Visualizing the “Greenhouse Effect” – Molecules and Photons

Guest Post by Ira Glickstein

This series began with a mechanical analogy for the Atmospheric “Greenhouse Effect” and progressed a bit more deeply into Atmospheric Windows and Emission Spectra. In this posting, we consider the interaction between air molecules, including Nitrogen (N2), Oxygen (O2), Water Vapor (H2O) and Carbon Dioxide (CO2), with Photons of various wavelengths. This may help us visualize how energy, in the form of Photons radiated by the Sun and the Surface of the Earth, is absorbed and re-emited by Atmospheric molecules. DESCRIPTION OF THE GRAPHIC

The animated graphic has eight frames, as indicated by the counter in the lower right corner. Molecules are symbolized by letter pairs or triplets and Photons by ovals and arrows. The view is of a small portion of the cloud-free Atmosphere. (Thanks to WUWT commenter davidmhoffer for some of the ideas incorporated in this graphic.)

  1. During the daytime, Solar energy enters the Atmosphere in the form of Photons at wavelengths from about 0.1μ (micron – millionth of a meter) to 4μ, which is called “shortwave” radiation and is represented as ~1/2μ and symbolized as orange ovals. Most of this energy gets a free pass through the cloud-free Atmosphere. It continues down to the Surface of the Earth where some is reflected back by light areas (not shown in the animation) and where most is absorbed and warms the Surface.
  2. Since Earth’s temperature is well above absolute zero, both day and night, the Surface radiates Photons in all directions with the energy distributed approximately according to a “blackbody” at a given temperature. This energy is in the form of Photons at wavelengths from about 4μ to 50μ, which is called “longwave” radiation and is represented as ~7μ, ~10μ, and ~15μ and symbolized as violet, light blue, and purple ovals, respectively. The primary “greenhouse” gases (GHG) are Water Vapor (H2O) and Carbon Dioxide (CO2). The ~7μ Photon is absorbed by an H2O molecule because Water Vapor has an absorption peak in that region, the ~10μ Photon gets a free pass because neither H2O nor CO2 absorb strongly in that region, and one of the 15μ Photons gets absorbed by an H2O molecule while the other gets absorbed by a CO2 molecule because these gases have absorption peaks in that region.
  3. The absorbed Photons raise the energy level of their respective molecules (symbolized by red outlines).
  4. The energized molecules re-emit the Photons in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.
  5. This frame and the next two illustrate another way Photons are emitted, namely due to collisions between energized GHG molecules and other air molecules. As in frame (2) the Surface radiates Photons in all directions and various wavelengths.
  6. The Photons cause the GHG molecules to become energized and they speed up and collide with other gas molecules, energizing them. NOTE: In a gas, the molecules are in constant motion, moving in random directions at different speeds, colliding and bouncing off one another, etc. Indeed the “temperature” of a gas is something like the average speed of the molecules. In this animation, the gas molecules are fixed in position because it would be too confusing if they were all shown moving and because the speed of the Photons is so much greater than the speed of the molecules that they hardly move in the time indicated.
  7. The energized air molecules emit radiation at various wavelengths and in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.
  8. Having emitted the energy, the molecules cool down.

DISCUSSION

As in the other postings in this series, only radiation effects are considered because they are the key to understanding the Atmospheric “Greenhouse Effect”. I recognize that other effects are as important, and perhaps more so, in the overall heat balance of the Earth. These include clouds which reflect much of the Sun’s radiation back out to Space, and which, due to negative feedback, counteract Global Warming. Other effects include convection (wind, thunderstorms, …), precipitation (rain, snow) and conduction that are responsible for transferring energy from the Surface to the Atmosphere. It is also important to note that the Atmospheric “Greenhouse Effect” and a physical greenhouse are similar in that they both limit the rate of thermal energy flowing out of the system, but the mechanisms by which heat is retained are different. A greenhouse works primarily by preventing absorbed heat from leaving the structure through convection, i.e. sensible heat transport. The greenhouse effect heats the earth because greenhouse gases absorb outgoing radiative energy and re-emit some of it back towards earth.

That said, how does this visualization help us understand the issue of “CO2 sensitivity” which is the additional warming of the Earth Surface due to an increase in atmospheric CO2? Well, given a greater density of CO2 (and H2O) molecules in the air, there is a greater chance that a given photon will get absorbed. Stated differently, a given photon will travel a shorter distance, on average, before being absorbed by a GHG molecule and be re-emitted in a random direction, including downwards towards the Surface. That will result in more energy being recycled back to the Surface, increasing average temperatures a bit.

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743 thoughts on “Visualizing the “Greenhouse Effect” – Molecules and Photons

  1. I like that animation. I have never been very good at that. :-)

    It is worth noting that gases can gain energy from collisions as well as by absorbing a photon. A molecule doesn’t care where it got the energy from. Once a molecule has more energy than the molecules around it, it will dissipate through releasing a photon or it could also lose it by another collision. Each molecule is having billions of such collisions per second under normal atmospheric conditions.

    You are correct in keeping it simple as the more complex the animation is, the harder it is to follow.

  2. Convection currents are the “great equalizer”…. I have a question I’m hoping someone can answer – I thought Al Gore got his graph backwards and the increase of CO2 followed warming – why are we still working with the assumption it creates it???

  3. The Ideal Gas Law PV=nRT quite accurately predicts the atmospheric temperature of Earth, Mars, Venus and Titan (at all altitudes) based on a simple pressure effect. Why add the unnecessary complication of the highly dubious Greenhouse Effect to an elegant and simple solution?

  4. Can we have one with the number of different GHG molecules represented proportionally for reference? We might not be able to see the CO2 one among all the others, though….

  5. Much of Newfoundland (“The Rock”) is at higher Lat than Minnesota. Though not as biting as “Hide the Decline,” tune-wise the M4GW guys have some professional competition…

  6. Let us use the back radiation according to the global warming science theory to warm our houses without burning fossil fuels.

    It is simple: paint a black square on your ceiling.

    Say the black paint covers 10% of the ceiling and let us compare the backradiation of the black square to the back radiation of the Co2 molecules in the air .

    Let us assume that the Co2 molecules in the airabove the ceiling are all present next to each other on the ceiling and radiating all the energy back to the floor . The backradiation of the Co2 molecules on the ceiling would be much stronger than the the real situation in the air.

    According to the global warming theory doubling the Co2 concentration would increase the temperature with one degree
    Instead of doubling the Co2 concentration we have a painted a black square.
    The surface of the black square is 250 times bigger than the Co2 surface
    The black paint will absorb and emit the whole infrared band from the floor
    whilst the Co2 surface will only backradiate from a narrow band in the spectrum.
    Because of its size the black square we can compare this to a (at least) 250 times increase in Co2 concentration.
    In other words we can expect an enormous increase in temperature in the room without using energy. Be carefull don’t paint the whole ceiling black !!!.

  7. Ira,
    I think a review of classical thermodynamics is in order here.
    You can not win
    You must lose
    You can not get out of the game.

  8. Very interesting discussion thread. I appreciate the simplification and the limitation of the argument, since this basic physical concept seems to confuse the waters in more cases than not.

    In the meantime, I’ve been pondering the question of atmospheric absorption of energy lately, and maybe you could pull out some interesting figures for the audience. Do you happen to have in your files the energy dependent photon absorption curves for various important gases in the air (i.e. N2, O2, H2O vapor, CO2, and why not Ar, H2, He, CH4, CO, and N2O)? I imagine the non-asymmetric molecules will not be very opaque, but it’s still a nagging question in my mind.

    Also, for the sake of education it my be instructive to show what the blackbody radiation spectra of a surface at, say 240K (Arctic), 285K (a normal summer night), and 310K (a hot day) look like.

    Thanks for the good work.

  9. Although there clearly is a greenhouse effect it is infinitesimal compared to the energy retaining effect of the oceans.

    Furthermore the non radiative processes seem to mostly or entirely negate its effect with the real reason for the Earth’s temperature being down to oceanic and not atmospheric energy retention.

    Really, we should regard the oceans as part of the atmosphere for energy budget purposes.

    The recent suggestion that removing CO2 would cause a snowball Earth cannot be right because there would still be solar shortwave into the oceans keeping them liquid and so a continuing water cycle.

    As far as I can see a removal of CO2 would just mean a little less energy in the atmosphere, lower the height of the tropopause a little for a shallower troposphere and redistribute the pressure patterns for a slightly slower water cycle for no significant temperature effect.

    The cooling effect of less CO2 would be all or mostly offset by the slowing of the water cycle just as the warming effect of more CO2 seems to be all or mostly offset by a speeding up of the water cycle.

  10. Sounds all very well BUT.

    This explanation completely ignores convection which is probably the most important heat distribution system. When this LW radiation adds energy to a GHG, ie warms up a CO2 molecule, it will immediately transfer this energy to molecules of lower energy levels, ie warm up the surrounding gasses. This is as per the 2nd law which states, through inference, that heat cannot be stored.

    When a parcel of gas becomes warmer than that surrounding it will convect due to density difference and rising air cools adiabatically. This cooling air will be warmer than that surrounding to continue convecting but cooler than the surface so energy will not be radiated to the surface because the 2nd law of thermodynamics forbids this to happen, energy only able to flow from a high to lower level.

    If it were possible to get energy to flow from cold to hot then we would have all our energy requirements solved. We would have a perpetual motion machine. This is impossible.

    So the above explanation may work in the laboratory but the atmosphere is a different place. If convection were not a major player in heat transport then we would see far less cloud than we do, most cloud being due to convection.

    When we look at geological history there were periods when atmospheric CO2 levels were many times today’s. There were ice ages and periods of warmth both independent of the atmospheric CO2 levels. Present day research shows that CO2 levels rise after rises in temperature which goes a long way to prove that CO2 does not drive temperature.

  11. bananabender. Go ahead, try to explain the temperature of the Earth using only the ideal gas law! It can’t be done, you see, if you actually start to think about what you do.

  12. “The Photons cause the GHG molecules to become energized and they speed up and collide with other gas molecules”
    I’m not really convinced about that, or better it could be not always true. In the particular case of the CO2 absorption band at 15um, as far I know, the energy of the photon is converted into molecular bending which just polarize the molecule. From my point of view, this shouldn’t increase the speed at all, but it should increase the probability of share energy with the surrounding molecules because the polarized molecule needs only two simultaneous collisions to share energy, while the unpolarized one needs three simultaneous collisions, one in one direction applied to the C atom and the other two applied in the opposite direction, or at least applied having angles on the O atoms which can bend the aligned O=C=O chain. If I’m right here, the CO2 molecules should be better as “photons absorbers” than “photon emitters”. So, it should have very little efficiency as “optical photon spreader”, but it should be a very efficient “thermodynamic photon sharer”.

  13. So a single CO2 molecule can both absorb photon energy and then re emit this energy in random directions? Given the fact that atmospheric CO2 only amounts to 0.039% of the atmosphere the net amount of re re radiated photons hitting the earth must be truly tiny, so small in fact as to be barely measurable.

    “The greenhouse effect heats the earth because greenhouse gases absorb outgoing radiative energy and re-emit some of it back towards earth.”

    The problem I have with this explanation is that the amount of re radiated energy directed downwards by such a trace gas as CO2 would be infinitesimally small and it is not constant and because it is not constant and the amount is so small this heating would be lost almost immediately within the far greater ‘noise’ signal of the other atmospheric gases which exist in far far greater amounts.

    The obvious question is one of proportions and percentages and given that atmospheric CO2 is present in such trace amounts just how much is CO2 heating the earth by? And given that global temperatures have not increased in the last decade and most of the 20th centurys small net warming occurred BEFORE the steep rise in atmospheric CO2 then I have strong reservations about CO2 being responsible for any measurable warming whatsoever.

    Here is the problem as I see it, the rise in atmospheric CO2 and the rise in global temperatures should be linked together and yet they are not, we have seen a fall in global temperatures even as CO2 increased and if CO2 was the driver of increasing temperatures we would have seen a greater increase in global temperatures at the same time as increasing CO2 levels yet the opposite has happened. CO2 rose quickly post war while global temperatures rose much more slowly post war and even as CO2 has been rising since 1998 global temperatures have not measurably increased at all.

  14. Careful – you’ll get Louis, Richard and the “Violation of the 2nd Law of Thermodynamics” crew on to you with outrageous statements like:

    “The greenhouse effect heats the earth because greenhouse gases absorb outgoing radiative energy and re-emit some of it back towards earth.”

  15. I am happy with this description overall but I am not sure about this part:

    “The Photons cause the GHG molecules to become energized and they speed up and collide with other gas molecules, energizing them”

    It is a long time since I studied absorption and scattering but this is how I see it.

    Photons interact with electrons in molecules. Photons with energies which excite any of the vibrational or rotational modes of the molecule will be absorbed, other wavelengths will generally not be absorbed. There are some inelastic scattering modes which result in photon absorbtion and re-emission at slightly different wavelengths but I do not think that this band broadening is what you are describing.

    A CO2 molecule colliding with, say, an N2 molecule, will pass on energy as well as receive energy during that collision. However the CO2 molecule has vibrational and rotational modes in the long wave energy band but N2 does not so any energy passed to the N2 molecule will not be radiated away whilst energy absorbed by the CO2 molecule might well be. On balance however I do not see this as being significant overall since the nett effect of the collisions should be small.

    Was it not you who showed the radiation plot looking up from the earth at night? This showed that there is virtually no atmospheric radiation outside of the absorption bands of the greenhouse gases. I do not understand why you feel a need to include these small scattering effects in your description while you exlude major effects like convection which change the altitude and therefore amount of radiation to space far more significantly.

  16. If all that is going on is absorption and re-emitting radiation there would be no heating of the atmosphere. Heating happens because of the re-distribution of energy through collisions between molecules. The “simple” visualization is thus too simple. The temperature of a gas IS the average kinetic energy of the gas molecules, this is something Roger Pielke Sr. has pointed out a number of times and for many years.

  17. Stephen Richards – that link that “disproves” theory of AGW – well for mine it is based on an incorrect understanding of the 2nd law and how it relates to warming. I see you beat me to my post at 2:06 though – damn moderation lag:)

  18. Jer0me says:
    March 29, 2011 at 12:43 am

    Can we have one with the number of different GHG molecules represented proportionally for reference? We might not be able to see the CO2 one among all the others, though….

    Like this one?

    ( In proportion: N2-green; O2-gray; Ar-brown; H2O-aqua (2.5%); CO2-red (400 ppmv) )

  19. Further to cal’s points, I don’t understand how CO2 and H2O are able to re-radiate in the way described.

    We are told by the AGW crowd that much of the expected (and so far undetected) warming occurs in the middle atmosphere, where the air temps are pretty low (-10C to -30C, up where the planes live).

    If a warm surface emits a photon that is captured at this height, and conduction results in almost instant passing on of the captured energy to surrounding gasses (primarily N2) then how do the H2O and CO2 molecules regain enough energy from the comparatively cold gasses to re-emit photons of similar energy?

    Surely they can only re-emit much lower energy photons, of similar energy levels to the black-body curve at -20C.

  20. So far the best write up.

    However, there is no mention of how much the re-emitted photon helps to warm the earth, and it is only a conjecture that it really does. For does it not hold true that if the re-emitted direction is random above a sphere the, in this case outer, space around it is the most likely direction?

  21. How much of this is theory, and how much observed, in the real atmosphere? Can we see it happening? Can we devise an experiment or observation which would work in the field? Personally, I’ve been struggling with radiative theory, I’ve seen stuff from all sides, and plainly theorists cannot agree. So what do we KNOW?

  22. Hank: thanks for your example. Although I am not a physicist, I thought that at the end in step 7 by Glickstein a little miracle occurred. If we have a perpetuum mobile working a tiny tiny bit, we can use that for heating our homes at least. Perhaps the next winter I will do your painting trick.

  23. Stephen Richards (2:05)

    the article in Climate Realists makes exactly the same mistakes as all those that think back-radiation somehow violates 2nd law thermodynamics. The last-but-one paragraph sums up his errors quite nicely – it’s basically completely wrong.

    Oddly, he describes a thought experiment with a black body and mirrors that he could easily try out, and which would show him he is wrong, Clearly he knows he is right, so doesn’t bother to actually test it.

    The other indicator that he is wrong is the massive introduction that is just waffle, before reaching any solid science. Why bother with all that if you are right?

  24. “where their energy is absorbed, further heating the Earth”

    That phrase is the crux of the difficulty that some have with the greenhouse effect.

    There is no ‘further’ heating.

    What happens is that the downward longwave radiation reduces the net upward flow by partially offsetting it until the temperature rises and a new equilibrium is reached.

    There is always a net energy flow from a warmer body to a cooler body in accordance with the Laws of Thermodynamics but in fact both bodies still radiate towards each other.

    A cooler body doesn’t stop radiating just because it is in the presence of a warmer body.

    It is the net rate of energy transfer between the two that changes with no need for the cooler body to effect any direct warming of the warmer body.

    The suggestions that the greenhouse effect somehow offends the Laws of Thermodynamics is a non starter and a hindrance to scepticism of the theory of AGW.

    Ira doesn’t seem to make that mistake but he has used a form of words that perpetuates the misunderstanding.

  25. Ira,

    Thanks for this, but for me the animation is way too fast to follow. Any chance you could slow it down?

  26. Ira,

    Good presentation!!!
    You’ve only touched partially on the tip of the iceberg of how complex this planets climate system is.
    Thermodynamics is garbage in being too simple and too broad in coverage so that simple minded scientists can grasp some hope of understanding.
    This planet is a globe and not a tube, that rotates and has gravity. From the equator to the poles, there are different factors going on besides the introduction of solar radiation, CO2, gases, water vapor, etc. Cold compression and superheated compression has very different areas in complexity as well as storing energy on a planet that is thrown through space at 300km/sec. We have yet to understand infused energy into the planet and solar system at creation and what that accomplishes in storing energy for slow release.

    Our current problem is that science can only grasp one area at a time and fails in understanding a multi-understanding of many areas into a super complex system.
    Following temperatures in a 150 year time period is ridiculous in the 4.5 billion year time frame and for what? To know what clothes to wear?

  27. Ira,

    I know you have the caveat paragraph “As in the other postings in this series, only radiation effects are considered because they are the key to understanding the Atmospheric “Greenhouse Effect”……

    But in comparison to the power of the hydrologic cycle your ‘radiation only’ approach is not unlike saying you are only looking at the retardation effects on the locomotive of hitting insects.

    The AGW CO2 warms the world hypothesis depends on the hydrologic cycle. It only works if they can show the ‘water vapor feedback’ which all the GCMs show as a tropospheric hotspot that in the real world does not exist therefore all the models are falsified. You cannot disregard the locomotive of the hydrologic cycle whose effects nobody has been able to quantify but is accepted as the main transport of heat from the surface to the tropopause and the ‘iris’ that increases albedo reducing incoming energy and only worry about the insect collisions of outgoing IR with CO2 in three small radiation bands.

  28. @Thomas says:
    March 29, 2011 at 1:57 am
    bananabender. Go ahead, try to explain the temperature of the Earth using only the ideal gas law! It can’t be done, you see, if you actually start to think about what you do.

    Try this:
    Timothy Casey B.Sc.(Hons.)

    http://greenhouse.geologist-1011.net/

    The temperature of the earth’s surface is often explained using the “Greenhouse Effect”. However, having refuted the “Greenhouse Effect”, we may wonder if it was necessary in the first place. The earth orbits the sun in the vacuum of space. There is no aether as Fourier, Tyndall and Arrhenius believed. Moreover, there is no heat capacity or thermal conductivity in space. The only way for heat to escape the planet is by emission to space. That makes the temperature of the absorbing mass of the earth a question of radiative heat transfer. Hereafter, I will refer to the that portion of the earth’s mass which absorbs solar radiation as the “solarsphere” because the atmosphere does not include the surface layer warmed by the sun on a day to day basis and there is no other term to encompass both. The method of calculation is to treat the solarsphere as an absorbing body subject to incident radiation from the sun.

    Given the solar constant of 1368 Wm-2 (Fröhlich & Brusa, 1981) and the fact that the cross-sectional area of solar radiation incident upon the earth is roughly one quarter of the earth’s surface area, it is unsurprising to observe that authors such as Kiehl & Trenberth (1997) arrive at 342 Wm-2 as the mean quantity of solar radiation that falls on the entire surface of the earth. Using this, we may calculate the expected geographical and altitudinal mean temperature of the earth’s solarsphere.

    Wm = σT4
    T4 = Wm/σ
    T = {Wm/σ}0.25

    Given Wm = 342:
    T = {342/0.000000056704}0.25 = 278.7ºK = 5.5ºC

    This figure, is an average or mean temperature for all times, latitudes, and altitudes of the the earth’s solarsphere. Just as the balance point or centre of gravity is found at the centre of mass, this average temperature may be found at the centre of heat capacity. In materials of similar heat capacity, this can be found near the centre of mass. Thus, in order to determine how well our 5.5ºC result -calculated above- corresponds to observed reality, we must first determine the average observed temperature at the barometric median in the part of the earth penetrated by solar energy.

    From the diagrams supplied by Vallier-Talbot (2007, pp. 25-26), we may roughly determine the centre of mass for a one square metre column extending from two metres below the surface to 50 kilometres above the surface. Soils and clays amount to roughly 2 tons per cubic metre, with the atmospheric column having to weigh 10 tons in order to yield a mean barometric pressure of roughly 1000 hectopascals at the surface. The total column weighs 14 tons with the centre of gravity corresponding to the barometric median at 700 hPa. Referring once again to Vallier-Talbot (2007, p. 26) we may determine that on average, this pressure corresponds to an elevation of roughly a mile or 1600m above the surface. Given the observed average atmospheric thermal gradient of -7ºC with every 1000m of elevation above the surface (Vallier-Talbot, 2007, p. 25), we may calculate the average absorbing mass temperature as it occurs at the altitude of the barometric mean for our absorbing column. No doubt you’ve worked out that the temperature drop over a tropospheric ascent is 11ºC per mile, and we all know that the average surface temperature is 15ºC (Arrhenius, 1896, p. 239; Burroughs, 2007, p. 124). Notwithstanding 100 years of apparently constant mean temperature from Arrhenius to Burroughs, we may determine that the observed temperature at the altitude corresponding to the centre of absorbing mass is 4ºC or 277ºK. This, via the reasoning above, extends to an observed average absorbing mass temperature for planet earth of 4ºC or 277ºK. This is slightly cooler than the mean absorbing mass temperature calculated above from the solar constant (278.7ºK, 5.5ºC) even if we do allow for 0.5º warming over the last century. However, if we were to consider the impact of convective cooling, I think we can agree that the temperature we derive from the Stefan-Boltzmann equation is well within the tolerance we must allow for such tests.

    Adding the tropospheric thermal gradient of 11ºC per mile we got from Vallier-Talbot (2007) above, our temperature (278.7ºK, 5.5ºC), calculated from the Stefan-Boltzmann Equation using the Solar Constant, yields a calculated surface temperature of around 16.5ºC. The fact that this is warmer than the observed mean surface temperatures of Arrhenius and Burroughs (15ºC) leaves no room for such dubious free energy mechanisms as Arrhenius’ “Greenhouse Effect”. The surface temperature of the earth can be much more simply explained without resorting to such complex and unverifiable entities as radiative amplification and power recycling via backradiation of the “Greenhouse Effect”. Absorptivity of any of the parts can vary, but that only alters the overall emissivity, which in turn leaves unchanged, the gross power flowing though the system. Once equilibrium is reached it is only the power flowing through a thermally isolated system that controls and maintains mean temperature. This is because power is required to offset the amount of heat that is lost spontaneously and continuously due to emission of radiation.

    Our calculation of mean surface temperature without the “Greenhouse Effect” above (16.5±0.5ºC corresponding to 16-17ºC) is made without considering the effect of carbon dioxide. According to Arrhenius (1906a, translated by Gerlich & Tscheuschner, 2009, pp. 56-57) the observed temperature should be 20.9ºC higher than that yielded by a calculation such as this, owing to the carbon dioxide in the atmosphere. The observed surface temperature of 15ºC (Arrhenius, 1896; Burroughs, 2007) is actually 1-2ºC lower than the calculated mean surface temperature of 16-17ºC. The lower atmosphere will always be warmer than the upper atmosphere because higher material density in the lower atmosphere dictates a much higher thermal conductivity, absorption and density of heat. In contact with an opaque surface warmed by the bulk of the heat absorbed from the sun, it is not difficult to explain why the surface is so much warmer than the altitude corresponding to the centre of mass in the solarsphere. Moreover, the Ideal Gas Law (PV = nRT) dictates that the temperature of a gas containing a given amount of heat invariably increases with pressure. As the highest atmospheric pressure is at the surface, it makes sense that the higher temperature is there, especially if obstruction to radiative outflow decreases with altitude.

    Turning our attention to the example of Langley’s greenhouse experiment on Pike’s Peak in Colorado (mentioned by Arrhenius, 1906b), we may be tempted to ask how it is that a greenhouse can reach such high temperatures. Qualitatively, we may attribute the difference between the 15ºC mean surface temperature and the 113ºC observed in Langley’s greenhouse to the fact that noon-time radiation at the surface is three to four times as intense as the mean radiation over the whole of the earth’s surface. Repeating our calculation method, this time for the midday conditions of a greenhouse:

    T = {Wm/σ}0.25

    Given Wm = 1368:
    T = {1368/0.000000056704}0.25 = 394.1ºK = 121.0ºC

    As you can see, our application of the Stefan-Boltzmann Equation predicts that incident Solar radiation at 1368 Wm-2 should produce a maximum daytime temperature of 394.1ºK or 121.0ºC in a greenhouse fully protected from heat losses to conduction. Although Langley’s temperature is lower by eight degrees, it is near enough and, allowing for conductive heat loss, remains a testament to the insulating effectiveness of double glazing.

    What is demonstrated in the above examples, is the fact that surface temperature and the temperature in a greenhouse can be explained without resorting to the extraneous entity called the “Greenhouse Effect”. This is significant in light of Ockham’s Razor, which states:

    Entia non sunt multiplicanda praeter necessitatem.

    This reads in English as:

    Entities are not to be multiplied beyond necessity.

    Although the terminology may seem unfamiliar in light of 20th century usage, if we look at the words for what they mean, we can nonetheless understand this statement. In short, William of Ockham is urging us not to hypothesise beyond what is necessary to explain the material evidence we possess. A hypothesis that does go beyond the support of material evidence violates this principle in that the evidence is already explained by a simpler theory.

  29. I agree with John Kehr says:
    March 29, 2011 at 12:11 am

    What he is describing is the Maxwell-Boltzmann kinnetic energy effect in the atmosphere.

  30. Hmm… from my second reading I note that no mention is made of energy absorbed being passed directly to nearby air molecules (collisions) and warming those before the absorbing molecule can fire off a photon and cool itself that way.

  31. [taunting is not an attractive color on you. go ahead and reply, some other moderator will not know why I deleted your little attempt at a humorous humorless polemic. ~ ctm]

  32. A quibble:
    Ira says:

    The energized air molecules emit radiation at various wavelengths

    I say:

    The energized air molecules emit radiation at various longer wavelengths

    The vast majority of downward radiation coming from the atmosphere is longer wavelength than 13 um. The area under the curve around 10 um looks like about 5 percent of the area under the curve greater than 13 um. In other words, the vast majority of the energy coming back at us is explainable as blackbody radiation. Note that the graphs I am attempting to display here are for arctic data.

    If I interpret the graphs correctly, even if CO2 absorbed all the energy around 10 um, and that energy was immediately passed to other molecules to re-radiate around 15 um, the energy absorbed by the CO2 wouldn’t account for much of the back radiation.

  33. This ‘GHG physics’ is bunkum. Let’s take the 15 micron band. 95% is absorbed in 1 m air. That’s why mirages exist – the air above a hot surface is warmed by IR absorption.

    OK, the air does re-radiate downwards but a lot of the hot air is convected upwards thus ensuring most of that original IR radiation from the ground never makes it back.

    What GHGs do is to increase the adiabatic lapse rate a bit, also raise the tropopause. And because more latent heat means more efficient precipitation, the upper atmosphere dries and it’s easier for the heat to radiate to space.

    So, GHG warming is controlled to a near constant level independent of [CO2]. it’s about time that ‘climate scientists’ and politicians realised it.

  34. bananabender. That calculation starts off with the simple mistake of forgetting to include albedo, that Earth reflect some of the sunlight, and thus gets a too high non-greenhouse temperature. (This temperature is measurable as the blackbody temperature of Earth as seen from space and is -19 C)

    Then the author come up with an altitude of 1600 meters that he for some reason thinks is important. Multiplying this arbitrary altitude with the temperature gradient and adding the (faulty) base temperature does give the right answer, but only because he picks that altitude to get the right answer.

    Now take a look at what the real atmosphere looks like:

    http://apollo.lsc.vsc.edu/classes/met130/notes/chapter1/vert_temp_all.html

    Still think you can explain it?

  35. John Kehr says:
    March 29, 2011 at 12:11 am

    “I like that animation. ”

    I find Dr. Glickstein’s animations (this isn’t his first) to be annoyingly distracting. Since it interfered with my focusing on the text, I decided to skip the essay.

    Did I miss an interesting and important post? Maybe. But it just wasn’t worth my effort.

  36. Ira,

    There are a few reasons that I do not trust “averaged out global mathematical calculations”.
    One time frames. Our current calculations fail due to the calculation do not include adjusting for planetary motion, slowdown or unknown surprising factors(such as solar flares, massive eruptions, salt changes, ocean current changes, impacts, etc.).
    Next, planetary positioning. Since the suns diameter at it’s equator is the greatest mass and the poles are much smaller to the drifting of the planet between the two.
    Thirdly is planetary shape of having a huge diameter equator and smaller diameter poles on a rotating planet.
    So, where on this planet or atmosphere does make a difference to any other place taking a measurement.

  37. Spartacusisfree, the amount of energy the Earth has to radiate to remain in thermal equilibrium is constant. Since we have GHG:s in the atmosphere only a small part of the radiation emitted into space comes from the surface, the rest comes from the atmosphere, and we can assign an “effective altitude” from which the average radiation is emitted. If you add more GHG:s radiation will have a harder time escaping and thus this effective altitude will increase, but since the amount of energy has to be constant its temperature has to remain the same. Now take the lapse rate into account. If the layer of constant temperature rises, the temperature at the surface has to increase as the lapse rate multiplied by the increase in altitude. If more GHG:s raises the tropause by 100 meters the temperature at the surface rises 0.7 degrees as a first approximation.

  38. Ira

    Carbon dioxide is a linear molecule. You show it with a bent structure like H2O
    Fix that mistake. It makes you look stupid.

    When a CO2 or H2O molecules absorbs an IR photon, its linear velocity does not increase. The absorbed photon causes an increase in the vibrational frequency of a bond or combination of bonds (e.g, bending). Fix that mistake.

    A vibrationally- excited CO2 or H2O molecule in the troposphere will not re-emit the absorbed photon but will undergo immediate collision deactivation with N2, O2, Ar or H2O. This causes an increase in the speed of these, i.e., they become slightly warmer.

    The collision frequency at 1 atm and room temp is about 100 billion collision per sec. Incidently, this is why nat gas in air explodes with great violence, i.e., the reaction goes at the collision frequency.

    FYI: CO2 is a weak absorber of IR because it does not have a permanent electric dipole like H2O.

  39. All of this is totally unnecessary. All we need is one valid graph of long-term temperature vs CO2 concentration, showing the lack of correlation. And we already have plenty of valid graphs like that.

  40. The greenhouse effect hypothesis, and specifically the warming caused by increasing CO2 in the atmosphere goes back to 1896, when Svante Arrhenius wrote an article and then a book that postulated the theory. All kinds of people ran with it afterwards although the theory was later found to be based on grossly erroneous spectra. These gross spectroscopy errors can be seen as conclusively proven wrong in MODTRAN(R) simulations. For a list of references to Arrhenius, including some by people that still believe it to be true look here:

    http://www.lycos.com/info/svante-arrhenius.html

    For the MODTRAN(R) simulations disproving the theory look here:

    http://members.casema.nl/errenwijlens/co2/arrhrev.htm

    Although I disagree with his conclusions (he is a warmist), Spencer Weart wrote a pretty thorough history of how this whole AGW thing evolved over the last century, ending with Charles (Dave) Keeling, before going into the modern research and model predictions that are so questionable.

    Regardless of his conclusions, I give Weart credit for being quite thorough and his book is worth a read because it encapsulates all the early prominent research that went into building AGW theory as we know it in it’s present form. This book is also a great place to look at all this greenhouse research in one place and to see where the all the holes and leaps of faith are. A pretty complete condensation of the book can be found here:

    http://www.aip.org/history/climate/co2.htm

    Sadly, and contrary to the whole point of the book, which should have been citations and discussions of the history of AGW research, Weart himself takes the ultimate leap of faith and joins the warmist camp, tainting a great read with his own conclusions. You can see that in his personal notes:

    http://www.aip.org/history/climate/SWnote.htm

    Best,

    Jose

  41. rusureuwant2know
    “I thought Al Gore got his graph backwards and the increase of CO2 followed warming – why are we still working with the assumption it creates it???”

    Greenhouse gasses definitely do warm the planet, there is no question of that. And, increasing the amount of CO2 in the atmosphere will increase the amount of heat retained here on Earth. The real question is, how much will the change be as a result of a doubling of the CO2? First order effects seem to point to a change of 0.8 to 1.5 degrees. All of the other warming the alarmists are trumpeting is from second order effects that are definitely not known. Will the H2O in the atmosphere increase significantly, thereby significantly warming the planet? or perhaps there is negative feedback and the H2O is reduced and the planet cools. I don’t know.

    As to your other point about the CO2 following temperature, there is definitely evidence of that too. Warmer oceans can’t hold as much CO2 so more of it ends up in the atmosphere. It’s a complicated system.

    Stephen Wilde
    Although there clearly is a greenhouse effect it is infinitesimal compared to the energy retaining effect of the oceans.

    The greenhouse effect and the energy retaining effect of the oceans are different things. The atmosphere and the ocean may retain heat, but the atmospheric heat retention is infinitesimal compared to the heat retention of the oceans. The greenhouse effect changes the amount of heat that flows out of the system.

    Perhaps you could think of the heat retention of the ocean as a bucket of water. Then, you have a stream of water flowing into the bucket and a hole in the bottom of the bucket that lets an equal amount of water flow out. The system is in equilibrium because the same amount of water is going out as is coming in, the water level remains constant. When you change the greenhouse gases you change the amount of water flowing out so the water level will raise of lower as a result.

    MikeEE

  42. To be thorough, there’s also an “anti-greenhouse” effect. See

    http://en.wikipedia.org/wiki/Anti-greenhouse_effect

    The atmosphere of Titan is transluctent to some wavelengths, and translucent to part of the surface radiation.
    This results in a surface cooler than a surface with NO atmosphere

    sun —–> 4 watts 2 watts 2 watts Titan surface 2 watts4 2 watts from atmosphere<——-
    2 watts from surface 2 watts from atmosphere
    4 watts 2 watts Earth
    The earth heats up, reradiating the 2 watts to the atmosphere, which will also heat up. The final
    balanc will be

    Sun —> 4 watts to atmosphere 4 watts to earth from atmosphere
    –>4 watts from sun, 4 watts from earth4 watts from atmosphere to earth
    <—4 watts to earth from atmosphere

    So the net effect of the surface warming from an infrared absorbing atmosphere is the same as the effect of NO atmosphere.

    See Trenbeth's figures here:

    http://stephenschneider.stanford.edu/Cl … lance.html

    Note that the zero greenhouse effect would apply for those 67 watts absorbed directly by the atmosphere.

    Caveats: Working out the temperature drop in a purely radiation
    cooled atmosphere, it can be shown that the temperature drop would be
    greater than the
    moist pseudoadiabetic lapse rate, making a purely radiation
    controlled atmosphere unstable. The adiabetic lapse
    rate places an upper limit on drop in temperature with height.

  43. Nice animation and explanation. But I think it is giving ony a passing thought to a very important component countering any heating effect of more CO2 — convection. As noted, a warmer molecule will be moving faster and hence have a faster Mean Free Velocity and it will rise. That warmer air will cool as it rises and condenses water pulled up with that rising mass, losing even more energy. Hence transporting ground air heat to the cooler higher atmosphere.

    Also missing from this, is that this affect in the animation is only when the sun it at noon in the location where the sun is directly over head. Daytime everywhere else gets lower energy from the sun because of the angle. Nighttime, more of that stored daytime heat is lost into space. Also, the warmed portions of the planet intermix that warm air to the colder regions of the planet (causing storms which disapates even more heat).

    In other words, the atmospheric system has a buffering affect and is why the planet have never “cooked” when CO2 was 20 times today.

    The planet is not suseptable to over heating, it’s suseptable to significant cooling, which is bad.

  44. Reply to Thomas: I agree that as the adiabatic lapse rate and the tropopause rise, surface temperature will increase. However, the reduction of [H2O] in the upper atmosphere will allow easier radiative heat transport to space. The interaction of this radiative heat transport with the ALR is actually quite interesting.

    The evidence of lower [H2O] is fairly conclusive. So, I expect GHG warming from CO2 to be very low.

  45. Igl
    What a lot of nonsense. Here is how it really works…

    Don’t believe everything you read. He says it’s not reradiation because “When the atmosphere emits radiation, it is not the same radiation”.

    What is ‘the same radiation’? that’s just nonsense. Any material (CO2 for example) that is not at absolute zero emits radiation to stay at equilibrium. If you raise the amount of energy that you put into that material you will raise it’s temperature and the amount of energy that it radiates. Whether it is the same of different energy is irrelevant.

    MikeEE

  46. Although this article mentions CO2 absorbing energy through collisions and radiating that energy (half to space) it does not mention that this is a cooling effect. And, when you add more CO2 you increase this cooling effect. So, what is the trade-off here. How much warming vs. how much cooling. I’ve never seen this tackled anywhere.

  47. Looking at Ira’s animation and Wayne’s proportional image, is it not clearly evident that CO2 does not have the volume required to have any dramatic impact on the Greenhouse effect?

    For example, the total Greenhouse effect represents approximately 33 degrees of warming for our planet. Simplifying the atmosphere as Wayne has done it would seem that CO2 can only add ~0.44 degrees (400ppmv CO2 / 30,000ppmv H2O x 33 degrees). This discounts other GHGs but I believe they are all in far less concentrations.

    MrC

  48. Stephen Richards: thanks for the link. I did not have enough time for reading the complete article by Postma. Perhaps tonight. My first impression is like yours: it’s over with the greenhouse theory and everything based on it.

  49. When you adjust for atmospheric pressure the absolute atmospheric temperatures of Earth, Mars and Venus only vary by 10-20%. This is despite Venus receiving about 8x as much solar radiation as Mars and 1.9x as much as Earth and the totally different atmospheric composition on all three planets.

    Measured Temperature vs Atmospheric Pressure:
    0.001Bar:
    Mars ~210K, Earth ~220K, Venus ~190K.

    1Bar:
    Mars (n/a), Earth 288K, Venus 350K.

    When you adjust for the different gas composition (relative molar mass) of each planet’s atmosphere the temperature differences are even smaller.

    This fact is explained far better by the Ideal Gas Law than any mythical Greenhouse Effect.

  50. Ira Glickstein says:
    “7 The energized air molecules emit radiation at various wavelengths and in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth …”

    As Ira states, heated air molecules radiate. So why are we so concerned about the ‘backradiation’ from CO2? As most of the heating of the atmosphere is by conduction and convection and all gases radiate in close proportion to their temperature, then all gases are greenhouse gases. If it were otherwise, N and O could not radiate away to the vacuum of space. Consequently even if it is accepted that there is backradiation from CO2 ‘further heating the Earth’ (impossible as this means the surface is heating up itself as it is the surface that has warmed the atmosphere in the first place) this effect would be dwarfed by the radiation from the remaining 99% of the atmosphere.

    Some commenters on this thread who deride those of us who keep on propounding the second law claim that we misunderstand the position. They say that the greenhouse gases ‘reduce cooling’ (rather than ‘further heating the Earth’). But reducing cooling cannot add heat. In the constant irradiance models that are used to demonstrate the greenhouse effect (e.g. Kiehl & Trenberth’s) the Earth is not cooling down. The most that can be said about the ‘reduced cooling’ theory of greenhouse gases (properly called an atmosphere effect) is that it will reduce heating during the day and cooling at night. Therefore the average temperature may be higher than it would be otherwise (e.g. on the Moon) as the possible minimum is 3K whereas the maximum is determined by the quantity of solar radiance. Those who think this is the greenhouse effect have not understood what real warmists (like K&T and Gavin Schmidt) really believe. Conventional greenhouse theory says that the downward radiation from greenhouse gases raises the Earth’s temperature higher than solar radiation can.

  51. 4. The energized molecules re-emit the Photons in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.

    This isn’t true for cold (below ionization energy threshhold) dense gases. They do not exhibit spectral emission lines they only have spectral absorption lines. The reason is that in a dense gas the molecules are packed so tightly that any increase in energy in an absorption line is immediately distributed to surrounding molecules through collisions. The emissions from the GHGs and all the other gases in the lower atmosphere are approximately continuous blackbody spectrum with a peak emisson frequency corresponding to the temperature of the air.

    7. The energized air molecules emit radiation at various wavelengths and in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.

    This is correct except for the part about “heating the earth”. That phrase justifiably causes people with a basic understanding of the laws of thermodynamics to balk. In the big picture this simple phrase (courtesy tallbloke) describes what’s happening: the sun warms the ocean and the ocean warms the air. So-called back radiation, which is what you’re describing, does not warm the surface except in limited and exceptional circumstances where a mass of air transported convectively (read winds) is warmer than the surface over which it blows and even then it is limited because the air has so little heat capacity compared to the surface – imagine trying to heat a cup of coffee by stirring it with a hot feather.

    What actually happens is the back radiation slows down the rate of surface cooling. This is because radiative transfer between two objects of different temperature is a two-way street. Say the warmer object is emitting 10 watts per square meter. The cooler object, unless it is at absolute zero, is also emitting energy. Say it’s emitting 5 watts per square meter. So the warmer object is emitting 10w/m and absorbing 5w/m while the cooler object is absorbing 10w/m and emitting 5w/m. The net transfer of energy is 5w/m from warmer to cooler.

    Absent greenhouse gases essentially all the radiation from the surface charges through the atmosphere at nearly the speed of light and is absorbed by the virtually infinite heat sink of the cosmic void. The cosmic void is 3 Kelvins – just a few degrees above absolute zero so the back-radiation from the cosmos is just about nil. Put some greenhouse gases in between the surface and the cosmic void and then you have some significant back radiation which slows down the surface cooling rate.

    At the end of the day because the ocean doesn’t cool as quickly at night with GHGs in play it’s a little warmer in the morning than it would be absent the greenhouse gases. The sun does its thing during the day warming the ocean and because the ocean was a little warmer at sunrise it’s a little warmer at the end of the day too. The cosmic void doesn’t change temperature but now that the surface is a bit warmer than it would be absent GHGs it cools more quickly at night – the higher the difference in temperature between two objects the faster warmer object loses energy to the cooler object. Thus a new equilbrium temperature is obtained where the energy received by sun during the day is emitted at night in perfect balance.

    In the real world equilibrium is moving target due to other factors like clouds (which reduce the amount of solar energy reaching the surface), convection (which mechanically transports energy from one place to another both horizontally and vertically), evaporation, and conduction. But the one thermodynamic fact is always operative in the real world – the farther out of equilibrium the system is the harder it tries to move back to the theoretical equlibrium temperate. So when everything is said & done and absent any feedbacks – GHGs raise the theoretical surface equilbrium temperature and for non-condensing (read everything except water vapor) GHGs which are well mixed and slow to change in concentration the change in theoretical surface equilibrium temperature is calculable and works out to about a 1C rise for every doubling of CO2.

    A 1C rise in surface equilibrium temperature per CO2 doubling is a very good thing especially when one considers that the bulk of the warming actually occurs at night, at higher latitudes, and in the winter. This has the beneficial effect of extending growing seasons when and where longer growing seasons are most needed. Add to that that higher CO2 concentration means plants grow faster/larger (provided sunlight, nutrients, and water are not limiting factors) and they use less water per unit of growth in higher CO2 as well.

    The whole brouhaha over scary global warming rests on the wholly fabricated idea that CO2 greenhouse warming puts more water vapor in the atmosphere and that, because water vapor is a powerful greenhouse gas, it constitutes a positive feedback that turns 1C per doubling into 3C per doubling. There isn’t a shred of empirical evidence to support that belief and IMO overwhelming evidence that the feedback is actually negative – the water cycle speeds up which serves to mechanically transport heat from the surface by evaporation and convection to the cloud deck where it then has an easier radiative path out to space. It also results in slightly greater cloud cover which reflects more sunlight back out into outer space before it can reach the surface to warm it. This prevents any chance of a runaway greenhouse and this is borne out by there never having been a runaway greenhouse in the earth’s history even though CO2 concentration in the past was as much as 20x greater than present. In fact the l paucity of CO2 today, relative to most of the earth’s history, is IMO undoubtedly a major factor in why the earth has been in an ice age for the past 3 million years – ice ages are rare but there is a tipping point where factors can combine that allow ice and snow to dominate the surface through positive feedback mechanisms – ice reflects a lot more sunlight than ocean or unfrozen land surface thus reducing the amount of solar surface heating (which makes conditions even better for glacier expansion) and because cold air holds less water vapor than warm air it also greatly reduces what’s normally our primary greenhouse gas in a viscious cycle of cold begetting even more cold. There is substantial but controversial evidence that glaciations in the remote past have frozen everything from pole to equator – appropriately called “snowball earth”. So global cooling, especially now with the Holocene interglacial overdue for its ending, should be our greatest concern and we should be welcoming whatever global warming we can get with wide open arms.

  52. re: Thomas says:
    March 29, 2011 at 5:39 am
    bananabender. That calculation starts off with the simple mistake of forgetting to include albedo, that Earth reflect some of the sunlight, and thus gets a too high non-greenhouse temperature. (This temperature is measurable as the blackbody temperature of Earth as seen from space and is -19 C)

    If you totally ignoreevery factor including albedo, atmospheric composition and incident solar radiation there is still a <10% temperature variation between the atmospheres of Mars and Venus at the same pressure. This is despite Venus receiving 8x as much solar energy as Mars.

    I prefer to accept an irrefutable Ideal Gas Law than the idea of "magical greenhouse gases" which defy virtually every basic principle of chemistry.

    The calculated value of -19C for Earth without an atmosphere is based on absurd reasoning.

  53. if I am reading it correctly (probably not) #7 violates the 2nd Law … the cooler air cannot heat up a warmer body …
    if you mean it slows down the rate of heat loss of the earth then I am with you but it cannot “heat” the earth …

  54. My impression is that there is confusion between thermal transfer between molecules (rotation, vibration etc) and radiation emission/absorption. The first depends on the thermodynamics, the second on quantum physics.
    As far as I remember, the absorption of a photon concerns only the electrons which can gain a higher (not stable) energy level. After some time the excited electron will re-emit the received excitation energy as a photon and fall back into the stable configuration. The radiation energy balance would be zero. But may be I have forgotten some essential laws of physics.

  55. Anders says:
    March 29, 2011 at 2:25 am
    If all that is going on is absorption and re-emitting radiation there would be no heating of the atmosphere. Heating happens because of the re-distribution of energy through collisions between molecules. The “simple” visualization is thus too simple. The temperature of a gas IS the average kinetic energy of the gas molecules, this is something Roger Pielke Sr. has pointed out a number of times and for many years.
    Your comments are as close to what makes me uncomfortable with this representation of the absorbtion and re-radiation of a photon. The kinetic energy seems like it could be represented as the mean free path and temperature. What I have a problem with is the absorbtion and re-radiation at the same wavelength. You can’t have that and warm the molecule as well. If warming occurs, then the re-radiation must be at a different wavelength, lower energy. Otherwise, energy is created from nothing.

  56. This article is incompetent, in the face of definitive experimental data that simply invalidates its premise:

    Venus: No Greenhouse Effect

    The simple analysis I did, within hours of finding the data on the internet, should have been done nearly 20 years ago. Every scientist who promulgates the greenhouse effect is incompetent. You will find the simplest, clearest discussions about the greenhouse effect on my website. My presentation of the definitive evidence, the first and only such presentation to date as far as I know, should be properly confronted and accepted by the entire science community; I have submitted it to “Physics Today” to get it before that community, but have received no response, which shows a complete lack of integrity and professional commitment to scientific self-correction — a fundamental failing of scientific institutions today.

    There is no longer any validity to arguments for the greenhouse effect. The science is settled, and what is needed is re-education of all those like Ira Glickstein who are promulgating false science.

  57. If the “greenhouse effect” had increased, a hot spot would be found over the tropics. This hot spot has not been measured.

    Now, let’s see where warming has been measured: mostly where you can find snow or ice. Maybe it could be good to evaluate the “greenhouse effect” inside of snow. First, it is wrong to think that snow reflects light like a giant mirror. Most of the light does enter inside of the snow where it is reflected in all directions by pieces of ice, and after a while, most of the light is reflected out of the snow. So it is similar to a “greenhouse effect”.

    Now, the impurities inside of the snow and the spectrum of light will probably have an effect on how much light is absorbed. Up to now, I have not seen any study about this.

  58. I think a more accurate animation will look like tsunami open sea wave models like this (where the islands represent GHG molecules):

    It certainly is not like a bunch of projectiles flying around.

  59. Thermodynamics gives us an idea of the direction of energy transfer. The transfer mechanism controls the rate. The rates of convection, evaporation/condensation, freezing/thawing are much slower than radiative energy transfer that can be considered “line of site, speed of light”. Model the energy input/output of a column of air in winter, over Arctic sea ice considering all these processes. Test your model by analyzing the reanalysis data. Take a guess as to the relative contributions of each process to slowing down the loss of energy to space. In your model design, don’t forget to include the insulating qualities of sea ice (thermal conductivity and thickness change). Energy is being transferred from water below the ice to the upper surface, as it freezes.

  60. Spartacusisfree, if you have any evidence for your claims I suggest you publish them. The general view is that H2O increases as temperature go up, and that seems much more reasonable to me. If you look at different areas of the Earth with different temperature the trend is definitely that warmer areas have more water vapor, so why should the trend be in the opposite direction if the warming comes from AGW? (But at least you don’t violate any obvious law of nature unlike most of the geniuses here)

  61. Thomas says:
    “bananabender. Go ahead, try to explain the temperature of the Earth using only the ideal gas law! It can’t be done, you see, if you actually start to think about what you do.”

    Ideal gas law PV=nRT can be rewritten as P/R=rho T . Ascending in altitude through the troposphere, the density decreases at a lesser rate than pressure

  62. steveta_uk says:
    March 29, 2011 at 3:15 am

    You need to explain in the same terms and manner as to why back radiation doesn’t violate black body theory not wave your arms around. I have read the entire doc twice and compared it to the BB theory and can find no problem with what he says. So, explain yourself using the known laws of physics and perhaps we can come to consensus :)

    Stephen Richards BSc Physic MSc Solid state Physics.

  63. bananbender, again I ask you to look at that figure of the real atmosphere. Where is the 0.001 bar altitude on Earth? It’s around the stratopause, and according to the standard atmosphere the temperature at that altitude is 270 K, not 220 K as you claim. You just make up your numbers to make them fit! I think I prefer the regular kind of science.

  64. steveta_uk says:
    March 29, 2011 at 2:56 am
    “…Surely they can only re-emit much lower energy photons, of similar energy levels to the black-body curve at -20C.

    Yes, however the temperature of a gas is actually defined as the average kinetic energy of the gas molecules, for example at -20C and not the amount of radiant energy it is emitting. In fact the laws of black body radiation only have a good fit to solid blackbodies which are in thermal equilibrium – something that seldom happens in our chaotic weather systems. The tiny effect the odd molecule of CO2 has is swamped by other processes which drive our turbulent, non-linear system.

  65. I think Cassandra King nailed it. To me, this is a problem of scale which, of course, is not represented in the original graphic.
    I would use the following analogy to visualise the scale:
    .

    Let us take a large sports stadium such as the Melbourne Cricket Ground. Capacity 100,000 seats. Although the concentrations of ghgs are by ppm by volume, let us approximate to 10.000 molecules per percentage volume (1 million divided by 100). This means that of the 100,000 spectators in the MCG, only 40 (wearing red for GHGs) are able to both hear the tannoy (LW radiation) and re-transmit what it says. Another 250 spectators (wearing blue for water vapour) can hear the tannoy but cannot re-transmit the message (because they are mute). The rest (wearing white) are deaf-mute. Let us now pretend the tannoy message was a very funny joke. The only way that the 99,710 spectators in white can even realise something funny has been said is by ‘conduction’ withthe 290 people, who are shaking with laughter (vibration). If they are touching the laughing people, they may well laugh in sympathy – but maybe with less conviction. As most of the ‘water’ spectators are in the lower stand, that is where most of the shaking is done. Back in 1850, the number of emitting spectators was only 28 (probably 29 if you include all the other dry ghgs). Therefore, according to the radiative forcing theory, the addition of 12 spectators (of any type of ghg) has increased the degree of overall shaking by 2.5% (0.8 C). According to the IPCC, doubling the number of emitters to 56 will lead to a ‘best estimate’ increase of 10% (3 C).
    Does anyone think the original 28 red people can have a significant effect, and will the addition of another 12, or even doubling the original figure, be likely to incur catastrophic warming (laughter)?
    .
    A lighthearted analogy, I admit…

  66. “The greenhouse effect heats the earth because greenhouse gases absorb outgoing radiative energy and re-emit some of it back towards earth”.

    To be true doesn’t this mean that CO2 must act like a mini sun with an energy multiplying effect.

    As an example if the earth contains 1 trillion units of energy. If 100,000 leave the earth’s surface as radiation. Somehow even though half or move does not return to the surface the energy that does return must be greater than 100,000 units. Otherwise there can not be a heating of the earth.

    But even this example is not correct as the radiation returning has a cooler energy impact than the earth and a cooler body can not heat a warmer one.

  67. Ira, you do keep pluggin’ away.

    1. If you are going to have H2O in the mix then you also need to recognize that it has a much higher specific heat than CO2. water vapor about 1.9 CO2 about .84. Water vapor can hold on to the heat. CO2 cannot.

    2. You should show the area under the curve. (this was mentioned before) It is about 8% for the 15 micro. People are only a small protion (3%) of the 8%. So our effect if any is small.

    3. You should define “surface”. If a proton of IR strikes me it disappears. If it hits a leaf or a blade of grass it is gone. So there is a huge portion of the downward IR that never gets to the surface. Again, the oceans cover 2/3 of planet. So your claim is that 1/3 of the surface area of the planet heats the other 2/3’s of the atmospher over water due to backradiation. I have my doubts.

    4. Heat transfer is totally and only dependent on temperature difference. q/A=(emissivity * sigma*(T^4-T^4)) If both e and sigma are the same no heat transfer but that does not cause either to stop radiating at its temperature. If I take a blow torch at 3000 C and add a second torch so the flame is exactly opposite I don’t get 0 at any point. If I combine the flames tips in the same direction I don’t get 6000 C. 3000 C is it.

    5. What is the emissivity of CO2 at partial pressure and the temperatures you list? Without that knowledge you cannot figure heat transfer and I bet it is very tiny. My heat transfer book does not go that low in temperature so I have to extrapolate to less than .1.

    6. PV=nRT for the air temperature at or near the surface is valid since the critical temperture of N2 is so far away. (Paul please note I said at or near. ;)) If this was not valid why do we have STP, standard temperture and pressure. Using standard pressure and 273 K the equation is in balance.

    7. Because of the equation in my #4 and the atmosphere under almost all conditions being at a lower temperature than the ground no heating of the surface by air can take place. Not to mention the mass diparity of both and the heat capacity difference.

    Please try again.

  68. “”””” Jan Oortwyn says:
    March 29, 2011 at 7:29 am
    My impression is that there is confusion between thermal transfer between molecules (rotation, vibration etc) and radiation emission/absorption. The first depends on the thermodynamics, the second on quantum physics.
    As far as I remember, the absorption of a photon concerns only the electrons which can gain a higher (not stable) energy level. After some time the excited electron will re-emit the received excitation energy as a photon and fall back into the stable configuration. The radiation energy balance would be zero. But may be I have forgotten some essential laws of physics. “””””

    In the lower regions of the atmosphere where weather happens, the mean time between intermolecular collisions, is much shorter than the mean lifetime of the molecular excitted states; so before an excited molecule gets to spontaneously decay to some lower state, it collides wiht another molecule or atom (of Argon) and energy is redistributed between the collision participants. That is what temperature is all about, so that original photon energy, becomes a melange of thermal energy of kinetic motion of the gas molecules. As a result of those collisions eventually some IR active species will receive enough kinetic energy from a collision, to put it into soe excited state from which it can radiate (before the next collision).

    Only in the rarified stratospheric regions, is the lifetime of the excited states shorter than the mean collision time.

  69. Arfur Bryant says:
    March 29, 2011 at 9:18 am

    I think Cassandra King nailed it. To me, this is a problem of scale which, of course, is not represented in the original graphic.
    I would use the following analogy to visualise the scale:

    Now that is a perfect and descriptive analogy and explanation of the facts, I really hope Ira reads your post and comments on it because it goes right to the heart of my concerns and I am not a scientist nor do I hold any scientific qualifications so your post really appealed to me and was a joy to read and understand and visualise. I say visualise because that is the key to understanding. IMHO you have created a masterpiece that should have a wider airing in order for us mere mortals to gain a better understanding of the issues.

  70. Reply to Thomas: 8.21 Stratospheric drying: http://www.npr.org/templates/story/story.php?storyId=123075836

    Solomon thinks it’s fewer thunderstorms: I think it’s more efficient precipitation in thunderstorms, part of the GHG control system that maintains near constant GHG effect in the atmosphere. Miskolczi has found from radiosonde data that the reduction of [H2O] in 61 years’ radiosonde data maintains constant IR optical depth/ 1.87 average IR photon absorptions in passing through the atmosphere [within 0,.1% of the theoretical value]: http://miskolczi.webs.com/

    My own work concerns the aerosol optical physics in the climate models. All the two-stream approximations are wrong because they fail to take into account direct backscattering you get from applying Mie theory to the first few scatterings as the light wave enters a cloud. What this means is that what is claimed to be ‘cloud albedo effect’ cooling is either neutral or heating, quite substantial in the palaeo-climate.

    Hence you don’t need CO2-AGW: it could well be net zero. The quid pro quo is that none of the models can predict climate.

  71. I am getting a bit depressed by the level of physics being discussed here and in some of the other web sites that we are being directed to. I am beginning to wonder if there is a strategy amongst the warmers to refute the greenhouse effect using bogus physics in order to quote these arguments, say how stupid they are, and thus justify the dismissal of all sceptic arguments as unscientific. Or I might just be paranoid!

    There are serious problems with the quantification of the greenhouse effect. There is a serious doubt about feedbacks. There is a serious problem that the “fingerprint” of an enhanced greenhouse effect cannot be identified.

    But there is no doubt that the atmosphere absorbs radiation emitted by the earth and eventually radiates it to space at a higher altitude which, being at a lower temperature, means a smaller energy loss. That reduction has to be compensated for by increased radiation losses elsewhere and that means a warmer surface.

    This is the overall energy balance scenario and Ira has tried to explain it at a mechanistic level to show what is happening within the climate system. I have problems with some details, as I have already commented, but overall I think it is a good effort.

    Some of the objections however are bizarre.

    For example some say the sky cannot radiate to the earth because it is warmer yet you can point an infrared detector at the sky at night and measure it! Anyway when a CO2 molecule radiates a photon downwards how does it know the earth is colder and it should not do it!? Of course if the earth is colder the nett energy flow would be upwards (from warm to cold) and that is the only necessary requirement thermodynamically.

    Then there are the arguments like CO2 is only a “trace gas” and H2O is a far more powerful absorber. If we a want to make the warmists accept the evidence of measurement we must accept the same rigour. One can measure the absorbtion by CO2 at say 1Km and do the same for H2O. By this altitude both have already trapped all the outgoing radiation at their characteristic absorption frequencies and re-radiated at least once. You can measure the radiation and see the big gaps in the black body spectrum associated with this absorption. You may think that CO2 (or H2O) at such low densities cannot do this but it does and there is already more than enough to do it totally. It is pointless coming up with simplistic arguments to deny these facts when we have data to confirm them.

    The greenhouse effect is very badly named but it is a real effect. It would do the sceptic cause a great deal of good for everyone to accept this and then focus on the real issue – what is the effect of further increments of CO2? This question is by no means obvious. When the warmists tackle this question they tend to use the same simplistic arm waving physics that I have been reading above. We need to do a better job not a worse one.

  72. Spartacusisfree, you brought up water vapor with this statement: “And because more latent heat means more efficient precipitation, the upper atmosphere dries and it’s easier for the heat to radiate to space.”

    Now you start to talk about the stratosphere which is clearly inconsistent with a statement about precipitation since there is far too little water vapor in the stratosphere to cause precipitation. Anyone brining up Miskolczi obviously have rather shallow understanding of the subject, so I think I pass on further discussion.

  73. Marc77 says:
    March 29, 2011 at 8:01 am

    If the “greenhouse effect” had increased, a hot spot would be found over the tropics. This hot spot has not been measured.

    Now, let’s see where warming has been measured: mostly where you can find snow or ice. Maybe it could be good to evaluate the “greenhouse effect” inside of snow. First, it is wrong to think that snow reflects light like a giant mirror. Most of the light does enter inside of the snow where it is reflected in all directions by pieces of ice, and after a while, most of the light is reflected out of the snow. So it is similar to a “greenhouse effect”.

    Now, the impurities inside of the snow and the spectrum of light will probably have an effect on how much light is absorbed. Up to now, I have not seen any study about this.

    Recommend you start here with snow absorption:

    http://www.google.com/search?sourceid=navclient&ie=UTF-8&rlz=1T4GGLL_enUS382US382&q=%22black+carbon%22+snow+albedo

    The so-called tropical upper troposphere hotspot I understand is the mark of the mythical water vapor amplification that turns 1C per CO2 doubling into as much as 3C per doubling. Supposedly the extra CO2 increases water vapor content of the upper troposphere more than lower troposphere resulting in upper troposphere (which is always far colder than the surface) warming more than the surface. “Hotspot” is an inaccurate name since it’s really differential warming between upper and lower troposphere.

    The increasing magnitude of the differential warming, if any, is difficult to measure and controversial. I believe notoriously poor GCM modeling of clouds is why the models are wrong. If the water vapor being delivered to the upper troposphere increases so does the amount of heat carried away from the surface by evaporation and convection and it condenses at an increased rate rather than giving the upper troposphere a higher absolute humidity. Thus evaporation, convection, and cloud formation is a negative feedback that negates the effect of more water vapor due to warmer surface temperature. There are several well developed hypotheses on this effect. The best IMO is

    which is basically saying that the atmosphere, with regard to water vapor greenhouse warming, is saturated and we can’t get any more greenhouse warming out of water vapor – we can get additional surface warming only from increases in non-condensing greenhouse gases like CO2 and methane. I am in fundamental agreement with the conclusion simply by considering that the earth has never experienced a runaway greenhouse effect which water vapor amplification must certainly entail. So regardless of whether Miskolczi is right something is putting a ceiling on the maximum surface temperature and this appears to me to explain why. CO2 levels in the remote past have been several doublings greater than present and the global average temperature was several degrees higher fitting perfectly with there being no water vapor amplification.

    My own pet theory is that CO2’s primary role is increasing the earth’s average temperature from below freezing to above freezing and this is accomplished by the first 100ppm in the atmosphere. CO2’s LWIR absorption curve is linear for low concentrations of the gas and becomes logarithmic in greater concentration (which has been known for over 150 years since John Tyndall experimentally measured it). It serves as “kindling” to get the water cycle going. Once the water cycle is going it takes over as the primary greenhouse gas and unlike CO2 is self-limiting because it’s a condensing gas. After that CO2’s role is limited to about 1C per doubling. There isn’t enough recoverable fossil fuel in the world to do more than two or three doublings which might possibly mean we could end the ice age and in a matter of thousands of years see the earth get green from pole to pole (which is its usual state – no polar ice caps at all). The downside of having a vastly increased amount of arable land is that ocean levels will return to their normal non-ice age level which is about 100 meters higher. That will take thousands of years to happen. No plants or animals will have any trouble migrating to higher ground. Humans have the bigger problem as we built a lot of immobile crap very close to the ocean under the mistaken impression that ocean level rises and falls only with the tides and is constant otherwise.

  74. Harry Dale Huffman,

    I agree with your analysis up to a point. In particular, the temperature ratio of the “effective radiating surface” of Venus should be 1.176 times the temperature of the “effective radiating surface” of earth (ignoring albedo effects). (I have also heard this called the earth’s “photosphere” since that is the regions from which photons are emitted from earth.)

    That “effective radiating surface” is a combination of the surface (warm), the clouds (cooler), and the GHGs (cooler yet since they radiate from close to the top of the troposphere). On Venus, almost no radiation comes from the surface, since it is blocked by continuous thick clouds and GHGs. On earth the surface DOES radiate a significant amount.

    To a zero-order approximation, you could say all the “effective radiating surface” is around the 500 mBar location on either, so this layer should have similar (within a factor of 1.176) temperatures on either planet.

    However, any more accurate estimate would need to include the actual locations of clouds and amount of cloud cover, as well as the location and emission spectrum of GHGs, since these all affect the “effective radiating surface”.

    (Once you get below this effective radiating surface, then the lapse rate is the key factor in determining the temperature for lower layers, so Venus with a much deeper atmosphere will have a much higher surface temperature.)

  75. “For example some say the sky cannot radiate to the earth because it is colder yet you can point an infrared detector at the sky at night and measure it! Anyway when a CO2 molecule radiates a photon downwards how does it know the earth is colder and it should not do it!? Of course if the earth is colder the nett energy flow would be upwards (from warm to cold) and that is the only necessary requirement thermodynamically”.

    Measure what though. See below that might be relevant. From the slaying blog

    And this has been discussed before but the issue is not whether radiation travels downwards. It is whether it has a warming impact if it returns to the surface.

    From slayingtheskydragoncom/en/blog/102

    Lord: “Back radiation can be simply demonstrated by pointing a simple infrared detector at the underside of a cloud. Try it.”

    Chorus: “My IR detector only cost $60! Simple! Agreed! Agreed!”

    Slayer: “Clouds do not absorb and re-radiate heat back to Earth. Clouds add THERMAL MASS which takes longer to heat and cool. Warmists ‘support’ this false hypothesis with IR thermometer readings, but the IR readings of a hot Barbie is the same from any distance; ENERGY is not. Your $60 REMOTE thermometer is not measuring the radiant energy you are receiving, it is measuring the resonance of the Barbie.”

  76. Yeah, yeah… that seems to be a very British concern to me. The country where they still have single glasing all over the place, and anyway, doors and windows never seal. I can totally see how they could save 90% heating by simply arriving in the 3rd millenium to that end. After many years, and by sheer luck, I finally ended up in a London house with double glasing for a change. And you can tell the difference – when friends come around for a visit, everybody goes like: ” Hey wow, you got double glasing…”

    It is also a national habit here that people simply won’t shut their car engines down in the street and let them run for ever, etc, etc, etc The truth is that the British are wastrels and being a bit more sensible alone would cut their energy consumption in half over night – no policy needed at all.

  77. Spartacusisfree says:
    March 29, 2011 at 5:16 am
    This ‘GHG physics’ is bunkum. Let’s take the 15 micron band. 95% is absorbed in 1 m air. That’s why mirages exist – the air above a hot surface is warmed by IR absorption.

    OK, the air does re-radiate downwards but a lot of the hot air is convected upwards thus ensuring most of that original IR radiation from the ground never makes it back.

    OK, Spartacusisfree, at what speed does the hot air convect upwards? 1 mile per hour? 10 MPH? 100 MPH? 1000 MPH :^) What?

    The downwelling radiation travels at … uh … the speed of light. That is over 600,000,000 MPH. So, what are the chances that convection will yank that heated air up and away before it has a chance to radiate energy in all directions, including back down towards the Surface of the Earth?

  78. @RJ: If the earth radiates 100,000 units of energy and then receives even 1,000 unit through back-radiation, then it’s still warmer than if there were no back-radiation at all. At the same time, the sun is radiating energy to the earth. The net effect is that the earth is warmer thanks to the back-radiation.

  79. I did not say precipitation occurs in the stratosphere, it’s much lower down. And it’s easy to prove the optical physics in the climate models, which predicts the cloud part of ‘global dimming’, is wrong by looking at clouds as droplets coarsen prior to raining: the base gets darker, exactly the opposite of what the models predict.

    After NASA discovered there was no experimental proof of this cooling, they published this: http://geo.arc.nasa.gov/sgg/singh/winners4.html

    Twomey’s physics, which he warned could not be extrapolated to thicker clouds, is based on Mie theory. There is no ‘surface reflection’ but it apparently fooled climate science, hence AR4 contained the imaginary cooling.

    So the high feedback hypothesis was always the artefact of a scientific mistake. Until it’s corrected, no model can predict climate. Correct it and out pops another AGW, ‘cloud albedo effect’ heating that explains palaeo-climate [probably via phytoplankton bio-feedback] far better than CO2-GW, which if not net zero, is much lower than claimed.

  80. polistra says:
    March 29, 2011 at 6:02 am
    All of this is totally unnecessary. All we need is one valid graph of long-term temperature vs CO2 concentration, showing the lack of correlation. And we already have plenty of valid graphs like that.

    I am on record as saying that Natural Cycles and Processes have more of an effect on raising and lowering average Global Temperatures than CO2. In the WUWT series that ended with Global Warming is a Pussy Cat I estimated that only 0.1ºC of the supposed 0.8ºC warming since 1880 is Human-Caused (CO2 and Land Use) . The averaged-results of WUWT commenters on my series put Human-Caused at 0.18ºC, almost twice my original value, but still smaller than Natural Cycles and Data Bias.

    Yes, for the past decade and a half, CO2 has continued its merry rise and there has been no net statistically significant Global Warming, and, as you point out, we have lots of charts to prove that. However, there are many factors that contribute to Global Warming and Global Cooling. Some, such as Natural Cycles, sometimes warm and sometimes cool. Others, such as rising CO2, have the effect of warming, but not as much as the official climate Team claims.

    The whole point of my Visualizing the “Greenhouse Effect” series is to help WUWT readers understand that the Atmospheric “Greenhouse Effect” (and I put “scare quotes” around it to distinguish it from a physical greenhouse) is a real effect. AGW (Anthropomorphic Global Warming) is real because we are partially responsible for average temperatures being higher than they would be absent human activities. Knowing that AGW is true, but has a minor effect, will (I hope) make you all look well-informed when you dispute the scam behind Catastrophic CAGW.

  81. The argument “backradiation warms the surface of the Earth” is often used when discussing the greenhouse effect. Is it fair to infer from this argument that an object’s temperature will always be higher in the presence of backradiation than it would be in the absence of backradiation? If so, I disagree.

    Consider two disjoint blackbody objects (a spherical active object, and an inert object) in the vacuum of cold space. Assume the internal thermal conduction properties of each object are such that the surface temperature is uniform. By blackbody I mean the surface of an object (a) absorbs all electromagnetic energy incident on it, and (b) radiates electromagnetic energy in accordance with Planck’s blackbody radiation law. By “active object” I mean an object with a constant internal source of thermal energy–e.g., radioactive decay. By “inert object” I mean an object that has no internal source of thermal energy.

    The active sphere in isolation will experience no back radiation–i.e., all radiation leaving the sphere will escape to space. The active sphere in the presence of the inert object will experience back radiation–i.e., some of the radiation emitted from the active object will be absorbed by the inert object–causing the inert object’s temperature to rise–causing the inert object to radiate energy, some of which will be directed “back” toward the active object. Provided there is no object-to-object thermal conduction or convection, I believe the steady-state temperature (the temperature when the rate of energy leaving an object is equal to the rate of energy entering the object) of the active object will be higher in the presence of the inert object than it will be in isolation. However, if conduction and/or convection between objects is allowed, then depending on the rate of energy transfer via conduction / convection, in the presence of the inert object and hence in the presence of backradiation the steady-state temperature of the active object can be either higher or lower than in the absence of the inert object. If true, then the argument that “backradiation warms (i.e., increases the steady-state temperature of) the active object” is not true.

    Ira explicitly stated that his arguments apply only in the absence of conduction and convection; and with that caveat, I agree. However, since conduction and convection are both present in the Earth/Earth atmosphere system, I am not convinced that “greenhouse gas backradiation” produces a net increase in the surface temperature of the Earth. Any attempt to discuss the “greenhouse effect” in the Earth’s atmosphere without including a discussion of conduction/convection is in some sense like trying to get your car to the top of a steep hill by pushing–you’ll waste a lot of time; and when you’re done pushing, your call will likely be farther down the hill than when you started.

  82. Ira this from Jennifer Marohasy blog. The calculation for the emissivity of CO2 is found in this thread.

    Comment from: Nasif Nahle March 30th, 2011 at 1:01 am

    ” The algorithm was developed (derived) from experiments by Hottel, Leckner, Lapp, etc., etc., etc.”

    Additionally, you can see the insignificant total emissivities of the carbon dioxide on the tables that you can find in any book on heat transfer by radiation. I’m not alone on this THEORY and LAWS. Mine is a humble calculation based on experimentation and observations made by many physicists.

    …” The total emissivity of the atmospheric carbon dioxide is 0.0017, whether you like it or not. …”

    :)Hottel found also that the total emissivity of the carbon dioxide in a saturated state was very low (Ɛcd = 0.23 at 1.524 atm-m and Tcd = 1,116 °C). [6]

    As Hottel diminished the partial pressure of the carbon dioxide, its total emissivity also decreased in such form that, below a partial pressure of 0.006096 atm-m and a temperature of 33 °C, the total emissivity of the carbon dioxide was not quantifiable because it was almost zero. [6] [7] [8]

    If has shown in ( http://jennifermarohasy.com/blog/2011/03/total-emissivity-of-the-earth-and-atmospheric-carbon-dioxide/#more-7739) that at STP the emissivity off CO2 is almost zero then no back radiation can take place.

  83. Matt

    All this might do is slow the rate of cooling. It will not warm the planet. And as radiation travels at the speed of light\(and the quantity of CO2 is small) the slower cooling if it occurs would be very minimal.

    But there is still the problem of a cooler body heating a warmer one. If radiation leaves earth’s surface at say 10 degrees and collides with CO2. Won’t this lower the potential energy in the radiation. So when it returns (if it does) its vibrational energy will be lower than the surface of earth. So can not warm the planet unless the planet has in the mean time cooled by more than the radiations potential energy.

  84. RJ says:
    in reply to March 29, 2011 at 10:31 am
    “For example some say the sky cannot radiate to the earth because it is colder yet you can point an infrared detector at the sky at night and measure it! Anyway when a CO2 molecule radiates a photon downwards how does it know the earth is colder and it should not do it!? Of course if the earth is warmer the nett energy flow would be upwards (from warm to cold) and that is the only necessary requirement thermodynamically”.

    Measure what though. See below that might be relevant. From the slaying blog

    And this has been discussed before but the issue is not whether radiation travels downwards. It is whether it has a warming impact if it returns to the surface.

    —————————–

    If it returns to the surface and does not warm it must be totally reflected. If it is totally reflected then the surface cannot emit at that wavelength either (basic law of thermodynamics). Therefore it is not possible for radiation to be emitted from the surface, absorbed by the atmosphere and re-radiated downwards without increasing the temperature of the surface.

  85. Ira

    I always had a problem with the logic of the GWG theory. And thankfully all the queries I had were answered and explained in the brilliant slayers book.

    I now debate with warmist in various ways. One is to dispute the GWG theory as per the slayers book. Another to dispute the extent of the warming assuming the GHG theory is correct. Another to point out the pointlessness of CO2 taxes and cap and trade etc.

    Assuming the flawed GHG theory is almost a fact (as you are doing) without understand how weak this theory is, and its flaws, was a hindrance not a help to me.

  86. Dave Springer (march 29@2:11pm)

    “So-called back radiation, which is what you’re describing, does not warm the surface except in limited and exceptional circumstances where a mass of air transported convectively (read winds) is warmer than the surface over which it blows and even then it is limited because the air has so little heat capacity compared to the surface – imagine trying to heat a cup of coffee by stirring it with a hot feather.” So let me get this straight you think backradiation is caused by a warm wind blowing across the surface, and that this is a limited and exceptional circumstance? You’re kidding, right?
    “What actually happens is the back radiation slows down the rate of surface cooling.”
    Does it now, and where are the experiments that quantify this? Objects above absolute temperature emit electromagnetic radiation sure, but HOW MUCH? How much IR does a block of iron at freezing emit. Is it sensible. Backradiation is a myth, go measure it and tell me how many btu’s or degrees kelvin it’s good for. I’d really like to know as it’s 3 weeks into spring, it’s freezing, and we have been below normal temps by 8-10 degs. How’s this, you or anyone else can’t measure or quantify it. To see what you’re up against see Dr. Roys Spencer’s amusing but failed “cavity experiment”.

    ” But the one thermodynamic fact is always operative in the real world – the farther out of equilibrium the system is the harder it tries to move back to the theoretical equlibrium temperate.” Does it now. You already invalidated any equilibrium by calling it a moving target. Any evidence that it “tries harder”.
    The bottom line is that there is no greenhouse effect. It is an atmosphere effect due soley to water vapor. Both Feedbacks and forcing are nothing but plausible fantasies.

  87. Ira, I really enjoy your contributions and the discussions they create. With so may ideas, one really has to think and ponder what is the most realistic explanation.

    After reading through the various contributions to this thread, it seems to me that the dualistic nature of electromagnetic radiation is not taken into account sufficiently. We talk of wavelength (the wave nature of electromagnetic radiation) and photons (the particle nature). Your diagrams are of molecules being hit by photons. So it is like a billiard ball situation.

    Someone said it was more like a tsunami wave. While I agree with that somewhat, radiation is not a wave front. It is more like a constant field. The billiard ball type diagram makes it seem as though the chance of a CO2 molecule being hit is small. And for one ray of radiation it is. But, with a constant field of radiation, the CO2 molecule is always being hit by some ray of radiation.

    Nevertheless, it is hard to understand how this could cause enough increase in the water vapor in the air for the water vapor to produce the increase in atmospheric temperature that the models show.

  88. Reed Coray says:
    March 29, 2011 at 11:20 am
    The argument “backradiation warms the surface of the Earth” is often used when discussing the greenhouse effect. Is it fair to infer from this argument that an object’s temperature will always be higher in the presence of backradiation than it would be in the absence of backradiation? If so, I disagree.

    Consider two disjoint blackbody objects (a spherical active object, and an inert object) in the vacuum of cold space. Assume the internal thermal conduction properties of each object are such that the surface temperature is uniform. By blackbody I mean the surface of an object (a) absorbs all electromagnetic energy incident on it, and (b) radiates electromagnetic energy in accordance with Planck’s blackbody radiation law. By “active object” I mean an object with a constant internal source of thermal energy–e.g., radioactive decay. By “inert object” I mean an object that has no internal source of thermal energy.

    The active sphere in isolation will experience no back radiation–i.e., all radiation leaving the sphere will escape to space. The active sphere in the presence of the inert object will experience back radiation–i.e., some of the radiation emitted from the active object will be absorbed by the inert object–causing the inert object’s temperature to rise–causing the inert object to radiate energy, some of which will be directed “back” toward the active object. Provided there is no object-to-object thermal conduction or convection, I believe the steady-state temperature (the temperature when the rate of energy leaving an object is equal to the rate of energy entering the object) of the active object will be higher in the presence of the inert object than it will be in isolation. However, if conduction and/or convection between objects is allowed, then depending on the rate of energy transfer via conduction / convection, in the presence of the inert object and hence in the presence of backradiation the steady-state temperature of the active object can be either higher or lower than in the absence of the inert object. If true, then the argument that “backradiation warms (i.e., increases the steady-state temperature of) the active object” is not true.

    If I follow the logic here you are suggesting that the loss of heat via convection and conduction to the inert body can be greater than the heat gained by back radiation. That means that the inert body is inside the atmosphere of the radiating body (or vice versa). In either case the inert body is in dynamic equilibrium with this atmosphere. In which case it would be the properties of the inert body relative to the atmosphere it replaces which would determine whether the surface warmed or not. If the inert body was thermally more conductive the surface would be cooler and if it were a better insulator the surface would be warmer. However I do not see how this is relevant to the situation we have where, in order to simulate the actual atmosphere, your “inert body” would have to have exactly the same composition and properties as the atmosphere itself.

  89. mkelly says:
    March 29, 2011 at 11:20 am

    If has shown in ( http://jennifermarohasy.com/blog/2011/03/total-emissivity-of-the-earth-and-atmospheric-carbon-dioxide/#more-7739) that at STP the emissivity off CO2 is almost zero then no back radiation can take place.

    You were doing fine until you said no back radiation takes place. CO2 has no line emission in a cold dense gas. It has an absorption line. After absorption it loses the energy kinetically via collisions primarily with oxygen and nitrogen. This thermalizes the gas mixture i.e. sensibly warmer. The air is ALWAYS emitting a continuous blackbody spectrum! Everything with a temperature above absolute zero emits a continuous blackbody spectrum with a peak frequency that rises with rising temperature. The higher the peak frequency the more energy in the emission. This is where the downwelling radiation comes from and its existence is without doubt.

  90. Brian W says:
    March 29, 2011 at 12:00 pm
    Dave Springer (march 29@2:11pm)

    “So-called back radiation, which is what you’re describing, does not warm the surface except in limited and exceptional circumstances where a mass of air transported convectively (read winds) is warmer than the surface over which it blows and even then it is limited because the air has so little heat capacity compared to the surface – imagine trying to heat a cup of coffee by stirring it with a hot feather.” So let me get this straight you think backradiation is caused by a warm wind blowing across the surface, and that this is a limited and exceptional circumstance? You’re kidding, right?
    “What actually happens is the back radiation slows down the rate of surface cooling.”

    Does it now, and where are the experiments that quantify this?

    “Heat: A Mode of Motion” by John Tyndall, 1859. This is old stuff. If you don’t know this you don’t know basic experimental physics.

  91. Dave Springer

    ” This is old stuff.” OH IS IT NOW! That’s funny NOWHERE in the 512 pages of Tyndall’s great experimental work does he EVER mention backradiation, nor does he mention the term greenhouse EVEN ONCE. You are blowing a lot of hot MOIST air and I’m calling you on it. Bring out some of that old stuff and let’s find out!

  92. George E. Smith says:
    March 29, 2011 at 9:42 am
    Where Argon ?

    Argone gone to lunch :^)

    As I mentioned, this simplified depiction is of a small (acually teeny tiny to use the technical term) section of the Atmosphere and, by luck, there was no Ar in that section.

  93. Dave Springer says: Everything with a temperature above absolute zero emits a continuous blackbody spectrum with a peak frequency that rises with rising temperature. The higher the peak frequency the more energy in the

    It’s not a peak frequency. It’s a median energy. Blackbody spectrum is a spectrum with a central energy and energys higher and lower thus forming the BB spectrum.

  94. “If it returns to the surface and does not warm it must be totally reflected. If it is totally reflected then the surface cannot emit at that wavelength either (basic law of thermodynamics). Therefore it is not possible for radiation to be emitted from the surface, absorbed by the atmosphere and re-radiated downwards without increasing the temperature of the surface”.

    My understanding is that this is not correct. As a cooler body can not warm a warmer body. Otherwise if for example a human was in a container of CO2 they would heat up rather than just cool at a slower rate. Or a human in a vacuum with a perfect radiation reflector would eventually cook to death. This surely is not possible.

    So my understanding is that radiation from a colder body can not heat a warmer body. As a teacher can pass and increase knowledge of a student. And a student to a younger student. But not increase knowledge (heat) by passing it back

    .

  95. Gerge E Smith says: My impression is that there is confusion between thermal transfer between molecules (rotation, vibration etc) and radiation emission/absorption. The first depends on the thermodynamics, the second on quantum physics

    I would lkie an euro for every time I’ve said this on this site and stil people continue.

  96. Surely they can only re-emit much lower energy photons, of similar energy levels to the black-body curve at -20C.

    This is a confusion between kinetic heat energy and quantum radiative energy.

  97. Dave Springer says:
    March 29, 2011 at 12:30 pm
    The air is ALWAYS emitting a continuous blackbody spectrum! Everything with a temperature above absolute zero emits a continuous blackbody spectrum with a peak frequency that rises with rising temperature. The higher the peak frequency the more energy in the emission. This is where the downwelling radiation comes from and its existence is without doubt.

    Of course it does not. Only a black body emits a black body spectrum. The atmosphere is not a black body. If it were a black body we would not be able to see the sun! Indeed most things are not even close to black bodies – that is why the world is colourful. Did you not see the spectral distribution looking up at the night sky that was posted about a week ago? It looked nothing like the black body curve. It was completely dominated by the greenhouse gas spectral lines as one would expect. The atmosphere is pretty transparent in the visible region and has an atmospheric window around 10 micron where there is not much absorption, so these wavelengths are absent in the downward radiation as well.

  98. This was a great post!
    Brilliant.
    Very very nice presentation.
    Thank you, now I will consider this in depth.

  99. The greenhouse effect and the energy retaining effect of the oceans are different things. The atmosphere and the ocean may retain heat, but the atmospheric heat retention is infinitesimal compared to the heat retention of the oceans. The greenhouse effect changes the amount of heat that flows out of the system.

    The oceans absorb the vast majority of the RADIATIVE ENERGY stiking the planet from the sun. This RADIATIVE energy leaves the planets surface by a route proportional to the density of the atmosphere (not just GHGs) and the speed of light. That’s why Stephen Wilde said what he said. ALL the energy on this planet comes from the sun ~ 99.9%. The Earth approximates to a blackbody in as much as it’s surface albedo modifys the amount of RADIATION that is reflected back into space. A blackbody cannot re-absorb radiation it has given up after equilibrium is reach. After equilibrium all radiation striking the BB is reflected away that’s why the earth and the moon cannot heat the sun by radiative reflection. After bombarding the earth with radiation for 5 billion years they are as near as damn it at BB equilibrium unless the sun’s radiative output changes.

  100. Dave Springer says:
    March 29, 2011 at 12:30 pm
    You were doing fine until you said no back radiation takes place.

    I was speaking very narrowly about CO2 per the drawing of Ira’s. If CO2 has at STP no emissivity then by Kirchoff’s Law it did not absorb.

  101. Reed Coray says:
    March 29, 2011 at 11:20 am
    … [excellent example of an inert object in the cold vacuum of space causing the surface of an active object to be warmer than it would have been absent the proximity of that inert object] …

    Ira explicitly stated that his arguments apply only in the absence of conduction and convection; and with that caveat, I agree. …

    Thanks Reed Coray for your clear example that works for you (and me) in the cold vacuum of space. Thanks also for agreeing with my argument that the Atmospheric “Greenhouse Effect” raises the temperature of the Surface under the conditions I stated, namely that I was not considering conduction and convection (although I also stated that they have significant effects on the Surface temperature of the Earth).

    However, since conduction and convection are both present in the Earth/Earth atmosphere system, I am not convinced that “greenhouse gas backradiation” produces a net increase in the surface temperature of the Earth. Any attempt to discuss the “greenhouse effect” in the Earth’s atmosphere without including a discussion of conduction/convection is in some sense … you’ll waste a lot of time …

    I hope I am not wasting my time or yours.

    Whether or not the Atmospheric “Greenhouse Effect” is present, the effects of convection and conduction (and precipitation and so on and on) depend upon the Surface and Atmospheric temperatures, humidity, and a bunch of other things. I think we both agree they have a net cooling effect.

    How could the effects of convection/conduction/precipitation more than counteract the effect of GHGs in the Atmosphere, which you agree, taken alone, would increase Surface temperatures?

    Are you proposing that a warmer Surface would cause more convection/conduction/precipitation? Well, yes, I think it would! However, if these net cooling effects were greater than the warming due to the GHGs, the Surface would cool, and the increased convection/conduction/precipitation would go away. Thus, it seems to me that convection/conduction/precipitation can only moderate the effects of GHGs, not cancel them out entirely.

  102. Inter-radiation of energy between molecules in all planetary atmospheres are random walks with an upward bias due to the curvature. The only place this does not occur is in the window where radiation can move from the surface to space directly.

    Cassandra King, you should be able to see this effect. It is what further reduces the effect of that 0.039% factor you mentioned for we live on a round Earth.

    Try reviewing this comment and the contained image: http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/#comment-629487

  103. cal says:
    March 29, 2011 at 12:12 pm

    Cal, thank you for your response.

    My argument has little if any relevance to the computation of the surface temperature of the Earth in the presence of greenhouse gases. However, it does have relevance to Ira’s statement: “That said, how does this visualization help us understand the issue of “CO2 sensitivity” which is the additional warming of the Earth Surface due to an increase in atmospheric CO2? Well, given a greater density of CO2 (and H2O) molecules in the air, there is a greater chance that a given photon will get absorbed. Stated differently, a given photon will travel a shorter distance, on average, before being absorbed by a GHG molecule and be re-emitted in a random direction, including downwards towards the Surface. That will result in more energy being recycled back to the Surface, increasing average temperatures a bit.

    In particular, his statement implies backradiation results in (a) more energy being returned to the Earth surface, and (b) an increase in average temperatures. If any situation can be constructed where (a) backradiation does NOT result in more energy being recycled back to the active object’s surface, and/or (b) backradiation does NOT increase the active object’s surface temperature, then in my opinion Ira’s statement requires additional justification.

    By applying Planck’s blackbody radiation law to the surfaces of (1) an active, uniform-surface-temperature sphere and (b) a thin, co-centered, disjoint, inert, planar, uniform-surface-temperature, circular annulus, it can be shown that in the presence of object-to-object conduction, the rate of inert-object to active-object backradiation, although present, can be made less than the rate of conductive energy transfer from the active object to the inert object. This will result in a lower active-object temperature. Furthermore, even in the absence of object-to-object conduction, increasing the outer radius of the inert annulus beyond a certain point, which provides a larger surface area from which backradiation can originate, results in less backradiation, not more. It may be true that greenhouse gas backradiation warms the surface temperature of the Earth, but Ira’s statements that (a) adding more greenhouse gas will increase the amount of backradiation, and (b) that the additional backradiation will warm the surface of the Earth do not convince me. I want to see an analysis of how and to what degree conduction and convection affect the amount of backradiation received by the Earth’s surface relative to the rate of energy transfer from the Earth’s surface by these thermal transfer mechanisms. For example, wouldn’t the increase in the amount of greenhouse gas increase the rate of conductive thermal transfer and possibly convective thermal transfer from the Earth’s surface to the atmosphere? If so, which effect is dominant, and under what conditions?

  104. cal says: March 29, 2011 at 10:12 am

    “I am getting a bit depressed by the level of physics being discussed here and in some of the other web sites that we are being directed to…. It is pointless coming up with simplistic arguments to deny these facts when we have data to confirm them.”

    AMEN!

    As to the original post
    * in (7) you say “The energized air molecules emit radiation at various wavelengths”
    The emission is still by the GHGs since other molecules are notoriously poor at emitting any sort of IR.
    * along with the previous point, the diagram in step 7 seems to show an O2 molecule emitting IR. This would not happen (to any significant degree).
    * Most importantly, you miss one major way that molecules can get the energy to emit IR — collisions with other molecules. Just as GHG molecules that have absorbed extra energy from IR photons can pass that energy on to other molecules, other molecules can collide with GHG molecules and give them extra energy from the collision that can then be emitted as IR photons.

  105. cal says:
    March 29, 2011 at 11:41 am

    If it returns to the surface and does not warm it must be totally reflected. If it is totally reflected then the surface cannot emit at that wavelength either (basic law of thermodynamics). Therefore it is not possible for radiation to be emitted from the surface, absorbed by the atmosphere and re-radiated downwards without increasing the temperature of the surface.

    Cal, I have to disagree. Consider a spherical active Earth surrounded by an “atmosphere” comprised of co-centered, thin, blackbody-surface (both sides), spherical shell with a vacuum filling the space between the Earth’s surface and inner surface of the shell. I know this isn’t a gaseous atmosphere, but in the sense that the shell surrounds the Earth, I will take license and call it an atmosphere. In isolation, the Earth’s surface temperature will be (a) directly proportional to the fourth root of the Earth’s internal rate of energy generation and (b) inversely proportional to the square root of the Earth radius. No backradiation exists for the Earth in isolation.

    When the shell is added, because in steady state the shell must radiate to space the energy generated internal to the Earth, the steady-state temperature of the shell (for a thin shell, the inner and outer shell surface temperatures will be very near the same) will be directly proportional to (a) the fourth root of the Earth’s internal rate of energy generation and (b) inversely proportional to the square root of the shell radius. Backradiation (radiation from the inner shell surface to the Earth) will exist. Since the radius of the shell is greater than the radius of the Earth, the shell’s surface temperature will be lower than the Earth surface temperature in isolation. If a thermal conduction path is provided from the Earth to the shell and if that path supports a rate of energy transfer greater than the radiative rate of energy transfer from the shell’s inner surface to the Earth, the Earth’s temperature in the presence of the shell with conduction will be lower than (a) the Earth’s temperature in the presence of the shell without conduction, and (b) the Earth’s temperature in absence of the shell. Since radiation is being emitted from the active object (the Earth) and backradiation from the inert object (the shell/atmosphere) exists whenever the shell is present, how can you claim backradiation from the shell/atmosphere must warm the surface temperature of the active object?

  106. Reed Coray says:
    March 29, 2011 at 3:16 pm
    … Consider a spherical active Earth surrounded by an “atmosphere” comprised of co-centered, thin, blackbody-surface (both sides), spherical shell with a vacuum filling the space between the Earth’s surface and inner surface of the shell. I know this isn’t a gaseous atmosphere, but in the sense that the shell surrounds the Earth, I will take license and call it an atmosphere. …

    When the shell is added, because in steady state the shell must radiate to space the energy generated internal to the Earth, the steady-state temperature of the shell (for a thin shell, the inner and outer shell surface temperatures will be very near the same) will be directly proportional to (a) the fourth root of the Earth’s internal rate of energy generation and (b) inversely proportional to the square root of the shell radius. Backradiation (radiation from the inner shell surface to the Earth) will exist. Since the radius of the shell is greater than the radius of the Earth, the shell’s surface temperature will be lower than the Earth surface temperature in isolation. If a thermal conduction path is provided from the Earth to the shell and if that path supports a rate of energy transfer greater than the radiative rate of energy transfer from the shell’s inner surface to the Earth, the Earth’s temperature in the presence of the shell with conduction will be lower than (a) the Earth’s temperature in the presence of the shell without conduction, and (b) the Earth’s temperature in absence of the shell. Since radiation is being emitted from the active object (the Earth) and backradiation from the inert object (the shell/atmosphere) exists whenever the shell is present, how can you claim backradiation from the shell/atmosphere must warm the surface temperature of the active object?

    Very clever Reed Coray! I have to agree that the shell will be at least a little cooler than the Earth Surface because it has a larger surface area. Given a nearly 100% conduction path (such as many metal struts from the Earth Surface to the shell), the shell could be a bit cooler than the Earth and more than compensate for the warming due to the “Greenhouse Effect”.

    Great theory, however there are at least three issues: 1) The small difference in surface area, 2) The large temperature increase due to the Atmospheric “Greenhouse Effect”, and 3) Convection/conduction/precipitation are not anywhere near 100% effective at transfering thermal energy from the Surface to the Atmosphere. Let us address each in turn:

    1 – If the shell is at an altitude similar to the effective altitude of the Atmosphere, which is a teeny tiny fraction (less that 1/1000) of the radius of the Earth, the shell’s surface area will be almost the same as the Surface area of the Earth, so the supposed temperature difference will be miniscule.

    2 – The effect of “Greenhouse” warming of the Earth is over 30ºC, which is many, many times the maximum possible temperature difference due to (1).

    3- Even if (1) was as large as 30ºC, the combined sum of convection/conduction/precipitation would fail to transfer anywhere near 100% from the Surface to your shell.

    Nice try though and THANKS for getting me thinking.

  107. Gravity is a force that is constantly pressurizing the atmosphere. Consider a warmed parcel of air at the surface. It will rise, no? When it rises, it eventually cools. When it cools and is pulled to the surface by gravity, its temperature increases because it is compressed.

    The constant compression of atmosphere by gravity results in higher pressure, density and temperature at the surface. No GHG’s are necessary for this..

  108. Reed Coray says: (in reply to my reply)
    March 29, 2011 at 2:44 pm
    …………. If any situation can be constructed where (a) backradiation does NOT result in more energy being recycled back to the active object’s surface, and/or (b) backradiation does NOT increase the active object’s surface temperature, then in my opinion Ira’s statement requires additional justification. ………It may be true that greenhouse gas backradiation warms the surface temperature of the Earth, but Ira’s statements that (a) adding more greenhouse gas will increase the amount of backradiation, and (b) that the additional backradiation will warm the surface of the Earth do not convince me. I want to see an analysis of how and to what degree conduction and convection affect the amount of backradiation received by the Earth’s surface relative to the rate of energy transfer from the Earth’s surface by these thermal transfer mechanisms. For example, wouldn’t the increase in the amount of greenhouse gas increase the rate of conductive thermal transfer and possibly convective thermal transfer from the Earth’s surface to the atmosphere? If so, which effect is dominant, and under what conditions?
    ————————————–
    Reed, I think it is totally reasonable to question whether there is a limit to the impact any change can have and whether there are compensating effects. Almost all natural effects have a limit and negative feedbacks. That is why the earth is ruled by cycles and, for example, does not get permanently hot or permanently cold. The greenhouse effect is logarithmic so we know that without positive feedback one has to double the concentration to get a 0.7C temperature rise and double again to get another 0.7C. Nothing to get concerned about. That is why positive feedback has to be postulated in order to scare us. With such small effects one has a duty to look at all other small effects that might create negative feedbacks which might partially compensate or even overcompensate and drive the temperature in the opposite direction as CO2 increases. You suggest this might be possible and while I do not follow your reasoning as to the specific mechanism I entirely agree that this is possible. So while I am convinced that back radiation warms the surface I am not convinced that further increases in CO2 will warm the surface further. I see these as two entirely different questions and is the reason why I believe sceptics should focus on this second question and not the first.

    You might be interested in one possibility that I raised (in person) directly with the UK met office. I have never had a satisfactory answer even though it was repeatedly promised. As you will probably be aware CO2 is the most significant molecule involved in the COOLING of the atmosphere. It radiates most of the energy from the thermosphere and stratosphere and together with H2O radiates almost all the energy from the top of the troposphere and tropopause. H2O is the dominant source of radiation from the mid troposphere and of course the surface radiates directly to space, mainly through the atmospheric window around 10 micron.

    So clearly there is the potential for a negative feedback through increased cooling as the concentration of CO2 increases. Which effect dominates? I believe the physics is too complicate to work this out. To find out the answer we have to do something outrageous. We have to measure what is actually happening!

    We need answers to questions like: what is happening to the position of the tropopause? what is happening to the temperature in the stratosphere immediately above the tropopause? If the tropopause and lower stratosphere are cooling then we might expect the surface to continue to warm but if there is no change to the tropopause or there is warming then (perhaps non intuitively) this means more radiation to space and a lower surface temperature. A couple of years ago there was a report of a measured increase in the altitude of the tropopause and implied cooling but since then it has gone very quiet, so I suspect the measurements are not yet providing the “right” answer!!

  109. Ian W says:

    The AGW CO2 warms the world hypothesis depends on the hydrologic cycle. It only works if they can show the ‘water vapor feedback’ which all the GCMs show as a tropospheric hotspot that in the real world does not exist therefore all the models are falsified.

    The water vapor feedback is now extremely well-verified, especially over timescales of months to a few years. See http://www.sciencemag.org/content/310/5749/841.abstract and http://www.sciencemag.org/content/323/5917/1020.summary

    Amplification in the tropical troposphere (the so-called “hot spot”) is also well-verified over these timescales. Over the multidecadal timescales, the data is simply too uncertainty to reach any definitive conclusions but it is difficult to imagine mechanisms by which the physics responsible for this would break down on these timescales without also breaking down on shorter timescales.

    Dave Springer says:

    The whole brouhaha over scary global warming rests on the wholly fabricated idea that CO2 greenhouse warming puts more water vapor in the atmosphere and that, because water vapor is a powerful greenhouse gas, it constitutes a positive feedback that turns 1C per doubling into 3C per doubling. There isn’t a shred of empirical evidence to support that belief and IMO overwhelming evidence that the feedback is actually negative – the water cycle speeds up which serves to mechanically transport heat from the surface by evaporation and convection to the cloud deck where it then has an easier radiative path out to space.

    The only way that radiation escapes to space is by stuff radiating and the amount of radiation that stuff emits is determined by the temperature. Hence, your statement amounts to the hypothesis that the “hot spot” ought to be more pronounced, not less pronounced, than the models predict.

    The models already take into account the negative feedback due to the lapse rate, which basically seems to be what you’re describing. But a literal interpretation of the claim that the hot spot is missing would be that the models are including a negative feedback that does not exist.

    If the water vapor being delivered to the upper troposphere increases so does the amount of heat carried away from the surface by evaporation and convection and it condenses at an increased rate rather than giving the upper troposphere a higher absolute humidity.

    Except that the satellite data seem to show otherwise, as I noted above.

    Thus evaporation, convection, and cloud formation is a negative feedback that negates the effect of more water vapor due to warmer surface temperature. There are several well developed hypotheses on this effect. The best IMO is

    which is basically saying that the atmosphere, with regard to water vapor greenhouse warming, is saturated and we can’t get any more greenhouse warming out of water vapor – we can get additional surface warming only from increases in non-condensing greenhouse gases like CO2 and methane.

    If that’s the best, I can’t imagine what the worst looks like! Miskolczi’s paper is basically nonsense with lots of extremely strange things like misapplying the virial theorem.

  110. I agree with Michael Larkin that the animation was much too fast. If you right-click on the picture and save it and open it in Irfanview you can save it as eight separate pictures and then move forward and backwards at leisure (Options – Extract all frames).

  111. BigWaveDave says:

    The constant compression of atmosphere by gravity results in higher pressure, density and temperature at the surface. No GHG’s are necessary for this..

    Your notion doesn’t even satisfy the 1st Law of Thermodynamics! In the absence of a greenhouse effect, the surface temperature of the earth is set by energy balance considerations. So, unless you are proposing a large source of energy…e.g., that the earth’s atmosphere is undergoing continual gravitational collapse, which would certainly be a novel hypothesis…your notion is completely in violation of known laws of physics.

  112. bananabender says:
    March 29, 2011 at 12:41 am

    The Ideal Gas Law PV=nRT quite accurately predicts the atmospheric temperature of Earth, Mars, Venus and Titan (at all altitudes) based on a simple pressure effect. Why add the unnecessary complication of the highly dubious Greenhouse Effect to an elegant and simple solution?

    You are wrong. John Tyndall showed that you are wrong in 1859, and his explanation of why the earth is as warm as it is has been accepted by scientists ever since. Even Fourier in 1829 knew you were wrong, but it was Tyndall who showed it conclusively.

  113. Two main mistakes as pointed out above.
    1. Absorbing by CO2 at 15 microns does not speed up the molecule, it raises its vibration state, and may change its rotation state (15-micron side-bands).
    2. Photons are not emitted by molecules other than GHGs and not at wavelengths other than those where they have spectral bands.
    Other than that, fine.

  114. Joel Shore says:
    March 29, 2011 at 4:51 pm

    Your notion doesn’t even satisfy the 1st Law of Thermodynamics! In the absence of a greenhouse effect, the surface temperature of the earth is set by energy balance considerations. So, unless you are proposing a large source of energy…e.g., that the earth’s atmosphere is undergoing continual gravitational collapse, which would certainly be a novel hypothesis…your notion is completely in violation of known laws of physics.

    You are mistaken. You can’t ignore that gravity is a force, and that a gas is compressible, and will increase in temperature as it is compressed. Unlike the GHG hypothesis, these are physical phenomena that are measurable and reproducible.

  115. “”””” cal says:
    March 29, 2011 at 10:12 am
    I am getting a bit depressed by the level of physics being discussed here and in some of the other web sites that we are being directed to. I am beginning to wonder if there is a strategy amongst the warmers to refute the greenhouse effect using bogus physics in order to quote these arguments, say how stupid they are, and thus justify the dismissal of all sceptic arguments as unscientific. Or I might just be paranoid!

    There are serious problems with the quantification of the greenhouse effect. There is a serious doubt about feedbacks. There is a serious problem that the “fingerprint” of an enhanced greenhouse effect cannot be identified.

    But there is no doubt that the atmosphere absorbs radiation emitted by the earth and eventually radiates it to space at a higher altitude which, being at a lower temperature, means a smaller energy loss. That reduction has to be compensated for by increased radiation losses elsewhere and that means a warmer surface. “””””

    Well there is one thing seriously wrong with your thesis, cal.

    The earth’s surface is mostly hotter than the air is at altitude. Therefore the surface emits at a higher emission level than the cooler atmosphere, and moreover it does so in a generally higher frequency (lower wavelength range; per wien’s Displacement Law.

    So the very hottest desert surfaces, are emitting plenty of radiation that goes right on by the CO2 15 micron absorption band right out into space unimpeded by CO2. And as you get higher, and the atmosphere gets colder and less dense, the absorption bandwidth of the 15 micorn CO2 band keeps getting narrower; each and every one of the high resolution lines in the CO2 band keeps get less broadened by doppler, and collisions, as you go up in the atmosphere, so it keeps on absorbing an even lower fraction of the surface emitted radiation, and escape gets easier, as you go higher. So you are not dependent on energy being transported to the highest least dense, and coldest upper reach4es of the atmospehre, and then expecting that material to radiate the entire heat load of the planet.

    And if you go in the opposite direction (atmospheric emissions are isotropic, so only half of it goes down, the downward path (that “back radiation thing”) sees a continuously warming, and denser atmosphere so the CO2 absorption spectrum gets more and more opaque for the downward radiation, instead of more and more transparent for the upward escape path.

    So the earth des not sit at some BB Temperature of 255 K due to some high altitude evanescent layer of colder rarified atmosphere. Certainly the energy borne aloft as heat by conduction, convection, and evaporation (latent heat) must eventually be radiated by higher altitude gases; but the very ground itself, is making a significant direct contribution to the radiative cooling that augments the other (thermal) processes.

    The earth is anything but Isothermal, and we should not expect it to radiate some isothermal radiation spectrum.

  116. BigWaveDave says:

    You are mistaken. You can’t ignore that gravity is a force, and that a gas is compressible, and will increase in temperature as it is compressed. Unlike the GHG hypothesis, these are physical phenomena that are measurable and reproducible.

    I am well-aware that gravity is a force and that gases are compressible. I, and other atmospheric and climate physicists are also aware of the constraints on lapse rates set by the decrease of pressure with height. However, that does not allow you to violate the Laws of Thermodynamics at will. And, these Laws tell us that the surface temperature of an IR-transparent atmosphere will go toward a steady-state value determined by the condition that the radiative emission balance the radiation absorbed from the sun (in the absence of any other significant energy source).

  117. Harold Pierce Jr says – – –

    Classic error in thinking and the modeling. The concept that once the CO2 molecule “absorbs” the IR, and becomes vibrationally excited, it “losses” that energy to the O2 and N2, and thus “traps” the energy in the atmosphere.

    NOT TRUE! There is a distribution of energies of the ensemble of molecules. There is EXCHANGE. Therefore, collisions between O2 and N2 molecules on the “high end” of the distributed energy, LOSE that translational kinetic energy to CO2 and H2O molecules.

    WHICH, can…with significant probability, re-emit an IR photon.

    That’s why the REAL question here is what is the “optical transparency” of the ensemble of the atmosphere, and has the increased CO2 since WWII changed that number significantly.

    Studying the work of Miskolczi, Ferenc M. 2010. “THE STABLE STATIONARY VALUE OF THE EARTH’S GLOBAL AVERAGE ATMOSPHERIC PLANCK-WEIGHTED GREENHOUSE-GAS OPTICAL THICKNESS”, Energy and Environment, 21, 243-262 , indicates that this is the case. I’m well aware of the critics of Miskolczi’s work. I’m also aware that based on “empirical observations” up to 1942, Dr. Elsasser came to the same conclusion as Miskolczi, and neglects CO2 in his “general radiation chart”, which has been used successfully to model day to day heat up and cool downs for “weather forecasting” since publication of his paper, “On the IR Heat Balance of the Atmosphere”, (Harvard Met. Series, 1942)

    Dr. Glickstein: I hope you are retired, and have the time to examine these matters in depth. I, alas, still have to work for a living.

    Max :)

  118. As soon as you put in convection, which completely dominates IR radiation in the troposphere, then I’ll be interested. Until then, it’s just a matter of “Given these conclusions what assumptions can a draw?” followed by “Assume there is no convection”…

  119. If backradiation was a phenomena that would heat up earth surface it would also heat up the sun up to the core.

    By using Stefan-Boltzmann the non-GHG surface temperature is found by using E = sigma T^4 (SB) where E can only originate from the sun. If now T would go up further because of GHG, this would mean more energy input would be needed. So where does this come from, are GHG capable of generating energy? Many people seem to think so, but what about the following?

    If we look at a CO2 layer as a plate with T2 above earth, you have something like the classic example of radiation between two plates here, where the sun would heat the lower one which is T1.
    So yes radiation will go both ways, there is ‘backradiation’ and I go with that. But read that in the state (equation 19.2) where also plate number two would be T1, Qnet will be zero. Surface 1 will never get any warmer by the (back)radiation, despite the accumulated (heat)energy in plate 2 and radiation going everywhere. Read from this that radiation is energy but never heat.

    So you can make calculations with multiple layers (like CO2 layers) and the result will be the same. T1 will not get any warmer with Qnet getting zero, and all the layers acting as insulation (like some people think CO2 does) and all plates actually permanently getting radiation from both sides. (Earth atmospheer would also be forced to become T1 from the surface up if it was’nt for gravity).

    Now think about multiple plates where the sun’s surface is the first plate and earth the second etc. So the greenhouse effect says that the second plate is supposed to get warmer by backradiation from the plates next in line? Then also the first one, the sun’s surface will get warmer (no matter how tiny) by this backwards phenomena. The sun has many more layers, but this process will go on up to the core! The source of the energy!

    Well, what is wrong here. The problem is that in our real universe the temperature of the sun in space is the driving force, telling all other layers down the line what temperature they must take on, in its path to the 3 Kelvin of space and not the other way around.
    With new layers somewhere in the line, the temperature of the layers behind the new one will simply go down to establishing a larger dT maintaining the same energy flux.
    Yes, it’s that pesky Second law, that in fact is constantly trying to lower the temperature of the sun.

    So what are the tricks with radiation that many people don’t see? For starters I want to say that the photons of EM radiation can be regarde as cold, only interaction with matter gives any thermal energy (to get a temperature) which is taken away (partly) when a(nother) photon leaves.

    Most greenhouse effect statements/sites claim an energy balance on earth surface because of the First law, the greenhouse theory makes fame by using energy in-out balances etc. convincing everyone who has no clue about thermodynamics. But this is dead wrong, the Second law rules here and this means simple energy conservation is not what determines radiative equilibrium temperature at the surface.
    The earth could conserve it’s energy by accumulating it until it would be as hot as the sun, there is nothing against the First law with this.

    So in GHG physics T of earth surface is supposed to go up with while receiving the same amount of energy from the sun. Now we know the Second law wants to establish a dynamic radiative equilibrium at the surface with a certain temperature but also that T = dE/dS, so this means that for T to be able to go up with backradiation Entropy must decrease during the absorption-emission process?? That will never happen.

    Looking closer it works like this:
    Why does the sun heat the surface to T? Because HQ radiation from the sun can leave earth as LQ IR radiation (to the cold space) in the process gaining entropy while raising kinetic energy as potential WASTE HEAT on the surface (and they don’t call this irreversible for nothing, the downgraded (photon)energy won’t be able to do the same trick at the surface again).
    The radiation Quality is expressed as theRadiation Energy Density
    Some LQ IR returns (backradiation), but it can never ever create more heat here. Because in that case it would have to leave with less energy so as even lower quality IR, and hey …… this can only happen at a LOWER surface temperature (think of the BB-spectra of earth next to the sun, all the energy flows from the high spectrum to the low one, anyone thinks some energy goes the other way here?).
    See the contradiction, that’s the mighty Second law of thermodynamics in action that says: it’s all wrong with this heating by backradiation philosophy.
    So backradiation will simply leave in a reversible proces (like reflection) without leaving waste heat, and with the same frequency as it had before and bounce whatever way it wants with no effect on the surface. It’s like in the Qnet = zero situation between the two plates (lotsa radiation, no more heating by the cold photons).

    So what is ‘the heat’ that can’t go from low to high temperature, warming the earth surface and why the confusion?
    Maxwell’s classic Theory of Heat states: heat is something which may be transferred from to another, according to the Second law of thermodynamics.
    In any case, heat transfer needs matter which has a temperature (as kinetic energy) and another piece of matter at lower T. Radiation has no temperature and can never ever represent heat. And so the photons of radiation can be regarded as cold.
    Heat is the waste from the irreversible radiation phenomena happening due to what the Second law dictates and is equivalent to the entropy gained and what is called dissipation of energy.

    Heat can be released when a photon interacts with matter. Many think ‘the heat’ has to do with all the energy (E= h x v) related to the cold travelling photon. But we have seen that this is not the case, this is imaginary heat (which is not heat but energy) that is never released as long as matter is’nt involved. Energy is only released when the SWR photon disappears in the matter and after that heat from this matter may flow if other colder matter is present, but this heat is not equivalent with the photon energy but is a function of the temperature difference between surface temperature and colder matter above it (like Q = k*A*dT). If there is no colder matter, there is no heat (dT=0).

    If earth had no atmospheer, the photons would warm earth surface to SB temperature but no heat would ever be exchanged. Then if there was a layer of CO2 around it at say 10 meters or 10 km, this would create backradiation, but the vacuum inbetween would be as cold as the 3K space, and the backradiation would not make the surface warmer, and there would be no heat despite all the radiation.

    Now because heat is flowing from earth surface we know we have a second irreversible process (waste heat, making entropy rise) we also know for sure the new lower energy photon can never release heat at this temperature where it came from ever again, so to think backradiation can go beyond that and even create a higher temperature is impossible by the Second law of thermodynamics.

    So the heat coming from the surface is the part of the solar radiation that did’nt get radiated away (directly) as IR (and the total energy of this IR is thus lower than the LWR that hits the surface). Heat is the part that has nothing to do with radiation, and the IR that left has nothing to do with heat.
    And so we get the surface balance: IR = LWR – HEAT. So the IR that leaves the surface does’nt only have a lower frequency, but the total energy of it is also lower than the incoming LWR. This is the second blow for the imaginary effect of backradiation.
    It is this HEAT only that warms the first layers of the atmospheer and gives the ground temperature’s measured (in air, so does not even represent the surface), and it can only take place by conduction from surface to air initially, it must be transported from the surface at T1 to matter at T2 and it finds its way up (conduction/convection – radiation) to space separated from the IR photons that left the surface.

    Papers about the sun that are not infected with the greenhouse virus go along the lines of my view. Photons jiggle around for tenthousands of years from layer to layer, and they do not heat the core. This degradation of high quality X-ray photons to relatively low quality optical photons, is only to be expected from the Second law of thermodynamics.
    Further: a photon can only travel a tiny distance before running into another hydrogen nucleus. It gets absorbed by that nucleus and the re-emitted in a random direction. If that direction is back towards the center of the Sun, the photon has LOST GROUND! It will get re-absorbed, and then re-emitted, over and over, trillions of times. The path it follows is called a “random walk”

  120. Max gets a good part of it, but there are a couple of things missing, the higher you go in the troposphere, the colder it gets, and the slower the emission from collisionally excited CO2, water vapor or any other greenhouse gas. Second, the distance a photon emitted from CO2 can go is not very far, so effectively, the height at which emission in the CO2 bands can reach space is about 10 km, where it is very cold. You can see this in the modtran spectra (just hit submit the calculation) which show what would be observed from say the space station. The emission between 600 and 700 cm-1 is much lower than it would be without CO2, because it comes from a layer 10km up where the temperature is 220 K.

  121. Joel Shore says:

    “I am well-aware that gravity is a force and that gases are compressible.”
    So, what happens to cooled air that descends?

    “I, and other atmospheric and climate physicists are also aware of the constraints on lapse rates set by the decrease of pressure with height.”

    If this is supposed to be some sort of appeal to authority, I’m not at all impressed. First, you have to say something that demonstrates you understand what you are talking about.

    You say: “However, that does not allow you to violate the Laws of Thermodynamics at will.”

    Nor does it allow you to ignore the system conditions with statements like this : “And, these Laws tell us that the surface temperature of an IR-transparent atmosphere will go toward a steady-state value determined by the condition that the radiative emission balance the radiation absorbed from the sun (in the absence of any other significant energy source).”; which totally ignores the influence of gravity on the initial state of the compressible gas that makes up your hypothetical IR-transparent atmosphere.

    But, where have I violated any laws of thermodynamics?

  122. When I got to the part where you demonstrated your complete lack of understanding the thermalization process I quit reading. The idea of an IR photon being absorbed and then quickly remitting with half going up and have down is absolute BS. Go back and study how the atmospheric thermal reservoir works.

  123. I’m stunned that there is debate about the ability of one radiating object being able to heat another radiating object. It is so numbingly obvious that this happens that it begs the question: What the hell would keep it from happening?!?

    As for which direction molecules radiate energy, try thinking of each of them as isotropic radiators (they’re not, but the principle applies). Except for geometry limitations of the molecule (asymmetric construction), they will behave like an isotropic radiator and radiate equally in all directions. In the atmosphere, where molecular orientation is assuredly random, even asymmetric molecules on average will not favor up vs down vs over there when it comes time to shed their packet of new energy.

    Here’s another mental widget to help get your head around this. We’re talking about radiation here, remember. Picture a flash bulb sitting in free space. Imagine it is lit off. Imaging a sphere of light radiating away from the bulb. Imagine that light hits other things. Imagine some of those things are other lightbulbs. Some lit, some not. Now try to understand – there is no intelligence delivered with that sphere of light. It simply illuminates any surface it touches. That includes other lit surfaces. The bulbs don’t know this is going on and don’t care. They are just following the rules of thermodynamics.

    Now the test – you put a very sensitive light sensor in the lab space and it has lenses so it sees but one single bulb. The entire field of vision of the sensor it filled with that bulb. Turn on the bulb and notice the sensor response. Now turn on another bulb that is conveniently located behind the sensor and so out of it’s sensing view. Note that the sensor detects that new light coming from the only thing in its view and it registers a higher light level than before. Some call it BS. In fact that just happened in this thread, but it is true that radiating objects light up other radiating objects in all parts of the spectrum including IR.

    Ever wonder why three logs stacked in a pyramid will burn to ash but one log alone will quickly stop burning and turn cold? It is because the logs radiate heat onto each other while they burn and heat is one of the three essentials for combustion.

    Nuther example because I sense a lot of ignorance here. You are sitting in your kitchen typing on your laptop, filling the WUWT blog with ignorance. It is dark out, and very cold – snow everywhere. You look at the outside thermometer and it is 5ºF – damn cold. You pull aside the drapes on your double paned sliding glass door so you can look out at the beautiful snowy wonderland. And after a few minutes you start to feel cold where before you were quite comfortable. How can that be? It is 72º in your kitchen – how can you feel cold? You granny comes over and pulls the drapes closed, calls you a dummy and that you’re going to freeze if you leave those drapes open. What does granny know? Did you notice too you were cold only on side of you facing the snowy winter wonderland?

    Granny knows your body is radiating the maximum possible heat out of the house and into that winter wonderland. If you pull the drapes closed that heat loss stops. Why? Think about that then send your granny a card. A real card, not a lame iCard, and let her know how smart she is.

  124. Ira Glickstein, PhD says:
    March 29, 2011 at 4:12 pm

    …there are at least three issues: 1) The small difference in surface area, 2) The large temperature increase due to the Atmospheric “Greenhouse Effect”, and 3) Convection/conduction/precipitation are not anywhere near 100% effective at transfering thermal energy from the Surface to the Atmosphere. Let us address each in turn:

    1 – If the shell is at an altitude similar to the effective altitude of the Atmosphere, which is a teeny tiny fraction (less that 1/1000) of the radius of the Earth, the shell’s surface area will be almost the same as the Surface area of the Earth, so the supposed temperature difference will be miniscule.

    2 – The effect of “Greenhouse” warming of the Earth is over 30ºC, which is many, many times the maximum possible temperature difference due to (1).

    3- Even if (1) was as large as 30ºC, the combined sum of convection/conduction/precipitation would fail to transfer anywhere near 100% from the Surface to your shell.

    Nice try though and THANKS for getting me thinking.

    I wish to thank you Ira. I too enjoy and benefit from these discussions. They often make me consider something I had completely missed. In that vein, I would like to retract my statement that it’s a “waste of time” to discuss Earth surface temperature without including a discussion of conduction and convection. Discussions of incorrect analyses of physical phenomena often lead to better understanding and to viable formulations. BTW–the immediately above statement applies to my analysis. If my analysis is incorrect, I will likely learn from my errors; and in the process maybe give the person who corrects my errors insight into an issue he didn’t have before the discussion.

    Now to address your three points.

    First, “The small difference in surface area,

    It’s true a shell that has the diameter equal to the atmosphere diameter will have at most a slightly larger area than the Earth. However, we’re discussing the simultaneous existence of backradiation and a cooler active object surface temperature. As such, there is no restriction on the radius of the shell. It’s also true that the larger the radius of the shell, the smaller will be the backradiation from the shell to the active sphere. The reason a larger shell radius results in less backradiation is as follows. For a thin shell, (a) the area of the inside shell surface is close to the area of the outside shell surface, and (b) for a highly conductive shell, the temperature of the shell inner surface will be close to the temperature of the shell outer surface. In a steady-state condition, the area of the shell outer surface must radiate energy away from the shell at a rate equal to that produced by the active-object internal thermal source. Since the inner shell surface area and temperature are close to the outer shell surface area and temperature, the rate energy is radiated from the inner shell surface will be close to the rate energy is radiated from the outer shell surface, and hence close to the rate of internal energy supplied by the active object. However, as the radius of the shell increases, more and more of the energy radiated from the inner surface of the shell will be directed toward another part of the shell inner surface, not toward the active sphere. As such, a corresponding drop in the backradiation to the active object will take place. Backradiation will still exist, but the rate of backradiation energy flow decreases as the sphere radius increases.

    Second, The effect of “Greenhouse” warming of the Earth is over 30ºC
    The approximate 30ºC number is obtained by assigning to the Earth a reception albedo of approximately 0.3 and an emissivity of approximately 1 [i.e., an albedo of approximately 0). If you treat the surface of the Earth as a graybody (i.e., you use a modified Planck blackbody radiation law with an emissivity (1 minus albedo) somewhere between 0 and 1], then you must use the same albedo for both reception and emission. For example, if you completely enclose an inert graybody object (non-zero albedo) within a surface that is maintained at a uniform fixed temperature, the steady-state temperature of the enclosed graybody object will equal that temperature. However, if to that scenario you apply Planck’s graybody radiation law with an emission albedo different from its reception albedo, the result will be that in radiation balance (rate of energy absorbed equals rate of energy emitted) the inert enclosed object’s steady-state temperature will be either higher or lower than the temperature of the surrounding material. The steady-state temperature of the inert object cannot both be “equal to” and “not equal to” the temperature of the surrounding (enclosing) material. I believe of these two choices (assuming one of them is correct), the correct answer is “equal temperature”. If true, then the use of different reception/emission albedos and Planck’s graybody radiation law are incompatible. As others have pointed out, I question the use of Planck’s graybody radiation law for a gas/surface interface. It may be valid for minute amounts of gas, but as the gas density increases, I have a feeling the law starts to break down. [As an aside, for a gas there is no clearly defined "surface". Since one term in Planck's blackbody radiation law is a differential area at a known temperature, any use of Planck's blackbody radiation law or a graybody version of that law for gases is inherently impossible.] If the same albedo is used for both reception and emission, the difference between the Earth’s actual temperature and it’s graybody temperature without an atmosphere is more like 7ºC, not 30ºC.

    Third, “Even if (1) was[n't] as large as 30ºC, the combined sum of convection/conduction/precipitation would fail to transfer anywhere near 100% from the Surface to your shell.” The rate of energy transfer via mechanisms other than radiation doesn’t have to be anywhere near 100%. The active object temperature will be decreased if the rate of energy transfer via non-radiative mechanisms is greater than the rate of backradiation. Since the rate of backradiation decreases with increasing shell radius, for a sufficiently large shell, it won’t take much conduction to produce a lower active object temperature. I believe experiments using eight or nine silver struts connecting the active object to the shell can be constructed that will likely transfer a sufficient rate of conductive thermal energy. If you don’t believe so, turn on your stove, place the blade of a looooong sterling silver knife on the element, and see how long you can hold onto the knife handle.

    Bottom line, I don’t claim greenhouse gases won’t increase the surface temperature of the Earth. However, I do argue that the backradiation argument is insufficient to make such a claim. This gets us back to my statement to the effect that any meaningful discussion of the Earth’s surface temperature must include a discussion of all thermal transfer phenomena, not just radiation.

  125. Well ok, I guess, although many here don’t seem to think it is ok. But why is the sensitivity of atmosphere to an additional molecule of CO2 logarithmic? Your model implies that it wouldn’t be.

  126. In this essay you have qualitatively described a small part of a complex nonlinear system which has huge amounts positive and negative feedbacks. Without numbers coming from theory and/or measurements and without describing connections of your radiation physics to the other parts of the system it is not possible to assess its validity and relevance. It is good that you did not have any conclusions or advice to the policymakers.

    Nice graphics, anyhow :-)

  127. @DP
    It’s a shame none of your examples support your argument.

    The first experiment is simply completely wrong.

    The three logs burn better because a pyramid allows much more air into the fire and increases the surface area exposed to the fire promoting combustion. Boy Scouts 101. Nothing to do with radiation.

    The reason it is cold near the window is because:
    a) The window is a much poorer insulator than the walls. This means the air near the window is cooled by conduction and radiation to the outside and convection on the inside surface of the window.
    b) The window is also further from the fire and receives much less thermal energy according to the inverse square rule of radiation.
    c) The window has a very low albedo so it doesn’t reflect heat back into the room.

    If you repeated the experiment with triple glazed windows it wouldn’t work.

  128. @DP
    It’s a shame none of your examples support your argument.

    The first experiment is simply completely wrong.

    The three logs burn better because a pyramid allows much more air into the fire and increases the surface area exposed to the fire promoting combustion. Boy Scouts 101. Nothing to do with radiation.

    How is it possible for a pyramid of logs to allow more air to the burning part of the log that a log standing alone without competition for that air from other logs? It really is the mutual heating that sustains the burn.

    FAIL

  129. If you repeated the experiment with triple glazed windows it wouldn’t work.

    Multiple panes of glass work against conduction. They are useless against radiated heat unless they are designed (coated) to radiate IR back into the room in which case it is exactly the same as drawing the drapes closed but much more expensive. Pick up a good science book and don’t put it down until you understand how radiated energy works. Then teach your kids.

  130. It’s more than 20 years since I did any spectroscopy so I’m pretty rusty. However we certainly weren’t told about any (alleged) heat trapping abilities of molecules by our PhD qualified chemistry lecturers.

    The author seems to have confused spectrophotometry which only involves the outer electron orbitals and IR spectroscopy which involves changes in the vibrational modes of molecular bonds.

    IR spectroscopy is primarily used for non-destructive testing of carbon rich solids such as coal, plastics and pharmaceuticals.

    It should be remembered that spectroscopy is an empirical tool for measuring structures, quantities or concentrations of a substance. It isn’t designed for determining the existence of a greenhouse effect.

  131. DP;
    How is it possible for a pyramid of logs to allow more air to the burning part of the log that a log standing alone without competition for that air from other logs? It really is the mutual heating that sustains the burn.>>>>

    Thank you thank you thank you thank you! What a wonderful, obvious, easy to understand example! Mind if I borrow it from time to time? I’ve been relying on my igloo example of cold things radiating heat to warm things and have become exasperated with the nonsensical objections.

    As for bananabender….
    If you’d actually taken Scouts 101 pretty much the first thing you learn is that any logs of a decent size simply will not burn well unless the burning faces are used to heat each other up. In your own example, if the pyramid of logs promoted surface area exposed to air, why wouldn’t the logs burn from the outside in instead of the inside out? If you took Scouts 201 you’d also know that arranging the logs in layers, each layer a set of parallel logs at right angles to the layer below, makes a fire that puts the pyramid to shame. It creates a whole series of surfaces radiating at each other to achieve much higher temps.

  132. George E. Smith says:
    March 29, 2011 at 5:38 pm
    I did not say that all the earth’s energy was absorbed by the atmosphere and re-radiated. It is only true for a small range of wavelengths associated with the absorbtion bands of the greenhouse gases. I have said this many times in different ways (including other comments of mine on this thread) but Eli Rabbet covers it pretty succinctly a few comments back. This is not a theory. It has been measured to be the way the world is.

  133. DavidM

    I’m just reading this article. This section below seems at variance to your very dubious Igloo example.

    tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf

    “Thermodynamics says that no object in the universe can heat itself by its own radiation, nor can heat flow from cold to hot”.

  134. BigWaveDave says:

    Nor does it allow you to ignore the system conditions with statements like this : “And, these Laws tell us that the surface temperature of an IR-transparent atmosphere will go toward a steady-state value determined by the condition that the radiative emission balance the radiation absorbed from the sun (in the absence of any other significant energy source).”; which totally ignores the influence of gravity on the initial state of the compressible gas that makes up your hypothetical IR-transparent atmosphere.

    I’m not ignoring gravity. I am just accepting the fact that gravity doesn’t magically create energy out of nothing. You write P/R=rho T and say that “density decreases at a lesser rate than pressure” in the troposphere and conclude that temperature must rise. However you fail to note that:

    (1) This is just an observed feature of the troposphere. It is not a necessary feature and, in fact, in other levels of the atmosphere (like the stratosphere) it is not the case that the temperature decreases with height.

    (2) The fact that temperature decreases with height does not imply that gravity somehow sets a constraint on the surface temperature. And, indeed, it does not…In the absence of an IR-absorbing atmosphere, the constraint on surface temperature is set by simple radiative balance considerations. (In the presence of an IR-absorbing atmosphere…i.e., the greenhouse effect, it is more complex but is still set by energy balance considerations, although one has to consider energy balance within the atmosphere too and convection is also involved.)

    But, where have I violated any laws of thermodynamics?

    You are proposing that the earth’s surface can be at a temperature where the earth would continuously emit much more energy than it absorbs from the sun, which is not possible for it to do without rapidly cooling down unless there is a significant other energy source that comes into play.

  135. RJ says:
    March 30, 2011 at 3:13 am
    DavidM
    I’m just reading this article. This section below seems at variance to your very dubious Igloo example.
    tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf
    “Thermodynamics says that no object in the universe can heat itself by its own radiation, nor can heat flow from cold to hot”.>>>

    NET heat cannot flow from cold to hot.
    If it didn’t then you couldn’t stay warm inside of an igloo anymore than you could standing outside the igloo. Igloos work. Period.

    Pour boiling hot coffee into a thermos at room temp. Which is hotter? The thermos of the coffee? The coffee obviously. Put a bowl out on the counter and fill it with an equal amount of the same coffee, cover it to prevent heat loss via convection. Wait six hours. Coffee in the thermos is still hot. Which is hotter? The coffee or the thermos? The coffee. What about the coffee in the bowl? Room temp. Huh.

    So what heat source was it that kept the coffee hot? Given that it was in a thermos that was at a lower temperature than the coffee and there were no other heat sources being put into the thermos, why did it stay hot?

    The answer is that it got some of its own heat back. It radiated heat to the thermos, which, as a consequence of warming up, radiated some back into the coffee.

  136. According to davidmhoffer an igloo is an example of ‘a cold thing radiating to a warm thing’.
    Snow is an excellent insulator as it contains pockets of air and, like a greenhouse, prevents convection. This means that a warm object (a human body, for example) will warm the interior of the igloo by conduction and radiation. The snow cannot add energy to a warmer body. The molecules of the warmer body are vibrating and this is the signal of its temperature. The radiation from a cooler body cannot vibrate the molecules more than they are already vibrating as it has less energy. The snow is not an extra radiator adding to the heat created by the warm body. If it were otherwise a an infra-red reflector (a mirror) would be a heater.

  137. davidmhoffer says:
    March 30, 2011 at 2:11 am

    As for bananabender….
    If you’d actually taken Scouts 101 pretty much the first thing you learn is that any logs of a decent size simply will not burn well unless the burning faces are used to heat each other up. In your own example, if the pyramid of logs promoted surface area exposed to air, why wouldn’t the logs burn from the outside in instead of the inside out? If you took Scouts 201 you’d also know that arranging the logs in layers, each layer a set of parallel logs at right angles to the layer below, makes a fire that puts the pyramid to shame. It creates a whole series of surfaces radiating at each other to achieve much higher temps.

    I was never a boy scout. That means I didn’t learn any of their incorrect explanations for how fires work.

    A pyramid or a “log cabin” fire forms a venturi drawing fresh air in from outside the base like a chimney. Hot air rises via convection up the the inside of the structure. This causes the evaporation of volatiles which then ignite. The outside doesn’t burn well because wood is a very poor conductor of heat and there is very little direct heating from the flames.

    In the case of a horizontal log the heat is carried upwards by convection so the wood below is not well heated. Charcoal is an exceptionally effective insulator so the wood below the flame doesn’t reach combustion temperature and the fire exhausts the fuel supply.

    Radiation is a negligible contributor to flame propagation in small fires. Heat flow is always from hot to cold.

  138. A number of burning logs in contact with one another will certainly aid combustion by aggregating their heat.

    However the jumble of logs will also obstruct and redirect and locally accelerate the flow of air between them so it is also true that accelerating the flow of air between the logs will fan the flames further.

  139. dp says:
    March 29, 2011 at 11:28 pm

    Multiple panes of glass work against conduction. They are useless against radiated heat unless they are designed (coated) to radiate IR back into the room in which case it is exactly the same as drawing the drapes closed but much more expensive.

    Curtains reduce heat loss primarily by preventing convection currents near the window surface. This in turn reduces the effectiveness of the glass as a conductive heat exchanger. Curtains have very little effect on radiative heat loss. Triple glazing is far more effective at preventing heat loss than curtains in a cold climate

    Triple glazed windows are effective primarily because they prevent conductive heat loss. Radiative heat loss across triple-glazed windows is low because glass is only partly transparent to IR radiation.

    Pick up a good science book and don’t put it down until you understand how radiated energy works. Then teach your kids.

    I’ve got three science degrees and have studied physics and chemistry at university. Unlike you I actually understand how radiation works.

  140. Igloos are temporary structures that work primarily by preventing wind chill and heat loss by convection. Snow is also a fairly good insulator reducing conductive heat loss to the outside air. The cold igloo doesn’t have any warming effect at all on the inhabitants.

  141. Ira says “2 – The effect of “Greenhouse” warming of the Earth is over 30ºC, which is many, many times the maximum possible temperature difference due to (1). ”

    This is only true if you disregard that STP is 0 C. If taken into account then GHG is 15 C max.

  142. A few things that stick out in this discussion:

    1 – The concept that the Earth surface warms by back radiation is a mis-understanding. The surface warms because of a rather large object 93 million miles away. The issue is that the OLR radiation leaving the Earth’s surface is slowed in its escape through the atmosphere because of absorption of certain wavelengths including the one about 14 microns that corresponds to an absorption band in CO2. This has the effect of increasing the energy and temperature of the near surface atmosphere, so slows the heat loss from the surface. If this is a violation of the 2nd law of thermodynamics, so are all forms of insulation…

    2 – There is a confusion between attempting to explain things at an atomic or quantum scale and in the macro world of classical physics.

    3 – the logarithmic effect of absorption (or at least that it is a reasonable approximation at the current CO2 concentrations) is easily demonstratable in absorption spectroscopy (i.e. can be verified by experimental observation). Absorption of a photon is probablistic, and dependent on how close the photon’s energy is to the absorption band of the CO2 molecule – this is why the argument that the CO2 wavelength is ‘saturated’ with regard to IR is a bit over-simplified, those CO2 molecules that have not absorbed a photon ideally suited to its absorption band can still absorb a photon at a markedly lower or higher energy, although the likelihood of absorbing this photon reduces the further from the ‘ideal’ wavelength the photon is.

    4 – Despite the above, all of this discussion is rather theoretical until adequate consideration is given to the dominance of convection as a bulk heat transfer mechanism in the lower atmosphere, and to the moderating effects of the oceans as a heat sink and source, remembering that the issue from a human experience perspective really is how much is the temperature increasing in the biosphere.

  143. Reed Coray says:
    March 29, 2011 at 8:59 pm
    … If the same albedo is used for both reception and emission, the difference between the Earth’s actual temperature and it’s graybody temperature without an atmosphere is more like 7ºC, not 30ºC.

    mkelly says:
    March 30, 2011 at 7:23 am
    Ira says “2 – The effect of “Greenhouse” warming of the Earth is over 30ºC, which is many, many times the maximum possible temperature difference due to (1). ”

    This is only true if you disregard that STP is 0 C. If taken into account then GHG is 15 C max/b>.

    from Wikipedia

    If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C. However, since the Earth reflects about 30% (or 28%) of the incoming sunlight, the planet’s effective temperature (the temperature of a blackbody that would emit the same amount of radiation) is about −18 or −19 °C, about 33°C below the actual surface temperature of about 14 °C or 15 °C. The mechanism that produces this difference between the actual surface temperature and the effective temperature is due to the atmosphere and is known as the greenhouse effect.

    While I certainly do not trust Wikipedia completely, since anyone may post almost anything to obscure topics, I know from personal experience that important scientific topics are monitored by domain experts who protect them from spurious information. Reed Coray, please cite a respected source for your “7ºC” estimate, and same for mkelly regarding your “15 C max”.

  144. “Ira says “2 – The effect of “Greenhouse” warming of the Earth is over 30ºC, which is many, many times the maximum possible temperature difference due to (1). ”

    Here’s a section from the document noted above

    It explains a reason why GHGs do not cause 30+ degrees of warming. This GHG theory is looking more and more shaky. (Is the GHG theory built on such poor science).

    tech-know.eu/upload/Understanding_the_Atmosphere_Effect.pdf
    [This link did not work for me. It brought up a plea for a donation, and I could not see any way to view the supposed .pdf document. It appears to be a scam of some sort. - Ira]

    The greenhouse theory states that the reason the ground is warmer than -18 C is because the atmosphere, via greenhouse gases like CO2, re-emits thermal radiation towards the ground and therefore amplifies the heat at the ground from -18 C up to +15 C.

    This is the same point as discussed just above, in using a projection factor of p=4. But to make the point more clearly: the entire surface of the Earth is not simultaneously illuminated by the incoming solar energy around all sides, but only half of the Earth is ever actually illuminated. When the solar input heating is incorrectly averaged over the entire Earth at once, there isn‟t a high enough radiative energy flux density to explain why the temperature ever gets above -18 C. Therefore a greenhouse theory must be proposed in order to explain why the ground temperature is +15 C.
    However, if the solar energy is correctly averaged over only the single hemisphere that actually physically receives sunlight, the heating temperature equals +30 C for that hemisphere (and much higher directly under the solar zenith, as we have seen). The +15 C average over both day and night, which is less than +30 C, is then easily understood as simply being due to the fact that the night-side is cooler. Of course, the night-side has to be cooler because it receives no solar energy, but it doesn‟t cool very fast because of the thermal capacity of the atmosphere and the ground. The average temperature of both the day and night hemispheres then comes out to +15 C, which is less than the input solar heating, not more.
    It is not surprising therefore that in the first case, we need to theorize a greenhouse effect which ends up violating the laws of physics,

  145. Ira Glickstein says

    …..” The effect of “Greenhouse” warming of the Earth is over 30ºC, which is many, many times the maximum possible temperature difference “…….

    Looking at Ira’s CV perhaps he comes from an electrical engineering background.
    Perhaps the degree options he picked did not include much thermodynamics.

    However he has to be congratulated for his attempts to make a moving pictorial representation of the atmospheric radiative effects.
    However these effects are greatly exaggerated such as the claimed 33K magnitude.

    Ira consider this analogy from electrical engineering.

    The suns effect on the earth surface as it spins resembles a half wave rectified AC signal (say after passing through a diode represents day/night) .
    The earth has a great capacity to store the energy it receives from the sun.
    This can be represented as three large capacitors connected in parallel.
    The largest by far represents the oceans.
    Another represents the atmosphere and a third represents the land surface say to a depth of two metres.
    Completing the circuit is a resistor in which represents radiation to space.

    Such a system would smooth out the day/night effect and would account for most of the 33K falsely attributed to the so called “greenhouse effect”.

  146. Thomas says:
    March 29, 2011 at 1:57 am
    “bananabender. Go ahead, try to explain the temperature of the Earth using only the ideal gas law! It can’t be done, you see, if you actually start to think about what you do.”

    Earth does not have a temperature, except in the fantasies of Climategaters. It also does not have a Santa Claus, an Easter Bunny, or a Thanksgiving.

  147. bananabender says:
    March 30, 2011 at 7:14 am
    Igloos are temporary structures that work primarily by preventing wind chill and heat loss by convection.>>>

    The Inuit lived in igloos for months at a time and if you knew the first thing about them you’d know they have a surprisingly large hole dead centre a the top. This is to promote fresh air exchange otherwise the inhabitants would suffer oxygen deprivation within days, let alone months. The design is promotes heat loss by convection and is a net cooling effect on the temp inside the igloo. As for wind chill, sorry, you’ve not done Scouts 101 nor any winter camping obviously. At -40 degrees, in the best sleeping bag you can buy, and with a wind chill of zero, you will most likely freeze to death sleeping outside beside the igloo. Don’t be daft, crawl into the igloo, deal with the -1 wind chill caused convection by you being a heat source inside the igloo and you will very likely be just fine.

  148. bananabender;
    I was never a boy scout. That means I didn’t learn any of their incorrect explanations for how fires work.>>>

    You claimed the authority of Scouting 101 and now admit the credential was false. You also don’t appear to have much experience with building fires.

    “A pyramid or a “log cabin” fire forms a venturi drawing fresh air in from outside the base like a chimney. Hot air rises via convection up the the inside of the structure. This causes the evaporation of volatiles which then ignite.”>>>

    So….the fresh cool air streaming into the centre of the fire would cool it….convection through the “venturi” is exhausting hot air OUT of the fire…this process sounds like net heating to you? The volatiles evaporate DESPITE the net cooling of the convection process because the combined heat of the various surfaces radiating at each other is high enough to overcome the cooling via convection. Don’t believe me? Build yourself a decent sized fire and turn a fan or leaf blower on it to put 10 or 20 times the air through and around the hole thing than convection could. Watch what happens and report back.

    “I’ve got three science degrees and have studied physics and chemistry at university. Unlike you I actually understand how radiation works.”>>>

    Really? What are these science degrees you possess? Clearly not physics or chemistry since you only “studied” them, and you no more understand how radiation works than you do Scouting 101. Oh yeah, you didn’t have that credential either.

  149. Stephen Wilde
    However the jumble of logs will also obstruct and redirect and locally accelerate the flow of air between them so it is also true that accelerating the flow of air between the logs will fan the flames further.>>>

    The air flow has a net cooling effect as cold air is pulled in while the hot air escapes out the top. Build yourself a small fire either teepee style or log cabin. Pull out your 3.5 Hrspwr leaf blower (mine’s hanging on the garage wall if you need to borrow it) and play it at idle on the flames. The illussion is that they get bigger. The volatile compounds burning at high enough temps to become incandescent get pushed farther and faster by the air, resulting in them travelling further before going out. The flames get longer, but the fire isn’t as hot. Now whip the leaf blower to max. Poof, fire went out.

    In fact you can prove these things with matches and candles. Bend ther wicks horizontal to the rest is easy to do. Strike a match and try and light a candle wick with the very tip only of the match flame. Now try with just the base of the candle flame. Now try with the side of the flame about 1/3 up from the bottom. You’ll find that is by far the hottest part of the flame. A long flame is a cool flame, its the short fat ones that are hot. A jumble of charcoal gets even hotter and it has no flame at all! And the purpose of closing your BBQ lid is to minimize heat loss by convection. Despite which the ones in the middle always turn to ash faster than the ones at the edge, because they heat each other up.

    For banana and stephen and richard, please be aware that I am a raging skeptic. CAGW is a total farce. But you can’t win the battle by arguing physics with warmists because on this issue they are correct and all you do is discredit yourself and any arguments that you might have that are correct. If any of you could falsify there being a positive flow of heat from cold to warm despite there being a NET negative flow of heat in total you will most likely win a Nobel prize. The laws of physics verified thousands of times through experimentation are now centuries old. They make it plain that any body radiates the exact same amount of heat at a given temperature no matter WHAT the temperature of the surrounding objects is. That heat impacts cooler things and warmer things and get absorbed by all of them. The warmer things radiate more heat than the cooler, and so the cooler things have a net heat gain relative to the warm things.

    If that is wrong, then the Stefan-Boltzmann Law http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law (1879!), Planck’s Constant, Wien Displacement Law, Raleigh-Jeans Law, and Kirchoff’s Law have all been disproven. Things like combustion engines, turbines, ramjets, and a whole long list of other things were therefore designed using falsified physics despite which they worked anyway.

    There is SO much the CAGW crowd has gotten wrong. Why attack them on a nitpicking little detail like this one where they just happen to be right? Order of magnitude, effect of feedbacks, THOSE are the places where their arguments fall apart.

  150. RJ says:

    The average temperature of both the day and night hemispheres then comes out to +15 C, which is less than the input solar heating, not more.
    It is not surprising therefore that in the first case, we need to theorize a greenhouse effect which ends up violating the laws of physics,

    Bryan says:

    Such a system would smooth out the day/night effect and would account for most of the 33K falsely attributed to the so called “greenhouse effect”.

    Both of these comments are simply incorrect. The energy balance constraint on the surface temperature of the earth in the absence of an IR-absorbing atmosphere is that the fourth root of average of T^4 over the earth’s surface must be equal to ~255 K. If the temperature on the earth was uniform, this would mean that the average of T would also be 255 K. However, to the extent that temperatures are non-uniform over the earth’s surface, the average of T is actually less, not more, since there is an inequality where will always be less than the fourth root of (where denotes average). This inequality is one of the few things that Gerlich and Tscheuschner get right in their paper (although they misinterpret what it means).

  151. In arguing (March 29, 2011 at 10:50 am) for the primacy of sensible radiative transport of heat based on its presumed near-light speed, Glickstein once again overlooks a crucial physical fact: the RATE of thermal energy transport by MOIST convection. At ~590 calories per gram of evaporated water, the latent heat transport is enormous. In fact, the Bowen ratio of sensible-to-latent heat transport rates has been shown in numerous experiments world-wide to be below unity in general, and invariably well below in marine environments that account for the great majority of the globe’s surface. Only in ultra-dry environments (Sahara, Antarctica) does it much exceed unity. Even Kiel & Trenberth’s “global energy-budget” cartoon reflects that fact–once the nearly null-net radiative exchange between surface and atmosphere is properly accounted for.

    Despite all the traffic generated by this pop-science series of posts on the ethereal notion of a “radiative greenhouse,” the lack of deeper comprehension of real-world physics does a disservice to WUWT readers.

  152. RJ:

    Here’s one link

    http://www.tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf

    The section posted above is on pages 25 and 26

    So now the link works but we are left wondering what your point is. If it is that one can find a lot of nonsense out on the internet, I agree with you. If it is something else, then I would not use a paper that is so clearly nonsense to try to make your point!

    Here’s a hint to you on where that paper goes wrong in all of its claims regarding thermodynamics and the greenhouse effect: The greenhouse effect does not say that (net) heat flows from hot to cold. (Some would say the word “net” is redundant since heat is usually a macroscopic concept describing the net energy flow.) In all models of the greenhouse effect, whether they are toy models or full-blown line-by-line convective-radiative transfer models, the heat flows from hot to cold.

    The greenhouse effect is simply stating this: In the absence of an IR-absorbing atmosphere, all of the terrestrial radiation would be emitted directly out into space. However, in the presence of an IR-absorbing atmosphere, some of the terrestrial radiation is absorbed and when the atmosphere subsequently emits radiation, some of this radiation comes back to the earth. This keeps the earth warmer than it would be in the absence of the IR-absorbing atmosphere.

    Also note that the Second Law of Thermodynamics is not magic: It does not state that colder objects magically detect the presence of a hotter object and refuse to emit any radiation towards said object. Rather, it is a statement derived from statistical physics that says that the amount of radiation that the colder object absorbs from the hotter will always be greater than the amount of radiation that the hotter object absorbs from the colder.

  153. davidmhoffer says:
    March 30, 2011 at 10:23 am
    Stephen Wilde
    However the jumble of logs will also obstruct and redirect and locally accelerate the flow of air between them so it is also true that accelerating the flow of air between the logs will fan the flames further.>>>

    The air flow has a net cooling effect as cold air is pulled in while the hot air escapes out the top. Build yourself a small fire either teepee style or log cabin. Pull out your 3.5 Hrspwr leaf blower (mine’s hanging on the garage wall if you need to borrow it) and play it at idle on the flames. The illussion is that they get bigger. The volatile compounds burning at high enough temps to become incandescent get pushed farther and faster by the air, resulting in them travelling further before going out. The flames get longer, but the fire isn’t as hot.
    In my experience with bellows at a forge, very little, and at a fireplace, a lot, adding a stream of air increases the temperature of the flame. This can be seen as the fire without the air cannot melt a given metal, but with the stream of air it can. Varying the amount of airflow controls the temperature lower for forging, higher for melting.

  154. Joel Shore says:

    However, to the extent that temperatures are non-uniform over the earth’s surface, the average of T is actually less, not more, since there is an inequality where will always be less than the fourth root of (where denotes average).

    Whoops…I guess the HTML markup doesn’t do well with the brackets that I used to indicate averaging. That sentence should read:

    However, to the extent that temperatures are non-uniform over the earth’s surface, the average of T is actually less, not more, since there is an inequality where the average of T will always be less than the fourth root of the average of T^4.

  155. davidmhoffer says:

    But you can’t win the battle by arguing physics with warmists because on this issue they are correct and all you do is discredit yourself and any arguments that you might have that are correct.

    David and I completely disagree on most things regarding AGW, but on this we completely agree. Trying to argue that the greenhouse effect doesn’t exist just discredits you in the eyes of any serious scientist. It is like arguments that the earth is only 7000 years old…It is just like wearing a sign around your neck that says, “I am an anti-science crackpot.” Better to argue about things like feedbacks and climate sensitivity, where there are at least some significant uncertainties in the actual science.

  156. Ira says:”However, since the Earth reflects about 30% (or 28%) of the incoming sunlight, the planet’s effective temperature (the temperature of a blackbody that would emit the same amount of radiation) is about −18 or −19 °C, about 33°C below the actual surface temperature of about 14 °C or 15 °C. “” and same for mkelly regarding your “15 C max”.”

    STP temperature is 0 C so -18 up to 0, the 15 C as actual surface temperature per your quote above. The difference between 0 C and 15 C would be GHG. You answered your own question.

    However, all this is based on the -18 C assumption.
    240 w/m^2 = SB X T^4
    240/SB = T^4
    T= 255 or -18 C

    1360/4=340
    340 times .7 = 240 rounded

    The 340 average may infact be incorrect for TOA as my heat transfer book shows 1063 w/m^2 at 90 deg sun angle (over head at equator) for a surface W/m^2.

  157. I have just realised that according to some of you I have transgressed terribly. I was tuning in my radio when I picked up some of the cosmic microwave background. Since this has a spectrum of a black body at 3K (that is 3 degrees above absolute zero) according to you I have clearly broken the second law of thermodynamics. I am currently 290 degrees warmer than 3K. I am sorry but my radio would only have warmed a little bit since the energy is very low.

    On the other hand maybe you do not have it quite right!

  158. mkelly says:

    The 340 average may infact be incorrect for TOA as my heat transfer book shows 1063 w/m^2 at 90 deg sun angle (over head at equator) for a surface W/m^2.

    On average, the sun is not at a 90deg angle overhead. The average power in W/m^2 at the top of the atmosphere is obtained by considering the solar constant and the ratio of the area of a disc to the surface area of a sphere. Needless to say, these geometric formulas are on a pretty solid footing!

  159. mkelly: Okay…My last post may have been in haste, as I now see what you might be trying to say. However, the 1063 W/m^2 is probably a value for a sunny day, i.e., it presumably neglects cloud albedo. (But, it also includes some things that lower the value from the solar constant, such as scattering from aerosols and other components in the atmosphere and absorption by the atmosphere.) [At any rate, even if we took 1063 W/m^2, divided it by 4 and used it in place of the 240 W/m^2, it would only raise the 255 K value to ~261.5 K.]

  160. Is “back radiation” really required to explain the greenhouse effect? It seems to me that it just confuses the issue.

    With a pure nitrogen atmosphere, almost all infrared radiation would pass through. The molecules of this atmosphere are primarily heated by direct contact with the surface of the sphere, and convection of this thin layer with the upper layers.

    Now replace a mass of that nitrogen with a like mass of CO2. Now you have layers of the atmosphere (top to bottom) heated by infrared radiation that was previously not heating any portion of the atmosphere. The atmosphere is warmer – period. No need to talk about whether or not back radiation will heat up the surface or not, because it is obvious that the atmospheric blanket itself is warmer than it was before. (If you really want to make it warmer, exchange H2O for either of the other two gases.)

    Do you feel warmer in a room with 15 degree air versus -15 degree air. Uh, yeah. But 15 degree air is below body temperature, so how could it possible be keeping you warm?! Well, neither is keeping you warm – that’s why you are cooling off. But you will cool at a slower rate in one versus the other.

  161. “This keeps the earth warmer than it would be in the absence of the IR-absorbing atmosphere.”

    But its a 30 degrees increase because of backradiation. Not keeps the earth warmer / reduces the amount of cooling.

    Some other points

    The article was very good. But then it does not challenge my strongly held beliefs.

    How much of the energy /heat leaves the surface by conduction. If the majority then the likelihood of the extra 30 degrees is surely even further reduced even if back radiation can add extra energy and heat the surface like the sun does.

    We are playing into the alarmists hands if we accept the GHG theory as a fact. It clearly is not so why not acknowledge this. Explain the theory but also explain the possible problems with this theory

  162. Joel Shore says:
    March 30, 2011 at 12:51 pm

    You can see by the earlier post I know where it came from and why. I could have said the chart in my heat transfer book shows from 90-5 deg and the range is 1063 to 41 w/m^2. I am starting to get an inkeling that the 1360/4 may not be correct for TOA that is all I said. I need to look into it further.

  163. RJ says:

    How much of the energy /heat leaves the surface by conduction. If the majority then the likelihood of the extra 30 degrees is surely even further reduced even if back radiation can add extra energy and heat the surface like the sun does.

    This diagram shows the relative transfers: http://www.cgd.ucar.edu/cas/abstracts/files/kevin1997_1.html And, yes, convection and evaporation / condensation play a very important role in heat transfer in the troposphere. However, that doesn’t affect the conclusion that the role of the greenhouse effect is to raise the average surface temperature by ~33 deg above what they would be in the absence of the effect. That is based on experimentally-observed surface temperatures. In fact, what I understand is that if one uses a purely radiative model to calculate the greenhouse effect, one predicts around twice that amount of warming of the surface…and it is the addition of convective effects to the model that then reduces that to something much closer to what is actually observed.

    By the way, although this back-radiation is a useful concept for explaining things, the technically-better explanation is to focus on the effect of greenhouse gases on the top-of-the-atmosphere radiation budget (because you know that the only significant heat transfer that occurs between the entire earth-atmosphere system and space is via radiation) and then to figure out what will happen on the surface in a way that accounts for the importance of convective transport in the troposphere. I believe that to the first approximation, what is expected to occur is that convective and radiative terms between the surface and the atmosphere will be such that the lapse rate is expected to remain about the same in an atmosphere with more greenhouse gases.

  164. Steve Keohane;
    In my experience with bellows at a forge, very little, and at a fireplace, a lot, adding a stream of air increases the temperature of the flame. This can be seen as the fire without the air cannot melt a given metal, but with the stream of air it can. Varying the amount of airflow controls the temperature lower for forging, higher for melting.>>>

    As with all real world examples, there’s no yes/no. In every situation, at every air flow, you get a curve and somewhere along that curve, yes, you can within a range on that curve control the heat of the fire. But at the far end of the curve, a large enough gust of air has the same result – fire goes out. Mini example – most people extinguish a match with a quick puff of air.

    But getting back to your forge….thanks for yet another fine example. Could you bring metal to the melting point with the same amount of fuel in an open fire? No, of course not, if you could there’d be no need to build a forge. Lots of heat going out by convection alone, and then along comes you pumping that bellows to get the maximum temperature. So with all that heat being pumped by convection PLUS the bellows, which is WAY more than an open fire would lose, it is still WAY HOTTER inside the forge…

    ’cause the walls of the forge absorb heat from the fire, and as they heat up they radiate heat. Some away to the outside…and some right back into the fire. Which is why it is so much hotter in there than in an open fire.

  165. RJ says:
    March 30, 2011 at 9:05 am
    Ira

    Here’s one link

    http://www.tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf

    Thanks, RJ, that link worked. I read up to page 6 and stopped, and I want my ten minutes back! The author goes on correctly and then provides an example:

    … imagine a blackbody which is absorbing energy from some hot source of light like a light-bulb, and it has warmed up as much as it can and has reached radiative thermal equilibrium. The blackbody will then be re-emitting just as much thermal infrared energy as the light energy it is absorbing. …

    The author is OK so far. Notice that the amount of ENERGY the blackbody is RE-EMITTING is EXACTLY the same as the amount of ENERGY it is ABSORBING. That is the very definition of thermal equilibrium, and absolutely correct. But, the author goes off the rails when he continues:

    However, because the blackbody doesn‟t warm up to a temperature as hot as the source of light, its re-emitted infrared light is from a lower temperature and thus of a lower energy compared to the incoming light that it is absorbing. …

    Say what? After correctly saying the RE-EMITTED ENERGY was EQUAL to the ABSORBED ENERGY, he now contradicts himself with the false claim that the RE-EMITTED ENERGY is of “a lower energy”. Nope! Thermal equilibrium is when energy in = energy out.

    Now, what is true is that the input energy was from a light bulb and was therefore in the shortwave region (around 0.2μ to 4μ) and the output energy was from a blackbody at a lower temperature and therefore in the longwave region (around 4μ to 20μ). But, BUT, the amount of ENERGY (say in Joules) is EXACTLY EQUAL.

    Having made that error, the author continues:

    Now here‟s the clincher: imagine that you take a mirror which reflects infrared light, and you reflect some of the infrared light the blackbody is emitting back onto itself. What then happens to the temperature of the blackbody? One might think that, because the blackbody is now absorbing more light, even if it is its own infrared light, then it should warm up. …

    Yep, it should warm up and (if it is a true blackbody in the longwave region) it will absorb that reflected longwave energy and, all else being equal, will warm up. (NOTES: (1) Indeed, some commercial incandescent light bulbs have an IR-reflective material on the inside to do just that. (2) In your example, we are dealing with longwave radiation and there is a problem with the idea of “reflecting” longwave radiation, so I will assume that the “mirror” proposed by the author is actually a thin surface that absorbs and re-emits longwave energy, about half back towards the source, and half outwards. Almost any metal painted black will do.)

    Given the incorrect science, the author continues with wrong conclusions:

    But in fact it does not warm up; it‟s temperature remains exactly the same. The reason why is very simple to understand but extremely important to physics: the blackbody is already in radiative thermal equilibrium with a hotter source of energy, the higher radiative energy spectrum light from the light-bulb. You cannot make something warmer by introducing to it something colder, or even the same temperature! You can only make something warmer, with something that is warmer! This reality is called the 2nd Law of Thermodynamics, and is so central and fundamental to modern physics it cannot be expressed strongly enough.

    The above is simply not true. It is a misunderstanding of the laws of physics.

    When you make a false assumption, you can “prove” false conclusions. For example, ASSUME I have a perfectly smooth elephant with neligible mass. Wow, can he do amazing circus tricks. Or, assume I have a frog who eats clay and has a rectangular anus. Wow, he will (make) bricks.

  166. RJ says:

    We are playing into the alarmists hands if we accept the GHG theory as a fact.

    It depends what your goal is. If your goal is just to sow confusion in the general public and hope that this helps keep policymakers from implementing the steps that you oppose, then you may be right…I don’t know. However, if your goal is to be scientifically-correct and to have an impact on the real scientific debate, then you are surely incorrect.

    When I see people challenging the basic facts of the greenhouse effect, what it tells me is that they are either misguided or (if they are knowledgeable enough to know better) that they are trying to actively deceive people and confuse them on the science, i.e., their real goal has nothing to do with science and everything to do with pursuing their policy objectives at the expense of science.

  167. Ira Glickstein says:

    When you make a false assumption, you can “prove” false conclusions.

    My favorite example of this, attributed to Bertrand Russell is showing that if 0 = 1 then I’m the pope: If 0 = 1, add one to each side to get 1 = 2. Then consider a room containing 2 people, myself and the pope. However, since 1 = 2, it can also be said that the room contains 1 person…and yet it must still be true that it contains me and the pope. Therefore, clearly I must be the pope!

  168. Steve says:
    March 30, 2011 at 1:07 pm

    Is “back radiation” really required to explain the greenhouse effect? It seems to me that it just confuses the issue.

    With a pure nitrogen atmosphere, almost all infrared radiation would pass through. The molecules of this atmosphere are primarily heated by direct contact with the surface of the sphere, and convection of this thin layer with the upper layers.

    Now replace a mass of that nitrogen with a like mass of CO2. Now you have layers of the atmosphere (top to bottom) heated by infrared radiation that was previously not heating any portion of the atmosphere. The atmosphere is warmer – period. No need to talk about whether or not back radiation will heat up the surface or not, because it is obvious that the atmospheric blanket itself is warmer than it was before. (If you really want to make it warmer, exchange H2O for either of the other two gases.)

    That’s a great point you raise.

    But Steve, sorry, you are wrong. If ignoring the tiny radiation bands in nitrogen, the pure nitrogen atmosphere would over time be very, very hot. During the daytime with no clouds and oceans the atmosphere would heat greatly at the surface and disperse upward by conduction and convection as you said. As the Earth turns and night comes, all convection would ceases and the atmosphere cannot lose any heat by radiation as you said. Day by day the atmosphere would get hotter and hotter until the entire atmosphere would reach almost the same temperature as the daytime soil. The only cooling occurring would be by strictly conduction near the surface at night in contact with a frosty soil for it would be radiating unimpeded.

    CO2 and all GHGs are a great fast conductors of heat. It is the over-half portion of radiation upward to space that keeps us from being literally cooked.

  169. Joel Shores simple world consists of an earth without oceans and a land surface that does not retain heat.
    Clouds, rain and snow (which give a much more continuous absorption/emission possibilities than gas molecules) apparently do not exist.

    The moon comes closer to Joel’s model but even there his simplistic assumptions do not give a realistic account of actual temperatures.

    According to Joel the so called 33K “greenhouse effect” is entirely due to CO2 and H2O gaseous molecules.
    Perhaps we should leave Joel with his simple fables.
    It seems pointless to explain how absurd his conclusions are as he is religiously attached to them.

  170. Probably already requested, but how about a page that has each graphic and each explanation below it? That way you can focus on the information in each graphic and not be distracted by the constant changing. Thank you.

  171. where is the low level boundary layer? Pretty much all IR radiation coming from the surface that can be absorbed by CO2 is absorbed by CO2 at ground level. The actual photons that CO2 can absorb beyond this must be generated from the heat in the atmosphere itself.

  172. Ira Glickstein, PhD says:
    March 30, 2011 at 8:13 am

    from Wikipedia

    If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C. However, since the Earth reflects about 30% (or 28%) of the incoming sunlight, the planet’s effective temperature (the temperature of a blackbody that would emit the same amount of radiation) is about −18 or −19 °C, about 33°C below the actual surface temperature of about 14 °C or 15 °C. The mechanism that produces this difference between the actual surface temperature and the effective temperature is due to the atmosphere and is known as the greenhouse effect.

    While I certainly do not trust Wikipedia completely, since anyone may post almost anything to obscure topics, I know from personal experience that important scientific topics are monitored by domain experts who protect them from spurious information. Reed Coray, please cite a respected source for your “7ºC” estimate, and same for mkelly regarding your “15 C max”.

    I believe the Wikipedia argument is correct. However, it compares apples to oranges. In particular, I’ll assume Wikipedia’s 30% number is correct. However, I then ask: “What part of the Earth reflects 30% of the sun’s energy?” From the papers I’ve read, it’s primarily the clouds and/or particulate matter in the atmosphere, not the surface of the Earth. When quantifying atmospheric effects on Earth surface temperature, shouldn’t you compare the measured Earth surface temperature (i.e., the temperature with an atmosphere) …to… an Earth model that is devoid of an atmosphere? Wikipedia’s model for the Earth surface temperature includes an atmosphere in that it’s the atmosphere that is reflecting 30% of the solar energy. So Wikipedia is comparing measured Earth surface temperatures to a model of Earth surface temperature that includes an atmosphere; and using the temperature difference to claim that the atmosphere raises the Earth surface temperature 33 °C . I call foul. It’s kind of like John Kerry saying he voted against it before he voted for it. (Sorry, I couldn’t resist).

    An Earth without an atmosphere will have an albedo much nearer 0 than 0.3. However, assume for a moment that the albedo of an atmosphereless Earth is “x”. Then as I discussed previously, if you’re going to use Planck’s graybody radiation law to determine the rate energy is emitted from that surface, the surface emissivity must be the same as one minus the surface albedo. If true, for an atmosphereless Earth albedo of “x”, the emissivity of that atmosphereless Earth will be “(1-x).” The energy absorbed by the Earth from the sun is equal to the product of (a) the solar energy density (Watts per square meter) at the Earth’s surface, (b) the surface area of the Earth perpendicular to the direction of radiation (pi times the radius of the Earth squared), and (c) one minus the albedo–i.e., (1-x). The rate energy is radiated from the surface of the atmosphereless Earth will be the product of (a) the surface area of the Earth (4 times pi times the radius of the Earth squared), (b) the Stefan-Boltzmann constant, (c) the Earth surface temperature to the fourth power, and (d) the emissivity of the Earth–i.e., (1-x). For steady-state conditions, these two rates are equal. Since the factor “(1-x)” appears on both sides of the Equation, the temperature of an “atmosphereless” Earth is independent of the Earth’s surface albedo (one minus emissivity). Thus, the temperature of an atmosphereless Earth is the same as the temperature of a blackbody Earth–using Wikipedia’s value: 5.3 °C, not -18 ºC or -19 ºC.

  173. “Then as I discussed previously, if you’re going to use Planck’s graybody radiation law to determine the rate energy is emitted from that surface, the surface emissivity must be the same as one minus the surface albedo. ”

    Not quite! The surface emissivity FOR VISIBLE LIGHT must be 1-albedo.
    The surface albedo FOR IR must be 1.

    There is no a priori reason the reflection for visible and IR must be the same.

  174. wayne says:
    March 30, 2011 at 2:44 pm

    “But Steve, sorry, you are wrong. If ignoring the tiny radiation bands in nitrogen, the pure nitrogen atmosphere would over time be very, very hot. During the daytime with no clouds and oceans the atmosphere would heat greatly at the surface and disperse upward by conduction and convection as you said. As the Earth turns and night comes, all convection would ceases and the atmosphere cannot lose any heat by radiation as you said.”

    I didn’t say that a nitrogen atmosphere cannot lose heat by radiation – is that what you are saying? I said that a nitrogen atmosphere would warm less than a CO2 atmosphere via the radiation emitted from a cooling earth surface. A molecule will radiate at wavelengths according to it’s emission spectrum, which is roughly equivalent to it’s absorption spectrum. Warming via conduction does not prevent cooling via radiation – the nitrogen in our atmosphere (most of the gas in it) cools every night. Are you proposing that the nitrogen gas in our atmosphere cannot cool off unless it collides with another type of atmospheric gas?

  175. Bryan says:

    Joel Shores simple world consists of an earth without oceans and a land surface that does not retain heat.

    No…Retention of heat is not relevant to the considerations that I mentioned. The way that it comes into it is that the ocean helps to keep the temperature more uniform, which means that the fourth-root of the average of T^4 is closer to the average of T, so the greenhouse effect is close to 33 K. If oceans were not there, then the temperatures would be more variable (between day and night, seasons, etc.) and the average surface temperature in the absence of the greenhouse effect would tend to be even lower.

    Clouds, rain and snow (which give a much more continuous absorption/emission possibilities than gas molecules) apparently do not exist.

    According to Joel the so called 33K “greenhouse effect” is entirely due to CO2 and H2O gaseous molecules.

    I never said that. Clouds are part of the reason why the atmosphere is not transparent to IR radiation and hence contribute to the greenhouse effect.

    Reed Coray says:

    Thus, the temperature of an atmosphereless Earth is the same as the temperature of a blackbody Earth–using Wikipedia’s value: 5.3 °C, not -18 ºC or -19 ºC.

    Sort of true…although really you are forgetting a couple of things. One is that I believe the albedo of the earth in the absence of clouds is still about 8% or so. The second is that such a colder earth would presumably have more ice on it and would therefore have even a higher albedo than this.

    At any rate, the full statement is that the atmosphere (including clouds) produces two effects: One is about 33 K of warming due to the greenhouse effect and the other is some cooling due to the albedo of the clouds. So, yes, the net effect of the atmosphere is not a full 33 K of greenhouse warming. However, again, the sort of calculations you discuss are under the assumption of no change in surface albedo [actually, you wrongly assumed 0 albedo]…In reality, if you removed CO2 from the atmosphere, not only would you remove a lot of the water vapor too, you would also increase the surface albedo of the earth.

  176. Reed –

    On re-reading what you wrote, I realize that you didn’t assume a zero albedo for the earth as I said in my last post but you did assume, as Tim Folkerts points out, that the emissivity is the same in the visible as it is in the far IR…which is indeed not a very good assumption. Emissivities of the earth’s surface in the visible vary widely with the type of surface, whereas emissivities in the far IR tend to be very close to 1 for nearly all surfaces. (I believe I’ve read that the only place that there can be any significant deviations from 1 is in some desert regions.)

  177. “”””
    Steve says:
    March 30, 2011 at 4:47 pm

    I didn’t say that a nitrogen atmosphere cannot lose heat by radiation – is that what you are saying? “”””

    Hi Steve. Yes, I was portraying that on what Phil. said in the prior post. I would agree with that. Nitrogen, except for some small lines near 3 µm, cannot radiate at all. I tend to agree with him though. There are many claiming that all matter radiates when above 0 K as you just said. If all matter radiates, and an atmosphere of pure N2 radiates, then this is no different than what a GHG is portrayed to radiate in all directions and all atmospheres and all molecules always cause a “greenhouse effect”.

  178. Tim Folkerts says:
    March 30, 2011 at 4:09 pm
    “Then as I discussed previously, if you’re going to use Planck’s graybody radiation law to determine the rate energy is emitted from that surface, the surface emissivity must be the same as one minus the surface albedo. ”

    Not quite! The surface emissivity FOR VISIBLE LIGHT must be 1-albedo.
    The surface albedo FOR IR must be 1.

    There is no a priori reason the reflection for visible and IR must be the same

    I agree. Real-world materials can exhibit emissivities (albedos) that are a function of frequency. For blackbodies, Planck’s radiation law contains a frequency dependent term of the form (f^3)*(df)/[e^(hf/kT) - 1]. When integrated over all frequencies (0 to infinity), the result is radiated power that is proportional to the fourth power of the temperature T. If to account for a different “visible light” emissivity and “IR” emissivity an additional frequency dependent factor g(f) is included in the law, then the integral over all frequency will only be proportional to the temperature to the fourth power if g(f) is constant. The T^4 rule may be a good approximation, but it won’t be theoretically correct. As such, if you’re going to compute the radiated energy using a temperature dependence of T^4, aren’t you implicitly assuming an emissivity (albedo) that is the same at all frequencies? If you’re going to use an emissivity that is a function of frequency, then shouldn’t you specify that dependence?

    I’m curious. What do you believe is the primary source of solar reflection: (a) the surface of the Earth, or (b) the Earth’s atmosphere and matter in the atmosphere, or (c) something else? Do you agree or disagree that if you’re going to determine the effect of the atmosphere on Earth surface temperature by (a) treating the measured Earth surface temperature as the “Earth with atmosphere temperature”, and (b) using a model to compute the atmosphereless Earth surface temperature, the model should be entirely devoid of atmospheric effects?

    If someone wanted to (a) model the position/frequency dependence of Earth surface absorption (no atmosphere) and the position/frequency dependence of Earth surface emission (no atmosphere), (b) compute the “average” Earth surface temperature using that model, and (c) argue that the effect of the Earth’s atmosphere is the difference between the average measured Earth surface temperature and the model average Earth surface temperature, I would have no objection. But that is not how the 33 degree number is computed. The 33 degree number is based on a reception albedo that, I believe, represents an “atmospheric albedo”, not an Earth surface albedo.

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

    “Having made that error, the author continues:

    One might think that, because the blackbody is now absorbing more light, even if it is its own infrared light, then it should warm up. …

    Yep, it should warm up and (if it is a true blackbody in the longwave region) it will absorb that reflected longwave energy and, all else being equal, will warm up.”

    Ira, so when will this cycle of selfheating stop?
    Lets slice a blackbody cavity which is at a equilibrium temperature of 288 Kelvin in half and replace the other half with a mirror. How much warmer will this cavity become?

  180. Nitrogen can’t radiate heat?

    Since when? Everything above absolute zero radiates energy.

    Atmospheric nitrogen has kinetic energy and therefore must radiate heat.

    The atmosphere is heated by molecular-kinetic effects i.e. molecular collisions. Atmospheric temperature is a function of pressure and density not radiative effects.

  181. bananabender says:
    March 30, 2011 at 7:08 pm
    Nitrogen can’t radiate heat?

    Since when? Everything above absolute zero radiates energy.

    Atmospheric nitrogen has kinetic energy and therefore must radiate heat.

    No that isn’t true for gases, N2 doesn’t have a dipole and therefore doesn’t emit in the IR.

  182. davidmhoffer says:
    March 30, 2011 at 9:39 am
    bananabender;

    So….the fresh cool air streaming into the centre of the fire would cool it….convection through the “venturi” is exhausting hot air OUT of the fire…this process sounds like net heating to you? The volatiles evaporate DESPITE the net cooling of the convection process because the combined heat of the various surfaces radiating at each other is high enough to overcome the cooling via convection. Don’t believe me? Build yourself a decent sized fire and turn a fan or leaf blower on it to put 10 or 20 times the air through and around the hole thing than convection could. Watch what happens and report back.

    For every phenomenon there is a simple and elegant explanation that is totally and utterly wrong.

    Your anecdotes are amusing folklore but they show absolutely no understanding of basic scientific principles.

    Have you ever used a bunsen burner? You vary the flame temperature of this burner by altering the amount of air mixing with the gas. The more air you add the more completely the gas burns and the hotter the flames. Fanning a fire is exactly the same principle.

    Air is an extremely poor heat conductor. The cooling effect of increased air flow into a fire is far less than the increased heat produced by more efficient combustion.

    You can hold your finger 1cm from the side of a bunsen flame and not get burnt. Gases have almost no mass so they radiate very little energy. A bunsen transfers energy directly from the flame by conduction not radiation.

    Really? What are these science degrees you possess? Clearly not physics or chemistry since you only “studied” them, and you no more understand how radiation works than you do Scouting 101. Oh yeah, you didn’t have that credential either.

    I was employed as chemist in the food and beverage industries after I graduated from university. I regularly used various analytical spectroscopy methods including AAS, GC-MS, spectrophotometry, IR and near IR spectroscopy . Spectroscopy is based entirely on radiative physics.

  183. During the daytime with no clouds and oceans the atmosphere would heat greatly…

    No need to talk about whether or not back radiation will heat up the surface or not…

    and therefore amplifies the heat at the ground from -18 C up to +15 C….

    Apparently, heat = “temperature”

    but the mechanisms by which heat is retained are different….

    that the temperature of a gas containing a given amount of heat…

    The atmosphere and the ocean may retain heat, but the atmospheric heat retention is infinitesimal compared to the heat retention of the oceans….

    Apparently, heat = “total thermal energy”

    Some would say the word “net” is redundant since heat is usually a macroscopic concept describing the net energy flow….

    As most of the heating of the atmosphere is by conduction and convection …

    ice reflects a lot more sunlight than ocean or unfrozen land surface thus reducing the amount of solar surface heating…

    Apparently, heat = “a flow of energy”

    As long as the discussion contains such different ideas about what is perhaps the single most important word in this discussion, I can’t see and hope of resolving anything. We will all be talking past each other as we each try to make out own points using our own understanding of the words.

    P.S. In thermodynamics, the 3rd definition (a flow of energy) is the accepted definition. If you see “Q” in an equation, it means how much energy was TRANSFERRED, not how much energy is CONTAINTED. If you see “Q” on a diagram, it is always associated with an arrow, not an object. Heat (and similarly work) are not “state variables” and do not describe the properties of any system.

  184. wayne says:
    March 30, 2011 at 6:09 pm

    “Nitrogen, except for some small lines near 3 µm, cannot radiate at all.”
    That determines the wavelengths at which it cools, it doesn’t limit the ability to cool off. It will still radiate according to it’s temperature. Nitrogen will radiate a lot of energy in a narrow band, instead of a moderate amount of energy across multiple bands.

    “If all matter radiates, and an atmosphere of pure N2 radiates, then this is no different than what a GHG is portrayed to radiate in all directions and all atmospheres and all molecules always cause a “greenhouse effect”.”

    The greenhouse effect is not a matter of the gases radiating, but absorbing. The surface of the earth radiates at a much larger variety of wavelengths than some gases can absorb, so some radiation goes directly to space without warming the atmosphere. If surface radiation is absorbed by an atmospheric molecule, it’s going to radiate that energy back out (within it’s emission spectrum), but not until it’s gained some vibrational/stretching/rotational energy in it’s bonds – it has to heat up to cool down.

    bananabender says:
    March 30, 2011 at 7:08 pm
    “The atmosphere is heated by molecular-kinetic effects i.e. molecular collisions. Atmospheric temperature is a function of pressure and density not radiative effects.”

    For convection to occur there must be a collision of masses. The vacuum of space doesn’t have enough mass for the upper atmosphere to collide with. The sun’s energy is radiated to the earth, and the earth radiates this energy back out into space as it cools.

    Atmospheric pressure is a function of temperature. Of course our atmosphere (as do most) behaves as a near ideal gas. There is no fixed volume – it can only expand against the force of gravity. The outward force of temperature balanced by the downward force of gravity determines the pressure and volume. There is no volume constraint, so changes in temperature generate changes in pressure. Do you think that the night side of earth is exposed to a lower pressure from outer space, so the temperature goes down?

  185. Steve says:
    March 30, 2011 at 4:47 pm

    Warming via conduction does not prevent cooling via radiation – the nitrogen in our atmosphere (most of the gas in it) cools every night. Are you proposing that the nitrogen gas in our atmosphere cannot cool off unless it collides with another type of atmospheric gas?

    I must admit I have been on both sides of this subject; does our atmosphere radiate as a whole, all molecules included, or only IR active molecules, and both according to temperature. My interests of decades has been in astrophysics but I have never seen a definitive spectrum of purely nitrogen or argon or for pure hydrogen to indicate that those symmetric gases in our galaxy are opaque to IR. Neither have I searched for them. I am like many and have only seen the lab spectrums of nitrogen that show no interaction in those frequencies but the one bump in then near IR.

    That is why I was careful to answer back and explain my assumption on that earlier comment.

    I have been to countless university sites on the subject and have found totally conflicting reports, usually because the word ‘gases’ is not separated out. So, you could flip me back so easily if you could just show me a spectrum with lines or continuum ranges that are identified to be nitrogen or argon for that matter. I just can’t find any so I stay in line with the apparent evidence, that whatever our atmosphere is radiating, it is only via IR active molecules.

    And surely you are not assuming Kirchhoff’s law does not apply in this fine point on a frequency by frequency basis. For if it radiates it must absorb equally. We all know for certain that solids and liquids radiate close to a gray body manner but I’ll ask you, why do think nitrogen gas is always radiating energy away in all directions?

  186. The whole “burning logs” thread seems to have too many variables. Let’s make it more controlled.

    Three separate identical vertical logs. The logs are each surrounded by identically-sized tubes a few inches bigger in diameter open at the top & bottom to allow convection.
    * Tube 1 is glass, which will not melt, but is mostly transparent to Visible and near IR
    * Tube 2 is oxidized metal, with an emissivity near 1.
    * Tube 3 is shiny metal, with an emissivity near 0 (ie almost a perfect mirror).

    I predict that Log 3 will burn fastest, due to all the reflected photons (which will look like a black body at the temperature of the glowing log) . These reflected photons will help raise the temperature of the log.

    I predict that Log 2 will burn next fastest, due to all the emitted photons at the temperature of the tube (which will be cooler than the log, but warmer than the room).

    I predict that Log 1 will burn slowest, since there will be very few photons striking it, so it will not get any boost in temperature.

  187. Hans says:
    March 30, 2011 at 7:01 pm
    Ira Glickstein, PhD says:
    March 30, 2011 at 2:02 pm

    “Having made that error, the author continues:

    One might think that, because the blackbody is now absorbing more light, even if it is its own infrared light, then it should warm up. …

    Yep, it should warm up and (if it is a true blackbody in the longwave region) it will absorb that reflected longwave energy and, all else being equal, will warm up.”

    Ira, so when will this cycle of selfheating stop?
    Lets slice a blackbody cavity which is at a equilibrium temperature of 288 Kelvin in half and replace the other half with a mirror. How much warmer will this cavity become?

    With longwave IR it would not be a mirror, but an absorbing/re-emiting thin, black-painted metal shell surface that would return about half the energy it absorbed (the other half radiating from the other side of the shell and away from the source). So, if the system was 100% efficient (which it could not be) it would be the sum of 1 + 1/2 + 1/4 + 1/8 and so on, which you must know has a limit at 2. Thus, if the longwave IR source was completely surrounded by the shell, the temperature could potentially increase to correspond to double energy, but, in the real world, it would reach equilibrium well before that.

    I am not familiar with the details of a blackbody cavity, but I can imagine something like an electric soldering iron. With a given power input, the tip, in free air, would reach a temperature of X. If you put the whole soldering iron in a metal box, and suspended the metal box in free air, the tip would reach a higher temperature, all else being equal. I include the last caveat because I believe that the resistance element in the soldering iron would increase its resistance as it heated up, which, at constant voltage input, would reduce the current somewhat, and therefore the input power. But, if you had a Variac (remember those – I used to own one) and you increased the voltage to get the exact same power input as before, the tip would increase in temperature compared to where it was in free air. You could calculate the new temperature using the Stefan Boltzmann (fourth power) Law. The new tip temperature would correspond to nearly twice the power.

    Note however that no extra energy is being created. At equilibrium, the metal box will radiate (and convect) away the exact amount of power as is being input to the soldering iron, energy in = energy out. The only difference will be that the tip of the soldering iron will be warmer when enclosed in the metal box than when it was in free air.

    Got it?

  188. BigWaveDave says:

    The constant compression of atmosphere by gravity results in higher pressure, density and temperature at the surface. No GHG’s are necessary for this..

    Your notion doesn’t even satisfy the 1st Law of Thermodynamics! In the absence of a greenhouse effect, the surface temperature of the earth is set by energy balance considerations. So, unless you are proposing a large source of energy…e.g., that the earth’s atmosphere is undergoing continual gravitational collapse, which would certainly be a novel hypothesis…your notion is completely in violation of known laws of physics.

    Gravity is a continual force and gas molecules are in motion, In any process where you pressurize a sufficient amount of gas, its temperature will increase. In processes where the gas is pressurized from one side, the temperature will be highest at the surface where where pressure is highest. This can be demonstrated with a centrifuge, or the leading edges of an aircraft. There’s nothing novel, or in violation of the laws of Physics here.

    Trying to argue that the greenhouse effect doesn’t exist just discredits you in the eyes of any serious scientist.

    This would be true if you replace “scientist” with “greenhouse hypothesis enthusiast”.

    However, in the presence of an IR-absorbing atmosphere, some of the terrestrial radiation is absorbed and when the atmosphere subsequently emits radiation, some of this radiation comes back to the earth. This keeps the earth warmer than it would be in the absence of the IR-absorbing atmosphere.

    This violates the second law of Thermodynamics.

    When I see people challenging the basic facts of the greenhouse effect, what it tells me is that they are either misguided or (if they are knowledgeable enough to know better) that they are trying to actively deceive people and confuse them on the science, i.e., their real goal has nothing to do with science and everything to do with pursuing their policy objectives at the expense of science.

    What you should realize, is that you have no physical explanation of what you believe to be “basic facts”. and that it is highly probable that the misguided one is yourself.

  189. From this thread I have discovered that:

    – I can heat my house for free by filling the roof space with CO2.
    – Pure N2 has a temperature of absolute zero because it can’t radiate energy.
    – Chimneys make fires cooler.
    – Igloos heat the occupants by radiating “cold” energy.
    – Two 50W light bulbs placed together are brighter than one 100W bulb.
    – CO2 always absorbs energy at ground level and radiates energy at the edge of space.
    – Chemists are all morons who think intramolecular bond stretching and orbital electron excitation are different things.

    Sarcasm off.

  190. wayne says:
    March 30, 2011 at 9:07 pm

    “My interests of decades has been in astrophysics but I have never seen a definitive spectrum of purely nitrogen or argon or for pure hydrogen to indicate that those symmetric gases in our galaxy are opaque to IR.”

    I think you misspoke there. Nitrogen gas is primarily transparent to IR, not opaque, just like it is primarily transparent to visible light. If nitrogen gas were primarily opaque to IR, imagine how difficult it would be to take a thermographic image in an atmosphere that is almost 80% nitrogen gas. It would be like trying to take a photograph in the fog.

    “So, you could flip me back so easily if you could just show me a spectrum with lines or continuum ranges that are identified to be nitrogen or argon for that matter.”

    In the infrared ranges? I can’t find those either, which is understandable. Typical infrared spectrometers use nitrogen gas as a purge for a very good reason – it doesn’t interfere with the readings. I have found a few articles, unfortunately behind paywalls, that reference it’s absorption/emission spectrum in the far infrared:

    http://link.aip.org/link/doi/10.1063/1.431648

    http://iopscience.iop.org/0022-3700/10/3/018

    http://www.informaworld.com/smpp/content~db=all?content=10.1080/00268978400102221

    “Why do think nitrogen gas is always radiating energy away in all directions?”
    Because outer space is much colder than the temperature of the atmosphere, and nitrogen gas makes up almost 80% of the atmosphere. I feel it get cold at night, so the heat is going somewhere. Since it doesn’t get continuously hotter every day, the heat must be leaving the atmosphere, right?

    You may be interested to read how satellite measurements pertaining to our temperature records are primarily readings in the microwave band, which correlates to heat radiated by oxygen gases (including ozone):

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

  191. bananabender says:
    March 30, 2011 at 7:51 pm
    Your anecdotes are amusing folklore but they show absolutely no understanding of basic scientific principles. >>>

    My anecdotes are an attempt to show via real world example how the laws of physics work. The equation to describe the amount of power in watts per square meter (P) that a black body will radiate at a given temperature (T) in degrees Kelvin is as follows:

    P = 5.67 * 10^-8 * T^4

    If you will take note Mr Banana, there is no variable in that equation pertaining to the temperature of the environment the black body exists in. In other words, the amount of energy being radiated by the black body is 100% governed by its temperature in degrees K. It matters not one whit if it is warmer or colder than its immediate environment, that is the amount of energy that it will radiate. The energy it radiates is in the form of photons, which travel at the speed of light (for purposes of this discussion they are light) and have a mass approaching a limit of zero. They exhibit the characteristics of both particles and waves, and are actually neither and both at the same time. Light exhibits a number of properties that demonstrate it exists physically outside of certain parameters we normally take for granted. Consider as an example a person on a high speed train reading a newspaper by the dome light above. If they were to calculate the speed with which photons emerge from the bulb and travel to the newspaper, the observer would arrive at a speed of exactly 1 times the speed of light. A stationary observer however would not see the light as going “straight down”. The stationary observer would see the light going down and sideways at the same time. If the stationary observer calculates the speed of the photons they will get… 1 times the speed of light. Before you start screaming that this violates simple pythagorean geometry and the laws of motion, please be advised that the speed of light is in fact immutable, the traveller on the high speed train does not pass through time as fast as the stationary observer, and hence they both see the light travelling at the same speed. If you want to argue with me as to this being correct or not, I shall cite one reference only:

    Mr Albert Einstein

    His theory of relativity provided for the quantification of precisely how much time a body in motion at a given velocity would lose in comparison to a stationary body, and this has been confirmed through experimentation multiple times. In brief, photons ain’t what you seem to think, they radiate from a surface according to the Stefan-Boltzman equation (above) and they don’t give a tinkers damn how fast you think they are going, they only have one speed, and that’s the speed they go until they collide with something, and they will collide with that something no matter what the temperature of that something is. Would you like me to explain Plank’s constant next?

    “Have you ever used a bunsen burner? You vary the flame temperature of this burner by altering the amount of air mixing with the gas. The more air you add the more completely the gas burns and the hotter the flames. Fanning a fire is exactly the same principle. “>>>

    Well, having discredited yourself as a physicist, chemist, and Scout, you now want to move on to Bunsen burners? Yes, I’ve used one. Many times. If you open the vent at the bottom all the way to let in as much air as possible, the flame will most likely blow out. At best you will get a crackling sound with the bottom of the flame starting about a quarter of an inch above the lip of the burner. Close down the vent until you get a tall silent flame instead. This results in the oxidizing molecules travelling as straight upward in a narrow column as possible in order to concentrate the heat best on the bottom of your flask or other object you happen to be heating. They’re very handy for heating up and bending glass tubing in this manner, but I recommend you keep the first rule of chemistry labs in mind when you try. Hot glass looks exactly the same as cold glass.

    “I was employed as chemist in the food and beverage industries after I graduated from university. I regularly used various analytical spectroscopy methods including AAS, GC-MS, spectrophotometry, IR and near IR spectroscopy . Spectroscopy is based entirely on radiative physics.>>>

    Congrats. I see you neatly side stepped the question, which I will repeat. You claim to have three degrees. What are they? Not physics or chemistry obviously or you wouldn’t be trying to build your credibility by quoting all the cool instruments you used at a job after university. Guess what? I can drive a car. That doesn’t make me an automotive engineer. I can use a computer, that doesn’t mean I can design one. Operating lab equipment doesn’t provide you with any understanding of the physics the equipment is based on. Unless you actually understand the physics, you’re just following the instructions in the manual and for all you know its just a big, fancy, thermometer.

    If you still think that operating a thermometer makes you a physicist, I may be willing to go another round. But at least come up with a tough one that doesn’t make you look foolish in the end, Mr not-a-Scout/chemist/physicist but I operated a spectrometer at work once and have three degrees.

  192. BigWaveDave says:
    March 30, 2011 at 10:28 pm

    Atmospheric pressure is a function of temperature. Of course our atmosphere (as do most) behaves as a near ideal gas. There is no fixed volume – it can only expand against the force of gravity. The outward force of temperature balanced by the downward force of gravity determines the pressure and volume. There is no volume constraint, so changes in temperature generate changes in pressure. Do you think that the night side of earth is exposed to a lower pressure from outer space, so the temperature goes down?

    I will rephrase my comment. The atmosphere can be sufficiently heated by molecular kinetic effects without needy any radiation-absorbing “GHGs”. Radiative energy transfer only becomes more important than conduction and convection at the outer edge of the atmosphere.

  193. I have just read through Nasif’s thread on Jennifer Marohasy’s blog and I haven’t had such an interesting and informative read or learned more about AGW issues in years. It will prompt me to read more too as there it would seem to show a possibility that not only is AGW not significant but that the whole greenhouse effect is not like it is viewed by the majority. The responses from the warmists sound like even they don’t believe it any more.
    I’m genuinely sorry that Nasif doesn’t get better coverage, he’s a bit eccentric but an excellent and thought provoking read

  194. Ira Glickstein,

    Thank you for your efforts to bring out the science of one isolated aspect, which I call the earth’s ‘atmospheric effect theory’. It is one out of many interacting aspects of our earth’s atmosphere. I understand you are doing so when there are many aspects (including the one you are discussing) of our atmospheres which do interact and which are not isolated in the actual complex atmospheric dynamic.

    The single isolated aspect (out of many interacting aspects) which you are presenting for discussion was purposely selected and isolated by you to aid in a clear presentation of just the one aspect that I call the earth’s ‘atmospheric effect theory’. I support your efforts to that extent with the understanding that what is not being presented by you is the total net effect (negative or positive) of the single isolated aspect you are discussing on the total complex earth dynamic system.

    This latest post is much better, in my view, than the previous posts in your series that used analogy. It is best to just present the science rather than use analogy. Analogies are always limited and have some misleading attributes.

    Things I find not at issue wrt the earth’s ‘atmospheric effect theory’ from the so-called consensus focused IPCC science:

    a) The gases in our atmosphere interact by collision with each other and in doing so transfer energy.

    b) Some of the gases in our atmosphere have the ability to absorb electromagnetic radiations of certain wavelengths. And in certain conditions re-emit it.

    c) Some of the gases in our atmosphere have varying interaction with the incoming solar electromagnetic radiation and in doing some energy is transferred to some of the gases.

    d) Some of the gases have varying interaction with the blackbody (or near blackbody type) radiation generated by all the surfaces of the earth and in doing so some energy is transferred to some of the gases.

    e) Some of the gases have varying interaction with the blackbody (or near blackbody type) radiation generated by the atmospheric gases themselves and in doing so energy is transferred to some of the gases and also sometimes to the earth’s surface.

    f) The electromagnetic radiation from the sun and also from the gases in our atmosphere has varying interaction with each of the earth’s various types of surfaces and in doing so some energy is transferred to the surfaces.

    g) Some of the electromagnetic radiation from the earth’s surfaces go directly to space.

    h) The electromagnetic radiation from all the gases in our atmosphere can, under certain conditions, go directly to space.

    Things I find at issue wrt the earth’s ‘atmospheric effect theory’ from the so-called consensus focused IPCC science:

    a) I would like to see the formal theory for what I call earth’s ‘atmospheric effect’. Show me the integrated formal theory. Has it been only premised that there is one, not shown that there is one? It cannot be uncoupled from other aspects of the climate system and remain meaningful.

    b) I would like to see the actual comprehensive /unbiased /multiple /corroborating observations that demonstrate that a theory of earth’s ‘atmospheric effect’ has a significant net effect on meaningful climate parameters. OK, on average GST too.
    Ira Glickstein,

    John

  195. @ Steve

    I’ll read those links if I can get to them. Thanks.

    One thing you might consider is that just because nitrogen by itself does not radiate it doesn’t mean nitrogen can not cool by radiation in the real atmosphere. It will do it through greenhouse gases by re-exiting them by a high enough collision and that greenhouse gas molecule will then radiate the energy away. I have also found that even though most references will show pure nitrogen with only a small band in the near IR and oxygen having bands in the microwave region, they both do radiate a small amount in the mid-infrared just above and below 20 µm due to N2-O2 collision electron interactions together. So, just because a pure nitrogen atmosphere might actually not be able to cool alone, it can once other species of molecules are put in the mix. Just found the paper last week describing that very combo with spectrums included. Well, that’s about where I stand at this point. And that is why I still think a pure nitrogen atmosphere with the same sea level pressure would be hot for sure, you have then taken away all of the ways that nitrogen can possibly lose IR heat. Two month’s ago I would have said exactly the same thing you said.

  196. Martin

    Thanks for posting this.

    What this GHG theory shows is how people will hold onto their beliefs no matter what. They are not acting as scientists should but hold onto their theory based on faith not science.

    The killer for me is that logically the GHG theory does not stand up. It leads to absurd possible outcomes. Like for example heating the house with CO2 in the roof space. Or ice in a Igloo heating the body by backradiation.

    Or an oven that heats a chicken without any outside power source.

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

    The GHG theory is flawed IMHO and its time to either move on or at least acknowledge the issues with this theory.

  197. Hans asks

    …..”Lets slice a blackbody cavity which is at a equilibrium temperature of 288 Kelvin in half and replace the other half with a mirror. How much warmer will this cavity become?”…..

    Ira Glickstein replies

    …..”With longwave IR it would not be a mirror, but an absorbing/re-emiting thin, black-painted metal shell surface that would return about half the energy it absorbed”…..

    My comment to Ira;

    This sounds as if you think that IR radiation cannot be reflected!

    When I was at school we did experiments to show that all EM radiation including Infra Red radiation exhibit the full range of wave properties of;
    Reflection, Refraction, Interference and Diffraction.

    Surely you did not mean to imply the opposite to what every school level physics pupil should know?

  198. I said

    Joel Shore’s simple world consists of an earth without oceans and a land surface that does not retain heat.

    Joel replies

    No…Retention of heat is not relevant to the considerations that I mentioned. The way that it comes into it is that the ocean helps to keep the temperature more uniform.

    My reply;

    The Oceans, covering 70% of planet surface, absorb solar radiation.
    Some shorter wavelengths penetrate the water to a considerable depth.
    In other words the effect is to increase the temperature of the water.
    Ocean currents(and tides) distribute the hotter water.
    E.g. the Gulf Stream effect keeps the west coast of UK much warmer than without it.
    Glasgow is at the same latitude as Moscow.
    The atmosphere, in particular clouds, help insulate the Earth surface.
    The gross error of IPCC dogma is to attribute a 33K increase effect as due solely to the radiative properties of CO2 and H2O in gaseous form.
    The early climate models did not even include clouds although some attempts are now being made to rectify this gross defect.

  199. Nah. You’ve all got it “wrong on the internet”!

    Here’s the Real Deal:
    Way back in Hypothetical History, the Earth had no CO2 and was in thermal and radiative equilibrium with Sol. But then one day the Three-Handed God of Gasses wiggled his middle pinkie and ~~BAZOWIE~~ CO2 joined the fun.

    This interrupted the normal escape routes and timing of surface radiation in a few LW slots, delaying it by hundreds, or even thousands, of microseconds, and it consequently warmed the air and surface slightly. But then the leading edge of the delayed LW escapee photons made it to the TOA, and fled the trap. Ever since, the surface and air temps have been nice and stable at the new minutely higher temperature.

    But, someday, the above-mentioned God of Gasses will wiggle his middle big toe, and ~~EIWOZAB~~ the CO2 will vanish! Then the “excess” will directly blow the joint and temperatures will return to pre-Pinkie levels.

    And now you know!
    ~~~~~~~~~~
    We Believers in the Gospel of the God of Gasses are called “The Laggards”! Join us now, even though it may already be too late.

    Or not.
    ;)

    Exegesis;
    The sequence above is actually slightly more complex and interesting. When the CO2 first appears, a drop in OLW occurs until the lagged photons hit TOA and the exit doors there. The energy they represent, as a function of flux times the lag duration, is the measure of the “excess” which hangs around and keeps temps a smidge above normal. When the CO2 vanishes, that drains away, and so does the temperature anomaly vis-a-vis the Pinky Event.

  200. Hans asks

    …..”Lets slice a blackbody cavity which is at a equilibrium temperature of 288 Kelvin in half and replace the other half with a mirror. How much warmer will this cavity become?”…..

    The cavity will remain at the same temperature! The phrase “at equilibrium” means “the photons in the cavity are at equilibrium with the walls of the cavity, all of which are at 288 K..” Putting a mirror down the center won’t change the temperature of either the photons or the walls. The left half of the cavity and the right half of the cavity will still be at 288 K.

  201. Hans says:
    March 30, 2011 at 7:01 pm
    Ira Glickstein, PhD says:
    March 30, 2011 at 2:02 pm

    “Having made that error, the author continues:

    One might think that, because the blackbody is now absorbing more light, even if it is its own infrared light, then it should warm up. …

    Yep, it should warm up and (if it is a true blackbody in the longwave region) it will absorb that reflected longwave energy and, all else being equal, will warm up.”

    Ira, so when will this cycle of selfheating stop?
    Lets slice a blackbody cavity which is at a equilibrium temperature of 288 Kelvin in half and replace the other half with a mirror. How much warmer will this cavity become?

    Good question. But I do not think the answer is what you think it is.

    If the total system is at equilibrium with its surroundings ( there is no other source of heat ) then it is at equilibrium and so the nature of the surfaces will have no effect on temperature.

    But let us take the more interesting example where the cavity has a power source. It does not matter how big but say it is 100 watts. Assume also that it is in vacuum so that it has no conduction losses and has a highly polished external surface so that it has no external radiation losses. If this is in dynamic equilibrium (that is it is losing heat at a constant rate but has a constant temperature) then the 100 watts will be dissipated through the opening of the cavity with a near black body spectrum characteristic of that constant temperature. Lets say for argument’s sake that this is 800K. If you silver the inside of part of the cavity it will be a slightly poorer black body so its spectrum will not be perfect and and therefore a small rise in temperature will have to occur in order to dissipate the same 100 watts. However the effect will be very small and there is of course no conflict with the conservation of energy.

    However an even more interesting experiment would be to take a sphere blackened on the outside with lamp black which, although not a perfect black body, is very close to perfectly black. Now heat this with 100 watts. Let us say that the sphere has a surface area of 100 units. The dissipation is therefore 1 watt per unit area (still in a vacuum so no conduction losses). If the surface area of this sphere was the same as the area of the apperture in the original black body cavity then the temperature would be close to but a little more than 800K because it is not quite perfectly black. Now place a semispherical shell a small distance from one half of the sphere. This shell has a black internal surface and a perfectly silvered outside. Energy dissipation on this side of the sphere is therefore absorbed by the shell but can only be radiated back to the sphere since the silvered external surface cannot radiate. Therefore, if we take the sphere and shell system as a whole, all the 100 watts has to be radiated from the 50 units of surface that is not covered. For this to happen the temperature has to rise to a level where the heat loss per unit area is double. If this temperature is T the fourth power of T is equal to 2 times the fourth power of 800. Thus T is 800 times the fourth root of 2. The shell will also be at this new temperature since the two facing surfaces must be in equilibrium if there is no nett heat flow (no dissiptation from the shell’s external surface).

  202. “The killer for me is that logically the GHG theory does not stand up. It leads to absurd possible outcomes. Like for example heating the house with CO2 in the roof space. Or ice in a Igloo heating the body by backradiation. ”

    The killer for me is that people could actually concluded that the GH effect predicts that you could heat a house with CO2 in the roof! No competent scientist would say that nor support that – the relative size and relative temperatures are wrong! Only poorly informed skeptics ever seem to come to this flawed conclusion (although poorly informed warmists can make it seem like a plausible conclusion based on their own faulty understanding of the GH effect).

    Since you are claiming the GH theory predicts that a house will warm up from C02 in the roof, perhaps you could explain the logic behind that prediction. How specifically would the GH theory lead to the conclusion that you could heat your house (or even limit cooling) simply from a layer of GHGs inside the roof space?

  203. Tim

    Why don’t you go to Jennifer’s blog and read Nasif’s work and the responses. It may surprise you and answer your question. It didn’t actually say you can heat your house with an attic full of CO2 but go and read and be ready to be surprised even.

  204. Tim

    I know no competent scientists would say this. But its what scientists seem to be saying with the GHG theory.

    CO2 supposedly adds heat to the earth due to backradiation. Why would the same not apply in a house (or if a person was enclosed in a container of CO2).

    Or as per the chicken in the oven example above.

    Or do you believe CO2 and backradiation can operate at higher levels but not at ground level. If so why not. If CO2 reflects back radiation to increase the temperature of the surface why can this not also apply at ground level.

  205. Bryan says:
    March 31, 2011 at 4:47 am
    … Ira Glickstein replies

    …..”With longwave IR it would not be a mirror, but an absorbing/re-emiting thin, black-painted metal shell surface that would return about half the energy it absorbed”…..

    My comment to Ira;

    This sounds as if you think that IR radiation cannot be reflected!

    When I was at school we did experiments to show that all EM radiation including Infra Red radiation exhibit the full range of wave properties of;
    Reflection, Refraction, Interference and Diffraction. …

    Common glass Mirrors that reflect visible light generally reflect near-IR (what we’ve been calling shortwave) at a fairly high level of reflectivity. However, their reflectiity generally falls off considerably in the Far-IR (what we’ve been calling longwave). But, you are correct, specially coated metal mirrors are available that reflect very well out to 10μ or even 20μ. For example:

    Metal mirrors are good general-purpose mirrors because they can be used over a very broad spectral range from 450 nm to 12 µm. They are also insensitive to polarization and angle of incidence, and provide a constant phase shift, making them appropriate for ultrashort-pulse applications. Their softer coating, however, makes them more susceptible to damage, and special care must be taken when cleaning.

    The point I was making was that a black-painted metal shell was all you needed, and that that was more like the Atmospheric “Greenhouse Effect” than more expensive, damage-prone, special far-IR mirrors.

  206. wayne says:
    March 31, 2011 at 3:24 am

    “And that is why I still think a pure nitrogen atmosphere with the same sea level pressure would be hot for sure, you have then taken away all of the ways that nitrogen can possibly lose IR heat.”

    There is still adiabatic cooling. The nitrogen gas will rise and cool, and then radiate it’s heat away at whatever wavelengths that cooler nitrogen gas radiates at, be it far infrared or microwave.

  207. RJ says:
    March 31, 2011 at 6:53 am
    Tim

    I know no competent scientists would say this. But its what scientists seem to be saying with the GHG theory.

    CO2 supposedly adds heat to the earth due to backradiation. Why would the same not apply in a house (or if a person was enclosed in a container of CO2).

    Or as per the chicken in the oven example above.

    Or do you believe CO2 and backradiation can operate at higher levels but not at ground level. If so why not. If CO2 reflects back radiation to increase the temperature of the surface why can this not also apply at ground level.

    CO2 in your loft space would be warming but the effect would be tiny since your ceiling and your roof already do a very good job of absorbing radiated heat and re radiating it back. Any downward radiation from CO2 would warm the floor of the loft and upward radiation would warm the roof. Since the CO2 would be at a slightly higher temperature than the roof my guess is that this warming of the floor would be slightly greater than it would have been if it were left to the roof alone to backradiate to the floor. Therefore there would be a slight drop in roof temperature and so a slight drop in total heat lost through the roof but it would be tiny. Silver foil would be more effective!

    Furthermore the scale is different. Energy radiated from the earth’s surface in the 14-18 micron band is initially radiated at around 290K and if there were no greenhouse gases this energy would be directly lost to space (just like the energy around the 10 micron band). The absorption by CO2 means that radiation to space can only take place at an altitude where there is little chance of further absorption. This is just below or within the tropopause ( around 8km at the poles and 16km at the equator). The temperature here is about 220K. At this temperature the energy radiated is about 10 times less than the original energy at the surface. This difference has to be made up because the incident radiation from the sun is fixed. There is no option but for the surface to warm in order to increase the radiation at the other wavelengths to compensate.

    In the case of your loft there is no radiation into space and even if there were the height of your loft would induce a temperature difference of about 1/50th of a degree.

  208. cal says

    …”The absorption by CO2 means that radiation to space can only take place at an altitude where there is little chance of further absorption. This is just below or within the tropopause ( around 8km at the poles and 16km at the equator). The temperature here is about 220K. At this temperature the energy radiated is about 10 times less than the original energy at the surface. This difference has to be made up because the incident radiation from the sun is fixed. There is no option but for the surface to warm in order to increase the radiation at the other wavelengths to compensate.”……

    Cal has given a good outline of the more sophisticated version of the Greenhouse Theory.(GT)
    This is quite a contrast to the cruder; …. atmosphere “toasts” the Earth Surface version favoured by Joel and others.

    In this version the famous experiment by R W Wood is not contradicted as the radiative effect of CO2 and H2O can still be very small for that particular volume.
    Ira likes analogies and the best one here is of a bath with an small open drain hole being filled with a bigger water supply from the tap.
    I’m not saying I agree with this version of GT, but at least its more plausable.

  209. martin mason says:

    I have just read through Nasif’s thread on Jennifer Marohasy’s blog and I haven’t had such an interesting and informative read or learned more about AGW issues in years.

    You mean the thread that starts out with this statement?

    Central to the theory of Anthropogenic Global Warming (AGW) is the assumption that the Earth and every one of its subsystems behaviors as if they were blackbodies, that is their “emissivity” potential is calculated as 1.0. [1]

    Since that statement is utterly wrong, I don’t really see why you would get anything out of his post. He then goes on to do a calculation for “total emissivity” of CO2 without even defining what he means by “total emissivity” (of an entire atmosphere’s pathlength?) … or to what wavelength his calculations apply.

    If you want to read about the science of calculating radiative transfer in the atmosphere, pick up a book on the subject or at least find a blog post by someone who has a clue what he is talking about.

  210. Bryan says:
    March 31, 2011 at 9:10 am
    cal says

    …”The absorption by CO2 means that radiation to space can only take place at an altitude where there is little chance of further absorption. This is just below or within the tropopause ( around 8km at the poles and 16km at the equator). The temperature here is about 220K. At this temperature the energy radiated is about 10 times less than the original energy at the surface. This difference has to be made up because the incident radiation from the sun is fixed. There is no option but for the surface to warm in order to increase the radiation at the other wavelengths to compensate.”……

    Cal has given a good outline of the more sophisticated version of the Greenhouse Theory.(GT)
    This is quite a contrast to the cruder; …. atmosphere “toasts” the Earth Surface version favoured by Joel and others.

    In this version the famous experiment by R W Wood is not contradicted as the radiative effect of CO2 and H2O can still be very small for that particular volume.
    Ira likes analogies and the best one here is of a bath with an small open drain hole being filled with a bigger water supply from the tap.
    I’m not saying I agree with this version of GT, but at least its more plausable.

    ————

    Thanks Bryan I have never been called sophisticated before!

    However I would point out that there is nothing contradictory about the two ways to describe the effect. I prefer this one (but have used both on this thread) because systems with many positive and negative feedbacks are difficult to predict. One can describe the greenhouse effect from bottom up, so to speak, as Ira has done, but it is hard to calculate the nett effect of small changes. However the top down approach (puns definitely intended) allows one to look holistically at the system and apply solid rules like energy balance equations to predict change on the basis of actual measurements. These are available at the macro level such as the temperature and position of the tropopause and radiation at different wavelengths leaving the atmosphere. It is because these measurements are readily available that I am sceptical of the warmists’ claims. The satellite data does not seem to provide evidence of the macro changes theory would predict. I can’t help feeling that if they had such data it would be spread all across the media. My conclusion is that rather than adapt their theory they are desperately trying to come up with an explanation. The issue is not whether the previous 10 doublings have had an effect. The issue is will this one. It is not obvious that it will because the effective level at which CO2 radiates into space is already at the coldest point in the atmosphere. For further reductions in radiation to occur the whole upper atmosphere has to cool. I will wait and see. I am sceptical, but everything is possible. I just think we would have seen at least a trend by now.

  211. Bryan says:

    Cal has given a good outline of the more sophisticated version of the Greenhouse Theory.(GT)
    This is quite a contrast to the cruder; …. atmosphere “toasts” the Earth Surface version favoured by Joel and others.

    I am not in favor of the cruder one. I have often pointed out that the picture described by cal is necessary for a fuller understanding. See for example, this post: http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/#comment-625057 and in general the discussion we had when I tried to explain to you why Wood’s experiment did not disprove the atmospheric greenhouse effect. I am glad to see you belatedly coming around to me point-of-view.

    However, the cruder picture is not without merit for getting some of the basic ideas down. Yes, it neglects some things like convection that eventually should be included in a more refined picture. But, simple models are always limited.

  212. “martin mason says: March 31, 2011 at 6:33 am

    Tim

    Why don’t you go to Jennifer’s blog and read Nasif’s work and the responses. It may surprise you and answer your question. It didn’t actually say you can heat your house with an attic full of CO2 but go and read and be ready to be surprised even.”

    Well, the very first paragraph is off to a bad start

    Introduction
    Central to the theory of Anthropogenic Global Warming (AGW) is the assumption that the Earth and every one of its subsystems behaviors as if they were blackbodies, that is their “emissivity” potential is calculated as 1.0. [1]

    http://jennifermarohasy.com/blog/2011/03/total-emissivity-of-the-earth-and-atmospheric-carbon-dioxide/#more-7739

    There is no such “central assumption” that “every one of its subsystems behaviors as if they were blackbodies.”

    The bulk of the paper deals with a rather involved equation that appears to be some empirical fit to CO2 absorption data. Without access to the original paper, it is very difficult to comment on the accuracy of the equation or Nasif’s application of of the equation.

    However, the conclusion states “the potential of the carbon dioxide to absorb and emit radiant energy is negligible”. His calculations clearly contradict various experimental results (http://www.skepticalscience.com/images/infrared_spectrum.jpg or http://brneurosci.org/spectra.png for example) which clearly show CO2 having a clear and significant impact on IR in the atmosphere. I have to go with experimental evidence and reject his calculations.

    Furthermore, Nasif’s comments in the discussion following the paper do not particularly impress me.

  213. RJ;
    The killer for me is that logically the GHG theory does not stand up. It leads to absurd possible outcomes. Like for example heating the house with CO2 in the roof space. Or ice in a Igloo heating the body by backradiation. >>>

    What is absurd is taking the explanations, which are based on 200 years of documented physics theory verified by thousands of experiments, and instead of learning how to understand the equations and the physics, instead make wild accusations that are completely out of proportion. Instead of responding with the ludicrous “so why can’t I heat my house by putting CO2 in the attic” as a proper question.

    Q; If I put CO2 in my attic, according to the laws of physics, would it heat my house?

    A; It would provide an amount of insulation so small that you likely could not measure it without some very expensive and accurate instruments, but it would have some effect compared to a vacuum for example. However it would still be positive compared to vacuum. A foot of fibre glass insulation would have a measurable effect, and by the way, works on the same physics. Does it generate “new” heat? No. It just slows down the escape of IR resulting in a warmer house for a given amount of heating input from the furnace, stove, people, etc. And BTW, if you take the temperature of the fibre glass insulation on a very cold day, you’ll find that it is indeed colder than the ceiling of the house it rests on. Despite which, it keeps the house warmer.

    Stop extrapolating actual physics to orders of magnitude higher claims than the physics actually makes, and the absurdity will go away with it.

  214. Joel
    I certainly do not believe you ever advanced a version of the greenhouse theory that explained it as being almost totally confined to above the tropopause.
    R W Wood and G&T are correct in saying that the radiative effects of CO2 are very small at atmospheric temperatures and pressures.

  215. In response to Hans, Ira Glickstein claimed that a radiating body would heat itself up with its own reflection.
    “Yep, it should warm up and (if it is a true blackbody in the longwave region) it will absorb that reflected longwave energy and, all else being equal, will warm up.”

    Bang go the laws of energy conservation then.

    If you believe in back radiation, then try this suggestion of Alan Siddons. Shine a beam of light on a surface which reflects some of the light, such as a wall. Shine the light at 45 degrees so that the part which is reflected bounces off at the same angle. Now take a mirror and position it so that it reflects back to the illuminated part of the surface the light that is reflecting off the surface. You will see that it makes absolutely no difference to the brightness of the surface. (It will, of course, illuminate the wall where it is not already illuminated by the beam.)

    Ira claims that the laws of thermodynamics are not broken because the energy exiting from inside his reflective shell is the same as that entering. But inside the shell (which in theory could be as large as a room, or a furnace) you are producing two units of energy out of one. This is energy creation, not conservation.

  216. @Steve

    If the entire pure nitrogen atmosphere could radiate sufficiently to overcome the cloudless sky daytime irradiance transferred by conduction, then you are, or course, correct. Just wish I could get some definitive proof and the answer to why professors in physics and spectrometry at various universities disagree with each other on that very point. Some would say that the ones saying “all matter always radiates if over zero K” are only referring to liquids and solids.

    I guess you do know what that would mean if you could prove that. That would mean that not only does nitrogen absorb and radiate adequately by itself but also oxygen and argon at 101325 Pa and there is no special attribute of GHGs at all over the atmospere as a whole, for the entire atmosphere then would be acting in exactly the same manner as they do. It’s just that the spectrums don’t seem to show that, or, possibly that their radiance is just so incredibly tiny and the hot atmosphere I mentioned above is still probably true though a bit cooler than with zero radiance.

    But I still would ask you on that hypothetical situation, if I could: If there was zero radiation from a pure nitrogen atmosphere, heck, lets make it pure argon as dense as ours, no oceans, no clouds, would it be cold or hot? What do you think?

  217. BigWaveDave says:

    Gravity is a continual force and gas molecules are in motion, In any process where you pressurize a sufficient amount of gas, its temperature will increase. In processes where the gas is pressurized from one side, the temperature will be highest at the surface where where pressure is highest. This can be demonstrated with a centrifuge, or the leading edges of an aircraft. There’s nothing novel, or in violation of the laws of Physics here.

    Yes, when you pressurize gas, it warms. However, if you pressurized some gas and then let it sit in equilibrium with its surroundings for billions of years, it will equilibrate with its surroundings. The earth’s atmosphere is not undergoing continual gravitational collapse.

    It is also true that the lapse rate, i.e., the decrease in temperature with height in the troposphere is determined by understanding what would happen if a parcel of gas rises up in the atmosphere and undergoes adiabatic expansion or sinks down and undergoes adiabatic compression. So, yes, the fact that the temperature decreases with height in the troposphere is due to this fact (coupled with the fact that the atmosphere is primarily heated from the bottom…since the adiabatic lapse rate only sets a stability limit on the lapse rate, which is why the temperature actually increases with height in the stratosphere).

    However, in the absence of an IR-absorbing atmosphere, the temperature at the surface would be set by the condition that the earth system must be in radiative balance with its surroundings. And, the temperature of the rest of the atmosphere would have to adjust accordingly.

    Trying to argue that the greenhouse effect doesn’t exist just discredits you in the eyes of any serious scientist.

    This would be true if you replace “scientist” with “greenhouse hypothesis enthusiast”.

    That is like me saying that trying to argue that the Earth is only 7000 years old discredits you in the eyes of any serious scientist and you responding that it would be true if I replaced “scientist” with “evolutionist”. All serious scientists understand that the greenhouse effect exists; you’ll notice that Roy Spencer or Richard Lindzen don’t dispute it, for example. It is based on well-understood laws of physics correctly applied.

    You may operate under the delusion that there is serious doubt about this…and, by all means, I very much encourage you to raise your lack of belief in the greenhouse effect with any serious scientist you might interact with; it will certainly help that scientist to quickly decide whether to take anything you say seriously.

    This violates the second law of Thermodynamics.

    No…You just clearly don’t have a clue what the Second Law actually says, as I have explained in posts above. The Second Law is not magic; it is a law deriving from statistical physics.

    What you should realize, is that you have no physical explanation of what you believe to be “basic facts”. and that it is highly probable that the misguided one is yourself.

    I and many others have endeavored to explain it to you. If you still cannot understand it, I don’t think it is all of our faults.

  218. Bryan says:

    I certainly do not believe you ever advanced a version of the greenhouse theory that explained it as being almost totally confined to above the tropopause.

    That is not what cal said. What he said is that the radiation that successfully escapes to space is mainly emitted from close to or a little below the tropopause (because radiation emitted from a lower layer will likely be absorbed again before escaping and higher layers have less emission). The greenhouse effect is not confined above a certain layer; it comes about from a holistic understanding of radiative transfer within an atmosphere where temperature decreases with height.

    R W Wood and G&T are correct in saying that the radiative effects of CO2 are very small at atmospheric temperatures and pressures.

    I don’t even know what that sentence means. In one sense it is exactly backwards…It is because CO2 absorbs so strongly at atmospheric temperatures and pressures that radiation emitted from near the surface is unlikely to escape to space without being absorbed again. However, if you mean that the fact that the temperatures in the troposphere decrease with height is vital in understanding the greenhouse effect, then this is correct. And, that is where jumping from Wood’s experiment to any conclusion about the atmosphere turns out to be wrong.

    Within the simple spherical shell models (which consider only radiative effects), one naturally finds that the temperatures of the shells decrease as one goes outward…i.e., the radiative equilibrium condition itself (coupled with the solar radiation mainly being absorbed at the surface) produces such a result. In the real atmosphere, convection prevents the lapse rate from being too steep…but you still get the temperature decreasing as you go up in the troposphere.

  219. AGW can never understand the objections re the Laws of Thermodynamics because AGWScience claims there is a “net” word added to the 2nd Law is because they have to pretend that Solar energy in the short-waves are absorbed by the Earth and so heat it, that the only Thermal IR in the energy budget comes from the thus heated Earth radiating into the atmosphere. They exclude the Thermal IR coming from the Sun that actually, really, heats the Earth, and us, that we can feel, to teach instead that non-thermal photons of light are in effect thermal, are actually felt as thermal. AGWScience has created another physics.

    I bring to your attention this old NASA page for children of the classic physics of Light and Heat energies, that they are different. This page is soon to disappear and the clear, straightforward descriptions will go the way of all AGWBSSscience mangling, so that AGWScience can continue to teach that cold molecules radiating back from the atmosphere heat the earth and that a cold interior wall of an igloo can cook a lump of meat if left there for a few hours, and the rest. So for the educational record, (the exclamation marks are in the original, this is for children):

    http://science.hq.nasa.gov/kids/imagers/ems/infrared.html

    Infrared light lies between the visible and microwave portions of the electromagnetic spectrum. Infrared light has a range of wavelengths, just like visible light has wavelengths that range from red light to violet. “Near infrared” light is closest in wavelength to visible light and “far infrared” is closer to the microwave region of the electromagnetic spectrum. The longer, far infrared wavelengths are about the size of a pin head and the shorter, near infrared ones are the size of cells, or are microscopic.

    Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared. The temperature-sensitive nerve endings in our skin can detect the difference between inside body temperature and outside skin temperature.

    Infrared light is even used to heat food sometimes – special lamps that emit thermal infrared waves are often used in fast food restaurants!

    Shorter, near infrared waves are not hot at all – in fact you cannot even feel them. These shorter wavelengths are the ones used by your TV’s remote control.”

    Because AGW doesn’t teach this, is why there are so many here who can’t understand convection or explanations by others here who do understand the real Thermodynamics Laws, for example:

    “If it were possible to get energy to flow from cold to hot then we would have all our energy requirements solved. We would have a perpetual motion machine. This is impossible.” (John Marshall March 29 @1:52 am)

    Contrast example: “There is always a net energy flow from a warmer body to a cooler body in accordance with the Laws of Thermodynamics but in fact both bodies still radiate towards each other.

    A cooler body doesn’t stop radiating just because it is in the presence of a warmer body.

    It is the net rate of energy transfer between the two that changes with no need for the cooler body to effect any direct warming of the warmer body.

    the suggestions that the greenhouse effect somehow offends the Laws of Thermodynamics is a non starter and a hindrance to scepticism of the theory of AGW.” (Stephen Wilde March 29 @ 3:45 am)

    The AGWScience energy budget, as Ira et al teach it, says that it is Solar energies which heat the Earth, that is, Visible Light and the two short waves either side of UV and Near Infrared. These are not hot. We cannot feel them. No matter how far Blue Light, for example, penetrates into the ocean, it will not heat it. UV may burn surfaces, as artifically intensified Blue light can also burn, but these do not raise the temperature of matter the way that Thermal IR does.

    Thermal IR heating the Earth is excluded from the AGW Energy Budget.

    Why?

    Where is the proof that Solar Light energies can heat the Earth, raise the temperature of the Earth?

    It’s Thermal IR that can cook food, Near Infrared is not hot.

  220. wayne says:
    March 31, 2011 at 1:01 pm
    “If there was zero radiation from a pure nitrogen atmosphere, heck, lets make it pure argon as dense as ours, no oceans, no clouds, would it be cold or hot? What do you think?”

    I’m still trying to understand what you mean. If there is zero radiation across an entire atmosphere, top to bottom, then it is extremely cold – absolute zero. So it wouldn’t even be an atmosphere, but a solid. I don’t think that is what you mean.

    If there is zero radiation in a particular wavelength, there are multiple explanations. The most obvious is that your atmospheric molecules don’t emit at the target wavelength. Or the atmosphere might be the wrong temperature, and it would radiate the target wavelength if it was hotter/colder.

    You also specify “as dense as ours”, and in a previous comment you specified “same pressure at sea level”. Since the atmosphere will approximately follow ideal gas laws, you have defined a priori the temperature of the atmosphere. Instead of asking a question, you have provided the answer via the assumptions.

    The question is “Given an atmosphere of equal mass to the current atmosphere, but consisting entirely of nitrogen (or argon), what would the density and pressure be at sea level? (and thus, the temperature)” Note that in this question “at sea level” refers to a height of a meter or two above the ground. The average pressure exerted by the atmosphere directly upon the ground’s surface at sea level will be unchanged, assuming the total mass of the atmosphere is the same.

    My hypothesis is that, given an equal mass of just nitrogen (or argon), the atmosphere will be denser at sea level because it will be colder. The atmospheric pressure at sea level (let’s say 2 meters above surface) will be lower, because the percentage mass of the atmosphere located below knee level will have gone up. The height of the top-of-atmosphere will also be lower.

    What I can’t hypothesize is how much denser/colder a pure nitrogen or argon gas atmosphere would be (at sea level) compared to the one we’ve got. And note that the radiation experienced by a person, such as us, standing in direct sunlight would skyrocket during the day and plummet during the night. The surface of the moon, with practically no atmosphere, hits over 115 Celsius at midday and drops below -170 Celsius at night (at it’s equator). Our atmosphere both reflects/absorbs harmful radiation during the day and radiates helpful energy down to the surface at night.

  221. Thanks Steve, interesting viewpoint. Seems almost every parameter of that is known to a close estimate. That just might deserve a quick simulation of such a scenario to see what that would show.

    You and I agree perfectly on the soaring temperatures during the day and frigid nights but I was including much weight on conduction and convection as the prime heater, so much that night would never be able to catch up with limited convection.

    The reason I perked up on that is it’s a perfect example to see what a lapse rate would be in such a scenario. Thanks for your viewpoint again. Very interesting.

  222. The Earth’s atmosphere is greenhouse regulated by the water cycle, it would be 67°C without water.

  223. Tim

    Nasif’s experiments and calculations and those of many others that he references show very much that he is correct. Real life observations would also show that he’s correct in the negligible contribution of CO2. I will accept any other views unreservedly if you would post on that site and show him where he is wrong. You can show I’m wrong easily but that’s different, it doesn’t make him wrong.

    I also believe his point that most scientists, even the best assume that the earth and CO2 molecules radiate as black bodies with an emissivity of 1.0

  224. Joel Shore says:
    March 30, 2011 at 5:47 pm

    Sort of true…although really you are forgetting a couple of things. One is that I believe the albedo of the earth in the absence of clouds is still about 8% or so. The second is that such a colder earth would presumably have more ice on it and would therefore have even a higher albedo than this.

    At any rate, the full statement is that the atmosphere (including clouds) produces two effects: One is about 33 K of warming due to the greenhouse effect and the other is some cooling due to the albedo of the clouds. So, yes, the net effect of the atmosphere is not a full 33 K of greenhouse warming. However, again, the sort of calculations you discuss are under the assumption of no change in surface albedo [actually, you wrongly assumed 0 albedo]…In reality, if you removed CO2 from the atmosphere, not only would you remove a lot of the water vapor too, you would also increase the surface albedo of the earth.

    Joel Shore says:
    March 30, 2011 at 5:55 pm
    Reed –

    On re-reading what you wrote, I realize that you didn’t assume a zero albedo for the earth as I said in my last post but you did assume, as Tim Folkerts points out, that the emissivity is the same in the visible as it is in the far IR…which is indeed not a very good assumption. Emissivities of the earth’s surface in the visible vary widely with the type of surface, whereas emissivities in the far IR tend to be very close to 1 for nearly all surfaces. (I believe I’ve read that the only place that there can be any significant deviations from 1 is in some desert regions.)

    Joel, taking your last statement first. If you’re going to use the rule that the rate radiation leaves a surface is proportional to T^4, then the emissivity must be independent of frequency. A T^4 rule with a frequency dependent emissivity may be a good approximation, but it is NOT theoretically correct. Thus, I contend that the use of a T^4 law implicitly implies a frequency independent emissivity.

    If the emissivity is independent of frequency and an inert surface is placed in a cavity whose walls are at a common temperature, the steady-state temperature of the surface will be equal to that temperature. If you assume the surface is at a single temperature and you use Planck’s graybody radiation law with constant emissivity to compute that temperature, I believe you will find that one minus the albedo must be equal to the emissivity. Thus, if you’re going to use a T^4 radiation rule to compute the rate energy leaves a surface, I believe you are implicitly assuming (a) both the emissivity and the albedo will be frequency independent, and (b) the emissivity will be equal to one minus the albedo.

    I agree, real world materials have emissivities/albedoes that are a function of frequency. And if you want to incorporate them into your temperature analysis, that’s fine. But don’t use the T^4 radiation law without clearly explaining why it is a good approximation. In addition, the rate of energy transfer via conduction is not proportional to T^4, but more nearly proportional to the difference in temperature between the two end surfaces of the conductive object. Since energy is transfered from the Earth surface to the Earth atmosphere via conduction and the Earth’s atmosphere radiates, don’t forget to include these effects as well. Then there’s convection, and evaporation, and … so on. Finally, I don’t know how to apply the T^4 law to radiation from a gas. The T^4 law is used in conjunction with a surface area, and a gas doesn’t have a definable surface. Bottom line, the complete computation of the atmosphere’s effect on Earth surface temperature is waaaaaay beyond my capabilities. I’m not against using simple models, but I want the use of a model to be internally consistent. In particular, I don’t want someone to use both a T^4 emission dependence and a frequency dependent emissivity; and I definitely don’t want someone to disregard atmospheric cooling effecs when estimating the effect an atmosphere has on surface temperature .

    Now to your first point. OK, I forgot to include the fact that a colder Earth would have significantly more ice–thereby increasing the Earth’s albedo. However, (a) I believe my omission is insignificant compared to Wikipedia’s omission that when computing the effect of the atmosphere on Earth surface temperature, Wikipedia completely ignores the fact that it’s the atmosphere that reflects most incoming solar radiation. Wikipedia is eager to give “credit” to atmospheric effects that warm the Earth’s surface, while completely ignoring atmospheric effects that cool the surface of the Earth.

    But even more important than that, (b) the increase in albedo that would accrue with more ice implies a corresponding decrease in emissivity. If the ice doesn’t exist everywhere, it will predominately exist in the polar regions (and I have to believe even Wikipedia would concede that point). As such, the effect of more ice on the surface temperature of a conductive Earth is to raise the Earth’s temperature not lower it. The reason for this is as follows. A one square meter of Earth surface area absorbs energy at a rate that is proportional to the size of the area (one square meter) and the COSINE of the angle between the surface normal and the direction of incoming radiation; but one square meter of Earth surface radiates energy at a rate that is proportional to the area (one square meter) without being modified (attenuated) by a COSINE function. At the poles, the angle between the incoming radiation and the normal to the Earth surface is large. As such, a given increase in albedo will have a smaller effect on absorbed energy than it will have on radiated energy. In particular, the conversion of one square meter of polar sea water to one square meter of ice will increase the albedo of that square meter. However, if that albedo increase causes one Watt less of energy to be absorbed, then for a steady-state Earth the reduction in the rate at which energy is emitted from that one square meter will be greater than one Watt. To counter this net negative change in the energy rate, the surface temperature will have to increase, not decrease.

    I have generated a PDF file with equations that show this to be the case. Specifically, I segmented the Earth into three regions (two polar regions of equal size, and the remainder of the Earth surface, which I call the equatorial region). I assume the albedo of the polar regions is higher than the albedo of the equatorial region–as would be the case for ice in the polar regions and sea water in the equatorial region. For a conductive Earth, the Earth surface temperature will be uniform. This means that the Earth surface will be all ice or all sea water. However, we can hypothesize an Earth whose polar and equatorial regions are comprised of differenct material where the Polar regions have higher albedos than the equatorial region. Since I want to use a T^4 law for radiation, I require (a) that both albedos be frequency independent, and (b) the both emissivities be one minus their corresponding albedo. For this Earth model we ask: “What will be the surface temperature of such an Earth compared to the surface temperature of an Earth with (a) the polar albedo everywhere, and (b) the equatorial albedo everywhere?” What’s interesting is that as long as the equatorial region albedo is not unity, independent of the equatorial region albedo, provided the polar region albedo is greater than the equatorial region albedo, the Earth’s surface temperature with mixed albedos will be higher than a uniform temperature Earth with a single albedo (either the polar region albedo, the equatorial region albedo or some other albedo). If you’re interested, I’ll Email the PDF file to you. You probably don’t have my Email address, and I’m reluctant to post it in an open forum. However, you can either post your Email address in a comment; or like me, if that’s not acceptable to you , you can tell Anthony that I give him permission to forward my E-mail address to you. You can send me an E-mail, and I’d be happy to respond.

  225. Ira – :)

    There’s an English expression, don’t know if it’s both sides of the pond, to get something ar*se about t*t. AGW has this wonderfully absurd comic book science which does that at every point of argument for its survival, it can best be seen as going through the looking glass with Alice as it creates impossible scenarios for our side of the mirror physical properties. Because, for this example, “greenhouse” means only warming in AGW the key role of the Water Cycle in our real greenhouse atmosphere of cooling has to be ditched and there are two outcomes of this.

    The normal physical well known real science facts are no longer understood because not taught by AGW and the greatly contorted explanations devised in an attempt to justify that greenhouse gases only heat produce people who believe impossible things are real and have no concept of actual physical reality.

    So, we end up with zilch AGW understanding of the Law of Thermodynamics and in its place a contorted explanation for the unneccessarily added word “net” in which is proposed non-thermal energies are thermal. As non-thermal Light energies are not radiating heat or creating heat capable of raising temperature (without burning surface, as does UV), then the “net” is irrelevant.

    To the extent that you et al end up believing that Solar energy, non-thermal Light, is thermal, you can believe impossible things like the cold insides of igloos radiate back energy capable of cooking lumps of meat, that back radiation from a source heated by the transfer of energy from a hotter source raises the temperature of the hotter source, etc. For you, generic, the atmosphere has no depth or weight or volume but is instead empty space with molecules zipping around in Brownian motion diffusing as if an imaginary ideal gas quite regardless that these molecules have non-ideal gas weight and volume and can’t do this for real. You miss all the funnies like filling one’s attic with CO2 to warm one’s house, because you, generic, have no sense at all of the real physical world around you.

    The Water Cycle cools the Earth by convection in taking up heat from earth via water vapour which is lighter than air and condensing out at higher colder levels of the atmosphere releasing heat away from the Earth.

    A hot little Carbon Dioxide molecule on the ground swept up in the movement of a hot volume of air, which is wind, will lose its heat as it rises, in this also cooling the Earth heated by the thermal energies from the Sun which are not Solar but Thermal IR. With a heat capacity of even less than Nitrogen and Oxygen, CO2 as quickly gives up its heat as it gets it, practically instantly. A cold CO2 molecule can not heat warmer matter around it because it is itself receiving heat and any heat it has from being in warmer matter, hot air, will be released the moment that air gets cold. As cold air, the volume of fluid gas, is not radiating energy to heat a warmer object (we can tell etc. see NASA description because we have an inbuilt ability to tell the difference between outside and inside body temperature), neither is cold CO2.

    So it’s not enough to say that “CO2 re-radiates in every direction, half downwards” and such, if that radiation is non-thermal it can do nothing to raise the temperature of the Earth.

    So, show me proof that Visible Light from the Sun is absorbed by the Earth and raises its temperature.

    In the real world in real science and not make believe through the looking glass with Alice AGW, we know your claim is nonsense, because we can use real science understanding in this our real physical world: http://www.ehow.com/facts_7868896_fluorescent-light-good-growing.html

    Benefits of Fluorescent bulbs – energy efficient and do not produce much heat, can place bulbs close to plants without risking burn damage, why? Because they produce mainly blue light. (Plants also need red light for growing on).

    Here: http://growlightuk.com/tabid/58/ProductID/17/Default.aspx?gclid=CJ_034CN-6cCFUEa4QodTmjiqA

    Oh what’s up? This lamp produces visible light of blue and red with virtually no heat. Does it somehow extract the heat these lights produce in your claim that Solar light is thermal or in your Solar budget heating the Earth are blue and red lights excluded because these colours do not produce heat? Are these bulbs energy efficient because it takes less energy to produce visible light?

    Come on Ira, everyone, let’s nail this once and for all. What is the physical reality here?

  226. Reed Coray says: March 31, 2011 at 10:15 pm

    “I require (a) that both albedos be frequency independent, and (b) the both emissivities be one minus their corresponding albedo. “

    Your whole set of calculations falls apart with this assumption. Snow has an albedo of close to 1 (for visible light), but also an emissivity of close to 1 (for IR).

    Google snow emissivity and the first two hits say:
    “Water, ice, and snow generally have a high emissivity, 0.94 to 0.99, across the thermal infrared region. Snow is unusual in that it has a high reflectance in the solar (visible) region where most of the downwelling energy is during the day, and a very high emissivity in the thermal region.”

    http://www.comp.glam.ac.uk/pages/staff/pplassma/MedImaging/PROJECTS/IR/CAMTEST/Icewater.htm

    “Snow emissivity is in the range 0.96-1.00 for all grain sizes and viewing angles. ”

    http://astrogeology.usgs.gov/Projects/PlanetaryIcesWorkshop/abstracts/swarren.html

  227. Reed Coray says:

    Thus, if you’re going to use a T^4 radiation rule to compute the rate energy leaves a surface, I believe you are implicitly assuming (a) both the emissivity and the albedo will be frequency independent, and (b) the emissivity will be equal to one minus the albedo.

    Not really. The conditions under which the T^4 rule will hold to a good approximation for not-too-large fractional changes in T (remembering that T is an absolute temperature scale!) are much weaker than the conditions under which you can assume that the emissivity in the far IR is equal to one minus the albedo in the visible. In particular, the T^4 rule (for relatively small changes in T) requires just that the emissivity doesn’t vary too rapidly with wavelength over the wavelength regime for which emission in that temperature regime is significant. Assuming that emissivity in the far IR is equal to one minus albedo in the visible assumes that the emissivity doesn’t change much when the wavelength varies by an order of magnitude or more!

    However, (a) I believe my omission is insignificant compared to Wikipedia’s omission that when computing the effect of the atmosphere on Earth surface temperature, Wikipedia completely ignores the fact that it’s the atmosphere that reflects most incoming solar radiation.

    I don’t think Wikipedia claims to be computing the total effect of the atmosphere on the Earth’s surface temperature. The computation is of the total greenhouse effect due to the atmosphere. I.e., we can divide the atmosphere’s effects into different contributions, rather than just looking at its net effect.

    But even more important than that, (b) the increase in albedo that would accrue with more ice implies a corresponding decrease in emissivity…

    Basically, everything beyond this point in your post relies on an assumption that the emissivity of ice is significantly different from 1 in the far infrared (and, in particular, that the reduction in emissivity is comparable to the albedo of ice in the visible). Here is the data of infrared emissivities showing that this assumption is incorrect: http://www.comp.glam.ac.uk/pages/staff/pplassma/MedImaging/PROJECTS/IR/CAMTEST/Icewater.htm (For comparison, here are some albedos for ice and snow: http://www.arcticice.org/albedos.htm )

  228. “So, show me proof that Visible Light from the Sun is absorbed by the Earth and raises its temperature.”

    Since you seem to not trust the most basic theories of light like E = hf, then lets just use a simple experiment.
    Dark clothing will make you warmer in sunlight than light clothing.
    –> dark clothing absorbs more energy from the sunlight
    The emissivity of cloth doesn’t change much with color.
    –> any color of clothing will absorb similar amounts of IR
    Sunlight consists almost entirely of visible light & IR.
    –> The visible light must be what makes the difference.

    “Oh what’s up? This lamp produces visible light of blue and red with virtually no heat.”
    My gracious! These are LED lights! They use a completely different process to produce light! Since they are not hot glowing objects, there is absolutely no reason to expect anything closer to a black body curve for the light produced!

  229. Joel Shore said:
    I and many others have endeavored to explain it to you. If you still cannot understand it, I don’t think it is all of our faults.

    Can you give a quantifiable, verifiable explanation of it then, please?

  230. martin mason:

    I also believe his point that most scientists, even the best assume that the earth and CO2 molecules radiate as black bodies with an emissivity of 1.0

    You believe his point based on what exactly? How extensively have you studied the literature or even a basic textbook discussing radiative transfer in the atmosphere?

    In fact, let’s test if that belief even makes sense. Can you define for me what it means to assume that CO2 molecules radiate as black bodies with an emissivity of 1.0 because frankly I am a bit confused on the concept being that emissivities is a macroscopically-defined quantity that doesn’t, as far as I know, even make sense to talk about on the scale of individual molecules. I think you have to talk about things like the optical depth of a certain thickness of atmosphere, but hey, if you know differently I’d be glad to hear about it!

  231. Myrrh says:
    April 1, 2011 at 3:43 am
    Come on Ira, everyone, let’s nail this once and for all. What is the physical reality here?

    Certainly not the crap you keep churning out!
    Anthony, this is the top Science blog?
    And we have to keep reading the nonsense this guy produces!
    The loss of temperature by an ascending gas volume is adiabatic i.e. no loss of heat!
    CO2 has a higher molar heat capacity than N2 or O2, ~28 vs ~20.
    Not to mention all his cockamamie ideas about ‘thermal’ vs ‘non-thermal’ radiation which have been rebutted here ad nauseam and his belief that the gas laws don’t apply!

  232. Joel Shore says:
    March 31, 2011 at 1:45 pm

    Yes, when you pressurize gas, it warms. However, if you pressurized some gas and then let it sit in equilibrium with its surroundings for billions of years, it will equilibrate with its surroundings. The earth’s atmosphere is not undergoing continual gravitational collapse.

    The atmosphere is constantly receiving energy from the surface via conduction, convection and latent heat via the phase changes of water. It hasn’t been in equilibrium with it’s surroundings for billions of years.

    To use an analogy the contents of a compressed gas cylinder will stay hot if you light a fire under the cylinder.

  233. wayne says:
    March 31, 2011 at 1:01 pm
    But I still would ask you on that hypothetical situation, if I could: If there was zero radiation from a pure nitrogen atmosphere, heck, lets make it pure argon as dense as ours, no oceans, no clouds, would it be cold or hot? What do you think?

    It would depend only on the reflectivity, absorptivity and emissivity of the surface and the distance from the sun.

  234. Myrrh;
    Come on Ira, everyone, let’s nail this once and for all. What is the physical reality here?>>>
    Phil;
    Anthony, this is the top Science blog?
    And we have to keep reading the nonsense this guy produces!>>>

    Sad is it not? The price of a forum in which contrary opinions are not suppressed is that people who don’t know what they don’t know get to distract and confuse everyone else with mind boggling mirepresentations of science. Sadder still, it is a mirror unto the real world where the contrary opinions have become the science and the scientists are accused of misrepresentation.

    Any sufficiently advanced magic is indistinguishable from science.

  235. Max Hugoson says:
    March 29, 2011 at 6:26 pm
    There is a distribution of energies of the ensemble of molecules. There is EXCHANGE. Therefore, collisions between O2 and N2 molecules on the “high end” of the distributed energy, LOSE that translational kinetic energy to CO2 and H2O molecules.

    WHICH, can…with significant probability, re-emit an IR photon.

    Yes but that probability is lower if the energy is acquired collisionally than if acquired radiatively. When excited radiatively the appropriate vibrational energy level is filled and is then capable of emitting however collisions reduce the lifetime of the state so that emission has a low probability. When excited collisionally that energy is unlikely to solely excite the vibrational level in fact it’s more likely to excite the translational energy of the CO2 molecule which has zero probability of emitting a photon.

  236. martin mason says:
    March 31, 2011 at 8:31 pm
    Tim

    Nasif’s experiments and calculations and those of many others that he references show very much that he is correct. Real life observations would also show that he’s correct in the negligible contribution of CO2. I will accept any other views unreservedly if you would post on that site and show him where he is wrong. You can show I’m wrong easily but that’s different, it doesn’t make him wrong.

    You can’t show Nasif he’s wrong, the reason he’s banned on here as I recall is because of his abusive attacks on other posters, Leif Svalgard for example.

    I also believe his point that most scientists, even the best assume that the earth and CO2 molecules radiate as black bodies with an emissivity of 1.0

    Believe it if you like but any physical scientist who assumes that gaseous CO2 molecules radiate as a black body with an emissivity of 1.o should have their degree rescinded.

  237. bananabender says:

    The atmosphere is constantly receiving energy from the surface via conduction, convection and latent heat via the phase changes of water. It hasn’t been in equilibrium with it’s surroundings for billions of years.

    To use an analogy the contents of a compressed gas cylinder will stay hot if you light a fire under the cylinder.

    Okay, so what source of energy do you propose is being ignored when it is concluded vi radiative balance that the earth’s surface temperature in the absence of an IR-absorbing atmosphere (with the same total albedo as present) would be approximately 255 K?

  238. Tim Folkerts says:
    April 1, 2011 at 3:59 am
    Reed Coray says: March 31, 2011 at 10:15 pm

    “I require (a) that both albedos be frequency independent, and (b) the both emissivities be one minus their corresponding albedo. “

    Your whole set of calculations falls apart with this assumption. Snow has an albedo of close to 1 (for visible light), but also an emissivity of close to 1 (for IR).

    Google snow emissivity and the first two hits say:
    “Water, ice, and snow generally have a high emissivity, 0.94 to 0.99, across the thermal infrared region. Snow is unusual in that it has a high reflectance in the solar (visible) region where most of the downwelling energy is during the day, and a very high emissivity in the thermal region.”

    http://www.comp.glam.ac.uk/pages/staff/pplassma/MedImaging/PROJECTS/IR/CAMTEST/Icewater.htm

    “Snow emissivity is in the range 0.96-1.00 for all grain sizes and viewing angles. ”

    http://astrogeology.usgs.gov/Projects/PlanetaryIcesWorkshop/abstracts/swarren.html

    Tim,

    My “whole set of computations” does NOT fall apart with these assumptions. Like Wikipedia, when computing the surface temperature of the Earth without an atmosphere, I assume a model. I admit my model does not match the real world–no model does; but I believe my assumptions are internally consistent and don’t violate any laws of thermodynamics. Do you believe (a) my model is internally inconsistent, and/or (b) violates a law of thermodynamics?

    Relative to the real world, I can make the argument that Joel’s and I guess your argument also “falls apart”. Specifically, how do you justify the T^4 rule? How do you justify using cavity radiation into a medium containing matter (i.e., an atmosphere)? How do you justify ignoring conduction and convection? How do you justify ignoring energy transfer via state change (water to vapor), movement of that vapor to various altitudes, radiation from that vapor, etc?

    Let me ask you. (1) Does the T^4 radiation law require an emissivity that is independent of frequency? If not, what is the frequency dependence that produces a T^4 law? (2) Will an inert surface, independent of the material that comprises the surface, when placed in the vacuum of a cavity whose walls are at a uniform temperature T, eventually attain the temperature T? (3) If you apply Planck’s graybody radiation law to that surface, will the rate of energy reception equal the rate of energy emission for any temperature other than T?

    But all my arguments that question the applicability of Wikipedia’s model pale when compared to my criticism of Wikipedia’s logic. To compute the effects of the atmosphere on Earth surface temperature, Wikipedia chose to assume for emission purposes a blackbody Earth, and for reception purposes assume an albedo that is a composite of atmosphere and surface. Wikipedia then attributes the difference solely to the atmosphere. I can’t buy this. For example, suppose the Earth’s surface was lampblack (not a perfect absorber, but close) with a highly conductive (thermally) interior. For such an Earth, without an atmosphere the surface temperature, T1, would come close to that of a blackbody in space, which I believe is approximately 278 K. To this surface add an atmosphere of clouds. The clouds reflect 30% of the incoming solar radiation. The lampblack surface will now “see” less solar energy. We compute the steady-state temperature, T2, of a lampblack Earth using the reduced radiation input value. We measure the surface temperature T3 in the presence of the clouds. We then make the claim that the atmosphere is responsible for increasing the surface temperature by the difference between of the T3 and T2. This is nonsense. Without the atmosphere the surface temperature was T1. With the atmosphere the surface temperature is T3. By any reasonable way of thinking, the atmospheric effect on the surface temperature is the difference between T3 and T1. How you come up with a model for T1 is open to disucssion. But whatever model you use, it should NOT contain an atmosphere.

    Joel Shore says:
    April 1, 2011 at 4:07 am

    I don’t think Wikipedia claims to be computing the total effect of the atmosphere on the Earth’s surface temperature. The computation is of the total greenhouse effect due to the atmosphere. I.e., we can divide the atmosphere’s effects into different contributions, rather than just looking at its net effect.

    I insert here a quote from Dr. Glickstein who quoted Wikipedia:

    Ira Glickstein, PhD says:
    March 30, 2011 at 8:13 am

    from Wikipedia

    If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C. However, since the Earth reflects about 30% (or 28%) of the incoming sunlight, the planet’s effective temperature (the temperature of a blackbody that would emit the same amount of radiation) is about −18 or −19 °C, about 33°C below the actual surface temperature of about 14 °C or 15 °C. The mechanism that produces this difference between the actual surface temperature and the effective temperature is due to the atmosphere and is known as the greenhouse effect.

    I don’t know how you interpret those words, but I interpret them as follows: (1) the atmosphere is responsible for the difference between the actual surface temperature and the “effective” surface temperature. (2) The “effective” surface temperature is based on the temperature of a ideally thermally conductive blackbody radiator in radiation rate equilibrium with the solar power density at the the top of the Earth’s atmosphere attenuated by 30% (28%). And (3) the name given to this effect (temperature difference) is the “greenhouse effect.” Based on Wikipedia’s words, I don’t see how you can argue that “…we can divide the atmosphere’s effects into different contributions, rather than just looking at its net effect“, and claim that Wikipedia meant to assign the name the “greenhouse effect” to one of those “contributions”.

  239. BigWaveDave says:

    Can you give a quantifiable, verifiable explanation of it then, please?

    Well, I don’t know what you are asking for that we haven’t already given or isn’t available in the literature. The atmospheric greenhouse effect can be thought of at a variety of different levels of detail, from simple toy “shell” models to detailed radiative-convective calculations. Ira Glickstein, Willis Eschenbach, Roy Spencer, and others have spent whole posts here explaining the effect, and even though they and I agree on rather little in regards to AGW, we are all in agreement on this aspect of the science.

    What about it exactly is tripping you up? Or maybe, to put it a different way, I said:

    However, in the presence of an IR-absorbing atmosphere, some of the terrestrial radiation is absorbed and when the atmosphere subsequently emits radiation, some of this radiation comes back to the earth. This keeps the earth warmer than it would be in the absence of the IR-absorbing atmosphere.

    Your response was that this violates the 2nd Law of Thermodynamics. So, explain to me exactly why you think it does. Do you believe it impossible for a colder object to even emit radiation toward a warmer object? Or, do you believe that it can but the warmer object won’t absorb any of the radiation? Or, do you believe that the warmer object can absorb the radiation but that it won’t cause it to have any additional thermal energy than it would have had otherwise?

    I am very confused about exactly what aspect of this process you believe violates the 2nd Law.

  240. Reed Coray says:

    Relative to the real world, I can make the argument that Joel’s and I guess your argument also “falls apart”. Specifically, how do you justify the T^4 rule? How do you justify using cavity radiation into a medium containing matter (i.e., an atmosphere)? How do you justify ignoring conduction and convection? How do you justify ignoring energy transfer via state change (water to vapor), movement of that vapor to various altitudes, radiation from that vapor, etc?

    Sorry…This is a false equivalence for a couple of reasons. First, all models or theoretical constructs of any kind involve making assumptions and approximations. However, not all such assumptions and approximations are equally good. What Tim and I have shown is that you made an assumption that is clearly not any good at all and your conclusions will quite clearly fail to apply to the real world because of it.

    By contrast, I gave you reason to believe that in many cases, it is in fact quite good approximation to assume the T^4 dependence holds. Heck, the dependence of radiative emission on temperature is so strong that at ANY frequency, the emission a that frequency is a monotonically increasing function of temperature. (People often have the picture of a shifting curve of emission power because you often look at normalized emission curves…but, in fact, it is a curve that grows rapidly with temperature that all of it lies above the corresponding curve for any lower temperature.)

    The second reason is that many of the issues that you mention are actually addressed by more complete models. So, yes, it is true that very simple models of the greenhouse effect neglect convection, assume simple emissive properties, etc. However, the justification for these assumptions not mattering to the final answer is that the more complex models give the same results at least for the things that we are using the simpler models to illustrate.

    That’s why I emphasize the importance of a hierarchy of models: A very complex model without simple models is simply a “black box” that gives very little insight to what the basic physics is. A very simple model without more complex models to back it up causes one to be unsure if the results shown by the model apply to the real world or are just an artifact of the approximations. However, if you take them all together, you can use the more complex models to back up what you find with the simple ones and then use the simple ones to get a simple picture of what is going on.

    I don’t know how you interpret those words, but I interpret them as follows: …

    When I read someone else’s work, I try interpret it in the most charitable way…i.e., the way that actually makes sense (assuming I can figure out a way in which it does make sense…which isn’t true, for example, for the work of Gerlich and Tscheuschner). If one chooses an interpretation that doesn’t make sense and then complains bitterly about it, I don’t really see what that accomplishes. Sure, maybe one could argue that the wording in that Wikipedia article could have been a little clearer. But, I don’t really see the need to make a federal case out of it. One just clarifies what they meant…that they were specifically talking about the magnitude of the atmospheric greenhouse effect, i.e., the magnitude of the effect due to the absorbance of the atmosphere in the far infrared part of the spectrum…and moves on.

  241. Reed,

    You certainly are welcome to do your computations and to see the effects of your model. I was perhaps a little harsh to say “it all falls apart”. But when you make an assumption that is almost completely the opposite of reality (assuming emissivity of snow ~0 when indeed is it ~1) then your conclusions will almost certainly be poor.

    To answer a few specific questions:
    How do you justify using cavity radiation into a medium containing matter (i.e.,
    an atmosphere)?

    While BB radiation was originally derived using a cavity, it has been shown that the same curve also works very well for other circumstances — ie some surfaces emit practically the same curve from a flat surface ie their emissivity is ~ 1 over the frequencies of interest. Experimentally, much of the surface acts much like a black body for IR wavelengths.
    The material into which the BB radiation travels does not matter — only the temperature of he emitting surface, so the properties of the atmosphere are not important here.

    This would seem to justify treating the surface as (approximately) a BB wrt IR.

    How do you justify ignoring conduction and convection? How do you justify ignoring energy transfer via state change (water to vapor), movement of that vapor to various altitudes, radiation from that vapor, etc?

    This is a little more subtle.

    Does the T^4 radiation law require an emissivity that is independent of frequency? If not, what is the frequency dependence that produces a T^4 law?

    As long as the emissivity is independent of the frequencies where (a significant amount of) energy is emitted, the T^4 law is (approximately) hold.
    For example, this table http://www.monarchserver.com/TableofEmissivity.pdf suggest that type 301 Stainless has an emissivity of 0.27 @ 24 C and 0.57 @232 C. This would mean that the IR power emitted by this surface would go up slightly FASTER than T^4 as the temperature rose from 297 K to 505 K.
    On the other hand, cuprous oxide decreases in emissivity as temperature increase, so it would be a little less than T^4

    Will an inert surface, independent of the material that comprises the surface, when placed in the vacuum of a cavity whose walls are at a uniform temperature T, eventually attain the temperature T?
    Yes. I’ve even seen it. I used to work with furnaces operating at > 1000 C. The uniformly heated walls and the small window in the door very closely approximate a cavity radiator. If you slide a sample in, everything – metal, ceramic, quartz – soon glows the same color.

    (3) If you apply Planck’s graybody radiation law to that surface, will the rate of energy reception equal the rate of energy emission for any temperature other than T?
    No. If the object is hotter than the chamber, is will cool; if the object is cooler it will warm up. Always.

    By any reasonable way of thinking, the atmospheric effect on the surface temperature is the difference between T3 and T1.

    I agree that there are many ways to interpret the effect of the atmosphere. And I agree with your calculations:
    * a=0; e=1 –> T1 = 278.7 (ideal BB)
    * a=0.3; e=1 perfect BB but clouds increase albedo to 0.3 –> T2 = 254.9
    * as above, but add GHG –> observed temperature T3 = 288

    So ONE EFFECT of the atmosphere (cloud cover) is to lower the temperature ~ – 23 K
    A DIFFERENT EFFECT of the atmosphere (known as the greenhouse effect) is to raise the temp ~ +33K
    The NET effect of the atmosphere is +10 K

    (And actually, the clouds are not the only thing reflecting visible light; this image from Wikipedia http://en.wikipedia.org/wiki/File:Albedo-e_hg.svg suggests that the earth as a whole reflects ~10-15 % already, so the clouds themselves only are responsible for PART of the cooling. This would leave maybe -10 K from the clouds and + 20 from the atmosphere as a whole.)

    So by a slightly different reasonable way of thinking, the atmospheric effect [specifically known as the green house effect that wikipedia is discussing] on the surface temperature is the difference between T2 and T1.

    Maybe we should go edit that page. :-)

  242. Bangs head against wall.

    Please listen to my question, as follows:

    Will you give me proof that Visible light from the Sun heats the Earth?

    Because until you do, you can keep on claiming it does and you can keep on describing the mechanism you say is doing this, but it remains an imaginary, unsubstantiated claim until you prove it. Just do it.

    And stop giving me daft meaningless ‘examples’ which mean nuttin. And which I’m now tired of explaining mean nuttin.

    You may well think that the knowledge you have about this is ‘standard’, but it is different from that which was previously taught.

    Since you are now teaching something different it is incumbent on you to provide proof that previous knowledge was wrong.

    Stop avoiding actually doing this.

    I realise, I’ve been looking at these arguments for some time now, that some might have difficulty believing there ever was any other understanding about this, because it’s very difficult to find any who admit it now with AGWScience takeover.

    However, after spending some considerable hours trying to find for myself the proof I’m asking you to provide, and unable to find it, I found this:

    “Many physics teachers traditionally attribute all the heat from the Sun to infrared light. This is inexact – visible light from the Sun accounts for 50 percent of the heating, and electromagnetic waves of any frequency will have a detectable heating effect if they are intense enough.” http://www.newworldencyclopedia.org/entry/infrared

    So let’s look at that together. Traditionally still holds good until you prove differently.

    The NASA page that I linked to is this same Traditional physics teaching.

    Please re-read what it says, I posted it above. It says that the heat we feel from the Sun is Thermal IR.

    Prove it is wrong, or shut the hell up about claiming you operate in the discipline of Science.

    Because until you actually show that the heat we feel from the Sun comes from Visible Light and is not the Thermal IR as by Standard Science Tradition, you are talking a load of bllcks.

    And therefore, your claim that 50% of heating comes from Visible Light is without foundation of proven fact.

    And, still looking at the extract I posted, this is followed by a statement which I’ve said often enough here, “electromagnetic waves of any frequency will have a detectable heating effect if they are intense enough.”

    For example, when I’ve explained that Visible Light is not thermal, but if concentrated in intensity it can burn.

    However, this scientific actual fact does not relate back to the claim in the previous firspart of the sentence. I hope you can see that.

    It is not confirming that “visible light from the Sun accounts for 50 percent of the heating.” I hope you can see that.

    Now, give me what I’ve asked for. This is a science blog.

  243. Bangs head against wall even harder.

    Perhaps you can tell me what would count as “proof” to you???

    ** I have given an example whereby sunlight raises the temperature of PART of the earth (a piece of cloth, but it holds as well for rocks of different colors or wood of different colors or cars of different colors). By extension, light could heat other parts of the earth (unless you can specifically show either that 1) dark objects are not warmer than light objects or 2) the earth absorbs lights in a way that is fundamentally different than the way cloth and rock and cars and wood absorb light)

    ** I have quoted the simplest possible equation for the energy of light: E=hf. Visible light has a frequency. Visible light has energy. The earth absorbs that energy (other than a bit which is reflected) Absorbed energy raise the temperature of objects. I even have mentioned several times that ~40% of the energy is visible and ~50 is IR.

    ** AND NOW — (don’t ask me why I even bothered, but) — I took a concave mirror ~ 0.4 m in diameter. The mirror is covered with glass, so little IR should get thru. The light from the mirror was focused to a fairly small area. I put my hand there and
    … PAUSE FOR A DRAMATIC DRUM ROLL ….
    my hand got warm (quite hot in fact if I got too close to the focal point)!

    I even put an extra pane of glass there to be doubly sure the IR would get absorbed. The light was passing thru 4 layers of glass. My hand STILL got warm.

    (And for that matter, since glass blocks most IR, then even a magnifying glass would have to be mostly focusing just the visible light, not the IR to char wood or burn ants).

    “You may well think that the knowledge you have about this is ‘standard’, but it is different from that which was previously taught. ”
    Yes, What I am discussing IS different from what was previously taught — IF YOU LEARNED PHYSICS before Einstein developed the idea of photons!

    “Traditionally still holds good until you prove differently. ”
    1) No — tradition is not what determines correct science — especially traditions among non-experts. Simple explanations that are traditionally presented by non-experts are often wrong. I’ve heard lots of poor explanations of the green house effect. And tides. And wings. And friction. And the 2nd Law of thermodynamics.
    2) Even so, NOT A SINGLE physics professor I have talked to has or would attribute all the sun’s heating to IR. Find the name of one physics professor who will go on record supporting your view.
    3) Even the source you quote says the tradition is wrong!

  244. Tim Folkerts says:
    April 1, 2011 at 1:42 pm
    Reed,

    You certainly are welcome to do your computations and to see the effects of your model. I was perhaps a little harsh to say “it all falls apart”. But when you make an assumption that is almost completely the opposite of reality (assuming emissivity of snow ~0 when indeed is it ~1) then your conclusions will almost certainly be poor.

    To answer a few specific questions:
    How do you justify using cavity radiation into a medium containing matter (i.e.,
    an atmosphere)?
    While BB radiation was originally derived using a cavity, it has been shown that the same curve also works very well for other circumstances — ie some surfaces emit practically the same curve from a flat surface ie their emissivity is ~ 1 over the frequencies of interest. Experimentally, much of the surface acts much like a black body for IR wavelengths.
    The material into which the BB radiation travels does not matter — only the temperature of he emitting surface, so the properties of the atmosphere are not important here.

    This would seem to justify treating the surface as (approximately) a BB wrt IR.

    How do you justify ignoring conduction and convection? How do you justify ignoring energy transfer via state change (water to vapor), movement of that vapor to various altitudes, radiation from that vapor, etc?

    This is a little more subtle.

    Does the T^4 radiation law require an emissivity that is independent of frequency? If not, what is the frequency dependence that produces a T^4 law?

    As long as the emissivity is independent of the frequencies where (a significant amount of) energy is emitted, the T^4 law is (approximately) hold.
    For example, this table http://www.monarchserver.com/TableofEmissivity.pdf suggest that type 301 Stainless has an emissivity of 0.27 @ 24 C and 0.57 @232 C. This would mean that the IR power emitted by this surface would go up slightly FASTER than T^4 as the temperature rose from 297 K to 505 K.
    On the other hand, cuprous oxide decreases in emissivity as temperature increase, so it would be a little less than T^4

    Will an inert surface, independent of the material that comprises the surface, when placed in the vacuum of a cavity whose walls are at a uniform temperature T, eventually attain the temperature T?
    Yes. I’ve even seen it. I used to work with furnaces operating at > 1000 C. The uniformly heated walls and the small window in the door very closely approximate a cavity radiator. If you slide a sample in, everything – metal, ceramic, quartz – soon glows the same color.

    (3) If you apply Planck’s graybody radiation law to that surface, will the rate of energy reception equal the rate of energy emission for any temperature other than T?
    No. If the object is hotter than the chamber, is will cool; if the object is cooler it will warm up. Always.

    By any reasonable way of thinking, the atmospheric effect on the surface temperature is the difference between T3 and T1.

    I agree that there are many ways to interpret the effect of the atmosphere. And I agree with your calculations:
    * a=0; e=1 –> T1 = 278.7 (ideal BB)
    * a=0.3; e=1 perfect BB but clouds increase albedo to 0.3 –> T2 = 254.9
    * as above, but add GHG –> observed temperature T3 = 288

    So ONE EFFECT of the atmosphere (cloud cover) is to lower the temperature ~ – 23 K
    A DIFFERENT EFFECT of the atmosphere (known as the greenhouse effect) is to raise the temp ~ +33K
    The NET effect of the atmosphere is +10 K

    (And actually, the clouds are not the only thing reflecting visible light; this image from Wikipedia http://en.wikipedia.org/wiki/File:Albedo-e_hg.svg suggests that the earth as a whole reflects ~10-15 % already, so the clouds themselves only are responsible for PART of the cooling. This would leave maybe -10 K from the clouds and + 20 from the atmosphere as a whole.)

    So by a slightly different reasonable way of thinking, the atmospheric effect [specifically known as the green house effect that wikipedia is discussing] on the surface temperature is the difference between T2 and T1.

    Maybe we should go edit that page. :-)

    Tim, thank you for your response. I agree with almost everything you said. My using an albedo that is one minus the emissivity is very likely a poor representation of reality, and therefore is likely to give poor conclusions. I wouldn’t quite say it that way. Rather, I’d say it this way: “My conclusions are valid, they just have little resemblance to the real world.” I’ll concede the point–my conclusions may have little relevance to the real world. It’s just that over time I became frustrated and irritated with what I considered to an illogical argument to make the claim that the atmosphere has a 33 degree warming effect on the temperature of the surface of the Earth.

    Regarding your point (3) above. I think there is some confusion. I also believe the temperature of the surface will trend to and eventually become the temperature of the cavity walls. If the inserted object’s temperature is higher (lower) than the wall temperature, the object’s temperature will decrease (increase). What I also believe to be true that when applying a graybody version of Planck’s blackbody law, the only temperature that will produce radiation rate equilibrium is the temperature of the cavity walls. Am I missing something?

    Now a new question, when integrated over all frequencies, the frequency-dependent portion of Planck’s blackbody radiation law, (f^3) / {e^[h*f/(k*T)] – 1}, not only produces a T^4 dependence, it also produces a multiplicative factor equal to (pi^4) * (k^4) / [15 * (h^4)], where k is Boltzmann’s constant and h is Planck’s constant. Planck’s blackbody radiation law in its integral form contains the multiplicative factors 2 * h / c^2, where c is the velocity of light in a vacuum. Combining these factors, one gets a compositve factor of
    2 * (pi^4) * (k^4) / [ 15 * (c^2) * (h^3)].
    pi times the value of this composite factor is called sigma, the Stefan-Boltzmann constant. The question I have, is to what degree does integration using a frequency-dependent emissivity affect the value of the composite factor, so that if the T^4 rule applies, can sigma still be used?

    As far as editing Wikipedia, I’ll leave that to you. ;) When it comes to anything to do with global warming, from what I’ve been told most Wikipedia editions that contradict the idea that global warming is real, manmade, and catastrophic don’t last long.

    Joel Shore says:
    April 1, 2011 at 12:15 pm
    Reed Coray says:

    I don’t know how you interpret those words, but I interpret them as follows: …

    When I read someone else’s work, I try interpret it in the most charitable way…i.e., the way that actually makes sense (assuming I can figure out a way in which it does make sense…which isn’t true, for example, for the work of Gerlich and Tscheuschner). If one chooses an interpretation that doesn’t make sense and then complains bitterly about it, I don’t really see what that accomplishes. Sure, maybe one could argue that the wording in that Wikipedia article could have been a little clearer. But, I don’t really see the need to make a federal case out of it. One just clarifies what they meant…that they were specifically talking about the magnitude of the atmospheric greenhouse effect, i.e., the magnitude of the effect due to the absorbance of the atmosphere in the far infrared part of the spectrum…and moves on.

    You elect to interpret scientific statements in a “charitable way.” That’s fine. I don’t. I don’t mind, in fact I believe in treating the statement’s author in a charitable manner. However, I choose to interpret scientific statements as written using my best understanding of the English language. I decline to defend/support/promulgate a logically-flawed scientific or mathematical argument because there is a “charitable” but illogical way to interpret what it claims.

  245. Joel Shore asks:

    Your response was that this violates the 2nd Law of Thermodynamics. So, explain to me exactly why you think it does. Do you believe it impossible for a colder object to even emit radiation toward a warmer object? Or, do you believe that it can but the warmer object won’t absorb any of the radiation? Or, do you believe that the warmer object can absorb the radiation but that it won’t cause it to have any additional thermal energy than it would have had otherwise?

    When you open a freezer door, can you feel all of that the 270°K (or so) warmth beaming at you? How long do you have to stand there before you get too hot?

  246. “When you open a freezer door, can you feel all of that the 270°K (or so) warmth beaming at you? How long do you have to stand there before you get too hot?”

    Compared to what, a freezer full of liquid nitrogen (77 °K)? Neither warms you up, but one will cool you off faster, and to a lower temperature, than the other. And if it’s a freezer door that opens a magic portal into the cold of space, you’re going to cool off REALLY fast.

  247. Joel says;

    ….. the way that actually makes sense (assuming I can figure out a way in which it does make sense…which isn’t true, for example, for the work of Gerlich and Tscheuschner)”..

    My reply;

    Its refreshingly honest for you to admit that you did not understand the work of G&T.

    As I’ve said to you in a previous post I dont think you even read the G&T paper.
    Perhaps you were asked to sign up in a political sense and your name appears as a of gesture of solidarity with Halpern.
    Anyone who has been anywhere near a thermodynamics book knows G&T are correct.

    I have made a point of asking people with a background in physics to find ANY mistake in the G&T paper.
    So far no mistakes have been found.
    Quite eminent physicists have agreed that there are no mistakes.
    A number have indicated in particular that their analysis of models and the difficulty of a climate solution for Navier Stokes equation is well founded.

    I have collected here a list of the relevant material.

    I am going on a short holiday so I will not be able to respond to any question you might raise.
    However after a week or so I expect you to have reviewed the material presented here.
    You seem to frequent WUWT and I will I hope you will be able to discuss in depth any faults you can find in the G&T paper.

    1] “Falsification Of the atmospheric CO2 greenhouse effects within the frame Of Physics” by Gerhard Gerlich and Ralf D. Tscheuschner; International Journal of Modern Physics B, Vol. 23, No. 3 (2009) pages 275-364.

    http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf

    [2] “Proof of the atmospheric greenhouse effect” by Arthur P. Smith; arXiv:0802.4324v1 [physics.ao-ph]

    http://arxiv.org/PS_cache/arxiv/pdf/0802/0802.4324v1.pdf

    In this paper Arthur Smith defends the current IPCC position and has the merit of taking issue with G&T for something that they did say.

    [3] “Comments on the “Proof of the atmospheric greenhouse effect” by Arthur P. Smith” by Gerhard Kramm, Ralph Dlugi, and Michael Zelger; arXiv:0904.2767v3 [physics.ao-ph]

    http://arxiv.org/ftp/arxiv/papers/0904/0904.2767.pdf

    Takes issue with Arthur Smith

    [4] 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.

    This must be the most embarrassing paper in history as it attacks G&T for things they didn’t say.
    Joel you must have a copy of the published comment

    [5] “Reply to ‘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” by Gerhard Gerlich and Ralf D. Tscheuschner, International Journal of Modern Physics B, Vol. 24, No. 10 (2010) pages 1333–1359.

    http://www.skyfall.fr/wp-content/Gerlich-reply-to-Halpern.pdf

    G&Ts reply to the absurd [4]

    (6.)
    Gerhard Kramm and others with a broader look at current climate science including the “greenhouse effect”

    http://www.benthamscience.com/open/toascj/articles/V004/137TOASCJ.pdf

  248. “Will you give me proof that Visible light from the Sun heats the Earth? Because until you do, you can keep on claiming it does and you can keep on describing the mechanism you say is doing this, but it remains an imaginary, unsubstantiated claim until you prove it. Just do it.”

    You mean AT ALL? OK, look up on a clear day at noon, and then look down. Is the ground beneath you as bright as the sun? No? OK, so it isn’t reflecting all of the incoming light. So where did the energy of the absorbed light go?

    That’s just the ground, which admittedly reflects so much visible light (unless you’re standing on asphalt) that I bet it is primarily heated by infrared wavelengths, which it reflects very little of. (I’m just guessing on that – ask a physicist) In terms of visible light significantly impacting the earth’s energy budget, repeat the experiment over an ocean, and keep in mind that this is the situation over most of the earth’s surface. Most of the visible light is absorbed by the upper layers of the oceans. Water has a high heat capacity, and the absorbed energy is distributed over a huge depth, so a diver isn’t going to feel a temperature increase at noon on a cloudy day just because the clouds break. The oceans are the primary heat reservoir of the energy of absorbed visible light from the sun – not the ground, and not the atmosphere.

  249. @ Steve

    You might find this intriguing. I located a paper related to the confusion of radiation of a pure argon atmosphere but guess what, article not available. Arggg.

    http://adsabs.harvard.edu/abs/2009APS..TSF.C3015B

    Under what conditions do accelerating charges radiate? An examination of recent literature

    Butterworth, Edward; Cox, Paul

    American Physical Society, Joint Fall 2009 Meeting of the Texas Sections of the APS, AAPT, and SPS Postdeadline, October 22-24, 2009, abstract #C3.015

    The process by which accelerated charges emit electromagnetic radiation remains surprisingly obscure: even at the advanced level, most textbooks do not treat it in detail, and published reports show a wide variety of descriptions of the process, some of which have led to paradoxes.

    Against published claims that uniformly accelerated charges do not radiate, Boulware (1980) and de Almeida & Suu (2006) propose that they do, but into a region of spacetime inaccessible to a comoving observer.

    Piazzese (2003) obtains the result that charges in uniform circular motion do not radiate, subject to particular constraints that limit orbital size; with the result that electrons in Bohr orbits do not radiate, while synchrotron radiation is allowed.

    No wonder I am confused !!! The radiation specialist are confused too!
    And this abstract once again does not even mention thermal collisions. There might be no hope right now in answering that question. But as I mentioned above, that is one unanswered question that AGW and CO2 promoters are hiding under, shaky ground. On one side CO2 is a cooler, not a warmer and on the other side CO2 is not special at all.

    @ Myrrh

    Noticed your note on 67 ºC, and I generally agree (that is unless an pure argon atmosphere really does radiate). That is an interesting question to follow in both astrophysics and spectrometry.

  250. BigWaveDave:

    Steve has answered your question. The alternative for the earth is to have an IR-transparent atmosphere, in which case it is basically a direct portal to the 3 deg background radiation of space. So, yes, being exposed to a relative cold atmosphere, even if we take the temperature at the tropopause of ~190 K, will feel like a freakin’ heat wave compared to being exposed to space!

    That is what you fail to understand: When we talk about the greenhouse effect, the comparison case isn’t having some nice warm surroundings radiating at the earth instead. The alternative is having all of the terrestrial radiation escape to space.

    A planet with no IR absorption in its atmosphere comes to an average surface temperature (really average of T^4) which is dictated by the radiative balance between what it gets from the sun and what it emits back into the coldness as space. A planet with an atmosphere that can absorb IR is able to maintain a warmer temperature at its surface. [As viewed from afar, the planet with its surrounding atmosphere still emits the same amount of radiation and acts as a body having the same average temperature (really average of T^4) as it would in the IR-transparent case; however, most of this radiation originates from higher up in the atmosphere where the temperature can be much colder than it is on the surface.] There is nothing particularly mysterious about this. It is basic physics, in perfect agreement with…in fact, dictated by…the Laws of Thermodynamics and radiative physics.

  251. Tim – re-read my post. I am asking specifically about one kind of light.

    I have already worked out that you’re not engaging with what I’m actually saying, hence my reference to daft meaningless examples which I’m no longer willing to waste my time on.

    The Traditional Physics is Still Being Taught – reference the newworldencylopedia, and NASA page.

    You, IRA are you listening?, generic, are teaching something different from AGWScience.

    You are saying that Thermal IR is NOT heating the Earth.

    This is A RATHER LARGE CLAIM TO BE MAKING, TO OVERTURN BASIC TRADITIONAL PHYSICS AS STILL BEING TAUGHT.

    Who’s claiming the Nobel Prize for Physics for this Revelation?

    Are you ever going to engage with my questions, IRA?

  252. BigWaveDave says:
    April 1, 2011 at 3:48 pm
    … When you open a freezer door, can you feel all of that the 270°K (or so) warmth beaming at you? How long do you have to stand there before you get too hot?

    When you open the door of a freezer, the cold air comes pouring out, cooling you, just as opening the door of an oven gets hot air streaming out, warming you. OK, get a relatively thin cylinder of plastic that passes both shortwave and longwave IR and stand inside of it, blindfolded.

    If you are in a room that is at 300 K (about 27ºC or 81ºF) you will be slightly warm, due to the 300 K radiation on all sides, plus the internal heat generated by your body.

    Now, your assistant rolls up an oven on your right and a freezer on the left (or vice-versa, she does not tell you which). The doors of the oven and freezer are opened, but, since you are enclosed in the plastic cylinder, you don’t feel the hot air on one side and the cold air on the other. However, you now have 250 K longwave IR radiation coming at you from one side and 350 K longwave IR radiation coming from the other side. I have no doubt you will be able to tell which side has the oven and which the freezer.

  253. Myrrh:

    You are presumably being ignored by Ira (and others including me up until now) because you have displayed no evidence of being teachable. It is a thankless task to try to teach someone who doesn’t want to learn.

    You have taken one line from one source and blown it totally out of proportion. What you are referring to is not “traditional physics”; rather, your source claims that it is something that “many physics teachers traditionally” have said that the source itself notes is not really correct. I am skeptical of the claim that many physics teachers traditionally say this, but who knows…I guess I was fortunate enough never to run into one of those physics teachers in the course of getting my PhD in physics.

    However, that it is not really correct is beyond dispute. If you want to believe otherwise, there is nothing that we can do to help you.

  254. Myrrh, you’re correct. There is a great amount of IR in the solar radiation and most of the albedo and refection off of the surface is strictly visible and above frequencies which must be subtracted from these higher frequency (left) portion.

    Use this: http://i56.tinypic.com/5wk13l.gif

  255. Myrrh says:

    You are saying that Thermal IR is NOT heating the Earth.

    No…He is not saying that. The intensity of solar radiation is found by integrating over all wavelengths.

    However, most of the intensity coming from the sun is in the UV, visible, and near IR parts of the spectrum. The “thermal IR”, which according to Wikipedia http://en.wikipedia.org/wiki/Thermal_infrared#Different_regions_in_the_infrared is used roughly for those IR wavelengths above 3 microns (although I don’t know how standard such a definition is), doesn’t make up that much of the solar spectrum…In fact, most plots of the solar spectrum don’t even bother to go out further than 2 or 3 microns. Furthermore, what little of it there is from the sun will get absorbed in large part before reaching the earth’s surface.

  256. Bryan says:

    Its refreshingly honest for you to admit that you did not understand the work of G&T.

    It is a little different “spin” to say that I don’t understand their work than to say that I cannot find any interpretation of what they are trying to say that makes any sense whatsoever. You, despite your sniping at me, have steadfastly refused to provide any such interpretation of your own. In fact, you here repeat your accusation that we “attack[] G&T for things they didn’t say” but continue to avoid my invitation to enlighten us on what they did say (that makes any coherent sense) regarding the greenhouse effect and the Second Law.

    You seem to frequent WUWT and I will I hope you will be able to discuss in depth any faults you can find in the G&T paper.

    My co-authors and I have discussed the faults in our comment on their paper. Arthur Smith on his comment alone has discussed other faults. The ball is now in your court, my friend. If you think that G&T is so sensible, then why don’t you tell us what it means? I would be happy to hear an interpretation of their work that both makes sense and is not in direct contradiction to things that they clearly state (like “The atmospheric greenhouse e ffect … essentially describes a fictitious mechanism, in which a planetary atmosphere acts as a heat pump driven by an environment that is radiatively interacting with but radiatively equilibrated to the atmospheric system. According to the second law of thermodynamics such a planetary machine can never exist.”). So far, you have been unwilling to provide one.

  257. wayne says:

    Use this: http://i56.tinypic.com/5wk13l.gif

    That shows the emitted radiation in W/m^2 at the surface of the sun and at the surface of earth (on a log-log plot, I believe). However, as you might not be aware, the earth isn’t located at the surface of the sun. To obtain the intensity of the solar radiation at the radius of the earth’s orbit around the sun, you have to multiply the emitted radiation by the ratio of the square of the radius of the sun to the radius of the earth’s orbit around the sun. After you’ve done that (and corrected for the factor of 4 between the area of a disc and the surface area of a sphere) and put it on a linear-linear scale, you’ll get something more like this: http://atoc.colorado.edu/wxlab/radiation/emissionspec.gif

  258. Myrrh says : “You are saying that Thermal IR is NOT heating the Earth.”

    No, I am saying that ALL wavelengths of sunlight are part of the energy going to earth, and therefore ALL wavelengths are providing heating to the earth. Please show ONE place where I said thermal IR is not part of the heating. (Thermal IR is a very SMALL part, but it is a part).

    On the other hand, you say “No matter how far Blue Light, for example, penetrates into the ocean, it will not heat it. UV may burn surfaces, as artifically intensified Blue light can also burn, but these do not raise the temperature of matter the way that Thermal IR does.”

    This seem pretty clear — you do not think blue light (or visible light in general) can help raise the temperature of the land or the oceans.

    “Thermal IR heating the Earth is excluded from the AGW Energy Budget.”
    Not sure how you came to this conclusion.

    One the one hand, thermal IR is effectively excluded from the energy budget when looking at sunlight. This is NOT due to some nefarious AGW scheme or poorly understood science. It is simply because thermal IR (specifically, I am referring to 4 um or longer), is less than 1% of of the incoming solar radiation and hence is only a minor player in the overall energy budget.

    On the other hand, thermal IR is BOLDLY included in the energy budget, since thermal IR is the 390 W/m^2 upward radiation from the earth and the 324 W/m^2 downward radiation from the atmosphere.

    Let me say one more time — the “Traditional Physics” you keep referring to is NOT traditional physics. There are several people in this discussion who have shown a pretty good level of scientific understanding — none of them seem to agree with your conclusions. I teach “traditional physics” and I know lots of other people who teach “traditional physics” and none of us teach that you can’t feel heat from visible light.

    Or perhaps I am still mis-understanding you.
    What SPECIFICALLY do you think “traditional physics” teaches about IR, visible light, and their heating effects that you agree with?
    What SPECIFICALLY do you think “AGW physics” teaches about IR, visible light, and their heating effects that is incorrect?

  259. Myrrh;

    You are some piece of work bud. Take a look at one of your own sentences:

    “For example, when I’ve explained that Visible Light is not thermal, but if concentrated in intensity it can burn.”

    So…. it isn’t thermal, but if concentrated…it is thermal.
    Magic.

  260. Is Glickstine’s ‘science’ and attempt to ‘debunk’ the man-made warming that we are experiencing at present? I do not see in here resume where she is even remotely qualified to comment on GHG’s.

    another reviewer wrote “Although there clearly is a greenhouse effect it is infinitesimal compared to the energy retaining effect of the oceans.”

    friggin DUH! how did the heat get to the oceans in the first place??!!! from the heat trapped IN THE ATMOSPHERE

  261. Certainly feels thermal and can heat up and burn the average body pretty badly. I believe thast UV light is the only part of the spectrum that causes sea temperature to rise.

  262. @ Cassandra King (March 29, 2011 at 9:59 am)

    Many thanks for your kind words, Cassandra! :)

  263. Richard Monror says……

    ……”friggin DUH! how did the heat get to the oceans in the first place??!!! from the heat trapped IN THE ATMOSPHERE”…..

    This is an unusual viewpoint.
    So we have no further use for the Sun!

  264. By the way, if you want to figure out the fraction of power in a certain wavelength range for a blackbody of a given temperature, here is an online Excel file that will do it for you: faculty.virginia.edu/ribando/modules/xls/HTTplnkslaw.xls

    Using it shows that for a blackbody at 6000 K (which approximates the sun quite well), less than 2% of the energy is emitted at wavelengths above 3 microns and less than 0.3% of the energy is emitted at wavelengths above 6 microns.

    By contrast, for a blackbody at 290 K (approximating the earth’s surface), ~99.4% of the energy is emitted at wavelengths above 3 microns and over 96% of the energy is emitted at wavelengths above 6 microns.

    This shows that there is very little overlap between the spectra of solar and terrestrial radiation.

  265. Backradiation acts as a perpetuum mobile.
    Thermodynamic theory already incorporates all directions of radiation in the surroundings in equilibrium with a radiative surface.

    Ira Glickstein, PhD says:
    March 30, 2011 at 10:10 pm

    “…..Thus, if the longwave IR source was completely surrounded by the shell, the temperature could potentially increase to correspond to double energy, but, in the real world, it would reach equilibrium well before that……

    I am not familiar with the details of a blackbody cavity, but I can imagine something like an electric soldering iron. With a given power input, the tip, in free air, would reach a temperature of X. If you put the whole soldering iron in a metal box, and suspended the metal box in free air, the tip would reach a higher temperature, all else being equal………You could calculate the new temperature using the Stefan Boltzmann (fourth power) Law. The new tip temperature would correspond to nearly twice the power.

    Note however that no extra energy is being created. At equilibrium, the metal box will radiate (and convect) away the exact amount of power as is being input to the soldering iron, energy in = energy out. The only difference will be that the tip of the soldering iron will be warmer when enclosed in the metal box than when it was in free air. ”

    This box would be a perpetuum mobile.

    Now lets look at the radiation energy density I mentioned earlier. Radiation energy density (RED) is the photon density (J/m^3) in the space surrounding a radiating surface in equilibrium with the Stefan-Boltzmann temperature, it’s simply related to the temperature as S-B * 4/c. And this RED will spread out over distance!!!

    So, we switch on the soldering iron that warms up to say 400 K, and it creates a RED corresponding to 400 K in the space around it. Then we put the box over it and the RED hits the box which is say 300 K, and the radiation from the iron will heat it up to 400 K, then the RED from these walls will also have increased from a density corresponding to 300 K initially to a level of 400 K. But that means the whole box now has a uniform RED, so what happens now? Nothing, because everything would be in equilibrium.
    In reality the walls won’t even become 400 K but 399,9…K at the most as long as the box radiates on the outside (and the RED inside spreads out), so the box will never get as warm as the energy source!
    Ira, thinks the RED will get equal and then magicly increase near the walls and backfire to the source, and that is impossible. The RED in the box depends on the source. The RED is the force acting on the walls and this results in the temperature of the wall which in return results in a RED from the walls and that will correspond to 400 K at the most.

    So many seem to forget that you need a gradient(delta) in temperature, pressure, concentration or density to get a flow. And if the gradient becomes zero the flow stops. This is the Second Law working, always trying to spread things out at the lowest energy level.

    So the die hards will say: because the walls also radiate I still believe in backradiation and this must do something.
    Well, look at the link where does the factor 4 come
    from?
    . Read and weep, radiation energy density already incorporates all directions of radiation. For the simple perpendicular radiation, half of the energy density in the waves is going toward the walls and half is coming out. But of course RED incorporates all possible angels of radiation because it describes a volume and you get an average factor of 4.
    Imagine a second wall on the right and one will see that nothing changes at all for the RED.

    You can also look at it as a pressure.Radiation
    pressure
    is RED/3. So you will have a pressure coming from the iron, what pressure will the walls receive? It is clear that they will never receive a higher pressure than the source and so never be able to establish a higher temperature.

    So heating by backradiation does not exist, it is radiative energy density pulled out of context.

  266. Hans,

    I have no idea why you introduced the concept of radiation density; it only complicates things to the point where you can make nonsense assertions that are far enough removed from simple understanding that perhaps some people won’t realize what nonsense it is.

    We’re not talking anything complicated here – It is really simple actually. Take an object, be it a soldering iron where electrical power is converted into heat or the earth that is absorbing radiation from the sun. If you have that object radiating into the vacuum of space, it is going to come into equilibrium by finding the temperature where the radiation it emits by the Stefan-Boltzmann Law matches the incoming power (provided by the sun in the case of the earth or the electrical power in the case of the soldering iron).

    Now instead, try surrounding that object by a box or shell that absorbs some of the radiation. That surrounding shell will heat up and by virtue of having a nonzero temperature, it will emit radiation. Now the object will be receiving not only the power it was originally receiving but additional power from the radiating shell; there is no way this can be avoided. Hence, in order to reach a state where energy in = energy out, it will have to emit more power. It does this by raising its temperature. It’s really that simple.

    That this is the way it works can be verified by simple toy models that can be solved trivially. I presented the blackbody shell model in another thread. You speculated about what would happen but I clearly showed you how it was easy to see that your speculation was dead wrong; the mathematical solution to the energy balance problem is trivial to solve and it confirms the picture that I have given above. End of story.

    It is embarrassing that there is really any serious argument here about this. If you can’t understand this stuff, how can you possibly hope to understand actual science that is anywhere near the leading edge of our current scientific understanding? It really shows a complete embrace of ideological dogma over reason.

  267. Hans says:

    Then we put the box over it and the RED hits the box which is say 300 K, and the radiation from the iron will heat it up to 400 K, then the RED from these walls will also have increased from a density corresponding to 300 K initially to a level of 400 K. But that means the whole box now has a uniform RED, so what happens now? Nothing, because everything would be in equilibrium.

    No…It is impossible for it to be in equilibrium. It was in equilibrium at 400 K when it didn’t have the walls at whatever elevated temperature that they are at emitting radiation at it. How can it possibly remain in equilibrium now that it is receiving more energy? That’s mathematically impossible. If it receives more energy, it has to emit more energy…and it does this by increasing its temperature.

  268. So many seem to forget that you need a gradient(delta) in temperature, pressure, concentration or density to get a flow. And if the gradient becomes zero the flow stops.

    Exactly…which is why your solution can’t be the steady-state. If there is no net energy flow away from the soldering iron, which is still plugged into the wall and converting electrical energy into thermal energy, what is it going to do? It is going to heat up until the heat leaving it does balance the thermal energy that it is generating! How could it do otherwise?

  269. Joel Shore says:
    April 1, 2011 at 7:02 pm

    Myrrh
    You are presumably being ignored by Ira (and others including me up until now) because you have displayed no evidence of being teachable. It is a thankless task to try to teach someone who doesn’t want to learn..

    That’s the best laugh I’ve had all week! AGWScience is nonsense, that’s an established scientific fact.

    You have taken one line from one source and blown it totally out of proportion. What you are referring to is not “traditional physics”; rather, your source claims tht it is something that “many physics teachers traditionally” have said that the source itself notes is not really correct.

    No, that is not what I have done. I have been arguing this for some time now on Ira’s threads and I posted it to show that what I have been saying is still traditionally taught physics – ongoing.

    “Many physics teachers traditionally attribute all the heat from the Sun to infrared light.”

    Many, means all those who still teach traditional physics and not this AGWScience, which mangles traditional physics in all aspects it uses to promote AGW.

    I posted the NASA page which shows clearly that this is still Traditional Science teaching.

    Read the NASA page, it is very clear what Traditional Science says about this. It says that the heat we feel from the Sun is Thermal IR. It says that Near IR is not hot.

    What I am sharing with you here is Traditional Science which makes sense, AGWScience does not make physical sense. My question to Ira is to prove that AGWScience claims are right and have really, actually, overturned Traditional Science.

    Really, please understand this, you are making a HUGE claim, altering all well known Traditional Physics in this AGWScience description of the Energy Budget. YOU NEED TO PROVE your claim. YOU NEED TO PROVE that Traditional Physics is wrong and you are right.

    Ira is ignoring me because I have ASKED FOR PROOF. I am still asking.

    Traditional Physics is exactly correct. Prove otherwise or get rid of teaching this AGWScience nonsense.

    I am skeptical of the claim that many physics teachers traditionally say this, but who knows…I guess I was fortunate enough never to run into one of those physics techers in the course of getting my PhD in physics.

    More’s the pity you didn’t, you wouldn’t have fallen for the AGWScience con, perhaps. I have already established that a PhD in physics by those who promote AGW doesn’t mean that the bearers have any real grasp of our physical world as described accurately by Traditional Science. One PhD I was questioning when wanting to discover how AGWScience could claim that CO2 could stay up in the atmosphere for hundreds and even thousands of years was convinced that a molecule of CO2 could move from the floor to diffuse into the atmosphere and there become thoroughly mixed without any work being done on it, by Brownian motion.. I don’t think you see how funny that is to someone who knows what Traditional Physics says here…

    However, that it is not really correct is beyond dispute. If you want to believe otherwise, there is nothing that we can do to help you.

    Cr*p! You really must get this straight in you mind, all of you here arguing for AGW, Traditional Physics says you’re talking nonsense! I am now DEMANDING that you prove your claim, and that means you too Ira.

    ######

    Wayne – http://wattsupwiththat.com/2011/02/28/visualizing-the-greenhouse-effect-atmospheric-windows/#comment-621497

    I wasn’t that surprised to see that Miskolczi had simply assumed that the Kiehl-Trenberth diagram meant downwelling Thermal IR. Why would anyone schooled in Traditional Science think otherwise? He thought they were real scientists..

    So, Joel, in your next post you say to my “You are saying that Thermal IR is NOT heating the Earth”

    No.. He is not saying that. The intensity of solar radiation is found by integrating over all wavelengths.

    Nope. I am saying that’s what he is saying because he is saying AGWScience which says that what heats the Earth is Solar energy and this is defined as being the Visible spectrum plus the two shortwaves either side, i.e., UV and Near IR. The AGW energy diagram says it is these which heat the earth, and that Thermal IR is what is radiated back from the Earth heated by these shortwave energies. These energies are not Thermal. Therefore, Thermal is excluded. Saying Solar energies heat the Earth is saying that Thermal IR does not heat the Earth. That is what he is saying.

    OK?

    Now, what I want proof of is that these energies defined as Solar on the AGWScience energy budget diagram actually heat the Earth.

    We are already aware from the NASA description that Near IR is not hot, think your remote control, you cannot feel it. What we feel as heat from the Sun, from fire, from the stove etc., is Thermal IR. Shortwave like the Visible are not Thermal. So, let’s keep this simple. Prove that Visible Light, heats the Earth, is capable of heating the Earth.

    Because until you, genericAGW, do so, that Energy Budget remains nonsense by the real Scientific Standard as still taught traditionally.

    When Ira was still talking to me, he said (in the discussion linked here to Wayne), March 1, 2011 at 6:11 am

    “Myrrh, you really need to get outside more and sit in the Sunshine and feel the warmth! That is how visible and near-visible (“shortwave”) light warms he Earth.”

    Has AGWScience also reversed the sex of Earth..?

    So, this is what I am asking y’all to prove. This is what AGWScience claims is fact in the Energy Budget. It is patent scientific nonsense. Attributing Thermal quality, heat, to shortwave energies we cannot feel, and which belongs to Thermal IR.

    It is Thermal IR which we feel as heat from the Sun – WE CANNOT FEEL SHORTWAVE SOLAR (“Sunshine”).

    Do you see the problem I’m having here? AGWScience is teaching rubbish according to traditional Physics.

    We cannot feel Visible light, or UV, or Near IR at all. We certainly do not feel them as the Warmth from the Sun! What we feel as warmth from the Sun, is Thermal IR. AGWScience has excluded Thermal IR from downwelling Solar energies.

    Here’s the diagram Ira posted in the linked-to discussion: “Sunlight Energy In = Thermal Energy Out http://wattsupwiththat.files.wordpress.com/2011/02/gw-spectrum-123a.gif

    So, two things immediately y’all have to grasp. The Energy Budget of AGW is make-believe according to traditional Science and therefore all calculations done on this AGW premise of Solar energy (“Sunlight” of Ira’s diagram) heating the Earth and excluding Thermal IR, will also be nonsensical. And secondly, that y’all are making one hellava Scientific claim here, counter all that Traditional Science understands and still teaches.

    I hope you can now see the enormity of the AGWclaim.

    Prove It.

  270. Myrhh says:

    Many, means all those who still teach traditional physics and not this AGWScience, which mangles traditional physics in all aspects it uses to promote AGW.

    Myrhh, you may think your use of the term “AGWScience” is cute, but frankly to those of us who actually know science, it is immature and more than a little paranoid. As has been pointed out to you, we are talking about stuff that has been understood at least since Einstein won the Nobel prize for the photoelectric effect, establishing the connection between the wavelength of electromagnetic radiation and the energy of the corresponding photons. (And, really, even before photons were understood, there was already a pretty good understanding of the energy associated with electromagnetic radiation…except for the quantization part. Heck, they even had empirically-determined the formula for the blackbody spectrum even if they could not explain why it didn’t blow up on the UV side [the so-called "ultraviolet catastrophe].)

    We are already aware from the NASA description that Near IR is not hot, think your remote control, you cannot feel it.

    Fine, how about I take a nice strong laser source and shine it on you? Heck, you won’t feel a thing! I’ll even be generous and use a visible laser, which you seem to think if anything has less ability to heat than the near IR!

    The reason you don’t feel your remote control is because the power involved is so small. If you have enough power, then believe me, you are going to feel it!

    By the way, the one thing that I have learned from you is that NASA really had some web pages written up by some science educator or engineer who didn’t have a clue what they were talking about. (Out of principle, I refuse to believe it was possibly a physicist who wrote that; however, given that Gerlich and Tscheuschner are physicists, I guess I have to admit that such disciplinary bigotry on my part is probably not justifiable.) I am surprised what a poor job NASA did on that website that you found…and now unfortunately there are gullible people like you who are thinking that such nonsense is actually the one example of unsanitized gospel truth before the great AGW conspiracy came through and swept all of the physical sciences clean of the correct understanding of electromagnetic radiation! To NASA’s credit, they do say that that site is obsolete and the new web page seems to have actually tried to get the physics correct, but no doubt that only adds to the nefariousness of the conspiracy in your mind.

    The only way that one can possibly interpret that outdated NASA web page as saying anything logical in regards to radiation is that it is true that most objects tend to have higher emissivities (and hence absorptivities) in the thermal IR than in the visible and near-infrared. So, you do have the possibility of reflecting more visible radiation than you do for the thermal IR; radiation that you reflect will indeed not heat you up. Also, visible light and near-IR photons actually have enough energy to do things like cause chemical changes, so there are more pathways for the energy than just direct conversion to thermal energy…although, the Second Law pretty much dictates that most of it is going to end up as thermal energy eventually, and often sooner rather than later.

    Therefore, Thermal is excluded. Saying Solar energies heat the Earth is saying that Thermal IR does not heat the Earth. That is what he is saying.

    Ira is just taking into account what I showed in this post: http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/#comment-634654 , namely that a negligibly-small amount of the solar energy is in the thermal IR because that is the nature of the blackbody spectrum of an object of about 6000K.

  271. Myrrh;
    We cannot feel Visible light, or UV, or Near IR at all>>>

    Myrrh, sorry, but your feelings have nothing at all to do with it. You can’t “feel” the IR from your television remote because is very low power. Low doesn’t mean zero, and how many w/m2 come out of your remote has nothing to do with the sun. The range of human senses (what you can feel) doesn’t define the facts. You can’t “see” UV or IR, does that mean they don’t exist? No more than you not being able to feel them means they don’t transfer energy.

    Ira’s ignoring you I’m thinking because what you’ve “proven” so far with all your rants is that you don’t understand Traditional Physics OR AGW Physics. You don’t even appear to understand your own arguments. I’m not certain if you understand arithmetic.

    What you should understand is that Joel Shore and I are on as opposite sides of the fence on AGW as one can get, and we’re both telling you stop spouting idiotic nonsense, and for the exact same reasons.

    Yes Joel, that was a compliment from me, don’t have a heart attack. Wanted to commend you on participating in a positive fashion. Much improved value from your comments as far as I am concerned.

  272. Tim Folkerts says:
    April1, 2011 at 8:41 pm

    No, I am saying that ALL wavelengths of sunlight are part of the energy going to earth, and therefore ALL wavelengths are providing heating to the earth. Please show ONE place where I said thermal IR is not part of the heating. (Thermal IR is a very SMALL part, but it is a part).

    I took it from your use of “Sunlight” as Ira has it, meaning the AGW Solar energies which exclude Thermal IR. So, right, if you’re not actually meaning this, but are using your own variation then my question still applies, because a “SMALL part” of Thermal IR, is for or all intents and purposes in this argument, discounting it. So, prove that the shortwave Visible, UV and Near IR which we cannot feel, are heating the Earth.

    On the other hand, you say “No matter how far Blue Light, for example, penetrates into the ocean, it will not heat it. UV may burn surfaces, as artificially intensified Blue light can also burn, but these do not raise the temperature of matter the was that Thermal IR does.”

    This seems pretty clear – you do not think blue light (or visible light in general) can help raise the temperature of the land or the oceans.

    “Thermal IR heating the Earth is excluded from the AGW Energy Budget.”

    Not sure how you came to this conclusion.

    See above my previous post. As per Ira’s description, it is the standard teaching of AGWScience in their Energy Budget, see Trenberth diagram. See also: http://en.wikipedia.org/wiki/Greenhouse_effect

    “the Earth receives energy from the Sun in the form UV, visible and Near IR radiation, most of which passes through the atmosphere without being absorbed. Of the total amount of energy available at the top of the atmosphere (TOA), about 50% is absorbed at the Earth’s surface. Because it is warm, the surface radiates far IR thermal radiation that consists of wavelengths that are predominantly much longer than the wavelengths that were absorbed. Most of this thermal radiation is absorbed by the atmosphere and re-radiated both upwards and downwards; that radiated downwards is absorbed by the Earth’s surface. This trapping of long-wavelength thermal radiation leads to a higher equilibrium temperature than if the atmosphere were absent.”

    So, as Ira has it, the AGW energy budget is “Solar” /”Sunlight” radiation – which is UV, Visible and Near IR, downwelling to Earth. This then heats the Earth’s surface which radiates out Thermal IR.

    One the one hand, thermal IR is effectively excluded from the energy budget when looking at sunlight. This is NOT due to some nefarious AGW scheme or poorly understood science. It is simply because thermal IR (specifically, I am referring to 4 um or longer), is less than 1% of the incoming solar radiation and hence is only a minor player in the overall energy budget.

    Which is as I said, you’ve excluded it. Technically picky add “effectively” if you want…

    On the other hand, thermal IR is BOLDLY included in the energy budget, since thermal IR is the 390 W/m^2 upward radiation from the earth and the 324 W/m^2 downward radiation from the atmosphere.

    Which is b*sh*t as you’ve ‘effectively’ excluded Thermal IR in heating the Earth, bear in mind it is what you feel as heat from the Sun, and you’ve yet to prove that the shortwave Visible, UV and Near IR which are not thermal are capable of and actually heating the earth.

    Let me say one more time – the “Traditional Physics” you keep referring to is NOT traditional physics. There are several people in this discussion who have shown a pretty good level of scientific understanding – none of them seem to agree with your conclusions. I teach “traditional physics” and I know lots of other people who teach “traditional physics” and none of us teach that you can’t feel heat from visible light.

    Then neither you nor they are teaching TRADITIONAL BOG STANDARD physics.

    AS per the NASA page, you cannot feel Near IR. If you are feeling heat, it is Thermal IR.

    You cannot feel UV. You can feel the effects when too much of it burns you, BUT, you cannot feel it burning you, because you cannot feel it. It is not Thermal. Do you understand that? Visible light in between these, likewise. They are not Thermal. You cannot feel them, they are not hot. See NASA page, Near IR is not hot.

    See NASA page, it is Thermal IR which is used to warm foods.. Near IR from the Sun will not warm you. Visible light will not warm you.

    Is that specific enough?

    Do you now understand my attempts to show this in real life real science physical application? For example, grow lights for plants?

    ##############

    davidmhoffer says:
    April 1, 2011 at 11:35 pm

    You are some piece of work bud. Take a look at one of your own sentences:

    “For example, when I’ve explained that Visible Light is not thermal, but if concentrated in intensity it can burn.”

    So…it isn’t thermal, but if concentrated…it is thermal.
    Magic.

    No, even when it burns it is not Thermal, it does not warm, it does not heat up matter as does Thermal IR energy.

    Thermal IR is absorbed by organic matter and heats it. Visible light is reflective rather than absorptive, it will penetrate organic matter only a little, say human skin, that’s why you cast a shadow. Near IR will penetrate further, UV hardly penetrates at all, you cannot feel it. If these non thermal energies of Visible and Near IR are artificially intensified they can burn, but they do not heat up organic matter naturally, they are not the heating mechanism for the Earth because they are incapable of it. UV is capable of burning because naturally intense. What is the difference? In the wavelengths. The even higher frequency shorter wavelength of UV is, like the friction of creating fire by drilling a stick into a log, it burns the surface, but it does not heat up organic matter; you will not feel warm from intense UV until you get sunstroke… Which is a secondary effect..

    What warms you up, absorbed by you and the Earth, is Thermal IR.

    The AGWScience Energy Budget is PHYSICAL NONSENSE.

    The Thermal IR radiated from the Earth is because the Earth has been heated by Thermal IR. Solar/Sunlight AGW’s UV, Visible, Near IR CANNOT raise the temperature of organic matter. At best, UV is able to burn the surface. If UV could penetrate further naturally we’d all be burned to a crisp inside.

    AGWScience has no concept of the real physical world around us. That’s a proven scientific fact, from observation of its claims..

    See again the NASA page and look at the description of size of IR, from the size of the pin head of far thermal to microscopic near non-thermal. Perhaps this will help?

  273. @ Myrrh

    See the enormity? I do. But, for you and I to get perfectly parallel you would have be willing to have a back and forth here for ten, maybe twenty steps. Staying on each point till we either agree or disagree. If I try to toss it all into one single comment it gets far to complex and leaves so many branches of possible misunderstanding. At the end I think we will both agree on many fronts, one that the energy budgets are misleading and in places a figment of someone’s imagination. Are you willing? Maybe we can both get our views so clear that others can see them.

    I’ll start out by telling you of an advantage I might have over you, on the subject of the higher frequencies of E/M. I have a major in physiology, though never followed, and I had to dig out of the garage my old books and read a bit. Seems the nerves that sense heat are mid to far-IR sensitive but not much to the shorter wavelengths. So, you might be getting severely burned and you won’t even sense it until the damaged cells start to pour out the histamine, then you will feel it, for sure, but more as a deep ache at first, the real pain comes later. Those frequencies go deeper than mid and far IR so the burn would be deeper. That’s from a pretty old book so you might look it up in a more recent one, but I don’t think the human body has changed that much but the knowledge may have.

    Now you know why with normal white light, UV, near-IR, you never feel it, the warm energy is spread vertically deep over many millimeters and the blood simply carries the warmth (energy) away too fast to matter. Focus that same amount of light, same type, with a magnifying glass on your skin and it does burn, and you do feel the warmth then, too much energy so fast that the heat then causes the nerves to fire, you feel not just warmth but immediate pain.

    So, I know from physics that these frequencies do carry energy and if ever absorbed or scattered and split this energy does warm. As shown above but too weak for us to “feel” it though it might be nearly the same intensity. But most of the strictly ‘albedo’ takes away from the mainly visible (or close to) frequencies so much of it is simply reflected back to space. Agree so far?

    Did you ever look at that tinypic mentioned above? It clearly shows the solar IR you keep speaking of, though that representation might magnify it’s influence too much. It’s hard to trust any of these representations as spectrum. More on that later. That’s topic two.

  274. Joel – that NASA page was written by someone who had more than a clue, had real grounding in the real science of our real physical world. ‘AGWScience’ is not some attempt to be cute, it is to make a real distinction between the real science of our physical reality and the junk that passes itself off as science by AGW which describes an imaginary, impossible world.

    I really hope, perhaps my last post to David will help, that you can begin to grasp there is a huge difference in concept between the two.

    But anyway, prove that visible light warms the Earth. I’m still waiting.

  275. Myrrh says:

    The Thermal IR radiated from the Earth is because the Earth has been heated by Thermal IR.

    Fascinating…So, like a few Watts/m^2 at most of thermal IR from the sun are enough to result in the emission of many hundreds of Watts/m^2 from the earth? How does this work exactly?

    But anyway, prove that visible light warms the Earth. I’m still waiting.

    Sorry…Science doesn’t work on proof; that’s the realm of mathematics. Science is inductive. For this reason, it is not possible to demonstrate science to someone who showed utter unwillingness to comprehend it. (Although I am not very confident I would even be able to “prove” something in mathematics to someone who refuse to abandon his mental blocks.) What’s the saying…”You can lead a horse to water but you can’t make him drink”? Those of us spanning a whole gamut of opinions on AGW as a whole have led you to vast lakes of fresh water. Now it’s your turn to demonstrate some ability to actually be worthy of our spending the time to even respond to your semi-coherent ramblings of scientific nonsense.

  276. Myrrh;
    No, even when it burns it is not Thermal, it does not warm, it does not heat up matter>>>

    Then….(choke, splutter)…Then how…(caugh, caugh, jeeez is he for caugh caugh real)…Then how does (ROFLMAO)…..pant, pant, tears in eyes…OK, got myself together again…. Then how does (LOL LOL LOL ohmigod I can’t breath wheeze wheeze wheeze)…OK, now then how does it burn the matter if it doesn’t heat up the matter? Please don’t answer, I don’t know if I can take anymore of this.

  277. Myrrh, somehow I missed a comment you wrote to me on Miskolczi back in #2 of this series, on Atmospheric Window at the bottom. You seem to understand all of this just fine, much deeper than many are giving you credit. I’m don’t think you and I need to go further, so skip that last comment and invitation to discuss I made. I’m already parallel to you on nearly everything you said.

  278. Myrrh

    “So, prove that the shortwave Visible, UV and Near IR which we cannot feel, are heating the Earth.”

    You really ARE unteachable — or more likely an intentional troll having fun getting smart people to engage your meaningless rants.

    It has been explained numerous times how these wavelengths CAN be felt and DO help heat the earth – warm sunshine (even after passing thru glass which will remove most of the IR) and E=hf and dark objects being warmer than light objects and lasers which cut things with heat. I for one am not going to continue to engage to other than to occasionally warn people about your lack of understanding if you try to post in other threads.

  279. David and Joel, it is totally pointless trying to reason with this man, he has absolutely no idea about science. The best policy is just to let him post this gibberish since he just responds with the same lunacy. Unfortunately any thread on GHE is going to be disrupted by this craziness, but Anthony lets it persist so what can you do?

  280. Myrrh, after a post here and in a previous forum you have yet to address my “proof”. If anything is “traditional physics”, it is the conservation of energy.

    So I repeat, look up at the sky and then look down at your feet. If you really want to get technical, look at a photo of earth from space. There is quite a bit of intense visible light coming from the sky (so much that it would blind you to stare at the sun), yet the ground isn’t nearly this bright. The visible light isn’t being reflected as visible light, and it isn’t passing through the earth, so what is happening to the energy of all that visible light? According to your “it isn’t heating the earth” hypothesis – poof, it disappears. How is that consistent with “traditional physics”?

  281. Phil;
    David and Joel, it is totally pointless trying to reason with this man>>>

    Well I for one have given up and am just laughing. Joel seems to have concluded the man is unteachable, said so, and then continues to try. Stubborn guy that Joel, gotta give him some credit.

    Reminds me of a conversation I had with a dear old lady in the family who insisted that man had never been to the moon, it was all faked. Said she could prove it. OK, so I was dumb enough to ask.

    “They can’t even get rid of mosquitos, how could they possibly get to the moon if they can’t even get rid of tiny little mosquitos!” she said triumphantly.

    Now she really was a very sweet old lady. And you have to admit, she won that argument hands down. I had….nothing.

    But Myrrh I can make fun of!

  282. RJ says:
    April 3, 2011 at 3:19 am
    DavidM
    Do you believe an oven like this is possible

    http://www.slayingtheskydragon.com/en/blog/111-a-pictures-worth-a-1000-words>>&gt;

    Absolutely Not!

    And if not why not if backradiation does add additional heat or energy to the earth’s surface.>>>

    This is a classic case of a ligitimate explanaton being twisted into something its not with the dilberate attempt to mislead. I find it sad that critics of AGW are stooping to the same tactics as the alarmists. Put a warm chicken on the counter and it will cool off. Put a warm chicken in that make believe oven and…. it will cool off. Not only will it cool off, but it will cool off to the temperature of the local surroundings, exactly like the chicken on the counter. BUT (pay attention now) the chicken in the make believe oven will cool off SLOWER than the chicken on the counter.

    The difference between that and the earth’s surface is that the chicken in the make believe oven has no in coming source of heat to off set what it is losing. The earth does. So let’s extend the analogy to something meaningful. Shine a 500 watt IR beam at both chickens. One on the counter and one through the glass door (special glass so IR could go through) of the make believe oven. The chicken on the counter heats up until it reached a new equilibrium temperature where it is radiating as much heat as it is abosrbing from the beam. The chicken in the make believe oven also heats up, but it heats up faster, and to a higher temperature than the chicken on the counter.

    Now if you want to be usefull at all in discussing AGW, put aside these childish, sarcastic, bullarky remarks that don’t represent a single thing Ira, or I, or Joel, or Phil or anyone of a number of people have said. The AGW debate is not if the chicken in the oven cooks itself, that stupendously rediculous. The AGW debate is about how much slower the chicken in the make believe oven cools off compared to the one on the counter, and how much hotter it gets when you fire a 500 watt IR beam at it.

    The CAGW crowd wants to claim a lot, the skeptics say bull, much smaller number, and with feedbacks included properly it may be negative. The battle ground is not IF the battleground is HOW MUCH and FEEDBACK NET EFFECT.

  283. Wayne – ta. You and I are obviously old school, with a sense of height and depth and width, and weight and volume..

    Unlike the opening reply here whose world is composed of an atmosphere of empty space with molecules rushing around having billions of collisions and so thoroughly mixing as they bounce off each other, as per ideal gas law.. Perhaps it’s now become practically impossible to convey a sense of the real physical world to AGWScience boffs who are so adept and ingrained in the misapplication of laws.

    I found this page earlier when I was trying to explain that molecules might well be moving at tremendous speed, but in the real un-ideal gas world they’re going nowhere fast..

    http://www.mediacollege.com/audio/01/sound-waves.html

    But still, they insist it’s Brownian motion moving the molecules around through our atmosphere, because no concept of the weight of the fluid gas Air above us in which these molecules have to move, no feel for volume or scale. So scent diffuses by Brownian motion through the room and Carbon Dioxide heavier than Air will by this motion diffuse through the atmosphere. Or sometimes, wind is like a wooden spoon constantly churning up Air and thoroughly mixing it… :) So of course, no explanation of convection gets through to them and the 2nd Law clearly proves to their satisfaction that a photon from a cold molecule of CO2 bounces back and heats the Earth, or, that CO2 sits up there accumulating, defying gravity and its own weight and heat capacity to trap Heat in an insulating blanket practically 100% holes.

    If you can believe that, and more absurdities, then of course it makes perfect sense to think of Solar energies being absorbed by the Earth and warming its organic matter by converting to heat energies, which are admitted as being Thermal IR…

    http://en.wikipedia.org/wiki/File:The_green_house_effect.svg

    Just don’t ask them to design a bright not hot light with neglible heat, not in their reading of the laws..

  284. Myrrh says:
    April 3, 2011 at 4:17 am
    Wayne – ta. You and I are obviously old school, with a sense of height and depth and width, and weight and volume..

    Unlike the opening reply here whose world is composed of an atmosphere of empty space with molecules rushing around having billions of collisions and so thoroughly mixing as they bounce off each other, as per ideal gas law.. Perhaps it’s now become practically impossible to convey a sense of the real physical world to AGWScience boffs who are so adept and ingrained in the misapplication of laws.

    Yes the real world as described by Einstein in 1905! Enough of this foolishness.

  285. Steve at April 2, 2011 at 11:57 pm

    Sorry, nearly missed your post twice.

    Visible light is easily reflected, so the blue sky as this spreads through the atmosphere.

    http://www.physicsclassroom.com/Class/light/u12l2f.cfm

    “The atmosphere is a gaseous sea that contains a variety of types of particles; the two most common types of matter present in the atmosphere are gaseous nitrogen and oxygen. These particles are most effective in scattering the higher frequency and shorter wavelength portions of the visible light spectrum. etc.”

    What we see as colour in our world is what is reflected. The green light is reflected by plants, which absorb blue and red light for photosynthesis, so in reflecting back at us we see green.

  286. davidmhoffer says:

    Well I for one have given up and am just laughing. Joel seems to have concluded the man is unteachable, said so, and then continues to try. Stubborn guy that Joel, gotta give him some credit.

    Alas, this phenomena is well-summarized by what might well be the best cartoon ever: http://xkcd.com/386/ But I think I should really try to adopt your approach in this case!

  287. davidmhoffer says:
    “The chicken in the make believe oven also heats up, but it heats up faster, and to a higher temperature than the chicken on the counter.”

    The temperature of the chicken cannot be higher than that determined by the quantity of thermal radiation it is receiving whether it inside an oven or not. This is similar to Ira’s misunderstanding when he wrote about a soldering iron in a box reaching a higher temperature than one in the open. If the flow of radiation is blocked by the box it does not result in the heated object achieving a higher temperature than that determined by the Stefan-Boltzmann constant (taking account of emissivity). If it did then, as Hans says, we have a perpetuum mobile. The plausible theory (first advocated by Fourier?) that if you block the escape of heat then this will add to the heat content is false as it contravenes the laws of the conservation of energy. Blocking the flow of radiation may reduce the rate of cooling (which is why there is an infra-red reflector inside a thermos flask) but it cannot add heat even where there is a source of heat inside the thermos. When the maximum temperature determined by the S-B constant is reached (let’s assume a blackbody chicken for simplicity) further heating stops. More energy is required to raise the temperature further – reflection or backradiation merely reflects exactly the same amount of energy so there can be no additional heating. (I wouldn’t recommend eating the chicken.)

  288. Phil – the atmosphere is the liquid gas Air, it has volume and it has weight, it is pressing down on us with a tremendous amount of pressure, around 14 lbs/square inch, around a ton pressing down on our shoulders. The molecules of Air, mainly nitrogen and oxygen, do not move very fast at all through all this.

    I haven’t checked how this compares with standing at the bottom of a swimming pool with 10 ft of water, if you know how much that weighs, but this is a description I’ve seen to explain our atmosphere. That going then and standing in the middle of a field in the open air is equivalent. We have a gaseous sea of Air above us. This is not empty space, because real gases have volume and weight and are subject to pressure and gravity, unlike the imaginary ideal gas which takes up no space at all..

    This is how sound manages to travel. The molecules of Air are a fluid moving ‘on the spot’ under all this pressure and gravity, sound travels by hitting a bunch of them setting them vibrating, still ‘on the spot’, they in turn bash into the next molecules which start vibrating hitting the molecules of Air next to them – it’s the energy of sound travelling, just like waves travelling in the sea. The water does not move very far from where it is en masse, it vibrates by the energy travelling through it and passes this wave along through the rest of the ocean to finally break on a shore somewhere.

    Read descriptions of the ideal gas, a good page on this will point out that it is an imaginary construct and describes no real gas and will explain that real gases have molecular weight and molecular volume, not attributes of this imaginary ideal gas. To scale up from this imaginary gas into ‘the atmosphere’, is creating an imaginary atmosphere.

    Ideal gas has no real gas interactions, there is no force of attraction or repulsion between them, they occupy no space, they do not have volume, they obey all gas laws at all temperatures and pressures. This is not a description of the REAL world. These may well be of use, if you understand the maths.., in calculations, but only by compensating for all the stuff that’s missing.

    Neither ideal gas laws nor Brownian motion, which is the movement of particles in a fluid (a gas or liquid), can be extrapolated without regard to real properties of weight and volume of real gases in the real atmosphere. (*)

    I’ll leave you to argue with others whether or not the 2nd Law actually breaks down on a molecular level..

    ..but as far as Solar energies being absorbed by the Earth heating it – I’m still demanding proof that Visible light from the Sun can actually do this, prove it can or take it out of the “Solar” energy budget in the AGWScience claim. Then do the same for UV and Near IR…

    (*) to grasp the concept of the tiny scale in which Brownian motion exists see the wiki page on Diffusion:

    “Under normal conditions, molecular diffusion is relevant only on length scales between nanometer and millimeter. On larger length scales, transport in liquids and gases is normally due to another transport phenomenon, convection.

    Therefore, some often cited examples of diffusion are wrong: If cologne is sprayed in one place, it will soon be smelled in the entire room, but simple calculation shows that this cannot be due to diffusion; the cause can only be convection. If ink is dropped in water, one usually observes an inhomogeneous evolution of the spatial distribution, which clearly indicates convection; diffusion dominates only in perfect thermal equilibrium.”

    CO2 cannot diffuse through our real world atmosphere, neither by giving it attributes of the ideal gas without volume or weight etc. and zipping through an atmosphere of empty space, nor by being diffused by Brownian motion. It has weight, the lighter molecules of nitrogen and oxygen are not strong enough to move it anywhere, they instead get displaced by the heavier than Air molecules of CO2 as these sink to the ground. It takes work from something stronger to move CO2, and it’s not always windy..

  289. Sorry, should be technically “fluid gas Air” in opening sentence. Liquids and gases are fluids in contrast to solids. I was thinking of the description “gaseous sea” from my previous post. But liquid Air as a description of the atmosphere above us is this “gaseous sea”, appreciating that as a concept is basic to seeing where AGWScience deviates from Traditional, by misapplying physical laws. By taking them out of context we get the impossible through the looking glass with Alice world where something heavier than air doesn’t sink..

  290. Myrrh says:
    April 3, 2011 at 7:54 am
    Phil – the atmosphere is the liquid gas Air, it has volume and it has weight, it is pressing down on us with a tremendous amount of pressure, around 14 lbs/square inch, around a ton pressing down on our shoulders. The molecules of Air, mainly nitrogen and oxygen, do not move very fast at all through all this.

    Most probable velocity of N2 at room temperature is 422 m/sec or 1520 km/hr,
    RMS speed is 517 m/sec or 1861 km/hr. He and H2 travel so fast that they can exceed the Earth’s escape velocity!

    Anthony do yourself and everyone on here a big favor and get rid of this anti-science nonsense!

  291. DavidH and RichardE

    Thanks for the replies.

    DavidH

    Isn’t this from Ira

    “Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth”

    Very different to

    “The chicken on the counter heats up until it reached a new equilibrium temperature where it is radiating as much heat as it is absorbing from the beam”.

    Is this new equilibrum similar to a CO2 changing the adiabatic heating capacity of earth. Rather than further additional heat or energy being added by backradiation.

  292. Joel Shore says:
    April 2, 2011 at 3:27 pm

    “Hans,

    I have no idea why you introduced the concept of radiation density; it only complicates things to the point where you can make nonsense assertions that are far enough removed from simple understanding that perhaps some people won’t realize what nonsense it is.”

    No, it gives a good insight on how to look at radiation coming from a surface. It’s not complicated but much easier as all these posts about the GHG effect. I did not make nonsense assertions, it’s all explained on the Hyperphysics page.

    “If you have that object radiating into the vacuum of space, it is going to come into equilibrium by finding the temperature where the radiation it emits by the Stefan-Boltzmann Law matches the incoming power”

    Following the gradient and cause and effect, the sun determines the surface temperature, then the surface will radiate.

    “Now instead, try surrounding that object by a box or shell that absorbs some of the radiation. That surrounding shell will heat up and by virtue of having a nonzero temperature, it will emit radiation. Now the object will be receiving not only the power it was originally receiving but additional power from the radiating shell; there is no way this can be avoided. Hence, in order to reach a state where energy in = energy out, it will have to emit more power. It does this by raising its temperature. It’s really that simple”

    The net power equals sigma *A (T1^4 – T2^4) so the power reduces when the shell gets warmer. You mix energy and power here. Energy in-out will always be the same, but not the power. Besides the in-out balance is First Law that just is what it is but tells you nothing about what will happen, it’s the Second Law that tells you what will happen.
    Without a gradient (delta T) nothing happens and in = out = zero.
    And think of cause and effect, it’s not power creating temperature but temperature creating power. Without temperature difference there will be no power, without power there will still be a temperature.

    “No…It is impossible for it to be in equilibrium. It was in equilibrium at 400 K when it didn’t have the walls at whatever elevated temperature that they are at emitting radiation at it. How can it possibly remain in equilibrium now that it is receiving more energy? That’s mathematically impossible. If it receives more energy, it has to emit more energy…and it does this by increasing its temperature.”

    You forgot this part: ‘In reality the walls won’t even become 400 K but 399,9…K at the most as long as the box radiates on the outside (and the RED inside spreads out), so the box will never get as warm as the energy source!’

    But if you insist, the limit would be where the surroundings outside this box would also become 400 K and the power becomes zero thus equilibrium.
    My point was that this unrealistic case would be the ultimate end, while no heating of the source by the box ever took place because for that to happen the box needs to exceed 400 K.

    ” So many seem to forget that you need a gradient(delta) in temperature, pressure, concentration or density to get a flow. And if the gradient becomes zero the flow stops.

    Exactly…which is why your solution can’t be the steady-state. If there is no net energy flow away from the soldering iron, which is still plugged into the wall and converting electrical energy into thermal energy, what is it going to do? It is going to heat up until the heat leaving it does balance the thermal energy that it is generating! How could it do otherwise?”

    The steady state is not my solution. It’s yours, by the idea of heating a power source by a shell that needs to create a steady state and then go beyond that and create a power flow in reverse.
    I explained that this is unrealistic, so don’t try to reverse things and make it my solution.

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

    Your quote of a quote:

    “”But in fact it does not warm up; it’s temperature remains exactly the same. The reason why is very simple to understand but extremely important to physics: the blackbody is already in radiative thermal equilibrium with a hotter source of energy, the higher radiative energy spectrum light from the light-bulb. You cannot make something warmer by introducing to it something colder, or even the same temperature! You can only make something warmer, with something that is warmer! This reality is called the 2nd Law of Thermodynamics, and is so central and fundamental to modern physics it cannot be expressed strongly enough.””

    Your reply to the quote of the quote:

    “Yep, it should warm up and (if it is a true blackbody in the longwave region) it will absorb that reflected longwave energy and, all else being equal, will warm up.”

    ———————————————————————————————————;
    False.

    The temperature of an ideal black body determines it’s cut off frequency.

    An ideal black body absorbs all incoming coherent electromagnetic radiation, re-emits the coherent electromagnetic radiation below the cut off frequency, and converts the coherent electromagnetic radiation above the the cut off frequency into incoherent high frequency white noise – or heat.

    Only electromagnetic radiation above the cut off frequency warms the black body.

    Also note, the black body model excludes the possibility of back radiation.

    I suggest you re-read the article.

    Two wrongs don’t make a right (but 3 lefts do.)

  294. RJ says:

    “The chicken on the counter heats up until it reached a new equilibrium temperature where it is radiating as much heat as it is absorbing from the beam”.

    Is this new equilibrum similar to a CO2 changing the adiabatic heating capacity of earth. Rather than further additional heat or energy being added by backradiation.

    No…It has nothing to do with changing the thermal energy of a gas by adiabatic compression, if that was what you are asking.

  295. Hans says:

    The net power equals sigma *A (T1^4 – T2^4) so the power reduces when the shell gets warmer. You mix energy and power here. Energy in-out will always be the same, but not the power.

    Power is just energy per unit time, so your last sentence is basically non-sensical. And, yes, the power does get smaller as the temperature of the shell increases…That’s the whole point. Now take that to its logical conclusion and you will understand why the temperature of the object must increase.

    Besides the in-out balance is First Law that just is what it is but tells you nothing about what will happen, it’s the Second Law that tells you what will happen.
    Without a gradient (delta T) nothing happens and in = out = zero.
    And think of cause and effect, it’s not power creating temperature but temperature creating power. Without temperature difference there will be no power, without power there will still be a temperature.

    Okay, so to interpret what you are saying charitably: You are saying if the temperatures are the same, there will be no heat flow…i.e., the power will be zero. So what, pray tell, is going to happen to the energy that the object is receiving (e.g., from the sun in the case of the earth or from the conversion of electrical energy into thermal energy in the case of the soldering iron)? You have just told me there is no energy flow away from the object. So, how is it not going to increase its thermal energy and hence its temperature?

    The steady state is not my solution. It’s yours, by the idea of heating a power source by a shell that needs to create a steady state and then go beyond that and create a power flow in reverse.

    I agree … Your solution is not a steady-state, which leads me to wonder why you have presented it to us as what will happen. What you are telling me is something that might occur at some instant of time for some contrived circumstances. However, that is not the subject of our discussion…We are talking about what state the system will find where its temperature is steady with time. The point being that such a temperature will be higher in one case than the other.

    Hans, you are really talking nonsense here and you are only embarrassing yourself by keeping it up. With Myrrh, I can understand that he just doesn’t have enough knowledge to educate himself, but I think in your case that you do. So, why don’t you just man up and admit that you are wrong? You’ll actually earn a little bit of respect that way.

  296. Richard E Smith;
    The temperature of the chicken cannot be higher than that determined by the quantity of thermal radiation it is receiving whether it inside an oven or not.>>>

    RJ;
    “Some of the re-emitted Photons make their way… back down to the Surface where their energy is absorbed, further heating the Earth”
    Very different to
    “The chicken on the counter heats up until it reached a new equilibrium temperature where it is radiating as much heat as it is absorbing from the beam”.

    Exactly the same, and for exactly the same reason. For simplicity we’ll use a pair of black body chickens at room temperature of 20 C. By SB Law they are radiating 418 w/m2. Put one chicken in the oven, which we’ll assume is also at room temperature. Chickens are radiating 418 w/m2 and the room and everything in it is radiating at 418 w/m2. Equilibrium.

    Turn on magical 500 watt beams, one aimed at Counter Chicken and one through the special window at Oven Chicken. Being black body chickens, they absorb 100%. Counter Chicken increases in temperature until it reaches 918 w/m2 or about 84 degC. Where does the extra 500 watts it is radiating go? Everywhere. It get’s absorbed by the environment the Counter Chicken is in, but the area in our fantasy kitchen of walls, ceiling, floor, is so large by comparison the effect is negligible.

    What of Oven Chicken? Oven Chicken is in an Oven made of…Oven Stuff, which is also a black body. Its not very much larger than the chicken. It starts out at 20 C, just like the chicken. So Black Oven is radiating 418 w/m2 and so is Oven Chicken, equilibrium. Until we hit Oven Chicken with that 500 watt beam. Oven Chicken rises in temp up to 918 w/m2 and 84 degrees. But that’s much hotter than Black Oven, which is at 20 degrees and only radiating 418 w/m2. So Black Oven is getting 418 w/m2 on the outside from the environment, but from the inside it is getting more because the Oven Chicken is at 84 degC. Now the area of the inside walls of the Black Oven is much larger than that of the Oven Chicken. Let’s say 10 times. So the inside walls of the Black Oven are getting an extra 500 w/m2, but that’s spread out over ten times as much area, so 50 w/m2. Since outside of Black Oven is receiving 418 and radiating 418, but inside Black Oven is receiving 468 but only radiating 418…Black Oven’s temperature must rise. How much?

    Black Oven has TWO surfaces that can radiate heat, one inside, and one outside. So being made of some magical Black Body Stuff so thin that the outside surface area is pretty much the same as the inside surface area, Black Oven must increase in temperature until it is radiating an extra 25 w/m2. That would be 418 w/m2 (room temp) plus 25 w/m2 (from Oven Chicken) a temperature of 24.3 degC. Now Oven Chicken has an equilibrium problem. Oven chicken is radiating at 918 w/m2, but stupid Black Oven is radiating at 25w/m2 more than room temperature. PLUS, it has an area 10 times that of Oven Chicken, so Oven Chicken is now getting ANOTHER 250 w/m2.

    Oven Chicken’s temp now goes up to 126 degC and it is radiating 1443 w/m2. But that’s 250 w/m2 going back toward Black Oven, with a 10 times surface area….so Black Oven is now getting ANOTHER 25 w/m2 more than before. But it has two surfaces to radiate from, so another 12.5 w/m2. So Black Oven has to get to 455.5 w/m2 or 26.4 degC.

    Now Black Chicken has a problem…. extra 125 w/m2 = 134.8 DegC.
    Now Black Oven is out of whack, extra 6.25 w/m2 = 27.4 DegC

    Now Black Chicken is getting another 62.5 w/m2….
    Then Black Oven get’s an extra 3.125 w/m2
    Black Chicken 31.25 w.m2
    Black Oven 1.66 w.m2

    And so on until the numbers get so small they don’t matter. Exactly the same process as Ira described for GHG’s warming the earth. Exactly the numbers from SB Law. Exactly in accordance with 2nd Law of Thermodynamics. And most importantly…

    Verified by experimentation. Build an apparatus like this, measure the effective black body characteristics of both Actual Chicken and Actual Oven, and you’ll be able to apply an Actual Heat Source and predict in advance the Actual Final Temperature with in fractions of a degree. Been done, many times, repeatedly, since 1879 when Stefan first formulated the equation. Theory and practical verfication match precisely.

    Sorry if that bursts your bubbles, argue cold things can’t heat warm things until you are blue in the face, the theory has been verified by experimentation thousands of times for more than a century. I consider the claims of the AGW Alarmongerers to be a total farce, but that has to do with a whole lot of other factors. On this issue they have it right. Stop getting your panties in a twist about this and hit them on the dozens and dozens of things they have wrong instead.

    At each step the amount of temp

  297. Richard E Smith says:

    The temperature of the chicken cannot be higher than that determined by the quantity of thermal radiation it is receiving whether it inside an oven or not. This is similar to Ira’s misunderstanding when he wrote about a soldering iron in a box reaching a higher temperature than one in the open. If the flow of radiation is blocked by the box it does not result in the heated object achieving a higher temperature than that determined by the Stefan-Boltzmann constant (taking account of emissivity).

    The flow of radiation is not “blocked” by the box. The box absorbs the radiation. Then, because it has a nonzero temperature, it emits radiation. You don’t have to call it “back radiation” if that term upsets you…In fact, there are people (e.g., this retired meteorologist http://www.ems.psu.edu/~fraser/Bad/BadGreenhouse.html ) who have recommended avoidance of that term because it tends to sow confusion. I always thought he was a bit too militant on this point, but I am beginning to see his reasoning now that I see how the term “back radiation” leads some people to say completely nonsensical things like “reflection or backradiation merely reflects exactly the same amount of energy so there can be no additional heating”.

    So when you say, “The temperature of the chicken cannot be higher than that determined by the quantity of thermal radiation it is receiving whether it inside an oven or not”, then you are correct. However, it is receiving more thermal radiation when it is in the oven because the oven walls are at a temperature higher than the surroundings (but lower than the chicken) and they radiate, just as any object does.

  298. Right on Phil – all these molecules travelling at escape velocity… we obviously live in a vacuum in which no sound is heard except the sound of one hand clapping.

    Well done. You’ve explained it nicely for me.

  299. Myrrh says:
    April 3, 2011 at 11:35 am
    Right on Phil – all these molecules travelling at escape velocity… we obviously live in a vacuum in which no sound is heard except the sound of one hand clapping.
    Well done. You’ve explained it nicely for me.>>>

    Myrrh, you clearly know nothing about physics. It takes two hands to clap. Traditional Physics. You can CATCH the clap, but that’s a biological clap not a physical clap. A physical clap requires two hands, and any assertion that a clap made with just one hand has any sound at all, not matter how quiet, is pure, unadulterated, claptrap. You’re making a fool of yourself by claiming otherwise. Look it up on any physics site, or any practical reference guide on clapping.

  300. Hans says:
    April 2, 2011 at 12:57 pm
    Backradiation acts as a perpetuum mobile …”

    I finally figured out where you have gone wrong here. You think that temperature, per se, is the same as energy.

    You think that I assume the metal box in which I hung that 400K soldering iron will get up to 399K or so. I assume nothing of the sort. The box is larger than the soldering iron and much larger than the tip. Indeed, if the box was the size of a tool box, you might be able to hold it in your hands and find it quite warm, but not burning. The tip, of course, would burn your fingers.

    When the temperatures stabilize, the radiation from the box to the free air in the room, we both agree, must exactly equal the electrical power in to the soldering iron. Since the box is so much larger surface area than the tip and shank and other hot parts of of the soldering iron, it can, acccording to SB, radiate that amount of energy at a much lower temperature than the 400K tip.

  301. There are a couple commenters in this and others of my Topic threads who I have judged to be incapable of reasoned cross-discussion. Either because they are so fixed in their prejudices or clueless in their science backgrounds. So, I just scroll through their comments without reply, for the most part, unless my eye catches something humorous or otherwise worth reading.

    I don’t get paid in cash for this blogging activity, but I do feel rewarded when my Topics score high in Page Views and Number of Responses, and when I rank high in the list of Guest Contributors. So (you know who you are) keep it up!

    Of course, the best reward is when a comment teaches me something important that I did not know before, which has occured several times in these Topic threads. THANKS! And thanks also to those WUWT readers who take the time to try to set those who honestly misunderstand on the correct track. And, of course, to those who take the time time to thank me.

    Every person adds joy to the world – some by arriving and some by leaving.

  302. Ira said that the warmth we feel from the Sun comes from the Solar energies which are the Visible and two shortwave either side, UV and Near IR, the downwelling energies of the AGW Energy Budget. None of these is Thermal. They cannot be felt as heat. What is felt as Heat is Thermal IR. This is what Ira is refusing to address in my questions. I can only conclude that it is a CON, and the refusal to engage with me on this is because it is knowingly a CON.

    Tim, I don’t believe this 1% for Thermal IR downwelling to Earth. The only thing so far I’ve been able to find on actually measuring the temperature of the light waves themselves was an experiment which said to first cover the thermometer bulb with black tape before doing the prism experiment. But, at every colour of light shown on the white paper Thermal IR is also being measured. The experiment however stopped there, and claimed that what was being measured was discrete colours visible and invisible. Nonsense, right? Now, mercury is practically a perfect reflector of visible light, so to get a true reading of each visible colour one would have to do the same experiment without the bulb being black taped, and then this result subtracted from the first. So many scientists here, do you all agree that would give me a true reading of each colour discretely?

  303. Myrrh says:
    April 4, 2011 at 7:54 am
    Ira said that the warmth we feel from the Sun comes from the Solar energies which are the Visible and two shortwave either side, UV and Near IR, the downwelling energies of the AGW Energy Budget. None of these is Thermal. They cannot be felt as heat. What is felt as Heat is Thermal IR. This is what Ira is refusing to address in my questions. I can only conclude that it is a CON, and the refusal to engage with me on this is because it is knowingly a CON.

    They are all able to warm an object when absorbed. The refusal to engage you is because you won’t listen and keep on trotting out the same garbage.

  304. Phil – what is the problem here? It is absolutely bog standard science fact that the warmth we FEEL is Thermal IR. That’s how we, scientists every one of us, can tell. The heat we feel from a fire is THERMAL IR. We each of are able to test for ourselves whether something is giving off heat. If it’s giving off heat it’s Thermal IR. It’s that bloody simple.

    Something doesn’t have to be hot enough to glow red to be able to feel the Thermal IR it’s giving off, a hot ring on a stove set to simmer won’t glow red, cooling embers of a fire.

    We cannot feel heat from UV. It can burn us if intense enough, but we do not feel it as heat, it is not Thermal IR. It doesn’t even penetrate the skin! It barely gets through to the edge of the first layer of skin (of three layers), the epidermis. UV may well be heating a small millimetre of the surface of the Earth, but that’s about it. If that’s “converting to heat energy, Thermal IR from the Earth upwelling”, then just how much is it?

    Visible light doesn’t penetrate that much further, we can’t feel it as heat. Near IR is not hot either.

    So, I’m still demanding proof that it’s these Solar energies of AGW that are actually heating the Earth and producing all that Thermal IR upwelling.

    Unless you can provide this, your AGW Energy Budget is pie in sky, fiction.

    And, I think it absolutely deplorable that you treat my questions about this with such arrogant disdain, vaunting yourselves as scientists and belittling me. This is AGW’s basic premise on which all other energy calculations are made. You have a scientific obligation to answer this. If you don’t, then you are either con artists or, not scientists.

  305. Myrrh says:

    So, I’m still demanding proof that it’s these Solar energies of AGW that are actually heating the Earth and producing all that Thermal IR upwelling.

    Unless you can provide this, your AGW Energy Budget is pie in sky, fiction.

    No, the energy budget is based on measured quantities. Whether or not you can “feel” visible light, which involves a lot of neurological and other complex issues in addition to physics is irrelevant, although there is no reason to believe you are right on that anyway.

    You have a scientific obligation to answer this. If you don’t, then you are either con artists or, not scientists.

    We don’t have any responsibility whatsoever. Out of some vague sense of wanting to help educate someone who is seriously confused, we have tried to engage you and we have been met with belligerent assertions of nonsense at every turn. You have made the choice to wallow in your own ignorance despite our best efforts to try to set you on a course to understanding. You are clearly too ideologically committed to what you believe to be disabused of that ignorance. It is ridiculous to claim that we have some sort of obligation to do something that you are making impossible for us to do.

  306. davidmhoffer says that a blackbody chicken in an oven emitting all the radiation it is absorbing will be heated up further by its own emissions returning to it from the inside of the oven it is sitting in. (David’s oven is assumed to have a thin blackbody skin.) So a cooler object (the oven walls) is heating a warmer object (the chicken).

    In his example the blackbody chicken receives 500 watts from a heat source, but the oven walls radiate half back to the chicken warming it up. The chicken is now emitting 750 watts and in David’s example, if I understand it correctly, half of this extra 250 watts is re-emitted back to the chicken and then half of this is re-emitted (250+125+62.5 etc) i.e. an extra 500 watts at no extra expense and the chicken is emitting twice as much energy as is being provided by the only heat source. (David’s decimal points and zeros are a bit wayward, but that is what I think he is saying. It certainly resembles standard greenhouse physics.)

    I have some observations about this.

    When the chicken is receiving 750 watts why does it not emit the whole of the 750 back to the walls? When half of this is returned we would then have an additional 375 watts. Add this to the 750 and we now have 1125. Half of 1125 is then re-emitted and so on. Such heating could go indefinitely.

    David says that his thought experiment has been proven thousands of times and the second law is not broken because radiative equilibrium is reached as 500 watts enter the oven and 500 go out. However, there seems to be the creation of energy inside the oven. The effect is the same as having two heating elements of 500 watts. Please point me to one of these experiments because it seems to me that if it is true, we only need to provide 500 watts to produce 1kw (or more?) and save a lot of energy in the cooking or heating process.

    The thought experiment seems to show that radiative insulation is impossible.

    Further, if instead of blackbody oven walls we had one that reflected all the radiant emissions from the chicken (like the solar oven that started this debate) the mutual heating between the chicken and the oven would produce so much thermal energy that it would explode making rather a sorry mess of the kitchen (not to mention the chicken). I know David does not think that this can happen (obviously) but it is not apparent to me why not, given that he believes that radiation from the surroundings of the chicken can heat the chicken up thereby adding more radiation from the chicken and so on – a perpetuum mobile.

  307. Myrrh;
    And, I think it absolutely deplorable that you treat my questions about this with such arrogant disdain, vaunting yourselves as scientists and belittling me.>>>

    Having attempted reasoned, factual, logical and detailed explanation founded upon theory verified by experimentation, and having failed utterly to get a single point through to you, we have concluded that you are protected by some sort of Magical Armour of Density.

    Our attacks on this seemingly impervious protection are now taking the form of sarcasm, belittling, and anything else anyone can think of to try an find a chink in the armour through which to slip a tiny sliver of knowledge.

    I believe myself that the effort may be futile, as it is well known that what you don’t know can’t hurt you, making some people invulnerable by default. That is how the Magical Armour of Density works. The Magical Cloak of Invisibility works in the opposite fashion, but has the same effect. Instead of giving the wearer such high Density that it can aborb nothing, the Magical Cloak of Invisibility simply allows everything to pass right through the wearer and so the wearer still absorbs nothing.

  308. “I know David does not think that this can happen (obviously) but it is not apparent to me why not, given that he believes that radiation from the surroundings of the chicken can heat the chicken up thereby adding more radiation from the chicken and so on – a perpetuum mobile”.

    Agree. I’m still waiting for an explanation on this point.

  309. Davidmhoffer -Avoiding giving me the specific information I have asked for is no reply.

    I am demanding a specific scientic proven fact, if you can’t provide it say so. Ad hominem attacks on me because of your own ignorance of the answer is getting tiresome.

  310. Myrrh, you are being ridiculed because your responses are lacking.

    For example, your response to my question was “Visible light is easily reflected…What we see as colour in our world is what is reflected. The green light is reflected by plants, which absorb blue and red light for photosynthesis, so in reflecting back at us we see green.”

    Which didn’t answer my question at all. What I asked you was what do you think happens to the energy of the visible light that isn’t reflected? Yes, everything around us reflects visible light. Now imagine every surface covered with a thin mirror, with the sun overhead. The difference in energy between that reflection and what you actually see when you look around is, approximately, the energy of the visible light that you need to account for.

  311. Richard E Smith says:
    April 4, 2011 at 10:28 am
    I know David does not think that this can happen (obviously) but it is not apparent to me why not, given that he believes that radiation from the surroundings of the chicken can heat the chicken up thereby adding more radiation from the chicken and so on – a perpetuum mobile.

    It’s feedback not a perpetuum mobile, the principle is used by engineers in radiation transfer calculation all the time!
    A classic example that I have given here before is of a thermocouple measuring the temperature in a gas turbine combustor exhaust. If a bare thc is used it will heat up and reach an equilibrium temperature (in radiative equilibrium with its surroundings). If a tubular silica radiation shield is put around it the measured temperature increases because it is now in Eq with the shield which is in Eq with the cooler surroundings. The radiation from the shield heats up the thc even though it is cooler than the thc.

  312. Joel Shore says:
    April 3, 2011 at 11:24 am

    “Hans says:

    The net power equals sigma *A (T1^4 – T2^4) so the power reduces when the shell gets warmer. You mix energy and power here. Energy in-out will always be the same, but not the power.

    Power is just energy per unit time, so your last sentence is basically non-sensical.”

    What is non-sensical? You have 100 J radiated away in 1 hour and get 100 J out, you have 100 J radiated in 10 hours and get 100 J out. So the energy in = out always no matter what and is just a given fact, the power is what results from the temperatures.

    “And, yes, the power does get smaller as the temperature of the shell increases…That’s the whole point. Now take that to its logical conclusion and you will understand why the temperature of the object must increase.”

    Less power is the conclusion. It seems to me that you think there exists a conservation of power principle that will dictate a temperature, and that doesn’t exist. First you have a temperature and the Second Law that tells you if any energy will flow and how fast and in which direction, then the First Law for bookkeeping. It flows from high to low, from source to receiver and the power will have to adjust to things happening at the end, not the other way around.

    Any restriction placed in a flow will reduce the flow and there is no physical effect that wants to restore it to the original flow.
    I have a pump pushing 2 m^3/s through a pipe, but it’s to much so I have a valve to reduce it to 1 m^3/s. But you want to tell me that this is impossible, because the pressure will go up because I had 2 m^3/s so I will always get 2 m^3/s because there is still electricity turning the pump.

    ” Besides the in-out balance is First Law that just is what it is but tells you nothing about what will happen, it’s the Second Law that tells you what will happen.
    Without a gradient (delta T) nothing happens and in = out = zero.
    And think of cause and effect, it’s not power creating temperature but temperature creating power. Without temperature difference there will be no power, without power there will still be a temperature.

    Okay, so to interpret what you are saying charitably: You are saying if the temperatures are the same, there will be no heat flow…i.e., the power will be zero.”

    You think there will be power?

    “So what, pray tell, is going to happen to the energy that the object is receiving (e.g., from the sun in the case of the earth or from the conversion of electrical energy into thermal energy in the case of the soldering iron)? You have just told me there is no energy flow away from the object. So, how is it not going to increase its thermal energy and hence its temperature?”

    You keep pulling things out of context don’t you? Am I describing the sun – earth situation with ‘Without a gradient (delta T)’ you think? Surely your smarter than that.
    So tell me now is there a power flux in Q = sigma *A (293^4 – 293^4)? Is there a temperature? So what drives what?

  313. Richard E smith says:

    Further, if instead of blackbody oven walls we had one that reflected all the radiant emissions from the chicken (like the solar oven that started this debate) the mutual heating between the chicken and the oven would produce so much thermal energy that it would explode making rather a sorry mess of the kitchen (not to mention the chicken). I know David does not think that this can happen (obviously) but it is not apparent to me why not, given that he believes that radiation from the surroundings of the chicken can heat the chicken up thereby adding more radiation from the chicken and so on – a perpetuum mobile.

    First of all, the walls are not reflecting the energy from the chicken. They are absorbing energy and they are (re-)emitting energy. They, like the chicken, will find the temperature at which the amount of energy they are emitting balances the amount that they are absorbing and that will be the steady-state.

    What david is describing to you is the iterative process by which one can figure out what is going to happen in this case. It is no more a “perpetuum mobile” than my describing the course of a runner in a 1 mile race as, “He first runs the first half mile, then runs the next quarter mile, then runs the next eight mile, …” It is a converging series that converges to the answer that satisfies the known laws of conservation of energy and of radiative emission from objects.

    One could just write down the equations that lead to the steady-state solution to this process without going through the iterative technique that he describes. However, the iterative technique is useful in that it describes how things actually evolve over time…i.e., how the chicken in the oven once it has heated up to the same temperature as the one on the counter is still receiving more energy than it is emitting and thus must heat up more over time. It also makes clear that, contrary to your claim that this somehow violates the conservation of energy, such a higher temperature is in fact ***REQUIRED*** by conservation of energy. When the chicken in the oven is only as hot as the one on the counter got then it is still emitting less than it is absorbing…and hence, unless energy was magically disappearing, it will continue to heat up further.

    However, there seems to be the creation of energy inside the oven. The effect is the same as having two heating elements of 500 watts.

    Again, it is actually required by conservation of energy that the chicken heat up higher. It is your proposed solution, not his, that violates conservation of energy, as it would require the destruction of energy to prevent the chicken in the oven from getting hotter than the one on the counter. Furthermore, there is no more “creation of energy inside the oven” than there is “creation of energy” when I put on a coat to keep myself warmer than I would be if I didn’t put a coat on.

  314. Ira Glickstein says:
    April 3, 2011 at 6:16 pm

    Hans says:
    April 2, 2011 at 12:57 pm
    Backradiation acts as a perpetuum mobile …”

    I finally figured out where you have gone wrong here. You think that temperature, per se, is the same as energy.”

    No I don’t, intensive versus extensive properties.

    “You think that I assume the metal box in which I hung that 400K soldering iron will get up to 399K or so. I assume nothing of the sort. The box is larger than the soldering iron and much larger than the tip. Indeed, if the box was the size of a tool box, you might be able to hold it in your hands and find it quite warm, but not burning. The tip, of course, would burn your fingers.”

    Agreed, but following your iron in a box idea you imply that the greenhouse effect heats up the core of the sun right?

  315. Steve – as in the example of the plants, visible energies are used by us. UV gives us Vitamin D, etc. There’s a whole lot more to life than heat in and heat out.. Near IR is used in medical therapies, it does not burn the cells targetted…

    Nor does it burn the finger when it is used in diagnosis – http://equipmentexplained.com/physics/respi_measurements/oxygen/oximeter/pulse_oximeter.html

    Visible light is not a good penetrator of the body, but the longer wave length of red can be used with Near IR to test oxygen levels in the blood. This uses red visible at approximately 650 nm and Near IR at 950 nm. It’s a fascinating read, real life scientific application our of lights real properties understood. So, Note Well, it makes adjustments for basic laws because real life is not an imaginary ideal, and if it didn’t make the necessary adjustments the results would be nonsensical.

    Carbon Dioxide does not rise up into the atmosphere to diffuse throughout it becoming thoroughly mixed by Brownian motion, nor does it travel through the atmosphere at superspeeds of the imaginary unreal ideal gas..

    ..only in Cloud Cuckoo land.

    Your ridicule is mis-directed, on par with your mis-application of the laws of physics. Step back through the mirror, come back to the real world.

  316. Myrrh says:
    April 4, 2011 at 11:30 am
    Davidmhoffer -Avoiding giving me the specific information I have asked for is no reply.>>>

    Avoiding the specific information that you have been given is proof of the existance of a Magical Armour of Density. Although it does raise the possibility of the existance of a Magical Armour of Avoidance.

  317. Let me see if I can answer the question — “which came first, the chicken or the heat”. To understand this, you need to look specifically where the energy goes.

    Let me make several assumption (none of which is absolutely critical, but which give us numbers to work with):
    * a spherical, ε = 1 chicken with a surface area of = 0.05 m^2.
    * a 500 W heater for the chicken which we can turn on or off (the exact nature doesn’t matter, but an electrical resistance heater would be simplest).
    * an un-insulated (and unheated!) oven with a surface area of 1 m^2 (ie 20x larger than the chicken). This oven has ε = 1 inside and out.
    * an ε = 1 room which is held at 300 K (room temperature).
    * the spaces between these are all well insulated, so there is no conduction of heat — only radiation.
    * all results are after equilibrium is reached.

    If the chicken was sitting in outer space far from any star or planet (ie not in the oven or room), it would eventually reach an equilibrium situation where it would radiate 500 W, which would be a temperature of 648 K (from BB radiation laws P = εσAT^4). If the heater was turned off, the temperature would head toward 0 K ( or 2.7 K since there is a LITTLE IR radiation in space).

    Suppose the chicken was placed inside the oven, and the oven is in outer space. With the heater running, the oven will have to eventually come to equilibrium radiating 500 W from its 1 m^2, which would require a temperature of 306 K for the oven surface. The oven at this temperature will radiate 25 W to the chicken. The chicken receives 25 W from the oven walls and 500 W from the heater; its temperature must be 656 K (ie 8 K warmer than if the chicken was “bare”).

    Suppose the chicken w/heater was placed inside the oven, and the oven is placed inside the room. The room will radiate 459 W to the oven. The oven needs to radiate this 459 W AND the 500 W from the heater = 959 W, which would be 361 K (ie the outside of the oven will be 61 K above room temp). The oven at this elevated temp will radiate 48 W to the chicken, so the chicken radiates 548 W, which requires a temp of 663 K (ie 7 K warmer than the oven in out space, and 15 K warmer than the bare chicken).

    Finally, suppose the chicken without heater was placed inside the oven, and the oven is placed inside the room. The room @ 300 K will radiate 459 W to the oven and the oven @ 300 K will radiate 459 W to the room. The oven 23 W to the chicken and the chicken will radiate 23 W to the oven. Everything will be @ 300 K

    So ..
    * Without a heater, the chicken will be at the temperature of the surroundings
    * With a heater, the chicken will have some temperature above the surroundings. How much depends on the situation, but in no case will the temperature grow without bounds, and in no case will the feedback grow without bounds.

    NOTE: I tried one more case — the “oven” wrapped closely around the chicken (but not touching). In this case at equilibrium the oven radiates 500 W outward from its 0.05 m^2 and 500 W inward (the maximum possible). The chicken then gets 500 W from the oven and 500 W from the heater. The oven reaches 656 K (he same as the chicken by itself in outer space), while the the chicken raises to a temperature of 771 K. This is the highest the chicken could reach without either powering up the heating element or adding more layers to the oven or adding other insulation.

    Of course, the original “powerless chicken oven” is absurd, violating basic physics. The powerless oven DOES help keep the chicken warm IF the chicken has some external power source, just like the powerless atmosphere helps keep the earth warm if the earth has some external power source (ie the sun).

  318. RJ; Richard E Smith;
    “I know David does not think that this can happen (obviously) but it is not apparent to me why not, given that he believes that radiation from the surroundings of the chicken can heat the chicken up thereby adding more radiation from the chicken and so on – a perpetuum mobile”.
    Agree. I’m still waiting for an explanation on this point.>>>

    Perhaps this will help? Add up the numbers in my model at each step. Remember that the Oven was modeled as having 10 times the surface area of the chicken, so the watts below are TOTAL not per meter squared.

    Chicken Oven
    0 0
    +500 => +250
    +250 => +125
    +125 => + 62.5
    + 62.5=> + 31.25
    + 31.25=>+15.625
    +15.625=>+7.8125
    +7.8125 => +3.906
    3.906 => +1.953
    1.953 =>.977
    .977 =>.488
    .488 => .244
    .244 => .122

    TOTAL WATTS OUT FROM OVEN = 499.76

    Continue the series for as long as you wish, you will eventually
    get to 499.9999…. watts coming out of the oven to the outside, matching exactly the watts going into the chicken in the first place.
    +2.1 => 1.0

    Beam of energy going into Oven = 500 watts
    250+125+62.5+31.25+16.5+8.25+4.1+2+1= 500.6

    If I’d bothered with the decimal places exactly it would have come

  319. Mods!
    Aw fiddlesticks, strike those last 4 lines, I didn’t run enough decimal places the first time got the wrong numbers, and then forgot to snip it off the end of the comment.

  320. “The heat we feel from a fire is THERMAL IR. ”

    This might be a big part of problem. The “established physics” typically defines “thermal IR” as 3 um – 15 um. A hot fire at 1/3 the temperature of the sun will emit only about 30% of its energy as “thermal IR” while the rest will be almost entirely “non-thermal IR”. Cooler fires will have a larger proportion of “thermal IR” but a significant part of the IR we feel will be other than thermal IR

    It is always important to have common definitions for terms in a discussion, like “thermal IR” or “heat”.

  321. Your getting closer Myrrh. You are admitting that there is non-reflected visible light responsible for… well, ultimately, just about every chemical bond in every organic molecule. So that’s some of your non-reflected energy of visible light, stored among all the biomass of earth (and it get’s released as heat when you burn it).

    Now, go back to imagining the difference between a mirror-covered landscape and the actual landscape for the entire earth. Plants don’t cover every surface, right? Have you bothered to try and calculate, just a little, the energy difference in reflected visible light between a mirror Pacific Ocean and the actual Pacific Ocean?

  322. Hans says:

    What is non-sensical? You have 100 J radiated away in 1 hour and get 100 J out, you have 100 J radiated in 10 hours and get 100 J out. So the energy in = out always no matter what and is just a given fact, the power is what results from the temperatures.

    No, Hans. We are talking about a case where power is being supplied constantly. If power going in at a rate of 100 W, i.e., 100 J / s and power is going out at a rate of 10 J / s, then after 1 second, the amount of energy would have increased by 90 J, after 2 sec by 180 J and so on. Hence, energy would be building up over time, which would result in the temperature of the object rising. You can’t just set the temperatures independently…If you are putting a fixed amount of power in, the temperatures will adjust until they find the point where the power being emitted equals that coming in.

    The statement of the problem is not that the sun sends a one-time contribution of 240 J/m^2 of energy to the earth. It sends (on average) 240 J/m^2 every second. If the earth doesn’t send back (out into space) energy at an average rate of 24o J/m^2 every second then the energy of the earth will increase, raising its temperature and (by the Stefan-Boltzmann Law) increasing the rate at which it emits energy until it is sending 240 J/m^2 every second back out into space.

    I have a pump pushing 2 m^3/s through a pipe, but it’s to much so I have a valve to reduce it to 1 m^3/s. But you want to tell me that this is impossible, because the pressure will go up because I had 2 m^3/s so I will always get 2 m^3/s because there is still electricity turning the pump.

    Okay…Nice analogy, Hans. Except you have made an error in solving the problem because you are mixing two things here, one being volume and one being the energy. This analogy, when looked at properly, actually allows you to understand why conservation of energy means that power in = power out.

    Power is energy per unit time. Likewise, volume flow rate is volume per unit time. So, the analogy with having conservation of energy in the systems we’ve been talking about is having conservation of volume in the water flowing in a pipe analogy. (It is NOT having a constant amount of power being used to drive a pump…that mixes energy and volume.) So, for an incompressible fluid completely filling a pipe, the volume flow rate must indeed be the same going in as coming out. If it is not, fluid will be piling up somewhere in between, which it can’t do for an incompressible fluid in a pipe. (If the fluid is compressible, then it is mass and not volume that must be conserved.) For example, if there was 2 m^3 /s of fluid going into a pipe in steady flow, there has to be 2 m^3 /s coming out. Thus, in your example, if you are forcing fluid to go into the pipe at 2 m^3 / s and you put a constriction in the pipe but continue to put 2 m^3 /s of fluid into the pipe, you will continue to get 2 m^3 /s of fluid out of the pipe.

    What your point-of-view would say, by contrast, is that you could be putting 2 m^3 / s into the pipe but only be getting 1 m^3 s out of the pipe. However, that would be impossible because the volume of fluid must be conserved. If you study this example and understand it, you will understand how Conservation of Volume implies that the volume flow rate in has to equal the volume flow rate out just as Conservation of Energy implies that the power in has to equal the power out.

    You think there will be power?

    No…At that particular instant of time there will not be any power being emitted. However, the situation that you describe cannot persist because the rate at which energy flowing in (i.e., the power) does not equal the rate at which it is leaving, so the temperature of the object will rise and then the temperature difference between the object and the shell surrounding it will no longer be zero.

    Hans: To be honest, I am beginning to doubt that you and Myrrh are really materially-different. Please show me otherwise by actually trying to open up your mind and learn something.

  323. Hans asks @ April 4, 2011 at 2:12 pm

    “Agreed, but following your iron in a box idea you imply that the greenhouse effect heats up the core of the sun right?”

    A point of clarification — are you asking about the green house effect of the sun’s atmosphere helping keep the core of the sun warm? In which case, yes, the insulating and “greenhouse properties” of the sun’s atmosphere definitely help keep the temperature of the core higher than it would be without those layers. (As opposed asking if the earth’s greenhouse effect and thermal radiation from the earth helping keep the sun warm, in which case it would in principle, but the effect would be such an infinitesimal amount that it would not be measurable.)

  324. Richard E Smith says:
    April 4, 2011 at 10:28 am
    davidmhoffer says that a blackbody chicken in an oven emitting all the radiation it is absorbing will be heated up further by its own emissions returning to it from the inside of the oven it is sitting in. …

    Let us do it this way to see if you can be convinced. Forget about emissions and consider cold cash instead.

    OK, we have three characters, Mr. Heat Source, Miss Chicken, and Mr. Wall.

    DAY #1: In the morning, Mr. Heat Source gives Miss Chicken $500. In the afternoon, she gives the $500 to Mr. Wall. (So Miss Chicken has zero cash in the afternoon.) That evening Mr. Wall blows half ($250) on booze and gives the remainder ($250) back to Miss Chicken. (So she has $250 in the late evening of that day.)

    DAY #2: In the morning, Mr. Heat Source gives Miss Chicken another $500. She has $250 left from the previous day, so she now has $750. In the afternoon, she gives the $750 to Mr. Wall. (So Miss Chicken has zero cash in the afternoon.) That evening Mr. Wall blows half ($375) on booze and gives the remainder ($375) back to Miss Chicken. (So she has $375 in the late evening of that day.)

    DAY #3: In the morning, Mr. Heat Source gives Miss Chicken another $500. She has $375 left from the previous day, so she now has $875. In the afternoon, she gives the $875 to Mr. Wall. (So Miss Chicken has zero cash in the afternoon.) That evening Mr. Wall blows half ($437.50) on booze and gives the remainder ($437.50) back to Miss Chicken. (So she has $437.50 in the late evening of that day.)

    DAY #4: In the morning, Mr. Heat Source gives Miss Chicken another $500. She has $437.50 left from the previous day, so she now has $937.50. In the afternoon, she gives the $937.50 to Mr. Wall. (So Miss Chicken has zero cash in the afternoon.) That evening Mr. Wall blows half ($468.75) on booze and gives the remainder ($468.75) back to Miss Chicken. (So she has $468.75 in the late evening of that day.)

    This goes on for many days. I hope you can see that Miss Chicken’s cash at the end of each day has been increasing. It went from $250 in the late evening of the first day to $468.75 on the late evening of the fourth day. Her cash in the morning, which she gives to Mr. Wall each afternoon, has gone from $500 the first day to nearly twice that, $937.50 on the fourth day.

    If we continue this tale long enough, things will stabilize, and Miss Chicken’s cash in the morning that she gives to Mr. Wall in the afternoon, will be around $1000. Mr. Wall, for his part, will be blowing about $500 on boooze each evening. So, money in = money out. Mr. Heat Source will give $500 each morning and Mr. Wall will blow the same amount each late evening. Miss Chicken will be have an account of $1000 each morning and early afternoon, which is twice what Mr. Heat Source gives her each morning.

    No cash had been created by the transactions between Miss Chicken and Mr. Wall. All meets with the approval of the accountants.

    Now do you get it? Mr. Heat Source is the Sun. Miss Chicken is the Earth, and Mr. Wall is the Atmosphere. I hope this convinced you. Richard E Smith

  325. David,

    One small comment — I think your oven infinite series above actually requires the “oven” to be close around the chicken. If the oven is 10x more area, then if the oven emits 250 W out into space, it will only send 25 W back to the chicken (since the chicken is 1/10 as much area as the oven. By the time the oven is emitting 500 W to space, it will be emitting 50 W to the chicken.

    The energy balance works in either case. If the system has reached equilibrium:

    10x AREA OVEN
    Chicken receives 550 W (500 W from heater; 50 W from oven)
    Chicken sends out 550 W (all to the oven)
    NET = 0 W

    Oven receives 550 W (from chicken)
    Oven sends out 550 W (500 to outer space; 50 to the chicken)
    NET = 0 W

    “outer space” receives 500 W (from oven)
    “outer space” sends out 500 W (electrical energy to chicken)
    NET = 0 W

    1x AREA OVEN
    Chicken receives 1000 W (500 W from heater; 500 W from oven)
    Chicken sends out 1000 W (all to the oven)
    NET = 0 W

    Oven receives 1000 W (from chicken)
    Oven sends out 1000 W (500 to outer space; 500 to the chicken)
    NET = 0 W

  326. Tim,
    Certainly!

    At day’s end, the oven must always heat up enough to emitt 500 watts to the outside to balance the 500 watts going in from the beam.

  327. Tim Folkerts says:
    April 4, 2011 at 4:12 pm

    Re: “The heat we feel from a fire is THERMAL IR.”

    This might be a big part of problem. The “established physics” typically defines “thermal IR as 3 um-15 um. A hot fire at 1/3 temperature of the sun will emit only about 30? of its energy as “thermal IR” while the rest will be almost entirely “non-thermal IR”. etc.

    Nope, the big part of the problem here is that non-thermal IR is not felt as heat. If you can feel heat you are feeling thermal IR. Adjust your um’s accordingly.

    AGWScience Energy Balance has UV Visible and Near IR – Near IR is not thermal. See NASA page for “ESTABLISHED PHYSICS”. You cannot feel Near IR, it is not hot. I repeat, in established physics, Near IR is not hot. You cannot feel it.

    You cannot feel Visible light, it is not hot. You cannot feel UV, it is not hot.

    It is always important to have common definitions for terms in a discussion, like “thermal IR” or “heat”.

    Damn right. Established physics says the heat we feel from fire and sun is Thermal IR.

    Adjust your thinking accordingly.

    Now answer my question.

    Ditto Steve for the above. And, The AGWScience Energy Budget says that Solar, UV, Visible and Near IR, heat the Earth. These are not Thermal energies. Show how these non-thermal energies produce the updwelling Thermal IR claimed. As Trenberth diagram I posted, and Ira; the claim is that it is these SPECIFIC energies which HEAT the EARTH and convert to THERMAL IR to upwell. Land and Sea both.

    Stick with the question I’m asking, don’t get distracted.

    I am demanding actual proof that these non-thermal energies do what it says in the AGWScience Energy Budget. I repeat. These are not Thermal IR in ESTABLISHED, TRADITIONAL PHYSICS.

    And, I don’t give a damn how many AGWScience reference you give to claim that traditionally non-thermal energies are thermal, they’re not. That AGWScience has corrupted traditional understanding is the REAL PROBLEM here. I am getting weary of explaining this to you. The NASA page describes the difference. The newworldencylopedia admits it is still traditional teaching.

    If you can feel heat, you are feeling Thermal IR. Full stop.

    Now, answer my question.

  328. Ira

    Re your money example above

    But does energy work this way. Energy is more like a teacher (sun) passing knowledge to a student (earth). The student can only take in so much knowledge and can not increase the teachers knowledge by passing the same knowledge back. But the student can increase the knowledge of another student (CO2) with less knowledge etc.

    IMHO the chicken and oven example in a very simple way demolishes the GHG theory as I understand it. This oven is impossible yet the GHG theory seems to be based on this concept.

    If the GH theory is correct and an 100% CO2 environment reflected back 100% of the radiation (rather than 50% or less). What then. The planet would continue heating even with only one days energy from the sun. This is clearly nonsense yet seems to be the concept the GHG theory is built on.

    Either CO2 reflects back radiation from the earth further heating the earth or it does not. In simple terms I can not see how the oven example is not correct. And complex explanations do not change this. Nor does the money example.

    I agree with the slayers book as others seem to on here. This thread has only reinforced my support for the sections on backradiation in this book.

  329. RJ says: “IMHO the chicken and oven example in a very simple way demolishes the GHG theory as I understand it. ”

    Then I humbly suggest that you do not, indeed, understand it. Analogies — “energy is like money” or “energy is like knowledge” — are just scratching the surface. Analogies are great for getting across some simple ideas, but you also have to know what is good and what is bad about the analogies to use them to make any conclusions.

    To really understand, you need to be able really use the equations. For instance, I gave some specific numbers for some specific cases a day ago based on my understanding of physics. The results agree with David’s approach using infinite series. They show clearly that the temperature of the chicken in the unheated oven is indeed warmer than the chicken outside the oven. They show clearly that the money analogy is better then the knowledge analogy.

    If someone wants to look for an error in the calculations and show how the calculations are wrong, I’d be happy to discuss it. That is how science moves ahead. Which of my numerical results are specifically wrong and what should they be instead? The simple truth is that you really need to be able to do these calculations to seriously enter into the conversation.

    If, instead, someone wants to simply say something like “I heard once that only thermal IR can be felt” even in the face of both experimental evidence and theoretical results — well — there is not much that can be done. If someone with such opinions wanted to calculate (based on any reasonable assumptions he wanted to make) how much energy he received on a sunny day from various wavelengths of the solar EM radiation and from various wavelengths of the earth’s EM radiation, and then show how only the wavelengths between 3 um and 15 um can elicit a response from temperature-sensing nerves, then I would be willing to take the conversation further.

  330. RJ says:
    April 5, 2011 at 2:24 am
    Ira

    Re your money example above

    But does energy work this way. Energy is more like a teacher (sun) passing knowledge to a student (earth). The student can only take in so much knowledge and can not increase the teachers knowledge by passing the same knowledge back. But the student can increase the knowledge of another student (CO2) with less knowledge etc. …

    RJ, IMHO you are wrong here on a couple of levels. (1) Energy/Matter cannot be created nor destroyed, so it is not at all like knowledge, and (2) I have learned a great deal from my students in the undergrad and grad system engineering courses I have taught at Binghamton University and currently teach online at the University of Maryland. Heck, I even learned from my students at the Brandeis University Summer Odyssey and they were “merely” bright High School students who elected to spend a month of ther summer vacations to get a taste of college. I learned from all of them despite the fact of my decades of successful experience doing real work as a System Engineer.

    I learn continually in my interactions with others – including especially here at WUWT. I have probably learned more in doing this “Visualizing” series than many of the WUWT readers, and I have thanked those who helped me understand better. It is never too late!

    I hope you and some others who have fixed, non-scientific opinions about the role of certain gases in the Atmosphere in keeping the Earth at a livable range of temperatures would take the opportunity to learn the accepted science. Accepting the truth of so-called “greenhouse” gases (GHG) and the small contribution of our burning of fossil fuels and land use changes to mean temperature levels does not mean that you subscribe to the outlandish Catastrophic CAGW theories of the official climate Team. The only way to fight darkness is with light. You can’t fight darkness with greater darkness.

    Good luck!

  331. Frankly, Myrrh, I have no idea what question it is that you want answered.

    So … phrase your question and/or hypothesis in precise scientific/mathematical language. Unless otherwise stated, I will assume you agree with wikipedia for definitions of any terms you use (eg “thermal IR” or “heat”) and you agree with wikipedia for equations (eg Stefan-Boltzmann law). [Not that wikipedia provides definitive knowledge, but at least it gives us a starting point for further discussion.]

  332. RJ says:

    But does energy work this way. Energy is more like a teacher (sun) passing knowledge to a student (earth). The student can only take in so much knowledge and can not increase the teachers knowledge by passing the same knowledge back. But the student can increase the knowledge of another student (CO2) with less knowledge etc.

    As Tim has pointed out to you, analogies are only useful to the extent that they illustrate actual laws of physics. To the extent that they illustrate imaginary laws of physics, they are worse than useless. Ira’s analogy illustrates the Law of Conservation of Energy.

    Your analogy, as near as I can tell, illustrates an imaginary physical law which, for lack of a better name, I’ll call the “Magical 2nd Law of Thermodynamics”. The “Magical 2nd Law” apparently says that if you put a cold object near a hot object, the cold object magically detects this and refuses to emit radiation toward the hot object. Or maybe it states that the cold object emits radiation toward the hot object but the hot object refuses to absorb it. The “Magical 2nd Law” has absolutely no experimental verification and, in fact, a lot of experimental evidence showing that it is nonsense. (It also represents a bizarre theoretical view of the universe, for whatever that is worth.)

    By contrast, the actual 2nd Law of Thermodynamics would say that the cold and hot objects both emit and absorb radiation but that the amount of radiation absorbed by the hot object from the cold object is always less than the amount of radiation absorbed by the cold object from the hot object.

    One fact it is perhaps worth making explicit: The laws of radiative transfer automatically satisfy the 2nd Law when applied correctly. If you find yourself applying the 2nd Law to somehow contradict something that the laws of radiative transfer have told you, this is an indicate that you are applying the Magical 2nd Law and not the actual 2nd Law (or, I suppose, that you applied the laws of radiative transfer incorrectly).

    IMHO the chicken and oven example in a very simple way demolishes the GHG theory as I understand it. This oven is impossible yet the GHG theory seems to be based on this concept.

    Since you have had it explained in gory detail why this is not the case, one can only guess that you cling to this belief because its suits what you want to believe rather than for any rational reason.

    If the GH theory is correct and an 100% CO2 environment reflected back 100% of the radiation (rather than 50% or less).

    Here, you have gone off the rails right from the get-go. The CO2 does not reflect radiation. What happens is that the CO2 in the atmosphere absorbs radiation and then as a result of its having a nonzero temperature, it emits radiation. While you can sometimes get away with imagining it being almost as if the CO2 reflected the radiation, such a point of view ultimately leads you astray, especially when you try to get quantitative about it.

    I agree with the slayers book as others seem to on here. This thread has only reinforced my support for the sections on backradiation in this book.

    Frankly, that is sad. What you are telling us is that you have no ability to distinguish between actual science and pseudoscientific garbage when the pseudoscience is something that you want to believe. That is something to be embarrassed about, not proud of.

  333. RJ;
    If the GH theory is correct and an 100% CO2 environment reflected back 100% of the radiation (rather than 50% or less). What then.>>>

    1. It doesn’t reflect. It absorbs and radiates.
    2. % of the environment which is CO2 has nothing to do with how much goes up and how much down. It absorbs some amount. It radiates that amount in a random direction. Random meaning about the same amount up as down. Sideways mostly a net cancellation. The % only has meaning in terms of magnitude, not ratio of up vs down.

  334. RJ;
    Either CO2 reflects back radiation from the earth further heating the earth or it does not. >>>

    It does not.
    As has been explained multiple times, it does not “reflect”.

  335. RJ;
    I agree with the slayers book as others seem to on here. This thread has only reinforced my support for the sections on backradiation in this book.>>>

    Then you have joined the ranks of the skeptics for all the wrong reasons. You’re as misguided by what you wish to believe as are your kindred spirits in the warmist community shouting that the sky is falling, not because the science says so, but because they wish to believe it.

  336. Tim Folkerts says:
    April 4, 2011 at 5:34 pm

    Hans asks @ April 4, 2011 at 2:12 pm

    “Agreed, but following your iron in a box idea you imply that the greenhouse effect heats up the core of the sun right?”

    …asking if the earth’s greenhouse effect and thermal radiation from the earth helping keep the sun warm, in which case it would in principle, but the effect would be such an infinitesimal amount that it would not be measurable.”

    Yes, infinitesimal. But as you say it’s the principle, and that is what I want to have confirmed.

    Ira, please comment.

    Joel, please comment.

  337. Myrrh;
    Ditto Steve for the above. And, The AGWScience Energy Budget says that Solar, UV, Visible and Near IR, heat the Earth. These are not Thermal energies. Show how these non-thermal energies produce the updwelling Thermal IR claimed.>>>

    As has been explained to you multiple times, all frequencies carry energy. Yet you continue to scream about frequencies being non-thermal. non-thermal doesn’t mean 0 energy. That you can’t “feel” outside of the range of thermal doesn’t mean it doesn’t carry energy. You cling to this concept, shout it repeatedly, because without this one piece of stupidity, your entire position collapses into a heap of smoldering b*****it.

    Is this a consequence of your Armour of Density, Invisibility, or Avoidance?

  338. What gives the CO2 molecule its ability to absorb and (as a consequence of physics) re-radiate radiation in infrared wavelengths that are important to the atmosphere and the climate is its number of atoms: three.’ “More than two” is the crucial concept here, because the other important infrared-capturing gases in the atmosphere also have three or more atoms in their molecules. When a gas has three or more atoms, it has modes of vibration inherent in its shape that can resonate with the frequencies of climate-affecting infrared waves. The matching enables the greenhouse molecules to intercept those waves and absorb their energy. Single atoms and two-atom molecules do not have those particular resonant modes. Just a few examples of other greenhouse gases make the point, if you count their atoms: water vapor (H2O)1 methane (CH4), nitrous oxide (N2O), and ozone (03). It is estimated that the earth would be 60f cooled without greenhouse gases. CO2 is the most variable of the greenhouse gases so it determines the tempurature of the biosphere. It is, in a natural situation, varied by other factors i.e. it follows global temp swings and that is because of the lag from the ocean’s heat collecting ability.

  339. By the way, although I (and, as I see now, David) have pointed out that CO2 absorbs and emits rather than reflects, it is perhaps worth considering what would happen if we surrounded a blackbody chicken with a perfect reflector.

    Case 1 – A chicken not being heated by any energy source: In this case, all the radiation emitted by the chicken would be reflected by the perfect reflector and end up being absorbed again by the chicken. So, if the chicken was at a temperature such that it emitted 500 W, it would also absorb 500 W. Since the power in and out are equal, the chicken is in radiative equilibrium and its temperature would not change. Of course, there is no such thing as a perfect reflector and to the extent that the reflector is imperfect (and assuming our chicken was initially at a higher temperature than the surroundings), there would be some escape of energy and the chicken would slowly cool off over time.

    Case 2 – A chicken being heated by an internal energy source (say, a radioactive chicken) at a constant power of, say, 100 W. In this case, all the radiation emitted by the chicken would be reflected by the perfect reflector and end up being absorbed by the chicken. The chicken would indeed get hotter and hotter indefinitely. However, in the real world, there are no perfect reflectors; rather, the reflector would either transmit or absorb some small fraction of the radiation. If it transmits, then eventually the chicken will reach a temperature such that the amount of radiation it is emitting is so large that the amount of this emission that is being transmitted through the reflector is 100 W. If the reflector absorbs, then by, Kirchkhoff’s Law (which is a requirement of the 2nd Law), it also emits and so the chicken and reflector will heat up until the reflector reaches a temperature such that the net amount it is emitting out into the surroundings (i.e., the amount it is emitting to the surroundings minus that it is absorbing from the surroundings) is 100 W.

    So, nothing is particularly bizarre here: An object with no internal heat source that is surrounded by a very good reflector just cools more slowly than one not surrounded by a good reflector. An object with an internal heat source can indeed get quite hot if surrounded by a very good reflector, because this reflector essentially insulates it, thus allowing the thermal energy to accumulate quite a bit until it reaches its steady-state. The latter, modulo the fact that greenhouse gases really absorb and emit rather than reflecting, is the rough picture of what actually is happening on Venus (and, to a much lesser degree, on Earth).

  340. davidmhoffer says:

    You’re as misguided by what you wish to believe as are your kindred spirits in the warmist community shouting that the sky is falling, not because the science says so, but because they wish to believe it.

    David, I hope you would agree on further reflection that this is a false equivalence. While you and I may disagree on whether the climate sensitivity is likely to be high enough to be a significant danger, I know that I for one would make a distinction between my belief that the scientific evidence points to at least a moderate climate sensitivity (> 2 K per doubling, say) and my belief that the greenhouse effect does not violate the 2nd Law. The former is still within the realm of reasonable scientific discourse, while the latter is basically an incontrovertible scientific fact.

    It is important to distinguish between what we essentially KNOW for sure and what we think the weight of the current scientific evidence shows.

  341. Ira,
    Thanks for your explanation with the cash analogy. Analogies can be useful as a way of explaining something, but does yours fairly represent what is actually happening in the real world?

    In your analogy Miss Chicken (clearly a relative of Chicken Little) gets $500 on each new day, gives all the cash she has to Mr Wall and gets half back. According to this analogy more heat (dollars) is coming in than is going out. Therefore the heat must add up until equilibrium is reached. It all sounds very plausible. (If it were not, then nobody would believe in it.)

    But back in the real world can you transfer energy like wads of cash? Can you put it in the bank? Can a body which is constantly irradiated give up all its thermal energy and then get half of it back? The fallacy is in comparing thermal energy, which flows according to a difference in temperature, to successive transfers of money.

    Argument by analogy can be misleading. On this and other threads the restriction of the flow of water out of a bath with the taps on is often used to demonstrate the principle that a blockage will raise the level of the water, so therefore a radiant blockage will raise the heat of the Earth. However, that analogy is incomplete because it does not take account of the energy in the flow of the water to the taps. If, for example, that water is supplied from an adjacent tank, the level of the water cannot by higher than the tank. Once that level is reached impeding the flow or even putting in the plug will make no difference.

    A better analogy, I would argue, is a pipe through which water flows from a reservoir. If the exit from the pipe is below the water level in the reservoir the water will flow through the pipe. High to low just like the transfer of heat from warm to cold. With no height difference, there is no flow and water level remains the same. If you partly block the pipe the water will still flow through, but none will be returned to the reservoir. If you completely block the pipe, nothing happens. None of the water returns to the reservoir. The water can only be returned to the source if energy is supplied either by raising the pipe above the level of the reservoir or pumping the water back up.

    In the constant irradiance models used to demonstrate the greenhouse effect, the Sun is pouring in a fixed quantity of watts m2 at all times. If the flow of heat from the Earth to space is blocked (as the water is blocked in the pipe) that heat cannot flow back heating the Earth even more because it does not have sufficient energy to do so – there is a temperature difference to overcome.

    Anyway – to return to Ira’s analogy.
    Mr Wall can hand back dollars to Mrs Chicken and the wealth can accumulate. But you cannot sum radiant energy like dollars. Ira says no cash has been created in his analogy – but it has been accumulated. In Ira’s analogy if Mr Wall returns all Miss Chicken’s money (reflection) she will accumulate it at $500 a day. But thermal energy does not accumulate like money that is not spent. It does not build up in the form of photons buzzing around under a radiative barrier until they burst through and restore a fictitious equilibrium. Heat cannot flow without a temperature difference. Thus the heated blackbody chicken in our make-believe oven cannot double its energy by its own emissions being absorbed and re-emitted from the oven walls. The chicken remains at a constant temperature which is as high or higher than the surrounding walls of the oven.

  342. One slight addition to Rich Monroe’s comments on IR active gases.

    Asymmetric diatomic molecules will also absorb/emit IR radiation. For example, CO, has significant IR absorption. N2 & O2 do not have significant absorption because they are symmtric (although molecules with different isotopes (eg N14-N15) are slightly asymmetric and have slight absorption)

  343. The smoking gun of CO2 induced warming is that the warming has been primarily within the troposphere and not higher.

  344. “Case 2 – A chicken being heated by an internal energy source (say, a radioactive chicken) at a constant power of, say, 100 W. In this case, all the radiation emitted by the chicken would be reflected by the perfect reflector and end up being absorbed by the chicken. The chicken would indeed get hotter and hotter indefinitely”.

    Thanks for the replies. I’m not an expert (far from it) but this topic just interests me. But I find it difficult to accept that a 100w power source could heat anything above 100w. But maybe I’m wrong. It would be great if this could be done as our energy problems would be solved.

    I will continue to debate alarmists in various ways. Few understand or are interested in the science even to a small degree so I tend to avoid this and focus on reduced living standards (due to higher energy costs) for no benefit whatsoever. If the science is discussed I use any tool available including the slayer viewpoints if appropriate.

  345. Richard E Smith says: April 5, 2011 at 10:32 am

    Analogies can be useful as a way of explaining something, but does yours fairly represent what is actually happening in the real world?

    You are quite correct that this analogy is fairly simplistic. We could improve on it by saying “Miss Chicken is charged a progressive tax based on how much total money she has (corresponding to progressively more energy being emitted as the temperature goes up). At some level — say $10,000 in her account — that tax rate is $500. Every day she is paid $500, so if her account is above $10,000 she will be losing money; if her account is below $10,000 she will be gaining money. …” But then we would find new limitations, and have to add on more and more “corrections”.

    Of course, at this point it would be better to discuss the ACTUAL energy flow. This analogy has already been pushed beyond the limit where is is providing insights and instead it is throwing up roadblocks when we try to use it.

    It’s kind of like expecting your globe to not only show where various countries are, but to also have oil reserves under the surface that you could find.

  346. Tim – look at the AGW Energy Budget as Ira has depicted it and as I linked to in Trenberth diagram. It claims that the the SOLAR energies of Visible, and the two shortwave either side of UV and Near IR are what heat the Earth, converting to heat energy, which is the upwelling Thermal IR.

    None of these are felt as heat, ergo, they are not in themselves heat energy, heat energy is Thermal IR. The AGW Energy Budget says that non-Thermal energies are heating the Earth and converting to Thermal IR. How?

    Remember, UV barely penetrates the epidermis on skin, it does not raise the temperature of the body further than the outer thin epidermis. Visible light is less intense, it is also not hot, but simple Light energy. It does not raise the temperature of the body, it is not felt as heat. Ditto Near IR. Near IR is not hot. It is not felt as heat, it is not thermal, it does not raise the temperature of the body, it penetrates the body further than Visible. Thermal IR, the heat we feel from the Sun, penetrates further, warming up the body. It is Thermal IR which warms up organic matter.

    If you are saying that Thermal IR is only 1% of downwelling, and therefore you say it can be ignored in the Energy Budget, I am saying that means that it is the 1% Thermal IR which is doing all the heating of the Earth, and the Solar energies UV,V,NR, are insignificant in the scheme of things.

    Either way, the AGWScience Energy Budget is a load of choose your favourite smelly mess, these Solar energies are incapable of raising the temperature of the Earth, of organic matter. They are incapable of converting into the heat energy as Trenberth states to upwell as this massive amount of Thermal IR.

    For example, UV may penetrate a little, a millimeter or two, any more intense and it would burn up everything. It’s energy is extremely limited in producing any heat. Visible light’s energy is not hot, it does not heat things up. Think of the lamp designed to produce mainly blue light for the initial growth of plants! Near IR is not hot, it is able to penetrate deeper in organic matter, it too is non-invasive, it does not raise the temperature of organic matter, it does not convert to Thermal IR. It is used in IR cameras because it reflects off the subject as does visible light, penetrating a bit deeper than Visible light, but still, not producing, not “converting to heat energy” of the Trenberth AGW claim.

    So, UV a little on the surface, Visible and Near IR not anything, then, how are these converting to heat energy, producing all that huge amount of Thermal IR claimed in the AGW EB? How??? So that’s the first most important question here because it establishes principle.

    Take a thousand or as many as you like remote controls, shine them into your bath full of cold water, tell me how long it takes before it heats up so you can take a bath. NASA page, Near IR is not hot.

    Just what is it contributing among the Solar energies of AGWEB in this claim that it is these energies which are converting to heat energy and upwelling in this huge amount heating the Earth globally? In traditional established science it is the Thermal IR energies which are converting to heat.

    Do you see the problem I’m having in this? There’s no scientific logic here, the properties of these energies are not able to do what is being claimed for them. What is being claimed for them according to Ira here, is that they are thermal energies. If they are thermal energies that means they are Thermal IR.

    So, that’s the first thing that has to be junked. The only thermal energy is Thermal IR, the heat energy of established traditional science.

    Therefore, for example, all the descriptions by AGW that these solar energies penetrate into the ocean and heat it up are b*ll*cks, these can’t even heat our bodies let alone hundreds of meters down in the ocean.

    So, this is my objection to the AGWScience Energy Budget. The claim that it’s these Solar energies of UV,Visible and NrIR which are converting to heat in organic matter, the Earth.

    Prove this. You can’t. Neither can all the AGWScience propaganda. What it is doing here is simply lying. By attributing to Solar the known physical properties of the IR that is capable of converting to heat. It’s this lie, this basic premise of AGW, that I am demanding to have proved.

    Until that is sorted, it is pointless to explore further what is happening in real life or in the depiction of the rest of the AGWScience Energy Budget, such as it grabbing more longer ums for its claim of Near IR.

    Traditionally then, by established science, it is the IR that by conversion creates heat which is called Thermal IR, or simply Heat, heat energy. Wherever it begins, the energies that do not convert to heat are not Thermal. Near IR is not thermal and this is established as fact in traditional science, that’s the cut off point, and likewise Visible light is not thermal and UV is not thermal – we do not feel these as heat, because they are not converting, do not convert to heat.

    I really don’t want to get distracted from this particular point, my objection and question how are these Solar energies converting to heat as claimed by AGW as per Trenberth, because this is crucial to the AGW claim and it goes against all established science, and I should be grateful if you would concentrate on it only.

    I hope that is better explained, if you’ve managed to work your way through it. No doubt if I had more experience I could have set it out more precisely mathematically/scientifically… but, sticking to English in describing concepts etc. should be more than adequate, as the adage goes, if you can’t explain your hypothesis to your cleaning lady then you don’t understand it. I’m working on improving my English. How am I doing?

  347. “But I find it difficult to accept that a 100w power source could heat anything above 100w. But maybe I’m wrong. It would be great if this could be done as our energy problems would be solved.”

    First of all, you are a little confused on the meaning of “100 W”. You heat things above a certain TEMPERATURE, not above a certain power. 100 W might heat a filament to 2000 C; it might heat an insulated tank of water to a 60 C; it might heat an un-insulated tank of water to 30 C.

    As to the second part, this can and is being done to help with energy problems — it is called adding insulation. With good insulation I can heat and cool me home with 1/2 or 1/10 the energy of an un-insulated home.

  348. “It claims that the the SOLAR energies of Visible, and the two shortwave either side of UV and Near IR are what heat the Earth, converting to heat energy, which is the upwelling Thermal IR. ”

    One step at a time. There is no point in looking at the rest of your post until we clarify what you mean here.

    When you say “heat the earth”, do you mean
    1) the whole earth on the diagram (the surface and the atmosphere and the oceans)?
    1a) I contend that 98% of solar energy traveling through space toward earth is between about 0.25 um and 4.0 um. Said another way, roughly 10 % is UV, roughly 40% is visible and 50% is IR (and only a small amount of that is >3 um which is traditionally called “thermal IR”).
    Further, I contend, that other power sources (notably geothermal & cosmic microwave background radiation & non-(visible/IR/UV) EM radiation) do not provide a significant amount of power to the earth as a whole.
    I agree that “SOLAR visible/UV/IR provide 99% of the energy that heats the earth.”

    DO YOU DISAGREE with any of this? If this solar energy is not what heats the earth as a whole, then what is your hypothesis?

    2) do you mean “the surface of the earth (excluding the atmosphere)”?
    2a) The diagram clearly shows that only ~168 W/m^2 of power to the surface is directly from the sun, while 324 W/m^2 is from “thermal IR” from the atmosphere.
    DO YOU DISAGREE with any of this? How then can you claim the diagram shows that solar energy is the only source of power to the earth?

    PLEASE choose (1) or (2) (or both) and answer, preferably with numbers. Or choose something else that you mean by “earth” and explain how the diagram relates to power to/from this “earth”

    Also, you say “heat energy is Thermal IR”.

    Googling “thermal infrared” gives these among the first hits:
    “Thermal infrared radiation refers to electromagnetic waves with a wavelength of between 3.5 and 20 micrometers. ”

    http://www.geog.ucsb.edu/~jeff/115a/remote_sensing/thermal/thermalirinfo.html

    “MWIR [3-8um] and LWIR [8-15 um] is sometimes referred to as ‘thermal infrared.’ ”

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

    Both of these suggest that “thermal IR” is only loosely defined, but applies above ~3 um.
    DO YOU DISAGREE? If so, give a definition of what wavelengths you think are “thermal IR”?

    A precise definition of “heat energy” would also be appropriate. Or is it identical to whatever definition of “thermal IR” you just gave?

  349. RJ says:

    Thanks for the replies. I’m not an expert (far from it) but this topic just interests me. But I find it difficult to accept that a 100w power source could heat anything above 100w. But maybe I’m wrong. It would be great if this could be done as our energy problems would be solved.

    I don’t know what it means to “heat anything above 100w”. The temperature that something equilibrates to is given by the balance of the energy in and the energy out. Watts is a measure of energy per unit time, so if you have an input power of 100 W, the total energy of the system will (in the absence of any losses) just continue to increase its energy indefinitely. If you input the energy for a million years, you would have input ~3.15 x 10^15 J. That’s a lot of energy!

    Of course, in the real world, objects do lose energy with time, e.g., via radiation. However, how quickly such energy loss occurs depends on how well-insulated the object is from its surroundings. I think you understand this intuitively. You certainly wouldn’t tell me that a person could walk out naked in -40 C weather and not get hypothermia. The person is generating thermal energy via metabolism (a typical rate being about 100 W). If the person bundles up, that 100 W will be sufficient to keep their body temperature at normal levels; however, if they are naked, it will not be sufficient. [The human body, of course, can adjust its metabolism to some degree as necessary to prevent hypo- or hyperthermia, as well as changing other things like dilation of the blood vessels and emitting perspiration in order to change the transfer of heat away from the body, but only within limits.]

  350. Hans says:

    …asking if the earth’s greenhouse effect and thermal radiation from the earth helping keep the sun warm, in which case it would in principle, but the effect would be such an infinitesimal amount that it would not be measurable.”

    Yes, infinitesimal. But as you say it’s the principle, and that is what I want to have confirmed.

    In regards to the thermal radiation from the earth: Yes, it does help warm the sun (relative to the case where the earth is not present to absorb and re-emit some of the radiation received from the sun). However, as was pointed out the effect is infinitesimal, both because the earth only intercepts a tiny fraction of the sun’s energy and because the sun then only intercepts a tiny fraction of the energy that the earth emits.

    As for how the greenhouse effect of the earth affects the sun: That is a more complicated question. At the end of the day (i.e., in radiative balance), the Earth system (meaning earth + atmosphere) must emit the same amount of energy back out into space as it receives, which (in the absence of a change in albedo for solar radiation) is the same whether or not there is a greenhouse effect on the earth. The greenhouse effect just means that the earth’s surface temperature has to be higher in order for this to be accomplished.

    So, my first reaction would be that the greenhouse effect of the earth has no effect on how much radiation the earth emits toward the sun. However, really the amount of energy that the earth emits in both cases has to be the same only when averaged over the various cycles like the seasonal cycle and the diurnal cycle. Take the diurnal cycle as an example: If the greenhouse effect changes the amount of radiation that the day side of the earth emits relative to the night side of the earth, then the amount of radiation that the earth receives from the sun could in fact be affected.

    But, all of these are ridiculously small effects in the grand scheme of things as far as the sun is concerned.

  351. Tim, says:

    I agree that “SOLAR visible/UV/IR provide 99% of the energy that heats the earth.”

    DO YOU DISAGREE with any of this? If this solar energy is not what heats the earth as a whole, then what is your hypothesis?

    I’m demanding that you prove it. You’re the one with the hypothesis…

    2a The diagram clearly shows that only ~168 W/m^2 of power to the surface is directly from the sun, while 324 W/m^2 is from “thermal IR” from the atmosphere.
    DO YOU DISAGREE with any of this? How then can you claim the diagram shows that solar energy is the only source of power to the earth?

    Because that’s the explanation. YOU’VE JUST SAID IT. That this SOLAR, WHICH YOU HAVE JUST SAID IS 99% OF THE ENERGY WHICH HEATS THE EARTH. It is this according to the diagram what is converted into heat energy, thermal IR, and radiated back into the atmosphere.

    You have said that no other sources are of any significance. Don’t you even listen to yourself?

  352. RJ;
    But I find it difficult to accept that a 100w power source could heat anything above 100w. But maybe I’m wrong. It would be great if this could be done as our energy problems would be solved.>>>

    You continue to be confused. In every example given to you, 100 watts goes into the system as a whole, and 100 watts comes out. Check the numbers. In my oven/chicken example, 500 watts goes into the chicken, and 500 watts comes out of the oven. Not only CAN this be done, it IS done, and in NO WAY solves our energy problems.

    You are confusing how much energy is flowing in the system at any one time with where it is CONCENTRATED at any one time.

    At equilibrium, energy out = energy in.

    The step I think you are missing is that brief time in between starting the beam of energy in, and REACHING equilibrium. Watts are actually joules/second. So using my original example, let’s do the steps again, but this time we’ll track the joules as well. Again, in reality this would be a smooth curve, not instant stepts, but the prinicipal is identical. (Calculus is just a way of reducing each step to a nearly infinitesimal size to get a more accurate answer)

    Start = Oven and chicken at room temperature.
    Beam = 200 watts = 200 joules per second.
    Each step takes 100 seconds.

    Step 1 (1 to 100 seconds)
    Chicken is radiating 0 watts = 0 joules /second
    Chicken (from beam) 200 joules/sec * 100 seconds = 20,000 joules
    Chicken (from oven) = 0 joules/sec * 100 seconds = 0 joules
    Joules retained in Chicken = (0 +20,000 + 0) = 20,000

    Step 1.5
    Chicken rises in temperature until it is radiating 200 watts. Joules/second being retained falls to 0. So 200 watts are going in, and 200 watts are coming out (of the chicken) but 20,000 joules of energy are IN THE CHICKEN.

    Step 2 (101 seconds to 200 seconds)
    Chicken is radiating 200 watts = 200 joules/second * 100 seconds = -20,000 joules
    Chicken (from beam) 200 joules/sec * 100 seconds = 20,000 joules
    Chicken (from oven) 100 joules/sec * 100 seconds = 10,000 joules
    Joules retained in Chicken = Step 1 + Step 2
    = 20,000 + (-20,000 +20,000 + 10,000) = 30,000 joules

    Step 2.5
    Chicken rises in temperature until it is radiating 300 watts.
    Oven absorbs 300 watts.
    Oven rises in temperature until it is radiating 300 watts, 150 inward and 150 outward.

    Step 3 (201 seconds to 300 seconds)
    Chicken is radiating 300 watts = 300 joules/second * 100 seconds = -30,000 joules
    Chicken (from beam) 200 joules/sec * 100 seconds = 20,000 joules
    Chicken (from oven) 150 joules/sec * 100 seconds = 15,000 joules
    Joules retained in Chicken = (Step 1 + Step 2) + Step 3
    = 30,000 + (-30,000 +20,000 + 15,000) = 35,000 joules

    Step 2.5
    Chicken rises in temperature until it is radiating 350 watts.
    Oven absorbs 300 watts.
    Oven rises in temperature until it is radiating 300 watts, 150 inward and 150 outward.

    So if you continue the infinite series:

    Beam 200w 200w 200w 200w…. 200w
    Oven 0w 100w 150w 175w…. 200w
    Energy Balance. Out = In

    But, while equilibrium of 200 out and 200 in was being established:

    Chicken (w) 0w 200w 300w 350w 375w… 400w
    Joules retained 20,000 30,000 35,000 37,500…. 40,000 joules

    Do you see now what has happened? There is a lag time for the oven to reach equilibrium and radiate 200 watts. As it rises in temperature, the joules/sec going into the chicken from the beam stay the same. So until the oven increases in temperature to the point it is radiating 200 joules/sec to the outside, the joules going into the system have to build up…in the chicken.

    If the chicken was sitting on an open counter, there would only be a Step 1. The chicken would retain 20,000 joules and then would be at equilibrium. From 101 seconds on, no more joules retained than radiated.

    After equilibrium is established, there’s more joules in the oven chicken than in the counter chicken. Now shut the beam off, let’s say after 1,000 seconds and let everything cool off to room temperature again. There is an imbalance at this point in terms of the joules and where they are. For both chickens, 200 joules per second has gone in from the beam for 1000 seconds. What has come out and been radiated to the local area?

    To local area from:
    Counter chicken = (1,000 * 200) – 20,000
    Oven Chicken = (1,000 * 200) – 40,000

    Now we shut off the beam, and let everything cool back down to room temperature. That means what ever joules of energy are in the chicken now escape to the local area until both are back at room temperature.

    To local area from:
    Counter chicken = (1,000 * 200) – 20,000 +20,000 = (1,000*200)
    Oven Chicken = (1,000 * 200) – 40,000 +40,000 = (1,000*200)

    No joules of energy are created, lost, destroyed, or hidden. During the time period when a new equilibrium is established, the joules build up in the chicken. Stop the input of energy and the joules leak back out.

  353. Richard E Smith says:

    Argument by analogy can be misleading.

    Indeed…See below.

    A better analogy, I would argue, is a pipe through which water flows from a reservoir. If the exit from the pipe is below the water level in the reservoir the water will flow through the pipe. High to low just like the transfer of heat from warm to cold. With no height difference, there is no flow and water level remains the same. If you partly block the pipe the water will still flow through, but none will be returned to the reservoir. If you completely block the pipe, nothing happens. None of the water returns to the reservoir. The water can only be returned to the source if energy is supplied either by raising the pipe above the level of the reservoir or pumping the water back up.

    Sorry…That analogy fails. It would work if when you “stopped up the pipe”, you were able to block the radiation coming from the sun to the earth. However, the radiation continues to come from the sun to the earth. (Hence, it is like a pipe where you are forcing a constant mass of water into the pipe each second and then asking how much water comes out the other end.) This causes the earth’s temperature to rise, effectively increasing the “pressure” in your water analogy until the flow out is again equal to the flow in. (If you could completely block the pipe with something that would not get dislodged ever, then the pressure would increase without bound. Of course, there is nothing that can withstand infinite force…and in the real system, there is no perfect insulation that would do this.)

    In the constant irradiance models used to demonstrate the greenhouse effect, the Sun is pouring in a fixed quantity of watts m2 at all times. If the flow of heat from the Earth to space is blocked (as the water is blocked in the pipe) that heat cannot flow back heating the Earth even more because it does not have sufficient energy to do so – there is a temperature difference to overcome.

    That last sentence is complicated. (Yes, heat…a macroscopic thing…can’t flow from hot to cold, but the heat is determined by a balance of radiation passing in both directions, and, yes, the radiation going back can increase.)

    But, if you are uncomfortable with radiation going back, just think of it this way: If you block some or all of the heat flowing out, you have more heat flowing in than flowing out, so the temperature rises until it is pushing the heat out so hard that you still get the same amount of heat flowing out as you have coming from the sun. So, in the final state, the heat flowing out is still the same as before but you need a higher surface temperature to obtain this flow. That is what we call the greenhouse effect.

  354. Tim:

    http://en.wikipedia.org/wiki/File:The_green_house_effect.svg

    “About half the solar radiation is absorbed by the Earth’s surface”
    “Absorbation solar radiation: 168 Watts per metre squared”
    “Radiation is converted to heat energy, causing the emission of longwave (infrared) radiation back to the atmosphere”
    Some of the infrared radiation [the upwelling after conversion] passes through the atmosphere and out into space”
    “Outgoing [into space] infrared radiations: 240 Watts per metre squared”
    “Some of the infrared radiation is absorbed and re-emitted by the greenhouse gas molecules.”

    Doesn’t say how much that is.

    However, amounts, for the moment, is a distraction to getting my questions answered, getting the proof I’m asking for.

    This is the point:

    So, prove that this Solar energy you say 99% heats the Earth is actually capable of doing so, actually capable of converting into heat energy, heating the Earth, and therefore capable of radiating out the thermal IR claimed.

    Also, still waiting for Ira to prove that it’s these Solar energies we feel as heat from the Sun, that these are thermal.

    These are two sides of one coin, asking proof from Ira is relevant to your proving that it’s these Solar energies which convert to thermal IR.

    From tradition physics these Solar energies do not convert to heat.

  355. Joel Shore said (in response to Hans):

    If the greenhouse effect changes the amount of radiation that the day side of the earth emits relative to the night side of the earth, then the amount of radiation that the earth receives from the sun could in fact be affected.

    Sorry…That should read

    …then the amount of radiation that the sun receives from the earth could in fact be affected.

  356. RJ says:
    April 5, 2011 at 2:24 am

    “But does energy work this way. Energy is more like a teacher (sun) passing knowledge to a student (earth). The student can only take in so much knowledge and can not increase the teachers knowledge by passing the same knowledge back. But the student can increase the knowledge of another student (CO2) with less knowledge etc.”

    ———————————————————————————————————;

    Claes Johnson’s student/teacher analogy in his comments on computational black body radiation:

    “We can also compare with a common teacher-class situation with an excited/high temperature teacher emitting information over a range of frequencies from low (simple stuff) to high (difficult stuff), which by the class is absorbed and re-emitted/repeated below a certain cut-off frequency, while the class is unable to emit/repeat frequencies above cut-off, which are in-stead used to increase the temperature or frustration/interest of the class. The temperature of the class can then never exceed the temperature of the teacher, because all coherent information originates from the teacher. The teacher and student connect in two-way communication with a one-way flow of coherent information.

    The net result is that a warm blackbody can heat a cold blackbody, but not the other way around. A teacher can teach a student but not the other way around. The hot Sun heats the colder Earth, but the Earth does not heat the Sun. A warm Earth surface can heat a cold atmospheric layer, but a cold atmosphere cannot heat a warm Earth surface. A blackbody is heated only by frequencies which it cannot emit, but has to store as heat
    energy.”

  357. On more comment…

    Richard E Smith says:

    Argument by analogy can be misleading.

    …Which is why those of us here who are arguing on the side of the correct science have actually solved the equations. We are not using analogies to try to determine what happens…We are using them to explain what we KNOW actually happens. This is a much safer way to use analogies.

  358. Joel Shore says:
    April 5, 2011 at 1:29 pm

    Hans says:

    …asking if the earth’s greenhouse effect and thermal radiation from the earth helping keep the sun warm, in which case it would in principle, but the effect would be such an infinitesimal amount that it would not be measurable.”

    As for how the greenhouse effect of the earth affects the sun: That is a more complicated question….The greenhouse effect just means that the earth’s surface temperature has to be higher in order for this to be accomplished.

    So, my first reaction would be that the greenhouse effect of the earth has no effect on how much radiation the earth emits toward the sun.”

    Joel, Ira’s example of the soldering iron in the box results in a temperature raise of the source. And that is what you stated also, to maintain the power flow.

    “Take an object, be it a soldering iron where electrical power is converted into heat or the earth that is absorbing radiation from the sun.
    Now instead, try surrounding that object by a box or shell that absorbs some of the radiation. That surrounding shell will heat up and by virtue of having a nonzero temperature, it will emit radiation. Now the object will be receiving not only the power it was originally receiving but additional power from the radiating shell; there is no way this can be avoided. Hence, in order to reach a state where energy in = energy out, it will have to emit more power. It does this by raising its temperature. It’s really that simple.”

    “No…It is impossible for it to be in equilibrium. It was in equilibrium at 400 K when it didn’t have the walls at whatever elevated temperature that they are at emitting radiation at it. How can it possibly remain in equilibrium now that it is receiving more energy? That’s mathematically impossible. If it receives more energy, it has to emit more energy…and it does this by increasing its temperature.”

    “If there is no net energy flow away from the soldering iron, which is still plugged into the wall and converting electrical energy into thermal energy, what is it going to do? It is going to heat up until the heat leaving it does balance the thermal energy that it is generating! How could it do otherwise?”

    ” And, yes, the power does get smaller as the temperature of the shell increases…That’s the whole point. Now take that to its logical conclusion and you will understand why the temperature of the object must increase.”

    Tim Folkerts made the logical conclusion that in principle this must also raise the temperature of the sun by an infinitesimal amount.
    And so do I because any restriction reduces the flow and this can only be restored by a higher temperature, so I’m very surprised that you don’t conclude this without doubts.

    I would also like Ira to give his thoughts on this, as he came up with the analogy.

  359. Claes Johnson’s student/teacher analogy in his comments on computational black body radiation:

    Claes is spouting nonsense, as seems to be his custom. This is my favorite thing that Claes has written – Look at Equation (10.3) here: http://www.csc.kth.se/~cgjoh/ambsblack.pdf He admits this equation is correct but then claims the problem is with the interpretation and spouts the standard talking points about no back-radiation, one-way transmission from hot to cold, etc.

    What is missing from all the mumbo-jumbo he writes is the fact that he has already lost the argument when he admits that Equation (10.3) is correct. It doesn’t matter how you interpret the terms in (10.3), e.g., if you consider the second term to represent “back-radiation” or not. It is a mathematical equation that gives the same mathematical solutions regardless.

    In particular, it says that if T_2 is the temperature of the earth and T_1 of the surroundings, then in the case where the earth has an IR-transparent atmosphere, T_1 is the 3 deg background temp of space; however, in the case where the earth has an IR-absorbing atmosphere, T_1 is the temperature of that atmosphere. And, the equation tells you that the heat flow away from the earth in the former case is greater than the heat flow away from the earth in the latter.

    Once you have that result, it is simply a matter of stating the steady-state condition for radiative balance at the surface (that this heat flow away equals the heat flow in from the sun)…and then what you find is a surface temperature T_2 that is higher as T_1 gets larger. I don’t care if you call the reason that this happens “back-radiation” or “mind control by aliens from the planet Xena”, the result of the mathematics is the same!

  360. >>I agree that “SOLAR visible/UV/IR provide 99% of the energy
    >>that heats the earth [surface & atmosphere as a whole].”

    >I’m demanding that you prove it. You’re the one with the hypothesis…

    Science develops a whole set of interconnected ideas. After a while these come to be accepted as “established science” (that presumably could be shown incorrect, but that would require significant upheaval).

    * the “solar constant” of ~1370 W/m^2 is “established science”. Divide this by 4 to get the average over the surface of the earth = 343 W/m^2. Do you disagree with this part of “established science”? http://en.wikipedia.org/wiki/Solar_constant

    * Planck’s law of BB radiation is “established science”. Do you disagree with this part of “established science”? http://en.wikipedia.org/wiki/Planck%27s_law
    * The sun’s surface temperature of 5800 K is “established science”. Do you disagree with this part of “established science”?
    If you accept these, then you have to accept the ratios given: ~10% UV; ~40% visible; ~48% “non-thermal” IR (0.7 – 3 um); ~2% “thermal IR” (>3um). [Or show by your own calculations where the error is.]

    * That the earth reflects ~ 30% of the incoming energy (from the clouds and from the ground) is “established science”. Do you disagree with this part of “established science”? http://en.wikipedia.org/wiki/Albedo
    * Conservation of energy is a cornerstone of science. Surely you can’t disagree with this.
    Since solar energy comes in (342 W/m^2) and only some solar energy leaves, the rest MUST be added to the earth (which eventually leaves as thermal IR). The exact numbers can be discussed (Kiehl & Trenberth gives 168 W/m^2 into the surface and 67 W/m^2 into the atmosphere). The relative magnitude, however, has been determined by the “established science” above (not to mention various other experiments I could reference).

    * Absorbed energy “warms” an object — once again “established science”.

    I don’t know how many other ways to say this! All the established science fits the hypothesis that solar energy in the proportions I quoted warms the earth in the manner I suggested. You now need to show which part of this established science is wrong and where else the energy might be coming from.

  361. Myrrh,
    So, prove that this Solar energy you say 99% heats the Earth is actually capable of doing so, actually capable of converting into heat energy,>>>

    What in heck makes you think it has to convert?
    Your Armour of Density or Your Armour of Avoidance?
    Energy is measured in joules.
    Solar energy is measured in joules.
    Heat energy is measured in joules.
    A watt is just a joule per second.
    UV can be measured in joules per second per area.
    IR can be measured in joules per second per area.
    Visible Light can be measured in joules per second per area.
    ALL energy is measured in joules, no matter what frequency, and power is measured in watts no matter what frequency. There is nothing to convert from an energy perspective. The only thing that gets converted is the frequency at which the energy gets transmited.

    Which has been explained to you about a thousand times.

    Proof positive that your Armour of Density (or Avoidance as the case may be) is effective.

  362. Joel Shore;
    …Which is why those of us here who are arguing on the side of the correct science have actually solved the equations. We are not using analogies to try to determine what happens…We are using them to explain what we KNOW actually happens. >>>

    And there lies the frustrating part of the discussion. We explain the math and get objections. We come up with analogies to try and explain the math that is beyond some people. More objections.

    But the truth of the matter is that the equations reign supreme. They are used every day by engineers and many other professions to design everything from combustion engines to rockets to boilers. They are used to design nuclear power plants, air conditioners and sex toys. They are used to calculate the optimum processes for making potatoe chips, plastic bags, and satellites.

    The equations work, they work precisely, and they work thousands of times every single day.

    Period.

  363. You continue to be confused. In every example given to you, 100 watts goes into the system as a whole, and 100 watts comes out. Check the numbers. In my oven/chicken example, 500 watts goes into the chicken, and 500 watts comes out of the oven. Not only CAN this be done, it IS done, and in NO WAY solves our energy problems.

    Thanks. I can see that 100 watts goes into the oven and 100 watts leave. But 500 watts goes into the chicken and also leaves The 100 watts increasing to 500 watts is the bit I still have a problem with. But I might have a block on understanding this.

    I still also have a problem with a colder body heating a warmer one. (although this has been discussed on these threads)

    But its not my area of expertise so I will just leave it for now and come back to it at some later point. An interesting thread though.

  364. davidmhoffer says:
    April 5, 2011 at 7:47 pm

    What in heck makes you think it has to convert?

    ? It’s a technical term, means what it says on the tin (established traditional science).

    As used here on the Trenberth diagram: http://en.wikipedia.org/wiki/File:The_green_house_effect.svg

    “Radiation is converted to heat energy”

    Conversion is the process by which radiation is converted to heat. This is the point I’m arguing about. The Solar energies of AGWScience’s Energy Budget do not convert to heat.

    Near IR does not feel hot. See NASA page. It is not converting to heat.
    Visible light does not feel hot. These wavelengths are not converting to heat.
    UV does not feel hot. Ditto.

    But, the Trenberth diagram says this is happening. That these wavelengths are converting to heat.

    The longer waves of IR convert to heat. Hence they are called Heat energies, or simply Heat. They heat things up because they convert to heat.

    Compare with, Light energies are not hot.

    What I want proved is that These Solar Energies which in traditional established science are known not to convert to heat, are said to be doing just that in AGWScience.

    How? How are these wavelengths heating the Earth? How are they heating the Earth to produce such a massive amount of Thermal IR radiating out from the heated Earth?

    Tim, will come back to this later, have a hedge to plant.

  365. Joel Shore says:
    “If you block some or all of the heat flowing out, you have more heat flowing in than flowing out, so the temperature rises until it is pushing the heat out so hard that you still get the same amount of heat flowing out as you have coming from the sun. So, in the final state, the heat flowing out is still the same as before but you need a higher surface temperature to obtain this flow. That is what we call the greenhouse effect.”

    I am sorry Joel, but this is non-physical.

    The blackbody object in the oven (chicken or whatever) is constantly receiving 500 watts. It heats up accordingly and its radiant emissions are a signal of its temperature – i.e. the vibrations of the molecules in the chicken. Now enclose it in an oven (to make it even simpler assume a vacuum) and its emissions are reflected back to it (or absorbed and re-radiated if you prefer to think about CO2). These emissions are at the wavelength determined by the blackbody’s temperature. When they return they are either at the same or a lower energy level. Therefore they cannot excite the chicken’s molecules to a greater level of energy than they were before. Your example of heat blocking confuses the return of a body’s own emissions with putting a lid on a saucepan. You cannot get more energy out than you put in.

  366. Tim – my reply to all that you wrote in your last post is the same as I wrote to David.

    I am asking how? this Solar downwelling, 99% of the energy which heats the Earth, actually converts to heat as per Trenberth diagrammatic.

  367. Myrrh;
    As used here on the Trenberth diagram: http://en.wikipedia.org/wiki/File:The_green_house_effect.svg
    “Radiation is converted to heat energy” >>>

    Of for g_d’s sake THAT’S what you’ve been yapping on about? I finally bothered to take a look at exactly where in the diagram those words appear. All you’ve got there is some poor wording on Trenberth’s part.

    OHMIGOD NOW I’M DEFENDING TRENBERTH!!

    There is no “conversion” per se other than the manner in which the energy is stored on a temporary basis. The energy is still measured in joules, and power is measured in joules/sec = watts. concentration of power = w/m2.

    The Sun emitts energy at multiple frequencies, each can be measured in w/m2.
    When the joules emitted by the Sun are absorbed by the earth, they are still joules. There is no conversion. But we don’t measure the w/m2 of the earth, we measure the temperature. Trenberth’s wording is sloppy as h*ll. What he meant was that joules of energy measured in w/m2 are absorbed by earth surface CAUSING the temperature to rise. The ENERGY did NOT get converted. Joules are joules no matter how they are stored! The only thing that got converted is how the joules are stored. As radiated energy they are
    (watts/m2)*time=joules.
    Stored as heat in the dirt its… Joel? Tim? its 2:30AM, I’m burnt out.
    Something along the line of joules/mass/volume
    In any event the concentration of joules in the dirt becomes denser, which we can measure as higher temperature. The reference to “heat” is just a reference to more joules of energy per mass/volume in the earth surface, which means exactly the same thing as higher temperature. And the result of a higher temperature is
    Power = 5.67*10^-8*T^4 where T is the TEMPERATURE in degrees Kelvin, and Power is in watt/m2 emitted by the body at that temperature.

    Sloppy wording on Trenberth’s part, that’s it. Nothing more. No conversion, just a reference to where the joules of energy are at a given point in time and how they are commonly measured. With Traditional Physics BTW.

  368. davidmhoffer says:

    Stored as heat in the dirt its… Joel? Tim? its 2:30AM, I’m burnt out.

    I swore off responding to Myrrh…but I guess I can respond to you about what he is saying. While others here have invented what I call the “Magical Second Law of Thermodynamics”, Myrrh seems to have invented the “Anti-Second Law of Thermodynamics”. One of the implications of the 2nd Law is that energy in more useful forms tends to be inevitably converted into thermal energy, which is a form of energy with high entropy (roughly speaking “disorder”…although it has a more precise mathematical definition). Myrrh’s anti-2nd Law appears to be that it is very difficult to convert energy in other forms into thermal energy. That is a rather novel view of the universe, to put it mildly.

    And, of course, it doesn’t change the fact that if one is going to claim that the incoming solar energy is not converted into thermal energy, one has to explain what form of energy it does go into. In what form is the ~2.07 x 10^7 J of solar energy that each meter-squared of the earth’s surface receives each day being stored?

  369. Richard E Smith says:

    These emissions are at the wavelength determined by the blackbody’s temperature. When they return they are either at the same or a lower energy level. Therefore they cannot excite the chicken’s molecules to a greater level of energy than they were before.

    Ah…Can you tell me what law of nature actually says this? It sounds like the “Magical 2nd Law of Thermodynamics” to me. While you are at it, can you tell me the experimental evidence in support of this law?

    A blackbody is a blackbody; it absorbs photons of all energies. (And, by the way, the energy distribution of the photons are statistical anyway: The distribution of photon energies for the two objects overlap and, indeed, when the temperatures are not too far apart, they overlap considerably.)

    You cannot get more energy out than you put in.

    You also cannot get less energy out than you put in. Could you demonstrate for us how my view leads to more energy out than in (which will be rather difficult since it is mathematically based on achieving radiative balance, as explained in words)? While you are at it, will you explain to us what your alternate view of the situation is and how this satisfies Conservation of Energy? Good luck!! (Your gonna need it!)

  370. Hans says:

    Tim Folkerts made the logical conclusion that in principle this must also raise the temperature of the sun by an infinitesimal amount.
    And so do I because any restriction reduces the flow and this can only be restored by a higher temperature, so I’m very surprised that you don’t conclude this without doubts.

    Because the geometry is completely different. The earth’s atmosphere is around the earth. It is not around the sun. What the atmosphere restricts is the flow of radiation out from the earth. Also, the sun’s radiation is mainly at UV, visible, and near-IR frequencies.

    I just don’t see where the equivalence is between these two different things. It is like asking why if my putting on a heavier winter jacket will help prevent me from getting hypothermia when I go out in the cold, why does my putting on a heavier winter jacket not help you from getting hypothermia.

  371. Myrrh says: April 6, 2011 at 2:38 am

    “I am asking how? this Solar downwelling, 99% of the energy which heats the Earth, actually converts to heat as per Trenberth diagrammatic.”

    So you are asking how – on an atomic level – that the energy of a photon can be transferred to a molecule in the ground?

    At some level, it doesn’t matter. You have already admitted a strong belief that longer wavelength IR can do this. In principle then, other wavelengths of light should be able to do this. Black paint is experimental confirmation — the photons go in, but don’t go out; as a result the painted object gets hot (where “hot” in this context is maybe 10 – 100 C warmer than a similar white object).

    Then the hot object will radiate thermal IR because that is what any hot object will do. The details of the atomic processes are probably not worth going into at this level (and I am not prepared to spend more time figuring out/explaining such details).

  372. Tim Folkerts says:

    Then the hot object will radiate thermal IR because that is what any hot object will do.

    Just to clarify (never can be too careful with Myrrh!): That would be that any object radiates and for an object at the typical temperatures that we are talking about here, most of the radiation will be in the thermal IR.

  373. No, no, no. Not all radiation converts to heat. What don’t you understand about Nr IR not being hot?

    AGWScience says that these Solar energies, UV, Visible and Nr IR, CONVERT TO HEAT to HEAT THE EARTH, WHICH THEN RADIATES OUT THIS HUGE AMOUNT OF THERMAL IR.

    I DEMAND THAT YOU PROVE THIS ON A WAVELENGTH BY WAVELENGTH BASIS.

    How much does UV heat the Earth? How much does Blue Visible Light heat the Earth, etc.

    And exactly HOW?

    This is the AGWScience BASIC ENERGY BUDGET. You are going to have to produce better explanations by detailed amounts than your current fudging.

    Prove it.

    Until you can prove it, your meanderings about possible ways it might be doing this, is of no interest to me whatsoever.

  374. I had to go doublecheck what I said …
    “… asking if the earth’s greenhouse effect and thermal radiation from the earth [would help] keep the sun warm, in which case it would in principle, but the effect would be such an infinitesimal amount that it would not be measurable.”

    I should amend that — the thermal radiation from the earth will infinitesimally help warm the sun, but the greenhouse effect will not matter. The earth emits some small amount of IR toward the sun. Consider a binary star orbiting close to the sun. That star would definitely provide a significant amount of energy to the sun and raise the temperature of the side facing the binary star. Now slowly shrink the star and cool the star and move it out as far as the earth is. The energy to the sun will keep decreasing, but there is always SOME small amount of energy and the near side would always be some infinitesimal amount warmer than the far side.

    The greenhouse effect, OTOH, will affect the spectrum of the outgoing light, but there will still be 343 W/m^s leaving the earth. Since the total energy heading toward the sun will be essentially unchanged as GHG’s increase, the effect on the sun will be unchanged.

  375. Myrrh says:
    April 6, 2011 at 11:39 am
    No, no, no. Not all radiation converts to heat. What don’t you understand about Nr IR not being hot? >>>

    What don’t you understand about it can be measured in w/m2? So its mildly warm instead of hot. Its not as hot as IR in other ranges. It still gets measured in w/m2. Some frequencies have more and some less, but none have zero.

    Never mind, Armour of Density, I forgot…

  376. Joel Shore says:
    April 6, 2011 at 5:34 am

    “Hans says:

    Tim Folkerts made the logical conclusion that in principle this must also raise the temperature of the sun by an infinitesimal amount.
    And so do I because any restriction reduces the flow and this can only be restored by a higher temperature, so I’m very surprised that you don’t conclude this without doubts.

    Because the geometry is completely different. The earth’s atmosphere is around the earth. It is not around the sun. What the atmosphere restricts is the flow of radiation out from the earth.”

    Of course, the atmosphere is just a small piece of the box but lets stick to the principle. The piece of box heats the source, and you with Ira tell me the source will then get warmer and warm the piece of box further to restore the flow. The smaller size of the area A just reduces the amount of extra heating of the source. Or do you want to say that the iron in the box is not how things work?

    “Also, the sun’s radiation is mainly at UV, visible, and near-IR frequencies.”

    Why do we have to split the radiation in sections now? Are there restrictions in frequencies coming from the box that re-heat the source. Some do some don’t?

  377. Ira?

    Your iron in the box analogy with heating of the energy source when shielding restricts the energy flow from the source, lead me to the conclusion that the sun must also raise it’s temperature to maintain the energy flow when GHG restrict the flow.
    Some here agree, but some don’t which questions your analogy. Please comment.

  378. Richard E Smith says:
    April 6, 2011 at 2:06 am

    “Joel Shore says:
    “If you block some or all of the heat flowing out, you have more heat flowing in than flowing out, so the temperature rises until it is pushing the heat out so hard that you still get the same amount of heat flowing out as you have coming from the sun. So, in the final state, the heat flowing out is still the same as before but you need a higher surface temperature to obtain this flow. That is what we call the greenhouse effect.”

    I am sorry Joel, but this is non-physical.”

    Indeed Richard, but as you’ll see a few comments above Joel seems to agree with you because he tells me here that the sun will not get hotter by the greenhouse effect.

  379. Hans says: April 6, 2011 at 2:08 pm

    “Your iron in the box analogy with heating of the energy source when shielding restricts the energy flow from the source, lead me to the conclusion that the sun must also raise it’s temperature to maintain the energy flow when GHG restrict the flow.
    Some here agree, but some don’t which questions your analogy. Please comment.”

    The outer part of the core of the the sun is about 1/4 the radius of the sun, so about 1/16 the area. To have the same power flow as the surface, it would need to only be 2x hotter since power is proportional to T^4. I.e. if the outer layers were removed and the energy stayed constant, the new surface would only need to be about 12000 K to radiate the same energy. Instead, the temperature is close to 7,000,000 at that part of the sun.

    The elevated temperature is due to the insulating/GHG effects of the radiative zone & convective zones.

  380. davidmhoffer says:
    April 6, 2011 at 12:55 pm

    What don’t you understand about it can be measured in w/m2?

    You presume too much. It is irrelevant to my question, I don’t care if it’s measured in cockle shells and pretty maids all in a row, but w/m2 will do fine.

    The AGWScience hypothesis, as shown graphically depicted by Trenberth, and constantly explained by AGWScientists, is that these particular Solar energies are the ones heating the Earth, which then sends out masses of Thermal IR, which then etc.

    I AM DEMANDING THAT YOU PROVE THIS.

    WHAT, AS YOU KEEP TELLING ME YOU ARE, DON’T YOU WITH YOUR SUPERIOR SCIENTIFIC UNDERSTANDING, UNDERSTAND ABOUT MY QUESTION?

    SOLAR IN, CONVERTING TO HEAT, HEATING UP THE EARTH, THERMAL IR OUT. THAT IS YOUR HYPOTHESIS. PROVE IT.

    How? How exactly how? Method how. Wave by wave: UV, the Visible Light, Near IR. Give me the facts and figures.

    Surely this must be somewhere on tap, as it were? This is the Main AGW Energy Budget Poster Child, on which all the rest of energies and runaway heating by CO2 is based. Fetch it ready made from AGWScience Stores, or explain it here in your own words. I don’t care one way or the other, but I demand to see this proved.

  381. davidmhoffer says:
    April 5, 2011 at 7:47 pm
    Myrrh,

    …ALL energy is measured in joules, no matter what frequency, and power is measured in watts no matter what frequency. There is nothing to convert from an energy perspective. The only thing that gets converted is the frequency at which the energy gets transmited.

    Which has been explained to you about a thousand times. …

    Dave: “About a thousand times?” I told you a million times, Dave, DON’T EXAGGERATE :^)

  382. Hans says:

    Of course, the atmosphere is just a small piece of the box but lets stick to the principle. The piece of box heats the source, and you with Ira tell me the source will then get warmer and warm the piece of box further to restore the flow. The smaller size of the area A just reduces the amount of extra heating of the source. Or do you want to say that the iron in the box is not how things work?

    For one thing, I don’t even understand the question anymore. Are you asking:

    (1) How will the greenhouse effect on the earth (i.e., the fact that some of the terrestrial radiation that is emitted by the earth comes back to the earth’ surface rather than asking the atmosphere) affect the sun? That is the question that I was answering.

    (2) How will the greenhouse effect of the atmosphere of the sun affect the temperature of the sun? [What Tim answered in his most recent post?]

    (3) How will the earth’s atmosphere affect the radiation coming from the sun…Will it cause some of it to go back to the sun? [What I now think you might be asking.]

    Besides the fact that you are talking about miniscule effects [at least for (1) and (3)], I also don’t really see the relevance of this to the actual geometry we are interested in. For example, the fact that the earth’s troposphere decreases with height is an important factor in the greenhouse effect. When radiation comes from the sun side, that aspect changes…besides the issues of frequencies below.

    But, basically, this is a question for the “advanced course”. Since you don’t even understand the elementary parts yet, I think we should stick to simpler things.

    “Also, the sun’s radiation is mainly at UV, visible, and near-IR frequencies.”

    Why do we have to split the radiation in sections now? Are there restrictions in frequencies coming from the box that re-heat the source. Some do some don’t?

    Well, again, I have lost what you are talking about. For the real earth, yes, the frequencies matter because it is the spectrum of the absorption lines that comes into play. The greenhouse gases absorb strongly in the thermal IR but not that much in the UV, visible, and near IR.

    Indeed Richard, but as you’ll see a few comments above Joel seems to agree with you because he tells me here that the sun will not get hotter by the greenhouse effect.

    I wouldn’t be so glib if I were you, given that you don’t seem to understand the elementary basics we are talking about here and both of you are spouting nonsense that clearly shows you have never even put a pen to paper and worked through any of the equations. Those of us who have are being very generous in trying to dispel your ignorance. We should probably just leave you hear to blather your nonsense and re-enforce to any scientists that you might interact with that “climate change skeptics” are utterly and completely ignorant of science.

    From my point of view, that might not be a bad outcome, if the scientist / physicist side of me didn’t have such an aversion to people spouting scientific nonsense. (And, I have a lot of sympathy for Ira and davidmhoffer who I imagine are horrified by the prospect of people like you representing to anybody outside of the skeptic community how climate change skeptics reason!)

  383. Tim Folkerts says:
    April 6, 2011 at 2:44 pm

    “The outer part of the core of the the sun is about 1/4 the radius of the sun, so about 1/16 the area. To have the same power flow as the surface, it would need to only be 2x hotter since power is proportional to T^4. I.e. if the outer layers were removed and the energy stayed constant, the new surface would only need to be about 12000 K to radiate the same energy. Instead, the temperature is close to 7,000,000 at that part of the sun.

    The elevated temperature is due to the insulating/GHG effects of the radiative zone & convective zones.”

    This is the worst analogy ever. It’s a well known fact that the temperature is that high in combination with pressure to balance the gravitational effects.

  384. Ira Glickstein, PhD says:
    April 6, 2011 at 3:54 pm

    ” Myrrh,

    …ALL energy is measured in joules, no matter what frequency, and power is measured in watts no matter what frequency. There is nothing to convert from an energy perspective. The only thing that gets converted is the frequency at which the energy gets transmited.
    Which has been explained to you about a thousand times. …”

    Exactly, that’s why I wonder why Joel Shore gave me this answer: What the atmosphere restricts is the flow of radiation out from the earth. Also, the sun’s radiation is mainly at UV, visible, and near-IR frequencies.

  385. Ira;
    Dave: “About a thousand times?” I told you a million times, Dave, DON’T EXAGGERATE :^)>>>

    Ira, is this your theory? I’m sorry, but “about a thousand times” seems a pretty good approximation to me. If you disagree, you will have to PROVE IT TO ME.

    YOU WILL HAVE TO COUNT EACH TIME, and submit a record of your findings. You will have to PROVE that I’m wrong, its YOUR THEORY why should I do any of the counting? And if you get less than a thousand, then all that proves is THAT YOU CAN’T COUNT. I’m demanding PROOF HERE! PROOF! IT’S BOG STANDARD COUNTING! We’re not talking AGW counting, we’re talking TRADITIONAL COUNTING. The way counting was done BEFORE THEY CHANGED IT. Look at any diagram of people’s hands. FOUR fingers on each hand, not FIVE! You don’t COUNT THE THUMB it isn’t an actual FINGER! But ask anyone and you’ll find out the truth, they’re teaching kids in class now that each hand has FIVE fingers. FIVE! You’re going to have to PROVE A HAND HAS FIVE FINGERS IF YOU WANT YOUR COUNTING TO BE ACCEPTED. I don’t mean just counting them off either, I mean proof, FINGER by FINGER! I want you to GIVE ME THE FINGER proof! You babble on about how scientific your counting is, but when I ask, YOU WON’T EVEN GIVE ME THE FINGER. Is the a SCIENCE BLOG? Or just a place to wave FINGERS at STOOPID PEOPLE?

    Back to counting the way it used to be.
    With only FOUR items at a time.
    And SINGING! You have to have SINGING!

    One of these people just doesn’t belong here…
    One of these people just isn’t quite sane.
    One of these people just doesn’t belong here
    So said Phil, tried to end the game.

    One of these people just doesn’t belong here
    Joel gave up and said the same
    Then Tim and Dave’s patience wained
    I think Myrrh has won this game!

    One of these people just doesn’t belong here…
    PROOF just isn’t about brains
    Its bullsh*t that matters, ‘cuz it baffles brains
    And Myrrh’s got lots of bullsh*t that matters
    Stuffed up there where we have our brains

    One of these people just doesn’t belong here…

  386. Joel Shore says:
    April 6, 2011 at 4:40 pm

    “I wouldn’t be so glib if I were you, given that you don’t seem to understand the elementary basics we are talking about here and both of you are spouting nonsense that clearly shows you have never even put a pen to paper and worked through any of the equations. Those of us who have are being very generous in trying to dispel your ignorance. We should probably just leave you hear to blather your nonsense and re-enforce to any scientists that you might interact with that “climate change skeptics” are utterly and completely ignorant of science.”

    So instead of answering a simple question, you keep making smokescreens and as a last resort a bunch of insults and personal attacks.
    Why do you suddenly come up with all those buts and iffs for the sun-earth situation after stating that an energy source heats up when restrictions come in play, by using the iron in a box analogy. I made it quite clear that I just wanted a simple confirmation that by the same principle the sun must get warmer, but somehow you don’t and kick in geometry, UV, visible an IR radiation while we both know neither makes a difference for the principle here.

    But then I read this:
    ” Richard E Smith says:
    April 6, 2011 at 2:06 am

    Joel Shore says:
    “If you block some or all of the heat flowing out, you have more heat flowing in than flowing out, so the temperature rises until it is pushing the heat out so hard that you still get the same amount of heat flowing out as you have coming from the sun. So, in the final state, the heat flowing out is still the same as before but you need a higher surface temperature to obtain this flow. That is what we call the greenhouse effect.”

    Now it can’t stop here because we learned from the iron in the box that a higher temperature of (the first appearing) restriction (now earth) heats the source (the sun). If not delta T sun-earth would now get smaller than before and the heat flowing from the sun would be less.

    Just a simple yes or no to my last sentence is all I’m asking for Joel.

  387. Myrrh says:
    April 5, 2011 at 11:30 am
    Tim – look at the AGW Energy Budget as Ira has depicted it and as I linked to in Trenberth diagram. It claims that the the SOLAR energies of Visible, and the two shortwave either side of UV and Near IR are what heat the Earth, converting to heat energy, which is the upwelling Thermal IR.

    None of these are felt as heat, ergo, they are not in themselves heat energy, heat energy is Thermal IR. The AGW Energy Budget says that non-Thermal energies are heating the Earth and converting to Thermal IR. How?

    Remember, UV barely penetrates the epidermis on skin, it does not raise the temperature of the body further than the outer thin epidermis. Visible light is less intense, it is also not hot, but simple Light energy. It does not raise the temperature of the body, it is not felt as heat. Ditto Near IR. Near IR is not hot. It is not felt as heat, it is not thermal, it does not raise the temperature of the body, it penetrates the body further than Visible. Thermal IR, the heat we feel from the Sun, penetrates further, warming up the body. It is Thermal IR which warms up organic matter.

    Check this out, it’s a laser power meter:

    http://www.newport.com/Laser-Power-Meter-Analog-407A/473001/1033/catalog.aspx

    It works by having an absorbing black surface in thermal contact with a thermopile which produces a voltage according to the temperature reached which is proportional to the amount of light energy absorbed. Note the following:
    “The thermopile detector can withstand 20 KW/cm2 average cw power density and is capable of measuring power from a few milliwatts up to > 20 Watts. The sensor delivers linear response over almost four decades of dynamic range and wavelengths ranging from UV to IR without bulky heat dumps. The head has been tested at up to 30 W laser power with no significant loss of linearity. Readings are extremely accurate since the detector’s absorbance varies only by ± 1 % between 400 nm and 1000 nm. ”
    Note that the same calibration applies over the whole visible range to the near IR.
    The one I used in my lab. gave the same results for 266nm, 355nm, 512nm and 1064nm, i.e. UV through near IR, so to a black body 1W of IR, vis or UV heats up equally. As far as the earth is concerned the same is true except for any wavelength dependent reflectivity. The effect on the earth’s surface does not depend on how light is absorbed by human skin.

  388. Myrrh;
    You presume too much. It is irrelevant to my question, I don’t care if it’s measured in cockle shells and pretty maids all in a row, but w/m2 will do fine.>>>

    That would be your magical Armour of Avoidance at work. You refuse to understand the meaning of the terms, and spout nonense as a consequence of your willful ignorance.

    How? How exactly how? Method how. Wave by wave: UV, the Visible Light, Near IR. Give me the facts and figures.>>>

    You want the exact figures? No problem. Start by reading this TRADITIONAL PHYSICS:

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

    When you are done that and understand it, then read this:

    http://en.wikipedia.org/wiki/Planck's_law

    Then this:

    http://en.wikipedia.org/wiki/Wien's_displacement_law

    And this:

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

    And this:

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

    By the end of which one of two things will have happened.

    1. You understood what you read and STFU.

    or

    2. One of the people just doesn’t belong here….

  389. Ira, David, Tim et al – do you have some kind of block here that’s preventing you from actually taking in my question? I’m asking about the specifics, how and how much does Solar actually heat the Earth.

    I’ll try again. AGWScience says that Solar energy, UV, Visible and Near IR, the Shortwave energies, actually heat the Earth which then radiates out Thermal IR (which then gets radiated out/trapped etc.). That’s the basic premise of the AGWScience Greenhouse. This should be bog standard understanding and the calculations at hand for any promoting this hypothesis.

    How and how much does UV heat the Earth? How and how much does Visible light heat the Earth? How and how much does Near IR heat the Earth?

    Prove it.

    I’m telling you, that these do not heat the Earth. These energies are incapable of heating the Earth to the extent that it gets hot enough to radiate out that amount of Thermal IR. These shortwave energies are not hot in themselves, that is, they are not Thermal Energies. Heat is Thermal Energy in transit from one location to another. Therefore, since they are not themselves thermal energy, they do not heat by their heat, because they have none.

    Show me the method and calculations, the detail, of how much these particular energies actually heat the Earth which is the BASIC PREMISE of AGW Greenhouse Hypothesis.

    If you can’t answer this – you have no scientific credibility in promoting this as real Science.

    What is it you don’t understand in my questions here? How and how much? Let’s limit this, for the sake of brevity, to Visible Light, which you, generic, keep stressing is the peak energy radiating out from the Sun heating the Earth (which then radiates out Thermal IR, etc.).

    Let’s make it even more simple, not only for the sake of brevity but also because you generic stress how deep it goes into the ocean compared with the other Visible wavelengths, how much does Blue Light from the Sun heat the Earth?

    Begin, by showing me how much blue light heats organic matter, solid and liquid, and take it from there.

  390. Myrrh, you said, “et al – do you have some kind of block here that’s preventing you from actually taking in my question? ”

    No, not at all in my case, in fact, you have really brought up a very curious question and experiment problem. How the heck could someone ever by experiment prove whether visible light warms and how much compared to unfiltered solar radiation.

    Well, it’s not summer yet but when it gets warm enough to play with some water I’m going to do such an experiment. I’m going to take a large black box with something like 1″ of water inside. Record the temperature, expose for a given time and get the delta T to compute the rate. Next is to cover this with a large enough plate of glass with a very thin layer of water flowing across it’s top to filter all but the visible frequencies. Repeat the sequence above.

    That is the only way I can come up without some expensive equipment that I no longer have. Does that sound like that should answer your specifications? (Would you believe it either way the results came out? I will. ) I’m going to pre-guess there will be some warming in the visible-only case but tiny compared to the unfiltered case and I’m trying to figure out some way to also filter out all but the lower infrared and repeat it there too.

  391. Myrrh, one more thing.

    I just figured out to do that very thing, fiter all but the IR, but I’m also going to need a small pump. If the water flowing over the glass to filter out the IR was not garden hose water but recycled water I can measure the rise of temperature of that water also, the water absorbing the IR. Sound about right?

  392. Hans says:

    I made it quite clear that I just wanted a simple confirmation that by the same principle the sun must get warmer, but somehow you don’t and kick in geometry, UV, visible an IR radiation while we both know neither makes a difference for the principle here.

    No,we do not know that, for the reasons I have stated in my previous posts if you actually bothered to read and comprehend them. I feel like I am talking to a brick wall. As for your question, I won’t even waste time starting to think about it until you make it clear whether you are asking about (1), (2), or (3) in this post: http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/#comment-637072

    Now it can’t stop here because we learned from the iron in the box that a higher temperature of (the first appearing) restriction (now earth) heats the source (the sun). If not delta T sun-earth would now get smaller than before and the heat flowing from the sun would be less.

    Statements like this make me wonder if you are comprehending anything that we tell you. Yes, it can stop there because I have described to you the final steady-state state that the system converges to. This is the solution you actually find if you solve the equations rather than just talk nonsense off the top of your head about what you think happens based on whatever misconceptions of reality you have.

    What davidmhoffer described in his post was how the system converges to this final steady-state. It is as if we wrote down the geometric sum 1 + 1/2 + 1/4 + 1/8 + … and David described to you how by successively adding each term you could see what it is converging to. And, I just noted the properties of the final result, which is 2 in this case.

  393. Joel Shore says:

    Richard E Smith says:
    ” ‘These emissions are at the wavelength determined by the blackbody’s temperature. When they return they are either at the same or a lower energy level. Therefore they cannot excite the chicken’s molecules to a greater level of energy than they were before.’ ”

    ‘Ah…Can you tell me what law of nature actually says this? It sounds like the “Magical 2nd Law of Thermodynamics” to me. While you are at it, can you tell me the experimental evidence in support of this law?’

    You can test it experimentally by pointing a spotlight at a wall in your house and holding a mirror so that as the light reflects off the wall it is reflected back to the wall by the mirror. You will find that the reflection from the mirror will illuminate the wall that is not already illuminated by the spotlight. If you direct it at the part already illuminated you will find that it does not brighten that part any more. The ‘extra’ light has no effect. Infra-red is a form of light and so an object cannot heat up itself by its own radiant emissions. Energy cannot multiply itself.

    I think a lot of confusion exists over the so-called greenhouse effect because people confuse light and heat. They transfer their experience of achieving greater warmth by the use of blankets and saucepan lids to so-called greenhouse gases. In fact a lid simply confines the area that is being warmed. This both restricts convection and reduces the volume of the area that is heated. It does not add heat to the source of the heat.

  394. Wayne and Myrrh: You are welcome to do all the experimentation that you like. However, I would also suggest you get a physics textbook and actually read it. Comprehend ideas like “conservation of energy”, “the photon model of light”, “blackbody radiation” that have come from centuries of experimentation and hypothesis testing by scientists. Your gaps in knowledge and misconceptions are way too large to be cured by us talking to you in this forum, particularly since you seem to have a mental block to absorbing anything that we say. Maybe you will have less of a block if you read it in a textbook. (The book that we use in the course that I teach is “College Physics” by Knight, Jones, and Field, which I would recommend except that it actually has a page on the greenhouse effect and global warming, so no doubt Myrrh would conclude it is part of the grand AGW conspiracy. Better to find an old textbook from before the time when you think this grand conspiracy overtook all of the physics textbooks.)

  395. Myrrh says: “I’m telling you, that these do not heat the Earth. These energies are incapable of heating the Earth to the extent that it gets hot enough to radiate out that amount of Thermal IR. ”

    As long as you are contradicting yourself within the course of 2 sentences, there is little change for any one else to explain anything to you! You are saying:
    1) ‘these wavelengths DO NOT heat the earth [at all]‘
    2) ‘these wavelengths DO heat the earth [just not enough to do the job]‘

    1) has been explained over and over.
    2) is true and is indeed the basis of the Greenhouse effect. Additional energy (thermal IR from the atmosphere) is needed to warm the earth.

    Exactly which wavelengths provide how much power would vary from location to location and from season to season. That is a question beyond the scope of a discussion here. If you are really keen to know, dig into the literature!

  396. Richard E Smith says:
    April 7, 2011 at 4:24 am
    You can test it experimentally by pointing a spotlight at a wall in your house and holding a mirror so that as the light reflects off the wall it is reflected back to the wall by the mirror. You will find that the reflection from the mirror will illuminate the wall that is not already illuminated by the spotlight. If you direct it at the part already illuminated you will find that it does not brighten that part any more. The ‘extra’ light has no effect. Infra-red is a form of light and so an object cannot heat up itself by its own radiant emissions. Energy cannot multiply itself.

    Well I’d suggest you conduct that experiment with a well controlled detector rather than the human eye (a square law detector with automatic brightness control)! In any case your experiment suggests that all of the reflected light from the mirror is absorbed when it hits the bright spot. Think about it.

  397. Myrrh says:
    April 7, 2011 at 12:09 am
    Ira, David, Tim et al – do you have some kind of block here that’s preventing you from actually taking in my question? I’m asking about the specifics, how and how much does Solar actually heat the Earth.

    It gets absorbed and heats up whatever absorbs it. A watt of UV heats up a black body the same as vis, ir or electric current for that matter. Evidence given above which you ignored!

    I’ll try again. AGWScience says that Solar energy, UV, Visible and Near IR, the Shortwave energies, actually heat the Earth which then radiates out Thermal IR (which then gets radiated out/trapped etc.). That’s the basic premise of the AGWScience Greenhouse. This should be bog standard understanding and the calculations at hand for any promoting this hypothesis.

    Which there is except for those who willfully ignore it.

    How and how much does UV heat the Earth? How and how much does Visible light heat the Earth? How and how much does Near IR heat the Earth?

    Prove it.

    Done, you are the one who makes the extraordinary statement that a watt of energy at a blue wavelength is incapable of heating an object which absorbs it. What do you think happens to that energy if it is not reflected, where does it go?

    I’m telling you, that these do not heat the Earth. These energies are incapable of heating the Earth to the extent that it gets hot enough to radiate out that amount of Thermal IR. These shortwave energies are not hot in themselves, that is, they are not Thermal Energies. Heat is Thermal Energy in transit from one location to another. Therefore, since they are not themselves thermal energy, they do not heat by their heat, because they have none.

  398. Phil. says:
    Correction to previous post.
    April 7, 2011 at 7:27 am
    Myrrh says:
    April 7, 2011 at 12:09 am
    Ira, David, Tim et al – do you have some kind of block here that’s preventing you from actually taking in my question? I’m asking about the specifics, how and how much does Solar actually heat the Earth.

    It gets absorbed and heats up whatever absorbs it. A watt of UV heats up a black body the same as vis, ir or electric current for that matter. Evidence given above which you ignored!

    I’ll try again. AGWScience says that Solar energy, UV, Visible and Near IR, the Shortwave energies, actually heat the Earth which then radiates out Thermal IR (which then gets radiated out/trapped etc.). That’s the basic premise of the AGWScience Greenhouse. This should be bog standard understanding and the calculations at hand for any promoting this hypothesis.

    Which there is except for those who willfully ignore it.

    How and how much does UV heat the Earth? How and how much does Visible light heat the Earth? How and how much does Near IR heat the Earth?

    Prove it.

    Done, you are the one who makes the extraordinary statement that a watt of energy at a blue wavelength is incapable of heating an object which absorbs it. What do you think happens to that energy if it is not reflected, where does it go?

    I’m telling you, that these do not heat the Earth. These energies are incapable of heating the Earth to the extent that it gets hot enough to radiate out that amount of Thermal IR. These shortwave energies are not hot in themselves, that is, they are not Thermal Energies. Heat is Thermal Energy in transit from one location to another. Therefore, since they are not themselves thermal energy, they do not heat by their heat, because they have none.

    False as can be proved by focussing light on any light meter (preferably a bolometer, the very existence of which proves you wrong, they’re traditional physics invented in 1878)!

  399. Joel Shore says:
    April 5, 2011 at 6:51 pm

    Claes is spouting nonsense, as seems to be his custom. This is my favorite thing that Claes has written – Look at Equation (10.3) here: http://www.csc.kth.se/~cgjoh/ambsblack.pdf He admits this equation is correct but then claims the problem is with the interpretation and spouts the standard talking points about no back-radiation, one-way transmission from hot to cold, etc.
    ——————————————————————————————————————;

    I think Johnson does an excellent job of presenting modern physics to the pseudo-sciences.

    There are no contradictions. It’s a physics problem.

    Johnson is simply reminding the readers that the Laws of Thermodynanics constrain the solutions of the differential equation to physically accessible states.

    The radiation portion of the Stefan-Boltzmann equation was provide by Stefan and the thermodynamic foundation was provided by Boltzmann.

    If you’re looking for detailed statistical mechanics justification of the constraint T_2 > T_
    1, then you should be reading Boltzmann (or Gibbs) instead of Johnson.

  400. Richard E Smith says:

    You can test it experimentally by pointing a spotlight at a wall in your house and holding a mirror so that as the light reflects off the wall it is reflected back to the wall by the mirror. You will find that the reflection from the mirror will illuminate the wall that is not already illuminated by the spotlight. If you direct it at the part already illuminated you will find that it does not brighten that part any more. The ‘extra’ light has no effect. Infra-red is a form of light and so an object cannot heat up itself by its own radiant emissions. Energy cannot multiply itself.

    And, you’ve actually tried this experiment before? Three things makes me pretty sure you haven’t:

    (1) If it were true, why would people design reflective luminaires for lights?

    (2) What happens to the energy in the reflected light? Does it just disappear and how is this consistent with conservation of energy?

    (3) You will presumably admit that two different spotlights shined on a wall will produce a brighter spot than one. So, the claim seems to be related to the light coming from one source. So, I would ask you, what makes the light from one source special? Do the photons carry little labels that say what source they came from? And, if so, what about a fluorescent lightbulb that is a meter long…Does light from the two ends constitute light from the same source or a different source? How far apart do the parts of the source have to be before it is no longer considered the same source?

    What I am trying to get you to do here is to think critically about your hypotheses.

  401. Richard E Smith;
    You can test it experimentally by pointing a spotlight at a wall in your house and holding a mirror so that as the light reflects off the wall it is reflected back to the wall by the mirror. You will find that the reflection from the mirror will illuminate the wall that is not already illuminated by the spotlight. If you direct it at the part already illuminated you will find that it does not brighten that part any more. The ‘extra’ light has no effect.>>>

    Don’t be daft. If you bother to pay close attention instead of just seeing what you want to see, you’ll SEE that you are wrong. Move the reflected spot from the dark part of the wall toward the illuminated part and you will see the dimmer edges of illuminated part show the outline of the reflection as it crosses into them. Just because at the centre the contrast is below what your eye is capable of detecting doesn’t mean it doesn’t exist.

    You can’t see air either. Does that mean it doesn’t exist?

    Stop contriving experiments based on the limits of human perception and attempting to present them as some sort of evidence. At least stop for a moment and ponder the fact that millions of engineers use these equations every day to design millions of things that WORK. If the principles being expained to you didn’t work exactly like they are being explained to you, then how is it that millions of engineers produce everything from refrigerators to turbines to nuclear submarines, all over the world, in dozens of languages, every single day, based on those exact same principles and manage to build things that WORK? Better still, work AS DESIGNED?

    Albert Einstein published a whole set of theoretical equations to describe photons, how they work, how they behave in 1905. If you think your spot light and mirror means anything consider Robert Millikan who was so convinced that Einstein was wrong that he built an experimental environment that took TEN YEARS to put together. His results? AN EXACT MATCH TO EINSTEIN’S EQUATIONS. A ten year experiment with the goal of disproving Einstein’s theories on light and photons, and all he accomplished was to prove Einstein right. In fact, he spent decades doing other experiments trying to find something wrong with Einstein’s equations and didn’t fully admit that his own work confirmed Einstein’s until 1958 when he said it “scarcely permits of any other interpretation than that which Einstein had originally suggested”.
    Millikan also conducted rigorous experimentation that confirmed Planck’s Constant.

    So if you want to put your silly flashlight and and mirror experiment up as evidence go right ahead. Or you can put aside your misguided notions about what you think you can conclude based on what you can see with the naked eye. Your flash light experiment is up against a ten year, meticulously documented experiment which confirmed Einstein despite being designed to refute him, by a man who won the 1923 Nobel Prize in Physics for that work (amongst others), who also confirmed Planck’s work which in term confirms Stefan-Boltzman.

    You want to refute the work of half a dozen Nobel Laureates all confirming each other’s work, all providing the basis for which millions of design engineers do their jobs every single day, repeatedly get the results they designed for, and expect to discredit them all because your personal pair of eyes can’t measure the brightness difference of reflected light from a flashlight and a mirror?

    I’ll go one better. If you are right, then laser pointers that you can buy for a few bucks don’t work either. Except they do. By bouncing light back and forth between two mirrors until the photons emerge in a very narrow beam all travelling the exact same direction, creating a bright point that you can see. Take the bulb out, look right at it, and you know what you will see staring straight at the bulb? Nothing, not even a glow, its below what your eye can detect. Put it back in the laser and presto, laser beam.

    The evidence that you are dead wrong is all around you, every day. The experiments have been done in mind numbing detail and they all confirm each other. All you have to do is read their work and understand it. Or shine flash lights and mirrors at walls and smugly tell yourself you must be right.

  402. Richard E Smith;
    In fact a lid simply confines the area that is being warmed. This both restricts convection and reduces the volume of the area that is heated. It does not add heat to the source of the heat.>>>

    Then could you please explain to me why a thick ceramic lid works better than a thin aluminum one? They both have the exact same effect in terms of restricting convections and reducing the volume heated. In fact, why does an aluminum pot heat up faster than a thick ceramic pot? But the ceramic pot eventually gets its contents to a higher temperature? Why is that? Where did the extra heat in the ceramic pot come from to make the contents hotter? Why on the exact same burner do the contents in the ceramic pot heat up slower, but eventually reach a higher temperature? Where did the extra heat come from? And why does it transfer heat to the contents slower, but still over time transfers more?

    And why is it that if you know the properties of the pots in enough detail, you can use all the equations we’ve been trying to explain to you to predict exactly how long each pot will take to reach maximum temperature and exactly what the temperature of the contents will be? If the equations are wrong, how does that work?

  403. davidmhoffer, joel and phil (I think) have all derided my experiment of reflecting the light from a spotlight back on to the wall where the spotlight is shining. They claim that the reflection from the mirror does add light to the bright spot but that it simply cannot be observed by the human eye. In other words they believe that reflecting light adds to the light from the source.

    Ok then. Assume that a torch is shining on a purely reflective surface – the opposite of a blackbody. 100% of the light will be reflected. Now get a mirror and hold it over the reflective surface. Let’s assume that the mirror reflects 10% of the light leaving the surface back to the surface. That’s now an extra 10% to add to the light from the torch. So according to david and joel 110% of the original supply is now being emitted from the surface. Add a few more mirrors and you can keep on increasing the brightness. Are you really suggesting that you can create energy like this? What about the first law of thermodynamics? – energy can neither be created nor destroyed.

    david uses the example of lasers to try to refute my observations, but a laser is not created by simply bouncing light between mirrors – the light has to pass through a gain medium with a supply of energy from an electric current. I do not doubt that light can be concentrated. You can do this with something as simple as a magnifying glass. I remember as a boy using a thick magnifying glass to focus the sunlight on to paper to set fire to it in the back garden in summer. Happy days!

    david also takes me to task about lids. He asks why a ceramic lid works better than an aluminium one. Simply because it is a better insulator. This is another way of saying it is not losing so much heat to the exterior so it needs less heat to reach the same temperature. It is nothing to do with backradiation. There is no extra heat.

  404. Agile Aspect says:

    I think Johnson does an excellent job of presenting modern physics to the pseudo-sciences.

    There are no contradictions. It’s a physics problem.

    Johnson is simply reminding the readers that the Laws of Thermodynanics constrain the solutions of the differential equation to physically accessible states.

    What Claes is doing is showing the derivation of the equation (correct as far as I have checked; at least I know the final result iscorrect) and then launching into a bunch of pseudoscience mumbo-jumbo to try to distract you from the fact that the equation itself shows the greenhouse effect and totally negates his point. In large part, he does this by flailing against strawmen (like the idea that heat can flow from cold to hot), which nobody is arguing about and partly he does this by launching into completely irrelevant (and incorrect) arguments about the interpretation of the terms.

    I explained to you in my last post on the subject why Claes’ interpretation is nonsense. To summarize:

    (1) There is good evidence supporting the notion that the colder object is actually radiating to the warmer one, not the least being that one can measure that radiation. (Of course, the warmer one is radiating toward the colder one to…and the 2nd Law tells us that the flow of heat, i.e., the net of the radiative flows, will be from the hot object to the cold object, as the equation itself explicitly shows.)

    (2) It doesn’t matter if you interpret one term as being “back-radiation” or not. Calling it “fred” instead isn’t going to change the fact that the equation itself shows how the heat flow from T_2 to T_1 is reduced as T_1 is increased while T_2 is kept constant. This is the origin of the greenhouse effect.

    The only slightly interesting question regarding Claes’ exposition is whether someone with his knowledge can really get so confused in the end or whether what his interpretation actually amounts to is deliberate deception. In the work I have seen on this subject, whether it is Claes’ or Gerlich & Tscheuschner’s, there seems to be this bizarre habit of mixing a lot of mathematics that is correct (although not really original) with this pseudoscience mumbo-jumbo interpretation. It is hard to imagine how this cannot be part of a deliberate attempt to deceive but maybe it is just testimony to how people can completely deceive themselves when their ideology prevents them from accepting what the equations are telling them.

  405. Richard E Smith says:

    Ok then. Assume that a torch is shining on a purely reflective surface – the opposite of a blackbody. 100% of the light will be reflected. Now get a mirror and hold it over the reflective surface. Let’s assume that the mirror reflects 10% of the light leaving the surface back to the surface. That’s now an extra 10% to add to the light from the torch. So according to david and joel 110% of the original supply is now being emitted from the surface.

    You are getting more and more ridiculous. First of all, there is no rule that says that light can’t reflect off of a surface multiple times. If it is an absorptive surface, some will be absorbed each time; however, for your imaginary, perfect reflector it is not. There is no problem with conservation of energy because none of the energy from the light beam is being removed from the beam by the perfect reflector!

    And, if you go far away from both reflectors and ask where the extra 10% comes from, it comes from the 10% that you took out of the original beam by the partial mirror. I.e., the partial mirror transmits 90% on the first pass, and the other 10% can then be transmitted on subsequent passes.

    At some point, what this all comes down to is this: For ideological reasons, you don’t want to believe in the greenhouse effect and you are willing to come up with pseudo-scientific nonsense to confirm your incorrect belief from now til the cows come home. There is no convincing someone whose has such religious beliefs prevent him from comprehending the science.

  406. Richard E Smith:

    david also takes me to task about lids. He asks why a ceramic lid works better than an aluminium one. Simply because it is a better insulator. This is another way of saying it is not losing so much heat to the exterior so it needs less heat to reach the same temperature. It is nothing to do with backradiation. There is no extra heat.

    And, this differs from the greenhouse effect how exactly? The greenhouse effect says that additional CO2 makes the atmosphere a better insulator. Hence, less heat is needed to reach the same surface temperature.

    All your mumbo-jumbo about “back radiation” doesn’t matter. Forget that anybody ever used the phrase “back radiation”. It is just a phrase sometimes used to explain at another level how the extra insulating effect works. If you don’t like it, don’t use it. (This retired meteorologist strongly counsels against it: http://www.ems.psu.edu/~fraser/Bad/BadGreenhouse.html ) However, what you call something doesn’t change the scientific reality coming out of the equations, which is exactly like what you have described for the ceramic lid case!

  407. Richard E Smith says:

    “Ok then. Assume that a torch is shining on a purely reflective surface – the opposite of a blackbody. 100% of the light will be reflected. Now get a mirror and hold it over the reflective surface. Let’s assume that the mirror reflects 10% of the light leaving the surface back to the surface. That’s now an extra 10% to add to the light from the torch. So according to david and joel 110% of the original supply is now being emitted from the surface. Add a few more mirrors and you can keep on increasing the brightness. Are you really suggesting that you can create energy like this? What about the first law of thermodynamics? – energy can neither be created nor destroyed. ”

    No excess energy is being created in your scenario.

    Let me make a slight variation of your situation. A beam of light shines on a mirror, and the beam bounces off. The beam hits a second mirror head-on and bounces straight back to the first mirror, so that 100% of the energy bounces back to the first mirror. This makes the spot 200% as bright!

    Let’s make a very special light — 1000W but it is only on for 1 ns, so it emits a 1 uJ pulse of energy. As the pulse reaches the mirror, 1 uJ of energy arrives at the mirror, and 1 uJ of energy leaves the mirror and 0 uJ of energy are absorbed by the mirror. Energy was conserved.

    Then a few ns later, the pulse returns to the mirror. 1 uJ arrives and 1 uJ leaves and 0 uJ are absorbed. Energy is again conserved.

    Send 1 pulse or 2 pulses or a billion pulses (1 second worth of light). Energy is conserved for each pulse, and hence conserved overall. The fact that one new pulse and one old pulse arrive at the same time in no way contradicts conservation of energy.

    (In fact, with a few modifications, I think this might be an effective way to explain the GH effect.)

  408. Apologies, I’m not going to able to concentrate on this until next Tuesday, and I haven’t been ignoring what y’all been saying, but nothing so far, and yes I have seen pages on your favourite laws etc., many, many, pages before, addresses my specific points, which I would like in English. So far, none of your explanation in English make any sense. A brief response to some of the points, I will go back over this from Tuesday on in more depth.

    Phil – the Earth is not a black body. Visible light get very easily scattered by the molecules of oxygen and nitrogen, I gave this explanation earlier in a link. If your contraption could be attached to real life organic matter..? And how does this laser, concentrated light, compare with that which reaches Earth from the Sun? Real life.

    A light meter measures how much light there is, not what I’m asking for, but could be useful in the package of explanation I’m waiting for.

    Please, I am still demanding specific answers to specific questions and you haven’t given them, much as you think you have.

    Done, you are the one who makes the extraordinary statement that a watt of energy at a blue wavelength is incapable of heating an object which absorbs it. What do you think happens to that energy if it is not reflect, where does it go?

    Good question, ask a plant.
    Then ask a rock.

    Generally, I’m beginning to tire of irrelevant or gobbledegook maths in responses, and to top it all, even saying that Trenberth didn’t know what he meant, and then being blamed for “not understanding”..

    Tim says:
    April 7, 2011 at 5:15 am

    2) is true and indeed the basis of the Greenhouse effect. Additional energy (thermal IR from the atmosphere) is needed to warm the earth.

    That’s not true, for all practically purposes their ability to heat anything is insignificant. That’s not at all saying just not enough to do the job. You said these Solar energies are 99% responsible for heating the Earth. HOw? And then you say all the thermal IR floating around is heating it. So UV drills a centimetre of heat into something, how much is that contributing to warming the Earth for it to radiate out the claimed Thermal IR? How intense does Blue light have to be to burn? Is that what we get from the Sun? A reference I gave earlier, I think in one of Ira’s discussions, says Blue light is Benign unless intensified, and then it can burn. But this is artificial.

    As AGW claim in the PREMISE of the AGW Energy Budget, it’s 100% Solar, so forget your extra IR needed. Quite specifically it says that it’s the SOLAR energies which Heat the Earth and the Heated Earth gives off x Thermal IR from this source. Whatever the figure that is, what is it?

    This, and only this, is what I’m talking about. The BASIC premise of the AGWScience Greenhouse Hypothesis. Remember the AGW greenhouse? It doesn’t give a damn about anything thermal IR heating the Earth, it’s only these Solar UV,Visible,NearIR which are said to heat the Earth which then radiates out Thermal IR. I’m not interesting in what happens next. Just the Solar in, Thermal IR out of the basic premise.

    Exactly which wavelengths provide how much power would vary from location to location and from season to season. That is a question beyond the scope of a discussion here. If you are really keen to know, dig into the literature!

    Find me some… It shouldn’t be beyond the scope of AGWScience explanation and I haven’t seen anything on this. That’s why I’m asking here of those who are promoting this idea…

    How much does Blue Visible Light heat the oceans? How much does Blue light heat a metre depth of water? I’m looking for actual proof that these heat the actual Earth as it is claimed. That this is not readily available is absurd.

    I’ve already given a reference to Blue Light being used in raising plant life; it is not hot, it can therefore be placed close to the plants. It doesn’t raise the temperature of the plants, it doesn’t cook them. They are doing something else with that energy, they’re not using it to keep themselves toasty warm.

    ..stored chemical energy until used by the plant for photosynthesis perhaps.

    Ira and Hans – The Trenberth Energy Budget is the AGWEnergy Budget, Ira, you use it. So how does Solar convert to Thermal IR? Miraculously? One second it’s Solar and then hey presto it’s Thermal IR?

    Wayne, I’m not sure what you meant in the first about running water filtering out all but the Visible frequencies. I would imagine that it is then highly reflective to short waves, scattering. Thermal IR is bigger, but I suppose it depends on how fast the water is flowing, but water is a very good absorber of it.

    Oh, as an aside to everyone here, I’ve recently read someone say that the Sun doesn’t emit infrared! The AGWEnergy Budget in all its glory of education.

  409. Richard E Smith;
    In other words they believe that reflecting light adds to the light from the source.>>>

    We only believe this because the math says it should be so and when we use instruments that measure these things it turns out that it matches the math. What do you think causes that? Pure coincidence?

    Add a few more mirrors and you can keep on increasing the brightness. Are you really suggesting that you can create energy like this?>>>

    What I’m suggesting is that you understand neither the explanation given to you, nor your own example.

    david uses the example of lasers to try to refute my observations>>>

    I’m not trying to refute your observations. I’m trying to point out that by carefull observation of your own experiment you will find out your observations are wrong.

    but a laser is not created by simply bouncing light between mirrors – the light has to pass through a gain medium with a supply of energy from an electric current.>>>

    Does it really? Have you ever built one? Do you know how? Do you even know what a gain medium is? I for one have no idea what you think that is. On the other hand I’ve actually built a low power laser powered by a flashlight battery with two mirrors and a cardboard tube lined with tinfoil and a 3 watt bulb.

    He asks why a ceramic lid works better than an aluminium one. Simply because it is a better insulator.>>>>

    GASP! You mean just like CO2?

    This is another way of saying it is not losing so much heat to the exterior so it needs less heat to reach the same temperature.>>>>

    GASP! You mean just like CO2?

    It is nothing to do with backradiation.>>>

    GASP! Call it something else if you want. The process is identical.

    There is no extra heat.>>>

    GASP! Right again! Just like with CO2!

    But per your own admission, an insulator (your word) means you need (your words) “less heat to reach the same temperature”. I presume you will also admit then that the same heat would result in a higher temperature? Or do you want to say no and look completely foolish?

    CO2 is just an insulator, albeit a very weak and over hyped insulator, but that is what it is. Since the amount of heat coming in doesn’t shange because of CO2, but it insulates against heat going out, higher temperatures result. You’ve just explained it with your own words!

  410. http://www.natscience.com/Uwe/Forum.aspx/optics/1023/Material-transparent-to-light-but-opaque-to-IR

    Found the above discussion, maybe of some use. Interesting last post about Lambert’s Law – not in real life, link I posted somewhere about plant life utilising IR for photosynthesis much deeper than that – according to Lambert they would be dead.

    I think this is the same recurring problem with ‘some of the laws’, people quote these as if they’re applicable in real life because they relate to a particular discrete context, and as the oximeter link explained, adjustments have to made if you don’t want gigo.

    http://equipmentexplained.com/physics/respi_measurements/oxygen/oximeter/pulse_oximeter.html

    “Early on, we discussed how the pulse oximeter uses Beer’s and Lambert’s Law (absorbance depends on concentration and path length) as part of its factors that it uses to compute oxygen saturation. Unfortunately, there is a problem. In physics, the Beer and Lambert law have very strict criteria to be accurate. For an example, the light that goes through the sample should go straight through like the lights rays in the image below. However, in real life this does not happen.”

    Anyway, as I’ve complained about before, giving me out of context laws as if they’re the answer to everything is nonsense in so much of AGW, carbon dioxide which can rise up through air even though it’s heavier, might make sense in a topsy turvy world of ideal gases, but not in real life.

    So, still asking for real information.

  411. Myrrh;
    So, still asking for real information.>>>

    No use. It just bounces of your Armour of Density.
    But I do note that having been given exactly what you asked for, the exact equations, the exact numbers, wavelength by wavelength, frequency by frequency, examples of the exact experiments, exactly documented, even the names of the people who did them and when, articles explaining in detail the exact answers to every question, you now babble on about something else entirely.

    I concluded that you either:

    1. Read what I told you to and didn’t understand it.

    Or

    2. You didn’t bother to read it.

    One of these people just doesn’t belong here….

  412. Once again, I give up!

    Myrrh, it is unrealistic t expect strangers to teach you all of physics and climate science for free via the internet.

    “I’m beginning to tire of irrelevant or gobbledegook maths”
    Math *is* the language of science. If calculus and infinite series and differential equations are goggledegook to you, you need to learn some more math before claiming you are right and everyone else is wrong. .

    “How much does Blue light heat a metre depth of water? ”
    There are plenty of references — go find one. Then phrase your question more precisely, like “how much energy is deposited in a square meter of ocean water by light between 400 nm and 450 nm from natural sunlight.” If you can’t calculate this yourself (or at least get a pretty good estimate), then you need to learn some more science before claiming you are right and everyone else is wrong.

    “This, and only this, is what I’m talking about. The BASIC premise of the AGWScience Greenhouse Hypothesis. Remember the AGW greenhouse? It doesn’t give a damn about anything thermal IR”
    The AGWScience Greenhouse Hypothesis is EXACTLY about the thermal IR — how thermal IR is emitted and absorbed by the surface and the atmosphere! You need to learn some more science before claiming you are right and everyone else is wrong.

    “So how does Solar convert to Thermal IR? Miraculously? One second it’s Solar and then hey presto it’s Thermal IR?”
    To paraphrase Clarke’s 3rd Law “Any sufficiently advanced science is indistinguishable from magic.” You need to bring yourself up several notches in scientific understanding until this conversion seem perfectly obvious, rather than a miracle!

    “Find me some [literature] ”
    We are not here as your personal research division. Find some and read them and discuss them. Do a little more of your own work.

    And with that — I am done with Myrrh in this thread.

  413. Tim Folkerts says:
    April 7, 2011 at 3:25 pm

    Let me make a slight variation of your situation. A beam of light shines on a mirror, and the beam bounces off. The beam hits a second mirror head-on and bounces straight back to the first mirror, so that 100% of the energy bounces back to the first mirror. This makes the spot 200% as bright!
    ———————————————————————————————–;

    False.

    You only had one light packet which you bounced off the mirror.

    Producing a light packet from where the Sun doesn’t shine violates energy conservation.

    But let’s assume there are two identical light packets and you send one towards the mirror which is reflected.

    Then it’s still false.

    In order to conserve energy, the light has to undergo a 180 degree phase change on the reflection boundary.

    Instead of adding adding the amplitudes, you have to substract them.

    It’s the a perfact candiate for the greenhouse warming conjecture.

  414. I started with one hypothetical packet, then considered a beam of light from a series of pulses. With enough pulses, the photons from an earlier pulse will reflect back from the second mirror when the photons from a later pulse are arriving. This gives twice the number of photons arriving compared to when a single pulse is used. It is a little inexact to call this “twice as bright” since brightness is a perception, but there is twice as much energy arriving as there was without the second mirror reflecting the light back.

    Phase change and coherence will affect details of constructive & destructive interference, but I really don’t think these details detract from the original supposition. If you prefer, consider the light source incoherent so we can average out all the interference effects. (Similarly, I’m not going to worry about diffraction effects. Classical “ray optics” will suffice here.)

  415. Richard E Smith says:
    April 7, 2011 at 1:26 pm
    davidmhoffer, joel and phil (I think) have all derided my experiment of reflecting the light from a spotlight back on to the wall where the spotlight is shining. They claim that the reflection from the mirror does add light to the bright spot but that it simply cannot be observed by the human eye. In other words they believe that reflecting light adds to the light from the source.

    That’s your fault for designing a crappy experiment, and you’re too dumb to realize that your hypothetical result shows that all the reflected light is absorbed which doesn’t exactly support your view of GHE does it? It’s not that the light can not be observed by the human eye, it’s that it’s incapable of the quantitative response you claim. Try covering the mirror with a red (or blue or green) filter, does the bright spot appear redder (or bluer or greener) when reflected light is directed on it? You claim not.

  416. For all those “cold things can’t heat warm things” people:

    How does a freezer work? Mine is pretty much room temperature on the outside, at the back there’s a pretty warm spot, and inside it gets very cold. Let’s say it hasn’t been running, it is room temperature inside and out. I put something in it (we’ll call it the Thing) at room temperature (20C), turn on the freezer, and after a while we check and the thing is frozen an at -5C.

    Now since the Thing started out at room temperature, and the outside of the freezer is still at room temperature, and the back is ABOVE room temperature, how did the Thing get cold? We know it is radiating energy, but where did it go? Since we “know” that cold things can’t transfer heat to warm things, and the outside of the freezer is clearly a lot warmer than the Thing, how did that happen? Did the warm outside of the freezer spray the inside with negatrons? Inversawatts? Snotohp (reverse of photons)? What? If cold things cannot radiate energy to warm things, how did the Thing get cold? Where did its watts go, since EVERYTHING AROUND IT IS WARMER THAN IT IS? If it had NOWHERE that it could send its watts because EVERYTHING around it was warmer, they would have had to stay in the Thing. But we know they didn’t, because the Thing isn’t just a little cooler than what it is surrounded by, it is -5C, 25 degrees colder. What did that cold thing do with all that energy it no longer has?

    Unless…. good gosh…. no, it isn’t possible… unless it transferred its heat energy to…something warmer than itself? No…not possible….must be another explanation….

    OOPS THERE ISN’T!

    Just askin’.

  417. “And, this differs from the greenhouse effect how exactly? The greenhouse effect says that additional CO2 makes the atmosphere a better insulator. Hence, less heat is needed to reach the same surface temperature.”

    Joel

    This is not as I understand the GHG theory.

    And isn’t there a big difference between CO2 reducing the rate of cooling. And this

    “The surface of the Earth is warmer than it would be in the absence of an atmosphere because it receives energy from two sources: the Sun and the atmosphere”

    I find your referenced web site confusing. And comments like this

    “Curiously, the surface of the Earth receives nearly twice as much energy from the atmosphere as it does from the Sun”.

    Surely this is complete nonsense. How can the sun heat the earth then the earth releases energy and twice as much comes back.

  418. Richard E Smith says:
    April 7, 2011 at 1:26 pm
    Ok then. Assume that a torch is shining on a purely reflective surface – the opposite of a blackbody. 100% of the light will be reflected. Now get a mirror and hold it over the reflective surface. Let’s assume that the mirror reflects 10% of the light leaving the surface back to the surface. That’s now an extra 10% to add to the light from the torch. So according to david and joel 110% of the original supply is now being emitted from the surface. Add a few more mirrors and you can keep on increasing the brightness. Are you really suggesting that you can create energy like this? What about the first law of thermodynamics? – energy can neither be created nor destroyed.

    No energy is being emitted from the surface since none is being absorbed! No energy is being created, just recycled, just like in multipass cells. By the way I’d be interested in where you propose putting the second (and subsequent mirror), interesting trick if you can pull it off. I guess you haven’t worked much with optics.

  419. Agile Aspect says:

    In order to conserve energy, the light has to undergo a 180 degree phase change on the reflection boundary.

    Phase changes don’t have to do with conservation of energy. They have to with satisfying the boundary conditions. Light can have a 180 deg phase change upon reflection or no phase on reflection depending on whether the material it is hitting has a higher or lower refractive index, respectively. Or, if the object is a metal and has a complex refractive index (meaning it absorbs light) then the phase change angle doesn’t have to be either 180 deg or 0 deg.

    And, as Tim noted, you can talk about incoherent light.

  420. RJ says:

    This is not as I understand the GHG theory.

    And isn’t there a big difference between CO2 reducing the rate of cooling. And this

    “The surface of the Earth is warmer than it would be in the absence of an atmosphere because it receives energy from two sources: the Sun and the atmosphere”

    No…It just means that there are different ways of looking at the same thing. The reducing the rate of cooling is the net result. The details about how this comes about involve the exchange of radiation.

    “Curiously, the surface of the Earth receives nearly twice as much energy from the atmosphere as it does from the Sun”.

    Surely this is complete nonsense. How can the sun heat the earth then the earth releases energy and twice as much comes back.

    You can’t just invent laws of physics because they sound plausible to you. You have to consider what the actual Laws of Physics say. A good example of this is a quiz question I recently gave my students. They had to choose which statement was NOT a consequence of the 2nd Law. The correct answer was: “A heat pump cannot pump more heat from a cold to a hot reservoir than the amount of work that is used to run it.” However, the majority did not choose this statement because it sounds like such a plausible consequence of the 2nd law, never mind that it is dead wrong. Nearly all practical heat pumps do in fact pump more heat than the work used to run them. In fact, as the two temperatures get closer together, the amount of heat one can pump from the cold to hot reservoir for a given amount of work gets larger and larger.

    What the Laws of Physics say for the situation we are discussing is that in steady-state, the rate of energy absorbed and emitted by an object have to be equal. If the earth is receiving power from both the sun and the atmosphere, then it can indeed radiate more than it received from the sun. In fact, on Venus, this is taken to an extreme…I think the amount of radiation that the surface receives directly from the sun is only a few percent of the amount that the hot surface emits.

  421. davidmhoffer says:

    For all those “cold things can’t heat warm things” people:

    How does a freezer work?

    I think I am going to have to argue with you on this one. A freezer is a heat pump. That is, it uses work (provided by electrical energy) to “pump heat” in the opposite direction than it naturally wants to flow, just as a mechanical pump can make water go uphill. So, the reason the object gets colder when you put it in the freezer is that the refrigerant pipes do cool the freezer so it is colder than the object and then heat flows from the object to the refrigerant.

    However, what is true about your example is that the object in the freezer is always radiating…It doesn’t magically start to radiate only once you turn it on and the refrigerant gets cold enough. It is just that the walls are radiating back as much or more than the object is radiating to them until the refrigerant gets cold enough. (There is also conduction and convection involved but I am simplifying the wording here…Imagine you have a perfect-vaccuum-freezer.)

    Unless…. good gosh…. no, it isn’t possible… unless it transferred its heat energy to…something warmer than itself? No…not possible….must be another explanation….

    It is actually true that in the absence of applied work, heat will not spontaneously go from a hotter object to a colder object. And, even within the freezer itself, the spontaneous flows from, say, the object to the refrigerant will be from hotter to colder. However, “heat” is a macroscopic concept and the reason heat flow must be from hotter to colder is due to the constraints imposed by the statistical physics. On a microscopic scales, the laws of physics involved are reversible. Applied to radiation, what this means is that a colder object will always absorb more radiation from a hotter object than the hotter object absorbs from the colder. Hence the heat flow will be from hotter to colder.

    What it does not say is that there is no radiation from the colder that is absorbed by the hotter. That is the “Magical Second Law of Thermodynamics” rearing its head.

  422. Joel Shore says:

    “The reducing the rate of cooling is the net result.”

    More anti-science. Just how would the imagined reduction of the rate of cooling be falsified? Or even empirically quantified?

    Once again, the inconvenient scientific method is ignored by pseudo-scientists.

  423. Smokey says:
    April 8, 2011 at 9:24 am
    Joel Shore says:
    “The reducing the rate of cooling is the net result.”

    More anti-science. Just how would the imagined reduction of the rate of cooling be falsified? Or even empirically quantified?
    Once again, the inconvenient scientific method is ignored by pseudo-scientists.

    Smokey;
    You know who I am and you know my opinions on CAGW, so please sit down before you read this:

    You haven’t been following this thread I presume, but Joel is 100% correct here.

    Joel – I think one of your responses was poorly worded. By saying the surface of the earth is heated by two sources, one the sun the other the atmosphere, I think you gave one of those technicaly right but gives the wrong impression answers. Better would have been something like the surface is heated directed by the sun, and directly by the atmosphere, but the atmosphere is really second hand sun. The second hand sun w/m2 are bigger than the direct sun w/m2.

  424. Smokey says:
    April 8, 2011 at 9:24 am
    Joel Shore says:
    “The reducing the rate of cooling is the net result.”
    Once again, the inconvenient scientific method is ignored by pseudo-scientists.

    Smokey;
    You know who I am and you know my opinions, so please sit down before you read this:

    You haven’t been following this thread I presume, but Joel is 100% correct here.

    Joel – I think one of your responses was poorly worded. By saying the surface of the earth is heated by two sources, one the sun the other the atmosphere, I think you gave one of those technicaly right but gives the wrong impression answers. The Sun heats directly with s/w, and indirectly with l/w. But the source of the watts in both cases is the Sun.

  425. Smokey says: April 8, 2011 at 9:24 am

    “More anti-science. Just how would the imagined reduction of the rate of cooling be falsified? Or even empirically quantified?”

    That’s a lot like asking “How would the imagined reduction of the rate of cooling for a covered pot compared to a similar uncovered pot be falsified?” Well, you could try do the experiment and finding a case where a covered pot cooled more quickly that a similar uncovered pot. Good luck with that.

    Similarly, the GHGs have been shown experimentally and theoretically to keep the earth warm. Good luck finding a counterexample to disprove this.

    That said, it is certainly possible that OTHER effects might tend to counteract this. If the uncovered pot was insulated on the other sides (perhaps a thick ceramic) then it might cool more slowly DESPITE the fact it was uncovered. If the earth had more clouds to reflect away incoming light, it might cool DESPITE the fact that GHGs are helping keep it warm.

    Everything else being equal, increasingCO2 will result in increased surface temperature” could in principle be falsified (but again, good luck with that). I don’t see how (everything else being equal), that more energy to the surface (in the form of IR from the atmosphere) could have no effect or a cooling effect.

    If other mechanisms are considered, increasing CO2 will result in increase the surface temperature” could be falsified, and you have a much better chance here because now other competing effects could come into play. Indeed, since CO2 has been monotonically increasing for decades but temperature has not, we can safely claim that there MUST BE other affects besides CO2. More CO2 could lead to more clouds, which reflect away more might and help to cool the surface. Upwelling cold water from the oceans could help to cool the surface. The sun could dim and help to cool the surface….

    Now we have LOTS of hypotheses to falsify, some where you have a good shot at success!

  426. davidmhoffer,

    You are right, I’ve not read nearly this entire thread. So are you saying, like Trenberth claims, that the reason there has been no CAGW is that there is empirical [ie: real world, testable, quantifiable, falsifiable and reproducible] evidence, per the scientific method, showing the amount of cooling that supposedly offsets global warming?

    If so, please demonstrate it. I’ve been wrong before, and if I’m wrong I will acknowledge it. Maybe I just missed that particular evidence.☺