Visualizing the “Greenhouse Effect” – Light and Heat

Guest Post by Ira Glickstein

Solar “light” radiation in = Earth “heat” radiation to Space out! That’s old news to those of us who understand all energy is fungible (may be converted to different forms of energy) and energy/mass is conserved (cannot be created nor destroyed).

My Visualizing series [Physical Analogy, Atmospheric Windows, Emission Spectra, and Molecules/Photons] has garnered almost 2000 comments, mostly positive. I’ve learned a lot from WUWT readers who know more than I do. However, some commenters seem to have been taken in by scientific-sounding objections to the basic science behind the Atmospheric “Greenhouse Effect”. Their objections seemed to add more heat than light to the discussion. This posting is designed to get back to basics and perhaps transform our heated arguments into more enlightened understanding :^)

Solar "light" energy in is equal to Earth "heat" energy out.

[Click on image for larger version]

As I’ve mentioned before, during my long career as a system engineer I’ve worked with many talented mathematical analysts who always provided precise results, mostly correct, but some precisely wrong, usually due to mistaken assumptions. I got into the habit of doing a “back of the envelope” calculation of my own as a “sanity check” on their results. If their results matched within reasonable limits, I accepted them. If not, I investigated further. In those days my analysis was really done using a slide rule and scrap paper, but I now use spreadsheets.

The graphic above is based on an excellent spreadsheet from http://serc.carleton.edu/files/introgeo/models/mathematical/examples/XLPlanck.xls. It uses Planck’s Law to calculate the black body radiation spectrum from the Sun, as observed at the top of the Earth’s Atmosphere. It also may be used to calculate the radiation spectrum from the Earth System (Atmosphere and Surface, see below for explanation) at any assumed temperature. (I will refer to this spreadsheet as “Carleton” in this posting.)

I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).

Since the mean temperature at the surface of the Earth is 288 Kelvins (+15ºC which is 59ºF), that leaves 33 Kelvins (33ºC which is 58ºF) to be accounted for. Guess how we acount for it?

The yellow curve (above left) shows that Solar radiation is in a tall, narrow “shortwave” range, from about 0.1μm (microns, or millionths of a meter) to about 4μm, which we call ultra-violet, visual, and near-infrared. The vertical axis is Intensity of the radiation, measured in Watts/m^2/μm, and the horizontal axis is Wavelength, measured in μm. If you divide the area under the yellow curve into vertical strips, and add up the total area, you get 240 Watts/m^2.

Since we humans sense the visual portion of this radiation as “light”, that is the name we give it, and that has led to the false assumption that it contains no “heat” (or “thermal”) energy.

The violet curve (above right) shows that, assuming a mean temperature of 255 K, Earth System radiation to Space is in a squat, wide “longwave” range, from about 5μm to beyond 40μm, which we call mid- and far-infrared. If you divide the area under the violet curve into vertical strips, and add up the total area, you get the same 240 Watts/m^2 as is under the yellow curve.

DETAILED EXPLANATION

Left: Actual Solar radiation spectrum observed at top of Atmosphere, compared to black body model. Right: Black body Earth System radiation spectrum out to Space.

The graph on the left shows the actual observed Solar radiation spectrum (in red) as measured at the top of the Atmosphere. It is superimposed on a black body model (in blue) showing very good correlation. Thus, while the Sun is not exactly a black body, it is OK to assume it is for this type of “sanity check” exercise.

If you calculate the area under the curve you get about 1366 Watts/m^2. That means that a square meter of perfect black body material, held perpendicular to the Sun, would absorb 1366 Watts.

However, the Earth is not a perfect black body, neither is it a flat surface perpendicular to the Sun! So, to plot the yellow curve at the top of this posting, I had to adjust that value accordingly. There are two adjustments:

  • The Earth may be approximated as a sphere, with the Sun shining on only half of it at any given time. The adjustment factor for this correction is 0.25.
  • The albedo (reflectiveness) of the Earth system, primarily clouds and light-colored areas on the Surface such as ice, causes some of the Solar radiation to be reflected back out to Space without contributing any energy to the Earth System. The adjustment factor for this correction is 0.7.

After applying these adjustments, the net Solar energy absorbed by the Earth System is 240 Watts/m^2.

The graph on the right shows the black body model for an Earth System at a mean temperature of 255 K, a temperature that results in the same 240 Watts/m^2 being emitted out to Space.

Of course, the Earth System is not a perfect black body, as shown by the graph in the upper panel of the illustration below, which plots actual observations from 20 km looking down. (Adapted from Grant Petty, A First Course in Atmospheric Radiation, Figure 8.2, http://www.sundogpublishing.com/AtmosRad/Excerpts/index.html.)

The actual measured radiation is the dark squiggly curve. Note that it jigs and jags up and down between the topmost dashed curve, which is the black body spectrum for a temperature of 270 K and a lower dashed curve which is the black body spectrum for 230 K. This data was taken over the Arctic, most likely during the daytime. The Petty book also has a graph looking down from over the Tropical Pacific which ranges from 300 K down to 210 K. Observations will vary by tens of degrees from day to night, summer to winter, and Tropical to Polar.

However, it is clear that my result, based on matching 240 Watts/m^2, is within a reasonable range of the true mean temperature of the Earth System as viewed from Space.

NOTE ABOUT THE ABOVE ILLUSTRATION

WUWT readers will notice some apparent inconsistencies in the graphs above. The top and bottom panels, from Petty, peak at 15μm to 20μm, while the purple, blue, and black curves in the middle panel, and the Earth System curves from the Carleton spreadsheet I used (see above) peak in the 9μm to 11μm range. Also, the Petty black body curves peak at a “Radiance” around 100 mW/m^2/sr cm^-1 while the black body curves from Carleton peak at an “Intensity” of around 14 W/m^2/μm. Furthermore, if you look closely at the Petty curves, the labels on the black body curves are mirror image! What is going on?

Well, I know some of the reasons, but not all. (I hope commenters who are more fluent in this than I am will confirm my explanations and provide more information about the differences between “Radiance” and “Intensity”.) I have Googled and Wikied the Internet and am still somewhat confused. Here is what I know:

  • The horizontal axis in Petty’s plots are what he calls “Wavenumber”, increasing from left to right, which is the number of waves that fit into a cm (centimeter, one hundredth of a meter).
  • This is proportional to the frequency of the radiation, and the frequency is the inverse of the wavelength. Thus, his plots are the mirror image of plots based on wavelength increasing from left to right.
  • The spreadsheet I used, and my previous experience with visual, and near-, mid-, and far-IR as used in military systems, always uses wavelength increasing from left to right.
  • So, when I constructed the above illustration, I reversed Petty’s curves, which explains why the labels on the black body curves are mirror image.
  • Fortunately, Petty also included a wavelength legend, which I faithfully reproduced, in non-mirror image, at the top of each plot.

But, that still does not explain why the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet and other graphics on the Internet. I tried to reproduce Petty’s blackbody curves by multiplying the Carleton values by the wavelength (μm) and that did not move the peak to the right enough. So, I multiplied by the wavelength again (μm^2) and, voila, the peaks agreed! (I hope some WUWT reader will explain why the Petty graphs have this perverse effect. advTHANKSance!)

ANSWERING THE OBJECTIONS TO BASIC ATMOSPHERIC “GREENHOUSE EFFECT” SCIENCE

First of all, let me be clear where I am coming from. I’m a Lukewarmer-Skeptic who accepts that H2O, CO2 and other so-called “greenhouse gases” in the Atmosphere do cause the mean temperature of the Earth Surface and Atmosphere to be higher than they would be if everything was the same (Solar radiation, Earth System Albedo, …) but the Atmosphere was pure nitrogen. The main scientific question for me, is how much does the increase in human-caused CO2 and human-caused albedo reduction increase the mean temperature above what it would be with natural cycles and processes? My answer is “not much”, because perhaps 0.1ºC to 0.2ºC of the supposed 0.8ºC increase since 1880 is due to human activities. The rest is due to natural cycles and processes over which we humans have no control. The main public policy question for me, is how much should we (society) do about it? Again, my answer is “not much”, because the effect is small and a limited increase in temperatures and CO2 may turn out to have a net benefit.

So, my motivation for this Visualizing series is not to add to the Alarmist “the sky is falling” panic, but rather to help my fellow Skeptics avoid the natural temptation to fall into an “equal and opposite” falsehood, which some of those on my side, who I call “Disbelievers”, do when they fail to acknowledge the basic facts of the role of H2O and CO2 and other gases in helping to keep temperatures in a livable range.

Objection #1: Visual and near-visual radiation is merely “light” which lacks the “quality” or “oomph” to impart warmth to objects upon which it happens to fall.

Answer #1: A NASA webpage targeted at children is sometimes cited because they say the near-IR beam from a TV remote control is not warm to the touch. Of course, that is not because it is near-visual radiation, but rather because it is very low power. All energy is fungible, and can be changed from one form to another. Thus, the 240 Watts/m^2 of visible and near-visible Solar energy that reaches and is absorbed by the Earth System, has the effect of warming the Earth System exactly as much as an equal number of Watts/m^2 of “thermal” mid- and far-IR radiation.

Objection #2: The Atmosphere, which is cooler than the Earth Surface, cannot warm the Earth Surface.

Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.

Objection #3: The Atmospheric “Greenhouse Effect” is fictional. A glass greenhouse works mainly by preventing or reducing convection and the Atmosphere does not work that way at all.

Answer #3: I always try to put “scare quotes” around the word “greenhouse” unless referring to the glass variety because the term is misleading. Yes, a glass greenhouse works by restricting convection, and the fact that glass passes shortwave radiation and not longwave makes only a minor contribution. Thus, I agree it is unfortunate that the established term for the Atmospheric warming effect is a bit of a misnomer. However, we are stuck with it. But, enough of semantics. Notice that the Earth System mean temperature I had to use to provide 240 Watts/m^2 of radiation to Space to balance the input absorbed from by the Earth System from the Sun was 255 K. However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.

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958 thoughts on “Visualizing the “Greenhouse Effect” – Light and Heat

  1. Ira, the basic bookkeeping of the simplest greenhouse calculations assumes that the re-emission of infra-red radiation by the greenhouse gas molecules is spatially symmetrical. I am well aware that spontaneous emission is equally likely to be in any direction so that approximately 1/2 will head downward to the surface. However, there is another kind of emission – stimulated emission – and it is massively biased in the same direction of travel as the direction of travel of the stimulating photon. Thats what creates laser light, but stimulated emission can occur without lasing. IF there were any significant amount of stimulated emission happening in the atmosphere then the basic 1/2 back down calculation would likely be wrong because most of the IR through the atmosphere comes from the ground up and, if it is stimulating atmospheric emission, would enhance outgoing IR. Has anyone ever looked for evidence of stimulated emission of outgoing IR in the atmosphere?

  2. Let’s add two more objections to this partial visualization of the Earth System.

    1. Where are the storages? Oceans store and release huge amounts of energy. Elaborate your views of the energy/radiation balance.

    2. Energy is not just radiation. Earth System is more complex than that. For example, absorbed radiation energy may result in increased cloudiness that has impact on who much radiation reaches the earth. Ignoring major part of the system does not help in believing your conclusions.

  3. CO2 vs Earth’s Atmosphere Screen Saver!
    1. Find out how many pixels there are on your computer screen. Using basic math calculate the proportional size in pixels of an image that represents 380 and 400 ppm.
    For example if your screen had exactly 1million pixels the image would be exactly 380 or 400.
    2. (Sticking with the above example) Go to Google images and search for images that are…380…400..or 20 pixels. Download and place in the centre of an empty screen.
    3. Show the images to people who Believe and ask them if they still think that CO2 is anything other than a trace gas and ask them if the eensy teensy baby image; 20 pixels (the increase in CO2) looks sufficient to destabilize the other 999,980 parts of the atmosphere!

  4. This posting is designed to get back to basics and perhaps transform our heated arguments into more enlightened understanding :^)

    A little humour goes a long way… Nice, Ira!

  5. Ira, the Second Law of thermodynamics works at every level and you cannot transfer heat from a colder body to a warmer body. A black body cannot also reabsorb and re-emit lower intensity radiation that it has already emited. I’d say that it is also possible to arrive at the temperature difference between the black body radiating temperature and the average surface temperature without any recourse to radiant heating (adiabatic compression by gravity) and same for the surface temperature on the surface of venus. Greenhouse heating by back radiation is surely wrong because it says that if you put a frozen steak inside a vacuum flask with reflective interior that the steak would cook, that if you stand in front of a mirror you will heat up from the reflected rays or that you can save on your heating bills by filling the loft with CO2?
    I agree that we can shoot down the dire predictions of the AGW whilst acknowledging that the basic greenhouse theory is correct but there are many now refuting that there is any greenhouse effect.

  6. It should be pointed out that the 33K difference referred to is the effect of the heat-trapping gases in the atmosphere. In the case of each gas, the first small amount has the greatest effect, with diminishing effect as more is added, as according to Beer’s Law. For CO2, it’s effect is 90-95% spent and, thus, the hypothetical doubling of atmospheric CO2 would have little effect, possibly 0.10–0.01 K.

    It should also be pointed out that, as CO2 partitions 50 to 1 into the oceans, we would have to emit 50 times more CO2 that required to simply double atmospheric CO2. There is not enough carbon available to do this. If we really tried, we might be able to do 20%.

    So, where is the majority of atmospheric CO2 coming from? Outgassing from the oceans with warming, as according to Henry’s Law. And, of course, with cooling, the oceans will soak up CO2. There is a lag period before increases reverse to decreases, but with 30–50 years of cooling ahead, we will be able to see it happen.

  7. Ira is just doing a gross energy balance, so it seems to me that JT’s and Cherry Pick’s objections all occur within the black box of the Earth-black-body emitter.

    I don’t like that Ira treats albedo as a constant. That sounds like a fudge factor you would throw into a GCM.

    “The adjustment factor for this correction is 0.7.”

    What is that? 70%? I don’t think so but I wait to be illuminated after Ira reflects on this.

    How much heat energy gets transferred to the Earth black box by the electromagnetic interactions between the Sun and the Earth?

    240 Watts/m^2. What’s the error margin on that for the Sun (solar “constant” anyone?) and for the Earth? Seems you could hide a great deal of heat in there.

  8. 3. Show the images to people who Believe and ask them if they still think that CO2 is anything other than a trace gas and ask them if the eensy teensy baby image; 20 pixels (the increase in CO2) looks sufficient to destabilize the other 999,980 parts of the atmosphere!

    Yes! and then replace “CO2″ with Cyanide and “atmosphere” with your body. If they don’t believe you, just ingest 2mg of cyanide for each kilogram of your body weight (1:500,000) and show those iteyucktuals what for!

  9. Ira; The surface air temperature is higher then theoretical because of the insulative mass of the air of the atmosphere. Just like the insulation in the wall of a cooler or a kiln.
    Energy out equals energy in. The air is heated at the ground level and looses energy to space. So the air at ground level has more energy per unit then the theoretical black body temperature. The thicker and less dense the insulation the slower the energy flow out. The thinner more dense the insulation (atmosphere) the faster energy lose.
    The solar radiation incoming is of wave lengths that are mostly transparent to the atmosphere and heat the oceans and land. Thermal energy as heat and that carried in water vapor is transported through the atmosphere towards space. The surface air temperature is just a single point of measurement in this energy flow. Solar radiation that energizes the atmospheric constituents change the density and thickness of the insulation and therefore the rate of energy flow to space. Ultraviolet and shorter wavelengths are more effective at energizing gas molecules to effect changes in the insulative value of the atmosphere. pg

  10. The basis for the adjustments to the total absorption due to geometric and reflectivity are based on???

  11. “The only rational explanation is the back-radiation from the Atmosphere to the Surface.”

    aren’t you overlooking the obvious? ever fry an egg on the sidewalk from back radiation of the atmosphere?

  12. One often meets the assertions that (a) the sea in polar and sub-polar regions is commonly several degrees warmer than the air above it; and (b) in these regions, incoming solar energy has heated the air which in turn heats the sea.
    Please expand?

  13. I still have issues with down welling or back radiation in the troposphere. It needlessly complicates the simple fact that clouds, air have a temperature that varies. Increased CO2 can change that temperature but doesn’t beam back little phaser blasts of photons. The layering of temperatures in the atmosphere just complicates the radiative cooling like walls of differing layers of insulation thickness. Without the cutesy back radiation into a warmer surface invention, it might be easier to explain the potential radiative impact of CO2 and with a better model of the atmospheric layers, understand why it is not living up to its potential. Other than that, nice post.

  14. Show the images to people who Believe and ask them if they still think that CO2 is anything other than a trace gas

    Remove that “trace” gas and all photosynthetic life of earth dies. And us with it.

    Small concentrations can have powerful effects.

  15. “I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).”

    When you say absorbed by earth system, this mostly means amount the surface warms up and the atmosphere itself is only absorbing a small percentage of the incoming solar flux?
    And assuming this is the case wouldn’t different material absorb different amounts. Or has this been simplified to mostly account for how much the oceans absorb energy, and since ocean absorb the vast majority of energy that reaches the earth surface, anything else can be mostly be ignored?

  16. Ira,
    Thanks for your visualization.
    I’d like to comment on your sentence and adress some of the comments here:
    “The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
    To start, here is my view that I already posted on Climate etc awhile ago..
    As I understand it: “backradiation” is the downwelling infrared radiation integrated over all directions that go into the surface.
    I am an experimentalist. For me a physical cause is something that I can switch on and off, at least in a thought experiment, and get an intended effect.
    If I switch off the sun in a thought experiment, I get cooling of the atmosphere and the surface. If I switch on the sun again, I get warming of the atmosphere and the surface.
    If I switch off ”back radiation” in a thought experiment on the night side I get a higher cooling rate.
    If I switch on “backradiation” on the night side I get a lower cooling rate.
    If I switch on “backradiation” on the day side I get a higher heating rate. Combining my thought experiments I deduct that if I switch on “back radiation” on the day side I get a higher heating rate due to a lower cooling rate because of the “back radiation”
    Moreover, I can deduct that with “back radiation” the earth system gets or stays warmer than a reference system without “back radiation”, because of a lower cooling rate.
    One might also say “back radiation” leads to a warmer surface compared with a reference system.
    However, I do consider sentences as incorrect with respect to physics that state only: “back radiation” is warming the surface.
    Of course in any private conversation we state things like: my new coat is warming me nicely. But in a scientific discussion, we need to employ the correct wording, especially in lectures or papers.
    For my opinion sentences like: “back radiation” is warming the surface
    cause therefore a lot of misunderstanding in the blogosphere. The Gerlich and Tscheuschner discussion is based on it.
    Thanks again for your article.
    Best regards
    Günter

  17. Ira, the radiance formula involves either a “df” or “dλ” at the end. Converting from frequency to wavelength thus involves replacing df by -cdλ/λ², which explains why multiplying the frequency data by λ² puts the peak back in the right place.

    This same issue arises in Wien’s Law. The “peak color” is different, depending on whether you use wavelength or frequency as a variable. Intensity, however, is integrated over all colors, and gives the same result for either variable.

    /dr.bill

  18. In re Objection #2:

    Gas molecules follow ballistic trajectories between collisions…

    … unless all the work on Gravity since Galileo is wrong, and doesn’t apply at the molecular level, of course.

  19. Would it make any difference if we doubled the average insolation for half a day, then dropped it to zero for the other half (we’ll call that half night)? We’re radiating outward the entire time, day and night. Messes up the simplicity of the model, but since the temperature varies on a 24 hour clock, the outward radiation does, too.

  20. “As I’ve mentioned before, during my long career as a system engineer I’ve worked with many talented mathematical analysts who always provided precise results, mostly correct, but some precisely wrong, usually due to mistaken assumptions. I got into the habit of doing a “back of the envelope” calculation of my own as a “sanity check” on their results. If their results matched within reasonable limits, I accepted them. If not, I investigated further. In those days my analysis was really done using a slide rule and scrap paper, but I now use spreadsheets.”

    During my long career as a nuclear engineer, I got into the habit of doing a “back of the envelope” calculation of my own as a “sanity check” on other’s results. If their results matched within reasonable limits, I accepted them. If not, I investigated further. I also encouraged new engineers to do the same.

  21. What happened to your previous thread. It was getting interesting until no more comments could be posted

    http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/

    This above thread convinced me even more that backradiation does not further heat the planet. Luke warmers are as scientifically wrong as alarmists if they believe this

    The average temperature might slightly increase due to more CO2. But the highest temperature is due entirely to the Sun. In no way can CO2 and backradiation further heat the surface. It can only slow the cooling rate and by so doing might raise the average temperatures

    But a colder object can never further heat a warmer one. And saying this time and time again will not make it true. So this objection is correct IMHO.

    “Objection #2: The Atmosphere, which is cooler than the Earth Surface, cannot warm the Earth Surface”.

  22. I’d like to comment the “However, it is clear that my result, based on matching 240 Watts/m^2, is within a reasonable range of the true mean temperature of the Earth System as viewed from Space.”

    Actually for an inhomogeneous body, and even more if it is not a perfect black body, there is nothing like “the true mean temperature”. You can define different averages, but not a single one; the important thing (never clearly stated by climate scientists, although they probably know it very well), is that the average surface temperature (the integral of TdS divded by the surface), is different from the effective temperature Teff which is such that Prad = S.sigma.Teff^4 , where sigma is the Stefan constant. And these temperatures would also be different from a “best fit Black Body temperature”.

    The important thing is that you can easily make one of the temperature vary , holding another one constant ! for instance, the average temperature can vary without varying the effective temperature, that is with the same energy budget – and conversely. This can be demonstrated very simply by noticing that the local temperature can be written as +∆T where is the average surface temperature and ∆T a surface (not temporal) “anomaly”. But the only condition is that = 0. Any repartition with a vanishing surface average gives the same average temperature, but NOT the same effective temperature, since on average the won’t vanish.

    This lets a lot of room for spontaneous variation even with no change of forcings. This is carefully hidden under the name of “unforced variability”. After all during El Niño/La Niña cycles, the average temperature changes by several tenths of °C (= several decades of observed trend ! ) without any significant change in the energy budget.

    The gospel of climate science is that the unforced variability is restricted to 30 years, which would mean that its power spectrum is cut-off for longer variations. But it can be easily seen that this cannot be true. Natural variability is necessary to explain that long term, secular variations, at century and millenium scales. If it were restricted to 30 years, for instance, the paleoclimate data should show that the average temperature should follow very closely the forcing (Milankovitch) curve, which is obviously not the case. And there are obvious physical reason why the Earth climate could oscillate naturally on millenium scale – for instance that’s the characteristic timescale of thermo-haline circulation and it can be expected to give cycles at this frequency. This is much overlooked by climate scientists, in my sense.

  23. (sorry for the italic tags, can you fix it Anthony. Sorry also for the possible bad english, that’s not my mother language :) ).

  24. Ira
    Congratulations, you have provided a good, visual explanation of the basic energy inputs and outputs for the earth. The devil is, of course, as I am sure you would agree, in the detail.
    When calculating surface temperatures, variations in the assumed constants in the equations need to be considered. To me the most obvious ‘variable’ constant is that assumed for the earth’s albedo. Spatial and temporal variations due, amongst other things, to the ability of H2O to exist as all three of its phases within the surface and atmospheric regions of the earth, coupled with the large amounts of energy emitted or absorbed at the transition temperatures result in variations in cloud and ice cover, affecting both incoming and outgoing radiation and provide major feedbacks.

    When these are coupled with the vast energy storage capabilities of the oceans and more subtle long-term variations of energy input imposed on the earth and solar system from ‘external’ sources known and unknown, the prediction of the earth’s surface temperature variation over an extended period is not within our present capabilities.
    In my view the attempts by the powers that be to convince us that the science is settled are the only significant source of ‘hot air’ in the debate!

  25. “How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”

    Since the earth’s core is molten (very hot), there must also be convection of heat from the core to the surface. How much of the extra 33K heat can be attributed to earth’s own heat source? Is it none?… a little?…. or perhaps a significant amount of the 33K? Just wondering.

  26. grrr I should’nt use > for average. The correct thing is that

    T = Taverage + ∆T

    with the condition ∆Taverage =0

  27. Mike McMillan, you seem to need something like these:

    Here are a couple of energy budgets using KT97 and TFK09′s data for the overall averages of reflection (albedo) per the papers, irradiance at the surface, and measured TOA LW upwelling radiation. These are somewhat rough, a first stab at a cosine weighted across 24 hour view but they give you a closer idea of what actually occurs with Trenberth and Kiehls’s numbers. You will notice the totals all match what you see on their graphics.

    http://i56.tinypic.com/avc5g.jpg : KT97

    http://i53.tinypic.com/ir6lie.jpg : TFK09

    They need some small adjustment I can see already:
    • Need a better estimate of the actual evaporation rate at nighttime compared to evaporation in the daytime.
    • The exact time that the minimum and maximum temperatures are reached, on the average globally, might need to be moved 30 minutes or one hour sooner.
    • Window radiation is strictly in those spreads gauged by the surface temperature and I don’t know if this is strictly correct since the cloud layer over 62% of the earth needs to be considered, but, the average should be close.
    • …

    These are built upon what is actually happening in our atmosphere. You will notice the lack of the mention of back radiation for back radiation is a mirage. A figment of you imagination brought to life by the way the Stefan-Boltzmann equation is applied. In a layered gravity held atmosphere there is no real backward movement of net radiation or any other energy for that matter of fact. When the Stefan-Boltzmann equation is applied correctly to each thin layer you will find this to be true.

    Also, radiation in the CO2 and H2O main bands which lie outside the radiative window frequencies cannot pass downward to any great distance to any great degree for the atmosphere is nearly completely “black” to these frequencies. Another way to put it is the mean free path (mean distance a radiation will travel before being absorbed) is simply to short in the lower troposphere to allow this imaginary backward propagation from high in the atmosphere to the surface to ever really occur. Both of these GHGs are simply fast and longer reaching conduction of parcels of energy just like conduction but conduction has to move molecule to molecule, this radiation can jump many meters. Above the tropopause these limits are no longer present enhancing the movement of radiation to space.

    So, that is why the familiar 390 & 396 W/m^2 huge red arrows of upward LW radiation and the 324 & 333 W/m^2 of downward IR radiation are missing. They do not belong there. However, they are real and can be measured by radiometers at the surface but this radiation that is being measured there is in the bottom few hundred meters or so except within the window frequencies. It is local, very local as to compared to vastness of the atmosphere. They deserve to be mentioned on those famous budget diagrams but would have more honestly portrayed as a footnote.

    Mike, fell free to use these to make you contention a bit stronger.

  28. Alcheson;
    Since the earth’s core is molten (very hot), there must also be convection of heat from the core to the surface. How much of the extra 33K heat can be attributed to earth’s own heat source? Is it none?… a little?…. or perhaps a significant amount of the 33K?>>>

    When I first became interested in climate science that was one of my questions. After much reading and research and my own back of the envelope calculations, the answer came out “insignificant”. Same for radioactive decay of elements in the earth, and I even calculated the energy released by burning of fossil fuels… in the end, all insignificant. Solar rules.

  29. wayne says:May 7, 2011 at 11:56 pm . . .

    Thanks, Wayne. Seems to me the atmosphere is more of a blanket than greenhouse, just slowing the outward movement of energy. The temperature rises down here to the point where the increased outward movement once again roughly equals the insolation, not that they’re ever exactly equal.

  30. “The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
    Has anyone actually calculated how much energy can be absorbed by the GHG molecules in the air? Radiation is not the only heat transfer mechanism.
    I have difficulty rationalizing how 2% of the molecules in the air (H2O and CO2) can heat themselves, the other 98% of the air molecules, and the earth surface, by 33K only through back radiation of IR. The GHGs absorb and emit IR but the other air molecules can’t so the heat transfer has to be through collisions. What is the probability of a non-GHG molecule to collide with an excited GHG molecule compared to the probability of colliding with the earth surface and have direct energy transfer from there? (Equally reasonable, we could have “back-collisions” as well, when an air molecule with higher energy than the earth surface, transfer its excess energy back when colliding). If the energy can be inhibited from instantly going out into space by being absorbed by 2% air molecules then it must also be inhibited by being absorbed by e.g. oceans, and released at a later moment. (Some of the energy is also stored in vegetation through the photosynthesis which uses energy from the sun to convert H2O and CO2 into oxygen and cellulose, but that is a longer time perspective.) As has been pointed out in previous postings, we can’t gain more energy than we receive from the sun, but we can delay its disappearance so the 33K energy that is missing has to originate from the sun anyway. Earth is not at instant equilibrium (never at equilibrium at all), and there are many variables that can affect the energy budget at any instant.

  31. “However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”

    Suppose you cover all the ocean of the world with plywood floating “floor”.
    You can paint the plywood any color- a nice blue, perhaps.
    Would this cause any significant change of global temperature?
    And would average global temperature increase or decrease?

    This “plywood ocean” could be hotter [though depends on type of paint- not necessarily on simply what color pigment it has] during the day. Since we are not sealing the ocean just covering it, the ocean water would still have some evaporation, but still the nite should be cooler.

    Of course such thing would cause a massive extinct event and have all kinds wild effects.

    But the point is, would it warm the planet or cool it?

    And my answer is it would cool it, significantly.

    It could increase global average daytime temperatures- especially if you included the area of the plywood oceans and have somewhat lower global average nite time temperatures. But the northern hemisphere [where most earth land mass is] would get colder- mostly nite time temperature. And other land masses would get colder, and the ocean [beneath the plywood] would get colder.
    And looking from space and measuring earth’s radiation would indicate that earth has lost a significant part of it’s “greenhouse affect”

    [And if painted the "plywood ocean" black, or put asphalt on it, the earth would still cool.]

  32. Ira,
    Very detailed explanation. Hard to argue with the fact that when one breaks the data down into that much detail, one gets only one possible conclusion. Either that or all the physics we know that engineers use every day to make planes fly, internal combustion engines that work, air conditioners that cool, furnaces that heat, are wrong. Since the planes they design do fly, the engines work as expected, the air conditioners cool exactly as designed and furnaces heat right to specification, either the physics is right or a few billion calculations based on the physics got it right by pure coincidence. I’ll put my money on the physics!

    That said, you can break this down to a much simpler calculation. The sun’s radiance doesn’t get absorbed 100% by earth surface. Some of it is absorbed by the atmosphere. Similarly, the earth’s radiance is not released 100% from the earth’s surface, some of it is released by the atmosphere itself. So the earth surface might be 288K, but as seen from space, any given outward bound photon could have come from the earth’s surface or from the atmosphere. Seen from space, earth is not a “surface” but a sort of fuzzy/murky/semi-transparent sphere. Take any “area” that you can “see” from space, and measure the w/m2 being emitted, and you get a total that is mostly originating from the surface, and some that originated from the atmosphere.

    If we assume that your 240 w/m2 is correct, we can use Stefan-Boltzman Law to calculate the “effective” black body temperature of the earth, which if you accept “average” as an argument, would occur not at earth surface, but somewhere between earth surface and the top of the atmosphere.

    SB Law being P=5.67*10^-8*T^4

    If we use P of 240 watts/m2 we get T of 255.07 degrees K or -17.93 degrees C.

    The same logic still applies. Since we can measure what’s going into the system at TOA (top of atmosphere) as well as what is coming out, and the various satellite measurements and experiments like ERBE and others all arrive at a MEASURED value of arounf 240 w/m2, then there is only one possible conclusion. Earth surface is 288K or +15C on average because the atmosphere retains heat that keeps the surface warmer than it otherwise would be. Call it “less cooling” or “back radiation” or “invisible pajamas” but the measurements and the math arrive at the same conclusion. 240 watts/m2 going in =240 watts/m2 going out = -18C. Since the earth surface is +15C, it can only be 33 degrees warmer because the invisible pajamas are making it warmer.

  33. Cherry Pick says:
    May 7, 2011 at 8:38 pm

    “Energy is not just radiation. Earth System is more complex than that. For example, absorbed radiation energy may result in increased cloudiness that has impact on who much radiation reaches the earth. Ignoring major part of the system does not help in believing your conclusions.”

    What you have just used as an example is one that would cause a cooling effect. I’m not sure if that was your intent, while trying to educate Dr. Glickstein.

    Good nickname, by the way.

  34. Thanks David. Seems the radioactive decay of elements is estimated to be ~30TW (thought to be ~80% of the earth’s internal heating based on Wikipedia) which would result in about 0.1K of the 33K difference. Wonder how well that value is actually known? Off hand, 30TW doesn’t seem like a lot to keep the earth’s core molten. I might have to spend some more time investigating how that 30TW was derived, and see if it makes sense given the thermal conductivity and temperature gradient of the earth to convince myself as well that it is totally insignificant.

  35. Ira

    You should read the article below.
    It will answer all of your misconceptions about the “greenhouse theory” which in fact is pure fiction.
    1. It is true that the radiative equilibrium temperature of the Earth is -18C.
    2. This figure is confirmed by Satellite measurement from space.
    3. It is NOT true that all the radiation emitted from the Earth surface.
    4. The emission is mostly from the cloud level.
    5. If we pick say around 5km as the average emission height to space we will be near enough.
    6. Now use thermodynamics expressed in the form of the lapse rate.
    7. An average moist lapse rate is -6.6K/km.
    8. The lapse rate has nothing to do with back radiation as even scienceofdoom admits.
    9. Work back down from -18C at 5km to the Earth surface and we get 5×6.6 = 33K.
    So now you have an Earth surface temperature of 15C without any “greenhouse effect”.

    http://www.ilovemycarbondioxide.com/pdf/Understanding_the_Atmosphere_Effect.pdf

  36. Alcheson says:
    May 7, 2011 at 11:54 pm

    Since the earth’s core is molten (very hot), there must also be convection of heat from the core to the surface. How much of the extra 33K heat can be attributed to earth’s own heat source? Is it none?… a little?…. or perhaps a significant amount of the 33K? Just wondering.

    =====

    From memory … about a tenth of a watt per meter squared. Big enough that it should probably be explicitly counted in really serious models, but probably not enough to really affect anything. As I recall, the number is in Wikipedia somewhere.

  37. I’d like to run a little thought experiment here, and ask a question. I’m not pushing an agenda. It’s a very long time since I had the maths to follow the physics above, and I’m too idle to re-learn.

    Given a warm radiating surface, and cooler atmosphere of, say, 1 molecule. The surface radiates 10 units of energy out. Nine escape, one unit is trapped by the molecule, and re-radiated back to the surface. Net result is a loss of nine radiation units from the surface. The 2nd law of thermodynamics is satisfied: the cool body is not heating the warm body, merely slowing the cooling rate. Effectively “insulating” the surface (but not in the sense of a “blanket” preventing convection).

    Isn’t this the effect of “back-radiation”?

  38. At a quick flick through I don’t think there was much discussion on the fact that the earth is a rotating sphere.
    The solar constant for Earth is commonly given at 1366 W/m^2, and to make a simple average of what is absorbed over the surface and atmosphere, and absorption rates etc, (and the outgoing emissions, spread over the entire sphere), when the wattage is proportional to the fourth power of T, introduces a few complications. In short, might I suggest that the effective radiative T of Earth as is popularly touted is uhm questionable.

  39. Gotcha Henry!
    He thinks CO2 is like a toxin in the atmosphere!
    An amazing toxin which is NOT TOXIC at 380ppm but suddenly BECOMES TOXIC at 400ppm. There in a single post is the fuzzy thinking of the Warmists! Fantastic.
    Textbook example of ‘Carbonaphobia’!

  40. Oh and by the way Henry, do bear in mind that every time you breath out you are breathing out that toxic gas…be careful not to inhale too quick after you breath out!!
    Intellectual…ha…like shooting fish in a barrel.

  41. JT says:

    I am well aware that spontaneous emission is equally likely to be in any direction so that approximately 1/2 will head downward to the surface. However, there is another kind of emission – stimulated emission – and it is massively biased in the same direction of travel as the direction of travel of the stimulating photon.

    This is I believe wrong in that you are assigning some kind of kinetic impact of the photon even though you call it stimulated emission. The photon has no mass, so when it is absorbed it is the electromagnetic properties of the photon that is acting on the molecular structure. In an instant the electromagnetic fields of its components are both attracted and repelled depending on the phase polarity of the photon at the time of absorption.

    I find wave theory a better example of the action. At rest the molecule is in balance with its competing fields. The magnetic wave is an oscillation of both positive and negative fields. Dependant on it’s phase, the approaching wave will attract all components of the molecule that are the opposite polarity and repel all those the are the same polarity. As the wave progresses, the polarity is reversed and the components that were repelled are now attracted and those that were attracted are now repelled.

    The energy of the wave has now been absorbed and for a brief time the molecule has kinetic energy as, notionally the competing fields struggle to reach equilibrium. In doing so the molecules components that are now vibrating disturb the electromagnetic balance of all the surrounding molecules electromagnetic fields which due to their state of equilibrium with the wider body of gas offer resistance but must acquiescent to the force of the field by propagating the energy though out medium.

    The above thought bubble only applies to an ideal gas.

  42. The average global surface albedo is about 0.121 to 0.124. The .3 figure includes the cloud contribution. The AR4 model albedos averaged about 0.14 per Roesch (2006), which amounts to about 3W/m^2 globally and annually averaged. This 3W/m^2 of absorbed radiation that the models currently under represent, chiefly from earlier observed spring snow melts while they allegedly “match” the 20th century climate, will be added as the models project the next 100 years, catching up with the snow melt, an addition of heat comparable to the projected CO2 forcing.

    Another way to think of the significance of the CO2 concentration is that it is about 2.66 grams per square inch, that is about the mass of a penny. Imagine that penny flattened into a square inch plate, and contiguous over the surface of the earth with all the other pennies. Instead of having the optical properties of copper, this contigous shell is opague to certain characteristic frequencies of CO2. Obviously this is thick enough for several absorption of photons in that range and to allow practically no chance of these photons escaping directly from the surface to space. I don’t see how one can just hand wave this away with no deeper analysis than a dismissive mention of the concentration.

  43. You state that the back radiation has been measured. Well it might, or at least some long wave IR that is more probably short wave IR that has lost energy traveling through the atmosphere. It would be impossible to tell the difference from measurements at the surface. It is also true that if a molecule of any so called ‘greenhouse gas’ adsorbs energy it must, by 2nd law rules, share this energy with the surrounding non ‘greenhouse’ gasses, ie it loses heat through conduction and radiation.

    I do not see in your calculations any adiabatic effect which is important. Taking this into account raises the temperature at the surface. Also, when any gas is heated it will convect. Not an easy effect to measure in the laboratory compared to the real atmosphere. Convecting air will cool.

    Jupiter, with an atmosphere of mainly hydrogen, has warming at the 1 atmosphere level similar to Earth. So there it must be an adiabatic temperature rise not ‘greenhouse effect’. Solar input out there is much lower than Earth’s.

    True, glass does not transmit IR energy but a greenhouse, transmits visible light and so doing allows energy losses inside, warming the interior by visible light. These losses mean that the lower wave light converts to IR which cannot be transmitted out through the glass. This, together with lack of mixing with cooler external air raises the internal temperature. At night the temperature falls rapidly. 2nd law dictates that the higher the temperature difference of two bodies the greater the heat flow between the two bodies. Entropy must increase.

    Taking into account the adiabatic effect, which is real and measurable, there is no need for the GHG effect.

    The greenhouse effect also has one fatal flaw. The warming in the mid troposphere in the tropics shown by all models has yet to be found in real observable data. I do not care what your calculations show (in reality another model), if this warm area shown by models is not there then the theory falls flat on its face!

  44. Martin Lewitt says:

    ……Another way to think of the significance of the CO2 concentration is that it is about 2.66 grams per square inch………

    Its an odd unit, but it seems much too high;
    Ive worked it out in g/cubic inch as 0.000074!

  45. Bob_FJ says:
    The solar constant for Earth is commonly given at 1366 W/m^2, and to make a simple average of what is absorbed over the surface and atmosphere, and absorption rates etc, (and the outgoing emissions, spread over the entire sphere), when the wattage is proportional to the fourth power of T, introduces a few complications. In short, might I suggest that the effective radiative T of Earth as is popularly touted is uhm questionable.>>>>

    One of… OK, my BIGGEST pet peave. There is no such thing as an average temperature or w/m2 that is all that meaningfull. The rough estimate commonly used in climate circles is 1366 x .5 (for day/night) = 683. 683 x .5 (cuz its a sphere) = 360. 360 x .7 (for albedo) = 250, pretty close to Ira’s 240.

    But that of course would be around 500 w/m2 at high noon at the equator, and pretty much 0 for six months at a time at the poles. So despite being relatively cold, the poles actually beam way more energy into space than they recieve from the sun, and the equator, despite getting very high insolation year round (in daytime at least) is a net absorber (the excess being moved away by wind and water currents).

    So trying to come up with an average insolation value and an average temperature value is meaningful in terms of the total of each. But since insolation varies with w/m2 and w/m2 varies with T^4… trying to track “warming” by following “temperature” is silly. 1 watt of extra insolation results in a much larger temperature anomaly at the south pole than it does at the equator. So when the IPCC says that the “average” temperature of the earth has increased about 1 degree “on average” since 1880, that is highly misleading.

    But if they said about 0.3 degrees at the equator and about 1.5 degrees at the poles, it would sound a lot less alarming. And before anyone starts screaming about the ice caps melting, let’s keep in mind that T^4 applies by latitude as well as season. So to get an even more reasonable expression of that “one degree average” they would need to say something like:

    Equator +0.3 degrees all year round.
    Poles +0.5 degrees in summer
    Poles +2 degrees in winter

    In other words, that scary “one degree” is mostly confined to the dead of winter in high latitudes. Sorry “the ice caps are melting we’re all going to drown” crowd, but if the poles go from -40 in the dead of winter to -38, the extra melting would be…. between almost zero and zero. And neither the polar bears nor the seals will notice. Or mind.

  46. I’m surprised that a “skeptical” site such as WUWT would print such nonsense:-

    “Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.”

    [b]No![/b] The earths surface is already emitting radiation in the Infra Red. It’s atoms and molecules are ALREADY excited sufficiently to be emitting IR. Therefore that IR scattering back to the surface CANNOT warm the surface any further because the atoms and molecules of the ground are already kinetically excited to this level. The IR will merely be scattered back to the atmosphere again- and eventually espace to space. The radiation from a body can ONLY transfer heat to a cooler body. The second law applies also to radiation.

    This sentence is true however:-
    ” Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.”

    Yes, and temperature of that surface is one of those characteristics. The radiation, does not need to “know” anything! If the surface is ALREADY emitting radiation that is “hotter” , i.e. more energetic then the “back radiation ” will in effect just be reflected again.

    Heat always passes from hot to cold, never the other way round, THERE ARE NO EXCEPTIONS!

    Ira Glickstein does not seem to understand how matter interacts with radiation.

  47. “I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).

    Since the mean temperature at the surface of the Earth is 288 Kelvins (+15ºC which is 59ºF), that leaves 33 Kelvins (33ºC which is 58ºF) to be accounted for. Guess how we acount for it?”

    We can account for the 33C difference without any need for an imaginary greenhouse effect. I urge everyone at WUWT to read this brilliant paper by physicist Joe Postma ;

    http://www.globalwarmingskeptics.info/forums/attachment.php?aid=327

    In this paper, Joe accounts for the 33C. The truth is that the radiative temperature of the Earth is measured from space IS -18C. And it is true that the average air temperature near the surface of the earth is 15C. However, if one includes the whole of the atmosphere right up to the edge of space and the surface of the earth in calculating the average radiative temperature of the Earths surface [i]and[/i] atmosphere, then we get to the -18C figure.

    There is no discrepancy, there is no need for an imaginary “greenhouse effect”.

    Surely the Phlogiston of our times!

  48. ….and just for the “cold things can’t heat warm things” crowd, here is a link to Denmark’s DMI which publishes average temperature in the arctic going back to 1958. The green line is the average and the red line is “this year”:

    http://ocean.dmi.dk/arctic/meant80n.uk.php

    Now take a close look at days 1 to about 60. That’s roughly January 1 to the end of February. The average is flat at about 247 K or -26 degrees C. Think about that for a moment. ITS DARK THE WHOLE TIME! NO SUNSHINE! NO INSOLATION!

    Despite which it drops down to around -26C by December… and then just stays there. Why doesn’t it keep cooling off? After all, the atmosphere above the arctic is even colder than the arctic itself… Heck, at 25,000 feet its colder than that even at the equator! So… with 0 watts/m2 from the sun, the arctic nonetheless sticks right around -26 degrees. Hmmmmm…..wonder what keeps it from cooling off more….

  49. Alcheson;
    Off hand, 30TW doesn’t seem like a lot to keep the earth’s core molten.>>>

    You’re thinking about it wrong way round. The 30 TW (actually I think it is 44 TW but no matter) is what ESCAPES from the molten core to the surface. OK correction, the outer core is molten, the inner core is actually solid. Fact is that dirt and rocks are pretty good insulators. In high latitudes where winter temps might be as low as -40, houses have to be constructed on a foundation that goes below the “frost line”. That line is a few feet below the surface and in even the harshest of winters, the dirt only reaches the freezing temperature by no more than an inch or so deeper than average. Get about 10 feet down and the temperature varies by almost nothing all year round. So figure that molten outer core is insulated by kilometers of dirt and rock…the heat gets out…but slowly.

    There’s a decent summary at wikipedia (not the best source for climate info but good enough in this case) http://en.wikipedia.org/wiki/Earth's_energy_budget

    Geothermal about 0.08 w/m2
    Tidal about 0.006 w/m2
    Fossil Fuel consumption about 0.025 w/m2.

    Compared to 250 w/m2 from solar… pretty minor.

  50. All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.

    But will be reflected by a body of a higher energy level.

    http://www.vermonttiger.com/content/2008/07/nasa-free-energ.html

    In fig. 3 a box states:

    CO2 & H2O ~15µ absorption; re-emit 7µ, 10µ, 15µ.

    But re-emission at a higher frequency (shorter wavelength) than the original absorption is not possible.

  51. Notice that the Earth System mean temperature I had to use to provide 240 Watts/m^2 of radiation to Space to balance the input absorbed from by the Earth System from the Sun was 255 K. However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.

    255K is the mean earth radiative temperature at around 4,000 metres altitude and above. Escaping IR radiation is from a volume of the atmosphere, not a flat surface.
    The mean surface temperature of 288K is simply the compression factor of gravity on the air. Air at the surface has the highest density and returns highest temperature. Think deserts, polar regions, equitorial regions. All very different suface temperatures but always the same adiabatic lapse rate, water vapour content allowing.

  52. There is a bathtub which can hold energy. (the earth’s atmosphere/surface/oceans)

    There is a energy tap flowing into it. (the sun)

    There is a hole in the tub where energy flows out. (radiative emission into space)

    Inflow = outflow, energy in = energy out.

    The quantity of energy the black-box “bathtub” is able to hold (wants to hold) can go up and down and is indicated by the temperature. When the variables which alter the level stored in the tub are changed (knobs turned) there will be an imbalance between inflow and outflow until the new equalibrium is reached – a equilibrium at which a different level of energy is stored in the tub. We expect a change in the total level of energy stored to affect the temperature at the bottom of the tub.

    I expect the IR interference to be one of those variables but I’m sure there are many others. The claim that 30 degrees celsius at the surface is all due to the greenhouse effect is rubbish.

    3 scenarious.

    A. Black body earth (-30 % albido).
    B. Earth with atmosphere and no IR interference.
    C. Earth with atmosphere and IR interference.

    A is cooler than B is cooler than C.

    C – B = greenhouse warming.

  53. Davidmhoffer,
    The ground and ocean warm the lower levels of the atmosphere in the polar winters.
    Does anyone else wonder what the rate of heat transfer to the surface is from conduction through the crust from the mantle?

  54. We can account for the 33C difference without any need for an imaginary greenhouse effect. I urge everyone at WUWT to read this brilliant paper by physicist Joe Postma ;

    http://www.globalwarmingskeptics.info/forums/attachment.php?aid=327

    Agree it was a brilliant paper. But this Postma paper was posted on Ira’s previous thread and Ira and others criticised the paper or would not read it all.

    http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/

    Ira Glickstein, PhD says March 30, 2011 at 2:02 pm

    This post by Ira has not really moving forward at all with Ira still posting this for example

    “How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”

    This is just total nonsense. (The only rational explanation). It is one explanation and not a very good one. Unfortunately Ira is just ignoring comments on previous thread so this series is not moving forward. He could summarise the main areas of contention without giving a viewpoint rather than posting again his beliefs that he knows many disagree with

  55. Engineering covers a considerable range of displines such as chemical, civil, electrical, mechanical etc. Heat transfer, thermodynamics, fluid dynamics are basic subjects for some of these disciplines but are not taught in all the disciplines. It is not surprising that some engineers do not understand the fundamental theory and experimental knowledge in these subjects. It is surprising, however, that qualified professional engineers who supposedly comply with a code of ethics which includes the requirement of competence should express opinions about a subject which they do not understand.
    Ira, you have been sucked in by climate scientists who have no understanding of heat transfer, fluid dynamics and thermodynamics. You make no mention of heat transfer by convection and phase change (evaporation and condensation). One can only assume that you have no knowledge of this subject. For a quick overview I suggest that you look at Perry’s Chemical Engineering Handbook. Learn about Nusselt and Reynolds numbers. Also, take note of the definition of a black body -it has a surface, and all absorbs energy of all wavelengths. The atmosphere is not a surface and the trace gas CO2 only absorbs radiant energy in very narrow wavelengths ie it is not a black body. The basic Stefan Boltzman equation is not applicable to the atmosphere. Engineers have carried out a huge amount of research on radiation involving gases in heat exchangers.
    Then read the following http://climategate.nl/wp-content/uploads/2011/02/CO2_and_climate_v7.pdf

  56. Dishman says:

    In re Objection #2:
    Gas molecules follow ballistic trajectories between collisions…
    … unless all the work on Gravity since Galileo is wrong, and doesn’t apply at the molecular level, of course.

    Ballistic trajectories are trajectories under the action of gravity. Note, he didn’t say they travel in straight lines.

    However: the approximate speed of molecules in the atmosphere is about 1,700 km/h. The path isn’t very curved, in other words. And the mean free path of a molecule in air is about 65 nanometres. You’d be pretty amazing if you could measure the deviation from a straight line of a path segment with that length and curvature.

  57. I’m reading the post by Joe Postma now. He appears to be saying my energy tub can only be filled to a maximum level and can’t hold more after that.

    It already occurred to me that climate scientists didn’t really have any firm understanding of the mechanism of action of the greenhouse “theory”, hence this very subject has been at the front of my mind lately, hence my bathtub theorizing. I’m glad to see this particular issue is slowly but surely coming to a head.

    If Joe is right they’ll just start saying that energy is being redistributed in a way to make the surface warmer.

  58. Bryan,
    I read the link you gave, and the guy got it partially wrong. He stated that a black body near a light bulb in equilibrium would not heat up if a reflector was then introduced to reflect back onto the black body some of the radiated heat. This is wrong. The black body would absorb the reflected radiation, and thus the equilibrium temperature would be increased. However, this is not related to the back radiation issue. The main remainder of yours and his points are correct, i.e., it is the effective average altitude of outgoing radiation and lapse rate that heats the surface, and backradiation is a consequence rather than cause of the heating. The reason is that buoyancy overcomes absorbed radiation heat transfer to prevent stagnant heating from the radiation.

  59. John of Kent says:

    I’m surprised that a “skeptical” site such as WUWT would print such nonsense:-

    “Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.”

    [b]No![/b] The earths surface is already emitting radiation in the Infra Red. It’s atoms and molecules are ALREADY excited sufficiently to be emitting IR. Therefore that IR scattering back to the surface CANNOT warm the surface any further because the atoms and molecules of the ground are already kinetically excited to this level. The IR will merely be scattered back to the atmosphere again- and eventually espace to space. The radiation from a body can ONLY transfer heat to a cooler body. The second law applies also to radiation.

    Firstly, I do wish that people who think they know something would just be polite about it; this is not a peer-reviewed journal (thank goodness). Anthony is offering a wonderful service here. The way I look at it is: the entire article, including comments, is the publication. If an article contains fallacies, we will learn from the more knowledgeable commenters.

    Having said that, you are not one of those more knowledgeable commenters. All radiators are absorbers. Contrary to your assertion, anything capable of emitting a certain wavelength at a certain temperature can absorb ditto. So yes, a photon radiated back towards the surface can indeed be absorbed by the surface. What this absorption will do, however, is raise the temperature slightly so that the surface now emits at a higher rate. Heat still travels on average from hot to cold. The temperature increase of the hotter surface, whilst happening only if the back radiation is happening, still comes about because of the impinging heat from the sun.

    Think of it this way: focus a flame upon a single cubic inch of steel that happens to be part and parcel of a massive steel girder. The flame’s heat will flow into the steel all around and you will find it very hard to increase the temperature of the cubic inch by very much. Now focus your flame on a cubic inch embedded in another surrounding cubic inch, embedded in a very poor conductor. Your heat will heat the first cubic inch, which will heat the second. The poor conductor will prevent escape of much heat and the molecules of steel will bump upon each other. Molecules in your first cubic inch will bump (heat) those in the second, those in the second will bump those in the first. The entire two cubic inches will quickly increase in temperature. The outer cubic inch will never become hotter than the cubic inch being directly heated, and yet the directly heated part will rise in temperature. So there is such a thing as back flow of heat and it is by no means contrary to the 2nd law.

    But having said that, the idea of back radiation as formulated by some warmists doesn’t hold much water. The most likely thing to happen to a CO2 molecule that has absorbed a photon and risen to a higher energy state is that it will collide with another molecule and transfer the energy into general heat of the whole gas. When that heats the gas enough, then the number of CO2 molecules undergoing the reverse phenomenon (getting knocked into a higher state by an impinging molecule – remember, these processes are always reversible) will rise enough to provide enough radiation that back radiation becomes significant. But there is a huge heat capacity to fill before that point is reached, since the CO2 is a trace gas heating a vastly huger bulk of N2, O2 etc. In the meantime, the heating of the gas starts convection, limiting the rise in lower air temperature to the adiabatic lapse rate. So once again, the back radiation has its effect auto-limited. Since all planets we see (Earth, Venus, Jupiter) are at close to maximum lapse rate, this is clearly a phenomenom that is quickly reached by almost any trace amounts of greenhouse gas in the atmosphere, and so the mechanism required for the AGW theory to work is already exhausted to within a few percent, and thus the phenomenon is basically harmless.

  60. Ira,
    I enjoyed your clear presentation of detail until I saw the kicker “How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”

    There have been several good presentations of the correct reason. The presence of IR absorbing gases acts as a partial radiation insulator. However, the atmosphere is unbounded, so if there were even a small amount of local heating occurs due to absorbing IR, the warmer air would rises and then cool by adiabatic expansion. In fact, the atmosphere mixes and rises so strongly, that noticeable radiation warming does not occur. The absorbed radiation passes energy into the surrounding air, but the more energetic air molecules (from the velocity distribution) impacting the absorbing molecules cause radiation, and the average is so in balance that the air is considered to be in LTE. However, the fact of the partial radiation insulation (i.e., prevention of direct radiation to space) results in the radiation to space occurring at higher altitude (average about 5 km), and it is the adiabatic lapse rate combined with the equilibrium temperature at 5 km that results in the ground temperature. The back radiation and ground up radiation are a result of the increased temperature, not the cause!

  61. I didn’t think he was right with the mirror example. If you shine a bright torch and a dim torch at a object it will get hotter than if you shone only the bright torch. If you shine the bright torch and reflect a mirror at the object it’s like adding another dimmer torch.

  62. “Heat always passes from hot to cold, never the other way round, THERE ARE NO EXCEPTIONS!”

    What about the sun corona? It’s a million degree and sun surface is cooler.
    Perhaps you referring to solids. Such as warm air can not heat up a warmer surface via radiant energy.
    Say, you had an object in space and it’s temperature was 100 K. This is a cold object, -173 C but compared to background temperature of Space 2.7 K it’s fairly warm. And it is certainly radiating energy into space. No part of earth is as cold as 100 K, if colder can not pass to warmer this means that with infrared telescope on Earth you could never detect such a cold object in space. It also means that your telescope mirrors would need to as cold or colder than any cool object you wanted to observe.

  63. * Blackbody – As you have stated, the inbound radiation spectrum from the Sun can be approximated as black body (at it’s temperature) and from the surface (at it’s temperature). The implication being, a blackbody emits in a radiant distribution dependant on it’s temperature. But you incorrectly indicate that Atmosphere is also a black body and it is NOT.

    “It also may be used to calculate the radiation spectrum from the Earth System (Atmosphere and Surface, see below for explanation) at any assumed temperature. ”

    If CO2 absorbs at 15 micron, it also emits at 15 micron (not a blackbody distribution of all wavelengths based on atmospheric temperature). So where your chart says CO2 ‘Re-Emit 7, 10, 15 micron’ – that is not correct. Similarly for H2O. The purple, blue and black curves in your atmosphere graphic have no meaning.

    The ‘looking down’ emission spectrum tells you where the energy was emitted if you look at the temperature curves. The atmospheric window wavelengths are like the surface temperature, cold from the antarctic, hot from the Sahara, with no emission from the intermediate atmosphere. The CO2 wavelengths always look like the tropopause temperatures no matter if they are measured over the desert or the poles, which tells us we are seeing emissions from high in the atmosphere, not re-emitted near the surface. And the H2O wavelengths seem to be in between, more like upper atmosphere and perhaps indicative of cloud tops or ice crystals.
    But ‘looking down’ distribution doesn’t look like a one-temperature blackbody distribution. Using a one size fits all blackbody emission from the atmosphere at any one temperature is probably not a good assumption.

  64. Radiant energy accounts for all of the outbound energy at the top of atmosphere, but only a fraction of the energy transfer at the surface. Understanding how the energy moves up through the atmosphere via non-radiative pathways is key in estimating potential GHG effects.
    If we use values from Kiehl & Trenberth 1977, they show 60% of the surface energy transfer to the atmosphere is due to Thermals (convection) plus Evapo-transpiration (water evaporation and release as latent heat). Leaving no more than 40% as Radiant Energy transfer, with 24% of that 40% as direct loss to Atmospheric Window. Whatever the GHG affects may be, they are happening in the 16% of the total that is left. (And I read somewhere that Trenberth used 40 W / m-2, knowing that the actual ‘window’ value was more like 66 W/m-2. That would increase the ‘atmospheric window’ from 24% to about 39%, leaving only about 1% of the total for the net of all other radiant emissions).

    How can that be, since we know the surface emits a blackbody distribution dependent on the surface temperature? One possible explanation is to look at what the energy ‘pathway’ might be. Sort of by definition, the radiant energy that is NOT in the atmospheric window wavelengths, is susceptible to being absorbed by one of the GHG, mostly CO2 and H2O. So the surface emits radiant energy, and the GHG susceptible wavelengths are absorbed fairly close to the surface. Next, we think that the absorbed photons are more likely to be ‘thermalized’ rather than re-emitted as photons (10,000 : 1 more likely thermalized). Thermalized means the GHG molecule collides with a non-GHG molecule in the air, converting the photon to kinetic energy, which can now be seen on a thermometer. Now all of the non-radiative pathways take over. In essence, the radiant emissions from the surface have ‘speeded up’ the energy transfer to get the energy into the other pathways. Hot air rises, clouds and thunderstorms happen when conditions are right, and generally energy moves up through the atmosphere at something like the wet or dry Lapse rate, as appropriate to conditions. When the energy gets high enough, the radiant pathways again become significant, and eventually take over, sending the energy out to space.
    Of course the devil is in the details, but that description seems to fit with what others have said earlier. It is not all about just the radiant component.

  65. “The only rational explanation is the back-radiation from the Atmosphere to the Surface.”

    There’s an objection you haven’t addressed yet – that if extra heat tried to build up near the surface, convection would immediately carry it away again because warm air rises.

    In an atmosphere without convection, it would indeed work as you suggest, and the Earth’s surface temperature would be about 60 C on average. When we stop convection artificially – with greenhouses and solar ponds – we do get much higher temperatures. But the mechanism doesn’t apply in a convective fluid, and the 60 C prediction (and other things like an exponential relationship of temperature with altitude) conflicting with observation falsify it.

    The back-radiation explanation (or the “shells model” as I sometimes call it) was the original one developed by De Saussure, Fourier, etc. and was overturned in the 1960s by the work of Manabe, Moller, Strickler, and others. The actual explanation is that extra greenhouses gases raise the average altitude of emission to space, raising the altitude of the level that equilibrates to -20 C, and the warmth of the surface relative to this level is due primarily to the adiabatic lapse rate (adjusted for water vapour latent heat).

    The surface temperature is the effective radiative temperature (-18 C) plus the average height of emission (5.5 km) times the (moist) adiabatic lapse rate (6 C/km).
    T_surf = T_eff + z_emiss * MALR.

    The CO2 greenhouse works by increasing z_emiss.

    The tropical hotspot, incidentally, is supposed to be caused by also decreasing MALR, through water vapour feedback.

    I’m very pleased to see a lot of people now picking up on the convection/lapse rate connection. Things have much improved over a few years ago.

  66. Leonard Weinstein

    ……”The black body would absorb the reflected radiation, and thus the equilibrium temperature would be increased.”……
    Yes I agree, he could have been clearer on that point.

    This aspect causes a great deal of confusion and some intellectual mischief.

    See if you would agree with this formulation.
    The reflected radiation is the reflective insulation component.
    If increased it will slow down the heat loss from the black body.
    If the black body has an unchanged energy input its equilibrium temperature would be increased.
    However if the energy input to the black body was cut off the temperature of the black body would drop despite the backradiation.

  67. Ira, explaining the 33K temp difference between surface and top of atmosphere in terms of LWIR backradiation AKA greenhouse gas effect is misleading; as it is already entirely explained by the adiabatic lapse rate.
    (see for instance William C. Gilbert here

    http://bit.ly/fIhWPS

    )
    This leaves us with the question – if the lapse rate explains the temp difference, why does the purported greenhouse effect contribute nothing – to which my answer would be: As Absorption and re-emission of LWIR is an extremely fast process the re-radiation process can only delay heat transport via LWIR radiation to space by minutes at best – it’s a ping pong game of photons, but no heat is “trapped”, no matter how often Dessler et. al. use the term “heat-trapping gases” ( see here:

    http://www.chron.com/disp/story.mpl/editorial/outlook/6900556.html

    )

    If this delay rises by a few percent due to increased CO2 levels then it is still only a short term delay.

    And that is why we see no warming related to CO2 (if there were such warming, it would correlate to the Keeling curve, which it doesn’t).

  68. And i see that others have mentioned Postma. A very good explanation of the lapse rate.

  69. The 33K is the result of the warming by gravitational potential energy of the atmosphere below the -18°C isotherm determined by equality of radiation out to radiation in. It’s about 5.5 km. Because it would be the same for an atmosphere of the same average Cp but without GHGs, it is unconnected to greenhouse warming.

    All that does is to provide extra heat in the lower atmosphere and it’s convected to the top where it can radiate away. In the absence of convection, we’d get another 44K warming so the convection/negative feedback is extremely efficient.

    If you increase [CO2], [H2O] ust decrease in the upper atmosphere, as observed in 61 years’ radiosonde data. The mechanism is whatever is consistent with the 2nd Law’s requirement of maximum rate of entropy production.

    Climate science’s confusion over greenhouse warming comes from the mistake made by Arthur Milne in 1922 when he solved a differential equation for IR absorption in the atmosphere using as boundary condition infinite thickness. The term it generated, extra ‘back-radiation’ is imaginary.

    This is not to say there is no back-radiation: that is essential to match exactly the IR radiation from the atmosphere below that point in specific spectral ranges consistent with the local deviation from the lapse rate, e.g from clouds. So, the spectral curve will vary depending on whether there are clouds for example.

    The problem is that most scientists haven’t a clue about practical conductive, convective, radiative heat transfer so are easy meat for the charlatans who have made a good career out of pretending there’s an effect of CO2. As an exercise they need to understand the UHI effect is caused by convective heat transfer. You get higher temperature because to get the required thermal equilibrium, you need more radiative flux from the Earth’s surface. Yet the sum of all the heat transfer components, equal to the short wave heating, remains the same once you have exhausted thermal storage effects.

    Similarly, at night, you can freeze water by digging a hole in the desert to restrict convection even though air temperatures can be 20°C+.

    Before anyone else contributes, they need to know some chemical engineering: http://rpaulsingh.com/teaching/LectureHandouts/Convection%20Heat%20Transfer_handout.pdf

  70. davidmhoffer says:
    May 8, 2011 at 3:19 am
    “Why doesn’t it keep cooling off? After all, the atmosphere above the arctic is even colder than the arctic itself… Heck, at 25,000 feet its colder than that even at the equator! So… with 0 watts/m2 from the sun, the arctic nonetheless sticks right around -26 degrees. Hmmmmm…..wonder what keeps it from cooling off more….”

    David, if the atmosphere above the arctic is colder than the arctic surface, do you want to imply here that this cold atmosphere is nevertheless capable of WARMING the surface? Are you sure you wanted to say that?

  71. Ira:

    As a sanity check you should do this calculation for the moon which receives exactly the same amount of solar radiation as the earth system and there’s no atmosphere or water to complicate the situation. Further, the average surface temperature of the moon was experimentally measured by two separate Apollo experiments where a borehole 3 meters deep was made and a probe with thermocouples at various depths was put into the boreholes. These experiments transmitted temperature data back for several years. The average temperature of the surface is that depth at which there is no daily, seasonal, or annual variation i.e. a constant year-round temperature. At two mid-latitude locations that temperature is -23C or -9F at all depths of 1 meter or more. The experiments were actually designed to measure the thermal conductivity of lunar regolith.

    So it would appear on the face of it that your theoretical figure for the earth of 33C warmer than it would be for two approximate black bodies is somehow in disagreement with experimental reality by about 5C.

    So before talking about the earth as an approximate black body, which we can’t measure because it ain’t one with a dynamic ocean/atmosphere wrapping it, you have to do a sanity check and see if your black body calculations agree with the measured temperature of the moon which actually is an approximate black body.

    I can’t possibly take seriously any discussion which hinges on this commonly number that the earth is 33C warmer than it would be sans ocean and atmosphere when it appears to be 5C smaller than experimental observations. Near as I can tell from the facts that number should be 38C. Normally when two numbers like that differ by something near 10% I shrug it off because it’s still pretty close to agreement. But in this case 5C of disagreement five times greater than the commonly claimed 1C of anthropogenic warming to date.

    This goes to the heart of the problem. Anthropogenic effects are so tiny as to easily buried in the noise of the real world and in the error margins of all these toy models of the real world.

  72. This makes sense. Absent CO2 and water vapor, we would freeze. How much warming does each contribute? If CO2 did it all, and mankind producses about 3% of the CO2, then we are responsible for 3% of the 33C warming or about 1C. Thus AGW is real. Except for water vapor which may cause 90% of the 33C, in which case we cause 0.1C warming. Big deal. Given the daily and seasonal variations in temperature, I find it very hard to believe that 0.1C will have any measureable effect. Or be measureable at all.

  73. “How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”

    Are we suggesting that the earth’s surface is one black body and the atmosphere another? And thus the atmosphere radiates to the surface? However, the atmosphere and the surface are part of the same black body at TOA. Treating them separately below TOA may lead to nonsensical results.

    A more reasonable explantion is that the missing 33 K is due to the weight of the atmosphere which compresses the gas nearer the surface, which we see as an increase in temperature.

    We know this by comparing the atmospheres of Venus (CO2), Earth(N2O2) and Mars (CO2). All three planets show that temperature varies as the pressure of the atmosphere and the distance from the sun, independent of the composition of the atmosphere.

    Venus is totally covered in clouds. If albedo has a 30% blocking effect on earth, then it should have a near 100% blocking effect on venus and the surface should be cold. But it is not.

    If CO2 has a warming effect, the mars, which has a greater partial pressure of CO2 at the surface than does earth, should be warm, but it is not.

    Climate science continues to build models. Why? Because the argument goes that their is only one earth, so we need models to tell us the effects of CO2 on climate.

    This is a rubbish argument. Our two nearest neighbors have CO2 rich atmosphere. These are perfect models for us to judge if the GHG theory is correct, but climate science continues to ignore this more accurate alternative.

    We hear about “run-away” greenhouse effect on Venus, without any explanation of how this is possible with 100% cloud cover if albedo and aerosols works as we think they do.

    The facts are that the surface pressure of Venus is 90 times. Yes 90 times greater than on earth. This is why the surface temperature is high, even though according to climate theories about clouds and albedo, it should be low.

    Why are the temperatures at the bottom of the oceans on earth not higher than the surface? After all this happens in the atmosphere, should it not happen in the oceans.

    The reasons is compression. The oceans do not compress, so a cubic meter of surface water has the same number of molecules as a cubic meter of water from the deep oceans. However, the atmosphere does compress. So, a cubic meter of atmosphere at the surface has more molecules than a cubic meter of atmosphere higher up.

    It is these molecules that carry the heat. If the energy of the molecules is unchanged, then the more you have per cubic meter, the greater the energy per cubic meter, which we will see as increased temperature.

  74. What is so bothersome about this topic is that people who take it up do nothing but rehearse the Great Assumptions, such as the claim that Earth should be treated as a black body. It’s like listening to discussions among Ptolemy and his colleagues. The solution is always the same: more epicycles and maybe an eccentric or two. Yet in setting forth the Great Assumptions, you mention known facts which contradict them. Earth has albedo. Suppose for a moment that Earth’s albedo became 100%, would that motivate you to change the Great Assumptions. Is an object with 100% albedo properly treated as a black body? Can anyone state a plausible account of Earth’s radiation budget that does not treat Earth as a black body? I don’t believe anyone can and I believe that fact reveals a limitation of imagination, not the firmness of accepted science. I guess we will just have to wait for the science to advance. Anyone know somebody named Kepler?

  75. Ted Dooley says:
    May 7, 2011 at 10:06 pm
    The basis for the adjustments to the total absorption due to geometric and reflectivity are based on???

    Ted Dooley: Thanks for asking.

    Simple geometry tells us that the surface area of a circle (the cross-section of the Earth) is Pi x radius^2. The equation for the surface of a sphere is 4 x Pi x radius^2. The ratio of surface area of circle/sphere is thus 1/4 = 0.25. Thus, the 1366 Watts/meter^2 Solar radiation intercepted by the Earth, when averaged and spread over the whole surface is 0.25 x 1366 = 341.5 Watts/meter^2.

    The albedo of the Earth System, including the reflectivity of clouds and of the surface is estimated to be 70% 30% by many experts. (Some a bit more, some a bit less, but all are close to 70% 30%.) Thus around 30% of the Solar energy that enters the Atmosphere is reflected back to Space without being absorbed by the Earth System. So, 341.5 x 0.7 = 239.05 which I rounded up to 240 because (as an engineer) I do not like to express results at a level of precision that grossly exceeds the accuracy.

    Now, one may object that there would be no clouds if the Atmosphere was pure nitrogen, and that is certainly true. Thus, my calculation assuming all else being equal is, strictly speaking, in error. However, I was doing an engineering “sanity check” on the basics of Atmospheric “greenhouse effect” science, and, in the real world, there are clouds and the Earth System does have an albedo. So, I think I was justified in my assumptions.

    Those who deny the reality of the Atmospheric “greenhouse effect” say that the water vapor and carbon dioxide and other so-called “greenhouse gases”, along with Atmosphere and Surface (as they are with clouds, rain, snow, storms, and ice and other light-colored items), taken as a whole, do not cause the Earth Surface to be warmer than it would be absent the back radiation from the Atmosphere.

    The point of my simplified calculation was to show that, absent back radiation from the Atmosphere, there is no good explanation for the approximately 33ºC (58ºF) temperature difference between the “light” input from the Sun and the “heat” output from an Earth that would have to be cooler than it is for the heat balance to work. In other words, if you take all else being equal except for the back radiation from the Atmosphere to the Surface, there is an unbalance which would cause the Earth Surface to cool below its actual measured mean temperature. So let us be thankful for the Atmospheric “greehouse effect”. Without it, life on Earth as we know it would be impossible.

  76. Ira (con’t):

    The first pass you need to make in accounting for this discrepancy is to take albedo into account. The moon’s albedo is constant and accurately measured at some 16% so that amount of insolation doesn’t get absorbed the surface. That will account for the theoretical difference between black-body absorption and reality. The moon would be about 5C warmer if the surface albedo was close to 0%.

    The problem doesn’t go away however because the earth has a non-zero albedo too and it isn’t known nearly as well as the moon’s with, depending on who you ask, an average albedo in the range of 32% which is primarily the result of some 70% being shrouded by clouds of some sort at any given instant. The problem here is there is no satisfactory agreement between experiments attempting to measure the earth’s average albedo and estimate range from as little as 30% to as much as 40%. All the experimental attempts to measure the earth’s average albedo do agree on one thing – it isn’t constant and varies by as much 1% year over year.

    One percent doesn’t seem like much variation but once again it’s a very large number when compared to anthropogenic warming. A 1% change in the amount of insolation actually reaching the surface and not being reflected directly back out into space is some 2.5 watts per square meter. The IPCC third assessment is “95%” confident that anthropogenic forcing lies in the range of 0.6 to 2.4 w/m2. Yet measured variations in earth’s albedo over just several years has it changing by more watts/m2 than the highest estimate of alleged anthropogenic warming. Worse yet, the actual albedo can’t be pinned down and estimates vary by about 7% which is 7 times greater than anthropogenic forcings.

    So how can we possibly start talking about anthropogenic forcings and surface temperature changes wrought by same when we don’t even know to +-5C what the average temperature of the earth should be due to our albedo measurements being so imprecise and having no bloody idea how, when, and why the earth’s average albedo varies.

    The earth being a dynamic water world makes all attempts to model it exercises in futility. The models are toys and should be regarded as precisely that – toys with zero credibility.

  77. 6. Now use thermodynamics expressed in the form of the lapse rate.
    7. An average moist lapse rate is -6.6K/km.
    8. The lapse rate has nothing to do with back radiation as even scienceofdoom admits.
    9. Work back down from -18C at 5km to the Earth surface and we get 5×6.6 = 33K.
    So now you have an Earth surface temperature of 15C without any “greenhouse effect”.

    Exactly, the earth and atmosphere are part of the same black body. Considering them as two different black bodies, based on observation of their combined effect at the TOA without consideration of lapse rate, convection and condensation seems unlikely to provide an accurate answer.

  78. Bryan stated that using clouds, you get a 33 C temperature difference between clouds and surface “without any greenhouse effect”. I’d say inseead, that a significant fraction of the greeenhouse effect is due to clouds.

    Ira Glickstein stated that the warming due to CO2 would only be about 0.2C. Here’s a way to get a rough estimate of the figure:
    We get an average of 342 watts/M^2 from the sun.
    As others have pointed out, excluding clouds, the average reflectivity of earth’s surface is about 0.124
    Earth is not a blackbody, most of the earth is covered by ocean, which has an emissivity between 0.92 and 0.96- I’ll use a 0.94 average. This non- blackbody emissivity means earth actually radiates away less than a blackbody at the same temperature.

    A non-greenhouse earth surface would receive an average of
    342 *(1-.124reflectivity)/0.94 emissivity would give an average surface flux of
    342*0.876/.94 = 318.7 watts. All of the positive and negative feedbacks to the
    greenhouse effect give an average temperature of 288K, or effectively 390.7 watts/m^2, for an effective magnification of 1.226

    From Trenbeth’s figures,

    http://content.imamu.edu.sa/Scholars/it/net/trenbert.pdf

    161 watts /m^2 hit the earth’s surface, with back radiation we get a total of 492 watts, not all in sensible heat, for a magnification of

    493/161 = 3.06,

    Putting the two formulas together, a theoretical additional factor of 2.06 multiplier due to the greenhouse effect becomes an actual multiplier of 0.226, thanks to negative factors like increases in albedo due to clouds, decreases in lapse rate, etc.
    A doubling of CO2 would supposedly increase the surface flux by something less than 3.7 watts/m^2, for a total of 496.7 watts, an increase to 335.7 from 332 over the original 161 watt/m^2 hitting the surface.

    Multiply that 335.7/332 * 0.226 and we get 1.2285* 318.7 for an effective
    wattage increase from 390.7 to 391.52 watts, or a temperature increase to
    (391.52/390.7)^0.25 = 288.15K from 288 K – Close to Ira Glickstein’s estimate of 0.2K- a drop in the bucket.

  79. Ira:

    “The only rational explanation is the back-radiation from the Atmosphere to the Surface.”

    This is the crux of the issue. It is the point of contention. You and others keep repeating the same GHE theory over and over again, with slightly different words. By now, I doubt that there are many people with any scientific background who do not understand what you are saying. The problem is that it has not been demonstrated empirically AND there are other ways to explain why the surface of the Earth is higher than -18C!

    The -18 C represents the average radiation temperature which comes from an altitude of about 5 km, NOT the surface. Due to the lapse rate, it is much warmer on the surface (viola, about 33C higher!).

    Now, that 33C “increase” may well be due to the GHE that you describe. However, it could also be explained simply by heat storage and the ideal gas law. So I’m still not sure.

    BTW, I hope you are not still trying to say that the surface is HEATED by the backradiation from the atmosphere. The atmosphere may manage to keep the surface from cooling so fast (due to the backradiation), but it is certainly NOT heating it. A colder body cannot heat a warmer one. Period.

  80. dr.bill says:
    May 7, 2011 at 10:38 pm
    Ira, the radiance formula involves either a “df” or “dλ” at the end. Converting from frequency to wavelength thus involves replacing df by -cdλ/λ², which explains why multiplying the frequency data by λ² puts the peak back in the right place.

    This same issue arises in Wien’s Law. The “peak color” is different, depending on whether you use wavelength or frequency as a variable. Intensity, however, is integrated over all colors, and gives the same result for either variable.

    /dr.bill

    Thanks dr.bill for trying to explain the reason the Petty curves place the peak “Radiance” at a longer wavelength than the “Intensity” of the Carleton spreadsheet and other Internet materials. However, I did not fully understand your explanation and I am not sure where you think the peak energy really is.

    When I use the Carleton spreadsheet for Solar radiation, the peak energy is at 0.5μm, which we see as “green” and which I happen to know is the most sensitive part of our visual system because biological evolution has “tuned” our visual system to be most sensitive where the light energy is maximized. Thus, I am sure the real peak for solar radiation is around 0.5μm, which would make the “Intensity” of Carleton the correct peak.

    So, I may be particularly dense on this topic, but I still do not understand why “Radiance” (per Perry) is used. I hasten to add that I know Perry obtained his data from real measurements from a NASA satellite sensor and I am sure that there is a good reason and rationale for using “Radiance”. I would just like to know what that rationale is. Anybody know?

  81. Ira Glickstein, PhD says:
    May 8, 2011 at 8:35 am

    You say the earth’s average is pretty close to 70%.

    First of all that’s the reciprocal of albedo. Albedo is a number ranging from 0-1 with 0 being dead black and 1 being a perfect mirror. Multiply by 100 for a percentage.

    Now I know wikipedia is frowned on as a reference but when it comes to global warming their distinct bias is always on the warmist side so if you find something in wikipedia that is contrary to warmist dogma you can bet your bottom dollar it’s a painful admission.

    I quote:

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

    Estimates of the Earth’s average albedo vary in the range 0.3–0.4, resulting in different estimated effective temperatures. Estimates are often based on the solar constant (total insolation power density) rather than the temperature, size, and distance of the sun. For example, using 0.4 for albedo, and an insolation of 1400 W m−2, one obtains an effective temperature of about 245 K.[25] Similarly using albedo 0.3 and solar constant of 1372 W m−2, one obtains an effective temperature of 255 K.[26][27]

    This is exactly the figure I gave you of 30% to 40% for estimated average albedo. However I didn’t get it from wiki. I got it from much more in depth reading of the experimental attempts to measure it. Wikipedia in this instance is accurate and for the warmista it is a fatal blow with regard to the credibility of their propaganda. We don’t know the average insolation that reaches the surface of earth any closer than +-25w/m2. To start talking about anthropogenic forcings in the range of -0.6 to 2.5w/m2 is ludicrous when the error margin in experimental attempts to measure average insolation is 25 watts.

    Before any toy model of the earth can begin to be given any credibility it must first have a accurate number for the earth’s average albedo down to a fraction of percent and then it must further model the observed variation in albedo down to a fraction of a percent. If it can’t do that it can’t even begin to seperate anthropogenic from natural forcing.

  82. Ira Glickstein, PhD says:
    May 8, 2011 at 8:35 am
    “The point of my simplified calculation was to show that, absent back radiation from the Atmosphere, there is no good explanation for the approximately 33ºC (58ºF) temperature difference between the “light” input from the Sun and the “heat” output from an Earth that would have to be cooler than it is for the heat balance to work.”

    Again: The Earth radiates from the top of the atmosphere and that’s where you find exactly the temperature needed to radiate enough. See also

    Harry Dale Huffman : No Greenhouse effect on Venus

    http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html

    where Mr. Huffman does the same computations for Venus.

  83. Probably should have read more comments before posting. fred burple already said about the same thing that I did. I think he is correct.

  84. jae –

    Someone who would benefit an understanding of ‘traveling waves'; incident and reflected waves (energy), S-Parameters (Scattering Matrix Parameters).

    Falling back on the simple ‘laws of thermodynamics’ does you a disservice …

  85. Ira Glickstein, PhD says:
    “The only rational explanation is the back-radiation from the Atmosphere to the Surface.”

    “absent back radiation from the Atmosphere, there is no good explanation for the approximately 33ºC (58ºF) temperature difference”

    As many comments above have pointed out, the adiabatic lapse rate alone can completely explain this. Thus, Ira you need to explain why in effect you consider this to be an irrational or poor explanation.

  86. I disagree with answer three. As a fellow systems engineer I appreciate and support your back of the envelope checks. But remember, we also have to explore all feasible answers, and the difference between black body and the actual measured radiation back to space does NOT require a ‘greenhouse’ effect.

    It simply requires a long time constant between heating and radiation, which is where heat sinks and non-solar heat sources come into play.

    The 33° K difference could simply be a result of a time delay between solar heating and thermal cooling. The heat capacity of the Earth could require the system to bank 33°K of heat before the there is sufficient flow of energy to the upper atmosphere to radiate.

    Moreover, the inner core and the active underwater volcanoes could easily be dumping a extra wedge of gravity-friction generated heat (akin to what drives the Io system). What our massive oceans provide is a heat store, not just a heat sink. Gravity-friction (which produce wave action, which transfers kinetic energy into heat) is a source of energy transfer at the surface, as is any flowing water and its passage to the sea.

    How much kinetic energy is in a river flowing to the sea? How was that energy stored? It was stored as water vapor evaporating from ground and bodies of water from solar heat, which rise to be COOLED, where the kinetic energy from gravity is then brought back to Earth with the rain. Friction supplies the transfer.

    You can identify numerous transitions where heat is captured, converted, released, transferred and held for days, weeks, months and years.

    What is wrong with ALL these models is they lack a time constant from which a joule of solar energy can travel through our system – over any one of hundreds of energy transformation paths – before it is made available for radiation. And we never remember to include the yet still unquantified heat source that is at the core of our planet. There is NO dissipation path for that heat except through radiation. It is a massive heat source which could EASILY account for the 33°K difference.

    We know so little, yet act as if we have all the answers. The minute that happens we stop being scientists and become zealots.

  87. DirkH says:
    May 8, 2011 at 7:56 am
    davidmhoffer says:
    May 8, 2011 at 3:19 am
    “Why doesn’t it keep cooling off? After all, the atmosphere above the arctic is even colder than the arctic itself… Heck, at 25,000 feet its colder than that even at the equator! So… with 0 watts/m2 from the sun, the arctic nonetheless sticks right around -26 degrees. Hmmmmm…..wonder what keeps it from cooling off more….”

    DirkH;
    David, if the atmosphere above the arctic is colder than the arctic surface, do you want to imply here that this cold atmosphere is nevertheless capable of WARMING the surface? Are you sure you wanted to say that?>>>>

    Dirk, I didn’t say the atmosphere made the surface warmer. I said, what keeps it from cooling off more? Why does the temp drop from 275K in July, to 240K in December, a drop of 35 degrees over 5 months, and then over the next 2-3 months, despite having 0 insolation, it drops by perhaps two degrees more. Since it isn’t getting any energy from the sun, where is the energy to keep it from cooling any further come from?

    There are several possibilities. The arctic is mostly ocean covered by ice, so warm water currents could be bringing in heat. Wind from warmer latitudes could be bringing in warm air. Reflection from clouds that are high enought in altitude to catch some sunlight cold contribute. All of these have been measured in various ways, all are part of the picture, but combined they are insufficient to keep the surface at -26.

    So why doesn’t it cool off more? Answer is (in part) that some of the outward bound radiance is in fact returned to the surface. In fact, the atmospheric window starts to close pretty fast right around that temperature, and by -40 it is shut tight. That’s why the very coldest nights at 80 N latitude are about -40 and the very coldest nights at 50 N latitude, where the “average” temp is MUCH higher, are around -40.

    Once the temp hits -25 not only is the “colder” atmosphere above the artic backradiating, reradiating, simulating pajamas, what ever term you want, it get’s more efficient at it. So it can’t EVER cause an INCREASING temperature trend, but it can certainly recycle, reradiate, pajamas, the surface radiance to the point where continued decreases in temp slow to a crawl.

  88. ferd berple says:
    May 8, 2011 at 8:22 am

    I don’t want to rain on your parade but the temperature at the top of the atmosphere is in the thousands of degrees. The TOA boundary for energy budget calculations is the top of the thermosphere. So the temperature is indeed much higher at TOA than it is at BOA.

    Your argument based on lapse rate is only valid to the top of the middle atmosphere and falls apart in the upper atmosphere where temperature increases with altitude.

    It’s very thin atmosphere but it’s still atmosphere and its temperature can be measured. So the temperature at the bottom of the atmosphere is indeed much colder than at the top.

  89. In my view, the explanation of the 33°C difference is somewhat tricky.

    If the atmosphere were totally transparent to radiation, the effective temperature of 280 K would be that of the ground (although, as I mentioned above, this effective temperature does not coincide with the average surface temperature). So the atmosphere must be opaque to thermal IR radiation to produce an effective temperature at the TOA different from the ground one. But absorption of IR radiation IS greenhouse effect. So “some ” greenhouse effect must be present to make the atmosphere partly opaque.

    BUT the open question is to compute the average difference between ground and TOA and this is much less obvious. In the absorption lines, the outgoing LWR intensity is determined by the local temperature at the last diffusion surface – so the upper atmosphere temperature. But the heat transfer in the troposphere is a complicated mixture of convection and radiation, and is indeed dominated by convection. So the difference temperature is mainly controlled by the adiabatic law between ground and and tropopause – and this can NOT be computed only with radiative physics. So in some sense, those claiming that the difference between the ground and the TOA is mainly due to gas thermodynamics are also right – this is a very complicated mixture of convective transport and radiative opacity.

    This can only be solve through detailed modeling – and it is quite possible that this modeling is inaccurate.

    A further complication is that as I mentioned , average ground temperature is NOT the same as effective temperature, so there is no clear relationship between the two: The Earth is NOT an isothermal copper sphere ! actually the average temperature depends strongly on meridional circulation that transports a lot of heat towards the high latitudes – especially with oceans that are, to say the least, not very well described and understood. Any variability in oceanic circulation could have strong effects on local, and hence average temperature, even with a fixed energy budget. This is again totally overlooked in “simple” isothermal , radiative arguments. I would say IMHO that this is actually the less known feature of current GCM – and the main reason for doubting about their accuracy and predictive power. And I am not really surprised that they struggle so much with the “lack of warming” .. :)

  90. ferd (con’t)

    The reason the middle and lower atmosphere get colder with height isn’t because of compression. It’s because in the middle and lower atmosphere the sun doesn’t do any appreciable heating of it.

    One must always begin with the big picture and work down to the smaller details not try to add up the details into a big picture.

    In the big picture the sun heats the ocean, the ocean heats the atmosphere, and the frigid cold of the cosmic background (3K) cools the atmosphere.

    The lower atmosphere is warmer than the middle atmosphere because the lower atmosphere is closer to the source of the ocean that heats it.

  91. Ira said:
    “Notice that the Earth System mean temperature I had to use to provide 240 Watts/m^2 of radiation to Space to balance the input absorbed from by the Earth System from the Sun was 255 K. However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
    No, by my point o f view there is another explanation. The incoming radiation from the Sun is truly measured from the radiometers because of the distance from the source we get all the parallel energy rays in the SW, while the emission of the Earth in the LW range is scattered in a spheric space around the planet, so the satellites’ radiometers can’t really measure all that energy because those rays are not parallel each other and many of them will never be perpendicular to the radiometer input slit.

  92. I suppose this is my segue into talking about the temperature of the ocean since it is responsible for the temperature of the lower atmosphere where we live and breathe.

    The average temperature of the ocean is 4C. 90% of the volume of the global ocean is at a near constant temperature of 3C. Only the surface layer comprising 10% of its volume gets any warmer (or colder) than that. It’s not a temperature/density thing either because seawater, due to its saline content, increases in density all the way down to its freezing point of about -2C.

    The only possible explanation for why the average temperature of the ocean is 4C is because that is the average surface temperature of the earth taken over a period of time long enough for convection and conduction to equilibrate the entire volume.

    Ostensibly I should think that period of time is about 120,000 years or once complete glacial/interglacial cycle.

    Any sane person armed with the facts should find their source of angst to be that huge volume of icewater lying just below the warm thin surface layer of the ocean. The current interglacial period is getting long in tooth so again by any sane measure we should be worried about an impending ice age and if there is any merit whatsoever in anthropogenic global warming we should be glad for it and try to get as much of it as we can in the hope that it might delay the inevitable return of glaciers a mile thick covering everything north of Washington, D.C.

    I wonder how many of the people I share the planet with are sane in this regard. The facts on the ground are incontrovertable. The only thing we have to fear is more ice not less ice. For this reason I call the warmista’s “ice huggers” in fond remembrance of their predecessors the “tree huggers”. The tree huggers morphed into ice huggers. Ain’t that a hoot? Trees don’t grow well in ice. They are lucky to survive where ice prevails much of the year much less thrive in such conditions. A warmer earth is a greener earth and a greener earth is what we all want, innit?

  93. Dave Springer:

    Please “debunk” the article at the link given by DirkH: http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html

    You folks are, in essence, saying that the lapse rate is CAUSED by the GHE. That is nonsense. The lapse rate is dependent upon only two variables, gravity (g) and the heat capacity of the air (Cp) and can easily be calculated from these two variables. GHE has no “place” in this formula. The lapse rate for ANY atmosphere (even pure nitrogen or oxygen) would be calculated the same way–without “allowance” for any GHE.

  94. [Update 7PM EST - I just noticed that I misunderstood the direction of Dave Springer's 5ºC delta. I am sorry about my error. The following is unchanged from what I originally posted. As you read it, please mentally reverse the sign of the 5ºC delta. Ira]

    Dave Springer says:
    May 8, 2011 at 8:08 am
    Ira:
    As a sanity check you should do this calculation for the moon which receives exactly the same amount of solar radiation as the earth system and there’s no atmosphere or water to complicate the situation. Further, the average surface temperature of the moon was experimentally measured by two separate Apollo experiments … At two mid-latitude locations that temperature is -23C or -9F at all depths of 1 meter or more. …

    So it would appear on the face of it that your theoretical figure for the earth of 33C warmer than it would be for two approximate black bodies is somehow in disagreement with experimental reality by about 5C. …

    Great to hear from you again Dave Springer and thanks again for putting me on to the radiation measurement curves in the Perry book in your comment on a previous thread. That information (from you and Tim Folkerts) was the basis for another one of my postings in this series. My only claim to special knowledge here at WUWT is as a system engineer and system scientist, and I appreciate it when those more familiar with climate science add to my knowledge.

    You make a great point about using the experimental reality of the Moon as a reference. That would reduce the 33ºC I calculated in my simplified “sanity check” by 5ºC, so it would be about 28ºC.

    How to explain the difference? Well, 28ºC is pretty close to 33ºC for a “sanity check”. Furthermore, the Moon’s albedo is about 0.12 compared to the albedo of the Earth which is about 0.37. If we assume all else is equal except the subject of this posting, namely the Atmospheric “greenhouse” effect, and assign the Earth System an albedo equal to that of the Moon, using my modified Carleton spreadsheet, that would increase my 255 K to about 272 K, reducing the 33ºC to be accounted for down to about 16ºC. OOPS, we have over-corrected your 5ºC on the Moon and we are now off by 12ºC in the opposite direction! Clearly there are other differences between the Moon System and the Earth System. If anything, your Moon example adds credibility to the reality of the Atmospheric “greenhouse effect” on Earth.

    I hasten to add that engineers are not as picky about precision of estimates as scientists and mathematical analysts are about their data. If we get estimates within 15% (33ºC vs 28ºC) that is usually close enough, since other uncertainties are likely to be in the mix. You may have noticed that I am unusually sensitive when mathematical analysts state their results to more decimal places than is justified by the accuracy of the assumptions that have gone into their models.

    I never claimed that my “sanity check” was more than just that, a way for a system engineer to double-check the rationality of the more exacting calculations of the supoosed experts. From that point of view, I think the 33ºC due to all Atmospheric effects (‘greenhouse” as well as others that are unknown at present to me) holds up pretty well. It is definitely in the ballpark of the truth.

    Thus, it seems you are correct that there are factors in addition to the Atmospheric “greenhouse effect” that account for the Surface being warmer than it would be absent that back radiation. Thanks for your input.

  95. No mention is made of ocean and atmospheric heat transport. With one side of the Earth exposed to near absolute zero temperatures and the other exposed to the Sun; any theory has to take into account the movement of vast amounts of heat energy from hot to cold and vice versa. Although the Earth may lose heat at night, there is still substantial heat transfer from the day side to the night side keeping the night side warmer than it would be otherwise. The true “greenhouse effect” is a perfect example. Without convective heat transport a greenhouse becomes much hotter than its unenclosed environment. Simplistic models with lots of fudge factors to make the numbers fit are insufficient to describe the climate.

  96. More on heating and cooling of gases due to compression and expansion.

    Heat and cooling only takes place as the gas pressure is changing. When the pressure stops changing so does the temperature. There is very little dynamic variation of atmospheric pressure at any constant altitude so there is very little heating or cooling taking place due to pressure variation.

    In a real world example I have an air compressor in my shop. If the air in the tank is not compressed and I run the compressor to bring up to 10 atmospheres the tank will get much warmer than the air in my shop and if I quickly bleed it back to 1 atmosphere the tank will get colder than the air in my shop.

    However (this is the key) if I leave the tank full of compressed air it won’t stay warmer than the air in my shop for long. That’s because the pressure is constant.

    Ferd’s assertion that atmospheres get warmer as you go deeper into them because of compression is quite wrong. If there were some mechanism that were taking volumes of air high in the atmosphere and transporting them lower in the atmosphere his thesis would have some merit but there is no such mechanism at work on Venus, Earth, or Mars. The pressure gradient is nearly static and if an actual volume of gas is neither expanding nor contracting there is no associated change in temperature.

  97. “A warmer earth is a greener earth and a greener earth is what we all want, innit?”

    It certainly is surprising that people fear the earth getting warmer, yet when it comes to spending their hard earned money to go on vacation, few it any head towards the poles. If warming is something to be feared, where did the term “tropical paradise” come from? Why isn’t it “polar paradise”?

  98. davidmhoffer says:
    May 8, 2011 at 9:49 am
    “Once the temp hits -25 not only is the “colder” atmosphere above the artic backradiating, reradiating, simulating pajamas, what ever term you want, it get’s more efficient at it. So it can’t EVER cause an INCREASING temperature trend, but it can certainly recycle, reradiate, pajamas, the surface radiance to the point where continued decreases in temp slow to a crawl.”

    David, this means that at -25 deg the upper atmosphere is no more colder than the ground; an equilibrium is reached. Of course, nothing happens from there; but your premise was to show that the people who say a cold object cannot warm a warm object are somehow wrong – your example shows that two equally cold objects are in equilibrium and does not contradict those people at all. In your example, the upper atmosphere is no more colder than the ground.

  99. I don’t want to rain on your parade but the temperature at the top of the atmosphere is in the thousands of degrees.

    Then it must be back radiating a fantastic amount of energy to the earth. At thousands of degrees this radiation would be in the visible or even UV.

  100. ” there is no good explanation for the approximately 33ºC (58ºF) temperature difference between the “light” input from the Sun and the “heat” output from an Earth”

    temperature is not heat – you can’t measure heat with a thermometer. degrees can not be converted to watts.
    the sidewalk doesn’t get hot because the air above it warmed it.
    the sidewalk stores heat and warms the air above it.
    incidentally, phase change from liquid to gas has no temperature change – but loads of heat is involved.
    incidentally, avogadro’s ‘laws’ trump co2 fetish fantasy – co2 molecules don’t get to be hotter than everything else around them – all molecules in a local region are basically at the same temperature via kinetic transfer.
    /me sick of co2 shrimpers. it’s like an old ed wood flick with fat ladies in a playpen and some old fart claiming to be the walrus. or maybe it’s like divine following that pooch.

  101. “It was stored as water vapor evaporating from ground and bodies of water from solar heat, which rise to be COOLED, where the kinetic energy from gravity is then brought back to Earth with the rain.”

    I recall years ago seeing a calculation that the amount of heat energy delivered to the earth by rainfall significantly exceed that delivered by bright sunlight.

  102. Ira:

    We seem to be out of synchronization. In a series of comments I left the second in the series explained why the moon is not at its theoretical black-body temperature i.e.. because it isn’t a black-body. It’s a gray body with an albedo around 12%. I found the missing heat for you in that comment. I then went on to explain that the earth is also a gray body with an albedo of 30% to 40% and pointed out the crux of the issue – the earth’s average albedo isn’t know to any better certainty than +-5% and +-5 uncertainty in albedo translates into 25 watts per square meter of uncertainty in how much insolation is actually absorbed at the surface. Moreover all actual attempts to measure albedo, while disagreeing by 10% (I actually only found 7% disagreement in actual studies but still…) all do agree that the average in any given year is not static and varies from year to year. One experiment using earthshine as a measure that went on for about five years found a 1.5% difference between year one and year five. Their methodology in that one was measuring the brightness of the new moon which is illuminated solely by light reflected from the earth. It’s a bit surprising how difficult it is to get an accurate measure of the earth’s average albedo in any one year but on the other hand it isn’t difficult to measure change from year to year. The earthshine measure might not be accurate but it is precise and consistent. Precision and consistency is all you need to identify trends.

    [Thanks Dave Springer and I appologize for not remembering that detail of your earlier attempts to clarify the issues. Please continue to share your special knowledge of climate science with me and others at WUWT. Your comments are most welcome here in my threads, even when they are corrections to my mistakes. In fact, especially when they help me improve my understanding and learn more about this topic. THANKS! - Ira 9:25PM EST]

  103. jae says:

    You folks are, in essence, saying that the lapse rate is CAUSED by the GHE. That is nonsense. The lapse rate is dependent upon only two variables, gravity (g) and the heat capacity of the air (Cp) and can easily be calculated from these two variables. GHE has no “place” in this formula. The lapse rate for ANY atmosphere (even pure nitrogen or oxygen) would be calculated the same way–without “allowance” for any GHE.

    The lapse rate is not caused by GHE’s per se. Your statement about the lapse rate is more or less correct…except that this sets a maximum lapse rate. That is, it is possible for an atmosphere to have a smaller lapse rate than you get from the stability calculation that you allude to. However, it won’ t have a steeper lapse rate because such an atmosphere becomes unstable to convection, with then brings the lapse rate down to the marginal stability value.

    So, the best way to put it is that the lapse rate is caused by the fact that the atmosphere is heated from below in concert with stability arguments based on buoyancy, adiabatic expansion, and the existence of convection.

    What the GHEs determine is where in the atmosphere the emitted radiation is able to escape to space. Radiative balance of the earth system then sets the temperature at this level in the atmosphere and the temperature at the surface basically follows from the lapse rate.

  104. Dave Springer says:
    May 8, 2011 at 10:23 am
    “For this reason I call the warmista’s “ice huggers” in fond remembrance of their predecessors the “tree huggers”.”

    Good idea. They really do love their Greenland ice and all that.

  105. Ira,
    I have been reading many posts here that discuss the blackbody radiation of the earth. You have claimed that the 33K difference is due to GHG back radiation. I don’t want to argue that with you… but I think it would be worthwhile to be very very clear on exactly what you are claiming.

    So… I can boil this down to a simple yes or no question…

    Are you claiming that if the atmosphere were replaced with a different set of gasses that do not contain GHG molecules, but still had all the other macro effects such as clouds etc in the exact same amount as our current atmosphere… are you claiming that this atmosphere would NOT cause (directly or indirectly) the surface of the earth to warm at all?

    You see… it does seem as if the answer to the above question is yes. It does seem as if you are making that claim. Are you? Or, are we just misunderstanding your argument?

    -Anton Eagle

  106. John of Kent says:

    We can account for the 33C difference without any need for an imaginary greenhouse effect. I urge everyone at WUWT to read this brilliant paper by physicist Joe Postma ;

    http://www.globalwarmingskeptics.info/forums/attachment.php?aid=327

    Bryan says:

    You should read the article below.
    It will answer all of your misconceptions about the “greenhouse theory” which in fact is pure fiction.
    1. It is true that the radiative equilibrium temperature of the Earth is -18C.
    2. This figure is confirmed by Satellite measurement from space.
    3. It is NOT true that all the radiation emitted from the Earth surface.
    4. The emission is mostly from the cloud level.
    5. If we pick say around 5km as the average emission height to space we will be near enough.
    6. Now use thermodynamics expressed in the form of the lapse rate.
    7. An average moist lapse rate is -6.6K/km.
    8. The lapse rate has nothing to do with back radiation as even scienceofdoom admits.
    9. Work back down from -18C at 5km to the Earth surface and we get 5×6.6 = 33K.
    So now you have an Earth surface temperature of 15C without any “greenhouse effect”.

    http://www.ilovemycarbondioxide.com/pdf/Understanding_the_Atmosphere_Effect.pdf

    Okay…So it is time to debunk the Postma nonsense again, even though we already did it in the last thread that Ira wrote.

    The problem with the Postma argument is that the “average emission height” depends on…you guessed it…the IR-absorbing substances, namely greenhouse gases and clouds, in the atmosphere. So, the Postma argument does not get rid of the greenhouse effect…He merely sweeps it under the rug by obfuscating the fact that this average emission height being above the earth’s surface is due to the greenhouse effect.

    Think about it. If the atmosphere contained no IR-absorbing substances, then all the IR emitted by the earth’s surface would escape into space and radiative balance would dictate that the earth’s average surface temperature (or really the average of emissivity*T^4 where T is the absolute temperature and the emissivity of most terrestrial materials in the wavelength range of interest is very close to 1) is set by the condition that the earth must radiate as much energy as it absorbs from the sun.

    Now…Slowly add IR-absorbing substances. What happens is that some of the radiation from the earth’s surface is absorbed in the atmosphere, which subsequently radiates it again. This absorption can even occur multiple times but once the radiation occurs high enough up in the atmosphere that the remaining IR-absorbing substances above that layer are unlikely to absorb the radiation, then it can successfully escape to space. Hence, the average emission height is what is determined by the IR-absorbing substances in the atmosphere.

    As the greenhouse gases in the atmosphere increase, this average emission height rises…and because of the lapse rate…this means less radiation is emitted back out into space. As a result, the earth system heats up until radiative balance is restored.

  107. Ira,
    I have a second question regarding the two emission diagrams you show.

    The upper graph, showing down-looking readings at the top of the atmosphere shows that approximately 1/2 of the 15 um radiation is absorbed. It looks as if it would be about 90 mW/m^2 without absorption, and instead is about 45 mW/m^2 or so with absorption.

    Then you show the lower graph showing the up-looking readings at the surface… showing about 90-100 mW/m^2 of radiation (at 15 um) down to the surface from the atmosphere.

    So, my question is… if the GH gasses are absorbing about 45 mW/m^2… how are the re-radiating almost 100 mW/m^2 back to the surface?

    It doesn’t look like the sun can be the source, since according to these curves the sun doesn’t radiate hardly anything at those wavelengths. So… where is the extra energy coming from?

    Thanks,

    -Anton Eagle

  108. Ira Glickstein

    Do you know what the adiabatic lapse rate is?.
    On page 21 of the Postma paper it is derived.

    This is not a controversial point, serious IPCC advocates include this derivation as part of their narrative.
    If you like I can give you a couple of such references.

    The clouds at 5Km are far better radiators than gaseous CO2 or H2O .
    Yet strangely enough it is the Latent Heat and high specific heat capacity of water in clouds that accounts for cloudy warmer nights rather than radiative effects.
    The convection mechanism is reduced if there is a smaller temperature difference between Earth Surface and clouds.

  109. John of Kent says:

    No! The earths surface is already emitting radiation in the Infra Red. It’s atoms and molecules are ALREADY excited sufficiently to be emitting IR.

    Yes..The molecules of the earth surface have kinetic energy because they are excited or are being exited with the absorption of radiation.

    This excitement distorts the magnetic field of the molecules components in a way that they are not in balance as they would be if the molecule was at rest. This leads the electromagnetic field of the molecule to send out waves of electromagnetic energy into space at decreasing rates if the source of the energy is turned of.

    When the source energy is turned off the emitting waves from the molecule don’t cease immediately because the molecules components have mass and it is the movement of these parts that cause the decreasing waves of electromagnetic disturbance as energy is lost.

    Similarly when the molecule is hit by a magnetic wave it will not reach its full kinetic and radiative potential until all its parts are moving in such a way as to propagate a field of disturbance equal to the constant of the energy input over time. At this point it is at radiative equilibrium with the incoming energy and no more work can be done.

    Let’s say we are on the moon so there is no overlaying gas and pick an arbitrary incoming value of 100Watts/m^2. The mass of the particles at the surface will eventually increase in temperature until the moving parts of the molecules are so excited they cause a electromagnetic disturbance that is equal to the incoming electromagnetic energy from the sun. At this point the waves of energy from the surface propagate outward and are lost to open space. 100Watts/m^2 in, 100Watts/m^2 out.

    This however is not the case with our planet. The over laying gas doesn’t impede the inflow of energy to the surface but the principle remains the same as the moons surface in that in its self the surface will move toward radiative equilibrium with the incoming energy.

    This time however the outgoing radiation is now blocked by a molecule of gas which of course also has mass and a state of rest.

    The propagating electromagnetic wave from the earths surface is now absorbed by the molecule which in turn adopts kinetic and thereby electromagnetic properties as the state of rest is now disrupted and its components vibrate, sending out electromagnetic waves as it struggles to balance its competing internal forces.

    Although free floating the rules for the gas molecule are the same as the molecules at the earths surface. It will absorb all the incoming energy from the earths surface until it radiates away the same amount of radiation it is receiving. It will do this in all directions, so as it moves to reach radiative equilibrium with the surface it will radiate half toward the surface and half away. The energy being radiated back to the surface is not kinetic, it is electromagnetic. It would be 50 Watts/m^2 in the first instance of time with the other 50 Watts radiating away. The surface doesn’t know where the 100 Watts or the 50 Watts are coming from it only knows it is receiving 150 Watts in the second instance of time.

    Think of this. If we had a closed system where no energy is lost and we were to put 100 watts a second into the system we would have a system with 100 joules after the first second. Now in the next second we only put 50 Watts a second into the system we get 150 joules in the system. The system doesn’t care that the that the last second had a lower value than the first second it only knows that over two seconds it has received 150 Watts of energy.

  110. John Marshall says:

    Taking into account the adiabatic effect, which is real and measurable, there is no need for the GHG effect.

    No…It does not as I have explained above.

    The greenhouse effect also has one fatal flaw. The warming in the mid troposphere in the tropics shown by all models has yet to be found in real observable data. I do not care what your calculations show (in reality another model), if this warm area shown by models is not there then the theory falls flat on its face!

    The prediction of tropical tropospheric amplification has nothing to do with the greenhouse effect. It has to do with the fact that in the tropics the temperature decrease with height is expected to closely follow the moist adiabatic lapse rate. Hence, this amplification is predicted for warming due to any mechanism.

    It is also worth noting that tropical tropospheric amplification is well-verified by the satellite and balloon data for the temperature fluctuations that occur on monthly to yearly timescales, e.g., due to ENSO. Where the data is more ambiguous is with the long-term multidecadal trends, where such data is easily contaminated by artifacts due to changes in instrumentation over time, switches from one satellite to another, etc. The different data sets…and various analyses of the data sets tend to differ not only from the model predictions (depending on which data set you look at) but also from one another.

    Even Richard Lindzen agrees that tropical tropospheric amplification follows from basic physics and has nothing to do with the greenhouse effect mechanism…and that the fact that is a problem with the observational data, not the models (and, at any rate, nothing to do with the mechanism causing the warming). His only point of departure is that he thinks the data more likely to be wrong is the surface data in the tropics whereas most others think it is the data at altitude.

    The specific fingerprint of warming due to an increased greenhouse effect (at least relative to, for example, the mechanism of heating by increased solar radiation) is that the greenhouse effect is predicted to cause cooling in the stratosphere. And, this is indeed what is seen. (Some of the cooling is due also to the decrease in stratospheric ozone but the amount and altitude distribution of the cooling apparently cannot be explained solely by this mechanism.)

  111. Ira Glickstein, PhD says:
    May 8, 2011 at 10:41 am

    “I hasten to add that engineers are not as picky about precision of estimates as scientists and mathematical analysts are about their data. If we get estimates within 15% (33ºC vs 28ºC) that is usually close enough, since other uncertainties are likely to be in the mix. ”

    Acceptable margins of error are established in context. Sometimes being within an order of magnitude is good enough and other times the proverbial nine nines isn’t good enough. But I understand your point. It’s illustrated by a trite expression that +-10% is good enough for government work. Left unsaid is that the government screws up a lot because what’s considered “good enough” by them often isn’t.

  112. Ira:

    A chain is no stronger than its weakest link. I think we can agree on that.

    When it comes to trying to sort out natural forcings from unnatural forcings we run into a link in the chain where we have a 25w/m2 range of uncertainty in surface forcing from natural factors. We might reliably know that unnatural forcings fall in the range 0.6 – 2.5w/m2. So that’s a strong link but it’s strength is meaningless when there’s a link in the chain with only a tenth of that strength. See what I’m saying?

  113. Martin Mason:

    Ira, the Second Law of thermodynamics works at every level and you cannot transfer heat from a colder body to a warmer body. A black body cannot also reabsorb and re-emit lower intensity radiation that it has already emited.

    You clearly do not know the basis for the Second Law, which is statistical physics. You seem to believe in some sort of magical version of the Second Law. The actual Second Law is based on the fact that although energy transfers occur in both directions, by simple statistics it becomes essentially astronomically improbable that the net flow of energy, what we call “heat” will be from the hotter body to the colder body for any macroscopic bodies.

    In particular, applied to two radiating objects at different temperatures, the Second Law simply states that the amount of radiation from the hotter body that is absorbed by the colder body is always larger than the amount of radiation from the colder body that is absorbed by the hotter body. The laws of radiative transfer automatically satisfy the Second Law when applied correctly and all models of the greenhouse effect, be they toy models or advanced radiative-convective codes, satisfy the Second Law.

    Greenhouse heating by back radiation is surely wrong because it says that if you put a frozen steak inside a vacuum flask with reflective interior that the steak would cook, that if you stand in front of a mirror you will heat up from the reflected rays

    No…That doesn’t happen because of simple conservation of energy. The steak is not producing thermal energy or receiving it from a hotter object. So, all that the reflective interior can do is slow the cooling of the steak.

    The earth is receiving lots of thermal energy from a hotter object, namely the radiation that it receives from the sun. Its steady-state temperature is set by the balance of what it receives and what it emits back out into space. The greenhouse effect, by affecting the rate at which the earth emits radiation back out into space for a given surface temperature, causes the earth’s temperature to warm in order to maintain radiative balance.

    or that you can save on your heating bills by filling the loft with CO2?

    Insulation in general does help you to save on heating bills. The reason that CO2 in particular does not is for a variety of reasons, one being that the heat loss from a house is due to many mechanisms besides radiation. (For the earth system as a whole, the only way it can lose heat is via radiation, although heat can be transferred between different parts of the system, e.g., surface to atmosphere by convection and evaporation/condensation.) Another is that, as discussed above, the full picture of how adding CO2 affects the greenhouse effect relies on the fact of the lapse rate in the troposphere, something that is not applicable in your example.

  114. Dave Springer says: May 8, 2011 at 10:23 am

    The only possible explanation for why the average temperature of the ocean is 4C is because that is the average surface temperature of the earth taken over a period of time long enough for convection and conduction to equilibrate the entire volume.

    Dave,

    Wouldn’t the explanation be that the surface temperature where there are down-welling currents must be around 4C?

    Currently (no pun intended) the major down-welling current is the gulf stream in the north Atlantic. As long as the north Atlantic is ~4 C, the currents along the bottom will be ~ 4 C. It is reasonable to speculate that down-welling currents would always be in cold regions, since that is where the surface water is most dense. (Add in the fact that evaporation of warm water as it heads pole-ward will increase salinity and increase density and we have a stronger reason for down-welling currents are in the polar regions.)

  115. Several commenters maintained that a colder body cannot heat a warmer body. They may have confused heat with temperature. Heat can be transferred from a cold body to a hotter body, but a colder body cannot increase the temperature or heat content of a hotter body. However, a colder body can decrease the rate of cooling of the hotter body by radiation to the hotter body. Electrons jumping orbitals and producing radiation don’t know if there is a warmer or colder body out there somewhere, therefore the total instantaneous radiation from any body is independent of the temperature of the surroundings.

    Something I did not see in the comments was mention of emissivity, which would have a direct impact on the sensible temperature of the air and surface of the earth. Good emitters (high emissivity) both absorb and emit energy better (grass, leaves, black bodies) than low emitters (most light colors, stainless steel). Emissivity will vary greatly from year to year depending on snow cover, clouds, crops, deforestation, etc.

    Dr. Glickstein’s discussion is interesting, but I think is too simplified to convince any warmist that CO2 is not cooking the earth.

    A couple of things that I am curious about that directly affect the emissions balance, and which have not appeared in any discussion I have read, are:
    1. Energy is absorbed and converted to chemical energy through photosynthesis. I have no idea how much energy that would be (probably not much), but with higher CO2 levels, vegetation is growing faster so it is logical that more radiation is being converted to plant material than in the past, which is stored energy that will not be re-emitted until the material is burned or otherwise reverts to it’s former state.

    2. The earth might correct temperature variations by moderating emissions, especially in the Arctic. When the northern oceans are warmer than normal, they freeze later in the season, and since they emit to an almost perfect black body (deep space with no sunlight) they release much more heat and take up very little, esp. in winter, helping to correct the global temperature. Ice and snow are poor emitters and also insulate the warmer ocean beneath whereas a dark ocean is a very good emitter. A change in ocean currents, for example diversion of a warm pacific current to the Arctic for a prolonged period, could cause global cooling by eliminating ice in the Arctic all year. I have no idea how significant this cooling would be, but suspect it is substantial. Perversely, an abnormally warm current to the Arctic would cause warming of the atmosphere while reducing heat content in the ocean, making it appear as if the earth is warming when in fact it is cooling. Do the GCM’s address this?

  116. As there are multiple periods in the geological record of tens of millions years in duration when CO2 levels were high and the planet was cold and when CO2 levels were low and the planet was warm, it appears there is a basic fundamental assumption in the model of atmospheric radiation that is incorrect or there is an omission of another mechanism from the standard models of atmosphere radiation.

    One possibility is the greenhouse mechanism saturates such that the initial CO2 causes most of the warming.

    Gas molecules transfer energy from molecule to molecule by collisions in addition to radiation.

    As one moves higher in the atmosphere there are more ions due to galactic cosmic rays striking the atmosphere. An ion radiates continuously due to motion of the ion in addition to the band specific radiation that is emitted when an ion captures an electron. It is possible that radiation from ions provides the leak to cause the CO2 mechanism to saturate.

    Atmospheric carbon dioxide levels for the last 500 million years

    http://www.pnas.org/content/99/7/4167.full

    Using a variety of sedimentological criteria, Frakes et al. (18) have concluded that Earth’s climate has cycled several times between warm and cool modes for roughly the last 600 My. Recent work by Veizer et al. (28), based on measurements of oxygen isotopes in calcite and aragonite shells, appears to confirm the existence of these long-period (_135 My) climatic fluctuations. Changes in CO2 levels are usually assumed to be among the dominant mechanisms driving such long-term climate change (29).

    Superficially, this observation would seem to imply that pCO2 does not exert dominant control on Earth’s climate at time scales greater than about 10 My. A wealth of evidence, however, suggests that pCO2 exerts at least some control [see Crowley and Berner (30) for a recent review]. Fig. 4 cannot by itself refute this assumption. Instead, it simply shows that the ‘‘null hypothesis’’ that pCO2 and climate are unrelated cannot be rejected on the basis of this evidence alone.

    http://www.nature.com/uidfinder/10.1038/nature01087

    Despite these successes in linking variations in greenhouse gas concentrations to climate change in the geologic past, the oxygen isotope palaeotemperature record from 600 Myr ago to the present displays notable intervals for which inferred temperatures and pCO2 levels are not correlated1. One of these occurred during the early to middle Miocene (about 17 Myr ago), a time well established as a warm interval (relative to today), but with proxy evidence for low atmospheric pCO2 (ref. 2). Moreover, whereas climate models predict tropical warming in response to elevated pCO2, geologic data — in particularly the oxygen isotope record — indicate muted warming or even cooling at low latitudes while higher latitudes warm (the ‘cool tropicsparadox’10–11).

  117. Bryan,

    The 2.66 grams per square inch figure is considering the whole air column.

    http://www.wolframalpha.com/input/?i=14.7+lbs+*+400%2F1000000

    Alan D. McIntire,

    The solar radiation hitting the earths surface is 184 W/m^2 per Trenberth 2009 or 198 W/m^2 per Trenberth 1997. The surface albedo must be applied to this to get the net gain. Back radiation adds another 333 or 324 W/m^2 respectively. Since Trenberth’s diagram don’t show any reflection at these wavelengths , it is unclear whether he has applied the albedo that corresponds to these wavelengths. The albedo of surfaces varies with wavelength, and for some surfaces like snow the variation is quite significant.

    If the emission from a surface is blocked from escaping in some parts of its spectral emission range, then, everything else being equal, it must increase in temperature until enough of its spectrum is in the ranges that do allow the energy to escape. That would be its equilibrium temperature. Of course, everything else isn’t equal, it will never have to reach that temperature, because conduction, evaporation, convection, poleward heat transport, etc. will assist in transporting the heat away.

  118. Energy in = energy out is not technically true though, the heat that reaches the surface is attenuated as the surface warms and cools, for example heating of the ocean surface and ground surface. When the sun first hits them in the morning they are not instantly at thermal equilibrium and radiating what they recieve straight back upwards, plus the deeper ocean can store and release energy – I know over time energy in = energy out, but it doesnt actually apply to any moment during a day, its why the coldest temps are night are the early morning, and why its warming in the afternoon, not when the sun is highest. I think the diurnal pattern of energy release is also something worth considering, as it does not match the incoming diurnal pattern.

    I think to further peoples understanding the impact of surface attenuation and the affect on the ocean on the global energy flow need to be discussed, as the greenhouse effect doesnt quite hold if these affects are not considered when looking at a point in time on the earth

  119. The issue at hand here is not that co2 can absorb energy, but what occurs and how does it contributes to AGW, and specific man’s co2 output.

    If I suspend in the middle of my room a 4 foot by 6 ft by 2 inch thick lead plate in the middle of my room (not touching the walls), then while that lead plate has huge ability to absorb energy (1000’s of times more than co2), that introduction of the lead plate will NOT cause the room to warm up. So, any energy absorbed by this lead plate will be re-emitted back into the room.

    If the above lead plate could heat up the room, the we would all purchasing big lead plates to suspend in the middle of the room for free heat.

    On the other hand, increasing the insulation of the WALLS of the room is a different matter.

    So, at the end of the day is not that co2 can absorb energy, but what occurs with that effect. As the lead plate example shows, just because physics says something can absorb energy does not necessary mean it affects the inflows and outflows of energy in a given system.

    And in fact it not the co2 that holding this energy anway, but the other substances such as ground and ocean and other parts of atmosphere that represents the bulk of the “mass” of the atmosphere to hold that heat energy.

    In fact, this explains why engine blocks and especially motor cycle heads are often painted black, not silver in color. The choice of black paint on the engine can cause the engine to cool better! In other words, placing black paint on an engine block INCREASES cooling effect yet black paint is known to absorb energy better!

    In effect, since black is a good absorber of energy it’s also thus becomes a good emitter of energy. Black objects emit more radiant energy (cools faster) than a white or silver to its cooler surroundings.

    The issue here never been that co2 can absorb energy, but the issue is not the slam dunk as to what effect this has in terms warming the overall system, or even as some claimed can in fact cause cooling of the system based on the same reasons why black paint on engines can help cooling!

    Super Turtle

  120. “If there were some mechanism that were taking volumes of air high in the atmosphere and transporting them lower in the atmosphere his thesis would have some merit but there is no such mechanism at work on Venus, Earth, or Mars. The pressure gradient is nearly static and if an actual volume of gas is neither expanding nor contracting there is no associated change in temperature.”

    Hadley cells transport volumes of atmosphere from high up to lower down. Don’t confuse the motion of the fluid with its pressure distribution. The fluid can move across a pressure gradient, expanding and contracting as it goes, while the pressure at every point remains constant.

    If there were no convection loops on Venus, it wouldn’t be so windy there.

  121. It seems that for incoming sunlight there is one surface (where sunlight strikes) but for the outgoing radiation there is another surface (at the top of the atmosphere for some wavelengths). One might say that the heat that consumes time migrating from the incoming surface to the outgoing surface represents the heat of greenhouse warming.

    I always wonder about the 15 degree C number quoted as the earth’s surface temperature. It seems that this is the just the air temp nearest the surface struck by sunlight (assuming I don’t misunderstand the source of the number). Why wouldn’t it be more appropriate to attempt to average all temps (air, sea, and land) inside the regions somehow touched by solar radiation (but excluding deep earth temps influenced by heat of radioactivity).

  122. Ron House wrote:

    Dishman says:

    In re Objection #2:
    Gas molecules follow ballistic trajectories between collisions…
    … unless all the work on Gravity since Galileo is wrong, and doesn’t apply at the molecular level, of course.

    Ballistic trajectories are trajectories under the action of gravity. Note, he didn’t say they travel in straight lines.

    However: the approximate speed of molecules in the atmosphere is about 1,700 km/h. The path isn’t very curved, in other words. And the mean free path of a molecule in air is about 65 nanometres. You’d be pretty amazing if you could measure the deviation from a straight line of a path segment with that length and curvature.

    Whether or not it’s directly measureable doesn’t matter. What matters is whether or not it’s happening.

    If molecules follow ballistic trajectories, the minimum KE will occur at zenith and be higher at all other points.

    A short MFP just means that equipartitioning will be rebalanced (on average) every MFP.

    Assuming air (a diatomic gas with 5 degrees of partitioning), for a given delta-h, the thermal energy will be reduced by 1/5th of the energy required to lift molecules by that delta-h.

    My point is that Objection #2 is not, in fact, correct.

  123. The solar radiation hitting the earths surface is 184 W/m^2 per Trenberth 2009 or 198 W/m^2 per Trenberth 1997. The surface albedo must be applied to this to get the net gain. Back radiation adds another 333 or 324 W/m^2 respectively.

    There seems to be a whole lack of common sense going around these days.

    So, according to T, the earth receives 184 w from the sun, radiates some of this back outwards, where some of it is absorbed by the atmosphere and re-radiated back to earth, where it provides 333 w.

    In other words, 184 w ends up as 666 w, because the back radiation is omni-directional at at most 50% reaches the earth.

    Of course, it is like two mirrors pointing in towards each other. Start out with 184w and after it reflects back and forth a bunch of times, you end up with 333w.

    Only one problem. If you start out with 184w from the sun, you can NEVER absorb more that that. Where is the excess energy coming from? The sun is the energy source.

    The atmosphere cannot manufacture 333 w of back radiation for a 184 w power source. The back radiation from the atmosphere must be less than the original 184 w from the sun.

    Otherwise we could put an ingenious array of one-way mirrors around a 100 watt light bulb, and generate millions of watts of power. What climate science has invented is a perpetual motion device, where the energy out is more than the energy in.

  124. “If there were some mechanism that were taking volumes of air high in the atmosphere and transporting them lower in the atmosphere his thesis would have some merit but there is no such mechanism at work on Venus, Earth, or Mars.”

    There is massive movement of air from high altitudes to low altitudes, as a result of an equally massive movement of air from low altitudes to high altitudes.

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

  125. Here is an experiment to disprove the existence of back radiation.

    CAUTION: Only perform with parental supervision.
    CAUTION: Treat any large Fresnel lens as if it were an open flame!
    CAUTION: Since this experiment is indoors, pay close attention to ANY light or infrared (IR) source, which could be focused by the lens. That includes sunlight through windows, hot kitchen stove filaments, etc.

    The only thing great expense in performing this experiment with be the Fresnel lens itself. One square meter Fresnel lenses are available through science supply sites. The lens need to be a plano-convex radial lens to focus parallel rays onto a one square centimeter spot (or even down to one square millimeter). I have heard an F-stop near one is best. This gives a one ten-thousand multiplication though that will to be reduced by up to ~50% due to refraction, scattering and optical imperfections in the facets.

    You will need a very small black metal cap to hold one cm³ (1 gram) of water or ice. This will be placed at the focal point of the lens. You will need an infrared (IR) heat source, a space heater, large wattage hair drier, something to add additional warmth to a wall later in the experiment. You will later need an ice cube.

    Similar to:

    http://www.modulatedlight.org/optical_comms/fresnel_lens_comparison.html

    http://www.scientificsonline.com/large-fresnel-lens.html?&cm_mmc=Mercent-_-Google-_-NULL-_-3052833&mr:trackingCode=CC6E1735-DB81-DE11-8C0A-000423C27502&mr:referralID=NA

    Now, ready? If back-radiation proponents are correct, we should be able to use this huge lens to focus back-radiation from any warm wall within your house, at the ambient room temperature of say 20 ºC (68 ºF), in some room to boil water in seconds.

    Use a small flashlight to illuminate the wall and so the exact focal point can be located. At this point place the black cap with the one gram of water. Wait. If they are correct the water should evaporate quickly, even boil, for that 70 ºF wall is giving off 5.67e-8*(273K+20K)^4 or 418 Watts per square meter that is going to focus down to 400 W x 5,000 multiplication factor or 2,000,000 Watts per square centimeter and we only need 2250 Joules to vaporize that gram of water. Wait, maybe there are some loses within your setup so be patient. Wait. Don’t know about your experiment results but mine is not working. Half an hour later I gave up, something mystical is blocking all of those back radiation photons from entering that lens. Watts up with that?

    Now replace the water and cap with an ice cube. it should develop a hole slowly at the focus. Here the temperature at the focus is less than the temperature of the wall and radiation flows.

    But I want to see the water boil, so cover the lens and warm up that wall in front of the Fresnel lens to a much higher temperature, let’s say 104 ºF (40 ºC). The watt flux should now be something near 5.67e-8*(313K-293K)^4 = 0.009 W/m^2 or 50% efficiency of 0.009 W/m^2 x 5,0000x = 45 W/cm^2. Uncover the lens. Well, it probably won’t boil but you should now evaporate that gram of water in about 2250 J/g ÷ 45 J/s or 50 seconds if you have a perfect setup with no loses. Let’s just say in a couple of minutes.

    Careful, but now fill the cap and point that heat source, directly into the lens from five or six feet away, this is why Fresnel lenses can be dangerous, the water should now boil and quickly in seconds.

    Thermal radiation heat going backwards does not exist in reality, at all, if you are a scientist type and not merely a AGWChurch member, run the experiment, prove it in reality.

    Sorry back-radiation lover’s, you love is just in your minds (and wrong application of Stefan-Boltzmann equation).

    That is also why Fresnel lenses are not dangerous in a very warm dark room at the focus. That room is at radiative equilibrium due to everything being at the same temperature and there is, in reality, no radiation at all, or you could focus it.

  126. Ferd Berple,

    You are just looking at the downward radiation, the upward radiation balances it, there is no energy problem. The mirrors on two sides of the bulb has some merits as an analogy, you wouldn’t doubt that the total flux of photons going back and forth between the mirrors sums to more than the 100 watts being input, but at equilibrium, only 100 watts escapes. A completely enclosed bulb is not a good analogy, because the energy is allowed to accumulate, the filament’s resistance will increase with temperature, reducing the amount of energy input, but still the bulb components will eventually melt.

    In the climate case, the system is very complex, because the increase in temperature in response to the back radiation, increases a number of other responses to increased temperature before equilibrium is reached.

    The idea that the CO2 effect is insignificant, and adiabatic explains all, fails for the common observation of another greenhouse gas, H2O. If it isn’t back radiation, how do you explain the difference in heat loss, on a clear calm humid night vs a clear calm dry night?

  127. DirkH;
    David, this means that at -25 deg the upper atmosphere is no more colder than the ground; an equilibrium is reached. Of course, nothing happens from there; but your premise was to show that the people who say a cold object cannot warm a warm object are somehow wrong – your example shows that two equally cold objects are in equilibrium and does not contradict those people at all. In your example, the upper atmosphere is no more colder than the ground.>>>>

    Read more carefully my friend. I said that at at 25,000 feet it is colder than that even at the equator. Two points:

    1. According to AMSU-A http://discover.itsc.uah.edu/amsutemps/ satellite data the global average at that altitude is in the range of -38C. It would be warmer than that over the equator, and colder… much colder, over the arctic.

    2. 25,000 feet isn’t “upper atmosphere”. In fact, it is only part way up the troposphere which is the lowest layer of the atmosphere. Troposphere ends at 11Km, then Stratosphere about 50Km, Mesosphere about 88Km and then the thermosphere about 500km and then the Exosphere.

    Lot’s of other examples by the way. Build a house with no insulation and measure the temperature in the house and between the inner and outer walls with the furnace running at a constant rate. Then insulate the wall cavity. With the furnace running at the same rate the inside of the house will become warmer. The thermometer between the inner and outer walls will also read warmer, but still cooler than inside the house. The insulation absorbs outbound heat and reradiates it back into the house. Pile snow up around a house, and despite being colder than the inside of the house, the house gets warmer with the exact same heat sources inside, that’s how the early pioneers in NA insulated their sod shacks in the winter. That’s how the Inuit survived centuries in the far north by living the winter in igloos where the cold ice walls nonetheless kept them warmer than they would have been in the open…It’s even the same principle that a freezer works on as does an air conditioner. A kiln gets much hotter than an open fire, but put a thermometer inside the wall of the kiln and you will find that it is cooler than the temperature inside the kiln. Same with a forge. And no, these examples have nothing to do with convection, you can build a forge or a kiln with thin walls or thick walls and have the exact same air circulation, but the thick walled ones have a higher temperature inside, and the temperature of the walls is cooler.

    I am a raging skeptic. An angry, they are misleading us with how they represent the results, producing results from tragically flawed and sometimes outright fraudulant “science” and are being used by lobbyists who are more interested in lining the pockets of their lobby than they are in what’s best for humanity, and I’m disgusted by the whole thing. But cold things radiate photons carrying energy and if they bump into something warmer, there are any number of factors that determine if they will be absorbed or not. But the relative temperature of the two bodies has zippo to do with it. The relative temperature of the two bodies only defines the magnitude and direction of the NET energy exchange, which can only be from warmer to cooler. But take the cooler body away, and the warmer body’s temperature must fall because it is no longer recieving any heat from the colder body.

  128. Martin Lewitt says:
    May 8, 2011 at 3:11 pm
    The idea that the CO2 effect is insignificant, and adiabatic explains all, fails for the common observation of another greenhouse gas, H2O. If it isn’t back radiation, how do you explain the difference in heat loss, on a clear calm humid night vs a clear calm dry night?
    ——
    Simply the difference in the temperatures of one layer to the other (clouds in this case). The air above the cloud tops will be cooler than normal shielded by the water mass in the clouds and the air between the surface and the cloud base warmer than normal. Please, pick up a calculator and calculate these layer’s flux, the overall flux from surface to TOA when all added remains the same, but, the inter-atmosphere profile of the temperatures causing the flux at each layer will be different in the two cases.

  129. Wayne,

    For how much of the infrared black body frequency range is your Fresnel lens able to focus at that one spot? Is your lens material transparent to infrared?

  130. Joel Shore says:

    …”Okay…So it is time to debunk the Postma nonsense again, even though we already did it in the last thread that Ira wrote.”……

    Joel, in case any readers are unaware, is one of the authors of one of the most imaginative pieces of fiction ever to have made its way into a science publication.

    Readers of Ira’s last thread will know that Joel postulated that;
    1. Heat can move from a colder surface to a warmer surface spontaneously.
    2. Invented statements and then proceeded to attack them.
    3. Has the Stephan Boltzmann equation applying to gases.

    Perhaps Joel is still a bit “rusty” about when and where the Stephan Boltzmann Equation applies.
    It applies to black bodies issuing a continuous spectrum centred around a characteristic temperature.
    It is the full integration of the Planck function.

    It does not apply to the line spectra issued by H2O and CO2 making up less than 1% of the atmosphere.

    Joels proof of the “greenhouse theory” involves getting rid of the Oceans and atmosphere.
    Postma is absolutely correct in stating that the long wave radiation produced by the Earth system is released at an average height.
    Some radiation from the ground leaves directly through the “window”.
    Radiation also leaves from different heights above the surface.
    The average height is about 5Km.
    The lapse rate which is completely independent of the radiative effects then determines the surface temperature .
    The greenhouse theory is like Joel’s paper, …… pure fiction.
    Bare Rock Earth is Joel’s favorite topic and he rarely strays from it.

  131. Charlie Foxtrot: exactly. Well-said.

    Dr. Glickstein: I appreciate your efforts to draw analogies to explain the CO2-induced warming of the Earth. However, I’m suspect the effort is doomed from the start because one cannot apply a steady-state model to a dynamic process. The Earth does indeed receive energy from the Sun, and radiates energy into space. But, the system is not at steady state due to Earth’s rotation around its axis, slight eccentricity of its orbit around the sun, greater or lesser cloudiness over time, absorption of energy into oceans or loss of energy from the oceans, and likely a host of other factors not mentioned in this list.

    I’ve done a fair bit of outdoor cooking, over a campfire with a chunk of meat rotating on a spit above the fire. One would have a tough time creating a valid model of how the meat cooks using a steady-state model. Same thing applies to the Earth’s energy balance.

  132. 1 Watt = 1 Joule/second

    1 Watt/m2 = 1 Joule/second/m2

    Now convert the whole concept into Joules and bring time into the equation – say a 24 hour period. It will look much different.

    The problem with all the calculations is that the definition of “Watt” has a time dimension already built into it so we tend to forget that all this has to happen in time.

    The Sun is beating down at 960 Joules/second/m2 at the height of the day but there is almost no energy coming at night. How come the average Earth temperature is only the equivalent of 418 Joules/second/m2 in the daytime and 364 Joules/second/m2 at night.

  133. ferd berple says:
    May 8, 2011 at 8:22 am
    “How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.” …

    A more reasonable explantion is that the missing 33 K is due to the weight of the atmosphere which compresses the gas nearer the surface, which we see as an increase in temperature. …

    Yes, ferd berple, when a gas is compressed its temperature increases. However, once compressed, that increase in temperature will radiate away.

    For example, when you pump up your bicycle tire, the pump and the air in the tire will be warmer than the ambient (uncompressed) air in the room. However, the air in the tire, although it is compressed to several times the normal ambient pressure, will not remain warmer for very long. The heat energy will radiate and convect and conduct away until the tire reaches equilibrium with the ambient temperature of its surroundings.

    The same is true of the Atmosphere. If we transported an Atmosphere’s worth of air to a planet that had no Atmosphere, the force of gravity of that planet would indeed compress the Atmosphere and warm it. However, before long, the Atmosphere would cool down to the temperature of the planet. If we then placed that planet in an Earth-like orbit around the Sun, short-wave Solar radiation would be absorbed by the surface of the planet and the surface would re-emit long-wave radiation to its Atmosphere where some would pass through and out to Space, while some would be absorbed by the H2O and CO2 and other “greenhouse” gases in the Atmosphere. The heated gas molecules would bump into other air molecules and warm them, and like any material above absolute zero, the Atmosphere would emit radiation at a variety of long-wave wavelengths in random directions, some of which would be absorbed by the surface of the planet, warming it further. Just like our Earth.

  134. Theo Goodwin says:
    May 8, 2011 at 8:31 am
    What is so bothersome about this topic is that people who take it up do nothing but rehearse the Great Assumptions, such as the claim that Earth should be treated as a black body. …

    Theo, please read before you comment. I wrote “… Of course, the Earth System is not a perfect black body, as shown by the graph in the upper panel of the illustration below, which plots actual observations from 20 km looking down. ..”

    So, I specifically said the Earth System is NOT a perfect black body and I presented a graph to prove how it differs from a black body! I said further: “… The actual measured radiation is the dark squiggly curve. Note that it jigs and jags up and down between the topmost dashed curve, which is the black body spectrum for a temperature of 270 K and a lower dashed curve which is the black body spectrum for 230 K. This data was taken over the Arctic, most likely during the daytime. The Petty book also has a graph looking down from over the Tropical Pacific which ranges from 300 K down to 210 K. Observations will vary by tens of degrees from day to night, summer to winter, and Tropical to Polar. …”

    So, I specifically said how the Earth’s radiation to Space differs from a black body and how it jigs and jags between black body curves as high as 300 K and as low as 210 K. How could I have been more clear?

    My use of a black body model at 255 K was only for what I said was a “sanity check” on the supposed experts who claim that 255 K is the temperature a black body Earth would have to be at to absorb and emit the same amount of Solar radiation the actual, non-black body Earth absorbs and emits. Got it?

  135. Martin Lewitt, it’s an experiment for you to perform, look on the specifications when your lense arrives, or better, mid-IR lens are also available, specify your choice when ordering. Let us know your results. ☺

  136. “However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”

    Maybe the effect of compressing the atmosphere might cause an increase in of 33 degrees C.

  137. I wish my fingers would learn that ‘lens’ is singular, ‘lenses’ are the plural. ☺

  138. “Yes, ferd berple, when a gas is compressed its temperature increases. However, once compressed, that increase in temperature will radiate away.

    For example, when you pump up your bicycle tire, the pump and the air in the tire will be warmer than the ambient (uncompressed) air in the room. However, the air in the tire, although it is compressed to several times the normal ambient pressure, will not remain warmer for very long. The heat energy will radiate and convect and conduct away until the tire reaches equilibrium with the ambient temperature of its surroundings.”

    A tire is an enclosed space where the air pressure is the same in all locations. The atmosphere is not. The air circulates. As it warms, it rises and cools, and as it cools it falls and warms. Go ahead and bat away a fact of science and try and shove your less than fully thought out process down some other people’s throats.

  139. And continuing.
    I would like to see the back radiation proponents do the following.

    Take a spherical black body object with a small heat source in its center and shine a light on it and read the temperature.

    Take the same spherical object and enclose it with a black body enclosure that does not touch the surface except at some required points to center it around it. use the same lamp at the same distance, and measure the temperature at the surface of the original sphere. Now make another black body enclosure and repeat the procedure. Do this until you have 12 layers of black body. If your argument is correct, then the temperature should sky rocket to the point of starting a fire. According to back radiation proponents statement of how it works.

  140. Ira Glickstein, PhD says:
    May 8, 2011 at 4:26 pm
    “The same is true of the Atmosphere. If we transported an Atmosphere’s worth of air to a planet that had no Atmosphere, the force of gravity of that planet would indeed compress the Atmosphere and warm it. However, before long, the Atmosphere would cool down to the temperature of the planet. If we then placed that planet in an Earth-like orbit around the Sun, short-wave Solar radiation would be absorbed by the surface of the planet”

    You’re right so far.

    ” and the surface would re-emit long-wave radiation to its Atmosphere where some would pass through and out to Space, while some would be absorbed by the H2O and CO2 and other “greenhouse” gases in the Atmosphere.”

    Here it starts going downhill. There are three modes of energy transport from the surface upward: LWIR Radiation, conduction and convection, LWIR radiation being the weakest. IOW, the oceans get warmed by visible and UV radiation and will give the energy to the atmosphere mostly by surface conduction; kinetically.

    ” The heated gas molecules would bump into other air molecules and warm them, and like any material above absolute zero, the Atmosphere would emit radiation at a variety of long-wave wavelengths in random directions, some of which would be absorbed by the surface of the planet, warming it further. Just like our Earth.”

    The gas molecules who get excited by LWIR photons would reradiate some immediately; in some cases they would give the energy to neighbouring molecules, thermalizing the energy. But Kirchhoff’s Law states that for any number of thermalization events there must be an equal number of dethermalization events as long as the gas is in local thermal equilibrium, which it is in the lower atmosphere. So no heat is trapped there.

    See

    http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/

    So, the consequence is that LWIR radiation is transporting energy to space. Some is re-emitted back to Earth from where it is re-emitted back upwards. Whether there is 0.03 % or 0.04 % of CO2 in the atmosphere only influences how often the photons get absorbed and re-radiated on their way to space – an increase in CO2 delays the process a little but does not change it fundamentally and *Does* *Not* *Trap* *Heat* any more than a sieve traps water.

  141. Dave Springer says:
    May 8, 2011 at 8:36 am
    Ira (con’t):
    The first pass you need to make in accounting for this discrepancy is to take albedo into account. … All the experimental attempts to measure the earth’s average albedo do agree on one thing – it isn’t constant and varies by as much 1% year over year.

    One percent doesn’t seem like much variation but once again it’s a very large number when compared to anthropogenic warming. A 1% change in the amount of insolation actually reaching the surface and not being reflected directly back out into space is some 2.5 watts per square meter. The IPCC third assessment is “95%” confident that anthropogenic forcing lies in the range of 0.6 to 2.4 w/m2. Yet measured variations in earth’s albedo over just several years has it changing by more watts/m2 than the highest estimate of alleged anthropogenic warming. Worse yet, the actual albedo can’t be pinned down and estimates vary by about 7% which is 7 times greater than anthropogenic forcings. …

    I agree completely Dave Springer! Anthropogenic forcing (human-caused warming due to land use changes that reduces albedo and burnign of fossil fuels that raises CO2 levels, etc.) is most proabably below the lower IPCC estimate of 0.6 W/m^2.

    In previous postings here at WUWT I have estimated the human contribution to net warming since 1880 at 0.2ºC, the natural cycles and processes contribution over which we humans have no control at 0.3ºC to 0.4ºC, with the remainder of the supposed warming of 0.8ºC due to data bias and cooking of the books by the official climate Team.

    My main purpose in this Visualizing series was not to justify any particular number for the warming effect, but rather to correct those of our fellow Skeptics who have bought into the false notion that the Atmospheric “greenhouse effect” is fiction. In some sections of the media and public opinion, outspoken, non-scientific Disbelievers make us look foolish by association.

  142. Ira,
    Please detail exactly why you think Swedish climatologist Dr. Hans Jelbring and chemical engineer William Gilbert are “foolish” “non-scientific Disbelievers” and “irrational” in their explanation of the so-called “greenhouse effect” based simply upon the adiabatic lapse rate:

    http://www.tech-know.eu/NISubmission/pdf/Politics_and_the_Greenhouse_Effect.pdf

    as well as physicist Joe Postma who explains the same

    http://www.globalwarmingskeptics.info/forums/attachment.php?aid=327

  143. Bryan says:

    Readers of Ira’s last thread will know that Joel postulated that;
    1. Heat can move from a colder surface to a warmer surface spontaneously.
    2. Invented statements and then proceeded to attack them.
    3. Has the Stephan Boltzmann equation applying to gases.

    Actually, what readers of that thread will know is that Bryan is a purveyor of pseudo-science. He has absolutely no serious desire to discuss the science and, like all purveyors of pseudo-science, tries to distract people by nitpicking words…even after the wording has been corrected to everyone’s satisfaction. (See here for more discussion specifically on the tactics that Bryan, Gerlich and Tscheuschner and other purveyors of such pseudo-science employ: http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/#comment-649274 ) The existence of people like Bryan within the skeptic movement, willing to actively engage in such deception, certainly helps to keep you guys marginalized in the scientific realm.

    3. Has the Stephan Boltzmann equation applying to gases.

    Joels proof of the “greenhouse theory” involves getting rid of the Oceans and atmosphere.

    More nonsense. Simple toy models of the greenhouse effect illustrate the concept and give one a qualitatively understanding of what is going on. Line-by-line radiations codes do the full calculations for the actual absorption lines as they exist in the atmosphere in order to nail down the effects quantitatively. Yes, the simple models don’t have all the gory details in them…That is the whole point.

    Postma is absolutely correct in stating that the long wave radiation produced by the Earth system is released at an average height.
    Some radiation from the ground leaves directly through the “window”.
    Radiation also leaves from different heights above the surface.
    The average height is about 5Km.
    The lapse rate which is completely independent of the radiative effects then determines the surface temperature .

    Fine…Everything you say is basically correct. The one thing that you and Postma are leaving out is what I said in my last post…that the average height of 5km is determined by the IR-absorptivity of the atmosphere, i.e., by the greenhouse gases in the atmosphere and clouds. (The technical statement is something like this: At any given wavelength, the height at which most of the radiation escapes to space is given by the height at which the “optical depth” [ http://en.wikipedia.org/wiki/Optical_depth ] of all of the atmosphere above that point to radiation of this wavelength is of order 1. More quantitative details can be found in books on atmospheric radiation, such as Ray Pierrehumbert’s book “Principles of Planetary Climate”.)

  144. I haven’t had time to read all of the comment but I haven’t seen any comment on land-use/botany. Even in the oceans there are plants. Much of the earth is a GREEN body that is converting solar energy into some form of biomass. How many watts does it take to produce a bushel of wheat or a 1,000 board feet of pine lumber or produce a nice turf (or Veg. Garden) at the White House? It isn’t being factored in!

  145. Anton Eagle says:
    May 8, 2011 at 11:32 am
    Ira,
    I have been reading many posts here that discuss the blackbody radiation of the earth. You have claimed that the 33K difference is due to GHG back radiation. I don’t want to argue that with you… but I think it would be worthwhile to be very very clear on exactly what you are claiming.

    So… I can boil this down to a simple yes or no question…

    Are you claiming that if the atmosphere were replaced with a different set of gasses that do not contain GHG molecules, but still had all the other macro effects such as clouds etc in the exact same amount as our current atmosphere… are you claiming that this atmosphere would NOT cause (directly or indirectly) the surface of the earth to warm at all? …

    WOW Anton Eagle, that is quite a question! My short answer is “yes, that is the essence of my claim.” However there are some caveats:

    1) The atmosphere would be composed of pure nitrogen and other gases that transparently pass both long-wave and short-wave radiation without any absorption at all.

    2) The albedo of the Earth System would be the same as it is now, around 30%. This could be accomplished by covering a portion of the Surface with reflective materials such that only about 240 Watts/m^2 would be absorbed by the Earth Surface, the remainer being reflected back out to space through the transparent Atmosphere.

    3) The effects of winds, storms, rain, snow, H2O phase change, ocean currents, and so on – including convection and conduction from the Surface to the Atmosphere and the reverse, would have to be simulated or corrected for.

    Under the above conditions, I believe the mean Earth Surface temperature would be closer to 255 K than to the present 288 K.

    I would appreciate hearing the answer to this question (and perhaps more caveats) by others who are more knowledgeable than I am.

  146. Ira has explained it here: http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655793

    I have explained it here: http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655686

    The only way that adiabatic compression could possibly explain the earth’s surface temperature would be if the atmosphere (and / or the earth itself) were undergoing continual gravitational collapse, which we know is of course not the case. Otherwise, such a hypothesis does not even satisfy the First Law of Thermodynamics (basically, conservation of energy): Without substances in the atmosphere that absorb terrestrial radiation, the earth’s surface at its present temperature would be emitting back out into space way more energy than it receives from the sun and hence would rapidly cool down.

    The temperature structure as a function of altitude, i.e. the lapse rate, in the troposphere is set by the considerations discussed regarding adiabatic expansion and compression (basically because a lapse rate higher than the appropriate adiabatic lapse rate for the given water vapor content is unstable and leads to convection until marginal stability is restored).

    However, this alone does not determine the temperature at the surface. That is determined by consideration of the absorption of the atmosphere of terrestrial radiation (and radiation emitted by the atmosphere), which essentially ends up determining at what altitude the temperature has to be determined via radiative balance between the Earth system (earth + atmosphere) and the sun and space [which for the earth system with its current albedo is ~255 K]. Then, by using the lapse rate, you can get the temperature at the surface.

    However, as one adds greenhouse gases, one raises this altitude and hence the required surface temperature. (In the simplest case, this is under the assumption that the lapse rate doesn’t change, whereas the more complicated calculations assume that there is somewhat of a decrease in lapse rate, by the well-understood physics of the moist adiabatic lapse rate in the tropics. This decrease in lapse rate is a negative feedback that is included in all of the climate models.)

  147. IRA: It is getting more and more conspicuous that you are completely ignoring the alternative explanations for the Earth’s temperature. You have completely ignored me and many others who have suggested some other theories to this point, unless I missed something. You need to respond. With some facts. Otherwise, my advice is to shut up.

  148. Thank you Hockey Schtick !!…

    William C. Gilbert’s words are much better than the ones I have been using and now I have someone to attribute them to. He’s right, it is merely not so simple, but understandable, physics.

  149. wayne says:

    William C. Gilbert’s words are much better than the ones I have been using and now I have someone to attribute them to. He’s right, it is merely not so simple, but understandable, physics.

    It is clear that Gilbert has never actually read any elementary textbook on atmospheric physics or climate science or he would be aware that the argument about adiabatic lapse rate that he presents is not new to anyone in the field…It is discussed in all of these books. (Alternatively, maybe Gilbert is aware of this but is confident that the readers who he wants to convince have not.)

    However, far from his claim that this temperature profile explains the 33 deg temperature difference between what is observed and what radiation balance requires, the fact is that it alone doesn’t explain any of it. You need to consider not only how the temperature varies with height but what then sets the constant that tells you what the absolute temperature is at some height in the troposphere. In the absence of absorption of terrestrial radiation by the atmosphere (and with the other caveats about still having the same albedo and such), that average temperature would have to be 255 K at the surface because of radiative balance and then the temperature would decrease with height at the lapse rate from there.

  150. jae says:

    IRA: It is getting more and more conspicuous that you are completely ignoring the alternative explanations for the Earth’s temperature. You have completely ignored me and many others who have suggested some other theories to this point, unless I missed something. You need to respond. With some facts.

    The errors in your explanations have now been explained to you multiple times in this thread. There is no excuse for you to continue to cling to them.

  151. Anton Eagle says:

    May 8, 2011 at 11:33 am (Edit)

    Ira,
    I have a second question regarding the two emission diagrams you show.

    The upper graph, showing down-looking readings at the top of the atmosphere shows that approximately 1/2 of the 15 um radiation is absorbed. It looks as if it would be about 90 mW/m^2 without absorption, and instead is about 45 mW/m^2 or so with absorption.

    Then you show the lower graph showing the up-looking readings at the surface… showing about 90-100 mW/m^2 of radiation (at 15 um) down to the surface from the atmosphere.

    So, my question is… if the GH gasses are absorbing about 45 mW/m^2… how are the re-radiating almost 100 mW/m^2 back to the surface?

    It doesn’t look like the sun can be the source, since according to these curves the sun doesn’t radiate hardly anything at those wavelengths. So… where is the extra energy coming from?

    WOW, another great question that I hope those more knowledgeable than I can answer. THANKS Anton Eagle – sometimes the right questions can lead to deeper knowledge.

    Here is my take:

    1) The Perry curves in this posting are based on NASA measurements taken in the Arctic, where the water vapor content of the Atmosphere is less than in regions closer to the Tropics. My earlier posting in this series at here reproduces more Perry curves in their full glory (and not mirror-imaged), including curves from the Tropics.

    2) The Perry curves plot what he calls “Radiance” and that seems to differ from “Intensity” (in the Carleton spreadsheet plots) in that the amplitude of longer wavelengths appears to be multiplied by the wavelength squared for reasons I do not yet understand. Thus, any readings we take in the ~15μm region are shown about four times higher than comparable readings in the ~7μm region, for example (because 15^2 is over four times 7^2).

    3) In comparing the amplitudes I think the best bet is to judge where they are with respect to Perry’s plot of the black body curves. Although it is not clear in the small versions I included in this posting, the full-resolution plots show that the topmost dashed curve in both plots represents a 270 K black body.

    4) As I understand it, the ~15μm radiation from the Surface to the Atmosphere is absorbed by H2O and CO2 molecules which, when excited, bump into nitrogen and oxygen and other air molecules, and heat the air. The result is emission of long-wave radiation at a range of wavelengths (not just in the ~15μm region, but including the ~7μm and ~10μm regions) in random directions, some of which finds its way back down to the Surface and is re-absorbed. That conversion of a portion of the ~15μm radiation into other wavelengths may explain why there is a “bite” out of the ~15μm radiation that seems to be missing from the radiation out to Space.

    5) In particular, notice that the ~10μm region (from around 8μm to 12μm) is very close to the highest black body curve, except for the little “bite” near 9.5μm which is where there is an oxygen/ozone absorption effect. In addition to the ~15μm “bite” that corresponds to the CO2 absorption region, notice that the H2O absorption areas, namely the ~7μm region and the part of the ~15μm region that extends from 18μm to 25μm and above is also somewhat below the highest black body curve.

    6) Thus, if we assume, as a first approximation, that the Surface approximates a blackbody at 288 K, with a spectrum something like the smooth blue curve in my illustration above, we see that the Atmosphere passes the ~10μm region (except for part of the ~9.5μm oxygen/ozone “bite”) and, from the Perry plot of Surface looking UP, re-emits much of the ~7μm and ~15μm region back down to the Surface. From the Perry Space looking DOWN plot, we see that the ~10μm region is the highest (except for the oxygen/ozone “bite”) and that the two H2O and the one CO2 regions are depressed. My take is that the radiation in the H2O and CO2 absorption bands shuttles back and forth between being absorbed and re-emitted by the Surface and Atmosphere (where about half heads in the direction of Space), all the while transforming a bit to the ~10μm region, which gets a free pass to Space, with each transaction.

    7) So, BOTTOM LINE, the readings taken from the Perry curves need to be interpreted properly to answer your question. There is no extra energy.

  152. Joel:

    “The only way that adiabatic compression could possibly explain the earth’s surface temperature would be if the atmosphere (and / or the earth itself) were undergoing continual gravitational collapse, which we know is of course not the case.”

    Thank you! I’ve had big trouble trying to explain this to people before. I tried using the example of a bike tire, which heats up when you compress the air but then doesn’t stay hot (even though it stays compressed, but the elasticity of the innertube, analgous to gravity, is not reducing the volume of gas). Success rate with this explanation is surprisingly low.

  153. Joel:

    “The errors in your explanations have now been explained to you multiple times in this thread. There is no excuse for you to continue to cling to them”

    ?? Can you please point me to those explanations, sir? I must have missed them, or else, perhaps, probably, most likely, they were balogne? Come on, fella, provide some tangible thing that I can relate to, instead of your Orwellian insistance that there is, “out there,” something that proves your case/argument/status/being/etc..

    WTF?

  154. “The only way that adiabatic compression could possibly explain the earth’s surface temperature would be if the atmosphere (and / or the earth itself) were undergoing continual gravitational collapse, which we know is of course not the case.”

    Typical CAGW-practitioner gobbledygook

    And the atmosphere does not behave as an enclosed bicycle tire. No wonder the “success rate” with that explanation is so low.

    Here’s Dr. Jelbring’s more detailed explanation:

    http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBYQFjAA&url=http%3A%2F%2Fruby.fgcu.edu%2Fcourses%2Ftwimberley%2FEnviroPhilo%2FFunctionOfMass.pdf&ei=9V3HTbH4IIP4sAOLzuTqAQ&usg=AFQjCNHTux_zuvY-oHVCehfC6OOmfzUafQ&sig2=p-LjGfmce1A-Vqk5ybrjZg

    Show me the exact error’s you claim, citing specific excerpts from his paper.

  155. ferd berple says:
    May 8, 2011 at 2:24 pm
    The solar radiation hitting the earths surface is 184 W/m^2 per Trenberth 2009 or 198 W/m^2 per Trenberth 1997. The surface albedo must be applied to this to get the net gain. Back radiation adds another 333 or 324 W/m^2 respectively.

    There seems to be a whole lack of common sense going around these days.

    So, according to T, the earth receives 184 w from the sun, radiates some of this back outwards, where some of it is absorbed by the atmosphere and re-radiated back to earth, where it provides 333 w.

    In other words, 184 w ends up as 666 w, because the back radiation is omni-directional at at most 50% reaches the earth……………………………………
    ///////////////////////////////////////////////////////
    There does appeart to be a complete lack of commonsense in these figures.
    If these figures were correct (ie., back radiation is approximately twice the power of solar), why are we wasting time with trying to capture solar energy rather than back radiation energy?
    Why don’t we construct a large array of mirrors collecting and focusing this back radiation much like the Californian Solar power station: see

    This not only would produce more energy (viz about 330 cf about 190 W/m^2; back radiation cf solar), it would additionally solve the storage problems inherent with present day green energy projects since back radiation is available day and night etc.
    The answer is simple, either this back radiation does not exist, or if it does exist it is incapable of doing any sensible work (ie., it lacks sensible energy).

    The lack of any serious research into collecting and utilising the back radiation energy source strongly suggests that leading energy scientists regard the back radiation set out in Trenberth’s energy diagrams to be complete and utter nonsence!

    Anyone who supports Trenberth’s diagram and considers that the back radiation theory is correct should be asked to explain why we are not utilising this fanastic energy source. After all, if it existed it should be the main focus of green energy projects.

  156. “The only way that adiabatic compression could possibly explain the earth’s surface temperature would be if the atmosphere (and / or the earth itself) were undergoing continual gravitational collapse, which we know is of course not the case.”

    Typical CAGW-practitioner gobbledygook

    And the atmosphere does not behave as an enclosed bicycle tire. No wonder the “success rate” with that one is so low.

    Here is a more detailed explanation from Dr. Jelbring:

    http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBYQFjAA&url=http%3A%2F%2Fruby.fgcu.edu%2Fcourses%2Ftwimberley%2FEnviroPhilo%2FFunctionOfMass.pdf&ei=9V3HTbH4IIP4sAOLzuTqAQ&usg=AFQjCNHTux_zuvY-oHVCehfC6OOmfzUafQ&sig2=p-LjGfmce1A-Vqk5ybrjZg

    Show me exactly where he is in error, citing exact quotes from the paper, without any hand waving.

  157. Luuuucy?????? Ira?????? we are missing some comments here. WTF? Are you THAT desperate????

  158. jae says:
    May 8, 2011 at 6:59 pm

    IRA: … You need to respond. With some facts. Otherwise, my advice is to shut up.

    jae says:
    May 8, 2011 at 8:10 pm

    … WTF?

    A courteous, well-bred person will not insult me, and none other can. [Paraphrasing William Cowper, ca 1763]

  159. Wow. Once again, misconceptions abound on both sides.

    Ira, just because it is the only seemingly rational suggestion that has been posited publicly does not mean that it actually describes the physical process accurately. If it did, the term “backradiation” would not have had to have been coined in the first place.

    It is a simple matter to increase the absorption in a gas above the level of the incident energy, without ever having to increase the relative concentration of the gas, and is done on a daily basis in my field, and has been for many decades, but we do not call it “backradiation”.

    Some (no, I don’t know exactly how much) of the reflected solar energy has already been absorbed by the atmosphere before being reflected. The statement that it contributes nothing to the atmospheric temperature profile is thus at least partly incorrect.

    So Ira, if you really are sincere about wanting to know where the 33K difference comes from, you could start by calculating the Mie backscatter cross section of a 10um dielectric sphere, and I would suggest calculating it with a 9.25um or 10.59um em wave incident on it. I suggest those 2 wavelengths because measurements have already been done with them, and you will have something to compare the calculated values to, although the measured values were normally given in steradians, which can also be calculated.

    You might also consider how it is possible to treat two entities that are physically coupled (think about it) as if they were two separate “blackbodies” radiating “at” each other.

    Cheers.

    BTW, N2 DOES have resonance lines in the solar UV region.

  160. A warm atmosphere will emit radiation. Some of it will be backradiation.

    Even without IR absorption the atmosphere will warm.

    http://earthobservatory.nasa.gov/Features/EnergyBalance/page6.php

    Convection.
    Evaporation.
    Incoming energy absorption (visible light).

    With this in mind, and accepting the backradiation mechanism, how can the “greenhouse IR” component be ascribed the full 33 C difference?

    Surely it can’t be?

  161. From what I see the proponents of the Greenhouse theory simply keep stating it as fact without explaining the observed phenomena that seemingly falsify it such as.

    – You cannot heat a hot body from a colder body which you have to be able to do.

    – The temperature at the surface and at height can be explained without recourse to radiative theory which to me means that the temperature is determined by the fact that we have an atmosphere not necessarily because we have an atmosphere with GHGs. The atmospheric column is nothing like a bicycle tyre in physical terms, that is a red herring.

    – In the past when we have had multiples of current CO2 levels we have not had runaway warming.

    – There is patently no runaway warming now while CO2 levels ramp up inexorably. There is no runaway warming on venus.

    – If back radiation is so powerful and it is a heat source, why don’t we use it?

    It’s the answers to the simple things that are important not the circular academic arguments. The consensus view is that AGW is happening because of a trace GHG, the only problem is that there is no unequivocally correct theoretical or observational data that supports it.

  162. John of Kent says:
    May 8, 2011 at 3:09 am
    Ira Glickstein does not seem to understand how matter interacts with radiation.

    On the contrary he seems to understand it fairly well, you however appear not to!

  163. Wayne,

    “The air above the cloud tops will be cooler than normal shielded by the water mass in the clouds and the air between the surface and the cloud base warmer than normal.”

    The problem is for those doubting back radiation to explain the greater night time head loss in dry clear skies vs humid clear skies. You don’t seem to be able to do it without clouds, are you doubting the humid vs dry clear sky difference exists?

    BTW, no huge lens covering a wide enough range of the infrared spectrum exists to conduct your hypothetical experiment.

  164. According to the more reasonable reconstructions of past temperature (Moberg, Loehle) the surface temperature has been rising since 1700. This is 230 years longer then co2 has been rising to any great degree.

    What is Ira’s explanation for this earlier warming?

    The Earth is a big heat engine. Most of the effective solar heating of the ocean occurs in the tropics, whereas most of the effective radiation of heat to space occurs near the poles. Most of the energy between the equator and temperate latitudes is shifted by the ocean. From the temperate latitudes to the poles, most of the energy is shifted by the atmosphere. This movement of energy is achieved mainly by convection of one sort or another, and since the temperature of the ocean is little affected by back radiation, changes in co2 concentration can’t have much effect.

    Why does Ira refer simply to the Earth’s surface as an amorphous entity instead of taking into consideration the fact that 70% of it doesn’t absorb much energy from back radiation?

    over 90% of downwelling radiation striking the ocean surface has been emitted a kilometre or less above it, because of re-absorption and re-emission. Since this re-emitted radiation is travelling at any random angle wrt to the ocean surface, most of it is travelling at angles which will either miss the surface or be reflected by it. The back radiation which is entrained into the ocean bulk by wave action is minimal, because wave action subducts water which is well below the level back radiation can penetrate to (little more than its own wavelength). Most of the radiation measured as coming from the ocean surface is actually either been reflected or is being emitted by molecules evaporated from the ocean surface.

    Since the ocean surface temperature changes precede surface air temperature changes by several months, and since the top two metres of ocean contain as much heat capacity as the entire atmosphere above it, it is clear that surface temperature and atmospheric temperature is strongly influenced by the ocean, which is heated by the sun, not by back radiation.

    Solar variation affects albedo, and that affects how much solar energy the ocean absorbs. This is why the number of sunshine hours correlates with surface temperature over the last 130 years much more closely than co2 level.

    In my opinion, surface temperature variation is due to the variation of the Sun. Co2 is along for the ride.

  165. Why is “the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet”?

    As dr.bill said, this is because the Planck equation is actually an integral. When you change the x-axis, the peak actually does shift. I have written a full explanation at

    http://mc-computing.com/Science_Facts/Blackbody/Blackbody_Equations.html

    There is also a plot showing the difference in the shape of the two 300K curves and their peaks.

    biological evolution has “tuned” our visual system to be most sensitive where the light energy is maximized

    Not quite, that is where the intensity per wavelength interval is brightest, not where the energy per frequency interval is largest.

    This demonstrates why people who believe in Global Warming always plot spectra versus wavelength. When you want to analyze the energy in the system, all plots should be versus frequency (normally expressed as a wavenumber) to show an honest plot. In that case, most of the energy from the Sun is in the IR. When plotted versus wavelength, the plots incorrectly imply that most of the energy is in the visible and UV spectra.

  166. The one big attractive feature of the “greenhouse theory” is its amazing elasticity.
    The climate getting warmer is a sure proof of CAGW
    But then so is the climate getting colder, say some IPCC advocates.

    Joel Shore has refined this even further to say Postema is a believer, only Postemo hasn’t quite realised it.
    Perhaps with the same elasticity Joel will prove that Gerlich and Tscheuschner also are really IPCC advocates.

    Joel goes on to say of me;

    …..”He has absolutely no serious desire to discuss the science”…..

    Quite the opposite Joel;

    On page 9 of your co-written Halpern et al comment paper you state that the Stephan Boltzman Law can be used to work out the thermal energy between atmospheric shells.
    I brought your attention to it in my post above;
    ….. ” 3. Has the Stephan Boltzmann equation applying to gases.”……

    Its no good trying to weasle your way out of it.
    Your comment was printed in a specialist Journal with readership almost entirely of Physicists.
    They would have no problem understanding your point.
    Do you still maintain this position or will this be your third major repudiation of this notorious document.
    In fact you should perhaps repudiate the document in its entirely!

  167. Ira, I’d really like some of the “back radiation” now. Just stepped outside my house in Kamloops BC and pointed an IR thermometer straight up into the sky and got a temperature reading of -10 F. When I pointed the thermometer horizontally down the street, got 34 F. Pointing it at the outside of my house gave a temperature of 44 F as did pointing it at my driveway. The airport temperature now is given as 48 F. The altitude of the airport is about 200′ lower than where I live. I should note that the sky is perfectly clear at this time with no clouds in sight (just wish it would do this in the daytime).

    The surface of the earth heats up considerably during the day and this heat is released during the night. Of course having heat buffers of variable capacity greatly complicates models. The -10 F temperature that I’m seeing with my IR thermometer is the average temperature of the atmosphere from 1600′ above sea level to space. What portion of this would be related to “greenhouse gasses” and what portion is the result of heat flux from sun-warmed surface and air to space?

  168. Martin Lewitt, you are right, I do not “believe” in “back radiation”, just plain radiation as taught in physics.

    Clouds make warmer nights because the clouds are usually warmer than the normal air temperature at that altitude and therefore the surface’s rate of loss by radiation upward will be less leaving you with a warmer than normal night. It is also that same radiation keeps those clouds warmer than te air at that altitude would normally be. Just using Stefan-Boltzmann will get you very close if you know the temperatures and emissivities involved. There are even rare cases when the clouds are actually warmer than the surface and then the clouds are radiating downward to the surface and also to space but as you should see, this condition is short lived since it has cooler on both sides unless continually replenished with warmth to maintain the temperature differential.

    BTW, that experiment does not require all frequencies to be valid. I’ll just leave it for each here to decide it’s merits, sounds like most here can handle high-school science correctly.

  169. From the comments here, a lot of people seem to believe that the greenhouse effect cannot exist because it violates the 2nd law of thermodynamics. To my shock, I am finding myself agreeing with Joel Shore on the elementary physics. I have just finished reading the Postma essay, and it contains the same strawman argument that crops up every time this matter is discussed.

    Postma described a black body warmed by a heat source and claimed that you cannot make it warmer by reflecting that emitted radiation back. Whilst acknowledging back radiation exists, he asserts that this energy cannot cause the black body to increase in temperature. He states that a body cannot raise its own temperature. Well of course not. The reason this is a strawman fallacy is because nobody is making that claim. The claim is that the back radiation reduces the rate of cooling of the earths surface because the net outflow in watts is slightly less than it would have been without the back radiation. It is the sun that causes the warming because energy is now coming in faster than it is leaving, so the temperature will rise until a new equilibriumis reached.

    An analogy is a tub with a ‘V’ cut near the top. The higher the water level is above the ‘V’, the faster it flows out. This is an analogy of the SB law of blackbody radiation. Suppose the sun is represented by a running tap. The water will rise above the ‘V’ just enough so that the rate of input is equal to the rate of overflow. Let us now represent the back radiation by blocking some of the ‘V’ which represents the reduced outflow of radiation from the earths surface. The water level (temperature) will now rise such that the outflow increases to balance the inflow. If you turned the tap off (remove the sun from the climate system), the water level (temperature) would start to decline just as you would expect from thermodynamics.

  170. Ira,

    Here is a schematic from a search, nothing special, merely from the first site that had a lapse curve, month of June, 45 degrees North latitude.

    Source was Bigg 2005, blog was http://scienceofdoom.com/2010/04/24/tropospheric-basics/

    The coldest temperature measured on earth is about -89 deg c at Vostok. This is colder than any part of the lapse curve below about 75 km above Earth.

    How can it get so cold? What stops heat getting to Vostok?

  171. “Anyone who supports Trenberth’s diagram and considers that the back radiation theory is correct should be asked to explain why we are not utilising this fanastic energy source. After all, if it existed it should be the main focus of green energy projects.”

    “- If back radiation is so powerful and it is a heat source, why don’t we use it?”

    Good questions. Although what about this free energy oven. Does it qualify?

    http://www.slayingtheskydragon.com/en/blog/111-a-pictures-worth-a-1000-words#comments

  172. Joel Shore says:

    You clearly do not know the basis for the Second Law, which is statistical physics.

    The basis for the second law (and the first) is careful and accurate measurement, mostly paid for by wicked 19th century capitalists, who wanted to know how to get more from their steam engines, and also how to avoid being sold snake oil.

    There is no need for a statistical ‘fudge’ unless you wish to introduce special pleading for “IPCC Approved” photons.

    If I were to dump a billion microscopic ice particles into a tank of hot water (di-hydrogen monoxide), your statistical approach would insist that a few of them may have warmed the water (hydrogen hydroxide), especially using a computer model.

  173. I use spreadsheets to calculate Earth IR emissions and absorption by CO2. Using mean beam lengths and CO2 partial pressures, I use PcL charts by Hoyt C hottel to obtain the emissivity of CO2 in the atmosphere. For current CO2 partial pressure or concentration I calculate a distance of 3600 m of traverse through the earth’s atmosphere by ‘IR radiation’ before the absorbable IR is ‘filtered out’ and reduced to zero. I then calculate that by doubling the atmospheric CO2 to 700ppm or so the ‘absorption distance’ is reduced to 2000 m but with no additional IR absorption by CO2. Therefore I conclude no global warming from increasing atmospheric CO2.

  174. Ira,
    “But, that still does not explain why the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet and other graphics on the Internet”

    Maybe I could help, or perhaps just confuse you further.
    If you plot the radiant energy against wavelength for blackbody around 287K, then you get a peak intensity at about 10 micron. The graphs reproduced from Petty show the peak intensity at about 18 micron. This apparent anomaly arises because the vertical axes of the 2 graphs use different units.
    The Plank distributions plotted against wavelength (the way I like to see them) have a vertical scale of ‘Watts per Square metre PER MICRON’. The graphs shown by Petty are plotted against wavenumber (not wavelength) and the vertical axes have units of ‘Watts per Square Metre per Steradian PER WAVENUMBER’.Note: graphs plotted in these different ways are NOT simply mirror images of each other. [Forget the Steradian bit, this is just a multiplying factor which doesn't effect the peak or shape]
    Say that the peak radiation on the wavelength plot occurs at 10.5 micron. Because the energy is plotted per micron it represents the total energy between 10 micron and 11 micron. In terms of wavenumber, 10 to 11 micron is equivalent to a range of 909 to 1000 per cm. So the energy for the 10.5 micron peak has to be dispersed over 91 different wavenumbers on the wavenumber plot. If we consider another point on the wavelength plot, say between 18 and 19 microns, this is represented by a wavenumber range from 526 to 555. So the energy around 18 microns has only to be shared between 29 wavenumbers whilst the energy around 10 microns has to be shared over 91 wavenumbers. As a consequence, the peak intensity no longer occurs at 10 micron, it now occurs at 18 micron. That is why the two methods of plotting give different peak intensities.

    Did that help, or just confuse everyone further?

  175. Boris Gimbarzevsky,

    Thanks for sharing your thermometer readings It would be interesting to know if on clear nights, sky readings taken at the same ambient temperature but different absolute humidities varied due to increased water vapor greenhouse effect.

  176. Dishman says:

    If molecules follow ballistic trajectories, the minimum KE will occur at zenith and be higher at all other points.

    A short MFP just means that equipartitioning will be rebalanced (on average) every MFP.

    Assuming air (a diatomic gas with 5 degrees of partitioning), for a given delta-h, the thermal energy will be reduced by 1/5th of the energy required to lift molecules by that delta-h.

    My point is that Objection #2 is not, in fact, correct.

    That may be your point, but nothing you say backs it up. You say “the thermal energy will be reduced by 1/5th of the energy required to lift molecules by that delta-h.”

    A quick calculation shows that the average molecular speed, assuming constant g, is sufficient to lift a molecule about 11 km. The mean free path of a molecule in air is about 65 nanometres. Dividing these, we see the energy needed to lift the molecule by your delta-h is about 1.7*10^-11 of its total energy. In other words, diddly squat. Gravitational effects make no difference to the result that could possibly be measured.

  177. Charles Nelson said: Gotcha Henry! He thinks CO2 is like a toxin in the atmosphere! An amazing toxin which is NOT TOXIC at 380ppm but suddenly BECOMES TOXIC at 400ppm. There in a single post is the fuzzy thinking of the Warmists! Fantastic. Textbook example of ‘Carbonaphobia’!

    I don’t read him that way. I think he’s making the point that the tiny size of the fraction, by itself, doesn’t tell us anything. If he’s making the point that you think he’s making, he’s doing a poor job of it.

  178. Ira,

    you do not consider conduction and convection, nor non-GHG absorbers/reflectors of radiation nor spatial issues. Simply saying that the difference in temperature between a perfect blackbody and the incoming radiation is due to “greenhouse” effects is disingenious. You can convince yourself by a thought experiment: would the atmosphere really be at zero degree Kelvin were it not for GHGs? I’d say obviously not and hence your simple difference doesn’t apply.

    best regards,

  179. Martin Mason said – If back radiation is so powerful and it is a heat source, why don’t we use it?

    Because the earth’s surface is already at the temperature to which the back-radiation has raised it.

  180. Allan M says:
    May 9, 2011 at 2:04 am

    “If I were to dump a billion microscopic ice particles into a tank of hot water (di-hydrogen monoxide), your statistical approach would insist that a few of them may have warmed the water (hydrogen hydroxide), especially using a computer model.”

    Some actually will warm the water. Things get more than passing strange when you leave to domain of bulk matter and look at what’s going on with individual atoms and molecules. Some of the water molecules in your tank of hot water are colder than ice and some of the water molecules in the ice are more than boiling hot. Those molecules are few and far between so for most practical considerations they are ignored. Temperature is an artifact of motion. In bulk matter where the molecules are all bumping into each other chaotically some small fraction will happen to be moving very fast and some very slow. A good analogy would be what happens to a set of billiard balls after the break. Initially all the motion is in the cue ball but when it hits the rack the energy of its motion is distributed amongst all the balls but not equally. The cue ball could stop dead in its track only to get whacked back into motion by one of the other balls a moment later or if it’s a glancing blow the cue ball might remain the fastest mover for a few moments. But because energy can be neither created nor destroyed the total kinetic energy in all the balls on the table will always equal the initial energy in the cue ball. Temperature is analogous to the total kinetic energy of all the billiard balls.

  181. Robert Stevenson says:
    May 9, 2011 at 2:16 am

    “Therefore I conclude no global warming from increasing atmospheric CO2.”

    This was disproven by expermental physicist John Tyndall over 150 years ago. Back radiation from gases that absorb IR is quite real. Suggest you read the original work here (it’s free):

    http://www.archive.org/details/heatconsideredas00tyndrich

  182. RJ says:
    May 9, 2011 at 1:41 am
    “Anyone who supports Trenberth’s diagram and considers that the back radiation theory is correct should be asked to explain why we are not utilising this fanastic energy source. After all, if it existed it should be the main focus of green energy projects.”

    “- If back radiation is so powerful and it is a heat source, why don’t we use it?”

    Good questions. Although what about this free energy oven. Does it qualify?

    We do use it all the time. One example is the smudge pots used to limit freeze damage in orange and avocado groves. The heat of the smoke from the smudge pots is inconsequential. The effect comes from the black carbon particles in the smoke which absorb upwelling IR from the ground and send a portion of it downwards and sideways. It works as an insulator. This is the same mechanism by which CO2 molecules insulate. Anyone who contests this is simply wrong and needs to take an introductory class in physics.

  183. “The water will rise above the ‘V’ just enough so that the rate of input is equal to the rate of overflow. Let us now represent the back radiation by blocking some of the ‘V’ which represents the reduced outflow of radiation from the earths surface. ”

    The problem is that Mr “T” says the energy from the sun is 184 watts and the amount being back radiated is 333 w. In other words, you are blocking more water than the maximum flow of the outlet, which is physically impossible in you analogy.

  184. Boris Gimbarzevsky says:
    May 9, 2011 at 12:52 am
    Ira, I’d really like some of the “back radiation” now. Just stepped outside my house in Kamloops BC and pointed an IR thermometer straight up into the sky and got a temperature reading of -10 F.

    You’d like more back radiation. You’re getting quite a bit of it now. The temperature of the cosmos is 3K and if it weren’t for back radation when you point your IR thermometer at the sky on a clear night you’d get a reading of -454F.

    Back radiation doesn’t come from just CO2. When a CO2 molecule absorbs upwelling IR it thermalizes surrounding non-GHG molecules via kinetic transfer. A common fallacy is that CO2 molecules emit a photon of the same frequency as that absorbed. That’s true in a very thin gas but in a cold dense gas the excited CO2 molecule almost instantly bumps into a neighboring molecule (likely a nitrogen molecule in the case of the earth’s atmosphere) and transfers some of its newly acquired energy to the neighbor.

  185. Ira Glickstein, PhD says:
    May 8, 2011 at 4:26 pm
    “If we then placed that planet in an Earth-like orbit around the Sun”

    Thanks Ira, what I read you to be saying is that if the atmosphere was pure N2O2, without any H2O, that it would have no effect upon the surface temperature of the earth. That since N2O2 do not absorb longwave radiation, then they cannot keep the earth warm.

    I’m saying that this is simply wrong. I’m saying that an atmosphere of pure N2O2 would change the surface temperature of the earth significantly as compared to an earth without an atmosphere.

    This is the calculation that must be undertaken first. How much of a difference would we see in the surface temperature of the earth due to N2O2.

    The next calculation is the effect of phase change of H2O in the atmosphere. Not the GHG effect of H2O, but the effects of evaporation and condensation in transporting heat vertically.

    Then, when you have completed those two calculations, what you have left is a candidate for the GHG effect.

    From what I’ve seen, when you apply these calculations to real models, such as venus, earth and mars, then the first calculation, the effects of the atmosphere without regard to the composition provides nearly the same amount of heating on venus, earth and mars.

    These are real observations, not theoretical models. And, observation trumps models, no matter how correct we think our models are. I don’t see this calculation in your work. I don’t see where you calculate the effects of a N2O2 atmosphere that does have any H2O or CO2.

    As to why the atmosphere of venus, earth and mars show the same amount of heating regardless of composition, that is a matter of theory. Put 10 experts in a room and you will have 20 opinions. None of these opinions will change the facts.

    The atmosphere of venus, earth and mars show a significant heating effect that is independent of the composition of GHG and this is not accounted for in your model.

    As to why this is, my theory is that it is due to compression. That the molecules i the atmosphere all have roughly the same amount of energy due to convection, but since there are many more of them per cubic meter at lower altitudes, there is more energy per cubic meter, which we register as increased temperature.

    For example, we could have 1 air molecule per cubic meter. When we went to measure the temperature of this meter of air, we would average out the energy of this one molecule over the volume and end up with a low temperature.

    Now, if we were to have a whole lot more molecules per cubic meter, all will the same energy as our first example, we would see this as an increase in temperature. However, in both examples the molecules are at the same energy level, yet the samples taken from lower down in the atmosphere will appear to have a higher temperature.

    However, this may not be the reason for what we observe. And it doesn’t really matter. What does matter is that FROM OBSERVATION the atmospheres of venus, earth and mars show similar warming at similar pressure, independent of their levels of CO2.

  186. Ira,
    Here’s why your bicycle tire analogy fails: you instead need a leaky tire that has to be inflated continuously. Would the air in that situation remain hot? Yes

    The open atmosphere essentially acts as a continuous ‘air conditioner’ or pump. Cold air continuously descends to the surface and is heated by the surface as well as from compression. That air then ascends and expands and cools until the process repeats. Thus, adiabatic compression is able to maintain a higher surface temperature on a continuous basis, the so-called GHE.

  187. We seem to have an unusually large number of comments from the anti-science brigade today who seem to have invented a whole new branch of previously unknown physics.

    Imagine an ice cube, at zero degrees Celsius, alone in the vacuum of space (which is at, say, absolute zero). What happens to that ice cube? Well, it radiates away heat (as everything above absolute zero does) and eventually cools to absolute zero itself, asymptotically. Now imagine an another ice cube, this time at -10 degrees Celsius, being placed near to the original ice cube. This colder ice cube also radiates heat; some of it will impact the first ice cube and ‘warm’ it. The original ice cube still continues to cool, but not as fast as it was doing before because it now has a heat input which was not there in the first case. The colder ice cube therefore helps to keep the original ice cube warmer. Of course, it never gets above its starting value of zero degrees Celsius, but that is how the presence of cold objects can keep hot objects warmer i.e. they are still warmer than what would be in their place if they were not there (in this case the cold vacuum of space at absolute zero).

    The 2nd Law of Thermodynamics? – Yes – it fully supports this – no contradictions. .

  188. martin mason says:

    From what I see the proponents of the Greenhouse theory simply keep stating it as fact without explaining the observed phenomena that seemingly falsify it such as.

    – You cannot heat a hot body from a colder body which you have to be able to do.

    In any model of the greenhouse effect, the net heat flows are from the warmer earth to the colder atmosphere. The atmosphere is not heating the earth itself…what it is doing is slowing down the cooling of the earth (for a given surface temperature). Since the steady-state temperature of the earth is determined by the balance between what it receives from the sun and what it emits back out into space, an IR-absorbing atmosphere will in fact cause the earth’s steady-state temperature to be higher than it would be if the atmosphere did not absorb IR. In a colloquial sense, one might call this “heating” the earth, although it is best to avoid such language since it leads to confusion and to instead state clearly what the effect of the atmosphere is, as I have explained above.

    – The temperature at the surface and at height can be explained without recourse to radiative theory which to me means that the temperature is determined by the fact that we have an atmosphere not necessarily because we have an atmosphere with GHGs. The atmospheric column is nothing like a bicycle tyre in physical terms, that is a red herring.

    No…Such a hypothesis does not even satisfy energy conservation! What can be explained without recourse to GHGs is the lapse rate in the troposphere. What can’t be explained is what the temperature at any altitude is. It is as if you told me the slope m of a line of the form y = mx + b and claimed that I could now compute y for any x. In fact, I would need the value of b to do so. The value of b is in essence what is determined by the absorption of IR radiation by the atmosphere.

    – In the past when we have had multiples of current CO2 levels we have not had runaway warming.

    – There is patently no runaway warming now while CO2 levels ramp up inexorably.

    This is a strawman argument. (Almost) noone is claiming that there will be “runaway warming”. [Hansen is talking about the possibility under certain circumstances and has specific arguments as to why the current case may be different from what the earth has experienced in the past, so let's leave him out of this.]

    What we are talking about is a climate that is quite sensitive, but not unstable, to perturbations. See, for example, here for a discussion of what the paleoclimate record tells us: http://www.sciencemag.org/content/306/5697/821.summary

    There is no runaway warming on venus.

    Really? You think the temperature on Venus is very pleasant? Did the runaway go on forever? No…An object that is linearly unstable will tend to find a point where stability is restored. However, a runaway did occur to produce Venus’s current very hot climate. Fortunately for earth, we are not as close to the sun and thus not subject to the runaway scenario that occurred on Venus.

    – If back radiation is so powerful and it is a heat source, why don’t we use it?

    How are you proposing that we harness it? As has already been noted by Smoking Frog, the back-radiation is already heating the earth. (Because most objects tend to be almost perfect blackbodies in the mid- and far-IR, almost all of the back-radiation that is received by the earth is absorbed.) Also, the back-radiation that we receive is diffuse (coming from all different angles) and so it cannot be focused in the same way that solar radiation (which is approximately from a point source) can. Furthermore, since the energy of an individual photon is proportional to its frequency, the back-radiation…unlike solar radiation…cannot be harnessed by photovoltaic devices, which rely on having photons of sufficient energy to cause electronic transitions (in particular, energies much larger than thermal energy of ~kT where K is Boltzmann’s constant). Finally, although the back-radiation we receive from the atmosphere is, in the global average (e.g., day and night at all latitudes), larger than the radiation we receive directly from the sun, the amount of radiation that we receive from the sun when it is actually shining and close to overhead is a fair bit larger.

  189. Ira – regarding your summary comment 4) at May 8, 2011 at 7:51 pm
    my comment – NO, the atmosphere does NOT emit LWIR across a distribution of wavelengths like a blackbody , see my earlier comment at Dave in Delaware says: May 8, 2011 at 7:00 am
    Ira Glickstein, PhD says:
    “4) As I understand it, the ~15μm radiation from the Surface to the Atmosphere is absorbed by H2O and CO2 molecules which, when excited, bump into nitrogen and oxygen and other air molecules, and heat the air. The result is emission of long-wave radiation at a range of wavelengths (not just in the ~15μm region, but including the ~7μm and ~10μm regions) in random directions, some of which finds its way back down to the Surface and is re-absorbed. That conversion of a portion of the ~15μm radiation into other wavelengths may explain why there is a “bite” out of the ~15μm radiation that seems to be missing from the radiation out to Space. ”
    ——————————————
    This is the part of your comment I disagree with –
    “The result is emission of long-wave radiation at a range of wavelengths (not just in the ~15μm region, but including the ~7μm and ~10μm regions) in random directions, …”
    My reply – If CO2 absorbs at 15μm, it emits at 15μm. The Nitrogen and Oxygen in the atmosphere neither absorb nor emit in the LWIR wavelengths of interest. That is why there is an Atmospheric Window – the atmosphere neither absorbs nor emits in that 10μm region.
    Think about the satellite-looking-down readings – the 10μm atmospheric window corresponds to the surface temperature. Over the Sahara, the 10μm readings show a surface temperature around 320 K. Over the Mediterranean, the 10μm readings show a surface temperature around 280 K. Over the Antarctic, the 10μm readings show a surface temp just below 220 K. If the atmosphere (mostly N2 and O2) emitted LWIR like a blackbody, the atmospheric window readings would look like some middle atmosphere temperature – not like the surface. There is an Ozone band in the middle of the atmospheric window, I expect from ozone in the upper atmosphere; if N2 was emitting blackbody LWIR there should be other N2 bands in the middle of the window, and there are not.
    ———————————-
    Ira comment – “That conversion of a portion of the ~15μm radiation into other wavelengths may explain why there is a “bite” out of the ~15μm radiation that seems to be missing from the radiation out to Space. ”
    My reply – Recall that the satellite-looking-down readings are Emissions; their temperature tells us something about where those emissions originated. Looking at the CO2 band satellite 15μm readings, the temperatures are around 220 K and nearly the same for the Sahara, the Mediterranean and the Antarctic. That looks to me like the Tropopause. It does not necessarily tell me how the energy got to the Tropopause, so I am not sure I could estimate a ‘bite’ value. That emitted energy could have arrived from lower atmosphere CO2 emissions, or from atmospheric lapse type convection, or possibly from the tops of thunderstorms.

    My comments are referenced to these satellite IR spectra at

    http://mensch.org/5223/IRspectra.pdf

    Slide (c) Western Tropical Pacific is particularly interesting because it compares ‘clear’ with ‘Thunderstorm Anvil’. Note that the CO2 band temperature is identical in the clear and thunderstorm curves. Also note that the atmospheric window 10μm region temperatures are MUCH DIFFERENT for clear vs thunderstorm anvil. This suggests to me that while the atmosphere does not emit like a black body, the thunderstorm (liquid water) does.

    Best regards to Ira – I hope my comments are constructive and instructive, and will help your ‘mental models’ and descriptions.

  190. Allan M says:

    The basis for the second law (and the first) is careful and accurate measurement, mostly paid for by wicked 19th century capitalists, who wanted to know how to get more from their steam engines, and also how to avoid being sold snake oil.

    There is no need for a statistical ‘fudge’ unless you wish to introduce special pleading for “IPCC Approved” photons.

    Experiment did provide the evidence for the Second Law. However, our modern (20th century) understanding of where the law comes from is based on statistical physics. To dismiss an entire field of physics that has successfully explained a huge wealth of empirical observations as “a statistical ‘fudge'” because one implication of it does not line up with your ideology is bizarre.

    This is essentially the project that Claes Johnson, one of the “Slaying the Sky Dragon” authors, has embarked upon. He wants to base things on the bizarre proposition that an artifact of numerical calculations of the differential equations actually governs the universe instead. The irony is that although he claims to have gotten rid of “back radiation” by coming up with a different interpretation of the terms in the equation governing radiative transfer between two bodies, he has in fact done nothing to change any numerical result using those equations…including all of the numerical results that support the existence of the greenhouse effect! What he has done is snookered a few people who want to believe the nonsense that he is peddling.

    If I were to dump a billion microscopic ice particles into a tank of hot water (di-hydrogen monoxide), your statistical approach would insist that a few of them may have warmed the water (hydrogen hydroxide), especially using a computer model.

    It depends what you mean by “microscopic”. It doesn’t take a very large particle size before the statistics make it vastly improbable for such a thing to occur. But, yes, if you particles were truly only a few molecules large, then you could detect behavior contrary to the 2nd Law. (Of course, the distinction between “ice” and “water” begins to lose meaning at such a small length scale since phases of matter are themselves a macroscopic concept.)

  191. Circa 1850 Tyndall experiments with greenhouse gases in a nutshell.

    Radiation flow:

    heat_lamp -> -> adjustable shade -> salt crystal window -> mirrored tube filled with test gas -> salt crystal window -> adjustable shade -> galvanometer

    Tyndal ran literally thousands of experiments with different gases and mixtures at different pressures, different infrared light sources, and different lengths of tubes. His setup took up a lot of space but I marvel at the cleverness he used to increase accuracy and precision using the technology of the day. His analog gear was so sensitive he read the reading from the galvinometer through a telescope because getting a warm body anywhere near the experiment mucked it up.

    His best most stable lamp was a black painted face of a vessel filled with boiling water. His galvanometer response is not linear so the primary purpose of the adjustable shades was keep the radiative energy input to the face of the galvanometer in a range where you got maximum needle deflection from minimal change in radiation.

    In lectures he used to take his galvanometer to the podium in the hall. He’d point it at a blank wall on the other side of the hall and ask someone from the audience to walk over to that wall. The galanometer needle would move dramatically when a person entered the scene.

    Anyhow Tyndall found a great many gasses, most famously water vapor, that would dramatically lower the galvanmeter reading when present in the tube versus the same mixture without it. He also used some rather ingenious ways of processing his gasses prior to filling the tube with them including means of completely drying a sample of the atmosphere.

    If one wants to dispute the concept of back radiation one must first explain why Tyndall’s galvanometer would read progressively lower as the absolute humidity of the atmospheric sample in the tube rose. No one will. Every attempt is fatally flawed in some way. Tyndall didn’t discover anything that wasn’t already predicted by the theoretical physics of the day. He confirmed the theory via experiment or in other words he was doing science the way it’s supposed to be done.

    Tyndall did discover some things that no one had worked through on paper at the time. One notable thing he discovered was the non-linear absorptive response as the partial pressure of an IR-absorbing gas went up. He found it to be linear at the lowest partial pressures then progressing to exponential or in other words when it comes to IR absorbing gasses increasing the amount of it is a case of diminishing returns after a certain amount.

    This is why the climate boffins talk about CO2 doublings and temperature rise. At the current atmospheric CO2 partial pressure its absorptive response is in the exponential range. The linear response range is all below the first 100ppm. It’s sort of like people picking low hanging fruit in an apple orchard. They can all pick as fast as they can for a while but as the number of people picking increases at some point they’ll be competing with each other for the same low hanging fruit. So putting more pickers in the orchard will always increase the rate at which fruit is picked but at some point it becomes a case of diminishing returns.

  192. Dave Springer

    You did not read what I had written – no additional heat absorbed- no temperature increase- no global warming. 150 years ago Tyndall new nothing about CO2 absorption bands

  193. Charles Higley says:
    May 7, 2011 at 9:33 pm
    “It should also be pointed out that, as CO2 partitions 50 to 1 into the oceans, we would have to emit 50 times more CO2 that required to simply double atmospheric CO2. There is not enough carbon available to do this. If we really tried, we might be able to do 20%.”

    It could also be pointed out that CO2 partitions 100,000 to 1 between the rocks and the air, so we would have to add 100,000 times more CO2 to just double atmospherics CO2…
    This is wrong, as the ocean comparison is wrong because of time-scales. It takes the oceans centuries if not millenia to absorb additional CO2 so that partitioning will take effect eventually, for rcok/air the timescales are more like millions of years…

  194. One little point about assumptions this is all based upon.

    Incoming energy is spread over HALF the earth’s surface.
    Outgoing energy is spread over the FULL surface of the earth, with an uneven distribution, probably (?) higher rate of outgoing radiation on the night side of the planet.

    So its entirely incorrect to say that incoming = 240W/m^2 and outgoing = 240W/m^2.

    Get the energy distributions right, then try again.

  195. Here is a reference showing that Jupiter has 60K higher temperature than expected.

    33K higher than expected on earth. 60K higher than expected on Jupiter. Co-incidence?

    http://burro.astr.cwru.edu/stu/advanced/jupiter.html

    •As seen in the table below, the average temperature of Jupiter is approximately 160 K. However, due to the equation for thermal equilibrium (below), it should only be about 100 K.

  196. Also one disputing back radiation must explain how many instruments that measure CO2 concentration manage to work. The basic design of these:

    15um infrared light source -> beam splitter

    one side of beam splitter -> sealed sample of air with known CO2 concentration -> IR photo transister

    other side of beam splitter -> sample of ambient atmosphere -> IR photo transister

    CO2 concentration is obtained by comparing the output of the two photo transisters.

    Pretty much the same as Tyndall’s experimental setup only tweaked for one particular gas in a narrow range of partial pressures and put into a space the size of a thimble using modern technology.

  197. Dave Springer says:
    May 9, 2011 at 5:57 am

    RJ says:
    May 9, 2011 at 1:41 am
    “Anyone who supports Trenberth’s diagram and considers that the back radiation theory is correct should be asked to explain why we are not utilising this fanastic energy source. After all, if it existed it should be the main focus of green energy projects.”
    “- If back radiation is so powerful and it is a heat source, why don’t we use it?”

    Good questions. Although what about this free energy oven. Does it qualify?

    We do use it all the time. One example is the smudge pots used to limit freeze damage in orange and avocado groves. The heat of the smoke from the smudge pots is inconsequential. The effect comes from the black carbon particles in the smoke which absorb upwelling IR from the ground and send a portion of it downwards and sideways. It works as an insulator. This is the same mechanism by which CO2 molecules insulate. Anyone who contests this is simply wrong and needs to take an introductory class in physics.

    ———–

    Dave, you and I structurally disagree, but end at the same conclusion. To me, and I am sure it is correct by physics, is the effect comes from the black carbon particles in the smoke which absorb upwelling IR from the ground increases the temperature at that altitiude (10 m) and by SB using delta T (LBL if you want) between the surface and these particles decreases the upward flux making it warmer than it would have been without the carbon particles in this thin layer where the smoke is.

    You do not need a special kind of “back radiation” that flows from cold to hot to explain it, it only scrambles minds away from what is actually happening. If you care about people’s minds at all, stop wording it like you just did. There is no special radiation called “back radiation” or “backward radiation”.

  198. “It takes the oceans centuries if not millenia to absorb additional CO2 ”

    This does not appear correct. CO2 absorbtion at the surface of the ocean is limited by the surface area. However, the surface area of H2O in the clouds is fantasitically greater than the surface of the ocean. In addition, the water in the clouds is cold, which increases its capacity to hold CO2.

    This is where CO2 is absorbed and returned to the oceans. High in the couds, not at the ocean surface. Near the equator, where the oceans are warm this CO2 will be released back into the atmosphere as the rainwater warms. In higher lattitudes where the oceans are cold this CO2 will not be released. The rain water will mix with the cold seawater and sink to the bottom of the ocean, carrying the CO2 with it. The polar oceans should be less caustic, more Ph neutral as a result, as compared to the tropical oceans. Which has been reported.

    Eventually this cold water will upwell in the tropics and release its CO2. Thus the CO2 being released during La Nina years was deposited at higher latitudes many years before. This accounts for the cause and effect lag observed between warming and CO2 release, which Al forgot to mention in his movie. This is strong evidence that temperature drives CO2. There is no similar observation that temperature lags CO2 in the paleo record, which is evidence that CO2 is not a driver of temperature.

  199. ferd berple says:
    May 9, 2011 at 8:11 am

    Here is a reference showing that Jupiter has 60K higher temperature than expected.

    33K higher than expected on earth. 60K higher than expected on Jupiter. Co-incidence?

    http://burro.astr.cwru.edu/stu/advanced/jupiter.html

    •As seen in the table below, the average temperature of Jupiter is approximately 160 K. However, due to the equation for thermal equilibrium (below), it should only be about 100 K.

    “This extra heat is generated due to gravitational contraction – the planet is slowly shrinking in diameter. This way, by compressing by only a few millimeters every year, it can generate heat by increasing the pressure of its constituent gas.”

    So Ferd, is it your assertion that the earth and venus are shrinking? I already explained to you that temperature only increases during the process of compression and the earth, Venus, and Mars (unlike Jupiter) aren’t shrinking.

    What part of that don’t you understand?

    Let me fix that for you by adding the explanation for the extra warmth on Jupiter from the article you quote:

  200. Robert Stevenson says:
    May 9, 2011 at 2:16 am

    For current CO2 partial pressure or concentration I calculate a distance of 3600 m of traverse through the earth’s atmosphere by ‘IR radiation’ before the absorbable IR is ‘filtered out’ and reduced to zero.

    —-

    Robert, I come up with a close conclusion, not exact, but close. However, the 3600 m is more than 36 times that is given in other conversations I have read speaking of it’s absorption very near the surface. Do you have an average absorption coefficient you are using, or a mean optical thickness for CO2 from which that is calculated? Maybe Hottel specifies emissivity direct then what equation are you using to get the 3600 m extinction distance?

  201. Wayne

    If you want a bit deeper understanding of the radiative energy flow one needs to understand that all matter above absolute zero radiates and where there are two bodies at different temperatures there’s a net transfer of energy between the two from warmer to colder. The net transfer rate is all the matters for most practical concerns but at the deeper level the radiative transfer goes two ways. The warm object doesn’t stop the cold object from radiating so energy from the cold object is indeed reaching the warm object just as energy from the warm object reaches the colder one. It’s simply a matter of the colder object receiving more energy from the warmer than the warm object receives from the colder.

    That doesn’t seem like proverbial rocket science to me but you may perhaps be right that it might as well be rocket science for many people.

  202. Explain why the surface of venus, which only receives some 2.5% of the sun’s energy due to the albedo effect of the clouds can have a temperature of 500C.

    Why do we use a figure of .7 when calculating the solar energy reaching the earth’s surface? If we did this on venus, then the surface temperature should be very cold indeed.

    The missing 33K comes from the assumption that only 70% of the solar radiation is available at the surface due to albedo. This is an incorrect assumption, as is clearly demonstrated by the surface temperature of venus.

    97.5% of the solar radiation never reaches the surface of venus, yet it has a high temperature. Thus, the .7 figure used when calculating solar radiation at the surface is refuted by observation.

    Consider 240k / 288K. This is .8. Thus, the missing 33K may not be missing at all. It may simply result from our asusmption that albedo is .7. If albedo is not what we assume it to be, for reasons that are apparent on venus, then the problem is not where is the missing 33K, but rather why temperatures are not hotter than they are.

    fully available in our assumption

  203. tallbloke says:

    May 9, 2011 at 12:07 am

    According to the more reasonable reconstructions of past temperature (Moberg, Loehle) the surface temperature has been rising since 1700. This is 230 years longer then co2 has been rising to any great degree.

    What is Ira’s explanation for this earlier warming? …

    Please re-read my main posting above where I say:

    The main scientific question for me, is how much does the increase in human-caused CO2 and human-caused albedo reduction increase the mean temperature above what it would be with natural cycles and processes? My answer is “not much”, because perhaps 0.1ºC to 0.2ºC of the supposed 0.8ºC increase since 1880 is due to human activities. The rest is due to natural cycles and processes over which we humans have no control. The main public policy question for me, is how much should we (society) do about it? Again, my answer is “not much”, because the effect is small and a limited increase in temperatures and CO2 may turn out to have a net benefit.

    Got it?

  204. “This extra heat is generated due to gravitational contraction – the planet is slowly shrinking in diameter.”

    What is the mechanism by which this is possible? Why has the planet not reached equilibrium? Why does the temperature of all the planets and the sun increase towards the center? If the temperature increase is a result of continued shrinkage, then this should long ago have ended on some, if not all planets.

    If this efect is unique to Jupiter, why is it so similar to what we see on other planets. Is is reasonable to suggest that a similar observation has unique and different causes on each different planet? Or is it more reasonable to assume that these unique explantions are a result of a lack of understanding. That they are modern day epicycles.

    We see a surface temperature of venus that is higher than expected if we used the same methodology for albedo as used on earth. We see a surface temperature for Earth that is higher than expected if we use .7 for albedo, but NOT higher than expected if we use for example .8 or .9. We do not have to increase this above 1, so it is possible the missing 33k is simply due to an error in our understanding about how albedo actually works.

    We see a surface temperature for Jupiter that is higher than we expect. This follow a similar pattern as with venus and earth. On each of the three planets science explains this by using 3 different methods, and proposes that these are unique to each planet.

    That may well be, but occam tells us that this is not the likely explanation. Since a simpler explanation exists, namely that we are not properly accounding for the effects of clouds in our calculations. The observations of venus suggest that clouds do not block solar radiation the way we think, or the surface of venus would be much cooler. When we apply this logic to earth, the”missing” 33 K is no longer missing. It is simply a result of us using .7 when a more accurate number would be .8 or .9. Jupiter is also covered with clouds. It could well be that the missing 60K on jupiter is due to the poorly understood clouds, not due to continued shrinkage.

  205. ferd berple says:
    May 9, 2011 at 8:28 am

    “It takes the oceans centuries if not millenia to absorb additional CO2 ”

    This does not appear correct. CO2 absorbtion at the surface of the ocean is limited by the surface area.

    The effective surface area is greatly increased or diminished by wind driven waves. It’s also limited by how far out of equilibrium it is. It’s also limited by how fast the surface water mixes with the deeper water which is governed by both winds and convective currents. The amount of CO2 that can be held in water increases with depth as well. In deep ocean trenches CO2 has been observed in liquid droplets. So the rate at which surface water is sequestered at depth is also very important.

    It’s been a bit of a surprise for the climate boffins that no matter how much faster humans introduce CO2 into the atmosphere only about half of it remains in the atmosphere. The mechanisms that serve as carbon sinks for the atmosphere (we didn’t even start talking about biological and chemical sequestration) are diverse and not well enough understood to predict whether this will remain true going forward.

  206. Dave Springer,

    You see, I know radiation in many cases can and does radiate in any and all directions, but that depends on the temperature and radiation from the surrounding matter. To radiate all directions the matter all around must be colder than the source that is radiating. The main difference you and I have is that radiation never actually flows from cold to hot matter, it doesn’t even flow from equal temperature pieces of matter or you could burn yourself with a Fresnel lens in a dark warm room as my little experiment should show you, if you will get curious and perform it yourself. Planets and stars radiate against basically nothing so you can let the other temperature in the Stefan-Boltzmann equation be zero, but always subtracting 2.7K would be more proper. You are one tiny step from moving your rather unrealistic view into what actually happens.

    Here is what happens because of the Stefan-Boltzmann equation that screws so many people up (sigma=5.6704r-8, epsilon=1):

    ε σ ( 288K^4 – 280K^4 ) = 41.572 W/m^2

    or

    ε σ ( 288K^4 ) – ε σ ( 280K^4 ) = 41.572 W/m^2

    The top one will always show you the correct and actual flow of radiation between two pieces of matter. The latter one implies that matter always radiates at all times in all directions even if the temperatures are identical and the last term gets this quirky term applied to it, “back radiation”. You will end up with the same numeric answer but it leaves you with the wrong impression that two things have occurred when in reality only one thing has occurred, radiation from the warmer to the cooler or not at all if the two temperatures are the same.

    This same logic applies to your split beams and it is why such instruments must be constructed to tightly control the temperatures inside. By the second set of equations temperatures of different pieces of the instrument do not matter at all, by the top equation surrounding temperatures of the pieces of the measuring instrument matters greatly. Go lookup an operation manual and see if I am not correct.

    Please, drop the “back radiation”, it really does not exist, use the top equation.

  207. Dave Springer says “I already explained to you that temperature only increases during the process of compression ”

    What you and Ira and Joel fail to consider is that there is continuous compression of cold descending air which heats that air until it becomes hot enough to then ascend, expand, and cool until the process repeats. The atmosphere thus acts as a continuous compressor (and expander).

    In addition to the 4 links I provided above, here’s 2 more peer-reviewed papers that explain this as well:

    http://www.climatephysics.com/PDFs/Chilingar%20-%20Cooling%20due%20to%20CO2.pdf

    http://docs.google.com/fileview?id=0B74u5vgGLaWoYTAyYTc4NmUtZDFkMC00ODg3LTgwYzAtOTg3OTJhZjI5MDVk&authkey=CKyD1r8H&hl=en

  208. ferd berple says:
    May 9, 2011 at 8:48 am

    So Ferd, is it your assertion that the earth and venus are shrinking?

    Earth’s Shrinking Atmosphere Baffles Scientists
    An increase in CO2 could be one reason why a layer of Earth’s upper atmosphere went through its biggest contraction in 43 years.

    http://news.discovery.com/earth/earth-atmosphere-shrinking.html

    Does this change air pressure at sea level significantly?

    That’s what’s happening on Jupiter. Pressure at depth is constantly rising. That isn’t happening on the earth. Atmospheric pressure at sea level rises and falls to some degree as horizontal pressure ridges sweep along but the average pressure remains the same at 1 bar.

  209. Jupiter is still undergoing gravitational contraction that began when the solar system was born. This process stopped for the earth billions of years ago. The sun stopped its gravitational contraction when temperature increase from ongoing compression lit off a sustained fusion reaction. The outward pressure from fusion counteracts the inward pressure from gravity and an equilibrium point was reached where they are equal and opposite in force and the sun has a relatively stable diameter that will persist until it starts running out of hydrogen fuel to sustain the outward pressure.

  210. Hockey Schtick says:

    What you and Ira and Joel fail to consider is that there is continuous compression of cold descending air which heats that air until it becomes hot enough to then ascend, expand, and cool until the process repeats. The atmosphere thus acts as a continuous compressor (and expander).

    We consider that…but it can’t change the result. You are talking about things that can only redistribute energy within the earth-atmosphere system. You still must satisfy conservation of energy when you look at the interaction of the Earth System with the sun and space. And, the combination of a surface temperature above the 255 K blackbody temperature and an atmosphere transparent to terrestrial IR radiation doesn’t do this: There is much more energy going out of the system than coming into it. The result of this would be rapid cooling.

  211. Ira
    I always enjoy reading your articles and the comments that they generate.

    I am sceptical as to how useful it is to consider just one element of how the atmosphere may work ignoring other factors which are known to exist and which in some circumstances are definitely more important factors (I have in mind convection, evaporation etc). Leaving that aside, I have problems in envisaging that the back radiation plays any significant role in the workings of the atmosphere.

    I have no problem with the argument on net energy flow and that if the atmosphere is warmer it slows down the heat loss from the warmer earth (or ocean). The problem I have is whether the atmosphere is warm simply because of the absorption of solar energy (a combination of absorption of incoming solar radiation and absorption of reflected and radiated solar energy from the bottom up) and adiabatic lapse, or whether in addition back radiation from GHGs plays a significant role.

    You will see the point I made at (richard verney says) at May 8, 2011 at 8:39 pm
    Smoking Frog sought to answer the point. He comments at May 9, 2011 at 5:21 am:
    “Martin Mason said – If back radiation is so powerful and it is a heat source, why don’t we use it?
    Because the earth’s surface is already at the temperature to which the back-radiation has raised it.”
    However, with respect that explanation cannot be correct. We know as fact that solar (possibly with the help of back radiation) heats the earth’s surface to the temperature to which solar energy has raised it, yet not withstanding this we can still extract work from the incoming solar energy. We can collect and focus the 184 w/m^2 of incoming solar energy and use it to melt salts etc in solar power plants. That being the case, it begs the question why can’t we collect and focus the 333 w/m^2 of back radiation and extract useful work from it.

    The fact that we cannot do this suggests that the ground is not being bomb barded with 333 w/m^2 of (back) radiation.

    I have a black and white marble path. On sunny days, this path is warm to walk on. The black marble slabs are considerably warmer than the white marble slabs. Not surprising since black is a better absorber than white and therefore the black marble stones absorb more of the incoming 184 w/m^2 of solar energy. IF back radiation existed and if there was some 333 w/m^2 of back radiation beating down on the ground, the black marble slaps should be significantly warmer than the white marble slaps at night or on cloudy days. However, they do not appear to be. In particular when left over night to bask in the supposedly 333 w/m^2 of back radiation, the black marble slaps do not heat up whereas they do heat up with only 184 w/m^2 of incoming solar energy!!

    Ira, I would like to see someone conduct a very simple experiment along the following lines;
    1. Get two identical metal slabs (same dimensions for area and thickness) save that one is painted brilliant white and the other matt black,
    2. Pre heat the slabs to the forecasted daily temperature (say picking a day when the estimated temperature is going to be 20degC).
    3. On a clear sunny day (ie the day estimated to be 20 degC) place the slaps in the sun and measure there respective temperatures every half hour.
    4. On a mild summer night (one estimated to be 20 degC), pre heat the slaps to the forecasted night temp of 20C and then leave them out after the sun has fully set and measure there respective temperatures every half hour.
    5. Compare the results.
    6. In particular does the matt black slab gain heat at night from the down welling back radiation. If it does not, why not?

    Ira, I would appreciate your comments

  212. Bryan says:

    Joel Shore has refined this even further to say Postema is a believer, only Postemo hasn’t quite realised it.
    Perhaps with the same elasticity Joel will prove that Gerlich and Tscheuschner also are really IPCC advocates.

    No…What I am saying is that Postma and G&T mix in a lot of correct science that is already known with some incorrect science…or just glossing over things…to reach completely incorrect and absurb conclusions. That tends to be how peddlers of pseudo-science operate.

    On page 9 of your co-written Halpern et al comment paper you state that the Stephan Boltzman Law can be used to work out the thermal energy between atmospheric shells.
    I brought your attention to it in my post above;
    ….. ” 3. Has the Stephan Boltzmann equation applying to gases.”……

    It is sad to see that the pull of pseudo-scientific arguments are just too strong for you to resist making them again and again. The shells that we consider are blackbody shells, so yes, the S-B equation applies by definition. This is a model. Is it a good model? Well, yes, it is quite a fine model to illustrate the basic features of the greenhouse effect. When you want to do detailed quantitative calculations, you adopt a better model. Is that concept so difficult to understand?

    It is not science to attack a model for being to simple when it is meant to be a simple representation and where more refined models are available and can be used to show that the qualitative predictions of the simpler models for which they are used are correct. That is instead the realm of pseudo-science, a realm that is apparently just too irresistible to you. That you cannot even distinguish the difference between scientific arguments and the pseudo-scientific nonsense that you peddle is very very sad for you, the people reading this thread, anybody who respects the scientific enterprise, and even AGW skeptics who don’t want to be considered complete crackpots. I actually pity people like Ira and David M. Hoffer for having to put up with people like you on their side of the AGW debate.

  213. Heh, and LOL. Unless I missed it, nobody has yet tackled the “venus thing” presented above by a commenter (which suggests that the GHE theory isn’t valid):

    http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html

    BTW, it holds for other planetoids with an atmosphere (I can provide links if anyone cares).

    Despite the Orwellian chants about “multiple lines of evidence,” the case for a GHE threat is really hurting these days, due to the total lack of any such lines of evidence. In fact the multiple lines of evidence appear to go the other way:
    1. No warming for 10-15 years, despite all the increases in the terrible GHGs (and not just OCO).
    2. Ice core records show no correlation between GHGs and temperature.
    3. The temperature swings during the past couple of millenia indicate the climate changes naturally without changes in GHGs. (We don’t need Mann’s proxies to know what happened during the MWP and LIA; it’s recorded in the history books, for crying out loud!)
    4. No “hot spot” in the mid-troposphere in the tropics, as predicted by ALL the climate models. (So much for treating their output as “data.”)
    5. No laboratory demonstrations or other real-world empirical demonstrations of any GHE effects.
    6. Areas with the most greenhouse gases (tropics) have maximum temperatures that are no where near as high as some desert areas with the lowest amount of GHEs (yeah, yeah, it’s the water evaporation and clouds; but that is precisely the point on Earth).
    7. Articles such as the one cited above, which show clearly that temperature in a gaseous atmosphere has nothing to do with the “greenhouse effect.”

  214. Dave Springer:

    “We do use it all the time. One example is the smudge pots used to limit freeze damage in orange and avocado groves. The heat of the smoke from the smudge pots is inconsequential. The effect comes from the black carbon particles in the smoke which absorb upwelling IR from the ground and send a portion of it downwards and sideways. It works as an insulator. This is the same mechanism by which CO2 molecules insulate. Anyone who contests this is simply wrong and needs to take an introductory class in physics.”

    Well, I question this. Since the backradiation is coming from cooler air above, it cannot be warming the orange grove.

    The Clean Air Act killed the smudge pots, so they now just blow air through the orchards or add heat by spraying water. Maybe they didn’t ever really need that backradiation?

  215. Hockey Schtick says:
    May 9, 2011 at 9:23 am

    “What you and Ira and Joel fail to consider is that there is continuous compression of cold descending air which heats that air until it becomes hot enough to then ascend, expand, and cool until the process repeats. The atmosphere thus acts as a continuous compressor (and expander).”

    Not in my case. I’ve discussed it in depth many times at WUWT. The following search reveals some of it.

    http://www.google.com/search?hl=&q=tunderstorm+heat+engine+springer+site%3Awattsupwiththat.com&sourceid=navclient-ff&rlz=1B3GGGL_enUS290US290&ie=UTF-8

    The big picture:

    The sun heats the ocean, the ocean heats the atmosphere, the cosmic void cools the atmosphere.

    This thread is about radiative transfer in and out which is a detail in the big picture. Convective transfer is a different detail but also very important. The water cycle in general is exceedingly important in basic understanding of atmospheric physics. Vast amounts of energy are lifted from the surface by evaporation in what’s called “latent heat of vaporization”. It’s called “latent” because it won’t register on a thermometer. When the vapor condenses the latent heat is released. As the earth gets warmer (for whatever reason) the speed of the water cycle increases. This is a negative feedback mechanism and it’s the most glaring flaw in the whole warmist narrative. There has never been a runaway greenhouse on the earth. Never. The hypothetical amplification of anthropogenic CO2 forcing by increased water vapor is a wholesale fabrication without a shred of evidence to support it and with mountains of contrary evidence. The water cycle is a negative feedback not a positive feedback. Most of the more informed global warmists and skeptics know that this amplification is the heart of the controversy. None dispute that when everything else is equal more CO2 will result in a warmer surface and that this is would be about 1.0C for each CO2 doubling starting from a base point of 280ppm. They also acknowledge that a warming surface will speed up the water cycle. The controversy is over whether a faster water cycle is a positive or negative feedback. The evidence says it’s negative. The alarmist must have it as a positive feedback because otherwise the only warming that’s gonna happen is a modest welcome amount that is of great net benefit to the biosphere. Living things don’t tolerate ice very well so the less of it there is the more life can flourish on this third rock from the sun.

  216. Climate boffins have pretty much conceded the debate over water vapor amplification. As this concession ocurred it manifested to the unwashed masses as “global warming” was first replaced by “climate change” and now is morphing yet again into “climate disruption”.

    The climate does nothing if not change and it has been doing so since long before humans entered the equation so “climate change” failed to elicit the needed level of fear response in the masses. Climate disruption has more fear factor in it.

    An inconvenient amount of climate disruption is a distinct possible result of anthropogenic activity but once again the evidence is all stacked against it. By the same token radically modifying anthropogenic activity to lessen any imagined disruption has an immediate negative consequence of slowing down the global economy which is driven in large part by fossil fuel consumption.

    Thus we have a risk/reward decision to make and in my view trying to reduce CO2 emissions is essentially all risk and no reward. Real pollutants that immediately and directly effect our health and well being are a much greater concern. Heavy metals in our water, particulate pollutants in our air, ozone, and lots of other things are legitimate red flag events but CO2 is plant food and the more of it we have the better off we are.

  217. “In deep ocean trenches CO2 has been observed in liquid droplets. So the rate at which surface water is sequestered at depth is also very important.”

    That is intersting. The ocean pressure goes up 1 atmosphere per 30 feet of depth. At 300 K CO2 becomes a liquid at 100 atmospheres. The deep oceans are all below 300K and 100 atmospheres is only 3000 feet – much less the the depth of the average ocean basins.

    So, what this means is that CO2 can exist as a liquid across most of the ocean bottoms. As liquid CO2 should weigh more than H2O, the question really should be why the bottom of the oceans is not all liquid CO2? Except that the oceans are not saturated, and thus liquid CO2 would be absorbed.

    http://en.wikipedia.org/wiki/File:Carbon_dioxide_pressure-temperature_phase_diagram.svg

    http://oceanservice.noaa.gov/facts/oceandepth.html

  218. http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html

    well worth the read, as well as the follow on explanation. what I find fascinating is that albedo has no effect on the temperature comparison between earth and venus, which accounts fully for the missing 33K in Ira’s analysis.

    quote from the article:
    For example, in the analysis, not only does the amount of CO2 not enter in (Earth has 0.04%, Venus a whopping 96.5%), but the albedo (from either cloud tops or the planetary surface) does not either (Venus has dense clouds that reflect much of the incident visible radiation, while Earth does not, and Earth’s surface is 70% deep ocean, while Venus is solid crust). The real atmospheres don’t care at all about these great differences in the two atmospheres and planetary surfaces, they only care, and quite precisely, about their distances from the Sun.

  219. This cannot simply be coincidence. How can two planets with completely different atmospheres, albedo’s and surfaces show almost identical temperatures as a function of pressure if the standard climate model is correct? Quite simply they can’t. These observations are either false or climate science has it fundamentally wrong.

  220. JAE says:
    May 9, 2011 at 10:02 am

    Well, I question this. Since the backradiation is coming from cooler air above, it cannot be warming the orange grove.

    Even without knowing about the exact situation you are discussing, I can tell you don’t understand the concept or the reality of back radiation.

    The air above is not warming the orange grove but slowing the heat loss by radiating some of the LW radiation back. If no more energy enters the grove from the outside it will continue cool over time. It is only the law of conservation of energy at work, so the back radiation is not extra energy but only energy that has had it’s ability to leave the system delayed.

  221. “”””” JT says:
    May 7, 2011 at 8:34 pm
    Ira, the basic bookkeeping of the simplest greenhouse calculations assumes that the re-emission of infra-red radiation by the greenhouse gas molecules is spatially symmetrical. I am well aware that spontaneous emission is equally likely to be in any direction so that approximately 1/2 will head downward to the surface. However, there is another kind of emission – stimulated emission – and it is massively biased in the same direction of travel as the direction of travel of the stimulating photon. Thats what creates laser light, but stimulated emission can occur without lasing. “””””

    I don’t believe that is true. Stimulated emission is still isotropic. The highly directional output of a laser beam, is a consequence of the resonant Optical cavity that the emission takes place in. Only wave modes, that are aligned with the axis of the resonator, can continue to propagate in the system. The off-axis modes are suppressed. The concept is quite simple. Take two parallel mirrors, and launch a geometrical ray normal to one of the mirrors, to wards the other. It will in the geometrical limit, bounce back and forth along the same line indefinitely. If the angle is even slightly off normal, the ray will walk over to the edge of one mirror, and get reflected into the side walls. Parallel flat mirrors are a completely unstable resonator configuration.

    Only certain mirror configurations produce stable multiple reflections that retain the energy within the resonator, except for small transmission losses that let light leak out of the business end of the laser. The other mirror is usually made as high a reflectance as technology allows, including the use of TIR mirrors.

  222. Joel Shore says:
    May 9, 2011 at 7:56 am

    Allan M says:

    The basis for the second law (and the first) is careful and accurate measurement, mostly paid for by wicked 19th century capitalists, who wanted to know how to get more from their steam engines, and also how to avoid being sold snake oil.

    There is no need for a statistical ‘fudge’ unless you wish to introduce special pleading for “IPCC Approved” photons.

    Experiment did provide the evidence for the Second Law. However, our modern (20th century) understanding of where the law comes from is based on statistical physics.

    If you you use statistics properly, the tail doesn’t wag the dog.

  223. Martin:

    “The air above is not warming the orange grove but slowing the heat loss by radiating some of the LW radiation back. If no more energy enters the grove from the outside it will continue cool over time. It is only the law of conservation of energy at work, so the back radiation is not extra energy but only energy that has had it’s ability to leave the system delayed.”

    Well, maybe so. But then aren’t you saying that the backradiation is affecting the lapse rate? That cannot be, since the lapse rate depends upon only Cp and g. ??

  224. wayne said:
    Clouds make warmer nights because the clouds are usually warmer than the normal air temperature at that altitude and therefore the surface’s rate of loss by radiation upward will be less leaving you with a warmer than normal night.

    This is nonsense. Typically, clouds are much colder than the ground. (When they are the same temperature, we call it fog.) Instead, they are more like styrofoam cups, they scatter IR radiation back toward the ground without affecting their own temperature. This is because the bottom of a cloud is a phase change boundary and, therefore, the temperature has almost no effect on the amount of energy released. Stefan’s equation does not apply.

    Bottom line, cold clouds make for warm nights. This is a part of the greenhouse effect.

  225. Robert Clemenzi says:
    May 9, 2011 at 12:26 am
    Why is “the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet”?

    As dr.bill said, this is because the Planck equation is actually an integral. When you change the x-axis, the peak actually does shift. I have written a full explanation at

    http://mc-computing.com/Science_Facts/Blackbody/Blackbody_Equations.html

    There is also a plot showing the difference in the shape of the two 300K curves and their peaks.

    biological evolution has “tuned” our visual system to be most sensitive where the light energy is maximized

    Not quite, that is where the intensity per wavelength interval is brightest, not where the energy per frequency interval is largest.

    This demonstrates why people who believe in Global Warming always plot spectra versus wavelength. When you want to analyze the energy in the system, all plots should be versus frequency (normally expressed as a wavenumber) to show an honest plot. In that case, most of the energy from the Sun is in the IR. When plotted versus wavelength, the plots incorrectly imply that most of the energy is in the visible and UV spectra.

    Robert Clemenzi, THANKS for answering my question and for the link to your full explanation which I read and appreciate.

    But, not surprisingly, I still have a problem understanding why multiplying the result of the Planck calculation by wavelength squared, thus moving the peak quite a bit to the longer wavelengths, makes physical sense.

    I mentioned human visual sensitivity as a “sanity check” on which method make physical sense, because biological evolution and natural selection has no “agenda” pro- or con-AGW. According to Wikipedia, human visual sensitivity peaks at about 553 nm (“green”).

    The Carleton spreadsheet comes with the Sun temperature set to 5880 K, and that value seems to best map to the actual measurements of the Solar spectrum as shown in Carleton and reproduced above as the leftmost panel of my second graphic. Carleton indicates maximum “Intensity” as 495 nm, which is 58 nm less than the value indicated by Wikipedia for human maximum color sensitivity. So far so good for Carleton.

    When I modify the Carleton spreadsheet calculation by multiplying the formula by wavelength squared, the Solar peak jumps to 889 nm, which is 336 nm more than human maximum sensitivity. Indeed, 889 is all the way into the near-IR range and totally outside the range of human vision, which ends around 700 nm.

    I have trouble believing that evolution and natural selection, which has had tens of millions (the Great Apes) if not billions of years to work on this problem, and an unlimited budget and “manpower”, could miss the actual peak power portion of Solar radiation by so much that it is out of the visual range. Therefore, I have to conclude that peak Solar radiation is closer to 553 nm than to 889 nm.

    You claim that “.. people who believe in Global Warming always plot spectra versus wavelength.” That is certainly true, but that fact alone does not make it scientifically incorrect to do so.

    You also claim that “… When you want to analyze the energy in the system, all plots should be versus frequency (normally expressed as a wavenumber) to show an honest plot. In that case, most of the energy from the Sun is in the IR. When plotted versus wavelength, the plots incorrectly imply that most of the energy is in the visible and UV spectra. …”

    How do you square that conclusion with the undeniable facts of evolution and natural selection which have no “agenda” other than the survival and reproduction of animals, and, in our case, of the Great Apes and humans?

    Again, I GREATLY APPRECIATE your work in this area, and your excellent write-up that I made clickable above, but I need some physical reason to accept that what you (and Perry) have done is a better representation of the energy in the Solar (and Earth) spectrum than what Carleton and Wikipedia and others have done.

    advTHANKSance for your reply!

  226. Bomber_the_Cat says:
    May 9, 2011 at 2:53 am
    Ira,
    “But, that still does not explain why the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet and other graphics on the Internet”

    Maybe I could help, or perhaps just confuse you further. …

    The Plank distributions plotted against wavelength (the way I like to see them) have a vertical scale of ‘Watts per Square metre PER MICRON’. The graphs shown by Petty are plotted against wavenumber (not wavelength) and the vertical axes have units of ‘Watts per Square Metre per Steradian PER WAVENUMBER’. …

    Say that the peak radiation on the wavelength plot occurs at 10.5 micron. Because the energy is plotted per micron it represents the total energy between 10 micron and 11 micron. In terms of wavenumber, 10 to 11 micron is equivalent to a range of 909 to 1000 per cm. So the energy for the 10.5 micron peak has to be dispersed over 91 different wavenumbers on the wavenumber plot. If we consider another point on the wavelength plot, say between 18 and 19 microns, this is represented by a wavenumber range from 526 to 555. So the energy around 18 microns has only to be shared between 29 wavenumbers whilst the energy around 10 microns has to be shared over 91 wavenumbers. As a consequence, the peak intensity no longer occurs at 10 micron, it now occurs at 18 micron. That is why the two methods of plotting give different peak intensities.

    Did that help, or just confuse everyone further?

    THANKS Bomber_the_Cat. The last part of your comment hit home. It reminded me of the guy who said his Black horses ate more than his Brown horses. It turned out that the guy simply had more Black horses than Brown ones!

    OK, so, the 10-11μm range has 91 “horses” (Wavenumbers) and the 18-19μm range has only 29 “horses”. If we give each range an equal amount of “hay” (energy) each 10-11μm “horse” will have less than each 18-19μm “horse”, so the peak “eating” of “energy” will occur with the latter group of horses, because there are fewer of them.

    In the “horses” example, if we give each “horse” and equal amount of “hay” the 91 “horses” in the 10-11μm range will, in total, consume about three times as much as the 19 “horses” in the 18-19μm range. Got it!

    Now, go one step further and please explain why you like to plot against wavelength.

  227. “Well, maybe so. But then aren’t you saying that the backradiation is affecting the lapse rate? That cannot be, since the lapse rate depends upon only Cp and g. ??”

    My question is why we assume that N2 in the atmosphere doesn’t back radiate. All objects radiate depending on their temperature. As the atmosphere is mostly N2, that must be the source of most of the back radiation.

    If we increased the atmospherice pressure 90 fold, such as on venus, then there would be 90 times as much N2 per cubic meter, and the back radiation would go up proportionally, as would the surface temperature. Thus, the surface temperature must be a function of atmospheric pressure, which is a function of gravity.

    The standard radiative model ignores gravity, which is why it cannot be applied to other planets to see if it is correct. Thus the reliance on computer models, which have the added bonus that there is no way they can be proven wrong in the short term.

  228. Joel Shore says:
    May 9, 2011 at 7:31 am
    … (Almost) no one is claiming that there will be “runaway warming”. [Hansen is talking about the possibility under certain circumstances and has specific arguments as to why the current case may be different from what the earth has experienced in the past, so let's leave him out of this.]

    I’ve appreciated and agreed with all you have said so far on my threads here at WUWT, until your comment on Hansen.

    The word on the street is that you are a Warmist. IMHO, there is nothing wrong with that. You appear to be totally honest, scientifically-oriented, intelligent, and well-informed. Your analysis of the available evidence leads to the belief that AGW is responsible for the majority of warming since 1880 and is likely to be dangerous for human civilization in the forseeable future. You conclude, therefore, that we (society) need to take action sooner than later. I’m a Lukewarmer-Skeptic whose analysis leads me to believe AGW has a real but minor effect compared to natural cycles and processes and data bias by the official climate Team, and that a bit of warming and higher CO2 levels may turn out to be of net benefit to humanity. Therefore we should do little about it. Our positions, while counter to each other, are both valid in the scientific sense, and deserving of mutual respect.

    However, I cannot agree to “leave him [Hansen] out of it.” I believe Hansen was one of the primary science advisors to Al Gore on his lecture and movie An Inconvenient Truth. He is therefore responsible for the consequences of the Catastrophic CAGW the sky is falling fiction bought into by the major media and opinion leaders. That led to Ethanol, Wind, and other unwise subsidies, the Cap & Trade scam, and other economy-wrecking policies.

    Gore is not and never has been a scientist, but Hansen, as a scientist and head of NASA GISS has misused his credentials and appears to have pressured his employees to cook the books, bending the temperature record by introducing “corrections” way after some of those measurements were old enough to qualify for Social Security, see Makiko Sato FOIA email and The PAST is not what it used to be.

    Should we interpret your “leave him [Hansen] out of it” as a repudiation of Hansen and his leadership of NASA GISS or as simple embarassment at the antics of a nice old uncle who means well but sometimes leaves his fly down by mistake?

  229. ferd:

    “My question is why we assume that N2 in the atmosphere doesn’t back radiate. All objects radiate depending on their temperature. As the atmosphere is mostly N2, that must be the source of most of the back radiation”

    N2 doesn’t radiate significally in the IR portions of the spectrum.

  230. Can you provide a reference to a recognized standard text which supports this contention that it is only the net difference which must respect the Second Law of Thermodynamics?

    The only instance I know of where a colder object can impart energy to a warmer object is in the entirely theoretical case of two objects being at negative Kelvin temperatures where the colder object must perforce be at a higher energy state the the warmer one. Yet in this case energy still flows from a higher state to a lower state and thus respects the Second Law.

    Heat is the manifestation of energy flowing from a higher state to a lower state. It is why heat pumps are efficient and air conditioners aren’t.

  231. Joel Shore

    You cannot weasel out of your gross error.
    You challenged me to discus the science.
    I am here to do so.
    If you prefer to avoid the science and instead issue smears instead then I will not be surprised.

    This item is directly addressed to Ira’s post and deserves an answer from you!
    On page 9 of your co-written Halpern et al comment paper you state that the Stephan Boltzmann Law can be used to work out the thermal energy exchange between atmospheric shells.
    There is a whole series of statements on pages 8,9,and 10 referring to the atmosphere, Stephan Boltzman Law and indeed including the SB formula.

    Gerlich and Tscheuschner specifically said that the Stephan Boltzman Law cannot be used for atmospheric gases.
    Do you now agree with them and withdraw your comments?
    Or will you persist in this unphysical assumption.

    Depending on your answer we can perhaps call on Ira’s spreadsheet to settle the matter.
    Your paper is shown below but I dont have a direct link.

    Comment on ‘Falsification Of the atmospheric CO2 greenhouse effects within the frame Of Physics’ by Joshua B. Halpern, Chistopher M. Colose, Chris Ho-Stuart, Joel D. Shore, Arthur P. Smith, Jorg Zimmermann.

  232. Dave Springer says:

    The hypothetical amplification of anthropogenic CO2 forcing by increased water vapor is a wholesale fabrication without a shred of evidence to support it and with mountains of contrary evidence. The water cycle is a negative feedback not a positive feedback.

    Climate boffins have pretty much conceded the debate over water vapor amplification.

    Actually, there is plenty of evidence of the water vapor feedback, both existence and rough magnitude, as discussed here: http://www.sciencemag.org/content/323/5917/1020.summary

    The only sense in which your argument for a negative water cycle feedback makes much sense is if you are grouping together cloud and water vapor effects in such a feedback (which I guess is not unreasonable when you refer to it as “water cycle” but becomes confusing when you refer to it as “water vapor feedback”).

    The cloud feedback is indeed associated with quite a large degree of uncertainty and basically the only way in which we could be spared from having a fairly significant climate sensitivity is to have a significantly-negative cloud feedback. Alas, besides not being what all of the models show (which admittedly could be a problem with the models), such a negative feedback makes it rather difficult to explain paleo-climate without somehow coming up with much larger forcings than are estimated right now (for. e.g., the glacial – interglacial oscillations).

    The controversy is over whether a faster water cycle is a positive or negative feedback.

    Limiting the controversy to specifically the feedback due to clouds, I essentially agree with this statement….which is why we both apparently find it so ridiculous to see people here wasting tons of time arguing about well-settled science that they have no hope of possibly overturning. (The most extreme cases being people like Allan M who apparently want to overturn all of modern statistical physics in order to come up with a version of the 2nd Law that is more toward their liking.)

    People should realize that, even from a purely tactical point-of-view, it would be wise to pick your battles. When you argue that the greenhouse effect doesn’t exist, the only cause your further, at least within the scientific community, is the cause of those who would like to show how ridiculously ignorant of science the AGW skeptics really are. I would invite those who want to help demonstrate that to continue to make their arguments here; the rest of you should consider carefully reading what Dave, David Hoffer, Ira, myself and others are writing here and actually trying to understand the correct science.

  233. “Bottom line, cold clouds make for warm nights. This is a part of the greenhouse effect.”

    Hmmm. Maybe not. Clouds affect convection and the lapse rate (decrease rates), too. Water (clouds) also store 4 times as much heat as air. Water vapor stores twice as much. How do you separate these effects from the “radiative greenhouse effect?”

  234. Ira Glickstein says:

    Should we interpret your “leave him [Hansen] out of it” as a repudiation of Hansen and his leadership of NASA GISS or as simple embarassment at the antics of a nice old uncle who means well but sometimes leaves his fly down by mistake?

    Ira,

    I think you may be over-interpreting the statement that I made. As far as I can tell, Hansen’s notion that if we really go to town using fossil fuels, we might trigger a true runaway greenhouse effect just seems rather vaguely-supported by any detailed argument or evidence at this point and does not seem to be the general belief within the climate science community. So, although I have a lot of respect for Hansen, who has the track-record of saying things that seem a bit far-out at the time but later become quite well-accepted by the scientific community (such as when he pronounced back in the late 1980s that the warming occurring was almost definitely due to greenhouse gases), I think it is at best premature to give much credence to this particular prediction of Hansen’s.

    So, I was simply saying that the generally-accepted projections for AGW are that positive feedbacks amplify the radiative effects due to greenhouse gases alone but do not lead to an actual “runaway” instability. For some reason that I don’t totally fathom, some people on the skeptic side (like martin mason above or Smokey in other threads) seem to like to group together this prediction of feedbacks amplifying things and producing an actual “runaway” instability, perhaps because it makes a better “strawman” argument to attack.

  235. Alleyne says:

    Can you provide a reference to a recognized standard text which supports this contention that it is only the net difference which must respect the Second Law of Thermodynamics?

    Any text on thermodynamics that treats it from a statistical physics perspective ought to do fine. In the context of radiative heat transfer specifically, any text that talks about the exchange of radiation between two objects or any object and its surroundings ought to give you the basic radiative transfer equations that are used in computing the greenhouse effect.

    Heat is the manifestation of energy flowing from a higher state to a lower state.

    Yes…So, heat (which is a macroscopic concept by its definition) always flows from hotter to colder (in the absence of work). However, note that the interpretation that the radiative energy flows in both directions, while having an abundance of empirical support, is not necessary to show there is a greenhouse effect. All that is really necessary to say is that the heat flow between two objects depends on the temperature of both objects and not just on the temperature of the hotter object.

    It is why heat pumps are efficient and air conditioners aren’t.

    I have no clue what you are trying to say in this last sentence. The term “heat pump” is sometimes used as a category that includes air conditioners and refrigerators. Other times, it is used to represent the subcategory of the “heat pump” in the above context that are actually used to heat a house rather than cool it. However, regardless of how the term is used, heat pumps used for heating and air conditioners operate by the same basic principle. Both use work to “pump” heat energy from cold to hot, i.e., in the opposite direction from which heat spontaneously “flows”.

  236. JAE says:
    May 9, 2011 at 11:28 am

    Well, maybe so. But then aren’t you saying that the backradiation is affecting the lapse rate? That cannot be, since the lapse rate depends upon only Cp and g. ??

    It’s not a question of back radiation effecting the lapse rate but rather the laps rate reduces the effectiveness of back radiation.

    As the air thins then there is an increased possibility that some of the electromagnetic wave may find windows that are not closed by a receptive molecule. This being the case then the wave propagates out into open space and so is not absorbed and not re-radiated.

    This is important with regard to H20, as humidity drops rapidly and the wave lengths that are absorbed by this molecule now escape at a greatly increased rate and so back radiation diminishes accordingly.

  237. Ira, Why are leaves green?

    It is because chlorophyll reflects green light. Leaves absorb and use the energy that is most abundant .. the near IR.

    People also usually ignore the fact that our retinas are highly sensitive to UV light. So sensitive, in fact, that there is a special UV filter associated with the lens to absorb it. This was discovered when people with new artificial lenses were able to see in the UV. Because our eyes use simple lenses, this caused significant chromatic aberration such that they were not able to focus when out doors. I mention this because it argues against the common idea that our eyes are most sensitive to green light because “that is the brightest part of the spectrum”.

    You should also consider that many people wear sun glasses because it is too bright near the equator. This implies that it is possible that our eyes developed to block the bright near IR and to use the significantly dimmer visible part of the spectrum.

    Therefore, I reject the argument that “using human eyes for a sanity check” makes sense. Using leaves (chlorophyll) makes a lot more sense since they have (it has) been around a lot longer.

    Remember, energy is expressed as frequency times Planck’s constant – E=hf. Therefore, when studying energy, frequency is the more appropriate x-axis.

    As for physical sense, the distance between 1 and 2 is the same as the distance between 9 and 10. However, the distance between 1/1 and 1/2 (0.5) is NOT the same as the distance between 1/9 and 1/10 (0.011). Since the energy is measured as the area under a curve (an integral), an adjustment has to be made to keep the areas the same when using the wavelength. Remember, energy is linear only in the frequency domain.

  238. Bryan says:

    You challenged me to discus the science.
    I am here to do so.

    What you (and G&T) are engaging in has ABSOLUTELY nothing to do with science, unless you include undermining and obfuscating science. I have explained the philosophy behind using a hierarchy of models here: http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655880 and here: http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-656374 . If you refuse to understand it, I don’t have any more to say to you.

    You are adding only useless noise to this thread.

  239. JAE says:

    Hmmm. Maybe not. Clouds affect convection and the lapse rate (decrease rates), too. Water (clouds) also store 4 times as much heat as air. Water vapor stores twice as much. How do you separate these effects from the “radiative greenhouse effect?”

    Your points are valid but clouds should not be appropriated by greenhouse enthusiasts

    Clouds, raindrops and hailstones are much better radiators than gases.
    Not because they are composed of so called “greenhouse gases”.
    ANY solid or liquid can provide a much larger range of radiation than the line spectra of gases.
    They don’t need to be composed of greenhouse gases to show this effect.

  240. Ira Glickstein,

    You have exactly conveyed my own world view regarding the effect and importance of CO2:

    …AGW has a real but minor effect compared to natural cycles and processes and data bias by the official climate Team, and that a bit of warming and higher CO2 levels may turn out to be of net benefit to humanity. Therefore we should do little about it.

    # # #

    Alleyne,

    Heat can flow from a colder to a warmer object; from higher to lower entropy. What is required is work. For example, a refrigerator does reverse entropy work. Work requires energy. Of course, total entropy in the universe only goes in one direction. It’s the exceptions that make it interesting.

  241. Robert Clemenzi says:
    May 9, 2011 at 11:34 am

    wayne said:
    Clouds make warmer nights because the clouds are usually warmer than the normal air temperature at that altitude and therefore the surface’s rate of loss by radiation upward will be less leaving you with a warmer than normal night.

    This is nonsense. Typically, clouds are much colder than the ground. (When they are the same temperature, we call it fog.)
    —-
    Respectable Robert, what you say I said is not what I said at all. Read what I said again. I never said clouds were warmer than the surface (well, except rarely and even that was pointed out later). I think you jumped the gun a bit. ☺

  242. Whew, I’ve read most of the comments over the last couple of days and yet I see nothing discussed about the cooling effect of GHGs.

    How can we discuss the temperature of the atmosphere without understanding the transfer of heat by collisions? When I see Trenberth’s energy chart it ignores heat transfer by collisions of gas particles. While I accept radiation flows both ways I understand collisions create a heat flow between the Earth’s surface and the atmosphere that also goes both ways.

    Therefore, the Earth loses energy at times to the atmosphere and gains energy at other times. However, just like in the case of radiation, the net flow is most likely to the atmosphere. Has anyone measured this heat flow? I’ve never seen it documented and yet it seems important if we are to understand the big picture.

    Now, why is this important? Well, consider the situation without GHGs and ignore the Earth radiating energy for the moment. The Sun would heat the surface and the atmosphere. Energy would move between them but would have no way to escaping. It would get pretty darn hot.

    Now, add GHGs to the system. The GHG-less atmospheric gases would start colliding with the newly added GHGs and heating them up. They would start radiating energy. Half would go down which would not change the heat in the system since none was escaping before adding the GHGs. However, half would go up and out of the system … thus cooling the atmosphere.

    Of course, we know the GHGs would actually intercept radiation from the surface as well. The so-called GHE. So, we have two conflicting processes. One delays the heat leaving the system and the other increases the heat loss that occurred as a result of heating due by other means (collisions with surface and solar).

    Without understanding both of these systems in detail how can we know the overall effect of GHGs? It seems to me only half the problem has been addressed by scientists, what am I missing?

  243. “In the past 100 years since the formation of the ‘greenhouse’ theory of atmosphere, we have discovered that the atmosphere is a MILES DEEP, FRIGID, compressible fluid, HEAT CONDUCTIVE, IMMERSION BATH, held in place around the earth, by gravity.

    The atmosphere possesses physical attributes like:

    A: a big, warm blankie
    B: a big, warm, greenhouse
    C: a MILES DEEP, FRIGID, HEAT CONDUCTIVE, IMMERSION BATH

    Someone who believes in a greenhouse effect will tell you the answer is A, but since it’s too complicated to talk about it as A, you can talk about it like it’s B, but not E.V.E.R. under A.N.Y. circumstances, is the answer C.

    T.h.a.t. is who you’re dealing with.

  244. N2 doesn’t radiate significally in the IR portions of the spectrum.

    How is that possible? Every object radiates depending on its temperature. It may not absorb at certain frequencies, but how can it avoid radiating in IR if that is its temperature?

  245. “The only sense in which your argument for a negative water cycle feedback makes much sense”

    Negative feedback makes sense when one considers how long life has existed on the planet. Under positive feedback anything that increased temperature (not just CO2) would have locked the earth into spiralling temperatures and the extinction of life.

    This doesn’t happen because the evaporation of liquid water is a super efficient cooling mechnism. As a result humans are able to survive the hottest conditions on the planet, so long as water is available.

    In contrast, humans cannot survive without protection temperatures under 82F for any length of time. Below this temperature we radiate more net energy than can be replaced through food and require technology to survive. Interestingly, 82F is the temperature of the tropical rainforests of the earth, which suggests that humans did not leave the jungles until after we developed technology.

  246. Smokey,

    Thanks, that was my point exactly.

    Where is the work, the external energy, which causes energy to flow from the atmosphere to the earth, from cold to warm?

  247. Knock Knock,

    The Earth is a rotating sphere, which emits radiative power from the surface according to the fourth power of T in all directions from all locations, varying with the time of day and the physical properties of the surface. Sunlight essentially has parallel radiation and is at a very much higher energy level with a “hot spot” notionally rotating around the planet. The outgoing stuff is very weak because it is spread around over the entire surface via atmosphere and ocean dynamics. That naughty T^4 thingy is roaming around everywhere, although it is in the minority in HEAT loss from the surface. (according to Trenberth).

    I think there are some meaningless averages touted around.

  248. ferd:

    “How is that possible? Every object radiates depending on its temperature. It may not absorb at certain frequencies, but how can it avoid radiating in IR if that is its temperature?”

    The electromagnetic radiation with wavelengths in the IR region is associated with changes in a dipole moment. See here:

    http://www.wpi.edu/Academics/Depts/Chemistry/Courses/General/infrared.html

    Key paragraph:
    “The fact that a molecule vibrates does not in itself insure that the molecule will exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the vibrational motion associated with that mode must produce a change in the dipole moment of the molecule. HCl, for example, with a center of positive charge at the H atom and a center of negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further, stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that cause changes in the value of the dipole moment from 0 to some non-zero value. An example is CO2, which has no permanent dipole moment because the individual bond dipoles exactly cancel. However, when CO2 undergoes a bending vibration, its dipole moment changes from zero to some non-zero value. This vibration produces a change in dipole moment and is IR active.”

  249. wayne says: May 9, 2011 at 9:21 am

    Here is what happens because of the Stefan-Boltzmann equation that screws so many people up (sigma=5.6704r-8, epsilon=1):

    ε σ ( 288K^4 – 280K^4 ) = 41.572 W/m^2

    or

    ε σ ( 288K^4 ) – ε σ ( 280K^4 ) = 41.572 W/m^2

    The top one will always show you the correct and actual flow of radiation between two pieces of matter. The latter one implies that matter always radiates at all times in all directions even if the temperatures are identical and the last term gets this quirky term applied to it, “back radiation”.

    Sorry, Wayne, but you are the one who is screwing it up. Either or both equations are correct. The object at 288 K will radiate ε σ ( 288K^4 ) independent of any other objects nearby. The other object at 280 K will radiate ε σ ( 280K^4 ) independent of any other objects nearby.

    If the second object was at 278 K, then both would radiate ε σ ( 288K^4 ), for a net transfer of zero. This is perfectly in accord with all the laws of physics.

    The first version highlights the correct and actual HEAT (ie net flow of energy).
    The second version highlights the correct and actual energy from each object.

    [Or more specifically, the net heat or energy per second per square meter].
    [And both rates of energy transfer depend on the geometry of the two objects. Just knowing how much each object radiates does not tell us how much is actually intercepted by the second object.]

  250. “Robert Clemenzi says: Ira, Why are leaves green? It is because chlorophyll reflects green light. Leaves absorb and use the energy that is most abundant .. the near IR.”

    Leaves reflect green because leaves usually contain a mix of chlorophylls (mostly types a and b), which predominantly absorb violet, blue, orange and red light. And as winter sets in, leaves become red, orange and yellow because the quantity of red absorbing chlorophyll is dropped in favor of violet absorbing chlorophyll (more bang for your metabolic buck). I have never read of any type of chlorophyll that absorbs in the NIR. In fact, chlorophyll meters specifically use the fact that chlorophyll doesn’t absorb NIR in order to take a measurement!

    “ferd berple says: How is that possible? Every object radiates depending on its temperature. It may not absorb at certain frequencies, but how can it avoid radiating in IR if that is its temperature?”

    Individual molecules are not black bodies. If nitrogen gas, which makes up 80% of the atmosphere, radiated “according to it’s temperature”, how would anyone be able to take an IR photograph that didn’t look like a photo taken in a thick fog?

    Radiation is due to the motions of atomic bonds. Temperature is a measurement of average kinetic energy. The temperature of a mass of nitrogen gas in our atmosphere is primarily due to entire molecules moving, not atomic bonds vibrating. So it predominantly cools by transferring this kinetic energy to neighboring molecules (conduction) or doing work (adiabatic cooling), which is the same way it was heated to that temperature in the first place (not by IR absorption).

  251. Smokey says:
    May 9, 2011 at 3:02 pm

    Said to Alleyne:
    Heat can flow from a colder to a warmer object; from higher to lower entropy. What is required is work. For example, a refrigerator does reverse entropy work. Work requires energy. Of course, total entropy in the universe only goes in one direction. It’s the exceptions that make it interesting.

    Ok Smokey, of course, the refrigerator and air conditioner and similar cases, of course, can and do work (sic ☺).

    But when viewing two objects as a whole, the atmosphere and the surface, that the atmosphere is performing work to warm the surface without the atmosphere equally cooling from doing that very work? Warmist tend to believe this, cake and eat it too. I can see isolated cases as compression at the poles and other curiosities but not on the average, and besides, even with those effects the temperature gradient is rarely actually inverted so in a net sense that is only slowing the cooling of the surface at the expense of equal cooling in the atmosphere which ends in a greater temperature gradient therefore a greater flux of energy upward to space. A local effect, may be the whole polar cap, but the net effect is the same.

    Do you really think this work by the atmosphere is happening and where, how? (Or am I missing something here?)

  252. Alleyne,

    Perfect question. Maybe those energetic little CO2 molecules are busily working away, heating up the planet and making it go all CAGW on us. ☺

    OTOH, maybe they just delay the transfer of heat, which would explain why Trenberth’s “missing” heat isn’t there – and never was.

  253. wayne says at 5:27 pm: [ ... ]

    Wayne, I probably didn’t make myself clear enough in the post you replied to, but as you can see from my reply to Alleyne above [which I posted before seeing your comment], I am in complete agreement with your view that CO2 does no work on its own.

  254. ferd berple says: May 9, 2011 at 4:30 pm

    “N2 doesn’t radiate significantly in the IR portions of the spectrum.”

    How is that possible? Every object radiates depending on its temperature. It may not absorb at certain frequencies, but how can it avoid radiating in IR if that is its temperature?

    Not only is it possible, it is REQUIRED! Any table of IR emissivity applies equally well to the object absorbing IR radiation or to it emitting IR radiation.

    Take two identical objects; one with a low emissivity and the second with a high emissivity. Warm them both above room temperature and set them out in the cooler room. The high emissivity object will cool faster because it is emitting more IR than the low emissivity object.

    Or do you think the low emissivity object will cool at the same rate because “but how can it avoid radiating in IR if that is its temperature?” In fact, here is someone who did this experiment and posted the results here. http://blog.khymos.org/2007/03/

    It is true for metal pots. It is true for gases in the atmosphere. How well they absorb is directly related to how well they emit.

  255. Richard M asks: May 9, 2011 at 4:16 pm

    Therefore, the Earth loses energy at times to the atmosphere and gains energy at other times. However, just like in the case of radiation, the net flow is most likely to the atmosphere. Has anyone measured this heat flow? I’ve never seen it documented and yet it seems important if we are to understand the big picture.

    The transfer of heat by collisions between the surface and the atmosphere is the “thermals” term. Air is a very excellent insulator and has a low heat capacity, so energy transfer by conduction is very limited. However, when the ground warms a layer of air near the surface, it expands and rises, carrying away energy from the surface.

  256. wayne says: May 9, 2011 at 5:45 pm

    Tim Folkerts, nothing new, I disagree.

    I agree that I said nothing new — I think it is all very standard, well-established physics. What specifically do you disagree with in the discussion of …

    εσ( 288K^4 – 280K^4 ) vs. εσ( 288K^4 ) – εσ( 280K^4 )?

  257. Joel Shore says:
    May 9, 2011 at 1:55 pm

    So, I was simply saying that the generally-accepted projections for AGW are that positive feedbacks amplify the radiative effects due to greenhouse gases alone but do not lead to an actual “runaway” instability. For some reason that I don’t totally fathom, some people on the skeptic side (like martin mason above or Smokey in other threads) seem to like to group together this prediction of feedbacks amplifying things and producing an actual “runaway” instability, perhaps because it makes a better “strawman” argument to attack.
    —————————————————–

    But Joel, you used to argue that on here, and quite vehemently as I recall. In fact I distinctly remember proposing the Joel Shore Uncertainty Principle whereby if what Joel Shore proposed was true, there would be no Joel Shore to observe it.

  258. Alleyne says: May 9, 2011 at 4:48 pm

    Where is the work, the external energy, which causes energy to flow from the atmosphere to the earth, from cold to warm?

    Work would indeed be required for a NET transfer of energy from cold to warm (eg an air conditioner) and you very correctly deduce this would not happen. However SOME energy can be transferred from cold to warm, as long as MORE energy is transferred from warm to cold (ie as long as the net transfer of energy is from warm to cool; ie as long as the heat is from warm to cool).

  259. philincalifornia: I am afraid you must be misremembering. What I have argued here is that the most likely range of the climate sensitivity is what the IPCC says the most likely range is…which corresponds to feedbacks amplifying the radiative effect due to CO2 alone by a factor of about 3 +/- 1.

    I have also consistently said that I am skeptical of Hansen’s claim that we could trigger a runaway effect. I know that because I have used it as an example to challenge Smokey to display some true skepticism by telling me what arguments on “HIS side” he is skeptical of. (Another example I gave where I remain skeptical is on the effect of AGW on hurricane intensity.) As I recall, he never took me up on it.

  260. Joel Shore,

    The water cycle feedback is about more than just the clouds and water vapor, it is about the evaporation and precipitation as well. Wentz (2007) documented that the models represented only one-third to one-half the observed increase in precipitation associated with the recent warming. Lindzen mentioned a paper which I wish I remembered that confirmed the increased precipitation with measurements of the latent heat flux (anyone save the cite?). The turning of the water cycle is like a big air conditioner with the water as the working fluid. At these new levels of relative and absolute humidity associated with the warming the air conditioner or heat pump turns faster. Wentz found it was turning faster in proportion to the humidity increase, while in the models it is one half to one third of that. While the water vapor feedback is positive, the net feedback of the whole water cycle may well be negative. This is the area in dispute in which the science and the models need to progress.

  261. Hi Ira.

    “Also, the Petty black body curves peak at a “Radiance” around 100 mW/m^2/sr cm^-1 while the black body curves from Carleton peak at an “Intensity” of around 14 W/m^2/mm.”

    As engineer to engineer. :)
    A “radiance” is a ‘comparison’ to an accepted ‘standard radiance’ (don’t ask me what frequencies this involves, as that’s defined by ‘the standard’), but an “intensity” is an ‘absolute measure’ of ‘energy’ (don’t ask me how the ‘Planck constant’ is considered there) and seems to be commonly used with a specific spectral wavelength. I hope this helps.

    It would be nice if someone were to construct a ‘table of resonant interaction’ for ‘atmospheric molecules/local EM field’. However, this would be a stratospheric altitude + project because LW photon emission hardly occurs in the troposphere (where it’s mostly molecular disturbance), meaning that this would come under the heading of insolation for the ‘tropo’. If only we could have the ‘unit measure’ the same.

    Best regards, Ray Dart.

  262. Dave in Delaware says:
    May 9, 2011 at 7:37 am
    Ira – regarding your summary comment 4) at May 8, 2011 at 7:51 pm
    my comment – NO, the atmosphere does NOT emit LWIR across a distribution of wavelengths like a blackbody …

    Thanks for your most welcome comments, Dave in Delaware. I did not claim that the Atmosphere, after being excited by radiation coming up from the Surface, emits down towards the Surface and up towards Space exactly like a black body, just that it emits at a variety of wavelengths.

    Please refer to my illustration above with the two Perry curves and the Atmosphere in between.

    Let us consider the Atmosphere as a “black box” (in the engineering sense as a box with a particular set of inputs and outputs, but unknown innards). I think we all agree that the Earth Surface emits pretty much like a black body, with a distribution of wavelengths from around 5μm to above 50μm. So the mean input to the bottom of the Atmosphere (BOA) is a more or less smooth, black-body-like curve at some temperature, probably around 270 K, more or less, since this is the Arctic. OK so far?

    Now, at the top of the Atmosphere (TOA) we see a jig-jaggy curve that skims along the 270 K black-body-like curve in the ~10μ region (between around 8μm to 13μm), except for a small “bite” in the 9.5μm O2/O3 region. (We know that corresponds to the absorption/emission band for oxygen and ozone.) You correctly interpret that as looking down from Space to the Surface through the ~10μm nearly clear pass-through window. This is in the Arctic region and the Surface may well be at 270 K, more or less.

    Looking from TOA at the ~7μm region (between around 5μm and 8μm) we see a depressed curve that coincides with the left-hand peak of the H2O absorption region indicated in gray in my illustration.

    You interpret that as looking down from Space at a middle of the Atmosphere (MOA) altitude which, due to the lapse rate, is cooler than the surface. Therefore, you say the depressed curve skims along the 260 K black body curve.

    Well, that is true, I guess. But, as an engineer looking at a “black box” I would say that a full blast of 270 K ~7μm radiation went in at the BOA and only a 260 K puff came out at TOA. So, imagining I do not know what is in that “black box”, and accepting conservation of energy, I would say the missing ~7μm radiation input from the BOA got transformed into some other region, most probably the ~10μm region!

    Similarly, looking from TOA at the ~15μm region (between around 12μm and 50μm plus) I see two distinct depressed curves. A very depressed curve (between 13μm and 16μm) coincides with the CO2 absorption region and the right-hand part of the H2O absorption region, both indicated in gray in my illustration. The second, less depressed curve (from 16μm and up) corresponds to the remaining right-hand part of the H2O absorption region.

    You interpret the very depressed curve as looking down from Space at a high-MOA altitude which, due to the lapse rate, is much cooler than the surface. Therefore, you say the very depressed curve skims along the 225 K black body curve. The less depressed curve would be interpreted as from about the same MOA altitude as the ~7μm we discussed in the previous set of paragraphs and skimming along the same 260 K black body curve.

    Well, that is true, I guess. But, as an engineer looking at a “black box” I would say that a full blast of 270 K ~15μm radiation went in at the BOA and only a 225 K piddle came out in the CO2 area and only a 260 K puff came out in the H2O area. So, imagining I do not know what is in that “black box”, and accepting conservation of energy, I would say the missing ~15μm radiation input from the BOA got transformed into some other region, most probably the ~10μm region!

    How could the absorbed radation get transformed? Well, while it is true that an excited molecule, in isolation, generally emits in or near the same wavelength where it absorbs. But, the H2O and CO2 molecules are not in isolation. They are colliding with each other and with N2 and O2 and O3 and Ar molecules and atoms. The temperature of a gas is something like the average velocity of the molecules that make it up and any material above absolute zero emits more or less (usually less in the case of a gas) like a black body at that given temperature. So, I think it is reasonable to assume that the H2O and CO2 and the other air molecules emit some fraction of energy at wavelengths outside their normal absorption regions. Furthermore, when some of the re-emitted radiation, even if it happens to be at 7 μm and 15μm happens to head back to the Surface, and get re-absorbed, the warm Surface will emit that energy at a variety of wavelengths, including ~10μm, roughly according to the black body curve at 270 K.

    So, that is my story of how the TOA jig-jag curve may be interpreted from a “black-box” standpoint. The initial 270 K, more or less, black body emission from the Surface is transformed, via multiple transactions, from a smooth black body curve to a jig-jag curve with proportionally more of it in the ~10μm region than went in, considerably less in the CO2 region, and a bit less in the two H2O regions and the O2/O3 region.

    Best regards to Ira – I hope my comments are constructive and instructive, and will help your ‘mental models’ and descriptions.

    I do consider your comments both constructive and instructive and they do help me better understand the physical models my blockhead brain insists on. THANKS for your contribution to a constructive dialog. (Oh, and by the way, your explantion may be entirely correct. Please try to tell your story as an energy-balance, physical way to get through my brain! :^)

    PS: Reminds me of the story of the engineer and the mathematician standing in front of an office building waiting for a bus. They see three people go in, and, several minutes later five people come out. The mathematician asks, “How many people are in the building?” The engineer says, “I don’t know, but I can tell you there must have been at least two people in the building before we got here.” The mathematician answers, “Silly engineers! Can’t you tell there are exactly minus-two people in that building!”

  263. Joel Shore:

    “philincalifornia: I am afraid you must be misremembering. What I have argued here is that the most likely range of the climate sensitivity is what the IPCC says the most likely range is…which corresponds to feedbacks amplifying the radiative effect due to CO2 alone by a factor of about 3 +/- 1″

    WHAT??

    Maybe I misunderstand this?? Is an “ampliyfing factor of MINUS ONE actually an AMPLYFIYING FACTOR?? Just how does this brand of logic work (outside of “climate science?”) AN “AMPLIFYING FACTOR OF NEGATIVE ONE????” WTF??? This reeks of “Kinetic Military Action,” “Overseas Contingency Operation,” “Climate Disruption,” and all the other blatant Orwellian Communistic propaganda crap from the “progressive” sickos.

    Are you sayin that IPPC actually says that there may be a …..NEGATIVE…..feedback? If so, just WHERE does that occur in the IPPC narritive? I must have missed that.

    BTW, you warmist “guys” are losing it, BIG TIME, IMHO. I detect some kind of desperado talk.

    Why don’t you just speak the native tongue?

    AND, BTW, it has now been shown that a very high percentage of the “literature” cited by the IPPC is Sierra Club -type political propaganda.

    Exit question: can any thinking person trust the IPPC?

  264. Tim Folkerts says:
    May 9, 2011 at 6:08 pm
    The transfer of heat by collisions between the surface and the atmosphere is the “thermals” term. Air is a very excellent insulator and has a low heat capacity, so energy transfer by conduction is very limited. However, when the ground warms a layer of air near the surface, it expands and rises, carrying away energy from the surface.

    Thank you, I believe that is the net transfer of energy. However, I suspect the total transfer of energy is much higher. The fact air works like an insulator may be due to energy going both ways easily, not that energy is not transferred. That is, if we measure the transfers in both direction we would see huge w/m2 going both ways but only a small difference which is identified as thermals. In any event the thermals term is bigger than the difference in total IR radiation.

    Now, the diagram does show the total IR radiation going both ways instead of just the small difference. Why are these treated differently? Well, most likely it is to highlight the back-radiation. But, think about this, if back-radiation heats the surface then thermal radiation toward the atmosphere should increase. If we don’t understand the total effect of thermal radiation then how can we understand the bigger picture?

    Also, I’m still waiting for anyone to explain why the cooling effect of GHGs is always ignored.

  265. Dammit, I screwed up in the last comment. I misunderstood the 3+/-1 figure. Please disregard all but the last half of the comment and the nasty attitude. The morons are now fighting a losing battle and are getting bolder and bolder and stupider and stupider, to their detriment. Brings tears to my eyes.

  266. IRA GLICKSTEIN:

    I consider it very strange, even weird, that you have not responded to or even acknowledged the other explanations about why the temperature on the surface of the Earth is 33 C higher than it “should be.” You are supporting the 19th century Ahrennius apostles, you know. Have you read and understood the articles that make mincemeat of your theory?

    Anyway, the silence is deafening. Do you not have any opinion about this? Are we beneath your contempt? WHAT? You act like a frigging Democrat.

  267. Tim Folkerts says:
    May 9, 2011 at 6:10 pm

    wayne says: May 9, 2011 at 5:45 pm
    Tim Folkerts, nothing new, I disagree.

    I agree that I said nothing new — I think it is all very standard, well-established physics. What specifically do you disagree with in the discussion of …
    εσ( 288K^4 – 280K^4 ) vs. εσ( 288K^4 ) – εσ( 280K^4 )?

    Thanks for the very civil reply. I was getting ready for a pie in my face!

    I need to let you know where I am now coming from, I keep getting acclimated to these areas of atmospheric physics which I had never spent much time getting in the depths, but it is all coming back slowly.

    When you get into spectrometry back comes the E/M waves and all of the associated equations governing the transfer of energy via radiation. The standing waves, traveling waves, Poynting complex vectors that I remember going over years ago. You see, I don’t remember matter being able to continually shed energy (continuous radiation) without a lose of temperature, energy will never transfer unless it should and then it must by temperature gradients always keeping in mind that we are speaking in an LTE environment along with the equipartition of the degrees of available freedom, that I will call these microstates for brevity.

    To my older knowledge there are standing waves in the complex plane that do not act as nodes on a vibrating string that most, in including me, tend to visualize it as, but that is flawed without including the imaginary portion. These standing waves, I have them right here in my room where all is at the same temperature, do not transfer energy unless one molecule has less microstates than the other molecules. That IS directly related to the temperature. The molecule with less free microstates will radiate to a molecule only if the destination molecule has more free microstates to accept the energy. Otherwise, you are just in a resonating state with no real energy being transferred as standing wave imply.

    Now that example is rather tinker-toy for in zillions of molecules there is always some misbalance of available microstates held by infrared enabled molecules (the GHGs) and so some energy is always being radiating but always in one net direction, here’s where we seem to disagree, to me in reality, always warm to cooler. Many professors at esteemed universities, MIT, Stanford, Cornell, Oxford, etc, do agree with my view of reality in energy transfer by radiation, some do not, they tend to be in the climate and ecology required courses.

    So, I have traced one big disagreement backwards to this very important point, does matter actually radiate energy that due to the temperatures does not have to violating the Principle of Minimum Entropy Production while at the same time maintaining the Maximum Energy Dissipation Principle always from warmer to cooler surfaces or matter at a microstate level.

    Prove me wrong, I’m having trouble doing that to myself.

    So: εσ( 288K^4 – 280K^4 ) vs. εσ( 288K^4 ) – εσ( 280K^4 )? I do see a disconnect at this very point in the way you visualize what is actually happening and that, to me, is why a huge Fresnel lens in a dark room no matter how hot it is will never be able to focus “back radiation” and cause damage. To me it is a manufactured concept and wrong, the traveling wave must manifest itself and they do when effective temperatures differ (the real description is very much deeper, see QED and why photons do not appear in space where logically you would say they should, or even must but don’t).

  268. Maybe I’m missing something but I thought the calculated black body temperature of the earth seemed low. I recalled the moon having extremes of temperature and a bit of Googling revealed peak equatorial temperatures on the moon of 380 K. Of course, given the lack of an atmospheric buffer, the non-sun exposed side of the moon can get close to absolute zero in shaded spots.

    So, if we assume that the peak temperatures on the earth would be the same as on the moon, it would appear that the primary role of the atmosphere is to cool the earth. I have to admit it’s been 35 years since I’ve taken a physics course but there appears to be something amiss here.

    To follow up with my previous post, daytime straight up in the air temperature -5.5 F when ground temperature was 65 F with readings done about 30 minutes ago.

  269. To wayne regarding the Fresnel lens experiment at room temperature.

    The critical factor you missed in your thought experiment is that the radiation from the wall is not all approaching the lens in parallel beams. The lens will only focus the heat radiation at the target that came from a spot on the all that is directly perpendicular to the surface of the lens. In order to focus the entire heat radiation from the wall onto the target, all of the radiation from the wall would need to be focused in parallel beams first, before it hits the lens. The best way to accomplish this would be to move the wall a long distance away, and heat it up so that the radiation from it reaching the plane of the lens is equal to the total radiation coming from the wall at room temperature. The water in the target would then heat up nicely, and the laws of thermodynamics are satisfied.

  270. Joel Shore;
    the rest of you should consider carefully reading what Dave, David Hoffer, Ira, myself and others are writing here and actually trying to understand the correct science.>>>

    Joel, I am going to take that as a compliment, and my hat is off to you, you and Dave Springer have been doing yeoman’s work in this thread. Your explanations have been bang on and I have to admit I’ve gotten a bit ill tempered in this particular topic of late. Your patience and clarity is appreciated to no end by me as a skeptic. Pardon me, a RAGING skeptic. For those skeptics who want to be part of winning the debate with the warmists, I have this suggestion:

    On the matter of “backradiation” and “greenhouse gases”, I recommend that you accept Joel Shore’s explanations at face value. CAGW is a farce of gigantic proportions, perpetrated by incompetant and/or fraudulent science and I have firm and specific reasons for saying that. Though a warmist he may be, I detect neither incompetance nor fraud in any of his explanations in this thread. And as I am certain Joel will tell you, if I did, he would hear about.

    Joel is explaining physics that is used and verified by design engineers by the thousands every single day, all over the world. As are Ira and Dave Springer and others. Some of the arguments that I’ve seen presented in this thread amount to no more than saying “the earth is flat, just LOOK at it”. Yup, I looked, and yup, it looks flat, and NOPE it isn’t.

  271. jae says:
    May 9, 2011 at 7:37 pm
    … “What I have argued here is that the most likely range of the climate sensitivity is what the IPCC says the most likely range is…which corresponds to feedbacks amplifying the radiative effect due to CO2 alone by a factor of about 3 +/- 1″

    WHAT??

    Maybe I misunderstand this?? Is an “ampliyfing factor of MINUS ONE actually an AMPLYFIYING FACTOR?? …

    jae, please! It is better to sit quietly in the corner and let some people think you are stupid, rather than open you mouth and prove it!

    Yes, you are missing something. Everyone but you knows that “3 +/- 1″ means a range from “3 + 1 = 4″ to “3 – 1 = 2″. In other words, the IPCC says the amplification is between 2 and 4.

    Joel accepts this IPCC estimated range. Some of us, including me, believe the amplification is less than 2, which would make the IPCC-specified range wrong.

    … blatant Orwellian Communistic propaganda crap from the “progressive” sickos….

    Now, please go back to the kiddie table or I’ll tell your mother and she’ll wash your mouth out with soap. :^)

  272. Boris Gimbarzevsky,

    Your reasoning from the lunar daytime temperature is wrong, that temperature has more to do with the heat capacity of the surface and the insulating properties of regolith. Even here on earth it gets hot enough on the hood of a dark colored automobile to fry eggs, even though the ambient temperature is much lower. See David Springer’s comment above for a better consideration of the lunar analogy.

    http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655571

    BTW, thanx again for your temperature readings. I think the ones at night will be the most interesting, especially if you can also report the humidity along with the ambient temperature.

  273. wayne says: May 8, 2011 at 3:01 pm

    “Now, ready? If back-radiation proponents are correct, we should be able to use this huge lens to focus back-radiation from any warm wall within your house, at the ambient room temperature of say 20 ºC (68 ºF), in some room to boil water in seconds.”

    No, Wayne, you are misunderstanding the physics involved. I can’t off-hand remember the name of the theorem involved , but you can’t focus the spot any hotter than the surface creating the light. (Similarly, the spot focused by a magnifying glass aimed at the sun can never be hotter/more intense than the light at the surface of the sun.)

    So no matter how thermal EM photons are focused, they can not make the surface they approach any warmer than the surface emitting the photons. The IR from the atmosphere could never (by itself) make the surface any warmer than the atmosphere. And the light from the sun could never make any part of the earth any warmer than 5700 K no matter how it was focused. Of course, the IR from the atmosphere PLUS some energy from the sun could warm the earth above the temperature of the atmosphere.

    (PS I don’t have time to address the other question of εσ( 288K^4 – 280K^4 ) vs. εσ( 288K^4 ) – εσ( 280K^4 ) .. i”ll have to see about doing that some other time.

    I’ll also have to see about addressing Jae’s concerns about Venus vs Earth vs Mars re the GH effect. Lapse rate is important, but I don’t believe it is the whole story — along the lines that some people have already mentioned.)

  274. JAE says:
    May 9, 2011 at 11:28 am
    the lapse rate depends upon only Cp and g. ??

    The Dry Adiabatic Lapse Rate (DALR) depends upon only Cp and g. The actual (environmental) lapse rate depends on other things and varies through out the day.

  275. ferd berple says:
    May 9, 2011 at 1:27 pm

    My question is why we assume that N2 in the atmosphere doesn’t back radiate

    Because it has been measured and N2 does not radiate at the temperatures normally found at the surface.

  276. wayne, I think you are right, I misread what you said.

    However, I stand by my comment that the temperature of the clouds has little effect on the surface temperature. On cloudy nights, the back radiation is almost the same as the radiation emitted from the surface. This is either because the clouds are the same temperature as the surface or because clouds simply scatter the surface radiation back toward the surface. Since clouds emit both up and down, they should be able to emit in each direction only half what they receive. In addition, since the water vapor mixing ratio generally decreases with altitude, if the surface temperature was the same as the cloud temperature, then there would be fog. Therefore, since the energy emitted toward the surface equals what is received, and since there is no fog, scattering makes more sense. Unfortunately, I have not found the measurements needed to be certain about this.

  277. Thanks to the following posters who talked about Latent Heat

    Dave in Delaware
    Nullius in Verba
    Bryan
    Dave Springer
    Martin Lewitt

    They’ve set me at ease for I now know I’m not the only one to have noticed that radiative models are missing a large piece of the puzzle.

    Even the climate keepers at Wikipedia have allowed it to give 23 % back to the atmosphere

    http://en.wikipedia.org/wiki/Earth's_energy_budget

    Sensible heat flux is mentioned too (convection) with 7 %.

    None of this has anything to do with greenhouse warming so comparing the blackbody example to earth now as a gauge of greenhouse warming is a pointless exercise.

  278. Joel Shore
    You challenged me to discus the science.
    I am here to do exactly that so you cant avoid the direct question posed below.
    If you prefer to avoid the science and instead issue smears instead then I will not be surprised, as insulting others, seems to be your default mode.

    Tell me which part of the query below is not science.
    This item is directly addressed to Ira’s post and deserves an answer from you!

    In your co-written Halpern et al comment paper you state that the Stephan Boltzmann Law can be used to work out the thermal energy exchange between atmospheric shells.
    There is a whole series of statements on pages 8,9,and 10(0r 1316 to 1318 of pdf) referring to the atmosphere, Stephan Boltzmann Law and indeed including the SB formula(constant x T^4).
    (The pdf below will allow readers to verify for themselves)

    However there is a major problem for you!

    A filtered spectrum does not vary with T^4.

    The Stephan Boltzmann Law cannot be used for the filtered spectrum of atmospheric gases.
    Do you now agree with me and withdraw your comments?
    Or will you persist in this unphysical assumption.

    Depending on your answer we can call on Ira’s spreadsheet to settle the matter.

    That is, to prove that the reduced Planck function does not vary with T^4.
    This has implications not just for your absurd paper but for ANY climate model with the T^4 assumption built in!

    Comment on ‘Falsification Of the atmospheric CO2 greenhouse effects within the frame Of Physics’ by Joshua B. Halpern, Chistopher M. Colose, Chris Ho-Stuart, Joel D. Shore, Arthur P. Smith, Jorg Zimmermann.

    http://scienceblogs.com/stoat/upload/2010/05/halpern_etal_2010.pdf

  279. Charlie Foxtrot, I saw Dave Springer’s comment and it may be that the average temperature of the moon is -23 C, but what I’m interested in is the immediate surface temperature of a planet, not an extremely low pass filtered temperature. I’ve worked in the NWT in areas of permafrost and it was very nice to be able to just dig a hole in the ground to keep our food cool when the outside temperature was 90 F during the longest days of the year. The temperature a couple of feet below the ground is below freezing but what we’re interested in is the surface temperature which is where we live. During a hot summer day I have to open all of my vehicle doors for 5 minutes before I can get into the vehicle and touch the steering wheel without getting burned. What I’m trying to get at is that the extremes of temperature get smoothed out by the low pass filter of the atmosphere and that heat transfer between sections of the atmosphere is probably far more important than any minimal effects of greenhouse gasses. The frequency response of the filters is what we’re interested in — I don’t care about the very long time constant of below ground temperatures when a relatively short pulse of heat will melt plastic items in my vehicle.

    As far as temperature of the chunk of air above my head goes, it seems to vary over a fairly small range. Just now stepped outside and found a minimum temperature of -12 F when the IR thermometer was vertical and humidity is given as 52% right now. Sky completely clear without a cloud to be seen and great night for stargazing but I have to work in a few hours:-(. When I’ve played around with this IR thermometer before I always seem to get a low value and it increases as one points the IR thermometer at angles of less than 90 degrees. If I was really obsessive about this I could mount the IR thermometer on a jig where I could precisely vary the angle and measure the temperature as a function of angle. I suspect this data could be used to compute a temperature profile of the atmosphere using similar math that one uses to generate an image with a CT scanner. The forecast is for rain tomorrow and will be interesting to see what kind of reading I get from the base of clouds.

  280. Wayne says:

    ‘Robert, I come up with a close conclusion, not exact, but close. However, the 3600 m is more than 36 times that is given in other conversations I have read speaking of it’s absorption very near the surface. Do you have an average absorption coefficient you are using, or a mean optical thickness for CO2 from which that is calculated? Maybe Hottel specifies emissivity direct then what equation are you using to get the 3600 m extinction distance?’

    Wayne,
    I read on a website (Gary Novak – Global Warming – not caused by carbon dioxide) that CO2 in air absorbs to extinction at its 15 micron peak in about 10m (Heinz Zug). I tried to confirm this figure using Plank – Hottel and got 3600m . I found that CO2 was too low at 0.038%. I used an absorptivity of 0.1905 for a hemispherical gas mass of radius L=3600m. Hottel gives CO2 emissivity direct from charts based on direct measurements of total emission. The precise method of calculating the effective absorptivity or emissivity is quite complex and is given in -(1) Trans. Am. Inst. Chem Engrs., 31, 517-549 (1935) Hottel, H.C., and H.G. Mangelsdorf. (2) Trans ASME, 57, 463-470 (1935) Hottel, H. C., and V. C. Smith.

  281. Wayne,

    When I said CO2 too low at 0.038% I meant too low for an extinction distance of 10m or so. However using the same technique (Planck- Hottel) for water vapour which has a partial pressure or concentration 100 x that of CO2 the extinction distance is much less at 120m (absorptivity 0.5734).

  282. Ferd berple says:
    ‘My question is why we assume that N2 in the atmosphere doesn’t back radiate. All objects radiate depending on their temperature. As the atmosphere is mostly N2, that must be the source of most of the back radiation’.

    Ferd

    The emissive power of N2 is small and insignificant compared with the emissive power of the entire spectrum of Earth radiation .

  283. David says: May 10, 2011 at 12:54 am

    …I now know I’m not the only one to have noticed that radiative models are missing a large piece of the puzzle [Latent Heat].

    David, I’m not sure why this came as a surprise to you. When trying to understand radiative effects (like Ira is doing here), naturally people concentrate on EM radiation. However, more advanced models include many more effects in much more detail.

    You note that “Even the climate keepers at Wikipedia have allowed it [latent heat] to give 23 % back to the atmosphere … Sensible heat flux is mentioned too (convection) with 7 %”. But, of course, those same terms were included in the Trenberth energy balance diagrams, so it is pretty clear that they have been considered important for a long time in all serious models.

    http://en.wikipedia.org/wiki/Global_circulation_model gives a quick overview of how the more advanced models work.

  284. Ira:

    Funny that you replied only to the comment which I already realized and acknowledged (next comment) was off-base. How about the other comments? Still that irritating silence from the warmers and luke-warmers.

  285. Robert says:

    “The Dry Adiabatic Lapse Rate (DALR) depends upon only Cp and g. The actual (environmental) lapse rate depends on other things and varies through out the day.”

    Yes. And your point is??

  286. Robert Clemenzi says:
    May 9, 2011 at 11:56 pm

    “Because it has been measured and N2 does not radiate at the temperatures normally found at the surface.”

    ALL matter with a temperature above absolute zero radiates. This is one of the most fundamental laws of physics. Either you are making things up out of thin air or the source you got it from doesn’t know his ass from elbow.

  287. wayne says: May 8, 2011 at 3:01 pm

    “Now, ready? If back-radiation proponents are correct, we should be able to use this huge lens to focus back-radiation from any warm wall within your house, at the ambient room temperature of say 20 ºC (68 ºF), in some room to boil water in seconds.”

    You can’t heat anything to a higher temperature than the temperature of the photons doing the heating. If you used a parabolic mirror aimed at a warm wall to heat water you could heat it to the temperature of the wall then the heating stops because the object being heated, now at the same temperature as the heat source, emits back as much radiation as it receives. It’s precisely because of back radiation that you can’t heat something hotter than the source of the heat.

  288. Retired Engineer says:
    May 8, 2011 at 8:22 am

    This makes sense. Absent CO2 and water vapor, we would freeze. How much warming does each contribute? If CO2 did it all, and mankind producses about 3% of the CO2, then we are responsible for 3% of the 33C warming or about 1C. Thus AGW is real. Except for water vapor which may cause 90% of the 33C, in which case we cause 0.1C warming.

    Things are more complicated than that. Water vapor is a condensing greenhouse gas. In frigid conditions most of it gets frozen out of the air. Antarctica has by far the dryest air on the planet’s surface. A second complication is that the IR absorption response of CO2 (and water vapor too) isn’t linear except at low concentrations (see my apple orchard vs. number of pickers in the field analogy) which is one of the surprising things John Tyndall discovered 150 years ago. Complicating things yet more is the albedo of water changes drastically between liquid and solid phase. The ocean absorbs almost all incident light while a glacier reflects almost all of it.

    So. Given the earth with an albedo similar to the moon (i.e. all rocks) it would have an average surface temperature well below freezing. But even your average rock is fairly dark and absorbs near 90% of incident light. But the earth is a water world so while the global ocean has an unfrozen surface it absorbs about 10% more light than rock. But what happens if the surface ever becomes almost completely frozen over? There would be hardly any water vapor in the atmosphere for a greenhouse effect and the snow & ice would reflect 90% of the sunlight. We would have a very frigid planet with no escape.

    But there is an escape hatch. If the earth is covered in snow & ice there are can be no green plants taking CO2 out of the atmosphere nor will there be any chemical processes (formation of carbonate compounds) removing CO2 from the atmosphere, nor will it be mixing with the ocean because the ocean is frozen over. In the meantime volcanoes will continue to vent CO2 into the atmosphere. CO2 level will keep on rising until the greenhouse effect from it kicks in and it starts melting ice first at the lowest latitudes and progressively higher latitudes. Water vapor again enters the atmosphere, ocean albedo changes from 90% to 1%, and the melt accelerates in one big hurry. Green plants bloom like crazy from the CO2-rich air/water and dissolved nutrients in the ocean accumulated over eons of nothing removing it.

    So basically CO2 greenhouse effect is what eventually stops a permanently frozen earth. Were we farther from the sun, like Mars, where it gets cold enough for CO2 to condense we wouldn’t have an escape hatch and the earth would be a cold dead rock much like Mars.

    Because CO2’s ability to absorb IR increases linearly at low concentrations (under 100ppm) a minimal amount even absent most water vapor serves to keep the earth just warm enough to prevent a snowball earth episode most of the time. It is somewhat controversial but it is generally believe that earth has had a few snowball episodes in the past and the CO2 hypothesis above is the only reasonable explanation for how it ever manages to melt once the ice takes over. The sun’s output has been gradually increasing over the billions of years and is about 10% higher now than in the distant past which is probably why there haven’t been any recent snowball episodes.

    My hypothesis taken from the above is that the first 100ppm of CO2 serves as a kind of kindling that fires up the water cycle i.e. it raises the average temperature just enough so that liquid water dominates frozen water. Evidently, at 280ppm, it’s just barely enough because the earth has been in an ice age for the past 3 million years with just some small cyclic variations in eccentricty & inclination being enough to transition between glacial periods and interglacial periods with the glacial ages lasting about 100,000 years and the interglacials about 15,000 years.

    Clearly the earth’s climate is at a tipping point but that tipping point is tipping back to a glacial period. Because CO2 at 280ppm is in the exponential part of its IR absorption curve it’s likely we can’t possibly enrich the CO2 content enough to end the 3 million year-old ice age but it would be great for all life on the planet if it did including us.

    In any case the indisputable testimony of the geologic column reveals that in the last 500 million years (since the Cambrian explosion when most of the modern phyla appeared and Ediacaran phyla disappeared) and animals started crawling out onto land the CO2 level in the atmosphere has, most of time, been 10 to 20 times greater concentration than today. Today’s atmosphere is CO2-depleted in historic terms and life is not nearly as abundant. The normal state of affairs for the earth is complete absence of any polar icecaps – it’s green from pole to pole. The living world today is a shadow of what it is during the good times with life struggling to survive in the nutrient-poor air and through the long cold winters. We should want to be moving away from the point where the climate tips back over into a glacial period. Evidently as a group we’re too stupid to know what’s good for us.

  289. Actually it appears that Mars too was once a liquid water world but it didn’t last long. Mar’s problem is insufficient gravity. Gravity is what gives us our air pressure of 14.7psi at sea level. This in turn raises the boiling point of water 212F above the freezing point. Mar’s lower gravity lowers the boiling point of water at the surface a lot so its oceans literally boiled away. For a time the excessive water vapor in the atmosphere would have given it an uber-greenhouse effect but slowly but surely it was electrolized by energetic radiation and blown off by solar wind – never to return. Mars is smaller than the earth so its liquid iron core cooled off a lot faster. A liquid iron core generates a magnetic field which deflects high energy radiation. The tipping point for Mars was probably when the iron core cooled off enough to solidify and its radiation shield vanished paving the way for it to be stripped of its water.

  290. >>Robert Clemenzi says: May 9, 2011 at 11:56 pm
    >>“Because it has been measured and N2 does not radiate at the
    >>temperatures normally found at the surface.”

    >Dave Springer says: May 10, 2011 at 6:32 am
    >ALL matter with a temperature above absolute zero radiates.
    >This is one of the most fundamental laws of physics. Either you
    >are making things up out of thin air or the source you got it from
    >doesn’t know his ass from elbow.

    I think the point is that N2 doesn’t radiate APPRECIABLE amounts of energy at typical atmospheric temperatures.

    The amount of energy radiated depends on the emissivity. Highly polished metal has an emissivity ~ 0.03 and will not radiate much energy. Rough, oxidized, dark surfaces often have an emissivity above 0.9. So it is easy for one material to emit 30x more radiation than another at the same temperature.

    It is easy to confirm that even many meters of N2 have little effect on IR – neither absorbing nor emitting appreciable amounts. The image at the top of the post — showing the IR spectrum looking up — shows quite clearly that for most wavelengths, there is almost no IR emitted by the atmosphere. The bulk of the IR that is seen is clearly attributed to H2O & CO2. Hence the entire column of N2 overhead emits no radiation (or close enough to zero for practical practical).

    The original statement is (for all practical purposes in climate modelling that I can imagine) correct.

    I presented evidence for my claim. Can you present evidence that shows IR emission from N2 is appreciable in any situations?

  291. David says:
    May 10, 2011 at 12:54 am
    Thanks to the following posters who talked about Latent Heat

    Dave in Delaware
    Nullius in Verba
    Bryan
    Dave Springer
    Martin Lewitt

    They’ve set me at ease for I now know I’m not the only one to have noticed that radiative models are missing a large piece of the puzzle.

    Even the climate keepers at Wikipedia have allowed it to give 23 % back to the atmosphere

    http://en.wikipedia.org/wiki/Earth's_energy_budget

    Sensible heat flux is mentioned too (convection) with 7 %.

    None of this has anything to do with greenhouse warming so comparing the blackbody example to earth now as a gauge of greenhouse warming is a pointless exercise.

    The toy global circulation models take latent heat flux into account so it isn’t missing although one might question the accuracy of the number or how it changes with temperature. The problem lies in how they model clouds. They believe that clouds have a net warming effect so as temperature rises from additional CO2 we get more clouds and more clouds cause even more warming. Clearly this is wrong because the end result is a runaway greenhouse and the earth has never in billions of years experienced a runaway greenhouse even when as recently as several million years ago CO2 levels were far higher than we could ever raise them through fossil fuel consumption. There just ain’t enough economically recoverable fossil fuel to get CO2 concentration into undiscovered territory. It’s questionable whether we can raise it high enough to cancel the effect of the Milankovich cycle which triggers glacial periods.

    Prior to 3 million years ago Milankovich cycles were not enough to trigger ice ages. It’s thought that the arrangement of continents are the other contributing factor. Depending on how the continents are arranged the global ocean conveyor belt changes and having a land mass over a pole blocks warm water from getting at the ice to melt it. Currently we have one large continent directly over a pole and one pole with none. The water sequestered at the south pole is a large amount which reduces the surface area of the global ocean and because land is higher albedo than water it raises the planetary albedo and thus cools it. If we had a continent over both poles at the same time we’d probably get a snowball earth episode that would last until either CO2 built up in the atmosphere to melt it or the continents drifted off the poles or some combination of both.

  292. CO2 extinction altitude can be seen in the spectrum looking down from 20 kilometers above the arctic ocean. IIRC the spectrum follows a 270K blackbody curve where the atmosphere is IR transparent i.e. it “sees” the temperature of the ocean surface but in the 15um region it drops down to follow a 250K blackbody curve. Going by dry adiabatic lapse rate (arctic air is fairly dry) of 1K per 100 meters the IR sensor is “seeing” the air temperature in the 15um range at a height of 2000 meters. I’m not an optics expert but I believe that altitude represents the optical depth of the atmosphere at 15um or in other words the extinction altitude. In any case there is no way in hell the extinction depth is in tens of meters.

    One of the concerns about increasing CO2 even if it doesn’t raise surface temperature all that much is that as it increases the extinction point will be lowered and near-surface adiabatic lapse rate will change as a result potentially causing changes in the weather but again I must go back to the indisputable testimony of the fossil record which reveals that the earth is increasingly friendly to life as CO2 rises even when it rises to 20 times the current level (greater than four doublings) so one might reasonably conclude it doesn’t cause climate disaster.

  293. Dave Springer:

    “ALL matter with a temperature above absolute zero radiates. This is one of the most fundamental laws of physics. Either you are making things up out of thin air or the source you got it from doesn’t know his ass from elbow.”

    Huh? I hope you are not saying that everything radiates in the IR frequencies. To repeat a comment I made above somewhere:

    http://www.wpi.edu/Academics/Depts/Chemistry/Courses/General/infrared.html

    The key paragraph:

    The fact that a molecule vibrates does not in itself insure that the molecule will exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the vibrational motion associated with that mode must produce a change in the dipole moment of the molecule. HCl, for example, with a center of positive charge at the H atom and a center of negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further, stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that cause changes in the value of the dipole moment from 0 to some non-zero value. An example is CO2, which has no permanent dipole moment because the individual bond dipoles exactly cancel. However, when CO2 undergoes a bending vibration, its dipole moment changes from zero to some non-zero value. This vibration produces a change in dipole moment and is IR active.”

  294. q = k A dT / s (1)

    q = σ T4 A (1)

    q = k A dT (1)
    Above from engineering tool box.

    Here are the basic heat transfer equations. Please note there are no inputs for back radiation. NONE. Your conclusion “The only rational explanation is the back-radiation from the Atmosphere to the Surface.” defies these equations.

    There are only two basic requirements for heat transfer 1. path 2. temperature difference. If either ar missing no transfer.

    I did not consider shape factor, emissivity differences, etc basics as you wanted to get back to.

  295. Martin Lewitt says:

    The turning of the water cycle is like a big air conditioner with the water as the working fluid. At these new levels of relative and absolute humidity associated with the warming the air conditioner or heat pump turns faster.

    Okay…Let me explain to you why I think this argument fails…at least from what I understand of the empirical data. Basically, you are saying that the evaporation / condensation increases, transporting more energy from the surface up into the atmosphere where it can be radiated back out into space. However, since the only way that energy can escape to space (in any significant amount) is via radiation and the amount of radiation depends on temperature, what this argument amounts to is the proposition that the upper portion of the atmosphere warms more than the surface.

    Now, to the extent that this is indeed expected to be true, it is already incorporated in the models via the lapse rate feedback (a negative feedback)…which says exactly this. The reason that the upper part of the troposphere is expected in the global average to warm more than the surface is that in the tropics is that one expects the lapse rate to closely follow the moist adiabatic lapse rate, which indeed implies more warming at altitude than at the surface. This is the so-called “hot spot” in the tropical troposphere.

    So, what you must be proposing is that this amplification of warming at altitude relative to the surface happens to a greater degree than the models predict. However, the data for the tropics currently show, if anything, just the opposite. I.e., people are complaining that the “hot spot” is missing. Whether it is really missing and the degree to which it is missing in fact depend on which analysis of the satellite data and radiosonde data one believes…and there are good reasons to believe there are problems with the data and that the models are basically correct. However, your picture would actually require that the “hot spot” really be more pronounced than the models predict, which seems rather unlikely based on what we do see in the data thus far.

    Alternately, I suppose you could argue that the models are correct (or overestimate the hot spot) for the tropics…but that they underestimate the warming at altitude (relative the surface) outside of the tropics by such a degree that the magnitude of the lapse rate feedback is still underestimated globally. However, as far as I know, the empirical satellite and radiosonde data, such as it is, don’t really support that idea either.

    So, that is my basic complaint about the argument that a faster water cycle somehow provides a negative feedback: To the extent that this expected to be true, it is incorporated into the models. If you want to argue that the models underestimate the effect, this has testable consequences for the temperature distribution in the atmosphere…And, at this point, the data for the temperature distribution do not seem to support this conclusion, if anything going the other way.

  296. mkelly,

    You needed to look a little farther and understand what you are reading. This is from the SAME source you quoted:

    Net Radiation Loss Rate

    If an hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as

    q = ε σ (Th^4 – Tc^4) Ac (3)

    http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html

    Please note there ARE inputs for back radiation (the Tc^4 term).

  297. Ref. mkelly:
    May 10, 2011 at 9:25 am

    There are no inputs for back radiation because it is taken care of by dT. No back radiation, then dT is simply the temperature of the emitter.

    It might be better to consider the back radiation from the atmosphere as a reduction in dT in order to use standard nomenclature. Using dT simplifies the calculation, esp. if emissivity is ignored. However, calling it back radiation clarifies the attribution to greenhouse gases, and allows the separate consideration of the back radiation from various concentrations of different gasses. Emissivity, of course, is a critical quality since it is this factor that makes CO2 and H2O significant contributors to the greenhouse effect, whereas N2 is not regardless of temperature (ignoring conduction, of course).

  298. Bryan says:

    A filtered spectrum does not vary with T^4.

    The Stephan Boltzmann Law cannot be used for the filtered spectrum of atmospheric gases.
    Do you now agree with me and withdraw your comments?
    Or will you persist in this unphysical assumption.

    This has implications not just for your absurd paper but for ANY climate model with the T^4 assumption built in!

    Okay, let’s try this one more time:

    When you choose what model you use, the model chosen is not independent of the purpose. If I were choosing a model to describe with as much quantitative fidelity as possible the greenhouse effect in the earth’s atmosphere, then the model I would choose would be a state-of-the-art convective-radiative transfer code using the actual composition and empirical absorption / emission lines for the atmospheric constituents. Such models most definitely do not just assume a the Stefan-Boltzmann Equation, as they do the calculation line-by-line (i.e., absorption line by absorption line).

    However, our purpose in our reply to G&T was not to model the greenhouse effect with as much quantitative fidelity as possible. Instead, it was just about as far as you could get from that in the other direction: We wanted to have the simplest models possible that illustrate the effect that G&T seemed to imply violated the 2nd Law of Thermodynamics, so simple in fact that nobody could seriously argue about whether or not we had solved them correctly because anybody could solve them on the back-of-an-envelope.

    So, that is what we came up with —A few very simple models, such as the one that involves 3 objects: one object A producing thermal energy and radiating energy at a fixed rate, two other objects B and C whose temperature is determined via radiative balance with object A and empty space, with a geometry such that the temperature of object B is higher than that of object C. And, what we wanted to illustrate is that the object C “warms” B in the colloquial sense of the word…i.e., that the presence of object C causes B to be at a higher temperature than if C is absent. By analogy, it then follows that if you choose A to be the sun, B to be the earth’s surface, and C to be the atmosphere, then one can immediately see the fallacy in claiming that “the greenhouse effect violates the 2nd Law because it implies heat flows from the colder atmosphere to the warmer surface” (that’s a paraphrase). In fact, the heat flows in our models are easy to calculate, so it is easy to verify that the 2nd Law is satisfied, but that, nonetheless, the presence of the colder object C causes the temperature of B to be warmer than in its absence.

    If you were worried about the issue of whether the T^4 dependence is critical in producing our result, it is not be hard to assume some more general dependence and show that as long as the amount radiated by the object increases with temperature, it doesn’t matter exactly what the temperature dependence is. Such are the advantages of simple models! (Note, by the way, that what is true for a radiating object is that the amount of radiation emitted AT ANY PARTICULAR WAVELENGTH is an increasing function of the temperature, a fact that is not always obvious because people often tend to normalize the emission curves when showing emission curves for different temperatures on the same graph.)

  299. Dave Springer says:

    The toy global circulation models take latent heat flux into account so it isn’t missing although one might question the accuracy of the number or how it changes with temperature. The problem lies in how they model clouds. They believe that clouds have a net warming effect so as temperature rises from additional CO2 we get more clouds and more clouds cause even more warming.

    This is not correct. The models…and empirical data…show that clouds have a combination of warming and cooling effects (depending on such details as the altitude and optical thickness of the clouds) but that the net effect of clouds on the earth is currently cooling. (I believe the magnitude quoted is on the order of 20 W/m^2.)

    Also, what the models show in terms of what happens with clouds is complicated. In general, to the extent that high clouds increase and low clouds decrease in the models, then the cloud feedback will be positive. You can find somewhere or other an analysis of the various effects due to clouds that are predicted by various models. They usually separate it out as the effect of the cloud feedback on shortwave radiation (i.e., radiation from the sun) and the effect of the cloud feedback on longwave radiation (i.e., radiation from the earth).

    Clearly this is wrong because the end result is a runaway greenhouse and the earth has never in billions of years experienced a runaway greenhouse even when as recently as several million years ago CO2 levels were far higher than we could ever raise them through fossil fuel consumption.

    A positive cloud feedback does not imply a runaway greenhouse effect. Ray Pierrehumbert would say that the reason why we cannot have a runaway effect on the earth with the sun at its current luminosity is actually quite well-understood. As I noted above, Hansen seems to feel differently, although the details of his reasoning remains obscure.

  300. R Stevenson says:
    May 10, 2011 at 1:58 am

    I tried to confirm this figure using Plank – Hottel and got 3600m

    To get a good figure, you must use the HITRAN data. At 15 cm-1, 99.7% is absorbed in less than 10 meters. At 14.97 cm-1, 99.7% is absorbed in less than 0.8 meters. Both of those measurements are with bandwidths of about 1 cm-1. With a bandwidth of 30 cm-1, 99.7% is absorbed in under 150 meters. To get 3.6 km you must be using a very wide bandwidth.

  301. [I'll just quickly add what I do know about why Hansen says that a runaway could potentially occur now while it hasn't in the past: One point he makes is that the sun has slowly gotten brighter, so CO2 levels back several hundred million years ago or more are not directly comparable to levels now with the current luminosity. Another point he makes regards the speed of the change in CO2 levels and that somehow overwhelming negative feedbacks that would otherwise occur if the change in CO2 levels happened over a longer timescale. I am not saying that he has compellingly shown that these two points materially change things, but just wanted to explain why he feels that something that never happened in the past could happen now.]

  302. Dave Springer says:
    May 10, 2011 at 8:55 am
    CO2 extinction altitude can be seen in the spectrum looking down from 20 kilometers above the arctic ocean. IIRC the spectrum……

    Dave Springer,

    What is IIRC?

    [Reply: IIRC generally means "if I remember/recall correctly," or similar. This is worth bookmarking. ~dbs, mod.]

  303. Tim Folkerts says:
    May 10, 2011 at 9:42 am
    mkelly,

    You needed to look a little farther and understand what you are reading. This is from the SAME source you quoted:

    Mr. Folkerts, you need to read what I wrote regarding the two conditions for heat transfer. Path and temperature difference or gradient. Temperature is not back radiation and trying to name it such does not make it so.

    Further please note I said “basic” per Ira’s want to go back to basics. Until you can show back radiation as an input for heat transfer then use the proper fromula and stop renaming things.

  304. Dave Springer you say at May 10, 2011 at 6:49 am

    “You can’t heat anything to a higher temperature than the temperature of the photons doing the heating. If you used a parabolic mirror aimed at a warm wall to heat water you could heat it to the temperature of the wall then the heating stops because the object being heated, now at the same temperature as the heat source, emits back as much radiation as it receives. It’s precisely because of back radiation that you can’t heat something hotter than the source of the heat.”
    ///////////////////////////////////////////
    Dave,
    What is the temperature of a CO2 re-radiated photon?

    What is the (average) temperature of the photons that supposedly downwell about
    330 W/m^2 (according to the Trenberth diagram)

  305. “”””” suricat says:
    May 9, 2011 at 6:58 pm
    Hi Ira.

    “Also, the Petty black body curves peak at a “Radiance” around 100 mW/m^2/sr cm^-1 while the black body curves from Carleton peak at an “Intensity” of around 14 W/m^2/mm.”

    As engineer to engineer. :)
    A “radiance” is a ‘comparison’ to an accepted ‘standard radiance’ (don’t ask me what frequencies this involves, as that’s defined by ‘the standard’), but an “intensity” is an ‘absolute measure’ of ‘energy’ (don’t ask me how the ‘Planck constant’ is considered there) and seems to be commonly used with a specific spectral wavelength. I hope this helps. “””””

    Well units of ” mW/m^2/sr cm^-1 ” are a little bit screwy; and need some brackets to properly set up.

    Izzat (mW) / (m^2/sr cm^-1); or is it (mW) /(m^2. sr. cm^-1). I think it is more likely to be (mW/cm^-1) / (m^2. sr). And the proper term is “spectral Radiance”, not “Radiance” because of that , /cm^-1 (per wave number increment) . And it represents Radiant energy in a small frequency (wave number) increment emitted from a small element of surface area (/m^2) into a small elemental solid angle (/sr) to some particular direction in space.

    And it is NOT equivalent to “Intensity” which is a property of a point source; so there is no per unit area of the source involved.

    We can reasonably talk about “intensity” when dealing with radiation for distant stars; but certainly not when referring to our local star. Well from a computational accuracy point of view, the measure of intensity is accurate to better than 1% so long as you are more than 10 emitting surface diameters away from a non-point source; but the point is there is NO per unit area section in an Intensity specification; but there is in Radiance, or Spectral Radiance, which these graphs properly are, and also in “emittance” which is simply W/m^2 without regard for directional properties, or wavelength or frequency properties, which would intorduce the spectral terms at least.

    “Radiance” without the “spectral” carries no information regarding frequency or wave number or wavelength or photon energy or any other such parameter, it simply measures total energy (Watts) emittted into some small solid angle (steradians) at any direction in space, from some small elemental surface area of the radiator. It’s photometric analog, for visible light conditions would be Luminance, or some like to use the term “Sterance” (not me). and “Brightness” would be a common colloquial and somewhat terrifying term, that we should all avoid like the plague; but we won’t, so we should only use it among friends; who can understand what it is we reqally mean. But the lay person will have no idea what exactly we mean.

    And the shift in the peak of the BB Spectral Radiance when we change from frequency or wave number over to a wavelength scale, is simply due to that switch from

  306. Robert Clemenzi says:

    To get a good figure, you must use the HITRAN data. At 15 cm-1, 99.7% is absorbed in less than 10 meters. At 14.97 cm-1, 99.7% is absorbed in less than 0.8 meters. Both of those measurements are with bandwidths of about 1 cm-1. With a bandwidth of 30 cm-1, 99.7% is absorbed in under 150 meters. To get 3.6 km you must be using a very wide bandwidth.

    Robert,

    I integrate between spectral band wavelengths of 12.5 to 16.5 microns for CO2.

  307. Joel Shore says:

    “…I am not saying that [Hansen] has compellingly shown that these two points materially change things, but just wanted to explain why he feels that something that never happened in the past could happen now.”

    If it were not for these endless “what if” scenarios, climate alarmists wouldn’t have much to say.

    And the models predicting a tropospheric “hot spot” have been falsified to the point that the backing and filling action now reverts to the stratosphere. But as Prof Richard Feynman makes crystal clear: if the models are contradicted by observations, the models are wrong.

    Joel Shore might want to take a few minutes off from writing his lengthy blog comments throughout the work day, and go look out the window. He would see that the climate is normal, and well within its past parameters. The models are wrong, AGW is wildly exaggerated, and the only effect from increased CO2 is on plant growth.

  308. Robert Clemenzi says:
    May 10, 2011 at 10:43 am

    R Stevenson says:
    May 10, 2011 at 1:58 am
    I tried to confirm this figure using Plank – Hottel and got 3600m

    To get a good figure, you must use the HITRAN data. At 15 cm-1, 99.7% is absorbed in less than 10 meters. At 14.97 cm-1, 99.7% is absorbed in less than 0.8 meters. Both of those measurements are with bandwidths of about 1 cm-1. With a bandwidth of 30 cm-1, 99.7% is absorbed in under 150 meters. To get 3.6 km you must be using a very wide bandwidth.
    —–
    Excuse me barging in but thanks Robert. I thought that very short mean path length strictly inc CO2’s frequencies was correct. Probably best just converted to tau which used the mean free path as the units maintaining the logrithmic nature of the absorbtion depths. Thanks again, that shores up everything I have been pointing out earlier in these threads in releation to that subject. If the 3600 m had been correct, I would have had to retract much I had claimed earlier, whew!

  309. mkelly,

    I’m not sure we are on the same wavelength. I agree with you that “path and temperature difference” are important parameters (especially for conduction). Thermal conductivity is also important, as is geometry. For radiation, emissivity is important, as are the specific temperatures (not just the difference). I think we are in agreement so far.

    You said “Until you can show back radiation as an input for heat transfer then use the proper formula and stop renaming things.”

    I believe this is indeed the proper formula (give or take a few details like you mention — eg the geometry of the objects and the emissivities of the two different materials.) (And this equation is actually the RATE of energy transfer, which would more commonly be labeled q/t)

    q = (ε σ Th^4 A) – ( ε σ Tc^4 A)
    or in words
    (net energy flow from radiation)
    = (“out radiation” energy from the object) – (“back radiation” energy to the object)

    I’m not renaming anything. “Back radiation” (which is a function of the temperature of the surroundings) IS an input for finding the heat transfer. It is 1/2 of the equation. I realize you said “basic”, but “basic” cannot involve dropping 1/2 of an equation where both halves are important.

  310. Joel Shore

    I said …” A filtered spectrum does not vary with T^4.
    The Stephan Boltzmann Law cannot be used for the filtered spectrum of atmospheric gases.”…..
    You now agree with me and withdraw this unphysical assumption.

    This has implications not just for your absurd paper but for ANY climate model with the T^4 assumption built in!

    Its of interest that you do not try to defend the Filtered T^4 atmosphere model that you presented in the article.
    Your talk of “simplification” and ” using heat in the colloquial sense of the word”.
    Your audience were physicists who must have cringed at the naive mistakes in your article.
    I would imagine that virtually all of them knew the thermodynamic meaning of heat.
    It is quite clear from your article that none of the Halpern et al team had any clue as to what heat meant.
    What was more disturbing is that none of the six of you could read properly.
    It is an almost unique example of group delusion!
    All of you came to the conclusion that G&T were of the opinion that colder objects could not radiate to hotter objects .
    Surely at this point one of you who actually read the paper should have said……
    “Hold on, there are a number of diagrams showing two way radiative interaction”.
    Another might have said “G&T have an extensive discussion on the Earths two way radiative response to the Sun”.

    Perhaps it is your view that your articles comments about HEAT and T^4 should not be taken seriously.
    I think your physics audience would take your remarks seriously.
    They would form a judgement on Halpern et al based on your comments.

    However from now on I will keep in mind that you should not be taken too seriously.

  311. Here is a thought experiment. If we hold everything else equal and double the amount of N2 in the atmosphere, because N2 does not participate in the back radiation, the surface equilibrium temperature should remain “unchanged” according to the radiative transfer model.

    However, comparisons of earth and venus suggest that the temperature of the atmosphere will only remain unchanged at 1 atmosphere pressure. As we now have more atmosphere, the pressure at the surface will be increased as compared to present. Under the gravitational model, the surface temperature would be expected to increase as a result of doubling the N2 in the atmosphere.

    Which of these two is correct? Adding N2 will not increase temperature, or adding N2 will increase temperature (holding all else equal)?

    The problem for me is the lapse rate. Doubling the N2 in the atmosphere will raise the height of the atmosphere. If the surface temperature does not increase, then the lapse rate dictates that since the atmosphere extends higher, that further up, past the present altitude it must get colder than it is at present. That seem nonsensical.

    It seems more likely that the top of the atmosphere will remain relatively unchanged, and due to the laspe rate the temperature at 1 atmosphere will be largely as it is now, and deeper down in the atmosphere towards the surface, temperatures will increase. This tells me that the surface temperature is a function of the amount of atmosphere, not (just) the specific gasses, as confirmed by comparison of the atmosphere of verus and earth.

  312. Ira,

    You haven’t answered so I’ll give it another try.

    Your attribution of the missing 33K from 240W radiation from earths system to “green house effect” assumes an atmosphere that is at 0K AND assumes no reflection (as I understand you, please correct me if I’m wrong). But even without GHGs the atmosphere must be rather warm (the dry adiabatic lapse rate is one lower limit, sensible heat and latent heat is another reason). The atmosphere is also reflective, Trenberth accounts close to 38% of longwave radiation from the ground to be reflected. Third, the atmosphere is heated by sunlight. Again according to Trenberth 78 W/m^2 is absorbed, i.e. 78 W/m^2 of SUNLIGHT is heating the atmosphere.

    Let’s assume (for the sake of argument) that the average optical depth is 7 km in the GHG free atmosphere (vs ca 5 km in our). DALR gives us a temperature of roughly 203 K at this point.

    If Trenberth is right on 38% reflection, 78 W is heating the atmosphere from above of which half will radiate downwards, and pure adiabatic (dry) lapse rate gives us 203K to begin with, we get 240W*(1-0,38)-78W/2- 5,67E-8*203^4=13,5 W.

    So, 13,5 W/m^2 is what is left to be explained by green house gases combined (if the above is correct). Do you disagree? Why so?

    Side note. Trenberths “Energy budget” http://mensch.org/5223/other/EarthsGlobalEnergyBudget.pdf has an implicit reflection of the atmosphere of 37% from below but 23% from above. Whenever I fly the clouds are bright white from above, but often dark from below. Should it not be the opposite of what Trenberth claims? Please enlighten me if you know…

  313. Ira, you say in your article above:
    “Since we humans sense the visual portion of this radiation as “light”, that is the name we give it, and that has led to the false assumption that it contains no “heat” (or “thermal”) energy.”
    You are making one assumption there that does not fit in with “Climate Science” at all. It has, ever since the days of Friedrich Wilhelm Herschel and Benjamin Franklin, been known to science that different colors absorb “solar light” at different rates.

    Having said all that, I too have been led to conclude that light, and indeed that all kinds of other radiation contains no heat. What light does contain however is “electronic energy”. Or to say it in a different way; light is always “on the move” and is transmitted as a part of the electro-magnetic wave-bands or spectrum, i.e. as radiation. – Only when that radiation is stopped or interrupted by “solids” i.e. atoms and/or molecules is that electronic energy either absorbed (transformed to kinetic energy) or reflected. (We can only see images as reflected light – or it’s source.)

    Kinetic energy is used to create molecular movements. – Molecular movements create friction. – Friction creates heat. – Heat can be measured as temperature. – So there it is – as far as I understand it. – Heat is the product of work done by energy provided by an energy source, in this case the Sun’s energy via the curtesy of radiation. Yes, I agree if there is radiation there must also be “back radiation” – Maybe I am not expressing myself very well, but I am hoping to keep my comment as short as possible.
    Furthermore, what I find curious with your “workings out” is the same as what I find unexplainable in any other “climate calculations” or models. They all “average” the solar W/m² input over the whole globe, which leads to an unavoidable bias in favor of solar irradiation. In your case, you say:
    “The Earth may be approximated as a sphere, with the Sun shining on only half of it at any given time. The adjustment factor for this correction is 0.25.”

    This 0.25 (or ¼) comes about because solar irradiation at the top of the atmosphere (TOA) which you say is 1366 W/m² will only retain it’s full force if it falls perpendicular onto a 2 dimensional disk. (Disk size does not matter as we are talking Watts for each individual, or any one square meter.) The difference between the square area of a disc and that of a sphere is ¼ – so if you divide 1366/4 you get 341.5, but the fact that the Sun is shining on only half of it at any given time, as you correctly state, is not taken into consideration. – According to what I once read about it in “Science of Doom” the thinking behind it was that if radiation out happens from the whole sphere’s surface then “to spread the solar constant” over the whole sphere seems –Fair enough.
    And maybe it is, – as the most likely probability is that nobody has a hope in hell of finding the exact average Wattage. But looking at it from the other angle: what if we averaged 0 solar radiation in (which is equally provided by the dark side? Then we would probably freeze to death before waking up in the morning. So of course if we have got sunshine 24/7 then we are going to get warmer! – CO2 or no CO2.

    Anyway, my fault is I waffle on, and on, and o – – – – .

    Your missing 33K could just be hiding in the old saying that this one is the “Goldilocks’ planet”. Your calculations only encompasses radiation in general and Infrared radiation in particular. It does not take into account anything else – like for example the Earth’s rotation, which is crucial if you wish to find the missing 33K. The Earth never warms “from scratch”, it keeps heat in during the night. In fact you would be hard pressed to register an overnight reduction in temperature in 70% of it (the oceans).

    Radiation my dear Ira is moving energy from one place to another – and back again – It may be a cause of “local warming – or cooling”, but is no more a cause of “global warming” than is conduction, convection and advection.

  314. My proof-reader has told me:
    Man, your last comment to Ira’s article is too heavy and therefore too dumb.

    In that case I shall just say; if you believe in AGW (or may it even just be a bit of warming by CO2), please show me an example where the product has reproduced its maker.

    Heat is the product of energy use, so please explain how heat can produce energy.

    If you believe CO2 is the producer of heat then please explain how CO2 is produced.

  315. Bryan says:

    [a bunch of nonsense and then...]

    However from now on I will keep in mind that you should not be taken too seriously.

    I would be perfectly happy if you were to just ignore me since you are clearly just interested in peddling your pseudo-science. There are other people here, such as David M Hoffer, who…while we may have strongly different opinions on the importance of AGW…do take me seriously and who I can have serious scientific discussions with. You are off in some make-believe land of your own devising, completely divorced from actual science…and in fact you are actively peddling pseudo-scientific nonsense that I think you are intelligent enough to know is such.

  316. George E. Smith says: May 10, 2011 at 11:28 am

    “””And the shift in the peak of the BB Spectral Radiance when we change from frequency or wave number over to a wavelength scale, is simply due to that switch from”””

    That looks to be an unfinished statement George. Would this finish it?

    “……, is simply due to that switch from” ‘its inverse form’.

    My time is short, so I thank you for your ‘expansion’ for the benefit of ‘Jo public’, but I think Ira understood the gist of my comment.

    Best regards, Ray Dart.

  317. JAE says:
    May 10, 2011 at 4:03 pm
    New calcs of radiative GHE shows the GHE is only 8-9 deg. C:

    http://activistteacher.blogspot.com/2011/05/radiation-physics-constraints-on-global.html

    //////////////////////////////////////
    Thanks for the heads up.
    The comment by Hans Shruder to that article is particularly interesting. I have repeatedly commented that we will be unable to understand how the atmosphere works as long as we continue to use average figures/average scenarios. This practice really disguises what is really going on. It will therefore be interesting to read the forthcoming paper to which Hans Shrudder refers.

  318. Further to my last post, I mispelt the name of the commentator. It should have read Hans Schreuder.

  319. Lots of discussion about the theory.

    But nobody is talking about what really happens on the surface of the Earth or what really happens up in the troposphere.

    The actual Earth surface and the troposphere does not react the way this theoritical framework indicates.

    Show me some actual radiation and temperature measurements of a location on Earth that meets this framework. Your backyard does not follow this framework – something else is happening.

  320. R Stevenson says:
    I integrate between spectral band wavelengths of 12.5 to 16.5 microns for CO2.

    From 12.5 to 13.3 microns, CO2 absorbs (and emits) almost no radiation. On the other side, the band goes to about 16.98 microns.

    When I integrate from 2 micron to infinity, assuming a constant pressure, 15c, and 1 km of thickness, CO2 alone absorbs about 73 of the available 390 W/m2. I suspect that there will always be a little more energy absorbed as the thickness is increased. However, once the amount remaining is “small enough”, it is common to consider the gas to be opaque. It is also important to note that about 54 of the 73 W/m2 overlaps with water vapor. Because of this overlap, doubling CO2 produces a net increase of only 4 W/m2 (according to my calculations).

  321. Smokey:

    And the models predicting a tropospheric “hot spot” have been falsified to the point that the backing and filling action now reverts to the stratosphere. But as Prof Richard Feynman makes crystal clear: if the models are contradicted by observations, the models are wrong.

    You might want to look at what Richard Lindzen has to say about this. Even you might admit that he knows a little bit more about atmospheric science than you do…and you might even believe him since he agrees with your ideologically-driven point-of-view on AGW in general. His wording that the data is wrong and the models / theory are right on this particular point is even stronger than mine ( wattsupwiththat.com/2011/01/17/richard-lindzen-a-case-against-precipitous-climate-action/ ):

    The dominant role of cumulus convection in the tropics requires that temperature approximately follow what is called a moist adiabatic profile. This requires that warming in the tropical upper troposphere be 2-3 times greater than at the surface. Indeed, all models do show this, but the data doesn’t and this means that something is wrong with the data.

    (The one difference between Lindzen and the consensus view is that Lindzen seems to think that the tropical surface data is suspect whereas the more standard view is that the problems are most likely mainly with the radiosonde & satellite data / analysis.)

  322. Phil: Maybe so, maybe not. You need to EXPLAIN WHY you disagree. With some kind of evidence, links, etc. Sorry, but I don’t know you and cannot take your word on this.

  323. Joel Shore,

    “So, that is my basic complaint about the argument that a faster water cycle somehow provides a negative feedback: To the extent that this expected to be true, it is incorporated into the models. If you want to argue that the models underestimate the effect, this has testable consequences for the temperature distribution in the atmosphere…And, at this point, the data for the temperature distribution do not seem to support this conclusion, if anything going the other way.”

    I mentioned the Wentz 2007 article in the journal Science that showed none of the models produced more than half the increase in precipitation seen in the observations, so how can you say the speedup is “incorporated into the models”. The presence or absence of a hot spot was never going to be decisive, even if it was as prominent as the models had it, it wouldn’t indicate whether feedback was positive or negative. The modelers were just scrambling to see the presence of something they claimed would be prominent. The lack of the temperature profile that you claim is expected from more turnover of the water cycle, can possibly be explained by the models under representing the efficiency with which greenhouse gases cool the upper troposphere, there is already good evidence that the stratosphere is cooler, that is attributed to this mechanism. But it is the models that aren’t matching the most basic observations of a faster water cycle.

  324. When Joel Shore says: “your ideologically-driven point-of-view on AGW,” you can be sure it’s merely Joel’s psychological projection. I personally have no such ‘ideological’ view. I simply point out the fact that there is zero evidence of global harm due to CO2, and that CO2 enhances plant growth; therefore CO2 is harmless and beneficial.

    Joel Shore always avoids trying to show evidence of global CO2 damage, because there is none. He is blinded by his own ideology. Alarmists like Shore have the onus of showing that their beliefs in CAGW are supported by empirical, testable evidence. Since they have failed, they project their own faults onto scientific skeptics.

    Someone please wake me when the alarmist crowd begins to follow the scientific method.

  325. JAE,

    Here is one specific flaw in http://activistteacher.blogspot.com/2011/05/radiation-physics-constraints-on-global.html

    But in fact, the evaluation of a 33 degree warming is in serious error. It is not physically reasonable to assume the true value of Earth’s average emissivity {ε} to be equal to 1. Indeed, it is physically impossible from usual known radiative processes for any opaque body to have a surface albedo (reflectivity) as high as 0.30 yet have an emissivity of 1. An ideal black body has a reflectivity of zero.

    This is not an elusive point in radiation physics. Emissivity has been studied since before the development of quantum mechanics and there exists a first-empirically-established relation between reflectivity (albedo) and emissivity. It is known as Kirchoff’s Law of radiation physics. Basically, it says that the larger the reflectivity the smaller the emissivity, in approximate linear relation to each other.

    As an equation, in this context, Kirchoff’s Law is:

    1 – {a} = {ε}. … (eq.5)

    It is a violation of Kirchoff’s Law to admit {a} = 0.30 yet use {ε} = 1 in calculating the Earth’s surface temperature, irrespective of the assumed amount of greenhouse effect.

    This reference (and many others like it) show that white paint can have an emissivity of ~ 0.9 for thermal IR, but white paint will have an albedo of ~ 0.9 for visible light. Here is empirical evidence that his interpretation of Kirchhoff’s law is wrong because absorption can and does vary with wavelength.

    {a} are weighted averages, where the weightings are completely different. There is no reason to expect the two values to be closely related.

  326. That last sentence should have been:
    {e} and {a} are weighted averages, where the weightings are completely different. There is no reason to expect the two values to be closely related.

  327. Dave Springer,

    High temperatures are achievable with infrared lasers:

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

    A high enough concentration of photons of any wavelength can cause heating. I think however, there are practical limitations to being able to perform Fresnel lens heating from wall radiation.

  328. jae says:
    May 10, 2011 at 6:46 pm
    Phil: Maybe so, maybe not. You need to EXPLAIN WHY you disagree. With some kind of evidence, links, etc. Sorry, but I don’t know you and cannot take your word on this.

    But you accept the unsupported statements in that paper?
    Try this for size:

    http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/albedo.html

    Includes this “The bond albedo is the total radiation reflected from an object compared to the total incident radiation from the Sun. The bond albedo for the Earth is given as 0.29 by de Pater and Lissauer, ”

    The Activist teacher paper asserts: “It is a violation of Kirchoff’s Law to admit {a} = 0.30 yet use {ε} = 1 in calculating the Earth’s surface temperature, irrespective of the assumed amount of greenhouse effect.”
    However as shown above the Bond albedo applies for the range of solar wavelengths, i.e. up to ~5 microns. It is therefore inappropriate to apply Kirchoff’s law to determine {ε} for the surface emission range (~5 – 50 microns), the value of 0.7 that Rancort derives is applicable to the UV to 5 microns range, not to the IR.

  329. Joel Shore;
    Hope you are still following this thread, could you hit my blog or otherwise provide your email address? I’ve got something I’d like to run by you.

  330. Ira Glickstein, PhD Reur May 9, 2011 at 9:01 pm

    You should retract your over-the-top insults to JAE; they give you a negative persona. You should have been able to work-out that JAE made a simple mistake, which BTW he shortly realized and corrected just before your rant. An apology is in order.

    I can only conclude that his suggestion that radiative hypotheses are the lesser in importance than other thermodynamics is inconvenient for you to discuss. I’ve followed JAE for several years and find his research and thinking on convection and thermalization etc to be of great interest. BTW Ira, HEAT is a different form of energy to EMR, regardless of wavelength. (the popular restriction of “thermal EMR” to IR is plain nonsense)

    JAE:
    You may be interested to know that I had an interesting Email exchange with Roy Spencer a while back where I asked in part if in a nominally warming world there would possibly be increased evapo-transpiration. If so, according to Trenberth, given that this is the largest HEAT loss from the surface, there ought be increased surface cooling, or in other words, a negative feedback. Roy eventually withdrew, but to summarise, he admitted that what he collectively called convection was indeed important, but that everyone is too busy working on the radiative stuff. (including himself)
    Ho hum.

  331. JAE,
    Further my post above, where I wrote:
    …everyone is too busy working on the radiative stuff…
    That might also translate to:
    “everyone is too busy competing on the radiative stuff”
    (including Roy Spencer)

  332. Joel Shore

    I said ….. ” However from now on I will keep in mind that you should not be taken too seriously. ”
    You replied
    …….”I would be perfectly happy if you were to just ignore me” …….

    If you wish to be taken seriously then avoid distorting words like HEAT which is inexcusable for someone with a Physics degree.

    When you do that your give the impression that;

    1. Your version of the IPCC position requires this distortion to be convincing.
    2. Or perhaps you really don’t know what your talking about.

    I am personally much more likely to be persuaded by the likes of Nick Stokes or Rodrigo Caballero who can present the IPCC case without departing from the traditional language of Clausius and Feynman.
    Read through Rodrigo Caballero’s Lecture Notes to see what I mean.

    http://maths.ucd.ie/met/msc/PhysMet/PhysMetLectNotes.pdf

    No wonder an educated person like Professor Gerlach “blew his top” when he found your paper full of elementary mistakes.

  333. Robert Clemenzi says:
    May 10, 2011 at 5:39 pm

    When I integrate from 2 micron to infinity, assuming a constant pressure, 15c, and 1 km of thickness, CO2 alone absorbs about 73 of the available 390 W/m2. I suspect that there will always be a little more energy absorbed as the thickness is

    Robert,

    We seem to be in close agreement on my spreadsheet model when I integrate between 0.5 microns and infinity at 15 C etc., CO2 absorbs 25.3 Btu/hft^2 (79.8W/m^2) of the available 124 Btu/hft^2 (391W/m^2) ie 20%.
    For water vapour alone, my spreadsheet integration gives 248W/m^2 absorbed of 391W/m^2 available or 63%.
    Using charts developed by Hottel I obtain absorptivities giving extinction distances of 3000 to 4000m for CO2 at 380ppm and 2000m when CO2 is doubled. Your extinction distance is less at 1km.

  334. Robert Clemenzi

    Further to above on my spreadsheet 97% of the absorbable IR is in 15micron band for CO2.

  335. I am impressed with Ira Glickstein’s post, and his clear willingness to share his logic and computations in order to invite criticism, correction, and further development. That speaks to me of scientific integrity, and it is refreshing to encounter it in the climate debate which is often dominated by what might be, at best, called ‘courtroom integrity’ in which antagonists vie with each other to present watertight cases immune to argument and contradiction.

    The whole idea of dealing with global mean temperatures and averaged-out energy budgets is by itself crude, and one commenter at least has noted that when an observed global warming of the order of 1C is small compared to the coarseness and sensitivity of such back of an envelope calculations, we need to look elsewhere to resolve disputes.

    I presume the answer lies in admitting more of the complexity of real case into the computations: if not the spinning, irregularly surfaced sphere, then at least the huge differential in solar heating ‘twixt the equatorial and the polar regions, the great daily poleward energy transfers which compensate thanks in large part to massive convective systems. I am with the commenter who sees the atmosphere primarily as a coolant – taking heat out of the tropical surface, and helping it escape to the temperate/polar zones and to space. Surely in there somewhere lies the possibility of crucial hypotheses to help clarify our differences of opinion and be capable of observational test/refutation/confirmation. The effects of additional CO2 seem, to me, to be likely to be so small that such hypotheses will be for correspondingly small, probably only regional or local, effects that might yet be measurable. I presume further that we would benefit from increased measurement of radiation environments and temperatures at various heights, coupled with measurements of moisture, cloud, and CO2 levels. A few years of good data there would, it seems to me (a mere outside observer of this science), produce a productive environment for scientific progress.

  336. Dave Springer says:
    May 10, 2011 at 8:55 am
    CO2 extinction altitude can be seen in the spectrum looking down from 20 kilometers above the arctic ocean. IIRC the spectrum follows a 270K blackbody curve where the atmosphere is IR transparent i.e. it “sees” the temperature of the ocean surface but in the 15um region it drops down to follow a 250K blackbody curve. Going by dry adiabatic lapse rate (arctic air is fairly dry) of 1K per 100 meters the IR sensor is “seeing” the air temperature in the 15um range at a height of 2000 meters. I’m not an optics expert but I believe that altitude represents the optical depth of the atmosphere at 15um or in other words the extinction altitude. In any case there is no way in hell the extinction depth is in tens of meters.

    That is the best contribution so far

  337. JAE says:
    May 10, 2011 at 8:56 am

    “Huh? I hope you are not saying that everything radiates in the IR frequencies. To repeat a comment I made above somewhere:”

    Solids, liquids, and dense gases emit continous blackbody spectra characteristic of their temperature. We were talking about atmospheric nitrogen which is a dense gas and its temperature, at least in the lower and middle atmosphere, puts its blackbody emission spectrum in the far infrared.

    What part of that don’t you understand?

  338. Smokey says:

    When Joel Shore says: “your ideologically-driven point-of-view on AGW,” you can be sure it’s merely Joel’s psychological projection …

    I just want to point out for the benefit of the readers what has transpired here. For the last several months, Smokey (often quoting me out of context) has hounded me for saying that I thought it was more likely that the models are correct and the data incorrect in regard to tropical tropospheric amplification.

    Now that it has been pointed out to him in my previous comment http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-657690 that Richard Lindzen says the same thing…except in even more unambiguous terms (stating flat-out that the data must be wrong), one might expect that a person with decent standards of personal responsibility would either

    (1) Apologize to me.

    (2) Rebuke Lindzen in equally-harsh words as he has rebuked me.

    What has Smokey done instead? He has avoided the issue entirely.

  339. Some links to comments on this thread which encouraged or informed my earlier one (3:07am)

    (1) ‘Too simple’ models:
    “This goes to the heart of the problem. Anthropogenic effects are so tiny as to easily buried in the noise of the real world and in the error margins of all these toy models of the real world.” Dave Springer May 8 8:08am
    Anthony Zeeman May 8 10:58am
    Roger Sowell May 8 4:02pm
    David May 10 12:54am
    richard verney May 10 4:03pm
    (2) Coolants:
    Richard M May 9 4:16pm
    Boris Gimbarzevsky May 9 8:38pm
    Radiation and/or temperatures at different heights and contexts:
    wayne May 7 11:56pm
    Bryan May 8 1:13am
    Martin Lewitt May 9 3:02am
    richard verney May 9 9:53am
    Robert Clemenzi May 10 12:25am
    Boris Gimbarzevsky May 10 1:51am
    (3) Importance of convection:
    Cematafried May 8 5:55am
    Leanard Weistein May 8 6:36am
    Dave in Delaware May 8 7:03am
    Nullius in Verba May 8 7:06am and May 8 12:53pm
    Alistair May 8 7:56am
    richard verney May 9 9:53am
    Dave Springer May 9 9:23am
    (4) meridional contrasts / transfers:
    davidmhoffer May 8 3:05am
    Gilles May 8 10:02am

  340. Bryan says:

    When you do that your give the impression that;

    1. Your version of the IPCC position requires this distortion to be convincing.
    2. Or perhaps you really don’t know what your talking about.

    In regards to our paper, you (and G&T) seem to be the only ones who are incapable of doing simple word replacements in a few places that have been described to you in detail. Besides which, I have been more careful in these threads to use the word “heat” more precisely; however, it seems that people peddling pseudoscience not only get inordinately hung up on word choices but then also don’t accept the concept of corrections.

    No wonder an educated person like Professor Gerlach “blew his top” when he found your paper full of elementary mistakes.

    While any actual physicists reading our paper may say we should have been a little more precise in our usage of the word “heat” in a few places, they will consider this to be “going a few miles above the speed limit” in comparison to Gerlich and Tscheuschner’s “serial killing”, i.e., huge scientific blunders in regards to the 2nd Law of Thermodynamics and the greenhouse effect. It is only people peddling pseudoscience who purposely try to misinterpret what other people are saying in order to confuse others.

  341. Typo: Need (3) added to give
    ‘(3) Radiation and/or temperatures ate different heights and contexts:’

    and then existing (3) -> (4), and existing (4) -> (5).

  342. John Shade says:
    May 11, 2011 at 3:07 am

    “I am with the commenter who sees the atmosphere primarily as a coolant – taking heat out of the tropical surface, and helping it escape to the temperate/polar zones and to space.”

    I try to make analogies to help people (myself included) understand things. In other commentary on WUWT I compared the poles to the radiator on an automobile. The equator is the engine. When we have a continent over a pole that’s like a blockage in the radiator limiting how well it can dump the heat (south pole today). When we have continents surrounding a pole with restricted openings that also, to a lesser degree, also limits the effectiveness of the radiator.

    I think heat transport in the atmosphere is almost all vertical with evaporation, convection, and condensation being the major players there.

    As far as heat capacity goes the ocean has roughly 1000 times as much as the atmosphere. So that relationship is akin to a dog and its tail. The tail doesn’t wag the dog of course. However if the dog is sleeping a tug on its tail can wake it up. The climate boffins I think are worried that anthropogenic CO2 is tugging the tail on a sleeping dog and believe it’s always best to let sleeping dogs lie. Or maybe it’s more like a tiger to them. :-)

  343. ferd berple says:
    May 10, 2011 at 1:43 pm
    Here is a thought experiment. If we hold everything else equal and double the amount of N2 in the atmosphere, because N2 does not participate in the back radiation, the surface equilibrium temperature should remain “unchanged” according to the radiative transfer model.

    However, comparisons of earth and venus suggest that the temperature of the atmosphere will only remain unchanged at 1 atmosphere pressure. As we now have more atmosphere, the pressure at the surface will be increased as compared to present. Under the gravitational model, the surface temperature would be expected to increase as a result of doubling the N2 in the atmosphere.

    Which of these two is correct? Adding N2 will not increase temperature, or adding N2 will increase temperature (holding all else equal)?

    Adding N2 will not increase temperature since it will have no impact on the radiative emission to space.

    The problem for me is the lapse rate. Doubling the N2 in the atmosphere will raise the height of the atmosphere. If the surface temperature does not increase, then the lapse rate dictates that since the atmosphere extends higher, that further up, past the present altitude it must get colder than it is at present. That seem nonsensical.

    Your assumption of a continuation of the lapse rate is flawed, the tropopause is the region where the positive lapse rate of the troposphere changes to the negative lapse rate of the stratosphere (heated from above). I see no reason why addition of N2 would change that very much.

  344. Someone hand Joel Shore a hanky.

    Blogrule #1: Never demand an apology. It is pitiful and impotent.

    Joel’s begging for an apology for his hurt feelings is simply his tactic to avoid answering the points I raised above. Here is the comment I made, which he still avoids answering:

    When Joel Shore says: “your ideologically-driven point-of-view on AGW,” you can be sure it’s merely Joel’s psychological projection. I personally have no such ‘ideological’ view. I simply pointed out the fact that there is zero evidence of global harm due to CO2, and that CO2 enhances plant growth; therefore CO2 is harmless and beneficial.

    Joel Shore always avoids trying to show evidence of global CO2 damage, because there is none. He is blinded by his own ideology. Alarmists like Shore have the onus of showing that their beliefs in CAGW are supported by empirical, testable evidence. Since they have failed, they project their own faults onto scientific skeptics.

    What is untrue about that comment? Truth doesn’t require apologies. I regularly ask Joel Shore and others to produce verifiable evidence of global damage caused by the rise in CO2. But rather than address that point, Joel has repeatedly accused me of being ideological; I am not. He is deliberately avoiding the central question in the whole trumped-up “carbon” debate.

    Joel Shore is wrong; I am only ‘biased’ against alarmist scientists feeding at the public trough, and who are trying to scare the public by spending a large part of their workdays wasting our tax money by writing long comments on blogs, instead of doing what they are paid to do.

    I am sorry Mr Shore’s boss allows him to misappropriate public funds in this manner. Gavin Scmidt and James Hansen have set a terrible precedent in this regard. There is no doubt that if someone like John Christy began blogging throughout the workday like Joel Shore does, the screams of outrage from the hypocritical alarmists would be deafening.

    If it is ‘hounding’ Joel Shore for any putative real world, testable evidence showing that rising CO2 – a tiny trace gas – causes measurable harm to the planet, he needs to stop prevaricating and produce it, or take being ‘hounded’ until he produces evidence, or admits that there is none. Calling others ideologically biased is just Joel Shore’s way of avoiding the question.

  345. I agree with much of what Dave Springer says, but I can’t agree with some of his conclusions about optical depth.

    Dave Springer says: May 10, 2011 at 8:55 am
    > IIRC the spectrum follows a 270K blackbody curve where the atmosphere is
    >IR transparent i.e. it “sees” the temperature of the ocean surface but in the
    >15um region it drops down to follow a 250K blackbody curve.

    From the graph at the top, the temperatures are closer to 265 and 225 K. That gives a difference of 40 K, not 20 K. Not a huge change, but significant.

    > Going by dry adiabatic lapse rate (arctic air is fairly dry) of 1K per 100 meters

    I would love a little more info here. The key is the relative humidity. Certainly cold arctic air has a very low absolute humidity, but is the relative humidity low?

    The observed environmental lapse rate is closer to 0.65 K/100 m (speaking in general — I don’t know details specifically for the arctic), and the temperature difference is closer 40 which would put the altitude closer to 40K / (0.65 K/100m) = 6000 m. So we have a very large uncertainty in the altitude here.

    >the IR sensor is “seeing” the air temperature in the 15um range at a height of
    >2000 meters. I’m not an optics expert but I believe that altitude represents
    >the optical depth of the atmosphere at 15um.

    I would not define the optical depth the way you seem to be doing.

    Optical depth, or optical thickness is a measure of transparency, and is defined as the negative logarithm of the fraction of radiation (e.g., light) that is not scattered or absorbed on a path. The optical depth is a measure of the proportion of radiation absorbed or scattered along a path through a partially transparent medium…
    Wikipedia

    In other words, of a region of the atmosphere has an optical depth of 1, then 70% of the incoming light has been blocked. (Or equivalently, 70% of the light we see would come from that region). If most of the light can pass from 2000 m up out of the atmosphere (passing thru well over 1/2 the total mass of the atmosphere) then, then most of the light passing up from the surface could presumable ALSO pass thru well over half the atmosphere. I.e. the optical depth looking up would be well over 2000 m. However as the altitude gets higher, the optical depth gets smaller.

    There is, however, another issue here. All we really know is that the photons come from a region where the temperature is about 220 K. First, I put the altitude closer to 6000 m, not 2000 m. But the temperature of 220 K is pretty close to the temperature of the tropopause, which typically extends for several km upward. The photons could come from pretty much anywhere over a range of several km of altitude and still appear to be ~ 220 K. The photons might be traveling a relatively short distance — for example, from space down to just 15 km altitude. So a very tiny part of the atmosphere could stop most of the photons — so a relatively short distance near the surface could stop the photons. We just can’t know very accurately using this method.

    (If someone has more details about the temperature and altitude range of the arctic tropopause, I would love to hear it.)

    >In any case there is no way in hell the extinction depth is in tens of meters.

    After all that, the simper way to determine optical depth near the surface is to look UP. From the graph at the top, the “temperature” of the 15 um IR radiation when looking up is practically the same as the temperature of the surface (certainly no more than 5 K different). The photons must come from no more than 500 m up (using the lapse rate argument), so the optical depth near the surface is well below 500 m.

    Whether the optical depth for 15 um photons near the surface is 3 m or 30 m or 300 m, I can’t tell from just this. But there is no way in hell it could be 2000 m.

  346. Dave Springer says:
    May 11, 2011 at 5:06 am
    JAE says:
    May 10, 2011 at 8:56 am

    “Huh? I hope you are not saying that everything radiates in the IR frequencies. To repeat a comment I made above somewhere:”

    Solids, liquids, and dense gases emit continous blackbody spectra characteristic of their temperature. We were talking about atmospheric nitrogen which is a dense gas and its temperature, at least in the lower and middle atmosphere, puts its blackbody emission spectrum in the far infrared.

    As far as blackbody radiation is concerned N2 is not a dense gas in the earth’s atmosphere and doesn’t emit as a blackbody. If it were a blackbody it would emit in the far IR.

  347. Heh. Come on, folks, I am not supporting the teacher’s different methods of calculatig the GHE; I only pointed it out. Don’t shoot the messenger. I have a strong hunch that NONE of these radiative calcs mean anything, anyway, because any of the effects are instantly overwhelmed by other phenomenon, such as convection and water evaporation/condensation.

    In that vein, I’m still really interested why there is so much silence surrounding this (and other related expositions):

    http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html

  348. Martin Lewitt says:
    May 10, 2011 at 8:28 pm

    Dave Springer,

    High temperatures are achievable with infrared lasers:

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

    A high enough concentration of photons of any wavelength can cause heating. I think however, there are practical limitations to being able to perform Fresnel lens heating from wall radiation.

    I’ll need to think about that some more. Shiva operates in the very near infrared (1um) which correlates to a temperature in the thousands of degrees. Moreover, it focuses 20 beams coming from different directions on its target and achieves heating to millions of degrees through compression (shock waves).

    Perhaps a more familiar technology is the microwave oven. Exceedingly cold matter emits blackbody spectra in the microwave region of the spectrum. The cosmic microwave background radiation is 3K for instance. That then raises the question of how a microwave oven can heat up a cup of coffee. It isn’t accomplished by microwave photons per se but rather by phase variation which causes dialectric molecules (primarily water in this case) to constantly re-orient with the magnetic phase. Friction then does the actual heating because microwave radiation in and of itself can’t raise the temperature of anything that isn’t within a couple degrees of absolute zero.

  349. jae says:
    May 11, 2011 at 6:30 am
    Heh. Come on, folks, I am not supporting the teacher’s different methods of calculatig the GHE; I only pointed it out. Don’t shoot the messenger.

    You asked my reasons for saying that his method was wrong, I gave them, that’s all.

  350. Phil. says:
    May 11, 2011 at 6:23 am

    “As far as blackbody radiation is concerned N2 is not a dense gas in the earth’s atmosphere and doesn’t emit as a blackbody. If it were a blackbody it would emit in the far IR.”

    A gas at 1 bar isn’t dense? In general a gas is considered dense if the molecules are close enough to each other to conduct heat. Nitrogen at 1 bar will certainly conduct heat. Try again.

  351. Joel quoted Richard Lindzen:

    “The dominant role of cumulus convection in the tropics requires that temperature approximately follow what is called a moist adiabatic profile. This requires that warming in the tropical upper troposphere be 2-3 times greater than at the surface. Indeed, all models do show this, but the data doesn’t and this means that something is wrong with the data.”

    I’m sorry Joel, but you need to get your sarcasm detector seen too. Lindzen’s tongue was so far in his cheek it should have been clearly visible to everyone from at least a mile away.

  352. Smokey says:

    Joel’s begging for an apology for his hurt feelings is simply his tactic to avoid answering the points I raised above.

    No…It is you who are desperately trying to avoid the fact that you have spent months attacking me (including in this thread) for a statement that Richard Lindzen makes in even more extreme terms than I do. That you are unwilling to address this when called out on it shows a completely lack of ethics and personal accountability on your part.

    As for what you ask: I have explained this stuff many times. It is also explained in many reports by reputable scientific organizations. I can’t convince someone who is unconvinceable.

  353. Joel Shore

    I said…… “No wonder an educated person like Professor Gerlach “blew his top” when he found your paper full of elementary mistakes.”……

    You said …”While any actual physicists reading our paper may say we should have been a little more precise in our usage of the word “heat” in a few places”…

    You have a particularly snide way of phrasing your replies.
    Are you here implying that Professor Gerlach is not a physicist?
    That a Professor from a German University is an impostor perhaps!
    That’s very impertinent from someone who finds coping with reading a big problem.
    For someone who is a bit “rusty” in the thermodynamics department yet was happy to add his name to a collection of elementary gaffs.
    Its too late now to withdraw you name from the Halpern et al embarrassment.
    I could recommend some useful textbooks to bring you back up to speed with Physics.

  354. David Springer, you need to slow down a little!

    “A gas at 1 bar isn’t dense? In general a gas is considered dense if the molecules are close enough to each other to conduct heat. “

    No. A gas is considered “optically dense” if it absorbs most of the photons at that wavelength.

    In spectroscopy, the absorbance A (also called optical density)[2][3] is defined as:[4]

    A_\lambda = \log_{10}(I_0/I)\,,

    where I is the intensity of light at a specified wavelength λ that has passed through a sample …
    Wikipedia

    N2 does not absorb much IR at any wavelength through earth’s entire atmosphere, so the entire depth of N2 in earth’s atmosphere is not considered “optically dense”. You have already been corrected once on this. You might consider doing a little research before being so sure of your own knowledge.

    In another post David discusses microwaves.
    “Perhaps a more familiar technology is the microwave oven. Exceedingly cold matter emits blackbody spectra in the microwave region of the spectrum. The cosmic microwave background radiation is 3K for instance. That then raises the question of how a microwave oven can heat up a cup of coffee. It isn’t accomplished by microwave photons per se ….”

    Once again, this is not right. It IS true that THERMAL microwaves would come from objects at a few Kelvins and that they could not warm an object above that same temperature. But microwave ovens don’t produce their microwaves by heating the walls of the oven to a few K. They use devices called “klystrons” or “cavity magnetron”. These devises are quite capable of producing very intense beams of microwave photons — several hundred watts of microwaves from a few square inches. It is exactly those microwave photons that heat the food.

  355. Let me, as a “thought experiment”, slow the radiation circuits down to a speed where we can see it happening (At least in “our mind’s eye”). And let’s also think of things as single units i.e. radiation = 1 ray, energy =1 unit (Watt) & CO2 = 1 molecule etc. Furthermore lets do what the scientists do on occasions like this; ignore everything other than radiation, i.e. adiabatic warming /cooling, Conduction, convection and advection etcetera. (after all the atmosphere is not needed for radiation.)

    Short Wave Radiation (SWR) comes in from space and warms the surface which begins to radiate Long Wave Radiation (LWR) back towards space, say at a constant energy rate of 1 W/m². Having sent this 1 W/m² away the surface must cool down respectively. CO2 intercepts and absorbs this energy and now CO2 warms up respectively and therefore begins to increase it’s radiation, – half of which (estimates may vary) returns to the surface and is now near to the end of it’s first circuit.

    Meanwhile a ray of SWR has been absorbed by the surface and the 1 W/m² that LWR took away earlier has been replenished, so now that one part, however small, of LWR returns for absorption it must mean that the surface will warm a little bit extra. (That is if this theory is correct)

    However what happens on the dark side of the Earth (or what we call at night)?
    Where the Sun does not shine and SWR therefore does not replenish the energy which LWR takes away, – there can be only cooling.

    To find out if the “cooling capacity” is great enough to bring the temperature right back to what it was the previous morning (i.e. heating starts again at “The zero line”) or if a part of the extra heating remains in the system and the heating starts again at a slightly higher temperature than yesterday it seems logical to me to bring back the Atmosphere (especially the Troposphere) and all it’s air-movements, pressures and lapse rates etcetera. And we must not forget the “Atmospheric Window”

    Do scientists have enough data to anything bar guessing?

  356. Dave Springer says:

    A gas at 1 bar isn’t dense? In general a gas is considered dense if the molecules are close enough to each other to conduct heat. Nitrogen at 1 bar will certainly conduct heat. Try again.

    We are not talking about conduction but radiative absorption. For N2, radiative absorption does not occur for the isolated molecule and can only occur because of collisional processes between molecules. Here is a paper discussing the measurement of such absorption lines at pressures of 0 to 10 atmospheres: http://www.opticsinfobase.org/view_article.cfm?gotourl=http%3A%2F%2Fwww.opticsinfobase.org%2FDirectPDFAccess%2FBC73981D-EC8E-6BE9-5F66346A49A16C1A_60399.pdf%3Fda%3D1%26id%3D60399%26seq%3D0%26mobile%3Dno&org=Rochester%20Institute%20of%20Technology

    Note that even for the strongest absorption line, the measurements rely on obtaining ultra-high purities of nitrogen because any small contamination by CO or CO2 overwhelms the measurement:

    At the path length used in this study, trace impurities in the sample presented a serious problem, and we outline in some detail our purification method. Both CO and CO2 have strong bands that fall near to or on top of the collisionally induced absorption band of N2. Concentrations of these molecules at the 10^-9 level cause interfering lines to be observed in the spectrum, and, in fact, commercial Ultra-High Purity nitrogen with a stated purity of 99.9995% contained enough CO and CO2 to render the measurements useless. Intensity measurements on individual impurity
    lines showed that the concentrations of CO and CO2 were 0.4 and 1 x 10^-6, respectively.

    So, apparently only 1 part per million of CO2 was, along with the CO, enough to render the measurements useless. Imagine what 380 parts per million does! Furthermore, as this graph shows http://www.learner.org/courses/envsci/visual/img_med/electromagnetic_spectrum.jpg , the absorption line of N2 that they are talking about, which is at 4.3 um, would already place it quite far out in the wings of the terrestrial radiation spectrum.

  357. Supplement to my previous comment:

    Oh yes, – “The Atmospheric window” – last time I looked at how it was once presented by “Climate Scientists” it showed that out of the 390 W/m² of long wave Infra Red (IR) radiation emitted by the surface 40 W/m² was let through that window “scott free”.

    Well, according to my rough calculations that is just around 10.25 – (or 10 %) – How can that be correct? – Yes, it may be correct in so far as they can say that; “around 10% of the wavebands emitted by IR radiation are made up of wave-lengths that cannot be absorbed by “Greenhouse Gases” (GHGs), but that cannot possibly mean that 0.04%, in the case of CO2 concentration but certainly less than 10% of the Atmosphere as a total has got what must be a “supernatural” ability to stop LWR. After all, according to scientific measurements the bulk of the Atmosphere, which is transparent to IR radiation, is made up from; Nitrogen (78%), Oxygen (21%) and Argon (0.9%) and the rest = 0.1% is made up from “trace gases” of which some are said to be “GHGs”. – Water Vapour (WV) does not mix evenly in with the other gases and atmospheric WV content varies from location to location but I believe it is estimated to be around 4 to 5%. – Of course the most dense WV concentration must be in the lower half of the Troposphere which may give the impression (by looking at Hydrometers) that is far more WV than a miserly 4-5%. However I doubt very much that the WV concentration near the surface amounts to more than say 40 – 50%. But of course I have got no official data as proof – and I could be wrong.
    So, as far as I can see, the case must be that whatever percentage of LWR passes through the “Atmospheric window” must be added to whatever passes through the “Transparent” part of the Atmosphere.

  358. Joel Shore:

    You say: “Silence?”

    Yeah, silence! Have you even read the article at the link? I don’t see how any of your links relate to it; all they do is repeat again the radiatitive GHE theory. What I want to know is why ALL planetoids with atmospheres, regardless of the gases (and solids) in those atmospheres, have the same temperature at 1000 bar, when distance to the sun is factored in. Are the articles wrong? If not, that simply cannot be due to any GHE. Just curious, man….

  359. Bryan says:

    [a bunch of nonsense and then...]

    I could recommend some useful textbooks to bring you back up to speed with Physics.

    That is very generous, but I think you had better look through the latest edition of such textbooks to make sure they pass through your “political correctness” filter. The last time that you quoted a physics textbook as a source, it turned out that the latest edition of that textbook had a whole page on the greenhouse effect and global warming (see http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/#comment-644823 ). To put it mildly, it did not provide any support whatsoever for your views on the subject!

  360. Several people are having trouble understanding the adiabatic lapse rates.

    Dry adiabatic lapse rate (DALR) means the change in temperature with a change in pressure (altitude) when the contained water vapor is NOT changing state.

    The Moist adiabatic lapse rate (MALR) (aka the Saturated adiabatic lapse rate – SALR) means the same thing except that the water vapor is condensing to form clouds.

    The difference has nothing to do with the amount of water in the atmosphere, but only with whether or not the water vapor is changing state.

  361. Bryan says:

    “You have a particularly snide way of phrasing your replies.”

    I would put it: whiny.

    Instead of misdirecting the conversation to Prof Lindzen, I’m still waiting for an answer based on the scientific method to the question: where is the putative global damage caused by CO2? In fact, if there was such evidence, it would have been trumpeted 24/7/365 by the CAGW crowd.

    Mr Shore, please provide testable, measurable, reproducible and convincing evidence demonstrating verifiable global harm specifically due to the rise in CO2 – you know: per the scientific method. Claiming that “I have explained this stuff many times” is a copout. We need solid evidence, because $trillions more in wasted tax dollars are at risk, based not on empirical evidence, but on GIGO models invented by people who can’t even get the definition of heat right.

    Alternatively, Joel Shore can be a man-up and admit that there is no such evidence, and then the whole baseless CAGW conjecture becomes clear to everyone: the purpose of the scare is to provide a free ride on the grant gravy train for these climate charlatans at taxpayer expense.

  362. R Stevenson,

    I would very much like a link to that spreadsheet. My attempts to locate Hottel’s method keep hitting paywalls.

    Using 1.0% water vapor, my program computes 282.68 W/m2 .. close to your value.

    For power absorbed, my program incorrectly assumes 1 km at constant temperature and pressure. However, it also plots the distance to absorb x% of the available radiation. As long as those values are under 100 meters, I feel that assuming constant temperature and pressure causes only minor errors.

    At any rate, I would like to investigate WHY the programs are giving different values to absorb the radiation.

  363. David Springer must won’t let go of this nonsense:

    “Solids, liquids, and dense gases emit continous blackbody spectra characteristic of their temperature. We were talking about atmospheric nitrogen which is a dense gas and its temperature, at least in the lower and middle atmosphere, puts its blackbody emission spectrum in the far infrared.

    What part of that don’t you understand?’

    I would have ignored this, except for the snide remark.

    Come on, David, I gave you a link to a very clear textbook exposition on the interaction of IR with gases, which states unequivocally that N2 is not “IR active,” and it provides specific detailed reasons why this is true (no dipole moment). Density probably has nothing to do with it, unless it is so dense it is a liquid (and that occurs at about 1600 psi at 25 C).

    I can see how we can debate something as complicated as the existence or amount of the GHE effect, but we should not be debating such basic, easily demonstrable, and widely accepted facts like the lack of signficant emissions of IR from N2.

  364. Smokey says:

    “Instead of misdirecting the conversation to Prof Lindzen, I’m still waiting for an answer based on the scientific method to the question: where is the putative global damage caused by CO2? In fact, if there was such evidence, it would have been trumpeted 24/7/365 by the CAGW crowd.”

    Come on, Smokey, you must not be listening: IPCC and the warmistas keep yelling to us that there are “multiple lines of evidence” for CAGW. That’s Orwell-speak for “we don’t have any direct clear evidence, but, considering all things that MIGHT happen…(whatever)…” When dealing with the “progressives,” you gotta learn a new vocabulary. Such phrases as “Kinetic Military Action” (war); “Overseas Contingency Operations” (war on terror); “man-caused disaster” (terrorist act), “climate disruption” (man destroying the world), “we must be civil” (the Conservative bastards must be civil). Ad nauseum.

    Problem with the CAGWers is that they can’t seem to locate one single clear line of evidence. An even bigger problem for them is that the “multiple lines of evidence” are actually demonstrating rather clearly that they are wrong, I’m sure to their inner horror (it’s a travesty…). Nothing seems to be working out for them, since the sea level is leveling off, the glaciers are not really melting everywhere, the temperature is not going up for the last 10-15 years, no “hot spot,” as predicted by models, etc., etc.

    On the other hand, I guess if you allow yourself to blame cold on warming, storms and any sort of malady on “climate change,” then you can allow yourself to say that, “multiple lines of evidence” show that we are killing ourselves through some sort of unclearly defined mechanism, which we will just call CAGW. LOL.

    It all just Brings tears to my eyes.

    But they are gonna fix the problem by setting up special forums and getting the liberal MSM to explain their (false) science better, so we morons can understand their brilliance. Or something like that.

  365. O H Dahlsveen says: May 11, 2011 at 8:45 am

    “but that cannot possibly mean that 0.04%, in the case of CO2 concentration but certainly less than 10% of the Atmosphere as a total has got what must be a “supernatural” ability to stop LWR. ”

    hmmmm … lets see try an order of magnitude estimation …

    A “typical” IR photon is ~ 10 um, or 2E-20 J

    The ground emits ~ 400 W/m^2, which implies around 400/ (2e-22) = 2E22 photons per second being emitted.

    Air applies 100,000 N/m^2 of pressure, so a square meter column of contains about 10,000 kg of air. CO2 is ~ 400 ppm by volume (and slight more than 400 ppm by mass), so there are about 10,000 kg x (400/1,000,000) = 4 kg =4000 g of CO2 in each square meter column of atmosphere.

    At 44 g/mole, that would be 100 moles of CO2, or ~ 6E25 molecules of CO2 per square meter.

    That makes 6E25 / 2E22 = more than 1,000 CO2 molecules for every IR photon passing by. CO2 only absorbs ~ 1/10 of the possible IR wavelengths, so that’s ~ 10,000 CO2 molecules per absorbable photon passing by each second. And the CO2 molecule only “holds” that energy for a faction of a second, so a given molecule could in principle absorb many photons every second.

    Each CO2 molecule absorbing 1 out of 10,000 photons passing doesn’t seem “supernatural” to me ….

    (I did this pretty quickly, so people are welcome to find flaws. I would not be surprised if my numbers are off by 10x or possibly even 100x, but that still doesn’t cahnge the fact that there are plenty of CO2 molecules around to absorb photons even at an “insignificant 0.04%” concentration.)

  366. First I apologise if I am rehashing something that has already been discussed, I have been out in the field and haven’t had a chance to catch up on all the posts in this thread as yet, however this one did.

    The equation cited does not, in my opinion, imply backradiation. If it did there would be terms for the emissivity of both objects as well as the surface area of both objects. There aren’t. What the equation does account for is the net difference in temperatures between the two objects or one object and its surroundings.

    The net difference in temperature is obviously one of the factors which determines the flow of energy from the hot body to the colder one. If this were water we would call it the hydraulic head, if it were electricity it would be the voltage. Furthermore, if the surroundings or other object did heat the hotter object through “backradiation” wouldn’t the equation not only require terms for the surface area and emissivity of the colder object, but also require integration of the effect of the backradiation? In other words, to consider just the colder object, as it emits energy to the warmer object it cools, but it also warms from the energy input of the warmer object which causes its temperature to rise as a function of the difference between its absortivity and its emissivity. The increased temperature would then cause it to backradiate more energy etc… While at the same time the hotter object would be behaving in a like fashion. So you would need to integrate the energy inputs to each from each as a function of not only emissivity but absorbtivity as the net temperature difference tended towards 0 ( assuming it could in fact tend towards 0)?

  367. Martin Lewitt:

    The lack of the temperature profile that you claim is expected from more turnover of the water cycle, can possibly be explained by the models under representing the efficiency with which greenhouse gases cool the upper troposphere, there is already good evidence that the stratosphere is cooler, that is attributed to this mechanism.

    I’m not sure what you mean by this. Are you trying to say that you don’t believe that the estimates of the forcing due to increases in CO2 is not around the value that nearly all scientists, including Roy Spencer and Richard Lindzen, accept it to be? And, if so, what evidence do you have to make the assertion? It seems to me that you are taking one rather slender piece of evidence (the Wentz paper, which is interesting but not completely definitive) and then extrapolating it like crazy to suggest all sorts of things that they haven’t suggested…and which don’t really seem in line with the empirical data that we have.

  368. Smokey says:

    Please provide testable, measurable, reproducible and convincing evidence demonstrating verifiable global harm specifically due to the rise in CO2 – you know: per the scientific method.

    How am I supposed to convince you of something that involves seriously weighing of scientific evidence when I cannot even get you to admit you have made in error when it is right in front of your face for everyone to see? I might as well convince a rock to do the tango! You are so far from a reasonable person able to objectively weigh scientific evidence that it is a fool’s errand to try to convince you of anything.

  369. ferd berple says:
    May 10, 2011 at 1:43 pm

    Here is a thought experiment. If we hold everything else equal and double the amount of N2 in the atmosphere, because N2 does not participate in the back radiation, the surface equilibrium temperature should remain “unchanged” according to the radiative transfer model.
    However, comparisons of earth and venus suggest that the temperature of the atmosphere will only remain unchanged at 1 atmosphere pressure. As we now have more atmosphere, the pressure at the surface will be increased as compared to present. Under the gravitational model, the surface temperature would be expected to increase as a result of doubling the N2 in the atmosphere.

    Which of these two is correct? Adding N2 will not increase temperature, or adding N2 will increase temperature (holding all else equal)?

    Adding additional N2 to doubling it’s contribution to total mass of the atmosphere will absolutely increase the temperature at the surface, quite a bit. Mainly water vapor and carbon dioxide will continue to be mixed and will absorb, thermalize and re-excite as normal, the lapse rate in a deeper atmosphere will do the rest.

    The problem for me is the lapse rate. Doubling the N2 in the atmosphere will raise the height of the atmosphere. If the surface temperature does not increase, then the lapse rate dictates that since the atmosphere extends higher, that further up, past the present altitude it must get colder than it is at present. That seem nonsensical.

    Your view of the lapse rate is right on the spot. At the one bar level temperatures will be very close to what we experience today.

  370. jae says:

    Yeah, silence! Have you even read the article at the link? I don’t see how any of your links relate to it; all they do is repeat again the radiatitive GHE theory.

    No…I also explained why the general claims (by people like Postma) that the lapse rate alone can explain the surface temperature are completely wrong.

    What I want to know is why ALL planetoids with atmospheres, regardless of the gases (and solids) in those atmospheres, have the same temperature at 1000 bar, when distance to the sun is factored in. Are the articles wrong? If not, that simply cannot be due to any GHE.

    That article just shows this as being true for Venus and Earth (assuming the data shown is accurate), which could very well be a coincidence, especially considering that the factoring in of the distance from the sun without taking the very different albedos into account seems rather unjustified. Can you show me the data for the other planetoids? Then maybe it would be worth thinking about why this might be the case.

    However, it would almost certainly have no bearing whatsoever on whether there is the GHE or not, since nobody has successfully told me how one can propose the earth and Venus have the surface temperature they do without invoking the greenhouse effect or abandoning conservation of energy.

  371. Tim Folkerts
    Perhaps you could shed some light on the CO2 photon capture aftermath.
    I did a calculation some time ago (and from memory) at atmospheric temperatures about 4% of CO2 molecules have the vibrational mode active.
    This means that 96% are available to absorb.
    So in the Earth surface up there will be a reasonable flux of 15um photons which can be absorbed to increase the active population.
    The relaxation time in this mode is relatively long compared with the chance of losing the energy by collision to probably N2, O2 molecules.
    This causes local heating and a local temperature rise.
    The fact that the re-emission of a 15um photon requires a CO2 molecule to reach a high and statistically unlikely translational speed means that the following seems more probable;
    1. Local temperature rise.
    2.15um Emissions are less than absorptions as a energy consequence of 1.
    3. Radiative emissions (when they happen) will favour the longer wavelength, lower energy, more probable and readily available H2O bands.
    Some people have trouble with 2 as Kirchhoff no longer holds.
    This is perhaps a needless concern as;
    1. LTE no longer holds – a condition for Kirchhoff
    2. Photon energy has been diverted along more probable paths.
    Thats my take – any comments?

  372. Joel Shore says:
    May 11, 2011 at 10:50 am
    “…since nobody has successfully told me how one can propose the earth and Venus have the surface temperature they do without invoking the greenhouse effect or abandoning conservation of energy.”

    Pressure, Joel, pressure. There I have successfully told you. Whether you accept is a different story.

  373. JAE, regarding some other comments made here, corrections are in order.

    Dr Richard S. Lindzen states: “If one assumes all warming over the past century is due to anthropogenic greenhouse forcing, then the derived sensitivity of the climate to a doubling of CO2 is less than 1C.” [my bold]

    None but the most radical alarmists claim that all of the warming over the past century is anthropogenic. Natural variability is always at work. Lindzen also points out:

    “There is ample evidence that the Earth’s temperature as measured at the equator has remained within ± one degree centigrade for more than the past billion years. Those temperatures have not changed over the past century.” [source]

    As one approaches the poles  the Earth’s temperature begins to vary more and more from one region to another, and from summer to winter, and day to night.  This is normal, and has always been so.  Greenland has very wide temperature swings, while Egypt is extremely uniform from year to year, from decade to decade, and from century to century.

    By selectively picking a base year for comparison, anyone can show either warming or cooling. Pick a warm base year such as 1997, and you can show cooling. Pick a cool base year like 2001, and you can show a warming trend. But the  Earth’s overall temperature is extremely constant, and a slight variation of a few tenths of a degree is routine and natural. Temperatures have fluctuated much more over the Holocene, sometimes by tens of degrees. Dr Lindzen points out how ridiculous the current arm-waving is, over the extremely *mild* natural variability of only 0.7°C over the past century and a half:

    “Future generations will wonder in bemused amazement that the early 21st century’s developed world went into hysterical panic over a globally averaged temperature increase of a few tenths of a degree, and, on the basis of gross exaggerations of highly uncertain computer projections combined into implausible chains of inference, proceeded to contemplate a roll-back of the industrial age.”

    Finally, regarding the question of the climate’s sensitivity to a doubling of CO2, the UN/IPCC preposterously claims it is 3°C or more. That is absurd. If sensitivity were that high, the temperature would closely track rises in CO2. It doesn’t.

    More knowledgeable scientists than the IPCC’s political appointees, including its head railroad engineer/bodice-ripper author, and its WWF advisors, provide much lower sensitivity numbers:

    Dr Chylek estimates the sensitivity at 1.4°C; less than one-half the IPCC’s lowest number. Dr Schwartz gives the sensitivity as 1.1°C. Prof Lindzen puts the sensitivity at under 1°C. Dr Spencer puts the sensitivity at 0.46°C, based on observations. Drs Idso, fils & pere, put the sensitivity at 0.37°C. Dr Miskolczi puts the climate’s sensitivity to a doubling of CO2 at 0.0°C.

    And there is still zero evidence of any global harm as the result of the increase in that beneficial trace gas. Conclusion: CO2, at current and projected concentrations, is both harmless and beneficial. And that answers the key AGW question: Is the rise in CO2 a problem? In fact, the increase in CO2 is, on balance, a net benefit to the biosphere.

  374. Bryan says: May 11, 2011 at 11:50 am

    “The fact that the re-emission of a 15um photon requires a CO2 molecule to reach a high and statistically unlikely translational speed …

    I agree with things before this line, but I disagree with this. The average KE of a gas molecule is 3/2 kT. At 250 K this works out to ~ 5e-21 J. This is the same approximate energy as photons with a wavelength of 15 um. Thus it seems that the KE due to thermal motion should be able to excite the modes needed to emit the IR photons. (it would take considerably more time to work out actual probabilities of having various specific energy.)

    Without thinking about it too deeply, it seems that peak in BB radiation would correlate quite naturally with the peak in kinetic energies of the molecules emitting the radiation. The fact that the name “Boltzmann” is associated with both ideas adds to the supposition that the two are related.

  375. >>Joel Shore says: May 11, 2011 at 10:50 am
    >>“…since nobody has successfully told me how one can propose the earth
    >>and Venus have the surface temperature they do without invoking the
    >>greenhouse effect or abandoning conservation of energy.”

    >mkelly says: May 11, 2011 at 12:12 pm
    >Pressure, Joel, pressure. There I have successfully told you. Whether you accept is a >different story.

    IMHO, pressure (and lapse rate) can indeed explain temperature differences for various layers within the atmosphere.

    But, mkelly, can you explain how pressure can determine either the location or the temperature of the “top of atmosphere”? What calculation will predict that a pressure of 1 bar should be ~ 15 C? Or that the temperature at 5 km altitude should be ~ -20 C?

    Until you can calculate such a temperature from first principles, you have not successfully told anything — you have simply made a bold, unsupported assertion.

  376. Tim Folkerts says:
    May 11, 2011 at 12:59 pm
    IMHO, pressure (and lapse rate) can indeed explain temperature differences for various layers within the atmosphere.

    Joel asked about the surface. I answered him. I believe your statement above would be valid for the layer that is at or immediately adjacent to the surface.

  377. vindsavfuktare says:
    May 10, 2011 at 1:46 pm
    Ira,
    You haven’t answered so I’ll give it another try.

    Your attribution of the missing 33K from 240W radiation from earths system to “green house effect” assumes an atmosphere that is at 0K AND assumes no reflection (as I understand you, please correct me if I’m wrong). …

    You ask some excellent questions and I was hoping someone else, more qualified than I in this thread would answer.

    My short answer:

    1) I do not attribute ALL the approximately 33ºC to the “back radiation” aspect of the Atmospheric “greenhouse effect”. However, if all else was equal except for the affect of so-caled “greenhouse gases” I am confident the mean temperature at the Surface would be about that much cooler.

    2) I actually used the Trenberth model to estimate how the 1366 Watts/m^2 should be distributed into the Earth System, which I specifically said included BOTH the Atmosphere and the Surface. As I wrote, we must first multiply by 0.25 to account for the fact that half the Earth System is not illuminated by the Sun at any given time, and the hemisphere that is illuminated is not all at right angles to the Solar rays. The second correction factor is for the albedo of the Earth System, and for that I used the 77 W/m^2 refleted by clouds plus the additional 30 W/m^2 reflected by the Surface. Doing that, and rounding (as I always do since I am an engineer :^) I got 240 Watts/m^2. Trenberth gets 67 Watts/m^2 absorbed by the Atmosphere plus 168Watts/m^2 absorbed by the Surface, which is total of 235 Watts/m^2, very close to my number for the mean rate of Solar energy input.

    3) Please keep in mind that the main purpose of this Visualizing series is to help those of our fellow Skeptics who (IMHO) have gone over the line to an equal and opposite error from the Alarmists, expressing total Disbelief in the very scientific basis of the ill-named “greenhouse effect” of the Atmosphere and in the total Earth System. I am completely open to those who do further analysis that casts doubt on Trenberth’s exact allocation and attribution of the Earth System energy budget.

    4) Perhaps the 33ºC attributed to the effect of the absorption and emission Atmosphere is too low or too high, but, again IMHO, that value is close enough for purposes of a “sanity check”. I’ll leave the exact value to the experts to argue about.

    5) Once we scientifically-oriented Skeptics accept the reality of the Atmospheric “greenhouse effect” we are, IMHO, better positioned to question the much larger issues which are: a) HOW MUCH does CO2 contribute to that effect, b) HOW MUCH does human burning of fossil fuels and land use changes that refuce albedo affect warming, and, perhaps most important, c) Does the resultant enhanced CO2 level and higher mean temperature actually have a net benefit for humankind?

  378. 5) Once we scientifically-oriented Skeptics accept the reality of the Atmospheric “greenhouse effect” we are, IMHO, better positioned to question the much larger issues which are: a) HOW MUCH does CO2 contribute to that effect, b) HOW MUCH does human burning of fossil fuels and land use changes that refuce albedo affect warming, and, perhaps most important, c) Does the resultant enhanced CO2 level and higher mean temperature actually have a net benefit for humankind?

    How about using real heat transfer equations to explain this rather than back radiation. There is no input for back radiation in any heat transfer equation so taking as if it does something cannot be shown via standard equations.

  379. mkelly says:

    Pressure, Joel, pressure. There I have successfully told you. Whether you accept is a different story.

    Sorry…Saying the world “pressure” does not get you around having to satisfy conservation of energy. As I have explained, those sorts of things explain the slope of the line (how temperature varies with height…or at least a limit on how steeply temperature can fall with height) but it does not determine the constant “b” in the equation “y = m*x + b”.

    The only way you can get around the requirement that the steady-state temperature be such that the energy in from the sun equals the energy out from terrestrial radiation is to have some other significant source of energy. For a gaseous planet undergoing gravitational collapse, that source of energy could be the conversion of gravitational potential energy to other forms of energy…but that is not happening here on the earth. Energy flow from the interior of the earth is negligible.

  380. O H Dahlsveen says:
    May 10, 2011 at 1:53 pm

    Ira, you say in your article above:
    “Since we humans sense the visual portion of this radiation as “light”, that is the name we give it, and that has led to the false assumption that it contains no “heat” (or “thermal”) energy.”
    You are making one assumption there that does not fit in with “Climate Science” at all. It has, ever since the days of Friedrich Wilhelm Herschel and Benjamin Franklin, been known to science that different colors absorb “solar light” at different rates.

    Having said all that, I too have been led to conclude that light, and indeed that all kinds of other radiation contains no heat. What light does contain however is “electronic energy”. Or to say it in a different way; light is always “on the move” and is transmitted as a part of the electro-magnetic wave-bands or spectrum, i.e. as radiation. – Only when that radiation is stopped or interrupted by “solids” i.e. atoms and/or molecules is that electronic energy either absorbed (transformed to kinetic energy) or reflected. …

    O H Dahlsveen, I have noticed (and I think replied) to your comments in my other threads and I very much appreciate them because they evidence a sincere, interested, well-educated person, probably oriented towards engineering and technology but not necessarily science, per se, who has done quite a bit of reading about the subject of climate science.

    However, when I read your statement I highlighted above, “I too have been led to conclude that light, and indeed that all kinds of other radiation contains no heat”, I get a disconnect!

    As you must know, all energy is fungible, and may be transformed from one form to another. The potential energy of water in an elevated lake that is rushing down a river does not contain “heat” per se, but, if we dam the river and divert the water pressure to run a hydroelectric generator, we can transform the river energy into electrical energy and then use that to power an electric heater. Or, we could use that hydro-generated electricity to power a pump and pump that water up into a water tower, turning the electrical energy back into potential energy.

    Thus, any form of energy contains “heat” and “potential energy” and “electrical energy” and so on. As I am sure you know, that water in the elevated lake got there because the Sun heated Surface water till it evaporated and rose high in the Atmosphere and came down as rain. You also know the energy in fossil fuels came about due to the Solar energy absorbed by long-ago living trees and sea life that got compressed and stored underground as coal and oil and natural gas. Thus, the Sun is the source of virtually all the energy on Earth (except for the relatively small amount we get from the hot core and atomic radiation).

    It ALL came in as “light” and was transformed into “heat” when absorbed by the Surface, into living things when photosynthesis transformed it into living things, and into potential energy when raised into the air by evaporation, and into wind and waves, and so on.

    That is what I mean when I say the Solar “light” contains “heat” or “thermal” energy. It is just a shorthand way of saying that energy is fungible and may be transformed into other forms.

  381. mkelly says:

    How about using real heat transfer equations to explain this rather than back radiation. There is no input for back radiation in any heat transfer equation so taking as if it does something cannot be shown via standard equations.

    This is utter and complete hogwash. The term “back radiation” is just a label given to radiation emitted by the atmosphere that happens to travel toward the earth. All actually calculations of the greenhouse effect are done with the standard heat transfer equations. If you don’t like that label, you can get rid of it and it will make not one iota of difference in what you get out of the equations.

  382. If the environmental lapse rate is larger than the dry adiabatic lapse rate, it has a superadiabatic lapse rate, the air is absolutely unstable — a parcel of air will gain buoyancy as it rises both below and above the lifting condensation level or convective condensation level.

    what happens to the black.body spectrum of water vapor as it condenses, joel and ira and other radiation freaks? say it loud, if you dare.

  383. mkelly says: May 11, 2011 at 1:55 pm

    How about using real heat transfer equations to explain this rather than back radiation. There is no input for back radiation in any heat transfer equation so taking as if it does something cannot be shown via standard equations.

    1) “back radiation” is radiation
    2) radiation is included in standard heat transfer equations
    Therefore “back-radiation” is included in standard heat transfer equations

    If you want more detail, here are some “standard equations” that specifically deal with thermal radiation from one object of arbitrary size, shape, emissivity, and temperature to a second object of arbitrary size, shape, emissivity, and temperature:

    Standard dQ/Dt (scroll about 2/3 of the way down).
    “The radiation heat transfer from one surface to another is simply equal to the radiation entering the first surface from the other [added: in other words "back radiation"], minus the radiation leaving the first surface.”

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

    Standard form factors to go with the above equation.

    http://www.engr.uky.edu/rtl/Catalog/tablecon.html

    Do you contend these are not standard equations?
    IF SO, what would you use instead?

  384. richard verney says:
    May 10, 2011 at 4:29 pm
    JAE says:
    May 10, 2011 at 4:03 pm
    New calcs of radiative GHE shows the GHE is only 8-9 deg. C:

    http://activistteacher.blogspot.com/2011/05/radiation-physics-constraints-on-global.html

    //////////////////////////////////////

    WOW! I just read the linked webpage and it seems fairly convincing. I hope Joel Shore and Dave Springer and others comment and either set me straight or agree that I and virtually everybody else on the Internet who claims the Atmospheric “greenhouse effect” contributes about 33º to the average temperature of the Surface. If the correct number is closer to a third of that amount, established science has got a lot of ‘splainin to do. (I am pretty sure there is an error in the new claims, but I cannot figure it out on my own.)

    In my “sanity check” I assumed that the Earth Surface emissivity was sufficiently high that, for all intents and purposes, I could use the black body model in the Carleton spreadsheet to estimate the temperature of the Surface absent the Atmospheric “greenhouse effect”. That is how I got an effective surface temperature of about 255 K which would emit about 240 Watts/m^2 and balance the mean input of Solar energy which is about 240 Watts/m^2 at the Surface.

    The linked website points out that the albedo of the Earth System is about 30% (the same value I used). Therefore, they say, since absorption and emission tend to be about the same for many materials, the emissivity of the Surface is bound to be less than 1.0, and may be around 0.7, if I read them correctly.

    Using the lower emissivity, they get 278.6 K (where I and almost all others get around 255 K).

    I have thought about where they are going wrong and have some ideas, but I will appreciate confirmation and correction by those who know more than I and the linked site:

    1) The albedo of the Earth System is indeed around 30%, but most of that is due to clouds. The albedo considering only the Surface is closer to 12% which would make emissivity (if it is equal to absorptivity) about 0.88 rather than 0.7. That change would decrease their 278.6 K by some amount, but not down to 255 K.

    2) The albedo of the Earth Surface is 12% based on absorptivity of visual and near-visual Solar radiation (what we have been calling short-wave). The radiative emission from the Surface, based on its black body temperature, is in the mid- and far IR (long-wave), so it is not clear to me that that emissivity at long-wave would be the same as the absorptivity at short-wave. For example, common glass has an absorptivity of close to 0.0 at short-wave but about 1.0 at long-wave. So, perhaps they are comparing apples and oranges and getting a fruit salad rather than the correct result.

    PLEASE EXPERTS: Comment on this very interesting link and argument and either confirm my conclusions or modify them or (Horrors!) accept the new argument as basically correct. advTHANKSance

    [UPDATE 6:20PM: I've just noticed that Tim Folkerts says: May 10, 2011 at 8:14 pm has confirmed my objection (2) with the example of white paint albedo reflecting 90% of visible short-wave light but absorbing 90% of long-wave IR radiation. Thus the linked website is wrong to assume that the 30% albedo of the Earth System in the short-wave region equates to only about a 0.7 emissivity in the long-wave. Using Tim's example, white paint would have an albedo of 0.9 in the short-wave but a an emissivity of 0.9 in the long-wave (rather than only 0.1 as the linked website would conclude. THANKS TIM and I appologize for not including you in my list of experts above. You certainly qualify, as do others here in this thread. - Ira]

  385. Ira,

    I have just a minor quibble with your wording here:
    “Thus, any form of energy contains “heat” and “potential energy” and “electrical energy” and so on. As I am sure you know, that water in the elevated lake got there because the Sun heated Surface water”

    I would change “contains heat” to “contains thermal energy”. In traditional thermodynamics terminology, “heat” is always energy being transferred, not contained in an object. An object doesn’t “contain heat” any more than it “contains work”. In many ways, it would be better if the language was more parallel. For example:

    dU = δQ – δW

    would be “the change in internal energy is equal to the ‘heat done to the system’ minus the work done by the system.” It sounds really awkward in English, but that is the traditional sense of the words in thermodynamics.

    It seems nitpicky to many, I’m sure. But when talking about laws of thermodynamics in discussions like this, much of the confusion comes from imprecise wording, like using “to heat” to as a synonym for “to warm”. Or using “heat” when one means “U” (as you did above).

    I do approve of your later use “the Sun heated Surface water”. There was indeed a transfer of energy from sun to water due to a temperature difference. :-)

    And I think you hit the nail on the head with:
    “5) Once we scientifically-oriented Skeptics accept the reality of the Atmospheric “greenhouse effect” we are, IMHO, better positioned to question the much larger issues which are: a) HOW MUCH does CO2 contribute to that effect, b) HOW MUCH does human burning of fossil fuels and land use changes that reduce albedo affect warming, and, perhaps most important, c) Does the resultant enhanced CO2 level and higher mean temperature actually have a net benefit for humankind?”

  386. Soft-science students of climate woul benefit greatly from performing a mass-weighted vertical integration of the standard atmospheric temperature profile. It nearly matches the theoretical expectation for the average graybody temperature of the Earth. This illustrates two vital points that are often overlooked: 1) the LWIR escaping into space comes NOT directly from the surface (except in the “window” wavelengths) but from a diaphanous mass of gases, and 2) there is NO additional thermal energy produced by the “greenhouse effect.”

    Air temperature is a nonconservative, intensive variable whose local value depends not only upon the radiative fluxes driven by thermalization of insolation, but upon upon the atmoshperic pressure, in accordance with Boyle’s law. Indeed, if a parcel of air is moved adiabatically from the surfaceto aloft, the temparature drops accordingly. The ~33K increase in temps that we enjoy is available only because we live at the bottom of the atmosphere. It is NOT the product of any additional thermalization, for that would require TOA LWIR planetary emissions of ~390w/m^2 to maintain a steady state. The fundamental flaw of the “radiative greenhouse” paradigm is the illusion that thermalization is a recursive process. In fact, it can occur only once. Like income, it cannot be reused.

  387. Ira Glickstein, PhD says:
    May 11, 2011 at 3:08 pm

    PLEASE EXPERTS: Comment on this very interesting link and argument and either confirm my conclusions or modify
    them or (Horrors!) accept the new argument as basically correct. advTHANKSance


    Since you have never responded to all of the real aspects I have supplied to you to think about, I don’t know why I am responding. Since you said please I guess. Seems you keep asking for an expert, which I will continue to maintain I am not, whether the surface of the Earth is a black body to long wave radiation peaking at 10 µm, the answer is no. It is very close to a gray body if the spectrum is taken merely meters from the surface. The is a paper by some Swiss scientists on the current radiation flux in the Alps that gives a great multi-altitude spectrum of the upwelling LW radiation. Search it out.

    Have you read Miskolczi’s papers yet as I suggested you do, by an atmospheric physicist, which explains plainly why this is so?

    Also, that is a rather cheap shot when you said if someone did not respond they will accept all you believe as laid out in this thread. That make you look so incredibly shallow.

  388. Tim Folkerts said….”I agree with things before this line, but I disagree with this. The average KE of a gas molecule is 3/2 kT. At 250 K this works out to ~ 5e-21 J.”

    I agree with this

    Tim Folkerts said….” This is the same approximate energy as photons with a wavelength of 15 um.”

    I dont agree with this.

    Energy of 15um photon is almost three times bigger.

    Redo your calculations and work out probabilities from Maxwell Boltzmann Statistics

  389. I am trying to figure out why a doubling of atmospheric N2 would necessarily cause a dramatic increase in temperatures at the surface, as I have seen some propose. N2 has little to no insulative properties, right?

    By ideal gas laws, PV=nRT. N2 is roughly 75% of the mass of the atmosphere. I know that n is moles, not mass, so let’s assume that we just increase N2 by “a whole helluva lot”, so that n(2) = 1.75n(1), in terms of the entire atmosphere.

    So we release a whole helluva lot of N2 gas into the atmosphere and then, much later, sample a cubic meter at sea level. What would we find is different about this volume V(2) as compared to V(1)? Is there any possibility that T(2) will have increased insignificantly compared to T(1)?

    Assuming n is the only change on the right side of the equation, sure. I can think of a pretty simple solution. P could increase by a factor of 1.75, and that’s it – equation balanced. We end up with a thicker atmosphere (a factor of 1.75 more moles per cubic meter), but the same surface temperature.

    I don’t think that is exactly what would happen, but I don’t see why increasing the mass of an atmosphere guarantees a dramatic increase in surface temperatures. It does guarantee an increase in surface pressures, but that just means a thicker atmosphere, not necessarily a hotter atmosphere.

  390. Bob Fernley-Jones says:

    May 10, 2011 at 11:26 pm (Edit)

    Ira Glickstein, PhD Reur May 9, 2011 at 9:01 pm

    You should retract your over-the-top insults to JAE; they give you a negative persona. You should have been able to work-out that JAE made a simple mistake, which BTW he shortly realized and corrected just before your rant. An apology is in order. …

    You criticize my reply to “JAE” (all upper case) but actually I was replying to “jae” (all lower case). I don’t know if both “JAE” and “jae are the same person, but “jae” has earned my reply, at least IMHO.

  391. A frigid fluid immersion bath, never was, isn’t, and won’t be tomorrow, a warmer of the object immersed therein.

    No matter how many times you try to claim you ‘know a guy who calculated’ it, and showed in a graph,

    the colder object,

    didn’t heat a warmer one.

    Thank you for calculating it but all we have to do, is sit two anvils down with thermocouples mounted in them.

    Not once has the cool one, subtracted it’s 75 degrees’ room temperature from itself and added it to the initially warmer one, creating a 150 degree anvil and a 0 degree one.

    Now; in case you think there’s some distantly possible way you know this is capable of happening, congratulations: you have made it to fundamental law number two, before you fell out of contention for the title ‘Grasper of Reality.’

    Thank you for playing, the attendant will see you to your perpetual motion-mobile.

    Your magical freezing and heating anvils look nice, with the little blinking thinglies alleged to be entropy violators.
    They do not however show us a cold object subtracted some of it’s own temperature, and added it to another object, already warmer. Oh no they did not, you did not, and if you claim you did, well you’re either a R.A.I.L. or another word using the same letters.

    Entropy 1.

    You, O.

    Your move.
    You don’t have one.
    That’s called checkmate.

    BECAUSE, when CALCULATING ENGINEERING of:
    ANY SYSTEM

    Y.O.U. WILL
    C.H.E.C.K.
    YOUR WORK,
    by ASKING YOURSELF
    if WHAT YOU ARE SAYING,
    AGREES
    with E.N.T.R.O.P.Y.

    If NOT,
    YOU WILL N.O.T.
    DETECT IT’S EXISTENCE in APPLIED systems, of ANY KIND, natural, or MANNMADE, because E.N.T.R.O.P.Y. WILL NOT BE DISREGARDED as the FAILSAFE for ANY MATHEMATICAL ENDEAVOR.

    And kids, PhD or not,

    you have tried to claim that ENERGY RESIDING in FINITE INTERSTITIAL VOLUME
    has VIOLATED entropy,
    and that a VOLUME of INTERSTICES
    containing a LOWER concentration of photons,
    sent that energy
    to a place with it’s interstices already MORE dense with them.

    Your claim is that *pOp* went the WEASEL, and magically, energy traveled the wrong way – from a miles deep, frigid, fluid bath,

    to the object submerged therein.

    Shiny blinkie spreadsheets or not, you did not warm the object submerged in the frigid gas bath,

    using that frigid gas bath.

    No, you did not, and that’s why you’re reduced to showing up without an instrument on earth that verifies the ***false claim*** you made.

    All we had to do was check and see if you’d wind up obeying entropy. And when you made net energy travel from the cooler to the warmer of two objects, y.o.u.
    were
    d.o.n.e.

    D.O.N.E. and OUT to PASTURE

    deep, D.E.E.P:

    in Area 51/Bigfoot/Psychic Spoon Bending/A.G.W. Country

  392. sky says:

    This illustrates two vital points that are often overlooked: 1) the LWIR escaping into space comes NOT directly from the surface (except in the “window” wavelengths) but from a diaphanous mass of gases,

    Yes…because some of the radiation from the surface gets absorbed and then radiation is emitted from the atmosphere…but less because the atmosphere is at a colder temperature.

    and 2) there is NO additional thermal energy produced by the “greenhouse effect.”

    Nonsense. What you have shown is there is NO additional thermal energy produced by the “greenhouse effect” other than that which is produced by the “greenhouse effect”. What do you think would happen if there were no IR-absorbing elements in the atmosphere? Would any of the radiation from the surface be absorbed?

    Indeed, if a parcel of air is moved adiabatically from the surfaceto aloft, the temparature drops accordingly.

    All that this does is tell you is the maximum rate that the temperature can drop with altitude, i.e., the lapse rate. It does not by itself tell you what the temperature at the surface is.

    The ~33K increase in temps that we enjoy is available only because we live at the bottom of the atmosphere.

    Yes…The bottom of an atmosphere that has IR-absorbing elements and hence a greenhouse effect.

    It is NOT the product of any additional thermalization, for that would require TOA LWIR planetary emissions of ~390w/m^2 to maintain a steady state. The fundamental flaw of the “radiative greenhouse” paradigm is the illusion that thermalization is a recursive process. In fact, it can occur only once. Like income, it cannot be reused.

    Not a clue what you’re trying to say here.

  393. OK, apparently I got the blockquote wrong and thus my previous post doesn’t make sense, so here goes again:

    “Tim Folkerts says:
    May 10, 2011 at 9:42 am
    mkelly,

    You needed to look a little farther and understand what you are reading. This is from the SAME source you quoted:

    Net Radiation Loss Rate

    If an hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as

    q = ε σ (Th^4 – Tc^4) Ac (3)

    http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html

    Please note there ARE inputs for back radiation (the Tc^4 term).”

    First I apologise if I am rehashing something that has already been discussed, I have been out in the field and haven’t had a chance to catch up on all the posts in this thread as yet, however this one did.

    The equation cited does not, in my opinion, imply backradiation. If it did there would be terms for the emissivity of both objects as well as the surface area of both objects. There aren’t. What the equation does account for is the net difference in temperatures between the two objects or one object and its surroundings.

    The net difference in temperature is obviously one of the factors which determines the flow of energy from the hot body to the colder one. If this were water we would call it the hydraulic head, if it were electricity it would be the voltage. Furthermore, if the surroundings or other object did heat the hotter object through “backradiation” wouldn’t the equation not only require terms for the surface area and emissivity of the colder object, but also require integration of the effect of the backradiation? In other words, to consider just the colder object, as it emits energy to the warmer object it cools, but it also warms from the energy input of the warmer object which causes its temperature to rise as a function of the difference between its absortivity and its emissivity. The increased temperature would then cause it to backradiate more energy etc… While at the same time the hotter object would be behaving in a like fashion. So you would need to integrate the energy inputs to each from each as a function of not only emissivity but absorbtivity as the net temperature difference tended towards 0 ( assuming it could in fact tend towards 0)?

    and just recently Tim said:
    “Standard dQ/Dt (scroll about 2/3 of the way down).
    “The radiation heat transfer from one surface to another is simply equal to the radiation entering the first surface from the other [added: in other words "back radiation"], minus the radiation leaving the first surface.”
    http://en.wikipedia.org/wiki/Thermal_radiation

    Sorry, but this is just a staight radiative energy transfer, no back radiation involved here at all. The energy transfered to an object is equal to the incoming energy minus the energy the object re-radiates. If back radiation existed the definition would need to be modified to say something like:
    ‘The radiation heat transfer from one surface to another is simply equal to the radiation entering the first surface from the other, minus the radiation leaving the first surface, plus the percentage of the energy leaving the first surface and entering the second object.’

    This would of course pose a problem because the energy transfer would never end and we would have created a perpetual motion machine, not to mention a really hot one :-) In other words you would never reach energy and temperature equilibrium between the two objects, which is obviously contrary to reality.

    Back radiation from a colder object cannot heat a warmer object, energy only travels from a warmer to a colder object, unless external work is applied.

  394. Dear Digitap,

    Which will cool off faster, and to a lower temperature:
    A 275K anvil in a 200K frigid bath…
    A 275K anvil in a 150K frigid bath…

  395. Whereas the condensation of water vapor liberates heat
    and whereas that phase change does not exhibit a temperature change
    therefore, black bodies from outer space are not telling much about the planet’s energy flux.
    We all know that on a sunny day it’s cooler to be in the shade.
    Gardeners know that at night, clouds will reflect heat and keep it warmer than a clear sky.
    It seems to go without saying enough, though, that all the while our water vapor clouds are moving the heat from the oceans to a place near you, CO2 sits there with it’s thumb up it’s butt, useless for shade or reflecting heat. It needs to be said and repeated, I guess.
    The amount of heat transported by the CO2 in any volume of our atmosphere is less than 1/50,000 that of the water vapor in that same volume. It just doesn’t count IRL.

  396. Ira:

    “You criticize my reply to “JAE” (all upper case) but actually I was replying to “jae” (all lower case). I don’t know if both “JAE” and “jae are the same person, but “jae” has earned my reply, at least IMHO.”

    Yes they are the same, but I think your reply was a little harsh. But I also don’t care.

  397. Steve:

    “Dear Digitap,

    Which will cool off faster, and to a lower temperature:
    A 275K anvil in a 200K frigid bath…
    A 275K anvil in a 150K frigid bath…”

    Answer obvious, and pertains only to anvils in frigid baths (or similar concepts), but it is completely irrelevant to the discussion at hand: it has NO relationship to atmospheric physics, due to “other factors,” such as convection and evaporation/condensation of water.

    Sorry.

    I am SO damn sick of the radiative effects being presented without any mention of all the other variables. It is, to put it as kind as I can muster, STUPID! This planet does NOT operate solely on the basis of some radiation cartoons! Please tell us how these radiative effects interact with convective and evaporative effects, and then you won’t have to deal with so many skeptics.

    Here’s the real question: Will an atmosphere of almost pure OCO warm the surface more than an atomsphere of 0.04% OCO?

    Evidently not, according to the links I provided above.

  398. Alleyne says:

    “Sorry, but this is just a staight radiative energy transfer, no back radiation involved here at all. The energy transfered to an object is equal to the incoming energy minus the energy the object re-radiates. ”

    I think we just have different definitions of the words. What you call “the incoming energy” (from radiation from some second object directed at the first object) is what I call “back radiation”. I know the name is poor and can lead to confusion.

    Then you say “the definition would need to be modified to say something like:
    ‘The radiation heat transfer from one surface to another is simply equal to the radiation entering the first surface from the other, minus the radiation leaving the first surface, plus the percentage of the energy leaving the first surface and entering the second object.’ ”

    I sort of see your point, but again, I think it is just a question of terminology. The “radiation back toward the first object from the second object” is certainly just a function of the temperature of the second object — as you say, no “percentage of the energy leaving the first surface and entering the second object” is needed in the equations. I agree 100% with you.

    However, I would ask what sets the temperature of the second object that is radiating back? Well, it could have many sources of energy, but one of them will be the energy it receives from he first object. That energy will raise the temperature higher than if that second object was not receiving energy from the first object. This elevated temperature will cause the second object to radiate more energy. In particular, the “radiation back toward the first object from the second object” will increase. In otherword, while the “back radiation” is completely determined by the temperature of the second object (along with geometry & emissivities), that temperature is at least in part a function of the “forward radiation” from the first object.

    I don’t expect the “divergent series problem you suggest here: “This would of course pose a problem because the energy transfer would never end and we would have created a perpetual motion machine, not to mention a really hot one :-) In other words you would never reach energy and temperature equilibrium between the two objects, which is obviously contrary to reality.”

    There will indeed be an equilibrium reached because the series will converge. There is a discussion in this thread if you want to see what was said before. Search for “chicken” to find the relevant parts of the discussion. http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/

  399. IAmDigitap says: May 11, 2011 at 5:20 pm

    A frigid fluid immersion bath, never was, isn’t, and won’t be tomorrow, a warmer of the object immersed therein.

    The earth is “immersed” in an “amazingly cold” fluid of 3 K microwave photons. With no energy source, the earth would cool to 3 K.

    The earth IS heated — it absorbs ~ 240 W/m^2 of sunlight. This is enough to raise the surface temperature quite a bit (the exact value depends on what assumptions you make about emissivity & albedo).

    Now immerse the earth in a “frigid fluid” of IR photons from the atmosphere. This fluid is indeed cooler than the surface. (For simplicity, we could leave a thin vacuum layer around the earth to prevent conduction & convection so that only radiation was important.)

    QUESTION. Will this surface, immersed in the “frigid” fluid of IR photons, be warmer than the surface immersed in the “amazingly cold” 3 K fluid of photons?

  400. Bryan,

    I agree that the energies involved in the MB distribution for room temperature gases would put the IR energies out toward the tail.

    It would be interesting to explore that issue more, but right now I don’t have time to dig into that. (There would also be messy details introduced my the fact that the energies of the vibrational modes are quantized). My hunch is that there will be fat enough tails for the molecules to emit the IR, but I could be wrong.

  401. It sounds like science doesn’t know what effect doubling the N2 in the earth’s atmosphere will have on surface temepratures, all else being held equal.

    As such, there is no way to know how much of the 33K “missing” in the radiative transfer model is due to N2/O2. Without that basic knowledge, the assumption that all of the 33K heating is due to GHG seems an enormous leap of faith. Especially as the main prediction of this model, the “hot spot” cannot be observed except through creative squinting. This in itself would in most science fields constitute falsification of the GHG theory of global warming.

    The fact that no hot spot has develped suggests that the effects of GHG are to increase the vertical convection rate, increasing the rate at which heat is returned to space. That gravity is responsible for the 33K in unexplained heating and contrary to the assumptions of the radiative transfer model, increasing the weight of N2O2 in the atmosphere will increase the surface temperature, as more and more molecules are packed into a smaller volume, resulting in a net increase in energy per cubic meter of atmosphere at the surface, which we measure as an increase in temperature.

  402. ferd berple says: “…increasing the weight of N2O2 in the atmosphere will increase the surface temperature, as more and more molecules are packed into a smaller volume, resulting in a net increase in energy per cubic meter of atmosphere at the surface, which we measure as an increase in temperature.”

    No, packing more molecules of a gas into a smaller space does not, in and of itself, raise the temperature of the gas. It raises the pressure. If what you are thinking were true, scuba divers would have one hell of a problem, given that tanks are typically filled to 200 bars. Those metal tanks are not insulated, and with internal temperatures 200 times the ambient air temperature… those would be some seriously hot tanks!

    jae says: “Here’s the real question: Will an atmosphere of almost pure OCO warm the surface more than an atomsphere of 0.04% OCO? Evidently not, according to the links I provided above.”

    Are you referring to the link regarding “Venus atmospheric pressure explains Venus surface temperature?” That link only demonstrates a misunderstanding of the ideal gas law.

  403. jae says:
    May 11, 2011 at 6:30 am
    In that vein, I’m still really interested why there is so much silence surrounding this (and other related expositions):

    http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html


    Presumably because it’s nonsense?
    An example, it starts with:
    “Venus is closer to the Sun, and gets proportionally more power from it. Earth is 93 million miles from the Sun, on average, while Venus is only 67.25 million. Since the intensity of the Sun’s radiation decreases with distance from it as 1 over r-squared, Venus receives (93/67.25) squared, or 1.91 times the power per unit area that Earth receives, on average.”

    It goes on to assert that any point in the Venusian atmosphere at the same pressure as the Earth’s would be at the same temperature with appropriate adjustment for the relative power received from the sun. This is said to be the 4th root of the ratio of the powers (∜1.91), and lo and behold they match!
    The trouble is that ~30% the power received at the Earth’s TOA is reflected and ~90% of the power received at the Venusian TOA is reflected, so by his argument Venus receives less power than Earth!

  404. Joel Shore says:

    “How am I supposed to convince you of something that involves seriously weighing of scientific evidence when I cannot even get you to admit you have made in error when it is right in front of your face for everyone to see?”

    Quit blustering. My comment was: “Please provide testable, measurable, reproducible and convincing evidence demonstrating verifiable global harm specifically due to the rise in CO2 – you know: per the scientific method.”

    I’m still waiting.

  405. Alleyne says:
    May 11, 2011 at 5:54 pm
    “Tim Folkerts says:
    May 10, 2011 at 9:42 am
    mkelly,

    You needed to look a little farther and understand what you are reading. This is from the SAME source you quoted:

    Net Radiation Loss Rate

    If an hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as

    q = ε σ (Th^4 – Tc^4) Ac (3)

    http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html

    Please note there ARE inputs for back radiation (the Tc^4 term).”

    The equation cited does not, in my opinion, imply backradiation. If it did there would be terms for the emissivity of both objects as well as the surface area of both objects. There aren’t.

    There certainly are, the energy balance is done at the surface of ‘h’, and the area is the same for both outgoing and incoming radiation, ‘Ac’, since ‘h’ is immersed in ‘c’.

    Strictly the equation should be:

    q =(ε σ Th^4 – a σ Tc^4) Ac
    however following Kirchoff’s Law a=ε
    so:
    q = ε σ (Th^4 – Tc^4) Ac

  406. Alleyne, very clearly and properly said. Can I use your words in a pinch?
    (I should have taken a writing course)

    You know, I searched for the first reference to the words “back radiation” and the only official place I could find it used is in I.P.C.C.’s AR4 WGI Chapter 3 – 3.4.4.2 Surface Radiation. It is not in any pre-1990’s physics books I can find. I sure would love to find out where and when this term and warped view of energy transfer came to be.

  407. ferd berple says: May 11, 2011 at 8:21 pm

    “It sounds like science doesn’t know what effect doubling the N2 in the earth’s atmosphere will have on surface temepratures, all else being held equal.”

    More specifically, it sounds like us amateurs posting here in our spare time don’t know the answer. I suspect there are calculations that have been done that could answer your question.

    The biggest question to me is how would that affect the distribution of CO2 & H2O since those are the key players in the radiation balance. Part of me says that more gas of any sort (ie the N2) would raise the whole atmosphere, including the CO2; part of me says each gas is independent and of the partial pressure of CO2 stays the same, then vertical distribution of CO2 would stay the same. That question is non-trivial and would require a little more thought for me to reach any firm conclusion.

    In the first case, I suspect the extra N2 would have a significant impact; in the second case, the N2 would have only a minor impact.

  408. Wayne,

    If you don’t like “back radiation”, you could always use “plain old thermal radiation from the molecules in the atmosphere that happens to be directed back down in the general direction of the surface”.

    Just like the light that happens to be heading west from a light bulb could be called “west radiation”. Just like microwaves aimed up and to the northwest could be called “(-i + j + k) radiation”. Just like a photographer’s flash that is aimed back toward the camera from the far side is called “back light”.

    Adding a designation to the direction of a flow of photons doesn’t create any new ideas or new laws of physics. It doesn’t invalidate any old laws of physics. It simply tells people the direction of the relevant EM radiation.

  409. Ira Glickstein might care to comment

    Tim Folkerts says …..”I agree that the energies involved in the MB distribution for room temperature gases would put the IR energies out toward the tail.”……

    Yes its all about probabilities for air at 250K.

    1. From temperature considerations 4% CO2 molecules in the active vibrational state.
    2. Now add a flux 15um radiation from Earth surface to further enhance the active numbers.
    3. The relaxation time of vibrational state is longer than the chance of collision deactivation
    4. The collision route will share the vibrational energy between CO2 and O2 molecules leading to local temperature increase.
    5. The radiative route takes two paths
    a. Collision activation of longer wave IR emissions from H2O is favoured for two reasons
    (i) Nearly 40 times more H2O molecules than CO2
    (ii) Plenty H2O bands at wavelengths > 15um => much more probable.
    6. Still about 4% chance of CO2 getting enough energy to be vibrationally active but even then back to 3 above.
    Net result of all this
    Photon energy from 15um Earth surface up going to increase in temperature of atmosphere and increasingly more radiation in wavelengths > 15um.

    Some might find a Kirchhoff problem here;

    But I think this worry is unfounded
    1. Photon energy being transformed into translational KE in N2,O2 molecules.
    2. Kirchhoff requires LT E to be valid and in this case this is not satisfied.

    My clincher evidence comes from the “bite around 15um ” seen in the satellite looking down spectrum.

  410. Tim Folkerts says

    …..”Wayne If you don’t like “back radiation”, you could always use “plain old thermal radiation from the molecules in the atmosphere that happens to be directed back down in the general direction of the surface”…….

    However Tim, climate science seems to have saddled itself with labels that are to say the least highly misleading.
    Back radiation includes a lot of solar radiation moderated by the clouds; in other words it NOT come BACK from the Earth surface.

  411. Robert Clemenzi writes:

    R Stevenson,

    I would very much like a link to that spreadsheet. My attempts to locate Hottel’s method keep hitting paywalls.

    Using 1.0% water vapor, my program computes 282.68 W/m2 .. close to your value.

    For power absorbed, my program incorrectly assumes 1 km at constant temperature and pressure. However, it also plots the distance to absorb x% of the available radiation. As long as those values are under 100 meters, I feel that assuming constant temperature and pressure causes only minor errors.

    At any rate, I would like to investigate WHY the programs are giving different values to absorb the radiation.

    Robert Clemenzi

    In my method I calculate the absorptivity of a hemispherical mass of gas (dry air for CO2 alone) of radius L (L= extinction distance) using graphs developed by Hottel and Mangelsdorf which can be found in Heat Trnsmission by McAdams .

  412. Bryan says:

    “Back radiation includes a lot of solar radiation moderated by the clouds; in other words it NOT come BACK from the Earth surface.”

    I think the preposition you chose is a big part of the misunderstanding. The radiation is back TOWARD the surface, not back FROM the surface.

    Beyond that, I’m not sure I agree with your wording that back radiation “includes a lot of solar radiation moderated by the clouds”. I interpret the Trenberth energy balance diagram as follows:
    * If the solar radiation is simply scattered/reflected AWAY from the surface, it is part of albedo, which is included in the 77 W/m^2.
    * If the solar energy is reflected/scattered TOWARD the ground, it is part of the 168 W/m^2.
    * If the solar energy is ABSORBED by the clouds, it is part of the 67 W/m^2.
    So the “solar radiation moderated by the clouds” is already all accounted for.

    The clouds do indeed provide part of the 324 W/m “back radiation”, with the rest provided by the GHGs. But that 324 W/m^2 of thermal IR comes from the energy OF THE CLOUDS (and air molecules) at a rate determined by the temperature OF THE CLOUDS (and air molecules).

    While there is a little more potential for misunderstanding from this term than from many others,with a small effort, anyone who knows some science should readily be able to understand the meanings.

    As another example, the phrase “for every action there is an equal and opposite reaction” is misleading to many people. That doesn’t mean Newton’s 3rd Law needs to be replaced. It just means you need to be a little more careful when teaching Newton’s 3rd Law.

  413. Tim Folkerts says:
    May 11, 2011 at 10:00 pm

    ferd berple says: May 11, 2011 at 8:21 pm

    “It sounds like science doesn’t know what effect doubling the N2 in the earth’s atmosphere will have on surface temepratures, all else being held equal.”

    More specifically, it sounds like us amateurs posting here in our spare time don’t know the answer. I suspect there are calculations that have been done that could answer your question.

    Or less specifically formulating an answer is an exercise in wool gathering.

    Let’s start off with one effect. It would increase atmospheric pressure at sea level to 1.7 bar. That would raise the boiling point of water to 250F or so. The three-minute boiled egg would become the two-minute boiled egg. There would be a zillion other things that change.

  414. Ira Glickstein, PhD says:
    May 11, 2011 at 3:08 pm

    New calcs of radiative GHE shows the GHE is only 8-9 deg. C:

    http://activistteacher.blogspot.com/2011/05/radiation-physics-constraints-on-global.html

    //////////////////////////////////////

    WOW! I just read the linked webpage and it seems fairly convincing. I hope Joel Shore and Dave Springer and others comment and either set me straight or agree that I and virtually everybody else on the Internet who claims the Atmospheric “greenhouse effect” contributes about 33º to the average temperature of the Surface. If the correct number is closer to a third of that amount, established science has got a lot of ‘splainin to do. (I am pretty sure there is an error in the new claims, but I cannot figure it out on my own.)

    Tim beat me to it but I wouldn’t use white paint as the example. I’d use the ocean.

    The ocean absorbs close to 100% of the visible spectrum that reaches it and the absorption continues to a depth of about 100 meters where there is effectively no more light.

    It doesn’t re-emit visible light. It emits in the far infrared and it sucks as an infrared emitter. It also sucks as an infrared absorber. Infrared radiation only penetrates the ocean to a depth of 1 micrometer. Instead of warming the water down to a significant depth it rather serves to increase the evaporation rate and the heat is carried off the surface as latent heat of vaporization. So the daytime radiative heating of the ocean isn’t followed by radiative cooling at night because water is quite opaque to IR. Only a thin 1 micrometer film at the surface can transmit IR. Convection does the heavy lifting. As the surface water cools at night through evaporation it gets denser and sinks while warmer water from below rises to replace it.

    The truism that good absorbers are good emitters has a caveat: it only applies to absorption and emission of the same wavelength. The nut of the greenhouse effect is dependent on this. CO2, like water, is largely transparent to visible light but opaque in the infrared spectrum. Energy enters the system as visible light but must escape as infrared because nothing gets hot enough to emit in the visible spectrum.

    [Update 3:30PM Florida time. THANKS Dave springer for this info. I fixed the blockquote stuff. - Ira]

  415. Speaking of caveats I owe Phil an explanation on atmospheric nitrogen thermal radiation. It must remain true that nitrogen radiates. ALL matter above absolute zero has thermal radiation which is a consequence of atoms being composed of oppositely charged particle in motion relative to each other. However, nitrogen is a poor absorber of infrared so it is also a poor emitter of infrared. So while it must have a thermal spectrum characteristic of its temperature the intensity of that spectrum is down in the dirt so it contributes nothing by itself, per se, to the infrared glow of the atmosphere. However it can still itself be thermalized kinetically and it can transfer kinetic energy to good IR emitters like water vapor and CO2 so it can’t just be ignored like it isn’t there at all.

  416. R Stevenson says:
    May 12, 2011 at 2:54 am

    My heat transfer book has a rough mean beam length formula of Le=3.6(V/A) where V is volume of gas and A is area.

  417. Tim Folkerts said ….”The clouds do indeed provide part of the 324 W/m “back radiation”, with the rest provided by the GHGs. But that 324 W/m^2 of thermal IR comes from the energy OF THE CLOUDS (and air molecules) at a rate determined by the temperature OF THE CLOUDS (and air molecules).”….

    The natural meaning of Backradiation is radiation that has come “back” from somewhere.
    The only place “back” can refer to in this context is the planet surface.
    Since it plainly does not all originate from the planet surface the term should be avoided.
    Downwelling long wavelength radiation is a much better title.
    If we were to accept your definition then from KT97 we have 350W/m2 up and 324 W/m2 back.
    The readily available spectra graphs show that this is wrong.

  418. Bryan says:
    May 12, 2011 at 12:57 am
    Ira Glickstein might care to comment

    Tim Folkerts says …..”I agree that the energies involved in the MB distribution for room temperature gases would put the IR energies out toward the tail.”……

    Yes its all about probabilities for air at 250K.

    1. From temperature considerations 4% CO2 molecules in the active vibrational state.
    2. Now add a flux 15um radiation from Earth surface to further enhance the active numbers.
    3. The relaxation time of vibrational state is longer than the chance of collision deactivation
    4. The collision route will share the vibrational energy between CO2 and O2 molecules leading to local temperature increase.
    5. The radiative route takes two paths
    a. Collision activation of longer wave IR emissions from H2O is favoured for two reasons
    (i) Nearly 40 times more H2O molecules than CO2

    True near the surface, becomes less so as you increase altitude or go near the poles or to deserts. So near the surface the energy absorbed by CO2 is likely to be shuttled through N2/O2 to be emitted at longer wavelength by H2O. It’s like how a Helium/Neon laser works, the He is excited by electric discharge collides with Ne thereby exciting it (3s level) and then Ne emits (3s->2p) the familiar red beam. As seen from the ‘down’ spectra though some CO2 is still emitted.

    (ii) Plenty H2O bands at wavelengths > 15um => much more probable.

    Again depends where you are in the atmosphere. Also depends on the relative radiation lifetime of the excited states of H2O, if shorter than the CO2 lifetimes or not. From what I recall from shocktube studies, H2O has a slightly shorter lifetime but that was at wavelengths 14 micron and below.

  419. mkelly says:
    May 11, 2011 at 12:12 pm

    “Pressure, Joel, pressure. There I have successfully told you. Whether you accept is a different story.”

    So if I compress air into tank until the compressional heating raises it above the boiling point of water then I can use it to run a steam engine forever, right? Because it will never cool off until I let the pressure escape.

  420. Bryan says: May 12, 2011 at 8:26 am

    “Downwelling long wavelength radiation is a much better title.”

    I think everyone agrees this is a much more descriptive and accurate name. So every time you hear “back radiation” just substitute in your mind “Downwelling long wavelength radiation”. The terminology is secondary, as long as everyone who really cares understands what is meant.

    “The readily available spectra graphs show that this is wrong.”

    Could you provide a link to specific spectra that you are referring to? What is wrong with them?

  421. Phil thanks for your input and your qualification of particulars in my post are well founded.

  422. Dave Springer says:
    May 12, 2011 at 8:51 am

    Sir, if you disgree that pressure causes an increase in temperature if volume is held steady that is well odd. PV= nRT discribes what is termed STP (Standard temperture and pressure). At 1 atmosphere the temperature of air is 0 C. On earth that accounts for 18 degrees of the so called 33 degrees of warming by GHG.

    Is it hotter in Death Valley or the top of Everest on the same day of the year?

    Further, I repeat that none of the heat transfer equations have an input for back radiation. So talking of something that does not exist in formulated heat transfer equations is strange. Please show me an equation that has an input for back radiation and is called back radiation in the formula. If you do I will speak no more of it.

  423. mkelly says: May 12, 2011 at 9:42 am
    “At 1 atmosphere the temperature of air is 0 C.”

    No, at 1 Atm the temperature of 1 MOLE OF GAS occupying 22.4 LITERS is 0 C. Go around the global at sea level and you will be right around 1 Atm the whole time, but I will guarantee the temperature will not be 0 C simply because the pressure is 1 Atm.

    “STP” is simply an ARBITRARY definition for uniform comparisons. It is in no way related to what temperature the air “should” be on earth or Jupiter or Pluto. Choosing a standard is only done to allow — well — standard measurements that can be compared easily.

    For example, in the US you buy natural gas by the cubic foot. But, of course, you would get a different number of moles in each cu ft depending on the temperature and pressure. The standard is 60 °F and 14.73 psi. This in no way says that natural gas “should” be 60 °F. It simply says that if you buy the natural gas at high pressure, you get charged more for the actual cu ft of gas because it would have more ‘standard cu ft’.

  424. mkelly says:

    Sir, if you disgree that pressure causes an increase in temperature if volume is held steady that is well odd.

    Nobody is disagreeing with that. However, the volume is not “held constant” in the atmosphere.

    You’ve had this all explained to you a million times: Considerations of the adiabatic lapse rate tell us the maximum rate at which temperature can decrease with height. However,

    (1) It does not tell us that the temperature must decrease with height at this rate, and in fact it doesn’t once you get beyond the troposphere. The actual temperature distribution depends on how the atmosphere is heated, with the adiabatic lapse rate just providing a stability limit…i.e., an upper bound on the allowed lapse rate. Since the bottom of the atmosphere (the troposphere) is heated strongly from below, it turns out that the actual lapse rate in this part of the atmosphere is usually close to the (appropriate dry or moist) adiabatic lapse rate.

    (2) Even once you know the lapse rate, that doesn’t determine the temperature at the surface alone. You must have an additional criterion, the temperature at some level in the troposphere, in order to determine the surface temperature. In particular, this criterion turns out to be that the average temperature at the effective radiating level in the atmosphere has to be equal to the “blackbody temperature” so that the earth system emits back out into space the same amount of energy it absorbs from the sun. This effective radiating level depends on atmospheric composition and, in particular, on those constituents that absorb terrestrial radiation.

    Further, I repeat that none of the heat transfer equations have an input for back radiation.

    If you repeat it a thousand times, it will be no less false than it was the first time. We have explained this to you. What you call it is totally irrelevant. If you don’t like the term, don’t use it. The greenhouse effect does not depend on calling the radiation that is downwelling from the atmosphere “back-radiation”. You can call it “Magical mystery radiation from the planet Zircon” and as long as you use the same equations, you will get the same result: namely, the greenhouse effect.

  425. [Perhaps this is obvious, but since nothing seems to obvious to state explicitly here, when I say "you use the same equations", I mean the standard equations used in all radiative transfer calculations, i.e., those used by scientists and engineers in practical calculations every day (as David M Hoffer has pointed out).]

  426. Joel Shore says:
    May 9, 2011 at 2:09 pm
    “Any text on thermodynamics that treats it from a statistical physics perspective ought to do fine. In the context of radiative heat transfer specifically, any text that talks about the exchange of radiation between two objects or any object and its surroundings ought to give you the basic radiative transfer equations that are used in computing the greenhouse effect.”

    Thank you Joel, but I’m afraid, not being as widely read as yourself, that I can’t find a textbook that supports your theory. Could you perhaps cite the document in question, and even better, quote the relevant passage(s)?  I am familiar with blackbody and greybody (to their surroundings) radiation equations and I confess that I find no evidence of backradiation in the equations.  Perhaps you could enlighten me?

    Joel Shore says:
    May 9, 2011 at 2:09 pm
    “Yes…So, heat (which is a macroscopic concept by its definition) always flows from hotter to colder (in the absence of work). However, note that the interpretation that the radiative energy flows in both directions, while having an abundance of empirical support, is not necessary to show there is a greenhouse effect. All that is really necessary to say is that the heat flow between two objects depends on the temperature of both objects and not just on the temperature of the hotter object.

    Of course the energy flow between the two objects depends on the difference in temperature between the two objects.  However that does not imply that energy flows from the cooler object to the warmer object.  That merely accounts for the difference in energy states.  The greater the difference in energy states, the more energy will flow from the higher state to the lower state.  The potential energy of an object 5 meters above the ground is greater than that of an object 1 meter above the ground.  The energy transfer from the higher object to the ground (via transformation to kinetic energy) will be greater than the energy transfer from the lower object to the ground.  However the energy transfer does not require that the ground move towards either object.  The energy transfer (current) in an electric circuit is unidirectional and proportional to the difference in the energy states between the source of the energy and its destination.  Energy is energy, it doesn’t change nature or characteristics, whether it is potential, kinetic or radiative.

    I don’t dispute that radiative energy can flow in both directions, I don’t think anyone does.  Any object above 0K will radiate some energy, just because you put a hotter one next to a colder one doesn’t mean the colder one stops radiating – but neither does it mean that the colder one will warm the hotter one.  

    The SB law of the radiation energy of a black body

    q = σ*T^4*A 

    states that the radiation energy of an object is proportional to the fourth power of its absolute temperature.

    The grey body equation for the radiation loss rate of an object to its surroundings 

    q = ε*σ*(Th^4 – Tc^4) Ah

    must correct for the temperature of the surroundings in order to correctly reflect the absolute temperature of the object, hence the term (Th^4-Tc^4).  This however in no way implies that the surroundings (which are radiating energy, not being at 0K) transfer energy to the object.

    Furthermore the equation has a term for the emissivity of the object and no term for the emissivity of the surroundings or the absorptiveness of the object, both of which would be required if the surroundings were transferring energy to the object.

    If you put a mirror a few feet away from an infrared source, say a heat lamp, the mirror radiates energy as visible white light and LWIR, both radiative energy, but that won’t increase the temperature of the heat lamp any.

    Joel Shore says:
    May 9, 2011 at 2:09 pm
    “I have no clue what you are trying to say in this last sentence. The term “heat pump” is sometimes used as a category that includes air conditioners and refrigerators. Other times, it is used to represent the subcategory of the “heat pump” in the above context that are actually used to heat a house rather than cool it. However, regardless of how the term is used, heat pumps used for heating and air conditioners operate by the same basic principle. Both use work to “pump” heat energy from cold to hot, i.e., in the opposite direction from which heat spontaneously “flows”.”

    I was using the term “heat pump” in the colloquial sense, meaning of a machine which heats a house by extracting heat from the surroundings and moving it into the house.  The reason I contrasted the two types of ‘heat pump’, heat pumps and air conditioners, was to illustrate the large amount of work (energy) required to circumvent the Second Law – ie to cool an object further by moving energy from it to a warmer one or warmer surroundings.

    I don’t necessarily disagree at this point with the concept that GHGs trap heat, I disagree with the idea contention that they heat the earth through backradiation.  They can’t, they don’t.

    But just for the sake of argument, tell me in your theory what the temperatures of the earth and atmosphere are, the amount of energy radiated from the earth to the atmosphere and backradiated from the atmosphere to the earth.  Please state all units in Kelvin if you don’t mind.  I’ll run the calcs and see if the numbers prove me wrong.

  427. Joel Shore says:
    May 12, 2011 at 10:42 am

    What you call it is totally irrelevant.

    Sorry I totally disagree. What a thing is named is important. What the equations call temperaturee you seem to want to call back radiation. Temperture is temperature and radiation is radiation. They maybe intertwined but they are not interchangeable.

    And since you did not supply an equation with a back radiation input I must declare I am correct.

    Alleyne says:
    May 12, 2011 at 11:29 am

    q = ε*σ*(Th^4 – Tc^4) Ah

    The above from an astute person shows clearly that if two objects of the same temperature are radiating at each other there is no (none, nada, zippo) energy exchange. No back radiation heating the other up as you claim back radiation heats the earth.

  428. MKelly says:
    My heat transfer book has a rough mean beam length formula of Le=3.6(V/A) where V is volume of gas and A is area.

    The original charts were for a hemispherical gas mass of radius L. Constants by which the characteristic dimensions of simple shapes are multiplied use 3.4(V/A) to obtain an equivalent mean hemispherical beam length L.

  429. Ira, thank you very much for your reply to my comment on your article “IraGlickstein, PhD says May 11, 2011 at 2:23 pm: “——-“

    I have not had a chance to reply until now. – Well, “until now” is not strictly correct as I finished an explanatory reply as to why you “get a disconnect”. However before I submitted my posting I read through some of the later comments and found that Tim Folkerts says it much more elegantly in his comment on May 11, 2011 at 3:12 pm, than I did in mine so I scrubbed my entry and recommend his instead.

  430. Mkelly,

    The above from an astute person shows clearly that if two objects of the same temperature are radiating at each other there is no (none, nada, zippo) energy exchange HEAT (ie no NET energy exchange). You even say specifically that they are radiating at each other; since radiation = traveling EM energy, you just admitted that energy is being exchanged in the very sentence where you say there is no energy exchange.

    As you just said, what you call things DOES matter — ESPECIALLY if you call them the wrong thing (ie saying that “q” is “energy” rather than “heat”)! There IS energy exchange when the temperature is the same — but the energy going each way is the same. There is no HEAT from one to the other.

    “What the equations call temperature you seem to want to call back radiation. ”

    This is just silly! Joel has always been way too careful to mistake temperature (measured in K) with radiation (in W in this case). Where does he do such an odd thing? Or is this another case where being careful to use the right words was not important?

  431. mkelly says: “Sir, if you disgree that pressure causes an increase in temperature if volume is held steady that is well odd. PV= nRT discribes what is termed STP (Standard temperture and pressure).”

    How are you applying pressure to your fixed volume? If I increase the pressure inside a scuba cylinder by pumping in more air, I get a proportional increase of n, not T, for my fixed volume.

    Are you trying to imagine some other way to apply pressure to a fixed volume without compressing it (not lowering V) AND without adding more gas (not increasing n)? That would indeed be an odd theoretical device – the non-compressing presser that heats a fixed volume. If I was trying to increase the pressure of a fixed volume, I would work the other way around – heat it to induce a temperature increase, leading to a pressure increase.

    In the real world, I do not think you would have “pressure causes an increase in temperature if volume (and moles!) is held steady”, I think you would have “temperature causes an increase in pressure if volume (and moles) is held steady.”

  432. Robert Clemenzi says:

    ‘Using 1.0% water vapor, my program computes 282.68 W/m2 .. close to your value.

    For power absorbed, my program incorrectly assumes 1 km at constant temperature and pressure. However, it also plots the distance to absorb x% of the available radiation. As long as those values are under 100 meters, I feel that assuming constant temperature and pressure causes only minor errors.

    At any rate, I would like to investigate WHY the programs are giving different values to absorb the radiation.’

    Robert,

    For water vapour I use a partial pressure of 0.231 atm. I calculate the emissivity/absorptivity of water vapour again using Hottel’s emissivity v absolute T graphs.

  433. This equation has shown up several times:

    q = ε*σ*(Th^4 – Tc^4) Ah

    Every equation applies in some specific set of circumstances (for example, d= vt only applies when v is constant). Looking at this equation, I would say it applies to radiation between two large, flat surfaces with area Ah with the same emissivity close to each other. It should work for a convex polyhedron of area Ah completely by the second object. It might well work for some other circumstances. (People are welcome to suggest other circumstances, or to suggest why I might be wrong, but that is not the point.)

    The given equation most certainly does NOT work for two arbitrary objects at arbitrary distances with arbitrary orientations with different emissivities. Expecting this simplified equation to work in general is a losing proposition.

    If you want an equation that applies to more general circumstances, you will need a more general equation. So the conclusion (furthermore the equation has a term for the emissivity of the object and no term for the emissivity of the surroundings or the absorptiveness of the object, both of which would be required if the surroundings were transferring energy to the object.) is more a statement about the limitation of the equation, not a statement about whether objects with different emissivities can transfer energy just one direction or both directions.

    PS As I pointed out once before, a more general equation is in Wikipedia http://en.wikipedia.org/wiki/Thermal_radiation . This equation DOES include the areas of the two objects, the emissivities of the two objects, and the geometry of the two objects. This removes the previous objection.

  434. Phil informs:

    “It goes on to assert that any point in the Venusian atmosphere at the same pressure as the Earth’s would be at the same temperature with appropriate adjustment for the relative power received from the sun. This is said to be the 4th root of the ratio of the powers (∜1.91), and lo and behold they match!
    The trouble is that ~30% the power received at the Earth’s TOA is reflected and ~90% of the power received at the Venusian TOA is reflected, so by his argument Venus receives less power than Earth!”

    Can you please share with us the source of your 90% reflectance number?

  435. Wayne,

    Sorry I missed this reply of yours.  Thank you for the kind words, you may of course use my words anytime you wish, for what they are worth :-)

    I too don’t remember the concept of “back radiation” occuring before AR4.  As with many things the IPCC claims, I find the actions they attribute to it unscientific and a misjudgement of cause and effect, as with the effect of clouds.  But I don’t want to open that can of worms at this time…

  436. Alleyne says:

    The grey body equation for the radiation loss rate of an object to its surroundings

    q = ε*σ*(Th^4 – Tc^4) Ah

    must correct for the temperature of the surroundings in order to correctly reflect the absolute temperature of the object, hence the term (Th^4-Tc^4). This however in no way implies that the surroundings (which are radiating energy, not being at 0K) transfer energy to the object.

    Well, yes, the 1st term is the transfer of radiant energy from the hotter object to the colder surroundings and the 2nd term is the transfer of energy from the colder surroundings to the hotter object. The heat q is the net radiative transfer and is clearly always positive when Th > Tc, meaning heat flows from the hotter object to the cooler object as the 2nd Law requires. [It is not necessary that you interpret the equation as two countervailing flows of energy in order to get the greenhouse effect, but since the actual radiation in each direction can be measured, it is worth noting that there really is a transfer of radiative energy back and forth.]

    Now, how does this lead to a greenhouse effect? What you have when Th is the temperature of the earth and Tc is the temperature of the atmosphere is a heat flow away from the earth that, while always positive, has a magnitude that depends on both Th and Tc. In radiative balance, this heat flow away from the earth will balance the heat flow to the earth from the sun. Now imagine that you start at Tc = 0 and then ramp it up. In order to maintain radiative balance, you have to ramp up Th. That is the greenhouse effect.

    [As Tim points out, the equation below has some simplifications in it already in regards to having a graybody object and blackbody surroundings and the surrounding completely surround the object...but for what we are using it to demonstrate, these simplifications arefine.]

  437. The -18 C “average temperature of the Earth” is the temperature where more radiation is leaving than is entering–about 5 km in the sky. As measured accurately by satellite. Because of the lapse rate (environmental lapse rate = -6.49 K/km), that means that the surface of the Earth has an average temperature that is about 33 C higher, or about 15C. We know that the lapse rate is not caused by a GHE, since it is simply a function of gravity and thermal capacity. I is caused by the relationship explained by the ideal gas law pV=RT. Therefore, it looks to me like there is no need to “explain” anything with a GHE.

    Of course, other planets with an atmosphere should show the same relationships, and it appears that they do. They all show the same type of increase in temperature with a decrease in altitude, such that the “black body temperature” of the planetoid is at a level that coincides with about 100 mb pressure (Earth is somewhat different, probably due to the presence of water). Doesn’t seem to matter what the gaseous constituents are. See the figure towards the end of this article:

    http://www.tech-know.eu/uploads/Greenhouse_Effect_on_the_Moon.pdf

    All right, now, tigers. Tell me why all this is wrong.

  438. Alleyne says:

    I was using the term “heat pump” in the colloquial sense, meaning of a machine which heats a house by extracting heat from the surroundings and moving it into the house. The reason I contrasted the two types of ‘heat pump’, heat pumps and air conditioners, was to illustrate the large amount of work (energy) required to circumvent the Second Law – ie to cool an object further by moving energy from it to a warmer one or warmer surroundings.

    Both a heat pump and an air conditioner use work to move heat from colder to hotter, so I don’t really see the distinction. The only substantive qualitative difference is that for the heat pump, the “hot reservoir” is your house and the “cold reservoir” is the outside and for the air conditioner, the “cold reservoir” is your house and the “hot reservoir” is the outside.

    I don’t necessarily disagree at this point with the concept that GHGs trap heat, I disagree with the idea contention that they heat the earth through backradiation. They can’t, they don’t.

    At some point, this comes down to terminology. It is most correct to say that because of back-radiation (in combination with the radiation from the sun), the earth is warmer than it would be if all the radiation from its surface escaped into space.

    But just for the sake of argument, tell me in your theory what the temperatures of the earth and atmosphere are, the amount of energy radiated from the earth to the atmosphere and backradiated from the atmosphere to the earth. Please state all units in Kelvin if you don’t mind. I’ll run the calcs and see if the numbers prove me wrong.

    This is an ill-defined question. The average flows of energy radiated and back-radiated have been measured to fairly good accuracy and are summarized in the diagram of Trenberth and Kiehl (e.g., http://www.nar.ucar.edu/2008/ESSL/catalog/cgd/images/trenberth9.jpg ). However, the atmosphere is not just at one temperature (nor is it a blackbody emitter). To compute what happens quantitatively, one must solve the equations for radiative transfer absorption-line by absorption-line through the atmosphere. This is quite an undertaking (although I think there are codes available on the web).

    To get a qualitative picture, you can use simpler blackbody (or graybody) shell models like the one we discuss in Section 2.3 of our comment on G&T: http://scienceblogs.com/stoat/upload/2010/05/halpern_etal_2010.pdf

  439. mkelly says:

    q = ε*σ*(Th^4 – Tc^4) Ah

    The above from an astute person shows clearly that if two objects of the same temperature are radiating at each other there is no (none, nada, zippo) energy exchange. No back radiation heating the other up as you claim back radiation heats the earth.

    Perhaps the most fundamental thing to understand about that equation is that if Tc was zero then there would be more heat going from the hotter object to the colder object than when Tc is not zero. That, along with the fact that the surface temperature of the earth is determined by balancing the heat it receives from the sun with the heat it radiates away tells you there is a greenhouse effect. In other words: the temperature Th necessary to radiate away the 240 W/m^2 of radiant energy that the earth receives from the sun depends on the temperature Tc.

    It is sort of amazing how such a simple concept seems to illusive for some people!

  440. JAE says:

    The -18 C “average temperature of the Earth” is the temperature where more radiation is leaving than is entering–about 5 km in the sky. As measured accurately by satellite. Because of the lapse rate (environmental lapse rate = -6.49 K/km), that means that the surface of the Earth has an average temperature that is about 33 C higher, or about 15C. We know that the lapse rate is not caused by a GHE, since it is simply a function of gravity and thermal capacity. I is caused by the relationship explained by the ideal gas law pV=RT. Therefore, it looks to me like there is no need to “explain” anything with a GHE.

    Are you reading anything we write here?!?! There is indeed something to explain: Why, from space, does the earth look like it has the temperature that is actually 5 km up in the sky? It certainly wouldn’t if all the radiant energy from the surface escaped to space. Then it would look like it had the temperature of the earth’s surface!

    For heaven’s sake, how many times do we have to say this?

  441. JAE says:

    http://www.tech-know.eu/uploads/Greenhouse_Effect_on_the_Moon.pdf

    All right, now, tigers. Tell me why all this is wrong.

    That paper has been debunked elsewhere. Basically, Hertzberg is clueless about how to calculate what the surface temperature has to be in the absence of the greenhouse effect and relate it to measured temperatures. What the blackbody temperature that you compute from the solar absorption actually constrains is the average of T^4 on the surface. It does not constrain the instantaneous temperature at a certain location, nor does it constrain the average temperature. (For a planet like earth where the temperature variations are not that large on an absolute scale, the difference you get from worrying about the average of T^4 vs the average of T is not that significant; for the moon, it is more significant).

  442. JAE says:
    May 12, 2011 at 1:56 pm

    The -18 C “average temperature of the Earth” is the temperature where more radiation is leaving than is entering–about 5 km in the sky. … Therefore, it looks to me like there is no need to “explain” anything with a GHE.”

    But the radiation that is leaving is coming in part from the GHGs, and also from clouds and also from the ground. The net radiation leaving is from all three added together. If there were no GHGs, they would not block the “warm” IR from the surface and they would not emit “cool” IR from the upper atmosphere. This would create a greater energy flow to space. To balance, the surface would have to cool off (or the cloud cover could change, or perhaps a few other things). You need to know about the GHE to understand this.

    Or put another way, why 5 km up? What physics leads to this? If the atmosphere was transparent to IR, all the IR would come from the surface, and the surface would be -18C. With a little GHG, the effective altitude might be 1 km — then the surface would -18 C + 6.5 C. With a lot, it might be 10 km and the surface would be (-18 C + 6.5C * 10) = 47.

    You NEED the GHGs & knowledge of the GHE to know the effective height. (Or equivalently, if you measure the effective height as 5 km , that tells you something about the actual GHE).

    The GHE sets the “effective altitude” where IR is emitted. The lapse rate sets how much warmer the surface is compared to this “effective altitude”. Both are needed.

  443. Sorry, Joel, but no entiendo. I’m reluctant to just accepting your words about “debunking,” “clueless,” etc. This talk reminds me too much of RC, which I can no longer bear to read. Sorry. Less armwaving, more specific info., and a link or two would be appreciated.

    But let’s go step-by-step. Earth first! What is wrong about my statements regarding this planet? Is it not true that the equilibrium radiation in/out is consistent with a temperature of about 255K, which is the average temperature at about 5 km elevation?

  444. “That paper has been debunked elsewhere. ”

    I think we need a “Snopes for Science”. Instead of looking into urban legends, it could look into scientific claims made on the internet. Then people could go there for info, rather than rehashing a topic every time it comes up in a discussion like this. :-)

    [Reply: WUWT is the 'Snopes for Science'.☺ ~dbs, mod.]

  445. Phil. said on Visualizing the “Greenhouse Effect” – Light and Heat
    May 11, 2011 at 9:12 pm

    “The equation cited does not, in my opinion, imply backradiation. If it did there would be terms for the emissivity of both objects as well as the surface area of both objects. There aren’t.

    There certainly are, the energy balance is done at the surface of ‘h’, and the area is the same for both outgoing and incoming radiation, ‘Ac’, since ‘h’ is immersed in ‘c’.

    Strictly the equation should be:

    q =(ε σ Th^4 – a σ Tc^4) Ac
    however following Kirchoff’s Law a=ε
    so:
    q = ε σ (Th^4 – Tc^4) Ac”

    Sorry Phil, you are correct, I should have been more accurate.  I should have said: the equation does not imply or support the claim that back radiation transfers energy to the hotter object.  All it says is that the energy transfer is equal to the difference between the radiative energy transfer per unit of time of the hotter object less the radiative energy transfer per unit of time of the cooler object.

    As I understand it Kirchoff’s Law states αλ = ελ, but only at thermal equilibrium and does not require that the absorbtivity and emissivity of the surroundings be the same as that of the emitting object, otherwise what you are suggesting is that everything has the same absortivity and emissivity and αλ = ελ = a constant.

  446. Tim Folkerts says:
    May 12, 2011 at 2:45 pm

    “Or put another way, why 5 km up? What physics leads to this?”

    The same physics that stops sunlight from penetrating the ocean more than 300 feet deep. It’s called optical depth.

  447. Tim:

    “With a little GHG, the effective altitude might be 1 km — then the surface would -18 C + 6.5 C. With a lot, it might be 10 km and the surface would be (-18 C + 6.5C * 10) = 47.”

    Well, the evidence does not appear to be supporting that concept. We have more GHGs now, but no “hot spot,” like your theory suggests.

    And is the level higher over Samoa (high levels of GHGs) than over the Sahara (very low levels of GHGs)?

    Was the average temperature 47 when OCO levels were at 2,000 ppm?

  448. Joel Shore says:
    May 12, 2011 at 2:15 pm

    “Why, from space, does the earth look like it has the temperature that is actually 5 km up in the sky? It certainly wouldn’t if all the radiant energy from the surface escaped to space. Then it would look like it had the temperature of the earth’s surface!”

    Actually there’s a fair size IR “window” where it does look the temperature at the surface. More or less water vapor can open/close other windows.

    Back radiation however is experimentally proven by pointing an IR spectroscope straight up in the air at night. If there were no back radiation it would only “see” the 3K cosmic background radiation. Instead it sees lots of energy at much higher temperatures up to and including the temperature of the dirt or water beneath it. The atmosphere glows (in all directions including downward!) in the infrared primarily around the emission frequencies of water vapor and carbon dioxide.

  449. FYI: an air conditioner is a heat pump. If you put one in your window to cool the house in the summer you can turn it around the other way to heat the house in the winter. More sophisticated models reverse themselves through plumbing so you don’t have to physically turn them around.

  450. mkelly says:
    May 12, 2011 at 12:00 pm

    Sorry I totally disagree. What a thing is named is important.

    A rose by any other name smells as sweet. ~Wm. Shakespeare

    Your lack of education evidently isn’t confined to science and extends into classic literature as well.

  451. The atmosphere gets colder as you move away from the surface because the surface is the source of the heat***. It’s works the same way as wearing layers of clothing in the winter. The layers are warmest nearest your body and get coldest nearest the outside air.

    ***This doesn’t apply once you reach the upper atmosphere where high energy radiation (which doesn’t penetrate the atmosphere very far) heats it up to thousands of degrees, ionizing it, and even causing weird stuff like the northern and southern lights. The international space station is orbiting about halfway through the upper atmosphere. The reason is doesn’t burn up is because the atmosphere is so tenuous at that height that it can’t transfer enough heat to the solid shell of the space station to have any effect on it. Nonetheless it is quite hot.

  452. The concentration of CO2 being very small compared to other gases is meaningless.

    It only takes a similar small percentage of pigment to turn white paint into dark paint. That tiny difference in pigment makes a huge difference in how hot the hood of your car gets on a clear day. CO2 is like a pigment – a little bit goes a long ways.

  453. Joel Shore says:
    May 11, 2011 at 5:45 pm

    It’s a small wonder that if you don’t understand the difference between PRODUCING energy, instead of merely spatially CONCENTRATING/STORING it, you’ll remain clueless about what I’m saying regarding thermal energy fluxes through the system.

  454. JAE says:

    But let’s go step-by-step. Earth first! What is wrong about my statements regarding this planet? Is it not true that the equilibrium radiation in/out is consistent with a temperature of about 255K, which is the average temperature at about 5 km elevation?

    Which part of the 25 times that Tim and I have explained this to you are you not understanding?!?

    Well, the evidence does not appear to be supporting that concept. We have more GHGs now, but no “hot spot,” like your theory suggests.

    You are just spouting meaningless nonsense talking-points that you pick up from garbage you read on the internet! The hotspot has to do with the moist adiabatic lapse rate; it says nothing to do about the mechanism causing the warming.

    Look, if you want to remain ignorant, then please just tell us and we’ll stop wasting our time (at least I will). In fact, I encourage you to go wherever you can and advertise that you are an AGW skeptic and believe the nonsense that you are defending here. You will just be confirming many intelligent people’s suspicions that AGW skeptics are a bunch of flat-earthers.

    I have better things to do with my time than waste it trying to teach others who don’t want to learn. David M Hoffer has already pretty much given up in hopeless disgust … and the poor fellow has more incentive than me to persist in that he has strong desire for the AGW skeptics not to look ignorant to the larger world. I actually feel quite a bit of pity for him and Ira and Roy Spencer who actually want to remain credible.

  455. David Springer:

    “Back radiation however is experimentally proven by pointing an IR spectroscope straight up in the air at night. If there were no back radiation it would only “see” the 3K cosmic background radiation. Instead it sees lots of energy at much higher temperatures up to and including the temperature of the dirt or water beneath it. The atmosphere glows (in all directions including downward!) in the infrared primarily around the emission frequencies of water vapor and carbon dioxide.”

    ??. I think I know what you mean, but you sure didn’t say it. It looks up and “sees backradiation.” OK, sure. But then you say “Instead it sees lots of energy at much higher temperatures up to and including the temperature of the dirt or water beneath it.” I guess you mean it sees radiation that is consistent with the temperature of the ….dirt or water. That part is nonsense, unless there is an inversion (which is why I think the Kiehl-Trenberth cartoon is also nonsense). The lapse rate dictates that the radiation “from the sky” will indicate a colder temperature than the dirt or water.

  456. Dave Springer says:
    May 12, 2011 at 4:05 pm

    >>Tim Folkerts says:
    >>“Or put another way, why 5 km up? What physics leads to this?”

    >Dave Springer says:
    >The same physics that stops sunlight from penetrating the ocean more than
    >300 feet deep. It’s called optical depth.

    EXACTLY! That optical depth is a result of the GHGs absorbing/emitting IR. This is exactly why GHGs and the GHE MUST be included!

  457. Dave Springer says:

    Actually there’s a fair size IR “window” where it does look the temperature at the surface. More or less water vapor can open/close other windows.

    Yeah…I meant in a total-integrated sense. But…you make a good point in noting that the earth doesn’t really look like some sort of blackbody object emitting at a certain temperature, corresponding to a certain level of the atmosphere. Rather, the emission varies strongly with wavelength due to the optical thickness of the atmosphere at these various wavelengths. The 5km “average (or effective) radiating level” is just that…an average that sweeps under the rug a lot of interesting spectral dependence.

    And, by the way, I should have also mentioned you in that list in my last post that included Ira, David M Hoffer, and Roy Spencer.

  458. “EXACTLY! That optical depth is a result of the GHGs absorbing/emitting IR. This is exactly why GHGs and the GHE MUST be included!”

    Eureka, eh?

    Possible logical error alert, I say.

    You are making the usual EXTREMELY big jump in saying that because there is absorption/emission by GHGs that there is a radiative GHE. That link HAS NOT BEEN ESTABLISHED EMPIRICALLY, JUST ASSUMED. You still don’t seem to get the FACT that there are all kinds of other things going on in the atmosphere. Convection might (and probably does) wipe out any GHE effects, just like with the “real greenhouse” effects.

    Science is NEVER settled until there is empirical evidence. And the empirical evidence, so fare, is against the link, as I pointed out many times above. Nice theory, though.

  459. Ira

    Your articles always generate much interest and comment, and this one is no exception.

    I am surprised that so many people hold such bullish views given the sparcity of empirical data and the complexity of the system and lack of understanding how it all fits together and works in practice. I remain unsure about many issues but I am reasonably confident of the following:

    1. That it is inappropriate to treat the Earth as if it were a blackbody (a rotating spherical object with oceans that act as heat sinks, swirling and variably heated atmosphere with variable latent heat characteristics, imprecisely known and changing emissivity etc is nothing like a blackbody). This fundamental misconception inevitably leads to a misconception as to what should be the ‘theoretical notional’ temperature of the Earth as if it had no atmosphere. Wrong model wrong answer.
    2. We have imprecise data with respect to albedo (Trenberth in his paper has albedo measurements varying between 27.1 and 35.8% and others have it is high as 40% and the albedo is constantly changing) and imprecise data with respect to the amount of solar energy absorbed by the atmosphere and imprecise data as to the amount of solar energy reflected by the atmosphere. These errors mean that we cannot accurately calculate the ‘theoretical notional’ temperature of the Earth as if it were a blackbody. These (potential) errors are not insignificant and afre such that they could lead to perhaps a 10 to 20K error in the assessment of the ‘theoretical notional’ temperature of the Earth (as if it were a blackbody).
    3. The water cycle alone, makes it inappropriate to assess ‘theoretical notional’ atmospheric temperature on the basis that the Earth is a blackbody.
    4. We do not have an accurate figure for the average temperature of the Earth. Indeed, this could be out by many degrees (say somewhere between 1 to 3K).
    5. That irrespective of the presence of GHGs, commonsense suggests that the atmosphere at or near the surface would be warmer than the ‘theoretical notional’ temperature of the Earth (assessed as if it were a blackbody). According to Trenberth some 78 W/m^2 of the incoming solar energy is absorbed by the atmosphere. Since the warmists case is that CHGs are transparent to incoming solar radiation, CHGs cannot account for this absorption. The absorption of this energy must have some heating effect on the atmosphere since energy is fungible and this incoming energy must have gone somewhere also not forgetting the latent heat content. The rotation of the Earth and the movement of the atmosphere (winds, convection etc) must inevitably lead to heating of the atmosphere (after all heat is a by product of work and considerable work is involved in moving the atmosphere). Most importantly, the oceans (which are heated by the sun during the day) are constantly heating the atmosphere day and night.
    6. If GHGs were responsible for all the warming of the atmosphere there would be some correlation between the concentration of GHGs and temperature both laterally and vertically but there does not appear to be such correlation.
    7. Adiabatic lapse/compression far better explains the temperature profiles of the atmosphere (although this is not a complete explanation in itself). You have likened it to a tyre and suggested that the temperature is radiated away. However, that analogy is incomplete. If you pump up a tyre and then drive on it (steering, braking, going over bumps etc), the side wall flexes slightly only slightly increasing the pressure as the tyre goes in and out of shape and this small movement is sufficient to maintain the temperature as long as the car is being driven (you see this in motorsport). The circulation of the atmosphere (hot air rising and cooling etc) is the equivalent and helps sustain the temperature of the atmosphere. The atmosphere is in constant flux and the work done inevitably generates heat.
    8. That Trenberth’s energy budget is wrong. It defies commonsence that DWLIR exceeds incoming solar energy given that the DWLIR can have been created only by the incoming solar energy (and energy radiated/released by the core which latter energy is said to be miniscule such that this can be ignored). The use of ‘average’ figures disguises and distorts what is truly going on. Just imagine for a moment what would be the position if the Earth’s speed of rotation was different. This would have little effect on the figures used in the Trenberth diagram but conditions on Earth would be radically different.
    9. That if DWLIR of the magnitude as indicated by Trenberth really existed, we would be able to extract energy from this and/or there would be considerable research into exploiting this natural green resource. After all, this DWLIR would have potential exceeding that of solar and wind.
    10. At most, the effect of GHGs is merely to delay heat loss being radiated. However, it is likely that at night (when there is no incoming solar energy) or at other times when atmospheric conditions are such that there is a temperature inversion, any LWIR that has been delayed has sufficient opportunity to radiate to space there by meaning that no excess heat is ‘trapped’ (ie., the heat in the atmosphere does not build up). Ira, see the article posted by E.M.Smith on Temperature Inversion titled Grostbite Falls January 23 2011 which gives a practical example of a ‘safety valve’ which serves to release any ‘excess’ temperature build up.

    I doubt that many people challenge the assertion that GHGs can absorb LWIR and re-emit some of this as potentially DLWIR (but bear in mind thermalisation which no doubt plays an important part). I doubt that many people would challenge the assertion that the warmer the atmosphere, the slower the heat loss from the surface of the Earth. The debate is not how gases behave in isolation in laboratory conditions but rather how the atmospheric system behaves as a whole with all its complexity and nuances, and it is because of this that many are skeptical as to whether in reality there is any significant ‘greenhouse’ effect.

    The upshot of the above is that I do not dismiss out of hand the contention that GHGs may contribute something to the warming of the atmosphere but I am very skeptical that the Earth is 33K warmer because of the presence of GHGs in the atmosphere. I consider that such a contention is far from proven both on figures and on principles involved.

  460. richard verney says: “According to Trenberth some 78 W/m^2 of the incoming solar energy is absorbed by the atmosphere. Since the warmists case is that CHGs are transparent to incoming solar radiation, CHGs cannot account for this absorption. ”

    The GHGs are transparent to incoming visible light, not all solar radiation. I think you have read an incorrect account of the “warmists” view.

  461. jae says: “You are making the usual EXTREMELY big jump in saying that because there is absorption/emission by GHGs that there is a radiative GHE.”

    “You still don’t seem to get the FACT that there are all kinds of other things going on in the atmosphere. Convection might (and probably does) wipe out any GHE…”

    How is convection going to “wipe out” a molecule’s ability to absorb/emit certain wavelengths of radiation? Are you saying that once energy is absorbed, a volume of gas is more likely to rise and expend all of that energy as work (adiabatic cooling), instead of emitting that energy as IR radiation? So the only GHE is an increased occurrence of rising warm air?

  462. Those advocates of the IPCC “greenhouse theory” should get real.
    If the radiative effects of H2O and CO2 are supposed to “heat” the atmosphere to 33K higher than “it would otherwise be” then they need to show significant radiative heating in a volume of the atmosphere.

    This they cannot do.

    Instead they distort equations and say look “there’s the proof”.
    If we went to the installer of our home central heating system with a complaint that it wasn’t working would we be satisfied with a couple of equations as an excuse.

    Call the bluff of IPCC “greenhouse theory” advocates!
    Ask for proof of radiative heating in a volume of the atmosphere.
    They will be stumped!

    The fact that the temperature profile of the troposphere can be derived from thermodynamics without reference to radiation sums it up.
    A good place for the IPCC advocates to start would be the radiative heating of a space the size of a real greenhouse.
    Lets get some real figures for say 30 cubic metres of air at STP on a KT average day.

  463. Tim Folkerts asks
    “The readily available spectra graphs show that this is wrong.”
    Could you provide a link to specific spectra that you are referring to? What is wrong with them?

    Tim, I have a spreadsheet of the Planck function similar to Ira’s
    So I can call up a blackbody spectrum for any temperature.
    This gives a surface up value.
    I then compared the atmosphere spectrum profile from Petty(2004) as given in page 127
    Physical Meteorology on line is Rodrigo Caballero’s Lecture Notes

    http://maths.ucd.ie/met/msc/PhysMet/PhysMetLectNotes.pdf

    With the surface continuous spectrum.
    My calculation is that all the radiation shown in this chart is reduced by 25% compared to the surface up radiation.
    So the maximum possible back radiation would be 75% of the surface radiation!
    K&T on the other hand have the reduction of only 8%.
    Thus they claim a figure of 92% of radiation is surface bound.
    I used the older K&T diagram .
    390 up (subtract 40 from ‘window’ as everyone agrees this radiation doesn’t come back).
    This leaves 350W/m2 up and 324 W/m2 “backradiation”.

    (350-324)/350 = 7.4%

    Caveat’s;
    1. A fairy rough and ready calculation
    2. The Petty spectrum might not be representative.
    However even with these reservations a reasonable conclusion can be drawn.

  464. jae says:

    Convection might (and probably does) wipe out any GHE effects, just like with the “real greenhouse” effects.

    Convection does not occur between the earth and space…and hence cannot get you around having to satisfy conservation of energy. Besides which, the convective-radiative models include convection. You are just spouting more nonsense.

  465. Bryan says:
    May 13, 2011 at 12:35 am

    If the radiative effects of H2O and CO2 are supposed to “heat” the atmosphere to 33K higher than “it would otherwise be” then they need to show significant radiative heating in a volume of the atmosphere.

    This they cannot do.

    There are some things that are easily provable yet there seems to be no dearth of skeptic who don’t accept what is proven.

    Point an infrared thermometer up at the clear sky at night. It shows a temperature much higher than the empty cosmos. That’s because the atmosphere glows in the far infrared. Point an infrared spectrometer up at the clear sky at night. It breaks the glow down into component frequencies and those are primarily the absorption/emission bands of water vapor and CO2. In the stronger absorption bands of water vapor when the absolute humidity is high the spectroscope will show temperatures in those bands closely approaching the ground temperature. The instrument in that case is seeing the temperature of the water vapor very close the ground.

    Emergency blankets commonly known as “space blankets” are made of exceedingly thin highly reflective mylar. A blanket of this material big enough to cover your body fits folded into a package the size of a teabag. It is made of a material with a mirror finish because that material reflects infrared radiation from your body back at your body and thus keeps you warmer than the same blanket made of non-reflective material.

    These things are indisputable proof of so-called “back radiation” and the FACT that back radiation slows the loss of heat from the body which it wraps whether that’s a human body on a cold night or the body of the planet.

    The person denying this just looks dumber than a fifth grader and makes it impossible to move on to the bits of the global warming story that aren’t well established facts like whether clouds have a net warming effect due to them being an effective insulator at night or a net cooling effect due them shading the ground underneath during the day. Or exactly how much daytime heating of the ocean gets lifted away from the surface by evaporation and convection. Or exactly how much heating of the equatorial ocean gets transported to the poles by ocean currents. Or exactly how much the albedo of the planet changes from year to year due to variation in type and amount of cloud cover and wind driven waves which break up the ocean surface and make it less able to absorb sunlight while at the same time increasing the evaporation rate which combine to make it much cooler than it would be if the winds were calm.

    There are a million details which are poorly characterized that are highly relevant to the anthropogenic global warming narrative. The basic physics of greenhouse gases are simply not one of those things that are not well-enough understood and if you don’t understand how greenhouse gases work you can’t possibly move on to any reasonable debate about other phenomena which can and do (IMO) largely negate the effects of increasing greenhouse gases and leave us in a situation where the modest increase in carbon dioxide has vast beneficial effect by warming the planet at high latitudes where warming is welcome, not warming it at low latitudes where it is already warm enough, increasing the growth rate of green plants, and decreasing the water needs of green plants at the same time.

  466. Steve says:
    May 12, 2011 at 10:13 pm
    richard verney says: “According to Trenberth some 78 W/m^2 of the incoming solar energy is absorbed by the atmosphere. Since the warmists case is that CHGs are transpa