Guest Post by Willis Eschenbach
Once again, the crazy idea that downwelling longwave radiation (DLR, also called infra-red or IR, or “greenhouse radiation”) can’t heat the ocean has raised its ugly head on one of my threads.
Figure 1. The question in question.
There are lots of good arguments against the AGW consensus, but this one is just silly. Here are four entirely separate and distinct lines of reasoning showing that DLR does in fact heat the oceans.
Argument 1. People claim that because the DLR is absorbed in the first mm of water, it can’t heat the mass of the ocean. But the same is true of the land. DLR is absorbed in the first mm of rock or soil. Yet the same people who claim that DLR can’t heat the ocean (because it’s absorbed in the first mm) still believe that DLR can heat the land (despite the fact that it’s absorbed in the first mm).
And this is in spite of the fact that the ocean can circulate the heat downwards through turbulence, while there is no such circulation in the land … but still people claim the ocean can’t heat from DLR but the land can. Logical contradiction, no cookies.
Argument 2. If the DLR isn’t heating the water, where is it going? It can’t be heating the air, because the atmosphere has far too little thermal mass. If DLR were heating the air we’d all be on fire.
Nor can it be going to evaporation as many claim, because the numbers are way too large. Evaporation is known to be on the order of 70 w/m2, while average downwelling longwave radiation is more than four times that amount … and some of the evaporation is surely coming from the heating from the visible light.
So if the DLR is not heating the ocean, and we know that a maximum of less than a quarter of the energy of the DLR might be going into evaporation, and the DLR is not heating the air … then where is it going?
Rumor has it that energy can’t be created or destroyed, so where is the energy from the DLR going after it is absorbed by the ocean, and what is it heating?
Argument 3. The claim is often made that warming the top millimetre can’t affect the heat of the bulk ocean. But in addition to the wind-driven turbulence of the topmost layer mixing the DLR energy downwards into lower layers, heating the surface affects the entire upper bulk temperature of the ocean every night when the ocean is overturning. At night the top layer of the ocean naturally overturns, driven by the temperature differences between surface and deeper waters (see the diagrams here). DLR heating of the top mm of the ocean reduces those differences and thus delays the onset of that oceanic overturning by slowing the night-time cooling of the topmost layer, and it also slows the speed of the overturning once it is established. This reduces the heat flow from the body of the upper ocean, and leaves the entire mass warmer than it would have been had the DLR not slowed the overturning.
Argument 4. Without the heating from the DLR, there’s not enough heating to explain the current liquid state of the ocean. The DLR is about two-thirds of the total downwelling radiation (solar plus DLR). Given the known heat losses of the ocean, it would be an ice-cube if it weren’t being warmed by the DLR. We know the radiative losses of the ocean, which depend only on its temperature, and are about 390 w/m2. In addition there are losses of sensible heat (~ 30 w/m2) and evaporative losses (~ 70 w/m2). That’s a total loss of 390 + 30 + 70 = 490 w/m2.
But the average solar input to the surface is only about 170 watts/square metre.
So if the DLR isn’t heating the ocean, with heat gains of only the solar 170 w/m2 and losses of 390 w/m2 … then why isn’t the ocean an ice-cube?
Note that each of these arguments against the idea that DLR can’t warm the ocean stands on its own. None of them depends on any of the others to be valid. So if you still think DLR can’t warm the ocean, you have to refute not one, but all four of those arguments.
Look, folks, there’s lot’s of good, valid scientific objections against the AGW claims, but the idea that DLR can’t heat the ocean is nonsense. Go buy an infrared lamp, put it over a pan of water, and see what happens. It only hurts the general skeptical arguments when people believe and espouse impossible things …
w.
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Tim_Folkerts,
So, water vapor DOES absorb in the near infrared where there is a rather large amount of insolation. (about 51% of solar is IR with most of it in the near infrared) Water vapor, the most important GHG, then is a moderator of insolation.
CO2 will cool or warm depending on the temperatures of what is providing the energy. CO2, at 390ppm, simply has too little mass in the atmosphere to make a difference as far as a storage media while its ability to absorb and either emit and/or transfer energy through collision is significant. Basically it is an energy distributor similar to water vapor, more efficient in a couple of bands but not covering as wide a spectra.
KuhnKat,
Water vapor and water droplets do still have one major difference that springs to mind in relation to “moderation”. The addition of droplets (ie clouds) reflects energy away from the earth, so that energy is completely removed, leading to a definite cooling effect . The addition of H2O gas absorbs the energy — less does indeed get to the surface, but more gets in the atmosphere. I suspect the net result for more H2O gas would would lead ti small changes in the surface temperature, so it is not nearly the same sort of “moderator” that clouds would be..
Bob, I am not sure what the purpose of this statement was: “BTW, did you notice that your hero Travesty Trenberth has been very naughty again, this time in the fracas over the Spencer and Braswell paper.”
Science is not about “heroes” or “villians”. When did I ever claim Trenberth or anyone else was a “hero”? The “fracas” seems to have a lot of losers and no winners — the biggest loser probably being the reputation of “science” in the eyes of the public. But eventually the science will sort itself out, independent of the egos and hot air on both sides.
And speaking of science sorting itself out — if there really was a glaring error in the magnitudes of the IR energy values in the Trenberth paper, don’t you think people like Spencer or Braswell or Lindzen would have pointed it out long ago? If you are going to make an “extraordinary claim” that numbers that are widely accepted even by “skeptical” scientists are wrong, then you will need “extraordinary evidence”, not simply gut feelings or analogies. Do the calculations and show us what the correct numbers should be, and submit the results to be published.
Tim Folkerts @ur momisugly September 11, 2011 at 3:47 pm
I thought my sarcasm over Travesty Trenberth would get you going, and I agree with your response. He does seem to excel in that sort of thing though, what?
Coming back to the paradox, there are some first principles that need to be established before dreaming about mathematical considerations. You did say earlier something like; first understand how it works in a transparent atmosphere before figuring out the effects of GHG’s etc. Now thence is where the initial surface emission gets modified and I guess will approach towards radiation normal to the surface, but very, very complicated, and that is a different argument to the claim that S-B radiation from the surface IS normal.
1) Do you agree that EMR waves pass through each other, without interference?
2) Do you agree that in a transparent atmosphere, the horizontal components of hemispherical radiation will escape to space approaching the tangential rather than normal?
3) Do you agree that an observer looking down, with a normal field of view, will not be able to see the tangential stuff?
4) Do you agree that if the observer shifts to a location where the first mentioned tangential stuff can be seen normally, there will be new tangential stuff from other locations that will then not be viewable?
5) Do you agree that the tangential stuff IS part of the S-B hemispherical emission, but is not normal as depicted by Trenberth et al?
Bob, I really am not quite sure what you are saying …
>1) Do you agree that EMR waves pass through each other, without interference?
Sure. (Well, there are some photon-photon interactions that are possible, but they are not important in this setting).
>2) Do you agree that in a transparent atmosphere, the horizontal components
>of hemispherical radiation will escape to space approaching the tangential
>rather than normal?’
What do you mean by “horizontal components of hemispherical radiation”? There are no purely horizontal photons. Certainly the more horizontal they are , the more tangentially they will approach some arbitrarily drawn sphere that we call the “top of atmosphere”. But all that is really important is if they do indeed escape somewhere, no the angle they are heading.
>3) Do you agree that an observer looking down, with a normal field of view, >
>will not be able to see the tangential stuff?
Do you mean normal = “perpendicular”? And what sort of “observer” do you have in mind. The only truly logical “observer” would be a meter that detects all radiation passing thru a given part of the surface in question. For example, i could put a small, horizontal black body there or a pyrgeometer pointing downward. Either of these would “see” radiation from any direction that happens to hit the “observer”.
It would be odd to only look at radiation that is heading straight up (ie normal = perpendicular). In fact, there would only be an infinitesimal amount of energy heading precisely vertically. So I guess, if you “put on blinders” so that only vertical photons were counted, you would get a much smaller number for the photon energy — way less than 390 W/m^2
>4) Do you agree that if the observer shifts to a location where the
>first mentioned tangential stuff can be seen normally, there will be
>new tangential stuff from other locations that will then not be viewable?
If I simply shifted my ANGLE (not position) and looked only at the photons heading straight in along that angle, then certainly you would see a different set of photons.
>5) Do you agree that the tangential stuff IS part of the S-B hemispherical emission,
>but is not normal as depicted by Trenberth et al?
I am quite sure Trenberth is not trying to depict the upward or downward IR as perpendicular (normal) to the surface.
I agree that the PARTIALLY-tangential stuff that is seen when the angle is shifted is indeed part of the SB emission. And when this is added up over all angles, it will indeed give the 390 W/m of SB radiation. (depending of course on the exact temperature of the ground/oceans. And then we would get to the details of the absorption by GHGs)
Maybe this is the crux: “However, with hemispherical radiation, the horizontal stuff is part of the 390 and in a transparent atmosphere would escape to space outside the field of view.”
There is no “outside the field of view”. The 390 from 1 m^2 of the surface would not all travel vertically thru the corresponding 1 m^2 directly above it. Some would “escape” at an angle thru other square meters. But energy from other square meters of the ground would head slightly sideways to pass thru the given upper square meter.
In the end, these must all balance (other than the slight decrease due to the curvature of the earth.)
Tim Folkerts @ur momisugly September 11, 2011 at 3:47 pm
Tim, you do raise a good point with your;
I think it is no secret that there has been a war going on between Dessler et al and Spencer et al concerning water vapour/cloud feedbacks*, what with their conflicting papers in GRL a few years ago etc. More recently, I was puzzled why so much emphasis in this war was concerning the feedback radiative effects, when, if we are to believe the Trenberth cartoon, by far the greatest surface cooling comes from thermals and evapotranspiration (~60%). I had several exchanges with Roy Spencer, suggesting that a small change in “convective factors” would intuitively have more negative feedback effect than was being researched in the feedback radiative effects. The outcome was that he agreed it was important, and that it was taken into account in the GCM’s, but, by reading between the lines, everyone was too busy trying to find the radiative stuff to worry about actually quantifying “convection”.
Perhaps the sceptics do not take the cartoon too seriously, and anyway the 390 W/m^2 is not all that important, it is the net heat loss that is, some of which goes directly to space BTW.
Here is a much more popular diagram, that is all over the NASA divisions etc, that does not even try to portray back radiation:
http://science-edu.larc.nasa.gov/EDDOCS/images/Erb/components2.gif
*That Dessler is desperate not to lose!
Tim Folkerts @ur momisugly September 12, 2011 at 4:06 pm
Groan, since what has been discussed only applies to a transparent atmosphere, let’s move on to what happens, per quantum theory, with opacity.
The S-B emission from the surface is hemispherical, but is instantaneously halted by the GHG’s etc in a rather complicated way. Photon free path length (FPL) will vary considerably according to the Planck energy distribution, the water vapour level, (0 to ~4%), lapse rate, and other stuff, together with heap big thermalization of O2 and N2. For consideration purposes, let’s hazard that the FPL varies between 100m and 1,000m. At the moment of initial absorption, the photons cease to exist, but others, not necessarily of the same energy level will be emitted, some as a consequence of collisions. Some might be from relatively high altitude, but the majority will be close to the surface, because as has been demonstrated earlier, most of the radiation is “sideways” through 360 degrees. Clearly at this point, S-B is no longer relevant, and “reemission” becomes spherical. All of this is a continuous process of course.
Because EMR is weird, (and not HEAT), all that S-B sideways stuff is part of the 390 W/m^2 but doesn’t transfer HEAT. A similar situation exists with the spherical reemissions; a significant proportion whizzes around sideways, doing nothing. So, let’s guess that from about 1m up from the surface, up (and down) flux has nothing to do with S-B. (which applies nominally to black bodies)
Bob,
I think we have gotten to (or perhaps gotten well beyond) the point where we can be productive in this setting. I agree with many things you said, but sorting out the precise details would be ineffective here.
Just to name a few details that are tough to hammer out:
* what does it mean to radiate energy from an area (eg “390 W/m^2”) when dealing with energy coming from a volume (eg radiation from GHGs)?
* in thermodynamics, “heat” is formally defined as a process of energy transfer, akin to “work”. You can’t “transfer work” and you can’t “transfer heat”. Often, heat is understood to mean “thermal energy”. Without more precise agreement on definitions, we will easily talk past each other.
* from outer space, you can detect some of the IR photons that left the ground, so SB has something to do with the radiation anywhere in or above the atmosphere. But deciding how much it has to do wit the radiation is more subtle.
* I don’t think we have really agreed on the simplified situation of a sphere at uniform temperature radiating through a transparent atmosphere. Trying to discuss more complex situations from a shaky foundation is a recipe for failure.