort”. Yes, Willis is also correct that the “smell” also changes and that’s because a little bit of Coal smoke in the House can go a long way in staying around odor wise. It’s a very “heavy Smoke”!

    Willis stated that I can smell the Heat…Well, that’s true to a degree but it’s when I can actually “feel” the temperature, does it feel warmer than other Fuels? Yes is my answer. I’ll put in order a list of the “best” heat to the “worst” heat IMO.

    1. Wood Heat- because you instantly feel warmth when it’s around you.
    2. Hydronic or Baseboard Heat- this heat produces a warm heat because it uses convection and doesn’t dry the air out as much. Type of Fuel doesn’t make any difference. Also depends on the efficiency of insulation in the House.
    3. Nat., Propane Gas, Fuel Oil – forced air to which I have Propane for warmer days. I live in the Rural area. Fuel Oil is slightly different but close.
    4. Electric Baseboard – As long it’s convection to which I have but do not use. It’s expensive to operate.
    5. Heat Pump- forced air obviously used with electric backup if the Heat Pump can’t keep up. Very dry heat..Hard to stay warm. Must Humidify.
    6. Electric Forced Air- this heat is the worst. If I’m in this for more than 5 minutes, I nose goes crazy because the RH is getting to 20% which is very dry and unhealthy for obvious reasons. If you have a very well insulated structure Humdity levels can be maintained. Wind blowing on a structure wicks the RH out if it’s not insulated properly.

    I hope this helps Anna…stay warm, dry and Healthy! :)

  422. Nick Stokes says:

    Ian Schumacher (19:35:21) :
    The Greenhouse Earth doesn’t absorb more energy than a black body – in these models it is in equilibrium, with no nett absorption at all. I think your issue is, how high can the surface temperature go. There is a limit. The GHE happens because part of the outgoing 235 W/m2 is emitted from the high troposphere, at a temperature of about 225K. If it weren’t for the atmospheric window emitting at higher intensity, this emitting layer would reach the snowball earth temp of about 255K, at which the outgoing IR balances incoming sun light. It can’t go higher.

    The surface could still get very hot if there were sufficient insulation – say several layers of “steel” plus vacuum. But the ability of the real atmosphere to insulate is sharply limited by the dry adiabatic lapse rate . This is about 9.8 K/km, and is a stability limit. If it were ever exceeded, the atmosphere would become very turbulent, and conduct heat upward very readily.

    So no matter how much GHG there is, the tropopause will be limited to about 255K, and the surface to an excess determined by that maximum lapse rate. That’s still hot.

  423. Willis Eschenbach says:

    par5 (20:58:51) :

    Willis, I think that I have solved your conundrum for you. I did not realise that the planet Cantobtainia was completely sealed with a thick layer of cantobtainium (which prevents outgassing). At first, I thought that my questions were valid, on topic and had merit. I was wrong and I do apologise. Anyway, I glanced at my atomic chart shower curtain today and realised that cantobtainium has to be some sort of radioactive isotope or trans-uranium element. Giggle. So you see, it is your encasement of cantobtainium that is producing your 235W/m^2.

    It is a win/win situation for both of us. You get to keep your planetary greenhouse without an atmosphere, and I have the pleasure of knowing that your planet isn’t going to last much longer due to its high rate of decay.

    Can we now shake hands, thank each other for the polite conversation and walk home happy and content?

    Done. Thanks.

  424. P Wilson says:

    carrot eater (18:00:48) :

    Thanks carrot eater but I regularly use IR thermal imaging devices outdoors

  425. P Wilson says:

    if an average man gives off 180 watts at night, though i record 228 watts sometimes, then how could earth be emitting 255w/m2?

    To calculate energy at this level you’d need to divide human surface area by 148, so 1.75/148=85w/m2, or 1.75/228=130w/m2

  426. carrot eater says:

    P Wilson (06:26:15) : Well, if you trust your thermal imaging device at all, then you’re using the very physics that you think are incorrect. The only trick is knowing the emissivity of a surface (and whether the emissivity is constant over the wavelength range in question), but that doesn’t really seem to be your difficulty.

    I can’t even figure out what you’re trying to say, otherwise.

  427. carrot eater says:

    Nick Stokes (00:45:31) : Thank you for your contribution. There are points here which got unfortunately totally lost in the discussion, because so many people were getting hung up on the basics of radiation. Namely, that either in the steel cartoon or with a real atmosphere, the final emission to space is about the same. What matters is the temperature where this emission takes place. Willis’s radiation-only no-atmosphere example cannot describe the actual lapse rate within our atmosphere; this has been touched on in other comments as well. But it becomes difficult to write a simple tinkertoy model that captures what actually happens in the atmosphere – the lapse rate, spectral absorption and emission, pressure effects, and so on. You quickly end up with a computer code of the sort that sceptics don’t like, because it’s complicated.

  428. lgl says:

    P Wilson

    Your body is emitting about 930 W (if you are 2 m2 and wearing clothes)
    but you are also receiving 830 W from the surroundings at 20 C.
    http://en.wikipedia.org/wiki/Black_body

  429. Nick,

    This idea that the outer atmosphere represents the blackbody is a common thing I hear. However it seems to me I can imagine a new sphere just inside the outer atmosphere. I can image the outer atmosphere as the source now, and our new smaller sphere the new body. Now we realize that the outer part of our new smaller sphere must also absorb and radiate like a blackbody … and so on, all the way to the surface.

  430. P Wilson says:

    wavelength expressed in colour as it accords to temperature is normally the way such devices are calibrated.

    I know all about the SB theory, though 7-14 microns is typical range of detection. At sunset- night, most outside objects are invisible to these wavelengths in a typical suburban environment -and humans are clearly visible across a quite a few wavelengths. A human emitting around at 8 microns average, corresponds to around 100w/m2, as the BMR is around 58w/m2 where radiation emitted is equalt to the BMR

    It leads to whether we should accept the SB for *normal* temperature matter at a constant and whether it should be reformulated to a coefficient, divided by as much as 12, to produce 30w/m2 for outgoing longwave from earth on average, since it’s clear that earth isn’t attempting to reach absolute zero in order to reach equilibrium, and emit as much as it possible can to achieve it.

    Lgl. I’m aware of the absurd figures produced from the SB constant. The only difficulty with absorbtion of radiation is that at 20C, matter is cooler than the human body, so its unlikely a human will absorb that much radiation. Experimental procedures show a recorded 85watts emitted at 27C.

  431. Nick,

    To address the second part. As in Willis model, the inside temperature can get to infinity with insulation and no holes. Holes make all the difference.

    Image a steel container of air, completely closed off. Now imagine there is a source of new air creating new molecules inside the container. The container pressure can build up to infinity. That is Willis model with a totally reflecting steel shell (or infinite absorbing shells).

    Now image a steel container that we fill up with air using an external source. We need a hole for the air to enter. Can we build up the pressure inside our steel container higher than the pressure of the source? No. This is the same with our energy flux from the sun. We can’t build up the energy density (a rough analogy to our air pressure) to be higher than the source.

  432. P Wilson says:

    correction: at 2m2 85/2=42.2w/m2 at 27C

  433. lgl says:

    “It leads to whether we should accept the SB for *normal* temperature matter at a constant and whether it should be reformulated to a coefficient”

    Yeah, and think of all the fun you can have if you do the same with the speed of light.

  434. DeWitt Payne says:

    carrot eater (08:12:54) :

    Nick Stokes (00:45:31) : Thank you for your contribution. There are points here which got unfortunately totally lost in the discussion, because so many people were getting hung up on the basics of radiation. Namely, that either in the steel cartoon or with a real atmosphere, the final emission to space is about the same. What matters is the temperature where this emission takes place. Willis’s radiation-only no-atmosphere example cannot describe the actual lapse rate within our atmosphere; this has been touched on in other comments as well. But it becomes difficult to write a simple tinkertoy model that captures what actually happens in the atmosphere – the lapse rate, spectral absorption and emission, pressure effects, and so on. You quickly end up with a computer code of the sort that sceptics don’t like, because it’s complicated.

    It’s not the complication, at least not for me. A line-by-line atmospheric radiative transfer program may be complex, but it’s truly physics based and produces results that conform to observation provided the input data (lapse rate, humidity profile, etc.) are correct. It’s the oversimplification or parameterization of complex processes and other kludges in GCM’s necessary to get the programs to run without blowing up and in a finite amount of time that are the problem. GCM’s don’t use line-by-line or even band models for radiative transfer. They take too much computing time. We know that even fairly recently some models used incorrect parameterizations of radiation transfer. Model cloud physics produces results that are wrong in both extent and location of cloud cover. Aerosol forcings, particularly the aerosol indirect effect, don’t seem to be based on observation, but are used as a fudge factor to produce a more or less correct hindcast whether the climate sensitivity of the model to ghg forcing is high or low. The list goes on and on.

  435. carrot eater says:

    DeWitt Payne (11:01:16) :

    Well, as you can see from this thread, you might be OK with radiative transfer code, but plenty of others are confused about radiation and the conservation of energy, in general. You can see how controversial this simple toy model was. As for how the radiation transfer is parameterised, I think that’s pretty well straightened out. Cloud parameterisations would be the topic of a whole other thread. Remember that no model is ever perfect – it just has to be good enough to be useful for whatever you are trying to learn.

  436. Nick Stokes says:

    Ian
    Now we realize that the outer part of our new smaller sphere must also absorb and radiate like a blackbody … and so on, all the way to the surface.
    Yes, you’re right. The top shell gets 470 W/m2 from below, and radiates 235W/m2 up and down. The next must radiate that 470 up, and so also 470 down; it gets 235 from above, and so must get 705 from below. And so it goes. Each shell successively has a nett flow through of 235 W/m2, as it must. And each one gets hotter, emitting in total 235 W/m2 more than the one above.

    This works because each is fully opaque, conducting (so both faces are at the same temp) and insulated (by vacuum) from each other. Just like the Multi-layer_insulation of spacecraft (read the description – it’s a steel greenhouse). But the atmosphere is not like this. The “layers” are only partly opaque, not fully conducting, and not separated.

    As to your point about the Sun – yes, the sun’s temp is another theoretical limit. In the air, the lapse rate is a much lower and more real limit. But OK, for Willis’s cartoon, your issue has some meaning. It’s true that for Willis’s postulate of a radioactive energy source, you could reach an arbitrarily high temp. But the analogy heat source he really means is where the “steel” layers, like GHG, allow sunlight in, but block outgoing IR through frequency selectivity, creating a 235 W/m2 heat source at the surface.

    As the temperature rises, the thermal outgoing radiation rises in frequency, and would approach the solar spectrum if the Earth got that hot. In that case, the shells can no longer exercise that frequency selectivity. The outgoing radiation would pass through as easily as sunlight got in. And since it goes out to all the sky, that exit radiation is far greater than the incoming, so in fact the bottom surface temperature could not rise to anything like the Sun’s.

  437. P Wilson says:

    After 10 exchanges (approx 2 minutes) it would be 2350w^m2.

    golly thats hot for a subzero source.

  438. Nick,

    Whether the layer is opaque or partially transparent doesn’t, we can still treat it as a source to the next layer.

    I’m not sure if you read my article yet (http://www.ianschumacher.com/greenhouse_effect_maximum.html) but in there I get a result which I ‘think’ is what you are thinking. A balance that is hot, less than the sun, but greater than blackbody equivalent, but then I discuss why this would not be the case because it assumes the outgoing energy spectrum is static, when in fact the energy will continously redistribute (trying to maintain the blackbody spectrum shape) to pour out the visible window.

    Anyways, that my thought.

  439. DeWitt Payne says:

    Fun experiments to try with your IR thermometer (strictly speaking a pyrometer), available for ~$50 at auto parts and kitchen stores (no home should be without one):

    1. Open your refrigerator and measure the temperature.

    2. Open your freezer and measure the temperature.

    Do you still think that colder objects do not emit IR radiation? Explain how the thermometer can get a reading if they don’t.

    3. Go outside and point it at the sky and measure the temperature. Don’t include the sun or moon in the field of view. Last night with a clear sky I measured a temperature of -27 C. This morning with cloud cover it was -4.9 C.

    Do you still think the atmosphere doesn’t radiate in the IR? Do you understand why it doesn’t feel as cold on a winter night under a cloudy sky compared to a clear sky? Now think how cold it would be if the atmosphere was completely transparent to IR so that the sky brightness temperature was -270 C.

    If after all that, you still don’t believe that there is a greenhouse effect from IR active molecules like water and CO2 in the atmosphere, then I give up. I don’t have Willis’ patience.

  440. DeWitt Payne says:

    carrot eater,

    I’m appalled by what seems to be willful ignorance on both sides. It’s truly depressing that physicists with advanced degrees could think that the greenhouse effect somehow violates the Second Law. I also find statements by people who should know better that Antarctica would be the only habitable continent if GMT goes up 6 C, which is still nowhere near as high as the Eocene Optimum 55 Mya. That doesn’t mean that a 6 C rise wouldn’t cause immense problems, but it wouldn’t make most of the planet unihabitable for a species that can live in climes ranging from the Arctic to the Arabian desert.

  441. P Wilson says:

    DeWitt Payne (10:22:23) :

    of course radiation emanates from the sun and heats earth surface, oceans and atmosphere, whilst parts of the atmosphere maintain whats known as a greenhouse effect.. Its the theoretical magnitude as opposed to the real measured magnitude that we’re fascinated by. Just that with certain constants taken from physics and applied to gases, which exagerrate their radiative effect by 12 times gives us a 2-6C rise.
    those who believe the SB will need to accept the validity of AGW

  442. DeWitt,

    You’re appalled because a couple of crackpots (out of thousands of non-crackpots that read the article) post crackpot ideas on a public forum?

    You don’t get out much do you :-)

    I assume most of the people here are laymen, not professional scientists. Really, assuming you are a real scientist, why are you here except to feed a superiority complex? Otherwise, gently correct, teach, help, and ignore. Simple.

  443. P Wilson says:

    DeWitt Payne (10:22:23)

    Ok. Here’s how to proceed. Open your freezer and line it with black paper leaving a 5 inch perimeter between the paper and the wall. Set the temperature to -19C. (what the constant says is the relationship between 235w/m2 and the temperature). Put in a black matt box, which will radiate at a theoretical 235w/m2, and see if the temperature is 10 times warmer than -19C after 1 day… Or something along those lines.

    Actually this thought experiment is replicable as an experiment even with black spheres and a black steel insulator

  444. P Wilson says:

    incidentally if it does what was said above somewhere – 470w/m2 – the SB constant puts that at 29C

  445. carrot eater says:

    Nick Stokes (02:22:27) : You beat me to it; it also occurred to me today that the theoretical limit in Willis’s cartoon would be the point where the Earth got so warm that it was emitting in wavelengths which would be transmitted by the ‘steel greenhouse’.

    DeWitt Payne (10:28:30) : If somebody posing as having an advanced degree in physics is making simple mistakes like those seen above, I would assume that person is making up credentials. Maybe they had a relevant undergrad class or two.

    What bothers me is the lack of humility and self-awareness. If you haven’t a clue what you’re talking about, you should realise this, instead of spewing attitude.

  446. Willis Eschenbach says:

    DeWitt Payne (10:28:30) :

    carrot eater,

    I’m appalled by what seems to be willful ignorance on both sides. It’s truly depressing that physicists with advanced degrees could think that the greenhouse effect somehow violates the Second Law. I also find statements by people who should know better that Antarctica would be the only habitable continent if GMT goes up 6 C, which is still nowhere near as high as the Eocene Optimum 55 Mya. That doesn’t mean that a 6 C rise wouldn’t cause immense problems, but it wouldn’t make most of the planet unihabitable for a species that can live in climes ranging from the Arctic to the Arabian desert.

    There is ignorance on both sides, but I think it is honest ignorance, not wilful ignorance.

  447. Willis Eschenbach says:

    P Wilson (08:48:58) :


    I know all about the SB theory, though 7-14 microns is typical range of detection. At sunset- night, most outside objects are invisible to these wavelengths in a typical suburban environment -and humans are clearly visible across a quite a few wavelengths. A human emitting around at 8 microns average, corresponds to around 100w/m2, as the BMR is around 58w/m2 where radiation emitted is equalt to the BMR

    It leads to whether we should accept the SB for *normal* temperature matter at a constant and whether it should be reformulated to a coefficient, divided by as much as 12, to produce 30w/m2 for outgoing longwave from earth on average, since it’s clear that earth isn’t attempting to reach absolute zero in order to reach equilibrium, and emit as much as it possible can to achieve it.

    P Wilson, I think by now everyone knows that you think a basic and long-established law of physics, the Stefan-Bolzmann Law, is incorrect. I note that the page I cited give two different ways to derive the law from fundamental principles, so to overthrow the SB Law you’ll have to show that both derivations are wrong …

    However, this is not the forum to argue that issue. This thread is for people who do believe the SB Law is correct, and are trying to understand the implications of that Law.

    I’m sure there is a forum on the web somewhere for like-minded people where you can discuss your ideas. I wish you the best of luck in overturning the SB Law. However, I fear I must I ask you to let it go here and refrain from further posting on the SB law, as it is far, far removed from what we are discussing here.

    Thanks,

    w.

  448. P Wilson says:

    Willis Eschenbach (15:00:35) :

    fine, as long as you can resolve the problems outlined aabove with the values taken from biology and the problem of the difference in temperature values from a black box at 235 increasing to 470w/m2 with no additional radiation input- along with quite a number of problems associated with its application rather than just censoring them.

    you’d also have to accept that heat as an entity were a fixed permanent constant, like solid matter that doesn’t change its form – like the ideal titanium spanner, whether it is an internal energy pressure caused through friction and/or whether heat is reduced and disippated from autonomous cooling, as Kelvin considered it as “heat loss”.

    Final comment prior to censorship: What happens to heat from a red hot poker from a hot furnace and exposed to normal temperature air? Does the heat disippate/disappear or does it have to be integrated into some other entity, or is it just a case that when objects cool and emit radiation they thermalise to the temperature set by the air, as the electrons and atoms lose their excited state?

  449. Willis Eschenbach says:

    P Wilson (18:17:03) :

    Willis Eschenbach (15:00:35) :

    fine, as long as you can resolve the problems outlined aabove with the values taken from biology and the problem of the difference in temperature values from a black box at 235 increasing to 470w/m2 with no additional radiation input- along with quite a number of problems associated with its application rather than just censoring them.

    you’d also have to accept that heat as an entity were a fixed permanent constant, like solid matter that doesn’t change its form – like the ideal titanium spanner, whether it is an internal energy pressure caused through friction and/or whether heat is reduced and disippated from autonomous cooling, as Kelvin considered it as “heat loss”.

    Final comment prior to censorship: What happens to heat from a red hot poker from a hot furnace and exposed to normal temperature air? Does the heat disippate/disappear or does it have to be integrated into some other entity, or is it just a case that when objects cool and emit radiation they thermalise to the temperature set by the air, as the electrons and atoms lose their excited state?

    I fear I can’t answer your questions, P Wilson, not because I don’t want to, not because I want to “censor” you, but simply because we have such radically different fundamental assumptions. As a result, we are talking past each other, rather than to each other, and anything I say in answer to your questions will be misunderstood. Or perhaps even “misunderstood” is wrong, but understood from a very different point of view, like a reflection in a fun-house mirror where everything is there but it is distorted in a host of ways that make discussion impossible.

    I wish it weren’t so, but there it is.

    Best regards,

    w.

  450. par5 says:

    Willis, I am glad to see that you are still here. I hope that I am not being a broken record, more like a 78 with a few cracks and chips. Since there is no material in a vacuum, how are you raising the temperature? There are no gasses to excite, no material to absorb and emit, no conduction or convection. How can you raise the temperature of nothing? If a steel shell surrounded a planet with no atmosphere, no pressure, complete vacuum- would not the space between them be zero?

  451. par5 says:

    Oh, I almost forgot- I laughed my @** off when I saw that pic of Foghorn Leghorn. I love that old bird. Good times…

  452. Willis Eschenbach says:

    par5 (23:00:31) :

    Willis, I am glad to see that you are still here. I hope that I am not being a broken record, more like a 78 with a few cracks and chips. Since there is no material in a vacuum, how are you raising the temperature? There are no gasses to excite, no material to absorb and emit, no conduction or convection. How can you raise the temperature of nothing? If a steel shell surrounded a planet with no atmosphere, no pressure, complete vacuum- would not the space between them be zero?

    There’s three ways that heat moves around – conduction (things touch), convection (a gas moves the heat), and radiation (electromagnetic waves). All of these, of course, have to act on something, so a vacuum has no heat and no temperature. There’s nothing there to heat up. The temperature of a vacuum is not zero, as many people think. It has no temperature at all.

    In the situation in Fig. 1, there is no conduction and no convection between the planet and the shell. But the radiation can still move through the vacuum to heat something at the other side of the vacuum.

    And Foghorn Leghorn? He’s my main man …

    Thanks for the question.

  453. par5 says:

    Over 450 comments so far- Anthony should give you a free subscription or something. Great job Willis. Hat tip to the mods…

  454. TomVonk says:

    Willis you said :
    .
    “First let me say that you are correct. Second let me say that the differences between what you say and the results I show above are trivial, on the order of 0.2%. This is because for all practical purposes, there is no difference between the situation I describe above and two infinite parallel planes.

    This is because the radii of the sphere and the shell are equal to within less than a tenth of a percent. As is common in analysing this particular situation, this tiny error is ignored, and they are treated as though they were infinite parallel planes. See e.g. the Kiehl/Trenberth diagram I show in Fig. 3.

    I think that answers your questions, if not, let me know.”
    .
    First I am glad that you saw now what I have been meaning since the beginning :)
    Second is that I had no questions but only a few comments .
    Third is that I said from the very beginning that I shared the opinion of L.Motl , e.g there were no substantial errors with your toy system even if it had nothing to do with the atmospheric dynamics and couldn’t simulate it .
    .
    I also said from the very beginning that P.r²/R² ~ P when r~R .
    So of course I agreed with your approximation and said so .
    What I did NOT agree with and what is still wrong is the use of the word CONSERVATION of W/m² .
    I hope you see the difference between a valid numerical approximation (e.g r~R) and a statement about a general law (e.g “W/m² are always conserved”) .
    While the first is valid , the second is wrong .
    .
    My second contribution was meant to be more pedagogical in nature .
    I have experience that many people consider the SB law as being a fundamental unique law what it is not .
    When I was teaching , those people always had a very hard time to understand why a 2 faced plate could radiate 2 times “more” than what it absorbed on 1 face .
    However the reality is that there is an infinity of SB laws and SB constants depending on the geometry .
    So experience showed me that the right approach to explain this “paradox” was to say that the SB law applying to 2 faced objects had an SB constant which was TWO TIMES bigger than the SB constant of the law applied to a 1 faced (convex) object .
    I believe that this explanation is short , correct and easy to understand .
    .
    Besides as the usual SB law formulation is derived from a plane surface radiating in a half space , people can then easily understand that as each differentiable surface can be approximated by tangent planes at each point , the (usual) SB law form can be generalised to any convex body like f.ex a sphere .
    .
    This insight avoids the HORRIBLE mistake to apply the (usual) SB law to a non convex cavity as a whole .
    I had a nice exercice asking students to compute the SB constant of a T form cavity :)
    Of course this has now nothing to do with your toy system because you did use only convex bodies .
    But it is a caveat that what you did cannot be generalised to ANY form of bodies .

  455. carrot eater says:

    TomVonk (06:49:02) :

    “However the reality is that there is an infinity of SB laws and SB constants depending on the geometry .”

    I think this is a horrific way to teach the concept. There is one SB constant; by definition it does not change with geometry. However, if you want to know how two blackbodies exchange radiation with each other, then geometry is taken into account by using the view factor (or shape factor).

    http://en.wikipedia.org/wiki/View_factor

  456. carrot eater says:

    P Wilson (18:17:03) :

    “Final comment prior to censorship: What happens to heat from a red hot poker from a hot furnace and exposed to normal temperature air? Does the heat disippate/disappear or does it have to be integrated into some other entity, or is it just a case that when objects cool and emit radiation they thermalise to the temperature set by the air, as the electrons and atoms lose their excited state?”

    Through the processes of radiation and convection, the thermal energy is transferred to the surrounding air and objects. The surrounding air and objects will end up with a higher temperature than what they began with. Energy does not disappear.

    I have a great deal of difficulty understand what you’re saying in most of your comments, but it seems to me that while you know the mathematical form of the SB law, you are having trouble applying it within an energy balance. Since you are not correctly writing the energy balance, you turn back and think the mistake is in the SB law itself.

  457. P Wilson says:

    carrot eater (08:17:40)

    No problem applying the constant and the results thereof.. nothing more to say on the matter as its upsetting the balance of the discussion. Out of respect of Willis

  458. P Wilson says:

    addendum *for respect of Willis’s request*

  459. David J. Ameling says:

    This has been the most thought provoking post I have seen. A steel greenhouse also illustrates how low level particulate pollution (especially black soot) has a greenhouse effect. The greenhouse affect of low level particulate pollution has been greatly under rated.

    You can argue cause and effect, but the warmest temperatures in the US occurred during the dust bowl years.

    Smudge pots prevent orchards from freezing.

    Over the last century, farmers have been tilling more and more land causing more soil erosion and low level particulate pollution. This has an affect.

    Over the last couple of decades farmers have been switching to no plow farming. This lessens erosion and low level particulate pollution. This has an affect.

    Climate models should pay more attention to these affects and a lot less attention to carbon dioxide.

  460. yonason says:

    Willis Eschenbach (20:41:08)

    “I can’t get to whatever is the nature of your complaint,…”

    Sorry if I was too vague. TomVonk says it much more clearly and concisely, than I was able to:

    “I hope you see the difference between a valid numerical approximation (e.g r~R) and a statement about a general law (e.g “W/m² are always conserved”) .
    While the first is valid , the second is wrong .”

    He also added that “your toy system … [has] nothing to do with the atmospheric dynamics and couldn’t simulate it .” (I hadn’t even gotten that far yet.)

    Finally, the temperature (in deg K) wouldn’t double, because the heat output is dependent on the fourth power of the temps, so, if I’m correct, it would at most increase by a factor of “2^(1/4)”, not “2” (for a sphere radius only slightly greater than that of the source sphere). The same for the temp of the penultimate sphere, with every additional sphere, which would consequently increase the heat source sphere’s temp, though not in a linear way, etc. After doing the calculation, I saw that there would be a substantial increase, but as TomVank said, it has “nothing to do with the atmospheric dynamics.”

    Anyway, thanks for getting me to remember a little bit of some course material from about 30 years ago.

  461. par5 says:

    Let’s try this another way. Let’s say that we find a rogue planet with the same conditions as you descibe- no atmosphere, total vaccuum and in the middle of nowhere. Now let’s say we don’t have any steel with us. Willis and I go down to the planet to do a survey. I’ve got the flashlight, Willis has the thermometer. Willis measures 235W/m2. Since we have no steel, I decide to take the outside skin of the planet, one meter thick, and raise it above our heads by three meters. So now we are standing on the new surface, in our pressure suits, in a vaccuum, with a one meter thick old surface over our heads. Willis measures the new surface we are standing on, and it is 236W/m2. And so is the roof three meters above. And the outside of the shell is still 235W/m2. Now we know the transfer rate of the planets surface material. It doesn’t matter how high we put the shell, three meters or three km, the outside of the shell will still be 235, and the inside will be 236. Let’s do another. We take the next one meter under our feet, and raise it three meters just like the last. So now we have two shells above the planet. Willis takes a new measurment of the new surface and gets 237W/m2. He then takes a measurment of the roof and gets the same.

    We have not added any exotic materials- everything is the same. The transfer rate, planetary suface, vaccuum- same. We can continue to do this, towards the core, until the material changes. This implies three things:

    1) The outside shell will always emit 235W/m2

    2) An outer shell will always be colder than an inner shell

    3) No matter what layer you are in, the roof and the floor will always measure the same

    Willis, if you agree to this, I can build your steel green house. However, there will be one more implication.

    Cheers!

  462. George Crews says:

    Hi Willis,

    Thanks for the interesting post. I think your clever double membrane model makes the greenhouse effect a lot easier to understand.

    The other reason for my comment: I think I see an asymmetry in Figure 4. The figure has a term (53 W/m2) that represents the upwelling longwave window from the surface through the troposphere membrane to the stratosphere membrane and beyond. However, the figure does not have a term for the downwelling window from the stratosphere to the surface. I would have thought that the clarity of the downwelling window would be the about the same as the upwelling window (ignoring temperature differences).

    If so, then this would mean the size of the downwelling term would be about 20 W/m2. I’m not a Windows person (pun intended) so I was not able to run your Excel model and check how adding this term would affect it.

    And I am sorry if this issue has already been addressed, but there are a lot of comments to this post and it would be easy for me to miss the issue.

  463. Willis Eschenbach says:

    George Crews (22:37:46) :

    Hi Willis,

    Thanks for the interesting post. I think your clever double membrane model makes the greenhouse effect a lot easier to understand.

    The other reason for my comment: I think I see an asymmetry in Figure 4. The figure has a term (53 W/m2) that represents the upwelling longwave window from the surface through the troposphere membrane to the stratosphere membrane and beyond. However, the figure does not have a term for the downwelling window from the stratosphere to the surface. I would have thought that the clarity of the downwelling window would be the about the same as the upwelling window (ignoring temperature differences).

    If so, then this would mean the size of the downwelling term would be about 20 W/m2. I’m not a Windows person (pun intended) so I was not able to run your Excel model and check how adding this term would affect it.

    And I am sorry if this issue has already been addressed, but there are a lot of comments to this post and it would be easy for me to miss the issue.

    You are correct. It is a term I left out of the drawing, although it is included in the model. However, if it is very large, the surface won’t warm, so …

    The model was written in Excel on my Mac, but it should run in a Windows version of Excel, I’d be surprised if it didn’t.

    w.

  464. stunned says:

    Everyone, read
    Smith (2007), Proof of the Atmospheric Greenhouse Effect [arXiv:0802.4324v1]
    It’s only nine pages, and it has actual physics in it.
    Sorry Willis, your physics fails.

  465. Gary Hladik says:

    stunned (09:54:49), out of curiosity, where does Smith’s paper disagree with Willis?

  466. NickB. says:

    stunned (09:54:49) :

    The article you referenced – http://arxiv.org/abs/0802.4324 – was published as a response to this “falsification” of atmospheric greenhouse theory – http://arxiv.org/abs/0707.1161

    I see no attempt to falsify anything with Willis’ work here, refine maybe but not falsify

  467. HankHenry says:

    My question: How would a thermometer read in a stevenson screen on the moon. Obviously there’s no need for the louvers, but how would a thermometer register in the shade of a box. My guess is that it would end up taking on about the same temperature as whatever the roof temperature is.

    I still think the greenhouse metaphor is broken. The air in our atmosphere tempers the earth’s temperatures in both directions (judging by the highs and lows of surface temperature seen on the moon). How does the tired out old greenhouse that everyone kicks around illustrate that point?

  468. HankHenry says:

    Also of interest:

    http://www.sciencedaily.com/releases/2009/09/090917191609.htm

    Paragraphs 9 and 10 indicate that there is something like regional climate on the moon based on the nature of the surface materials. I suppose that would be things like heat capacity, conductivity, depth of material, and blackness. This seems suggestive of the difficulties of modeling climate. One needs to account for various kinds of surface material. I also want to think that somewhere one needs to account for the length of a day in any good model since we know that the extreme heat of a day does not coincide with high noon.

  469. B Louis says:

    I like the steel greenhouse, because everyone knows that a steel shed gets hot inside. Still can’t tell if the radiative greenhouse is real or falsified.

    Can someone please explain why it is that climate scientists measure “energy” in W/m2? If any of us tried that in high school physics, we would flunk. All the math/physics that follows falls over because conservation of energy is robust but conservation of radiative flux does not exist.

    I note that the Kiehl/Trenberth Global Energy Budget does not seem to make any account of earth-generated thermal pollution. Neither does IPCC. Hence climate scientists have trouble understanding Nordell’s analysis of energy, which claims to account for 74% of observed warming. I think there are some missing assumptions in Nordell and Gervet’s work but not sure of their magnitude – ie how much energy is stored in industrially generated materials and not turned to heat.
    Paper via

  470. Willis Eschenbach says:

    B Louis (13:32:04), thanks for your note. you say:

    I like the steel greenhouse, because everyone knows that a steel shed gets hot inside. Still can’t tell if the radiative greenhouse is real or falsified.

    Can someone please explain why it is that climate scientists measure “energy” in W/m2? If any of us tried that in high school physics, we would flunk. All the math/physics that follows falls over because conservation of energy is robust but conservation of radiative flux does not exist.

    I note that the Kiehl/Trenberth Global Energy Budget does not seem to make any account of earth-generated thermal pollution. Neither does IPCC. Hence climate scientists have trouble understanding Nordell’s analysis of energy, which claims to account for 74% of observed warming. I think there are some missing assumptions in Nordell and Gervet’s work but not sure of their magnitude – ie how much energy is stored in industrially generated materials and not turned to heat.
    Paper via

    While radiative flux is not conserved, W/m2 is. If a body in space is receiving X W/m2, if it is in equilibrium it is also radiating X W/m2.

    I was interested by your description of “earth-generated thermal pollution”. What is that? Do you have a reference?

  471. B Louis says:

    Nordell and Gerbett – see 2009 paper at bottom of page + some responses and replies by authors: (Sorry must have forgotten to paste the link.)
    http://www.ltu.se/shb/2.1492/1.5035?l=en

    W/m2 is radiative flux (also J/m2/s). Energy is Joules or kWh.

    (Laws of physics include conservation of mass, energy, momentum, charge.)

    So a body in space in thermal equilibrium must not retain energy. If the body was a steel sphere in the sun, the energy flux it receives is spread over its lit surface area. The heat it radiates if it conducts is over its whole surface with a different area and a different flux. Physics of a rotating earth is a bit more complex, since it never really reaches equilibrium it is always heating daytime and cooling at night. The response to the CO2 greenhouse falsification contained complex physics of that. (http://arxiv.org/abs/0802.4324). not sure how to tell who is right.

  472. Willis Eschenbach says:

    B Louis, here’s why I find Nordell’s explanation, that the warming is from human use of energy, totally unconvincing.

    In the BP Statistical Review of World Energy June 2004, I find the following world energy consumption including all sources (fossil fuels, hydroelectric, nuclear) in Tonnes of Oil Equivalent (TOE)

    9.7E+09: TOE for world, 2003

    From the same source, the energy content of a TOE is:

    4.2E+10: joules/TOE

    Multiplying these gives us a total of:

    4.1E+20: joules/yr from human fuel use.

    Since a watt is a joule/second, dividing joules/year by the number of seconds in a year gives us:

    1.3E+13: watts

    And then we have the area of the earth:

    5.1E+14: sq. metres planetary surface

    Dividing one by the other, we get:

    0.03: W/m2 from “thermal pollution”

    Given that the total downwelling radiation at the surface (short and long wave) is on the order of half a kilowatt per square metre, this is a trivially small amount from “thermal pollution”, far too small to make any discernible difference.

    Please check my numbers, but I believe that they are correct.

    w.

  473. B Louis says:

    Sorry Willis, the numbers may be correct but the assumptions have nothing to do with the reality.

    The issue is: how does one convert energy to temperature? It has nothing to do with W/m2. There is a robust physical relationship between energy, specific heat and mass. The equations used by Nordell appear reasonably robust. One can compute energy from W/m2 time m2. What is the relevant specific heat? How does one account for thermal conduction? Nordells work accounts for all of that in a more complicated but apparently robust and verifiable manner. He has checked his numbers.

    Climate scientists on the other hand defy the laws of physics. The magical formula to convert radiative flux to temperature change is (ΔTs): ΔTs = λRF, where λ is the climate sensitivity parameter

    None of that formula has any robust basis in fundamental physics. Mass is unaccounted. Specific heat is unaccounted. Lambda is extremely wooly and is derived by observation of flawed temperature data and with a range that varies by an order of magnitude. Talk about junk science. W/m2 is not a measurement of energy. m2 from a sphere is not constant when the radius of interest changes. In fact m2 has nothing to do with the planetary surface, since in the radiative balance is represents area irradiated by the sun (assuming a flat earth), hopefully correcting for the earths curve and atmospheric effects and rotation. Climate math is a lot easier if the earth is flat! Conservation of radiative flux is great for perpetual energy.

  474. B Louis says:

    If we assume the earth is in long-term equilibrium for a moment. (Gerlich and Tscheuschner argue that a mean temperature for an earth is nonsense.) Any particular patch of earth warms in the daytime and cools at night. The net energy gain is positive positive in daytime, negative at nighttime, positive in spring/summer and negative in autumn/winter more or less. So for a whole year on average, despite the high radiative flux that hits the earth on the sunny side, net energy gain is near zero because of losses on the dark side. So large radiative flux numbers on average have not a lot to do with long-term warming.

    The fact that heat is generated by burning fossil fuels is absolutely proven. The magnitude of this energy gain over the mass of the earth has been calculated. One could argue that this will be lost as radiation or convection, but that doesn’t help the flux warmist cause since a warmer earth resists further heating. (Unless one doesn’t believe in the second law of thermodynamics.)

    I read somewhere that liquids don’t emit blackbody radiation, so that puts a big dent in the radiative flux theory.

  475. B Louis says:

    If one is having trouble with Gerlich and Tscheuschner, try this:
    http://ilovemycarbondioxide.com/pdf/DEFINITIVE_DEATHKNELL_to_CLIMATE_ALARMISM.pdf

  476. Dan says:

    I disagree with the multiple steel ball idea where the inner temperature is doubled each time another steel shell is added.
    The first shell will recieve 235w/m2 and heat to some temperature where it will emit a total of radiation that will keep it in equilibrium with the surroundings. This radiation will be in inward and outward direction. Let’s say half goes inward and half goes outward. This means that an extra shell added outside the steel sphere will recieve 50% of the radiation recieved by the first shell. This outer shell will, due to the radiation of 117.5 w/m2 heat to some lower temperature than the inner shell. The inner shell will due to this recieve on its outer surface 50% of 117.5 w/m2 and heat to some degree where a new equilibrium of energy is established.
    A third shell will in analogy also recieve 50% of 117.5 w/m2 and heat to its equilibrium temperature.
    All this does not mean that the inner temperature is doubled each time we add a shell, it means that each shell added will have a diminishing effect on the inner temperature. At some number of shells, the outermost shell will have a negligble effect on the inner temperature.
    My 2c
    Dan

  477. B Louis says:

    If you think Smith’s refutation of Gerlich and Tscheuschner is successful, read this:
    Comments on the “Proof of the atmospheric greenhouse effect” by
    Arthur P. Smith
    Gerhard Kramm1, Ralph Dlugi2, and Michael Zelger2
    http://arxiv.org/ftp/arxiv/papers/0904/0904.2767.pdf

  478. Willis Eschenbach says:

    Dan (01:14:49) : edit

    I disagree with the multiple steel ball idea where the inner temperature is doubled each time another steel shell is added.
    The first shell will recieve 235w/m2 and heat to some temperature where it will emit a total of radiation that will keep it in equilibrium with the surroundings. This radiation will be in inward and outward direction. Let’s say half goes inward and half goes outward. This means that an extra shell added outside the steel sphere will recieve 50% of the radiation recieved by the first shell. This outer shell will, due to the radiation of 117.5 w/m2 heat to some lower temperature than the inner shell. The inner shell will due to this recieve on its outer surface 50% of 117.5 w/m2 and heat to some degree where a new equilibrium of energy is established.
    A third shell will in analogy also recieve 50% of 117.5 w/m2 and heat to its equilibrium temperature.
    All this does not mean that the inner temperature is doubled each time we add a shell, it means that each shell added will have a diminishing effect on the inner temperature. At some number of shells, the outermost shell will have a negligble effect on the inner temperature.
    My 2c
    Dan

    Dan, there is a misunderstanding here. The temperature is not doubled with each additional shell. Instead, with each shell the multiplier goes up by one. With one shell, you get a surface which radiates at twice the source radiation (not temperature but radiation). With two shells you get three times the radiation, with three shells you get four times the radiation, and so on.

    Sorry for my lack of clarity. Draw it up and do the math and it should be clearer.

    My best to you,

    w.

  479. DirkH says:

    “Willis Eschenbach (20:15:26) :

    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, you argued that a one-shell model does not suffice to explain earths temperature of about 15 degrees C so you went for a 2 shell model. So you’ve already left the pure world of thinking about radiation balances and start to take real world temperatures into account. So it would be only fair to take real world emissivity into account as well.

    I took the values of Alan D. McIntire (16:36:32) : for seawater emissivity and computed the changes in temperature with the SB law. When we lower the emissivity of the surface from 1 (blackbody) to 0.9425 we need an increase of surface temperature of approx. 5 degress C to achieve the same level of radiation.

    (Even though the absolute temperature appears as the 4th power in the SB law, we only have to do relatively small changes in the temperature – from, say 288 K to 293 K to compensate for this big drop in emissivity)

    I think it makes your model more realistic – it feels strange talking about blackbodies on one hand and using the real world estimates from the K/T diagram on the other hand.

    But we can say that we get a temperature change from this change in emissivity. The next question follows logically and i haven’t throught it through by now: How do cloud albedo, cloud cover, icecap albedo and land surface albedo affect the emissivity and the necessary temperature?

  480. DirkH says:

    Willis, forget my last post.

    Reading through the thread i recognized that you’re only using the numbers from the K/T diagram to have the order of magnitude right for your Gedankenexperiment with the lone planet in space surrounded by two steel shells.

    But it’s really fun to play with the SB law. I found this link above:
    http://climateaudit.org/2006/04/09/hansen-and-schmidt-predicting-the-past/
    where they point to this breakthrough proof of Hansen and Gavin Schmidt that the earth “receives more energy than it radiates” with a difference of 0.85 W/m^2.

    Now with all the latency and stuff the earth would probably heat up mightily until it balances that out… i computed it with the SB law and end up with a warming of 0.15 degrees C until it’s balanced. Impressive.

  481. BLouis79 says:

    Please note that the net difference of 0.85W/m2 is the number concocted in the computer model. Trenberth et als Global Energy Budget based on sound measurement does not balance for reasons that are unclear. It is balanced by adjustments made with reference to Hansen’s computer model. Trenberth in the CRU emails admits the budget doesn’t balance. Nordell and Gervet claim they can explain a large chunk (55-74%) of observed warming, which leaves even less atributable to “radiative forcing”. A recent attempt to account for thermal pollution does not quote Nordell in the powerpoint references I have seen but simply concludes the phenomenon can be ignored because the W/m2 is small. If we correct for airport heat island bias and thermal pollution according to Nordell and Gervet, then the net energy addition remaining to be found could easily be zero or even negative. Given the bounds of the errors in the Global Energy Budget, such an answer is not inconsistent with the measured data.

    To quote Trenberth’s latest attempt at the Global Radiative Energy Budget:
    http://www.cgd.ucar.edu/cas/Trenberth/trenberth.papers/10.1175_2008BAMS2634.1.pdf

    “There is a TOA imbalance of 6.4 W m-2 from CERES data and this is outside of the realm of current estimates of global imbalances (Willis et al. 2004; Hansen et al. 2005; Huang 2006) that are expected from observed increases in carbon dioxide and other greenhouse gases in the atmosphere. The TOA energy imbalance can probably be most accurately determined from climate models and is estimated to be 0.85±0.15 W m-2 by Hansen et al. (2005) and is supported by estimated recent changes in ocean heat content (Willis et al. 2004; Hansen et al. 2005). A comprehensive error analysis of the CERES mean budget (Wielicki et al. 2006) is used in Fasullo and Trenberth (2008a) to guide adjustments of the CERES TOA fluxes so as to match the estimated global imbalance. CERES data are from the Surface Radiation Budget (Edition 2D rev 1) (SRBAVG) data product. An upper error bound on the longwave adjustment is 1.5 W m-2 and OLR was therefore increased uniformly by this amount in constructing a “best-estimate”. We also apply a uniform scaling to albedo such that the global mean increases from 0.286 to 0.298 rather than scaling ASR directly, as per Trenberth (1997), to address the remaining error. Thus the net TOA imbalance is reduced to an acceptable but imposed 0.9 W m-2 (about 0.5 PW). Even with this increase, the global mean albedo is significantly smaller than for KT97 based on ERBE(0.298 vs 0.313).”

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