Guest post by Bob Fernley-Jones by Bob Fernley-Jones AKA Bob_FJ
CAUTION: This is written in Anglo-Oz English.
Here is the diagram as extracted from their 2009 paper, it being an update of that in the IPCC report of 2007 (& also 2001):
The unusual aspect of this diagram is that instead of directly showing radiative Heat Transfer from the surface, it gives their depiction of the greenhouse effect in terms of radiation flux or Electro-Magnetic Radiation, (AKA; EMR and a number of other descriptions of conflict between applied scientists and physicists). EMR is a form of energy that is sometimes confused with HEAT. It will be explained later, that the 396 W/m^2 surface radiation depicted above has very different behaviour to HEAT. Furthermore, temperature change in matter can only take place when there is a HEAT transfer, regardless of how much EMR is whizzing around in the atmosphere.
A more popular schematic from various divisions around NASA and Wikipedia etc, is next, and it avoids the issue above:
- Figure 2 NASA
Returning to the Trenberth et al paper, (link is in line 1 above), they give that the 396 W/m2 of EMR emitted from the surface in Fig.1 is calculated primarily by using the Stefan–Boltzmann law, and global year average conditions. Putting aside a few lesser but rather significant issues therein, it is useful to know that:
1) The Stefan-Boltzmann law (S-B) describes the total emission from a flat surface that is equally radiated in all directions, (is isotropic/hemispherical). Stefan found this via experimental measurement, and later his student Boltzmann derived it mathematically.
2) The validity of equally distributed hemispherical EMR is demonstrated quite well by observing the Sun. (with eye protection). It appears to be a flat disc of uniform brightness, but of course it is a sphere, and at its outer edge, the radiation towards Earth is tangential from its apparent surface, not vertical. It is not a perfect demonstration because of a phenomenon called limb darkening, due to the Sun not having a definable surface, but actually plasma with opacity effects. However, it is generally not apparent to the eye and the normally observed (shielded) eyeball observation is arguably adequate for purpose here.
3) Whilst reportedly the original Stefan lab test was for a small flat body radiating into a hemisphere, its conclusions can be extended to larger areas by simple addition of many small flat bodies of collectively flat configuration, because of the ability of EMR waves to pass through each other. This can be demonstrated by car driving at night, when approaching headlights do not change in brightness as a consequence of your own headlights opposing them. (not to be confused with any dazzling effects and fringe illumination)
4) My sketch below demonstrates how radiation is at its greatest concentration in the lateral directions. It applies to both the initial S-B hemispherical surface radiation and to subsequent spherical radiation from the atmosphere itself.
5) Expanding on the text in Figure 3: Air temperature decreases with altitude, (with lapse rate), but if we take any thin layer of air over a small region, and time interval, and with little turbulence, the temperature in the layer can be treated as constant. Yet, the most concentrated radiation within the layer is horizontal in all directions, but with a net heat transfer of zero. Where the radiation is not perfectly horizontal, adjacent layers will provide interception of it.
A more concise way of looking at it is with vectors, which put simply is a mathematical method for analysing parameters that 
possess directional information. Figure 4, takes a random ray of EMR (C) at a modestly shallow angle, and analyses its vertical and horizontal vector components. The length of each vector is proportional to the power of the ray, in that direction, such that A + B = C. Of course this figure is only in 2D, and there are countless multi-directional rays in 3D, with the majority approaching the horizontal, through 360 planar degrees, where the vertical components also approach zero.
6) Trenberth’s figure 1 gives that 65% of the HEAT loss from the surface is via thermals and evapo-transpiration. What is not elaborated is that as a consequence of this upward HEAT transfer, additional infrared radiation takes place in the air column by virtue of it being warmed. This initially starts as spherical emission and absorption, but as the air progressively thins upwards, absorption slows, and that radiation ultimately escapes directly to space. Thus, the infrared radiation observable from space has complex sources from various altitudes, but has no labels to say where it came from, making some of the attributions “difficult”.
DISCUSSION; So what to make of this?
The initial isotropic S-B surface emission, (Trenberth’s global 396 W/m2), would largely be absorbed by the greenhouse gases instantaneously near the surface. (ignoring some escaping directly to space through the so-called “atmospheric window”). However, a large proportion of the initial S-B 396 surface emission would be continuously lateral, at the Trenberth imposed constant conditions, without any heat transfer, and its horizontal vectors CANNOT be part of the alleged 396 vertical flux, because they are outside of the vertical field of view.
After the initial atmospheric absorptions, the S-B law, which applied initially to the surface, no longer applies to the air above. (although some clouds are sometimes considered to be not far-off from a black body). Most of the air’s initial absorption/emission is close to the surface, but the vertical distribution range is large, because of considerable variation in the photon free path lengths. These vary with many factors, a big one being the regional and more powerful GHG water vapour level range which varies globally between around ~0 to ~4%. (compared with CO2 at a somewhat constant ~0.04%). The total complexities in attempting to model/calculate what may be happening are huge and beyond the scope of this here, but the point is that every layer of air at ascending altitudes continuously possesses a great deal of lateral radiation that is partly driven by the S-B hemispherical 396, but cannot therefore be part of the vertical 396 claimed in Figure 1.
CONCLUSIONS:
The vertical radiative flux portrayed by Trenberth et al of 396 W/m^2 ascending from the surface to a high cloud level is not supported by first principle considerations. The S-B 396 W/m^2 is by definition isotropic as also is its ascending progeny, with always prevailing horizontal vector components that are not in the field of view of the vertical. The remaining vertical components of EMR from that source are thus less than 396 W/m^2.
It is apparent that HEAT loss from the surface via convective/evaporative processes must add to the real vertical EMR loss from the surface, and as observed from space. It may be that there is a resultant of similar order to 396 W/m^2, but that is NOT the S-B radiative process described by Trenberth.
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ADDENDUM FOR AFICIONADOS
I Seek your advice
In figure 5 below, note that the NIMBUS 4 satellite data on the left must be for ALL sources of radiation as seen from space, in this case, at some point over the tropical Pacific. The total emissions, amount to the integrated area under the curve, which unfortunately is not given. However, for comparison purposes, a MODTRAN calculator, looking down from 100 Km gives some interesting information for the figure, which is further elaborated in the tables below. Unfortunately the calculator does not give global data or average cloud/sky conditions, so we have apples and pears to compare, not only with Nimbus, but also with Trenberth. However, they all seem to be of somewhat similar order, and see the additional tabulations.
| Compare MODTRAN & “Trenberth”, looking down from 2 altitudes, plus Surface Temperature | ||||
| Location | Kelvin | 10 metres | 100 Km. | (Centigrade) |
| Tropical Atmosphere | 300K | 419 W/m^2 | 288 W/m^2 | (27C) |
| Mid-latitude Summer | 294K | 391 W/m^2 | 280 W/m^2 | (21C) |
| Mid-latitude Winter | 272K | 291 W/m^2 | 228 W/m^2 | (-1C) |
| Sub-Arctic Winter | 257K | 235 W/m^2 | 196 W/m^2 | (-16C) |
| Trenberth Global | 288K ? | 396 W/m^2 | 239 W/m^2 | (15C ?) |
| Compare MODTRAN & “Trenberth”, looking UP from 4 altitudes: W/m^2 | ||||
| Location | From 10 m | From 2 Km | From 4Km | From 6Km |
| Tropical Atmosphere | 348 | 252 | 181 | 125 |
| Mid-latitude Summer | 310 | 232 | 168 | 118 |
| Mid-latitude Winter | 206 | 161 | 115 | 75 |
| Sub-Arctic Winter | 162 | 132 | 94 | 58 |
| Trenberth Global | 333 Shown as coming from high cloud area (= BS according to MODTRAN) | |||
wayne, why would you think absorption features of the gases is not important? Sun energy reaching, number of particles (which, if we are talking about similar gases, correlates to first order with mass), and ability to absorb are all key ingredients. [Even the curvature of the planet is important.] Do you have specific evidence to suggest absorption is not an issue?
Mass is one tool we can use within the context of a particular system (where a particular “attenuation coefficient” might have a specific value), but generally mass in one atmosphere would not directly match mass in another without some scale factor to account for these other variables.
Keep in mind that near the ground, with lots of hills and mountains, there will be a lot more chances for “re-radiation” to hit the surface.. and there is no need to abandon re-radiation (which includes a downward slope).
A mostly flatter (temp) lapse rate curve results from the homogeneity of Mars atmosphere. Do you have samples of those curves that lead you to think absorption characteristics are not important?
>> the talk in my comments above of NET energy never returning to the surface once it has entered and been absorbed by the atmosphere
Saying “net” and “not returning” just means that we get a (net) distribution profile that is most intense at the excitation boundary (the planet) and trails off towards the other end.
Just think of putting your hand against a heating coil. The closer you are, the hotter it feels. This is true regardless of the contents of the atmosphere; however, this doesn’t negate the concept of greenhouse “trapping” (absorption and emission in all directions).. ie, raising the overall profile higher the more greenhouse effect there is.
Also, because of the curvature of the planet, the higher you get, the lower is the percentage of your potential re-emission direction that intersects the planet. Combine this effect with the lower density higher up (meaning that the photon is less likely to be intercepted). These two things (combined with absorption/radiation) are consistent with the table Spector provided a little higher up. [Note that if the planet was flat and infinite, a particle would always have half of its potential emission directions intercepting the planet no matter how high the particle was.]
Concerning climate models, I suspect they model particle interactions (or wave equations) at a fairly low level. Solving physics time-dependent differential (or short time step difference) equations at many subdivisions of the planet’s atmosphere (with the power of supercomputing) takes the need for intuition or for coarse approximations out of the picture. That “lateral” effect (and absorption, density, and everything else) would be factored in automatically.
I want to point out a few things
>> At some altitude like 2 km, I would suggest that very close to ALL of the SB surface emission in certain bands has been absorbed.
Whether 2 km is accurate or not would depend on the density of ghg in the atmosphere that could absorb in those bands. I mention this since Mars was mentioned. On some planets (maybe within our solar system or not), there could be no such value where one could say “close to ALL” because of the thinness of the ghg. Put differently, on one group of planets, we might be able to say “close to ALL” even as high up as several kilometers, but on other planets we might not be able to say “close to ALL” even when on ground level. This effect may play a role in the temp distribution in the atmosphere and help explain increasingly strong “skewing” of the profile beyond (or some bending at) some height. We have some form of “saturation” and then a trailing off much more quickly.
Additionally, as stated in an earlier comment, the curvature of the planet (aided by decreasing gas density with height) suggests that being higher in the atmosphere increases the chances significantly that the next emission will direct a photon essentially away from the planet even if the net direction is partially “downward” (the photon straight line path may initially approach the planet a little but then hit minimum nearness and diverge and keep diverging towards infinity).
>> Conversely, outside the bands where GHGs absorb, very nearly 100% of the surface IR photons will still be heading generally upward.
Note on “generally upward”: In the limit towards infinite distance, all such out-of-band surface emitted photons that don’t run into dust, mountains, etc, will essentially follow a path that approximates a radial path from the planet. They will all eventually be seen as mostly moving “up” from the center of the planet. [Eg, Bob’s discussion of some sun rays leaving from the edge of the sun laterally is approximated by saying that the ray is moving up from the center of the sun.]
Note: The chances of emitting in such a band is a probability calculated within the quantum mechanics framework (I have some but not too much experience here). The narrower are the absorption bands (and taking into account LTE), the greater the likelihood the next emission from a molecule will be “upwards”. To relate to the Mars example, having enough total ghg band coverage (aka, distinct ghg) greatly improves the odds of an emission _from the surface or other solid body_ being absorbed and not just passing through into space unimpeded.
>> For the sake of argument, .. GHG’s absorb 1/3 .. 264 W/m^2 out of 264 of surface .. photons .. will pass upward
Note: 264 is 2/3 of 396. This just says that at the initial “hemisphere” emission on the surface (assuming all such angles have equal probability) a full 2/3 will simply leave the planet unable to be impeded (ie, absorbed).
Note: Of the other 1/3 absorbed in the near atmosphere, a fair number of photons will then later be in a “re-emission” mode where we look at the full spherical direction, of which approximately half will be right back at the planet and will be absorbed essentially no matter the frequency.
Note: To improve this simple model, we would have to factor in the distance and direction the gas molecule would travel (including, on average, losing some of that energy with LTE) before the next photon release. So a “second” round of releases, on average, would be some distance from the planet that may not coincide with the average distance of (first) absorption of planet surface photons.
Note: FWIW, the “lateral” effect plays a role to the extent it lowers the average distance from the earth of a further collision. If we could somehow create a field that skews emissions to be mostly “up” then the average distance to next absorption would be higher.
>> On the other hand, if you look at MODTRAN output looking down from 2 km, you do not see 100% missing from any bands. Why is that? Simply because those are also the bands where GHGs emit IR photons.
Note: So even though certain bands get almost entirely caught in the lower 2 km. They eventually get re-emitted _always_ at that same band range, but, being higher up, now have an improved change to make it out of the 2 km (enabling it to be measured ..so to “fill up” that would-be missing band).
>> By the time you are 2 km up, the gas is getting cooler, and the power is a little less in the GHG absorption/emission bands.
Note: Although these re-emissions just mentioned are in the “same band”, because of average lower energy higher up, they will be redistributed within that band towards the lower energy ranges of that band as compared to the distribution we’d see nearer to the ground.
>> Note: 264 is 2/3 of 396. This just says that at the initial “hemisphere” emission on the surface (assuming all such angles have equal probability)
The angle probability part at the end is irrelevant.
>> To relate to the Mars example, having enough total ghg band coverage (aka, distinct ghg) greatly improves the odds of an emission _from the surface or other solid body_ being absorbed and not just passing through into space unimpeded.
What I mean here is that if a planet mostly has just one type of ghg, as Tim Folkerts’ explanation implies, a larger fraction of all surface photons would simply leave unimpeded into space in comparison to a planet that had more energy band coverage because the atmosphere had a wider array of ghg in significant numbers. This may address a complaint wayne mentioned recently about Mars’ mostly CO2 atmosphere. So for Mars, losing half their CO2 but doubling up on some other gases (even though the new number of total gas particles would still be much less) might lead to a greater greenhouse effect. [Maybe we would need to more than just double up on some of these trace gases, but we might still remain with many fewer total gas particles for the same gh effect.] And if we reverse the argument with methane playing the role of dominant gas, the same logic would apply (even though CO2 has a weaker gh effect). This invokes the principle of “diminishing returns”.
Jose_X November 13, at 11:51 pm
Jose, you wrote in part:
[1] Actually, both Trenberth and the MODTRAN model calculator do not show that, and if the 396 W/m^2 S_B surface emission did make it to high altitude/space, then there would be more energy leaving the Earth than was received from the Sun.
[2] All parts of the atmosphere emit photons isotropically. In successive layers approaching the TOA, some photons can escape more readily to the normal because of shorter path lengths, whereas the more horizontal are absorbed. When closer to the TOA, even those approaching the horizontal will also escape to space. Thus an observing instrument needs a wide angle field of view to pick-up all the radiation. (it is not all normal)
Regarding the MODTRAN calculations presented above, they seem perfectly logical and understandable.
I) The final column should clearly be labeled “Net IR Energy Flow Out”
II) The only 2 numbers on this chart that Trenberth is dealing are the first row (428.610 up – 365.496 down = 63.114 up at the surface) and the last row (292.334 up at the TOA). These numbers don’t quite match his numbers, but his numbers global averages and these are clear sky tropics, so there is no reason to expect them to match exactly.
III) The true Net Flow should be 0 W/m^2 (or ~ 0.9 W/m^2 if you want to trust Trenberth’s estimate of net warming) at any level. At TOA, the ONLY way to send energy up is IR radiation, so this should (approximately) match the net inflow. The global average inflow is 341 -102 = 239. This is less than the MODTRAN results, but the inflow at the tropics would be larger, so 292 vs 239 seem reasonable.
At the surface, there are many other ways to send energy up, so the net incoming solar (maybe 200 – 250 in the tropics), would be matched by the net outflow (the 63 IR outflow + evaporative outflow + convection outflow).
The higher you go, the less able the other energy flows are (evaporation & convection), so IR must take up the slack.
Wayne says “Dr. Richard Feynman explained all of this some forty years ago…quantum electro-dynamics…”
We seem to be having considerable difficulty with real photons; I don’t think getting into virtual photons is going to make things clearer. 🙂
Jose_X, I think I agree with pretty much all of your additional notes on my comments. You seem to be preemptively addressing a variety of issues that others might have. It is always tough to know how much detail to put in, and how much to wait until people have additional questions. And of course, things that are obvious to me (like 2/3 x 396 = 264) may not have been as obvious as I thought.
Jose_X, I read your tips and don’t even want to consider such complexity at this point. Simplicity first. But thanks anyway. After all, if I don’t already know to consider what you mentioned plus some, there’s probably no hope. ☺
Bob,
[1] You are right that saying “all” didn’t make sense there. “All” might only apply if we were just releasing photons from surface straight into space.
[2] The instruments cover approx a full hemisphere, yup.
Tim Folkerts, yes, I wanted to anticipate questions and take advantage of connecting various ideas/points with each other. Also, I too am shaping and refining my understanding, so the comments were part of Jose_X answering Jose_X’ [I have multiple internal identities. The debates can get really heated sometimes.]
wayne, some of those analysis “tips” were not intended for mere humans. I don’t blame you, but don’t blame me for trying to indulge a moment in the heavens.
Wayne,
Thanks for your suggestions and your offer to collaborate on a new article, but things have become complicated for me in the near term.
I had planned to re-write the article to include some pre-emptive stuff WRT various queries and criticisms in this thread of over 400 comments. I would then offer it to JoNova and/or WUWT. Oddly, if I tie-up all the loose ends it may result in not much traffic. On the other hand, if Anthony agreed to run it, it would be good if R. Gates and Willis Eschenbach among others would comment. (I’m disappointed they have not joined-in). Then if Joanne Nova ran it, it would hopefully stimulate a different audience, and it might provoke the interest of David Evans, her husband.
I was also thinking along the lines of a part 2, concerning some other issues with the Trenberth cartoon, the biggest of which seems to be the alleged 40 W/m^2 radiated directly to space from the surface, which is somewhat more than the remaining 23 of the 63 net leaving the surface absorbed by the atmosphere, and….
However, currently I’m hopping mad that the ACMA (Australian Communications & Media Authority) have sent me a 29 page report rejecting my appeal on complaint originally rejected by the National Radio, the ABC, concerning their so-called “Science Show”. They produced a whitewash that did not actually address the issues I raised. This process has taken over one year. I suspect that it may be political because it covertly criticises junk science from the AAS. (Australian Academy of Science)….. and boy it took them about 6 months to come-up with their 29 page report which they say they wil publish on their website.
I need to do something about this, and to raise a new complaint to ACMA showing how else the “Science Show” is knowingly biased.
I’m also planning to have a break in Tasmania of about a month, and where I’m going, radio contact may not be available.
It’s a bit off-topic, but there is more on my complaint here if you are interested:
http://bobfjones.wordpress.com/2011/03/01/abc-radio-australia%e2%80%a6-misleading-%e2%80%9cscience%e2%80%9d-%e2%80%a6-no-3/
Click on Complaint No.3 for the original complaint but there is also much correspondence.
Any comments I make from here-on for a while are likely to be brief and to the point!
Bob_FJ: “from here-on for a while are likely to be brief”
Hey, no problems. I mainly study science, scientists, and scientist’s logic and methods. That simple. And there is a pot-lot of work left to do in the realm of planetary atmospheric science. Addressing Earth, it is suck in a rut.
Thank goodness there are individuals like yourself willing to keep open eyes, such as admit there is even a 3-D aspect, mostly unaddressed, and bravely analyze it, and for goodness sake not merely follow “the script”.
Later Bob. Hope you endeavors to better that science show are successful. ( You mention Tasmania, enjoy! If I ever got that far south it would include a stop in Adelaide Yacht Club to give young Jessica Watson a big pat on the back. (youngest girl to solo circum-naviagate last year) )
Jose_X @ur momisugly November 14, at 7:20 pm
Very quickly:
So you now agree that radiation from earth is not all normal to the surface?
And earlier:
Erh from memory the average diametr of the Earth is about 12,500 Km. Tim Folkerts suggested that the curvature effect amounted to about 1%, which is trivial compared with the other stuff. MODTRAN suggests that most of the radiative action in the atmosphere is almost over at an altitude of around 10 Km in the tropics. (presumably lower towards the poles). So how about you do a calc for 12,510 squared / 12,500 squared?
Tim Folkerts @ur momisugly November 14, at 1:43 pm
Quickly;
You wrote in part:
Me no understand; the surface emission is isotropic!
Bob, the spectral gap I was referring to is with respect to the wavelengths, not locations or angles. So yes, I agree that the surface emission is indeed isotropic.
Refer to Figure 5 in your post. The envelop for the data seems to be a BB curve with a temperature ~ 295 K. So at ground level, the curve would (very nearly) follow the 295 BB curve. In particular, the scale would be ~ 140 (in the units of the graph) at 15 um.
As the photons travel thru the atmosphere, some will get absorbed by CO2. The deep “bite” around 15 um is (I am pretty sure) due to CO2. In the first km, a big fraction of the 15 um photons would get absorbed. For the sake of argument, lets assume that 99.99 % get absorbed.
If I understand your argument correctly, you would say that only ~ 1/6 of those 15 um photons (about 25 units on the graph) would be heading mostly upward, so the observed result should be ~ 25 for the “brightness” of 15 um photons at 1 km according to your model.
According to my model, about 140 of the 140 units of 15 um photons should be observed heading upward. Even though ~ 0% of the 15 um surface IR photons survive, I say the missing 15 um surface photons are replaced by an approximately equal amount of 15 um photons emitted by the atmosphere.
In other words, I predict approximately no “gap” in the IR spectrum at 15 um when looking down from 1 km. And (I believe) you predict a deep “gap” at 15 um. (Feel free to correct me if I am misinterpreting your model).
>> So you now agree that radiation from earth is not all normal to the surface?
Bob, I’m not sure where you think I said every emission is normal. I said (a) normal appears to be an accurate way to represent (what I anticipate is) the net average effect at ground level; (b) normal is accurate representation if we are doing simple diagrams (engineers use simple box models all the time); and (c) far from earth, as the earth can be approximated by a point, the photons we do find that originated there will be very near a radial line from the center of the planet.
I also told wayne I understood his explanation of your graph and agreed (in different words that) the majority of area on the surface of a sphere is not directly up or even mostly up.
I can understand you misinterpreted Trenberth’s diagram (as I don’t think Trenberth was trying to suggest every emission from the ground was exactly up at 90 degrees or only in the general “up” direction). Yes, that diagram appears to be an oversimplification and not great for learning about the process.
I haven’t changed my mind on any of this.
>> So how about you do a calc for 12,510 squared / 12,500 squared?
When I mentioned the curvature issue, I was not referring to anyone else’s context. It was partly to state that high up it becomes extra unlikely (assuming random direction emissions) a photon will head on a course to hit the planet. The times I mentioned that I was not necessarily thinking up of a figure for altitude, especially since the conversation has been general and even included talk of other planets; however, to demonstrate the point, we would not look at the 10km section since there it wouldn’t be much of an issue. Looking at wikipedia, the exosphere goes up till about 10,000 km. Whatever calculations you wanted me to do for 10 km would not be necessary now. 🙂
Point one:
“Up arrow” can certainly signify the _entire_ upper hemisphere rather than what you (Bob) are calling simply the upper portion of the upper hemisphere. That is a reasonable interpretation for an up arrow, especially in the context of an audience that knows obviously not all emissions are 90 degrees up.
Further, this interpretation makes even more sense as explained in the next point.
Point two (and clarification):
>> (a) normal appears to be an accurate way to represent (what I anticipate is) the net average effect at ground level;
A sensor would likely accept approximately anything from the full half space (think hemisphere) towards which it is oriented. This means that a photon coming quasi-laterally from far away would on average (at a given height, say, very close to the ground) make up for a photon directly under the sensor which was inclined laterally as well.
So by “net” I was looking at this from the perspective of a measuring device, where what matters is the number of photons hitting in a certain (eg) square meter and not the angle any particular photon might have.
I understand you were bothered by this, Bob, but think of the context around this paper and consider no longer thinking there were evil intentions behind it. I am sure many laypeople would go bonkers thinking you are trying to deceive them if they see some of your diagrams intended for other (mechanical) engineers.
Tim Folkerts @ur momisugly November 16, at 4:08 am
So I guess you did not mean to say the following in your previous post; my bold added:
Tim Folkerts @ur momisugly November 14, at 1:43 pm
Quickly,
You wrote in part:
Bob says “So I guess you did not mean to say the following .. ”
Actually, I think I do mean what I said. Or I mean what I thought I said.
Near some wavelengths — for example near 11 um (near 900 cm-1) — the atmosphere is quite transparent. Nearly 100 % of the 11 um photons emitted from ground level (“surface IR photons”) will reach a surface 1 km up.
Near some other wavelengths — for example ~15 um (~ 700 cm-1) — the atmosphere is quite opaque. Nearly 0 % of the 15 um photons emitted from ground level will reach a surface 1 km up.
So the spectrum observed (a la Figure 5) due to photons emitted directly from the surface would be “bright” near 11 um and “dark” near 15 um. The “dark bands” are the “gaps” I was talking about in the spectrum of the IR photons emitted from the ground.
Does that clarify my position?
~~~~~~
And I would really enjoy hearing your reply to my post:
“In other words, I [Tim] predict approximately no “gap” in the IR spectrum at 15 um when looking down from 1 km. And (I believe) you [Bob] predict a deep “gap” at 15 um. (Feel free to correct me if I am misinterpreting your model)”
Ie. i think the spectrum looking down from 1 km up will be ~ identical to the black body spectrum looking down at ground level even if the GHGs in between absorb a lot of IR photons and send them out “mostly sideways”.
Jose_X says:
November 16, 2011 at 11:20 am AND 11:38 am
You wrote in part:
I feel you should either retract that statement or clarify what you meant to say.
I don’t have time to respond to the rest of it.
Tim Folkerts, it seems you are saying at the top half of your last comment (“[a]ctually, I think I do mean what I said”) that if we could hypothetically track the photons to filter out all except the surface photons, then we would get gaps. Then at the bottom, it seems you are saying that actually measuring quasi-near the ground would reveal little distortion away from blackbody of ground.
Here is what I am thinking now, and it offers a way to explain figure 5.
You did say earlier (according to my interpretation) that the varying temperatures across the atmosphere would distort the observed in band spectrum from high above. I would agree with this statement but would add that we also get distortion from the following effect. [Here, “in band” and “out-of band” refer to energy ranges relative to ghg.]
Assume, to get a first order result that we can ignore night and day differences, clouds, etc (I’d have to think more to see what effects these might have on average).
The earth surface receives in band and out-of band radiation X from the sun and essentially only in-band backradiation Y. [For the record (but not used much in the analysis): the X is from immediate sunlight shine while the Y includes leftover atmospheric effects going back into the distant past (negative infinity in time).] This X+Y input gets output as the in band and the out-of band of blackbody spectrum at surface temp.
A) The out-of band makes it to satellite measurement.
B) However, only some of the in-band energy leaving the surface at time t=0 will eventually make it upward as in-band. A part of this surface radiated in band energy at t=0 will make it back to earth at various future times via the Y stream. This means that some of the photon energy released in band at t=0 never make it to space in band because it detours by the earth and gets redistributed to _full_ in surface temp blackbody spectrum. If we were keeping track of in-band energy released at t=0, we would find that some of it makes it to satellite as in band energy at numerous future times, but a large part of it becomes, in small chunks, part of out-of band radiation (at surface temp), each chunk at some future point in time.
So, (A), as out-of band energy radiated from the surface at time t, fairly evenly fills out the out-of band blackbody curve at surface temp as measured by the satellite at about time t. (B), as in band energy radiated from the surface at time t, helps build the out-of band (A) of many future points in time, with the leftover helping to fill out the in-band satellite measurements taken at many future points in time.
Alternatively, we can look at the picture from the point of view of the satellite reception: Out-of band satellite received photons at time t were radiated from the surface at about time t and accounts for energy from sun and that had previously in time been generated as in band from the surface but returned to the surface. In band satellite received photons at time t were radiated as in band in the atmosphere and/or the surface of the earth with “pre-cursor” photons having been generated at a distribution of times in the past (also from the surface and/or atmosphere). [“pre-cursor” is an approximate term that might make sense if we were accounting carefully for energy. Of course there is LTE, etc, to muddy the picture.]
The out-of band at satellite is fairly even and corresponds to surface temp. The in band at satellite is varied (due to radiating at varying altitudes/temps) and falls significantly short of the out-of band curve. This description qualitatively matches the satellite curves shown in fig 5.
I’ll have to think a bit more about these details. What do you think?
@Bob: sorry if that comment upset you. After posting I realized I was carrying over from your comment at your blog site (to Jose Testx), where you are upset at the IPCC. So what you quoted of my reply here doesn’t apply here. Sorry. [I may not get a chance any time soon to read any of the books you mentioned on that other thread at your site. I may reply there later on after I try to read some of what you mentioned.] I thought about clarifying but got distracted with my prior comment to Tim Folkerts.
@Bob: OK, this prior comment only applied to the “there were evil intentions behind it” part.
The part about your engineering diagrams was my way of saying that Trenberth’s diagrams were probably understood by much of his intended audience. A person not an expert in climate science might very reasonably misinterpret the meaning of the arrows in that diagram. If you read the entirety of that comment I wrote, you’ll read that I now think the up arrow more likely than not refers to upper half plane rather than to a region around the surface normal direction (as I think was your interpretation in this article).
>> The earth surface receives in band and out-of band radiation X from the sun and essentially only in-band backradiation Y.
Ooops, X is mostly just out-of band (since in-band would largely be blocked on way down).
And I am talking about _directly_ receiving the radiation from the sun in the sense that the photons were generated in outer space (likely at the sun).
So in my hypothetical energy accounting, I start off by partitioning the earth received radiation into what comes immediately from the sun and what comes from elsewhere (ie, from the atmosphere).
Tim Folkerts @ur momisugly November 16, at 2:55 pm
Tim, look, I’m sorry but you are asking for discussion on something which is a tad complicated and speculative that I don’t have time for, as I explained above.
For instance, the IR radiation seen from space by Nimbus must include that originating from the surface a la S-B, together with thermals, evapotranspiration, atmosphere, and clouds. These things all have difficult gradation issues with altitude. They don’t have labels attached, and there are other bothersome things like lapse rate causing lengthening of wavelength with altitude, and varying H2O vapour level and, and, is that enough?
Bob says: “Tim, look, I’m sorry but you are asking for discussion on something which is a tad complicated … ”
I agree they are complicated. But remember you are the one who started this by questioning the Trenberth diagram and the physics of IR radiation. It is more complicated than drawing a colored circle with an “X” and guesstimating how much IR should go different directions.
A quote from Lord Kelviin comes to mind:
Until we are discussing actual numbers and actual 3D integrals, any discussions we have are destined to be “meagre and unsatisfactory”.