Energy Balance at the Tropopause

The IPCC defines radiative forcing at the tropopause. However, nowhere do they provide a diagram showing energy balances above the tropopause and below the tropopause – something that seems like one of the first things to do. Instead, they show the Kiehl and Trenberth cartoon which treats the atmosphere as a whole without distinguishing balances above and below the layer said to be critical to radiative forcing calculations Kiehl and Trenberth 1997 online here. (Kiehl and Trenberth 1997 is a very good and interesting article and deservedly is widely cited and relied on.) Willis Eschenbach has attempted to make these estimates and has produced a very interesting calculation and diagram, doing exactly this. His calculation also sheds some interesting light on the IPCC/Houghton explanation of the enhanced greenhouse effect as being due to higher effective radiation from the troposphere.

First here is the IPCC AR4 diagram (see FAQ and chapter 1) based on Kiehl and Trenberth. A similar diagram was used in TAR. This version varies from Kiehl and Trenberth by a few wm-2 here and there, but nothing material. As you see, there is no breakout of absorption above and below the tropopause.

balanc3.jpg
FAQ 1.1, Figure 1. Estimate of the Earth’s annual and global mean energy balance. Over the long term, the amount of incoming solar radiation absorbed by the Earth and atmosphere is balanced by the Earth and atmosphere releasing the same amount of outgoing longwave radiation. About half of the incoming solar radiation is absorbed by the Earth’s surface. This energy is transferred to the atmosphere by warming the air in contact with the surface (thermals), by evapotranspiration and by longwave radiation that is absorbed by clouds and greenhouse gases. The atmosphere in turn radiates longwave energy back to Earth as well as out to space. Source: Kiehl and Trenberth (1997

Next here is Willis’ corresponding diagram first posted here distinguishing between the troposphere and the stratosphere, breaking out these energy balances (assumptions to be discussed below.)

greenhouse.bmp

Willis observes of this diagram:

1) The majority of the radiation emitted to space (147 w/m2 out of 237 w/m2) comes from the lower stratosphere, and not from the troposphere (50 w/m2) or the surface (40 w/m2).

2) The predicted temperature of the lower stratosphere (147 w/m2 and -48°C [226K] ) agrees well with the actual temperature of the lower stratosphere. This is the “cold, high” emission you [Steve] referred to in your head article.

3) All layers (surface, stratosphere, troposphere) are balanced in that they emit what they absorb. The atmospheric layers are balanced in that they emit the same amount both up and down.

4) The emission from the troposphere is at about 0°C (321 w/m2 = ~1°C). This makes sense to me, although I’m not sure I could explain why. It also puts the emission in the lower troposphere where we’d expect it to be.

6) The “temperature sensitivity” is highly dependent on how much of the increased downwelling radiation is turned into evapotranspiration/convection. If 3/4 of the increase comes back out as sensible/latent heat, it takes a whole lot more radiation to raise the global temperature than if only say 1/4 of the increase comes back as sensible/latent heat.

A few comments of my own on this. It is my understanding that, in the Pacific Warm Pool and hot tropical areas that, in a sense, are “heat engines” for the planet, incremental energy is nearly all used for evaporation and that it is hard to raise SST. So it’s not unreasonable to think that this is a very important effect in the tropics at least.

Secondly, let’s use this diagram to re-consider Houghton’s explanation of how enhanced AGW works, previously discussed here. Houghton said:

Absorbing gases in the atmosphere absorb some of the radiation emitted by the Earth’s surface and in turn emit radiation to space. … Radiation is emitted out to space by these gases from levels somewhere near the top of the atmosphere – typically between 5 and 10 k high (See Fig. 2.3) … Let us imagine that the amount of CO2 in the atmosphere suddenly doubled, everything else remaining the same. What would happen to the numbers in the radiation budget? The solar radiation budget would not be affected. The greater amount of carbon dioxide in the atmosphere means that the thermal radiation emitted from it will originate on average from a higher and colder level than before (Fig 2.3).

Below is my re-plotting of Houghton’s Figure 2.3:

As Willis observes below, if one takes a weighted average level for radiation-to-space, the average is in the troposphere. However substantially more (147 wm-2) radiation-to-space originates from the stratosphere as from the troposphere and surface combined (50 wm-2 + 40 wm-2). As we previously noted [see this image], in the stratosphere, temperatures are increasing with altitude. If Willis’ estimates are correct, then a larger proportion of radiation-to-space originates in a region with increasing temperatures with altitude (stratosphere) than in a region with decreasing temperatures with altitude (troposphere). This would make Houghton’s heuristic completely unusable. Note that Houghton’s heuristic was endorsed in TAR as follows (see ref in prior note):

Note that it is essential for the greenhouse effect that the temperature of the lower atmosphere is not constant (isothermal) but decreases with height.

Willis’ spreadsheet for calculating these results is here. When you open the spreadsheet, you’ll probably receive a diagnostic that there is a circular reference. As Willis states in his notes, you have to set the Options for iterative calculation. [I got wrongfooted on this my first try and thought there was a problem.] Here are Willis’ spreadsheet notes:

1 The model is currently coupled so that some of the change in downward radiation is offset by a change in sensible + latent heat loss. This coupling is adjusted by cell E4 (orange). The zero point (in total downwelling radiation at the surface) is set by cell E5 (yellow)

2 Stratospheric and tropospheric absorption are currently coupled, so that a change in tropospheric absorption (cell G25) causes a change in stratospheric absorption.

3 If an impossible value is entered, the model will go off the rails, and I know of no way to get it back on except to close it and re-open it.

4 For the model to work, set the “Calculation” (Mac menu item “Excel/Preferences/Calculation”, PC menu item “Tools/Options/Calculation) to “Iterative” with a limit of 100.

5 Only cells which are colored yellow or orange should be changed. Messing with the others is instant disaster.

6 Arrows /\ | indicate upwelling and downwelling energy flows
| V
7 The calculation of the changes (∆) are made from a copied and pasted “comparison scenario”, located below the model and the “∆” change calculatations, starting at cell A71. If you wish to set up a scenario for comparison, adjust the variables, then do a copy and a “Paste Special/Values Only” in the comparison scenario area.

Again, I’m not saying that Willis’ calculations disproves Houghton’s argument. I’m just feeling our way through the data right now and trying to understand the provenance of various data. It’s quite possible that Willis has got something wrong in his calculations. However, at a minimum, Willis’ diagram showing the energy balances above and below the tropopause seems like a mandatory diagram for people who rely on radiative forcing at the tropopause as a mechanism for analyzing climate change and the apparent absence of such a diagram from peer reviewed literature seems astonishing. If Willis’ diagram is wrong in some respect, it would be instructive to see a Trenberth version and then see exactly how the differences arose.

Reference:
Kiehl and Trenberth 1997 online here).


236 Comments

  1. LadyGray
    Posted Jan 10, 2008 at 9:24 AM | Permalink

    As I look at these diagrams, they always seem to show energy just flowing through various steps, with whatever amount of energy that goes into something also coming out of the something. However, in real life this does not happen. The amount of energy that gets absorbed by molecules is usually an order of magnitude greater than the amount of energy that is emitted by those same molecules. This is easily illustrated by a gas laser, where the amount of energy required to excite the molecules (the energy is absorbed) is much greater than the energy emitted, in the form of light and heat, from those molecules.

    Is there something basic that I’m missing here?

    Steve: Yes.

  2. Gunnar
    Posted Jan 10, 2008 at 9:48 AM | Permalink

    >> Further, as we previously noted [see this image], in the stratosphere, temperatures are increasing with altitude.

    You are of course correct, but I just wanted to point out this nitpick. The image here is a text book cartoon meant to teach the concept. In real life, the temperature rise is only slight. Long ago in a thread far, far away, I posted an empirical version.

  3. Steve McIntyre
    Posted Jan 10, 2008 at 9:55 AM | Permalink

    #2. Gunnar, I’d be interested in a “more empirical” version. The version here is certainly a representative public presentation.

  4. Dennis Wingo
    Posted Jan 10, 2008 at 10:15 AM | Permalink

    Steve

    I am really skeptical of that entire cartoon. They use the radiative average of 342 w/m2 for the incoming radiation when the atmosphere and incoming solar radiation simply do not operate in that fashion.

    When the sun is overhead and the full 1361 w/m2 is burning down from zenith there is one response from the atmosphere that is akin to the cartoon. However, what about the mornings and late afternoons? With a longer path length through the atmosphere the dynamics change. If you doubt that just look at the color of the sun at these times. There needs to be an integration done over the full solid angle of the sun’s path that also takes into account the variation in re-radiation back to space during non zenith conditions.

    Is this wrong?

    Steve: Probably. I have no objection to showing this type of overall balance sheet in a cartoon and I have no reason to presume that the Kiehl and Trenberth numbers are materially incorrect.

  5. Gunnar
    Posted Jan 10, 2008 at 10:22 AM | Permalink

    I found the link that I previously posted:

    http://www.ux1.eiu.edu/~cfjps/1400/atmos_struct.html

    However, I was wrong. I misinterpreted the graphs on that page (not noticing that one graph had miles on the RHS). So, the graph you linked to is correct. The empirical version is still interesting because it shows how it changes with latitude. Of course, it’s quite different at night, when I believe the thermosphere temperature collapses.

  6. Gunnar
    Posted Jan 10, 2008 at 10:39 AM | Permalink

    >> When the sun is overhead and the full 1361 w/m2 is burning down from zenith there is one response … Is this wrong?

    I say you are not wrong. I think it would be helpful to draw one cartoon for daytime, and one for night. The average output of a system is not equal to the output of the system with average inputs. The average temp is determined from the high and low. The highs and lows are determined from alternating direct exposure to the sun and space. If one considers the earth as one object, then the sun is always shining on it. Imagine a pool that is rotating, being filled with a hose that is fixed. The rotation doesn’t change the fact that the hose is adding water to the pool.

  7. SteveSadlov
    Posted Jan 10, 2008 at 10:50 AM | Permalink

    It is exceedingly important, what is raised in this post. Thanks Steve M for putting this up. Held’s models appear to assume more or less a closed loop, in terms of energy flows in the “upper limbs” of the Hadley Cells. In this construct, energy flux is near vertical in / near the ITCZ, then top out at the “high points” of the tropical tropopause, then undergo a slow descent along the poleward “slope” of the tropopause (of course spiraling around clock wise ever more toward the east) then subsides in the mid latitudes. Then, energy coupling to Ferel cells eventually moves the energy to the polar regions. However … if there is a “fan out” of energy flux in the tropical latitudes of the tropopause, then such models would be incorrectly posed. If a significant quantity of energy flux is shunted aside in the tropics, up into the stratosphere, into the area of “preferred” flux into space, then we must reconsider the “classical” notions regarding poleward transport. All of the above diagrams appear to depict all sensible and latent heat being absorbed by the troposphere and yet, it has been known for at least 50 years that strong tropical convection results in bubbles / spikes air passing through the tropopause into the lower stratosphere. This can occur both in and around the ITCZ as well as nearby the subtopical jet streams. (The corresponding structures of such events make for some truly eerie and beautiful cloud photos of the upper portions of these thunderheads).

  8. SteveSadlov
    Posted Jan 10, 2008 at 10:58 AM | Permalink

    Clarification – I am referring purely to the “circulatory” energy flux related to Hadley cells, not radiation. So, I referring to a component or small set of them.

  9. AK
    Posted Jan 10, 2008 at 11:06 AM | Permalink

    Re #7, this was the point I was going to make in a more general fashion. There is a good deal of exchange of air between the troposphere and the stratosphere with accompanying forced convection. Thus the assumption of separate radiative balance for troposphere and stratosphere is not really valid. The question is whether this exchange can be considered negligible.

    IMO the tropopause should be regarded not as a barrier to non-radiative heat transport but rather a region of qualitatively different kind of transport.

    IMO there should be a difference between explanatory cartoons and models (which can be criticized for things other than clarity).

    As for energy balance, IMO it should be considered separately for each wavelength, since the effective radiative height varies with wavelength. This is especially true for understanding the effects of CO2 changes, since these (effects) also differ by wavelength.

  10. MarkW
    Posted Jan 10, 2008 at 11:15 AM | Permalink

    Gunnar,

    For that matter there would be different diagrams for the tropics and the poles.

    The tropics get more sunlight (especially in polar winter). The poles get heat transported in which they then radiate away.

  11. Cliff Huston
    Posted Jan 10, 2008 at 11:28 AM | Permalink

    RE: #3 Steve McIntyre said:

    I’d be interested in a “more empirical” version. The version here is certainly a representative public presentation.

    In another thread Geoff Sherrington posted a link to sides for a tropopause lecture which included this slide of atmosphere temperatures:

    For Geoff’s link to the whole set see this post: http://www.climateaudit.org/?p=2572#comment-193772
    (If the above image does not show, it is on page 4 in the document that Geoff linked)

    Cliff

  12. D. Patterson
    Posted Jan 10, 2008 at 11:39 AM | Permalink

    SteveSadlov says:

    January 10th, 2008 at 10:50 am
    [....]All of the above diagrams appear to depict all sensible and latent heat being absorbed by the troposphere and yet, it has been known for at least 50 years that strong tropical convection results in bubbles / spikes air passing through the tropopause into the lower stratosphere. This can occur both in and around the ITCZ as well as nearby the subtopical jet streams.[....]

    Most attention regarding transfers of chemical species and heat between the troposphere and the stratosphere has been directed towards overshooting tropical convective activity. However, there are also the much neglected transfers occurring as a consequence of the atmospheric wave activity exploiting the differentials in altitude and baroclinic properties between adjacent tropopauses. Also unclear is the extent to which researchers do and do not understand the role of jet stream vorticity in the entrainment of tropospheric air into the strotosphere.

  13. John Creighton
    Posted Jan 10, 2008 at 12:37 PM | Permalink

    Averages behavior in terms of energy flows is a perfectly valid way to describe the system because energy adds linearly.

  14. AJ Abrams
    Posted Jan 10, 2008 at 1:21 PM | Permalink

    Can anyone give me a reason that the polar trop has less temperature variation than other latitudes in the above graphic? Are we talking large convection differences in the area of 300-100 hPa?

  15. SteveSadlov
    Posted Jan 10, 2008 at 1:27 PM | Permalink

    RE: #12 – D. Patterson, thanks for pointing this item out, which I had overlooked. There have been transects done by R&D aircraft, attempting to characterize the tropopause, E to W, across CONUS. Especially interesting where the jet stream passes over the Rockies.

  16. Gary Gulrud
    Posted Jan 10, 2008 at 1:28 PM | Permalink

    Do you really expect this gambit to work?

  17. SteveSadlov
    Posted Jan 10, 2008 at 1:36 PM | Permalink

    Gambit? Is that all it is to you? Far be it for anyone to actually want to improve scientific understanding of the atmosphere. If it is a gambit, then why does even NASA, who I routinely lambaste, nonetheless continue to devote research dollars to this area? Please explain that.

  18. xtronics
    Posted Jan 10, 2008 at 2:09 PM | Permalink

    From Willis’ spreadsheet notes:

    The atmospheric layers are balanced in that they emit the same amount both up and down.

    Is this exactly true? Yes and no – It is important to note that if you are in the stratosphere the “sky is bigger than the ground” as an angular field. Not sure it adds up to much, but it would introduce an error. Should be easy to run the numbers.

    Steve:
    this issue has been raised elsewhere and is immaterial at the level of this cartoon.

  19. Willis Eschenbach
    Posted Jan 10, 2008 at 2:38 PM | Permalink

    Steve M., you say:

    According to Willis’ calculations (which I have not verified), nearly three times as much (147 wm-2) radiation-to-space originates from the stratosphere as from the troposphere (50 wm-2). If this is correct, then Houghton’s statement that radiation-to-space from these gases is typically “from levels 5 to 10 km high” is incorrect. It seems implausible that Houghton would goof on something like this. But it’s worth checking. If Houghton’s number is correct, then Willis has goofed somewhere and this would be worth figuring out. Willis’ spreadsheet documenting his calculation is here; Houghton provided no documentation, support or reference for his assertion.

    I don’t see any discrepancy between my figures and Houghton’s. It appears that Houghton has averaged the outgoing radiation into one single radiation level.

    My calculation, on the other hand, splits up the outgoing radiation into stratospheric and tropospheric, with the stratospheric about 3/4 of the total. The contribution from the stratosphere is at 147 w/m2, while the the contribution from the troposphere is at 321 w/m2. The weighted average of these is 190 w/m2, which equates to about -32°C, This in turn implies an average emissions altitude of about 7,500 feet, which is in agreement with Houghton.

    w.

    Steve: good point. I’ll amend the post to reflect this.

  20. AK
    Posted Jan 10, 2008 at 2:52 PM | Permalink

    Re: #14

    A brief description of the tropopause behavior may be found here. Note that latitudinal variation wrt height in both cartoons above (as well as the one in the linked text) is averaged. The actual height varies over latitude, longitude, season, and the location of the jet stream.

  21. SteveSadlov
    Posted Jan 10, 2008 at 3:02 PM | Permalink

    http://acd.ucar.edu/~andrew/papers.html

    One person at UCAR. This barely scratches the surface. There still so much that we do not know.

  22. John G. Bell
    Posted Jan 10, 2008 at 3:03 PM | Permalink

    If you ignore refraction, given a random orientation, your chances of being pointed at the earth from some point in the stratosphere is about 47%. That is if I’ve done the math right :). Please check!. So shouldn’t the radiation emitted by the stratosphere to the sky be 155 w/m2 and the back radiation from the stratosphere be about 138 w/m2?
    This with a cloudless sky.

  23. AJ Abrams
    Posted Jan 10, 2008 at 3:13 PM | Permalink

    AK, thanks for the link. So in short I was correct, but not for the reasons that came to mind at first. Convections, but not from specific mixing with the strat, but instead from expansion. Great link.

  24. Posted Jan 10, 2008 at 4:02 PM | Permalink

    #19 Willis, however, given your model asserts the NECESSITY of more than
    one shell, and Houghton/IPCC does not, there seems to be opportunity for
    a falsification of either theory. One could imagine a list of
    predictions of the 2 shell theory and a corresponding list of
    predictions of a one shell theory and put them in a table for
    comparison with a third column, the observed measurements.
    So Houghton could quite will be proved wrong, by ommission of
    a essential feature.

  25. AK
    Posted Jan 10, 2008 at 4:16 PM | Permalink

    I must admit I skimmed the original post, and missed one of the assumptions:

    3) All layers (surface, stratosphere, troposphere) are balanced in that they emit what they absorb. The atmospheric layers are balanced in that they emit the same amount both up and down.

    In fact, the surface and troposphere have enormous conductive heat flow between them (both sensible and potential), and cannot by any stretch of the imagination be regarded as only radiatively coupled.

    Re: #24, I don’t think you could get worthwhile predictions with either “model”, these should only be used as simple explanations of more complex models.

  26. DocMartyn
    Posted Jan 10, 2008 at 4:20 PM | Permalink

    Could you prepare a pair of diagrams showing what happens at 1 O’clock in the morning and in the afternoon. The idea of “averaging” the day and night cycle makes no sense to me at all; in the same way that an average of the pressure/temperature of the gas in the cylinder of an internal combustion engine makes no sense.

  27. Willis Eschenbach
    Posted Jan 10, 2008 at 4:34 PM | Permalink

    AK, you say:

    In fact, the surface and troposphere have enormous conductive heat flow between them (both sensible and potential), and cannot by any stretch of the imagination be regarded as only radiatively coupled.

    Yes, they do. Note that these flows are specifically included in both the graph and the spreadsheet.

    w.

  28. AK
    Posted Jan 10, 2008 at 5:12 PM | Permalink

    Re: 27

    Willis, you seem to me to be using two different definitions of the word “emit”. You can’t have it both ways with one definition: either the troposphere is not emitting the same amount as it absorbs, or the surface is not.

    I tend to think of “emitting” as radiative (excluding conduction), but either definition is OK as long as it’s consistent. As far as I can tell, it’s not in this assumption.

  29. AK
    Posted Jan 10, 2008 at 5:16 PM | Permalink

    Re: #27,28

    Never mind, I get it. It works as long as you assume only conduction from the surface to the troposphere, which is only true for a global average.

  30. Willis Eschenbach
    Posted Jan 10, 2008 at 5:23 PM | Permalink

    David Stockwell, you raise an excellent point:

    #19 Willis, however, given your model asserts the NECESSITY of more than one shell, and Houghton/IPCC does not, there seems to be opportunity for a falsification of either theory.

    The question seems quite simple to me. The part that people seem to have missed is that a single shell greenhouse can only heat up the planetary surface to a maximum theoretically possible temperature of 2 * S, where “S” is the incoming solar radiation after albedo. For the Earth, S is about 235 w/m2, and twice that is 470 w/m2. That’s all of the energy theoretically available with a perfect system with no losses.

    Since the temperature of the earth is about 390 w/m2, this means that only 80 w/m2 is available for all possible losses. But we know that we have the following approximate losses in the system:

    68 w/m2 of incoming solar absorbed in the atmosphere
    22 w/m2 of sensible heat lost from the surface to the atmosphere
    76 w/m2 of latent heat lost from the surface to the atmosphere

    This is a total of about 166 w/m2 in losses. Since with a single shell greenhouse there is only 80 w/m2 available for all possible losses, this means that either our estimates of the losses is way, way too large (very doubtful), or that a single shell greenhouse system is not a feasible model for the earth’s greenhouse system.

    I say that it is very doubtful that the estimate of the losses is way too large because there is either good experimental or theoretical data for the numbers. The amount of solar absorbed in the atmosphere has been measured directly in a variety of places, times, and manners. The total evaporative heat loss can be estimated through worldwide estimates of rainfall or evaporation, which give similar answers. The ratio between the amount of sensible and latent heat (the “Bowen ratio”) has been measured experimentally. See the Kiehl Trenberth (K/T) paper for the derivation of these estimates.

    So overall, while these loss figures might be out by 10-20%, it is extremely unlikely that they have been overstated by more than 100%. This makes the single shell model physically impossible.

    w.

  31. Willis Eschenbach
    Posted Jan 10, 2008 at 5:33 PM | Permalink

    Steve Sadlov, you say:

    All of the above diagrams appear to depict all sensible and latent heat being absorbed by the troposphere and yet, it has been known for at least 50 years that strong tropical convection results in bubbles / spikes air passing through the tropopause into the lower stratosphere.

    You are correct, the diagram is a bit simplified in that regard. However, in the spreadsheet itself, I include the sensible/latent transfer from the surface through the tropopause to the stratosphere.

    w.

  32. SteveSadlov
    Posted Jan 10, 2008 at 6:24 PM | Permalink

    RE: #31 – Willis, very interesting indeed. This is highly productive work!

  33. Posted Jan 10, 2008 at 6:53 PM | Permalink

    #30. Willis, so you could say roughly that the probability of Houghton’s
    description being correct is a 5-Sigma event or over 1 in 3 million
    chance of being correct.

    So overall, while these loss figures might be out by 10-20%, it is extremely unlikely that they have been overstated by more than 100%. This makes the single shell model physically impossible.

  34. Black Wallaby
    Posted Jan 10, 2008 at 7:04 PM | Permalink

    Trenberth Cartoon funnies(Part 1)

    If we do a HEAT transfer calculation from the surface, it is simple: 390 -324 = 66. Checking that result: (via HEAT in = HEAT out): 168 -24-78 = 66…OK.
    Thus in this calculation the back radiation, (EMR), is a negative quantity, BUT it is treated as a positive. That is to say, Trenberth shows EMR of 324 from a cold source, HEATING the warmer surface!

    We also know that if there is free passage from any source, EMR, (which is not HEAT), radiates in all directions, including straight out of the cartoon and into your face. So what happens to all the other stuff apart from the up and down? Is it merrily HEATING everything?

    The heat loss from the surface via EMR of 66 is also confirmed by a NASA diagram, which is fundamentally the same as Trenberth’s, except that the “simplistic” depiction of the greenhouse effect is omitted.

    When I say “simplistic”, I mean it in the sane way that some climatologists brazenly state that the greenhouse effect is 33C!

    Does anyone disagree with the above, part 1?

  35. Willis Eschenbach
    Posted Jan 10, 2008 at 7:07 PM | Permalink

    DocMartyn, you say:

    The idea of “averaging” the day and night cycle makes no sense to me at all; in the same way that an average of the pressure/temperature of the gas in the cylinder of an internal combustion engine makes no sense.

    I would respectfully disagree entirely. The average temperature and pressure of the gas in the cylinder of an internal combustion engine can tell us a lot about the engine. The average cylinder pressure tells us how much energy is being transferred to the piston, while the average temperature and pressure taken together give us a measure of the thermodynamic efficiency of the engine. In addition, certain combinations of average temperature and pressure will indicate problems with the engine (loss of cooling fluid, pressure leaks, bad fuel, etc.) In fact, there is a host of valuable information in the averages provided we approach them with an open mind.

    Certainly it is possible to do much more complex evaluations and investigations of the climate, some of which would necessitate a day vs night analysis. However, that does not mean that less complex analyses are not also valuable.

    w.

  36. Black Wallaby
    Posted Jan 10, 2008 at 7:18 PM | Permalink

    Some funnies in the Trenberth Cartoon

    If we do a HEAT transfer calculation from the surface, it is simple: 390 -324 = 66. Checking that result: (via HEAT in = HEAT out): 168 -24-78 = 66…OK.
    Thus in this calculation the back radiation, (EMR), is a negative quantity, BUT it is treated as a positive. That is to say, Trenberth shows EMR of 324 from a cold source, HEATING the warmer surface!

    We also know that if there is free passage from any source, EMR, (which is not HEAT), radiates in all directions, including straight out of the cartoon and into your face. So what happens to all the other stuff apart from the up and down? Is it merrily HEATING everything?

    The heat loss from the surface via EMR of 66 is also confirmed by a NASA diagram, which is fundamentally the same as Trenberth’s, except that the “simplistic” depiction of the greenhouse effect is omitted.

    When I say “simplistic”, I mean it in the sane way that some climatologists brazenly state that the greenhouse effect is 33C!

    Does anyone disagree with the above, part 1?

  37. Willis Eschenbach
    Posted Jan 10, 2008 at 7:18 PM | Permalink

    Black Wallaby, welcome to the discussion. You say:

    That is to say, Trenberth shows EMR of 324 from a cold source, HEATING the warmer surface!

    Umm … don’t know why this rates an exclamation point. If we have two objects of different temperatures, they will both heat each other by radiation. The sun warms the earth by radiation, but at the same time, the earth also warms the sun by radiation. If the earth did not exist, the sun would be cooler … seems counterintuitive, I know, but it’s true. The energy that strikes the earth from the sun would otherwise be lost to space … but with the earth there, some of the sun’s energy returns to the sun as radiation from the earth, and makes the sun warmer.

    It is only the NET flow of energy that goes from hot to cold. Individual radiation flows, on the other hand, add energy to whatever they strike regardless of the temperature of the receiving substance.

    w.

    PS – for a most curious form of energy transport, consider that it is not always heat that flows in natural systems, sometimes it is cold that flows in natural systems. Not only that, but cold it can flow in either direction (warmer to colder, or colder to warmer). Go figure … nature is full of surprises.

  38. Larry
    Posted Jan 10, 2008 at 7:32 PM | Permalink

    Hey guys, conduction doesn’t amount to diddly squat in the atmosphere. Convection and radiation dominate in different parts, but the thermal conductivity of air (or any gas) is so small, that if you could get the stuff to stay still, it wouldn’t amount to anything. Why do you think foam is such a good insulator?

  39. Posted Jan 10, 2008 at 8:16 PM | Permalink

    Powerpoint which illustrates the irregular nature of the tropopause, for background.

  40. Ian McLeod
    Posted Jan 10, 2008 at 8:39 PM | Permalink

    Willis, you said in #37

    PS – for a most curious form of energy transport, consider that it is not always heat that flows in natural systems, sometimes it is cold that flows in natural systems. Not only that, but cold it can flow in either direction (warmer to colder, or colder to warmer). Go figure … nature is full of surprises.

    What are you saying here? Surely your not saying cold can flow to hot. Are you referring to radiation. Sorry, I’m not following.

  41. John Creighton
    Posted Jan 10, 2008 at 9:58 PM | Permalink

    Willis Eschenbach, I’m looking at the spreadsheet and it doesn’t look like the energy which is absorbed is re-emitted. How come?

  42. John Creighton
    Posted Jan 10, 2008 at 10:03 PM | Permalink

    Oh, sorry, Ignore post #41. I found where it sees emitted.

  43. John Creighton
    Posted Jan 10, 2008 at 10:06 PM | Permalink

    I think I found something missing though. I don’t see anywhere in the spreadsheet where a percentage of the radiation reflected off the troposphere is absorbed in the stratosphere.

  44. Willis Eschenbach
    Posted Jan 10, 2008 at 10:06 PM | Permalink

    Ian, I wondered when someone was going to ask that. Congratulations for being the first, the willingness to ask is a valuable asset.

    And indeed, cold can flow to hot.

    It involves that most curious of all weather phenomena, hydrometeors. These include rain, snow, graupel, sleet, hail, and the like. With these, Nature does a most bizarre trick. By enlisting gravity, hydrometeors can move frozen water from aloft to the ground. As we know, the snow or hail will often melt when it hits the ground … and thus we have cold flowing instead of heat, and naturally flowing from a colder region to a warmer region.

    Go figure …

    w.

    PS – I consider this the gaping hole in the K/T energy budget … but I haven’t figured out any way to estimate the amount of heat moved (or cold removed, or however you want to say it), so I have ignored it. Any ideas on amounts of energy moved gladly accepted.

  45. John Creighton
    Posted Jan 10, 2008 at 10:41 PM | Permalink

    PS – I consider this the gaping hole in the K/T energy budget … but I haven’t figured out any way to estimate the amount of heat moved (or cold removed, or however you want to say it), so I have ignored it. Any ideas on amounts of energy moved gladly accepted.

    I’m not sure what you are saying.

  46. John Creighton
    Posted Jan 10, 2008 at 10:44 PM | Permalink

    Well, here’s one conclusion from Willis’s model. If I set the percentage of the IR radiation by the troposphere to 100% then the temperature goes up 4.59
    degrees.

  47. Ian McLeod
    Posted Jan 10, 2008 at 10:53 PM | Permalink

    Willis #44

    I accept what you say but I am struggling with the thermodynamics. I suppose when it hails, cold precipitation travels from a cold region in the troposphere to the surface where it is warmer. And you say that cold is flowing to hot … Damn, I’m not going to sleep tonight Willis. Thanks a lot.

  48. Black Wallaby
    Posted Jan 10, 2008 at 10:53 PM | Permalink

    Hi Willis, Reur 37
    1) I’m not sure if you understand the difference between EMR and HEAT. Just to clarify that a bit, are you saying that two bodies at the same T, and hypothetically isolated from any other effects, and radiating at each other as a consequence of their T, are HEATING each other. If so, why does their T not increase? (You might also contemplate a layer of absorptive gas at a constant T)
    2) I was a bit stunned by your concept of COLD flowing like HEAT. For instance, temperate ice contains almost as much heat as tepid water. It does not contain any energy known as COLD, or negative HEAT. It became a bit clearer where you were coming from with your 44, but there is nothing mysterious about it. Yes of course, water can be pumped uphill, and similarly HEAT can be pumped with additional energy to where it cannot naturally flow.
    Regards Black Wallaby

  49. John Creighton
    Posted Jan 10, 2008 at 11:00 PM | Permalink

    #48, anyway, my response to my observation #48 is that the energy emmitted shouldn’t be symmetric. More on this later.

  50. Andrey Levin
    Posted Jan 10, 2008 at 11:22 PM | Permalink

    Willis Eschenbach:

    I admire your efforts to make sense from IPCC radiation flowchart. It is important to clean it up, however limited its utility is.

    Couple of thoughts.

    On Trenberth and your charts it appears that solar irradiation (depicted in yellow) is in visible and UV specter alone. However, it is not the case. 45.2% by energy of solar constant of 1366 W/m^2 (averaged over Earth surface 1366/4=342 W/m^2) is in IR, about 44.8% in visible, and about 10% in UV. As such, solar and surface IR radiation appears to receive very different treatment in atmosphere in both charts.

    Take a look at Solar Radiation page at Wiki, for example:

    http://en.wikipedia.org/wiki/Solar_constant#Solar_constant

    and pay attention to Solar Radiation Spectrum Fig. Measured at clear sky and averaged over globe and year amount of radiation in all specter received by Earth surface (depicted in red) is substantially less than averaged total solar irradiation of 342 W/m^2, as it should be. It is important to note, that it is total measured value from all sources, Sun and back radiation of atmosphere altogether. Yet on presented charts average total irradiance received by Earth surface with shielding effect of clouds is 169+30+321=520 W/m^2. This discrepancy is, in fact, root of all grudging over violation of 2-nd law. Couple of times I asked if somebody have data on measured radiative specter at Earth surface facing up at cloudless tropical night, to put this discrepancy to rest, but to no avail…

    Couple of thoughts about applicability of this flowchart to purposes different from educational.

    1. Physical laws describing black-body radiation are not applicable to gases at temperatures found in atmosphere. One example is fluorescent bulb, having effective radiative temperature of 4500K, while actual temperature is barely 350K. Effective radiative temperature of troposphere or stratosphere you calculate are abstractions, and do not directly correspond to actual air temperatures.

    2. All parameters found in your chart are highly averaged. Solar irradiance at apogee/perigee (1321/1412 W/m^2), surface and air temperatures over night/day, over summer/winter, over desert and ocean, over tropics and polar regions. Differential equations and formulas describing heat transfer and radiation fluxes are applicable only to actual physical bodies having defined temperature and other parameters, and do not work on abstract “averages”. Averaged heat flows, temperatures, and radiation fluxes in the global flowchart you are assembling could not be plugged afterwards into such equations and formulas to derive some conclusions.

  51. Black Wallaby
    Posted Jan 10, 2008 at 11:28 PM | Permalink

    Ian,
    Fret not. Just as a refrigerator pumps heat from cold to hot, so precipitation can pump heat of a lower level to the warmer surface below. The driving force in this particular pump is gravity
    Regards, Black Wallaby

  52. Ian McLeod
    Posted Jan 10, 2008 at 11:45 PM | Permalink

    Willis,

    A couple quick questions about your model. You use the Stefan-Boltzmann equation for calculating temperature, the constant is 5.675×10^-8 W/(m^2.K^4), you convert to degrees C with -273.15 K, and so forth. What about emissivity? Are you saying Kirchhoff’s Law applies here so the surface emits what it absorbs? Just wondering.

    I see that you divided by 2 in the latent heat cell and then multiply by 0.8. I understand the division by 2, but not the .8.

    Okay, I see you also multiply the sensible heat by .2. So, 20% sensible, 80% latent?

  53. Ian McLeod
    Posted Jan 10, 2008 at 11:50 PM | Permalink

    Thanks Black, Willis caught me off guard and its getting late. In fact, too late to wrestle with thermodynamic semantics.

  54. John Creighton
    Posted Jan 11, 2008 at 12:13 AM | Permalink

    All right it’s late. My latest thoughts for non symmetric equations of emitted radiation.
    Let
    y1=the upward IR radiation from the surface through the troposphere
    y2=the downward radiation to the surface through the troposphere
    (1-exp(-a))*100%=The percentage of the upward radiation absorbed via the troposphere

    Then

    dy1/dz=-0.5*a y1 + 0.5*a y2
    dy2/dz= 0.5*a y1 – 0.5*a y2

    y1(0)=The upward radiation from the surface
    y2(0)=The downward radiation hitting the surface
    y1(1)=The amount of radiation re-emitted into the stratosphere
    y2(1)=The amount of downward radiation re-emitted from the stratosphere into the troposphere

  55. Posted Jan 11, 2008 at 12:15 AM | Permalink

    #52 Andrey, I disagree with the view that simple models are limited because they exclude details, even night/day seasonality etc. Its up to people who complain about this to show that inclusion of this feature would change the AVERAGE result. Even if it does change it, a well constructed average model will only change minimaly as you add temporal and spatial fluctuations. This demonenstrates its usefulness even more, because it provides an abstract framework against to measue deviance due to other factors.

    That is, a model is not wrong or limited simply BECAUSE it doesn’t include some variability or feature of the real world. This view has lead to a lot of the element simulation silliness that pervades climate sciences. E.g. the view that GCM are the minimal model complexity needed to prove anything. It should be shown that such detail is necessary, and still the abstract model that presents the minimum necessary provides conceptual understanding.

    For example, Einstein always started with a very simple abstract model of the phenomena that he could explain to a child, such as chasing a beam of light, and then built up from there. It got complicated fairly fast, sure, but the basic model was always simple.

  56. Willis Eschenbach
    Posted Jan 11, 2008 at 12:39 AM | Permalink

    Ian, you say:

    Willis #44

    I accept what you say but I am struggling with the thermodynamics. I suppose when it hails, cold precipitation travels from a cold region in the troposphere to the surface where it is warmer. And you say that cold is flowing to hot … Damn, I’m not going to sleep tonight Willis. Thanks a lot.

    You are more than welcome, anything to assist a friend. Black Wallaby, on the other hand, said:

    I was a bit stunned by your concept of COLD flowing like HEAT. For instance, temperate ice contains almost as much heat as tepid water. It does not contain any energy known as COLD, or negative HEAT. It became a bit clearer where you were coming from with your 44, but there is nothing mysterious about it. Yes of course, water can be pumped uphill, and similarly HEAT can be pumped with additional energy to where it cannot naturally flow.

    Bro’, we’re moving ICE. Now to me, ice is not HEAT. It is COLD. (I am perfectly aware of the fact that there is no such thing as “cold energy”.) It is the cold sleet coming down in rivers. Now, you can consider that a flow of heat in the other direction if you wish. I consider it a flow of cold.

    And yes, water can be pumped uphill, and refrigerators can cool rooms. But how often does water flow uphill in nature? About as often as cold flowing from a colder to a warmer place … except in the marvelous solar powered global air conditioning machine that we call the climate.

    An interesting thought, the global airconditioning machine. The greenhouse effect is efficient at warming the surface. Without losses, we’d be cooking. All of the phenomena we usually call weather, like storms and wind and clouds and rain and snow, all of those are parasitic losses whose effect, one and all, is to cool the earth’s surface. Wind cools us. Rain and snow cool us. The clouds cool us. Storms are cold. And all of these are driven by some local/regional/temporal temperature differential (∆T). As soon as a hot spot forms, the air rises, and the wind comes in. This wind is cooler air from outside, plus it helps evaporate any water. All of this cools the hot spot down. Some of the excess heat is radiated away, but much more is moved as evaporation and convection. This is especially true in the tropical ocean where I live.

    And if that spot continues to heat up, poof! A cloud forms above it, and the fun begins. A very efficient heat pipe is soon formed by a pillar of cloud that extends vertically. Outside it looks placid. Inside, air is rising at up to 1000′ (300m) per minute. Hot moist light air is taken up from the surface and sped upwards, first condensing into cloud, and then rain, and then perhaps freezing, on the way up. It is like a hose, which is picking up wet air at the surface and spewing out cool reduced humidity air near the top of the troposphere, after wringing out all the moisture and using it to cool the surface. Each cloud is a heat engine that is providing a parasitic loss path for surface energy to escape to high altitude, while simultaneously cooling the surface with refreshing showers. To me, that’s a wonderfully ingenious heat engine. Two completely different and complementary ways to cool the surface – move the heat up, and at the same time move the cold down.

    Now the observant will have noted that running a heat engine off a hot spot on the earth’s surface, while simultaneously cooling that same surface is … umm … won’t work. And you’re right. In calm weather, when a cloud tower forms perfectly vertical, the cold rain falling directly underneath soon cools the hot spot, and the cloud dissipates. The cloud can only grow large when the cloud tower is slanted, so that the rain falls away from the hot spot that is driving the heat engine. But I digress. I was speaking of hydrometeors of all types being part of the global cooling system that we call “climate”.

    I was standing on my porch the other evening here at 9 South, listening to the roar of the rain coming from a nearby ridge. Shortly after the rain started, the “rain wind” arrived. This is the cooled (by evaporation) air that is entrained with the rain and plunges to the earth. On a muggy tropical evening, the rain wind is wonderfully cool … and again I marvel at the solar powered air conditioner, it’s blowing cool air in my face in the middle of the tropics.

    Gotta run, more later …

    w.

  57. DeWitt Payne
    Posted Jan 11, 2008 at 1:38 AM | Permalink

    I’m not sure there is anything critical about the choice of the tropopause to calculate forcing other than it’s where the optical density of the atmosphere for thermal IR becomes low enough that the majority of radiation emitted upward escapes to space and radiative heat transfer dominates. Also, the atmosphere above the troposphere has a very short equilibration time constant (on the order of months) since there isn’t much thermal mass and radiative heat transfer is rapid.

    Another important feature of the energy balance is that the tropics absorb more heat from incident solar radiation than is emitted to space by the surface and atmosphere while the higher latitudes (greater than 45 degrees) emit more heat than is absorbed. So a significant amount of heat is constantly being transferred to higher latitudes by ocean and air circulation. This (and the areas covered by liquid water year round) makes the system behave almost like an isothermal body (and raises the average temperature) even though the temperature difference between the poles and the equator is still quite large.

    Looking at the measured IR emission spectra, I would say that the majority of radiation emitted to space comes from the upper and middle troposphere, rather than the lower stratosphere. Radiation intensity from the 15 micrometer CO2 band and 10 micrometer ozone band is consistent with emission from the tropopause or lower stratosphere (220 K), but much of the rest of the emission has a higher brightness temperature implying a lower altitude. The radiation from water vapor, for example, will be emitted much lower because the scale height (the altitude where the pressure is 1/e of the surface pressure) of water vapor is only 2 km compared to 8 km for non-condensible gases like CO2.

    I think some knowledge of Physical Meteorology in addition to radiative transfer physics is necessary to begin to understand greenhouse theory. Some understanding of classical thermodynamics is, unfortunately, unavoidable in order to understand Physical Meteorology.

  58. Black Wallaby
    Posted Jan 11, 2008 at 4:02 AM | Permalink

    Hi Willis, Reur 58
    1) I’m not sure if you understand the difference between EMR and HEAT. Just to clarify that a bit, are you saying that two bodies at the same T, and hypothetically isolated from any other effects, and radiating at each other as a consequence of their T, are HEATING each other. If so, why does their T not increase? (You might also contemplate a layer of absorptive gas at a constant T)
    2) I was a bit stunned by your concept of COLD flowing like HEAT. For instance, temperate ice contains almost as much heat as tepid water. It does not contain any energy known as COLD, or negative HEAT. It became a bit clearer where you were coming from with your 44, but there is nothing mysterious about it. Yes of course, water can be pumped uphill, and similarly HEAT can be pumped with additional energy to where it cannot naturally flow.
    Regards Black Wallaby

  59. Gord
    Posted Jan 11, 2008 at 4:21 AM | Permalink

    Why are Vectors not taken into account in the Energy Budget Diagrams?

    1. Vector addition of fields…

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html#c3

    2. Radiation from a dipole source..and cancellation of sound..and an animation

    http://www.kettering.edu/~drussell/Demos/rad2/mdq.html

    3. Field Lines for a Magnetic Dipole…

    http://www.earthsci.unimelb.edu.au/ES304/MODULES/MAG/NOTES/fieldl.html

    The field is zero along the plane of the dipole.

    4. Cancellation of Light

    http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html

    http://webexhibits.org/causesofcolor/15F.html

    http://micro.magnet.fsu.edu/optics/lightandcolor/interference.html

    All these links demonstrate constructive and destructive Vector additions of fields.

    This phenomena is taught throughout ALL areas of Physics and Engineering…is Climate Science any different?

  60. Geoff Sherrington
    Posted Jan 11, 2008 at 4:26 AM | Permalink

    Re Willis # 58 and DeWitt Payne # 59

    Willis, some months back on CA I introduced an analogy with the refrigerator which is similar to your air conditioner. The essential feature was that work was done, with opposite effect, in two (or more) places that were separated. The compressor on one hand and the evaporator/freezer on the other. There can be various types of connectors between them, some with large loss and some with little loss. The separation is the important factor, otherwise you have an entity sitting still, stewing in its own juice. Along the way I started to remember the fundamentals of energy, work, force, power, heat, temperature, conduction, convection, radiation – some of which terms are used rather loosely on CA. Then I started to fell embarrassed at some of my earlier comments, too much shooting from the hip.

    DeWitt, about a year back, I was pointing to both the lack of accepted definition of “tropopause” and the complexity of it. I don’t think that much understanding or modelling of the tropopause is possible until you have studied its characteristics (as known) in great detail. I have browsed but not studied. Several events important to temperature change occur there; they vary in importance with time, latitude, convection, ozone chemistry etc. There can be huge discontinuities in the tropopause – its geometry cannot be modelled by an oblate spheroid of fairly even surface, it is extremely complex, affected by jet stream motion and other perturbations. It should not be defined as the height at which radiation in = radiation out. I sincerely hope that I am not teaching Grandma to suck eggs. I do not know your level of comprehension, training or experience (which all seem high in general) but after a year of reading I have concluded that I am not qualified to talk about it in CA and I suggest it is a topic best left to those for whom it is a life’s work.

    The enthusiasm of writers on CA is to be applauded because out of that type of melting pot can come the sudden insight that leads to simplification, understanding and accuracy. I don’t want to dampen that, but I, like many others, have strayed a little beyond my bounds by writing when I should have been reading. Steve writes beautiful introductions then I and others rapidy go OT. My apologies.

  61. Black Wallaby
    Posted Jan 11, 2008 at 5:09 AM | Permalink

    Gord,Reur 62:

    Yes, “climate science” as I observe it, is DIFFERENT.

  62. Gord
    Posted Jan 11, 2008 at 5:16 AM | Permalink

    Black Wallaby, regarding 64:

    Can you describe HOW?

  63. Pat Keating
    Posted Jan 11, 2008 at 7:30 AM | Permalink

    62 Gord
    A short answer is that energy IS additive.

    The examples you quote are all linear fields, where vector addition is required if the two components are correlated. Energy is a quadratic function of a linear field and a phase is not defined.

    Moreover, the different fields that contribute to emitted radiant energy are normally uncorrelated, the only exception being two or more beams from a single source when the path difference is less than the correlation length of the beams (which is just about never met for broad-band radiation).

  64. jae
    Posted Jan 11, 2008 at 7:44 AM | Permalink

    One of the things that intrigues me about these radiative models is the effect of the other 98+ percent of the atmosphere. The N2 and O2 molecules have to have temperatures in equilibrium with their environment, and that energy has to come from somewhere. Do these molecules “steal” that energy from the GHGs in thermalization reactions (collisions), thereby affecting these radiation schemes? I wonder if the whole thing doesn’t boil down to a big bunch of convection loops.

  65. Phil.
    Posted Jan 11, 2008 at 8:15 AM | Permalink

    Re #63

    DeWitt, about a year back, I was pointing to both the lack of accepted definition of “tropopause” and the complexity of it. I don’t think that much understanding or modelling of the tropopause is possible until you have studied its characteristics (as known) in great detail. I have browsed but not studied. Several events important to temperature change occur there; they vary in importance with time, latitude, convection, ozone chemistry etc.

    As I understand it the tropopause arises because the atmosphere is heated from both above and below, solar UV and surface IR. Without O3 you don’t get a tropopause, see Mars for example

  66. buck smith
    Posted Jan 11, 2008 at 8:19 AM | Permalink

    That Russian scientist that just predicted cooling due changes in Solar irradiance said that convection is more important than radiation in moving heat from lower atomosphere to where it can be radiated out to space. That does not seem to represented in these diagrams at all or am I missing something?

  67. Jan Pompe
    Posted Jan 11, 2008 at 8:22 AM | Permalink

    jae says:
    January 11th, 2008 at 7:44 am

    Do these molecules “steal” that energy from the GHGs in thermalization reactions (collisions)

    No jae the GHGs share willingly. If the other 98% of the atmosphere is warmer than 0K it will radiate warmth if it is warmer than the ~2.7K of space there will be net radiation to space.

  68. Jan Pompe
    Posted Jan 11, 2008 at 8:29 AM | Permalink

    buck smith says:
    January 11th, 2008 at 8:19 am

    Hi Buck,

    That Russian scientist that just predicted cooling due changes in Solar irradiance said that convection is more important than radiation in moving heat from lower atomosphere to where it can be radiated out to space. That does not seem to represented in these diagrams at all or am I missing something?

    Yes the thermals (22 w/m^2) and the latent heat (76 w/m^2).

  69. Posted Jan 11, 2008 at 8:46 AM | Permalink

    The 1970s climate regime shift included a tropical tropopause temperature shift . The pattern is similar, but of a smaller magnitude, in the extratropical tropics.

    This is based on NCEP reanalysis data.

    I offer no conjecture as to why this apparent shift happened. The pressure-height of the tropical tropopause did not shift at the same time, which adds to this little puzzle.

  70. Tom Vonk
    Posted Jan 11, 2008 at 9:14 AM | Permalink

    Such diagrams are average energy balances , have the merits that making averages has and there is a factor 2 between what is emitted by high latitudes compared to low latitudes .
    Hwoever they don’t depend on frequency and don’t pretend that they do .
    As IR absorbing gases are all about frequency dependent processes , their effects don’t appear in the diagram and can’t be deduced from it .
    Indeed it doesn’t make much sense saying that x W/m² “come from” this or that altitude .
    Especially then not when (unless I am mistaken) some people convert a total radiative energy in a gaz temperature by SB law (a gaz = black body ?) and then by looking at the altitude that happens to have a similar temperature conclude that this radiation “comes” from that altitude .
    The absorption/emission depends on frequencies and this relationship changes with altitude – f.ex going up the troposphere , the H2O concentration decreases and therefore absorption in H2O absorption frequencies decreases too .
    Averaging (in this case integrating) energies for all frequencies is good for energy balances but it makes disappear all the spectral information and I don’t really see how such a diagramm can help to show or to explain the effect of a variation of IR absorbing gazes .

    The very positive point I see here is that it takes explicitely the stratosphere in account and as LTE holds about to the top of the stratosphere and a bit beyond , we are still in the domain where collision processes rule .
    Things change dramatically higher because then we have no more LTE and radiation rules .

  71. Gord
    Posted Jan 11, 2008 at 9:33 AM | Permalink

    Pat Keating 68:
    Correct..except that vectors have a magnitude and a direction.
    Addition of fields that have even slightly opposing directions result in some form of cancelation and/or addition when added.
    This is absolutely shown to be true by Fourier Analysis of more than one waveform.
    The superposition principal also applies for this type of analysis.
    This is why at least some form of noise reduction is available in communication circuits.

  72. Gord
    Posted Jan 11, 2008 at 9:45 AM | Permalink

    Pat Keating 68:

    I forgot to mention this link which shows light cancelation over the entire visible light spectrum.

    http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html

    This link demonstrates how a wideband spectrum of light can be completely canceled.

  73. Jeff A
    Posted Jan 11, 2008 at 10:06 AM | Permalink

    It does not contain any energy known as COLD, or negative HEAT.

    It was my understanding that there really isn’t any such thing as “cold”, only varying levels of heat or lack thereof.

  74. Sam Urbinto
    Posted Jan 11, 2008 at 12:46 PM | Permalink

    Usually, humans set hot/cold at the freezing point of water. Just a convention. Hotter vs colder. (more heat versus less heat) But I suppose that specifically, cold would have to be below absolute zero (anti-heat?).

  75. SteveSadlov
    Posted Jan 11, 2008 at 1:01 PM | Permalink

    RE: #44 – somewhat related – last night, the cloud deck came down onto my ridge. Clouds were dropping drizzle and light rain. I observed activity using the beam of one of our flood lights out back. The motion of the droplets making up the cloud mass was reminiscent of brownian motion (although, the driver of that motion was turbulent flow of air. Collisions of droplets were constantly occuring. Some would eventually combine into drops and fall. This was a mostly kinetic (with obvious other factors such as surface tention, dipole moment, other forces) process. Certainly, it was not strongly exothermic other than whatever heat may come from collisions. A far cry from the condensation component of the process which had earlier led to droplet formation. Those falling drops brought liquid phase H20 to the ground, at cloud temperature.

  76. DeWitt Payne
    Posted Jan 11, 2008 at 1:14 PM | Permalink

    Re: #65 and #60

    Geoff,

    It’s not my life’s work, but unless I risk exposing my ignorance by posting stuff, I can’t really prove I understand it.

    Phil,

    As I understand it the tropopause arises because the atmosphere is heated from both above and below, solar UV and surface IR. Without O3 you don’t get a tropopause, see Mars for example.

    Strictly speaking, it’s the presence of oxygen. You won’t have ozone without oxygen. But yes, that is indeed the case, as I understand it. Without the warming of the stratosphere caused by the absorption of short wavelength UV by oxygen and ozone, there would be no temperature inversion and no tropopause, however it’s defined.

  77. SteveSadlov
    Posted Jan 11, 2008 at 1:17 PM | Permalink

    RE: #74 – The convention for what constitutes a “cold” environment is the 10 deg C mean annual isotherm. Interestingly, the cloud temperature in #74 was 10 deg C.

  78. SteveSadlov
    Posted Jan 11, 2008 at 1:19 PM | Permalink

    Sorry, typo, I meant to write “the cloud temperature in #75 …”

  79. Willis Eschenbach
    Posted Jan 11, 2008 at 3:55 PM | Permalink

    Several people have postulated about the existence of the tropopause, and what defines it.

    For me, the tropopause is defined as defined and shown by Cliff Huston above. Take a look at his graphic, it is where the temperature “kinks”, where it stops cooling and starts rising.

    However, this raises an interesting question – why does the temperature bottom out at that point, and then start to rise as you ascend further? I’ll give my answer, and then you guys can tear it apart …

    I say that it is because in a multi-layer greenhouse system, there is one layer whose temperature is fixed by the physics of the situation. This is the outer layer, which is constrained to emit the difference between the total incoming (downwelling) energy and the upwelling energy which passes through the final outer layer to space. Or to put it another way, the upwelling radiation that passes through the outermost layer, plus the amount the outermost layer emits, must equal the total incoming radiation. The other layers are free to take up various intermediate temperatures. The outmost layer is not. Whatever energy does not pass through it, it has to emit.

    This means that the temperature of the outermost layer is fixed. My model shows that on average it is fixed at somewhere around -50°C. So the answer to the question of why the temperature of the atmosphere bottoms out at about -50° or so at somewhere around 10 – 15 km up in the air is that it is constrained to do so by the presence of an atmospheric layer that has to be a given temperature to maintain the energy balance.

    So, that’s my story and I’m stickin’ to it … let the scientific process begin.

    w.

  80. DeWitt Payne
    Posted Jan 11, 2008 at 4:27 PM | Permalink

    Willis,

    The atmosphere of Mars has no similar kink. The temperature decreases all the way up. Well, there is sort of a kink at 23,000 feet, but rather than an increase in temperature, the temperature decreases faster with altitude. It’s oxygen. Besides, a real atmosphere has effectively an infinite number of layers. You can approximate the behavior with fewer, but it’s not going to be completely accurate.

  81. Posted Jan 11, 2008 at 5:57 PM | Permalink

    compare this with

    this modtran spectrum

    and this derived effective emission height per wavelength at the tropopause

    I’d say the bulk of the outgoing radiation at the tropopopause comes directly from the surface (logically because that is the most effective emitter).
    There is a clear tripartition in the outgoing radiation:
    1) up to 7 micron from Mid troposphere water vapour (second)
    2) 7 to 13 micron from surface radiation and low ozone (bulk)
    3) Mid and high troposphere CO2 (third)

  82. Gary
    Posted Jan 11, 2008 at 6:13 PM | Permalink

    Surely when completing a radiative balance the object is to move up through the atmosphere until there is no change in the emitted radiation flux. Whether this occurs at 10km or 70 km is irrelevant as long as we have reached steady state emittance. The physical properties of the various layers just determines the parameters of the equations.
    To work out a flux and say this is equivalent to a temperature at 7000 m is so UHH!
    If the temperature was equivalent to the minimum winter temperature in Calgary (we dont have these temps in Aussie) is this relevant???
    Willis, when someone says “its counter-intuitive, its true” the missing words are”trust me”. Evidence and experimental results is what drives science. What you describe makes for an unstable universe spiraling out of control. The fact that you say that the temperature of the sun will rise is a transfer of heat from cold to hot without work being done. Reductio ad absurdum: if I have two infinite planes ie each has the other as field of view and both at same temperature you infer that they will both heat up due to the radiation from each other. What you are describing is a positive feedback with no negative which is unstable. I would put my house on the fact that both surfaces will stay exactly the same temperature.
    Also your cold flow confuses because everyone thinks of thermodynamic heat transfer (convection, conduction, radiation) and what you are describing is transfer of matter with the then subsequent heat transfer. The transfer is still hot to cold it is just that you moved the cold mass to mix with the hot mass.

  83. Posted Jan 11, 2008 at 6:19 PM | Permalink

    Reductio ad absurdum: if I have two infinite planes ie each has the other as field of view and both at same temperature you infer that they will both heat up due to the radiation from each other.

    One plane alone cools faster than two planes facing each other! Please do the sums before you start handwaving that it’s impossible.

  84. jae
    Posted Jan 11, 2008 at 6:22 PM | Permalink

    I’m still troubled by the fact that the areas within 30 deg. N. Lat and 30 deg. S. Lat with the most greenhouse gases (water vapor) are nowhere near as hot as the areas with the least amount of greenhouse gases. I know that’s due, at least in part, to subsistence and adiabatic heating in the dry areas, but wouldn’t one expect all that “back radiation” from the GHGs to at least equal the effects of adiabatic heating?

  85. DeWitt Payne
    Posted Jan 11, 2008 at 6:41 PM | Permalink

    Re:#81

    Hans,

    For clear sky conditions, which I assume is what you used in the MODTRAN input, I agree that most of the radiation comes from the surface. BTW, what were the input parameters, 1976 standard atmosphere? I’d like to see the equivalent plot for the tropical atmosphere. Also, a substantial fraction of the sky isn’t clear. IIRC, you have to assume significant coverage by multi-layer clouds to be able to get the correct overall energy balance.

  86. DeWitt Payne
    Posted Jan 11, 2008 at 6:45 PM | Permalink

    Read the graph title dummy, tropical atmosphere. The comment about cloud cover still applies though

  87. Posted Jan 11, 2008 at 6:47 PM | Permalink

    #79 Willis, that there exists a constraint due to a requirement to conserve energy is explanation enough for me. Thats why it doesn’t seem necessary that an increase in a forcing due to CO2 density increase say, necessarily leads to a great change in temperatures or balances. However, I would love to see it in equations without the hand waving.

    Some other things that would be good, are an explanation for the kink above the stratosphere, and a differential form that accounted for the thickness of the layers, rather than a thin shell.

  88. Gary
    Posted Jan 11, 2008 at 7:10 PM | Permalink

    #83, I agree Hans, but that is not what Willis is saying. The difference between one plane and two is availability of a heat sink. I dont think it is handwaving at all. I am questioning a physical assumption that a body radiating at a certain wavelength will absorb IR at a longer wavelength. My thoughts would be that the radiation bounces around the universe like a billiard ball until it finds a cold pocket to fall into. I just need to be convinced of the reality thats all. Simple

  89. Jan Pompe
    Posted Jan 11, 2008 at 7:11 PM | Permalink

    Gord says:
    January 11th, 2008 at 9:33 am

    Correct..

    Of course he is he knows his onions.

    except that vectors have a magnitude and a direction.

    That’s the whole point behind the statement:

    Energy is a quadratic function of a linear field and a phase is not defined.

    direction isn’t defined either. Squaring the velocity component for example (0.5mv^2) kind of takes care of the direction problem and squaring the sinusoid takes care of the zero average amplitude, to determine energy or power in the wave, too. Recall that power is generally expressed in terms of RMS power

  90. Pat Keating
    Posted Jan 11, 2008 at 7:13 PM | Permalink

    72 Gord

    I refer you to what I said in my post:

    the different fields that contribute to emitted radiant energy are normally uncorrelated, the only exception being two or more beams from a single source when the path difference is less than the correlation length of the beams (which is just about never met for broad-band radiation).

    The bubble example is why I said “just about never”. This is due to the fact that the bubble skin is so thin that the two beams (from front and back surface) have a path-difference less than the coherence-length.

  91. DeWitt Payne
    Posted Jan 11, 2008 at 7:35 PM | Permalink

    Re: #88

    I am questioning a physical assumption that a body radiating at a certain wavelength will absorb IR at a longer wavelength.

    But a body, as opposed to in isolated atom or molecule, doesn’t radiate at just a certain wavelength. It emits a continuum of radiation described by the Planck function and the emissivity (as a function of wavelength and viewing angle) of the body. By Kirchhoff’s law, if it emits, it absorbs, so even a very long wavelength photon can be absorbed by a warm body because there is no cutoff on the long wavelength tail of the Planck function. A warm body with the same emissivity as a cooler body will emit more energy at any wavelength than the cooler body. This leads to the conclusion that a warm body radiating to space will cool less rapidly if there is another body that is above 2.725 K close by.

  92. Willis Eschenbach
    Posted Jan 11, 2008 at 7:45 PM | Permalink

    Gary, you say:

    What you describe makes for an unstable universe spiraling out of control. The fact that you say that the temperature of the sun will rise is a transfer of heat from cold to hot without work being done. Reductio ad absurdum: if I have two infinite planes ie each has the other as field of view and both at same temperature you infer that they will both heat up due to the radiation from each other. What you are describing is a positive feedback with no negative which is unstable. I would put my house on the fact that both surfaces will stay exactly the same temperature.

    Here’s a though experiment for you. Imagine a sun in space, no planets around it. It is at some equilibrium temperature Ts.

    Now, we plunk down a large planet near the sun. What happens to the temperature of the sun?

    Well, the rays of the sun that used to go to deep space are now intercepted by the planet. The planet warms, and begins to radiate.

    Now, some part of that radiation from the planet hits the sun, and raises the sun’s temperature. Doesn’t raise it much, to be sure, unless the planet is huge and close in … but it raises it nonetheless, to Ts + a.

    This is because objects just radiate. All the time. Depending on their temperature. That’s why infra-red night vision goggles work. You radiate. I radiate. So do rocks, and water, and trees, and CO2.

    This radiation travels through space, and transfers energy to (warms) whatever it hits. The temperature of the object it hits doesn’t matter. Whatever infra-red hits it warms.

    w.

  93. DeWitt Payne
    Posted Jan 11, 2008 at 8:14 PM | Permalink

    Willis,

    Now I’m going to have to go and look up Dyson Spheres to see if he included the increase in solar surface temperature caused by building a sphere around the sun.

  94. John Creighton
    Posted Jan 11, 2008 at 8:28 PM | Permalink

    Off topic but I think a Dyson sphere could actually cool the sun, as I don’t see IR radiation helping fusion and if it heats up the sun then the sun could expand which would reduce the pressure where the thermo nuclear reaction takes place.

  95. Black Wallaby
    Posted Jan 12, 2008 at 1:04 AM | Permalink

    Gord 62,

    You asked how is “climate science” different?

    If you go back to my 36, “Some funnies in the Trenberth Cartoon”, I asked if anyone disagreed on some initial observations. Apart from Willis, who responded in part, (as further discussed elsewhere) there has been a deafening silence.
    I mentioned that the 324 down-EMR has to be negative in the HEAT transfer calculation from the surface, yet Trenberth/IPCC clearly shows it to be positive, being absorbed as HEAT, and adding, (not subtracting) to the HEAT-loss from the surface. Pray how can the 324-down be both positive and negative at the same time. (when viewed from the same ralativity?) What other branch of science has such weird mathematics? There is more besides. Regards, Black Wallaby

  96. W Robichaud
    Posted Jan 12, 2008 at 1:14 AM | Permalink

    Willis Thank you! A Veil as been removed.

  97. Black Wallaby
    Posted Jan 12, 2008 at 1:26 AM | Permalink

    Hi Steve Mc,
    It looks like two posts have been deleted, without reference….OK, fine, your call I guess. Problem is, when we refer to each other’s posts by number, it gets a bit confusing, when they become two numerals out.

    With the utmost of respect, Black Wallaby

  98. Willis Eschenbach
    Posted Jan 12, 2008 at 1:37 AM | Permalink

    Wallaby, use a link to the post instead of the post number. The link to the post is under the large numeral at the right side of the post, right click on that, copy the line, and use it in your reference.

    w.

  99. Black wallaby
    Posted Jan 12, 2008 at 1:57 AM | Permalink

    Hans Erren 83,

    Run that by me slowly again! You need to define the conditions each side of the planes surely!

  100. Posted Jan 12, 2008 at 2:24 AM | Permalink

    You need to define the conditions each side of the planes surely!

    Of course, for an excact calculation you need to define the geometry, but it’s really not rocket science.
    See also dewitt’s post

  101. Gord
    Posted Jan 12, 2008 at 4:07 AM | Permalink

    Pat Keating 90:

    The bubble example just means that the thickness of the bubble surface just has to be less than about 1/10 wavelength of
    the highest frequency emmission. All longer wavelengths will be reflected 180 deg out of phase. I don’t know if this
    phenomena is prevalent in the atmosphere or not.

    Can you comment?

    I do know that this link..
    Heat Radiation

    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2

    Shows that the equation..
    P = e*BC*A(T^4 – Tc^4)

    Where P = net radiated power (Watts), e = emissivity, BC = Stefan’s constant, T = temperature of radiator and Tc =
    temperature of the surroundings.

    ..when rearranged gives

    P/A = e*BC*T^4 – e*BC*Tc^4 in (Watts/m^2)

    and this shows that this is an obvious subtraction of EM fields.

    Can you comment?

  102. Gord
    Posted Jan 12, 2008 at 4:16 AM | Permalink

    Black Wallaby 95:

    Thanks for your response…I share your views.

  103. Jan Pompe
    Posted Jan 12, 2008 at 4:49 AM | Permalink

    Gord

    I don’t know if this
    phenomena is prevalent in the atmosphere or not.

    For wavelengths longer than 6m since the IR radiation from earth surface is much much shorter than that it’s not affected that way.

  104. Black Wallaby
    Posted Jan 12, 2008 at 5:27 AM | Permalink

    Willis and ALL,
    Let me introduce you to Andrew Lacis, an impressive atmospheric scientist in GISS, and a link to his many publications:

    http://pubs.giss.nasa.gov/authors/alacis.html

    As an expert reviewer of the IPCC 2007 report draft, he had this to say in part, with my capitalization emphasis added:

    `…obvious that any gas that absorbs thermal radiation is a greenhouse gas, and that its ability to contribute to the greenhouse effect depends on the strength of its absorption bands, its atmospheric concentration, AND ITS LOCAL TEMPERATURE DIFFERENCE WITH THE GROUND SURFACE. [Andrew Lacis]” The arbitrating authors accepted.

    Elsewhere on this blog, Gary wrote: “Reductio ad absurdum: if I have two infinite planes ie each has the other as field of view and both at same temperature you infer that they will both heat up due to the radiation from each other”.
    And also he separately wrote, “My thoughts would be that the radiation bounces around the universe like a billiard ball until it finds a cold pocket to fall into. I just need to be convinced of the reality that’s all. Simple”

    Elsewhere on this blog, I wrote twice, following your first “oversight”:
    “I’m not sure if you [Willis] understand the difference between EMR and HEAT. Just to clarify that a bit, are you saying that two bodies at the same T, and hypothetically isolated from any other effects, and radiating at each other as a consequence of their T, are HEATING each other. If so, why does their T not increase? (You might also contemplate a layer of absorptive gas at a constant T)”

    Will YOU or ANYONE make a clear denial to Gary, Andrew Lacis, and me, concerning the rate of HEAT transfer being proportional to T difference?.
    Remember, EMR is not HEAT, but can be converted to HEAT by molecular absorption, if “willing” molecules can be found!
    For instance, we subjectively experience sunlight warming our skin, but sunlight (EMR), is not HEAT. What we actually experience is its conversion to HEAT via absorption by molecules in our skin being at a LOWER temperature than the EMR source; the Sun.

    BTW, when you say that Planet Earth HEATS the Sun, I’m sorry, but you have the cart before the horse! Our dear ole planet SLOWS the escape of heat from the Sun, much like a blanket thrown over a bed slows heat from escaping. The blanket does not heat the bed. In heat transfer calculation terms, the rate of heat loss, in the normal, from the Sun is S-E, where S is the solar radiance and E is the back radiation from Earth. (think of it like a back pressure) In more detailed terms there are also some vectors involved, outside the parallel or normal.

    Willis, divergent thinking in science is very good and has achieved great things, but in the above, I fear you do indeed have the cart before the horse! Sorry, whilst I sincerely admire your enthusiasm.

  105. Posted Jan 12, 2008 at 6:04 AM | Permalink

    Black Wallaby:
    Please do not capitalise. Any body that has a temperature radiates, even the universe. Also any body that radiates, will cool because it looses energy, unless more energy is received than emitted. Now with these fundaments of science look at sun earth interaction: The sun radiates, but also generates, so the sun doesn’t cool. All the bodies in the universe also radiated and heat the sun but the amount is neglecatable compared to the amount that the sun generates.

    So, it’s not a question if the sun is heated by exernal bodies, but how much the sun is heated by external bodies.

    The answer: “not much”.

  106. Black Wallaby
    Posted Jan 12, 2008 at 6:28 AM | Permalink

    Hans Erren,
    Please do not Bold for emphasis, please use Caps, it’s easier.
    You raise some points which I agree with, but why you do so is a puzzle. Otherwise:
    Soreeeee Hans, but the simple primary heat transfer calculation is S-E where S is the solar radiation, and E is the back radiation from the Earth. There are also vectors to consider outside the parallel or normal.

  107. Black Wallaby
    Posted Jan 12, 2008 at 6:39 AM | Permalink

    OH Hans, what’s more,

    It was Willis who introduced the Sun-Earth thing not me, and the magnitude of the heat transfer is irrelevant and blindingly screamingly obviously miniscule. It is the PRINCIPLE that is under debate, not its scale

  108. MrPete
    Posted Jan 12, 2008 at 11:37 AM | Permalink

    (Black Wallaby– online convention has it that extended all caps looks like someone “yelling”. If you don’t want to use fancy text, acceptable alternatives include _underlining_ and *emphasis*.)

    Skimming this conversation, it would appear we’re discussing another one of those non-obvious realities. Reminds me of the discussion about Let’s Make A Deal :)

    What some seem to be forgetting is that everything above zero K is radiating — cooling — as well as absorbing. Thus, those two hypothetical sheets don’t need to get hotter as long as their radiation exceeds absorption.

  109. Richard Sharpe
    Posted Jan 12, 2008 at 12:00 PM | Permalink

    Thus, those two hypothetical sheets don’t need to get hotter as long as their radiation exceeds absorption.

    Indeed. It seems to me that these two infinite sheets were a thought experiment.

    They should radiate the same amount of energy from each side and thus be losing temperature. However, on the sides facing each other they would absorb the same amount of energy they are radiating on that side, and so would cool down at a lower rate than if they were the only objects radiating into an otherwise empty space.

    I am sorry if I am simply restating the obvious, but it seemed worth saying.

  110. Willis Eschenbach
    Posted Jan 12, 2008 at 4:02 PM | Permalink

    Black Wallaby, you say:

    BTW, when you say that Planet Earth HEATS the Sun, I’m sorry, but you have the cart before the horse! Our dear ole planet SLOWS the escape of heat from the Sun, much like a blanket thrown over a bed slows heat from escaping. The blanket does not heat the bed. In heat transfer calculation terms, the rate of heat loss, in the normal, from the Sun is S-E, where S is the solar radiance and E is the back radiation from Earth. (think of it like a back pressure) In more detailed terms there are also some vectors involved, outside the parallel or normal.

    and

    For instance, we subjectively experience sunlight warming our skin, but sunlight (EMR), is not HEAT.

    You say electromagnetic radiation (EMR) is not heat. You say that when EMR hits a “willing molecule” it turns into heat. I assume that a “willing molecule” is one which can absorb the EMR. We are in agreement so far.

    Since we are in agreement, then consider the following, regarding my example of a sun without a planet and a sun with a planet.

    1) The EMR from the sun strikes the planet and turns into heat.

    2) As the planet warms, it begins radiating EMR itself.

    3) Some of the planet’s EMR strikes the sun.

    Now, at this point we have a few possibilities:

    1) The EMR is reflected by the sun. This can’t be true, because the sun is radiating at all frequencies, so it must be absorbing at all frequencies.

    2) The EMR is absorbed by the sun, and in the process removes energy from the sun (cools the sun) … but this can’t be true either, because as you have pointed out, when an object is struck by EMR, it warms the object, it doesn’t cool the object.

    3) The EMR is absorbed by the sun, and in the process adds energy to the sun (warms the sun).

    It seems like we have a problem of terminology. You say in effect that ‘the planet is not warming the sun, it is slowing the rate of cooling of the sun’ …

    But, as I specified in my post, the sun starts out at equilibrium, at a temperature “Ts”. And it ends up at a new higher temperature equilibrium, Tx + a, because it is now absorbing heat from the planet.

    Now on my planet, when an object starts out at one temperature, and ends up at a warmer temperature because it has absorbed heat from a second object, we say that the second object has “warmed up” the first object. Perhaps on your planet you say the second object has slowed the cooling of the first object, but here on this planet, we look at the starting temperature (Ts) and the ending temperature (Ts + a). If it ends up warmer than it started, as in my example, we call that “warming up”.

    The sun ends up warmer than it started out. You are certainly free to call that “responding to a reduction in the amount of net radiative cooling”, or whatever circumlocution you want to use.

    You are not free, however, to say it is not warming if it ends up warmer than it started out.

    So … in an effort to put this question to bed, let me propose the following terminology:

    1) Adding heat to an object by any means can be called “warming” the object.

    2) Removing heat from an object by any means can be called “cooling” the object.

    If you don’t like those terms, fine. Mentally replace “warming an object” with “removing an opportunity for the object to cool”, and replace “cooling an object” with “allowing the object to assume a more relaxed and nuanced thermal position”, or whatever terms you might prefer. I will use “warming” and “cooling”, and I will not discuss it further.

    Now, can we get back to the question of the energy balance and budget of the Earth?

    w.

    PS – anyone who seriously thinks that when electromagnetic radiation hits a warm object, it bounces off and continues bouncing around the universe until it hits something cool and is finally absorbed, needs more professional help than I can provide.

    PPS – anyone who thinks that they will gain credibility on this blog by “introducing” Andrew Lacis, a Hansen co-author and participant in the IPCC mummery, needs to spend some time reading the back pages of the blog. In my opinion, the fact that the IPCC accepted whatever it was that Lacis had to say increases the chances of it being false … or to use your terminology, I should say that it “reduces the rate at which the object is radiating EMT (electromagnetic truth)” …

    But beyond that, the cited passage by Lacis has nothing to do with what we are discussing.

    PPPS – Perhaps another example will help. You and I leave a house which is at say 70°F, and go outside where it is freezing cold. After a while, I say, “Let’s go in the house and warm up”. Now, here’s the relative temperatures:

    House – 70°F

    Outdoors – 30°F

    You and I – 98.6°F

    How on earth can a house a 70° warm us up? We’re at 98.6°F or so, all the house can possibly do is cool us down … but we still say “Let’s go inside and warm up”. Why? Because inside the house, we will receive more heat from our surroundings than we do outdoors. So like I say, you’re free to exclaim “boy’s, lets go in side and cool down less” … just don’t expect people to understand what you mean.

  111. Gary
    Posted Jan 12, 2008 at 7:45 PM | Permalink

    Willis, I really don’t know where your coming from. I think you are just trying to pull everyones chain. In the last example you feel warm inside the house because the rate of heat loss from the body decreases due to the lower dT between the house and body (98-70). You feel cold outside because the rate of heat loss is much greater and in case of hypothermia the body may not be able to generate heat to keep up.
    In the example of the sun and planet why did you stop at Ts+a. Why not continue along the infinite series which is therefore as the sun now radiates more the planet warms up to Tp+b which reheats the sun to Ts+a+a1 which then reheats the planet etc, etc, etc where does it stop? we would all be star material now if this was the case.

  112. Willis Eschenbach
    Posted Jan 12, 2008 at 8:05 PM | Permalink

    I just noted that Steve M. said

    As Willis observes below, if one takes a weighted average level for radiation-to-space, the average is in the troposphere. However substantially more (147 wm-2) radiation-to-space originates from the stratosphere as from the troposphere and surface combined (50 wm-2 + 40 wm-2). As we previously noted [see this image], in the stratosphere, temperatures are increasing with altitude. If Willis’ estimates are correct, then a larger proportion of radiation-to-space originates in a region with increasing temperatures with altitude (stratosphere) than in a region with decreasing temperatures with altitude (troposphere). This would make Houghton’s heuristic completely unusable.

    I now see my use of the term “lower stratosphere” as the location of the radiating layer is not entirely clear. I meant at and slightly above the “kink” in the temperature gradients visible above. This “kink” at the top of the troposphere is the signal, the marker, of the lower edge of the upper radiating layer of the atmospheric greenhouse. On average that layer is energetically constrained to a very low temperature(-50° to -70°C). Both above and below that level, the temperature rises. At the layer, there is often a region where there is no temperature change with altitude. That layer is cold because it is radiating much more than the atmosphere above or below it.

    Here is the nub of the matter. Some of the radiation emitted from either the surface or the troposphere passes right through the outer layer (at and above the tropopause). That radiation is not absorbed in the outer layer. Let me call that escaped radiation “R”.

    On average, the temperature of the outer layer has to be about 235 w/m2 minus R.

    This is because as a whole, on average the planetary system cannot emit more or less than it absorbs from the sun. And it absorbs 235 w/m2 from the sun. That, less R, gives the temperature of the outermost layer. My model puts it at about -50°, which seems a bit warm. However, my toy model is indicative of relationships more than actual numbers, and deals in averages. There are some conclusions to be drawn nonetheless.

    One is that, regardless of the number of layers, the outermost layer of a greenhouse system has a specifically constrained radiation temperature. This is the amount of energy received, minus the amount of energy that passes through the outmost layer. That is to say, if S is the incident solar energy after albedo (in w/m2), the radiation temperature of the outermost layer is S times the transmissivity (Tr) of the outermost layer.

    T = S * Tr

    The second insight is that as the transmissivity increases (more radiation escapes to space), the outer layer temperature drops. And as more radiation is absorbed, the outer layer temperature rises.

    So this energy balance calculation says that as co2 increases, more energy should be absorbed in the stratosphere, and that as a result, the temperature of the outer layer (located at the “kink” in the temperature profile) should increase.

    Don’t know if we could measure that, though.

    w.

  113. Willis Eschenbach
    Posted Jan 12, 2008 at 8:42 PM | Permalink

    Gary, you say:

    Willis, I really don’t know where your coming from. I think you are just trying to pull everyones chain. In the last example you feel warm inside the house because the rate of heat loss from the body decreases due to the lower dT between the house and body (98-70). You feel cold outside because the rate of heat loss is much greater and in case of hypothermia the body may not be able to generate heat to keep up.

    My point is not to do with the net change in energy. It is to do with the nature of radiation. If an object radiates energy, it warms whatever it hits. It doesn’t matter whether what it hits is warmer or colder. That only affects the net energy change.

    Next, you are partially correct when you say:

    In the example of the sun and planet why did you stop at Ts+a. Why not continue along the infinite series which is therefore as the sun now radiates more the planet warms up to Tp+b which reheats the sun to Ts+a+a1 which then reheats the planet etc, etc, etc where does it stop? we would all be star material now if this was the case.

    You are correct that there is a slight feedback effect. However, you are totally incorrect when you claim that any feedback always leads to runaway feedback.

    In fact, as the steel greenhouse example shows, even wrapping a sun totally in a steel shell does not lead to runaway warming.

    w.

  114. Black Wallaby
    Posted Jan 13, 2008 at 12:24 AM | Permalink

    Willis,
    Thank you for your lengthy dissertation. Rather than respond on the many things you raise at this time, I’d like to get back to 3 basic points that you have not addressed.

    1) I asked TWICE for your explanation of what happens in the following hypothetical: “…are you saying that two bodies at the same T, and hypothetically isolated from any other effects, and radiating at each other as a consequence of their T, are HEATING each other. If so, why does their T not increase? (You might also contemplate a layer of absorptive gas at a constant T)”
    Gary asked a similar hypothetical question without success, and to help you understand it, here is another hypothetical presentation of it. Two identical stars at surface T of 6000C approach each other from an immense distance, and go into a binary system. What happens to their T? Oh, and I walk into a room at the same T as my body T. Does the room heat me?

    2) Concerning my analogy: “…much like a blanket thrown over a bed slows heat from escaping…” Can you not see that the blanket does not HEAT the bed?

    3) Re my 36, do you agree that the net heat transfer from the surface in Trenberth’s cartoon is 390 – 324 = 66

  115. Willis Eschenbach
    Posted Jan 13, 2008 at 3:10 AM | Permalink

    Wallaby, thanks for your questions. You say:

    Thank you for your lengthy dissertation. Rather than respond on the many things you raise at this time, I’€™d like to get back to 3 basic points that you have not addressed.

    1) I asked TWICE for your explanation of what happens in the following hypothetical: are you saying that two bodies at the same T, and hypothetically isolated from any other effects, and radiating at each other as a consequence of their T, are HEATING each other. If so, why does their T not increase? (You might also contemplate a layer of absorptive gas at a constant T)
    Gary asked a similar hypothetical question without success, and to help you understand it, here is another hypothetical presentation of it. Two identical stars at surface T of 6000C approach each other from an immense distance, and go into a binary system. What happens to their T? Oh, and I walk into a room at the same T as my body T. Does the room heat me?

    I don’t want to get into a discussion without a clear definition of terms. In my terms, if an object is radiating, and some other object absorbs that radiation, that second object is heated by the first, in the exact amount of heat absorbed. So to take your examples in turn:

    1) Are two radiating bodies heating each other? Yes.

    2) Their temperature does not change because the net of the heat gains and losses is zero.

    3) If two identical stars approach each other, they will heat each other equally. If they form a binary system, their new equilibrium temperature will each be higher than it was when they were isolated from each other. Are you saying that as two stars approach each other, they do not heat each other?

    2) Concerning my analogy: much like a blanket thrown over a bed slows heat from escaping - Can you not see that the blanket does not HEAT the bed?

    Depends on whether we are talking the main function of the blanket (preventing convection/conduction through the air) or the radiative function of the blanket. The part you don’t seem to get is that when an object radiates, it just radiates. Doesn’t depend on the temperature of the objects around it.

    And when that radiation strikes another object, it warms it.

    3) Re my 36, do you agree that the net heat transfer from the surface in Trenberth’s cartoon is 390 – 324 = 66

    Not sure what you mean. Since the system is in equilibrium, the amount of energy entering the surface has to be equal to the amount of energy leaving the surface. The surface is receiving 324 + 168 = 492
    w/m2. It is emitting 390 + 24 + 78 = 492 w/m2. The net heat transfer is zero.

    w.

  116. fFreddy
    Posted Jan 13, 2008 at 3:24 AM | Permalink

    Re #114, Black Wallaby

    are you saying that two bodies at the same T, and hypothetically isolated from any other effects, and radiating at each other as a consequence of their T, are HEATING each other.

    He is saying they are passing heat from one to the other. He is not saying they are causing a net increase in heat in either body.
    If it needs clarifying: when body A passes heat to body B, then body B gets a bit hotter and body A gets a bit cooler. Body B then returns the favour, and we are back to where we started.

  117. fFreddy
    Posted Jan 13, 2008 at 3:27 AM | Permalink

    Whoops, sorry Willis. Slow update.

  118. Dave Dardinger
    Posted Jan 13, 2008 at 5:24 AM | Permalink

    BTW, since not too many people have been weighing in here, I’d like to support Willis 100% in this discussion. I don’t understand why this cooler vs warmer argument comes up every time the subject is discussed. But his expanations are extremely correct and all I can suggest to anyone who doesn’t get what he’s saying is to reread his messages and work through his examples until the aha moment arrives.

  119. Gary Moran
    Posted Jan 13, 2008 at 6:18 AM | Permalink

    Heat, by definition, is the transfer of energy from one body to another body which have different temperatures, and which results in a change in the internal energy of both. A cooler body can only heat a warmer body through the action of a heat pump. It is a trivial exercise to validate this definition by using Internet search engines. From this definition we can see that a body radiating at another body, can only be said to have heated it, if: its internal energy has diminished and the other bodies internal energy has increased. In the case of the Earth and Sun, only the Earth is being heated up, even though both are radiating at each other.

    From this it would appear that Willis Eschenbach has a different definition of heat. He should state it and provide supporting references. Or Willis is talking about a different phenomena?

  120. lgl
    Posted Jan 13, 2008 at 6:25 AM | Permalink

    I can’t get this spreadsheet running (there’s a lot I can’t find in Excel 2007 :-( so can someone please tell me how much of a 1 W/m2 increase in latent heat loss from the surface will reach the TOA and escape to space?

    There has been a significant increase in water vapor the last decades so it should be interesting to know the cooling effect of this. I assume 1% increse in water vapor means 1% increse in latent heat loss from the surface.

  121. fFreddy
    Posted Jan 13, 2008 at 8:57 AM | Permalink

    Re #119, Gary Moran

    From this it would appear that Willis Eschenbach has a different definition of heat.

    Your definition of “heating” refers to total net heat flow. It seems pretty clear to me that Willis is talking about the components of that flow.
    For example, if the Sun shines 100J of heat on the Earth and the Earth reflects 1J of heat back at the Sun, then both Earth and Sun have heated the other, in Willis’ terms.
    In your terms – of net transfer – the Sun has heated the Earth by 99J.

    While I am all in favour of precision of terminology, this is really not worth arguing about …

  122. Gunnar
    Posted Jan 13, 2008 at 9:26 AM | Permalink

    >> While I am all in favour of precision of terminology, this is really not worth arguing about …

    fFreddy, since most of the confusion is caused by differences in semantics, it’s a huge waste of bandwidth not to simply agree on the terminology first. Gary Moran in #119 clearly expressed the absolute truth. He said it was trivially verified, but yet you felt arrogant enough to skip the verification and claim that Gary Moran expressed his own definition of “heating”, rather than agreed upon standard scientific definition. I’ll help you look it up:

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

    In short, the earth is radiating towards the sun, not heating it.

  123. beng
    Posted Jan 13, 2008 at 9:28 AM | Permalink

    I think Willis’ simplified model could be a submitted paper, considering that Trenberth’s diagram is apparently quite “popular” & Willis’ is an improvement over that, especially the 2-layer aspect. The tropopause seems a ready-made “boundary” to seperate the layers.

    Realize that Willis makes some major simplifying assumptions in his model, namely the “shells” (steel or otherwise) are of negligible thickness/heat capacity, and 100% radiation-opaque, and are perfect “conductors” — same temperature on both sides. I don’t see how this changes the overall heat flows in/out of the control volumes, tho. Willis has reduced the “greenhouse” effect to a much more easily comprehended (to me) “insulation” effect.

    Perhaps the next steps, in addition to Dave Dardinger’s “holes” permitting the proper layer radiational opacities (not 100%), would be to expand the shells to proper thicknesses to determine temp “profiles”.

  124. fFreddy
    Posted Jan 13, 2008 at 9:47 AM | Permalink

    Re #122, Gunnar

    since most of the confusion is caused by differences in semantics

    Agreed. I hope my trivial example in #121 will assist in reaching mutual understanding.

    … [remainder] …

    How frightfully rude.

  125. Gunnar
    Posted Jan 13, 2008 at 9:55 AM | Permalink

    >> How frightfully rude

    I’m sorry for being rude. I overreacted. I guess I’m offended when standard scientific definitions and concepts are painted as your personal definition or theory.

  126. jae
    Posted Jan 13, 2008 at 10:13 AM | Permalink

    Willis: Any idea why average temperatures in very humid areas (tropics, e.g.) seldom, if ever, get above 32-33 C, even on clear days; whereas, average temperatures in dry areas can be much hotter?

  127. kim
    Posted Jan 13, 2008 at 10:31 AM | Permalink

    #124, how rudely ironic.
    ================

  128. Steve McIntyre
    Posted Jan 13, 2008 at 10:56 AM | Permalink

    this sort of elementary discussion of the meaning of heat cold and heat is exactly the sort of discussion that I want to discourage on the blog.

  129. Dave Dardinger
    Posted Jan 13, 2008 at 1:26 PM | Permalink

    Steve,

    It may be the sort of thing you want to discourage, but until the issue is settled, no progress can be made. If you would be so kind as to put forth a definition of “HEAT” to be used on this site, then fine, we’ll live by it. Frankly I vote strongly for Willis’ definition and pace Gunnar and the rest, strongly believe it’s the standard scientific definition.

  130. Willis Eschenbach
    Posted Jan 13, 2008 at 2:32 PM | Permalink

    Steve, I agree, and I have attempted to cut through the semantic question. However, it is a difficult question to put to bed. Gunnar, for example, says:

    I’ll help you look it up:

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

    In short, the earth is radiating towards the sun, not heating it.

    However, when you look further down on the page that Gunnar recommends, you find:

    Whenever EM radiation is emitted and then absorbed, heat is transferred.

    Now, I’m describing this process wherein EMR is emitted by one object and absorbed by another and heat is transferred between them by saying “the first object is heating the second object”. And as I said before, you can call this process what you want. You don’t like the term “heating” to describe a situation where, as the reference puts it, “heat is transferred”? That’s fine, call it “calorification” or whatever you want.

    But the process is real, and the heat transferred is real, and I will continue to call it “heating”.

    Now perhaps we can get back to whether my toy model is of any value in understanding the world.

    w.

  131. Black Wallaby
    Posted Jan 13, 2008 at 4:29 PM | Permalink

    Willis, 115
    Thanks for that. The most illustrative point of yours was:

    “If two identical stars approach each other, they will heat each other equally. If they form a binary system, their new equilibrium temperature will each be higher than it was when they were isolated from each other. Are you saying that as two stars approach each other, they do not heat each other?”

    I’m sorry Willis, but although the two stars will reach a higher equilibrium T, it is not because they HEAT each other, but because they loose less HEAT to space. Two objects at the same T cannot HEAT each other, (Cannot raise each other’s T) per the 2nd law of thermodynamics etc.

    To try and explain this, let’s look at equivalencies in energy transfer in other systems. A good example is a reservoir of water. In this equivalency, the potential hydraulic energy is “HEAT”, (in a body at constant T), and the water pressure is “EMR”. If we take any level in the reservoir and divide it into A & B, they are exerting pressure on each other, (“EMR”) but there is no change in hydraulic PE, (“HEAT”). All forms of energy require a potential difference to flow. The bigger the difference, the higher is the rate of flow. The lower the water outlet is, the higher is the KE released and the pressure (“EMR”) If there is any back pressure to the hydraulic flow, the equivalency would be back radiation, and the net water pressure would be equivalent to net “EMR”

    Coming back to our two remote identical stars, they were loosing HEAT via EMR over their entire surfaces, but when coming together, each was strongly radiating at the other, equally. Whilst their net EMR PD was zero, with zero HEAT transfer, a considerable surface area that was previously radiating to space has been neutralized. Thus they get hotter, yet they do not HEAT each other.

  132. fFreddy
    Posted Jan 13, 2008 at 5:25 PM | Permalink

    Methinks I smell some trolling …

  133. Willis Eschenbach
    Posted Jan 13, 2008 at 5:44 PM | Permalink

    Wallaby, I’m not going to argue terminology. If you want to say that two stars don’t warm each other, that’s fine by me. The mathematics of it is clear, and we agree, both stars end up warmer than they started.

    My main point is that, contrary to your assertion, a cooler object can radiate energy to a warmer object. If we want to find out the net heating or cooling, it is necessary to calculate the energy emitted by an isolated body, as well as the energy absorbed by an isolated body.

    Thus the atmosphere, which is cooler than the surface, nonetheless radiates energy (by whatever name you want to call it) to the surface. And the surface also radiates (and conducts/convects) energy to the atmosphere, by whatever name you want to call that process. To calculate the net heat transfer, we must first calculate the individual heat transfers.

    These individual heat transfers are what I have enumerated in my cartoon of the global energy balance above, and in my spreadsheet that calculates that balance.

    Call it what you want, I don’t care. Please note, however, that you are not correct when you say that there is “zero HEAT transfer” between the stars. They are transferring huge amounts of heat between them, but the heat transfers are equal. You are confusing HEAT transfer (which is non-zero) with net HEAT transfer (which is zero). As the page recommended by Gunnar says,

    Whenever EM radiation is emitted and then absorbed, heat is transferred.

    Since each star is emitting radiation which is then absorbed by the other star, it is obvious that heat is being transferred, even though the net heat transfer is zero.

    All the best,

    w.

  134. Cliff Huston
    Posted Jan 13, 2008 at 6:33 PM | Permalink

    RE#131 Black Wallaby

    Step 1: Go to ht tp://web.mit.edu/lienhard/www/ahtt.html (remove space in http)
    Step 2: Down load ‘A Heat Transfer Textbook,  3rd edition’
    Step 3: Open the file and go to page 32 and read the section titled Radiant heat exchange.
    Step 4: Tell Willis you are sorry for trashing his thread with your confusion.

    Cliff

  135. John Creighton
    Posted Jan 13, 2008 at 6:52 PM | Permalink

    Willis, I was wonder if in your calculation of temperature you should use emissivity:

    acording to:

    http://en.wikipedia.org/wiki/Kirchhoff's_law_of_thermal_radiation

    objects emit what they absorb. The modified Stephan Boltzmann equation is:
    F=E /sigma T^4

    http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

    Your stratosphere temperature calculation gives:

    =(I20/0.0000000567*100/G17)^0.25-273.15=-53
    G17=70
    I20=308

    Since you say the stratosphere absorbs 70% of IR radiation I’ll take the emissivity to be 0.7 which gives:

    =(I20/0.0000000567*100/G17)^0.25-273.15=-33 degrees celsius.

  136. Ian McLeod
    Posted Jan 13, 2008 at 7:35 PM | Permalink

    John #135, I had the same question about emissivity in an earlier post.

    http://www.climateaudit.org/?p=2581#comment-194480

  137. Willis Eschenbach
    Posted Jan 13, 2008 at 7:44 PM | Permalink

    John C, you ask a good question:

    Willis, I was wonder if in your calculation of temperature you should use emissivity:

    acording to:

    http://en.wikipedia.org/wiki/Kirchhoff’s_law_of_thermal_radiation

    objects emit what they absorb.

    I didn’t use emissivity because I don’t have good numbers for the various surfaces involved. If you have them, please plug them in.

    I’m not sure that your comment that “Since you say the stratosphere absorbs 70% of IR radiation I’ll take the emissivity to be 0.7″ is correct, given that we are talking about a mixture of gases … but then, there’s lots about the climate that I’m not sure about. Clarification is always welcome.

    w.

  138. jae
    Posted Jan 13, 2008 at 8:17 PM | Permalink

    Now perhaps we can get back to whether my toy model is of any value in understanding the world.

    Yes. Now, are you saying that less than 2% of the atmosphere (the GHG portion) is backradiating 321 watts/m^2 energy? My rudimentary understanding of thermo is that LTE requires that the N2 and 02 share the energy absorbed by the GHGs. Are these molecules helping to produce the 321 watts, or are the GHGs doing this all by themselves? Are you assuming blackbody radiation at all levels in the atmosphere?

    Another question, regarding the two stars. If there is total saturation of all the radiation bands in the “atmospheres” of both stars, then what happens to that back-radiation from each one?

  139. Willis Eschenbach
    Posted Jan 13, 2008 at 9:22 PM | Permalink

    jae, thanks for the questions.

    Now, are you saying that less than 2% of the atmosphere (the GHG portion) is backradiating 321 watts/m^2 energy? My rudimentary understanding of thermo is that LTE requires that the N2 and 02 share the energy absorbed by the GHGs. Are these molecules helping to produce the 321 watts, or are the GHGs doing this all by themselves? Are you assuming blackbody radiation at all levels in the atmosphere?

    The O2 and N2 molecules do absorb/emit a small amount, but the heavy lifting is done by the GHGs. I answered above that I am assuming blackbody throughout.

    Another question, regarding the two stars. If there is total saturation of all the radiation bands in the “atmospheres” of both stars, then what happens to that back-radiation from each one?

    I believe it still gets absorbed. My understanding is that “saturated” merely means that the layer is absorbing all of the impinging radiation of a given frequency. It does not mean that the layer turns into a reflector, or that it cannot absorb more energy. It just means that it is absorbing everything that is hitting it from the inside.

    However, I’ve never thought about this question, so it’s just an educated guess. There’s only three possibilities, however (absorption, transmission, or reflection). Of those, I think that impinging radiation from the outside would be absorbed rather than transmitted all the way through the star, or reflected from the star. Neither of those possibilities seem very likely.

    w.

  140. John Creighton
    Posted Jan 13, 2008 at 10:07 PM | Permalink

    Another note is that I think the energy dissipated through convection should be more a function of the difference between surface and tropopause temperatures then it should be a function of the incident radiation. Convective flows attempt to equalize unstable temperature gradients. Well the lapse rate over a large scale is constant I don’t think this applies at very small heights above the surface. For instance 1 cm above the pavement it seems a lot hotter then the surrounding air. Thus I conclude there should be a steep temperature gradient near the surface that drives convection.

  141. Black Wallaby
    Posted Jan 14, 2008 at 2:34 AM | Permalink

    Willis, keeping it brief, you wrote:

    “Since each star is emitting radiation which is then absorbed by the other star, it is obvious that heat is being transferred, even though the net heat transfer is zero”

    If there is zero net heat transfer, (to which I agree), then the HEAT content of the two stars is not increased by this process, and thus no increase in T. Elsewhere you have said that the identical stars will increase in T as a consequence of them heating each other, via their EMR. Would you please explain for me and others, how you figure this?
    Regards

  142. Chris Harrison
    Posted Jan 14, 2008 at 5:14 AM | Permalink

    Black Wallaby,

    If there is zero net heat transfer, (to which I agree), then the HEAT content of the two stars is not increased by this process, and thus no increase in T. Elsewhere you have said that the identical stars will increase in T as a consequence of them heating each other, via their EMR. Would you please explain for me and others, how you figure this?
    Regards

    The stars are not static but part of a dynamic system: each star is continuously generating heat and radiating it away. The star’s temperature is determined by the equilibrium point where the heat generated matches the heat radiated. The second star reduces the rate at which heat is being radiated away from the star, causing the temperature of the star to increase until a new equilibrium point is reached. One could replace the second star with a mirror in this thought experiment if that makes things clearer.

  143. MarkW
    Posted Jan 14, 2008 at 5:18 AM | Permalink

    Expanding the sun would not decrease the pressure at the core.

  144. Dave Dardinger
    Posted Jan 14, 2008 at 5:46 AM | Permalink

    Chris,

    The second star reduces the rate at which heat is being radiated away from the star,

    This depends on the definition of “away” Certainly radiation is leaving each star. It’s just that instead of seeing 3 deg k radiation coming back from space, each is seeing several thousands of degrees of radiation coming back from a small region of space. The real problem with working with net changes is that it’s mathematically rather difficult to work with, especially if you have more than two objects. It’s easier to deal with each closed system object as a point, so to speak, and then the diagrams are a lot more useful.

  145. Tom Vonk
    Posted Jan 14, 2008 at 8:34 AM | Permalink

    Willis Eschenbach # 130

    Now perhaps we can get back to whether my toy model is of any value in understanding the world.

    I don’t think that it is .
    I have already written why so I’ll repeat it .
    A gaz is not a black body . A spectrum of a gaz is not the one of a black body . The emission and absorption of a gaz is not described by the Planck formula .
    From that wrong assumption come then several dubious statements like :

    The O2 and N2 molecules do absorb/emit a small amount, but the heavy lifting is done by the GHGs. I answered above that I am assuming blackbody throughout.

    This is a contradiction . If the atmosphere is a black body then the radiated power/m² depends only on the temperature . So as the atmosphere can be approximated by an N2 and O2 mixture , the GHG can be neglected . We would have a black body constituted by N2 and O2 that stand at radiative equilibrium with their surrounding . On the other hand if frequency dependent processes are important (aka selective absorption/emission) then we have no black body . One can’t have it both ways at the same time .

    This is because as a whole, on average the planetary system cannot emit more or less than it absorbs from the sun. And it absorbs 235 w/m2 from the sun. That, less R, gives the temperature of the outermost layer.

    No it doesn’t . As the atmosphere is not a black body , the absorbed or radiated power doesn’t depend on temperature only . It depends also on the frequency , pressure and composition . Measuring only the total power gives no information about temperatures , compositions and geometry .
    Besides that , this “outer layer” is not even in LTE let alone black body . The measured power is an integral of everything over the whole space and it doesn’t come from any particular location .

    Looking at the measured IR emission spectra, I would say that the majority of radiation emitted to space comes from the upper and middle troposphere, rather than the lower stratosphere. Radiation intensity from the 15 micrometer CO2 band and 10 micrometer ozone band is consistent with emission from the tropopause or lower stratosphere (220 K)

    Same like above . In addition precisely absorption bands like 15µ are horribly non black bodyish because the absorbed energy (at 15µ) is not equal to the emitted energy (at 15µ) . The Planck s law or worse , the SB law , are of no use for such frequencies .

    Last but not least . There is an unsaid assumption and the same is in Trenberth .
    The radiation absorbed and emitted in the troposphere and bottom stratosphere is so through 2 processes : the gazes absorb/emit and the clouds absorb/emit .
    If the former are no black bodies , the latter are much more so .
    Follows that those processes are very different . Taking a constant for the sum of both consists to make an assumption that a particular geometry/structure of cloudiness (f.ex some yearly average or whatever) is not only relevant but conserved .
    There is no law of “cloudiness structure conservation” and the impact on radiation exchanges is huge if that parameter changes .
    Therefore this assumption is at best without justification and probably wrong .

  146. Gunnar
    Posted Jan 14, 2008 at 11:51 AM | Permalink

    Tom, great post!

    >> This is because as a whole, on average the planetary system cannot emit more or less than it absorbs from the sun.

    I think it’s been clearly established that this is simply false. Jupiter…

    Analogy: the water level of lake ontario is dependent on how much water comes in (Niagara falls, rivers) versus how much is going out (St Lawrence, evaporation).

    Your statement above is like saying the lake cannot emit more out the St Lawrence than comes in from Niagara. This is simply false. There is a huge reservoir. The physical processes that govern the inflow and the outflow are completely independent. The lake could be filling or draining. Reality doesn’t have a preferred level, and there is no physical law that is striving for balance.

    Steve: Gunnar, please take this to BB.

  147. Jonathan Schafer
    Posted Jan 14, 2008 at 12:20 PM | Permalink

    Sorry if this is a dumb question.

    The cartoon depiction at the top of this post shows reflection and emission by clouds, as well as greenhouse gases. My simple mind has a difficult time understanding the difference between a ghg and a cloud. A cloud is nothing more than a visible collection of water vapor and water vapor is a ghg. So, I guess my question is what is it about the massing of water vapor into a “cloud” that makes its properties different than water vapor that isn’t in a “cloud”, and why would that need to be modeled differently? In other words, the behavior of water vapor I would assume is the same no matter what. It’s the density of the water vapor at the location being measured that has an effect, not the fact that it’s a “cloud”. Am I just way off base here, or is there something specific about “clouds” that require different parameterizations and characterizations vs the density of water vapor in the atmosphere.

  148. Gunnar
    Posted Jan 14, 2008 at 12:33 PM | Permalink

    Jonathan, someone else can answer your question in a substantive way, but I’d just like to point out that a cloud is not water vapor, it’s floating liquid water. Quite different I think.

  149. Larry
    Posted Jan 14, 2008 at 12:44 PM | Permalink

    From an albedo standpoint, a cloud is very different from vapor. They’re white, because the droplets refract, and bend a certain amount of light to the point where it returns to space. From a radiative standpoint, I’m not sure. Each droplet is so dense, that it wouldn’t behave according to the logarithmic rule; the odds of a photon being absorbed in the droplet would be very high.

    Clouds, by everyone’s admission (well, maybe not the Goracle), are the great unknown.

  150. AK
    Posted Jan 14, 2008 at 1:34 PM | Permalink

    Re 147

    So, I guess my question is what is it about the massing of water vapor into a “cloud” that makes its properties different than water vapor that isn’t in a “cloud”, and why would that need to be modeled differently? In other words, the behavior of water vapor I would assume is the same no matter what. It’s the density of the water vapor at the location being measured that has an effect, not the fact that it’s a “cloud”. Am I just way off base here, or is there something specific about “clouds” that require different parameterizations and characterizations vs the density of water vapor in the atmosphere.

    Water vapor is a gas with a specific absorption/emission spectrum. Liquid and solid water are effectively black bodies at IR wavelengths.

  151. Gunnar
    Posted Jan 14, 2008 at 1:39 PM | Permalink

    >> Steve: Gunnar, please take this to BB.

    But other people can discuss the same topic ad nauseam? An interesting psychological question: What is different or controversial about #146? compared to #145?

    Steve:
    Please set an example. It applies to others as well.

  152. Jonathan Schafer
    Posted Jan 14, 2008 at 1:40 PM | Permalink

    #147,

    Gunnar,

    Well, it’s true that it’s different in that it’s a liquid (or solid depending on height) vs a gas, but does the form change the ir qualities of the molecules. H2O is H2O, regardless of what state it is in, so it would seem that from a physics standpoint, it’s IR capabilities would not change.

    #148

    I agree that clouds can affect any number of things, including blocking ir from the surface, blocking ir from space, albedo changes, etc. But I’m always a bit confused when it is said that we don’t know this or that about clouds. But from a physics standpoint, I assume we know many of the properties of water in all three states. Unless those states change the ir absorption/emission, I don’t understand why they are a great mystery.

    Again, I apologize if I am obtuse about this, but I am really trying to understand the graphs and the literature.

    • AK
      Posted Jan 14, 2008 at 2:09 PM | Permalink

      Re: #152

      But from a physics standpoint, I assume we know many of the properties of water in all three states. Unless those states change the ir absorption/emission, I don’t understand why they are a great mystery.

      The key unknown is how the cloud cover responds to changes in the temperature profile caused directly by CO2 or through water vapor feedback. Those cloud cover changes, in turn, cause changes to the radiative profile that feeds back into the temperature profile. The effect of any specific cloud cover condition can be estimated, and any specified change can be introduced into the temperature profile, but without knowing what impact changes to the temperature profile have on the cloud cover conditions, it’s impossible to know what the net feedback on the overall temperature profile is.

      (I’m assuming this is on topic, Steve. It seems elementary to me, but very relevant.)

  153. Black Wallaby
    Posted Jan 14, 2008 at 1:42 PM | Permalink

    Chris H, 141
    Yes, precisely,when to two identical stars approach each other they have less surface area to radiate to space, and thus they loose less heat and T increases. This is what I tried to explain to Willis in my 131. However, Willis still insists that their T increases because they are heating each other via their EMR. I like your analogy of a mirror.

    Steve:
    Please take this elementary discussion to BB.

  154. Black Wallaby
    Posted Jan 14, 2008 at 2:10 PM | Permalink

    pFreddy 116,
    Sorry for delayed reply, you wrote:

    “He [Willis] is saying they are passing heat from one to the other. He is not saying they are causing a net increase in heat in either body.
    If it needs clarifying: when body A passes heat to body B, then body B gets a bit hotter and body A gets a bit cooler. Body B then returns the favour, and we are back to where we started.”

    Or, if I can translate what you say, two bodies at the same T radiating at each other are HEATING each other equally, and at the same time are COOLING each other equally by the same amount, and thus nothing happens. The end result is zero.

    So why does Willis claim that their T increases as a consequence of them doing nothing to each other?

  155. Gunnar
    Posted Jan 14, 2008 at 2:15 PM | Permalink

    >> Steve: Please set an example. It applies to others as well.

    Does it apply to Willis?

    >> … elementary discussion

    Yes, when speculative ideas are contradicted by elementary basic physics, it makes fools of the promoters and those that can’t recognize it.

  156. Black Wallaby
    Posted Jan 14, 2008 at 2:16 PM | Permalink

    pFreddy further my 155,

    Sorry, with respect to the binary stars, I should have added:
    “…as a consequence of them doing nothing to each other,”
    , other than preventing full surface area radiation to space, re my 131?

  157. Phil.
    Posted Jan 14, 2008 at 2:57 PM | Permalink

    Re #154

    It’s not so much the absorption properties of the water that is at issue rather the size, shape and concentration of the drops or crystals that are formed. Those parameters greatly effect the light scattering properties of the clouds formed, high cirrus clouds contain long needle-like crystals as I recall, lower clouds contain drops of varying sizes, different wavelengths will be effected differently (the key parameter is d/lambda, lambda changes by a factor of 40+ and drop size probably by a factor of 100)!

    • AK
      Posted Jan 14, 2008 at 3:48 PM | Permalink

      Re: #158

      Chapter 2 of Marshak and Davis has a good basic discussion of the effect of cloud parameters on radiative (absorptive) properties. See especially the figures on pp. 121-122.

      All of this can be calculated, and confirmed by laboratory and field observation. What creates the “mystery”, though, is the unknown effect of changed temperature and water vapor profiles on droplet size distribution and density, as well as cloud size and location.

      Clouds affect the radiative profile in many ways. They modify the shortwave (SW) albedo, absorb SW radiation that would otherwise have reached the ground, absorb LW as black bodies (effectively), and radiate LW as black bodies at their current temperature. Thus high clouds have a pronounced greenhouse effect, while low clouds do not. High clouds can also scatter much more SW radiation than they absorb, which changes the effect of lower clouds on the radiative profile.

      Clouds don’t end abruptly, they are usually surrounded by a “halo” of air just at the dew point with a much lower density of droplets. Changes to the typical size and droplet density of this halo can have important effects on the radiative profile.

      AFAIK there’s no good understanding of the effect of changing the temperature profile on any of these “parameters”.

  158. fFreddy
    Posted Jan 14, 2008 at 3:12 PM | Permalink

    I’m setting an example.

  159. Sam Urbinto
    Posted Jan 14, 2008 at 4:28 PM | Permalink

    Short answer.

    Liquid
    Solid
    Gas

    These behave differently to IR. Clouds, ice and snow reflect it (unless other agents change their albedo). Water as a Vapor and Liquid absorb it, heat and emit in various ways.

    All I know is I’m all for allowing an object to assume a more relaxed and nuanced thermal position. Heck, who’s not?

  160. Black Wallaby
    Posted Jan 15, 2008 at 12:11 AM | Permalink

    Willis, 110; your important question:
    “…You and I leave a house which is at say 70?F, and go outside where it is freezing cold. After a while, I say, “Let’s go in the house and warm up”. Now, here’s the relative temperatures: House – 70?F Outdoors – 30?F You and I – 98.6?F
    How on earth can a house a 70? warm us up? We’re at 98.6?F or so, all the house can possibly do is cool us down … but we still say “Let’s go inside and warm up”. Why? Because inside the house, we will receive more heat from our surroundings than we do outdoors. So like I say, you’re free to exclaim “boy’s, lets go in side and cool down less” … just don’t expect people to understand what you mean.”

    ANSWER~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    The reason one feels warmer is simply that the rate of heat loss from a body is proportional to the difference to the (colder) sink. The body is not heated by the cooler room. The body simply looses less heat than if it were outside.

    PROOF~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    Surface area of my body radiating outwards 100 m square, all of which radiate equally in all directions hemispherically. Thus my body receives maybe 50x more EMR from the room, than I give to it, even when the room is cooler, yet I do not get “hot”.

    Note to Steve: I am answering an important question from Willis, that is neither trivial or off topic. May I also answer his other questions and issues? My 155/7 crossed your criticism added to my 153.
    Regards, Black Wallaby

  161. Black Wallaby
    Posted Jan 15, 2008 at 12:21 AM | Permalink

    Willis/Steve,

    Sorry, my 162 should read in part:

    PROOF~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    Surface area of my body radiating outwards is less than 2m square, and typical area of walls, ceiling and floor more than 100 m square, all of which radiate equally in all directions hemispherically. Thus my body receives maybe 50x more EMR from the room, than I give to it, even when the room is cooler, yet I do not get “hot”.

    Regards, Black Wallaby

  162. Andrey Levin
    Posted Jan 15, 2008 at 4:20 AM | Permalink

    So, clouds cover about 50% of the Earth surface, and act like black body for IR radiation, even at “atmospheric window” where even nasty GHG let surface heat radiation escape into space unmolested. As such, clouds could be viewed as steel shell, covering 50% of Earth, according to virtual experiment of Willis. How much back radiation clouds return to the surface, and why this supposedly first degree effect is not depicted in the cartoons?

  163. Willis Eschenbach
    Posted Jan 15, 2008 at 2:18 PM | Permalink

    Andrey, thank you for your interesting question, viz:

    So, clouds cover about 50% of the Earth surface, and act like black body for IR radiation, even at “atmospheric window” where even nasty GHG let surface heat radiation escape into space unmolested. As such, clouds could be viewed as steel shell, covering 50% of Earth, according to virtual experiment of Willis. How much back radiation clouds return to the surface, and why this supposedly first degree effect is not depicted in the cartoons?

    The forcing by the clouds, as you point out, is an important factor in the greenhouse. It is included implicitly in the cartoon, where the effect of the clouds is lumped in with the action of the atmospheric layers. To give the results shown in the cartoon, the absorption of each of the layers has to be quite a bit greater than 50% (see the spreadsheet), and part of that is from the clouds.

    It should be noted in this regard that clouds also have a strong effect on the incoming radiation, in three ways. One is that they reflect the incoming sunlight. The second is that they diffract and refract incoming sunlight. The final way is that they absorb incoming sunlight.

    As a result, even in my toy model (which ignores diffraction/refraction), a 1% increase in cloud reflection/absorption leads to a 0.75° drop in surface temperature. A 1% change in atmospheric IR absorption, on the other hand, only leads to a 0.25° change at the surface.

    It is also worth noting that a 1% change in cloud reflection/absorption gives a 1.4 w/m2 change in forcing at the tropopause layer, while a 1% change in IR absorption gives a 2.8 w/m change at the tropopause layer.

    If this toy model is even approximately correct, this means that the “climate sensitivity” figure so beloved of the AGW crowd does not have a physical meaning. The albedo/absorption sensitivity is 0.75°/1.4 w/m2 = 0.5°/w-m2 tropospheric forcing change, whereas the sensitivity for IR absorption changes is 0.25°/2.8 w/m2 = 0.2°/w-m2. The IR sensitivity is only about 40% of the cloud sensitivity.

    While this model is assuredly only a toy model, it still points out an important fact. Depending on what is happening at the lower layer and the surface, a given change in forcing at the tropopause can lead to very different changes at the surface, and thus the “climate sensitivity” doesn’t have a single value.

    w.

  164. lgl
    Posted Jan 15, 2008 at 2:57 PM | Permalink

    Great fun Willis, thanks

    But the only way I can get this match the observations is by reducing stratospheric SW absorption, not by increasing tropospheric absorption for instance. Wonder where I derailed.
    I have taken the liberty to copy a couple of runs, please tell if you don’t allow.
    http://www.virakkraft.com/volcanoproposal.ppt (2,5 MB)

  165. Sam Urbinto
    Posted Jan 15, 2008 at 3:00 PM | Permalink

    Black Wallaby. You’re arguing semantics. We get your point. Losing less heat is not the same as providing more heat. But you’re like a person looking at a quarter on the tails side arguing with somebody looking at it from the heads side.

    Most people think of it this way. When you’re outside, and losing heat, you feel cold. When you’re inside not losing as much heat you feel warm. When you’re not losing as much heat, you’re warmer, right?

    Or even think of it this way.

    It’s 32 F outside. It’s 70 F inside. If you walk inside, part of the heating you feel is real (insulation + heat source keeping the inside 38 F warmer) and some is perceived (because you’re keeping more heat inside your body, which is the same as losing less heat).

    So if you lose less heat, are you warmer? Can it be 70 F inside when it’s 32 F outside without some source of heat and the insulation of the structure?

  166. Posted Jan 15, 2008 at 3:04 PM | Permalink

    Willis, breaking the toy model is that while it uses a black body relationship for the steel shell, gasses do not act like black bodies. If they did, they would absorb all spectral frequencies (they don’t) and would emit red when heated (they don’t, though some flouresce).

    Your model is still very useful though. Regards

  167. MarkW
    Posted Jan 15, 2008 at 3:28 PM | Permalink

    Sam,

    Yet another way of looking at it is the fact that the body generates heat that it needs to loose.
    If it is loosing less heat than it generates, it feels hot.
    If it is loosing heat at about the same rate it is being generated, it feels comfortable.
    If it is loosing heat fast enough that it has to turn up the metabalism to compensate, it feels cold.

  168. Willis Eschenbach
    Posted Jan 15, 2008 at 4:07 PM | Permalink

    David S, thanks for your comment. You say:

    Willis, breaking the toy model is that while it uses a black body relationship for the steel shell, gasses do not act like black bodies. If they did, they would absorb all spectral frequencies (they don’t) and would emit red when heated (they don’t, though some flouresce).

    Your model is still very useful though. Regards

    Actually, while what you say is true, it doesn’t affect the model at all, much less break it. This is because the model merely adjusts how much is reflected or absorbed of IR and shortwave. It does not depend on the frequency of the radiation, nor on whether the bodies are black bodies, gray bodies, or any other kind. It just sets the absorption percentage.

    w.

  169. Sam Urbinto
    Posted Jan 15, 2008 at 4:20 PM | Permalink

    Very good MarkW, nicely put. I had not been thinking about that indeed we do create heat/warmth, and the environment we’re in causes us to be comfortable in it (or not).

    80 feels pretty good, because we are in a basic equilibrium; unless we have just gotten out of a cold shower (too cool), or just finished running 10 miles on a treadmill (too warm).

    But really, if we talk about temperature, what makes things “hot” or “cold”? What we compare them to. Is an ice cube cold? Compared to boiling water, yes. Compared to liquid nitrogen, no.

  170. Black Wallaby
    Posted Jan 15, 2008 at 4:29 PM | Permalink

    Sam Urbinto 167, MarkW 169,

    You’re missing the point guys: Trenberth and Willis make a fundamental claim that EMR from a cold source is absorbed as HEAT by a warmer body. I am offering PROOF at 163, that this is not the case. If any of you do not believe that every square millimetre of the walls, ceiling, and floor area in the room radiates equally towards any object in the room, take a look at the Sun. It appears to be a disc of uniform brightness, but being spheroidal, it is obvious that the sunlight coming at us from the circumference is tangential to its surface, and must trend progressively to the normal projection at the centre.
    Back to the room: The relatively huge amount of lower level EMR at ~ 290K from a vastly greater surface area radiating towards a small warmer body radiating back at ~309K clearly does not HEAT the body. EMR is not HEAT, and is only converted to HEAT when striking absorptive material at a lower T.
    Per the 2nd law of thermodynamics.

  171. Posted Jan 15, 2008 at 5:14 PM | Permalink

    Willis, Wouldn’t the material affect the model? E.g. bodies are wrapped in reflective foil because it keeps them warmer that a dark substance. So what the shell is made out of should affect the equilibrium result. Just trying to test out the limits of the model.

  172. Sam Urbinto
    Posted Jan 15, 2008 at 5:55 PM | Permalink

    BW, if electromagnetic radiation hits a surface that can absorb the energy, if any work is done the object heats. Why do you need to make it so complicated?

  173. Gunnar
    Posted Jan 15, 2008 at 6:50 PM | Permalink

    >> thus the “climate sensitivity” doesn’t have a single value.

    A fact which is extremely obvious.

  174. Gunnar
    Posted Jan 15, 2008 at 7:05 PM | Permalink

    >> EMR is not HEAT, and is only converted to HEAT when striking absorptive material at a lower T.

    BW, you are certainly correct, but many people here don’t want to be told that basic science contradicts their idiosyncratic theories. They want to define their own control system theory, their own thermo, their own basic physics. Tom Vonk completely shatters their illusions, so they ignore him.

    >> thus the “climate sensitivity” doesn’t have a single value.

    But when I explained this months ago, it was advocating my own idiosyncratic theory, rather than stating a basic fact learned from years of experience and knowledge as an applied scientist.

  175. Willis Eschenbach
    Posted Jan 15, 2008 at 7:49 PM | Permalink

    David, you say:

    Willis, Wouldn’t the material affect the model? E.g. bodies are wrapped in reflective foil because it keeps them warmer that a dark substance. So what the shell is made out of should affect the equilibrium result. Just trying to test out the limits of the model.

    What you say would definitely affect the absorption … but the absorption is a variable that’s set by the user. The model doesn’t care how you arrived at the figures for e.g. absorption, it just uses the figure (10% or 90% or whatever) you give it.

    w.

  176. Posted Jan 15, 2008 at 7:58 PM | Permalink

    Willis, But then the ‘steel greenhouse’ model provides no constraints on the surface temperature.
    At one extreme, a radioactive orange wrapped in a perfectly reflective foil would rapidly increase in temperature until the whole thing melts. A perfectly transmitting layer would act as if it wasn’t there, wouldn’t it? Or are you saying that only the component that is absorbing acts in some invariant way. Even so the reflectivity would affect the surface temperature, wouldn’t it? Cheers

  177. AK
    Posted Jan 15, 2008 at 8:05 PM | Permalink

    Re:#172, #176

    I’ve responded at the bulletin Board. Why not continue the discussion there as Steve requested?

    (Who knows, he might delete any subsequent posts on this subject.)

  178. jae
    Posted Jan 15, 2008 at 8:06 PM | Permalink

    Gunnar and Black Wallabe, I think you are right here. Here’s an experiment someone should do. Put 20 45 watt bulbs in a circle around a 100 watt light bulb and see if the 100 watt bulb increases in temperature with the 45 watt lights on. Simple experiment. I would do it if I had a decent thermocouple for measuring the temp. of the 100 watt bulb. I will bet a tip to the tip jar that the 100 watt bulb does not get hotter.

  179. John Creighton
    Posted Jan 15, 2008 at 8:12 PM | Permalink

    #178, does it matter. Even 100% IR absorption in Willis’s model only gives about a four degree temperature change.

  180. Willis Eschenbach
    Posted Jan 15, 2008 at 8:40 PM | Permalink

    Gunnar and Wallaby, let me echo Steve’s request that you please take this elementary (and error ridden) discussion of basic radiation physics to the bulletin board. Since you both agree that (emphasis mine)

    EMR is not HEAT, and is only converted to HEAT when striking absorptive material at a lower T

    it is clear that both of you would benefit greatly by reading page 32, and then all of Chapter 10, of the excellent MIT textbook recommended above by Cliff Huston before further posting. Wallaby, they discuss the exact situation you describe above … but they come to a very, very different conclusion than yours …

    w.

  181. Willis Eschenbach
    Posted Jan 15, 2008 at 8:59 PM | Permalink

    David S., another good question:

    Willis, But then the ’steel greenhouse’ model provides no constraints on the surface temperature.
    At one extreme, a radioactive orange wrapped in a perfectly reflective foil would rapidly increase in temperature until the whole thing melts. A perfectly transmitting layer would act as if it wasn’t there, wouldn’t it? Or are you saying that only the component that is absorbing acts in some invariant way. Even so the reflectivity would affect the surface temperature, wouldn’t it? Cheers

    If the foil were perfectly 100% reflective, yes, it would heat on and on.

    But if it is only 99.999999999% reflective, the maximum radiation temperature is exactly twice the initial temperature of the orange. See the steel greenhouse discussion for the math.

    w.

    • AK
      Posted Jan 15, 2008 at 9:42 PM | Permalink

      Re: #183

      But if it is only 99.999999999% reflective, the maximum radiation temperature is exactly twice the initial temperature of the orange. See the steel greenhouse discussion for the math.

      Willis, let me see if I’ve got your example right. Take two oranges, both with enough radium in them to produce 1 watt of energy. One is effectively a black body. The other is wrapped in aluminum with an emittance of 0.0001 (1/10000). Put them both in space with a temp of 2.5K. The surface temperature of the aluminum around the wrapped orange will be ~10 times that of the surface of the unwrapped orange. Given no conduction(/convection) the surface temperature of the wrapped orange is 2^1/4 the temperature of the wrapping. Agreed so far?

      Does this map to your model? What in mine maps to the “maximum radiation temperature?” What is “exactly twice?” What maps to “the initial temperature of the orange” (i.e. which orange in my example)?

      Do you perhaps mean “the maximum thermal radiation is twice that of the unwrapped orange?”

  182. Posted Jan 15, 2008 at 10:22 PM | Permalink

    John, I thought Willis’s example was that the shell raises earth surface by ~30K (2S).

  183. Willis Eschenbach
    Posted Jan 15, 2008 at 10:23 PM | Permalink

    AK, thanks for the reply:

    Willis, let me see if I’ve got your example right. Take two oranges, both with enough radium in them to produce 1 watt of energy. One is effectively a black body. The other is wrapped in aluminum with an emittance of 0.0001 (1/10000). Put them both in space with a temp of 2.5K. The surface temperature of the aluminum around the wrapped orange will be ~10 times that of the surface of the unwrapped orange. Given no conduction(/convection) the surface temperature of the wrapped orange is 2^1/4 the temperature of the wrapping. Agreed so far?

    Does this map to your model? What in mine maps to the “maximum radiation temperature?” What is “exactly twice?” What maps to “the initial temperature of the orange” (i.e. which orange in my example)?

    Do you perhaps mean “the maximum thermal radiation is twice that of the unwrapped orange?”

    The radiation temperature of the wrapped orange will be twice the radiation temperature of the unwrapped orange. What the physical temperature equilibrates at depends on the emissivity of each of the objects (orange and foil).

    w.

  184. Jan Pompe
    Posted Jan 15, 2008 at 10:31 PM | Permalink

    Willis Eschenbach says:
    January 15th, 2008 at 8:40 pm

    it is clear that both of you would benefit greatly by reading page 32, and then all of Chapter 10, of the excellent MIT textbook recommended above by Cliff Huston before further posting

    Willis could you check the link? I don’t know how the to read the 10th chapter of a six chapter book by a different author (John Lienhard lV an0d V) seems to be the right topic though.

  185. Cliff Huston
    Posted Jan 15, 2008 at 11:16 PM | Permalink

    RE#187 Jan,

    That is a six part book with eleven chapters and an appendix. Chapter 10 is in section IV (4), starting at page 525.
    And yes, A Heat Transfer Textbook, Third Edition by John H. Lienhard IV and John H. Lienhard IV – I provided the link, I did not write the book.

    Cliff

  186. Jan Pompe
    Posted Jan 16, 2008 at 12:37 AM | Permalink

    Thanks Cliff,

    I looked at the author’s name and really looked no further than the index. I think Willis and Gunnar have semantic problems and I’m not inclined to buy into it here.

  187. Andrey Levin
    Posted Jan 16, 2008 at 2:30 AM | Permalink

    Re#165, Willis Eschenbach:

    Thank you for detailed reply. It is really important to keep in mind how powerful are sunlight reflective properties of clouds. However, I was throwing stones in different window.

    As I see it, the basic problem of presented flowcharts is incorrect application of Kirchhoff’s law. It could be applied to radiative balance of whole Earth/atmosphere system averaged over long term, but it is not applicable to any smaller part of the system, even at thropopause, where latent heat/convection heat transfer is supposed to be negligible.

    Take for example my example of clouds, assumed to behave as blackbody for IR radiation. Clouds do not re-emit all of adsorbed thermal radiation, as Kirchhoff law would suggest. Clouds emit amount of thermal radiation as defined by SB law, according to 4 power of their absolute temperature, regardless of amount of IR adsorbed. The only way to increase back radiation from clouds is to heat them. I think that it would be impossible: clouds just will form higher and push thropopause higher, if initial moist air will be heated near surface. Again, I am talking strictly about IR, because in visible light clouds reflect, not just adsorb, re-emit, and thermalize.

    Now, gases compounding atmosphere are quite different, but I suspect general idea will hold: backradiation from air will increase only if we will manage to heat-up the atmosphere.

    All in all, I strongly suspect that backradiation numbers presented in the cartoons are order of magnitude overestimated.

  188. MarkW
    Posted Jan 16, 2008 at 5:18 AM | Permalink

    EMR is converted to heat when it is absorbed. Period. The relative temperature of the emitter and absorber doesn’t factor into the equation.
    The hot object emitts IR. The cold object emitts IR. The hot object emitts more than the cold object.

    It is possible for IR from the cold object to hit the warm object, and warm it further. At the same time, a greater volume of IR is travelling from the hot object, to the cold object, and warming it.

  189. Tom Vonk
    Posted Jan 16, 2008 at 6:08 AM | Permalink

    David Stockwell #168

    Willis, breaking the toy model is that while it uses a black body relationship for the steel shell, gasses do not act like black bodies. If they did, they would absorb all spectral frequencies (they don’t) and would emit red when heated (they don’t, though some flouresce).

    You are right even if what breaks the model is not the absorption .
    In this model , if I understood it well , absorption is whatever you want – it doesn’t check the physicality or unphysicality of the figures as it obviously can’t .
    What breaks it is the emission because this figure is a constraint .
    Yet this figure is calculated by the good old SB law that means that the atmosphere is considered as a perfect black body .
    It is not .
    For instance when Willis says that the troposphere emits “308 W/m² at 0°C” , he is actually saying that it emits as a black body (emissivity = 1) at 0°C .
    Well it doesn’t .
    Same remark is true for the stratosphere which doesn’t emit like a black body at some – 50°C and I have already pointed out the contradictions in statements saying simultaneously “N2 and O2 neither absorb nor emit” and “the atmosphere emits like a black body” .

    That’s why I don’t think that this toy model helps much to understand what happens when GHG concentration varies .
    It gives an average energy balance based on an assumption that atmospheric layers are black bodies what allows to play with a planet whose atmosphere could be approximately described as a black body .
    But that planet is not Earth .

  190. Jan Pompe
    Posted Jan 16, 2008 at 6:32 AM | Permalink

    MarkW

    A little addendum

    It is possible for IR from the cold object to hit the warm object, and warm it further. At the same time, a greater volume of IR is travelling from the hot object, to the cold object, and warming it.

    and the hot object is cooling.

  191. MarkW
    Posted Jan 16, 2008 at 6:38 AM | Permalink

    Jan,

    To go a little further.
    When an object emits IR, it cools.
    When an object absorbs IR, it warms.

    In the two body example, with one hot body and one cold body.
    Both objects are emitting and absorbing IR.
    The cold body absorbs more than it emits.
    The hot body emits more than it absorbs.

    As a result, the hot body cools, and the cold body warms.

  192. MarkW
    Posted Jan 16, 2008 at 6:41 AM | Permalink

    I may have gone too far there.

    The rate at which the bodies change temperature depends on the total geometry of the experiment. Which affects how much of the emitted IR is absorbed by the other body.

    This is my last post on the subject because I think I hear the zamboni warming up.

  193. Posted Jan 16, 2008 at 7:23 AM | Permalink

    Hmm… Another issue I just thought of is that the toy model assumes
    perfect conductive isolation between the shell and the earth. I think
    that conductivity (not radiation) between
    them would lead to equilibration. So it needs to be imperfect
    in terms of absorbion, and perfect in terms of insulation.

  194. Jan Pompe
    Posted Jan 16, 2008 at 8:28 AM | Permalink

    MarkW

    What I think is happening is that people are confusing heat with energy (radiation is energy in transit) if your were to do a dimensional Energy has dimension ML^2 T^-2, radiation ML^2T^-3 and heat has only temperature (theta). To determine what temperature (how hot) a body should be we need to know how much energy (vibrational for solids) it contains it’s mass and the empirically determined specific heat.

    Those who are saying the absorbing bodies convert the absorbed energy to heat have it right. There is nothing in Lienhard and Lienhard chapter 10 to suggest otherwise.

  195. Willis Eschenbach
    Posted Jan 16, 2008 at 1:09 PM | Permalink

    Jan Pompe, you say:

    Thanks Cliff,

    I looked at the author’s name and really looked no further than the index. I think Willis and Gunnar have semantic problems and I’m not inclined to buy into it here.

    ?!!?!!?!?

    Perhaps if you actually took the trouble to read the book, your opinion might change …

    w.

  196. Sam Urbinto
    Posted Jan 16, 2008 at 1:32 PM | Permalink

    Now. Planet. Molten core. Heat from the planet goes out. 10 KM of what used to be space. Steel shell. Space.

    1. How much heat is released from the crust.
    2. How does the heat transfer from the crust to the shell.
    3. What temperature does the used to be space non-atmosphere become.
    4. How much does the steel shell heat.
    5. How much heat does the steel shell transmit back into the now closed loop system.
    6. How does the heat transfer from the shell to the crust.
    7. How much more does the crust heat.

    It depends. What is the crust releasing, what is it made of. What is the shell absorbing, how thick is it? What “cooling” influence upon the shell does the outside have compared to the “heating” influence of the inside, given the crust and the closed loop?

    I give up.

    As far as the other matter, again, I ask why this is so complicated.

    When electromagnetic radiation hits a surface that can absorb the energy, when work is done the object getting the energy heats and re-emits it.

    Okay.

  197. Willis Eschenbach
    Posted Jan 16, 2008 at 5:09 PM | Permalink

    Sam U, you say:

    Now. Planet. Molten core. Heat from the planet goes out. 10 KM of what used to be space. Steel shell. Space.

    1. How much heat is released from the crust.
    2. How does the heat transfer from the crust to the shell.
    3. What temperature does the used to be space non-atmosphere become.
    4. How much does the steel shell heat.
    5. How much heat does the steel shell transmit back into the now closed loop system.
    6. How does the heat transfer from the shell to the crust.
    7. How much more does the crust heat.

    It depends. What is the crust releasing, what is it made of. What is the shell absorbing, how thick is it? What “cooling” influence upon the shell does the outside have compared to the “heating” influence of the inside, given the crust and the closed loop?

    I give up.

    1) The radioactive centre of the planet releases 235W/m2 from the crust.

    2) The heat transfers by radiation.

    3) There is no atmosphere. Space has no temperature.

    4) The steel shell heats until it is releasing 235 W/m2 from its surface outward to space. At that point, the system is in equilibrium.

    5) Since it is radiating 235 w/m2 outward, it is also radiating 235 w/m2 inward. It is not a “closed loop” system, since 235 w/m2 is going to space.

    6) Radiation again.

    7) The crust heats by the 235 w/m2 absorbed from the shell, giving it a total radiation temperature of 470 w/m2.

    w.

  198. Larry
    Posted Jan 16, 2008 at 5:19 PM | Permalink

    Space has no temperature.

    Actually, it does. The background radiation of space is equivalent to something like 2.7k (??). They have to use this number instead of 0k, because there’s that much microwave radiation in all directions.

  199. Reference
    Posted Jan 16, 2008 at 5:36 PM | Permalink

    #201 Willis Eschenbach

    According to this site the internal heat flow at the surface of the Earth is about 0.075 Watts/meter2

    This global heat map may also be of interest:

  200. Gunnar
    Posted Jan 16, 2008 at 6:23 PM | Permalink

    >> According to this site the internal heat flow at the surface of the Earth is about 0.075 Watts/meter2

    That’s not hard coded. It just means that there is not much delta T.

  201. Jan Pompe
    Posted Jan 16, 2008 at 6:35 PM | Permalink

    Willis Eschenbach says:
    January 16th, 2008 at 1:09 pm

    Perhaps if you actually took the trouble to read the book, your opinion might change …

    I did take the trouble to look at it and it isn’t very different from texts I already have and have had to work through in the past though a while ago, and I see no reason to change my opinion. You do get some things wrong so perhaps you need to work through it rather than just read it. For example in 201 you say:

    giving it a total radiation temperature of 470 w/m2.

    470w/m^2 does not have the dimension of temperature but of ML^2T^-3 (mass, length squared divided by time cubed) temperature is it’s own dimension. A black body which may be treated as a massless Bode gas (think of space that way) that radiates 470w/m2 will have a temperature of 301.7K. For a grey body the temperature will be higher for the same radiation you can calculate it by dividing it by the emissivity (

  202. Jan Pompe
    Posted Jan 16, 2008 at 6:48 PM | Permalink

    me

    Note to self: Watch those special characters.

    Now completing:

    For dappled coloured earth take a wild guess or integrated over space and spectrum.

  203. Willis Eschenbach
    Posted Jan 16, 2008 at 9:34 PM | Permalink

    Jan Pompe, thank you for your post. I had said:

    Perhaps if you actually took the trouble to read the book, your opinion might change …

    To which you replied

    I did take the trouble to look at it and it isn’t very different from texts I already have …

    My apologies for the misunderstanding, I had actually believed you when you said:

    I looked at the author’s name and really looked no further than the index.

    You are right that I used the term “radiation temperature” in an incorrect manner, please replace it with the terminology of your choice.

    w.

  204. jae
    Posted Jan 16, 2008 at 9:58 PM | Permalink

    1) The radioactive centre of the planet releases 235W/m2 from the crust.

    2) The heat transfers by radiation.

    3) There is no atmosphere. Space has no temperature.

    4) The steel shell heats until it is releasing 235 W/m2 from its surface outward to space. At that point, the system is in equilibrium.

    5) Since it is radiating 235 w/m2 outward, it is also radiating 235 w/m2 inward. It is not a “closed loop” system, since 235 w/m2 is going to space.

    6) Radiation again.

    7) The crust heats by the 235 w/m2 absorbed from the shell, giving it a total radiation temperature of 470 w/m2.

    w.

    I’ve beem biting my tongue, so as not to bug Steve Mc. But I just have to ask a question. 235 w/m^2 POWER is equivalent to 324 watt-hr/m^2 ENERGY. Are you saying that I can double the energy output of the furnace in my house by putting a steel shell around it? Or am I really missing something. I would be happy with 1/2 the energy back.

  205. Jan Pompe
    Posted Jan 17, 2008 at 12:16 AM | Permalink

    Willis

    thanks for your response. I had actually looked at it by the time I posted 205

  206. DeWitt Payne
    Posted Jan 17, 2008 at 1:21 AM | Permalink

    I’ve beem biting my tongue, so as not to bug Steve Mc. But I just have to ask a question. 235 w/m^2 POWER is equivalent to 324 watt-hr/m^2 ENERGY. Are you saying that I can double the energy output of the furnace in my house by putting a steel shell around it? Or am I really missing something. I would be happy with 1/2 the energy back.

    The energy output of the furnace doesn’t change by putting a shell around it, only the temperature. If you put a kerosene heater in an open field, the air temperature won’t change much. Put it in an enclosed space and the air temperature in the space will rise. If you draw an imaginary sphere around the heater that includes the enclosed space, the heat flux at the boundary of the sphere will be the same whether the walls are there or not.

  207. jae
    Posted Jan 17, 2008 at 8:11 AM | Permalink

    OK, I forgot about the equilibrium that is reached (BTW, I meant 235 w/m^2 power is equiv. to 235 watt-hr energy). Now, what happens when you turn off the heater (night time)? No addition of heat, so no more radiation (assuming a massless system). What is the average for the day?

  208. Gunnar
    Posted Jan 17, 2008 at 9:38 AM | Permalink

    >> BTW, I meant 235 w/m^2 power is equiv. to 235 watt-hr energy

    you actually mean 235 w/m2 results in 235 watt-hrs, after one hour. Just like driving 60 miles/hour results in being 60 miles away, after one hour. Driving 60 mph is not equivalent to being 60 miles away.

  209. jae
    Posted Jan 17, 2008 at 9:56 AM | Permalink

    OK, Gunnar.

  210. John Creighton
    Posted Jan 17, 2008 at 9:55 PM | Permalink

    Willis, it still really bugs me that none of the reflected IR radiation in your model gets absorbed. I’m trying to expand on your model and be consistent with it but that part of it just looks wrong to me.

  211. John Creighton
    Posted Jan 17, 2008 at 9:55 PM | Permalink

    Correction

    Willis, it still really bugs me that none of the reflected UV radiation in your model gets absorbed. I’m trying to expand on your model and be consistent with it but that part of it just looks wrong to me.

  212. DeWitt Payne
    Posted Jan 18, 2008 at 3:49 PM | Permalink

    John Creighton,

    The reflected UV doesn’t get absorbed (or better, can be neglected) because the shortest wavelengths where oxygen and ozone absorb have already been removed by the initial passage through the stratosphere and upper troposphere.

  213. DeWitt Payne
    Posted Jan 18, 2008 at 4:03 PM | Permalink

    Re: #211

    jae,

    A massless or zero heat capacity system is pretty unrealistic, but assuming radiation to space, the night time temperature would be 2.725 K so the average would be pretty low. Assume equal time on and off. Heater on blackbody temperature at the outer shell is about 255 K and off 2.7 K, which gives 129 K average temperature. Any finite heat capacity raises the heater off temperature drastically. The minimum surface temperature of the dark side of the Moon is well over 100 K, for example.

  214. jae
    Posted Jan 18, 2008 at 4:28 PM | Permalink

    DeWitt:

    Any finite heat capacity raises the heater off temperature drastically.

    Yes, definitely. How much effect do you think the simple heat capacity of the atmosphere has on temperature?

  215. DeWitt Payne
    Posted Jan 18, 2008 at 6:58 PM | Permalink

    jae,

    A lot. Dry ground has fairly low thermal conductivity which limits the effective heat capacity. For ground above the frost or dew point, the drop in temperature pretty much stops when condensation starts. Even when the temperature is below the dew point, conduction/convection keeps the temperature up when the temperature gradient gets high enough. If you model the heat transfer, with no convection, the temperature gradient only affects the first few centimeters of the ground while going up a couple of meters in the air.

  216. Pat Keating
    Posted Jan 18, 2008 at 7:14 PM | Permalink

    210 DeWitt

    This is somewhat off-topic, but do you have any references or links to IR earth radiance data for the 100-600cm-1 (16-100u) region, or any part of it?

  217. jae
    Posted Jan 18, 2008 at 7:54 PM | Permalink

    219: DeWitt: I am not asking about the ground. I am asking about the air, which has a heat capacity of about 1,033 joules/kg/K. There is about (10,000 kg/m^2 of surface area) of mass (STP) of air that is heated up every day and cooled every night. I’m curious about how important that heated mass is to maintaining the temperature of the planet?

  218. DeWitt Payne
    Posted Jan 18, 2008 at 10:01 PM | Permalink

    jae,

    Radiative cooling from any one kg of air is not very large compared to the heat capacity of that kg of air. That 10,000 kg column of air is only radiating on the order of 200 Watts to space total, assuming the Kiehl and Trenberth number for direct radiation to space from the ground is accurate, for a maximum cooling of 4.5 degrees in 12 hours and nets less than that, probably equal to twice the 24 hour absorption of incident solar radiation by the atmosphere or 134 W because the ground still radiates at night.

  219. DeWitt Payne
    Posted Jan 18, 2008 at 11:10 PM | Permalink

    Re: #220

    do you have any references or links to IR earth radiance data for the 100-600cm-1 (16-100u) region, or any part of it?

    No. I think it may be a detector problem. I’m not at all sure there are any good detectors for that range or if there are, they’re really expensive or not flyable or both. There is a version of the ARM AERI instrument that goes to 25 microns, but that’s about it AFAIK.

  220. Pat Keating
    Posted Jan 19, 2008 at 7:47 AM | Permalink

    223
    Thanks, DeWitt. Is there any earth-radiance data from that AERI instrument?

  221. beng
    Posted Jan 21, 2008 at 9:05 AM | Permalink

    What’s interesting to me about Willis’ simplified model is that it could suggest upper “limits” to potential greenhouse effects. Since the “real” atmosphere including GHGs doesn’t absorb all the radiation like the pure blackbody “shells”, that would indicate that the model will produce max effects & the real atmospheric warming will be less.

    Where/how much the greenhouse effects will occur in the atmospheric profile is of course the big question — the surface being of most concern.

  222. Sam Urbinto
    Posted Jan 21, 2008 at 3:41 PM | Permalink

    Larry: We decided to drop the background radiation since it’s very small. On the other hand, if the shell is blocking it, it’s not getting inside anyway.

    Willis: That’s why I called it used to be space non-atmosphere.

    Maybe I get it. (maybe not…) A light bulb in a vaccuum sphere made of steel. If the bulb has a radiant emittance of 235, then the sphere has an irradiance of 235. Are you saying that the sphere then has a radiant emittance of 235 that goes back to the bulb, at which time the watts per square meter at the bulb is 235 of radiant emittance and 235 of irradiance for a total of 470?

  223. Willis Eschenbach
    Posted Jan 21, 2008 at 7:13 PM | Permalink

    Sam U., you say:

    Maybe I get it. (maybe not…) A light bulb in a vaccuum sphere made of steel. If the bulb has a radiant emittance of 235, then the sphere has an irradiance of 235. Are you saying that the sphere then has a radiant emittance of 235 that goes back to the bulb, at which time the watts per square meter at the bulb is 235 of radiant emittance and 235 of irradiance for a total of 470?

    Exactly. Which is why a single shell greenhouse can only double the incoming energy and nothing more, and why the earth must perforce be modeled by at least a two shell greenhouse, with minimal sensible/latent heat transfer between the shells.

    w.

  224. Gary
    Posted Jan 22, 2008 at 3:43 AM | Permalink

    Willis,
    The max doubling is an outcome of the model not a physical reality. The max doubling only applies to an isothermal shell. Replace your shell with one with some thickness and a thermal conductivity. There will exist a temperature gradient across the shell from the inner to the outer surface. Analagous to the lapse rate. The inner surface has a radiant flux greater than the outer surface. The back radiation in this case is higher than the outer surface. The radiation flux from the inner sphere will be equal to the incoming plus the higher back radiation. The total irradiance can be more than doubled as the thermal conductivity falls.

  225. John Creighton
    Posted Jan 27, 2008 at 4:58 AM | Permalink

    In my model presented here:

    http://www.climateaudit.org/phpBB3/viewtopic.php?f=4&t=53

    The effect of more radiation being emitted towards the earth then away is shown. However, although I tried to improve on Willis’s model I sill don’t think the model I arrived at is very realistic.

  226. Willis Eschenbach
    Posted Jan 28, 2008 at 6:03 AM | Permalink

    Gary and John, thank you for your comments.

    Gary, you are correct that you can wrap a blanket of some kind around a source of radiant heat and achieve any temperature. In addition you can do it with or without radiant heat exchange. Just cut down on the conduction, no problem.

    However, in the atmosphere, there is very little heat transferred by conduction (other than the initial transfer by conduction from the surface). Almost all of the heat is transferred by convection (or radiation). This is because air is a very, very poor conductor of heat, viz:

    THERMAL CONDUCTIVITIES
    copper___________224
    aluminium________117
    steel_____________27
    ice______________1.3
    water___________0.32
    air____________0.014

    Thus, we are justified in ignoring conduction in the atmosphere.

    John, I took a look at your model, and I can only shake my head in awe. That way lies madness … but a stupendous work of madness, indeed. I salute you.

    For both of you, I was pushing the envelope the other direction — what is the minimum toy model sufficient to investigate the effect of a change in radiative properties? What can we do without? How simple a model can I make that is realistic enough to be useful?

    w.

  227. Black Wallaby
    Posted Feb 2, 2008 at 2:37 AM | Permalink

    It seems to have gone quiet here so, I ask for a discussion, which is relevant to the Trenberth cartoon.

    There’s this controversy where a minority of heretics have difficulty in accepting the theory that EMR from any source is always absorbed when it strikes an absorbent material, even when the target is hotter than the source. This appears to offend the second law of thermodynamics, but the standard explanation is that law 2 does not apply to EMR, because there are two different opposing flows, not 1. (in the simplest case)

    1) Despite this claimed uniqueness, “opposition of flow” has equivalences in all translatable forms of energy, without offending law 2, and here follow some examples. However, before listing them, and you dismissing it with “so what”, there is this strongly related consideration:

    It is not necessary to invoke in the calculation of net radiative heat transfer, that the radiation from a colder source is first absorbed by the warmer body. It is simply a matter of difference in potential, and is similar to conductive heat transfer, a matter of T1 – T2, and all of the translatable energy transfer calculations contain a similar term. (in EMR typically Q1 – Q2….or T to the fourth can be substituted for Q).

    Analogy 1: Consider a simple electrical circuit with a DC power supply of 12v. Apply an opposite supply of 3v, and the net voltage will be 9v. There is no transition point between 3v and 12v; implying that the 3v passes through the 12v, and subtracts from it. The 3v is not absorbed by the higher potential source, and, the 12v source continues to exert 12v. The resultant of V1– V2 is translatable to power, W1 – W2, and is also converted to heat across a resistor. (“absorbed” as heat just like EMR and translatable as Q1 – Q2).

    Analogy 2: Consider a water reservoir with a hypothetical perfect flow outlet pipe. The rate of flow will be proportional to the fluid pressure, or the difference in height between the water surface and the outlet. (potential) If we then place say a restrictive grille across the outlet, a back-pressure will be created, and we end-up with the familiar calculation P1 – P2 where P = pressure, in opposite directions, which is the equivalent of EMR. The PE of the water that is converted to KE, is the equivalent of heat.
    This example can also be adapted to show equivalence of EMR between matter of equal T: If we now close the outlet and divide the reservoir into A & B, the pressures each side of the divider are equal but opposite. Remove the divider, and there is no flow of water, with equal opposing pressures. This is the equivalent of no flow of heat between matter at equal T, or equal opposing EMR

    Other analogies may be derived, (assume hypothetical perfection in each case), such as:

    Analogy 3: Consider a block of material sliding down an inclined plane uniformly …(F1 – F2) x d = KE

    Analogy 4: Consider processes within opposing magnetic fields.

    Analogy 5: Consider opposing heat flows in a conductive material at equilibrium.

    As step 1 in the discussion, (more to come), do you agree that these ARE equivalences to EMR, and its translation into heat.

  228. Willis Eschenbach
    Posted Feb 2, 2008 at 3:24 PM | Permalink

    Black Wallaby, thank you for posting. Steve M. has created the “CA Forum” for discussing (inter alia) these types of physics questions, so I have answered your question there.

    All the best,

    w.

  229. Colin Davidson
    Posted Mar 3, 2010 at 6:49 PM | Permalink

    I posted in Unthreaded #43, but my comment really belongs in this (very old) thread. [Steve ticked me off for advancing a personal theory, but it really was a quibble about where the IR Radiation from CO2 comes from, so I thought I had better repost it on a more apposite thread.]
    It has been stated that the IR radiation from to space from CO2 in the 15um band probably originates in the troposphere. I disagree.

    The standard absorption tables for CO2 imply that at most 10% of the atmosphere is required for complete absorption. So any 15um band radiation from below the 90% height of the atmosphere, will be absorbed by the overlying CO2 and not make it out into space. I make this firmly above the tropopause. Even in the wings of the band, the majority of the radiation emanates from above the tropopause.

    This implies that any increase in concentration of CO2 will tend to increase the temperature of the radiation in this band – ie a NEGATIVE forcing.

    [The table I consulted for absorption by CO2 is as follows:
    (Figures are % transmitted)

    Wave Number Concentration, atm cm
    0.2 0.5 1 5 10 100 1000

    600 98/ 96/ 93/ 78/ 69/ 28/ 2
    650 74/ 61/ 48/ 16/ 8/ 0.1/ 0
    700 86/ 76/ 64/ 33/ 22/ 2/ 0
    750 99/ 98/ 97/ 88/ 82/ 45/ 10 ]

    • Dave Dardinger
      Posted Mar 3, 2010 at 7:12 PM | Permalink

      Re: Colin Davidson (Mar 3 18:49),

      One thing you need to keep in mind is that radiation from CO2 will arise from the temperature of the atmosphere itself as opposed to the surface temperature. That is, a CO2 molecule which absorbs IR or a water vapor molecule which absorbs IR will be hit by other molecules and lose the energy to the bulk of the atmosphere. Only the top layer of the atmosphere will be sufficiently thin to allow the back IR radiation to be appreciably less than the emission to space.

  230. Colin Davidson
    Posted Mar 4, 2010 at 12:49 AM | Permalink

    I agree with Dave Dardinger, and this is also what John Nicol so cogently points out – nearly all absorbed radiation is transferred as atmospheric heat to the bulk of the air due to collisions (only about 1 in 10,000 excited molecules can re-radiate before being hit in a collision.)

    It is also collisions which provide the energy for emissions, which is why the intensity is temperature and concenntration dependant. So for the majority of emissions the process is:
    1. collision(s) excites molecule
    2. If molecule is not again collided with, it may emit a photon. In general this is a low probability outcome – usually the molecule is stripped of its energy by another collision before it has time to radiate. However there are lots of molecules and lots of collisions, so emissions do occur.

    At the top of the atmosphere, there are fewer potential absorbers between an emitter and space. So if a CO2 molecule emits a photon, it is more and more likely that it will leave the planet the higher up you are (ie the less CO2 there is to absorb it).

    The question I posed was – what depth of CO2 will cause total extinction (or near enough to it) – and the answer I got was about one tenth of the atmosphere (mass) for the 15um band. So the lowest level from which 15um radiation can escape the planet is above 90% of the atmosphere, ie well into the Tropopause.

    For the wings (eg Wavenumbers 600, 750) over 50% seems to be coming from above this level as well.

    So virtually all the radiation from CO2 originates from the Tropopause and above…if my use of the absorption tables is correct…

  231. Methuselah
    Posted Mar 24, 2010 at 8:40 PM | Permalink

    Long ago I was taught that the only reason the Earth avoids a runaway greenhouse effect is that convection carries water vapour above the radiation traps. When the vapour turns to ice, the latent heat is liberated and mostly radiated away into space. This is a negative feedback, of course. Without any calculation at all, it seems obvious that a hotter sea surface means warmer, wetter air which is (on both counts) lighter, and this implies that more heat will evade the radiation traps by this process. Currently it appears from the diagrams that 80 W/sq m escape in this way, compared to 23 by reflection and 40 by surface radiation, so this is the principle mechanism. Of course it could be that positive feedbacks, such as more effective trapping of radiation due to water droplets, might outweigh the negative feedbacks (latent heat effect, cloud reflection effect). But if that were so, wouldn’t the Earth necessarily have net positive feedback, and a runaway greenhouse effect. Or, as I learned equally long ago, if A then B; not-B, therefore not-A.

    I have a nasty feeling I’m going to get a science lesson. Serve me right…

  232. Posted Dec 11, 2011 at 12:50 AM | Permalink

    To Whom It May Concern,
    We are General Education Taiwan(GET) Program performed by National Taiwan University, Taipei City, Taiwan,R.O.C., and have found the 1 image (FAQ 1.1, Figure 1. Estimate of the Earth) from the website http://climateaudit.org/2008/01/10/energy-balance-at-the-tropopause/ is quite beneficial in our educational program.
    We are sincerely requesting for your permission to use the image on our non-commercial teaching website (http://get.nccu.edu.tw:8080/get/ ; http://ocw.aca.ntu.edu.tw/). The image will be available for viewing on our teaching website and will be excellent teaching aids for the course ”Investigating the Earth” and visual references for our students. By granting this request, you would enable the educational personnel and the students at our institute as well as others in the collegiate community in Taiwan to learn more about our earth.
    If granted permission, we will be sure to note the original source of the data as well as provide a link to your homepage.
    Thank you for your time, and look forward to your reply and permission to post the image on the Website.
    Sincerely yours,
    Kelly Chang
    Assistant of the General Education TW
    Room 805 the Institute of Zoology, the college of LifeScience National Taiwan University Lifescience No.1, Sec. 4, Roosevelt Rd., Da’an Dist.,Taipei City 106, Taiwan (R.O.C.).
    TEL:+886-2-3366-9863
    FAX:+886-2-2363-8179

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