Can a gaseous atmosphere really radiate as a ‘black-body’?

Posted: January 21, 2016 by tallbloke in climate, Electro-magnetism, methodology, physics, Temperature

Commenter ‘USteiner’ asks this question on suggestions 16 and explains the reason for asking.

blackbodies

Let’s put aside the conclusion that CO2 makes the antarctic cool the earth. In the Schmithüsen paper (http://onlinelibrary.wiley.com/doi/10.1002/2015GL066749/full) they claim to have made the calculation, and it shows that. However, what is new? This had been measured with – gosh – real data some 45 years (!) ago. See here the Nimbus data from 1971 in Fig. 12d (http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19710026041.pdf )

Further, look at Schmitthüsen’s Fig 2. You see the spike on the CO2 pimple at the South pole at a Spectral radiance of 2.5. And exactly at the same level for the US Standard atmosphere. This is at least consistent with an interpretation that CO2 radiates the same everywhere on earth, but becomes only visible, when the background black body radiation from the ground is low enough. And this requirement is fulfilled only at the very cold south pole. So, what is left as excitement?

My “excitement” , however, is elsewhere. Right in eqn1 of Schmitthüsen they define the atmosphere as a black body radiator, by:

“the emission of the atmosphere ε(index atm) x σ x T(index atm)4”

Gases are no black body radiators. Never. And never were! Not even “greenhouse gases”. So when you do start with a nonsensical assumption, what value does the rest have?

And since they do not give an explanation for that strange assumption, I think it is worth some discussion. In particular since I think this to be a standard assumption in climate modelling, though I have not seen that assumption expressed so explicitly.

Comments
  1. A. Ames says:

    tallbloke:

    Thanks for the topic.

    I venture that if you asked the average person if GHGs radiate heat to a far greater extent than they absorb it, they would think you were daft. So I give credit to Schmitthüsen et. al. for a conventional publication showing that green house gasses can actually cool. Several non-physics academic friends of mine have expressed great surprise at the notion but now accept the idea.

    Energy exchange between photons and gas molecules are described by the Einstein A and B coefficients. I don’t remember exactly, but after all the Boltzmann excitations and collisions have been included I think you get either a T^3 or T^4 dependence for the (now collision broadened) individual lines. It is possible to get from Einstein to Planck but I don’t remember how.

    I think the spike you noted is the non local thermodynamic equilibrium radiation from excitations and radiation in excess of the equilibrium rate.

    As I recall GISS E did use the correlated K to produce a grey body emissivity. Perhaps stating this will encourage someone who knows to help.

  2. oldbrew says:

    Wikipedia does ‘cloud forcing’ here but fails to convince IMO.

    ‘Clouds remain one of the largest uncertainties in future projections of climate change by global climate models, owing to the physical complexity of cloud processes and the small scale of individual clouds relative to the size of the model computational grid.’
    http://en.wikipedia.org/wiki/Cloud_forcing

  3. A. Ames says:

    tallbloke

    From the quite long discussion here
    http://www.drroyspencer.com/2015/06/what-causes-the-greenhouse-effect/
    it appears many people do not understand the difference between the properties of gasses, which have more or less isolated transitions, and solids which have interactions such that the isolated transitions are broadened into the bands that give rise to black body radiation.

    The many times most active player in the green house effect is water (not mentioned in Schmitthüsen et. al ) for which the isolated lines are partially coupled through (I argue) electrostatic interaction to form a solid like continuum, which (I argue) should behave sort of like a grey body.

  4. p.g.sharrow says:

    This argument over the actions of gasses in the Atmosphere is like fingernails on a chalkboard to me.
    CO2 and water are NOT greenhouse like gasses! They are effective energy transporters, water and water vapor is the refrigerant that transports vast amounts of energy from the surface to the Stratosphere. Oxygen, Nitrogen as Argon, are selective insulators that let radiation in and restrict the loss of energy to space, REAL GREENHOUSE LIKE GASSES!…pg

  5. A. Ames says:

    oldbrew
    There are references showing that skyward emission of thunderheads is that of a black body at the temperature of the top. But as noted we cannot predict the incidence of such events.

  6. A. Ames says:

    pg.
    The classic paper
    http://www.john-daly.com/forcing/hug-barrett.htm
    shows directly that CO2 does interact with radiation in a way that N2 etc. do not. This is the phenomenon on which the EU and USA base their redirection of some large part of their economies — quite startling when you think about it.

  7. AlecM says:

    Every GHG band in a planet’s OLR has a spectral temperature defined by the altitude at which the emission sites (GHG molecules) are space filling when seen from a radiometer in the vacuum of Space, pointing towards the surface.

    That is the definition of a black body. The part of OLR in the 8-14 micron atmospheric window has a partial OLR emissivity of ~0.97, because it is from the surface!

    CO2 15 microns band has a spectral temperature of ~-30 deg C because it is in the zone of constant temperature with altitude just above the tropopause. I have not yet worked out why it has a constant temperature, known by all airline pilots who fly in that zone!

    When polar temperature falls to ~-30 deg C, the fixed CO2 special temperature equals surface temperature.

    Readers might wish to see the Big Radiative physics’ mistake made by R D Cess in 1976, which with a ‘fudge’ by GISS in their first modelling paper in the same year apparently created the same 1.2 K CO2 sensitivity. Both papers were wrong but it set up the present IPCC scam because the papers were not rejected in peer review. Result: 40 years of progressively more expensive science fraud.

  8. Re Schmitthüsen’s Fig 2: The graphs have different vertical scales. However, note the similarity between 380ppm and 1000ppm CO2 curves.

  9. gbaikie says:

    **Gases are no black body radiators. Never. And never were! Not even “greenhouse gases”. So when you do start with a nonsensical assumption, what value does the rest have?

    And since they do not give an explanation for that strange assumption, I think it is worth some discussion. In particular since I think this to be a standard assumption in climate modelling, though I have not seen that assumption expressed so explicitly.**

    I believe plasma does radiate as blackbody but plasma is different state of matter than gas.
    Gases of course emit radiation at specific wavelengths and blackbodies emit entire spectrum,-limited by their temperature. Also it seems common blackbody emits in one direction [though I guess not with plasma] as compared to a gas which emit a specific wavelength in a random direction.
    Water and ice would radiate as blackbody, but the size of particle of liquid or solid would limit
    the spectrum absorbed and also seem to effect the spectrum emitted.
    Or when dealing with microns of scale of thickness, liquids and solids are transparent [to all wavelengths]. Or very thin and small liquids and solids would behave more like gases.

  10. Thermodynamics and heat transfer are engineering subjects which few or possible no scientists understand. In 1967 the book “Radiative Transfer”, by Profs Hoyt Hottel and Adel Sarofim, was published by McGraw Hill. It is a compilation of many journal articles and actual practical research by the two authors who were Professors of Chemical Engineering at MIT. Hoyt Hottel authored the sub-sections on radiation heat transfer in Marks Mechanical Engineering Handbook and Perry’s Chemical Engineering Handbook where radiant heat absorption and emission of gases are mentioned. Here is section 4 (Heat) from Marks http://www2.hcmuaf.edu.vn/data/phamducdung/thamkhao/Mark%27s%20Standard-Handbook/Heat.pdf treats gases on a volume basis and uses partial pressures and path length. So called “climate scientists” who do not understand this should not be commenting or publishing.

  11. Peter Shaw says:

    USteiner –
    This topic produces much confusion. I suggest a change of POV, which may clarify things.

    1. The Stefan-Boltzmann law applies strictly:
    a) in a warm cavity with pinhole;
    b) to a warm sphere in an empty universe.
    Neither of these is realistic, so S-B is a mathematical abstraction, to be used with due precaution.

    2. In the real universe, light always goes somewhere; IOW it is a form of transaction, having a source and a destination of equal importance.

    3. Where S-B applies, nett heat flux varies as the difference of 4th powers (Kirchoff). This always has a factor of the temperature difference, so heat flow is everywhere the familiar heat-transfer (2nd-Law-compliant) – with the complication that the heat-transfer coefficient may be a strong function of *average* temperature. Heat transfer from Earth to space can ignore the 3K CMB in practice, but not conceptually.

    4. The Planck Equation is fundamental. This has an “emissivity” term for that particular frequency (= colour), Where this term is constant, Planck sums to S-B. The absorption spectrum of a gas is essentially a plot of “emissivity” vs frequency. which indicates that S-B is a very poor approximation except in particular contexts (such as a narrow band).
    This can lead to an important error: If you separately S-B two greenhouse gases having no common colours, you’ll combine their effects in series, where you should combine them in parallel. Colour matters.

    5. Antarctic cooling: There’s a better spectrum graphic in a slide from Prof. Happer 2014 (sorry, link mislaid; doc name is UNC-9-8-2014)**.
    This needs careful reading. The CO2 “hump” (600-750 /cm) is *not* “on top of” the ice-sheet black-body curve of ~180K. Consider the total area under the curve between those limits. This (scaled) is the heat-transfer to space from the *whole* air column via CO2. The area below the 180K isotherm is the *reduction in heat loss* from the surface via CO2. So, the surface cools less but the air more than without CO2; the nett is increased cooling. The 667 “spike” is stratospheric cooling by CO2.

    6. The Sun produces a near-BB spectrum from hydrogen plasma. It appears that in the cooler surface regions where protons and electrons begin recombining to hydrogen atoms, a few go one further to hydride ions. Because of weak binding, the lines of hydride are “smeared-out” to a near-BB spectrum. You’ll read of “continuum absorption” of water-vapour, very probably due to weak dimers -(H2O)2 ; I surmise that what water dimers do for molecular bands (IR), hydride does for electronic bands (UV-VIS).

    ** TB – I have the slide as a picture in Office2010/Word. Can I post such?

  12. William Astley says:

    The so called one dimensional greenhouse warming calculations ignored (assumed there was no water vapor in the atmosphere which is goofy) the infrared spectrum absorption of water vapor and CO2 overlap, I repeat overlap which is a physical fact not a theory.

    As the earth is 70% covered with water there is a great deal of water vapor in the atmosphere, particularly in the tropics. The overlap of absorption of water vapor and CO2 reduces the forcing for a doubling of atmospheric CO2 by roughly a factor of 4, so the warming for a doubling of atmospheric CO2 with no ‘feedbacks’ is less than a watt/m^2 rather than 3.7 watt/m^2 which will result in warming of roughly 0.15C, not including the offsetting increase in convection cooling.
    The warming of 0.15C is so small the warming without feedbacks will be the same as with feedbacks.

    Check out figure 2 in this paper.

    http://journals.ametsoc.org/doi/pdf/10.1175/1520-0469%281982%29039%3C2923%3ARHDTIC%3E2.0.CO%3B2
    ” Radiative Heating Due to Increased CO2: The Role of H2O Continuum Absorption in the 18 mm region
    In the 18 mm region, the CO2 bands (William: CO2 spectral absorption band) are overlapped by the H2O pure rotational band and the H2O continuum band. The 12-18 mm H2O continuum absorption is neglected in most studies concerned with the climate effects of increased CO2.”
    The IPCC’s general circulation models have more than 100 different parameters that than can and have been adjusted to create model output warming to push the cult of CAGW’s agenda. A key variable that is adjusted is wind speed which is reduce with warming to reduce evaporation cooling. The reduction in wind speed accounts more than a 1/3 of the warming.

    Secondly the one dimensional greenhouse gas ‘warming’ calculations (done by different authors that all followed the party line to create a warming issue where there is none) ignored the fact that an increase in ‘greenhouse’ gas in the atmosphere will cause an increase in convection cooling, a change in the lapse rate (hot air rises which causes cold air to fall which almost completely which is also a physical fact not a theory) which almost completely offsets the greenhouse gas warming due to a doubling of atmospheric CO2.

    The warming in the last 150 years is almost completely due to solar cycle changes not the rise in atmospheric CO2. The same regions that warmed in the last 150 years are the same regions have warmed and cold cyclically in the paleo record correlating with solar cycle changes.

    The recent Antarctic cooling is due to a reduction in high latitude clouds not the increase in atmospheric CO2. The albedo of clouds is less than the Antarctic ice sheet so an increase in clouds in that region causes cooling not warming which explains the so called polar see-saw where the Antarctic ice sheet cools when the Greenland Ice sheet warms and vice versa.
    The following paper by Svensmark explains the mechanism.

    P.S. The solar cycle has been interrupted. The sun spots are shrinking and then disappearing which is different than a reduction in the number of sunspots. The last gasp of warming is due solar wind bursts which are caused by coronal holes. The coronal holes now starting to also disappear.

    “The Antarctic climate anomaly and galactic cosmic rays

    Borehole temperatures in the ice sheets spanning the past 6000 years show Antarctica repeatedly warming when Greenland cooled, and vice versa (Fig. 1) [13, 14]. North-south oscillations of greater amplitude associated with Dansgaard-Oeschger events are evident in oxygenisotope data from the Wurm-Wisconsin glaciation[15]. The phenomenon has been called the polar see-saw[15, 16], but that implies a north-south symmetry that is absent. Greenland is better coupled to global temperatures than Antarctica is, and the fulcrum of the temperature swings is near the Antarctic Circle. A more apt term for the effect is the Antarctic climate anomaly.”

  13. gymnosperm says:

    The reason the usual “shark bites” are inverted as “teats” on a satellite spectrum of Antarctica is that the co2 and a bit of water are radiating at a higher temperature than the surface. Antarctica is the only place this happens because it is the only place surface temperatures are regularly colder than the tropopause.

    There is another “teat” in the ozone bands (to the left at um scale they use) in the atmospheric window. It is the only blemish in an otherwise perfect Planck curve where the satellite reads surface temperature. The ozone is important.

    Satellites see co2 and a bit of water radiating at the temperature of the lower stratosphere in the 15 um vicinity all over the planet. This is because the solar resonating ozone thermalizes Nitrogen and Nitrogen has a special relationship with co2 in the following transitions:

    Below the tropopause there is zero absorption and zero transmission in these bands until you get down to the first some meters above the surface where co2 and water absorb, radiate, and thermalize all this light to extinction.

    Does increased atmospheric co2 cool the planet from Antarctica? The co2 that leaks into the lower stratosphere does. Just like it does everywhere else. Down near the surface additional co2 just lowers the extinction altitude.

  14. AlecM says:

    Just to emphasis the points in my previous post, OLR is a ‘black body effect’, but it is mostly comprised of GHG IR bands which, when self-absorption ends, radiate to Space from a wide variety of altitudes hence spectral temperatures.

    This is what MODTRAN shows, and that is as absolute as any Atmospheric Science computer programme. The only parts of the spectrum that do not behave this way are the atmospheric window (8- 14 microns) from surface and clouds, also the central spikes in the CO2 15 micron and O3 bands, caused by emission from the hotter (than lower stratosphere) upper stratosphere.

    The failure to understand that OLR is primrily a black body spectrum, but from a range of temperatures, has confused 1000s of people including the best of the warmists, like Pierrehumbert.

    Treat it as a range of partial OLRs ad you can’t go wrong. However, R D Cess in 1976 did go wrong by claiming OLR/surface exiting is the planet’s emissivity, a catastrophic failure of physics’ understanding because it implied 40% more atmospheric warming than reality. The rest of the IPCC fraud is, shall we hope, History!

  15. Richard111 says:

    Peter Shaw, thanks for bringing up the subject of water dimers. When do water dimers become water molecules? Or is the water dimer considered already a water molecule? I assume the dimer does not have the reactive effect of the molecule and will also have very different radiative characteristics.

  16. A. Ames says:

    richard111
    I believe the only evidence for a dimer is that the strength of the H2O continuum is proportional to the concentration squared. If there were only short range forces this would conclusive. But the water molecule has a vary large dipole, and most of the IR spectrum would be perturbed by dipole-dipole or electrostatic perturbation. So far the MOPAC program does not indicate any particularly stable versions of H2O4 but maybe there are some, or some other interaction. For now, I would bet on perturbation by random electrostatic interactions.

  17. AlecM says:

    @A Ames: the effect of the hydrogen bond is high in solid and liquid water. However, in the gas phase the molecule’s thermal energy is so much larger, that he probability of dimers existing is vanishingly small.

    If there were an effect. it would have to be from ions or elections in the atmosphere, for which the electrostatic attraction is still strong. This might contribute to the Svensmark effect.

  18. Peter Shaw says:

    Richard111 –
    Sorry, I didn’t explain the jargon.

    Genuine dimers exist in some vapours, eg acetic acid. They’re unstable (then re-form), but last long enough to measure directly.
    Not so, water.
    Kinetic theory has gas molecules bouncing off each other freely. Imagine water molecules as slightly “tacky”, giving lingering contact between them.
    Consider the CO2 “spike” in the NIMBUS spectra at 667 /cm (15 um). This is from the middle stratosphere, where CO2 molecules emit in isolation, producing near-monochromatic light. Encroaching other molecules compromise this, broadening (and lowering) the peak. The spectrum of dilute water-vapour has many sharp peaks, broadened by this process; however the water:water effect is extreme, such that those peaks overlap indistinguishably to a “continuum”. Other than that. we measure only individual molecules.
    This effect applies to triple collisions (which happen) etc, but the whole is lumped into “water dimers”.

    Happer presentation is at
    http://www.sealevel.info/Happer_UNC_2014-09-08/UNC-9-8-2014.pptx

  19. wayne says:

    Peter Shaw, William Astley, thank you for some very insightful point of views to ponder and I agree with much of what you both have laid out. Just had to be said, thanks again.

  20. A. Ames says:

    AlecM
    Thank you for pointing out that the OLR must be a composite spectrum. I agree it must be but it is seldom noted. As for dimers, Shaw’s sticky model should work just fine in all cases.

    tallbloke
    Do you have an answer yet? I think the answer is yes and no. Yes wrt. temperature specific to location, but with lines not continuum, and possibly reduced emissivity to grey not black. We might have a talk about the cartoon, which could be improved.

  21. If one has any experience with heat transfer rather than reading texts which are likely to be wrong, then it will be known that no gas can be a black body which absorbs and emits radiation in all wavelengths. In fact it is doubtful if there are any black bodies other than in mathematical limits. The sun is not a black body. A number of papers have challenged the notion of a cavity being a black body eg “On the equation which governs Cavity Radiation II” by Pierre-Marie Robitaile, Progress in Physics Vol10 July 2014 . In the summary he states “Arbitrary cavities do not emit as blackbodies. A proper evaluation of this equation reveals that cavity radiation is absolutely
    dependent on the nature of the enclosure and its contents”
    further many people do not understand Kirchhoff. The equality of absorption and Emission ONLY applies “in its surrounds”. A body can absorb radiation from another body which is at higher temperature but it will emit radiation only at wavelengths at or below its own temperature. Liquid water can absorb radiation from the sun which is estimated to have an average emission temperature around 5800K but it can only emit radiation at its temperature (in the range 273 to 373 K.)
    Few seem to understand the Stefan-Boltzmann equation and the limits of its application. As I said above most scientists do not understand Heat and Mass transfer and that applies particularly to those calling themselves climate scientist who should be dismissed as incompetents.

  22. AlecM says:

    @Cementafriend: few understand that Hottel’s self-absorbed GHG IR emissivity data (corrected by Leckner) are for the inside of the gas column. At any optical heterogeneity, e.g. the planet’s surface, or ToA, any self absorbed GHG band’s partial emissivity is by definition at the black body amplitude. For CO2, the ratio, optical heterogeneity/internal is about 5:1.

    This has been completely missed by Climate Alchemists and Astronomers, right back to Schuster and Schwarzschild. I like the German word Dumbkopfen.

  23. Trick says:

    tallbloke top post – “Right in eqn1 of Schmitthüsen they define the atmosphere as a black body radiator, by: “the emission of the atmosphere ε(index atm) x σ x T(index atm)4”

    For an ideal BB radiator, ε(index atm) term would be 1 but the paper Chapt. 1&2 reads as if that term is not 1: “one atmospheric layer at temperature Tatm with LW emissivity εatm”.

    The paper is actually written: “assuming the emissivity of the surface to be 1”. Thus the paper Chapt.s 1 and 2 are not assuming the atm. gas as a BB radiator, only the L&O surface is assumed to be ideal BB (no reflection or transmission) reasonably rounding measured L&O 0.96 or so up to 1. Thus the paper is not starting with a nonsensical assumption.

    Chapter 3 Fig. 1 is from observations so include the natural ε of both surface and atm. gas so the paper moves on from the earlier assumptions to reality.

  24. tallbloke says:

    Trick, welcome back.🙂

  25. Gail Combs says:

    Peter Shaw mentions Dr Happer.

    Here is Dr Happer’s presentation where he stomps on the CO2 log curve so beloved of the Warmists and luke warmers.

    (I had the privilege of getting to audit this seminar at UNC thanks to prodding by David Burton. )

    Slides and video and e-mail conversations available here:
    http://www.sealevel.info/Happer_UNC_2014-09-08/

  26. USteiner says:

    @ Trick:
    @Gail Combs:

    The eqn 1 in Schmitthüsen has 2 terms: one is the BlackBody radiation of the surface – that is ok, and it does no matter, what shades of grey they assume – and the other is for the atmosphere. That latter one is the Boltzmann equation for a BB radiator! And thus they assume the N2, O2, H20, Co2 gases to be radiating like a BB. That remains nonsense.

    The discussion of Happer referred to by Gail Combs, is just one more discussion that repeats that those gases are line radiators, albeit there can be many lines due to the spin +/-1 …+/-infinity around the center line, but with strongly diminishing intensity.

    Nobody, to my knowledge, has ever shown – nor even claimed to have shown – that the energy emitted by a line radiator can be approximated by a BB radiation determined as sigma*T^4 !

    In fact, as it is also discussed by Happer here (http://www.sealevel.info/Happer_UNC_2014-09-08/Another_question.html) the lifetime of an excited state of a CO2 molcule is of the order of 1 second, while the time between molecular collisions is in the order of 1 nanaosecond. So, in 99.9999999% of the time the CO2 molecule will loose its energy by collisions, i.e. make the other molecules faster, aka warm the environment. The chances for de-excitation by radiation does increase with thinning atmosphere, and finally result in a cooling of the stratosphere. Whether the net effect of cooling from stratosphere minus warming in the troposphere is positive or negative remains unclear. A hiatus in global warming while CO2 is significantly increasing does not suggest the effect to be strongly positiv!

    So far, I have no problems. But what justifies naming the remaining emission a BB radiator?

  27. Gail Combs says:

    Dr Happer also mentions that CO2 does not start radiating IR until 11 kilometers up (very little) and radiates most strongly at ~47 kilometers up. This is above the tropopause (black dotted line) so both water vapor AND water droplets are below the radiating CO2.

  28. Peter Shaw says:

    Gail Combs –
    On any of Dr Happer’s NIMBUS spectra, the black-body curves provide a temperature scale. If you assume the black-body-in-narrow-band approximation, you can read off the “average” temperature of an emission peak, which from the air-temperature profile gives the “effective emission height” (remembering that the stratosphere warms with height, so choose the right option).
    Inspection shows that “average EEH” is a concept of very limited use.

    Whether emission is above the weather or not depends on the particular line strength; at which point you need a clear-air radiative model.

  29. A. Ames says:

    USteiner

    wrt. T^4 dependence of radiation by a gas, by definition temperature is only defined at equilibrium.
    By definition the heat flow between isolated bodies is the same at equilibrium. Radiation which does not interact is irrelevant to the molecule.

    So if a gas molecule is isolated from collisions and is in equilibrium with a black body, it must transmit and receive the same amounts, thus must vary as T^4.

  30. A. Ames says:

    Gail Combs

    Thank you for the Happer video. It ties up several loose ends. I am having much trouble hearing it, but at 1:15 I believe he says it is not possible to get cooling from CO2, which is the thesis of the Schmithüsen et al paper. A bit later he does emphasize that the most significant unknowns are the properties of H2O and its phases. I would like to be corrected on these points if I misheard.

  31. suricat says:

    Guys, water precipitation ‘scrubs’ CO2 from the atmosphere! Surely we should be aiming our sights at the ‘differential’ between CO2 in the ‘strat : tropo’ to realise a ‘warming : cooling’ forcing for CO2.

    It’s not ‘just’ about its effect, its about ‘what it does’, ‘where it’s present’! Moreover, why is its ‘presence’ (of import to this discussion) in its current ‘form/systemic configuration’ relevant?

    Best regards, Ray.

  32. USteiner says:

    @Gail 3:08pm: Interesting graph, can you provide a link/reference to the article, from which this graph was taken? Thx

    The concept of an “Effective Emission Height” for the atmosphere is the same misleading assumption of the gases being BB radiators, and arguing with it is kind of a circular reasoning.

    Taking from the Nimbus data ( http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19710026041.pdf ) e.g. fig. 11c (lat/lon=14.8N/4.7W is from tropical Mali, Africa) one can argue that the spectrum between wave number 800-1000 and 1150-1250 is in agreement with a blackbody radiation from the ground at a temperature of 320K or 47°C (wow! Does the soil in the tropics really get that hot? I’d thought that evaporation would keep it cooler).

    Now looking at the CO2 absorption band of 15µm equivalent to the (satanic) 666 wave number, the radiation is at a level which the BB radiation of the soil would have, if the soil were actually radiating at roughly 220k (-53°C).

    However, this is now interpreted as being the temperature of the atmosphere at the altitude of maybe 10km where the gases radiate as a BB radiator.

    Nothing justifies that assumption. The CO2 (and H20, O3) gases can be excited by absorbing IR radiation and also by collisions, though they do loose energy by radiation and by collisions. As collisions are less frequent with a thinning atmosphere, I do not know what radiation to expect. But it does still seem to be nonsense to relate this to the temperature of a gas radiating as a BB!

  33. Roger Clague says:

    A. Ames says:
    January 23, 2016 at 11:43 pm

    Yes wrt. temperature specific to location, but with lines not continuum, and possibly reduced emissivity to grey not black

    What is a gray body?
    In my opinion, there is no scale of blackbodyness. A radiator is either a black body or it is not.

    A. Ames says:
    January 26, 2016 at 8:28 pm

    So if a gas molecule is isolated from collisions and is in equilibrium with a black body, it must transmit and receive the same amounts, thus must vary as T^4.

    Gas molecules are not isolated from collisions. It is collisions that cause gas properties.

    How can a molecule of gas be in equilibrium with a black body?

  34. Roger Clague says:

    Trick says:
    January 24, 2016 at 10:02 pm

    For an ideal BB radiator, ε(index atm) term would be 1 but the paper Chapt. 1&2 reads as if that term is not 1: “one atmospheric layer at temperature Tatm with LW emissivity εatm”.

    A black body has only one temperature. The lower atmosphere has a range of temperatures from 300K to 220K.

    I don’t think dividing the atmosphere into layers is acceptable way to avoid the restriction.

  35. A, Ames says:

    USteiner

    As I understand, AlecM above argues that OLR looks like BB “by accident”, and one cannot see the surface.

    On the matter of molecular radiation and its relation to BB radiation,
    https://en.wikipedia.org/wiki/Einstein_coefficients
    particularly the sections on “equilibrium” and “balancing” are worthwhile.

    I think part of our problem is the inappropriate use of temperature derived from LTE radiation as a global temperature. If the system is not at equilbrium various modes will be at different temperatures and heats calculated with the LTE fit will be wrong.

    Molecular radiation at its relevant wavelengths may or may not be exactly equal to BB radiation depending on the state of equilibrium. I grant that in the atmosphere it is most likely not and LTE phenomena dominate.

    For anyone who has not done so, I recommend Petty’s book “A First Course in Atmospheric Radiation”.

    Thanks to Gail Combs for referring http://rtweb.aer.com/lblrtm_frame.html

    While interesting, it is worth remembering that the uncertainties in CO2 properties are small compared to uncertainties the effect of water and its phases.
    https://www.researchgate.net/publication/257864234_Emissivity_of_the_main_greenhouse_gases
    shows powers of 10 and does not even get to condensed phases.

  36. USteiner says:

    Looks like some of this discussion had been had on talkshop before :-): https://tallbloke.wordpress.com/2014/09/12/hockey-schtick-co2-does-what-exactly/comment-page-1/

    Looking at this graph http://rtweb.aer.com/lblrtm_frame.html, already referred to by Gail and Ames, one finds that CO2 does NOT warm the troposphere – except for only a little bit around the tropopause – but does significantly cool the stratosphere, greatly exceeding ozone’s contribution to cooling.

    Assuming this graph shows radiation only, then I understand not seeing the warming from CO2 in the lower atmospheric layer, as most energy absorbed by CO2 would be lost by collisions (and thereby warm the air). In the thin atmosphere at 1mbar (~48km) there would be few collisions and hence radiation would be the primary process of de-excitation of CO2.

    Once the ground temperature is low enough – as it may be in the Antarctic – then the stratospheric CO2 radiation would be higher than the BB radiation from the ground, and form a bump on the IR spectra. So, now we have measurements from 1976, a calculation (this graph) from 1995 – what is the newness of a 2015 dissertation to show that very effect?

    Still does not answer the question why you can use a BB model for the atmosphere.

  37. A. Ames says:

    USteiner: “Still does not answer the question why you can use a BB model for the atmosphere.”

    I am no expert, but I believe that current models for atmospheric radiative transfer are evolved
    from Schwartzchild’s theories for radiation transfer. This model uses the models of Beer, Planck, and Kirchoff and is summarized here.
    http://www.geog.ucsb.edu/~joel/g266_s10/lecture_notes/chapt04/oh10_4_3/oh10_4_3.html
    And I do recall reading that correlated K is used to combine isolated lines into greybody bands.

    The Schwartzchild model is criticised here
    http://claesjohnson.blogspot.com/2015/04/unphysical-schwarzschild-vs-physical.html

    I have just now discovered, and so have not read
    http://www.hindawi.com/journals/ijas/2013/503727/
    which apparently derives Schwarzchild’s equations and uses Plancks model.
    Section 4.3. “Schwarzschild Equation in a Thin Gas looks interesting”.

    I have not yet backtracked from the state of the art summary.
    https://www.aer.com/science-research/atmosphere/radiative-transfer

    Here is where we agree. I strongly believe that the validity of Plankian laws as grey bodies depends on having continuum radiation. My argument is that If there are spectral holes then energy modes are decoupled from the local thermal equilibrium and global thermal equilibrium cannot be achieved. Continuum radiation depends on having the transitions broadened to the extent of overlap at least part of the time.

    If such turns out to be the case, your intuition and the thesis of this blog entry was right on.
    And once again, its all about the water.

  38. Tim Folkerts says:

    ” Right in eqn1 of Schmitthüsen they define the atmosphere as a black body radiator, by:

    “the emission of the atmosphere ε(index atm) x σ x T(index atm)4”

    Gases are no black body radiators.”

    No, they don’t define the atmosphere as a blackbody radiator. In fact, they clearly define the atmosphere as NOT a blackbody radiator!

    For a blackbody, the emissivity is ε = 1 and is typically left out of the equations. The fact that they include a variable for the emissivity specifically means they are assuming the atmosphere is NOT a blackbody.

    They even discuss how the emissivity is not exactly 1, stating “ε(atm) depends on the concentration c and increases with increasing c in certain absorption bands.

    If ε=1 in their equation (ie if the atmosphere were indeed a BB), then all the TOA radiation would come from the atmosphere.
    If ε=0 in their equation (ie the atmosphere were perfectly transparent), then all the TOA radiation would come from the surface.
    But since 0< ε < 1, then the TOA radiation is a combination of surface radiation and atmospheric radiation. The equation may be a bit of a simplification, but it should work pretty well. Increases in GHGs would lead to slight increases in ε (ie the atmosphere becoming a better radiator/absorber) and slight decreases in TOA radiation, exactly as would be expected.

  39. USteiner says:

    I think there is a misunderstanding in some posts: At issue is NOT the fact that the emissivity epsilon is less than 1, because some emitting object is not an “ideal” blackbody. This is normal, because there simply aren’t any ideal blackbodies!

    There are other cases in physics, where „ideal“ formulations are made, which never are really true. However, they may oftentimes be helpful. Like the ideal gas law PV=nRT is never true, because there aren’t any ideal gases. Even Hydrogen fails that gas law at pressures as low as 50bar (relevant for fuel cell cars, which use pressure tanks at 700bar).

    The emissivity may be anywhere from 0 to 1, and yet it may be justified to call such an object a “gray body”, depending on its emission characteristic.

    Many solid bodies are well determined in their greyness, like soil, grass, snow, ice, water. Sometimes it depends on the wavelengths considered, like snow: in visible light snow is highly reflective, i.e. “white” , in infrared almost black. See more data here, e.g. fig 2-20: https://books.google.de/books?id=z_anVNTmQLUC&pg=PA56&hl=de#v=onepage&q&f=false

    At issue here is that the the proportionality to absolute temperature to the power of 4 (~T⁴) is an indication that the underlying assumption for the spectral emission of the emitter is following Planks’s Law ( https://en.wikipedia.org/wiki/Planck's_law ), and the key point here is that it means a CONTINUUM radiation. However, a gas, like CO2 in the infrared spectrum, is a line radiator and does NOT emit a continuum. Other gases, like N2, O2 don’t radiate at all in IR!

    But if the claim is that the integral for the energy emission of a line radiator is also proportional to T⁴, then I would like to see that derived.

  40. Chic Bowdrie says:

    AlecM said:

    “Readers might wish to see the Big Radiative physics’ mistake made by R D Cess in 1976, which with a ‘fudge’ by GISS in their first modelling paper in the same year apparently created the same 1.2 K CO2 sensitivity.”

    This reader certainly would like to know the story of the “fudge.” I found the R D Cess paper and can probably track down the GISS paper, too. Would it be possible to get a reference for the latter to save time? Also, if there are discussion about this online, I would appreciate a link or two.

  41. A. Ames says:

    USteiner
    Consider the fate of a gas molecule loose in a black body cavity. By Kirchoff’s law it must emit as much as it absorbs, and what it absorbs varies as T^4, does it not? I grant there are interesting issues with respect to quantization of the exchanged energy.

  42. tjfolkerts says:

    A. Ames asks: “… and what it absorbs varies as T^4, does it not?”

    No, not specifically. The intensity at a given wavelength does not vary as T^4 — only the integral does. It turns out that a long wavelengths (ie much longer than the wavelength of the peak), the intensity increases linearly with temperature. But at short wavelengths, the intensity increases much faster than T^4. (If every wavelength increased in proportion to T^4, then the shape of the curve and the location of the peak would not change).

    For example, radiation from a 200 K BB radiator peaks near 15 um (ie an absorption band of CO2). If a BB cavity doubled from 50K (peak near 60 um) to 100 K (peak near 30 um), the intensity of 15 um radiation would increase much faster than T^4. If a BB cavity doubled from 400K (peak near 8 um) to 800 K (peak near 4 um), the intensity of 15 um radiation would increase much slower than T^4.

  43. Trick says:

    tallbloke – “Trick, welcome back.”

    Thx tb, I get interested enough & motivated enough to post up comments in limited basic subjects. Some responses and comments from this thread:

    Peter Shaw: “S-B is a mathematical abstraction, to be used with due precaution…that S-B is a very poor approximation…”

    Broadband S-B emissivity term is found from test, S-B is far from a mathematical abstraction. This is justified by experiment as, so far, whenever I measure the brightness temperature of natural objects with my $30 Ryobi IR002, the laser guided digital readout agrees with the object’s Weber digital thermometer temperature. More generally iterative LBLRTM method works particularly well for earth atm. temperature profiles.

    ——

    USteiner – ..”the other is for the atmosphere. That latter one is the Boltzmann equation for a BB radiator! And thus they assume the N2, O2, H20, Co2 gases to be radiating like a BB.”

    Eqn. 1 first term is the L&O surface natural emissivity ε ~0.96 rounded up to 1.0 for convenience, so the surface is set to BB radiator in part 1 (for convenience only).

    Eqn. 1 second term applies the atmosphere emissivity ε to the surface temperature! Strange, but they explain this with a formula showing their estimate for atm. transmitted surface radiation which is less than emissivity 1 since earth atm. has an optical depth; our atm. is not nearly transparent from the orbit of satellites.

    Eqn. 1 third term is for the atm. with an emissivity less than 1 as shown by the ε term as measured. Thus eqn. 1 does NOT assume “N2, O2, H20, Co2 gases to be radiating like a BB”

    11:15am: “the same misleading assumption of the gases being BB radiators”..

    Nowhere I have seen; there is never ever assumption of gas as BB radiator, only the rounded up L&O surface, see Tim’s 7:39pm comment below. The EEH is not misleading, it is measured, and limited in application.

    10:41am: “Other gases, like N2, O2 don’t radiate at all in IR!”

    Your eyes are not a spectrometer, point a real spectrometer up at the night time atm. and it will record a non-zero emission intensity at all light wavelengths (including IR; all means all, no hedging here) down to its particular sensitivity level.

    ——

    Gail Combs 3:08: The Clough and Iacono JGR 1995 graph you show has the right scale in K/day-cm^-1) of cooling. Positive numbers are cooling, negative numbers are warming. Thus the band labeled CO2 below the dotted line is warming (up to about -1K/(day-cm^-1) and the areas above the dotted line to be strong CO2 cooling up to about +100K/day-cm^-1.

    The blue legend +400K/day-dm^-1 is spurious as it does not appear in the original paper. That 400 square of blue crept in behind somewhere.

    ——

    Roger Clague: Dividing any atm. into layers for energy balance iteration is an accepted method as the analytical iterated temperature profile results always properly compare reasonably close (less than 0.1K difference) compared to that found from testing by balloon and rocket thermometry.

    ——

    A.Ames 7:30pm: The Schwarzschild model derives from his own test, Claes J. has not a single test of his own supporting his analytical work that I have ever tracked down. If you know of one, please advise.

  44. Trick says:

    USteiner 10:41am: “The emissivity may be anywhere from 0 to 1..”

    Not necessarily, actually I have read of a certain Univ. Prof. that would give his entering meteorology grad. students a lab assignment to find the emissivity of various materials to prove that they had properly learned about emission/absorption as undergrad. meteorology students.

    One particular material was designed (by him) to test out with emissivity greater than 1.0. This result gave him much entertainment as he observed various grad. students being alarmed at the test results showing ε ~ 1.1 and seeking ways to repeat their testing to all hours of the night so as to get the number to report below 1.0. The means and lengths some would go to & try to avoid turning in what they considered as an embarrassment endlessly amused him. Probably the students did not vote him #1 in popularity but they learned more than the usual student.

    Anyone want to write up how he prepared the specimen with ε ~ 1.1?

  45. tjfolkerts says:

    Trick, I can imagine a variety of tricks to make a surface “test out” with an emissivity greater than 1.000, but all involve some miscalibration or some incorrect temperature or some spurious light source. For instance, if you are using an IR thermometer and comparing the actual temperature to the IR reading, you could trick the IR meter by reflecting some light from an even hotter source off of the surface.

    But using proper physics & proper engineering? Not a chance. You can’t possibly absorb more than 100% of the incoming light. And the 2nd Law prevents a surface from emitting better than it absorbs.

  46. Trick says:

    Tim – I see that Prof. has one more student. He designed the experiment so the students had to calculate emissivities of very small particles (powder) at IR wavelengths. The students, to their horror and dismay, were obtaining emissivity values greater than 1. In undergrad. of course, they had learned (brainwashed?) this was not possible as even you write. Not so.

    Supposed upper limits on emissivities hold (approximately) only for objects much larger than the wavelength of interest. In order to determine all the radiation incident on a body, what is meant by “incident” has to be well defined, but it is not. There are always departures from geometrical optics, although for sufficiently large bodies these departures (e.g. diffraction) are negligible.

    In Planck’s time, testing had already run across the issue these students were again finding (at the expense of and with the expertise of a certain Prof.). Thus in the opening paragraphs of his writings, Planck restricts his law to objects much larger than the wavelength under consideration.

    I mention this because of its application to the moon. Much (~50%) of the moon’s surface is powder, particle diameters measured to be at IR wavelengths. Strictly Planck Law is thus inapplicable. Diffraction is not negligible as Planck required for light incident on the moon surface and needs to be accounted.

  47. tjfolkerts says:

    Yeah, i wasn’t considering quantum effects an objects with sizes similar to the wavelengths. If you can’t give the “size” accurately due to various quantum effects, then I could imagine that you might calculate an impassivity greater than 1.

    I would still expect that an collection of these could not emit more thermal radiation from a macroscopic surface (say at least 1cm x a cm) that a blackbody.

    Do you have a link to details?

  48. Trick says:

    No details Tim, it is just a story from Craig Bohren’s et. al. “Fundamentals of Atm. Radiation” 2006 p. 13. Your 1cm by 1cm of powder still has the rays of light diffracted by the individual particles being the same diameter as wavelength, the size of physical sample will not matter. Scale up to moon size, still have the issue with diffraction that needs be accounted.

    Bohren also writes the seemingly heretical emissivity greater than 1 would be considered almost trivial when cast in the language of antenna engineers. Any antenna engineer knows that the effective area of a receiver can be much larger than its geometrical area.

  49. tjfolkerts says:

    Trick,

    There are two different ideas here.

    1) I accept that you could spread the powder out sparsely over 1 cm^2, covering, say, 50% of the area but have it absorb/emit as if it were covering 55% (Effective emissivity = 1.1).

    2) But if you covered 100% of the 1 cm^2 with this powder, it would only absorb/emit as well as 1 cm^2 of blackbody surface.

    In any case, it is a fascinating (if extremely esoteric) idea.

  50. Trick says:

    If 1) there would be no story Tim, the students would have immediately seen only 50% of the area was covered & would not have been a bit suicidal. The story of ε greater than 1 is from 2) due to diffraction of the light bending around the powder particles and coming back to source more than if negligible bending from particle size much larger than wavelength where it goes on to interact with the next particle. This should also teach rays emitted do not come from the solid surface but within an object.

    Planck drops out diffraction ASAP from consideration, like a hot potato, rules out particle size of the wavelength of interest because it has such a big effect, very noticeable. Not negligible even in 2). The story comes about as an effort to get students to painfully find limits in application of various laws before embarrassing themselves in practice. Not a perfect method but reduces chances.

  51. tjfolkerts says:

    Let’s just say I am skeptical, as a true emissivity greater than 1 would allow the surface cool naturally below the temperature of the surroundings. A sphere covered in such a material inside a cavity (both at the same temperature) would absorb all the the incoming radiation and emit MORE — thereby cooling below the temperature of the surroundings (violating the 2nd Law of Thermodynamics).

  52. Trick says:

    “..as a true emissivity greater than 1 would allow the surface cool naturally below the temperature of the surroundings.”

    Not so Tim, the emissivity greater than 1 comes from diffraction, the small particle powder the students tested would have been in equilibrium with the surroundings, room temperature, diffraction included. The moon’s powder, a sphere covered ~50% in such material, is in equilibrium with the surroundings, long term steady state, diffraction included. The non-negligible diffraction is part of the mix, simply means the small particle emissivity is measured greater than 1, entirely physical & all 2LOT compliant. The analysis simply must account for the diffraction, as nature does.

  53. USteiner says:

    I read the Schmitthüsen dissertation ( http://elib.suub.uni-bremen.de/edocs/00104190-1.pdf ) and found this statement (p48):

    “The emitted long-wave flux at the top of atmosphere FTOA can be separated into the transmitted surface radiation (1 – ­ α) σ Tsurf^4 and the emission of the atmosphere itself α σ Tatm^4, with α being the emission coefficient of the model’s atmosphere, σ the Stefan-Boltzmann constant, Tsurf the surface temperature, and Tatm the temperature of the atmosphere.”

    Huh? On top of the atm being a BB radiator, the ground emissivity now depends on the atmosphere? Where does that come from?

    Later on, p76ff, he discusses the CMIP5 models, and finds them all over the place with respect to antarctic radiation and temperature, but by some odd magic, the average of all models also shows CO2 cooling similar to his (fig.2.27).

    At the end he presents a thought experiment (p85) which I actually find worth considering:

    “3.1 A thought experiment on negative greenhouse effect

    The term negative GHE might seem to sound odd, as we think of GHGs to act like a blanket for the planet, shielding terrestrial radiation from being emitted to space. “Anti-shielding” does not make sense. The following thought experiment demonstrates that GHGs can actually help the planet to lose energy, that would not be emitted without them: Say, there were no GHGs in the Earth’s atmosphere. Clouds shall be neglected as well, to make things easier. The planet gains energy over the tropics (positive budget) and loses this extra energy over the poles (negative budget). The energy transport inbetween is carried out by the atmosphere. The ocean, of course, also contributes to this meridional transport of energy, but this is not of importance here.

    The energy gained over the tropics, which is then transported to the poles, must enter the ground in the polar regions before it can be emitted to space. This is because no GHGs and no clouds, also no aerosol, shall be contained in this hypothetical atmosphere. The atmosphere cannot emit energy directly to space, as it lacks longwave emitters. Consequently, any “imported” energy that shall leave the Earth-atmosphere system in the polar regions, must be transported via sensible heat flux into the ground. From there it can then be emitted to space.

    Now, GHGs shall be introduced. Sure, they have a “shielding” effect over the tropics by causing long-wave downwelling radiation to heat the surface. The same happens, to some smaller extent though, in the polar regions. In addition to that, GHGs give the atmosphere the ability to emit energy directly into space, without the need to transport it through the surface first. This increases the ability of the planet to get rid of energy at the poles, which has been collected over the tropics. In essence, this helps the atmosphere to perform its “task” of meridional energy transport; GHGs help to balance the radiative imbalance between the tropics and the poles.

    The conditions in central Antarctica, being a high-altitude plateau and having a continental climate, are such, that the “shielding” effect of GHGs is excelled by the “helping in losing energy” effect. This, one can name negative greenhouse effect.”

    Leaving the odditiy of finding such a paragraph in an otherwise party-line PhD thesis aside, this is my thinking, too. But how can this be quantified?

  54. tjfolkerts says:

    ME: “I accept that you could spread the powder out sparsely …”
    TRICK: “… the students would have immediately seen only 50% … ”
    ME: ” A sphere covered in such a material …”
    TRICK: “a sphere covered ~50% in such material …”

    You seem to be switching your position. You started with a claim that it would work for 100% coverage (which I still think is impossible) , but now seem to be backtracking to say it would work for 50% coverage (which I can vat least imagine being possible).

    Once again, it is conceivable that a single powder particle emits as if it had a larger effective area than its physical area. This could give an emissivity greater than 1 if you use the physical size of the particle in your calculations.

    However if you dump a layer of these on a macroscopic surface — say 1 m^2 — so that the surface is completely covered, the absorption from the square meter cannot be greater than 1, and the emissivity of that square meter can also not be great than 1 without violating the 2nd Law of thermodynamics (see https://en.wikipedia.org/wiki/Kirchhoff's_law_of_thermal_radiation).

  55. tallbloke says:

    Usteiner quoting Schmitthüsen:
    Now, GHGs shall be introduced. Sure, they have a “shielding” effect over the tropics by causing long-wave downwelling radiation to heat the surface.

    Errm, how does downwelling radiation from a colder body heat a warmer body without violating 2LOT? Or does Schmitthüsen mean the DWR slows the rate of heat loss from the surface? If he does mean that, then he really ought to say that.

  56. Trick says:

    tb – Correct, his use of the word heat is wrong, should have written slow the cooling. This seemingly slight issue has generated more blather in this universe than all of creation.

    ——

    USteiner – “But how can this be quantified?”

    Using Planck distribution & S-B with some labor. However Schmitthüsen should have received some criticism from his committee over this:

    “GHGs can actually help the planet to lose energy, that would not be emitted without them..no GHGs…The atmosphere cannot emit energy directly to space, as it lacks longwave emitters.”

    Of course there would be less energy emitted from the atm. with less IR active gas ppm but atm. LW would still be emitted (feebly) from N2, O2 the other 99%. Schmitthüsen should have realized there would be no “energy gained over the tropics” so no energy to lose in his hypothetical atmosphere since the sun in this hypothetical also “cannot emit energy directly to space”.

    ——

    Tim – I am not backtracking, the 50% coverage would have been a clue to the students as to why that particular sample emissivity was out of the ordinary. I am guessing the Prof. presented all the samples with 100% coverage and the powder (small particle size) was not unusual sample in any obvious way. He was teaching them the limits of particle size vs. wavelength of interest. An innocuous sample would be required.

    “..the emissivity of that square meter can also not be great than 1..”

    The testing showed emissivity greater than 1, Tim, without any violation of 2LOT because the sample was physical, this is not a thought experiment. Non-negligible diffraction caused the testing to measure as it did.

    You seem to be arguing that particle size does not matter, that Planck was incorrect excluding diffraction from his law by forcing diffraction to be negligible in correct applications of the law. That the students really should have found an emissivity of less than 1. The story would not exist if that were the case.

    Consider my Ryobi IR002 fixed at 0.95 emissivity. Point it at a lab glass of boiling water reads same as thermometer, 212F, an ice water lab glass, reads 32F. Point it at the room temperature lab glass filled with powder (particle size IR wavelength) surface it will now read different brightness T than the thermometer inserted in the powder. One would have to adjust the emissivity above 1 to get the correct room temperature brightness reading to equal the thermometer, the diffraction causing the issue. By the student test results.

  57. A. Ames says:

    tallbloke
    Thanks again. At the outset I was unaware of the conclusion below
    which I think is accurate.

    Our language calls use of a coupon which reduces spending a “saving”,
    so, sadly, reducing a cooling can be termed heating.
    ———–

    tjfolkerts
    I agree that you answered exactly. It depends on the molecule. As for the emissivity argument don’t forget the magic word “equilibrium” where the empirical “emissivity” is defined . With lasers tuned to molecular transitions one can produce all sorts of anomalies, even spontaneous cooling as I remember.
    ————-

    USteiner and tb

    The complete answer for the isolated molecule in a BB sphere starts with Planck’s equation 282 for the space density of monochrome radiation, u(nu,temperature). This is the function which when integrated over nu gives the Stefan-Boltamann law.

    Harde, eq. 49 http://www.hindawi.com/journals/ijas/2013/503727/ shows a directly parallel equation assuming kinetic equilibrium but not radiative equlibrium.

    So it appears that if an isolated molecule had lines that uniformly sampled frequency space (a finite sum approximation to the integral) and was in kinetic equilibrium, its radiation would have T^4 dependence.

    Thus, thermal equilibrium is not needed for T^4, just kinetic equilibrium or LTE.

    More deeply, we don’t even require kinetic equilibrium, just some approximation of the Boltzmann phase space near the energies involved in the transitions, and we will observe black body radiation
    at the approximate temperature of the phase distribution.

    My conclusion is that BB radiation is such a good model that it is useless for diagnosis of physical states as nearly everything looks like a BB, a BB band, or a BB line radiator, though perhaps at the wrong temperature.

    ——
    Trick

    Your question on emissivity did not provide enough details to comment. Opals, snow flakes and flourescent plastics all came to mind, and one can imagine nanoparticle flourescence based lasers.

    I am not supporting Claes J, just noting the argument. I do believe in photons and “back radiation”.

    Once again I call attention to the differences between the exchange of photons with solids having bands and with gasses having only electronic levels. Organic molecules and linear aggregates fall in between.

  58. USteiner says:

    tallbloke: This guy Schmitthüsen is a German native, writing his thesis in English. Don’t be overcritical over wording! In particular when A. Ames tells you how the English language is already messed up.

    Otherwise I move to defend Schmitthüsen and declare this topics to be a “moot point”, then lean back and watch with joy how the British and American speakers go after each other! (I learned this when I myself made this mistake:-/ )

    But as to the underlying question re the 2nd Law, Harde (from the link provided by Ames, http://www.hindawi.com/journals/ijas/2013/503727/ ) has a comment (p26), which I found helpful:

    “Thermal radiation is electromagnetic radiation and no
    heat. Therefore, in the same way, as radio waves can propagate
    from a colder antenna to a warmer receiver, microwaves
    can be absorbed by a hot chicken, or CO2 -laser radiation
    (10.6 μm) can be used for welding and melting of metals up
    ∘to several thousand C, so any back radiation from colder and
    higher atmospheric layers can be absorbed by the lower and
    warmer layers, and this back radiation can also be absorbed
    by a warmer surface of the earth without violating the 2nd law
    of thermodynamics. As long as the surface is assumed to be
    a black or gray absorber, it does not filter any frequencies of
    the incoming radiation, in the same way as it does not reject
    any frequencies of the broad Planck spectrum of a thermal
    radiator, independent, if it has a higher or lower temperature
    than the earth. Radiation converts to heat after an absorption,
    followed by an emission in accordance with a newly adjusting
    thermodynamic equilibrium, which only requires that the net
    energy transfer is in balance.”

    I think a partial answer to the current question may actually be found in Harde’s paper.

  59. USteiner says:

    @Trick says:
    January 31, 2016 at 4:15 pm

    Of course there would be less energy emitted from the atm. with less IR active gas ppm but atm. LW would still be emitted (feebly) from N2, O2 the other 99%.

    No, no, no, no, no! N2, O2, Ar and other single-atomic or di-atomic molecules do not – repeat: NOT – emit or absorb radiation at the IR wavelengths in discussion! That is the whole point for me bringing up this issue: the vast majority of the gases making up the atmosphere are NOT RADIATING (at those IR wavelengths)! Not to mention they are not radiating as a blackbody.

    You need to wiggle at least 3 atoms to see IR; hence you see IR from CO2, H2O, O3, N2O, CH4, …

  60. USteiner says:

    @A.Ames
    “Harde, eq. 49 http://www.hindawi.com/journals/ijas/2013/503727/ shows a directly parallel equation assuming kinetic equilibrium but not radiative equilibrium.”

    I guess this is getting us on the right track. Following equation 49 Harde is saying:


    This derivation differs insofar from Einstein’s consid-
    eration leading to (16), as he concluded that a radiation
    field, interacting with the molecules at thermal and radiation
    equilibrium, just had to be of the type of a Planckian
    radiator, while here we consider the origin of the thermal
    radiation in a gas sample, which exclusively is determined
    and rightfully defined by the spontaneous emission of the
    molecules themselves. This is also valid in the presence of
    molecular collisions. Because of this origin, the thermal
    background radiation only exists on discrete frequencies,
    determined by the transition frequencies and the linewidths
    of the molecules, as long as no external radiation is present.
    But on these frequencies, the radiation strength is the same
    as that of a blackbody radiator.

    My take is now: if the CO2 molecule is absorbing a photon, it is unlikely to emit it also, as the lifetime of the excited state is so long (orders of milliseconds to even second) that it more likely will be loosing the energy through a collision, thereby converting the original radiation energy into heat (because the colliding molecule now has more kinetic energy, which it may spread through bumps with other molecules, and at the end the whole ensemble is faster, a.k.a has a higher temperature, aka the atmosphere has been warmed).

    However, even with no radiation energy – e.g. from a hot surface – being available, the molecules will fly around and bump into each other. Some of those collision will result in an excitation, from which a de-excitation occurs mostly by another collision, but sometimes also through radiation. Thus even molecules with no external radiating source do emit radiation, as long as they are not at zero Kelvin.

    And because the kinetic energy of the gas molecules does depend on temperature, then the radiation will also depend on temperature.

    Ok. But why should that dependence a function identical to the planck curve?

    And even if that is the case, the SB Law is the result of integrating the planck curve. But with an IR emitting gas, like CO2, we have 3 lines in IR (4µm, 15µm, and a 3rd at ???µm). Now integrating over a planck function with gaps before, after and between these lines cannot possibly give the same SB law of sigma*T^4 as if all gaps were filled???

  61. tallbloke says:

    USteiner: if the CO2 molecule is absorbing a photon, it is unlikely to emit it also, as the lifetime of the excited state is so long (orders of milliseconds to even second) that it more likely will be loosing the energy through a collision, thereby converting the original radiation energy into heat

    For me this is a key point. We see energy balance diagrams with 340W/m^2 of DWLR coming from sky to ground, but this ignores convection. Since excitation energy is likely to be shared with the 98% of the atmosphere which is non-radiative in short order, you’d expect that with such short path lengths, little of energy from the DWLR emitted at 8km altitude is going to get back to the surface. This surely is part of the reason enhanced GH theory says the mid troposphere should be warming at a faster rate than the surface. It conspicuously isn’t, according to a comparison of satellite and radiosonde mid trop data and surface thermometer data.

  62. Trick says:

    tb – “We see energy balance diagrams with 340W/m^2 of DWLR coming from sky to ground, but this ignores convection.”

    The balance diagrams do not ignore downward convection (downdrafts).

    Convection up from the warmer ground to colder atm. in a planet’s gravity field has its energy dumped into the atm. by radiative, conductive & convective transfer indistinguishable from whence it came. Because of that, the balances do not show separate streams of downward thermals & LH, they lump all these together into an “all-sky” downward returning energy balance (here “all” means cloudy & clear). Convection does not get to space, CERES does not need to measure passing (upward & downward) windiness to/from deep space.

    If you prefer, draw each all-sky return arrow, separating the one big down arrow into its constituent parts; presumably your 340 comes from Stephens 2012, 345+/-9 “all-sky” to surface balance, you would then show 4 down arrows labeled (in place of the one really big one) returning in steady state:

    24 downdrafts (SH) + 88 evaporation/transp. (LH) + 75 SW solar absorbed + 158 LW sky = 345 +/- 9 LW (LW also known as terrestrial radiation).

    It is unfortunate the standard way these cartoons are drawn does not do so.

  63. Trick says:

    USteiner – “..German native, writing his thesis in English..”

    Translation issues are not the root cause of the problem.

    Native American English, Canadian, Queen’s English, Australian (you name it) speakers exhibit confusion over misuse of “heat” term and tb’s point about “slows the rate of heat loss”. Whole blogs exist, whole books have been written over this unneeded confusion.

    Look at Planck’s Treatise title “Theory of Heat Radiation” word written in German “Waermestrahlung” and translated. I recall but can’t pinpoint discussion over the actual translation of the word “Waermestrahlung”. Modern physics would take his title and translate to “Theory of (that which doesn’t exist in the universe) Radiation” or maybe “Theory of (kinetic energy in the thermal sense) Radiation” or the best “Theory of (light ray) Radiation”.

    In the translation of Planck 1913, Prof. Planck makes an immediate attempt in the General Introduction to explain his use of the translated “heat” term, Planck’s translated words: “The term “heat radiation,” then, will be applied to all physical phenomena of the same nature as light rays. Every light ray is simultaneously a heat ray.”

    Heat doesn’t physically exist anymore separate from kinetic energy and light rays. To avoid confusion whenever you see the word “heat radiation” substitute in english : “light ray radiation” to help ascertain if the writer makes sense or not.

    What about heat radiation makes sense? Nothing. Hot bodies radiate (heat), the implication being that cold bodies do not? All bodies radiate, be they hot or cold. There is no such thing as heat radiation being a special radiation with the unique capability of heating matter. Radiation of any frequency is capable of heating matter given sufficient power or the matter illuminated is suitably chosen.

    Catch my drift? Get my meaning?

  64. Trick says:

    USteiner – “No, no, no, no, no! N2, O2, Ar and other single-atomic or di-atomic molecules do not – repeat: NOT – emit or absorb radiation at the IR wavelengths..”

    You will need to inform Prof. Planck’s followers of this new revelation. They will point out tests show pure as we can make N2 – the gas does attenuate IR signals down to the intensity we can physically measure contrary to this assertion.

    “You need to wiggle at least 3 atoms to see IR..”

    What is the translation source word for wiggle? Humans cannot “see” IR so even with 3 atoms we cannot see the atm. water vapor or liquid (or N2) glowing at night. Point a sensitive enough spectrometer at the night all-sky though and it will output non-zero intensity at all light ray wavelengths (all means all, no hedging here). Look up an atm. spectrum on the internet.

    N2 absorbs and emits light of non-zero intensity at all frequencies (if you buy N2 is matter). If you look up the relevant generalized Hamiltonian energy list you will find N2 does so by incident IR energy stored/released in translation, rotation, bond extension and bending. Not as long a list as water but sufficiently long to attenuate IR signals.

    I have seen some argue that you can’t call radiation visible if your eyes can’t detect it in the darkness. Fair enough. Just give me a sensitive enough spectrometer though in the dark and I’ll measure you some intensity in the 400-700nm range. Call it what you will.

    ——

    A. Ames – ”.. with gasses having only electronic levels.”

    The spectrum of even a gas is continuous. Yes there are lines at certain frequencies that absorb (dark) and emit (bright) which coincide with certain natural modes of vibration/electron level in each molecule, these form the basis of identifying species at long range – you know across the universe once the line shift is accounted. A spectrometer though will measure an intensity at each frequency down to its sensitivity level as per Planck’s ideal distribution (made real with S-B). Plug in a frequency of light, you get non-zero emission intensity.

  65. A. Ames says:

    USteiner
    Perhaps the way to look at it is that T^4 results from Maxwell-Boltzmann, as reflected by BB. The common term between all of these is the [nu^3 dnu/Bose-Einstein] which Planck integrates to T^4.

    Any integral can be approximated by the sum of a finite comb of samples to arbitrary accuracy. Perhaps if the comb happened to be in a region of high emissivity, like the equal spaced lines
    of the parabolic oscillator, it would look like T^4 for a short interval of temperature?

    ——————

    Tallbloke, Trick
    I too would very much like to see a better portrayal of the heat movement in all directions.
    I was of the impression that the mid-tropo heat bump is part of the clear sky radiative transfer
    model. Is this wrong?

    Even if we define heat, the momentum of society will crush us. I the meantime would you get me my sweater? I need to warm up a bit.

  66. tallbloke says:

    Trick: pure as we can make N2 – the gas does attenuate IR signals down to the intensity we can physically measure

    True, but it’s a tiny fraction compared to H2O and CO2 is it not? Is there any estimate of the amount of radiative energy absorbed/re-emitted by the N2 and O2 in the atmosphere?

  67. A. Ames says:

    USteiner
    Perhaps the way to look at it is that T^4 results from Maxwell-Boltzmann, as reflected by BB. The common term between all of these is the [nu^3 dnu/Bose-Einstein] which Planck integrates to T^4.

    Any integral can be approximated by the sum of a finite comb of samples to arbitrary accuracy. Perhaps if the comb happened to be in a region of high emissivity, like the equal spaced lines of the parabolic oscillator, it would look like T^4 for a short interval of temperature?

    ——————
    Tallbloke, Trick
    I too would very much like to see a better portrayal of the heat movement in all directions. I was of the impression that the mid-tropo heat bump is part of the clear sky radiative transfer model. Is this wrong?

    Even if we create words to carry all the needed subtleties,the momentum of society will crush us. In the meantime would you get me my sweater? I need to warm up a bit.

    I will take the position that many molecules in isolation have portions of the spectrum where they are completely transparent. Because of quantum mechanical coupling, solids and liquids and even dense gasses may not. I reserve my definition of “completely”.

  68. tjfolkerts says:

    Trick, it is probably time to drop the “emissivity >1” discussion. It is fun but
    1) At the moment, we only have your paraphrase of the situation.
    2) It really isn’t germane here, anyway.

    On the question of absorption of IR by N2 or O2, any absorption is minuscule to the point of being ignoble. Check out the spectra at the website below — I am sure you can figure out how to play with the controls. You will find that the strongest IR peaks for N2 & O2 — even scaled atmospheric abundance — are many orders of magnitude weaker than the peaks for CO2 or H2O. And outside those peaks, the IR absorption is mostly too small to even include in the dataset.

    http://www.spectralcalc.com/spectral_browser/db_intensity.php

  69. Trick says:

    “Even if we define heat, the momentum of society will crush us.”

    No kidding. The best cure is to use the “heat” term as little as possible, I will guaran-damn-tee your writing clarity will improve. Bury the “heat” term will full honors. Just like we have buried slide rules, now use calculators and iPads (of any size), watches, phones, etc. Slide rules and “heat” both had their day, did what was asked of them, but have been superseded.

    The caloric theory is officially dead, thus should receive a decent burial and remain below ground. Yet as you imply – I note most textbooks, blogs, wiki do acknowledge the death of caloric, they then proceed to do everything possible to breathe life into the corpse. So many people write about “heat” that by hook or crook a corporeal form for “heat” must be invented to save appearances.

    In everyday language you will hear “heat rises”? WTF? Heat was buried, can’t rise. They say “heat is added to the system” or “heat transfers”? WTF again. Heat is not a substance that can be “added” or “rise” OR “transfer”. Kinetic energy transfers, light ray energy transfers.

    Keep “heat” buried. Join the revolution! Out with the 1800s, in with the new! Vive light rays!

    ——

    tb 9:19pm – Very true, the intensity of atm. N2, O2 emission is of the order “feeble” as I wrote but non-zero. Internet searches and texts will be more specific. If I see something, I’ll post up.

    ——

    A. Ames – ”I will take the position that many molecules in isolation have portions of the spectrum where they are completely transparent.”

    This will be in opposition to Planck’s law. You will need extraordinary proof. All matter radiates at all frequencies at all temperatures all the time according to his testing ref.s.

    ——

    Tim – ”(emissivity >1) isn’t germane”

    Well, I responded to OP USteiner 10:41am: “The emissivity may be anywhere from 0 to 1..”, to point out the range is not that limited. Looks like Tim might have answered tb’s question with spectralcalc link.

    The problem of the feebleness of N2, O2 emission arises when it is set to zero with non-physical conclusions then being drawn as in Schmitthüsen dissertation “The atmosphere cannot emit energy directly to space”. This is never true no matter the atm. composition.

  70. A. Ames says:

    Trick
    Planck does not place a lower limit on emissivity, and does admit of a “diathermous” (as I remember) or transparent material. His cavity contains a speck of black material to assure mode swapping just for this case. He also did not know about many phenomena since discovered like the heat capacity of hydrogen. http://employees.csbsju.edu/cgearhart/pubs/H2_Utrecht08.pdf
    No levels means no radiation.

    Do you really believe that the N2,O2 continuua outside the Hitran data base are significant? If so, by how much, and what interactions do you propose? I think there is a lot left to say about H2O, and I do wonder if there might be H2O-CO2 events but these won’t change the overall picture.

  71. Trick says:

    “His cavity contains a speck of black material to assure mode swapping just for this case.”

    Not in the real cavities tested in the ref.s he cites. They did not employ a speck of carbon though one could argue that certainly negligible impurities existed.

    Planck reserves the bit of carbon is needed only for discussing non-real ideally perfectly reflecting cavities which present a singularity he gets around in theory by introducing that minute particle of carbon “just as small as we please” compared with the radiation in the cavity of arbitrary magnitude. In this way, an impurity will always exist to make the anomaly disappear no matter the polished perfection of the cavity. Much later, LASERs made use of this concept.

    Significant is relative. I will stick to my N2,O2 emission being “feeble” for Earth STP. This eliminates the singularity for zero emission from N2,O2 similar to Planck’s “feeble” emission of the carbon particle eliminating the polished cavity singularity.

    Thanks for the discussion.

  72. Brett Keane says:

    What is the likely effect and outcome for a completely 3N2/O2 atmosphere under our radiance? Stephen’s idea of expansion and loss to gain equilibrium is the only one I’ve seen so far?

  73. USteiner says:

    tallbloke says:
    January 31, 2016 at 6:11 pm
    … you’d expect that with such short path lengths, little of energy from the DWLR emitted at 8km altitude is going to get back to the surface. “

    My thinking, too. In fact, we can do some calculation for the 15µm line. The extinction coefficient (taken from here: http://www.eike-klima-energie.eu/fileadmin/user_upload/Bilder_Dateien/HUG_THE_ein_Artefakt_II/Hug-EIKE-Artefakt_II-1.pdf ) is 20.2 m²/mol. At 400ppm CO2 (=0.0179 mol/m³ near ground) the Absorbance A for a path length of 1m is 0.362. The amount of radiation absorbed is then:
    Path length 1m: …. 57%
    Path length 2m: …. 81%
    Path length 10m: .. 99.98%

    So, put on your 15µm goggles and look down – you will barely be able to see your toes!

    The typical depiction for greenhouse warming of some layer up in the atmosphere, perhaps where the airplanes fly, which radiates like mad down to the earth, does not exist. All “Back Radiation”, which impinges on the ground, originates in the first few meters of the atmosphere from the ground up.

    And what I have learned here is this “Back Radiation” is not from the CO2 molecules sending a previously received photon back to earth – because this will be quickly thermalized by collisons – but rather from the CO2 molecules emitting photons simply due to them being at a temperature greater than 0K. And such radiation would happen even if the molecules received no ground-radiation at all.

    Lastly, not to forget: Harde also concludes (p25):

    … should only contribute to a temperature increase of
    about 0.3 K. This is in clear contradiction to the IPCC, which
    issues an official climate sensitivity (temperature increase at
    doubled CO2 ) of CS = 3.2 K [25]. But even a more sophis-
    ticated consideration,… only gives a
    climate sensitivity of 0.6 K [20], which is still 5x smaller than
    the IPCC value.

    It still leaves open the possibility that the climate sensitivity may be even negative, as Schmitthüsen mulled over in his thought experiment (see above).

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