the Air Vent

Because the world needs another opinion

Radiative Physics Simplified II

Posted by Jeff Id on August 6, 2010

Radiative physics of CO2 is a contentious issue at WUWT’s crowd but to someone like myself, this is not where the argument against AGW exists.  I’m going to take a crack at making the issue so simple, that I can actually convince someone in blogland.  This post is in reply to Tom Vonk’s recent post at WUWT which concluded that the radiative warming effect of CO2, doesn’t exist.  We already know that I won’t succeed with everyone but when skeptics of extremist warming get this wrong, it undermines the credibility of their otherwise good arguments.

My statement is – CO2 does create a warming effect in the lower atmosphere.

Before that makes you scream at the monitor, I’ve not said anything about the magnitude or danger or even measurability of the effect. I only assert that the effect is real, is provable, it’s basic physics and it does exist.

From Tom Vonk’s recent post, we have this image:

Figure 1

Short wavelength light energy from the sun comes in, is absorbed, and is re-emitted at far longer wavelengths.  Basic physics as determined by Planck, a very long time ago.  No argument here right!

Figure 2 below has several absorption curves.  On the vertical axis, 100 is high absorption.  The gas curves are verified from dozens of other links and the Planck curves are verified by my calcs here.  There shouldn’t be any disagreement here either – I hope.

Figure 2 – Absorption curves of various molecules in the atmosphere and Planck curve overlay.

What is nice about this plot though is that the unknown author has overlaid the Planck spectrums of both incoming and outgoing radiation on top of the absorption curves.  You can see by looking at the graph (or the sun) that most of the incoming curve passes through the atmosphere with little impediment.  The outgoing curve however is blocked – mostly by moisture in the air – with a little tiny sliver of CO2 (green curve) effective at absorption at about 15 micrometers wavelength (the black arrow tip on the right side is at about 15um wavelength).  From this figure we can see that CO2 has almost no absorption for incoming radiation (left curve), yet absorbs some outgoing radiation (right curve).  No disagreement with that either – I hope.   Tom Vonk’s recent post agrees with what I’ve written here.

Energy in from the Sun equals energy out from the Earth’s perspective — at least over extended time periods and without considering the relatively small amount of energy projecting from the earth’s core.  If you add CO2 to our air, this simple fact of equilibrium over extended time periods does not change.

So what causes the atmospheric warming?

Air temperature is a measure of the energy stored as kinetic velocity in the atoms and molecules of the atmosphere.  It’s the movement of the air!  Nothing fancy, just a lot of little tiny electrically charged balls bouncing off each other and against the various forces which hold them together.

Air temperature is an expression of the kinetic energy stored in the air.  Wiki has a couple of good videos at this link.

“Warming” is an increase in that kinetic energy.

So, to prove that CO2 causes warming for those who are unconvinced so far, I attempted a thought experiment yesterday morning on Tom Vonk’s thread.   Unfortunately, it didn’t gain much attention.  DeWitt Payne came up with a better example anyway which he left at tAV in the comments.  I’ve modified it for this post.

Figure 3- Experimental setup. A – gas can of air with all CO2 removed at ambient temp and standard pressure. B – gas can of air diluted by 50 percent CO2, also at ambient temp and standard pressure. C ultra insulated laser chamber with perfectly transparent end window and a tiny input window on the back to allow light in from the laser. Heat exit’s the single large window and cannot exit the sides of the chamber.

Figure 4 is a depiction of what happens when  C contains a vacuum.

Figure 4 - Laser passes straight through the chamber unimpeded and a full 1000 Watt beam exits our perfect window.

The example in Figure 5 is filling tank C with air from tank A air (zero CO2) at the equilibrium state.

Figure 5 – Equilibrium of hypothetical system filled with zero CO2 air from canister A.

Minor absorption of the main beam causes infrared absorption and re-emission from the gas reducing the main beam from the laser. This small amount of energy is re-emitted from the gas through the end window and scattered over a full 180 degree hemisphere.

What happens when we instantly replace the no-CO2 air in chamber C with the 50% CO2 air mixture in B?

Figure 6 – Air in C is replaced instantly with gas from reservoir B

From the perspective of 15 micrometer wavelength infrared laser, the CO2 filled air is black stuff.  The laser cannot penetrate it.  At the moment the gas is switched, the laser beam stops penetrating and the 1000 watts (or energy per time) is added to the gas.  At the moment of the switch, the gas still emits the same random energy as is shown in Figure 5 based on its ambient temperature, but the gas is now absorbing 1000 watts of laser light.

Since the beam cannot pass through, the CO2 gains vibrational energy which is then turned into translational energy and is passed back and forth between the other air molecules building greater and greater translational and vibrational velocities.  —- It heats up.

As it heats, emissions from the window increase in energy according to Planck’s blackbody equation.  Eventually the system reaches a new equilibrium temperature where the output from our window is exactly equal to the input from our laser – 1000 watts. Equilibrium! – (Figure 7)

Figure 7 – Equilibrium reached when gas inside chamber C heats up to a temperature sufficient to balance incoming light energy..

The delay time between the instant the air in C is switched from A type air to B air to the time when C warms to equilibrium temperature is sometimes stated as a trapping of energy in the atmosphere.

“CO2 traps part of the infrared radiation between ground and the upper part of the atmosphere”

So from a few simple concepts, two gasses at the same temp, one transparent the other black (at infrared wavelengths), we’ve demonstrated that different absorption gasses heat differently when exposed to an energy source.

How does that apply to AGW?

The difference between this result and Tom Vonk’s recent post, is that he confuses equilibrium with zero energy flow.  In his examples and equations, he has a net energy flow through the system of zero, which is fine. Where he goes wrong is equating that assumption to AGW.

What we have on Earth, is a source of 15micrometer radiation (the ground) projecting energy upward through the atmosphere, exiting through a perfect window (space) – sound familiar?   Incoming solar energy passes through the atmosphere so we can ignore it when considering the most basic concepts of CO2 based warming (this post), but it is also an energy flow.  In our planet, the upwelling light at IR wavelengths is a unidirectional net IR energy flow (figure 2 – outgoing radiation), like the laser in the example here.

Of course adding CO2 to our atmosphere causes some of the outgoing energy to be absorbed rather than transmitted uninterrupted to space (as shown in the example), this absorption is converted into vibrational and translational modes (heating). Yes, Tom is right, these conversions go in both directions.  The energy moves in and out of CO2 and other molecules, but as shown in cavity C above, the gas takes finite measurable time to warm up and reach equilibrium with space (the window), creating a warming effect in the atmosphere.

None of the statements in this post violate any of Tom’s equations; the difference between this post and his, is only in the assumption of energy flow from the Sun to Earth and from Earth back to space.  His post confused equilibrium with zero flow and his conclusions were based on the assumed zero energy flow.   The math and physics were fine, but his conclusion that insulating an energy flow doesn’t cause warming is non-physical and absolutely incorrect.

Oddly enough, if you’ve ever seen an infrared CO2 laser cut steel, you have seen the same effect on an extreme scale.

————-

So finally, as a formal skeptic of AGW extremism, NONE of this should create any alarm.  Sure CO2 can cause warming (a little) but warmer air holds more moisture, which changes clouds, which will cause feedbacks to the temperature.   If the feedback is low or negative (as Roy Spencer recently demonstrated), none of the IPCC predictions come true, and none of the certainly exaggerated damage occurs. The CO2 then, can be considered nothing but plant food, and we can keep our tax money and take our good sweet time building the currently non-existent cleaner energy sources the enviro’s will demand anyway.  If feedback is high and positive as the models predict, then the temperature measurements have some catching up to do.

Even a slight change in the amount of measured warming would send the IPCC back to the drawing board, which is what makes true and high quality results from Anthony’s surfacestations project so critically important.

This is where the AGW discussion is unsettled.


172 Responses to “Radiative Physics Simplified II”

  1. BillyBob said

    Isn’t a very large portion of the CO2 absorption curve also the same as the H20 curve?

  2. Jeff Id said

    #1 sure is.

  3. Gary said

    Now that’s an explanation a caveman can understand [even the ones in southern Oregon, JAE ;-).]

  4. JAE said

    Don’t yell at me, but I think you are still not “trapping heat with CO2,” but you are simply storing heat in all the air molecules in the form of kinetic energy, in accordance with their specific heat values (water vapor stores twice as much, BTW–about 2 joules/g/K). You would store exactly the same amount of heat in your “magic can” if you had only a tenth of the CO2 molecules in there.

  5. Jeff Id said

    #4 I agree the terminology ‘trapping’ isn’t perfect. DeWitt had a better word for it once, but I can’t remember.

  6. Steve Fitzpatrick said

    Jeff,

    More power to ya! You seem to have the patience of a saint.

  7. Jeff Id said

    As far as storing the same heat with less CO2, what happens in reality is a portion of the beam always escapes so as the concentration drops more beam escapes – less heating. In this example, I’ve shown no beam escaping to simplify it.

  8. Jeff Id said

    #6 I offered the post to Anthony, that’s when the stuff will really hit the fan. That may be the first time anyone compared me to a saint. hehe.

  9. BillyBob said

    If you subtract H2O absorption from Co2 absorption, isn’t O2+O3 as much as greenhouse gas than Co2?

    And is there any energy left for Co2 to absorb after O2+O3 absorbs the outoging energy considering that O2+O3 is 21% of the atmosphere and Co2 is .039%.

  10. Steve Fitzpatrick said

    #8. Great patience probably does not by itself confer sainthood, but it may be required for serious consideration. ;-)

  11. Carrick said

    Nice post, Jeff. Now write down the mathematical equations and post the R code. ;-)

  12. Michael Moon said

    What is not discussed in any of these exchanges is the Second Law. 15 microns corresponds to a temperature well below zero F which means that, typically, the only CO2 that can absorb this energy is high in the atmosphere several miles up, where it is colder. Heat up there does not drop down here! Photons or no photons, the cold surface of the Earth cannot transmit heat to warmer atmospheric gases.

    CO2 lasers cut like crazy, but the wavelength is not 15 microns, but much shorter. 15-micron is far infra-red/microwave band. This has been rather difficult to intuit ever since I started paying attention. Somebody, help please…

  13. […] A reply to Vonk: Radiative Physics Simplified II Posted on August 6, 2010 by Anthony Watts Radiative Physics Simplified II […]

  14. JAE said

    Before you introduce the 15 micron beam, a certain number of the CO2 molecules (5%?) are already in the “excited” state that corresponds to absorbing a photon with a 15 micron wavelength (call them “hot” molecules). I guess you are saying that the beam will increase the proportion of hot CO2 molecules, which will then react with other air molecules, making the whole air mass hotter. It looks logical, but I would like to hear Vonk’s take on it.

  15. YFNWG said

    Perhaps it would confuse the “picture” but Figures 5-7 should have radiative emission out from the laser side of the canister in order to properly depict CO2 IR emissions.

  16. John Eggert said

    Excellent post. I agree that people who dispute the fact of radiant heat absorption (and others who speak of the second law of thermodynamics or the overwhelming influence of water absorption) detract from the strong case against AGW. I question the doubling hypothesis myself, but am willing to be talked into it. If anyone cares, Tallbloke’s talkshop (see link to the right) was kind enough to host three papers I wrote regarding this. My view is that the doubling is an oversimplification. It is pretty close from 1 ppm to 100 ppm, but there is a case to be made that this drops off after that. Could be wrong, often am.

    JE

  17. Steve Fitzpatrick said

    #12,

    The 15 micron wavelength does not have any associated temperature… it is not “cold”, it simply carries energy. Yes, if you have a black-body at -20C its emission curve has amaximum emission near 15 microns, but the energy content of that black-body radiation at 15 microns (number of photos times energy per photon) is quite low. Jeff’s 1000 watt laser has lots of photons… enough to sum up to 1000 watts, even though the energy per photon is low. Don’t put your hand in front of that laser beam, or it will burn a hole in your skin… fast!

    Microwaves in your microwave oven carry a lot of energy (they cook the food!) even though the wavelength corresponds to a cryogenic black-body temperature. The microwave oven sure doesn’t suck heat out of the food due to the low black-body equivalent temperature of the microwaves!

    If you still have doubt after reading Jeff’s (very clear) explanation and the above comments, then I suggest you find a course in radiative physics.

  18. josh said

    Jeff, are you really sure that a CO2 filled box is “black stuff”. Emission and absorption are equal probability events, and much more likely than thermalization. The CO2 should just scatter the 15 micrometer light, not stop it dead. I think thay all you’ve done is create a 15 micrometer light bulb from a 15 micrometer laser, even immediately after you turn it on – there is no need to wait for any heating.

  19. JAE said

    Where I have hesitation is here: At the initial temperature of the gas before you introduce the 15 micron beam, the Boltzmann Distribution says you have only a certain proportion of atoms at each energy level. Now, you add a beam of photons at a single energy level and tell us that you will suddenly change that whole distribution of energies. How does this happen? You can change the distribution only by changing the temperature of the gas. Chicken/egg???

  20. Steve E said

    Please forgive me, I’m a layman. This has been one of best simple explanations I’ve seen.

    Is it fair to say that what is at issue in AGW is the changing equilibrium point resulting from the rising concentration of CO2 plus feedbacks–positive or negative–and just how much heat is retained at each subsequent point of equilibrium?

  21. Carrick said

    #17, that is a nice explanation.

    IMO, one of the chief disservice’s people are given in their science education is introduction to “classic” thermodynamics before (if ever) they are introduced to the microscopic (“statistical”) theory.

    There is no such thing as heat, period. It is a 19th century concept of a “caloric” fluid, which has every bit as much reality as aether, which is to say none.

    Heat does not “flow” from warm bodies to cool bodies. Kinetic energy associated with individual molecular vibrations gets transmitted either from collisions between molecules or via the exchange of particles (e.g. thermal photons).

    In the exchange of a thermal photon. in particular, the atom doesn’t first take a poll of the space surrounding a distant molecule (which it must some how know it going to absorb in the future) before it decides whether it can irradiate the photon or or not, so as not to violate the second law of thermodynamics!

  22. josh said

    I am still having trouble with this Jeff, I think you are misunderstanding the physics badly. The CO2 transmits the light just fine, it just scatters it. So you will no longer have a coherent beam, most of the light will just go off at some other angle having been absorbed and re-emitting many times along the way. Some of this light, a small fraction, will result in additional translational energy of the CO2 molecules, but most of it will merely be slowed down a little bit and sent off in a different direction, resulting in no net heating.

    Now, if the sides of your can are IR absorptive, you would heat up, as some fraction of the IR beam is scattered into the side, but you could accomplish the same result with a mirror in the middle of the can placed at a 45 degree angle to the beam.

  23. Alan D McIntire said

    Another way to look at the issue of earth cooling and the earth/atmosphere relationship is to consider Newton’s law of cooling:

    http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html

    T(t) = Tambient + (Toriginal- T ambient)*e^-kt
    Start, say with a temperature of 294 K, plug in the temperature of a clear atmosphere, say 255 K, figure T(t) after 12 hours and you’re able to compute k.

    Plug in a different ambient temperature, say 3K, the temperature of the background radiation in space, and you get a lot quicker cooling-

    Plug in the temperature for a cloudy sky, say about 270K, and you get a lot slower cooling

  24. Jeff Id said

    #22, CO2 absorbs and re-emits the light. It is black in color to the beam – no transmission. This is caused by the quantum vibration modes of the molecule being centered on the 15um laser.

  25. josh said

    “#22, CO2 absorbs and re-emits the light. It is black in color to the beam – no transmission. This is caused by the quantum vibration modes of the molecule being centered on the 15um laser.”

    What exactly do you think “transmission” is? Water droplets in a cloud absorb and re-emit light – are clouds black? No, they are white.

    Basically Jeff what you are claiming is that the entire energy of the beam is trapped by the CO2 and not re-emitted as IR at the same wavelength, which is patently non-physical. Or does this light somehow get stuck in a circular path of absorption and re-emission that never intersects the boundaries of the container?

    In reality, the majority of the light will pass right through the gas, being absorbed and re-emitted by successive CO2 molecules until it leaves the gas. A tiny fraction of the energy will be converted into heat (translational energy).

  26. Dave Springer said

    Why not better simulate reality?

    The energy source would be the top of a column of water several hundred meters deep at 62F. If we make it one square inch (for convenience) that should be in the neighborhood of 300 milliwatts of continuous gray body. Our air column would be vertical at 14.7 psi at the bottom and be near 0 psi at the top some 25 miles long. Let’s make it standard atmosphere minus all GHG’s (including water) except for 380ppm CO2. The far end of our column of course would be 3K blackbody (cosmic background).

    We’ll need to keep in mind that water vapor is present at about 3% or more over the ocean surface in the real world and has an absorption band that overlaps CO2 so our CO2 induced temperature stratification will be exagerated near the surface. We’ll also be ignoring latent of vaporization which in the real world, where water surface will be evaporating, can carry a substantial amount of non-sensible energy upward through convection and release it much higher in the column when the vapor condenses, bypassing GHG absorption between point of evaporation and point of condensation. Lack of accounting for evaporation will also exagerate CO2 effect in the lower portion of the air column.

    Is it too much to expect that pretty much all PhD physcists should be able to crunch the numbers (that task is beyond me) and agree about what the temperature profile of our air column from bottom to top at say one hour intervals for 12 hours starting from time zero will look like?

  27. Joel Heinrich said

    “The gas curves are verified from dozens of other links and the Planck curves are verified by my calcs here. There shouldn’t be any disagreement here either – I hope.”

    Well, hope for the best, but expect the worst. :-)

    The Planck curve of the sun in the image here has a maximum at 350 nm (UV) while your “verification” has a maximum at ~ 900 nm (IR) while the real sun has its maximum at 500 nm (green).

    And since the x-axis is logarithmic while the curves of sun and earth looks identical, it makes the area under the curves different. The earth in this figure 2 and in your “verification” is emitting some 80 times more energy then the sun!

  28. Dave Springer said

    Oh, and no fair keeping the water temperature the same using a heat source. In the real world it will be cooling down as the night progresses. So I guess it would be nice to include a temperature profile of the water column along with the air.

    Then of course remove the CO2 from the equation and compare the result side by side with the previous.

  29. Jeff Id said

    “What exactly do you think “transmission” is? Water droplets in a cloud absorb and re-emit light – are clouds black? No, they are white.”

    Now I’ve been working as an optical engineer for the past 20 years so I could be wrong, but from memory, water is clear.

    Transmission is different from absorption.

  30. Jeff Id said

    #27, Nice, it’s the shape that matters. If you look at the link where I said I verified it, you will see I confirm your results.

  31. Dave Springer said

    his conclusion that insulating an energy flow doesn’t cause warming is non-physical and absolutely incorrect

    100% agreement here on that. I’m always quick to say that greenhouse gases should be called insulating gases. Imagine how much confusion that would clear up! Even a caveman knows what the effect of adding layers of clothing will do to help keep a body warm.

  32. Steve Fitzpatrick said

    #21,
    “the atom doesn’t first take a poll of the space surrounding a distant molecule (which it must some how know it going to absorb in the future) before it decides whether it can irradiate the photon or or not, so as not to violate the second law of thermodynamics!”

    LOL.

    Damn, and I was thinking those molecules had studied thermo! Not fair that they don’t suffer like the rest of us.

  33. Earle Williams said

    Josh,

    Jeff’s example is a taken from a place called Physics Wonderland. It is a place where you can have perfect spheres, frictionless surfaces, and perfect insulators. People don’t live there, they only visit to conduct Thought Experiments.

    You’re having trouble grasping Jeff’s model because you inhabit The Real World and apparently do not visit Physics Wonderland often enough to appreciate its elegance and simplicity.

    Thought Experiments in Physics Wonderland can be very useful. For a simple pendulum the difference between the the predicted motion in Physics Wonderland and the observed motion in The Real World is very small. This is because the Physics Wonderland model is a close approximation of The Real World.

    Jeff’s Physics Wonderland model is not so useful because it is a poor approximation of The Real World. It attempts to isolate specific elements of radiative physics to prove a point.

    As has been noted here and at WUWT Jeff sabotages his own model in Figure 6 where he portrays the radiated energy exiting the left side of the High Tech Gas Chamber but no energy exiting the right side of the chamber. Perhaps the chamber was ill-specified and the right-hand pane was intended to be perfectly transmissive for the incoming laser energy and perfectly reflective for the radiated energy resulting from CO2 absorption.

    One can argue that at equilibrium there must be conservation of momentum, and attempt to explain the oddity presented in the figure. However then you are leaving Physics Wonderland for its suburb, Approximationville. Things get pretty messy there, as there are no zoning ordinances and what does get constructed is frequently shored up with struts made of armwavium.

    It’s illustrative to me that something as ‘simple’ as radiative physics and the greenhouse effect has traps laying in wait for the unwary. Hopefully it is instructive as well that when lecturing others on simple physics that one doesn’t perpetuate further misunderstanding.

  34. Jeff Id said

    #32, It seems you were right Steve.

  35. Jeff Id said

    “As has been noted here and at WUWT Jeff sabotages his own model in Figure 6 where he portrays the radiated energy exiting the left side of the High Tech Gas Chamber but no energy exiting the right side of the chamber. Perhaps the chamber was ill-specified and the right-hand pane was intended to be perfectly transmissive for the incoming laser energy and perfectly reflective for the radiated energy resulting from CO2 absorption.”

    The input window is a tiny hole to let the beam in.

    “C ultra insulated laser chamber with perfectly transparent end window and a tiny input window on the back to allow light in from the laser.”

    Too small to let much out.

  36. Steve Fitzpatrick said

    #29,

    Well at least you can maintain your sense of humor. :-)

    #25,

    Woa! Optical transmission (water, glass, etc) is most certainly not the same as absorption. Finely divided, optically clear (non-absorbing) particles do indeed just scatter light.. like the small droplets/crystals in clouds via refraction and diffraction. But transmission within the bulk phase of a homogeneous clear material is coherent and essentially without loss. Interaction of the photons with the clear medium shows up as a loss in velocity relative to the speed of light in vacuum (AKA refractive index >1.0), not a loss in intensity.

    That process is nothing like absorption, where the photon ceases to exist when it is absorbed, and it’s energy is converted to the energy of a molecular vibration. Re-emission (should this take place prior to quenching of the excited vibrational state) is in random directions, not in the same direction as in transmission. Since re-emission can’t possibly be 100% efficient, and some fractional quenching will take place, so it is inevitable that the gas in Jeff’s cavity must heat up.

    What do you think causes the absorption lines (specific frequencies with low emission intensity) in the outgoing infrared radiation from Earth’s atmosphere? (hind: infrared absorbing gases in the air that have those corresponding absorption lines)

  37. josh said

    Jeff, you will find many explanations of how CO2 scatters IR on the Internet. This is the very claimed mechanism of warming. Instead of the IR light leaving to space, it has a random chance of being back-scattered to the surface where it can strike an object that will absorb the energy and convert it to heat – thus the CO2 in the atmosphere is “glowing” in IR. CO2 is not “black” to IR as you claim it is. Further, the claimed mechanism of CO2 warming does not involve the CO2 itself heating up, though the CO2 can heat other molecules if the energy absorbed from an IR photon is given to a different molecule in a collision, but this is a very small effect. Mostly it just re-emits the IR in a random direction.

    As for your appeal to authority. Please. It’s unbecoming.

  38. Dave Springer said

    22.josh said
    August 6, 2010 at 2:40 pm

    I am still having trouble with this Jeff, I think you are misunderstanding the physics badly. The CO2 transmits the light just fine, it just scatters it. So you will no longer have a coherent beam, most of the light will just go off at some other angle having been absorbed and re-emitting many times along the way. Some of this light, a small fraction, will result in additional translational energy of the CO2 molecules, but most of it will merely be slowed down a little bit and sent off in a different direction, resulting in no net heating.

    My emphasis above.

    That depends on the initial temperature of the insulating gas. Say I throw a cold blanket over a hot rock at night. The blanket can’t get any warmer than the rock but it can certainly warm up closer to the temperature of the rock. And the rock will be warmer in the morning than it would have been without the blanket.

    At least conceptually everyone should be able to understand how insulators work, even a caveman – I’m sure one of them probably noticed that if he skinned half of a freshly killed wooly mammoth on a cold winter afternoon, came back to finish the job in the morning, that the meat would be a lot cooler where the wool was missing. But none of it would be as warm as when the animal first fell. It ain’t complicated.

  39. Brian H said

    The energy imparted to a CO2 molecule by interception of a 15um wave is not special energy. Any and all vibration it experiences, whether from radiative or thermal contact, is equally likely to be re-radiated. Heating of surface air and subsequent convection will thus warm the lowest layers of CO2 and cause it to finger-print re-radiate, just as reception of IR from ground molecules or other CO2 will. Hence, CO2 will radiate more IR than it receives.

    Which will cool it.

  40. Brian H said

    correction: “finger-print radiate”, not “re-radiate”.

  41. Earle Williams said

    #35

    My bad, didn’t read that caption. That would force all energy to be radiated out the exit chamber. And it makes your model even more contrived.

  42. Jeff Id said

    #37 Sorry Josh, It’s just a little funny to have someone make the statements you did. I’ll answer again because you are missing some of the concepts of absorption.

    The molecular bonds resonate at the frequency of the laser light. This means they are ready to become excited at this mode. In transmission the resonance is between the electrons and the light, in certain modes of absorption, the molecular bonds vibrate with the entire mass of the atoms. It’s a form of kinetic energy despite what some physicist wrote on WUWT. Once it’s turned into motion energy it can be turned into linear motion by collisions. All very different from a cloud.

    Now re-emission which is the same as emission, happens several ways, one is a release of an equal wavelength photon. This represents an output on the same wavelength after some time delay. However, the collisions of the gas guarantee that the net re-emission spectrum will drift to the planck blackbody curve (or some close approximation). This means that the CO2 does emit light, but this is entirely different from the diffraction/reflection/transmission of the cloud – which is what made it funny – sorry.

    This ‘glowing’ as you call it, is exactly the point of the post. The glowing CO2 doesn’t transmit light either, it absorbs it and is forced to emit it agian later. It’s still absorption, it’s still opaque black to the laser. Think of it like shining a light on a filament (a really dim one in the example), the filament is white hot, but the light you shine on it will not pass through. It will add to the heat of the filament and be re-emitted though.

  43. Brian H said

    About the radiation source graph. Someone (don’t have link to hand) has pointed out that the nominal 5780°K temperature of the sun has no basis. The peak radiation from the sun is in the green band, at a rather higher temperature. Not sure what the consequences of that are, other than to shift the Plank curve left, overlapping fewer of the “absorption curves”.

  44. Jeff Id said

    #41, If you think of the atmophere which is in equilibrium and define a 1 meter cylinder boundary from ground to orbit. Energy which escaped from the side of the cylinder will come back in from adjacent air on the other side of the cylinder. (Infinite insulation)

    If you think of energy re-striking the ground, an equal energy is coming back up, also infinite insulation.

    So you have a perfectly insulated cylinder with a perfect emitting window at the top. — not so bad I think. DeWitt Payne get’s credit for it though.

  45. Steve Fitzpatrick said

    Really Jeff, how can you imagine CO2 could cause the Earth’s surface to warm? It just re-radiates everything it absorbs…. and so is effectively transparent, even at it’s (incorrectly named) “absorption peak”.

    OK, no need to answer that.

    On a more serious note, does anybody have estimates of quench rate for CO2 versus emission rate from the excited (bending mode) state at different pressures? This might help clarify what is going on. DeWitt? Carrick?

  46. Earle Williams said

    Jeff,

    It would help your dissertation to describe your cylinder accordingly. I assumed you were describing a hypothetical bench experiment.

    Josh,

    If you add enough water to clouds they do indeed look black. Well, blackish.

  47. Pat Frank said

    #22, Josh, you’re right that the first process is IR scattering. The central issue of the whole business is in Jeff’s statement that, “Since the beam cannot pass through, the CO2 gains vibrational energy which is then turned into translational energy and is passed back and forth between the other air molecules building greater and greater translational and vibrational velocities. —- It heats up.

    The excited vibrational modes of CO2 only transfer to translational energy (heat) by collision with other gas molecules. Otherwise, the vibrational energy is just re-emitted off somewhere through 4 pi radians; i.e., scattered.

    The lifetime of a vibrationally excited state of CO2 (tau) is given approximately by 1/tau = [8 x pi x freq^2 x S]/c, where S = 10^-16 cm [1]. For the 15 micron band, tau is about 30 milliseconds. This is a long time relative to the collisional rate of CO2 at 1 atm (~10^8/sec).

    The total thermalization rate of CO2 is (the collisional rate) x (the probability of energy transfer), which varies with the mass of the gas molecules. But, at 1 atm of pressure, most of the radiationally excited CO2 will likely collisionally decay before it decays by re-radiation.

    The extinction coefficient of CO2 at 15 microns is about 2.5 cm^-1 atm^-1 [2]. If you plug 400 ppm (4×10^-4 atm) of CO2 into that, you get an absorbance of about 1 per 10 m of air, which means that 99.9% of the 15 micron band is absorbed in the first 30 meters of air above the surface.

    The R and P branches of the 15 micron band are about 3x less intense at their maximum, but much broader. So, they’ll max out their absorbance between, say 90 m above the surface at their center and at about 900 m above at their 10x less intense wings.

    The same is true at the 1900 CO2 level of 290 ppmv, with a 15 micron absorbance of 1 at 13.8 m, and 99.9% absorbance of 15 micron radiation at 41.4 m above the surface. In short, all of the effect of CO2, since 1900 and no matter how many doublings we achieve, will always exert virtually its entire radiative-to-collisional heating within the first km of the troposphere above the surface.

    There’s an interesting discussion of this at the late John Daly’s excellent, but unfortunately moribund, “Still Waiting for Greenhouse” website, here.

    With all this, most of the dissipation of the heat from CO2 absorption near 15 microns will be by turbulent convection and by conversion into the latent heat of water evaporation. Any increased water vapor re-condensation in the upper atmosphere will change the cloudiness and be a source of long-wave IR radiation into space. These two processes — convection and cloud formation — are exactly the processes most poorly modeled by GCMs. So, the climate sensitivity of CO2 that is built into the climate models is not reliable.

    However, all the processes that were already established within the climate of earth to remove the collisional atmospheric heating by CO2 at 290 ppmv (turbulent convection from the surface and water evaporation/re-condensation) are also available to remove the added collisional heating of CO2 at 400-600 ppmv. If the response rates of these processes were unsaturated at 290 ppmv CO2, which seems likely, then one will expect the response rates to increase with added CO2. If the two response rates increase only linearly with added CO2 forcing, the effect of any new CO2 on net surface temperature will be about zero.

    At the end, Jeff is right. For all the purity of radiation physics, no one knows whether any of the thermal energy induced by added CO2 will be detectable as heat or be beyond natural variability. It may show up as slightly stronger updrafts and/or slightly increased rainfall. One might end up seeing a counter-intuitive result from added CO2: an increase in the long-wave radiation from the upper atmosphere without any apparent increase in the surface air temperature.

    [1] J. T. Houghton (1967) “Fluorescence from the v3 vibration of carbon dioxide” Proc. Phys. Soc. 91, 439-448

    [2] C.W. Schneider, et al. (1989) “Carbon dioxide absorption of He-Ne laser radiation at 4.2 [micrometers]: characteristics of self and nitrogen broadened cases” Applied Optics 28, 959-966, and the NIST spectrum of CO2 at http://webbook.nist.gov/chemistry.

  48. Pat Frank said

    Jeff, “Think of it like shining a light on a filament (a really dim one in the example), the filament is white hot, but the light you shine on it will not pass through. It will add to the heat of the filament and be re-emitted though.

    If the filament is white hot then there’s been a large population inversion. Absorption efficiency is a function of the probability of absorption, which is small for an inverted population. If the ground state population given by the Boltzmann distribution is already tiny, there’s only a tiny absorption of light and no detectable effect on the filament.

  49. Stilgar said

    Good job with the plain explanation. People either don’t know or don’t realize that vibrational energy is heat. If you have more vibrating things, you have more heat. Fairly simple concept.

    It is funny that some of those who are looking at the example are saying yes, but your simple experiment does not account for real world complexities. No kidding, what part of simple was it that you didn’t understand? Was it the part that means leaving off more complex details so that it is actually simple?

    Another funny reaction are the ones that say, C02 doesn’t actually warm because H2O (or other feedbacks/processes like convection) provides overall cooling. It still doesn’t change what more CO2 does. More CO2 means more heat, even if the chaotic atmospheric reaction to that heating means after all is said and done that the atmosphere will actulally be cooler, the net cooling does not change the fact that CO2 did the initial heating that triggered the cooling. You don’t need an atmospheric model to know a CO2 increase will increase heat, that is classroom stuff. What is not understood is the reaction of the atmosphere to that increased heat. Uncertainty of atmospheric interaction does not change the basic physics of CO2.

  50. Jeff Id said

    #49, Exactly my thoughts.

    #47 really nice comment.

  51. Jeff Id said

    #48, Pat,

    Now this is an interesting problem. Reflection is an inverse resonance of electrons, transmission is a resonance, but what happens when a filament is highly energized. If you have 10 watts of laser, can the light from a laser simply pass through the metal of an illuminated filament, does it reflect? My opinion is that it simply is absorbed into a higher average state and that the filament just warms more.

    The saturation (negative temperature) effect which can exist in a laser cavity, is far away. Plasma far in my opinion.

  52. Steve Fitzpatrick said

    #47,

    Thanks very much for the numerical data. Those numbers make it clear that most absorbed energy will be converted to heat before an excited CO2 molecule has a chance to re-radiate, except very high up in the stratosphere.

    Or as Jeff said, the CO2 in the cell appears pretty much like “black stuff” to the 15 micron laser beam.

  53. Steve Fitzpatrick said

    #48, #53,

    Then if the filament surface is very hot, does this imply it becomes a very good “mirror” that just reflects arriving photons? That doesn’t sound right to me, but I have not thought about it much. Does an ideal black-body really have the possibility of a population inversion? That doesn’t sound likely either; isn’t an ideal black-body one where any size energy transition is allowed (that is, not limited to transitions between specific energy levels)?

  54. Jeff Id said

    “The saturation (negative temperature) effect which can exist in a laser cavity, is far away. Plasma far in my opinion. ”

    I don’t agree with myself anymore. Now that I’m older and wiser, I think we can keep heating until plasma state and beyond (whatever that is). The saturation of a laser cavity – inversion – is a different beast.

  55. Pat Frank said

    #51, Jeff, if the target population is strongly inverted and the laser is at the resonant frequency of the target, then it will likely stimulate emission from an excited state. What one sees amounts to reflective scattering. Remember the photoelectric effect applies. Absorption occurs only at resonance. You can further pump an excited state, but only with higher energy photons. Two-photon processes are QM forbidden and have very small resonant cross-sections.

  56. Jeff Id said

    #55 So we get spontaneous re-emission of the light energy and some residual heating right? Not bad.

  57. Pat Frank said

    #56, Jeff, I guess one could get further heating if the intensity of the laser was increased to be above the intensity of the radiation emitted by the target, per unit surface area. That could increase the population inversion, if it were not already saturated.

  58. freespeech said

    I assume your cylinder is long enough to actually block the laser, i.e. a very short cylinder with 100% CO2 won’t block all of the energy (i.e. reduce length to 1 atom to see that most will pass through)

    So let’s assume it is 10km long (the approximate height of the atmosphere if it was all the same denisty).

    At what concentration of CO2 would the contents of your cylinder become opaque to the laser?
    I assume that for concentrations below this level, the temperature will rise with some proportionality to the concentration?
    What happens to this temperature/concentration relationship when the CO2 concentration exceeds this opaque level? Is it different?

    My understanding is that this is a closer model for our atmosphere, and that the current (and previous pre-industrialisation) concentrations of CO2 were already opaque. So it seems that the saturated/opaque heating formula is the one we need to use, assuming it is different to the heating relationship prior to saturation.

  59. gallopingcamel said

    Opting for a 15 micron source was a good choice. It takes less than a meter of CO2 at 1 Bar to absorb >99% of the incident energy.

    DeWitt Payne:
    Did I drop a decimal point somewhere?

  60. gallopingcamel said

    Jeff,
    I have this very nice 1,000 W laser but it runs at 10.6 microns. Can you use it in your experiment?

  61. gallopingcamel said

    Jeff,
    In figure 5 you have CO2 free air scattering 15 micron radiation. While that will happen due to Rayleigh scattering the effect is insignificant.

    My “back of the envelope” calculation says it would take more than 1,000,000 km of air to scatter 50% of the 15 micron radiation

  62. gallopingcamel said

    Pat Frank (#47),
    Absorption and scattering are not the same.

    When CO2 absorbs radiation it is stored in excited states that eventually decay and radiate photons. The radiated photons are at lower energies than the incident photon. Some of the incident energy may be retained in the CO2 as kinetic energy, thereby raising the temperature.

    When scattering occurs (e.g. Rayleigh scattering) no energy is absorbed so there is no local heating and the wavelength of the scattered radiation is identical to the incident radiation.

  63. Pat Frank said

    #62, Absorption and re-emission is also a scattering phenomenon; it’s just not Rayleigh scattering.

    Following electronic absorption, the radiated photon energy is the same as the energy of the absorbed photon, unless the initial electronic excited state has decayed to a lower energy excited state through some alternate decay channel.

  64. Geoff Sherrington said

    Josh said at 2.40 pm “Now, if the sides of your can are IR absorptive, you would heat up, as some fraction of the IR beam is scattered into the side, but you could accomplish the same result with a mirror in the middle of the can placed at a 45 degree angle to the beam.”

    Like I noted before, toy models can get a bit complex and sometimes confuse the argument rather than clarify it. I suppose in an open system, you have to know the average time a CO2 molecule stays in its excited state, its typical velocity (related to temperature and the density of other atoms & molecules around it) and how far it would travel before emitting its photon when returning to a lower state.

    At the extreme, what is the temperature in space? Is it determined by the microwave radiation said to be left over after the Big Bang, or is it in some way, difficult to measure, related to the kinetic energy of the odd atom or molecule or photon that strays through the void? How do you measure temperature as gases become more dilute, to the stage where their probability of interacting with a sensor becomes vanishingly small? What is really meant by the temperature changes reported to exist above the tropopause?

  65. anette said

    i am a bit confused over the concept temperature. when i took phys chem i learned that temperature is only the translational kinetic energy (TKE). rotations (R) and vibrations (V) are internal energy. but in other sources i have read also rotations and vibrations are included in the temperature concept? TKE can be understood from classical mechanics, that a collision at any speed can transfer any amount of energy to another molecule, but for R and V only specific quanta can be transferred. can such a quantum be transferred to TKE or only to new R or V? i hope someone can give me an answer.

  66. steveta_uk said

    I think a lot of the confusion is because many folk use the pool ball model in their minds when thinking about a gas. What they don’t think about is the quintillions of pool balls required for this image to work.

    Several of the comments on WUWT seem to assume that the chances of a collision and thus an energy transfer prior to re-radiation are small enough to ignore. That may be true for a specific photon absorption-emission event, but there are a heck of a lot of these events going on – so the chances of any given photon (don’t get into photon identities!) reaching up 30m into the air then coming back down to the surface must be negligible.

    One of the posters on WUWT assumed that a given IR photon might take perhaps 10 times longer to reach space than one not absorbed. It reminds me of the high-school physics question, of “how long does an individual electron in a flashlight take to complete the circuit thru the bulb and back to the battery. The answer, that it probably never does in the lifetime of a battery, surprises a lot of folk.

    Does anyone know the real lifetime of a given photon leaving the surface before reaching space?

  67. jeff id said

    Geoff,

    Assume mirrored zero heat transfer walls.

    The example is just supposed to show people the mechanism by which CO2 Traps, delays, warms etc. It’s supposed to present it in as conceptually simple way as possible so that people don’t roll around in blogland claiming these other weird effects eliminate the warming. One of my favorites is at the WUWT thread where people claim the warming causes convection which causes release in the upper atmosphere and a net cooling. They never seem to realize that the gas that warmed now has more energy to release and has to cross back through the zero point (where it was before) to cool more.

    There are probably a trillion complications but all they do is start with a warmer gas. The most powerful (stable) negative feedback will still sit slightly warmer than no CO2 initial point. The warming could be 0.000001 C, but it is still warming. I don’t believe the effect is this weak.

    —–
    #60 My wife says no. Something about shooting my eye out. Thanks for the offer though. :D

    The scattering was supposed to be from other elements in the atmosphere absorbing like CO2 does — not Rayleigh, my wording wasn’t clear enough. Especially for those who already have this all figured out.

  68. Colin Aldridge said

    excellent post especially the input and output spectrum which I hadn’t seen before. I am constantly amazed and depressed that sceptics are equated with people who don’t think CO2 warms the earth. Any sensible analysis shows that it does. The real issue is “what about all the other stuff”, feedbacks, forcings etc which is where the real debate is… or should be

  69. Paul said

    After equilibrium is reached in cans(A,B,C)– when the laser is turned off, how long is the ‘captured’ energy maintained in each?

  70. GHowe said

    #68 I agree. Settle the feedback arguments ( which should take 20 more years or so) and then the radiative physics will be obvious, maybe. i.e. kick the cans down the road. ;)

  71. Paul Linsay said

    #47 Pat Frank makes a most important point here. The CO2 absorption is already saturated for most of its IR lines, i.e., all IR generated by the heated surface is absorbed and thermalized by the atmosphere within less than 1 km of the surface, mostly within 30m. Adding extra CO2 only pushes the level at which the heating occurs closer to the surface, e.g., doubling CO2 drives 30m down to 15m. It does not add any more energy into the atmosphere. Extra CO2 causes extra heating if there are unsaturated lines left that allow IR to escape to space.

    The real question is whether it matters that the heating from CO2 takes place within 30 m of the surface or 15m of the surface. I’d be surprised if there was much difference.

  72. Jeff Id said

    #71, Even though the energy is already being captured before it escapes, what happens is the pathlength shortens so the amount of heating still increases – by ever decreasing amounts. I freely admit the amount could be exceedingly small, I’ve seen absolutely no evidence which proves the magnitude of warming. The 1.2 C for doubling CO2 with no feedback stuff is a bunch of rubbish IMO – and that sort of thing is what is incorporated into models.

    Does someone have a good article on the physical basis of the zero feedback magnitude of CO2 warming?

  73. Mark F said

    71: Um, I don’t think doubling will halve the thickness of the (very non-uniform density) layer, for starters. Consider also that the atmosphere in total isn’t being compressed into half of its thickness – CO2 is not exactly the largest component.

  74. Pat Frank said

    Shoot, well, I misread C. Schneider, et al.1989, and the extinction of CO2 at 15 microns is 5 cm^-1 atm^-1, not 2.5 cm^-1 atm^-1. So, all the distances must be halved, i.e., at 400 ppmv, the 99.9% saturation altitude is 15 m, not 30 m, etc.

    Sorry about that.

    #65, Anette, your initial understanding was correct. Heat is a measure of translational kinetic energy, only.

    There are quantized translational kinetic energy states, of course, but the splittings are so small and the density of states so large, that they may as well be a continuum.

  75. Jeff Id said

    “There are quantized translational kinetic energy states, of course, but the splittings are so small and the density of states so large, that they may as well be a continuum.”

    Exactly, that’s what bugged me about Vonk’s post. Ya can’t blast energy at something and not expect warming. If you could, it would make a pretty cool wire.

  76. Jeff Id said

    #74, I wonder if you would be kind enough to write a short post on the radiative decay time vs interaction time for CO2 in the atmosphere. This is a commonly discussed issue and were it documented better the confusion could be reduced. It would need equations and the sources of numbers used but I really think it would be helpful for the general understanding.

  77. It amazes me that such a simple physics problem is so difficult for people to understand. Jeff, that was a pretty dang good job of simplification. Still, there has to be a better way to convince some people that there are simple laws of physics that are just the way it is. If Wille E. Coyote stands under the anvil when the rope comes untied he gets a headache. BTW, 1 Watt per meter squared or approximately 0.75 degrees C seems to be the most reasonable estimate for a doubling of CO2 alone.

  78. Al Tekhasski said

    Pat wrote: “The extinction coefficient of CO2 at 15 microns is about 2.5 cm^-1 atm^-1 [2]. If you plug 400 ppm (4×10^-4 atm) of CO2 into that, you get an absorbance of about 1 per 10 m of air, which means that 99.9% of the 15 micron band is absorbed in the first 30 meters of air above the surface.”

    I have a different number. Instead of using unquantified formula for undefined coefficient, I use
    http://www.spectralcalc.com/calc/spectralcalc.php .

    The calculator says that for a cavity length of 10cm and 400ppm of CO2 at normal pressure and 296K temperature, the transmittance is 0.65 at 14.98um, which means to me that 35% will be absorbed and 65% will pass through. The next 10cm will do the same, so the transmittance across 2x10cm layers will be 0.65 * 0.65 = 0.42. Your number of “30m” is equivalent to 300 10-cm cavities in sequence, therefore the residual beam will be attenuated by a factor of (0.65)^300 = 7.48^10-57 after passing acros 30m od air. Which is quite less than 0.1% you arrived to.

    What am I missing? :-)

  79. Al Tekhasski said

    gallopingcamel said: “Jeff, I have this very nice 1,000 W laser but it runs at 10.6 microns. Can you use it in your experiment?”

    I would think he can. However, CO2 has much less absorption at 10.6 um lines.
    At 10.6um a 10cm-long cavity with 50% of CO2 will be 99% transparent to the laser beam, again as per spectralcalc.com.
    Still 1% of 1000W is 10W, which will make quite measurable effect: the internals of chamber “C” will “warm up”.
    How much?
    Again, assuming the output window as 10cm2 (or 1/100 of m2) and all other walls well insulated, it gives the output flux of 10W*100 = 1kW/m2,so the gas in cavity will warm up at least to 365K before it stabilizes. I can assure you would feel it with bare hands, although I would recommend this measurement technique only for CAGW promoters :-)

    Also, before doing this experiment in your garage, be aware that there could be run-away effect, because at 400K the cavity will be only 96% transparent (as I just checked with spectralcalc), so the absorbed power goes to 40W and corresponding rise in temperatures must be expected. Exact degree of warming would require solving some equations with more details on chamber insulation, properties of windows, inner coatings, etc.

    Also make sure that your chamber is made from something stronger than a Coca-Cola can, such that it will not blow up due to pressure build-up.

    Good luck.

  80. #72 It all depends on how much convection is generated by the heating, how quickly it starts, and how fast it carries the warmed air upwards.

    #73. I’m not sure I understand. The extinction coefficient is the product of the photon absorption cross-section [area] and the CO2 concentration [#/volume] Doubling the concentration doubles the extinction coefficient. The absorption length is the inverse of the extinction coefficient, hence is halved with a doubling of the concentration.

  81. Pat Frank said

    #78, Al, using your site, I got a transmittance of ~0.96 at 15 microns for 1 cm of 400 ppm CO2 (1000 mbar total pressure).

    Absorbance is log(1/T) => 1.8×10^-02/cm x 1000 cm = an absorbance of 1.8 per meter. That’s about 10x larger than is produced by C. Schneider’s extinction coefficient as given in #74, so I’m not sure where the problem is.

    Schneider’s extinctions were measured with CO2 in a balance of nitrogen to make 1 bar total, and so could be different from the value for the pure dilute gas. I.e., I’m guessing your spectroscopic site doesn’t include collisional broadening.

    If you go to the NIST link and search on CO2, you’ll find the spectrum as mixed with N2. Look at the 15 micron band (667 wave numbers). It’s very broadened compared to the well-resolved rotational/vibrational peaks in the spectrum of your spectroscopic site.

    The spectrum of 357 ppmv CO2 in wet air shown at John Daly’s site is also broadened and gives an absorbance of 3 at 10 meters, compared to about 1.8 in dry N2 at 10 m if the extinction is 5 atm^-1 cm^-1 (extrapolated from C. Schneider’s work).

    All of these results imply that virtually the entire effect of CO2 heating by reflected IR is restricted to the near-surface atmosphere, however.

    #76, Jeff, I’m not sure what you mean by “interaction time for CO2 in the atmosphere.” But sorry to admit that I haven’t much time for any more independent research just now. :-)

  82. Jeff Id said

    #81, I was just describing the emittence time for 15um energized CO2 vs collision Xfer time. It’s a very big issue in blogland and some headway could be made.

  83. Suibhne said

    Accepting that your graphs are correct, there seems to be significant absorption of the incoming Solar Radiation.
    H2O several bands;CO2 two bands around 4um.
    On the upward Earth Thermal Radiation H2O is shown to absorb from 10um to 20um;
    CO2 absorbs in 15um area.
    Given that water vapour has about 30 molecules to each CO2 molecule how much of the heating effect around 15um is due to CO2?.
    We then move on to the backradiation from the atmosphere.
    Very few dispute that the atmosphere radiates to the Earth Surface.
    What is in doubt is the magnitude of the radiation due to CO2
    What is not in doubt is that the backradiation cannot raise the temperature of the Earth Surface.
    That is if the second law of thermodynamics has any meaning.

  84. Jeff Id said

    #83,

    Accepting that your graphs are correct, there seems to be significant absorption of the incoming Solar Radiation.there is some
    H2O several bands;CO2 two bands around 4um. there isn’t much energy at 4um
    On the upward Earth Thermal Radiation H2O is shown to absorb from 10um to 20um;
    CO2 absorbs in 15um area.
    Given that water vapour has about 30 molecules to each CO2 molecule how much of the heating effect around 15um is due to CO2?. The pro’s say 1/25th
    We then move on to the backradiation from the atmosphere.
    Very few dispute that the atmosphere radiates to the Earth Surface.
    What is in doubt is the magnitude of the radiation due to CO2
    What is not in doubt is that the backradiation cannot raise the temperature of the Earth Surface.I disagree
    That is if the second law of thermodynamics has any meaning.In the experiment above, the gas warms from the laser. In the case of the earth, the warmer gas would be sitting against the surface of the earth. According to the second law, the surface must warm.

    I’m not sure why people claim there is some violation. No magic energy is created, it’s just prevented from flowing out as easily.

  85. Suibhne said

    What is not in doubt is that the backradiation cannot raise the temperature of the Earth Surface

    Jeff Id……”I disagree”……

    Take the example of night time so as not to confuse the issue with Solar radiation.

    An atmosphere at say -30c cannot raise the temperature of the Earth Surface which is perhaps at +10c.

    What some people say is that if there were no atmosphere; then…….
    But that’s a whole different ball game!

  86. Jeff Id said

    #85, If the average day and night temps rise by .000001, so nightime temperature has risen to 30.00001 from the standard 30, does the summation of day and night warm the surface?

    Did we violate the second law?

  87. Suibhne said

    During a typical night the Earth surface and the troposphere above are losing temperature.
    The troposphere at -30c can radiate to the surface and at best produce a result of -30c.
    Meanwhile the surface is radiating far more energy to the troposphere.
    The flow of HEAT is always from hot to colder.
    I don’t think anyone expects the Earth Surface at 30c to increase to say 32c because of backradiation.
    However this is exactly implication of the misinformation that the IPCC circulates.
    Diagrams showing backradiation Heat leaving the atmosphere and being absorbed by the surface are unfortunately quite commonplace.
    Very few don’t accept that CO2 and H2O absorb and thermalise the energy of IR photons.
    The heat transfer from The Sun to the Earth Surface and Atmosphere is perfectly well explained by 80 year old physics.
    The atmosphere insulates the Earth, conduction convection and radiation as well as phase change all play their predictable parts.

  88. TomVonk said

    Hello Jeff

    In the case that you have not read it , I have just posted on WUWT the following comment on your post :

    ” I have no issues with this post and agree with almost everything .
    The only (minor) issue is when you state : Where he goes wrong is equating that assumption to AGW.

    I do not think that I used the term AGW once and I neither analysed radiative transfer nor considered transient regimes . The purpose of my post was very exactly and exclusively to give arguments for the statement written at the beginning .
    Like you rightly say I considered a volume in LTE what , per definition , excludes transients at microscopical level .
    If you wish , in the frame of your laser experiment , my post would cover what happens inside the cylinder for all t > t0 where t0 is the time at which the cylinder emits by the transparent side 1000 W .
    It is for those t that the “CO2 does not heat the N2″ even if it continues to absorb the whole 1000 W .
    My purpose was to say that in those conditions there can’t be a net energy transfer from the vibrational modes to the translational modes because else the cylinder would continue to heat and emit less than it absorbs .

    A minor technical point is that like Merrick says , temperature is indeed not merely an average of the kinetic energy but an average of energy what includes non kinetic energy too .”

    Going a bit farther . The point of my post was that there could not be a “natural law” saying that all vibrationnal energy must be transferred to translationnal energy (of N2) what one very often hears and reads .
    On the contrary if that was true , then there could not be equilibrium . Ever .
    If the above statement was true following infinite process would take place :

    1) CO2 absorbs an IR photon
    2) CO2 gives immediately the vibrationnal energy to N2
    3) N2 heats up , collides with CO2 (T-T process) and heats it up to establish equilibrium
    4) But … go to 1 .

    That’s why what actually happens is that the stage 2 of the above loop is not the same in every loop . It converges .
    In the first loop almost all energy goes to N2 and none comes back (to the CO2 vibrational energy)
    In the middle loop some energy goes to N2 and some comes back (to the CO2 vibrational energy)
    In the last loop the exchange is balanced and equilibrium may exist .

    Does that seem trivial ? I think so , yet you will still see often saying that CO2 heats N2 in all cases (e.g also in equilibrium) because there MUST always be a net energy transfer V (CO2) -> T (N2) . The reverse is forbidden .
    It is this exact point that triggered my post . Not AGW .

  89. Steve Fitzpatrick said

    TomVonk,

    Of course the reverse is not forbidden. Any CO2 that reaches an excited state by random thermal process can emit a 15 micron photon, just a as a CO2 molecule that is not is that excited state can absorb a 15 micron wavelength photon.

    But please note that Jeff’s 1 KW laser beam is very bright, and so the beam will in fact continue to be absorbed and generate intense local heat, even after the container as a whole warms enough so that the gas emits the same total energy as the laser beam provided. The absorption will take place in a tiny fractional volume of the gas in the container, and that tiny absorbing volume will always be MUCH hotter than the gas as a whole. Heat will be continuously (physically) transported from the very hot absorbing region to the cooler surrounding regions. So viewed as a whole, the system at your time >t0 is in energy equilibrium, but that doesn’t mean there is not a continuous conversion of 15 micron photons into thermal energy which then migrates into the remainder of the gas by diffusion, convection, and some re-radiation, before escaping from the transparent window. At no time or place will the interior of the container be in energy equilibrium, even though the container, considered as a whole, is in equilibrium.

  90. JAE said

    Tom Vonk:

    “Going a bit farther . The point of my post was that there could not be a “natural law” saying that all vibrationnal energy must be transferred to translationnal energy (of N2) what one very often hears and reads .”

    Hmmm, I’ve never read that. Are you subtly trying to say some other things, also?

  91. suricat said

    Hi Tom, Hi Jay.

    It’s a while since I enjoyed your dialogues on the (now defunct) CA forum. :)

    I think this post needs a response that I gave on the Sciencefile site recently:

    “Miscolczi took many spatial observations spanning ~61 years and made LBL (line by line) calculations to conclude that during this period the depth to extinction for CO2 associated emission remained unchanged.

    The kinetic theory of gasses that I linked to doesn’t include the ‘bendy, stretchy, spiny thing’ that CO2 undergoes, and because this is the model used to claim that CO2 ‘heats the atmosphere’ I’ve persisted looking for a link (besides the fact that I’m an engineer and don’t explain science too well):

    http://webphysics.davidson.edu/alumni/jimn/co2/pages/co2theory.htm

    However, as you can see, this model is based on CO2 gas LASER theory! CO2 LASERs need a power input and a lot of ‘tweaking’ (population inversion at the base level) to ‘stimulate’ the ‘emission’ of multiple photons. Insolation levels met in the upper atmosphere may well provide this energy requirement, but both insolation levels and ‘collisional kinetic’ at near surface temperatures are unlikely to excite an emission. After all is said and done, GHE works both ways! Perhaps this is why the IPCC attribute an average altitude of 5.5 km for CO2 emission (or was that for something else)? [smiley=TIC.gif]

    Best regards, Ray Dart.”

    I’ve not read all the comments here, but it seems that CO2 is expected to emulate a LASER simulation within Earth’s atmosphere. This may be true for high altitudes, but I think that low altitudes prohibit this.

    Your thoughts?

    Best regards, Ray Dart.

  92. kevoka said

    I second Steve Fitzpatrick’s request for the quench rate, or the equations used to calculate it.

  93. Brian W said

    Don’t we all love a good dose of pseudoscience. My dear Jeff you are just full of it. I love how you guys like to use abstraction and thought experiments to set up your playing field.
    Figure 3 ” Heat exit’s the single large window and cannot exit the sides of the chamber”. Not so, simple thermal heat will be retained and will not exit easily. The only trapping going on is the volume of gas inside your fantasy chamber. No air exchange is possible. Since I live in the real world there is no perfect anything, then all the sides are subject to thermal conduction.

    Figure 4 “From the perspective of 15 micrometer wavelength infrared laser, the CO2 filled air is black stuff. The laser cannot penetrate it”. I hope you are not serious! Who says,you do? What do you mean by black stuff? Co2 molecules act like little black bodies absorbing all radiation striking said molecules. Wow, got me. The gas will NOT absorb all 1000w! Pure nonsense. Prove me wrong.

    Whats up with the switching thing? You are merely taking A and B up from an ambient temperature of say 75oF. Your starting point in this case is 75oF. You say “At the moment of the switch, the gas still emits the same random energy as is shown in Figure 5 based on its ambient temperature”. Yeah, so, whats your point?

    Figure 7 “As it heats, emissions from the window increase in energy according to Planck’s blackbody equation.” This would only be true if the window was the emitter. Blackbody physics is aimed at solids not gases. Gases do not behave like solids and I’ve never seen a gas with a surface. In traditional studies the walls of a cavity are heated not the air inside. Gases emit from all points within completely unlike a solid. This is called an inappropriate use of the Stephan Boltzmann constant. You say “The delay time between the instant the air in C is switched from A type air to B air to the time when C warms to equilibrium temperature is sometimes stated as a trapping of energy in the atmosphere.“CO2 traps part of the infrared radiation between ground and the upper part of the atmosphere”. The time it takes to heat a gas to a particular temperature has nothing to do with trapping anything. Tell me how do you trap an electromagnetic wave in the atmosphere?

    Ready for more pseudoscience. You say “Oddly enough, if you’ve ever seen an infrared CO2 laser cut steel, you have seen the same effect on an extreme scale.” A Co2 laser does not operate because of high insulation! The walls do not insulate and typically use water to remove heat. The complete opposite. Sealed tubes need helium for cooling or they will burn up. In addition water cooled units have a continous flow of gas through the tube or dissociation occurs. Co2 lasers work because of mirrors,focussing and a high voltage power supply not from any insulation factor.
    Oh, I almost forgot is that little arrow in the atmosphere showing reflection or absorption/reemission.

  94. Geoff Sherrington said

    Brian W

    Agreed. The trick is in the focussing.

    We once had a wheat farmer who wanted to use our CO2 Megawatt lasers to harvest his wheat crop instead of cutting it by machine. He took a while to grasp the importance of focus. The helium in the gas mix of a CO2 laser does a few interesting transition things, as well as heat conduction, as I suspect you know.

  95. TomVonk said

    Jae
    “Going a bit farther . The point of my post was that there could not be a “natural law” saying that all vibrationnal energy must be transferred to translationnal energy (of N2) what one very often hears and reads .”

    Hmmm, I’ve never read that. Are you subtly trying to say some other things, also?

    I am not aware of saying something else than what I said .
    As for not ever having read that , the thread at WUWT contains it and this thread too . You can look it up .
    The classical and most frequent form is : “CO2 cannot radiate any 15µ radiation BECAUSE the average time to emit is too long . As the excited CO2 molecule undergoes a large number of collisions during this time , every single absorbed photon’s energy is transferred to the translationnal energy of N2 before it has the time to be reemitted as IR .”
    There are variants of that but this is the idea .
    The corrolary is that CO2 can indeed never radiate at 15µ because it gives this energy irreversibly to N2 instead .
    The rebuttal is easy , look f.ex at the downwelling IR spectrum during the night . There is 15µ radiation there .
    The purpose of my post at WUWT was to explain why there is 15µ radiation as observed and why the statement that ALL vibrational CO2 energy must be transferred to N2 is wrong .

  96. TomVonk said

    Steve Fitzpatrick
    At no time or place will the interior of the container be in energy equilibrium, even though the container, considered as a whole, is in equilibrium.

    This statement is self contradictory . You need to elaborate on what you understand as “being in equilibrium” . The LTE works exactly the other way round : “Even if the “whole” is not in equilibrium , there exists a neighborhood of every “interior” point which is in equilibrium “.
    LTE is very strict requirement to be able to define a local temperature . Without LTE , you cannot define a local temperature .
    LTE being a weaker requirement than TE , if a system is in TE , it is necessarily in LTE too .

  97. TomVonk said

    Brian W
    CO2 traps part of the infrared radiation between ground and the upper part of the atmosphere”. The time it takes to heat a gas to a particular temperature has nothing to do with trapping anything. Tell me how do you trap an electromagnetic wave in the atmosphere?

    You are right here but it is just a non conventional use of the word “trapping” .

    As it heats, emissions from the window increase in energy according to Planck’s blackbody equation.” This would only be true if the window was the emitter. Blackbody physics is aimed at solids not gases.

    This is right too . However I didn’t consider that as being a major problem . Jeff could (and should) have dropped the reference to BB Planck law all teh while keeping the spirit of the experiment .
    He could have written “There will be “some” radiation from the window and it will increase so that …. etc ” .

  98. Suibhne said

    TomVonk said

    ……”The rebuttal is easy , look f.ex at the downwelling IR spectrum during the night . There is 15µ radiation there .”………

    How are you so sure that the 15u is from CO2 rather than say H2O ?

  99. Jeff Id said

    Tom,

    First, your post was well done as I said in my first article on the topic. You seem to have a thick enough skin anyway, but this is what I said.

    Tom really did do a good job, so I’m not disparaging his work, what it does do though is create a lot of confusion in the crowd which is something Tamino accuses every non-extremist warmer of doing.

    Statements like this:

    What we have proven is that the CO2 cannot heat the atmosphere in the bulk but the whole system cannot be reduced to the bulk of the atmosphere .

    and this

    If you search for “greenhouse effect” in Google and get 1 cent for statements like…

    “CO2 absorbs the outgoing infrared energy and warms the atmosphere” – or – “CO2 traps part of the infrared radiation between ground and the upper part of the atmosphere”

    …you will be millionaire .

    Even Internet sites that are said to have a good scientific level like “Science of doom” publish statements similar to those quoted above . These statements are all wrong yet happen so often that I submitted this guest post to Anthony to clear this issue once for all.

    I don’t see anything wrong with saying CO2 absorbs outgoing infrared energy and warms the atmosphere. I know there is nuance to the statement but calling it wrong sowed a lot of confusion – which you can read in the WUWT comment thread. When you combined that direct statement with the example of a sphere with no net energy transfer, and the title, it very much seemed to me (and a whole lot of other people at WUWT) that the CO2 warming effect was being discounted.

    There are a group of people out there who flatly beleive CO2 doesn’t capture, delay or cause any warming whatsoever. I’m enough of a skeptic to say maybe it doesn’t even cause 0.00001C from doubling, because of feedback. It doesn’t look like that ridiculous low level of warming is true from what I’ve read, but I’m not studied or convinced enough to say it is wrong. This group however, does itself no service to go around claiming the basic physics are wrong, it causes them to be immediately discounted on other valid points.

    I think it would be helpful to WUWT to clarify your position on this, whether it agrees with me or not. – if Anthony is interested – if not you can put it as a post here.

  100. Jeff Id said

    Don’t we all love a good dose of pseudoscience. My dear Jeff you are just full of it. I love how you guys like to use abstraction and thought experiments to set up your playing field.

    Nice Brian.

    Figure 3 ” Heat exit’s the single large window and cannot exit the sides of the chamber”. Not so, simple thermal heat will be retained and will not exit easily. The only trapping going on is the volume of gas inside your fantasy chamber. No air exchange is possible. Since I live in the real world there is no perfect anything, then all the sides are subject to thermal conduction.

    Of course some things are more perfect than you realize. See number 44 for example.

    If you think of the atmophere which is in equilibrium and define a 1 meter cylinder boundary from ground to orbit. Energy which escaped from the side of the cylinder will come back in from adjacent air on the other side of the cylinder. (Infinite insulation)

    If you think of energy re-striking the ground, an equal energy is coming back up, also infinite insulation.

    So you have a perfectly insulated cylinder with a perfect emitting window at the top. — not so bad I think. DeWitt Payne get’s credit for it though.

    This I need to answer inline.

    This would only be true if the window was the emitter. Blackbody physics is aimed at solids not gases. Not true. Gases do not behave like solids and I’ve never seen a gas with a surface. In traditional studies the walls of a cavity are heated not the air inside. I’ve seen no ‘traditional studies of this effect. Gases emit from all points within completely unlike a solid. Not true either, solids also emit from all points, it just gets reabsorbed. This is called an inappropriate use of the Stephan Boltzmann constant. You say “The delay time between the instant the air in C is switched from A type air to B air to the time when C warms to equilibrium temperature is sometimes stated as a trapping of energy in the atmosphere.“CO2 traps part of the infrared radiation between ground and the upper part of the atmosphere”. The time it takes to heat a gas to a particular temperature has nothing to do with trapping anything. So tell me, where did that energy go?Tell me how do you trap an electromagnetic wave in the atmosphere? You would use an absorbing medium to transfer the electromagnetic energy to heat

    Ready for more pseudoscience. not really You say “Oddly enough, if you’ve ever seen an infrared CO2 laser cut steel, you have seen the same effect on an extreme scale.” A Co2 laser does not operate because of high insulation! the heat is created when an absorbing medium turns electromagnetic energy to mechanical (heat) at the surface of the steel The walls do not insulate and typically use water to remove heat. The complete opposite. Sealed tubes need helium for cooling or they will burn up. In addition water cooled units have a continous flow of gas through the tube or dissociation occurs. Co2 lasers work because of mirrors,focussing and a high voltage power supply not from any insulation factor.
    Oh, I almost forgot is that little arrow in the atmosphere showing reflection or absorption/reemission. it shows the NET direction of IR flow.

  101. suricat said

    Jeff.

    This may be an inopportune moment for me to add a comment, but CO2 is alleged to ‘warm’ in the lower tropo and ‘cool’ in the upper tropo in the ‘real world’.

    This being the case, how is it that when CO2 exhibits a more ‘diatomic’ property than ‘triatomic’ it is expected to emit photons (LASER level stuff) in the lower tropo (where energy levels are low), in comparison to the upper tropo and above where the energy levels from insolation are able to provide the required energy for CO2 to emit photons?

    There’s also the ‘duality’, or co-dynamic, of observation. As Suibhane said, “How are you so sure that the 15u is from CO2 rather than say H2O ?”

    This is a real muddler!

    Best regards, Ray Dart.

  102. Jeff Id said

    #101, I’m not sure how to answer you. As far as muddling the result above, it makes no difference. As far as muddling how much warming, I fully agree.

    As far as what you’re saying here:

    This being the case, how is it that when CO2 exhibits a more ‘diatomic’ property than ‘triatomic’ it is expected to emit photons (LASER level stuff) in the lower tropo (where energy levels are low), in comparison to the upper tropo and above where the energy levels from insolation are able to provide the required energy for CO2 to emit photons?

    I’m not really sure what you mean, some concepts seem to be mixed.

    What I do know is that more CO2 means at least some warming – predicting how much is up to someone else.

  103. Brian W said

    Jeff Id post #100

    Your response certainly deserves a reply! Out to the back woodshed we go! Previously I stated that “I love how you guys like to use abstraction and thought experiments to set up your playing field.” So what do you do in number 44? Precisely that. First off the atmosphere is never in equilibrium. Ever. Second you want me to consider a 3ft diameter tube made of ?? running from the ground to space? #44 is a complete load of abstract nonsense. Its literally perfect gibberish. Its certainly not real physics. No wonder you don’t take credit for it. I certainly wouldn’t have either(lol).

    Blackbody physics was conceived with the idea of investigating the radiating behaviour of a planar solid exhibiting the characteristics of total absorption at all wavelengths. Its 100% TRUE. Get over it.

    “Traditional study” was poor wording lets use “lab experiment” instead. Blackbody behaviour is investigated using a cavity whereby the sides are electrically heated hot enough to radiate. If you want to study it go buy one.

    You say “Not true either, solids also emit from all points, it just gets reabsorbed.”
    I said from all points within. A solid is governed by intermolecular forces or covalent bonding. The much tighter bonding promotes radiation of a solid as a single or whole unit. Gas molecules on the other hand are not constrained by such forces. A gas is not a solid. One does not behave like the other. You are wrong and what is up with the reabsorbed thing.

    You say “So tell me, where did that energy go?”. I wasn’t aware any was missing please elaborate.

    You say “You would use an absorbing medium to transfer the electromagnetic energy to heat”. Its not a transfer its a conversion and simply heating something doesn’t trap anything since it will thermalize with the surrounding air once the source of heating is removed.

    You say “the heat is created when an absorbing medium turns electromagnetic energy to mechanical (heat) at the surface of the steel”. This meaningless statement is only your attempt at extricating youself from a difficult position. The ability of a laser to cut metal is not a function of insulation taken to an extreme. If Co2 was a superinsulator it would have been discovered long ago and proudly announced to the world.

    The arrow shows reflection what do you say? Good luck with the pseudoscience.

  104. TomVonk said

    Jeff

    I don’t see anything wrong with saying CO2 absorbs outgoing infrared energy and warms the atmosphere. I know there is nuance to the statement but calling it wrong sowed a lot of confusion

    I think that this is the major (only ?) point where we differ . I see everything wrong with saying that and this was precisely the reason why I wrote the post I wrote .
    As it indeed appeared to have sowed confusion in a small but non negligible part of comments , I must elaborate . I might also submit an extensive clarification post at WUWT depending on time available .

    1)
    There is a difference between saying “CO2 heats the atmosphere” and “IR radiation may heat the atmosphere” . While the latter is correct , the former is not . Can you see the difference ?

    2)
    Another variation .
    I substitute to your statement quoted above another one .
    “I don’t see anything wrong with saying CO2 absorbs outgoing infrared energy and reemits it and doing that warms and cools the atmosphere.”
    This statement is again correct . Can you see the difference between this and the quoted statement ?
    3)
    Why the 2 statements “CO2 heats the atmosphere” and “I don’t see anything wrong with saying CO2 absorbs outgoing infrared energy and warms the atmosphere.” are wrong ?
    If they were true , one would have to propose and prove a mechanism that shows a net energy transfer (e.g “heating”) from CO2 molecules towards N2 molecules . Please note the absence of words “radiation” , “infrared” , “radiative equilibrium” , “AGW” , “Greenhouse effect” , “lapse rate” and such .
    This mechanism is indeed proposed f.ex on the Science of Doom pages and elsewhere .
    It is the following “The increase of the vibrational energy of CO2 due to the absorption of IR photons is immediately and irreversibly transferred to the N2 molecules by collisions .”
    Clearly this mechanism has indeed for consequence that “CO2 heats N2 and the atmosphere” .
    My point is that this mechanism is wrong and therefore the statement “CO2 heats the atmosphere” is wrong too .

    Beside the theoretical arguments in my post , the experience with the hollow sphere is also an obvious evidence that this mechanism is wrong .
    According to the collisional energy transfer mechanism , if it was true , it would also take place within the sphere . You surely agree with that .
    There would be 15µ radiation emitted by the sphere , there would be CO2 so there would of course be also absorption . All the conditions being fulfilled , according to the “collisional transfer mechanism” , there would be :

    a) heating of the N2 within the sphere
    b) no 15µ radiation observed by a detector on the surface of the cavity (because it all goes to N2 via collisions) .

    Both conclusions are falsified by observation .
    Therefore the proposed mechanism is wrong and therefore the statement “CO2 heats the N2″ is wrong too .

    Specific remark :
    This statement would be correct for a limited time in a violently out of equilibrium system with strong transient like f.ex a gas laser .
    The troposphere which we consider here is in LTE and I specifically precise that my argument only applies on systems like the troposhere which are in LTE .

  105. TomVonk said

    Suricat and Suibhne

    There’s also the ‘duality’, or co-dynamic, of observation. As Suibhane said, “How are you so sure that the 15u is from CO2 rather than say H2O ?”

    Because the 15µ radiation is a signature of the first vibrationaly excited state of the CO2 molecules that is relaxing to the ground state .
    H20 has not this strong signature .
    So if you see a strong 15µ line in a Galaxy 1 billion light years away , you are sure that there is CO2 present .

  106. Suibhne said

    TomVonk said

    ……”Because the 15µ radiation is a signature of the first vibrationaly excited state of the CO2 molecules that is relaxing to the ground state .
    H20 has not this strong signature .”………

    I said in post 83 with reference to the graphs that Jeff Id supplied at start of article.

    “On the upward Earth Thermal Radiation H2O is shown to absorb from 10um to 20um;
    CO2 absorbs in 15um area.
    Given that water vapour has about 30 molecules to each CO2 molecule how much of the heating effect around 15um is due to CO2?.”

    CO2 radiates strongly in the 15um band
    H2O radiates to a lesser extent around the 15um band

    But since in the atmosphere there is much more H2O than CO2, your firm conclusion escapes me.

  107. Jeff Id said

    Tom,

    There is a difference between saying “CO2 heats the atmosphere” and “IR radiation may heat the atmosphere” . While the latter is correct , the former is not . Can you see the difference ?

    The statement doesn’t say CO2 heats the atmosphere, it says CO2 absorbs IR which heats the atmosphere. Which is what it does. I’m not sure about your ‘may’ qualifier in IR may heat. Of course it does.

    “I don’t see anything wrong with saying CO2 absorbs outgoing infrared energy and reemits it and doing that warms and cools the atmosphere.”

    I don’t agree with this statement on a subtlety of temperature which I think must be acknowledged for clarity. Temperature is a bulk property, while the transfer on an individual particle scale is back and forth, the net effect of additional CO2 is warming through absorption of IR therefore saying warming and cooling in the context of particulate atoms makes no sense. Please note that this is again different from increased emissivity, convection etc. because the net effect is higher energy level and warming at the surface.

    It is the following “The increase of the vibrational energy of CO2 due to the absorption of IR photons is immediately and irreversibly transferred to the N2 molecules by collisions .”
    Clearly this mechanism has indeed for consequence that “CO2 heats N2 and the atmosphere” .
    My point is that this mechanism is wrong and therefore the statement “CO2 heats the atmosphere” is wrong too .

    You are confusing the bulk property that CO2 will cause additional IR absorption, with the particulate property of energy transfer. I have no idea why it matters if the heat is irreversably transferred or not, there is still more heat captured. Shorter path lengths between absorption guarantees some additional energy absorption and warming – again, how much is another story.

    Beside the theoretical arguments in my post , the experience with the hollow sphere is also an obvious evidence that this mechanism is wrong .
    According to the collisional energy transfer mechanism , if it was true , it would also take place within the sphere . You surely agree with that .

    I agree with your experiment and result, what I don’t agree with is the conclusion you draw from it. In your sphere there is no net energy flow through the gas. The temperature of it stays constant no matter which gas is inserted. No flow of energy, no effect on temperature. If you were to insert CO2 at a temperature colder than the sphere walls, the CO2 would absorb energy slightly faster than inserting equally cold N2.

    If you put a jacket on an icecube in your freezer it does nothing – this is equivalent to your CO2 in a sphere example.

    This statement would be correct for a limited time in a violently out of equilibrium system with strong transient like f.ex a gas laser .
    The troposphere which we consider here is in LTE and I specifically precise that my argument only applies on systems like the troposhere which are in LTE .

    I don’t agree with either of these points. What you are missing is that we have a system in equilibrium(ish) yet it has a net flow of energy. Heat is coming in at one wavelength and being let out at another – completely different from your sphere. The outgoing radiation is captured in the CO2 bonds and transferred to other molecules. Changing the CO2 concentration in a flow, creates the warming effect. Changing CO2 concentration with no net flow, makes no difference.

    We can’t equate equilibrium exclusively with zero energy flow as you demonstrate in your sphere example. Zero energy flow in and out of a system is a subset or type of equilibrium. It does not represent the lower troposphere which has a 15um flow from ground to space.

  108. Carrick said

    Brian:

    Blackbody physics was conceived with the idea of investigating the radiating behaviour of a planar solid exhibiting the characteristics of total absorption at all wavelengths. Its 100% TRUE. Get over it.

    Oh really?

    That’s a remarkable claim given that Planck developed his theory using Boltzman’s kinetic theory of gases.

  109. Jeff Id said

    #108, He makes a number of remarkable claims. I can’t even begin to address the nonsense.

  110. DeWitt Payne said

    I haven’t read all the comments in detail, but I see several general concept problems. First there’s the concept of local thermodynamic equilibrium (LTE) and all that implies. Here’s the opening paragraph of an article by H. Roos, “On The Problem of Defining Local Thermodynamic Equilibrium”

    There are two statements which may come to your mind when reading the title of this paper: “Everybody knows what local thermodynamic equilibrium means”, and: “Nobody has ever given a really satisfactory definition of local thermodynamic equilibrium”. I can defend both of them under suitable conditions; and I shall do so in the following concentrating mainly on the second one. This article will not give answers, let alone a precise definition of local thermodynamic equilibrium; my aim is to bring the related problems into sharp focus.

    A system does not have to be in thermal equilibrium everywhere at all time scales for it to be in local thermodynamic equilibrium. Local thermal equilibrium is important for atmospheric radiative transfer because Kirchhoff’s Law (absorptivity equals emissivity) only applies under LTE. An operational definition of LTE in the atmosphere is that the vast majority of energy transfers between molecules are by collision rather than radiation. If I understand the question about quenching time correctly, it has no meaning. It does not matter how long any individual molecule is in an excited state. Emission is determined by the fraction of molecules in an excited state at any given time. Under LTE, this fraction is calculated using the Boltzmann distribution with a degeneracy of two because there are two identical (degenerate) modes for the 15 micrometer CO2 vibrational transition (in plane and perpendicular to the plane). For Earth atmospheric pressures, the effect of increasing pressure is to broaden the line width. A broader line has higher total absorptivity.

    The other problem is the statement that increasing CO2 concentration doesn’t matter because it just shortens the photon mean free path. The answer to this is too long for a comment. A one line reply would be: That’s true for the center of the band, but not the wings. For more detail, I refer you to The Science of Doom series of articles “CO2 – An Insignificant Trace Gas? (starting with Part One) and specifically to Part 8 – Saturation. Additionally, the atmosphere is neither isotropic nor isothermal and that makes a big difference when doing the radiative transfer calculations for the entire vertical column of the atmosphere.

    Finally, CO2 and other molecules that absorb in the thermal IR spectrum do warm the atmosphere above where it would be if those gases were not present.

  111. suricat said

    #105 Tom Vonk said.

    “H20 has not this strong signature .”

    That’s ‘brilliant’. :)

    So if no CO2 is present the brilliance from 10 um through 20 um is ~constant, but if there’s a brilliance peak at 15 um CO2 is present. That makes sense, thanks Tom.

    However, any ideas on how an increasing altitude above the temperature inversion point would affect a CO2 emission?

    As an engineer, I’d expect an increasing temperature to negate (to some degree [pardon the ‘pun’]) the extra distance to extinction by increasing molecular velocity. Thus, the time period for energy gained before emitting a photon is about, or ‘near’, average. Though, emission at higher altitudes is greater. Could this be due to higher insolation levels in the upper atmosphere? Are you unsure of this?

    After all is said and done, ‘you can’t get out more than you put in in the first place’!

    Best regards, Ray Dart.

  112. suricat said

    #102 Jeff Id said.

    “What I do know is that more CO2 means at least some warming – predicting how much is up to someone else.”

    Why “up to someone else”?

    We already know that the photon energy that is released by CO2 to the EM (electromagnetic) system was provided by an equivalent energy input to the molecule from kinetic and/or EM systems so there’s no ‘amplification’ factor, only ‘unity’ (discreet and measured ‘packets of energy’)! It can’t be this!

    So, if the ‘radiative properties’ of CO2 can’t be a part of this ‘warming’ influence of CO2, what can?

    IMHO, the fact that CO2 has a relatively low Cp (thermal capacity) WRT other atmospheric constituents, CO2 is less able to ‘convect’ energy away from the surface, and because CO2 is a ‘heavy’ gas WRT other atmospheric constituents, it also proposes a ‘negative’ influence to ‘convection’.

    Your thoughts?

    Best regards, Ray Dart.

  113. Jeff Id said

    “Why “up to someone else”?”

    ‘Cause I don’t know.

    “We already know that the photon energy that is released by CO2 to the EM (electromagnetic) system was provided by an equivalent energy input to the molecule from kinetic and/or EM systems so there’s no ‘amplification’ factor, only ‘unity’ (discreet and measured ‘packets of energy’)! It can’t be this!”

    Unreadable. Mixing of topics.

    “So, if the ‘radiative properties’ of CO2 can’t be a part of this ‘warming’ influence of CO2, what can?”

    Can’t understand you.

    “IMHO, the fact that CO2 has a relatively low Cp (thermal capacity) WRT other atmospheric constituents, CO2 is less able to ‘convect’ energy away from the surface, and because CO2 is a ‘heavy’ gas WRT other atmospheric constituents, it also proposes a ‘negative’ influence to ‘convection’.”

    I’ve never bothered to look up the heat capacity of CO2, despite that falling within my engineering formal education. It has no bearing to this post

  114. Suibhne said

    suricat said
    August 11, 2010 at 7:28 pm

    #105 Tom Vonk said.

    “H20 has not this strong signature .”

    That’s ‘brilliant’.

    I take it you mean the little spike in the “total” slope around 15um.
    This should also give an indication of the H2O/CO2 radiative flux ratio around this wavelength.

    However Jeff Id’s position(as I understand it) does not exclude a certain amount of CO2 emission at this frequency along with a larger amount of thermalisation.

  115. TomVonk said

    deWitt Payne

    A system does not have to be in thermal equilibrium everywhere at all time scales for it to be in local thermodynamic equilibrium. Local thermal equilibrium is important for atmospheric radiative transfer because Kirchhoff’s Law (absorptivity equals emissivity) only applies under LTE. An operational definition of LTE in the atmosphere is that the vast majority of energy transfers between molecules are by collision rather than radiation. If I understand the question about quenching time correctly, it has no meaning. It does not matter how long any individual molecule is in an excited state. Emission is determined by the fraction of molecules in an excited state at any given time. Under LTE, this fraction is calculated using the Boltzmann distribution with a degeneracy of two because there are two identical (degenerate) modes for the 15 micrometer CO2 vibrational transition (in plane and perpendicular to the plane). For Earth atmospheric pressures, the effect of increasing pressure is to broaden the line width. A broader line has higher total absorptivity.

    Thanks ! I was beginning to wonder if there was somebody who knew what LTE meant . I agree 100% with the above quote . Most of what I wrote in my WUWT post and attempts at clarifications here is contained in teh quote . I recommend everybody to read it several times and to make sure that there is no doubt about what the words and sentences really mean .

    Jeff
    I agree with your experiment and result, what I don’t agree with is the conclusion you draw from it. In your sphere there is no net energy flow through the gas. The temperature of it stays constant no matter which gas is inserted. No flow of energy, no effect on temperature.

    There are only 2 possibilities – either you have not read what my conclusion is or you have misinterpreted it .
    Here is my conclusion again and I will reword it in 5 variants which are STRICTLY equivalent to make sure that it is clear what my conclusion really is :
    1)
    In LTE the energy transferred from vibrationnal decay of CO2 towards kinetic energy of N2 molecules is exactly balanced by the energy transferred from the kinetic energy of N2 molecules towards vibrational excitation of CO2 molecules
    2)
    In LTE there is no net energy transfer from the vibrational degress of freedom of CO2 towards translational degrees of freedom of N2 .
    3)
    Regardless whether there is IR radiation absorbed by CO2 or not , there is no net energy transfer from the vibrational energy of CO2 towards the translational kinetic energy of N2 in LTE
    4)
    In LTE CO2 does not heat N2 by energy transfer fromm its vibrational energy towards N2 kinetic energy regardless whether there is or is not IR radiation
    5)
    In LTE saying that CO2 doesnot transfer energy from its vibrational levels towards N2 is equivalent to saying that any IR radiation absorbed by CO2 must be reemitted .”

    So now as all these statements are equivalent , you either agree with ALL of them or with none .
    The only forbidden way is to agree with some and disagree with others .
    For the sake of clarity , I remind for the Nth time that the low atmosphere is in LTE even if my statement applies to any gas mixture in LTE . Low atmosphere or the sphere is just an example of a mixture in LTE .

    So now do you agree with all or disagree with all ?

  116. Jeff Id said

    Tom,

    As I said before, and you quoted from me in 115. I don’t disagree with any of your points in bold even in the slightest way. NOT even a little. Nothin’ to disagree with there. When you first wrote LTE, I knew what is was and even read up on it to make sure. No arguments here.

    Try to read what I’m saying, your interpretation of these statements that the lower troposphere cannot warm from CO2 is where I disagree.

    You wrote this at WUWT:

    “CO2 absorbs the outgoing infrared energy and warms the atmosphere” followed by These statements are all wrong

    If my experiment above reaches equilibrium, with an atmospheric level of CO2 where 50% of the laser passes through unmolested, wouldn’t it also reach a warmer LTE when the CO2 concentration was increased?

    Your experiment has a problem in comparison to the atmosphere, while it is definitely in LTE, your experiment exhibits no flow through the volume.

    Compare that to the troposphere. Let’s take a 1m column of air from ground to space. Divide the column in 1m cubes. Each cube is basically in LTE at any moment. Ignoring other much stronger IR absorbing gasses, the bottom cube captures radiation from the earths surface in it’s CO2 this is transferred into and out of N2 however in the case of our column, the net flow is from ground to space. We have a lapse rate so that despite LTE where energy in is equivalent to energy out, the out is on the top and the in is on the bottom. We have flow.

    The IR from the surface of the earth absorbs into the first block. This block gains energy from the photon, it’s transferred to other molecules which transfer it back to CO2, but according to the second law of thermo, the NET energy goes one way into the next block up which is slightly cooler. The process goes on and on until the top blocks emit the IR into space.

    By adding CO2 you capture the IR closer to the ground, the mean difference in path length for an IR emission shortens. The time taken to transfer that photon of energy from one block to the next increases. – maybe very slightly but it increases because it has more CO2 plant food in the way to capture that occasional IR emission. The delay time means more energy in that volume, which equates to a higher temperature.

    Therefore, this statement ““CO2 absorbs the outgoing infrared energy and warms the atmosphere”

    Is true.

    Even in LTE

  117. TomVonk said

    Now let’s continue and suppose that a reader agrees with one of the above formulations of my statement what means that he agrees with all of them .
    Instead of a theoretical gas mixture in LTE we will take a volume in the real atmosphere .
    For example a sphere of volume 1 cm^3 .
    This volume will be in TE .
    This is not a typo , I dropped the L because the atmosphere being in LTE , I know that there exists a neighborhood of every point which is in TE (definition of LTE) .
    If you think that 1 cm^3 is too big , take 1 mm^3 it doesn’t matter .
    There will also be an IR flow of some 30W/m² in the 15µ band irradiating this gas volume – that’s what the ground is radiating at room temperatures .
    Of course the CO2 molecules in our sphere will absorb some of that IR .
    Will the gas sphere of 1 cm^3 taken in the real atmosphere heat ?
    No as the formulation N°4 with which you agreed is telling .

    To finish let’s suppose that you ask “But what happens if I suddenly changed the IR souce so that it emits now 100 W/m² ?”
    Well first what happens is that you destroy the TE in all small volumes .
    Second what happens is that my statement is no more correct because it requires LTE .
    Third is that for an extremely short time the CO2 molecules will absorb more IR than what they did before and their vibrational energy increases above what it was . This additional vibrational energy will be extremely fast spread out among all the molecules by collisions and there WILL be a net energy transfer from CO2 to N2 .
    The mixture heats up untill a new TE is achieved in all small volumes and it will happen very fast .
    At this point LTE will be restored and my statement will be correct again despite the fact that the IR is now 100 W/m² instead of the 30 W/m² it was before – in other words the CO2 no more heats up the N2 once this point is reached .

  118. Steve Fitzpatrick said

    TomVonk,

    Do you agree with this comment from DeWitt?

    “CO2 and other molecules that absorb in the thermal IR spectrum do warm the atmosphere above where it would be if those gases were not present.”

    A clear answer one way or the other would maybe help.

  119. TomVonk said

    Jeff

    You wrote this at WUWT:

    “CO2 absorbs the outgoing infrared energy and warms the atmosphere” followed by These statements are all wrong

    Yes I did and for me it is more a matter of logics than physics .
    If the statement you quoted is indeed wrong as I submit , then it means that the negation of this statement is true .
    What is the negation ?

    Well it is “CO2 cannot heat the atmosphere regardless whether it absorbs or not IR radiation”
    And this statement is nothing else than the Variant N°4 written above with which you agree .

    So if we stay consistent , if you agree with :
    “CO2 cannot heat the atmosphere regardless whether it absorbs or not IR radiation” is true
    Then you necessarily also agree with :
    “CO2 absorbs the outgoing infrared energy and warms the atmosphere” is wrong .
    Both formulations are exactly the same statement .

    It’s really just that . There is no need to talk about radiative transfer or extinction lengths etc – they have nothing to do with the statement .

  120. Jeff Id said

    #118, Your first comment on the last thread on this topic made the same point I’ve been trying to make. I thought people might get caught up in equilibrium discussion at that time but it only took comment #1 to find the error in his post.

  121. TomVonk said

    Steve Fitzpatrick

    “TomVonk,

    Do you agree with this comment from DeWitt?

    “CO2 and other molecules that absorb in the thermal IR spectrum do warm the atmosphere above where it would be if those gases were not present.”

    A clear answer one way or the other would maybe help.

    Yes .
    (Can’t be clearer , can it :))

  122. Jeff Id said

    Tom,

    So if we stay consistent , if you agree with :
    “CO2 cannot heat the atmosphere regardless whether it absorbs or not IR radiation” is true I don’t agree with this for the above reasons, which you have failed to grasp

    Then you necessarily also agree with :
    “CO2 absorbs the outgoing infrared energy and warms the atmosphere” is wrong .I don’t agree with the first so cannot agree with the second

    Both formulations are exactly the same statement . agreed

    Now try and read what I wrote.

  123. Jeff Id said

    Here’s one,

    Does the total time taken for the IR emission from the earths surface to outer space change when CO2 is added to the atmosphere.

    Is that time change caused by the absorption spectrum of the CO2 molecule.

    Does that time change cause warming?

  124. TomVonk said

    Btw I think that I see now where all the misconceptions come from .
    Some , not all , people equate “CO2 heats the atmosphere” with “the temperature of an atmosphere containing CO2 is higher than what it would be without CO2″
    .
    Those statements are clearly not equivalent , at least not for a physicist .
    The first statement is a dynamical statement , it says that there is a NET FLOW of energy from CO2 towards N2 . This needs a precise mechanism of energy transfer .
    The second statement is comparison of 2 static equilibrium states which have nothing in common .
    More specifically the second statement doesn’t need a mechanism transferring directly energy from CO2 to N2 .

    I thought that it was clear for everybody that I have been only talking about the dynamical statement which is the natural sense of the sentence “X heats Y” . I hope it is clear now .

  125. Steve Fitzpatrick said

    TomVonk,

    The two statements

    “CO2 cannot heat the atmosphere regardless whether it absorbs or not IR radiation.”

    and

    “CO2 and other molecules that absorb in the thermal IR spectrum do warm the atmosphere above where it would be if those gases were not present.”

    appear to me to be mutually exclusive, yet you seem to have agreed both are true. Can you please explain why you think both are true?

  126. Re: TomVonk (Aug 12 07:47),
    Tom,
    Do you think that ozone, by absorbing UV, transfers heat to N2 and O2 in the stratosphere?

  127. Richard111 said

    Tom Vonk wrote in 117 above:

    “”At this point LTE will be restored and my statement will be correct again despite the fact that the IR is now 100 W/m² instead of the 30 W/m² it was before – in other words the CO2 no more heats up the N2 once this point is reached .””

    We have a change from 30 W/m² to 100 W/m². Is all this energy seen by the CO2? Surely the 15 micron band energy level is much less???

    Sorry for a confused question from a confused layman but I am trying to learn.

  128. TomVonk said

    Sorry Jeff I stop here .
    I can’t make sense of the logical contradictions you accumulate .
    You said in 116

    As I said before, and you quoted from me in 115. I don’t disagree with any of your points in bold even in the slightest way. NOT even a little

    N°4 of these points was :
    In LTE CO2 does not heat N2 by energy transfer fromm its vibrational energy towards N2 kinetic energy regardless whether there is or is not IR radiation

    Then in 122 you say :

    CO2 cannot heat the atmosphere regardless whether it absorbs or not IR radiation” is true I don’t agree with this for the above reasons, which you have failed to grasp

    It is obvious that those are same statement with which you simultaneously agree and disagree . Ok i forgot to specify “… cannot heat the atmosphere in LTE regardless …” but since the beginning of this thread I hope that it is clear for everybody that all my statements require LTE .

    So you are right , I have difficulty to handle statements that are right and wrong at the same time and prefer to stop here .
    Good luck .

  129. Steve Fitzpatrick said

    TomVonk,

    Well, thanks for #124.

    So it appears the argument is basically one of semantics.

    OK, with everything defined as you choose to, you can draw a distinction. But in a practical sense (like, what is the expected effect of adding CO2 to the atmosphere?), you are mainly confusing the discussion.

    If you believe that adding CO2 to the atmosphere will cause some (exact amount unknown) warming of the Earth’s surface, then it may be best to clarify that with the readers at WUWT, since I believe they drew exactly the opposite conclusion from your post. If you do not believe that adding CO2 to the atmosphere will warm the Earth’s surface (even a little), then however you are doing your analysis, the result is simply wrong.

  130. Jeff Id said

    Tom,

    What I think you are missing is in your balanced equation you are ignoring which N2 molecule and which CO2 molecule gets the transfer. You are treating LTE it as though everything is in a trapped volume. In fact due to the atmospheric lapse rate, the probabilities are such that a higher altitude N2 or CO2 is cooler and the Co2 <> N2 balance which occurs, has an increased probability of exciting a higher altitude molecule which causes an upward flow – conduction, second law and all that. LTE is maintained by incoming thermal radiation – absorbed by the CO2.

    As I’ve tried to explain, It’s this flow which makes the statement that CO2 can warm the atmosphere true.

  131. Steve Fitzpatrick said

    TomVonk,

    One final comment: if you believe adding CO2 to the atmosphere will warm the Earth’s surface by some amount, and if you believe that the radiating level of the atmosphere has to remain effectively constant in temperature by the requirement of radiative loss equal to solar gain, then it logically follows that the atmosphere must (on average) be warmer as well. Or in other words, more CO2 causes some warming of the atmosphere.

  132. Steve Fitzpatrick said

    Jeff,

    A brave effort, but I fear not completely successful.

    As I said in the other thread, I think Anthony needs some kind of informal review/advise on what gets posted.

  133. Jeff Id said

    #132. I agree.

    Tom, I wish you would point out where my statements contradict instead of this cop out:

    So you are right , I have difficulty to handle statements that are right and wrong at the same time and prefer to stop here .

    I fear he may have figured out that I was right and decided to bail.

  134. Suibhne said

    The particular method by which the 15u photon transfers its energy to the other molecules be it translational, vibrational or rotational KT is not for me the most important aspect.
    As Merrick says in his posts at WUWT getting the molecular KE to change back to an IR emission is much less likely.
    There appears to be is an asymmetry between absorption and emission.
    I tried to make the same point over at SoD’s site but all I got back was a truncated statement on Kirchoff’s Law

  135. Jeff Id said

    #134, It’s a different topic, but also interesting

    “As Merrick says in his posts at WUWT getting the molecular KE to change back to an IR emission is much less likely.”

    I don’t think it’s being considered properly. The 30ms number calculated by Pat Frank above for IR emission is an average probability. The molecule doesn’t just wait for 30ms every time. Since the CO2 is re-energized as often as it is de-energized in the lower atmosphere CO2 <>N2, I believe there is still the same average time between emissions of CO2 IR. IR emissions are therefore not less likely.

  136. Carrick said

    I like DeWitt’s contribution here.

    The other thing I’m going to throw into the mix is classical, macroscopic thermodynamics is a phenomenological theory whose language is still highly colored by the caloric theory of heat., one should not hang too much hay on this.

    When considering for example of whether an emission of a photon by an atom a high altitude and its absorption by another atom at the surface, the issue of whether the atoms on the ground have a higher mean kinetic energy than the atoms at the high altitude does not come in to play. What matters for this process are time scales on the same order of magnitude as the emission and absorption times… which are typically small fractions of a nanosecond, and the distance scales that are important are again fractions of a micron.

    The correct direct for developing a physical picture involves integration of atomic physics processes to a macroscopic environment, with “allowed” versus “forbidden” having to do with quantum mechanical selection rules, not classical arguments.

    Beyond that for classic physics hanger-ons, I can give you a physical example where net positive heat energy is transferred from a colder object to a warmer one when the two regions are in thermal contact (search “negative temperature”), and where adding heat energy makes them colder not warmer (search “negative heat capacity”).

    I’m left with no clue what TomVonk was originally trying to argue. As it stands now, it appears to be a confused mess.

  137. Steve Fitzpatrick said

    #136,

    I think TomVonk just is struggling with the transition between local equilibrium and macroscopic energy flows. As a result, he ends up saying things that are contrary to clear observational evidence (AKA “a confused mess”).

    Would you be willing to look over proposed posts at WUWT for basic consistency with physical reality? I’m not talking about if the UHI adjustments for a set of weather stations are correct, just if there are obvious errors in basic physics. I could do this for any chemistry/physical chemistry related content.

  138. Suibhne said

    Jeff Id said

    …..”Since the CO2 is re-energized as often as it is de-energized in the lower atmosphere CO2 N2, I believe there is still the same average time between emissions of CO2 IR. IR emissions are therefore not less likely.”…….

    The energy E=hf for the 15u photon is over twice the average translational KE of the molecules at STP.

    If you believe in the thermalisation route, this excess KE is quickly shared out amongst neighbouring molecules, because of the ten to the power of ten collisions per second.

    So for absorption all CO2 molecules would appear to be equally likely.

    However for emission only those CO2 molecules with translational KE> twice the average are eligible.

    Hence the apparent asymetry!

  139. Jeff Id said

    #138, “So for absorption all CO2 molecules would appear to be equally likely.” – don’t forget that the energy comes back equally CO2<>N2. I would say that an equal proportion of CO2 is ready to absorb the 15um. With an already excited CO2 molecule we might get spontaneous emission or something – Tom would be better than I to answer that question.

    You can look at the blackbody curve to judge both the intensity and amount of emission at each wavelength. It’s dependent only on temperature.

  140. Steve Fitzpatrick said

    Carrick,

    I wanted to also mention that I agree with you (#21) about the contrasts between classical “thermo” and statistically correct (microscopic) treatments. Unfortunately, most people who are exposed to some thermo (and there are lots!) never get to see, or maybe see but don’t really understand, that the equations of classical thermodynamics drop out of a more fundamental description of the statistical nature of matter at the molecular level. Newton’s laws are more accurately described by Einstein’s equations, even though Newton’s laws are useful approximations in practice… just like thermo is useful in practice, but does not address reality as as fundamental level, or like the classical description of covalent bonds is useful, but quantum chemistry offers a more fundamental description of covalent bonding.

  141. Carrick said

    Steve, I’m not sure I have time right now to do any sort of review of Tom’s work. (Meaning there is no way I would have time, lol)

    I won’t state what is true for him, but it is clear that for many one of the problems is as you stated a confusions over ” local equilibrium and macroscopic energy flows”. Whether you have a 30-minute overturning rate of the atmosphere plays no role in the computation of photon emission rates from e.g., CO2 (other than how this 30-minute over turning affects local temperature and other thermodynamic properties).

    That and a confusion regarding placing a primacy on macroscopic, phenomenological “laws” over the actual fundamental underlying microscopic physics.

  142. Suibhne said

    Jeff Id said

    …..”You can look at the blackbody curve to judge both the intensity and amount of emission at each wavelength. It’s dependent only on temperature.”…..

    If emission is just as likely as absorption then TomVonk is correct – no net heating effect from CO2.
    Absorption, the 15um photon energy is transferred to KE of neighbouring molecules, local heating is the result.
    Emission, a CO2 molecule gets the statistically less likely (but possible) over twice average KE by collision with neighbouring molecules and emits 15um photon.
    Result local cooling.

    However I think that if photon energy transfer can be changed into something other than radiation(in this case thermalisation) then Kirchoff’s Law no longer applies.

  143. Jeff Id said

    “If emission is just as likely as absorption then TomVonk is correct – no net heating effect from CO2.”

    Nope, there is a flow, and the probability of gaining/loosing energy, balances for an upward transfer of energy. It’s more likely that a the unenergized molecule for the xfer is at a higher altitude. So at any point in the troposphere equilibrium, there is an equal probability of absorption and emission in LTE, however the direction of that probability balances upward toward space.

    If you slow the flow, you capture the heat.

  144. Re: Jeff Id (Aug 12 12:40),
    “So at any point in the troposphere equilibrium, there is an equal probability of absorption and emission in LTE,”
    I don’t agree. Mostly the air is getting IR from somewhere warmer. That determines the probability of absorption – probability of emission is determined by local temperature.

    That was the point of my query to TV about ozone. The probability of absorption is determined by sunlight. The probability of emission is determined by local temp, and the probability of UV emission is nil. O3 can emit thermal IR, but would have to emit many photons to balance absorption of 1 UV.

  145. Steve Fitzpatrick said

    #144

    I think a fair amount of the energy from absorbed UV in the stratosphere goes into production of ozone, that is, the UV energy is converted to chemical bond energy. That “chemically bound” part of the UV energy would not be re-radiated as IR energy.

    #141
    Carrick, I was not suggesting a review of TomVonk’s efforts. I meant a quick read-over of potential posts, just to make sure really crazy, not close to right stuff is not posted (like the post last year which claimed solid CO2 might be forming in Antarctica). I do not know if Anthony would be interested in any kind of screening/reality check, but it seems to me you have competence in physics that would be very helpful. Of course if you are too busy, I can certainly understand that.

  146. kevoka said

    Dewitt:

    “If I understand the question about quenching time correctly, it has no meaning. It does not matter how long any individual molecule is in an excited state.”

    The reason the quenching rate is interesting is due the nature of the Boltzmann distribution. There is a % (3-6?) of CO2 in the excited state that can emit at the 15um. That leaves 94% or more at some other energy level – unable to emit a 15um photon. Some of these are in the zero state (what %?)and ready to absorb another 15um. But there are is also % that are excited, but not enough to emit, and too much to absorb (AFAIK collisions do not transfer energy as quantum units).

    The quenching question is (for me) essentially what is the mean time that the CO2 molecule remains in this “half baked” state? Because while it is in this state, the molecule can neither absorb or emit at 15um.

    Is the mean quenching time greater than the average re-emission time (30ms from Pat Frank)? Less than? Equal?

  147. kevoka said

    #139 Jeff

    “I would say that an equal proportion of CO2 is ready to absorb the 15um.”

    Equal proportion to what? Those excited enough to emit?

  148. Douglas Hoyt said

    The problem with the radiative transfer thought experiment lies in the assumption that an instantaneous doubling in CO2 is a reasonable assumption. It is not and the radiative forcing calculated using that assumption is not correct for any planet that is wet and has moisture in its atmosphere. The instantaneous doubling assumption will only give a valid answer for a completely dry atmosphere over a surface with zero emissivity. The other problem with an instantaneous doubling of CO2 is that the temperature of the atmosphere and radiation field are far from LTE and will never occur in the real world.

    With a wet atmosphere and a surface with an actual emissivity close to one, the additional radiation absorbed by more CO2 will be converted to heat which will then heat the atmospheric moisture and the surface and that will then re-radiate at all wavelengths, much of which will escape to space through the atmospheric windows. The net result is that the assumed forcing for an instantaneous doubling should be multiplied by exp(-1), where 1 is the optical depth one. Only a portion of the initially absorbed radiation is retained in the system and ultimately heats it.

    In short, the highly simplified thought experiment used to justify catastrophic warming leads to an overestimation of the greenhouse forcing by a factor of about three, which happens to be what McKitrick has recently found.

  149. DeWitt Payne said

    Re: kevoka (Aug 12 23:41),

    That’s not how it works. If 6% is in the excited state then 94% is in the ground state. The time it takes to change states is very short so the fraction of molecules that are in the process of emitting or absorbing at any given instant is vanishingly small. If it weren’t, it wouldn’t be possible to build a laser with a femtosecond scale pulse duration.

  150. Jeff Id said

    #149, I was trying to explain to someone else that the fact that the collision time is so much shorter than the radiative decay time doesn’t reduce the amount of emissions due to radiative decay because in thermal equilibrium, there is always a similar amount of energized CO2. It’s more complex than I can write because of the spread of energy levels due to the collisions but the claim has been made that CO2 15um emission is vanishingly rare in the lower troposphere whereas I say it’s not.

  151. Suibhne said

    Jeff Id said

    …..”CO2 15um emission is vanishingly rare in the lower troposphere whereas I say it’s not.”…

    The 15um emission is not so much rare as more unlikely than absorption.
    However the CO2 4um band in that small but interesting overlap between solar and terrestial IR can be absorbed but since emission requires over eight times the translational KE of STP molecules it seems that it must be pretty rare in the “backradiation ” spectrum.

  152. DeWitt Payne said

    Re: Jeff Id (Aug 13 11:12),

    The life time of an excited CO2 molecule is actually on the order of a microsecond rather than a nanosecond even though the mean time between collisions at STP is ~1 ns. Most collisions are elastic and don’t change the state of the vibrational mode. Even if the time constant for spontaneous emission for a CO2 molecule is on the order of 10 microseconds or higher, some fraction of excited molecules will radiatively decay much more quickly. The overall effect is that the total flux is not a strong function of pressure.

    The other thing is that you can measure the atmospheric emission spectra at high resolution in the CO2 band. There are places where this is done every day. The observed spectra are in good agreement with calculated spectra. The total upward and downward flux is also measured and is just what is expected from R/T theory. How some people can ignore this fact and still claim that the emissivity of CO2 is small is beyond me.

  153. Jeff Id said

    #152, Thanks DeWitt. I asked Pat Frank to do a post on the decay rate of energized CO2 and a comparison with the collision time. It’s a very often misunderstood topic by those who consider themselves skeptics of the science. I wonder if you wouldn’t mind?

  154. Jeff Id said

    You know write a couple of equations out, describe their significance and fill in a couple of values. Explain what the whole thing means.

  155. DeWitt Payne said

    Re: Jeff Id (Aug 13 18:48),

    The math would be the same as for radioactive decay. In principle we know the rate and the concentration. From that we should be able to calculate the decay time constant. The self absorption makes it more complicated, but it shouldn’t be too much trouble. The time constant should be inversely proportional to the line strength. 238U, for example, has a half life of ~4.5E9 years, but as long as enough atoms are present, the decay rate is measurable.

  156. RuhRoh said

    Hey Jeff;

    I was wondering if this effect would be expected on Mars.

    I was looking at a NASA site that seemed to imply there was more total CO2 around Mars than around the earth, but that the Martian surface temperature was equal to the blackbody temperature.

    I did a big post with all the links at the time but it went poof.
    Probably cockpit error here.

    But I still wonder why there is no hint of warming there.

    TIA
    RR

  157. RuhRoh said

    The Martian example would seem to be almost ideal;

    No oceans
    No Clouds
    ~constant albedo
    No water vapor
    No Evaporative heat transfer
    No Crappy, drifty thermometers
    Minimal AGW/C
    No Volcanic aerosols
    No UHI…

    So, does the effect only work in more dense atmospheres?

    TIA
    RR

  158. DeWitt Payne said

    Re: RuhRoh (Aug 15 21:27),

    The depth of the atmosphere has a strong effect on the magnitude of the greenhouse effect. Because the Martian atmosphere is 95% CO2 and the average surface pressure is 7 millibar, the effective emission height of the Martian atmosphere is close to the surface. The value of g on Mars is smaller and the adiabatic lapse rate is 4.5 K/km compared to 9.8 K/km on Earth. As a result the magnitude of the greenhouse effect is smaller on Mars, approximately 5 C, even though the total amount of CO2 in the Martian atmosphere is greater than the amount in the Earth’s atmosphere. You can see a similar effect on Earth where the forcing from doubling CO2 at the poles is 1/3 that at the equator because the atmosphere is colder and denser at the poles. If you replaced half the CO2 in the Venusian atmosphere with nitrogen, the surface temperature would drop by only about 9 C.

  159. Ryan O said

    Regardless of all of the other radiative physics basics above, the Tom Vonks of the world make one grand assumption in their no-net-heating argument: that the re-emitted photon is lost to the system. While most of them are eventually lost, because the re-emission process shows no (or little) directional dependence, the residency time of photons with wavelength X within the atmosphere is increased as the absorption of wavelength X increases. As the residency time increases, so does the chance for the energy to be lost via KE transfer from a collision rather than re-emission.

    It doesn’t matter that this increase is small (because we already knew that). All that matters is that some of the incident radiation is transferred to KE as a function of optical depth.

  160. Gary Moran said

    Regardless of all of the other radiative physics basics above, the Tom Vonks of the world make one grand assumption in their no-net-heating argument: that the re-emitted photon is lost to the system.

    Tom is very precise with his wording and at no point does he say what you claim. He clearly states he believes the system will be warmer because of the presence of Co2, a nuanced but important point.

  161. Jeff Id said

    #160, Tom did a great job, my only problem with his work was the over generalization of LTE to a system without any energy flow. In a system with an energy flow, the statements about heat trapping he claimed were incorrect – are in fact all true.

  162. Ryan O said

    Gary, LTE assumes that all incident photons are lost to the system. To quote Tom:

    “Because the number of excited molecules in a small volume in LTE must stay constant , follows that both processes emission/absorption must balance . In other words CO2 which absorbs strongly the 15µ IR , will emit strongly almost exactly as much 15 µ radiation as it absorbs . This is independent of the CO2 concentrations and of the intensity of IR radiation.”

  163. Suibhne said

    To quote Tom:

    …“Because the number of excited molecules in a small volume in LTE must stay constant , follows that both processes emission/absorption must balance . In other words CO2 which absorbs strongly the 15µ IR , will emit strongly almost exactly as much 15 µ radiation as it absorbs . This is independent of the CO2 concentrations and of the intensity of IR radiation.”….

    LTE is not something that can be imposed.
    Its more of a statement of fact, about the thermodynamics of the locality.
    A sample of air at STP may well have LTE.
    If however CO2 molecule absorbs say a 4um photon the KE of the molecule jumps to over 9 times the average.
    Most people think that this thermalisation is quickly shared out amongst neighboring molecules.
    This process means that in the locality of the excited CO2 there is temporarily no LTE.
    Similarly for the 15um photon.

  164. RuhRoh said

    Re 158;

    OK, I think I understand your text.

    But where would I see the effect of the greenhouse ~5C warming?

    i.e., would the surface be warmer than blackbody predictions?
    or is the warming only ‘atmospheric’?

    Thanks for your kind efforts thus far.

    RR

  165. RuhRoh said

    Mr DeWitt Payne;

    So, when I check the martian data at

    http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html

    it says that the blackbody temperature is 210.1K and the average atmospheric temperature is ~210K.

    Where would I expect to find that 5 K that you said is happening there?
    TIA
    RR

  166. Douglas Hoyt said

    Thsi paper by Spencer (http://www.uah.edu/news/newspages/campusnews.php?id=564) is consistent with my comment number 148 above in that the radiative forcing is overestimated because the radiative transfer calculations are erroneous.

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  169. DeWitt Payne said

    Re: RuhRoh (Sep 8 11:00),

    Sorry I missed this earlier.

    You have to understand the consequences of Holder’s inequality. The blackbody temperature is an upper bound. Given the large range in Martian surface temperature, the measured mean temperature, Tphys, would be lower than the blackbody temperature, Teff. See for example Earth’s moon. The lunar Teff, corrected for an albedo of 0.12 is 288 K. The actual average surface temperature is ~200 K. Mars spins faster and has an atmosphere so Tphys in the absence of a greenhouse effect would be about 205 K.

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  171. Hello! I know this is somewhat off topic but I was wondering
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  172. AVes said

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