People Living in Glass Planets

Guest Post by Willis Eschenbach

Dr. Judith Curry notes in a posting at her excellent blog Climate Etc. that there are folks out there that claim the poorly named planetary “greenhouse effect” doesn’t exist. And she is right, some folks do think that. I took a shot at explaining that the “greenhouse effect” is a real phenomenon, with my “Steel Greenhouse” post. I’d like to take another shot at clarifying how a planetary “greenhouse effect” works. This is another thought experiment.

Imagine a planet in space with no atmosphere. Surround it with a transparent shell a few kilometres above the surface, as shown in Figure 1.

Figure 1. An imaginary planet surrounded by a thin transparent shell a few kilometres above the surface (vertical scale exaggerated). The top of the transparent shell has been temporarily removed to clarify the physical layout. For our thought experiment, the transparent shell completely encloses the planet, with no holes. There is a vacuum both inside and outside the transparent shell.

To further the thought experiment, imagine that near the planet there is a sun, as bright and as distant from that planet as the Sun is from the Earth.

Next, we have a couple of simplifying assumptions. The first is that the surface areas of the planet and the shell (either the outside surface or the inside surface) are about equal. If the planet is the size of the earth and the transparent shell is say 1 kilometre above the surface, the difference in area is about a tenth of a percent. So the simplification is warranted.

The second simplifying assumption is that the planet is a blackbody for longwave (infra-red or “greenhouse”) radiation. In fact the longwave emissivity/absorptivity of the Earth’s surface is generally over 0.9, so the assumption is fine for a first-order understanding. You can include the two factors yourselves if you wish, it makes little difference.

Let’s look at several possibilities using different kinds of shells. First, Fig. 2 shows a section through the planet with a perfectly transparent shell. This shell passes both long and shortwave radiation straight through without absorbing anything:

Figure 2. Section of a planet with a shell which is perfectly transparent to shortwave (solar) and longwave (“greenhouse”) radiation.

With the transparent shell, the planet is at -18°C. Since the shell is transparent and absorbs no energy at all, it is at the temperature of outer space (actually slightly above 0K, usually taken as 0K for ease of calculation). The planet absorbs 240 W/m2 and emits 240 W/m2. The shell emits and absorbs zero W/m2. Thus both the shell and the planet are in equilibrium, with the energy absorbed equal to the energy radiated.

Next, Figure 3 shows what happens when the shell is perfectly opaque to both short and longwave radiation. In this case all radiation is absorbed by the shell.

Figure 3. Planet with a shell which is perfectly opaque to shortwave (solar) and longwave (“greenhouse”) radiation.

The planet stays at the same temperature in Figs. 2 and 3. In Fig. 3, this is because the planet is heated by the radiation from the shell. With the opaque shell in Fig. 3, the shell takes up the same temperature as the planet. Again, energy balance is maintained, with both shell and planet showing 240 W/m2 in and out. The important thing to note here is that the shell radiates both outward and inward.

Finally, Fig. 4 shows the energy balance when the shell is transparent to shortwave (solar) and is opaque to longwave (“greenhouse”) radiation. This, of course, is what the Earth’s atmosphere does.

Here we see a curious thing. At equilibrium, the planetary temperature is much higher than before:

Figure 4. Planet with a shell that is transparent to shortwave (solar) radiation, but is opaque to longwave (“greenhouse”) radiation.

In the situation shown in Fig. 4, the sun directly warms the planet. In addition, the planet is warmed (just as in Fig. 3) by the radiation from the inner surface of the shell. As a result, the planetary surface ends up absorbing (and radiating) 480 W/m2. As a result the temperature of the surface of the planet is much higher than in the previous Figures.

Note that all parts of the system are still in equilibrium. The surface both receives and emits 480 W/m2. The shell receives and emits 240 W/m2. The entire planetary system also emits the amount that it receives. So the system is in balance.

And that’s it. That’s how the “greenhouse effect” works. It doesn’t require CO2. It doesn’t need an atmosphere. It works because a shell has two sides, and it radiates energy from both the inside and the outside.

The “greenhouse effect” does not violate any known laws of physics. Energy is neither created nor destroyed. All that happens is that a bit of extra energy is temporarily trapped at the surface of the planet. This warms the surface.

So yes, dear friends, the “greenhouse effect” is real, whether it is created by a transparent shell or an atmosphere.

And now, for those that have followed the story this far, a bonus question:

Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?


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422 thoughts on “People Living in Glass Planets

  1. Well, how about day & night, water vapour (incl stratospheric) & clouds, heat transport by atmosphere & water currents to the poles, albedo changes, biosphere reactions, oceanic heat storage, solar variations (incl Milankovitch cycles), cosmic ray effects, aerosols, soots, volcanoes – and I guess a few more things.
    I don’t pretend to be an expert, btw.

  2. It only deals with radiation and does not include the major energy storage and transfer mechanisms in our atmosphere water and how the maintenance of an equilibrium in the amounts of each state of water (solid – liquid – vapor) stores and releases energy independent of temperature at each phase change.

  3. TomB thanks for going first; I’m no scientist but I would guess that’s the answer; from my understanding the earth’s atmosphere is not uniform, for example there are oceans over about 70%; there are variations in cloud cover, vegetation, terrain and all these things make a difference to the way radiation is absorbed and reflected. I know this is pretty simplistic and I’m looking forward to being educated by all you clever people out there; thanks Willis for such an interesting puzzle to start off my Sunday

  4. So, if only those few molecules of water/CO2 are the reason why Earth night is warmer than night on Moon, why on Sahara, with its negligible humidity, today midnight temperature will be +19°C? Overall “greenhouse effect” there is by magnitude weaker than over equal, but humid area.

    How long the simple heat-keeping capacity of remaining >99% of the atmosphere will be totally ignored?

    Btw, real experiment was done by well known physician Woods.

    http://neighbors.denverpost.com/blog.php/2009/02/04/greenhouse-theory-disproved-a-century-ago/

  5. Here comes the -18°C again. In their excellent paper “Falsification Of The Atmospheric CO2 Greenhouse Effects Within The Frame Of Physics”, the two german physicists have clearly explained that -18°C comes from a wrong way of applying Stefan-Boltzmann law. On the non rotating and atmosphere free planet, the real average temperature would have been -129°C, which is more or less what we have on the moon. They have also explained why this glass approach is not to be used, since the so called greenhouse gazes do not behave as a physical or reflecting barrier. This post is unfortunately pointless.

  6. In response to my question, Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?:

    TomB says:
    November 27, 2010 at 12:01 pm

    Because we’re not in a vacuum inside the shell?

    True, but not the answer. Many simplified models of systems do not contain many physical attributes of the system. There is a more serious problem with using the above model to represent the earth.

  7. aarrggh!

    Some time ago this site had a GREAT graphic showing the actuality of the ‘shield’. It aint a light filter over a petri dish!

    It was a global shell AROUND a globe, not a flat shell with some translucence. I fully accept that the carbon based molecule absorbs long wave radiation and re-emits it. However that radiation emits in all directions, to be absorbed by the next molecule it happens to come across. Which then re-emits, and so on…

    But here’s the kicker: Does, or does not, that radiation travel in a straight line?

    And what happens to ANYTHING projected in a straight line from, or above, the surface of a globe?

    Now forgive me, I’m just a poor dumb engineer {and electrical, at that}, son of a poor dumb farmer, but how the hell does it not occur to most science and physics wonks to take that MAJOR factor into account?

    So, SURE, to extrapolate to water vapor.. a cloudy night after a cloudless day is gonna be warmer than a cloudless night after the same cloudless days. And maybe the day after too.. but EVENTUALLY an equilibrium is reached. Regardless of the amount of vapor on a given night.

  8. Convection and heat transport from the tropics to the poles, I’d say. That completely changes the dynamics of the whole system.

  9. The real situation is far more complex, with a continuum of temperatures through a continuous medium, not through a shell, with conduction and convection, with a water cycle, with a spectrum of IR frequencies, with the SR and IR from the sun cycling down to zero every 24 hours, with etc, etc.

  10. Because the shell is not opaque to all long-wave radiation, only that at certain wavelengths, nor is it transparent to all short-wave radition, a proportion of it is reflected, mainly by clouds.

    This is where things start getting complicated…

  11. Off the top of my head…

    1) The “shell” is a gas column of varying temperature
    2) Convection transports engery through “the shell”
    3) The shell is is a multi bandgap filter, not opaque
    4) The “shell” bandgap filters are a non-linear function of the temperature, given the that critical gas components (water vapor) undergo phase change within the range of

  12. The earth is not a black body insofar as radiation and absorption of heat is concerned. The equation-confusing factors of gases and vapours between the shell and the ‘not-black-body’ planet are conveniently missing as well.

  13. Because the atmosphere has many distinct layers, because CO2 is not the only gas in the atmosphere, because all of the gases have different properties that must all be considered, and because cloud cover is not being considered?

  14. Your results are correct, but your diagrams are sure messed up. They imply that the earth is an energy source.

  15. Where are the feedback mechanisms? No polar ice caps, no water vs. land specific heat capacity, no cloud formation.

  16. The Earth IS a long wave infrared source. Taking the ‘vacuum under the opaque shield’ instance mentioned above, the earth’s crust would not be at absolute zero because of magma temperatures bleeding off through the crust.

  17. Willis!

    Well ive had this question in m head for a long time and I think you have to calculate the atmospfere as a part of the “blackbody`s surface” to get the energy balance right.
    “The atmospfere is no shell its a part of the surface” thats my answer.
    And I am a total layman in fysics so its just a guess. :)

  18. You cannot use the word ‘equilibrium’ to describe a steady state, the thermodynamics of steady states and equilibrium systems are quite different.

    You models suck. The Earth rotates. At least 50% of the time the Earth is radiating into space and is not absorbing light. However, your greenhouse material is a superconductor and must be at the same uniform temperature all over.

    (You will also note that we can establish the (internal) heat radiated from the actual planet from the mid-winter Antarctica, -89°C or 184K)

  19. Variability of cloud cover thickness and distribution; atmospheric air flow from tropics to poles; transmission of internal core heat; there’s no “plastic shell” in reality

  20. pettyfog says:
    November 27, 2010 at 12:36 pm

    aarrggh!

    Some time ago this site had a GREAT graphic showing the actuality of the ‘shield’. It aint a light filter over a petri dish!

    It was a global shell AROUND a globe, not a flat shell with some translucence

    Dude, take a deep breath. This is a thought experiment designed to show how a planetary “greenhouse” works, not an accurate drawing of said “greenhouse”. And as Figure 1 show, it is a global shell AROUND a globe. The later figures just show a small part of that larger system.

  21. Vorlath says:
    November 27, 2010 at 12:49 pm

    Your third graph doesn’t work if the shell is opaque.

    Huh? Why not?

    Sam Hall says:
    November 27, 2010 at 12:51 pm

    Your results are correct, but your diagrams are sure messed up. They imply that the earth is an energy source.

    In all three of the diagrams, the amount of energy emitted by the earth is the same as the amount of the energy absorbed. Thus, the earth is not an energy source. It merely absorbs and then emits energy, it is not a source of energy.

  22. The alititude of the effective radiating surface of the earth is a function of concentration. Below this altitude heat is transported mainly by convection/conduction. Above this altitude it is transported via radiation. Glass greenhouses don’t do this.

  23. In response to my question, Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?, a number of answers have been proposed. These have all revolved around the idea that some part of the actual system has not been included, or not all details of some part have been included.

    But that is always true of models. They are, after all, models and not the real thing. So there will always be parts left out, or parts that don’t contain all of the real world details. In general, people have been correct, many things are not considered by my simplified model.

    But there is a much more fundamental reason why that model is inadequate to represent the Earth’s climate system in even the most simplified way … and no one has mentioned it yet.

  24. Willis Eschenbach says:
    November 27, 2010 at 1:21 pm

    Vorlath says:
    November 27, 2010 at 12:49 pm

    Your third graph doesn’t work if the shell is opaque.

    Huh? Why not?
    ———————

    Because if the shield is opaque to longwave, then the only way you can have longwave emitted from said shield is if it came from outer space. This would also mean that no longwave could escape from inside and your calculations are wrong. What your third graph is describing is something *like* a black hole with respect to shortwave radiation.

    To fix it, you’d need something that allows half of the longwave radiation through.

    Also, the planet acts as an energy source in all three graphs. Where’s that energy from?

  25. am not so sure about the left hand up arrow in Fig 4 being at 480 W/m2… I would have thought 240 W/m2…

    However, for the earth, it all depends on how opaque (or transparent) the atmosphere is… and we know cloud cover is variable… and the sun might encounter snow, ice, water, plants, sands, rocks etc on the surface of the earth… we also know that the heating effect of the sun is not instantaneous – it take times… as does the cooling in the evening… and then we have the other half of the day when the sun has gone to bed… and we know it gets cooler at night.

    Now my thought experiment is more of a real experiment… I park my sedan in the middle of the car lot at my local Wall Mart in the early morning… and I sit in the car… as the sun comes up I get hotter… and by midday it is hotter in the car than it is outside the car… and by midnight it is older inside the car than it is outside… it might even be frosted over while the parking lot is frost free. Now being tired and hungry I go home.

    So the next day I decide to change my experiment a little bit… I take my girl friend’s convertible to the car lot… now when it gets hot I open up the roof to let the heat out…. and I don’t close it at night and the car stays frost free.

    Therefore, my conclusions are:
    1) The greenhouse effect only works when the air is trapped inside the car… after all hot air rises.
    2) If the sun stayed up all night then it would be too hot all the time.
    3) If the sun stayed in bed all the time then it would be too cold all the time.
    4) It takes time for everything to warm up in the morning.
    5) It takes time for everything to cool down in the evening.

    So my guess is that the earth has a very clever mechanism called CLOUDS… they work a bit like a convertible… they form when it gets hot and stops the earth from getting too hot… and they tend to disappear when it gets very cold so the earth has its best chance of being heating up the sun… but as to pinning down the precise science of the heat flows… dream on… everything is variable and subject to time delays.

  26. The shell is best represented by a “venetian” blind, or slat blind – with the slats vertical to the surface & gaps between the slats – as the “greenhouse” is only absorbing a small portion of the radiation – the rest escapes directly to space. The slats are also “opaque” as the absorb all they can – and making the slats longer in the vertical direction (more greenhouse gases) has little additional effect.

  27. Just looked into this post and comments and I am puzzled at the lack of mention at the fact there is no real equilibrium in the real earth. solar in and energy out are highly unlikely to ever be in equilibrium, the atmosphere and oceans, planetary surface, etc are all just giant storage heaters, storing energy at daytime and releasing at night. some days the charge is hotter, some days its less hot (due to albedo etc). but the time period (lag if you like) it takes for the thermal ‘mass’ to equalise around the planet and then emit out to space and this will always define the actual climate temperature as rising or falling. (I am ignoring the different ‘spheres’ upper trop, etc).

  28. Come on this is stupid. The greenhouse effect doesn’t exist for the simple reason that there is no “greenhouse” effect in a greenhouse – it would be like using the theory of the ether to explain radio waves and talking about the “ether effect” and not expecting most people who knew what was being talked about not to roll with laughter on the floor.

    And … even putting it in quotes as in the “greenhouse” effect doesn’t explain why CO2 has to result in warming when it has a cooling effect due to the greater emissivity of CO2.

    To summarise: the greenhouse effect is Noddy science and even when it is “Noddy” science it still doesn’t follow CO2 causes warming.

  29. All right I went back and read your Steel Greenhouse post to get a clue, is the answer that upward radiation does not equal downward radiation?

  30. well, the model system doesn’t have climate- no atmosphere, water, land, or biomass- all of which constitute climate, or perhaps more correctly, weather. So of course it doesn’t explain anything about climate or why or how the energy transfer through the atmosphere to the ground and back affects climate.

    the model system is a good explanation of why the earth is in energy balance and perhaps the limits to the temperatures involved, as opposed to the current AGW hypothesis that CO2 in the atmosphere has a positive feed back that increases the response of temperature.

  31. PS
    The other big problem with Fig 4 is the right hand down arrow… although you argue that 240 W/m2 is radiated back towards earth this does not mean that all this energy is received by the earth… there is a problem with distance… remember if you stand close to the fire you get hot… and as you step way from the fire you cool down… so size is important too :-)

  32. Too many assumptions that are not real world. Kind of like thinking a sports team will always win because on paper they have the better players. Unfortunately, they still have to play a real game.

  33. BTW – is there not a shift in emission spectrum due to higher BB temp? and of course, this would constantly change slightly, resulting in permanent non equilibrium condition?

  34. But there is a much more fundamental reason why that model is inadequate to represent the Earth’s climate system in even the most simplified way … and no one has mentioned it yet.

    No thermostatic regulation in the model?

  35. PPS
    Simple models simply don’t work for complex situations!
    Just look at the accuracy of the AGW computer models if you need confirmation….

  36. A transparent shell would not be at 0K. If it were perfectly transparent (which is physically unreal, but never mind) it would retain whatever temperature it had when you first put it there, because it is neither absorbing nor losing any heat. If it were merely very nearly transparent (which is the physically realisable case) , then in this example it would be at a temperature ~ 255K. The temperature would be somewhat higher or lower than this, depending as the emissivity were higher or lower in the visible or thermal.

    This model implicitly assumes uniform insolation, which is not valid for a sphere illuminated from one direction. Steady state requires that the planet does not rotate (relative to the sun), or has negligible heat capacity; the temperature would then range from a factor 4^1/4 higher (ie 361K) at the subsolar point, through ~73K around the terminator, to ~0K on the dark side. There would be no meaningful average or global temperature at all. The greenhouse effect of the shell would be the same factor of 2^1/4 (for the visibly transparent, thermally opaque case). Add diurnal rotation and finite heat capacity and thermal conductivity, and things become much more complicated. There is no steady state, and the temperature curve for a given latitude is messy to compute. The thermal capacity of the shell has to be considered too. The general effect of thermal capacity and rotation is to transfer (some) heat between the day and night sides.

  37. The comments that dispute Willis’ excellent post highlight some of the problems with the AGW sceptic arguments. Those who argue that the greenhouse effect does not exist do the sceptics’ side a disservice, by arguing an untenable position. It creates a weakness in the sceptical position, which warmists focus on to try to discredit all sceptic arguments.

  38. In other words, the atmosphere “denies” entry of a lot of what the sun etc tosses our way, but not all, and that changes all the time for various reasons.

  39. Willis – I can’t be sure what model inadequacy you have in mind, but perhaps the answer is that climate involves much more than just temperature. Another thought is that climate (and weather) vary with time and location on the Earth’s surface, but your model seems incapable of representing either of these aspects.

  40. DocMartyn says: The Earth rotates. At least 50% of the time the Earth is radiating into space and is not absorbing light.

    Have I missed a response to this? The graphics appear to assume a Flat Earth, which reinforces what the RealClimate folks have been saying about climate skeptics… ;-)

  41. I guess that a couple of relevant factors would be the extinction coefficient or saturation of the long wave absorption capacity of the shell and the permanence or otherwise of the shell (residence time of GHG in our case).

  42. Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?

    Too hot?

    Third picture shows equilibrium of ~30 C at the surface. Average for Earth is lower than that.

  43. Here is my short explanation of why it is wrong (this could be expanded into several books).

    At the real Earth surface, the atmosphere has a molecule every 3.7e-26 square metres. The longwave photons, therefore, have to travel by/through 5.4e+30 (54 with 29 Zeros behind it) layers of atmospheric molecules (opaque shells) on its way to space.

    About 20% of shortwave photons from the Sun hits one of those 5.4e+30 molecules on its way to the surface so a longwave photon only needs to hit a few more than 20% to cause just enough lag in the time between the two types of photons so that the Earth’s surface is warmer than it should be. Then factor in the fact that the Earth’s land and ocean surface then absorb 85% of the 85% of the shortwave photons that reach the surface and those land molecules hold onto them for a little while and the surface is a nice comfortable 288K (for some of us).

    All physics equations should have “Time” included in them. Given that 1 watt/m2 = 1 joule/m2/second, the shortwave photons actually come in 960 joules/m2/second at the height of the day (which is 3e+21 individual sunlight photons per square metre per second) and 0 joules/m2/second at night (but don’t forget the Cosmic Background Radiation).

    The surface temperature at 2 metres, however, only changes by +0.001 joules/m2/second during the day and loses 0.0017 joules/m2/second through the night despite the large difference in shortwave energy coming in during that 24 hour period. Given the longwave photons are at lower energy levels, there are still almost as many thermal photons emitted per second at night as the sunlight coming in at the height of the day. Lots of layers and molecules have to be navigated by that unbelievable number of photons even though they travel at the speed of light and could escape the atmopshere in 0.0003 seconds if they weren’t intercepted so often.

    The greenhouse effect description needs to move to the level it occurs at, the quantum level. Technically, each shortwave photon from the Sun ends up in about 8 billion different molecules before it escapes from the Earth system – that is a lot of random walking for a lot of individual photons.

  44. Vorlath says:
    November 27, 2010 at 1:30 pm

    Willis Eschenbach says:
    November 27, 2010 at 1:21 pm

    Vorlath says:
    November 27, 2010 at 12:49 pm

    Your third graph doesn’t work if the shell is opaque.

    Huh? Why not?
    ———————

    Because if the shield is opaque to longwave, then the only way you can have longwave emitted from said shield is if it came from outer space. This would also mean that no longwave could escape from inside and your calculations are wrong. What your third graph is describing is something *like* a black hole with respect to shortwave radiation

    Vorlath, the shell (shield) absorbs shortwave and emits longwave … as does almost everything on the earth. We are warmed by the shortwave radiation of the sun, and re-radiate that energy as longwave (infra-red) radiation.

    Energy can change from one form to another (chemical to mechanical, mechanical to heat, etc.). It cannot be created. So the Figures above show that at equilibrium the total energy entering something equals the energy leaving it, although the energy may leave in a different form from the way it entered the system.

  45. Malaga View says:
    November 27, 2010 at 1:30 pm

    am not so sure about the left hand up arrow in Fig 4 being at 480 W/m2… I would have thought 240 W/m2…

    In Fig. 4, the surface receives 240 W/m2 from the sun, and 240 W/m2 from the shell … how much should it radiate at equilibrium, if not 480W/m2?

  46. Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?

    Earth is a sphere?

    Your picture illustrates one point on the surface. But if you “zoom out” to show whole earth, you realize it’s receiving radiation on one hemisphere but radiating to space from a full sphere.

  47. Kev-in-UK says:
    November 27, 2010 at 1:37 pm

    Just looked into this post and comments and I am puzzled at the lack of mention at the fact there is no real equilibrium in the real earth. solar in and energy out are highly unlikely to ever be in equilibrium, the atmosphere and oceans, planetary surface, etc are all just giant storage heaters, storing energy at daytime and releasing at night. some days the charge is hotter, some days its less hot (due to albedo etc). but the time period (lag if you like) it takes for the thermal ‘mass’ to equalise around the planet and then emit out to space and this will always define the actual climate temperature as rising or falling. (I am ignoring the different ‘spheres’ upper trop, etc).

    The equilibrium condition is an average condition, not an instantaneous condition. It merely means that over some longish time period the temperature of the planet doesn’t change. Since the temperature is not changing we know that on average the planet is in equilibrium.

    And while instantaneous equilibrium certainly doesn’t exist, as you say, long-term equilibrium exists for any planet which doesn’t change average temperature over some longer period.

    Since the average temperature of the Earth hasn’t changed even a single degree in a century, we can see that although there are always spatial and temporal imbalances, overall, the earth is at something very near to an equilibrium state.

    In my thought experiment, of course, such daily and hourly temporal and spatial imbalances don’t concern us.

  48. Mike Haseler says:
    November 27, 2010 at 1:39 pm

    Come on this is stupid. The greenhouse effect doesn’t exist for the simple reason that there is no “greenhouse” effect in a greenhouse – it would be like using the theory of the ether to explain radio waves and talking about the “ether effect” and not expecting most people who knew what was being talked about not to roll with laughter on the floor.

    And … even putting it in quotes as in the “greenhouse” effect doesn’t explain why CO2 has to result in warming when it has a cooling effect due to the greater emissivity of CO2.

    To summarise: the greenhouse effect is Noddy science and even when it is “Noddy” science it still doesn’t follow CO2 causes warming.

    Mike, it is merely called the “greenhouse effect”. I put it in quotes and referred to it as the “so-called greenhouse effect” so that foolish people wouldn’t assume that it had anything to do with a real greenhouse … guess I’ll have to re-think that plan.

  49. Malaga View says:
    November 27, 2010 at 1:46 pm

    PS
    The other big problem with Fig 4 is the right hand down arrow… although you argue that 240 W/m2 is radiated back towards earth this does not mean that all this energy is received by the earth… there is a problem with distance… remember if you stand close to the fire you get hot… and as you step way from the fire you cool down… so size is important too :-

    You are missing a couple of things. One is that as you move away from a fire, you absorb less energy because you intercept less of the radiation. But in the figures above, all inward radiation from the shell is intercepted by the earth.

    The second thing is that air absorbs some of the heat from the fire, just like the atmosphere absorbs some of the heat from the earth … but my thought experiment is in a vacuum.

  50. Willis Eschenbach says:
    November 27, 2010 at 2:36 pm

    Vorlath, the shell (shield) absorbs shortwave and emits longwave … as does almost everything on the earth. We are warmed by the shortwave radiation of the sun, and re-radiate that energy as longwave (infra-red) radiation.
    ———————-

    I’m talking about Figure 4, the third graph.

  51. Your model is inadequate because the earth rotates and so the incoming radiation will oscillate. In other words, your model is static, while the real world i dynamic.

  52. My question was:

    Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?

    A number of people have proposed that the problem with the model is that it is missing something – clouds, atmosphere, water, feedbacks, the list is long. But that’s not the main problem with the model. Even if those were fixed, this model wouldn’t work

    There is a reason that this model cannot physically represent the Earth, even in a simplified manner. And unless I’ve missed it, no one has mentioned it yet.

  53. I think it’s erroneous to add the Wm2 from the sun to the Wm2 from the shell (atmosphere) which effectively doubles the energy i,e, the original 240Wm2 becomes 480 Wm2.

    Take a basic strip heater of say 1000W, the above fallacy suggests all we have to do is add a well conducting bar across the front of the element, say an inch away (unwired and not receiving any power, i.e. benign) and voila’ we now have a strip heater putting out twice the energy.

    Can someone help me with patents please?

  54. The greenhouse effect is the most abused non-existing item on this planet.
    Let me explain:

    CO2 does have an absorption in the far IR. However, H2O does also. CO2 sits in a window of H2O vapour.

    Both of them absorb the black body radiation of the earth. The fact is, earth is NOT a perfect black body. It is a reflecting, colourful body, which means that it reflects a lot of its incoming energy. When emitting, it will only emit radiation with a max temp of 300K. Which means that it does nothing to heat the atmosphere. All atmospheric gasses emit the same amount of radiation at their black body temperatures. Since there is no boundary layer, emissions cancel out.

    Ther is no green house effect.

  55. The shortcoming in the model is that each molecule absorbs heat emitted from the earth, the re-radiates it out spherically, meaning that half of the energy is returned to the earth, and that this process repeats itself until vacuum is reached. So the surface will be warmer than the top, though the earth will always find equilibrium and emit its 240 W/m2. Or did I not get the right answer?

  56. My use of the greater than symbol has messed up the html! Again:
    Day/night cycles are not considered. Day: outgoing is less than incoming. Night: outgoing is greater than incoming. The change (raising) of surface temperature is the difference between these two budgets.

  57. Dear Folks,

    There is a simple way to understand this effect. It is a case of: Radiant Power in = Radiant Power out

    The radiant power in is simply (S alpha piR^2), where S is the solar constant (watts/meter^2), alpha is the absorption coefficient, and piR^2 is the area of the Earth’s disk in terms of its radius R.

    The radiant power out is given by the grey-body version of Stefan’s Law, equal to epsilon signma T^4 (4piR^2), where epsilon is the emission coefficient, sigma is the Stefan-Boltzmann constant, T is the surface temperature of the Earth, and the expression in parentheses is the surface area of the sphere of the Earth.

    If we equate them, we get

    S alpha piR^2 = epsilon sigma T^4 (4piR^2)

    By cancellation and rearrangement of terms, the temperature is solved as

    T = (S alpha / 4 epsilon sigma)^1/4

    Now we are in a position to understand the influence of the various parameters. S and sigma are physical constants and are not subject to variation (in this analysis). The absorption coefficient, alpha, may vary from zero (perfect reflection) to unity (perfect absorbtion) and characterizes the Earth’s absorption of radiant power from the sun. If alpha = 0, the Earth cannot absorb any power and will cool to zero temperature. Likewise, the emission coefficient, epsilon, may vary from zero to unity and characterizes the Earth’s emission of thermal power from Stefan’s Law. If epsilon = 0, the Earth cannot emit any power and the temperature will increase infinitely.

    The crux is this: if alpha = epsilon (both close to unity), then a certain equilibrium temperature will be reached that will be close to the “airless planet” temperature (for S = 1400 w/m2 and sigma = 5.67 x 10–8 w/m2K4 this will be T = 280.3 K or 7.15 C).

    However, if alpha > epsilon, as is the case when there is imperfect emission (grey-body conditions, epsilon < 1), then the equilibrium temperature will be higher (if alpha/epsilon = 1.3 then T = 299.3 K or 26.15 C). This is what happens with a re-radiative atmosphere; it is simply equivalent to the Earth’s emission coefficient being reduced by the thermal radiation backscatter from the atmosphere.

    Contrariwise, if alpha is reduced (as by higher albedo from increased clouds), the equilibrium temperature will be lower. As this equation makes clear, albedo has direct leverage over the result, whereas CO2 has very indirect leverage.

    The mechanics of heat transfer within the atmosphere and to and from the oceans and land masses are details that affect the values of alpha and epsilon, but are not otherwise involved in the equilibrium radiative thermal balance among the Earth, Sun, and outer space.

  58. Willis Eschenbach says:
    November 27, 2010 at 2:39 pm
    In Fig. 4, the surface receives 240 W/m2 from the sun, and 240 W/m2 from the shell … how much should it radiate at equilibrium, if not 480W/m2?

    Because a Thermos Flask helps keep things hot… but it doesn’t make them hotter… your only input is 240 W/m2 from the sun!

  59. Willis, you have had the same answer twice; your diagram needs to show
    Sun
    ————–Shield

    EEEEEEEEE Earth

    ————–Shield

    The shield radiates at the same temperature all over the sphere, but half the sphere is in darkness. This is why the heat flux budget diagrams are such bollocks, the Earth rotates. We do know that the Earth generates heat from internal radioactive decay; something missing from most heat budget.

  60. Willis, I must be missing something too…
    but I do not quite understand your query…the reason, in basic terms that the planetary surface is ‘warmer’ is because the rate of thermal energy (or rather just energy) transfer is reduced or ‘slowed down’ by the shell.
    In terms of light transmission through a substance, the light is bent and deviated, resulting in slower (altered) transmission (i.e. refraction – which is different for different wavelengths too!) and this means the light travels a longer path. This could be analagous to the thermal ‘storage’ in that the atmosphere acts as a deviation for the thermal ‘light’ energy (ok – its radiation, but it is an analogy) – thereby lengthening its stay in the atmosphere, which is not a vacuum and its contents can be excited (vibrated) by the passing radiation, thereby allowing it to warm the atmosphere.
    Mind you – it’s 11pm here in UK, and after a very nice red wine, with nothing on the TV – this may not be making much sense! LOL

  61. Willis,
    Your model does show some essential features that determine planetary “greenhouse effects”. However, it does not tell the full (still simplified) story of what actually happens. The following is a somewhat more complete version. The point is that it is actually the combination of the fact that a lapse rate becomes established in combination with the movement of the outgoing radiation to a high altitude that is required.

    Atmospheric greenhouse effect:
    Once a sufficient amount of atmospheric long wavelength IR absorbing gases (called atmospheric greenhouse gases by convention) are present, they reduce the direct radiation transmission to space enough so that convection and relayed radiation becomes the dominant modes to transport absorbed energy from the surface to a high altitude where it is radiated to space. The only way the energy can be radiated to space from the high altitude is from the absorbing and radiating gases (or clouds, but that is another story). Once this situation is encountered, adding more absorbing and radiating gases (but not enough to significantly increase the total mass of the atmosphere) do not change the fact that convection is still the dominant heat transport mechanism within the atmosphere. In fact, this reduces the radiation conduction, so that convection carries even more of the total energy to the high altitude. However, adding more absorbing gas does raise the altitude of outgoing radiation somewhat (not because of the tiny added mass, but because the height where the radiating gas concentration is suitable to radiate to space is increased). It is this increase in altitude of the outgoing radiation that results in the slight temperature increase.

    Once the dominant mode of heat transport is convection (buoyancy, winds, and turbulent mixing), the atmosphere will form and maintain (on the average) an adiabatic lapse rate. This lapse rate is a temperature gradient due to the cooling effect of rising gas in a dropping pressure (due to gravity). Evaporation from the surface and condensing water vapor in the atmosphere change the level of the lapse rate from the dry air value, and this is called the wet adiabatic lapse rate. The outgoing radiation has to equal the incoming absorbed radiation unless the temperature is changing, but here we consider the case where the temperature has leveled off for simplicity (as it has on Earth for the last decade or so). In that case, the match of radiation out to space, from a particular effective altitude, to absorbed input radiation, determines the effective temperature of the gas at that altitude. This temperature is then added to the lapse rate times the effective altitude of outgoing radiation, and this gives the ground effective temperature. The combination of moving the location of the effective level of the atmosphere, where the radiation to space occurs, to a higher altitude, and adding the effect of the lapse rate times increased altitude is the source of higher surface and low altitude temperatures.

  62. Your model assumes a transparent shell and a vacuum within. These are not true for the earth since clouds can reflect or scatter the short wave radiation and the atmosphere is not a vacuum.

  63. I am intrigued by this thread.
    Perhaps I have missed some glaring fundamental principle but the only answer to the bonus question that I can come up with is the scale of complexity of the model.
    For any realistic calculation to be made the model would have to include all the goings on of the many layers of the shield as well as other factors like convection , cloud formation etc.
    I think it is safe to say that once the radiation has arrived on Earth it can never again be emitted as short wave so the question is whether the long wave emissions can maintain the energy balance. It occurs to me that the long wave opacity of the atmosphere might have gaps in it which will affect the numerical calculations but I do not see any holes in the principle.
    In summary: My guess is the inability of the basic model to cover all necessary parameters.

  64. Molecules, like air, do not reflect energy. If they absorb, and then retransmit, it is not directional, but in all directions from the molecule, as a tensor, not a vector.

  65. It’s the impervious nature of a shell. Like the glass in a real greenhouse. It traps and elevates heat. Our atmosphere does not resemble glass or a shell.

    If it did we would all be cooked.

  66. Willias,
    I forgot to add: The higher value of the ground temperature results in a higher radiation out level from the ground than if the ground were not as hot. However, the atmospheric radiation from just above the ground (due to the local atmospheric temperature acting on a radiating gas) is omnidirectional, and the downward component, called back radiation, almost equals the upward radiation from the ground, resulting in a greatly reduced NET radiation heat transfer. It is convective heat transfer that carries most of the energy from absorbed solar radiation in the ground to the upper atmosphere. Thus the higher radiation level up and back radiation do not cause the heating, they are a result of it.

  67. I think Paul Birch got it. The earth is a sphere, with a “point” source of energy in the sun a single 1D model can’t capture the system.

  68. I’ll just say it again as clear as I can since I and others have mentioned it several times. The last figure. Figure 4. The math is wrong. If the shell is opaque to longwave, then you have two possibilities for the 240W/m2 radiating out into space from the shell.

    1. Longwave is coming from space or the Sun.
    2. Shortwave is coming from the surface of the Earth and converting to longwave as it passes through the shell.

    Neither of those are shown in your graph. You have all of 240W/m2 going to the surface of the Earth. So option #1 is no good. Second, you have blue lines indicating longwave radiation on the inside of the shell. So option #2 is no good. Option #2 has another issue that makes it wrong is that the line of radiation from the Sun is shown to go all the way through to the surface of the Earth without any conversion. So if there is no conversion going in, then there is no conversion possible going out either.

    IOW, the math is critically wrong in figure 4. Please fix this before continuing.

  69. Well this looks as if it will be an interesting thread.

    I am not sure what the most important missing factor is, but would vote for the fact that the earth rotates. This means that on average the incoming radiation only warms some of the time, but the outgoing is present all the time. However this only affects the average temperature, and the basic principle that warming occurs is still correct.

  70. The Earth is vastly more complicated than the model, and has many known and unknown feedback mechanisms.

    Most of these must be negative feedbacks, or we’d have had a runaway greenhouse or freezer effect eons ago and life would not have arisen.

  71. Willis – Roy Spencer claims that there is a net negative feedback which more than counters the CO2 effect.
    He claimes that this is not just theory but that he has measured it from observations.
    Please comment.

  72. The “greenhouse effect” is, if I remember, only a PR definition that was used to simplify stuff for the policy makers or something.

    Secondly the greenhouses isn’t used to trap heat per se (if they were they’d be padded down with loads and loads of insolation) but to control the inside environment for maximum growth using sunlight, artificial light and heat (for night time and winter), humidity, CO2 and Nitrogen levels, pesticides, and of course water flow and nutrients. CO2 is of course not used to trap heat or to add to the heat because if it were that much of a problem plants getting heat stroke would be a real problem and you’d end up spending more money on cooling systems, so the trace gas is simply used as nutrient in a controlled environment.

    So the analogy that gases acts with a “greenhouse effect” is just bad since the greenhouse is to control a balanced environment.

  73. I get the feeling I’ll be slapping my forehead and exclaiming; Doh! I shoulda thoughta that one!

    Let me be the first (?) to state that I am in fact a blathering blowhard, and a stuffed shirt! I take what little I know and make up the rest! I pass wind at your silly little mystery question! Hahahahahahahahahahaha… I’m dressed like a turnip!

  74. Willis,
    I see the following as a being a problem with your model:

    The incoming energy (from the sun) you express in w/m^2, lets simplify it even more and say that energy is delivered in truckloads.

    Lets say we get 2 truckloads per hour.

    With your transparent shell, we end up with 2 truckloads in and two truckloads out, things are balanced.

    But when we come to your semi-transparent shell, you are still getting two truckloads per hour, but you say that these two truckloads are delivered to both the earth and to the shell — that makes 4 truckloads/hr.

    Where did the extra two truckloads come from?

    You then make things even worse, by saying that the magical two extra truckloads that are delivered to the shell are again magically multiplied, with two going out to space, and two more going to the earth.

    Where do these magical extra truckloads of energy come from?

  75. “There is a vacuum both inside and outside the transparent shell. ”
    =======
    A vacuum inside ??
    Not my kind of planet.

  76. Willis I’m having trouble with fig.3

    If the shell is radiating 240Wm2 to space AND 240Wm2 to the planet, it must therefore be receiving a total of 480Wm2, which is what fig.3 shows.
    IF it is receiving 480Wm2, how can it be the same temp as the planet surface i.e. 255K or -18C? The planet surface is receiving only 240Wm2.

    Maybe I’m not comprhending this situation too well. To put it another way..
    Take an energy source radiating 240Wm2 and point it at a black body object. This object will reach 255K.
    Now, if we add a SECOND energy source of 240Wm2 from the opposite direction, according to fig.3 this object will stillbe at 255K???

  77. If it is not the cloudy thingies, could it be that the basic theory is a load of dung beetles raison d’etre?

  78. “Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?”

    Because the oceans also have their own “greenhouse effect”. They allow downward shortwave radiation (visible light) to warm the oceans to depths of 200 meters (with the warming diminishing with depth), but only release heat at the surface. John Daly made the argument that a planet that was all ocean would be warmer than a planet of all land. Refer to the following post inder the heading of “The oceans also behave this way”:

    http://www.john-daly.com/deepsea.htm

  79. “Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?”

    When your shell absorbs the blackbody radiation from the planet, it heats up to a lesser temperature than the planet’s surface, so it will emit a lower frequency blackbody radiation than it received. CO2 re-emits the same frequencies it receives.

  80. Wait. Your figure 3 is wrong. 480 W go up to the shell, it must reach the same temperature as the planet’s surface, otherwise it can’t re-emit 480 W. (Again, this is not the case with CO2 in the atmosphere, as it re-emits the frequencies it receives due to absorption bands, not blackbody spectra)

  81. Observed from a distance, Mr Eschenbach’s imaginary planet and shell would appear to be in equilibrium at 240 w/m2 in and 240 w/m2 out in the first part of his explanation. In the latter part of his explanation, with the shell opaque to LW, oberved from a distance the planet would appear to be in equilibrium at 240 w/m2 in and 240 w/m2 out. In other words, the distant observer would see both situations as the planet being the exact same temperature.

    The only thing the observer might notice out of the ordinary is that if the imaginary shield went from “clear” to “opaque” to LW in an instant, there would be a perturbation (spelling not my strong point at best of times, so I’m pretty certain that’s wrong) to the outgoing energy until equilibrium was re-established. The distant observer would conclude that “something happened” but that the temperature of the planet as observed from space was the same before and after.

  82. The energy conversion between the different forms of energy is not perfect, there is loss, entropy. No one has mentioned that yet.

    “In physics, the term energy describes the amount of work which may potentially be done by forces or velocities (kinetic energies) within a system, without regard to limitations in transformation imposed by entropy. Changes in total energy of systems can only be accomplished by adding or subtracting energy from them, as energy is a quantity which is conserved, according to the first law of thermodynamics. According to special relativity, changes in the energy of systems will also coincide with changes in the system’s mass, and the total amount of mass of a system is a measure of its energy.

    Energy in a system may be transformed so that it resides in a different state. Energy in many states may be used to do many varieties of physical work. Energy may be used in natural processes or machines, or else to provide some service to society (such as heat, light, or motion). For example, an internal combustion engine converts the potential chemical energy in gasoline and oxygen into heat, which is then transformed into the propulsive energy (kinetic energy that moves a vehicle.) A solar cell converts solar radiation into electrical energy that can then be used to light a bulb or power a computer.

    The generic name for a device which converts energy from one form to another is a transducer.

    In general, most types of energy, save for thermal energy, may be converted to any other kind of energy, with a theoretical efficiency of 100%. Such efficiencies may even occur in practice, such as when potential energies are converted to kinetic energies, and vice versa. Conversion of other types of energies to heat also may occur with high or perfect efficiency.

    Exceptions occur when energy has already been partly distributed among many available quantum states for a collection of particles, which are freely allowed to explore any state of momentum and position (phase space). In such circumstances, a measure called entropy, or evening-out of energy distribution in such states, dictates that future states of the system must be of at least equal evenness in energy distribution. (There is no way, taking the universe as a whole, to collect energy into fewer states, once it has spread to them).

    A consequence of this requirement is that there are limitations to the efficiency with which thermal energy can be converted to other kinds of energy, since thermal energy in equilibrium at a given temperature already represents the maximal evening-out of energy between all possible states. Such energy is sometimes considered “degraded energy,” because it is not entirely usable. The second law of thermodynamics is a way of stating that, for this reason, thermal energy in a system may be converted to other kinds of energy with efficiencies approaching 100%, only if the entropy (even-ness or disorder) of the universe is increased by other means, to compensate for the decrease in entropy associated with the disappearance of the thermal energy and its entropy content. Otherwise, only a part of thermal energy may be converted to other kinds of energy (and thus, useful work), since the remainder of the heat must be reserved to be transferred to a thermal reservoir at a lower temperature, in such a way that the increase in entropy for this process more than compensates for the entropy decrease associated with transformation of the rest of the heat into other types of energy.”

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

    How this applies to the alleged greenhouse effect is an interesting question.

    How it applies to the actual real planet Earth is an even more interesting question.

    Is the effect significant? If so by how much and where does it take place?

  83. I do believe that there are errors in figures 3 and 4. It is stated that the shells are at a temperature equivalent to 240 W/m^2, but the sum of the energy emissions is actually 480 W/m^2, because there is 240 W/m^2 towards earth and 240 W/m^2 out into space for both. This might be a source of confusion.

    I’m interested in hearing where the failure is in the model. Nothing really sticks out to me.

  84. I’m not sure what deficiency Willis is thinking of, but the one that occurs to me is that the main transfer of energy between the surface and the ‘shell’ in the real planet is not radiation but convection. There is enough GH gas in the atmosphere to absorb all the photons in the absorption bands within about ten metres. The main radiation layer is where the H2O ceases (the upper atmos. is dry) and the H2O radiation band gets a clear line of sight to outer space – that is the height corresponding to Willis’ shell. Hot air rising from the surface layer to the upper atmosphere is the main energy transport mechanism, and so the extra 240 Wm-2 is not needed to get the energy up there.

    Also, people keep saying “multiple shells”. The above is the reason why that is not the reason.

  85. Am going to hit the sack now but I just wondered if Willis’ query is perhaps related to the enthalpy of the system?

  86. I’m going to have to agree with those who point out that the shell is not at thermal equilibrium. Our atmosphere is not made of a shell of Maxwell’s daimon’s radiating with a directional variance like that.

  87. A further thought experiment…

    I take a hot rock and place it outside where it’s zeroDeGC. The hot rock will cool down in a given period of time.
    If I repeat the experiment, but this time place a 2nd hot rock alongside the first one, the first hot rock will take some time longer to cool down compared to the first experiment, BUT IT’S TEMPERATURE WILL NOT RISE. This we know to be true.

    Hence this is the problem facing the proponents of the Enhanced Greenhouse Effect (EGHE) hypothesis. The EGHE should manifest itself in higher minimum temps. The proponents need to explain why we should rearrange our energy use and economies just because of higher minimum temps.

    If they believe higher min T’s will eventually lead to higher max T’s, they then need to first prove this, and second, demonstrate empirically how long it will take before we reach these higher max T’s. Is it 1yr? 5yrs? 20yrs?

    So far as I know, this hasn’t been done

  88. Firstly, your link to Judith Curry’s site is not correct (wrong format)

    Now let try this by the Sherlock Holmes principle of eliminating the unnecessary.
    Things which only affect the actual magnitude of the effect, and not its existence, are not necessary for a proof-of-concept model. So, I earlier suggested that the earth’s rotation was needed, but it is not, as that only affects the actual temperature, not the existence of warming. Likewise clouds – they affect albedo, but that does not matter. Oceans likewise – who says this model planet has oceans. Is an atmosphere needed? Well this provides extra mechanisms for heat transport i.e.convection but that does not seem to be essential.
    Objections to the name “greenhouse effect” on the grounds that it is not a real greenhouse are just silly, and objections to the energy balance in figure 4 are also silly (figure 4 is correct).
    What else – the shell in the model has to be infinitesimally thin otherwise you need convection to get heat through – is that it? You need several concentric shells of finite thickness in a real system?

  89. I think the clue is given away in the question and that the simplification of a thin shell is the weak point. If you allow for a thick shell with a temperature gradient you can find a heat balance that allows different temperatures for the planetary surface. For example a planetary surface temperature of 35C (510 W/m2) and a temperature of -10C for the lower surface of the shell (270W/m2) would balance. The key then would be to find how the heat gradient can be calculated through the shell.

  90. One thing for certain, is that there is no “barrier” separating the atmosphere from space as in the model presented.

  91. I have observed that women like Dr. Curry have demonstrated that they can break through the gas ceiling. ;>)

  92. I take back my remarks about the temperatures of shells. I missed the fact that the shell has twice the area since it has a top and a bottom, and so the per-area radiation is still just 240 W/m^2 for the shell.

  93. Lets do a thought experiment on these shells of yours.

    What happens if the shell is further away or closer. Does the temperature increase or decrease?

    What happens if there are several shells? Does each lower level increase in temperature? Could we build a device that uses solar energy with say 500 shells that creates a temperature of say 25300 kelvin?

    So if we look at what happens with the “greenhouse effect” we currently have, does moving the distance from say 10 feet off the ground to only 5 feet off the ground increase the temperature, because the bandwidth of the spectrum in question is fully absorbed within 10 of leaving the ground surface by CO2.

    Does the fact that we are not in a vacuum change the temperature? My understanding is that as pressure increases, so too does the temperature. It has been shown that the density and weight of the Venusian atmosphere captures 100% the temperature one would expect without CO2 methane and other “greenhouse gasses” in the atmosphere.

    To point, the 255 kelvin theoretical temperature of the Earths surface is as a black body with nothing surrounding it allowing the heat to escape back to space. But our Earth does not end at the surface, there is an ocean of air that extends this surface outward several miles. Thus, the temperature of 255 kelvin would be at the central point in the system from which the radiation is emitted, several miles above the surface of the Earth. Points below this position would increase in temperature and those beyond it decrease in temperature. So, is 30 degrees kelvin beyond the scope of this?

    Basically, your whole idea is a joke original author, and while it can make sense, it is not compatible with what science shows.

  94. AusieDan says:
    November 27, 2010 at 3:51 pm

    Willis – Roy Spencer claims that there is a net negative feedback which more than counters the CO2 effect.
    He claimes that this is not just theory but that he has measured it from observations.
    Please comment.

    Roy generally knows what he is talking about. I say something similar, that there is both net negative feedback and also an active governor system that regulates the earth’s temperature. However, neither is the subject of this thread.

    Thanks,

    w.

  95. @Eschenbach
    > Finally, Fig. 4 shows the energy balance when the shell is transparent to
    > shortwave (solar) and is opaque to longwave (“greenhouse”) radiation.
    > This, of course, is what the Earth’s atmosphere does.

    The Earth’s atmosphere is not completely opaque to longwave. There is a transparent transmission band extending from 8 to 13 microns, in the center of the terrestrial thermal black body curve. This allows up to 30% of the longware to escape.

    The rest of the longwave is absorbed by the atmosphere, mostly by water vapor.

    Note that CO2 has 4 absorption bands, but the 2,3 and 5 micron bands are not in the main part of the thermal curve and don’t have much effect. The 15 micron band is the largest and is centered in the 310K thermal curve.

    CO2 is a powerful absorber of longwave energy. Though present in only “trace” amount (380 ppm), it completely absorbs 100% of the 13-16 micron longwave emitted from the surface, all within the few hundred meters above the surface.

    But adding more CO2 does not make this band “more opaque”. Water (in all of its states) is doing all of the “heavy lifting” here on Earth, in terms of warming and regulating Earth’s climate. CO2 has a relatively small “greenhouse effect”.

    To see this, consider Mars, whose atmosphere is 95% CO2. Though much thinner, it contains almost 30 times as much CO2 per surface area unit, than Earth. Yet it has virtually no greenhouse warming effect: the mean surface temperature is very close to the theoretical black body temperature of 210 Kelvins.

    Mars Facts http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html
    Visual geometric albedo 0.170 (Earth 0.367)
    Solar irradiance (W/m2) 589.2 (Earth 1367.6)
    Black-body temperature 210.1 K (Earth 254.3 K)
    Average temperature: ~210 K (Earth 287 K)

  96. Thierry says:
    November 27, 2010 at 12:36 pm

    Here comes the -18°C again. In their excellent paper “Falsification Of The Atmospheric CO2 Greenhouse Effects Within The Frame Of Physics”, the two german physicists have clearly explained that -18°C comes from a wrong way of applying Stefan-Boltzmann law. On the non rotating and atmosphere free planet, the real average temperature would have been -129°C, which is more or less what we have on the moon. They have also explained why this glass approach is not to be used, since the so called greenhouse gazes do not behave as a physical or reflecting barrier. This post is unfortunately pointless.

    I’m glad no one else has commented on this, and I probably shouldn’t too. I thought this had been put to bed long ago.

    Lunar conditions don’t apply due to many things beyond the lack of CO2. Lunar soil is a poor heat conductor, in part because its dry. That’s one reason why some deserts, e.g. the high deserts of Oregon or Arizona, cool off so quickly at night. (Basalt walled canyons in Oregon will bake you all night.) Worse, a lunar night is 14 days long – that’s a lot of time for radiational cooling to cool!

    Also, the effect of real greenhouse is achieved by locking convection so that warmed air doesn’t rise and leave the greenhouse. There’s no convection in a real greenhouse. As Willis noted, he’s talking about the “Greenhouse Effect,” which is unfortunately very different from real greenhouse physics.

  97. Some folks upthread have said that what is missing is rotation. OK, let’s assume that the sun is some form of luminous cloud that illuminates the planet evenly with 240 W/m2 of solar radiation.

    Even with that, there is still something that prevents this single-shell planetary “greenhouse” in our thought experiment from representing our real planet in even the simplest way. I still have not seen anyone point to it.

  98. Sorry, but figure 4 is incorrect. The 240 going into space comes from nowhere. I don’t know what kind of games Willis Eschenbach is playing, but this flaw is really trivial. And the outgoing energy is not the only flaw. It’s not about energy balance. It’s about the description. It’s wrong. You cannot have outgoing energy without that energy coming from somewhere. But nowhere in the graph is it shown how that energy can pass through a shell that doesn’t allow that kind of energy to pass through it.

  99. > … I still have not seen anyone point to it.

    Are you perhaps thinking of “convection”, which is a major transporter of heat in the Earth’s atmostphere?

  100. The atmosphere is the shell. For climate we meaurse the temperature of the shell not whats is encased in the shell. That is why the greehouse effect does not reflect the real world.

  101. Our average radiation is based on our average temperature. I don’t believe 390PPM of CO2 contributes anything measurable to our average temperature. I think the difference between the calculated black body temperature and our actual temperature is because we don’t live on a black body disk, we live on a gray, spinning sphere with lots of integrating thermal mass and a homogenizing atmosphere.

    Want to see some negative feedback? Try increasing radiation by the fourth power of a temperature difference. Mother Nature does not want us to be warm!

  102. Jimmi says:
    November 27, 2010 at 5:24 pm

    Firstly, your link to Judith Curry’s site is not correct (wrong format)

    Thanks, Jimmi, fixed. WordPress changed their default, I missed it.

  103. There is a “greenhouse” effect in a greenhouse, although it is less important than trapping the warmer air inside. Like CO2, glass is transparent to the sun’s shortwave energy. Also like CO2, the glass absorbs IR and then reemits it in all directions. The half that goes back into the greenhouse helps heat the greenhouse. Having two panes of glass means, in a very crude sense, that only 1/4 of the IR makes it out.

  104. Vorlath: The 240 W/m^2 outwards is thermal radiation from the shell, not the radiation from the earth somehow passing through the shell.

  105. Willis Eschenbach;
    Even with that, there is still something that prevents this single-shell planetary “greenhouse” in our thought experiment from representing our real planet in even the simplest way. I still have not seen anyone point to it.>>

    There are so many defficiencies in this model Willis that you’re probably going to have to throw out a hint. You’ve so oversimplified things that there are dozens of simple things wrong with it. What perspective are you looking at this from?

    o An outside observer would see the temperature of the planet as identical if the shell is opaque or clear to LW. Both radiate 240 w/m2.
    o For the shell to radiate 240 w/m2 it must have some sort of mass that heats up to emitt 240 w/m2. Depending on what material you presume the shell to be made of, that could be a very large amount of energy, or very small.
    o For the model to be vaguely accurate, it cannot presume a shell of 0 thickness. Per above, the shell must have mass to function, and therefore some sort of thickness. If we were to change the model to presume the shell surface is one kilometer above the earth surface, and that the shell thickness is 1 km, then the model breaks down entirely. Both the clear and opaque to LW shells would then have a temperature gradiant between earth surface and TOS (Top of Shell). The distant observer would still see 240 w/m2, but it would be EFFECTIVE black body temperature that exists somewhere between planetary surface and TOS. The difference between the clear and opaque models would then be that the effective black body temperature is identical, but the altitude at which it occurs is different. Further, the earth surface in this case would NOT double its radiance as it must in a o thickness 0 mass shell model.
    o The sun emitts considerable LW though mostly SW. An opaque shell would absorb and re-emitt LW from the Sun, and I expect this would be measurable.
    o Though you postulate the Sun as some sort of cloud emitting constant radiance to earth surface at all times, the earth surface is not constant. Even assuming constant 240 w/m2 constant over entire surface, land and water will react differently. In the case of a presumed vacuum between shell and earth surface, I presume also that we have oceans that don’t boil, but otherwise have identical properties. In that case, LW only penetrating a few microns of water, the water surface would heat up enormously with the hot water rising to the top and becoming a barrier to mixing. Land would heat up less as it does not flow and conductance would carry some heat to some depth. This is the reverse of the real world where land would heat up faster than ocean surface. If we presume instead that the shell is again 1 km thick and touches the earth surface, then the top few microns of ocean would heat up and transfer the bulk of the energy back into the shell without appreciable changing the heat content of the ocean at all.

    I could go on, but I think you’ve got to narrow the scope somewhat.

  106. Sorry Willis… but you have just added to the rather large chunk of climate science junk.
    You said “In the situation shown in Fig. 4, the sun directly warms the planet. In addition, the planet is warmed … by the radiation from the inner surface of the shell. … As a result the temperature of the surface of the planet is much higher than in the previous Figures.”
    NO, No and No.!!! The colder inner surface of the shell can not warm the warmer surface of the Earth (Second Law of Thermodynamics). However, if the colder shell surface was slightly less cold (due to the absorbed infra-red radiation from the surface of the Earth) than the loss of heat from the surface of the Earth would be less due to the lower temperature differential. This in effect would lead to increase in the minimum temperature readings at the Earth’s surface (as stated by Baa Humbug above). I think there is evidence that there is an increase in minimum temperatures leading to an increase in the average surface temperatures ( Can someone verify this?). If so then this may be evidence of an increased “greenhouse” effect?

  107. Willis,

    Add one more variable to your thought experiment. Add a layer of 100% IR transparent atmosphere to the model. Assume very high gravity so that even with a thin layer, the atmosphere is extremely dense at the surface, say, 100x earth’s, yet near zero at the outer diameter that you considered, like ours. Now some energy will be transferred by conduction and diffusion from the surface to the atmosphere at the lowest levels so that the lowest layer will be the same temperature as the surface. The temperature profile will decrease via the usual adiabatic lapse rate for whatever gravity we choose (this might take some time as conduction and diffusion are both slow, but we will wait for equilibrium). The highest level will be near absolute zero, due to near zero pressure. What happens? Now what happens when the density at the surface is 1000x? 1 million x? By failing to consider temperature simply due to pressure, and the atmosphere whose molecules are also emitting IR (even though they are IR transparent), the greenhouse effect by greenhouse gases is overstated, while the effect simply due to pressure rise with depth is understated or ignored. Forget convection for this small addition to your model.

  108. Willis,
    I normally find your posts very interesting, but this ‘bonus question’ lark is a bit too juvenile. There are a whole host of reasons why your model is inadequate to represent the climate system of the earth – earth has an atmosphere, no hard shell, with convection, water that changes state, and with oceans, and with a hot core, etc, etc, etc, etc, etc.
    This strikes me as a bit like a weird game of ‘one of these things is not like the others’ where the question-poser eventually reveals the fact that the odd one out is obviously the one which has a different penultimate letter, and no other differences count.

  109. I get accused of believing some odd things, but I’m not the one denying that N2 and O2 molecules have temperatures. 781,000PPM of N2 has no thermal capacity or thermal mass? I don’t care how many degrees of freedom your fantasies have, the temperature of 781,000PPM N2 has nearly nothing to do with anything 390PPM of CO2 can do. Stefan-Boltzmann says things radiate based on their temperature, not whether they’re made of CO2 or water vapor.

  110. The earth is round and it’s not a black body.
    NASA Abandoned Flawed Climate Calculations in 1960’s
    “Lunar Temperatures Cast Doubt on Climate Theory
    NASA had found that daytime temperatures on the lunar surface were lower than expected because planetary bodies also conduct heat to their inside rather than radiating it all into space – an empirical fact that challenges the GHG theory. Computer models supporting GHG theory had predicted that such heat energy would be ‘blanketed’ above a planet’s surface.”

  111. Baa Humbug says:
    November 27, 2010 at 4:27 pm
    Willis I’m having trouble with fig.3

    If the shell is radiating 240Wm2 to space AND 240Wm2 to the planet, it must therefore be receiving a total of 480Wm2, which is what fig.3 shows.
    IF it is receiving 480Wm2, how can it be the same temp as the planet surface i.e. 255K or -18C? The planet surface is receiving only 240Wm2.

    The key to understanding the effect is to remember that the shell has two sides. It radiates energy from both sides equally.

    The planet has one side. If it is receiving 240 W/m2, it must radiate at 240 W/m2 to stay in balance.

    A shell, on the other hand, has two sides. Suppose it is heated from one side at a given rate, say 480 W/m2. With twice the surface area, the must radiate at half the rate per square metre to maintain the balance. Half goes out to space, half goes in to the planet. And the rate of radiation is what determines the blackbody temperature.

    Not sure if that is clear. If not, ask in a different way.

    Thanks,

    w.

  112. Morris Minor:
    The second law of thermodynamics is about the net flow of heat. In Figure 4, the net heat flow between the warm planet and cooler shell is still towards the shell, so there is no violation.

  113. You’re killin’ me Willis. People say, it’s missing this or it’s missing that and you say, well, all models miss stuff. The obvious rejoinder being, if it’s missing something important, than it’s not much of a model, is it? And you say, even if it included everything, it still wouldn’t provide a workable model of the “greenhouse effect” on our planet.

    I was hoping by the time I got to the end of the comments, you’d spill the beans, but not yet. So out with it! Put an end to this infernal guessing!

  114. In our world the atmosphere has a temperature, that’s why it radiates some photons to earth. The model has but a vacuum and a shell of presumable infinite thinness so there is nothing there to have a temperature and thereby radiate energy.
    Do I get some sort of prize?

  115. I think it’s the fact that you are showing us a simplified model of a “perfect” system, sort of in the spirit of the ‘ol college physics 101 “assume there is no friction”.

    In your model, the shells operate at 100% efficiency and the “earth” is a perfect blackbody for exactly what you are modeling.

    Nothing is perfect or works at 100% efficiency. Iskandar hit on this a little earlier up in the posts.

  116. Did anyone mention the bottom line? Two additional watts of forcing at the surface from all anthropogenic GHGs together for a total of 242 watts per square meter temporarily until equilibrium is reestablished. New equilbrium temperature will be close to 2 degrees higher if nothing else changes.

    There’s barely a third of that surface warming actually observed if we can trust the instrumental record. Therefore Trenberth’s “missing heat”.

    I suspect the missing heat is being rejected by higher speed water cycle which means more convection and higher albedo from more clouds. Humans still emit aerosols along with their CO2 and methane and aerosols have a surface cooling effect so some of the 2 watt GHG forcing is negated by those. Nothing is missing. Trenberth should relax about it.

  117. Earth does not receive ALL of it’s energy from the sun. Their is an additional source from within the earth radiating heat away from it’s active core.

  118. Morris Minor;
    NO, No and No.!!! The colder inner surface of the shell can not warm the warmer surface of the Earth (Second Law of Thermodynamics).>>

    No violation of 2nd Law occurs. Consider a similar thought experiment. A house is heated by a 10,000 watt heater and exists in a -40 C atmosphere. At equilibrium, the temperature inside the house is higher than -40, and the house radiates 10,000 watts to the oustide. Double the thickness of the walls from say R2o to R40. At equilibrium, the house radiates exactly 10,000 watts to the outside. But the temperature in the house is higher. Why? Because the layer of extra insulation added must heat up until the house as a whole is radiating 10,000 watts to the outside. But since it radiates in two directions, not one, the interior temperature of the house must increase until the amount of extra energy it radiates balances the energy absorbed by the extra insulation and radiated back into the house.

    I’m a hardcore skeptic and I don’t have a single problem with this. My problem is with all the other factors in a chaotic system that physics requires to be a mitigating factor that the climate alarmists conveniently ignore. Well I have two problems. The second is trying to stay awake long enough to find out what the heck Willis is getting at.

  119. The blue down arrow on your figure 4 disappears in the case of a vanishingly thin transparent sphere. This is because the mass of the transparent sphere affects its capacity to capture and reradiate energy.

  120. The shell as proposed is a constant, and does not allow for any feedbacks due to, for example, that pesky water vapor.

    Interestingly, I think that both Miscolskzi and Hansen would agree with me on this.

  121. Your opaque shell example seems nonsensical. If the shell is opaque at all wavelengths there is no albedo and the shell absorbs near total TSI and would radiate half that to the surface, But if the shell is opaque to LW, the energy reradiated by the planet couldn’t escape and would build up in the interstice between shell and planet.

  122. Matthew Sullivan says:
    November 27, 2010 at 6:35 pm

    Vorlath: The 240 W/m^2 outwards is thermal radiation from the shell, not the radiation from the earth somehow passing through the shell.
    ————-

    Thermal radiation from what energy source? And why doesn’t fig 3 show this if what you say is true? Also, Willis Eschenbach disagrees with you. He’s saying it does come from the Earth and it’s passing through the shell.

    I’m seeing a lot of goalpost movement here. The math is flawed. Plain and simple.

    —————
    Willis Eschenbach says:
    November 27, 2010 at 6:55 pm

    Baa Humbug says:
    November 27, 2010 at 4:27 pm
    Willis I’m having trouble with fig.3

    If the shell is radiating 240Wm2 to space AND 240Wm2 to the planet, it must therefore be receiving a total of 480Wm2, which is what fig.3 shows.
    IF it is receiving 480Wm2, how can it be the same temp as the planet surface i.e. 255K or -18C? The planet surface is receiving only 240Wm2.

    The key to understanding the effect is to remember that the shell has two sides. It radiates energy from both sides equally.
    —————————–

    If so, then it’s no longer opaque. Here’s the definition of opaque as it relates to energy.

    2. not transmitting radiation, sound, heat, etc.

    So please stop changing the goalpost, or at the very least stop using better terms to describe your models.

  123. Willis. My answer [guess] to your question.

    The posited globe receives much less energy at the higher latitudes than at the equator.
    This would make for an energy imbalance and [consequently] an energy flow between the low latitudes and the poles.

  124. Why The Thought Experiment Can’t Represent Earth

    It is not powerful enough.

    What I have illustrated above is the energy balance in a theoretically perfect single shell planetary “greenhouse”. It shows that the maximum amount that such a system can produce is a doubling of the input. Not one Watt per square metre more. It is not a magical system. At best, it can double the input.

    Since the real Earth receives about 240 W/m2 (after albedo reflections), the most the system could theoretically produce is 480 W/m2.

    At first glance, that seems like plenty. I mean, blackbody temperature for 480 W/m2 is 30°C, and the Earth only averages around 15°C. But in the real world there are losses.

    On the incoming side, about 70 W/m2 of the incoming solar energy is absorbed by the atmosphere. In addition, there are losses from convection and from evapotranspiration. These losses move energy directly from the planetary surface to the shell. These are estimated at another 100 W/m2. Finally, about 40W/m2 goes straight to space from the surface. Total loss is about 210 W/m2.

    So out of the 480 W/m2 from our theoretically perfect system, we end up with only about 270 W/m2 for the surface temperature, and the rest goes up in losses.

    And 270 W/m2 is a blackbody temperature well below freezing … the system pictured is simply not powerful enough to heat the surface given the real-world losses.

    This means that at a minimum a system to model the earth must have two separate and distinct shells. One shell is not enough to concentrate enough energy to satisfy the known constraints.

    With two shells, you end up with a theoretical maximum of three times the incoming W/m2. For the earth this is 720 W/m2. That is enough to allow for known losses and to still have enough to warm the surface to the Earth’s 15°C.

    People may have gotten the erroneous idea that I am opposed to models. To the contrary, I use them and write them myself. I am particularly interested in what I call “Tinkertoy models”. These are the models which contain the absolute minimum information necessary to model a system.

    For radiation/convection/evapotranspiration models of the earth, the absolute minimum configuration is two shells. My iterative Excel model of the two-shell system is here (I hope, server hassles lately). All models are wrong. Some models are useful.

  125. As always Willis, a great, thought-provoking post.
    Judging by the responses, so far, you’ve hit the odd raw nerve!
    Nice demonstration of just how shaky the application of the adjective “Greenhouse” is to the”consensus” that “underpins” CAGW.

  126. typo in my last message: stop = start

    Here’s the corrected version:
    So please stop changing the goalpost, or at the very least *start* using better terms to describe your models.

    Also, if you’re letting the shell transmit energy, is this happening in figure 3 as well? So it’s not truly opaque. But even so, you’ve got a delay that doesn’t show up even if the totals do add up eventually with a continuous supply of energy. If half the energy from the beam goes back into space, then that’s 120W/m2 leaving 120W/m2 going to the planet surface and back to the shell where half again goes to space. So we’re at 60W/m2. On and on until all of it is sent back into space. It will add up to 240W/m2. But the description of opaque is completely off.

  127. ‘Willis Eschenbach says:
    November 27, 2010 at 7:50 pm
    Why The Thought Experiment Can’t Represent Earth’

    Well throw another log on the fire, I’ll get the beer and popcorn. This sholud get interesting.

  128. Well, I’ll take a stab at the ‘bonus’ question: I would say in short because the earth’s climate system of predominately the amosphere and the oceans move heat around the earth (convection, latent heat of vaporization from the oceans, and ocean currents) to locations they can more easily escape, in essence ‘negating the shell to some extent.

    The oceans absorb heat, and ocean currents move the heat towards the poles, where it can radiate the heat into space, without being ‘reheated’ the same as at the equator because of lower angle of incidence of the sun (thus the poles are acting as a ‘heat outlet’ of the earth.

    The atmosphere transports a lot of heat in water vapor from the oceans (the latent heat of vaporization) higher in the atmosphere, where when the water vapor condenses, it gives up its heat higher into the atmosphere, where it is radiated into space quicker (due to the thinner atmosphere) than heat radiated at a lower level in the atmosphere.

    . . is this explanation on the right track . . ?

  129. Willis Eschenbach;
    For radiation/convection/evapotranspiration models of the earth, the absolute minimum configuration is two shells.>>

    So when I said one shell of o mass and o thickness was unrealistic, that much better would be a shell of 1 km thickness resulting in a temperature gradient from earth surface to TOS, that would be different how? You proposed two shells as being the minimum and I proposed (in effect) a near infinite number of shells of near 0 thickness stacked to a depth of one kilometer. Your two shell minimum is better than one shell, but an infinite number of shells is far superior to both, easily described by calculus, and much much more accurate than two shells. Or three. Or 10.

    Great thread by the way, provoked a very interesting discussion. But your two shell minimum isn’t much better than your original one shell.

  130. Here goes Willis again wandering off into the never never land of naively simplistic science. This post if I can understand it correctly deals with Greenhouse effect which most people will interpret as the greenhouse GAS effect. It has already been pointed out correctly that there is no shell but a continuum of atmosphere from the surface to the TOA.
    Now the facts are that CO2 can only interact with about 7% of the outgoing radiation and H2O somewhat more than that but the majority of the incoming and outgoing radiation interacts with the particulate in the atmosphere which is not resonant frequency sensitive as are the gasses. Particulate is defined as water droplets and solid particles such as volcanic ash carbon and fly ash etc.
    Clearly the vast majority of the particulate is water droplets or clouds. Therefore the so called greenhouse gas effect is a very small contributor to the general warming produced by our atmosphere.
    The debate over whether clouds have a net positive or net negative forcing effect has raged for years IPCC claim net positive thus the effect of an increase in temperature is an increase in humidity which produces more cloud therefore warms the earth Drs Lindzen, Christie and Spencer feel the reverse is true and that clouds have a net negative forcing effect which creates an equilibrium.
    The cloud cover according to satellite data has varied from 62% to 69% over the last 30 years which has had a significant effect on temperatures.
    To get back to Willis’s childish illustrations there is never a 100% transparent or 100% opaque atmosphere and to cite a “global average” radiation is even more ridiculous than trying to compute a “global average temperature” from a bunch of computer generated statistics.
    Basically the clouds do tend to smooth out the average radiation budget although they will reduce the average as they increase and increase the average as they decrease.

  131. Paul Birch says:
    November 27, 2010 at 2:00 pm
    A transparent shell would not be at 0K. If it were perfectly transparent (which is physically unreal, but never mind) it would retain whatever temperature it had when you first put it there, because it is neither absorbing nor losing any heat. If it were merely very nearly transparent (which is the physically realisable case) , then in this example it would be at a temperature ~ 255K. The temperature would be somewhat higher or lower than this, depending as the emissivity were higher or lower in the visible or thermal.

    This model implicitly assumes uniform insolation, which is not valid for a sphere illuminated from one direction. Steady state requires that the planet does not rotate (relative to the sun), or has negligible heat capacity; the temperature would then range from a factor 4^1/4 higher (ie 361K) at the subsolar point, through ~73K around the terminator, to ~0K on the dark side. There would be no meaningful average or global temperature at all. The greenhouse effect of the shell would be the same factor of 2^1/4 (for the visibly transparent, thermally opaque case). Add diurnal rotation and finite heat capacity and thermal conductivity, and things become much more complicated. There is no steady state, and the temperature curve for a given latitude is messy to compute. The thermal capacity of the shell has to be considered too. The general effect of thermal capacity and rotation is to transfer (some) heat between the day and night sides.

    Useful analysis. So even this (supposedly) extremely simplified scenario results in thermal complexity and a non-equilibrium thermal state, with all the non-equilibrium implications of nonlinear dynamics etc.

    Does this not make it even more absurd to make any attempt to analyse earth’s real climate in terms of linear, equilibrium physics? OK there is always a heat in – heat out equation. But a complex chaotic system can respond in complex including adaptive ways…

  132. There is no greenhouse effect because the greenhouse effect assumes all energy is transfered radiatively. Your model shows only radiative energy transfer.

    I highly doubt that it is even possible to reliably model all the seperate modes of energy transfer in the earth-atmosphere system, conduction, convection, radiation, evaporation, condensation and mass transfer(i.e. wind and precipitation)

    Here’s a couple thought experiments on heat transfer:

    1. You have a really hot cup of hot chocolate and you want to cool it down do you A. Drop an ice cube in it (conduction), B. blow on it (convection) or C. hold it up to the clear night sky (radiate)

    2. Thermoses use a vacuum for their insulation, because purely radiative energy transfer is slower then conduction and convection when a gas is present.

    3. Argon gas is used in insulated windows becuase of its low thermal conductivity and co2 isn’t used because the greenhouse effect isn’t real.

  133. Vorlath: The shells aren’t transmitting the radiation. All of the radiation leaving the shell is thermal radiation originating from the shell, the same way the energy going up from the modeled planet is thermal radiation originating from there.

  134. Bill Illis says:
    November 27, 2010 at 2:36 pm

    Bill, can you give a quantum explanation of why the CO2 “greenhouse” or IR radiative heat trapping effect does not saturate in a few hundred meters?

    The mean free path and radiative balance scenario that you describe could surely be turned on its head by missing out just one subtle feedback somewhere.

  135. Just another comment on Willis’s article.
    Go to Climate4you.com select global temperature and go down to the top of atmosphere radiation section.
    You will see a wild variation in TOA radiation from 216 W/m2 to 238 W/m2 with a mean around 230 W/m2, not exactly what Willis was promoting, The CO2 curve v’s TOA radiation is also very instructive.

  136. Willis Eschenbach says:
    November 27, 2010 at 6:55 pm
    Baa Humbug says:
    November 27, 2010 at 4:27 pm
    Willis I’m having trouble with fig.3

    Yes thnx Willis.

  137. The model does not work in that if you make the ‘shield’ distance to the earth surface thinner and thinner until it is actually AT the surface you are in fact at the same condition as either figure 2 or 3. This cannot be if the model is to be consistent.

    Roger

  138. The glass sphere is uniform but greenhouse gasses(water vapour,clouds and co2) are not uniform around the earth,they are variable over time.

  139. davidmhoffer says:
    November 27, 2010 at 8:20 pm
    Willis Eschenbach;

    For radiation/convection/evapotranspiration models of the earth, the absolute minimum configuration is two shells.>>

    So when I said one shell of o mass and o thickness was unrealistic, that much better would be a shell of 1 km thickness resulting in a temperature gradient from earth surface to TOS, that would be different how? You proposed two shells as being the minimum and I proposed (in effect) a near infinite number of shells of near 0 thickness stacked to a depth of one kilometer. Your two shell minimum is better than one shell, but an infinite number of shells is far superior to both, easily described by calculus, and much much more accurate than two shells. Or three. Or 10.

    Great thread by the way, provoked a very interesting discussion. But your two shell minimum isn’t much better than your original one shell.

    I posted the thought experiment to provoke discussion, so success on that front.

    I’m looking for the simplest model that will give me earthlike results. That model contains two shells. Yes, you can model it with four, or four hundred. I’m looking to simplify, not complicate.

    Thanks,

    w.

  140. Barry Moore says:
    November 27, 2010 at 8:22 pm

    To get back to Willis’s childish illustrations there is never a 100% transparent or 100% opaque atmosphere and to cite a “global average” radiation is even more ridiculous than trying to compute a “global average temperature” from a bunch of computer generated statistics.

    Barry, my planet with a 100% transparent shell is not the real world. It is a THOUGHT EXPERIMENT. These have been used throughout the history of physics. In a thought experiment, one can imagine such things as a perfectly clear shell. It is useful because it helps us to understand our mysterious universe, not because it is correct in the details.

  141. Willis,
    You still haven’t replied to several posts that say you can’t add the two 240 W/m2 downward radiation element and get 480 W/m2 upward.

  142. Geoff Sherrington says:
    November 27, 2010 at 7:36 pm

    The blue down arrow on your figure 4 disappears in the case of a vanishingly thin transparent sphere. This is because the mass of the transparent sphere affects its capacity to capture and reradiate energy.

    Thickness will only affect the response time for a change in input. Near 0 = fast response, thick = slow response. As small at 1 atom thickness would still be in equilibrium.

  143. It would be difficult to have an earth model with only two shells. You would need one for the gasses (which are mostly uniform), one for the water vapour (which is not uniform, although can be persistant in some areas), and one for the clouds (which are not uniform, although can be persistant in some areas). These ‘shells’ overlap, for the most part, but are distinctive in their responses. You could also include the ozone layer and thermosphere. But I would not calculate the earth as a black body.

  144. Vorlath,

    I don’t see what’s wrong with the third diagram. Shortwave radiation comes through shield (transparent to shortwave) hits planet (opaque to SW) and heats it, where upon planet emits LW. LW radiation hits shield (opaque to LW) heating it, where upon it begins to emit LW in both directions, further heating the planet surface, and losing LW to space. Just because a material is opaque (non transparent) to a wavelength doesn’t mean it can’t emit that wavelength. I would suggest that white light won’t pass from a source behind the tungsten filament in an incandescent globe to the other side, even when the incandescent light bulb is on.

  145. Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?

    Without reading the comments, I would venture to guess that the energy in the system you are postulating doesn’t do any work.

  146. Okay, my guess. A lot of incoming shortwave radiation is absorbed by LIFE on the planet, it is not immediatly reradiated as longwave radiation?

  147. Braddles says:
    November 27, 2010 at 2:13 pm

    The comments that dispute Willis’ excellent post highlight some of the problems with the AGW sceptic arguments. Those who argue that the greenhouse effect does not exist do the sceptics’ side a disservice, by arguing an untenable position. It creates a weakness in the sceptical position, which warmists focus on to try to discredit all sceptic arguments.

    True, then on the other hand, warmists occupation with this self-created strawman does a devastating disservice to its side. It exposes its rage as being unfocused, not centering on the strong skeptic gems in there. Informed minorities on both sides should ignore both the untenable and the rage.

  148. In for a penny, in for a pound. Had to read the lot and hope I did not miss any points. Several made great store on their observation that the Earth was not an emitter.

    Really

    Its a humungus ball of molten rock at about 2,000 degrees with a very thin layer of colder rock on top. Space is about zero degrees and if i am correct hot moves to cold, I.E. the earth emits. Volcanoes, hot rocks, hot springs, geothermal vents, it emits all the time and is a very effective means of warming the oceans and thus the air.

    It is not a black body like the moon

  149. Ok, Willis, so I have a really stupid question relating to the thought experiment.

    You posited a shell and the implied assumption was that the shell was unchanging over time. Obviously this was to restrict the experiment to simply defining the upper boundry of radiation.

    The real shell (atmosphere) is created by life, which is constantly replenishing it. We’re now at 390 ppm of CO2. An increase in CO2 ought to result in an increase in O2, although the climate community doesn’t seem to have accounted for this. As such the shell ought to be somewhat thicker now than 1890. Maybe not by much, but thicker all the same. Moreover, the assumption of CO2 being (n)% of the atmosphere and growing seems off. If plants are also emitting more O2 then the (n)% isn’t growing as much as it may seem at first blush. As such this would imply that much of the apparent temp increases are more related to land use etc than CO2 (or natural recovery from the LIA, etc.) since it appears that the AGW argument is that CO2 is rising whereas all else is purely static.

    So… my question. Am I missing something really basic here? Thanks…

  150. Hi Willis; these types of thought experiments are interesting and serve a purpose. If I may point out an aspect to do with your figure 4; in the first instance only 240 w/m2 reaches the surface in the form of SW so only 240 w/m2 is reemitted to the disc in the form of LW; of that 240 w/m2 isotropic effect causes 120 w/m2 to be reemitted from the disc back to the surface and 120 w/m2 outwards to space; so in the next instance the surface has 240 w/m2 SW + 120 w/m2 LW for a total of 360 w/m2 and so on; in short this is, of course, a limiting sum geometric series:

    Sn=a/1-r where a=1 and r=0.5; or 1/n^2 for n=1 to infinity is 2.

    In otherwords the surface will only receive 480 w/m2 at infinity.

  151. Willis Eschenbach says:
    November 27, 2010 at 2:51 pm
    but my thought experiment is in a vacuum.

    So standing your Glass Thermos Flask in the sun at mid-day will slow down the cooling during the daytime… but by mid-night the contents will be cool… and probably cold just before dawn…

    On a more constructive note it would be wonderful if you could create, based upon the very intelligent comments you have received so far, a revised version of your diagram that is technically correct and isn’t based upon a vacuum… because I remember being taught: Nature abhors a vacuum

  152. Willis:

    Please forgive the abruptness of this post which O am fitting in between duties I have today.

    At November 27, 2010 at 8:20 pm davidmhoffer said to you:
    “Your two shell minimum is better than one shell, but an infinite number of shells is far superior to both, easily described by calculus, and much much more accurate than two shells. Or three. Or 10.”

    You have replied to that at November 27, 2010 at 10:12 pm saying:
    “I’m looking for the simplest model that will give me earthlike results. That model contains two shells. Yes, you can model it with four, or four hundred. I’m looking to simplify, not complicate.”

    OK. I understand that, but it can be misleading to simplify beyond the physical process being modelled: an analogue is not a model of a mechanism but may be an accurate model of the behaviour of that mechanism.

    GHG molecules in the atmosphere absorb IR photons from the Earth’s surface then (ignoring collisional energy exchange) release that energy by emitting photons in all directions. So, the molecules emit half the energy in upward directions and half in downward directions.

    Thus, the GHG molecules act like half-silvered mirrors transmitting visible light. They accept radiation from below, then emit half of that energy upwards and half downwards.

    Ignoring effects of collisional energy exchange, the number of ‘half-silvered mirrors’ to model the observed GHG effect is 7.

    So, I suggest that your “simplest model that will give me earthlike results”is 7 spheres each with 50% transmittance and 50% reflectance. And a more realistic model would include energy extraction by molecular collisions from each shell.

    Richard

  153. I also found the following passage very interesting…
    see http://www.newton.dep.anl.gov/askasci/env99/env223.htm

    The flow of heat through the sandstone and limestone overlying Wind Cave’s passages is extremely slow. Temperature fluctuations of over 60°F between day and night are not uncommon on the surface, but if we were to monitor the temperature only two feet below the surface on such a day, the fluctuation in temperature would be only about 1°F. Therefore, it does not stay warm long enough during the day, nor does it stay cool long enough during the night, to significantly change the temperature
    of the rock only two feet underground. The same principle holds true for seasonal temperature fluctuations as well, although the depth at which temperatures begin to stabilize is greater. Seasonal temperature fluctuations of 80°F are reduced to only l°F at a depth of about 50 feet.

    So does the temperature gradient look like from the earth’s core up to ground level?
    Is the earth internally generating heat or just simply cooling down?
    Sorry to be throwing another spanner into the works….

  154. You can solve this with one shell. No problem.
    But You must add little bit of reality.
    The energy balance between surface and your shell is NOT a radiative balance.
    In real world is it a heat balance driven by convection. Then You get the height of the real world “shell”. 33K/6 Kk/km = 5,5 km
    6 K/km are average lapse rate.

  155. I’m no scientist, but how about uh… barbecues, campfires, air-conditioners, billions of cars, motorcycles, ships, airplanes, heaters, stoves, buildings on fire, industries, power stations, nuclear stations, Borneo, Russia & Sumatra forest fires, geysers, electricity, volcanoes, lava streams, E=MC2, energy and heat produced by nature, earth and man? Doesn’t that add substantially and explain a fraction of a degree rise of temperature perhaps? Do I get at least a bonus?

  156. I am puzzled by the long wave radiation out to space of 240 W/m2 (top RHS of figure 4)

    If the boundary does not allow long-wave radiation out then where does this radiation come from?

    Under steady state conditions (equilibrium)

    Outflow must equal Inflow. Outflow=Inflow (1)

    Under these conditions storage has no effect on the result

    i.e. S dT/Dt is zero where S is storage and T temp and t time.

    As far as I can see the components don’t obey equation 1 in Figure 4.

  157. “Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?”
    ================================
    Assuming you wanted the simplest answer and noting that you said ‘planetary’, I posted at 2:58 pm;

    “Day/night cycles are not considered. Day: outgoing is less than incoming. Night: outgoing is greater than incoming. The change (raising) of surface temperature is the difference between these two budgets.”

    You first need two conditions, not two shells, for your construct to work. When I was at school (1960′s), a greenhouse was explained to work because glass is semi-transparent to IR. Therefore, when the sun is shining, its internal temperature increases as less IR escapes than is received. When the sun is not shining, its internal temperature decays to ambient more slowly (than without glass) because of the same phenomenon – the semi-transparency. Convection losses are present at all times but vary as delta T changes – applicable to a greenhouse but not a planet.

    I claim my $5!-)

  158. Willis Eschenbach says:
    November 27, 2010 at 7:50 pm
    For radiation/convection/evapotranspiration models of the earth, the absolute minimum configuration is two shells. My iterative Excel model of the two-shell system is here (I hope, server hassles lately). All models are wrong. Some models are useful.

    Damn. I started to reply that the atmosphere is made of an almost infinite number of ‘shells’ but didn’t submit because you said it was a simple model.

    I’d like you to expand though. How do you think your insight relates to the real world? The path lengths of longwave radiation go from about 10^-27m near the surface (due to the density of the atmosphere) to around 1m at the top of the ionosphere. Radiation becomes more important than convection as a transporter of heat somewhere near the tropopause.

    Humidity near the tropopause has been falling since 1948 on the average. I think this will have negated any rise in co2. Specific humidity near the tropopause mysteriously mimics solar activity levels: http://tallbloke.files.wordpress.com/2010/08/shumidity-ssn96.png

    So what caused the warming? Reduced cloud cover and high solar activity. ISCCP data shows a drop in tropical cloud cover from 1980-1998: http://tallbloke.files.wordpress.com/2010/11/isccp-temp.jpg My calcs show this was causing an extra 4W/m^2 forcing on the surface in the 1993-2003 decade.

    Simples. :)

  159. More junk science.

    The problem with thought experiments is that there are always factors left out to simplify. Climate Science has already left out too much science. Leaving out more simply proves the thought experiment is worthless for this application.

    Try your thought experiment again with a reasonable figure for the energy flux within the earth. Then keep adding the other factors left out and viola, no NET greenhouse effect!!!

    Capiche??

  160. Robert Weber says:
    November 27, 2010 at 10:25 pm

    Willis,
    You still haven’t replied to several posts that say you can’t add the two 240 W/m2 downward radiation element and get 480 W/m2 upward.

    Not sure what you are referring to. The energy flows are exactly balanced, at all levels and in all three figures (Figs. 2, 3, 4). Perhaps you could explain what you see as the problem. There have been several posts that I did not understand, such as the one that said:

    The shortcoming in the model is that each molecule absorbs heat emitted from the earth, the re-radiates it out spherically, meaning that half of the energy is returned to the earth, and that this process repeats itself until vacuum is reached.

    Sometimes I think I understand that, but then it slips from my grasp again.

    In any case, if you could restate what you think the problem is with my thought experiment, I will see if I can explain it.

    w.

  161. Between Willises post here and J Currys “Skeptics make your best case”, my brain is frazzled.

    If I’ve got it right, Willis says at 7:50pm “It’s not powerful enough…..This means that at a minimum a system to model the earth must have two separate and distinct shells. One shell is not enough to concentrate enough energy to satisfy the known constraints.”

    At 2:42 pm I gave a simple short answer to the simple question thus:

    Baa Humbug says:
    November 27, 2010 at 2:42 pm

    Layer upon layer upon layer

    So, do I get a cyber prize or a cyber kick up the backside?

  162. G.L. Alston says:
    November 27, 2010 at 11:26 pm (Edit)
    Ok, Willis, so I have a really stupid question relating to the thought experiment.

    The only stupid question is the one you don’t ask.

    … An increase in CO2 ought to result in an increase in O2, although the climate community doesn’t seem to have accounted for this. As such the shell ought to be somewhat thicker now than 1890.

    I don’t follow that. If anything, it seems like making more CO2 should decrease the amount of O2 in the atmosphere. Oxygen makes up about 21% of the atmosphere, or about 210,000 ppmv. Every CO2 requires one O2. So for the CO2 to go from 280 to 390 ppmv, the O2 might go down to 209,890 ppmv. Of course, this doesn’t include any feedbacks, sources, or sinks … just a rough number. In any case the change in O2 concentration is on the order of five hundredths of a percent (0.05%), so the difference would be trivial.

  163. kuhnkat says:
    November 28, 2010 at 1:53 am

    More junk science.

    The problem with thought experiments is that there are always factors left out to simplify. Climate Science has already left out too much science. Leaving out more simply proves the thought experiment is worthless for this application.

    Thought experiments have a long and distinguished history in science. They are used to explore things that are difficult to explore experimentally.

    Try your thought experiment again with a reasonable figure for the energy flux within the earth. Then keep adding the other factors left out and viola, no NET greenhouse effect!!!

    Capiche??

    I have done exactly that in my Excel model referred to above. I still find a net warming effect. Here is a more complete view of a two-shell system, with the other factors included. The two shells are shown as horizontal gray bands:

    Finally, your claims would have more weight if they were accompanied by numbers.

  164. Willis, see

    Oxbridge Prat says:
    November 27, 2010 at 1:12 pm

    Multiple concentric shells.

    Can I have a virtual banana please?

  165. The Greenhouse Effect is a ridiculous name for an effect that has been used as blame for climate. Greenhouses do not act like the atmosphere. Greenhouses are heated by visible light and the energy loss of that light to become IR energy. Glass does not transmit IR, or UV for that matter, but greenhouses still get warmer than ambient. Incoming visible light is transmitted and this warms the interior and suffers energy loss so that some if that light becomes IR which remains inside the greenhouse because the glass traps it inside. The temperature will build and build until someone opens a vent to let in cooler air and expel the hot air.
    The atmosphere is transparent to IR, and some UV light, as well as visible light. Yes water vapour and CO2, as well as other misnamed GHG’s, do warm up by adsorbing this energy, but this heat is not stored to be re-radiated later because the 2nd law of thermodynamics does not let this happen. The adsorbed heat is immediately conveyed to the other atmospheric gasses by convection, conduction or radiation to cooler areas. (Heat will only be transferred from hot to cold never in the other direction). So some of this heat will be transferred back to space because the lower atmospheric layers will be warmer!
    The whole idea that something can ‘store’ heat violates the laws of thermodynamics and so makes the GHG warming hypothesis a failure.

  166. To all those who think that the Figure 4 is wrong:

    If we say we only absorb long-wave radiation in this example, then we get only 240 W/m^2 to the earth. In the first time step, we emitt 240 W/m^2 to the shielding shell (because Heat Capacity is 0). This shell then emits 120 W/m^2 into space and the same back to the planet. In the next step. The planet emits 360 W/m^2 and half of that gets remitted to the planet: 180 W/m2. In the next time step we emit 420 W/m^2 to the shell, the shell divides it and reemitts 210 W/m^2 back to earth. Then we get 450 W/m^2 that is emitted by the planet and thus 225 W/m2 back to earth. This continues until we reach the equilibrium.

    E(t) = Solar rate + S’(t)
    S’(t) = E(t) /2 (half of it into space)
    S(0) = 0
    Solar rate = 240 W/m^2
    Shell can’t emitt in the initial condition, because it has no energy.

    You can easily reiterate this in Excel and you will see that E(t) converges to 480 W/m^2.

  167. Oxbridge Prat says:
    November 28, 2010 at 2:27 am (Edit)
    Willis, see

    Oxbridge Prat says:
    November 27, 2010 at 1:12 pm

    Multiple concentric shells.

    Can I have a virtual banana please?

    Of course. Have two.

  168. Max says:
    November 28, 2010 at 2:28 am
    Max, as I said earlier, there cannot be an equilibrium, defined as per Willis’s Figure 4 where the surface receives 480 w/m2, until infinity:

    cohenite says:
    November 28, 2010 at 12:27 am
    Hi Willis; these types of thought experiments are interesting and serve a purpose. If I may point out an aspect to do with your figure 4; in the first instance only 240 w/m2 reaches the surface in the form of SW so only 240 w/m2 is reemitted to the disc in the form of LW; of that 240 w/m2 isotropic effect causes 120 w/m2 to be reemitted from the disc back to the surface and 120 w/m2 outwards to space; so in the next instance the surface has 240 w/m2 SW + 120 w/m2 LW for a total of 360 w/m2 and so on; in short this is, of course, a limiting sum geometric series:

    Sn=a/1-r where a=1 and r=0.5; or 1/n^2 for n=1 to infinity is 2.

    In otherwords the surface will only receive 480 w/m2 at infinity.

  169. I assume that the amount of heat humans create is known to some degree and accounted for in climate modelling (unless it is considered trivial)?

  170. cohenite says:
    November 28, 2010 at 12:27 am

    Hi Willis; these types of thought experiments are interesting and serve a purpose. If I may point out an aspect to do with your figure 4; in the first instance only 240 w/m2 reaches the surface in the form of SW so only 240 w/m2 is reemitted to the disc in the form of LW; of that 240 w/m2 isotropic effect causes 120 w/m2 to be reemitted from the disc back to the surface and 120 w/m2 outwards to space; so in the next instance the surface has 240 w/m2 SW + 120 w/m2 LW for a total of 360 w/m2 and so on; in short this is, of course, a limiting sum geometric series:

    Sn=a/1-r where a=1 and r=0.5; or 1/n^2 for n=1 to infinity is 2.

    In otherwords the surface will only receive 480 w/m2 at infinity.

    Yes. However, remember that radiation happens at the speed of light, so for all practical purposes the system will equilibrate in a finite time.

    This reminds me of a story my high school math teacher used to tell about the meaning of “for all practical purposes”. You may be familiar with “Zeno’s paradox”. This paradox states that if we move half the distance to our goal, and then half the remaining distance to the goal, and half again and so on, we will never reach our goal.

    Mr. Hegji, my math teacher, gave us a thought experiment. He said “Suppose we lined up all of the boys along one wall of the classroom, and all of the girls along the other. Then suppose each time I clap my hands, both groups move forward until they are half as far apart as before.”

    “Now you know”, he said, “that according to Zeno the two groups will never reach each other. However, before I have clapped my hands too many times, the boys and girls will be close enough for all practical purposes …”

    So yes, you are correct. We will only reach 480 at infinity. But starting at 240, and halving the distance to 480 each time as in your math, we get to within a tenth of a percent(0.1%) of 480 in just nine steps, and within a hundredth of a percent (0.01%) in 13 steps … close enough for all practical purposes.

  171. TimM says:
    November 28, 2010 at 3:24 am

    I assume that the amount of heat humans create is known to some degree and accounted for in climate modelling (unless it is considered trivial)?

    Globally the heat humans create (mostly through burning fuels of all types) is not large. From memory it’s in the hundredths of a watt per square metre.

    However, in some densely populated areas like say the Netherlands, again from memory, it can be a couple of Watts per square metre. It is also makes more of a difference the further north you go, and in wintertime.

    Whether this is accounted for in the climate models I don’t know. My guess is no for most models.

  172. Max,

    Then how do you answer Baa Humbug who says:
    November 27, 2010 at 2:56 pm Take a basic strip heater of say 1000W, the above fallacy suggests all we have to do is add a well conducting bar across the front of the element, say an inch away (unwired and not receiving any power, i.e. benign) and voila’ we now have a strip heater putting out twice the energy. Can someone help me with patents please?
    ……………………………
    I try to make the same point, that the shell has to have some rather special properties to do anything. Also, there is a big mix-up (as usual) between static analysis and dynamic analysis. This shows with what cohenite says:
    November 28, 2010 at 12:27 am – In otherwords the surface will only receive 480 w/m2 at infinity.

    It’s actually worse than that because less than 50 % of isotropic emission will return to Earth from the shell because of geometry.

    Similarly where Michael D Smith says:
    November 27, 2010 at 10:29 pm – Thickness will only affect the response time for a change in input. Near 0 = fast response, thick = slow response. As small at 1 atom thickness would still be in equilibrium.

    That’s dynamic analysis, not static. You have to wonder how a thickness of 1 atom interacts with a wavelenth of IR light of say 10 microns. It would take a rather long time to show. Logically, a tickness of 1 atom would imply a separation between atoms, so the sphere is not merely transparent to IR, it has holes in it. Ad absurdum, if the mechanism works with holes in the sphere , it will work with one atom. (Don’t think so).

  173. The issue is surely not the existence of a “greenhouse effect”, but its magnitude. Errors abound in calculations for the “non-greenhouse” earth, for example the usual assumption that the surface acts as a perfect “black body”. If it did, radiation scientists wouldn’t need to struggle to devise perfect”black body” reference sources for measurements, they’d just need a handful of dirt or a bucket of sea-water.

    The relationship between temperature and radiation is non-linear. Averaging a wide range of one element to calculate the other introduces an error – the wider the range, the larger the error. Kiehl & Trenberth show that for small increments in either radiation or temperature, the relationship can be considered to be linear. However, they do this AFTER averaging wide ranges in radiation to calculate average temperature and vice-versa, and assuming that surface and atmosphere behave as perfect “black bodies”. Also water-vapour is NOT uniformly distributed over the earth’s surface, therefore neither are clouds, and the heat transfer involved in evaporation and condensation is very large, and not properly included in the internal atmospheric “budget” These factors mean that the averaging process used invalidates the “energy-balance” diagrams by as much as several degrees K. For example, averaging the radiation deriving from temperatures of 0°C and 40°C results in a figure which gives 22.1°C rather than 20°C. That’s not insignificant when temperatures are calculated to a tenth of a degree.

  174. Willis Eschenbach says:
    November 27, 2010 at 7:50 pm
    “Why The Thought Experiment Can’t Represent Earth
    It is not powerful enough.
    What I have illustrated above is the energy balance in a theoretically perfect single shell planetary “greenhouse”. It shows that the maximum amount that such a system can produce is a doubling of the input.”

    This is silly.
    1) Plenty of people upthread have pointed out that there isn’t just a single shell; you poo-pooed them for adding complexity.
    2) Doubling the input is ample for a simple model of Earth’s climate; you have only made it apparently inadequate by adding additional shells or absorption mechanisms that cut down the energy reaching the surface in places beyond what you assumed, which is an unnecessary complication.
    3) Even a single shell can in principle produce any degree of greenhouse warming required. Off hand, I can think of at least three mechanisms:
    (i) if the ratio of the thermal emissivities of the lower and upper surfaces is E then the temperature of the shell is enhanced by a factor E^1/4, and the flux re-radiated to the planet is increased by the same factor of E. One could easily engineer a factor of up to ~100 (using, eg, aluminium film), which would push the Earth’s temperature to almost 1000K! In the real atmosphere, with its thick shell, those emissivities do not differ by anything like so much, but they are not identical.
    (ii) if the shell is reflective (as you have implicitly assumed by setting the albedo greater than zero) then the amount of energy radiated up and down from the shell can be much less than the amount of thermal energy simply reflected or scattered back down. Again, an aluminium film could boost the heat reaching the surface a hundredfold!
    (iii) if the shell is not perfectly conductive, there is a temperature difference between top and bottom, and there is no fundamental limit on how large that temperature difference can be. In the real atmosphere, there is indeed such a temperature gradient.

    You could argue, I suppose, that (i) and (iii), in using a shell with different properties for the top and bottom surface, are really two shells; but that pedantic objection would not apply to (ii).

    Note that in principle one is not even limited to the 6000K temperature of the Sun; if one were to allow through only the blue or “hot” end of the spectrum, the planetary temperature could theoretically exceed the source temperature. This does not contravene the second law of thermodynamics, because there is still a net entropy gain (more photons leave the planet+shell than reach it).

  175. Willis Eschenbach says:
    November 28, 2010 at 2:26 am
    in my Excel model referred to above. I still find a net warming effect. Here is a more complete view of a two-shell system, with the other factors included. The two shells are shown as horizontal gray bands:

    Willis, that’s an interesting reworking of the Kiehl-Trenberth energy budget sketch. Have you been able to produce an estimated sensitivity using it?

  176. The so called “greenhouse-effect” does not exist.
    Radiation of heat through a body only depends on the temperature of that body and its coefficient for its emissivity. Back radiation cannot influence this. It only could transfer heat back to that body.
    But transfering heat energy from “cold” to “warm” violates the second law of thermodynamics.
    There is no way around that.

  177. tallbloke says:
    November 28, 2010 at 3:57 am
    Willis Eschenbach says:
    November 28, 2010 at 2:26 am

    in my Excel model referred to above. I still find a net warming effect. Here is a more complete view of a two-shell system, with the other factors included. The two shells are shown as horizontal gray bands:

    Willis, that’s an interesting reworking of the Kiehl-Trenberth energy budget sketch. Have you been able to produce an estimated sensitivity using it?

    Well, the sensitivity cannot be estimated from that budget. This is because of the great unknown, the power of the feedbacks and the size of the losses.

    For example, you can say that 390 W/m2 is emitted by the surface as radiation, and 100 W/m2 is lost from the surface through convection and evapotranspiration. So somewhere around a fifth of the energy is not going into warming the surface.

    But that is in the aggregate. It does not tell us what the net effect of a change in the forcing might be. The Clausius-Clapeyron relationship, along with the linear dependence of evaporation on wind speed, would indicate that for an additional forcing above the current temperature, much more than a fifth of the incoming energy would go into evaporation and convection, and correspondingly less would go into heating the surface. This fits with our experience — we know that as heat engines run hotter, parasitic losses increase rapidly.

    So while we can’t calculate the sensitivity, it certainly appears just from an examination of the losses that there is not much chance that the net feedback is positive. The losses are too large for the net feedback (which of course includes the parasitic losses) to be a net gain.

  178. Geoff Sherrington says:
    November 28, 2010 at 3:41 am

    Max,

    Then how do you answer Baa Humbug who says:

    November 27, 2010 at 2:56 pm Take a basic strip heater of say 1000W, the above fallacy suggests all we have to do is add a well conducting bar across the front of the element, say an inch away (unwired and not receiving any power, i.e. benign) and voila’ we now have a strip heater putting out twice the energy. Can someone help me with patents please?

    ……………………………
    I try to make the same point, that the shell has to have some rather special properties to do anything.

    No. Take a basic strip heater. Put it in a vacuum. Measure the temperature. Then surround it with a metal shell. Measure the temperature. The temperature will go up, because of the so-called “greenhouse effect”. The shell radiates some of its heat back to the strip heater, making it warmer.

    Also, there is a big mix-up (as usual) between static analysis and dynamic analysis. This shows with what cohenite says:
    November 28, 2010 at 12:27 am – In otherwords the surface will only receive 480 w/m2 at infinity.

    See my answer to this above.

    It’s actually worse than that because less than 50 % of isotropic emission will return to Earth from the shell because of geometry.

    You are correct. Again, this is being ignored as a simplifying assumption because the effect is small. With the radius of the earth being over 6,000 km, when you are a km or two above the earth the difference between a flat plane and a sphere is not large.

    Similarly where Michael D Smith says:

    November 27, 2010 at 10:29 pm – Thickness will only affect the response time for a change in input. Near 0 = fast response, thick = slow response. As small at 1 atom thickness would still be in equilibrium.

    That’s dynamic analysis, not static. You have to wonder how a thickness of 1 atom interacts with a wavelenth of IR light of say 10 microns. It would take a rather long time to show. Logically, a tickness of 1 atom would imply a separation between atoms, so the sphere is not merely transparent to IR, it has holes in it. Ad absurdum, if the mechanism works with holes in the sphere , it will work with one atom. (Don’t think so).

    That I don’t know about, I have no idea if it would work if the shell were one atom thick … not sure why or if that is relevant.

    Will the mechanism work with holes in the sphere? Yes, they just introduce losses where some amount of the radiation goes out to space. This makes the surface less warm, but the mechanism is still working.

    w.

  179. MostlyHarmless says:
    November 28, 2010 at 3:44 am

    The issue is surely not the existence of a “greenhouse effect”, but its magnitude. Errors abound in calculations for the “non-greenhouse” earth, for example the usual assumption that the surface acts as a perfect “black body”.

    AAARGHH! This is a thought experiment. I used a simplifying assumption, that the earth was a black body for IR. This introduces an error in the conversion of radiation to temperature on the order of about 5%, acceptable for a first-order understanding of what is happening. I said if you want to include it, do so …

    However, it makes no difference to the radiation calculations themselves. If 240 watts/m2 is absorbed by a planet, 240 watts/m2 will need to be radiated if the planet is to stay at equilibrium, regardless of whether the planet is a black body or not.

  180. The thermodynamics of this model neglects entropy.

    The Earth is not at equilibrium and it while the energy balance may be that the net energy flux integrated over a surface surrounding the Earth may be zero, the system is not at equilibrium, in which the entropy production is zero, but is in an approximate steady state. There are gradients of pressure, temperature, velocities of air and water, i.e.: decreased entropy, which are driven by an energy flux through the system. These would decay if there was an opaque shell surrounding the earth. A simple radiative transfer model striles me as somewhat simplistic.

  181. Here’s the problem that I see. In any of your “fully opaque” scenarios, the 240 units of energy from the sun should be REDUCED because the energy is the sum total of all wavelengths. So if none of the longwave energy can pass from the sun to the earth, then it’s not 240 that hits the earth, it’s 170 (or whatever… just an example.)

  182. Willis — I don’t follow that. If anything, it seems like making more CO2 should decrease the amount of O2 in the atmosphere.

    Sorry I was not more clear. Your shell is the atmosphere. The atmosphere is always replenished in that there’s leakage into space. The atmosphere is (re)generated by life, so it’s not a steady state where increase of A necessarily means B decreases.

    Specifically I was thinking of photosynthesis. If the number of plants is a steady state, then more CO2 = decrease in O2. On the other hand more CO2 = more plant growth (more plants?), hence more O2 released. More O2 = slightly more atmosphere. All things being equal there ought to be more plant activity with more CO2 meaning there ought to be more O2 emitted as well. In real life the volume of the atmosphere, which is a living system of sorts, ought to change slightly as parts are lost and subsequently replenished, so we know it’s being created. But enough to be measureable? I don’t know.

    (Mental model: CO2 concentration goes up, plant volume up. More O2 created. Atmospheric thickness goes up. )

    I also don’t know how the ppm # is created. If machinery counts up a million molecules and derives ppm, then OK, I’m dead wrong. OTOH if the machinery is calibrated against a steady state and ppm = derivation from this state, then what’s being measured is an absolute CO2 molecule count vs pctg of atmosphere, hence this count may not mean what it ought to mean.

    Is it far off enough to make a difference? No idea. I’m simply wondering if we’re measuring what we think we’re measuring.

    No I’m not questioning the greenhouse effect or offering alt theory here. Mostly it’s a nagging question regarding the reality of the hard numbers, and I’d thought of your list of known losses etc in your thought experiment.

    Thanks for listening.

  183. Willis

    “For example, you can say that 390 W/m2 is emitted by the surface as radiation, and 100 W/m2 is lost from the surface through convection and evapotranspiration. So somewhere around a fifth of the energy is not going into warming the surface.”

    “there is not much chance that the net feedback is positive”

    So there must be 390+100 watts absorbed by the surface. There is only about 164 watts solar absorbed + lets say 36 watts solar re-emitted from the atmosphere, i.e there is an 490/(164+36) amplification. Why wouldn’t an additional 4 W/m2 (~1% change) see about the same 2.45 amplification?

  184. The greenhouse effect was shown to be bogus by R W Wood.

    He showed that;

    1. A glasshouse only worked by stopping convection.

    2. The remaining real residual radiative effect was so small that it could be almost ignored.

    See page 17

    http://www.friendsofginandtonic.org/assets/hutton%20-%20climate%20change.pdf#page=17

    Here is a paper that more people should read.
    Especially as it comes from a source with no “spin” on the AGW debate.

    The way I read the paper is it gives massive support for the conclusions of the famous Woods experiment.

    Basically the project was to find if it made any sense to add Infra Red absorbers to polyethylene plastic for use in agricultural plastic greenhouses.

    Polyethylene is IR transparent like the Rocksalt used in Woods Experiment.

    The addition of IR absorbers to the plastic made it equivalent to “glass”

    The results of the study show that( Page2 )

    …”IR blocking films may occasionally raise night temperatures” (by less than 1.5C) “the trend does not seem to be consistent over time”

    http://www.hort.cornell.edu/hightunnel/about/research/general/penn_state_plastic_study.pdf

  185. OMG, I really got desperate while reading some of the comments. Perhaps 30%. Is the general commenter here THAT clueless?

    The next time I see a post which explains something very basic and/or proposes readers to solve a problem, I promise not to read any of the comments. It puts my morale down. Twice was I tempted to answer with expletives… It is incredible the authority with which absolutely clueless people can talk about things that they cannot understand.

    Willis, really, just give up. Whoever cannot understand the GHE, after the several posts that have appeared here in the last months about the subject, will never do. It’s not worth the effort.

  186. Following on from my post at 1:32 am.
    Received average W/m^2 for half a cycle = 240 = Wr
    Transmitted average W/m^2 for a full cycle = Wt

    For 1 cycle: Wr+2Wt = 0 (steady state). Therefore Wt = Wr/2 = 120

    This is true with and without a ‘greenhouse’ effect. So what lifts average temps from -18 deg? The answer is in the semi-transparent nature of the atmosphere to solar radiation, particularly IR, which creates a lag or hysteresis. With hysteresis, the work done on the system is the area under the hysteresis graph. In this instance, the work done is the raising of average temperature from -18 deg, an increase in potential energy. To understand lag, imagine yourself as a photon travelling into the atmosphere along-side a group of others: as it is semi-transparent, some, including yourself, pass through but on the way out some get out but you don’t. Your eventual departure has been delayed: you are lagging behind the others.

    How you would show that in a diagram, I don’t know but that diagram will have to show 1/2 cycle input and time lag.

  187. Willis asks, “Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?”

    Willis answers, “It is not powerful enough.”

    ——-

    Willis E.,

    Your answer is not the most fundamental answer to your own question.

    So, let’s play the game with you instead of your game with us!

    MY COUNTER QUESTION: “What is the most fundamental reason why the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?”

    John

  188. People,
    lots of interesting stufff here, but we are all overcomplicating and over critising this simple thought experiment to explain the fundamental mechanics of the green house effect. Not the earths atmosphere, BUT the greenhouse effect.

    My GUESS is that the thin shell is described only in its characteristics of being transparent or opaque (as to whether or not it passes or blocks shortwave or long wave radiation.) It does not take into account its mass (in this example not stated so taken as zero) and as such does not account for the heat that it itself would store, how long it would heat up and also cool down….
    depending on the temparature differential from one side to the other of the shell.

  189. @Geoff:

    Well, Willis posted a good reply for the heater system up there, so I won’t repeat it here a second time.

    But when it comes to static between dynamic systems, we have a steady-state system as long as all boundary constraints stay the same, as Willis showed in his figures. He only displayed the steady-state, because (and I only give my opinion here) of time constraints and simplification.

    What I added was merely to show how you arrive from the initial condition of only solar radiation influx to the steady-state over a certain amount of time (read: iterations). As you can see the pure mathematical model NEVER actually arrives at 480 W/m^2, which is the convergation limitation of the dynamic model AND not a fault in the thinking. From there on, the model is in a steady-state until the influx from the solar radiation changes.

    There is no violation of any kind of Energy law. The Earth in this model is only an intermitting energy source, BUT not the ultimate energy source, that is still the sun. In reality, the Earth heat capacity would further delay the arrival of a steady-state (along with several other factors which would help avoid any steady-state for a longer period). If you want to check the energy balance, you have to cut the whole dynamic system free at the shell. Then you have as much energy going away from the shell as is coming in (480 comes in, 240 goes into space and 240 is reemitted back to earth). The same you can do for the Earth side of the equation: 240 + 240 is coming in (from the sun and from the shell) and 480 is emitted to the shell. There is no “new” energy or energy creation on earth, it is just blind power, which you can also find in hybrid drive trains (as any engineer can tell you).

    In those drive trains you have to ways to transmitt power and thus torque from the engine to the wheels. First you might have a mechanic transmission and a hydraulic transmission (for the sake of simplicity). The power is divided at a planetary gear just behind the combustion engine and then part of it is transmitted via the hydraulic transmission and the rest through the mechanic transmission. It is then added to a single shaft by three gears. We there have an addition and/or subtraction of torque and it might be that we actually have power that is only “circulating” between the two transmission and IS NOT leaving it due to their relative rotation speeds. I could go more into detail, but I don’t have the time to design some accompanying drawings that would help illustrating the point.

    Also, similar effects can be observed in electrical systems (alternate current): http://en.wikipedia.org/wiki/Reactive_power#Real.2C_reactive.2C_and_apparent_power

    It is sometimes a difficult to grasp concept, but nonetheless valid and proven.

    quote:
    Max,

    Then how do you answer Baa Humbug who says:
    November 27, 2010 at 2:56 pm Take a basic strip heater of say 1000W, the above fallacy suggests all we have to do is add a well conducting bar across the front of the element, say an inch away (unwired and not receiving any power, i.e. benign) and voila’ we now have a strip heater putting out twice the energy. Can someone help me with patents please?
    ……………………………
    I try to make the same point, that the shell has to have some rather special properties to do anything. Also, there is a big mix-up (as usual) between static analysis and dynamic analysis. This shows with what cohenite says:

  190. Greenhouses work by containing convection and not by preventing the re-transmission of radiant energy.

    Whatever the “effect” on the Earth’s climate it has little to do with what happens in greenhouses.

    It should more properly be called ” the insulation effect” as the atmosphere limits the transmission of heat energy by its attenuation within a material.

  191. Willis, I’m curious about the effect of a high mass, 100% IR transparent atmosphere as posed in my first question above… Not sure if you saw it. Thanks, Mike S.

  192. Willis

    How does the glass shell radiate twice as much energy as it receives? Inquiring minds would really like to know.

  193. Ken Coffman says:
    November 27, 2010 at 6:53 pm

    “I get accused of believing some odd things, but I’m not the one denying that N2 and O2 molecules have temperatures.”

    Individual molecules will have wildly varying temperatures if you could somehow find a way to measure them individually.

    “781,000PPM of N2 has no thermal capacity or thermal mass?”

    Of course it has thermal mass and as a group of molecules will have a measurable temperature which is the average of the group.

    “I don’t care how many degrees of freedom your fantasies have, the temperature of 781,000PPM N2 has nearly nothing to do with anything 390PPM of CO2 can do.”

    This is where you’re wrong. Imagine two thick sheets of copper sitting out in the sun. On one we put a put a very thin coat of black paint and on the other we put a very thin coat of white paint. The paint weighs next to nothing compared to the weight of the metal. Will the paint cause a difference in the temperature of metal as it sits out in the sun? I think you know the answer to that.

    “Stefan-Boltzmann says things radiate based on their temperature, not whether they’re made of CO2 or water vapor.”

    That is correct. But what governs absorption? The key is that CO2 absorbs infrared radiation at characteristic infrared frequencies while N2 is completely transparent in infrared. In a dense mixed gas (air is “dense” in this context) of N2 and CO2 when a CO2 molecule absorbs infrared radiation it translates to kinetic energy – the molecule is moving and vibrating faster as a result. The motion is what registers on a thermometer. Since it’s in a dense gas it almost immediately bumps into a neighboring molecule and transfers some of its kinetic energy to the molecule it bumps into. In that manner CO2 acts to absorb infrared energy emitted from the earth’s surface and raises the temperature of all the molecules surrounding it. Without the CO2 the radiation would have passed straight though the N2 and the N2 would not have been made warmer.

  194. Hello all,

    To Ric Werme’reply, November 27, 2010 at 6:03 pm

    1/ Everytime you see a reference to -18°C in the climate debate, you have to remember this value comes from the use the other way around of the Stefan-Boltzmann’s law as related in the peer-reviewed paper “Falsification Of The Atmospheric CO2 Greenhouse Effects Within The Frame Of Physics” (Gerlich & Tscheuchner). Therefore, this -18°C value means absolutely nothing and should never be used in the climate debate.

    2/ Since the wrong calculation is usually done for a non rotating and atmosphere free planet, the comparison with the moon do apply here since in the same situation, earth would end up with the same desert-like surface anyway. The moon low rotation motion (14 days) is slow enough to reach a surface temperature steady state.

    3/ The whole point is that a pure radiative approach of earth climate is meaningless as pointed out by G&T.

  195. @Ken (con’t)

    The “greenhouse effect” (a poor term but we have to live with it) is physics that was known theoretically 200 years ago. About 150 years ago John Tyndall demonstrated it expermentally in literally thousands of different experiments with various gases under various pressures irradiated at various frequencies. He published his results in a seminal tome titled “Heat: A Mode of Motion”. You can read the book in its entirety on Google Books:

    http://books.google.com/books?id=eZEAAAAAMAAJ&dq=john%20tyndall%22%20heat%20motion&pg=PP1#v=onepage&q&f=false

    For ease of reading I suggest downloading the PDF file.

    The bottom line is the greenhouse effect was demonstrated experimentally over 150 years ago. Anyone claiming it isn’t real is sorely lacking in understanding of introductory level physics.

  196. Others may be interested in the related discussion starting here and particularly the following comment on the shell model, and also here. The general idea was covered further at greater length here.

  197. Willis

    A simple question for you. Recognizing that your analogy is extremely simplistic to prove the existence of the “greenhouse affect”, what happens if you make the shell only 80% opaque to long wave radiation and then increase that gradually to 80.1%?

    My guess, it results in an unmeasureably small change in the planets temperature!

    Please feel free to enlighten me at 2billyarber@Gmail.com

    Bill Yarber

  198. It stands to reason that heating a globe who’s core has been cooling overall since it’s formation would be an intensive matter. And because of our size and shape, we don’t capture much of the Sun’s infrared. If the Earth fails to capture and hold onto longwave, we roast during the day and freeze at night.

    The premise here is to layer up. One blanket does not trap heat nearly as well as multiple blankets. And a single opaque blanket may lead to cold damp conditions. So it’s a good idea to have a bit of flow-through in each of these layers. Which equates to the idea of the shells being not absolutely opaque but nearly opaque to longwave.

  199. TimM says:
    November 28, 2010 at 3:24 am

    “I assume that the amount of heat humans create is known to some degree and accounted for in climate modelling (unless it is considered trivial)?”

    Yes and no. It isn’t trivial when a thermometer is located in close proximity to an artificial heat source! Averaged over the entire surface of the earth it is indeed a trivial amount of heat. Also trivial is the amount of heat from the molten hot core of the earth that is emitted from the surface due to the crust being a very good insulator.

  200. Willis: “OK, let’s assume that the sun is some form of luminous cloud that illuminates the planet evenly with 240 W/m2 of solar radiation.”

    Then the planet heats up to the same temperature as the luminous cloud. Luminous cloud is a simplifcation too far, principles need to be argued from point source.

    Flow of energy is then crucially determined by geometry. A disc (facing into the point source) has much the same issue as luminous cloud. I’d say it is essential to stick to a sphere.

    Even a highly idealised spherical model needs to be detailed enough to integrate incoming radiation distributed over the sphere (or one side of a sphere at any time), conduction over the surface, and differentials of OLR at different coordinates.

    I’m not going to try to do that because it relies on too many assumptions (that I’m not in the best position to deal with). However I supect that the combination of assumptions, integrals and application of S-B will introduce some significant numerical factors not evident in the simple “vertical line” diagrams above.

    (BTW, before you answered the question set at the top of the thread, my answer was going to be “no angles”.)

  201. It seems there should be some time considerations that should play into a “greenhouse gas” model. I imagine that on the gasless moon radiation strikes the surface, heat is created in the atoms at the surface and some short time later the heat is available to be re-radiated (at a longer wavelength), but if there were no “short time later” you would have a reflective surface and presumably there would be no heating. Likewise the time needed for radiation to be absorbed and re-radiated within the atmosphere should have a bearing on what equilibrium temp is achieved. In other words the atmosphere can be thought of as hindering or slowing the passage of radiation from the surface, and how quickly that happens helps determine how much warming occurs – with less warming occurring when radiant energy passes more quickly.
    Does the atmosphere have an R-factor? Shouldn’t a blanket or insulation be part of this awkward, inapt greenhouse metaphor? Maybe an electric blanket where the electricity coming in on a wire in an electric blanket is akin to the shortwave sunlight passing freely through the atmosphere to warm the surface?

  202. @Dave Springer says:
    “The bottom line is the greenhouse effect was demonstrated experimentally over 150 years ago. Anyone claiming it isn’t real is sorely lacking in understanding of introductory level physics”

    Nobody is saying that there is no backradiation. The point is that backradiation emitted by a layer with -55 Celsius cannot transfer heat to a surface much warmer that -55 Celsius. This is the second law of thermodynamics. This second law is true for all of us: Being realists, physicist or oekofascist…

  203. Willis,

    If it is the simplest possible model you want that is USEFUL then I think I already suggested it. A shell one km thick being in contact with earth surface and a resulting temperature gradient from earth surface to Top Of Shell. Still no where close to reality, but sufficient to model re-radiance of energy based on varying amounts of CO2.

    As for those who are jumping on Willis because they don’t get how the shell winds up doing what it does, I’m sad to say that this hard core skeptic agrees 100%, that is EXACTLY what happens. In the real world the “effect” is distributed across the thickness of the atmosphere, and it seems counter intuitive when you look at the drawing, but that is what happens. As upward bound photons are absorbed by CO2 molecules and then re-emitted in a random direction, the amount of energy they absorb gets re-emitted 1/2 upward and 1/2 downward. For conservation of energy to take place, the earth surface mut heat up enough to create upward bound photons that balance the re-emitted downward bound photons.

    Which is why a “shell” of a given thickness with a temperature gradient from bottom to top is usefull. It isn’t the surface temp per se we are interested in, it is the change in the gradient. If you do it that way you will find that doubling co2=3.7watts=+1 degree AT SOME POINT ON THE GRADIENT and that surface temp of earth only goes up about 0.6 degrees and the observed temperature of earth from space goes up…. O.

  204. Sorry, model doesn’t work. In figure 2 the temperature of the shell is given as 0 Kelvin. Since energy is either transmitted perfectly or reflected perfectly, that temperature doesn’t change. As a result in figure 4 it has to radiate at 0 Kelvin.

    Another way of going about it, after going opaque we change the suns output from 240W/m^2 to, say, 500W/m^2. What happens on the inside? The answer must be that nothing changes. The shell is opaque and all energy is reflected. Inside the numbers stay at 240W/m^2 and on the outside the numbers are now 500W/m^2. Does the shell now have two temperatures?

  205. Dave Springer says:
    November 28, 2010 at 6:41 am

    The “greenhouse effect” (a poor term but we have to live with it) is physics that was known theoretically 200 years ago. . . .[edit] . . .

    The bottom line is the greenhouse effect was demonstrated experimentally over 150 years ago. Anyone claiming it isn’t real is sorely lacking in understanding of introductory level physics.

    . . . [edit] . . . The bottom line is the greenhouse effect was demonstrated experimentally over 150 years ago. Anyone claiming it isn’t real is sorely lacking in understanding of introductory level physics.

    - – - – - -

    Dave Springer,

    I agree. The name ‘GHG effect’ is a terribly misleading and very unscientific name. So let us forthwith not use it. I suggest something like Atmospheric Effects of Molecular Dipole Moment Gases is much much better.

    Also, regarding the GHG effect being proven, yes, but a simplified demonstration in laboratory does not mean the earth atmosphere effect when mixed with actual atmospheric processes has always the same effect as the laboratory at all altitudes and in all regions. It is simple in lab principle is all we can conclude.

    John

  206. The actual surface temperature does not follow the Radiation Theory Explanation.

    The Earth should get much, much hotter during the day and much cooler at night. Basically, the Earth absorbs just an extremely tiny amount of the solar energy received during the day and releases back that tiny amount at night.

    It absorbs only 0.00026% of the energy received by the Sun during the 12 hours of sunshine and then releases back that 0.00026% at night (give or take some small daily seasonal change).

    In other words, 99.99974% of the solar energy received during the day is sent back up to the atmosphere and eventually to space every second that the Sun is up.

    A given square metre of Earth Surface receives 20,736,000 joules of solar energy during the day but the surface temperature only rises by 54 joules during that period.

    You could think of it as, over time, the Earth has accumulated a tiny differential in outgoing versus incoming radiation and if the Sun didn’t come up tomorrow, it would all be gone in 86 hours (give or take a lag from the ocean and land – it might take a few decades to fall to -270C but it would fall to -75C in just a few days).

    This is a fairly typical 24 hours of measured Radiation Flows for a given location. The explanation does not match reality. Table Mountain Colorado on November 17th, 2010.

  207. Good post! You addressed both greenhouse effect and zero greenhouse effect. You overlooked a third possibility. On Titan, the atmosphere absorbs some frequencies directly from the sun, and is transparent to some of the longwave frequenices from the planet, resulting in a possible ” anti- greenhouse effect.

  208. How about this fundamental problem…

    In figure 4 the blue arrow is radiating energy from a low energy black body (the glass) towards a higher energy black body (the earth). Isn’t this a violation of entropy?

  209. Furthermore, this can’t happen unless work is somehow applied to the model and there isn’t any in the model is there?

  210. bessokeks,

    “But transfering heat energy from “cold” to “warm” violates the second law of thermodynamics.
    There is no way around that.”

    There is a lot of misunderstanding surrounding the second law of thermodynamics. The original formulation of this law says approximately what you attribute to it. This was applied to heat transfers in conducting bodies. The modern formulation is in entropy and states that in a closed system, the total entropy must always increase. The application to heat transfer has been changed to refer to net heat transfers. The implications to conduction of heat have not changed. However, we know that radiative transfers do indeed take place in both directions.

    In order for the second law to be obeyed, the net transfer must always be from warmer to cooler. This should be intuitive – even a cool body emits photons of energy which must travel at the speed of light regardless of the fact that they may impinge on surfaces that are warmer. The net transfer argument should make it plain that more photons will be emitted from the warmer surface.

    A modern version of Maxwells demon thought experiment is to imagine a demon sitting between the cool and warm surfaces. The demon is supposed to allow the photons from the cooler surface to reach the warmer surface, but prevent the flow in the opposite direction. The action of the demon would cause the cool surface to become cooler and the warm surface to become warmer. This is clearly in violation of the second law. The Maxwell demon paradox was solved when it was realised that the demon needs information to sort the photons and this prevents the system from being closed.

    In practice, although photons from the cool surface carry energy to the warm surface, a higher flow occurs in the opposite direction. The result is that the warm surface cools and the cool surface warms. Ie, they equilibriate and their entropy increases. Thus, back radiation does not violate the second law.

  211. Thinking a bit further. in figure 4 the temperature of the planet has to rise until it radiates 240W/m^2 at and above the short wavelength where the shell is transparent. The radiated spectrum will look a bit unnatural but that’s to be expected when working with miracle matter.

  212. John Bowman says:
    November 28, 2010 at 6:02 am

    “It should more properly be called ” the insulation effect” as the atmosphere limits the transmission of heat energy by its attenuation within a material.”

    Exactamundo!

    That should be enough said and it’s a shame when it isn’t. Most people at least intuitively understand the effect of insulation if not the physical mechanism by which it works.

  213. willis,
    can you explain why the temperature of the shell in figures (3) and (4) is only 255K rather than 303K? It seems to be absorbing and radiating at 480W per m2 – the same as the planet in figure (4).

    thanks
    nige

  214. I agree that thought provoking hypothesis can generate discussion amongst discerning individuals but they must be accompanied with caveats which clearly state that the thought presented is a special case and on its own has no practical implications in the real world except it is one of a long list of interdependent variables which make up the final result.
    My concern is that some will take such a presentation too literally and form a wrong impression.
    The assumption that 100% of the absorbed radiation is reemitted back from the earths surface is incorrect since the plant life of the world absorbs radiation energy to generate carbohydrate from CO2 and H2O.
    The grossly oversimplified diagrams showing radiation from the gasses or particulate in the atmosphere has a free ride back to the earths surface is nonsense since radiation downward will have the same degree of absorption as radiation upwards. Further to this point with regard to gasses they absorb resonant frequencies very quickly ( try doing a Beer’s law calculation) however the close neighbors to the resonant frequencies are absorbed much slower so reach higher into the atmosphere, when a molecule reradiates however it does so at exactly its resonant frequency so its return path is much shorter. Thus none of this energy can return to the surface. I note that in one comment an exception was made to the transfer of energy by collision this is not a viable argument since the time taken by a molecule to reemit IR energy is about 10 000 times longer than the time taken to collide with another molecule thus there is virtually no reemission.
    One last point, to assume the atmosphere of the entire world contains some mix of components which forms a 100% opaque screen to all the outgoing radiation is a worthless hypothesis I think a number of shells representing all the different components with their contribution to the warming effect of our atmosphere would be far more instructive.

  215. Willis, I believe you figure 3 and 4 are misleading. Here why. Instead of using energy, use dollars with a checking account (the shield) and a saving account (black body). You must do this in cycles.
    We have $240 going into the checking account half of which going out, $120. The other half, $120, goes into the saving account then comes out into the checking account. Half is goes out $60 the other half goes back into the saving account and so on. The most you are working with is $360 ($120 in the saving account plus a new cycle of money, $240.)
    Figure 4 It doesn’t make any difference if the money hits the saving account first then is transferred to the checking account. Half, $120, goes out, half, $120, is going back into the saving account before starting a new cycle and another $240 is added. You are never working with $480. The most you are working with is $360.

  216. the “greenhouse effect”

    Maybe this would better be explained by the use of “S-Parms” as they are used in the characterization of uWave (Microwave) components and devices; S-Parms more formally known as Scattering Matrix Parameters.

    A system by which the vector sums of the forward (incident) and reflected (back scattered) waves or energy are characterized and allows for the characteristics of the ‘system’ to have different behavior at different wavelengths (as uWave components characterization often uses frequency swept measurements).

    Now, if one considers that incoming solar insolation and outgoing earth LW IR encounter elements in three-space (including an optional 3-dimensional atmosphere of some thickness) that have reflective and scattering properties at IR wavelengths then a measurable ‘flux’ exists between the earth, the sun and the infinite ‘sink known as deep space, and each element will have some associated temperature above absolute zero given ‘forward’ and ‘reflected’/re-radiated IR/thermal energy … and the earth will have _with_ an atmosphere some temperature above that what it would have without an atmosphere … especially true if some of the elements encountered exhibit different wavelength characteristics to the spectra of the incoming and outgoing energies (the EM waves, the IR etc.)

    I don’t expect anyone exc a couple of uWave RF engineers or physicists to comprehend this … MODTRAN et al (as they account for CO2 and WV) are kinda examples of this.

    S-Parms: http://www.google.com/search?hl=en&client=opera&hs=yf6&rls=en&q=s-param+scattering+matrix&aq=0&aqi=m1&aql=&oq=s-parmsscattering+matrix&gs_rfai=

    .

  217. Jordan says:
    November 28, 2010 at 7:49 am
    Willis: “OK, let’s assume that the sun is some form of luminous cloud that illuminates the planet evenly with 240 W/m2 of solar radiation.”

    “Then the planet heats up to the same temperature as the luminous cloud.”

    Only if the cloud is optically thick. If you make the cloud of a hot (6000K) but highly rarefied plasma, it could provide the same amount of “sunlight” as the sun, but stop only 1/160,000 of the planet’s thermal radiation. More simply, if you want a model with effectively uniform insolation (which may indeed be a “simplification too far”), then you can surround the Earth with a cloud or constellation of miniature suns, allowing the thermal radiation to escape through the gaps.

  218. I read through this post and realized – this post exemplifies exactly what is wrong with serious AGW arguments (note – serious does not include the alrmist fringe – there is no explaining their arguments).

    Serious AGW arguments always seem to start off with a “pretend Earth is a Terrarium” tangent and then try to build off of it. Well, Earth is not a terrarium. Most (all?) serious AGW skeptics understand that the Earth will not behave like a simple system. It isn’t a black body, and it doesn’t have a glass shell. It has a very dynamic atmosphre and multiple non-linear response patterns to incoming energy.

    Even if we tried to start fixing your “pretend a planet is surrounded by a shell” thought experiment – it soon becomes so complex as to lose its power to explain anything. For example – lets first give your planet 3 concentric shells representing high, medium, and low atmosphere. Each responds differently to temperature becoming more or less reflective (representing clouds) over part of their surface in a very dynamic way. Oh, and the planet has a range of colors and refelctivity, and two main materials that store and reflect heat differently, and also can change refelctivity due to temperature (snow). Now we have a simple (!) model that accounts for reflectivity and absorbtion. We are merely missing heat transport (air convection, ocean currents), pollution effects (not all man-made), and variable light input (solar changes, orbital patterns, tilted rotating and orbiting planet). Does this model help explain anything? No, it makes people bug-eyed.

    If atmosperic science could be studied in a controlled environment, we could tease apart the various interactions. Unfortunately, all that can be studied in a controlled environment are tiny bits and pieces (like what happens if we add CO2 to the terrarium?). People with science degrees in specific areas, but no understanding of large complex (often chaotic) systems then run off with the bits and pieces and make grand predictions pointing at their “facts”. (I am also ignoring the people that make up their facts – I am talking about serious AGW supporters here)

    This is where the two sides can never seem to find common ground. AGW supporters want to over-simply to the point its “easy to understand”, and AGW skeptics can’t help but wonder how many of the gazillion detials left out are actually important.

    My experience tells me that natural systems are complex and dynamic. Climate science is not going to “fit” into a simple realtionship between CO2 and temperature response. If it did, we would have been warming for the last 20 years at a minimum of a linear response, and more likely an increasing response. It hasn’t happened – therefore AGW supporters needs to re-think their side of the argument.

  219. Answering your Test-Question:

    -The wavelengths which water and CO2 absorb are (mostly) saturated. So this model would have to include multiple shells for certain wavelengths. The number of these shells depends on the average distance between radiation and absorption within the atmosphere. This distance is correlated to the concentration of the greenhouse gases.

    - Heat transport through convection between these shells and between the shells and the the surface.

    - Varying opacity due to clouds.

  220. The Greenhouse Effect is not ill-named. The principle is quite simple and quite general; whenever you make it easy for sunlight to get in, and hard for heat to get out, you get greenhouse warming. This is true no matter how the outwards heat flow is impeded, whether as convection, conduction, radiation, or something more exotic. The Earth’s atmosphere impedes convection, conduction and radiation. The glass of a garden greenhouse impedes convection, conduction and radiation. They are not fundamentally different, though of course the relative importance of each transfer mechanism differs.

    I should also point out that garden greenhouses do impede thermal radiation, more than is sometimes thought, because the water condensing on the glass at night is a strong absorber.

  221. Bryan.

    The woods experiment has nothing to do with radiative transfer, especially in the column of gases, miles high, that we call the atmosphere. The properties of transfer change as you go higher in the atmosphere and are very different in the dry stratosphere. So, just very simply, woods test fails to test the real issue. His experimental set up does not replicate the conditions in the atmosphere, and thus it says nothing about the radiative properties of C02 in the atmosphere. We know from measurements of back radiation that outgoing longwave is reflected back. We know from physics which molecules are responsible and why. we know the same from TOA measurements. The question, the only open question worthy of debate (worth our time) is what sort of feedbacks are present. every doubling of C02 results in a warming of say ~1C. First order estimation based on known working physics. everything else beyond this or below this can only be estimated based on models and by looking at transient responses of the system. Someone like Willis would argue that the system is damped ( naturally regulated) and climate science argues that it is undamped ( in the short term at least).

  222. With all due respect, Mr. Springer, you mix up a thin varnish with 781,000 microdots of clear epoxy and 390 molecules of opaque microdots and tell me you can see the difference in visible light, particularly when I’m doing everything I can think of to distract you, like reflecting and slopping on the varnish unevenly and modulating the incoming white light and waving banners between you and the surface you’re trying to study. I accept the resonance on CO2 (and water vapor molecules), but I don’t accept that you can measure the effect of it. If you can’t measure it, then I want you to stop it with the AGW nonsense.

  223. Willis, I like it. It is what it is. A thought provoking model that does not and was not meant to mimic reality.
    My physics mentor gave me something similar. A 1Tonne Bull charges a lump (very large) of putty. Calculate the energy absorbed by the deformation, converted to heat and it’s location (the energy’s that is).

    It’s a very simplistic model that when iterated to infinity will give you the answer to everything climatic. Just one thing, I might have missed it, but won’t the shell reradiate in all directions. 240w absorped, ideally, 240 radiated over 4/3 pi r²? 120 w up and 120w down from both sides of the opaque shell. In simplistic terms.

  224. “So, just very simply, woods test fails to test the real issue. [...] We know from measurements of back radiation that outgoing longwave is reflected back.”

    But the reflection of outgoing longwave radiation isn’t what controls the temperature, because of convection.

    Take the analogous case of a pan of boiling water on a stove. The temperature of the water is 100 C. We turn up the heat. Double the power is entering the system at the base of the pan. But the temperature that results from all this extra heat is… 100 C.

    The reason it doesn’t get hotter, even though more heat is entering the system, is that convection and evaporation speeds up the rate at which the heat rises and escapes.

    Logically, you cannot leap from “more heat” to “higher temperature” without talking about how convection responds, any more than you can say the pan of water gets hotter when you put more heat in. And in a convective atmosphere below the tropopause, convection dominates in determining the temperature profile. The downwelling longwave radiation is still larger in magnitude, but the radiative effects discussed here are ‘short-circuited’ by the non-linearity of convection, and forced to match the adiabatic lapse rate. The adiabatic lapse rate is entirely unaffected by the radiative properties of ‘greenhouse’ gases.

    There is a greenhouse effect, and the Willis’s description is how it would work in a non-convective atmosphere, but with convection present it’s effectively an entirely different mechanism and radiative balance is not what determines surface temperature.

  225. Willis Eschenbach says:
    November 28, 2010 at 4:13 am
    you can say that 390 W/m2 is emitted by the surface as radiation, and 100 W/m2 is lost from the surface through convection and evapotranspiration. So somewhere around a fifth of the energy is not going into warming the surface.

    Hi Willis, thanks for your reply. Although the figures for downwelling and upward radiation are big in comparison to other forms of energy transport, I think we need to step back and consider their relative importance from a wider perspective. The LW flux results in a net cooling of the surface of around 71W/m^2 according to your modified K-T diagram. This is less than the combined convection and evapotranspirationof 100W/m^2 also cooling the surface.

    I thin we need to be careful with terminology to avoid confusion. You talked about downwelling LW ‘heating’ the surface. Since the net flow of down and up LW results in a cooling of the surface, perhaps it’s better to say that downwelling LW slows the rate of heat loss from the surface rather than saying it warms or heats it?

    But then it’s clear that if there wasn’t as much downwelling LW there wouldn’t be as much upward flow either, and the other point to make here is that downwelling LW doesn’t penetrate water (70% of the surface) so contributes to evaporation by superheating surface water molecules, a cooling effect.

    So it is apparent that the only way the greenhouse effect operates is by raising the altitude of the outward radiating shell. The additional co2 will have raised it around 1-1.5% or about 100 – 150m.

  226. PJP says:

    “… lets simplify it even more and say that energy is delivered in truckloads. Lets say we get 2 truckloads per hour. … when we come to your semi-transparent shell, you are still getting two truckloads per hour, but you say that these two truckloads are delivered to both the earth and to the shell — that makes 4 truckloads/hr. … Where did the extra two truckloads come from?”

    OK, say the Sun is shipping two truckloads of orange juice to the Earth each day through a Shell that is transparent to orange juice – all the orange juice gets through unchanged and the two truckloads of the stuff are absorbed by the Earth.

    Whatever incoming juice the Earth absorbs is converted to blueberry juice and put on trucks that head back to the Shell. So, two truckloads of blueberry juice head to the Shell. The Shell absorbs the two truckloads of blueberry juice and then (pay attention here!) sends one truckload of blueberry juice out to Space and one truckload back to the Earth where it is absorbed.

    The next day, the Sun has sent another two truckloads of orange juice to Earth and Earth has absorbed and transformed it to blueberry juice. So, on the second day, the Earth has THREE truckloads of blueberry juice to send up to the shell (the truckload sent back by the Shell yesterday and two new truckloads from today). OK, those THREE truckloads get to the Shell and half is returned to Earth (1.5 truckloads) while the other half goes to Space.

    On the third day, the Sun has sent yet another two truckloads of orange juice to Earth, which is absorbed and converted to blueberry juice. That, along with the 1.5 truckloads returned from the Shell yesterday, is sent back to the Shell, for a total of THREE AND A HALF truckloads today.

    As you should be able to see, as the days pass, the traffic in blueberry juice from the Shell to the Earth and back will continue to increase asymptoticaly until there are FOUR truckloads of blueberry juice a day going from Earth to the Shell and TWO coming back.

    So, in the steady state, each day there are TWO truckloads of orange juice and TWO truckloads of blueberry juice going from the Shell to the Earth, FOUR truckloads of blueberry juice from the Earth to the Shell, and TWO truckloads of blueberry juice going from the Shell to Space. Exactly as Willis shows in his Figure 4. QED!

  227. Steven Mosher

    …..”The woods experiment has nothing to do with radiative transfer, especially in the column of gases, miles high, that we call the atmosphere.”…….

    The part of the Woods experiment that most people pass by is that for the remaining real radiative effect the result is almost negligible.

    Some people have argued that this is explained because the Woods experiment box had very little height.
    Hence my post on the much larger polyethylene tunnels which bear out the conclusions of R W Wood.

    The radiative effect of CO2 and H2O seems to be of little consequence in large greenhouses even though the IR blocking was 100%.
    If this concentrated “greenhouse effect” is of such little consequence that the recommendation of the study group is not to bother adding it to polyethylene what does that tell us about the atmospheric greenhouse effect?

    The only effect that the “greenhouse gases” seems to engender is speculative posts and thought experiments.
    For bulk effects of the atmosphere such as;
    The polytunnel study in post above.
    Another bulk experiment is obtained by pointing parabolic mirrors at the dark night sky.
    This is capable of freezing water at the mirrors focus even when the planet surface temperature is above zero.
    I realise of course that the focus is not perfect for an extended source such as the atmosphere however a concentration effect seems to be evident.

    On the other hand is there any hard physical evidence of a bulk effect to show that the large claimed magnitude of the “greenhouse effect” is real?

    I discount the claimed assertion that it amounts to a 33C surface temperature increase unless backed with a plausible cause.

  228. H2O, CO2, O3, CH4, N2O absorb outgoing infrared radiation, promoting these molecules to higher-energy excited states. The excited states collide inelastically and repeatedly with N2 and O2 molecules, increasing the kinetic energies of those N2 and O2, raising the average temperature of the upper atmosphere and returning the excited greenhouse molecules to their ground states. Outgoing radiation is therefore prevented from radiating into space but instead is absorbed and transferred to the atmosphere.

    The warming of the atmosphere reduces the thermal gradient between the surface and upper atmosphere and therefore acts to insulate.

    Is this explanation correct?

  229. Paul Birch says: “hot rarified plasma” around the planet.

    To be relevant to AGW discussion, this thought experiment needs to relate to our situation, or it should move in that direction when elaborated. You are running in the wrong direction.

  230. @ Mike Haseler:

    “Come on this is stupid. The greenhouse effect doesn’t exist for the simple reason that there is no ‘greenhouse’ effect in a greenhouse – it would be like using the theory of the ether to explain radio waves and talking about the “ether effect” and not expecting most people who knew what was being talked about not to roll with laughter on the floor.”

    Are all those gardeners who have greenhouses stupid? Have you never warmed yourself on a cold but sunny day by standing with your back to a window through which the sun is shining?

  231. “Are all those gardeners who have greenhouses stupid?”

    Nope. Greenhouses work by preventing heat loss via convection, not by the “greenhouse effect”.
    ~And how does CO2 in the atmosphere prevent convection?

  232. Ok for those of you who keep claiming fig 4 is wrong, here is why its not:

    Solar energy to the planet: 240 w/m^2
    Energy reradiated by the atmosphere back to the planet: 240 w/m^2
    Add the two together: 480 w/m^2, which is what the planet radiates back.

    The atmosphere intercepts half of the 480 and reradiates it back to the planet in a continuous cycle that is the equilibrium, while the atmosphere radiates half the 480, or 240 w/m^2, to space, which is the same amount the sun sends to the planet.

    For those of you who keep saying convection: that does eliminate the interior vacuum, so you will get, in addition to reemission of heat from one molecule to another from the surface to the edge of space, you will get some degree of convection as molecules rise when they are warm, and drop when they are cool, but this is limited to certain layers of the atmosphere, there is no actual convection from the surface to the edge of space.

    For those of you who point to the earth being round with one side in darkness, as Willis says, the model is time averaged. The peak solar insolation is far higher than 240 w/m^2, 240 is a 24 hour average, which includes the time the sun isn’t shining on the surface. Peak solar insolation, with no clouds, is about 1000 w/m^2 at the surface at the equator, and zero at the poles, while about 400 w/m^2 is blocked/reflected by the atmosphere from the start (more when there are clouds involved).

    For those of you who keep saying the molecules in the atmosphere radiate in all directions: Yes, this is true, which is why the model shows 240w/m^2 going up, and 24ow/m^2 going down, but not quite exactly…..

    The flaw in Willis’ model, however is that you wont have an equal division of 240 up and 240 down, because the earth is round, the surface curves away convexly from the molecule’s perspective, so there will always be slightly more heat radiated to space than reabsorbed by the ground.

    There is also a factor of variance that involves how the solar cycle’s variation in solar wind and UV and other factors causes the earth’s atmosphere to vary in diameter. When it is swollen up during solar maximum, it has a larger diameter so it will intercept more radiation than during solar minimum, when it is a smaller diameter.

  233. “For those of you who keep saying convection: [...] but this is limited to certain layers of the atmosphere, there is no actual convection from the surface to the edge of space.”

    Good answer. But there is convection throughout all the layers that actually radiate to space. Most ‘greenhouse’ radiation to space is from the lower 10 km of atmosphere, the troposphere, and it’s all convective.

    The exception is the ozone layer in the stratosphere, which absorbs UV and is non-convective.

    Incidentally, since everybody knows that hot air rises, why are the tops of mountains so cold? Why is the top of the troposphere 11 km up at a temperature of -54 C? What maintains the temperature gradient?

  234. Jordan says:
    November 28, 2010 at 12:04 pm
    Paul Birch says: “hot rarified plasma” around the planet.

    “To be relevant to AGW discussion, this thought experiment needs to relate to our situation, or it should move in that direction when elaborated. You are running in the wrong direction.”

    I was not running in this direction. On the contrary, it was I who pointed out that the Earth is not subject to a uniform insolation. However, your objection to Willis’s fix was erroneous. The uniform insolation simplification is workable. How useful it is may be debatable; but those who cannot or will not understand even this simple model are unlikely to profit from more realistic ones.

  235. Willis,

    Sorry, but I liked some of the other answers (like Leonard Weinstein’s) better than yours. There are all kinds of reasons why a single shell model separated by a vacuum would not be very representative of real Earth temperature, such that if it was anywhere close it would have been a fluke. The question is whether a given model is informative about the impact of higher greenhouse gas concentrations. I don’t think the two shell model is much more informative than the one shell model. Sorry to be critical, but you’re the one who asked a fuzzy question and then held out for a particular answer that isn’t particularly interesting.

    I think that the rest of the post is fine, I just think the final question and answer is a bit silly. Next time maybe you should drop a couple of hints to move the conversation in the desired direction.

  236. Ira,

    That was a brilliant explanation. I’ve tried explaining the physics a dozen different ways and never got to a concise explanation that was easily understood and still reasonably accurate. thanks, and with your permission, soon to be repeated to many.

  237. Willis, your model is what I like to call the “Planet Dewar” model of the GHG effect. Of course, the air really is in contact with the surface, and it really has an exponential temperature gradient as altitude rises caused by gravity, at least up to the tropopause, but it sounds like these are not the problems you have in mind.

    One paradox that arises from this model is when it is extended to what I call the “Planet Dante” model, with more multiple crystalline heavens. The same reasoning that leads 2E to reach the surface when only E is incoming with one heaven leads 3E to the the sruface with 2 heavens, etc. With multiple heavens, the surface can be made arbitrarily hot by this reasoning, at least until the surface begins to incandesce and emit visible light and short-wave IR which can escape through the assumedly transparent shells.

    So other than this, what is wrong with the good old Planet Dewar model?

  238. >>Willis Eschenbach says:
    >>November 27, 2010 at 6:05 pm
    >>Some folks upthread have said that what is missing is rotation. … I still have not seen anyone point to it.

    You assume a shell with 2 sides but a one sided earth. The rotation of the earth carries heat from the hot to the shaded side, where it radiates back to the shell and out to space. The shell itself cannot be assumed to be a constant temperature because it receives less solar heating at the poles.

    Your model needs gravity and a circulating atmosphere to create a lapse rate, giving different temperatures between the shell and the earth’s surface.

  239. Robert of Texas says:
    November 28, 2010 at 10:42 am

    I read through this post and realized – this post exemplifies exactly what is wrong with serious AGW arguments (note – serious does not include the alrmist fringe – there is no explaining their arguments).

    Serious AGW arguments always seem to start off with a “pretend Earth is a Terrarium” tangent and then try to build off of it. Well, Earth is not a terrarium.

    Oh, please. When someone claims that some huge group of arguments “always start off with” something or other, we’re into a world of exaggeration.

    Some discussions of the climate use thought experiments. Some of these are good, some not. My point was to stimulate discussion of a simplified system that still exhibits the “greenhouse effect”. I’d say I was successful …

  240. Lets start with the infamous woods experiment:

    As with all climate science I like to review the ACTUAL paper and get the data. Wood’s “paper”

    R. W. Wood: Note on the Theory of the Greenhouse
    The following text is from the Philosophical magazine

    XXIV. Note on the Theory of the Greenhouse
    By Professor R. W. Wood (Communicated by the Author)

    THERE appears to be a widespread belief that the comparatively high temperature produced within a closed space covered with glass, and exposed to solar radiation, results from a transformation of wave-length, that is, that the heat waves from the sun, which are able to penetrate the glass, fall upon the walls of the enclosure and raise its temperature: the heat energy is re-emitted by the walls in the form of much longer waves, which are unable to penetrate the glass, the greenhouse acting as a radiation trap.

    I have always felt some doubt as to whether this action played any very large part in the elevation of temperature. It appeared much more probable that the part played by the glass was the prevention of the escape of the warm air heated by the ground within the enclosure. If we open the doors of a greenhouse on a cold and windy day, the trapping of radiation appears to lose much of its efficacy. As a matter of fact I am of the opinion that a greenhouse made of a glass transparent to waves of every possible length would show a temperature nearly, if not quite, as high as that observed in a glass house. The transparent screen allows the solar radiation to warm the ground, and the ground in turn warms the air, but only the limited amount within the enclosure. In the “open,” the ground is continually brought into contact with cold air by convection currents.

    To test the matter I constructed two enclosures of dead black cardboard, one covered with a glass plate, the other with a plate of rock-salt of equal thickness. The bulb of a themometer was inserted in each enclosure and the whole packed in cotton, with the exception of the transparent plates which were exposed. When exposed to sunlight the temperature rose gradually to 65 oC., the enclosure covered with the salt plate keeping a little ahead of the other, owing to the fact that it transmitted the longer waves from the sun, which were stopped by the glass. In order to eliminate this action the sunlight was first passed through a glass plate.

    There was now scarcely a difference of one degree between the temperatures of the two enclosures. The maximum temperature reached was about 55 oC. From what we know about the distribution of energy in the spectrum of the radiation emitted by a body at 55 o, it is clear that the rock-salt plate is capable of transmitting practically all of it, while the glass plate stops it entirely. This shows us that the loss of temperature of the ground by radiation is very small in comparison to the loss by convection, in other words that we gain very little from the circumstance that the radiation is trapped.

    Is it therefore necessary to pay attention to trapped radiation in deducing the temperature of a planet as affected by its atmosphere? The solar rays penetrate the atmosphere, warm the ground which in turn warms the atmosphere by contact and by convection currents. The heat received is thus stored up in the atmosphere, remaining there on account of the very low radiating power of a gas. It seems to me very doubtful if the atmosphere is warmed to any great extent by absorbing the radiation from the ground, even under the most favourable conditions.

    I do not pretend to have gone very deeply into the matter, and publish this note merely to draw attention to the fact that trapped radiation appears to play but a very small part in the actual cases with which we are familiar.”

    Why is his second to last paragraph wrong?
    Comments and correction to this section are welcome!

    Firstly, note that unlike the experiments described earlier, this paragraph merely expresses his opinion.

    1. he expresses an OPINION
    2 He neglects the stratosphere which is not subject to convection.
    3. the troposphere is largely opaque to IR. Thus convection plays a large role
    in moving heat up from the surface, but since the stratosphere has very little water vapor and is largely transparent to IR it is instrumental in the propagation of heat via radiation to space.

    and we would need to discuss line broadening at some point

    So, willis needs another shell,

  241. ‘Mike Lorrey says:
    November 28, 2010 at 12:33 pm
    Ok for those of you who keep claiming fig 4 is wrong, here is why its not:
    Solar energy to the planet: 240 w/m^2
    Energy reradiated by the atmosphere back to the planet: 240 w/m^2
    Add the two together: 480 w/m^2, which is what the planet radiates back.’

    Sorry, You can’t do that in accounting. That violates accounting principles. You deposit $240 into your checking account. If you transfer $240 into your saving account then transfer back into your checking account. You still only have $240 not $480. It does not make a difference if you do it in one lump sum or a dime at a time. Part or all of the money is going to be in of three places. your checking account, cyberspace or saving account. The only way to reach $480 is to add $240 to the system.

  242. “Of course, the air really is in contact with the surface, and it really has an exponential temperature gradient as altitude rises caused by gravity, at least up to the tropopause, but it sounds like these are not the problems you have in mind.”

    You mean a linear gradient. It would only have an (approximately) exponential gradient if the purely radiative model was correct.

    “So other than this, what is wrong with the good old Planet Dewar model?”

    That it neglects convection and the adiabatic lapse rate?

  243. Chico sajovic said “3. Argon gas is used in insulated windows becuase of its low thermal conductivity and co2 isn’t used because the greenhouse effect isn’t real.”

    True on thermal conductivity. Putting CO2 between glass would be like putting glass between glass. But CO2 is exactly like having shards of glass in the atmosphere, it absorbs heat and spreads it to the rest of the atmosphere (including O2 and N2).

  244. John Marshall said “…light becomes IR which remains inside the greenhouse because the glass traps it inside.”

    Glass absorbs IR and re-emits it. CO2 does exactly that and conducts heat to the O2 and N2 around it.

  245. Steve Mosher’s post @2:50 pm above is paraphrased from Jan Schloerer, who states: “Caveat: This is not my field.”

    Schloerer does a credible job, but at the end his preconceived conclusions show that his mind was already made up:

    “It is an open question how soon the uncertainties can be narrowed down, and whether climatologists will be able to predict details reliably before they start to happen in the real world [IPCC 95] [Morgan]. There is a natural inclination to wait and see until we know what we shall have to face. By then it may be too late.” [my emphasis]

    Schloerer wrote that almost 14 years ago, and it is another good example of why the scientific method must be followed, rather than panicking over scary what-ifs. Fourteen years on, and after spending many billions of dollars on a non-problem, it is clear that the climate is, if anything, even more benign.

  246. “If you transfer $240 into your saving account then transfer back into your checking account. You still only have $240 not $480.”

    You pay the $240 from your employer and another $240 from your savings account into your checking account. At the same time, you transfer $480 from your checking account to your savings account, and $240 from your savings account to the grocery store.

  247. Hu McCulloch says:
    November 28, 2010 at 1:34 pm

    Willis, your model is what I like to call the “Planet Dewar” model of the GHG effect. Of course, the air really is in contact with the surface, and it really has an exponential temperature gradient as altitude rises caused by gravity, at least up to the tropopause, but it sounds like these are not the problems you have in mind.

    One paradox that arises from this model is when it is extended to what I call the “Planet Dante” model, with more multiple crystalline heavens. The same reasoning that leads 2E to reach the surface when only E is incoming with one heaven leads 3E to the the sruface with 2 heavens, etc. With multiple heavens, the surface can be made arbitrarily hot by this reasoning, at least until the surface begins to incandesce and emit visible light and short-wave IR which can escape through the assumedly transparent shells.

    So other than this, what is wrong with the good old Planet Dewar model?

    Thanks, Hu, good to hear from you. In theory there is no limit to how much you can concentrate energy (slow down its loss) in this manner. I don’t see why people think there should be a limit. Take an electric heating element. Put it in a Dewar flask. The heating element reaches a higher temperature. Put the Dewar in a larger Dewar. Higher temperature. Repeat ad lib., you can achieve a very, very high temperature … why is that seen as being so impossible?

  248. old construction worker says:
    November 28, 2010 at 3:06 pm

    ‘Mike Lorrey says:
    November 28, 2010 at 12:33 pm

    Ok for those of you who keep claiming fig 4 is wrong, here is why its not:
    Solar energy to the planet: 240 w/m^2
    Energy reradiated by the atmosphere back to the planet: 240 w/m^2
    Add the two together: 480 w/m^2, which is what the planet radiates back.’

    Sorry, You can’t do that in accounting. That violates accounting principles. You deposit $240 into your checking account. If you transfer $240 into your saving account then transfer back into your checking account. You still only have $240 not $480. It does not make a difference if you do it in one lump sum or a dime at a time. Part or all of the money is going to be in of three places. your checking account, cyberspace or saving account. The only way to reach $480 is to add $240 to the system.

    If I understand you, you seem to be confusing total (net) flows with individual flows. The net flow from the planet to the shell in Fig. 4 is 240 W/m2. The individual flow from the planet to the shell is 480 W/m2. Fig. 4 shows only individual flows, and not net flows. It is the individual flow from the planet to the shell that determines the blackbody temperature.

    PS – among my other skills, I am an accountant. I assure you that no accounting principles were violated in any of the Figures.

  249. old construction worker;
    Sorry, You can’t do that in accounting. That violates accounting principles. You deposit $240 into your checking account. If you transfer $240 into your saving account then transfer back into your checking account. You still only have $240 not $480.>>

    Physics isn’t accounting. But if it were, it would go more like this.

    I send You a check for $240.
    You send Me a check for $240.
    Every day for many days.

    Insert Willis.

    I send you a check for $240.
    You send Willis a cheque for $240.
    Willis sends Me a cheque for $120 and you a cheque for $120.
    You now have an extra $120, so the next day…

    I send you a cheque for $240.
    You send Willis a cheque for $360.
    Willis sends me a cheque for $180 and you a checque for $180.
    First day you wound up with $120, extra and second day $180, so on 3rd day…

    I send you a cheque for $240.
    You send Willis a cheque for $420.
    Willis sends me a cheque for $210 and you a cheque for $210

    And so on and so forth until every day (at equilibrium) you are sending Willis a cheque for $480 every day, and he is sending you and me a cheque for $240 each every day. My cheque to you still goes straight to you for $240. The books still balance. The only difference is that Willis started out with $0 and now on any given day has $480.

    If we were back to physics, that would be the “shell” and it would be retaining how ever much energy is required to make it radiate 240 w/m2 in each direction.

  250. Hi, Willis. You seem to have some German gene inherited. The one that makes Germans say,”why make things simple if we can make it complicated”, a gene that those who worked in electronics are quite aware of. :-)

    Your explanation of why the Climate Model you devised doesn’t work is OK, technically. But I see it as a close look to the tree. Walking away from the tree and having a better view of the forest, I would say that given the fact that there is a tremendous amount of variables working in the climate system, and that feedbacks are positive as well as negative, and because there is no way to calculate how interaction of ALL those feedbacks will work our, the Climate System is CHAOTIC.

    So the simplest explantion would be: “The model is useless because we cannot model chaos”.

    Still in Salomon islands? Please scubba down and bring me one bolt from a Japanese Zero sunk off the coast… :-)

  251. Willis Eschenbach says:
    November 28, 2010 at 3:47 pm
    ….
    Thanks, Hu, good to hear from you. In theory there is no limit to how much you can concentrate energy (slow down its loss) in this manner. I don’t see why people think there should be a limit. Take an electric heating element. Put it in a Dewar flask. The heating element reaches a higher temperature. Put the Dewar in a larger Dewar. Higher temperature. Repeat ad lib., you can achieve a very, very high temperature … why is that seen as being so impossible?

    So are you saying that in my Planet Dante example, the surface really could be heated indefinitely (until incandescence started leatting the heat through)?

    On another matter, did you get my e-mail of 10/23 regarding our Tanganyika piece? I first sent it 10/19, but it didn’t seem to go through so I sent it again, both times using your taunovobay address. Please e-mail me either way.

  252. Willis Eschenbach says:
    November 28, 2010 at 3:35 am
    Hi Wll, thanks for the reference to Zeno; however I think your reference to the real world confounding the application of a a limiting sum geometric series to Figure 4 so that the final amount of flux is never received misses the point that this effect in the real world prevents a runnaway situation from occuring; to that extent your comment that “In theory there is no limit to how much you can concentrate energy (slow down its loss) in this manner.” I think is wrong and it is the mistake the IPCC makes with its climate sensitivity formula. There is no evidence for this. The IPCC attributes ACO2 as being the forcing agent, F, for this scenario, with water vapor the feedback, f, and temperature, t, the parameter for the change; the interaction of these variables is measured by the state vector, S, which would itself change if F has the effect the IPCC alleges. IPCC represents this dynamic thus:

    dS/dt=S/f+F

    IPCC assumes that f is +ve so if we intergrate by dividing both sides by fS+F, and multipling both sides by f*dt we get:

    (S2+F/f)/(S1+F/f)=exp(f*(t2-t1))

    The problem with this is because it predicts that as the final value of t, t2, approaches infinity, the value of S2 becomes infinite. This is wrong because if there is a climate forcing in operation, at infinite time, the temperature anomaly should approach its finite equilibrium value even if there is positive feedback. This is shown by Venus which is paraded by AGW supporters as being the inevitable result of AGW; but, if there was any greenhouse effect on Venus it has now stopped despite high levels of CO2 and obviously its equilibrium was less than infinity. The correct formula for measuring feedback is done by Spencer and Braswell:

    http://www.drroyspencer.com/Spencer-and-Braswell-08.pdf

    Their equation 2 is:

    Cp*T/*t=-^T+N+f+S

    The difference with S&B’s equation is that it introduces a term for the stochastic properties of clouds, N and breaks F into -^T and f; f is ACO2 and -^T is a total feedback term which must be negative so that an infinite equilibrium is impossible.

    On another point with your figure 4 it seems to me that not just Zeno’s arrow would apply but Zeroth’s law; if you assume that between the shell and the surface there will be an equilibrium then an isothermal state must be reached; that is if the flux is equivalent everywhere then temperature must be equal everywhere. Of course even in a hypothetical situation like yours that can’t be because gravity would produce a gradient with pressure higher at the bottom; even allowing for a vacuum that would impact on temperature so that a disequilibrium between the surface and the shell temperature would occur?

  253. Smokey,

    The woods experiment is a farce. Now to be sure until the 1950s we did not understand that the stratosphere was dry. It took the airforce to figure that out. The results of that research and the effects of C02 at that altitude are now gratefully better understood. As I have explained to you before, if we did not understand how C02 interacts with IR throughout the air column (not in an enclosure) we would not have been able to build satellites that actually work in sensing the earth and clouds.
    Stealth aircraft would not be stealthy. IR missiles would not work as they do.

    If you want to argue about feedbacks, that’s fine. That question is open for debate. If you want to argue that known working physics doesnt work, well then there is no debate. TWO entirely different questions. I would suggest that you will get more respect from people if you read what Anthony, Willis, Lindzen, Spencer, Christy, Monkton, say about C02 and radiation. None of them deny the basic physics. I would guess because they either have worked with the physics or understand it. Once you do, then you’ll stand on BETTER skeptical ground.

  254. On reading the comments in more detail, I see that I was overlooking your proposal of a multi (2) shell Dante model, on the 27th at 750PM.

    But this doesn’t make the 1-shell model wrong as an insight into how the GHG effect works — as you say, it just isn’t strong enough, given leakages.

  255. I like the example of a heating element inside a Dewar. But when it comes to the atmosphere the heating element (the sun) is outside. What does that do to the results?

    I have greatly enjoyed this thread which tries to answer a question. I would like to see a similar thread that seeks to answer the question as to why the warmest temperatures on earth occur where there are the least amount of green house gases. That is arid areas. I do not have the answer, but I would enjoy the attempts to answer that question.

  256. Willis Eschenbach says:
    November 27, 2010 at 7:50 pm

    “Why The Thought Experiment Can’t Represent Earth

    It is not powerful enough.

    What I have illustrated above is the energy balance in a theoretically perfect single shell planetary “greenhouse”. It shows that the maximum amount that such a system can produce is a doubling of the input. Not one Watt per square metre more. It is not a magical system. At best, it can double the input.

    Since the real Earth receives about 240 W/m2 (after albedo reflections), the most the system could theoretically produce is 480 W/m2.

    At first glance, that seems like plenty. I mean, blackbody temperature for 480 W/m2 is 30°C, and the Earth only averages around 15°C. But in the real world there are losses. …”

    There was a reply that you seem to have skipped over, which was on the money. The shells you are talking about don’t model the lapse rate.

    Will Nelson says:
    November 27, 2010 at 12:33 pm

    “Adiabatic lapse rate?”

  257. Hi Steven Mosher,

    If you re-read my post you’ll see that I wasn’t commenting on Wood’s experiment at all, nor about feedbacks, although your response indicates tht’s what you believed. I simply pointed out where your own comment came from, and the fact that when preconceived assumptions were made by Schloerer, they were typically alarming – but wrong. None of what he feared and expected happened. In fact, just the opposite.

    Maybe you were inadvertently responding to Bryan?

    And your comment…

    “Why is his second to last paragraph wrong?
    Comments and correction to this section are welcome!”

    …was apparently lifted directly from Jan Schloerer’s very similar statement here.

    Just sayin’, that’s all.☺

    Carry on.

  258. Steven Mosher;
    If you want to argue about feedbacks, that’s fine. That question is open for debate. If you want to argue that known working physics doesnt work, well then there is no debate.>>

    Oh yes there is. There is a huge debate. Massive in fact. Because the known physics is not what is being argued by the alarmists. The IPCC reports gloss over the fact that CO2 is logarithmic. Known physics. They gloss over the exponential negative feedback response (T to the power of 4) of the planet surface. Known physics. They extrapolate temperature increases from economic scenarios that would require 10 to 100 times current consumption to achieve. Known physics. They predict +1 degree from CO2 doubling, but fail to mention that it is the effective black body temperature they are calculating against, not the much lower average surface temperature. Known physics. They highlight large temperature increases at the pole as if they are a harbinger of things to come planet wide while they know full well that increases at the equator will miniscule. Known physics. They present the temperature increases at the high latitudes as of they are uniform over the course of the year when they know full well that the bulk of the rise is in the depths of winter and the summers are very little warmer. Known physics. They talk incessantly about daily averages, glossing over that the day time highs change little but the night time lows increase. Known physics. They predict higher total energy in the system resulting in more severe weather though the result of most of the warming happening in the coldest places at the coldest times results in more uniform temperatures globaly, and it is temperature differentials that cause extreme weather, meaning less extreme weather in a warmer world, not more. Known physics.

    They misrepresent the physics repeatedly, and when one attempts to engage them in a proper discussion of the physics, what happens? Well look at this temperature record. Well look at these tree rings. Well look at these tidal gauges. Well look at this ice extent. Anything, absolutely ANYTHING to get the topic off the physics and turn the discussion to secondary or even more removed proxy data that they then claim proves the physics. But discuss the physics? Not a chance. They haven’t the kahonies to have a real discussion of the physics. If they did, they would have to admit:

    o CO2 is logarithmic and the amount of oil we would need to burn to get to the catastrophic temperature increases they predict is nearly astronomical.
    o Using effective black body exagerates the actual surface temperature change
    o Averages are meaningless because most of the temperature increases that make up the average will be in the coldest climates, in the depths of winter, and at night time lows.
    o A warmer world is more uniform temperature wise, and hence less severe weather.

    No debate about the known physics? Are you kidding me? The arguments of the alarmists rest on everything EXCEPT the “known physics”. They want to debate temperature records and species extinction and temperature reconstructions from 7 trees in Siberia to prove the physics. Except it isn’t the “known physics” that they are trying to prove, it is “their physics”, a “special physics”, physics proven by proxy, mythology, and photoshop. The “known physics” says, point blank, NO PROBLEM, so they have set about replacing it with tree rings, polar bears on ice floes, and tidal gauges on sinking islands.

    The known physics you speak of and the known physics of the alarmists are two different things, which is why there IS in fact a debate. If you were being honest Mr Mosher, you would excoriate the alarmists for glossing over the actual physics and when being critical of statements like Smokey’s you would include the physics that mitigates and limits the effects of CO2 along with your argument that the warming effects of CO2 are known and not debatable.

    One variable in isolation does not a physics equation make.

  259. Miswording above:
    They predict +1 degree from CO2 doubling, but fail to mention that it is the effective black body temperature they are calculating against, not the much lower average surface temperature.

    Should have read “…not the much higher average surface temperature (meaning much lower change at surface)”

  260. Sorry smokey, I thought I was responding to him.

    I think william connelly is the one who actual found the document in the stacks. Could be wrong about that. We discussed it a while back on lucia’s site.

  261. Willis,

    the real problem with these models is that they use the wrong form of the energy equations. When you have two bodies radiating against each other you have to use the form that takes into account the temperature of the body being radiated against. In the case of GHG’s, since their radiation is based on molecular bond instead of temperature, how would you do that??

    Using the form of the RTE which computes energy radiated to 0K simply does not do it and is one of several reasons why climate science is so unphysical. Simply put, it ignores the fact that photons also have wave properties and waves can interact. These interactions can cancel and interfere with wave propagation so that your 120 down does NOT all get to the ground and your 240 up does not all get up!!! What happens to the energy?? Maybe you can find some physicist to tell us!!

  262. Willis Eschenbach says on November 28, 2010 at 3:47 pm – In theory there is no limit to how much you can concentrate energy (slow down its loss) in this manner.

    Well, there is, as I have shown by arguing in extremis. There has to be some property of the shell (or the walls of the Dewar) that allow mechanisms to operate in a realistic and probabilistic time. For example, your shell will take forever to warm up on the inside if it is hugely massive and it will not warm up much at all if it is so thin that energy transfer is of low probability. Your shell has to have properties that allows it to act as a shell and I suggest that that one property is appropriate mass.

    To the extent that mass is relatable to density in a real life situation, I think that this is part of the reason for the decrease of temperature with altitude to the tropopause. We are argiung about how much importance that “part” has.

  263. Steven Mosher says:

    ……….”The woods experiment is a farce.”………

    R W Woods has been often described as the best experimental Physicist that America ever produced.
    Yet you describe his experiment and comment as a farce!

    His experiment still stands.
    It has been verified for much larger structures such as the polyethylene polytunnel.

    The IR radiative effects of CO2 and H2O exist but in the context of the troposphere they play an insignificant role.
    You can call this insignificant role the “greenhouse effect” if you like.

    Leonard Weinstein with William C Gilbert set out a much more coherent explanation of heat transfer in the troposphere on SoDs site.
    It involved convection modifying the gravitationally established lapse rate.
    and in my opinion this explanation is correct.

    The radiative effect becomes more important above the tropopause where it radiates away the much longer end of the IR spectrum.

    My point about raising the question of Woods experiment was to underline the fact that the radiative effect is so small, in contrast to the convective heat transfer, that it is almost unnoticeable.

  264. The Climate System is CHAOTIC.

    Another experiment,

    I’ll hang up the sun for 1000 years.

    And in the distant future (3010).

    The sun will be turned on again.

    If AGW theory is correct. November 4010 will be the hottest year in history.

  265. ‘davidmhoffer says:
    November 28, 2010 at 4:18 pm
    Insert Willis.
    I send you a check for $240. You balance is $0. My balance is $240
    You send Willis a cheque for $240. Your balance is $0, My Balance is $0, Willis is $240
    Willis sends Me a cheque for $120 and you a cheque for $120.’ Willis balance is $0 Your and my balance is $120 each , So where is the extra $120
    You now have an extra $120, so the next day…

  266. This is a topic that is very frustrating to me. Clearly the atmosphere causes the temperature to be warmer. I was baffled when I initially found people objecting to the GHE as a whole.

    I do disagree with the diagrams posted though. I have re-done the Energy Balance by Trenberth (08) and by focusing on the NET transfers I get a solid GHE with the surface transferring 120 W/m2 to the troposphere. That is enough to warm the first km of the troposphere up 9.3C.

    The energy balance and the GHE need to have discussion picked up a notch. :-)

    John Kehr
    The Inconvenient Skeptic

  267. Vince Causey says:
    “But transfering heat energy from “cold” to “warm” violates the second law of thermodynamics.
    There is no way around that.”

    “There is a lot of misunderstanding surrounding the second law of thermodynamics.”

    This is right ;-)

    “The original formulation of this law says approximately what you attribute to it. This was applied to heat transfers in conducting bodies. The modern formulation is in entropy and states that in a closed system, the total entropy must always increase. The application to heat transfer has been changed to refer to net heat transfers. ”

    There is no such thing like “net-heat”. How should a flow of energy know that an other one exists and it has to accept to get netted??? The rule is simple: there is NO transfer of heat from cold to warm. Full stop.

    “The implications to conduction of heat have not changed. However, we know that radiative transfers do indeed take place in both directions.”

    You know that? WOW !!!

    “In order for the second law to be obeyed, the net transfer must always be from warmer to cooler. This should be intuitive – even a cool body emits photons of energy which must travel at the speed of light regardless of the fact that they may impinge on surfaces that are warmer. The net transfer argument should make it plain that more photons will be emitted from the warmer surface. ”

    Wron question! Right qestion is: can the photons, emitted by the colder body transfer their energy to the warmer body? Answer: NO

    “A modern version of Maxwells demon thought experiment is to imagine a demon sitting between the cool and warm surfaces. The demon is supposed to allow the photons from the cooler surface to reach the warmer surface, but prevent the flow in the opposite direction. The action of the demon would cause the cool surface to become cooler and the warm surface to become warmer. This is clearly in violation of the second law. The Maxwell demon paradox was solved when it was realised that the demon needs information to sort the photons and this prevents the system from being closed.”

    The second law applies to open AND closed systems

    “In practice, although photons from the cool surface carry energy to the warm surface, a higher flow occurs in the opposite direction. The result is that the warm surface cools and the cool surface warms. Ie, they equilibriate and their entropy increases. Thus, back radiation does not violate the second law.”

    This is voodoo-physics

  268. pwl says:
    November 28, 2010 at 7:40 pm

    “How can the troposphere, the imaginary shell(s), at -65 c warm the surface at +15 c?”

    It cannot. But it can slow down how fast the surface can give up heat. The energy transfer is moving from surface to the black of space which sits at 3 degrees above absolute zero. Throw anything in the way which isn’t completely transparent to the infrared frequencies involved and the transfer rate slows.

    The problem for the warmists is that things can also be thrown in that speed up the transfer where that “thing” is evaporation and convection which skips around the radiative transfer and mechanically transports a massive amount of energy as latent heat of vaporization in the H2O molecules and where that latent heat is released high in the troposphere or in some stronger convective cells even in the stratosphere.

    After this surface energy is mechanically transported thousands of meters above the ground and released the underlying greenhouse gases now make it more difficult for that energy to radiate back towards the surface and thus make the radiative path upwards to space even easier than it would be absent any greenhouse gases.

    Heat transfer from surface to space occurs by convection, conduction, evaporation, condensation, and radiation. All must be considered simultaneously because all three happen together. Isolating only the radiative transfer only works in a vacuum and doesn’t tell much of the story on a planet with a thick atmosphere and a surface covered with liquid water. In the end, as long the water cycle is running (temperature above freezing and below boiling), the water cycle is the dominant player in how the earth warms during the day and cools during the night. If the earth is covered by glaciers and sea ice then and only then do non-condensing greenhouse gases like CO2 play a dominant role.

    The only important thing CO2 does is raise the average surface temperature above freezing then the water cycle takes over and limits how warm it can get. If more non-condensing greenhouse gases are added the extra energy in the system simply speeds up the water cycle and negates (evidently) about 75% of the potential surface temperature increase of the non-condensing greenhouse gases. It does this, as described above, by mechanically transporting heat away from the surface in the form of latent heat of vaporization in water molecules.

  269. davidmhoffer says:
    November 28, 2010 at 7:59 pm (Edit)
    Steven Mosher;
    If you want to argue about feedbacks, that’s fine. That question is open for debate. If you want to argue that known working physics doesnt work, well then there is no debate.>>

    Oh yes there is. There is a huge debate. Massive in fact
    1
    2
    3
    4
    5
    6

    Well said.

  270. old construction worker,

    the problem with trying to apply accounting analogies is that they depict an instantaneous position whereas the situation is one of flows. I prefer the following analogy:

    Imagine a line of people filing into a building through the in-door, at the rate of 240 people per hour, and filing out the other end through the out-door. This represents the energy flow arriving and leaving the earth. Now suppose that half of the outgoing people pair up with those on the incoming line. You now have 360 people per hour entering the building. These extra people haven’t materialised out of thin air – they came in the original line.

  271. bessokeks says:
    November 29, 2010 at 6:24 am

    “There is no such thing like “net-heat”. How should a flow of energy know that an other one exists and it has to accept to get netted??? The rule is simple: there is NO transfer of heat from cold to warm. Full stop.”

    Not by conduction at any rate. Radiative transfer is an exchange. That’s why we can see a satellite from the ground and the satellite can see us looking at it. Radiation from the satellite is reaching us and radiation from us is reaching the satellite. The satellite might be much colder or hotter than the ground observer yet each can still observe the radiative energy from the other.

    Whether or not there is any net transfer of energy it will be from the more energetic source to the lesser. If they are both exactly the same temperature the exchange is exactly even but in no case does the radiation from one not fall upon the other.

  272. bessokeks says:

    “There is no such thing like “net-heat”. How should a flow of energy know that an other one exists and it has to accept to get netted???”

    It does not need to know. The term net energy flow means that more energy will flow from the warmer body to the cooler body than the other way round. This is a simple consequence of the warmer body having a higher radiative flux as per Stefan Boltzman.

    “You know that? WOW !!!”

    Yes.

    “Wron question! Right qestion is: can the photons, emitted by the colder body transfer their energy to the warmer body? Answer: NO”

    Let us consider the question in stages. Can a cooler body emit a photon? Clearly yes. Can that photon impact a cooler body? Obviously yes. Can it also impact a warmer body? Yes. Does a photon add energy on impact? Yes – a photon absorbed by an atom will raise an electron to a higher energy state. Therefore, ergo demonstratum, the body has gained energy.

    “The second law applies to open AND closed systems.”

    Yes, but it makes no difference to the argument.

    “This is voodoo-physics.”

    Sorry to disappoint. The modern form of the second law is about entropy, that must increase. Any closed system in which entropy decreased would be in violation of the second law. As long as the cooler body gets warmer and the warmer body gets cooler as they approach the same temperature, then entropy has increased to a maximum.

  273. @Vince Causey says:
    “Does a photon add energy on impact? Yes – a photon absorbed by an atom will raise an electron to a higher energy state. Therefore, ergo demonstratum, the body has gained energy.”

    This is an often found mistake: raising an elektron has nothing, absolutely nothing to do with transfer of heat. When, in our case, heat is transferred, the molecules will raise the frequenzy of the intermolecular binding. The energy to raise electrons is much higher than what we get from our infrared radiation.
    Please also be aware, that the second law ONLY applies on heat, not on energy in general. The frequencies which make you “see” the satellite are far from the infrared spectrum emitted by a warm body

  274. @Vince Causey says: (2)
    “The term net energy flow means that more energy will flow from the warmer body to the cooler body than the other way round. This is a simple consequence of the warmer body having a higher radiative flux as per Stefan Boltzman”

    Imagine a small, warm body allmost completely surrounded by a big, rel. colder body.
    The amount of energy radiated from the colder body to the warmer body will be much bigger than the amount of energy that is radiated by the warmer body to the colder one. If the energy of the colder body would be absorbed by the wormer one, it would heat up. But it does not…

  275. @Dave Springer says:
    ““How can the troposphere, the imaginary shell(s), at -65 c warm the surface at +15 c?”

    It cannot. But it can slow down how fast the surface can give up heat.”

    NO! Nothing can influence the surface to give up heat. It only a function of the temperature of that surface and its factor of emissivity

  276. Well Willis, your model does what it claims to do; demonstrate that a “Greenhouse effect” as we understand that term (which is not the way greenhouses work) can raise the temperature of a glass enclosed planet.

    Too bad your model is of an infinitely conductive thermally earth that does not rotate.

    Your glasss earth does not explain how climate works; but it does show that a greenhouse effect is possible.

  277. bessokeks, Vince Causey,Dave Springer

    There seems to be a great deal of confusion when the word “heat” is used.

    There is a huge difference between the vernacular use of “heat” and the thermodynamic use of “heat”

    Vernacular use tries to convey the idea of “hotness” and “temperature” and even the sensation of taste of a curry.
    Thermodynamic use of “heat” is very precise and it deals with the process by which thermal energy is transferred from a higher temperature object to a lower temperature object.

    Heat always travels from hot to cold never the reverse.
    Thermal energy can be radiated and absorbed by both hot and cold objects.

    Heat (thermodynamic use) has the capacity to do WORK which again had a precise meaning.
    If you are in any doubt about whether “heat” is involved use the “WORK” test it always sorts out the confusion.
    Have a look in a thermodynamics text book particularly the chapters dealing with the Carnot cycle and the second law, there any loose use of the word “heat” would render the chapters uninteligable.

  278. Vince Causey and bessokeks:

    The waters surrounding the issue of a second law violation by the “back radiation” are muddied by a widespread misuse of the word “heat” by climatologists. The radiative energy flux that is incident upon a surface at a space point is not a heat flux; it is properly referenced as a “vector irradiance.” The radiative energy flux that is reflected, emitted or transmitted through the surface at the same space point is not a heat flux; it is properly referenced as a “vector radiosity.” Climatologists err in referencing the vector irradiance and vector radiosity as “heat fluxes.” As neither is a heat flux, neither the vector irradiance nor the vector irradiance is bound by the second law.

    By its definition in thermodynamics, the “heat” is the energy that crosses a boundary. It is the heat flux that is bound by the second law. A component of the heat flux may be the radiative heat flux. At a space point, the radiative heat flux is the vector difference of the vector radiosity and vector irradiance.

    The entity which, in a Kiehl-Trenberth diagram, is labelled as the “back radiation” should be relabelled as a “vector irradiance” with specified spectral characteristics. Relabelling it would help to eliminate the appearance of a second law violation.

  279. The Maxwell demon paradox was solved when it was realised that the demon needs information to sort the photons and this prevents the system from being closed.

    Then what prevents information from playing a role in all energy transfers? Why would information only become important in this transfer? And what is the decision making process of the energy when given this information?

  280. bessokeks says:
    November 29, 2010 at 10:06 am

    @Dave Springer says:
    ““How can the troposphere, the imaginary shell(s), at -65 c warm the surface at +15 c?”

    It cannot. But it can slow down how fast the surface can give up heat.”

    NO! Nothing can influence the surface to give up heat. It only a function of the temperature of that surface and its factor of emissivity

    By the same token, nothing can influence how well a surfaced absorbs photons except the emissivity of the surface. So if there are lots of incoming photons, the surface will absorb more energy; if there are few incoming photons, the surface will absorb less energy. This extra absorbed energy will slow the cooling of surface.

    Consider a 303 K rock in outer space (and assume the rock is a perfect black body & it is far from any stars). The outgoing IR radiation is 480 W/m^2. The incoming radiation from the 2.7 K cosmic background is going to be around 0.000005 W/m^2 = 0 W/m^2 for all practical purposes. The net loss is 480 W/m^2

    Put the same rock (in a vacuum container to avoid conduction & convection) in a freezer at 255 K. The rock still radiates 480 W^m2. But the container radiates 240 W/m^2 back to the rock (again assuming the container is a blackbody). The net loss is 240 W/m^2. The rock will cool half as fast as before (at least initially).

    Put the same rock (in a vacuum container again) in a 303 K room. The rock still radiates 480 W/m^2. But the room radiates 480 W/m^2 back to the rock. The net loss is 0 W/m^2. The rock will stay 303 K.

    However you want to say it, the net result is that the cooling rate of the rock is slower based on the surrounding environment.

  281. bessokeks says: November 29, 2010 at 10:01 am

    Imagine a small, warm body allmost completely surrounded by a big, rel. colder body.
    The amount of energy radiated from the colder body to the warmer body will be much bigger than the amount of energy that is radiated by the warmer body to the colder one. If the energy of the colder body would be absorbed by the wormer one, it would heat up. But it does not…

    Sorry, but you are wrong again. The real consideration is the solid angle subtended by the radiating source. If the cooler container completely surrounds the hot object, then the rate that energy arrives at the hot object is the same, whether the container is a little larger or much larger.

  282. Willis’ model is ridiculous because planets rotate, allowing a heated side to radiate to space at night. His model would only apply to a body that had zero rotation and had a surface that constantly faced the sun.

  283. The rule is simple: there is NO transfer of heat from cold to warm. Full stop.”

    Bassokeks must have a very inefficient air conditioning system if it stops working as soon as the temperature inside the house is less than that outside.

    “Wron question! Right qestion is: can the photons, emitted by the colder body transfer their energy to the warmer body? Answer: NO”

    Let us consider the question in stages. Can a cooler body emit a photon? Clearly yes. Can that photon impact a cooler body? Obviously yes. Can it also impact a warmer body? Yes. Does a photon add energy on impact? Yes – a photon absorbed by an atom will raise an electron to a higher energy state. Therefore, ergo demonstratum, the body has gained energy.

    The second part of this quote (from Vince Causey) is correct except that an photon in the infra-red region excites the molecular vibrations, not the electronic levels, which means the nuclei move with increased kinetic energy, which means ‘hotter’ at the molecular level. And work has to be done to achieve this, so it is heat by the classical thermodynamic definition as well.

    so Bessokeks reply to this

    This is an often found mistake: raising an elektron has nothing, absolutely nothing to do with transfer of heat. When, in our case, heat is transferred, the molecules will raise the frequenzy of the intermolecular binding. The energy to raise electrons is much higher than what we get from our infrared radiation.
    Please also be aware, that the second law ONLY applies on heat, not on energy in general. The frequencies which make you “see” the satellite are far from the infrared spectrum emitted by a warm body

    is in fact, nonsense. There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction. By the way, this can apply to conduction as well – if you think carefully about what happens when you allow two gases at different temperatures to mix, you will realize that some of the flow of both molecules and energy has to be from what was ‘colder’ towards what was ‘hotter’. Again it is just the net flow that is hot to cold.

  284. Vince Causey says:
    November 29, 2010 at 9:12 am
    old construction worker,

    the problem with trying to apply accounting analogies is that they depict an instantaneous position whereas the situation is one of flows. I prefer the following analogy:

    Imagine a line of people filing into a building through the in-door, at the rate of 240 people per hour, and filing out the other end through the out-door. This represents the energy flow arriving and leaving the earth. Now suppose that half of the outgoing people pair up with those on the incoming line. You now have 360 people per hour entering the building. These extra people haven’t materialised out of thin air – they came in the original line.

    I agree with you 100%. But if you look it the “flow charts”, they do not indicate a “content flow”. They indicate one unit of 240 W/m2 going from point a to point b to point c to d (Fig 4#). To correct the misinterpretation, not Willis’s explanation, I would add “content flow rate of ” above 240 W/m2 where it enters the system.

  285. “”””” bessokeks says:
    November 29, 2010 at 6:24 am
    Vince Causey says:
    “But transfering heat energy from “cold” to “warm” violates the second law of thermodynamics.
    There is no way around that.”

    “There is a lot of misunderstanding surrounding the second law of thermodynamics.”

    This is right ;-) “””””

    Well I like the Clausius form of the Second Law.

    No cyclic machine can have no other effect, than to transport heat from a source at one Temperature to a sink at a higher Temperature.

    Note the restriction to “cyclic” machines.

    Machines of any kind involve MATTER. Without matter there is NO HEAT; it is simply the mechancial kinetic energy of matter.

    Electromagnetic radiation exists even in the total absence of matter. Whether you consider it a photon stream or an electromagnetic field, it can and does transport energy from anything to anything.

    We get too wrapped up in the quantum nature of EM radiation; so it raises questions like; if CO2 molecules are only one in 2500 of the total atmospheric molecules; how do they ever get hit by a photon ?

    But if you think of thermal radiation as being an electromagnetic field; then it can propagate throughout space, and encounter any CO2 molecule in its path.

    It helps if you understand some radio-physics; specifically antenna theory. Antennas can radiate EM energy because Voltages, and currents in their elements generate fields, and those fields cannot appear instantaneously. An ordinary dipole antenna consisting of two wires pointing in opposite directions and driven at their close ends by an AC Voltage can have a current flow along those wires, because the distant ends are still connected together by a small capacitance, which is capable of carrying an AC current; and because of the flow of that AC current, there also is a magnetic field set up around those wires. The direction of the flux lines of the magnetic field are always perpendicular to the lines of the electric that is due to the varying Voltage along the wires. Because the resultant EMf ield takes time to set up and start to fill all of space, it is possible for the direction of the current and or Voltage to reverse, before the filed gets enough time to collapse; there’s that propagation delay thing again. as a result of those propagating EM fields, free space; behaves somewhat like a transmission line with a characteristic impedance that is known exactly; in fact it is 120 Pi Ohms or 377 Ohms. If you know how to make 377 Ohm paint then you can make yourself invisible by just dunking yourself in it. Any EM wave that strikes you will be totally absorbed with NO reflection, and it will apear to a viewer that the radiation just went right on through you to the end of the universe.

    Well don’t get too self confident; because if by definition you do absorb all radiation that hits you (you’re a BB) then you likely will undergo a Temperature rise; and ultimately I guess you might have to emit thermal radiation; but don’t quote me on that.

    I like to think of it as something like a kid’s wire loop that she sticks in the soap solution and blows bubbles. The air pressure bellows out the stable flat water film , and surface tension tries to pull it into a sphere attached to the wire loop. Somehow the bubble necks down near the loop and pinches off the bubble so it is free to propagate to the edges of the universe..

    The antenna EM waves do the same thing; you get a close in (near field) behavior within about a wavelength or so of the antenna, and then there develops disconnected closed loop fields of mutually perpendicular magnetic, and electric fields that are no longer constrained to the antenna, and they propagate at a velocity c (in free space) and in a direction that is also perpendicular to both the electric and magnetic fields.

    Well atoms or molecules that emit “resonance” types of EM radiation are just bloody small antennas; well they are if they have a dipole moment.

    Some have argued here that a molecule requires three or more atoms to emit Infrared radiation or absorb it; diatomic molecules do not absorb or emit in the infrared.

    Well your should try telling that to an HF molecule which has almost as big a dipole moment as does H2O; and it most certain can and does absorb and emit infrared radiation.

    But it is generally true of things like N2 and O2 that they are NOT infrared active, as far as resonance radiation goes; because they are symmetrical.

    But it simply is NOT TRUE, that these gases or gases in general DO NOT emit THERMAL RADIATION.

    The distinction is a fourth grade science question for “Are You Smarter than a Fifth Grader”.

    Resonance emission or absorption is a line spectrum that is characteristic of the specific material (molecule species), and is essentially independent of any Temperature. Well experiment shows that the width of these resonance radiation lines is somewhat Temperature dependent, and also pressure dependent ; due to the Doppler effect (Temperature) and the molecular collision frequency (Temperature and pressure related). But the basic frequencies stay the same.

    Thermal radiation as its name implies is a consequence of the Temperature of the emitting material and is essentially independent of the nature of the material; and depends ONLY on the Temperature.

    Its immediate Physical cause is the simple fact that the MOLECULE contains electric charges; and those charges undergo motion and acceleration as a result of the translational motion of the molecule itself which is the direct result of the Temperature of the material. Classical Physics requires that accelerated electric charges constantly radiate EM radiation and energy.

    As evidence that atomic or molecular structure is unrelated to the thermal radiation; we have the obvious case of a free electron or proton; which don’t exactly have a whole lot of orbitals associated with them; but which are well known to lose energy by radiation in the process of being accelerated in particle beam accelerators. Particularly in circular orbital accelerators, electrons eventually radiate more energy during one revolution than the machine can replace next time the packet returns to the kick site.

    That is the whole reason for the Stanford, 2- mile Linear Electron accelerator; and also for the whacking great size of the LHC machine that covers most of Europe (underground). Going in circles is constant acceleration (rate of change of momentum) so constant energy loss.

    Charged particles at near light speed crashing into materials at greater than the velocity of light in the material undergo rapid deceleration, and during that time they emit radiation (usually a blue light); well the Cerenkov radiation is usually of that nature. I’m not exactly sure of the distinction between Cerenkov radiation, and Bremsstrahlung (which I gather is German for “Braking Radiation.” ) And I don’t do umlauts so if you need one grab one.

    Bottom line is that whereas N2 and O2 may not absorb or emit infrared characteristic line spectra, they most certainly DO emit thermal radiation in a BB like spectrum dependent only on the Temperature.

    My Quantum mechanics timed out at about the Master’s level, and I haven’t much kept it up; so I don’t have a good grasp of the quantum mechanics of thermal radiation; but I have seen suggestions that it is during the collision times between molecules, when all those electrons get in each other’s hair, that the quantum selection rules like the Pauli principle and the like, and assist in the thermal emission; but I am out on thin ice on that; so you need to ask the PhDs like maybe Phil; if you really want to delve into that. I’m happy to accept that the Planck derivation of BB radiation didn’t really need any deviation from the Classical physics (statistical mechanics ) of the Wien and Raleigh Jeans derivations; other than to insist that the energy levels couldn’t be closer together than kT/2; whereas the classical partition rules had no minimum energy difference.

    And if I got this a bit too screwed up; well you should be able to go and look someplace for the fair dinkum scoop on it; but don’t go to wiki; because they are more screwed up than I am.

  286. I have one more quation Willis,

    Why the few km height of the glass shell; why not a few cm or a few atto metres for that matter ?

    We know that spectrally selective absorbers do work quite well in collecting solar energy; by having high solar spectrum absoptance; but much lower thermal sepctrum emissivity. Well they work well enough to take the chill off your swimming pool water in winter.

  287. “jimmi says:
    November 29, 2010 at 2:19 pm

    There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction.”

    Can this be demonstrated/verified via an experiment?

  288. George E. Smith says: November 29, 2010 at 4:49 pm

    Thermal radiation as its name implies is a consequence of the Temperature of the emitting material and is essentially independent of the nature of the material; and depends ONLY on the Temperature.

    This is not even close to right. The material makes a huge difference. Look at the table here: http://www.bacto.com.au/downloads/IR%20&%20Emissivity.pdf

    The energy radiated by an object is not simply P = AσT^4.
    Rather it is P = AεσT^4 where ε = emissivity.

    The IR emissivity can vary greatly depending on the material (and the surface finish). Polished metal is often below 0.05. Soot is up around 0.95. A painted metal object (ie high IR emissivity) will cool off much faster than a similar size/shape/temperature shiny metal (ie low emissivity) object because the low emissivity object simple cannot radiate IR effectively.

    Since N2 & O2 don’t absorb IR well (very low emissivity), they also don’t emit well. If you are waiting for N2 to cool simply by radiation, you will presumably be waiting a very long time.

  289. ,blockquote>There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction.”

    Can this be demonstrated/verified via an experiment?

    To be honest it is self-evident, but if you know someone with night vision glasses (the thermal imaging type, not the light amplification variety) you could try the following:
    Turn out all the lights, turn off the power to your fridge (so the internal light does not come on) put on the glasses and open the fridge door. Do you see anything? If so you have proved it – the glasses are at room temperature, the contents of the fridge are at about 5C, and those imaging glasses should work for objects down to about -5C (according to the specs). I haven’t tried this myself (no night vision glasses) so if anyone can try it I am sure some people would be interested.

  290. Can this be demonstrated/verified via an experiment?>>

    Well, how do you suppose an igloo works? Or a quinzey? Do you know how the first settlers insulate their houses? By piling snow up against them. Now how could that cold cold snow POSSIBLY make it warmer inside the house? How could you POSSIBLY warm yourself up by hollowing a hole out of a snowbank and sitting inside it? Ever done any winter camping? Sit on top of a pile of snow at -30 and you will die before dawn. Dig a hole in the side of the exact same pile of snow and crawl in. You’ll still die unless you remember to poke a small hole inthe top to allow for circulation of the air. If you remember, and you’ve dug the hole no larger than necessary, you’ll likely want to take your jacket off to cool down after a couple of hours.

  291. Tim,

    Oh Oh. You’re right, but now you get to reprise my argument with Svalgaard. :)

    I’m still trying to figure out the problem with George’s 377ohm paint. I suspect it has something to do with the characteristic impedance of air not being the same as vacuum, but that’s not entirely satisfying.

    George?

  292. Willis,

    Thank you for the response to my first post. You did misunderstand my poorly phrased suggestion. I was not saying that you would not find any warming. I tried to say you would not find any warming from GHG’s!!!

    BIG difference!!

  293. jimmi and davidmhoffer,

    “There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction.”

    Well, it may seem self-evident, which is not always the case with quantum physics, but real experiment is needed to verify that a photon from a colder object is indeed absorbed (meaning energy transfer?) and that many more [photons] are going in the opposite direction. I am not sure your proposed test with fridge will demostrate it, perhaps I will see reflected photons from outside? Igloo example is also not clear regarding the specific statement above.

    And speaking of “many more,” if we were to use a focusing lense or parabolic mirror to increase number of photons from a colder object so that there are more going in this direction from colder to warmer, will the warmer object absorb them and increase its temperature?

  294. from NY says:
    November 30, 2010 at 3:26 am

    “Well, it may seem self-evident, which is not always the case with quantum physics, but real experiment is needed to verify that a photon from a colder object is indeed absorbed”

    How about you look outside next time it’s snowing and explain how your eyeball at 98.6F can see a snowflake at 32F.

    If you can see it you’ll have proved to yourself that photons from colder objects fall upon warmer objects.

    If that doesn’t convince you I really don’t know what will.

  295. @Bryan says:
    “Heat always travels from hot to cold never the reverse.”

    o.k.

    “Thermal energy can be radiated and absorbed by both hot and cold objects.”

    of course – but thermal (IR) radiation of a colder body cannot be absorbed by a relatively warmer body. This does not mean, that a colder body cannot radiate to a warmer body. Think of radiation as a flow of energy which can be described with two parameters: quantity (amount) and quality (frequency). To transfer heat by radiation you need beside quantity also a certain quality

  296. @Terry Oldberg says:
    “The entity which, in a Kiehl-Trenberth diagram, is labelled as the “back radiation” should be relabelled as a “vector irradiance” with specified spectral characteristics. Relabelling it would help to eliminate the appearance of a second law violation.”

    The second law gets violated when radiation should transform into heat.

    Think of this:
    Kiehl-Trenberth shows back radiation with 324 W/qm, radiation from sun with 168 W/qm. Why do we build sun collectors and not back radiation collectors. Back radiation is available round the clock, 365 days a year and is delivering double as much energy as radiation from sun. Has anybody ever tried to look at back radiation as a source of energy? Ask yourself why!

  297. @Tim Folkerts says:
    “Put the same rock (in a vacuum container to avoid conduction & convection) in a freezer at 255 K. The rock still radiates 480 W^m2. But the container radiates 240 W/m^2 back to the rock (again assuming the container is a blackbody). The net loss is 240 W/m^2.”
    Thermodynamics know three kinds of systems:
    1.open (exchange of energy and matter)
    2.closed (exchange of energy)
    3.self-contained (no exchange)

    “The rock will cool half as fast as before (at least initially).”

    Really???

    the second law applies for 1 and 2

  298. @Tim Folkerts says:
    “The real consideration is the solid angle subtended by the radiating source. If the cooler container completely surrounds the hot object, then the rate that energy arrives at the hot object is the same, whether the container is a little larger or much larger.”

    please rethink…

  299. @jimmi says:
    “There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction. ”

    This means a large cold objekt can warm a small, warmer object, because it is only a question of quantity…

    “By the way, this can apply to conduction as well – if you think carefully about what happens when you allow two gases at different temperatures to mix, you will realize that some of the flow of both molecules and energy has to be from what was ‘colder’ towards what was ‘hotter’. Again it is just the net flow that is hot to cold.”

    think about Brown…

  300. @George E. Smith says:

    “Well I like the Clausius form of the Second Law”

    Well, I love it… But where is the point?

  301. from NY says:
    November 29, 2010 at 6:31 pm
    ” “jimmi says:
    November 29, 2010 at 2:19 pm

    There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction.”

    Can this be demonstrated/verified via an experiment?”

    nice question…
    ;-)

  302. @davidmhoffer says:
    “Well, how do you suppose an igloo works? Or a quinzey? Do you know how the first settlers insulate their houses? By piling snow up against them. Now how could that cold cold snow POSSIBLY make it warmer inside the house? How could you POSSIBLY warm yourself up by hollowing a hole out of a snowbank and sitting inside it? Ever done any winter camping? Sit on top of a pile of snow at -30 and you will die before dawn. Dig a hole in the side of the exact same pile of snow and crawl in. You’ll still die unless you remember to poke a small hole inthe top to allow for circulation of the air. If you remember, and you’ve dug the hole no larger than necessary, you’ll likely want to take your jacket off to cool down after a couple of hours.”

    what do you want to prove by that?

  303. @from NY says:
    “And speaking of “many more,” if we were to use a focusing lense or parabolic mirror to increase number of photons from a colder object so that there are more going in this direction from colder to warmer, will the warmer object absorb them and increase its temperature?”

    Hey, welcome to the club ;-)
    This “netting”-thing is just bullsh… They call it “modern” physics. Makes me wonder where “fashion” will go in the future…

  304. bessokeks says:
    “… thermal (IR) radiation of a colder body cannot be absorbed by a relatively warmer body. This does not mean, that a colder body cannot radiate to a warmer body…”

    I’m not sure how you can seriously say those two sentences — they contradict each other! A “colder body radiating to a warmer body” IS thermal radiation of a colder body being absorbed by a warmer body.

    Or how about this.
    * The sun (5800 K) emits some 15 um IR as part of its thermal radiation.
    * The earth (300 K) emits some 15 um IR as part of its thermal radiation.
    * The atmosphere (250 K) emits some 15 um IR as part of its thermal radiation.

    Can an object that is absorbing the 15 um photons know where those photons came from? Of course not! Whether the object absorbing the light is 200K or 275K or 350K or 3000K, it will absorb the 15 um photons from any temperature source exactly the same.

    “Why do we build sun collectors and not back radiation collectors? ”

    Because heat engines require a temperature DIFFERENCE to operate. The thermal radiation from the cool atmosphere can slow the cooling of the surface, but could never be used to operate a heat engine where the “cool side” is the warmer earth.
    Because photoelectric cells require HIGHER ENERGY visible light photons to operate — focused low energy IR photons (no matter how bright or well focused) will not work.

  305. “And speaking of “many more,” if we were to use a focusing lense or parabolic mirror to increase number of photons from a colder object so that there are more going in this direction from colder to warmer, will the warmer object absorb them and increase its temperature?”

    A clever attempt, but ultimately it is wrong.

    Consider using a lens to focus sunlight and burn word. Normally the sun provides about 160 W/m^2. If you use a 5x magnifying glass, what you have done is make the sun look 5x bigger from the perspective of the focused spot on the wood surface(=25x more area). That 25x “larger sun” will radiate 25x more light to the focus = 4000 W/m^2. That is a lot – enough to char the wood. But it is still way less than the 60,000,000 W/m^ the sun radiates from its surface.

    Even if you focused sunlight from all direction, you can’t focus more photons to an object than you could get by surrounding the object completely by a thermally radiating object of the same temperature.

    The same goes for the example above. Focusing photons from a cold object to a second object will never do more than surrounding the object by the colder object — which will cool the second object as expected. So the supposition “so that there are more going in this direction from colder to warmer” is never possible. (Or you WOULD violate the laws of thermodynamics and win a Nobel Prize.)

  306. bessokeks

    ….”two parameters: quantity (amount) and quality (frequency). To transfer heat by radiation you need beside quantity also a certain quality”…..

    I don’t think were in a big disagreement here.
    The colder body cannot raise the temperature of a warmer body above its own colder temperature.
    Which means the absorption of photons by the warmer has no effect.
    The vector dealing of the radiations as outlined by Terry Oldberg above gives the best approach.

  307. bessokeks

    I was told in another thread that you can’t “use a focusing lense or parabolic mirror to increase number of photons” of LW because it comes from all directions (unlike the radiation from the Sun).

  308. Bryan says:
    November 30, 2010 at 6:27 am

    “The colder body cannot raise the temperature of a warmer body above its own colder temperature.”

    A room temperature blanket can be used to raise the temperature of a warm bodied person. I understand the main article, but it seems like a lot of extra work over saying “The atmosphere acts like a blanket.”

  309. Steve

    Yes insulation best describes the radiative interaction between hot and cold bodies.
    The presence of the cold body reduces the heat loss from the warmer one.
    But even this does not always happen.
    For instance if the hotter body on its own was in thermal equilibrium with its surroundings and colder object was brought near .
    The colder object could reduce even faster the temperature of the hotter body if the colder body was colder than the surroundings.

  310. Which means the absorption of photons by the warmer has no effect.

    I’m sorry, but this is quite simply wrong. Please find any statement in any physics book which says that photons can be absorbed by an object in such a way as to destroy energy, because that is what you are proposing – you have now modified the 1st law of thermodynamics as well as the 2nd. Where does the energy go if it is not absorbed? Do photons carry a little sign saying “please don’t eat me?” How does the absorbing body know the photons are from a colder object?

  311. @Tim Folkerts says:
    “A “colder body radiating to a warmer body” IS thermal radiation of a colder body being absorbed by a warmer body”
    NO, because the photons do not need to be absorbed. There is no “thermal radiation”.
    There is only radiation. A certain kind of radiation (dep. on frequency) can be absorbed and transferred into heat
    “it will absorb the 15 um photons from any temperature source exactly the same. ”
    Right, dep. on the temperature of its surface. What temperature do you need to emitt 15 um photons?
    ” “Why do we build sun collectors and not back radiation collectors? ”

    Because heat engines require a temperature DIFFERENCE to operate. The thermal radiation from the cool atmosphere can slow the cooling of the surface, but could never be used to operate a heat engine where the “cool side” is the warmer earth.
    Because photoelectric cells require HIGHER ENERGY visible light photons to operate — focused low energy IR photons (no matter how bright or well focused) will not work.”
    According to Kiehl/Trenberth 324 W/qm get smashed against the surface of the earth.
    You are saying, the energy of that 324 W/qm can be absorbed by earths surface, but not by a collector (black surface, water inside, covered by thermal insulation glas which got a coating letting IR in, but not out) producing hot water. Are you serious????

  312. @lgl says:
    “I was told in another thread that you can’t “use a focusing lense or parabolic mirror to increase number of photons” of LW because it comes from all directions (unlike the radiation from the Sun).”
    Please do not think of parabolic mirrors, think of collectors producing warm (hot) water. These collectors work even when the sun is not directly shining on them.
    Radiation from all directions (background radiation) can be parallelized by bundels of tubes and then also be used in parabolic mirrors reaching much higher temperatures

  313. “”””” Tim Folkerts says:
    November 30, 2010 at 6:15 am
    “And speaking of “many more,” if we were to use a focusing lense or parabolic mirror to increase number of photons from a colder object so that there are more going in this direction from colder to warmer, will the warmer object absorb them and increase its temperature?”

    A clever attempt, but ultimately it is wrong.

    Consider using a lens to focus sunlight and burn word. Normally the sun provides about 160 W/m^2. If you use a 5x magnifying glass, what you have done is make the sun look 5x bigger from the perspective of the focused spot on the wood surface(=25x more area). That 25x “larger sun” will radiate 25x more light to the focus = 4000 W/m^2. That is a lot – enough to char the wood. But it is still way less than the 60,000,000 W/m^ the sun radiates from its surface. “””””

    Well Tim, I would say you need to take a course in Optics; because your above “explanation” is quite at variance with the laws of Geometrical Optics.
    For a start, what you describe as a 5x magnifying glass, is only 5x in the sense that when used to view a real object such as a postage stamp, produces a 5x magnified VIRTUAL

  314. @Steve says:
    ““The colder body cannot raise the temperature of a warmer body above its own colder temperature.”

    A room temperature blanket can be used to raise the temperature of a warm bodied person. I understand the main article, but it seems like a lot of extra work over saying “The atmosphere acts like a blanket.”

    Please do not forget the following:
    -the body is a source of heat ( meaning the sun is inside the blanket…)
    -the blanket reduces the loss of heat by preventing konvection which is the main reason for the loss of energy

  315. “”””” Tim Folkerts says:
    November 30, 2010 at 6:15 am
    “And speaking of “many more,” if we were to use a focusing lense or parabolic mirror to increase number of photons from a colder object so that there are more going in this direction from colder to warmer, will the warmer object absorb them and increase its temperature?”

    A clever attempt, but ultimately it is wrong.

    Consider using a lens to focus sunlight and burn word. Normally the sun provides about 160 W/m^2. If you use a 5x magnifying glass, what you have done is make the sun look 5x bigger from the perspective of the focused spot on the wood surface(=25x more area). That 25x “larger sun” will radiate 25x more light to the focus = 4000 W/m^2. That is a lot – enough to char the wood. But it is still way less than the 60,000,000 W/m^ the sun radiates from its surface. “””””

    Well Tim, I would say you need to take a course in Optics; because your above “explanation” is quite at variance with the laws of Geometrical Optics.
    For a start, what you describe as a 5x magnifying glass, is only 5x in the sense that when used to view a real object such as a postage stamp, produces a 5x magnified VIRTUAL image at a comfortable apparent distance of say 25 cm or ten inches .

    But the ability of that “magnifier” to raise the W/m^2 of sunlight by 25 x is a total fiction. It is a fundamental theorem of Optics, that NO OPTICAL SYSTEM can produce an image (real or virtual) that is of higher radiance (or luminance) than the image produced by an “Aplanatic” system; which is an imaging system that is simultaneously corrected for both Spherical Aberration and Coma, those being the two simplest of the “Seidel Aberrations” of imaging systems. And another theorem says that an aplanatic system cannot produce an image that has a higher radiance (or luminance) that the object.

    This latter theorem in fact was deduced by Clausius from the second law of thermodynamics. If you can design such an imaging system, you simply use it to image the exit aperture of a black body radiation source, at some Temperature (T) onto the entrance aperture of a second black body. And if the image has higher radiance than the source, then the Temperature of the second body will be raised to a higher value than the Temperature of the first; and that violates the second law.

    The reason why “magnifying glasses” can set fire to paper or wood or something else, is because they gather solar energy over a large area , and compact it onto a smaller image area. If the lens is aplanatic for the infinity focus conjugates, then that image will be a real image of the sun, having a size governed by the angular diamter of the sun (approximately 1/2 degree) and the focal length of the lens.
    So for example a one diopter reading glass lens, would have a one metre focal length, so it forms a sun image about 8.73 mm in diameter. The amount of solar energy that the lens can push into that area would depend on the diameter of the lens clear aperture.

    So to get the most energy into the smallest spot you need the “fastest” lens ; which means the smallest “f-number”. Of course following the principle of cussedness, the faster you try to make the lens in that sense, the worse the aberrations are likely to be, so keeping the design aplanatic may be a problem. As for the actual magnification; that is nearly zero, since you only got about a 9 mm image of the 860,000 mile diameter sun.

  316. Timing is fairly good for this. I am doing a series about the energy balance and I have tweaked it to discuss the Greenhouse Effect as well.

    The GHE is very real. Without it the Earth would be cold. Trenberth cheated and used radiative flux instead of energy transfer. That is why his balance is all funny looking.

    John Kehr
    The Inconvenient Skeptic

  317. “You are saying, the energy of that 324 W/qm can be absorbed by earths surface, but not by a collector (black surface, water inside, covered by thermal insulation glas which got a coating letting IR in, but not out) producing hot water. Are you serious????”

    I am completely serious, but you misunderstand what I am saying, and misunderstand the situation.

    That “black surface” will indeed absorb the 324 W/m^2 of incoming thermal IR, just like the earth does. But that black surface also radiates 380 W/m^2, just like the earth does. (Give or take a little depending on the exact temperatures.)

    The heating you are supposing is due to other sources besides the IR. Here is a simple thought experiment. Put out your “solar collector” at night. It is still collecting the same IR from the atmosphere (at least roughly) as during the day. Now, however, the “collector” will cool off (jsut like the earth cools off at night. There is no way the collector can use radiation from a cooler source (no matter how you try to “focus” it) to create a net heating of some other object. (It will, of course, still slow the cooling of the collector or earth compared to having the collector facing an even cooler object.)

    BTW, “glass which got a coating letting IR in, but not out” can’t exist. If a particular wavelength passes one way thru glass, that same wavelength will pass the other way thru the glass. So called “one way mirrors” let light thru equally in both directions. You just have to have a dark room on one side so their is not much might available to get out and the reflections from the other side are brighter.

  318. George E. Smith says:
    November 30, 2010 at 12:25 pm
    Well Tim, I would say you need to take a course in Optics;
    Thanks, but I already have. And after shifting thru all the talk of “aplanatic” and “coma” and “f-stops”, it seems we agree 100% on the main point.

    For a start, what you describe as a 5x magnifying glass, is only 5x in the sense that when used to view a real object such as a postage stamp, produces a 5x magnified VIRTUAL image at a comfortable apparent distance of say 25 cm or ten inches .
    OK, I was a bit sloppy here – but that is not at all the main point and I wasn’t really planning on getting into an optics lesson. I should have left out the numbers and focused on the concept (or worked thru the numbers more carefully) — my bad.

    The point here is that the lens is gathering more light and getting more energy to the spot on the surface than without the lens (or parabolic mirror). Others were suggesting that this could be used to create a “concentrator” to use the cool IR radiation in some sort of solar collector.

    “… an aplanatic system cannot produce an image that has a higher radiance (or luminance) that the object.”

    Ah ha! This sentence is exactly my main point. We seem to be in complete agreement on the main point. No optic system can have a higher radiance than the original source. In other words “even if you focused sunlight from all directions, you can’t focus more photons to an object than you could get by surrounding the object completely by a thermally radiating object of the same temperature” as I said before.

    ** There is no way to produce more than 60,000,000 W/m^2 by focusing sunlight (since the sun @ 5800 K is radiating 60,000,000 W/m^2 at its surface).
    ** There is no way to produce more than 240 W/m^2 by focusing 255 K bb radiation (since the bb @ 255 K is radiating 240 W/m^2 at its surface).

    So the original plan that “if we were to use a focusing lens or parabolic mirror to increase number of photons from a colder object so that there are more going in this direction from colder to warmer” is impossible. The warmer bb object (> 255K) will always be emitting more than the 240 W/m^2 and hence the cooler object (the whole sky in this case) will be always be cooling the warm object if only the IR is included.

    “… Clausius form the second law of thermodynamics. … and that violates the second law.”

    Yep. We once again are in agreement – we both conclude it will violate thermodynamics to do what they are suggesting. It will be interesting to see what bessokeks will have to say next ….

  319. bessokeks says:
    November 30, 2010 at 11:59 am

    “Please do not forget the following:
    -the body is a source of heat (meaning the sun is inside the blanket…)”

    The Earth’s surface is a (reflective) source of heat, inside the blanket of the atmosphere.

    “-the blanket reduces the loss of heat by preventing konvection which is the main reason for the loss of energy”

    Yes, exactly, although “main reason” isn’t necessarily true. Is the air dead or do you have a fan blowing?

    The molecules of the blanket are relatively fixed in comparison to the molecules of the surrounding air. Air around the body which is heated by radiative transfer is free to flow away from the body, so the heated air need not radiate much energy directly back to the surface of the body. The blanket keeps heated molecules close to the body (lower convection), so heat that was radiated away from the body is more likely to radiate back to it. With a blanket, at any point in time the surface of the body is being heated by the energy of the constant biochemical reactions plus the radiative transfer of heat that was captured by the blanket. Air captures heat too, but since heated molecules flow away (convection) that heat is more likely to go to the surrounding room than back to the surface of the body.

  320. “”””” bessokeks says:
    November 30, 2010 at 11:42 am
    @Tim Folkerts says:
    “A “colder body radiating to a warmer body” IS thermal radiation of a colder body being absorbed by a warmer body”
    NO, because the photons do not need to be absorbed. There is no “thermal radiation”.
    There is only radiation. “””””

    Well I wouldn’t bet my paycheck on that assertion. Particularly since thermal radiation is by far the most common in the entire universe; every single thing in the universe emits “Thermal Radiation”.

    Thermal radiation is electromagnetic radiation that is emitted by all matter in the real universe solely as a result of its Temperature. That’s why it is called “thermal”. Physicists (like me) tend to use the term as an alternative to saying “Black Body” radiation; since a Black Body is a theoretical construct which doesn’t really exist anywhere. But it is relatively simple to build “nearly black bodies” which emit a radiation spectrum wherein most of the output energy fairly closely matches that of an ideal black body; and most real objects that emit thermal radiation have real spectra that are not too difernet from that ideal. No body emits a broad spectrum of EM radiation as a result of its Temperature, where that radiation at ANY wavelength exceeds that emitted by a black body at the same Temperature. So it is very useful to do preliminary thermal radiation calculations in the context of a BB, and then apply corrections as needed. This results in the total radiation emitted by any body above zero K approximating that calculated from the Stefan-Boltzmann formula (sigma T^4); and havingg a spectrum roughly that calculated from the Planck Radiation formula.

    Theremal radiation results solely from the kinetic energy of motion of complete molecules, as a result of the acceleration of the electric charges in those molecules. It involves no intra-molecular atomic motions. Those changes internal to the molecule are constraiend by chemical bonds, and bond strengths and other structural details, which have resulting resonant types of oscillations, which can and do result in the emission (or absorption) of very specific, and very molecular species specific narrow line spectra. these radiations, whether emitted or absorbed, are NOT a function of the molecular Temperature; only its structure. There are secondary Temperature effects; which are a result of the Doppler effect, which will broaden the width of those resonances over and above the intrinsic line width; which depends on quantum effects.

    Then of course there is the so-called “nuclear radiations” of which gamma radiation is clearly electromagnetic in nature, while others involve actual material particles such as +/-elctrons, or alpha particles and all the more exotic ones. Those “radiations” which are the ones that send layfolks into apoplexy, are mostly Temperature independent; so they aren’t “Thermal Radiations”.

  321. “lgl says:
    November 30, 2010 at 8:18 am
    bessokeks

    I was told in another thread that you can’t “use a focusing lense or parabolic mirror to increase number of photons” of LW because it comes from all directions (unlike the radiation from the Sun).”

    Why exclude radiation from Sun? There was nothing in jimmy’s post to exclude Sun from consideration.

    “There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction.”

    —-

    Dave Springer says:
    November 30, 2010 at 4:16 am
    from NY says:
    November 30, 2010 at 3:26 am

    “Well, it may seem self-evident, which is not always the case with quantum physics, but real experiment is needed to verify that a photon from a colder object is indeed absorbed”

    How about you look outside next time it’s snowing and explain how your eyeball at 98.6F can see a snowflake at 32F.

    I believe jimmy was talking about emitted as opposed to reflected light. I think you would agree with me that when I look next time at a snowflake at 32F (in presumably sunlight conditions) more photons go to my eyes at 98.6F from snowflakes than from my eyes to snowflakes. I’d be appearing as a Terminator were it the case according to jimmy’s claim.

    Jimmy, any response from you to expect in the near future?

  322. The snowflake example (which was not mine) is a poor one, as you see the snowflake by reflected visible light not by emitted long wavelength infra red. As far as emitted light goes you are giving out more light than the snowflake. That is not a claim, it is basic physics and if you used a camera sensitive in the thermal wavelength (= far infra red ) part of the spectrum, it would be obvious. There are lots of images on the web of objects taken using thermal imaging. Here is a whole bunch of animals

    http://www.seedgen.com/thermallondon/animals.htm

    Note how hotter objects, i.e. the animals, are emitting more radiation than cooler ones, i.e. the background.

    Do you think there is something which prevents a photon from a colder body being absorbed by a warmer one? If so what? How does the warmer body know where that photon came from?

  323. To those who are interested in an experiment to demonstrate cold body radiation being absorbed by a warmer body, I propose the following as a valid (but quantitatively limited) measure of this effect. I have not seen any reference to this technique anywhere but I think it is a poor man’s pyrgeometer.

    At least some of the so called downwelling radiation (LWR) can be observed with a relatively inexpensive device. I have used a hand held IR thermometer that detects LWR in the range of 8-14 microns (the most commonly available in hardware stores) to convince my skeptical self that downwelling radiation is real. The IR thermometer measures the voltage generated by a thermopile due to photons entering through the optics. The principle is of course well established for temperature measurement as long as the emissivity compensation is made and good practices are observed. For this experiment, I set the emissivity to the maximum setting so that I can use the table mentioned below.

    I am not suggesting that the temperature reading is a true temperature (it is not). It can; however, be used to back into the radiation that the detector is seeing. Figure on page 4 at

    http://www.hartscientific.com/publications/pdfs/3187781_A_w.pdf

    can be used to estimate the watts/sq m. The straight line is a representation of Wien’s distribution law.

    In a cloudless area of the sky and shielded from sunlight, I have found that the observed “temperature” is a function of dew point much more so than ambient temperature. It drops off rapidly at dew points less than 60 deg F becaause the water vapor in the atmosphere (aka water column) is falling off exponentially. At about 56 F dew pt, the reading is off scale at -40 F (-40 C). I’ll leave the total sky downwelling radiation for later and/or for others. The viewing angle is relatively small and needs to be integrated (ugh) over the entire sky. Numerous issues I see with that.

    I’ve also observed that the reading in a cloudless sky increases as the angle of sight decreases toward the horizon but it hardly matters whether it is daytime or night time- a real surprise to me. I presume that at lower angles above horizon, there is more water vapor that is radiating back. Just a guess though.

    Results are quite a bit different on cloudy days. The “temperature” of a cloud is much higher. So downwelling radiation (at least in the 8-14 micron LWR range) from a cloud is much greater than that from CO2+ water vapor. It is also not a function of viewing angle.

    The 8-14 micron range was selected by the developers to avoid CO2 interference in temperature measurements so it does miss most (but not all) of the CO2 radiation around 16 microns.

    Comments are welcome.

    Regards,

    PB

  324. buntche70

    Good post
    Most people accept that colder objects radiate to hotter ones.
    But what is the result of this radiation arriving at a hotter surface?
    This paper shows that radiation from a nighttime clear sky can freeze water at the parabolic dish even when the ambient temperature is well above zero c

    http://solarcooking.org/research/McGuire-Jones.mht

  325. @jimmi says:
    “Do you think there is something which prevents a photon from a colder body being absorbed by a warmer one? If so what? How does the warmer body know where that photon came from?”

    Where that photon came from is irrelevant. The only thing that counts is the “quality” of the photon.
    How does the wormer body know that he has to accept the energy of the photon, because he belives he does not get warmer?
    Remember: the warmer body radiates according to the rule of Max Planck. If you radiate back energy, you warm it up again. This means you raise temperature again above the present temperature. There is no way to let the warmer body cool down slower

  326. @George E. Smith says:
    “Thermal radiation is electromagnetic radiation”

    …and therefore not heat. It can be converted to “heat” IF the photons get absorbed…

  327. @Steve says:
    “The Earth’s surface is a (reflective) source of heat, inside the blanket of the atmosphere”

    I guess I should have phrased it clearer: earth does not generate heat, like your body

  328. @John Kehr says:

    “The GHE is very real. Without it the Earth would be cold”

    Thre is no valid evidence for that.

  329. @Steve says:
    “Air around the body which is heated by radiative transfer is free to flow away from the body, so the heated air need not radiate much energy directly back to the surface of the body”
    please do not forget that “air” (O2, N2) is blind to IR, so radiation playes a very tiny role. konvection gets propelled by air, which is heated by conduction at the surface of the body

  330. bessokeks,

    You make some very good points. Your thought experiment where a large cooler shell surrounds a small warmer body is a valid one, since it postulates a situation where the greater flow is of photons from the cooler body impacting the warmer body. According to the net flow idea, the warmer body would get warmer, which, of course, does not happen, so the net flow idea is wrong.

    The conclusion I would draw is that a photon from the cooler body is incapable of raising the energy of a particle in the warmer body to make it warmer still. Yet, what is happening to this extra energy? Where has it gone? Perhaps it is re-emitted. This would conform to the radiation budgets. Ie, the inner body now radiates at a greater flux density than before. The problem is, everyone then applies the Stefan-Boltzman formula and concludes the temperature must have increased. But maybe the equation cannot be used because it is not behaving like a black body. If this is the case, then the GHE could not work to raise the temperature of the inner body. All it could do is stop the temperature falling below that of the outer shell.

  331. bessokeks says:
    December 1, 2010 at 6:04 am

    “please do not forget that “air” (O2, N2) is blind to IR, so radiation playes a very tiny role. konvection gets propelled by air, which is heated by conduction at the surface of the body”

    So the air in a typical room has absolutely no IR absorption in your universe? And apparently it is 100% O2 and N2! Why don’t you check out this graph. Note that the IR band begins just before the 1, and “thermal IR” is commonly considered to be the range of 3 to 15 micrometers.

  332. Hi Willis,
    I read over this post and the comments a couple of days ago but it keeps nagging me — a simple model to support the Green House effect.

    It occurred to me that you really don’t need to look much further than the turn of each day. The Green House effect rises in the East and sets with the Sun in the West. This is the part Climate “Science” fails to embrace (its a natural dynamic event).

    Until Climate “Science” embraces and communicates the dynamic relationships as an integrity, its unlikely the Science will accurately predict anything other than the ongoing waste of more tax dollars.

  333. Vince Causey says: December 1, 2010 at 7:02 am

    bessokeks,

    You make some very good points. Your thought experiment where a large cooler shell surrounds a small warmer body is a valid one, since it postulates a situation where the greater flow is of photons from the cooler body impacting the warmer body. …

    Arrg. No, this point is not valid. It is not valid at all! There will still be a greater flow of photons out from the hot surface than in to the hot surface!

    This picture gets at the heart of the problem:

    If the surrounding surface gets twice as big in radius, the area increase by 4x. But the intensity of light from any part of the surface decreases by 1/4. The net effect is that the surrounding surface provides the same intensity of light (= the same number of photons = the same W/m^2) to an object inside no mater how far away the surrounding surface is.

    Put another way, the surrounding surface emits more and more light as it gets bigger (more total photons), but fewer and fewer of them hit the object inside (ie more and more miss the object and simply return to the surround surface at some other location). The two effects exactly cancel out.

    If you still don’t get it, then I give up.

  334. So apparently the warmer body always reflects photons from a cooler one?

    Imagine you have 3 objects at temperatures 10C, 11C and 12C. So the theory is that the object at 11C absorbs photons from that at 12C, but reflects photons from the object at 10C. In other words the emissivity of an object changes sharply according the temperatures of remote objects? This is Phantasy Physics not real physics at all.

  335. “jimmi says:
    November 30, 2010 at 11:19 pm

    Do you think there is something which prevents a photon from a colder body being absorbed by a warmer one? If so what? How does the warmer body know where that photon came from?”

    I don’t know. But if we follow your approach and manage to get more photons from a colder body to a warmer one, which should be possible, as the colder body can be much much larger, while a temperature difference may be rather small, then a warmer body should become even warmer according to your reasoning, and that would be against 2LOT.

    Your thoughts?

  336. No, even if you completely surround the warmer object you can never supply more inward photons than there are outgoing ones. Imagine a warm object radiating photons into the vacuum of space. The flux (power per unit area) would be given by the Stefan-Boltzmann equation.
    flux = 5.67e-8 * T&#8308
    Now completely surround that object by another, but with a tiny gap so that energy has to radiate from one to the other. The flux from the second object will also be given by the Stefan-Boltzmann equation, but if the temperature is less, so will be the flux. You cannot make the second object any bigger, or closer, than one that completely surrounds the first, so you can never have the inward flux greater than the outward flux no matter what the geometry. Clearly though, if the outside object had a temperature only slightly greater than the inner one, the inner one would cool very slowly.

  337. Um, that funny looking formula was suppose to be T-to-the-power-4, ie T superscript 4, but it has not come out properly

  338. jimmi, do you think 2 glowing light bulbs produce about twice the number of photons or the same number as just one glowing light bulb?

  339. jimmi, do you think 2 glowing light bulbs produce about twice the number of photons or the same number as just one glowing light bulb?

    You must have some odd view of physics if you think that question reveals anything. Of course they have approximately twice the output of one bulb. If you think I was suggesting otherwise you have not understood anything. Anyway, I trust my previous comment shows that the idea of heating a small object with a very large colder one is invalid.

  340. You have an omission in the description of Figure 3. It is correct only assuming that the shortwave albedo of the shell is the same as that of the earth.

    Perhaps you should also emphasize more clearly that the LW emissivity of the shell is 1. The full opaqueness does not imply 100% emissivity as it allows for a lower albedo.

    By these comments, I do not contradict your example.

  341. @jimmi, I just wanted to verify this point. Here is a setup that I think will produce more photons from lower temperature (LT) objects to a higher temperature object (HT). Lets select an HT object with a temperature slightly higher than LT, so that it produces 10% more photons. Put an LT object in the focus point of a parabolic reflector, and then use a lense to focus reflected photons on HT object. This way we should be able to capture about 60% of all photons emitted by the LT object*. Put another LT2 object in a similar setup with output also directed to HT object. As a result we have total number of photons coming from LTs to HT at 120%, while HT is 110% of this number. Of course, if needed we can use more LT objects.

    Comments?

    Note: I used for reference this site for transmission/reflection losses:

    http://www.telescope-optics.net/functions.htm

  342. from NY says: December 2, 2010 at 6:36 am
    … Put an LT object in the focus point of a parabolic reflector, and then use a lens to focus reflected photons on HT object….

    It certainly is a clever idea, but ultimately it will fail for reasons that have been discussed before.

    As a slight aside, the 60% from the website you listed is 60% of the light entering the telescope, not 60% of the light leaving the object. You could certainly focus 60% of the light leaving an object onto some other object, but it would take a little different set of lenses/mirrors. It is easy to focus 60% of the sunlight entering the telescope into your eye; it is impossible to focus 60% of the light leaving the star into your eye!

    Here is one slightly new argument. If you really wanted to focus light from the cooler object, you should put it at one focus of a perfectly reflecting ellipsoid and put the other (warmer) object at the other focus. Then you could get 100% of the light from the cooler object focused on the warm object — not just the 60% you suggested.

    However:
    1) this completely surrounds the warm object, so you can’t possibly add a second similar source. You could “cut a hole” in the ellipsoid so only 60% of the energy was focused on the warm object. But the size of the hole would limit you to another 40%, never an extra 60% if you added a second cool source.
    2) By symmetry, all the light from the warm object is focused on the cool object. Since the warm object is emitting more energy, we will always have more energy from warm to cool than from cool to warm.

    I’m afraid any more discussion would require being able to draw pictures and/or talk in person and/or perform integrals in spherical coordinates. Since none of these are feasible in this setting, I think it is time for me to drop out of the discussion.

  343. “Tim Folkerts says:
    December 2, 2010 at 11:06 am”

    Tim, you constructed a straw man “ellipsoid” argument, bearing no relation to what I wrote. And wrong – 60% is for the light leaving the LT object. If you think this is too high – note, I wrote “Of course, if needed we can use more LT objects.”

    If you have anything specific against what I described – please post.

  344. NY,

    I don’t think we will be able to resolve this issue in this venue – the tools are just not available. Maybe with webcams and mics we could get our points across, but short of that i don’t see how we can communicate our ideas clearly enough here.

    I will just say that as I understand it (with degrees in math and physics) — the fields of optics, geometry, and thermodynamics all conclude that you cannot get more energy from the cool object to the hot object than from the hot object to to the cold object. I can sort of imagine the set-up of mirrors and lenses you want, but no matter how I picture it, I see physical and/or fundamental optics limitations that limit the results to no better than surrounding the hot object by a sphere of material at the low temperature.

    With all due respect, since you are the one making an “extraordinary claim”, it is up to you to give extraordinary evidence — not just “this seems plausible to me”. I (and others) have already given standard responses.

    (BTW, the elliptical mirror is not a strawman. It is actually a simplification of what you propose — it collects the light like your parabolic mirror AND focuses it like your lens. The parabolic mirror is a limiting case of an elliptical mirror where the second focus is at infinity.)

  345. From NY,

    I think I can deal with this without maths provided you get the idea of a limiting case, which is a standard way of dealing with physics problems. Suppose you were wanting to work out whether a given process was large enough to have a certain effect. You take the limit where you make that process as large as it possibly can be , and if it is still not large enough then there is no point in looking at other configurations.
    So in the case I gave, you make the cooler object completely surround the warmer object with just a tiny gap between them. This means that absolutely all the energy leaving the cooler object must strike the warmer one, and you cannot do better than 100% of the energy. Even in this case the flux (power per unit area) from the cooler object towards the warmer one is not sufficient as it is less than the flux going the other way. i.e a cooler object surrounding a warmer one, with a small gap, will always result in the warmer one eventually cooling down – it just takes longer than if the warmer one were left to itself (congratulations you just invented the thermos flask)

    Now I did not say what the cooler, surrounding object was made of, and that is because it does not matter – you can aggregate any number of smaller objects, mirrors etc in any configuration you like and you will never be able to come up with a case where more than 100% of the available energy hits the enclosed object. Using two or more heat sources makes no difference, provided they are all at the same temperature. Nor does it matter whether the objects are pure Black Bodies – again think of limiting cases e.g what if one object was a prefect mirror which only reflected radiation and neither absorbed nor emitted it – if the outer, surrounding object were a perfect mirror it would just reflect all radiation from the inner object back on itself, and nothing would change.

    What would happen if the cooler object did not completely surround the warmer one? Well there is no problem there either, because in that case you have indirectly introduced a third object – call it “space” or “the vacuum” i.e if there are holes in the outer object radiation can leak away into the vacuum. The effect? Well now both objects are cooling – some of the energy will radiate directly into the vacuum, some will come from one object (does not matter which), and take an indirect route out as a result of being reflected and/or absorbed and re-radiated from the other object – either way the energy eventually leaks away. So in this case both objects are cooling and there is no problem w.r.t LOT2.

    The model presented at the top of the thread cannot be a quantitative description of the GHE, in the sense that it is too simple to give a value for the temperature increment, but it illustrates one property that can give rise to a greenhouse effect, and that is that the incoming radiation is unhindered, whereas the outgoing radiation has difficulty leaving. The model has no problems with LOT2, and neither does the actual GHE, which is a real and entirely naturally occurring effect (which may or may not have been altered by human activities – but that is a separate discussion)

  346. @jimmi, December 2, 2010 at 7:28 pm …

    Energy radiation (Stefan–Boltzmann law) is proportional to 4th power of temperature per unit of time per unit of surface area. In the case you describe when colder object surrounds warmer object with a tiny gap, the radiating surfaces are about the same, thus colder object radiates less than the warmer one.

    This is why I asked you in a previous post if you think 2 glowing light bulb radiate approximately twice as one, apparently it was not clear why I asked the question, but you agreed that twice the output is approximately right. So, unlike your example, which essentially is one colder bulb and one warmer bulb, I constructed 2 colder bulbs (objects) with the output directed with some losses against one warmer bulb (object). I hope it is clear now; your example is invalid.

  347. From NY

    Sorry, but my example is as many light bulbs as you want so as to surround the warmer object – not just 2 – as many as you want as long as each is cooler than the central object and as long as the central object is completely surrounded so that every square metre of the central object is irradiated. I actually said that previously, “using two or more heat sources makes no difference “. And if you do not surround the warmer object it does not work due to losses to the vacuum.

  348. @NY

    The crux here is the difference between
    a) “radiating more from it’s own surface” and
    b) “radiating more
    to the other surface“.

    No one is refuting (a). A large cool object can certainly radiate more total energy than a small warm object. If the cool object is 10% cooler but twice as big, then it will indeed radiate more energy (about 30% more) than the slightly warmer, smaller object.

    But (b) is believed to be impossible (based on very fundamental optics and thermodynamics). In my example above, the cool object was radiating 30% more total energy. BUT some of that energy is not going to hit the warm object, since there must be a gap between the two. Some of the photons will “miss” the warm object and get reabsorbed by the cool object. On the other hand, all of the photons from the warm object will get absorbed by the cool object, since the cool object completely surrounds the warm object.

    You would get something like
    * Warm object emits 100 J per second — all of which gets absorbed by the cool object

    = 100 J per second from warm to cold

    * The cool object emits 130 J/s — 65 hits the warm object and 65 hits the cold object object thru the “gaps”

    = 65 J per second from cool to warm (and 65 J/s from cool to cool)

    No matter how you arrange it, the “gaps” will let more energy “miss” than you gain by making the surrounding cooler object bigger. And no amount of mirrors or lenses can change that, either.

  349. Steve says:
    “So the air in a typical room has absolutely no IR absorption in your universe? And apparently it is 100% O2 and N2! Why don’t you check out this graph. Note that the IR band begins just before the 1, and “thermal IR” is commonly considered to be the range of 3 to 15 micrometers.”

    Absorption: YES
    Reemission: NO

    Up to 10km height, animated molecules give away the extra energy they absorbed before by collision

  350. @Tim Folkerts says:
    “If the surrounding surface gets twice as big in radius, the area increase by 4x. But the intensity of light from any part of the surface decreases by 1/4. The net effect is that the surrounding surface provides the same intensity of light (= the same number of photons = the same W/m^2) to an object inside no mater how far away the surrounding surface is.

    Put another way, the surrounding surface emits more and more light as it gets bigger (more total photons), but fewer and fewer of them hit the object inside (ie more and more miss the object and simply return to the surround surface at some other location). The two effects exactly cancel out.

    If you still don’t get it, then I give up.”

    Have patience, Tim ;-)

    Imagine two coins opposite to each other. one small and warm, one 10 times bigger and slightly colder. Then put a lens between an direct the photons of the slightly colder coin exactly onto the surface of the smaller coin. You would transfer roughtly 10 times the energy the warm coin radiates onto the warm coin…

  351. Imagine two coins opposite to each other. one small and warm, one 10 times bigger and slightly colder. Then put a lens between an direct the photons of the slightly colder coin exactly onto the surface of the smaller coin. You would transfer roughtly 10 times the energy the warm coin radiates onto the warm coin…

    No, this is one of the cases I was discussing with FromNY – you actually have three objects here – the two coins and “space” – both coins are cooling because they are both radiating to space – the fact that some of the energy from the larger, cooler object goes via the smaller one is irrelevant.

    Also, regarding FromNY’s light bulbs – note that we have accidentally changed the example from the initial one, which was just considering warmer and cooler objects – if we now start considering powered objects, then we also need to consider what is happening to the ultimate power supply – in other words FromNY needs to think about what happens when he puts all these light bulbs in place, waits from them to warm up, and then switches off the power supply

  352. bessokeks says:
    “Imagine two coins opposite to each other. one small and warm, one 10 times bigger and slightly colder. Then put a lens between an direct the photons of the slightly colder coin exactly onto the surface of the smaller coin. You would transfer roughtly 10 times the energy the warm coin radiates onto the warm coin…”

    Last I checked, lenses work in both directions. If the lens captures light from he cool object and focuses it on the warm one, then it will simultaneously focus light from the warm object onto the cool one. If you focus 10x more light from cold to warm, you will be focusing 10x more light from warm to cold.

    Or to clarify what you were trying to say before …”You would transfer roughly 10 times the energy AS YOU DID WITHOUT THE LENS FROM THE COOL COIN onto the warm coil. You would also transfer roughly 10 times the energy AS YOU DID WITHOUT THE LENS from the warm coin onto the cool coin.” If the Warm coin had been “winning” before, it will still be “winning” if all the numbers are multiplied by 10.

    Following up on jimmi’s comment, the lens reduces the energy that is going to “to space” and increases the energy going to “the other coin”. BOTH coins get more energy than before from the other. The coins will stay warmer longer, but they will not violate the 2nd Law.

    Since you are the one trying to violate the 2nd Law of Thermodynamics, I leave it to you to propose a specific lens arrangement and coin sizes, and then do the calculations that show us that your system truly focus light in such a one-sided way as to send more photon energy from cool to warm than from warm to cool. Give us the hard numbers of energy flux from cold to hot and from hot to cold.

    If you can do that, you will have yourself a Nobel Prize. If you can’t do that, then you might as well stop now trying to come up with “this seems like it should work” scenarios.

  353. Tim Folkerts says:
    December 4, 2010 at 4:36 pm
    “this seems like it should work” scenarios.

    Here’s a very simple one. A massive self-gravitating planet is surrounded at a distance by an unbroken shell. The outer shell is warmed by the sun, the planet in turn is warmed by the shell. Photons emitted from the shell gain energy falling down the gravitational well onto the planet. Photons emitted from the planet lose energy climbing back out. Then, in the steady state, the planet is hotter than the shell (as measured by thermometers on each). The temperature ratio is equal to the gravitational blue shift. The time rate is also higher on the planet, in the same proportion. In all, the brightness of the shell is boosted by the fourth power of the blue shift, the rest coming from a compression in the apparent angular size of the shell overhead (aberration of light). This does not violate the second law, even though the hotter object (the planet) has been heated to that temperature by a cooler object (the shell).

  354. Tim Folkerts — December 4, 2010 at 6:28 am

    Tim, (a) follows from law, and (b) is demonstrated, for example, in example by bessokeks:

    “Imagine two coins opposite to each other. one small and warm, one 10 times bigger and slightly colder. Then put a lens between an direct the photons of the slightly colder coin exactly onto the surface of the smaller coin. You would transfer roughtly 10 times the energy the warm coin radiates onto the warm coin…”

    Tim says, “Last I checked, lenses work in both directions. If the lens captures light from he cool object and focuses it on the warm one, then it will simultaneously focus light from the warm object onto the cool one. If you focus 10x more light from cold to warm, you will be focusing 10x more light from warm to cold.”

    It is interesting to follow your argument, for now you directly dispute jimmi’s claim:

    “jimmi says:
    November 29, 2010 at 2:19 pm

    There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction.”

    Perhaps Tim would now address jimmi’s questions — November 30, 2010 at 11:38 am

    “Where does the energy go if it is not absorbed? Do photons carry a little sign saying “please don’t eat me?” How does the absorbing body know the photons are from a colder object?”

    In another formulation,

    “jimmi says:
    November 30, 2010 at 11:19 pm

    Do you think there is something which prevents a photon from a colder body being absorbed by a warmer one? If so what? How does the warmer body know where that photon came from?”

  355. To those who remain unconvinced of the greenhouse effect, focus on Willis’ claim: “It doesn’t require CO2. It doesn’t need an atmosphere.” His statements zero in on the operational feature which troubles many; how can a cold object (atmosphere) warm a warmer object (earth)?

    Consider an airless planet orbiting an isolated star, one side always facing the star.

    The night-side of that planet transfers less heat to space than is received by the day-side (the bulk difference is reflected or radiated back to the sun). Therefore, for that sector of space, the planet is a restriction in the rate of heat transfer from the star. With less heat leaving the star’s surface and presumably no decrease in heat arriving from the interior, the star’s surface temperature will be higher.

    Add to that planet an atmosphere with Tyndall gases (‘greenhouse’ gases) and the planet will be warmer for the same reason the star was warmer – restriction in outgoing heat transfer.

  356. All the explanations for the “Greenhouse Theory” I have ever seen or read about rely on the atmosphere being able to stop – or restrict convection. This “Steel Greenhouse” is no different. Well no, – it is different as it has got no atmosphere but a vacuum, so in stead a presence of air, the shell (or replacement atmosphere) this time is opaque to outgoing long wave radiation. Stopping outgoing long wave radiation should work just as well, after all this model is not warming an atmosphere just the planetary surface
    This, by the way is not the first time I have seen this kind of planetary contraption and the first time I saw it, yes – it had many shells (all imaginary ones) but that just made the whole thing even more unrealistic and I did wonder at the time; “Why do these people make things so complicated?”
    – Well if you have a vacuum then of course you do not have to explain away convection. – Could that be the reason?

    In this “Steel Greenhouse” there are lots of numbers, some are relating to energy in W/m² and some are relating to temperatures in °K and some in °C. – I would like to know how they worked any one of those numbers out. And by the way as there seems to be 240 W/m² hitting the planetary surface from the Sun and a further 240 W/m² from the inside of the shield, why not carry this “thought experiment” a bit further? What is there to stop the now 480 W/m² from bouncing down to the planetary surface again and again ad infinitum doubling it’s square wattage with each bounce?
    And lastly, can anybody work out how 240 W/m ² can possibly leave the outside of the shield?

    How did that wattage get through the shield?

  357. Paul Birch says: December 5, 2010 at 3:22 am

    Here’s a very simple one. A massive self-gravitating planet is surrounded at a distance by an unbroken shell. The outer shell is warmed by the sun, the planet in turn is warmed by the shell. Photons emitted from the shell gain energy falling down the gravitational well onto the planet. Photons emitted from the planet lose energy climbing back out. Then, in the steady state, the planet is hotter than the shell (as measured by thermometers on each)….

    You are one of the few people I have ever heard of speaking general relativity as “simple”. :-)

    We are getting far afield, but I don’t think your example invalidates anything I said.
    In relativity all the measurements are — well — relative. Suppose the the system has reached equilibrium. The temperature of the two objects as measured in any given reference would be the same.”
    * Mr P. on the planet would measure the same spectrum (and total energy flux) for the upgoing radiation as the downgoing radiation –> he concludes the planet and the shell are the same temperature.
    * Mr S. on the shell would measure the same spectrum (and total energy flux) for the upgoing radiation as the downgoing radiation –> he concludes the planet and the shell are the same temperature (although this temperature is a little less than Mr P would measure for the two)

    These two numbers for the temperatures (as measured by Mr P vs Mr S) are different because they are measuring the temperature in different frames. In relativity it is perfectly OK to get different numbers in different frames. In any specific reference frame, however, the temperatures will indeed be equal if the radiation in and out is balanced.

  358. O H Dahlsveen says: December 6, 2010 at 3:11 pm

    All the explanations for the “Greenhouse Theory” I have ever seen or read about rely on the atmosphere being able to stop – or restrict convection.

    Then with all due respect you have been reading the wrong explanations. The main problem is that there are two very different processes that are BOTH called “the greenhouse effect.”

    * For actual greenhouses, “the greenhouse effect” does indeed depend primarily on the restriction of convection.

    * For planets, “the greenhouse effect” depends on the IR that is shining down from the atmosphere. Atmospheres without GHG’s are very ineffective at radiating IR, so they cannot provide any significant IR toward the surface.

    Anything that warms the atmosphere will increase the downward IR, which depends on a) the temperature of hte atmosphere and b) the concentrations of various GHG’s. Convection warms the higher levels a bit. Evaporation/condensation warms it more. IR from the surface warms it even further. So this warm atmosphere will radiate IR, some of which will radiate toward the surface and hence provide extra energy for the surface.

  359. @ From NY

    I don’t quite follow what you are objecting to — I don’t think I am contradicting anything jimmi said.

    Let me try one more time. Lets suppose as a starting point, two blackbody discs with an area of 1 m^2. The cool one radiates 100 W/m^2 and the warm one radiates three times as much: 300 W/m^2. The discs are placed 10 m apart facing each other. The rest of space is very cold and provides no appreciable radiation.

    Only a small bit of the energy from the cool disc actually reaches the warm disc, since most of the radiation “misses” the warm disc. I estimate the actual flux reaching the warm disc as 0.16 W/m^2, but the actual number is not important. What IS important is that the cool disc will receive 3 times as much power per meter (0.48 W/m^2). Thus cool disc receives more power than it loses. There is a net flux from the warm disc to the cool disc

    If you put TEN cools discs side-by-side (or a single disc of area 10 m^2), then EACH cool each square meter radiates 0.16 W/m to the warm disc = 1.6 W/m^2. But the warm discs radiates 0.48 W/m^s to each cool disc = 4.8 W/m^2 total. Thus cool disc receives more power than it loses. There is still a net flux from the warm disc to the cool discs.

    If you put a lens between the 1 warm disc and the 10 small discs, you could focus more of the power from the cool disc that would missed the warm disc. Say the lens is big enough to focus 5 times more power from the cool discs. There is now 10 x 1.6
    W/m^2 = 16 W/m^2 hitting the warm disc. But as I said, the lens works in both directions, it will also focus 5x as much energy from the warm disc = 48 W/m^2. There is still a net flux from the warm disc to the cool discs.

    Not matter how you arrange cool discs, lenses or mirrors, I challenge you to find a specific example that either has a) more energy from cool to warm than from warm to cool or b) more than 100 W/m^2 from the cool surfaces to the warm surface.

    Back to my paying job. :-)

  360. “Tim Folkerts says:
    December 6, 2010 at 7:47 pm

    I don’t quite follow what you are objecting to — I don’t think I am contradicting anything jimmi said.”

    OK, let me point you to exact words:

    “Imagine two coins opposite to each other. one small and warm, one 10 times bigger and slightly colder. Then put a lens between an[d] direct the photons of the slightly colder coin exactly onto the surface of the smaller coin.”

    Would the photons from the colder coin be absorbed by the warmer coin or not?

  361. Yes, the photons from the cooler object will be absorbed by the warmer object, just like photons from the warmer object will be absorbed by the cooler object. Indeed, if we are assuming that the objects are blackbodies, then temperature is really immaterial — any blackbody at any temperature will absorb any photons from any source at any temperature.

    (Of course, no object is a true blackbody. How well they actually absorb photons will depend on the wavelength of the incoming light. For example, the shell in the original model was assumed to be a blackbody (emissivity = 1) for LWIR, but a “clearbody” (emissivity = 0) for visible wavelengths.)

  362. Tim Folkerts says:
    December 6, 2010 at 6:52 pm
    “We are getting far afield, but I don’t think your example invalidates anything I said.
    In relativity all the measurements are — well — relative. Suppose the the system has reached equilibrium. The temperature of the two objects as measured in any given reference would be the same.”

    Strictly speaking, it is not possible to measure the temperature other than in the rest frame of each object. One can infer the temperature by observing the emitted radiation, but to do so one has to make certain assumptions – such as that the radiation one observes is thermal and has not been doppler shifted – that are not always valid. The temperature difference in my example is a real or absolute one. We can see this by imagining that the two objects are made of a single material that has a melting point above the temperature of the shell but below that of the planet. The shell will remain solid; the planet will melt.

    The significance of this thought experiment is that it is unwise to rely upon archaic formulations of the second law – like “heat cannot flow from a cooler object to a hotter object” – based on temperature or heat flux, because they hide assumptions that occasionally bite back. They are very nearly – but not quite – correct. The correct basis of the second law is entropy, and the key question whether entropy is non-decreasing.

  363. “The temperature difference in my example is a real or absolute one. ”
    I see where you are coming from, but I disagree with your terminology. There is no “absolute” reference frame in relativity. These might be called the “proper temperature” is suppose.

    “The shell will remain solid; the planet will melt. ”
    OK. In different frames lengths are different and clocks run differently. And I strongly suspect materials melt at different temperatures. But hashing out relativistic thermodynamics here is beyond my expertise or motivation. :-)

    But I completely agree that “The significance of this thought experiment is that it is unwise to rely upon archaic formulations of the second law – like “heat cannot flow from a cooler object to a hotter object””

  364. “Tim Folkerts says: December 7, 2010 at 10:33 am
    Yes, the photons from the cooler object will be absorbed by the warmer object”

    OK, so what happens to the warmer object? Does its temperature go up? After “Then put a lens between an[d] direct the photons of the slightly colder coin exactly onto the surface of the smaller coin.”

  365. @ From NY

    The warmer object gains the energy, but its temperature does not go up. Specifically, the warm object absorbs the energy from the photon, but thermodynamics/blackbody radiation laws tells us that the warm object will necessarily be losing more energy from its own outgoing photons than it is gaining from the incoming photons. Or put another way, the photons from the cool object will make the warm object warmer than it would have been without the photons from the cool object, but the warm object will still be getting cooler.

    It’s kind of like I give you $10 every minute and you give me $5. Do I get richer from your $5? Well, I don’t get poor as quickly, but I am still getting poorer.

    And the lens still will not be able to focus more from cold to hot than from hot to cold. At this point I fear you will either have to trust me (and jimmi), or do some heavy-duty analysis with principles from BB radiation, optics, thermodynamics, and surface integrals until you can show a specific counterexample that violates the 2nd law. For over 150 years clever people have been trying to come up with ways to violate the 2nd law — but as far as I know, no one has succeeded. So I am quite confident you will not be able to come up with such a violation.

  366. “Tim Folkerts says: December 7, 2010 at 6:06 pm

    Specifically, the warm object absorbs the energy from the photon, but thermodynamics/blackbody radiation laws tells us that the warm object will necessarily be losing more energy from its own outgoing photons than it is gaining from the incoming photons.”

    How can the warm object absorbing energy stay at the same or lower temperature and radiate more? This would be against the Stefan–Boltzmann law.

  367. Specifically, the warm object absorbs the energy from the photon, but thermodynamics/blackbody radiation laws tells us that the warm object will necessarily be losing more energy from its own outgoing photons than it is gaining from the incoming photons.”

    By the way, this statement is factually incorrect. I sure observed warm objects getting warmer when placed in the sunlight.

  368. Tim Folkerts says:
    December 7, 2010 at 4:00 pm
    PB: The temperature difference in my example is a real or absolute one.
    “I see where you are coming from, but I disagree with your terminology. There is no “absolute” reference frame in relativity. These might be called the “proper temperature” is suppose. ”

    It is a misconception that there are no absolutes in relativity. Temperature is such an absolute (that is, the absolute temperature of an object, not the apparent brightness temperature of a radiation source), as is the number of sides of a plane figure. All observers will agree on the latter; and all observers will agree the point at which phase changes etcetera take place, and on the position of chemical equilibria. Proper temperature would also be a proper term.

    “OK. In different frames lengths are different and clocks run differently. And I strongly suspect materials melt at different temperatures. But hashing out relativistic thermodynamics here is beyond my expertise or motivation. ”

    No, no, no. Time runs differently. Clocks run exactly the same. The local physics is independent of the local rest frame (it is impossible to determine its location, velocity or potential by purely local measurements).. Materials melt at the same temperature/pressure wherever they are. There is a slight caveat here in that neither the shell nor the planet is in an inertial frame; since pendulum clocks obviously need the same gravity to work correctly, this would have to be supplied (easy enough with a centrifuge).

  369. “By the way, this statement is factually incorrect. I sure observed warm objects getting warmer when placed in the sunlight.”

    But the sun is warmer than the “warm object”! So the “warm object” you are thinking of has become the “cool object” in our discussion, and the sun has become the “warm object”. And it is still completely impossible to have a net heat transfer from a cool object on the earth to the warmer surface of the sun.

  370. “It is a misconception that there are no absolutes in relativity. Temperature is such an absolute … ”
    First of all, I said there are no absolute reference frames, which I do beleive to be correct.
    Secondly, google turns up several papers discussiong effects of relativity on temperature, so I am not convinced that it is “absolute.”

    In any case, this seems like it has become a discussion for a physics/relativity blog, not a climate blog. I am sure there are people more learned in this topic than either of us who could shed much more light on the topic.

  371. Tim Folkerts says:
    December 8, 2010 at 8:38 am
    “First of all, I said there are no absolute reference frames, which I do beleive to be correct.”

    I was talking about the absolute temperatures of the shell and planet. This has nothing to do with absolute frames of reference; and even then it needs to be understood that although velocity is relative, acceleration or gravity is not.

    “Secondly, google turns up several papers discussiong effects of relativity on temperature, so I am not convinced that it is “absolute.” ”

    I don’t know what alleged effects you are referring to, but that local physics is invariant is the fundamental principle of relativity (the constancy of the speed of light in vacuo being just one aspect of that).

    “In any case, this seems like it has become a discussion for a physics/relativity blog, not a climate blog. I am sure there are people more learned in this topic than either of us who could shed much more light on the topic.”

    Speak for yourself! As for its relevance: quite a few previous threads here have misapplied relativity (and quantum mechanics) to climate issues (I’m thinking of various solar, solar neutrino and cosmic ray threads, greenhouse effect discussions and others). In this thread, the last week’s discussion has been based upon people talking past each other with muddled notions of heat and temperature, which a grasp of this simple relativistic scenario could have prevented, since it demonstrates how an object at one temperature can heat another to a higher temperature; so this cannot be what the second law prohibits.

  372. @Tim Folkerts December 8, 2010 at 3:34 am

    Ok, you meant the warmer object, not just warm. Now, will you answer my question posted December 8, 2010 at 1:54 am ?

  373. At this time Tim Folkerts, you really have got into a deep discussion from all sides, haven’t you?
    I’ll just let you know that I have, of course come across both of the “greenhouse theories” you mentioned earlier . What I meant was that the only one that makes any sense, to me – is the one that relates to greenhouses. But I was not clear about that, nor was I clear about what puzzles me about this particular one.

    Look at fig.4 with a shell which blocks long wave radiation but lets short wave in. My question should have been; HOW CAN LONG WAVE (BLUE) RADIATION RETURN AS ENERGY TO SPACE IF THE SHELL IS OPAQUE TO IT?

    If one speculates that the heat is conducted from the shells inside to the shells outside before being radiated back to space, then there will be a delay and one must ask how long does that conductive delay take?
    This is an important question indeed as a correct answer will enable us to calculate how many times the long wave radiation wave (BLUE ARROW) will bounce, at the speed of light, between the shell and the planet’s surface. If one can (as indicated in the drawing Fig.4) double the W/m² for every bounce one may be able to make a very reliable estimate of that planet’s greenhouse warming.

    Then again: is the first delay the only delay?

    Oh and about the answer you gave me, you say; “Atmospheres without GHG’s are very ineffective at radiating IR, so they cannot provide any significant IR toward the surface. “

    Could you please give me one, just one example of such an atmosphere? I.E. an atmosphere devoid of “greenhouse gases” as the comparison may/will prove – or disprove everything concerning AGW.
    You see I believe in most that is so far been discovered about radiation, but I do not believe that radiation from an average temperature of 15 °C, or less is powerful enough to have any impact, even if the basic “Greenhouse Effect Theory” is correct.

    I am looking forward to an answer from you because, yes – some of your answers out there are pretty good.

    Lastly, think about this one: In nature everyting increases or decreases one step or one unit , whatever that unit is, at the time. Now think about temperature (T) and an increase to the power of 4.

    Or as it is often portrayed: “The temperature goes up with 2 °C which means that radiation increase by 2 x 2 x 2 x 2 = 16 times. “Oh yes somebody’s constant is involved as well, – I know -. But it hould be 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1…….. Do you follow? No, well think about it some more.

    But when all is said and done I still cannot accept we know all there is to know about radiation. Of course everybody’s working outs concur. they will every time if everybody are using the same flawed theories!

  374. “At this time Tim Folkerts, you really have got into a deep discussion from all sides, haven’t you?”

    You can say that again! I seem to have hit the trifecta of heavy-duty physics!
    * quantum mechanics (IR absorption by polyatomic molecules)
    * relativity (redshifted photons)
    * thermodynamics (second law/energy transfer from warmer to cooler surfaces)

    “HOW CAN LONG WAVE (BLUE) RADIATION RETURN AS ENERGY TO SPACE IF THE SHELL IS OPAQUE TO IT? ”
    I think the key idea is that the energy does not directly “return to space”. If the shell is at 255 K, it will radiate 240 W/m^2 into space in the form of long wavelength infrared (LWIR). This is calculated with the the Stefan-Boltzmann law. This LWIR is “created” by the surface independent of how the surface got to that temperature.

    Of course, if there was no other energy source, then the shell would start to cool off since it is losing energy. So we need a heater. We could heat the shell with an electric heater. We could heat it with a star radiating 240 W/m^2 of LWIR onto the surface. Or in Fig 4, we could heat the shell indirectly by having the planet absorb 240 W/m^2 of SWIR/visible light. The planet then heats the shell from below with its own LWIR. This LWIR from the planet is absorbed and ceases to exist. Instead it simply becomes thermal energy of the shell. Then the shell can create its own LWIR as in the first paragraph.

    “… will enable us to calculate how many times the long wave radiation wave (BLUE ARROW) will bounce, at the speed of light, between the shell and the planet’s surface. ”
    Again, I don’t think this is the right picture of what is happening. The LWIR doesn’t “bounce back and forth” If we are assuming the surfaces are blackbodies, then the LWIR is completely absorbed when it reaches the surface.

    If the shells are not perfect blackbodies, then some of the light will indeed reflect at each surface. But objects that are not BBs will also emit less light. So there would not be lots of extra IR bouncing around anyway. Changes in emissivity could adjust the equilibrium temperatures a bit, as Paul Birch said November 28, 2010 at 3:56 am

    “If one speculates that the heat is conducted from the shells inside to the shells outside ”
    Yes, the original model does indeed postulate that the shell is a perfect thermal conductor and that it will have the same temperature everywhere. The LWIR heating the inner surface spreads out evenly thru the shell (inside and out, sunny side and dark side). If this is not the case, then there is some “delay” in the energy getting thru the shell and the analysis will get more complicated, which Paul also addressed in the post i just mentioned.

    “Could you please give me one, just one example of such an atmosphere [without GHG's]?”
    I can’t think of any real planet or moon that has such an atmosphere. GHG’s like CO2, H2O and CH4 are common around the universe. But we can certainly measure the IR absorption of various gases and extrapolate what an entire atmosphere of that gas woudl be like as it relates to IR absorption. I would suggest looking here: http://scienceofdoom.com/2010/07/24/the-amazing-case-of-back-radiation-part-two/

    “Or as it is often portrayed: “The temperature goes up with 2 °C which means that radiation increase by 2 x 2 x 2 x 2 = 16 times. “Oh yes somebody’s constant is involved as well, – I know -. But it should be 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1…….. Do you follow? No, well think about it some more.”

    The RATIO of the power increases with the RATIO of the temperature.
    For example, if the temperature goes up with 2 °C from 300K to 302K (1.0067 times hotter), the power increases by a factor of (302^4 / 300^4) = 1.0067^4 = 1.027 times more energy. You would have to DOUBLE the temperature (for example, 300K to 600K) to get 16x power. Google “Stefan-Boltzmann Law” for more info.

    “But when all is said and done I still cannot accept we know all there is to know about radiation. Of course everybody’s working outs concur. they will every time if everybody are using the same flawed theories!”
    Certainly we don’t know all there is to know about radiation (or anything for that matter!). But ultimately the theories have to match observations. We have LOTS of observations of of glowing objects to confirm the basics of Stefan-Boltzmann. (Applying it to a complex system like the atmosphere becomes more of a challenge).

  375. from NY says: December 8, 2010 at 4:38 pm

    Ok, you meant the warmer object, not just warm.

    Sorry if I was not clear. In discussions like this I assume that “warm” means “warmer than the other object”.
    * A “warm” object at -200 C will radiate net energy to a “cool” object at -250 C
    * A “warm” object at 20,000 C will radiate net energy to a “cool” object at 15,000 C

    “How can the warm object absorbing energy stay at the same or lower temperature and radiate more? “

    I think we are facing the same sort of semantics here. There are a couple issues that are interconnected.
    * A warm object will radiate more thermal EM energy per m^2 per sec than a cooler object with the same emissivity. This is the stefan-boltzmann law.
    * A warm object will radiate more thermal EM energy per sec to a cool object than the cool object radiates to the warm object (independent of size, location, emissivity, or mirrors/lenses). This is the 2nd Law of Thermodynamics (ignoring relativistic effects :-) ) .

    So the warm object can absorb energy and cool because it is radiating even more energy than it receives. Even if both objects cool, the warm object is “warmer than the cool object” but “cooler than before”. So it radiates “more energy than the even cooler object” but it radiates “less energy than it did before”.

    Keeping track of precisely what is meant by “warmer” or “more than” is critical, and I was perhaps making too many assumptions that others were interpreting the word the way I was intending in a particular context.

  376. Paul Birch says: December 8, 2010 at 11:35 am
    “In this thread, the last week’s discussion has been based upon people talking past each other with muddled notions of heat and temperature, which a grasp of this simple relativistic scenario could have prevented”

    With all due respect, I disagree. From my view, most of the “muddled notions” are much more basic. For example:
    * “warm” vs “warmer”
    * Multiple claims that Fig 4 is completely wrong in its energy balance.
    * muddled analogies of “truckloads of energy” and hot rocks and electric heaters.
    * misunderstandings about the nature of GHG’s.
    I don’t think relativistic thermodynamics will help clarify those issues!

    I have probably spent too much time trying to explain things in this thread. Part of the problem is background — from PhD’s to “soundbite scientists”. (I just coined that phrase! :-) ) . Another problem is motivation — some are eager to learn while others already have their minds made up. And finally, what gets resolved here will really not matter to more than a handful of people — the same issues will likely come up again next week in a different entry at a different blog with different people.

    But the teacher in me hates to see poor science go by without some discussion …

  377. Tim, you are dodging the question. The Stefan-Boltzmann law states, that “the total energy radiated per unit surface area of a black body per 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.” Look up Wiki for example.

    A body radiates in a vacuum too. The situation we are discussing is having a coin with certain temperature radiating according to its temperature and surface in a vacuum, and then having photons from a colder coin directed to this coin.

    You claim that 1) “Yes, the photons from the cooler object will be absorbed by the warmer object”, 2) “The warmer object gains the energy, but its temperature does not go up.” 3) “Specifically, the warm object absorbs the energy from the photon, but thermodynamics/blackbody radiation laws tells us that the warm object will necessarily be losing more energy from its own outgoing photons than it is gaining from the incoming photons.”

    This contradicts the Stefan-Boltzmann law. The temperature does not go up, yet the warmer coin radiates more than before. That is what I am questioning and you are dodging.

    The problem starts with your acceptance of absorption of photons from the colder body. You don’t explain how this extra energy is going out at the same temperature.

    There is no problem if colder photons are not absorbed.

  378. From NY,

    Are you postulating that the warm object and/or the cold objects have some sort of other heaters? I was assuming the objects were simply warm and slowly cooling as they lost heat. That might be the source of our miscommunication.

    Just to make it more concrete, lets use some numbers. Suppose we have a blackbody sphere with a surface area of 1 m^2 @ T = 303K radiating 480 W/m^2 (the “warm object”) We have available a thin shell blackbody with a surface area of 4 m^2 (twice the radius of the warm object) at a temp of 255 K which would radiate 240 W/m^2 = 960 W total (the “cool object”). (The inner surface will also radiate 960 W, but all of this would be absorbed by other parts of the inner surface.)

    CASE 1: Neither is heated.
    1A) The warm object is in the middle of nowhere. It loses -480 W to space, absorbs 0 W from other sources. It starts at 303 K but cools as it loses -480 W.

    1B) The warm object is surround by the cool object. The warm object still radiates -480 W, all of which goes to the cool object. The warm object absorbs 240 W from the outer shell (the inner surface of the shell radiates 960 W, but 720 W would go back to the shell, while 240 W would go to the warm object. The warm object now has a net loss of only -240 W so it will cool, but not a quickly as before.
    For the cool object, the outer surface radiates away -960 W. The inner surface radiates -960 W as well, but 720W goes back to other parts of the inner surface. The warm object radiates 480 W to the inner surface of the cool object. The cool object has a net flow of -960 W (to space), -240 W to the warm object, and +480 W from the warm object = -720 W net loss. (so it cools more slowly than if it was by itself radiating -960W to space.)

    Over time, all the temperatures and radiation will drop, but there will always be more power warm –> cool than cool –> warm.

    CASE 2: Both have thermostats.
    1A) The warm object is in the middle of nowhere. It loses -480 W to space, so the thermostat causes the heater to provide 480 W. It stays at 303K

    1B) The warm object is surround by the cool object. The heater will provide 240W to the warm object to keep it from cooling. The heater provides 720 W to the cool object. The temperatures of all objects stay the same and the analysis of CASE 1 applies.

    CASE 3: Both have heaters (480W for the hot object; 960 W for the cool object, which would independently maintain the current temperatures).
    1A) The warm object is in the middle of nowhere. It loses -480 W to space, but the heater to provide 480 W. It stays at 303K

    1B) The warm object is surround by the cool object. The heater will provide 480 W to the warm object to keep it from cooling. The net flow for the warm object is -480 W from radiation + 480 W from the heater and + 240 W from the cool object. The warm object starts to warm up.
    The net flow from the cool object is -960 W to space + 960 W from the heater – 240 W to the warm object +480 from the warm object. The net power is +240 W. It also warms up.
    Eventually, the cool object will warm up until it radiates away all the power from the heaters = 960 W + 480 W = 1440 W = 360 W/m^2. The warm object will be getting 360 W from the cool object, and 480 W from the heater, so it will warm until it radiates 840 W/m^2.

    CASE 1B specifically matches my claims that you quoted. Stefan-Boltzmann is satisfied for all cases. 2nd Law of Thermodynamics is satisfied for all cases. There would be a HUGE problem if the photons from the cooler object were not absorbed by the warmer object.

    And if you use just part of the outer shell (eg cut out a coin-shaped section), the results will be somewhere between (A) and (B) but we will still have the appropriate laws satisfied. You could also change the shape of the center object to a coin — the calculations would just be a little more complicated.

    I *think* that covers pretty much all the options and hopefully this no longer “dodges” which ever specific circumstances you were thinking of.

  379. Tim, this is not what we discussed. The case was suggested by bessokeks December 4, 2010 at 9:14 am

    “Imagine two coins opposite to each other. one small and warm, one 10 times bigger and slightly colder. Then put a lens between an direct the photons of the slightly colder coin exactly onto the surface of the smaller coin. You would transfer roughtly 10 times the energy the warm coin radiates onto the warm coin…”

    This is what we were talking about all the time, coins, not spheres. Just read my last post again, please:

    “The situation we are discussing is having a coin with certain temperature radiating according to its temperature and surface in a vacuum, and then having photons from a colder coin directed to this coin.”

    And read the bessokeks’ description for details. Thanks.

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