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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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) | |||
Tim Folkerts @ur momisugly November 9, at 6:52 pm
Tim,
You wrote:
I am pretty sure the 80 W/m^2 _is_ the evaporative cooling you speak of. The surface is losing those 80 W/m^2, cooling the surface. When the water re-condenses (in clouds), the air is warmed by that amount. I’m not sure what additional cooling of the surface you are thinking of, or where that energy comes from/goes to.
Tim, I’m shocked! Try this experiment: put on an old t-shirt or something and douse yourself with water. Notice that you will feel colder than if you were dry. It’s called evaporative cooling. In the same way, evapotranspiration cools the surface of the earth, (removes surface heat), and there is a convective warming of the atmosphere. Then, as the water vapour ultimately meets the ambient conditions that promote it, it condenses in clouds, thus releasing latent heat. These are two different things.
Evaporative cooling is exquisitely explained in quantum theory, as I explained above somewhere, and in fact I’m happy to call this example: Quantum mechanics. (rather than theory).
RW and Tim,
Straight up is the shortest distance to space. Any vector other than straight up will encounter more atmosphere. Out of 390 w/m^2 only 40 manages to make it all the way out of the atmosphere and most of that will be going straight up. Why?? Because the densest part of the atmosphere is next to the ground. Any almost horizontal vectors will extunguish fairly quickly due to the density. As the angle from horizontal increases the density decreases as does the distance to space.
There are actually 2 computations for SB:
http://www.spectralcalc.com/blackbody_calculator/blackbody.php
On this page for 288k with 1 for emissivity it gives both radiant emittance at 390 and radiance at 124. RE is for the full hemisphere. R is for a specific amount of angle.
http://en.wikipedia.org/wiki/Irradiance
Notice a more realistic .99 emissivity is only 386, not 390!! How much was Trenberth missing again?? What a sad little man. Then there is the issue of the atmosphere’s emissivity. Probably down around .7 which would only give 270 instead of 333 coming back down. As the atmosphere ISN’T as warm as the surface usually, even that number is inflated!!!!
Tim, less than half will be going in the direction of earth as soon as you clear any obstructions on the horizon. Remember the earth is round and finite, not an infinite flat surface to intersect the radiation. The higher up the less is going in the direction of earth in the cone.
RW @ur momisugly November 9, at 7:04 pm
Hi RW,
The issue I’ve tried to explain is that in an absorptive atmosphere, only part of the initial S-B surface emission and its progeny escapes directly to space. Well, other than those recalcitrant naughty photons escaping through the so-called window by virtue of their particular wavelengths. The stuck-at-home guys are repetitively annihilated by absorption in the GHG’s, at various layers of altitude. A very large proportion of them are annihilated in the lateral directions, at all levels until approaching the TOA, and by proper vector analysis there are both vertical AND horizontal components. The greater horizontal components of the 396 W/m^2 and its progeny are constantly there, and cannot be part of the vertical flux, yet they ALL evolve as PART OF the S-B initial surface emission in the context of radiative emissions from the surface.
To that must be added the radiative effects of evapotranspiration and thermals. (per Trenberth terminology)
Bob is shocked: “Try this experiment: put on an old t-shirt or something and douse yourself with water. Notice that you will feel colder than if you were dry. It’s called evaporative cooling. ”
Yes, I know about evaporative cooling. I did exactly what you suggest one summer while driving through the desert in a car with no air conditioning.
I am simply saying that “evaporative cooling” is synonymous with the 80 W/m^2 of energy leaving the surface. The oceans are the ‘wet t-shirt around the earth’, cooling the surface of the earth by 80 W/m^2 (well, you do need to also include transpiration from plants). You seem to be saying that there is some OTHER evaporative cooling beyond the evaporative cooling already in the diagram.
“In the same way, evapotranspiration cools the surface of the earth, (removes surface heat), and there is a convective warming of the atmosphere. “
These are two different things, each of which is counted in the Trenberth diagram. Evapotranspiration removes ~ 80W/m^2 from the surface and deposits 80 W/m^2 into the atmosphere. Convection removes ~ 20 W/m^2 from the surface and deposits ~ 20 W/m^2 into the atmosphere.
Then, as the water vapour ultimately meets the ambient conditions that promote it, it condenses in clouds, thus releasing latent heat. These are two different things.
I’m not exactly sure which two things you are calling different.
Latent heat and convection are certainly two different things. Evaporation and condensation are two aspects of the same thing. If evaporation removes 80 J of energy, then conservation of energy requires that condensation deposits 80 J of energy somewhere else, namely the atmosphere.
Kuhnkat exclaims: “Probably down around .7 which would only give 270 instead of 333 coming back down. As the atmosphere ISN’T as warm as the surface usually, even that number is inflated!!!!
Once you calm down from your 4 exclamation points, you might want to switch to contemplating a few scientific points. In particular, the emissivity of clouds is very close to 1 and clouds cover ~ 2/3 of the earth. If the cloud base was 1 km up, the average temperature would be about 6 K cooler than the surface. Such clouds would radiate on the order of 5.67e-8 * (288K- 6K)^4 = 358 W/m^2.
Clouds come in all sorts of types and altitudes, but some of them clearly radiate well above 333 W/m^2, while areas of clear sky radiate less than 333 W/m^2. It sure SEEMS the average of numbers above 333 and below 333 could average to 333. Do you have clear support for your contention that the global average is indeed below 333 and that the number is inflated as you claim?
Tim Folkerts @ur momisugly November 9, at 7:39 pm
Tim,
So 66 goes up and 330 goes sideways?
You can only sum two different sources of EMR when they have the same sign. If you add -66 to + 66 you have no heat transfer.
BTW, the 396 from the surface is isotropic, so a single square metre of ground will not radiate into just the cube above it.
Oh and your analogy was for a black body anyway Tim
Bob,
“The issue I’ve tried to explain is that in an absorptive atmosphere, only part of the initial S-B surface emission and its progeny escapes directly to space. Well, other than those recalcitrant naughty photons escaping through the so-called window by virtue of their particular wavelengths. The stuck-at-home guys are repetitively annihilated by absorption in the GHG’s, at various layers of altitude. A very large proportion of them are annihilated in the lateral directions, at all levels until approaching the TOA, and by proper vector analysis there are both vertical AND horizontal components. The greater horizontal components of the 396 W/m^2 and its progeny are constantly there, and cannot be part of the vertical flux, yet they ALL evolve as PART OF the S-B initial surface emission in the context of radiative emissions from the surface.”
I know what you’re saying, but I think you’re overlooking that the average flux is still really perpendicular to the surface even though on a photon by photon basis it’s going all over – much of it in a more horizontal upward direction as you describe. Remember, the surface to space direct transmittance or ‘window’ is the amount of surface radiative flux that passes straight through the atmosphere to space as if the atmosphere wasn’t even there. There is no distinction or stipulation that it is only from perpendicular or directly upward emitted photons. It is true that a more horizontally upward emitted photon has longer path to the TOA, resulting in a greater chance of being captured by the atmosphere, but this in and of itself does not affect what percentage of the flux ends up going straight to space.
Let me ask you this. The upward emitted half of isotropic emission from the atmosphere that ultimately escapes to space – is it incorrect to show or calculate this as an all upward flux from the atmosphere to space? Some of the more sideways emitted upward photons have to travel through more atmosphere in order to get to the TOA, right? But they ultimately still get there, right? How is this really any different from the emission from the surface?
I think what you’re arguing would have some validity if the Earth’s surface was a flat plane, but the fact that it’s a sphere will no ‘ends’ in any direction makes it moot.
Bob,
I wrote:
“It is true that a more horizontally upward emitted photon has longer path to the TOA, resulting in a greater chance of being captured by the atmosphere, but this in and of itself does not affect what percentage of the flux ends up going straight to space.”
What I meant was the so-called ‘window’ transmittance accounts for the equally multi-directional emission from the surface.
Bob,
You do know that the so-called ‘window’ transmittance or direct surface to space transmittance does not just include the parts of the emitted spectrum that are mostly transparent to the atmosphere (i.e. between about 8-13u), but also includes the non-transparent components, a portion of which still passes directly to space in the non-saturated bands, right?
RW @ur momisugly November 10, at 3:39 pm
RW, you wrote in part:
Let me ask you this. The upward emitted half of isotropic emission from the atmosphere that ultimately escapes to space – is it incorrect to show or calculate this as an all upward flux from the atmosphere to space? Some of the more sideways emitted upward photons have to travel through more atmosphere in order to get to the TOA, right? But they ultimately still get there, right? How is this really any different from the emission from the surface?
No, they do not ultimately still get there. In fact, most of them are annihilated quite close to the surface. If you study Trenberth’s diagram you will see that 161 W/m^2 of descendants of surface heat loss is claimed to leave the TOA. (239 less 78 solar absorbed by atmosphere = solar absorbed by surface).
So, whatever happened to the 396 + 80 + 18 leaving the surface?
Have a play with the MODTRAN calculator looking down from various heights.
http://geoflop.uchicago.edu/forecast/docs/Projects/modtran.html
Bob,
“No, they do not ultimately still get there. In fact, most of them are annihilated quite close to the surface.”
I’m aware that most of the absorption happens close to the surface, but are you trying to say it’s not possible for a more sideways emitted photon from the surface to travel directly to space without being capture by the atmosphere? I realize of course the chances a photon of the same wavelength getting captured increases the longer the distance to the TOA, but that’s separate issue relative to the global average.
“If you study Trenberth’s diagram you will see that 161 W/m^2 of descendants of surface heat loss is claimed to leave the TOA. (239 less 78 solar absorbed by atmosphere = solar absorbed by surface).”
All of the post albedo of 239 W/m^2 gets to the surface, albeit not all directly. The 78 W/m^2 designated as ‘absorbed by the atmosphere’ does not mean that of the 239 W/m^2 flux leaving, 78 W/m^2 are from that designated as being ‘absorbed by the atmosphere’. This is the fundamental problem with his diagram – it doesn’t show net energy flux from the TOA to the surface and from the surface to the TOA, which is what really matters. All the energy that enters the atmosphere, from the Sun or the surface, is either on a path back to the surface or out to space.
While I certainly agree that not all the post albedo gets to the surface as SW radiation, and some of it is absorbed in the atmosphere (mostly by clouds). While it’s entirely possible for some of these absorbed photons to exit at the TOA without ever reaching the surface, all that is happening in this case is post albedo energy absorbed in the atmosphere from the Sun is being traded off for surface LW flux absorbed in the atmosphere that would otherwise have to be leaving the planet and subsequently getting to surface indirectly in this way as a result. The bottom line is by definition all the post albedo absorbed in the atmosphere enters the thermal mass of the planet, which is mostly surface but part atmosphere as well.
“So, whatever happened to the 396 + 80 + 18 leaving the surface?”
It’s all accounted for when the full post albedo is getting to the surface. 396 + 80 + 17 = 493 and 161 + 333 (78 + 80 + 17 + 157) = 494. Subtract out Trenberth’s phony extra watt and you have 493 W/m^2 entering the surface from the atmosphere and 493 W/m^2 leaving the surface into the atmosphere.
Tim,
you accept that only 40 w/m^2 makes it past the dense low atmosphere and eventually straight to space. The clouds do a really good job of irradiating mostly that same dense low band of atmosphere. Remember, over half of everything absorbed goes back up before getting to the ground!! Your point should have been that I should have taken into account that it would be somewhere in between 270 and 358. Of course, 1k or below clouds do not necessarily WARM as anyone who is around water, fog, and low overcast will tell you, they COOL!!
So, my 270 is only appropriate where there is no cloud and your 358 is only appropriate when the cloud temps are really as high as 288!! The averages are really poor. Of course, my 270 is based on the same temp as surface so is an overestimate and probably not as far from reality as you are trying to make it!! Or NOT!! The .7 or so is an AVERAGE so includes clouds!!
http://www.xylenepower.com/Emissivity.htm
Tim Folkerts @ur momisugly November 10, at 8:56 am
Tim, you wrote in part:
* Evaporative cooling from the surface results in warming of the air at the interface and consequent additional convective activity.
* The process involved is that higher energy water molecules escape, leaving a greater proportion of lower energy molecules behind. (BTW there is a mass transfer to the atmosphere)
* The water molecules eventually end their convective travel upwards when the ambient conditions result in their condensation in clouds. This results in further warming.
* The latter process involved is latent heat loss, due to material phase change.
These are two entirely different processes, and there are also at least two other associated processes which are probably relatively trivial.
* Cooling of the surface by precipitation from a colder region.
* Warming of the air and surface by air friction and impact of precipitation.
Also, as a bit of a “David Karoly”, shortly after the climategate scandal, over at RealClimate, Gavin Schmidt ran a period of allowing many sceptical comments to be posted. (long-since abandoned). I raised the topic of evaporative cooling and latent heat of condensation there, and my recollection is that no one, zilch, disputed the two different processes. However, the message I got, very strongly was that latent heat loss in the clouds is the big deal, and that evaporative cooling was small by comparison. Of course those assertions were without any data or relative numerical foundation. (and I have no idea of their relative scales).
But actually, this is all wandering away from the paradox that I raised.
RW @ur momisugly November 10, at 6:57 pm
A quickie:
Thankyou RW for your long post, but right at the moment I don’t want to grind through the detail. Please try to understand that the Trenberth alleged 396 W/m^2 surface emissions a la S-B are isotropic. As I understand you, you intuitively think that this is somehow converted to a normal (vertical) presentation. But, IF the atmosphere were indeed transparent, then such an integration that you visualize would be arguable, what with the horizontal stuff escaping directly to space. (if viewed from space with an encompassing wide-angle instrument). However we have an absorptive atmosphere, and a lot of stuff don’t make it to space.
Furthermore, my purpose in asking “what happens to the 396 from the surface” was to get you thinking that indeed it substantially dies, and does not head-out vertically to space as you seemingly wish to say it does.
Starting from a one layer model, going to 2, 3..and finally a multilayer model of chicken wire.
Applying it to the atmosphere gave coherent results.
Applying it to K&T and the like gave different results.
http://www.ilovemycarbondioxide.com/pdf/IRabsW27102011.pdf
Let’s go back to the top …
Bob, you claim “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.”
Could you provide a summary of what Trenberth actually claims from some paper? As far as I know, he only claims the surface radiates 396 W/m^2, which is (as far as I know) exactly in accord with SB calculations. I doubt he claims that the radiation higher up is exactly SB in nature (since it involves radiation from many places that that are many different emissivities, rather than from a single opaque surface). I suspect this is a bit of a strawman, attributing to Trenberth conclusions that he never specifically made.
Bob, thank you much for the link the paper by Joseph Reynen in France. Seems others have come to the same conclusion as being discussed above, and, in much more detail. That will take a while to absorb. Great reference, will use it. Also noticed his immediate jump to net transfers. Bravo!
BTW, this was one great post. I rank it at the very top being the real science.
Tim Folkerts @November 11, at 12:54 pm
Tim,
You are welcome to read the 2009 paper a link for which is in the first paragraph of the article. The diagram shows the surface emission as a vertical arrow into the high clouds other than the 40 directly to space. The earlier 1997 version is similar, and is used in the IPCC reports of 2001 & 2007
You are also welcome to look at the following sketch again:
http://bobfjones.wordpress.com/2011/10/29/quick-sketch-for-trenberth-cartoon/
Bob, love this statement in the paper:
“Back radiation is just an algebraic manipulation of the SB law which
happens to give the same result as for the present model when the
atmosphere is taken as one single slab.”
Could not agree more. That is a great way to put it.
Just a note: also found in multiple astronomy sites is the figure of how much more atmosphere mass radiation passes through when near horizontal in direction. It is about 38 times the zenith mass. This is closely related to “Mass Attenuation Coefficient” or “Mass Extinction Coefficient”, logarithmic in nature. Large telescopes greatly rely on such figures to calculate the dimming of radiation at various frequencies and vectors through the telescopes. This has a huge effect in most IR frequencies compared to visible frequencies.
Also took the time to write a numeric integration program over a sphere’s surface of random radiation from a point as seen in three dimensions. Just wanted to make sure some of my statements above might not be mathematically correct. Horror! As I suspected it is a bit harder than you would think at first. You cannot take random directions in +/- x, y, & z and you cannot do the same evenly distributed in theta and phi if in spherical coordinates for the integration will report more radiation going up and down than tangent. The trouble is these distributions both ignore the area weighting on a sphere being much smaller at the poles. Thanks to Wolfram Research for the proper corrections and it is as stated above, 1/6 exactly in all of the six possible half-axes when normalized to one. Just had to check that. Even though it seems so logical, I was saying it above with no real proof.
JWR @ur momisugly November 11, at 12:00 am
wayne @ur momisugly November 11, at 2:32 pm
Wayne, we have JWR to thank, for the link not me
JWR, thanks for the link. I’ve not had time to properly read it, but it looks interesting
Bob,
“Thankyou RW for your long post, but right at the moment I don’t want to grind through the detail. Please try to understand that the Trenberth alleged 396 W/m^2 surface emissions a la S-B are isotropic. As I understand you, you intuitively think that this is somehow converted to a normal (vertical) presentation. But, IF the atmosphere were indeed transparent, then such an integration that you visualize would be arguable, what with the horizontal stuff escaping directly to space. (if viewed from space with an encompassing wide-angle instrument). However we have an absorptive atmosphere, and a lot of stuff don’t make it to space.
Furthermore, my purpose in asking “what happens to the 396 from the surface” was to get you thinking that indeed it substantially dies, and does not head-out vertically to space as you seemingly wish to say it does.”
I’m a bit confused, as I’m not quite sure where it is we are disagreeing.
For the record, I do not think that the atmosphere is transparent to the more horizontally emitted photons from the surface. I agree that most of them would be captured by the atmosphere relatively close to the surface and relatively few would pass directly to space. I also agree that of the more horizontally emitted photons, a higher percentage of them would be captured by the atmosphere than those emitted more perpendicular to the surface, as a result of the longer distance between the surface and the TOA.
I also agree that of the 396 W/m^2 emitted radiatively at the surface according to Trenberth, not all of this makes it to the clouds before much of it is absorbed by the atmosphere and subsequently re-emitted by the atmosphere. No doubt that there are multiple absorptions and re-emissions along the way (either out to space or back to the surface).
So tell me, where is it we are disagreeing?
Ditto the thanks JWR. ( seems a maximized window would help ☺)
To follow-up on my previous post, I am thinking that much of the discussion has been focused on a side-track.
Bob says: “Furthermore, my purpose in asking “what happens to the 396 from the surface” was to get you thinking that indeed it substantially dies, and does not head-out vertically to space as you seemingly wish to say it does.”
I think we can all agree that (give or take a few W/m^2 in all the following numbers) ~ 396 W/m^2 of IR photons get radiated by the surface of the earth, and that these specific photons head out pretty nearly isotropically from the surface.
Beyond that, I think we agree that 356 W/m^2 of that 396 W/m^2 gets absorbed before escaping to space (either by GHGs or by clouds), while 40 W/m^2 gets emitted to space.
To a large extend, that is all that really matters — energy leaves the surface and energy enters the atmosphere. The claim (made by Bob in his post)
is, in my opinion, a red herring. There is no claim of 396 W/m^2 continuing up through the atmosphere. There is no “alleged 396 W/m^2 of upward flux” anywhere but at the surface. Any details withinthe atmosphere are glossed over in the Trenberth diagram. This is a “black box” approach (not “black body”) that ignores the details of the atmosphere.
Specifically, the diagram makes absolutely NO attempt to describe the energy flows by specific means at specific altitudes (other than the very generic claim that the NET flow at the TOA is ~ 0.9 W/m^2). It is easy to hypothesize/assume/conclude that the higher you go, the fewer LWIR photons from the surface you will find, but many of them will be replaced by LWIR upward from the atmosphere. To the extent that the air is the same temperature as the ground, the net change in upward LWIR will be rather small.
If I had to make a stab at the flow 100 m above the ground, I would say something like
The GHGs absorb LWIR isotropically; the GHGs emit LWIR isotropically. To the approximation that the atmosphere and the ground are the same temperature, the absorption of upward LWIR by GHGs and the emission of upward LWIR by GHGs will very nearly cancel out.
RW @ur momisugly November 11, at 6:10 pm
Sorry RW if I misunderstood, but I thought you inferred somewhere that the isotropic radiation from the surface was vertical.