Does the Trenberth et al “Earth’s Energy Budget Diagram” Contain a Paradox?

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)
0 0 vote
Article Rating
669 Comments
Oldest
Newest Most Voted
Inline Feedbacks
View all comments
Ken Methven
October 26, 2011 5:31 pm

“Leave a Reply”?….if only I could!

Bob Fernley-Jones
October 26, 2011 5:34 pm

Unfortunately, an old draft of the article has been posted in errror. I’ve Emailed Anthony, asking for the intended version to be posted. It contains additional information.

R. Gates
October 26, 2011 5:46 pm

There is lot’s to chew on here, but please clarify this statement:
“…temperature change in matter can only take place when there is a HEAT transfer..”
____
If by “temperature” you are referring the average translational kinetic energy, then of course your statement is erroneous, for the temperature of matter (as measured by average translational kinetic energy) can be changed by both heat transfer, AND/OR having work done on it. In the end, the results are indistinguishable. A perfect example is the compression of air when it flows between zones of different pressure…this is work done on the air molecules, their temperatures will rise as they are compressed and the average kinetic translational energy rises from the work done.

October 26, 2011 6:07 pm

Well, in a word: YES!
In a transparent atmosphere and ignoring the curvature of the earth, the average 396 W/m^2 would be constant all the way up. The energy leaves the spherical surface of the earth. By conservation of energy, the energy MUST go somewhere. The energy does not get absorbed (by definition in a transparent atmosphere). Therefore it must leave the from any arbitrary spherical surface above the earth. In the approximation that the earth is not curved (or that we have not gone high compared to the radius of the earth) the same energy must leave upward thru every m^2 of the upper surface as left every m^2 of the surface. Unless you can find a good reason to disagree with euclidean geometry or conservation of energy, the average upward flux of thermal IR from the surface is constant in these circumstances.

Bob Fernley-Jones
October 26, 2011 6:12 pm

R. Gates @ 5:46 pm,
Yes, you are correct, that work can result in heating, or cooling. An oversight on my part, where I was trying to show the difference between heat and EMR

Billy
October 26, 2011 6:17 pm

If i followed your presentation correctly, then it seems to me that you left out an important fact. The radiation lost “sideways” above one point, should be exactly made up by the radiation “sideways” from some other point on the globe. I’m sure that, if you consider the earth as a sphere, these effects all average out and your concern is misplaced. I think that, to the extent that the earth is not a sphere, more surface radiation is absorbed in the atmosphere than in the spherical case.
Billy

October 26, 2011 6:21 pm

“The length of each vector is proportional to the power of the ray, in that direction, such that A + B = C.”
Energy is not a vector! You cannot take components of energy. If a photon with energy 10^-19 J is heading north and at a 45 degree angle above the horizontal, it does not have 0.707 * 10^-19 J of “northward energy” and 0.707 * 10^-19 J of “upward energy”. It simply has10^-19 J of energy. If it passes upward through a surface, it carries 10^-19 J of energy upward thru the surface, not 0.707 * 10^-19 J.

davidmhoffer
October 26, 2011 6:32 pm

An interesting observation about R. Gates. Has anyone else noticed that he pops up almost immediately in any thread where Trenberth’s work is being scrutinized? On other threads he is noticeably absent, but pick a hole in Trenberth’s work and POOF! there’s R. Gates defending it with one argument or another, usualy a tengential one intended to hijack the thread.
BTW R. Gates, you welched on your bet with me.
And I repeat my question from another thread…Do you have on official relationship with Trenberth, and if so, what is it?
Excellent article Bob Fernley-Jones.

wayne
October 26, 2011 6:34 pm

Bob Fernley-Jones, I owe you a debt of gratitude. I have no idea if you just stumbled upon this though on your own or if it is that I have been harping on this on and off here since this spring when I personally realized it. But one thing I can say either way is you explained this hugely better than I have been able to explain it.
Stefan-Boltzmann is a three dimensional law, not one dimensional as so many of the climate related sciences is applying it. That is why the 396 Wm-2 is pure fantasy when speaking of energy net transfer as you have eloquently explained. This one fact in science is why global warming as we speak of it is now over, to me anyway. The maximum affect, or sensitivity, is 1/6th (numerically exact) of what so many of the calculation have been applied using S-B law. Pure radiation loss of energy from the surface is a minor, I’ll repeat, a minor player in Earth’s loss of the solar radiation gathered on the lit side every second of every day, not a major
Please help me to not let this thought go stale. It is one of the saddest states I could every imagine of how the physic branches of science have let this notion not only exist, but to become an accepted view of how energy transfers, always in one dimension, up and down, never accounting horizontally which is 4/6th of what actually occurs with symmetrical cancellation of effects. Any deviation from exactly 4/6th is due to the fact our world is a sphere, not a flat plane and the cooling and thinning of the atmosphere with altitude.
Excellent, excellent, article (paper). To me, the notion of global warming from CO2 died on this 3D thought. Radiation cannot do what they have led us to accept as truth… it isn’t.

doug s
October 26, 2011 6:34 pm

I must say I don’t get any of this really, but what I have always found odd about this line of thinking is, what about kinetic and potential energy transfer of the heat. It would seem to me, that the sun shining and warmth associated converts into water moisture, which rises, forms clouds and falls. It creates wind from pressure variances. So when they start accumulating forcings and adding it all up, in fact that little extra just becomes one days little bit extra rain, just a little bit extra wind. It doesnt add up over the years, it just turns out to be a little more movement day to day.
When someone says climate engine, or weather engine, this is what I think of.

wayne
October 26, 2011 6:44 pm

Tim Folkerts says:
October 26, 2011 at 6:07 pm
Well, in a word: YES!
In a transparent atmosphere and ignoring the curvature of the earth, the average 396 W/m^2 would be constant all the way up. […]

Wrong.
It decrease with temperature with an increase in altitude.

jimmi_the_dalek
October 26, 2011 6:49 pm

“…temperature change in matter can only take place when there is a HEAT transfer..”
Sorry , but no. Temperature is directly affected by absorption of radiation – so how and why are you distinguishing HEAT (in capitals!) from EMR.

Richard Keen
October 26, 2011 6:52 pm

Along with the paradox, the Trenberth et al. diagram also contains a statistical fantasy, which is the obscenely precise value of 0.9 W/m2 for the net absorbed. Considering the multiple W/m2 uncertainty in some of the other numbers (including cloud reflection), that 0.9 W/m2 could only have been extracted from some authorial orifice. A couple of years ago Vincent Gray and I had a short discussion of this statistical uncertainty, along with the unreality of the flat earth, on ICECAP
http://icecap.us/index.php/go/icing-the-hype/the_flat_earth
Thus, the “missing heat”, aka the “net absorbed” on the diagram, could be positive, negative, or zero.

wayne
October 26, 2011 7:13 pm

Tim Folkerts says:
October 26, 2011 at 6:21 pm
“The length of each vector is proportional to the power of the ray, in that direction, such that A + B = C.”
Energy is not a vector! You cannot take components of energy. […]

Energy transfer is a vector and you must take it’s dimensional components when analyzing it in a real world. First study some real three dimensional physics.

Gail Combs
October 26, 2011 7:18 pm

I am no physicist or mathematician so let me see if I have this correct.
1. Energy from a point is radiated in all directions not just up or down. – no problem
2. the “apparent” vertical component of any line that is not straight up is represented by the side of a right triangle and is alway less than the hypotenuse.
However the energy from the point of view of an atom of gas can be thought of as a “packet” Upon absorption it kicks the atom to a higher energy level. when the atom falls back to the original energy level the exact same amount of energy “packet” is emitted.
I view it as sort of a pinball machine with the energy “packet” the ball getting whacked back and forth from atom to atom until it finally “escapes” The energy “packets” value does not change only its direction and the amount of time of travel.
If the energy “packet” is not the right “flavor” the atom will not “eat it” and therefore will not be absorbed and re-radiated. So it bounces off the atom. If it does not hit any atoms it just head out to space even if it is at a very long tangent with a very low angle to the ground.
Have I blown the physics somewhere?
(I really hate defending that model by the way)

DocMartyn
October 26, 2011 7:19 pm

One would assume that there would be some work term. The conversion of saline into airborne fresh water, then its transport onto land would be included, but no.
There is really no point in attempting to use such diagrams as this to calculate anything. The average temperature and average emission(s0 of a rotating planet that orbits a star every 365.25 days is a nonsense.
The difference between total radiative influx between the summer and winter solstices is 3.5%. If global is global then 3.5% of 341, 12 W/m2 should drive the difference between the global temperature around June 22nd and December 22nd. Sadly, this gives 12W/m2 = -0.17 degrees.
The problem is the heterogeneous distribution of different types of water and of land.
Trenberth ignores this complexity and presents box diagrams that were abandoned by everyone else some decades ago.
The Earths average temperature, see the BEST June+July vs Dec+Jan, is lower when it is closer to the sun than when it is further away.

jimmi_the_dalek
October 26, 2011 7:32 pm

“Energy is not a vector…etc”
But an energy flux is.
However once you start doing vector algebra, you have to integrate over all angles. So consider that little vector diagram in the article, and put next to it a similar diagram representing the neighbouring point on the surface, but think of the vector pointing the other way (ie same vertical component, opposite horizontal component). The horizontal fluxes will cancel. Once you integrate over all spherical angles, and over the surface area, all components except the vertical will cancel. This I think is why these diagrams only ever show a vertical component.
You could profitably read the discussion on the Trenberth diagram at ScienceOfDoom
http://scienceofdoom.com/2011/06/21/whats-the-palaver-kiehl-and-trenberth-1997/
SoD is not a warmist or an alarmist or even a “sceptic” site in the sense used here – it is that unusual thing in this field, a “pro-science” site (and if you doubt that, just consider how much they talk about fundamental physics, and how little about computer models…)

October 26, 2011 7:32 pm

>>Tim Folkerts says:
>>In a transparent atmosphere and ignoring the curvature of the earth,
>>the average 396 W/m^2 [of upward thermal IR] would be constant
>>all the way up. […]
>wayne says:
>Wrong
>It decrease with temperature with an increase in altitude.
The temperature of the atmosphere certainly decreases — ie the energy of the molecules decreases as you get higher. But the energy of the PHOTONS does not change. If 396 J worth of photons leave the surface, those photons will still have 396 J of energy when they are 10 m or 10^9 m from earth.
(OK — there is actually a very small gravitational redshift of he photons, but that is immaterial here and almost certainly not what you were intending.)

CRISP
October 26, 2011 7:33 pm

The 333W/m2 back-radiation supposedly heating the Earth is nonense.
You CANNOT transfer heat from a colder body (the upper atmosphere) to a hotter body (the lower atmosphere and Earth surface) without doing work.
The 2nd Law of Thermodymanics avbsolutely forbids it.

Jim Masterson
October 26, 2011 7:38 pm

Interesting post.
One of things that I’ve noticed over the years is that Climatologists don’t seem to differentiate between surface temperature and surface air temperature (SAT). GISS seems to understand this problem (see this for example). Trenberth’s diagrams treat both temperature types as one and the same.
There’s another problem. The Earth’s surface isn’t a perfect black body radiator. It does seem to radiate as a gray body, but then you have to use a modified S-B law that includes emissivity. Now emissivity is wavelength/frequency dependent, so an average emissivity is somewhat problematical. I’ve seen estimates anywhere from 0.90 to 0.99.
KT 97 assumes a surface temperature of 15 °C and an emissivity of 1.0. This gives us his surface radiation of 390 W/m². Trenberth’s 2009 update uses a surface temperature of 16 °C and an emissivity of 1.0 to give us a surface radiation of 396 W/m².
If we use a more realistic emissivity of say 0.95, then the KT 97 surface radiation drops to 370.6 W/m² and his 2009 surface radiation drops to 375.8 W/m². That’s almost a 20 W/m² difference. Trenberth needs add more reality into his diagrams instead of wringing his hands over a missing 0.9 W/m².
Jim

October 26, 2011 7:43 pm

jimmi_the_dalek,
“Temperature is directly affected by absorption of radiation – so how and why are you distinguishing HEAT (in capitals!) from EMR.”
because EMR can be reflected with no net heat transfer among other issues. Then there is the issue of frequency. Not everything absorbs energy from every frequency whether it reflects or just scatters the frequency or simply ignores it.
This is the root of a lot of uncertainty in the so-called settled radiative transfer equations. The climate types are telling us that every bit of IR going down IS absorbed 100% by whatever composition of the earth is there and nothing else can happen to it. Wanna bet?!?!?!?!

KevinK
October 26, 2011 7:46 pm

Mr. Fernley-Jones;
With respect, I do understand that this particular post is an “initial draft” and perhaps additional details are forthcoming. I await these with interest.
In the meantime I would just like to point out a few observations relative to your (or more accurately NASA’s) Figure 2;
All of the following energy flows are travelling at the speed of light; “incoming solar energy = 100%”, “reflected by atmosphere = 6%”, “reflected by clouds = 20%”, “reflected by Earth’s surface = 4%”.
The following energy flows that are travelling at “close” to the speed of light after a slight delay (caused by absorptions and remissions by gases with thermal capacities) include; “absorbed by atmosphere = 16%”, “absorbed by clouds = 3%”.
Some of the remaining energy flows are made up of various flows that travel through the system at a combination of speeds; “radiated to space from clouds and atmosphere = 64%” which travels in part at close to the speed of light after absorptions and emissions but also contains energy flows that are “absorbed by land and Ocean’s = 51%” and “carried to clouds and atmosphere by latent heat in water vapor = 23%” which travel through the system at a close to the speed of heat.
One interesting thing to note is the ABSENCE of any text that indentifies the source / destination / and magnitude of the lower rightmost red arrow which starts as “absorbed by land and Oceans = 51%” (after removing 23% + 7% (or 30%) this would equal 21% and seems to leave the Earth as “radiated to space directly from Earth = 6%”. Somehow this 21% gets 15% “deleted” from to it and it becomes 6%. Seems to make sense, HOWEVER upon careful observation the arrow labeled “absorbed by atmosphere = 16%” MAGICALLY becomes “radiation absorbed by atmosphere = 15%” after passing behind the BIG 64% red arrow. Perhaps a typo, perhaps an accounting error?
In summary, while these nice little graphics do a reasonably good job of describing the first “order” understanding of the energy flows through the Sun / Atmosphere / Earth / Atmosphere / Universe system they have MANY faults. The largest of which is totally discarding any consideration of the speed at which each of these energy flows travel through the system.
In electrical engineering this is considered as a “DC” (direct current) analysis. Once the “speed” / “lag time” / “response time” / “delay time” is incorporated into the analysis a totally different prediction of the systems response usually results.
Hey, a percent here and a percent there, it probably cancels out and we are STILL WARMING FROM THE “GREENHOUSE” EFFECT RIGHT ???
Cheers, Kevin.

October 26, 2011 7:49 pm

Wayne,
I agree that you can define a direction that energy is flowing, just like you can define a direction that the mass in a stream is flowing. But that does not make energy or mass a vector. If 10^19 photons of energy 10^-19 J pass thru a 1 m^2 surface oriented perpendicular to the earth’s surface, then 1 J passes thru the surface independent of the direction that the photons are moving. If the photons are all traveling at 45 degrees to the normal, the flux is still 1 W/m^2, not 0.707 W/m^2 “upward” and . 0.707 W/m^2 “sideways”
Figure 4 in the post seems to suggest that only the “upward component” of photon energy counts, which is NOT the case. My comments fit in with Billy’s comment October 26, 2011 at 6:17 pm. The total upward energy flux from the IR photons from the surface (in a transparent atmosphere) will remain constant. A proper integral over all directions will confirm this.
Now when the IR absorption of the atmosphere is added in, this will complicate the analysis a bit. But I have not been addressing that yet.

Ray
October 26, 2011 8:08 pm

What about evaporation, condensation and changes of state? Are those really equal in time and balancing each other?

jorgekafkazar
October 26, 2011 8:14 pm

Let’s see an update. This doesn’t move me.

October 26, 2011 8:21 pm

Mr. Watts,
The earth is a globe. Horizontal radiation will leave the atmosphere at an angle if it makes it that far. Shallower angles would leave at angles approaching vertical. Of course, with so much more atmosphere to traverse the likelihood of it leaving on that path is much smaller than straight up, the shortest path!!! As the emission point rises the angle that can irradiate the earth is increasingly small and the radiation has a larger area that points to space through increasing amounts of atmosphere (your cone).
The IR from the surface is most likely going to be absorbed by GHG’s close to the surface anyway. At the pressures and density there I am told collisions will transfer much of that energy to oxygen or nitrogen and be convected away as collisions happen oftener than the emission of IR.
As we go up in the atmosphere that gradually changes until the Strat. where the density apparently favors emission.
GHG’s therefore speed the heating and cooling of the bulk of the atmosphere which has a magnitudes lower absorption and emission rate in the far IR. This also brings into question just how much IR there is to actually get back to the surface if the energy is being preferentially transferred though collisions in the far IR opaque lower trop.
Another interesting bit is that GHG’s absorb in slightly wider frequencies near the ground due to the line broadening. It would seem if they emit at the same frequency the IR going up would have a better chance of escaping as the pressure/temp decreases narrowing the bands and not absorbing as much as the lower.
This would all seem to be especially apropros as to why Venus does NOT have a huge Greenhouse. The effect drops off so quickly only near the surface is it possible to be so massive, yet, the collisional transfer even though it is from CO2 to CO2 must be almost total under such temp/pressure/density!! It would almost seem to be more like water, conducting instead of radiating internally.
I must be wrong, but I don’t know enough to see why.

Bob Fernley-Jones
October 26, 2011 8:25 pm

Tim Folkerts @ 6:07 pm,
I don’t have an issue with your transparent atmosphere discussion, if I understand it correctly. (there do seem to be a few typos). However, we do have in our atmosphere GHG’s, particulates, clouds, and precipitation. Note that the MODTRAN calculations in the tables in the addendum, suggest that up-down radiation diminishes with increasing altitude.
See also Wayne @ 6:44 pm below and that ol’ T^4 thingy

October 26, 2011 8:30 pm

jimmi_the_dalek,
“The horizontal fluxes will cancel. ”
The surface is a sphere. The adjoining points would NOT cancel. Surface irregularities will confuse this so that the surface is irradiating itself just as the surface of the water does. One of the requirements for S-B to apply is that the surface geometry does NOT allow it to irradiate itself. OOOPS!!!!

Bob Fernley-Jones
October 26, 2011 8:32 pm

Billy @ 6:17 pm,
I think my item 5) in the article should answer your concern, if not, please elaborate.

Ed_B
October 26, 2011 8:38 pm

jimmi_the_dalek says:
“Sorry , but no. Temperature is directly affected by absorption of radiation – so how and why are you distinguishing HEAT (in capitals!) from EMR.”
Since long wave radiation is absorbed and then re transmitted, how can there be a NET absorbtion to affect temperature?

Jim Masterson
October 26, 2011 8:41 pm

>>
KevinK says:
October 26, 2011 at 7:46 pm
In summary, while these nice little graphics do a reasonably good job of describing the first “order” understanding of the energy flows through the Sun / Atmosphere / Earth / Atmosphere / Universe system they have MANY faults. The largest of which is totally discarding any consideration of the speed at which each of these energy flows travel through the system.
<<
These diagrams are “steady-state.” That means the transients have had time to damp out or stabilize.
>>
In electrical engineering this is considered as a “DC” (direct current) analysis. Once the “speed” / “lag time” / “response time” / “delay time” is incorporated into the analysis a totally different prediction of the systems response usually results.
<<
That is unless you’re measuring flows that have had time to stabilize. These diagrams represent such measurements. I admit that these diagrams have a multitude of problems, but bad DC analysis isn’t high on the list of those problems.
Jim

harry
October 26, 2011 8:46 pm

Billy said:
I’m sure that, if you consider the earth as a sphere, these effects all average out and your concern is misplaced.”
I agree that in theory at any point in space all the scattered rays would cross, “averaging out” to be the total energy. However, don’t need to test the measuring device to see if it properly registers the energy from a beam scattered on a very shallow angle?
Its design could reflect those rays.
An interesting analogy is a PV cell on my roof. Calculating the angle of the sun and its cosine allows me to produce the expected electricity production from the cell. Effectively it is the ratio of the area exposed to the sun, compared to the area of the PV cell. But because of the design of the cell when the sun is low in the sky, most of the rays reflect off the surface of the protective cover of the PV cell. So whenI should see ~300W, I see ~30W.
I wonder if the measuring devices have been calibrated or designed to measure emissions at low angles of incidence. The height of satellite orbit would also affect this, the higher the better for reducing this problem.

Bob Fernley-Jones
October 26, 2011 8:57 pm

Tim Folkerts @ 6:21 pm
I can understand that you find the vector consideration a tad obscure, but nevertheless, vectors are very useful for analysing parameters having directional information. In fact I wondered if someone might come up with your concern, and contemplated instead showing the classic inclined plane with a mass upon it, but decided it was too tedious and possibly thought to be off-topic.
If you place a mass upon an incline, where it overcomes the coefficient of friction, assuming it, and flatness, to be constant, and ignoring elasticity in the materials, then it will exert a perpendicular force upon the reacting surface according to the vertical component of its mass. If you complete the vector loop there is also an unseen horizontal component vector. This is pure vector maths, but nevertheless the mass can only slide down the direction of the incline

Ask why is it so?
October 26, 2011 9:06 pm

I’m not sure I understand this article but just to clear a few things up, radiation is not heat. Heat is the result of the absorption of radiation by a surface or molecule. 2nd Law of Thermodynamics does state that heat (without work) will only travel from higher to lower temperature, that is correct, however, radiation is not heat. Radiation can travel in any direction and because of this radiation reflected or re-emitted can also travel in any direction, up, down, sideways. Is the question here how much of the radiation is actually absorbed by the system, surface and atmosphere, and how much is lost (unused) out to space, leaves the system without being absorbed? I just don’t know how any of these figures are even calculated. If the earth reflects 4% that presumes that every day, the same amount of energy and direction of radiation and absorbing surfaces never change. This presumption is preposterous. Climate scientists and Physicists cannot do their calculations until they have an average figure but the real world is an every changing system that doesn’t know what an average is. I’m still trying to work out how long wave radiation can produce more heat than short wave radiation thereby causing global warming. I suspect it just ain’t so.

Bob Fernley-Jones
October 26, 2011 9:09 pm

Wayne @ 6:34 pm,
Thanks for your comments. I’ve actually been ruminating over the Trenberth/IPCC cartoon for several years, and wondering why I’ve not seen anything in the literature critiquing it. (and I have not seen your comments on it). Oh well, I thought just recently; why not be brave and challenge the great authority.

grayman
October 26, 2011 9:15 pm

Myself i am a layman, but i know that the sun is a flaming band and it ebbs and flows. So my question is who came up with the figure on the amount of radiation coming in and is that always constant. Can anyone tellme how it is measured and is it the same amount in the UK as in the USA or Australia, or is it what is measured at the Eqautor and considered the measurement for all of Earth? I have been told by warmist that a specific radiation amount sqaured is constant, on paper yes, but IMHO, the real world, i do not see it.

grayman
October 26, 2011 9:17 pm

Sorry, Flaming BALL not Band

Bob Fernley-Jones
October 26, 2011 9:28 pm

jimmi_the_dalek @ 6:49 pm
Jimmi, you are describing ONE of the ways of transporting HEAT. In the case of radiation this ONLY happens when there is a potential difference between two sources of radiation.
Check – out Wikipedia on heat transfer, for a start;
http://en.wikipedia.org/wiki/Heat_transfer

ferd berple
October 26, 2011 9:34 pm

A large block of ice and a small candle flame both emit the same amount of EMR. However, only one is capable of warming a human being.
Trenberth’s analysis ignores this simple observation and assumes that since two object emit the same EMR, they contribute the same amount of warming.

Jeff D
October 26, 2011 9:41 pm

I would love to see this in a 3d ray traced model animation. With the multiple freq bands it would have to be broken up into several animations to visualize it. I really enjoy these type threads even though I only understand about 2/6th of the content! Thanks for the puzzle.

October 26, 2011 9:42 pm

I have several problems with these types of simplistic diagrams, and the first of which was mentioned by others that there is no way in hell that you can provide an average measurement with a granularity of 0.1 w/m2. Not possible.
I do a pretty good bit of engineering and space science. We have done a lot of work on the Nimbus II HRIR data sets and clouds are a lot more of an influence than what these simplistic diagrams indicate. In the Nimbus data the temperature of clouds were all the way down to below the 210 kelvin calibration of the sensor over large areas. This indicated a very high altitude for these clouds and a LOT of reflection.
I have measured a 90% decrease in insolation at the surface from cloud cover. This is a broad spectrum measurement with solar panels. Multiply this by the average global cloud cover and it is a hell of a lot more than is indicated.
I also hate using this funky 396 w/m2 average number. The vertical number is 1366 w/m2 and so the second graph does a better job of fixing this problem.

ferd berple
October 26, 2011 9:46 pm

R. Gates says:
October 26, 2011 at 5:46 pm
A perfect example is the compression of air when it flows between zones of different pressure…this is work done on the air molecules, their temperatures will rise as they are compressed and the average kinetic translational energy rises from the work done.
Under that argument, air rising and falling also has work done on it. Sort of like the explanation that the brakes on a car stop the car by turning the motion of the car into heat.

Bob Fernley-Jones
October 26, 2011 9:51 pm

Gail Combs @ 7:18 pm
What may not be commonly realized is that the GHG molecules are not the sole carriers of thermal energy as a consequence of their absorption of photons from EMR energy. There are countless molecular collisions between them and the N2 and O2 molecules etc that comprise the vastly greater bulk of the atmosphere. It is commonly called thermalization, of non-greenhouse gasses. These collisions result in a change of kinetic energy (heat) of individual molecules, and just because a GHG molecule absorbs a photon, it does not mean that it will re-emit it, because there is a lot of other stuff going on.

R. Gates
October 26, 2011 9:56 pm

Bob Fernley-Jones says:
October 26, 2011 at 6:12 pm
R. Gates @ 5:46 pm,
Yes, you are correct, that work can result in heating, or cooling. An oversight on my part, where I was trying to show the difference between heat and EMR.
____
Thanks for your reply. I assumed it was an oversight on your part. But to be clear, in a pure sense, heat is a measurement of energy in transit, and should not be taken to be a measurement of the average translational kinetic energy (normally called temperature) nor the total internal energy of the object. Again, heat is a energy in transit flowing between objects. An excellent resource on this can be found at:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
and on the whole issue of heat and thermodynamics, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
To your point about the difference between heat and EMR, They are completely different things,as one is the measurement of the flow energy, and one is a form of energy.

ferd berple
October 26, 2011 9:57 pm

How is it that solar radiation is partially reflected by the surface, but back radiation is 100% absorbed? How is it that the is 0% reflection of back radiation?

George E. Smith;
October 26, 2011 10:00 pm

Well I have always been critical of the Trenberth “cartoon” earth energy budget, or whatever they currently call it in your new posting. And I should add, that I do not use the word”cartoon” in any derogatory sense; simply a descriptive label for a diagram that is not very scientific.
So tell me; why would anyone (including me) take any notice of anything written by a “scientist” apparently with a Doctorate Degree (Dr Laura has a PhD, and she doesn’t know beans about climate either) who uses “Watts” or “Watts per metre squared” as units of ENERGY.
When I went to school and studied Physics, Watts, was a unit of POWER, and “Watts per metre squared” would be a unit of AREAL POWER DENSITY.
And at the position of earth’s orbit (average) the arriving power density is 1362 Watts per metre squared; it certainly is NOT 341 W/m^2.
The response of real materials to an incident power density of 1362 W/m^2 (irradiance) will be quite different to the response to 341 W/m^2. For example the former might melt or vaporize a material whereas the latter may not. Such a change might be irreversible, in that the altered material may then become widely dispersed.
An example would be, a somewhat attenuated proportion of each of those purported radiances being incident on the surface of say an Arctic (of Greenland) ice mass.
The former larger radiance might result in warming the ice surface sufficiently for it to melt, whereas the latter could do no such thing.
Remember that WATTS is a rate of supply (or loss) of energy, and if the supply rate does not exceed the loss rate from other processes, such as simple thermal radiation; or conduction, then melting will NEVER occur, no matter how long that condition is maintained.
And since the earth rotates, under that incident power density (radiance) of 1362 W/m^2, if melting does not occur, during the irradiation portion of the daily cycle it never will melt, and it certainly won’t melt under 341 W/m^2 irradiance.
So Trenberth’s cartoon is unscientific gobbledegook.

R. Gates
October 26, 2011 10:04 pm

ferd berple says:
October 26, 2011 at 9:34 pm
A large block of ice and a small candle flame both emit the same amount of EMR. However, only one is capable of warming a human being.
_____
Depends how cold that human being might be. Energy will flow from a block of ice to a human if that human is chilled below the temperature of the block of ice (by floating in interstellar space for a few minutes for example). Also, on what basis do you contend that a “large” block of ice and “small” candle flame emit the same “amount” of EMR? How can you know this?

Roy Clark
October 26, 2011 10:38 pm

I have just published a little monograph on this subject titled ‘The Dynamic Greenhouse Effect and the Climate Averaging Paradox’ . The links are:
Paperback:
http://www.amazon.com/Dynamic-Greenhouse-Climate-Averaging-Paradox/dp/1466359188/ref=sr_1_2?s=books&ie=UTF8&qid=1319675042&sr=1-2
Kindle:
http://www.amazon.com/Dynamic-Greenhouse-Climate-Averaging-ebook/dp/B005WLEN8W/ref=sr_1_1?s=books&ie=UTF8&qid=1319675042&sr=1-1
I put my dynamic or time dependent version of the K-T diagram on the front cover. Just copy it from Amazon. This may be challenge to Trenberth that you are looking for. Journals like Nature and Science wouldn’t dare touch this one. Once the dynamics are understood then the whole global warming problem disappears into the noise of the daily surface energy transfer. The ‘average’ solar heating and convective cycle is 12 hours not 24.
The funamental issue is that there is no such thing as a climate equilibrium state. The whole radiative forcing approach is just plain wrong. There are no ‘forcings’ or ‘feedbacks’. The climate models have been wrong at least since 1967 when Manabe and Wetherald started down the wrong path.
There is more info on my little web site at http://www.Venturaphotonics.com for those who are interested.

anna v
October 26, 2011 10:42 pm

What he said:
George E. Smith; says
October 26, 2011 at 10:00 pm
Since delving in the mire of IPCC climate modeling I came to the opinion that some climate “scientists” think that physics is a la cart: pick and choose what you like, feel free to redefine what you like and create new terminology that sounds like physics. They have managed to dominate their field so even quite solid climate scientists use some of their terminology, making life very hard for physicists :). It feels like wading through hair.

George E. Smith;
October 26, 2011 10:46 pm

There is a lot more examples of contention even in the second NASA version. While it is headed ENERGY budget, it diesn’t give units so that is ok.
BUT !! just let’s start with the INPUT at the left edge in yellow. Starting at 100%, 20% is reflected by clouds. Well average global cloud cover is supposed to be in excess of 60%, so if only 20% is reflected from 60% coverage, then the cloud reflectance must only be 33%. And the reflectance of clouds over the solar spectrum is well known to be much higher than 33%, so that 20% number doesn’t fly.
Now look there at the cloud ABSORPTION; only 3%
Now if the clouds REFLECT 20% and ABSORB 3%, what does that say about how much of the solar energy the clouds TRANSMIT ?
Well anyway, that is just an aside; the real point is that in CLEAR AIR there are NO clouds; so the clear air losses of solar energy are then 6% REFLECTED by the atmosphere; 16% ABSORBED by the atmosphere and 4% REFLECTED by the surface (oceans plus land). So assuming that is all true, then the ground level solar insolation (clear air) is 26% less than the TSI or 74% of 1362 which comes out to 1007 W/m^2, and that seems like a believable number since 1,000 W/m^2 is the usually accepted maximum power density for solar energy availability.
The 6% reflectance of the atmosphere is news to me; I’ve never been aware of it. Is it possible that they are using the term REFLECTANCE to describe processes like Raleigh scattering, that produces the blue sky color, and results in a like amount of solar energy being scattered to space.
The incoming solar rays are certainly not obeying the reflection laws of geometrical optics, so it is incorrect to describe Raleigh or Mie scattering as REFLECTANCE.
As to the 4% reflected from oceans and land, about 75% of the earth surface in the major solar input latitudes, is ocean, and the normal relfectance is 2%. That leaves another 2% to be reflected from the land, which comprises only 25% of that surface area. a simple calculation then shows that the land reflectance must only be 10%; and that would be a maximum, because, while 2% is the normal incidence reflection coefficient of water, over all angles, the total reflectance is more like 3%, since the 2% number holds relatively constant up to teh Brewster angle (57 deg incidence), but then increases rapidly beyond that.
So in that case, the land must only contribute 1% to the total reflectance and that reduces the land reflection coefficient down to 7%.
Whether 7% or 10%, both numbers are way too low for the reflectance of land areas.
But at least it looks as if they may be closer than Trenberth, and they avoid the incorrect units problem, and its consequences

George E. Smith;
October 26, 2011 10:54 pm

“”””” R. Gates says:
October 26, 2011 at 10:04 pm
ferd berple says:
October 26, 2011 at 9:34 pm
A large block of ice and a small candle flame both emit the same amount of EMR. However, only one is capable of warming a human being.
………………………………………………..
Also, on what basis do you contend that a “large” block of ice and “small” candle flame emit the same “amount” of EMR? How can you know this? “””””
Well clearly ferd berple’s statement is correct, if we agree on say total WATTS of power output, and then agree on how large a block of ice is necessary to achieve equality, assuming we know the flame temperature or otherwise know its output power.
But R Gates’ objection is valid too, since the optics of illumination by a large block of ice, and a near point source candle, are quite different, and never could produce the same result; even if they were spectrally identical; which they are not.
So I agree with R on this and ferd’s example is not instructive.

charles nelson
October 26, 2011 11:00 pm

I know this is going to sound really dumb to a lot of you out there,
but I’m just a humble individual whose job it is to control, temperature and humidity…so could someone explain to me please exactly how or more precisely where heat would accumulate in the Warmist model.
Given that the air temp at 300kPa is minus 30 degrees C over the equator and minus 50 over the poles…(and we all know from traveling in planes that air reguarly rises to this height then cools and falls) I’ve just never understood WHERE the warming is meant to BE and when it is there what stops it from rising up to the frigid void at the top of our atmosphere?
Be gentle with me!

George E. Smith;
October 26, 2011 11:11 pm

“”””” CRISP says:
October 26, 2011 at 7:33 pm
The 333W/m2 back-radiation supposedly heating the Earth is nonense.
You CANNOT transfer heat from a colder body (the upper atmosphere) to a hotter body (the lower atmosphere and Earth surface) without doing work.
The 2nd Law of Thermodymanics avbsolutely forbids it. “””””
Well CRISP, nobody (but you) is transporting “heat” from a colder body to a hotter body, with or without doing work; so the second law of thermodynamics is not invoked. What IS being transported from the colder body to the hotter body, and can do so with impunity is ELECTROMAGNETIC RADIATION which is a form of energy, which is quite unrelated to “heat”; which is a different form of energy requiring physical matter to be present. EM radiation needs NOTHING in order to go where it chooses to.
EM radiation knows absolutely nothing about either “heat” or “Temperature.”, and it can go wherever it darn well pleases .

October 26, 2011 11:28 pm

For me, thermals seemed always greatly underestimated.
Second, all atmospheric molecules radiate IR, not only so called “greenhouse gases”. Those fantasy diagrams being obsessed with wrong greenhouse analogy ignore 99% of the atmosphere and things like simple heat retention.

Leg
October 26, 2011 11:29 pm

As I look at Trenberth’s work, I have to wonder if some of these climatologists have a fundamental grasp of how radiation (electromagnetic/photonic*) interacts with matter. Anyone who works with gamma, X-ray, IR, light, radio, and et cetera radiation quickly recognizes the 3D issues presented as photonic radiation interacts either by full or partial adsorption of the energy. Perhaps a primer on how radiation interacts with matter would be useful to readers of this article?
* Of course there is a large amount of energy transferred to the earth via particulates (muons, pions, et al) but much gets converted to photonic energy in the atmosphere.

October 26, 2011 11:42 pm

kuhnkat says:
October 26, 2011 at 8:21 pm

GHG’s therefore speed the heating and cooling of the bulk of the atmosphere which has a magnitudes lower absorption and emission rate in the far IR.

Sorry, but it seems to me that that assertion was refuted by the replication of Al Gore’s high-school science experiment. CO2, at least, is a green-house gas that actually retards the speed of heat transfer through the atmosphere.

Man Bearpigg
October 26, 2011 11:54 pm

.. and they call us skeptics ‘Flat Earthers’ when their calculations are based exactly on that notion.

J.H.
October 26, 2011 11:59 pm

Walter H. Schneider….. During the ATTEMPTED replication of Al Gore’s high-school science experiment….. AKA, Science 101…..;-)
Anthony Watts showed quite clearly that Al’s experiment was busted….. But did prove to be informative of the fact that pure, or high concentrations of CO2 have a different thermal conductivity than that of air….

jason
October 27, 2011 12:24 am

Very interesting discussion, which demonstrates clearly that the view that sceptics are politically rather than scientifically motivated is false.
Good to see R Gates removing more of his mask. He starts off posting in the arctic threads arguing with steve m, then reveals he is a buddy of Trenberth, and now demonstrates in this thread that his understanding of the subject goes way above a layman. R Gates is I believe someone far closer to the heart of the debate than he has let on until now. Genuinely sceptical mind turned, or trojan horse?

Erik Ramberg
October 27, 2011 12:31 am

I think everyone is making this way too complicated. Physics is all about symmetries. For the purpose of explaining Figure 1 (which, after all, is a cartoon) imagine the processes shown are equal across the sphere of the Earth. Since there is no preferred direction in that case, there can be no net radiative or convective transport, except in the direction normal (up or down) to the surface. There is no net horizontal energy flow.
Obviously the real Earth breaks this symmetry and has a preferred direction – namely the axis of rotation. That is why calculations are done for energy transfer as a function of latitude. When you do that you get Hadley cell circulation north and south, which then produce the trade winds, etc., by the Coriolis effect.
Suffice it to say that Trenberth understands this, and that Figure 1 is a very useful and informative graphic.

Gail Combs
October 27, 2011 12:47 am

Bob Fernley-Jones says:
October 26, 2011 at 9:51 pm
Gail Combs @ 7:18 pm
What may not be commonly realized is that the GHG molecules are not the sole carriers of thermal energy as a consequence of their absorption of photons from EMR energy. There are countless molecular collisions between them and the N2 and O2 molecules etc that comprise the vastly greater bulk of the atmosphere…..
_____________________________________________
Thanks,
That is sort of what I was getting at. (bouncing being collisions) however I did forget that during a collision energy can be split with some transfered as heat or added momentum without absorption or did I mix that up? (my physics training is over forty years old)
A collision unlike absorption and emission of a photon does not have to be in “discrete packets” that is at specific wavelengths.
So this means there are at least 4 separate types of activities happening to the radiation leaving the earth.
1. absorption of a photon
2. emission of a photon
3. Collision (bounce) with no transfer of energy (reflection)
4. Collision with some transfer of energy to the atom.
I also specified atom (or molecule to be more precise) because air is a mix of gases and absorption can take place with CO2, H2O, CO and possibly others. Collisions with all the different molecules.
Hopefully my fumbling around will help clarify this for others who are lurking. WUWT can proved to be one heck of an education in physics!

David
October 27, 2011 1:07 am

DocMartyn says:
October 26, 2011 at 7:19 pm
“Trenberth ignores this complexity and presents box diagrams that were abandoned by everyone else some decades ago.
The Earths average temperature, see the BEST June+July vs Dec+Jan, is lower when it is closer to the sun than when it is further away.”
Sunlight, falling on the Earth when it’s about 3,000,000 miles closer to the sun in January, is about 7% more intense than in July. Because the Northern Hemisphere has more land which heats easier then water most people state that the Earth’s average temperature is about 4 degrees F higher in July than January, when in fact they should be stating that the ATMOSPHERE is 4 degrees higher in July. In January this extra SW energy is being pumped into the oceans where the “residence time” within the Earth’s ocean land and atmosphere is the longest. There are also other factors, such as the Northern hemispheres winter increase in albedo exceeds the southern hemisphere’s winter albedo due to the far larger northern hemisphere land mass. So at perihelion we have a permanent loss to space of ? W/2m SWR due to increased albedo and a loss of SWR to the atmosphere as at perihelion the SWR is falling on far more ocean, where it is absorbed into the oceans for far longer then if that SWR fell on land. Do these balance (unlikely) or is the earth gaining or losing energy during perihelion??? The TOA flux should tell us and climate models should accurately predict the observation.

Gail Combs
October 27, 2011 1:09 am

Leg says:
October 26, 2011 at 11:29 pm
As I look at Trenberth’s work, I have to wonder if some of these climatologists have a fundamental grasp of how radiation (electromagnetic/photonic*) interacts with matter. Anyone who works with gamma, X-ray, IR, light, radio, and et cetera radiation quickly recognizes the 3D issues presented as photonic radiation interacts either by full or partial adsorption of the energy. Perhaps a primer on how radiation interacts with matter would be useful to readers of this article?
* Of course there is a large amount of energy transferred to the earth via particulates (muons, pions, et al) but much gets converted to photonic energy in the atmosphere.
_________________________________________________
Go for it.
The uninformed like me would love to see the information and Anthony has asked the WUWT community to pinch hit for him for a while so this would be a really good time.

Kelvin Vaughan
October 27, 2011 1:16 am

Just a thought! anti phase sine waves of the same frequency cancel out. Does this happen at infra red wavelengths?

Bob Fernley-Jones
October 27, 2011 1:47 am

R. Gates @ 9:56 pm
I was hoping this matter of semantics would not arise, and which I discussed in an earlier draft of this article which was not approved by Ric Werme, (For Anthony Watts)

There are conflicts in terminology between physics and engineering, (quantum theory and thermodynamics), but since the latter field is closer to common understandings, I lean towards that terminology. For instance if we heat a pot on a stove, we say that it gets hotter, but a physicist may prefer to argue no; it has obtained more thermal energy, and HEAT is a transient condition. Surface radiation is also known as infrared light, or EMR, or Electro-Magnetic Radiation, or upwelling radiation, or infrared radiation, or long wave, and more, and is sometimes confused as thermal energy or heat radiation. Some climatologists refer to thermal radiation as being restricted to the infrared, whereas it is a proven fact that solar visible light is also thermal, and more strongly so.

In the classical meaning of HEAT, it can be calculated within any particular material phase from its specific heat (factor) and temperature per unit mass, but can also be described as sensible heat. If there is phase change involved then latent heat needs to be added or subtracted. BTW, doesn’t Trenberth/NASA mention sensible and latent heat?

Leg
October 27, 2011 2:00 am

Gail @ 12:47
You are getting at the basics of how radiation interacts with matter, which makes it a little easier to understand Trenberth’s graph and its refutation in this article. Adding to your list of potential radiation interactions: for incoming cosmic radiation, which may be >10E6 eV/photon, there is enough energy in the photon to disrupt the nucleus of an atom. This is how we get C-14 from nitrogen in our atmosphere as one example. Just looking at the atomic level, energy distribution from radiative forces in our atmsophere is clearly complex.
One thing I have not seen is how all the different forms of electromagnetic energies are measured for determining the relative amounts of energy transfer to earth and space (once measured and summed, conversion to potential energy such as watts/meter square is relatively easy). I know we have instruments that measure incoming radiation for a lot of spectrums (maybe the entire spectrum), but do we have instruments measuring outgoing radiation? Is outgoing measured or just calculated? I’d love to see some references that talk to how they measure outgoing radiation if someone has this handy.

Thomas U.
October 27, 2011 2:20 am

@ Eric Rambert
The key word in your post is : IMAGINE. That seems to be what too many climate “scientists” are doing. The article of Bob Fernley-Jones raises important questions, as do many of the posts, those of George E. Smith being one example. And no, it is not true “that everyone is making this way too complicated.” The fact is, REALITY is much more complicated than the simplistic “models”, “cartoons”, “theories”, “assumptions” employed in climate “science” suggest.

October 27, 2011 2:48 am

This is really disappointing – both the article and the meandering, and largely confused, blog responses about 3-D geometry. It casts no light at all for me on the important central questions:
1. Does back radiation to the Earth’s surface occur at all or is it “unphysical”?
2. If it does, what proportion of that back radiation is due to CO2?
3. What proportion of the back radiation due to CO2 is due to man-made CO2?
Any answers to the above questions should then be tested for credibility against the real-world data which indicates that near-surface average temperatures between 1850 and 2010 increased at the decidedly unalarming long-term trend rate of about 0.4degC per century. (See: http://www.thetruthaboutclimatechange.org/tempsworld.html.)
Cmon guys – please let’s get focussed!

Michel
October 27, 2011 2:58 am

Few points:
– both diagrams, Trenberth and NASA, show the summary of simple energy balance models over an idealized 24h day and from poles to equator for the whole globe, assuming that any other parameters are in steady state. The diagrams are useful to grasp orders of magnitude but not to explain the underlying physical processes. Each flux is the results of many parameters that cannot be discussed at that coarse summary level.
o Trenberth diagram: to emit 396 W m-2 with an assumed emissivity of 1.0, the irradiating Earth surface (a mix of land, ice and sea) should have a temperature of 16°C (289K).
o NASA: if 70% (240 W m-2) of the incoming solar radiation is re-emitted to the outer space by Earth surface and atmosphere, then either the average surface temperature should be -18°C (255 K) or the average emissivity should be 0.604, or something in between.
But average temperatures or average emissivities don’t have any physical meaning. They are just calculation aids for these simplistic models.
– The NASA diagram is more understandable because it does not introduce the “back radiation” concept
– In the Trenberth diagram I could not understand why the 333 W m-2 back radiation is only going back and not also forth? Has anyone an idea?
– Ground and sea surfaces, as well as dust particles and droplets suspended in air emit radiation at their surface temperature following Planck’s law. At Earth temperature levels it is in the infrared range (IR) of roughly of 2 to 100 micrometers. The Stefan Boltzmann equation is the integral of Planck’s law over the whole spectrum. Gases do not irradiate (or only at very high temperature in the plasma state like in a flame or at the surface of the sun).
– IR absorption takes place at finite wavelengths. Each IR absorbing gas like CO2, H2O has an own absorption spectrum calculated line by line from the possible rotational and vibrational transitions that the given molecule allows. The absorption spectra can be downloaded from the HITRAN Database with a 0.1 cm-1 resolution from http://spectra.iao.ru/ .
– The radiated energy would not “largely be absorbed by the greenhouse gases instantaneously near the surface”, as writes Bob Fernley-Jones.
In a line by line calculation over a 2000 metre layer the absorbed energy (also called forcing) is approx. 10% of the total in dry air, and 28% in very wet one.
See graphs: http://dl.dropbox.com/u/6905434/Air-Full-Spectrum.png and http://dl.dropbox.com/u/6905434/Air-Full-Spectrum-tropic.png
– The CO2 forcing can be calculated: results are in line with the typical forcing of 3.7 W m-2 for any doubling of the CO2 concentration, as published by Myhre et al. in Geophysical Research Letters, vol. 25, no.14, p. 2715-2718, July 15, 1998.
That should no more be a debated issue, neither in its nature, nor in its extent.
– However, looking again at the numbers: 3.7 W m-2 are only approx. 1% of the Earth re-emitted IR energy, one order of magnitude less than what can be related to changes of humidity. The Trenberth and NASA diagrams are helping us grasping these sizes.
– This is why we should concentrate on the consequences of the increase of CO2 in atmosphere:
o Water displacement? Change of emissivity?
o Resulting in Warming? Cooling?
o and combined with other driving forces?
– And let’s not forget that neither weather nor climate have ever been in a steady state that would need to be preserved.

Mydogsgotnonose
October 27, 2011 3:03 am

Trenberth’s ‘Back Radiation’ is the biggest scientific cock-up in History. He has confused it with ‘Prevost Exchange Energy’, exactly offset by IR from the ground plus a bit from lower, hotter gas. A Dutch PhD student recently shinned up an 800 foot radio mast and showed that the [up-down] signal decayed exponentially to zero, Beer’s Law.
Because the modellers include this imaginary energy in their CO2-AGW myth, they have to hide it by cloud cooling. The first thing they do is to overestimate low level cloud albedo by >2. The second is to claim the 1st AIE applies to thicker clouds. It doesn’t, so the sign is wrong.and it’s the real AGW which heated the World as Asian industrialisation spewed out aerosols which switched off a direct backscattering mechanism. It’s why Asian low level clouds look dull from the top compared with 30 years’ ago; they transmit more energy. It’s self-limiting hence the oceans stopped warming in 2003, and Trenberth’s ‘missing heat’.
Because they also overestimate present greenhouse warming by a factor >3, the IPCC ‘consensus’ CO2 climate sensitivity is a factor of at least 9 too high.
So long as Trenberth continues his elementary mistake [he clearly wasn’t taught the correct physics] , he is condemning himself to be considered by history as a failure. As for Hansen with his recent claim of over double 1st AIE to keep his 4.2K climate sensitivity despite no warming, he is an expert on aerosol optical physics so has no excuse for what is almost certainly fraud, the claim by NASA from 2004 of imaginary ‘surface reflection’ purporting that clouds with small droplets have high albedo,.
Look at any rain cloud – it’s dark underneath because albedo is high, and it’s a complex large droplet phenomenon. Yet workers in climate science believe the surface reflection myth.

October 27, 2011 3:18 am

The answer to the “paradox” is contained in the two emission curves depicted. The flux emitted through the “atmospheric window” (approx.750-1150 cm-1) can be seen to approximate to a black-body curve for about 298°K, or about 25°C, consistent with the “Tropical Pacific” source, and an emissivity close to 1. There is no paradox.
Secondly, a correction – K&T 2009 does not compute surface radiation for a “surface temperature of 16 °C and an emissivity of 1.0”. They did so in K&T 1997:
“For example, in KT97, we used a single column model constrained by observations, to represent the average fluxes in the atmosphere. We compared results at TOA with those from the NCAR CCM3 and found good agreement, so that the spatial structure was accounted for. At the surface, the outgoing radiation was computed for blackbody emission at 15°C using the Stefan–Boltzmann law”
In K&T 2009:
“To compute these effects more exactly, we have taken the surface skin temperature from the NRA at T62 resolution and sampling and computed the correct global mean surface radiation from (SB) as 396.4 W/m² . If we instead take the daily average values, thereby removing the diurnal cycle effects, the value drops to 396.1 W/m² , or a small negative bias. However, large changes occur if we first take the global mean temperature. In that case the answer is the same for 6-hourly, daily, climatological means at 389.2 W/m² . Hence, the lack of resolution of the spatial structure leads to a low bias of about 7.2 W/m² . Indeed, when we compare the surface upward radiation from reanalyses that resolve the full spatial structure the values range from 393.4 to 396.0 W/m² .
The surface emissivity is not unity, except perhaps in snow and ice regions, and it tends to be lowest in sand and desert regions, thereby slightly offsetting effects of the high temperatures on LW upwelling radiation. It also varies with spectral band (see Chédin et al. 2004, for discussion). Wilber et al. (1999) estimate the broadband water emissivity as 0.9907 and compute emissions for their best-estimated surface emissivity versus unity. Differences are up to 6 W/m² in deserts, and can exceed 1.5 W m² in barren areas and shrublands.”
I see plenty to argue about in the detail of K&T 2009, but let’s get the facts straight.

wstannard
October 27, 2011 3:31 am

According to Trenberth’s figure the downward radiation from the GH gasses in the atmosphere. is 333 W/m2. But there must be an equal amount upward. Does’nt the atmosphere radiate infrared in all directions. If 333 W/m2 were radiated from the atmosphere to space we would have far more energy leaving the Earth then received! Surely this backradiation figure is a nonsense.
Can some one answer another question. All up 239 units of IR radiation are emitted to space, 40 directly from the Earths surface, 199 from the atmosphere (including clouds). How is this 199 units radiated to space? – It has to be by Greenhouse Gases in the atmosphere. If these gases were not present this energy could not be emitted. By increasing GH gas concentrations the ability for the atmosphere to radiate energy to space is increased. Where does warming come from?

wayne Job
October 27, 2011 3:33 am

FJ a most profound analysis of an old graph, all imputs to the discombubilation of the hockey team are a step forward to reality and common sense. Well done mate.

Robert Clemenzi
October 27, 2011 3:38 am

MODTRAN looking down will almost always be different than satellite measurements because a significant amount of the available heat is moved from the equator toward the poles via the wind. Thus, at the equator a satellite measures less heat than computed, and at the winter poles, much more.
Because the energy coming from clouds is sometimes much more than predicted by Stefan’s equation, it makes more sense to treat them as mirrors than as blackbody emitters. Yes, I have measured the energy using an IR thermometer. Very interesting.

October 27, 2011 4:13 am

Bob Fernley-Jones says: October 26, 2011 at 8:57 pm
“Tim Folkerts @ 6:21 pm
I can understand that you find the vector consideration a tad obscure, but nevertheless, vectors are very useful for analysing parameters having directional information.

Actually, I don’t find vectors particularly obscure, I work with vectors regularly, and I have done the vector surface integrals that show the upward flux will indeed remain constant (in the idea case: uniform temperature, transparent atmosphere, not TOO high above the surface). Furthermore, these integrals are indeed overkill, as Erik Ramberg said earlier: “I think everyone is making this way too complicated. Physics is all about symmetries.” The symmetry of the idea case makes it obvious that the net upward flux is indeed constant.

Bomber_the_Cat
October 27, 2011 4:20 am

Fred Berple raises the old ‘ice cube and hot object fallacy’. He says at 9:38 PM,
“A large block of ice and a small candle flame both emit the same amount of EMR. However, only one is capable of warming a human being.”
Normally we think of an ice cube as being cold, because most things around us are normally warmer than that. Thus the analogy seems reasonable at first sight. How could an ice cube warm us? We all make the implicit assumption that if the ice cube wasn’t there, there would be something else, warmer, in its place – this is our normal experience.
But lets say you are in the cold vacuum of space, close to absolute zero. Now the ice cube is relatively warm, and you will receive much more radiation from it than you were previously receiving from ‘nothing’. If the energy received from it is the same as from the candle flame (which was the predicate) then It will provide the same heating effect as the candle flame. A black body absorbs all the radiation falling on it, whether from an ice cube or a candle.
.

October 27, 2011 4:22 am

wstannard says: October 27, 2011 at 3:31 am
According to Trenberth’s figure the downward radiation from the GH gasses in the atmosphere. is 333 W/m2. But there must be an equal amount upward. Does’nt the atmosphere radiate infrared in all directions. If 333 W/m2 were radiated from the atmosphere to space we would have far more energy leaving the Earth then received! Surely this backradiation figure is a nonsense.
The diagram is indeed “nonsense” to the extent that it is intended to be about the simplest possible diagram. One vast simplification is that it shows the atmosphere as one object, when in fact the calculations and observations that support it must look at the atmosphere as layered.
Willis Eschenbach created a great “slightly more realistic” diagram that splits the atmosphere into a lower layer and and upper layer. This helps explain why the atmospheric radiation is not the same up into space as down to the surface.
http://homepage.mac.com/williseschenbach/trenberth_mine_latest_big.jpg

son of mulder
October 27, 2011 4:38 am

A question inspired by this article though only slightly connected. If there is some warming from the increased CO2 in the atmosphere and so the air gets warmer, but because it is an open system the atmosphere will expand slightly and increase the mean free path of the photons leaving the surface. This would mean the window to space would get slightly larger and so act as a negative feedback. Has anyone, anywhere quantified the magnitude of this and is it significant?

October 27, 2011 4:58 am

“This is not science, it is an attempt to understand a climate system so complex that it can be discussed, modelled, hypothesised, even analysed, but can never confidently be understood and defined in scientific certainties. Not even attaching statistical degrees of uncertainty resolves this issue. The number and degree of uncertainties far outweigh our ability to make meaningful conclusions. In particular conclusions that require or support changes to our civilization such as are currently being FORCED upon us.”
I suggest the bigger picture is, by far, the more important aspect. The above comment is extracted from my blog http://tgrule.wordpress.com/2011/10/27/global-warming-more-questions-than-answers/, where this post is acknowledged.
Thanks to Anthony for the usual ‘on the ball’ posts.

Myrrh
October 27, 2011 5:15 am

For one heart-stopping moment I thought someone had really got what was wrong with ” EMR is a form of energy that is sometimes confused with HEAT.”
Ask why is it so? says:
October 26, 2011 at 9:06 pm
I’m not sure I understand this article but just to clear a few things up, radiation is not heat. Heat is the result of the absorption of radiation by a surface or molecule. 2nd Law of Thermodynamics does state that heat (without work) will only travel from higher to lower temperature, that is correct, however, radiation is not heat.
Thermal radiation, thermal infrared, is heat. It is the thermal energy of the Sun. It is heat on the move. Not all electromagnetic radiation is heat.
What is wrong with this energy budget is that LIGHT from the Sun, shortwave, non-thermal, has been given the properties of HEAT from the Sun, which is thermal energy on the move, thermal infrared from the Sun direct to us, thermal radiation.
For example, visible LIGHT cannot heat water because it doesn’t have the mechanism to do so, water is a transparent medium for visible LIGHT and so visible light is transmitted through without being absorbed, visible light cannot move the molecules of water into the vibrational state which is heat.
HEAT on the other hand, does have the mechanism to heat water, and does. This is the real energy which is heating the land and oceans directly from the Sun.
Not only has the KT taken out the world’s real energy budget the energy direct from the Sun to the surface which is the real source of heating the Earth, the Sun’s thermal energy, thermal infrared – it has given the properties of thermal infrared to visible, it has swapped HEAT to LIGHT around. It is saying that shortwave is thermal energy.
Not so much a paradox, more junk science fiction produced by the AGWSF department to sell its wares.
I have just had a long discussion on this and do not have the time to go through it all again here, but for any interested in telling the difference between this junk energy budget in order to understand what is really going on in our world, this has to begin with disentangling the science fiction memes about heat and light from the real physical properties of energy and matter.
You can start here with the difference between Heat and Light:
http://wattsupwiththat.com/2011/10/18/replicating-al-gores-climate-101-video-experiment-shows-that-his-high-school-physics-could-never-work-as-advertised/#comment-778960
=======================
Bob Fernley-Jones – your link to HEAT takes to a page of junk physics.
Here’s real world physics on HEAT.

http://thermalenergy.org/
“What is thermal energy ?
Thermal Energy: A specialized term that refers to the part of the internal energy of a system which is the total present kinetic energy resulting from the random movements of atoms and molecules.
The ultimate source of thermal energy available to mankind is the sun, the huge thermo-nuclear furnace that supplies the earth with the heat and light that are essential to life. The nuclear fusion in the sun increases the sun’s thermal energy. Once the thermal energy leaves the sun (in the form of radiation) it is called heat.
Heat is thermal energy in transfer. Thermal energy is part of the overall internal energy of a system.
At a more basic level, thermal energy comes from the movement of atoms and molecules in matter. It is a form of kinetic energy produced from the random movements of those molecules. Thermal energy of a system can be increased or decreased.
When you put your hand over a hot stove you can feel the heat. You are feeling thermal energy in transfer. The atoms and molecules in the metal of the burner are moving very rapidly because the electrical energy from the wall outlet has increased the thermal energy in the burner. We all know what happens when we rub our hands together. Our mechanical energy increases the thermal energy content of the atoms in our hands and skin. We then feel the consequence of this – heat. Laws of Thermodynamics [link]”
[And]: http://thermalenergy.org/heattransfer.php
“Thermal energy and heat are often confused. Rightly so because they are physically the same thing. Heat is always the thermal energy of some system. Using the word heat helps physicists to make a distinction relative to the system they are talking about.”</blockquote

Bill Illis
October 27, 2011 5:19 am

The radiation budget graphic should be re-done in various scenarios:
– at noon when it is cloudy;
– at noon when it is clear;
– middle of the night when it is cloudy;
– middle of the night when it is clear.
– for the surface;
– for the tropopause
It will be a far less cloudy picture and be much more clear in that case.
Incoming solar irradiance is 1366 W/mw at noon and Zero at night. All the numbers are vastly different when it is cloudy versus clear. The tropopause energy levels barely changes at all from day to night, from clear to cloudy conditions, everything happens below that level.

alex
October 27, 2011 5:40 am

Anglo-Oz “physics”.
It would be better, WUWT abstains from such “contributions”.

October 27, 2011 5:51 am

Sorry, but the Earth has no lid, and not having a lid impedes heat confining, as the so called “green-house effect” it is not other thing than confined heat:
http://www.scribd.com/doc/28018819/Greenhouse-Niels-Bohr
What would you choose for keeping your feet warm, a bottle filled with warm air or, instead, a bottle filled with hot water?….That´s simply because its Volumetric Heat Capacity is:
Air=0.001297 joules cm3/kg
Water=4.186 joules cm3/kg
This means that water holds heat 3227 times more than air.

mkelly
October 27, 2011 6:48 am

Why is it that the 161 W/m^2 incoming from the sun can heat my sun tea, but the 333 W/m^2 down from the atmosphere (CO2) cannot?

MartinGAtkins
October 27, 2011 7:08 am

wstannard says:
October 27, 2011 at 3:31 am
According to Trenberth’s figure the downward radiation from the GH gasses in the atmosphere. is 333 W/m2. But there must be an equal amount upward.
I have a similar problem with the first graphic. On the left hand side it has 161 W/m^2 being absorbed by the surface as the initial forcing but on the right hand side 396 W/m^2 as out going surface radiation.
If this figure is derived from near surface back radiation then the value in an open system can only be twice the constant input value of the black body (BB) radiator. This is because the near gas molecule can only accumulate enough energy to radiate (in this case) 161 Watts back too the BB and 161 Watts away. The BB and molecule will then be at radiative equilibrium of 322 Watts. I don’t understand where the 396 W/m^2 comes from.

R. Gates
October 27, 2011 7:11 am

jason says:
October 27, 2011 at 12:24 am
Very interesting discussion, which demonstrates clearly that the view that sceptics are politically rather than scientifically motivated is false.
Good to see R Gates removing more of his mask. He starts off posting in the arctic threads arguing with steve m, then reveals he is a buddy of Trenberth, and now demonstrates in this thread that his understanding of the subject goes way above a layman. R Gates is I believe someone far closer to the heart of the debate than he has let on until now. Genuinely sceptical mind turned, or trojan horse?
_____
Really, aspersions as to my intentions, relationships, etc. are quite unecessary and, more importantly, completely inaccurate. I comment here when I see something that sparks my interest, I have no relationship to Dr. Trenberth other than the fact that we live in the same state, and I think for many of those frequenting WUWT our knowledge of the subject at hand go well beyond the “layman”.
In terms of my own position of AGW. I remain more convinced than not that it is happening (I put the percentage at 75% just to show it is more than 51/50, but not 99%). Finally, I am currently not a believer in catastrophic AGW.

H.R.
October 27, 2011 7:58 am

@R. Gates says:
October 27, 2011 at 7:11 am
“[…]
In terms of my own position of AGW. I remain more convinced than not that it is happening (I put the percentage at 75% just to show it is more than 51/50, but not 99%). Finally, I am currently not a believer in catastrophic AGW.”

I’ve known known your position stated in the first portion of that last paragraph, as anyone who’s read WUWT for a while has seen you post that a number of times phrased a number of different ways. I don’t recall you ever posting a statement as direct about CAGW as that last sentence, though you could have and I missed it. (P.S. Thanks for your comments. WUWT is not an echo chamber and it’s because you are one of several who helps keep it that way with some good points that you throw into the mix. Thanks again.)

davidmhoffer
October 27, 2011 8:18 am

R. Gates;
Really, aspersions as to my intentions, relationships, etc. are quite unecessary and, more importantly, completely inaccurate.>>>
REPLY: If you would provide straight forward answers to questions that wouldn’t be the case. But you don’t.
R. Gates;
I comment here when I see something that sparks my interest, I have no relationship to Dr. Trenberth other than the fact that we live in the same state,>>>
REPLY: You proudly proclaimed yourself as the “go between” who arranged for an invitation for 20 people from WUWT to meet with Kevin Trenberth. You accomplished this merely by living in the same state as Trenberth? I call BS.
R. Gates;
and I think for many of those frequenting WUWT our knowledge of the subject at hand go well beyond the “layman”.>>>
REPLY: Putting aside the royal “our”, excuse me, but you barely qualify as a layman. Consider the bet with me which you jumped into and then welched on. You agreed to wager that Al Gore’s experiment, if conducted as illustrated, would show the results as illustrated. You failed to notice that Al Gore’s experiment as illustrated made use of infrared heating lamps, making it IMPOSSIBLE for the experiment to demonstrate the greenhouse effect in the first place. Then, to compound your so called “well beyond” the knowledge of a layman, you suggested taking the globes OUT of the jars as they were superflous to the experiment. Really? Tell me please, with nothing in the jar to absorb SW and re-radiate it as LW, exactly how could the “greenhouse effect” be demonstrated? That thread alone demonstrated how limited your understanding of the physics is, and I could draw many more examples from many more threads. You constantly parrot the words of others, and when people with knowledge start asking pointed questions, you get backed into a corner and simply stop responding. Or welch on the bet. Whatever.
R. Gates;
In terms of my own position of AGW. I remain more convinced than not that it is happening (I put the percentage at 75% just to show it is more than 51/50, but not 99%). Finally, I am currently not a believer in catastrophic AGW.>>>
REPLY: Really? Can you point to a single skeptic comment about any part of the CAGW debate anywhere in the WUWT forum? Ever? Just one? As for not believing in “C”AGW, shall I cut and paste from all those comments of yours about chaos theory and sand piles collapsing under the weight of a single grain of sand?

Matt G
October 27, 2011 8:23 am

The difference between solar radiation and back radiaiton is that the latter is hardly absorbed at all by a volume of water. ( 65 percent solar) Water in the energy buget diagram is completely ignored is this situation and despite the oceans being the most important with regulating global atmospheric temperatures. This is easily shown by comparing how a volume of water warms during one day in the sun and in the shade. Therefore the heat transfer is demonstrated to be huge by solar, but negligible with back radiation. I always new the energy buget diagram was wrong and didn’t reflect observed reality.

October 27, 2011 8:24 am

Myrrh says:
October 27, 2011 at 5:15 am
For example, visible LIGHT cannot heat water because it doesn’t have the mechanism to do so, water is a transparent medium for visible LIGHT and so visible light is transmitted through without being absorbed, visible light cannot move the molecules of water into the vibrational state which is heat.
Yet it does. 100 meters of water absorbs all the visible light falling on it, so heats it. Visible light move water molecules into overtones of symmetric and anti-symmetric vibrations.

David
October 27, 2011 8:24 am

R Gates says…In terms of my own position of AGW. I remain more convinced than not that it is happening (I put the percentage at 75% just to show it is more than 51/50, but not 99%). Finally, I am currently not a believer in catastrophic AGW.
Well there is wisdom in that, would you care to quantify the warming thus far between natural, UHI, and AGW?
One more question if you are in the mood. If a hypothetical GHG were to increase the LWR in the atmosphere, but decrease the SWR entering the oceans by an equal amount, which would have a greater effect on the earths energy budget and would that affect be warming or cooling?

JPeden
October 27, 2011 8:28 am

Richard Keen says:
October 26, 2011 at 6:52 pm
Along with the paradox, the Trenberth et al. diagram also contains a statistical fantasy, which is the obscenely precise value of 0.9 W/m2 for the net absorbed.
http://icecap.us/index.php/go/icing-the-hype/the_flat_earth

Thanks. I’ve always had a hard time even looking at Trenberth’s representation, because it’s really too much like a Cartoon out of some pre-Kindergarten book.

Matt G
October 27, 2011 8:38 am

R. Gates says:
October 27, 2011 at 7:11 am
“Finally, I am currently not a believer in catastrophic AGW.”
That’s quite a statement because the main sceptic argument has been this, not that a little AGW likely doesn’t have at least a little influence.

Matt G
October 27, 2011 8:44 am

Matt G says:
October 27, 2011 at 8:23 am
The format doesn’t like “” close together with numbers.
Therefore the bracket in this previous highlighted post should read ” (absorbance less than 1 percent LWDR v greater than 65 percent solar)”

October 27, 2011 8:44 am

mkelly askes:
“Why is it that the 161 W/m^2 incoming from the sun can heat my sun tea, but the 333 W/m^2 down from the atmosphere (CO2) cannot?
Perhaps you are looking at it incorrectly. The “natural” temperature of the universe is 3K. Absent some input of energy, objects would radiate away energy until they reached this temperature. If you took your sun tea far from the earth and sun, the sun tea would cool to 3K. Bring that sun tea back to the earth in a transparent, insulated container that removed any heat transfer by conduction. Even if all sunlight is blocked, the IR radiation would heat the sun tea to ~ 300 K.
Conversely, if you kept the tea far from the earth’s IR, but exposed it to 161 W/m^2 of sunlight, it would warm up, but it would end up well below 300 K (around 230 K for a blackbody).
So it is easy to argue that the earth’s IR is BETTER at warming your sun tea than the sun. The sun can only effectively warm the tea if the IR is ALREADY present.

October 27, 2011 9:04 am

Myrrh,
Without going in to the rest what you wrote, let me simple state that the source you choose as “real world physics” gives two completely contradictory definitions of “heat”.
1) “Heat is thermal energy in transfer.”
2) “Heat is always the thermal energy of some system.”
“Heat” cannot be both always within a system and always between two different systems.
If you base your posts on a source that is clearly inconsistent, then there is a good chance that any of your conclusions will also be inconsistent.

October 27, 2011 9:04 am

This is a most welcome article. Many thanks to WUWT for having the cahones to showcase it.The comments are also a delight as they reveal that more and more skeptics are waking up to the ‘Slayers’ science. The paradigm shift is upon us: there is no greenhouse gas effect and the K-T energy budget is utterly bogus.
Astrophysicist, Joe Postma says it all here:
http://principia-scientific.org/publications/The_Model_Atmosphere.pdf

mkelly
October 27, 2011 9:09 am

Tim Folkerts says:
October 27, 2011 at 8:44 am
mkelly askes:
“Why is it that the 161 W/m^2 incoming from the sun can heat my sun tea, but the 333 W/m^2 down from the atmosphere (CO2) cannot?
Mr. Folkerts says: “Even if all sunlight is blocked, the IR radiation would heat the sun tea to ~ 300 K.”
Balderdash. The IR given off by CO2 cannot do this. The 15 micro emission has an associated temperature via Wien’s Law of -73 C. The best that could happen is the tea would attain the air temperature via conduction.
Also, please let us talk of the earth and the atmosphere and not of space. I don’t live in space beyond the atmosphere. So what is important is what does or can happen where I live. I was looking at it correctly. The KT diagram says 161 W/m^2 at the earth’s surface that is where I put my sun tea and the sun heats it, CO2 cannot.

Robert Clemenzi
October 27, 2011 9:12 am

David Socrates asks:
1. Does back radiation to the Earth’s surface occur at all or is it “unphysical”?
Yes it does. It is easy to measure it with an IR thermometer. It can be seen in lapse rate plots. Back radiation is what causes the morning temperature inversion over land.
2. If it does, what proportion of that back radiation is due to CO2?
That is the important question. No one knows. My analysis indicates that it is close to zero at the surface and about 100% in the stratosphere.
To all, the fact that the amount of energy toward space is less than the amount toward the surface is proof that the atmosphere is IR opaque over a significant part of the spectrum.

Editor
October 27, 2011 9:15 am

R Gates
I have asked you this many times and perhaps you have answered but it got lost within the deluge of comments.
When do you believe the globe started warming? (we shall leave aside the one third of stations that are cooling that we pointed out in our article ‘in search of cooling trends’
I believe it started warming from 1607/8 albeit it has been in fits and starts with numerous serious reversals and astonishing advances (such as around 1700.)
Berkely appear to partially support my view with a rise from their start date of 1800.
Supplementary question; Who do you believe, Michael Mann or the Berkely study?
tonyb

October 27, 2011 9:22 am

We are just waiting for a very “cool” cooling….enjoy it!, it will be better than in the Maunder Minimum 🙂

October 27, 2011 9:24 am

You might like more writing on the paradox summarized quite nicely here:
http://www.tech-know.eu/uploads/Copernicus_Meets_the_Greenhouse_Effect.pdf
and a more lengthy version here:
http://www.tech-know.eu/uploads/The_Model_Atmosphere.pdf

David
October 27, 2011 9:28 am

Leif Svalgaard says:
October 27, 2011 at 8:24 am
Myrrh says:
October 27, 2011 at 5:15 am
For example, visible LIGHT cannot heat water because it doesn’t have the mechanism to do so, water is a transparent medium for visible LIGHT and so visible light is transmitted through without being absorbed, visible light cannot move the molecules of water into the vibrational state which is heat.
“Yet it does. 100 meters of water absorbs all the visible light falling on it, so heats it. Visible light move water molecules into overtones of symmetric and anti-symmetric vibrations.”
A great deal of the ocean has greater clarity and a small percentage, about 1% goes deeper. (sometinmes called the bad light zone) However this 1% may be in a spectrum that varies more then most over solar cycles. The intensity of visible light transmitted through water decreases rapidly with
increasing depth. Roughly 60 percent of incoming light is attenuated in the first meter (3.3 ft), 80
percent in the upper 10 meters (33 ft), 99 percent at a depth of 150 meters (500 ft) in
very clear water, and there is essentially no light penetration below 1000 meters.
http://www.google.com/url?sa=t&rct=j&q=solar%20radiation%20penetrates%20bad%20light%20zone%20oceans%20800%20meters&source=web&cd=3&ved=0CDQQFjAC&url=http%3A%2F%2Fwww.mhhe.com%2Fearthsci%2Fgeology%2Fduxfund4e%2Fstudy%2Fchap5.pdf&ei=dYapTuCMHoKhsQLUtonmDw&usg=AFQjCNGnVOSvWJbH6orsFOLUGmjjve5hMQ

David
October 27, 2011 9:35 am

Leif, a second reference for you…
http://www.google.com/url?sa=t&rct=j&q=bad%20light%20zone&source=web&cd=10&ved=0CHUQFjAJ&url=http%3A%2F%2Fwww.scienceclarified.com%2FMu-Oi%2FOcean-Zones.html&ei=CIepTrzlLsHnsQKqy52vDw&usg=AFQjCNHtNuybbpchX9S35Pp8IywjkMq8hQ
From 660 to 3,000 feet (200 to 900 meters), only about 1 percent of sunlight penetrates. This layer is known as the dysphotic zone (meaning “bad light”). Below this layer, down to the deepest parts of the ocean, it is perpetual night. This last layer is called the aphotic zone (meaning “without light”). At one time, scientists thought that very little life existed within the aphotic zone. However, they now know that a variety of interesting organisms can be found living on the deepest parts of the ocean floor.
Read more: Ocean Zones – body, used, water, process, Earth, life, plants, chemical, form, energy, animals, carbon, oxygen, parts, primary, plant, surface http://www.scienceclarified.com/Mu-Oi/Ocean-Zones.html#ixzz1c06eJJzu

Lars P.
October 27, 2011 9:55 am

Bob, thank you for the article! It is clear that you “ruminated” some time on it give us some time to digest the information. Here my two cents:
As you correctly showed – in divergence to T&All (F1 diagram) – the radiative exchange does not happen between the “floor” and a “ceiling”, but between “floor” and all levels of the atmosphere. The 333 W/m2 “back radiation” from a high ceiling is an erroneous number leading to misunderstandings – this is clearly showed in the tables – high cloud area is above 7 km – looking up from 6 km in the table we have between 58 and 125 W/m2 pointing down. There is no “back radiation” but energy transfer through radiation in all directions.
Figure 2 is a much more clear and without bias overview of the energy budget.
Furthermore the second point was that radiation results also in heat transfer between various strata of the atmosphere, especially with warm air raising up.
The third point I retained was that the outgoing radiation seen by Nimbus 4 is a result of radiation from several strata. The use of S-B formula is of limited help and should be used very carefully.
Finally I think there are bigger problems with the F1 diagram as pointed out by Mydogsgotnonose says: October 27, 2011 at 3:03 am, Myrrh says: October 27, 2011 at 5:15 am (is any radiation equally transferred into heat?), George E. Smith; says:October 26, 2011 at 10:00 pm (bogus flat world asumption etc) and others.

October 27, 2011 10:36 am

David says:
October 27, 2011 at 9:35 am
Leif, a second reference for you…
You are missing the point completely. The point is that the oceans eventually absorb all the visible light falling on them [minus that small part that is reflected] and hence is heated by that.

October 27, 2011 10:38 am

Mkelly, we are a bit off topic, but let me say a couple quick things
>>Mr. Folkerts says: “Even if all sunlight is blocked, the IR radiation
>>would heat the sun tea to ~ 300 K.”
>Balderdash. The IR given off by CO2 cannot do this.
The ~ 333 W/m^2 of IR we are talking about comes partly from CO2, but also from H2O (both gas and liquid). (In addition, the sides of the sun tea will be exposed to both the atmosphere and the ground, further raising the incoming IR). On a cloudy day, the IR spectrum will be nearly black body, and would warm the sun tea to ~ room temperature without any input from the sun or from conduction.
>The 15 micro emission has an associated temperature via
>Wien’s Law of -73 C.
No, Wien’s Law specifically applies to BB radiation. You can’t conclude that a non-BB like CO2 emits most strongly at 15 um has an effective temperature of (2.90 E-3 K/m) / (15 E-6 m) = ~ -73 C.
>Also, please let us talk of the earth and the atmosphere and not of space.
>I don’t live in space beyond the atmosphere.
The connection to space cannot be ignored. In fact, that is precisely the reason that CO2 by itself could not heat the sun tea to ~ 300 K as you correctly concluded above. CO2 by itself only emits in certain bands, leaving large “windows”. These “windows” connect space to everything on the ground, so we are all “connected” to space, just as we are connected to the sun. This is why the tops of cars are often frosted when the sides are not.
>The KT diagram says 161 W/m^2 at the earth’s surface that is where
>I put my sun tea and the sun heats it, CO2 cannot.
Actually, that is not what the diagram says. It says that during an average 24 hour period, the average solar energy is 161 W/m^2. During the night the average is 0 and during an average the day the average is 2 * 161 W/m^2 = 322 W/m^2. Assuming you are making tea not too early or late on a sunny day, the solar energy at that time could be up to about 1000 W/m^2. This is one other reason that sunlight is more effective.

rbateman
October 27, 2011 10:45 am

The Sun does not surround the Earth, illuminating and heating it from the vertical all day long at all points on the surface. The Sun illuminates/heats the dayside of the Earth as if it were 1/2 a circle the size of the Earth….from the vertical. So, the Earth’s dayside gets, at most, 1/2 of the 1366 W/m^2. 1366/2 = 683.

October 27, 2011 11:11 am

rbateman says:
October 27, 2011 at 10:45 am
The Sun does not surround the Earth, illuminating and heating it from the vertical all day long at all points on the surface. The Sun illuminates/heats the dayside of the Earth as if it were 1/2 a circle the size of the Earth….from the vertical. So, the Earth’s dayside gets, at most, 1/2 of the 1366 W/m^2. 1366/2 = 683.
And because the Earth is round, you can cut that number in half as well, getting to the 342 W/m2.

Rich
October 27, 2011 11:14 am

Anyway, shouldn’t it be A^2 + B^2 = C^2 for that little vector diagram, not A + B = C?

October 27, 2011 11:22 am

A black body emits W/m2 AT A POWER according to it’s temperature.
A gray body emits AT A LOWER POWER than a black body (at the same temperature) for many well known reasons.
THAT, is NOT an energy flow, for a gray body IT IS the power the energy flow is emitted at…….
The volume of the energy flow IS NOT DESCRIBED.
So, as the gray bodies depicted in the K&T budgets are all different sizes and temperatures,
the diagram CAN NOT be correct (for power or volume),
UNLESS THEY ARE ALL (EQUAL) BLACK BODIES….
This also applies to all present explanations of GH “theory”.
Did you realise K&T and GH is ALL explained in “black body”???
It has taken me ages to realise the above, and it’s importance. It’s all imaginary.
http://www.globalwarmingskeptics.info/forums/thread-1071-post-10470.html#pid10470
(Particularly post 21 onwards.)
I doubt I am alone in that.
K&T depicts a black body world, in all it’s parts, AND, with no life, as does GH “theory”.
Makes me wonder how they “measure” the figures supposedly from (gray body) “reality”….
It’s a scam, plain and simple. A politically convenient, imaginary hobgoblin.
The world ain’t flat, and CO2 does not drive, nor even measurably influence, climate.
Wake up people.
There can not possibly be a “greenhouse effect” as presently hypothesized,
IN ANY OF IT’S PRESENTLY TOUTED “FORMS”.

October 27, 2011 11:28 am

Rich says:
Anyway, shouldn’t it be A^2 + B^2 = C^2 for that little vector diagram, not A + B = C?
If you assume he meant that as a vector equation (with the little arrows over the letters), then (vector A) + (vector B) = (vector C).
But if they represent the magnitudes, then you are right, the Pythagorean theorem would apply.

Myrrh
October 27, 2011 11:30 am

Leif Svalgaard says:
October 27, 2011 at 8:24 am
Myrrh says:
October 27, 2011 at 5:15 am
For example, visible LIGHT cannot heat water because it doesn’t have the mechanism to do so, water is a transparent medium for visible LIGHT and so visible light is transmitted through without being absorbed, visible light cannot move the molecules of water into the vibrational state which is heat.
Yet it does. 100 meters of water absorbs all the visible light falling on it, so heats it. Visible light move water molecules into overtones of symmetric and anti-symmetric vibrations.

Utter codswallop. Water is a transparent medium to visible light, visible light is transmitted through without being absorbed because that’s what what transparent means. Visible light isn’t capable of moving molecules of water into vibration, it ain’t big enough and doesn’t have the power to do so, it works on electron scale, electronic transitions, and doesn’t even get in to play with electrons in water (as it does in the atmosphere where the electrons of the molecules oxygen and nitrogen absorb it. It takes real power, the power of heat, thermal infrared, to move molecules of water. And this is what it does, direct from the Sun, the land and oceans are heated by thermal infrared.
The AGW/KT/NewScienceFiction is a nonsense. The Science Fiction department of AGWInc has given the properties of heat, the thermal energy from the Sun, to light, and, then says that the Sun’s real thermal energy thermal infrared, doesn’t get to the surface to heat it, and takes it out of the budget! To claim the actual heat from the Sun doesn’t heat the world is bad enough, but for anyone here calling calling himself scientist who thinks visible light is capable of doing this, needs to go back to primary school, in the real world, there are still some teachers left…
….
Tim Folkerts says:
October 27, 2011 at 9:04 am
Myrrh,
Without going in to the rest what you wrote, let me simple state that the source you choose as “real world physics” gives two completely contradictory definitions of “heat”.
1) “Heat is thermal energy in transfer.”
2) “Heat is always the thermal energy of some system.”
“Heat” cannot be both always within a system and always between two different systems.
If you base your posts on a source that is clearly inconsistent, then there is a good chance that any of your conclusions will also be inconsistent.

No contradiction – “Using the word heat helps physicists to make a distinction relative to the system they are talking about” It’s all of those things in context, because “thermal energy and heat are the same thing.”
What I am talking about here in context is the Sun’s heat, the Sun’s thermal energy on the move, thermal infrared which is this heat energy, this heat from the Sun travelling to us here and heating land and oceans and us. Or simply, heat from the Sun is thermal infrared, it is invisible, but we can feel it as heat because it warms us up from the inside. Because it is heat and that’s what heat does.
Visible light cannot heat us up.
You have excluded the primary means of heating of the Earth, it is missing from your energy budget, this is the real travesty of the missing heat.

Myrrh
October 27, 2011 11:52 am

Man Bearpigg says:
October 26, 2011 at 11:54 pm
.. and they call us skeptics ‘Flat Earthers’ when their calculations are based exactly on that notion.
A blast from the past:
“NASA covered up for forty years proof that the greenhouse gas theory was bogus. But even worse, did the U.S. space agency fudge its numbers on Earth’s energy budget to cover up the facts?”
http://sppiblog.org/tag/stefan-boltzmann-equations
“What ignited this latest Climategate-linked rumpus is a sensational new research paper, ‘A Greenhouse Effect on the Moon?’ otherwise called the ‘Moon Paper.’
Researchers for the paper scientifically proved that since at least 1997 climate scientists knew that guesswork was underpinning the whole greenhouse gas theory. In fact, so flaky are these numbers that they can be rendered to show a GHG effect on Earth’s moon, where no greenhouse gases exist! Thus, skeptics argue, the burning embers of political heat generated by the discredited theory should now finally and unequivocally be extinguished.
But more sinisterly, it turns out that NASA climate scientists, with access to better climate equations used for the Apollo Moon mission, forsook those in favor of dodgy Dr. Schmidt’s ‘back of an envelope’ numbers.”
http://usactionnews.com/2010/06/nasa-charged-in-new-climate-fakery-greenhouse-gas-data-bogus/
NASA knew how to calculate it properly, it had to junk Stefan-Bolzmann to get accurate figures for the moon landing.

mkelly
October 27, 2011 12:03 pm

Mr. folkerts says: “Assuming you are making tea not too early or late on a sunny day, the solar energy at that time could be up to about 1000 W/m^2. This is one other reason that sunlight is more effective.”
This I fully agree with the 1000 W/m^2 is why the diagram is not correct. However, you obfiscate on other issues, but on this we are in total agreement. You are coming along my little padawan.

R. Gates
October 27, 2011 12:04 pm

climatereason says:
October 27, 2011 at 9:15 am
R Gates
I have asked you this many times and perhaps you have answered but it got lost within the deluge of comments.
When do you believe the globe started warming? (we shall leave aside the one third of stations that are cooling that we pointed out in our article ‘in search of cooling trends’
I believe it started warming from 1607/8 albeit it has been in fits and starts with numerous serious reversals and astonishing advances (such as around 1700.)
Berkely appear to partially support my view with a rise from their start date of 1800.
Supplementary question; Who do you believe, Michael Mann or the Berkely study?
tonyb
_____
Tony, by your question, I take it you mean when did human activity (specifically the burning of fossil fuels) begin to warm the climate? I think the issue would be one of measureability of the signal through any natural background fluctuations. Certainly, IMO, most of the decadal, century scale, and even 1500 year cycles in climate (i.e. non-Milankovitch) can be traced back to the sun. They may be reflected in ocean cycles, but the oceans themselves are not the cause of the cycles, but only a resultant effect. Then there are the shorter term cycles in climate linked to events like ENSO, volcanoes, etc. So in trying to find an anthropogenic signal amongst all this other natural background climate noise is a difficult proposition. For example, there can be times when the natural cycle is toward cooling (i.e. a Maunder or Dalton Minimum), but there could be human influences which make the cooling less severe or deep than it might have been. Identifying that type of effect (a less intense cooling) as in indication of warming is certainly quite a challenge. However, not to dodge your question, in my opinion the first really visible signs of human warming of the planet would be early in the 20th century, with the signal growing progressively stronger throughout (even though there were periods of natural cooling). I think for certain by the early 1980’s we could see temperatures begin to deviate from the natural solar influenced cycle, meaning that the influence of human GH forcing became greater than the solar influence. As to who I believe, Mann or Berkeley, I find that an odd choice. For example, I think the MWP was probably a bit warmer and bit more global than Mann might like to posit, but I also think that in general, the trend of global temperatures increases over the past century as reflected by Berkeley et. al, are pretty much on target, give or take some inconsequential adjustments. Based on natural cycles plus human forcings, we are due for continued warming (despite the AGW skeptics excitement over the flattening of the upward trend over the past decade). So we are due for warmer, not cooler times ahead…and the next glacial period by Milankovitch cycles is not due for more than 10,000 years.
One final note, of course human influences on the climate go both ways , with some leading to cooling and some leading to warming. So even these can be competing signals against the background of natural forcings.

October 27, 2011 12:05 pm

Bob_FJ,
As to the last section where you claim “333 W/m^2 Shown as coming from high cloud area (= BS according to MODTRAN)”
First, the 333 W/m^2 is coming from the atmosphere as a whole, not “high clouds”. It includes CO2 and humidity and high clouds and low clouds.
And following up on that point, the calculations you have are all for clear skies. If you model a cumulus cloud base, the numbers are much higher. For example, the first line for looking up from 10 m in a clear tropical sky was 348 W/m^2 in your table, but 418 W/m^2 when cloudy. Arctic winter goes from 163 W/m^2 to 243 W/m^2. All the other numbers will also be higher for cloudy weather. As such, you would have to average not only the different zones, but also cloudy and clear for each zone.
Given that the downward radiation with cloudy conditions is so much higher than the clear conditions you quoted, it is now plausible that the average from MODTRAN (all areas, all cloud covers) would agree with trenberth’s 333 W/m^2.

Kelvin Vaughan
October 27, 2011 12:27 pm

Leif Svalgaard says:
October 27, 2011 at 11:11 am
rbateman says:
October 27, 2011 at 10:45 am
The Sun does not surround the Earth, illuminating and heating it from the vertical all day long at all points on the surface. The Sun illuminates/heats the dayside of the Earth as if it were 1/2 a circle the size of the Earth….from the vertical. So, the Earth’s dayside gets, at most, 1/2 of the 1366 W/m^2. 1366/2 = 683.
And because the Earth is round, you can cut that number in half as well, getting to the 342 W/m2.
And surely the same argument goes for backradiation, only 1/4 of it will reach the surface of the earth?

October 27, 2011 12:30 pm

mkelly says:October 27, 2011 at 12:03 pm
“This I fully agree with the 1000 W/m^2 is why the diagram is not correct.
This seems like a non-sequitur. We seem to agree that sometimes sunlight is bright (1000 W/m^2 on a sunny noon); sometimes it is sort of bright (~200 W/m^2 later in the afternoon); sometimes it is gone (0 W/m^2 at night or under clouds). A weighted average that comes out to 161 W/m^2 would certainly be possible — how does this invalidate the Trenberth diagram?

October 27, 2011 12:34 pm

Kelvin Vaughan says:
October 27, 2011 at 12:27 pm
And surely the same argument goes for backradiation, only 1/4 of it will reach the surface of the earth?
No. The IR comes from all directions, not just from one direction. There is no “night side” where IR from the atmosphere does not shine.

jae
October 27, 2011 1:14 pm
richard verney
October 27, 2011 1:22 pm

@Leif Svalgaard says:
October 27, 2011 at 8:24 am
Just to be clear, are you talking about pure water, or the soup that is sea water?

davidmhoffer
October 27, 2011 1:26 pm

Tonyb;
Congrats on getting a response from R. Gates… well, sort of.
Me he continues to ignore entirely because the only answer he has left is to admit that he was totaly wrong and welched on his bet with me.

Editor
October 27, 2011 1:29 pm

R gates
Thanks for your answer. So the observed 400 year warming trend only became a warming trend influneced by humans in the last 20 years or so? So for 380 years it was natural and then it switched to being man made. Is that correct?
tonyb

davidmhoffer
October 27, 2011 1:32 pm

Myrrh;
Utter codswallop. Water is a transparent medium to visible light, visible light is transmitted through without being absorbed >>>
In various threads you demanded experimental proof, and when it was provided to you, declared it impossible. You demanded text books, and when reffered to same, declared them irrelevant. You demand that others read and understand your blather, but when confronted with facts, you scream “codswallop”.
Tell me please. Since it gets dark at about 100 meters in the ocean, where did the light go? If it wasn’t absorbed, where did it go? Did it make a U turn and head back to the surface? One would think the casual observer would notice light shining up from the bottom of the ocean? Did it come to a stop and just sit their in suspension in the water? Did it turn into flying pigs? Where does the light go if it wasn’t absorbed by the water? Maybe it just disappeared?

Bart
October 27, 2011 1:43 pm

Tim Folkerts says:
October 26, 2011 at 7:49 pm
“If 10^19 photons of energy 10^-19 J pass thru a 1 m^2 surface oriented perpendicular to the earth’s surface, then 1 J passes thru the surface independent of the direction that the photons are moving.”
Unh-uh. If that were true, we would not need to steer solar arrays to point at the sun to get maximum power. You have to integrate over the projected area, which is proportional to the cosine of the angle between the (-) surface normal and the direction of propagation.

Jeff D
October 27, 2011 2:19 pm

davidmhoffer says:
October 27, 2011 at 1:32 pm
Myrrh;
Utter codswallop. Water is a transparent medium to visible light, visible light is transmitted through without being absorbed >>>
_______________________________________
I don’t have any proof to the unhindered transmission of light threw pure water. However the sea is not pure. Particles and dissolved organics will decrease penetration of light in the sea. How these effect temperature I don’t have a clue.

October 27, 2011 2:24 pm

Myrrh says:
October 27, 2011 at 11:30 am
Utter codswallop. Water is a transparent medium to visible light
You have been shown several times that this is not true. Are you not embarrassed by displaying such ignorance? and repeatedly.
Not only did you get a reference to such a textbook, but also a condensed explanation that should be understandable to the general reader: “A water molecule consists of an oxygen atom with two hydrogen atoms sitting on ‘hydrogen bonds’ [like little springs sticking out from the oxygen atom]. The two springs form an angle of some 105 degrees. There are three main modes of vibrations: mode 1, where the hydrogen atoms vibrate in and out in unison [‘symmetric vibration’]; mode 2 where the bonds bend back and forth, changing the angle;and mode 3 where the hydrogen atoms vibrate in opposite directions [when one goes in, the other goes out – called ‘asymmetric vibration’]. Because of the dense packing in liquid water, mode 2 does not happen [there is not enough room], so in liquid water only mode 1 and 3 occur. As with any vibration, there is a fundamental frequency [which is activated by far infrared], but here are also overtones [or harmonics – this is what makes a note, like ‘A’, sound differently on a violin and a trumpet]. The overtones that combine mode 1 and 3 [combined to 6th and 5th harmonics] correspond to the visual frequencies of light of 511 nm [green] and 606 nm [yellow], so visual light can and does excite vibrations. Overtones are normally of much less amplitude [strength] than the fundamental vibrations, so the visual light absorption is up to million times smaller than that of far IR. That infrared is absorbed [and thus heats] in the first few millimeters of the water column, while it takes 100 meters or more of water to absorb [and be heated by] visual light. But the ocean is deep enough for that, so even visual light gets absorbed [otherwise the ocean floor would be bathed in light – which it is not] and thus heats the oceans.”

October 27, 2011 2:34 pm

Myrrh says:
October 27, 2011 at 11:30 am
Water is a transparent medium to visible light
Here you can learn more about absorption of visible light by water:
http://www.lsbu.ac.uk/water/vibrat.html
” Water is very slightly blue in color [131]c as overtone and combination vibrational absorption bands (albeit far less intense, see above [130]) extend through the red part of the visible spectrum with a small peak at 739 nm and shoulder at 836 nm, both varying somewhat with temperature [268] plus a smaller fourth overtone of the v1:v3 stretch at 606 nm, and very small fifth overtone (at 514 nm) and combined overtone (at 660 nm) bands. This absorption spectrum of water (red light absorbs 100 times more than blue light), together with the five-times greater scattering of blue light over red light, contributes to the blue color of lake, river and ocean waters.”

Bob Fernley-Jones
October 27, 2011 2:48 pm

MostlyHarmless @ October 27, at 3:18 am
You seem to have misunderstood the article in two ways:
1) The problem is expressed in the absorptive spectra, which is why I wrote: (ignoring some [radiation] escaping directly to space)
2) The problem is concerning the S-B isotropic radiation when affected by an absorptive atmosphere. It is not chrystal clear to me what they did with their initial S-B calculation in 2009, but it doesn’t matter how the final number of 396 was derived, and the same argument would apply if they stayed with their 390 number of 1997. That is why I wrote: 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. And why I also put a question mark against the temperature in the tables

October 27, 2011 2:55 pm

Nice Job Bob FJ,
A couple interesting points,
1. The 63 Wm-2 (396-333) is a good bit different than the NASA value of 83Wm-2 (21% at surface) and 59Wm-2 absorbed by the atmosphere versus only 23Wm-2 absorbed in the K&T.
2. The 396Wm-2 from the surface is via S-B with emissivity = 1, perfect black body. If the 333 is assume as a real energy source, via S-B emissivity = 1, The clouds where the DWLR would be originating are at approximately 277 degrees K, I have always found that interesting.
3. The assumption of only up or down radiant energyflux is perfectly fine, until the tropopause roughly. Then the horizontal window should be considered.
The cartoon needs a proper funeral so things can move on.

Bob Fernley-Jones
October 27, 2011 3:13 pm

Michel @ October 27, at 2:58 am
What I wrote in full was: 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”).
The reason I claim this is that most of the radiation is in lateral directions, and even if the photon free path lengths are long, their initial absorption would be near the surface. Of course those streams of a more vertical aspect would mostly go much higher.

October 27, 2011 3:22 pm

>>Tim Folkerts says: October 26, 2011 at 7:49 pm
>>“If 10^19 photons of energy 10^-19 J pass thru a 1 m^2 surface oriented
>>perpendicular to the earth’s surface, then 1 J passes thru the surface
>>independent of the direction that the photons are moving.”
>Bart says: October 27, 2011 at 1:43 pm “Unh-uh. If that were true,
>>we would not need to steer solar arrays to point at the sun to
>>get maximum power. You have to integrate over the projected area… ”
If you read carefully what I said, then we are not in disagreement. Whatever angle a photon hits a solar array, if that photon gets absorbed, it deposits all of its energy. It does not deposit Ecos(theta) or any such thing. This deposits E. Bob_FJ seems to be saying that only the ‘perpendicular component of the photon energy” gets deposited.
I fully agree that if you change the angle of the surface, then the number of photons hitting the surface will change. That is separate issue.

Berényi Péter
October 27, 2011 3:22 pm

Well, It can be put this way:
According to Trenberth heat transfer by thermal radiation from surface to atmosphere is 23 W/m², while thermals + latent heat (related to phase transitions of water) carry off 97 W/m². As 40 W/m² is supposed to escape directly from surface to space through the atmospheric window by thermal radiation, it is only 14.4% of the entire heat flux leaving the surface that can be connected to the greenhouse effect.
More importantly in Table 2a of their paper they say global land area has an overall 15.6 W/m² heat flux deficit, that is, it radiates 5.57 PW more to space than it receives from the sun. This figure matches exactly the latent heat carried by the amount of water vapor carried by winds over land that is seen again as global river runoff to the ocean, glacier discharge added (slightly more than a million cubic meter per second).
Their value of 80 W/m² for the latent heat of evaporation also matches global average precipitation closely (as it should). However, water, once up in the atmosphere in the form of vapor, seldom returns to the surface upon first condensation. Most droplets and ice crystals, up to 90% of them evaporate again as they descend to lower layers, well before reaching the surface. That vapor is then raised again until it re-condenses, and so forth.
That is, even that 14% of heat flux that left the surface as thermal radiation and was re-absorbed by the atmosphere at some (pretty low) level, does not have to proceed painfully upwards as radiation, but is aided in its progress by the water cycle, until it makes up to a level where the air is clear & dry & cold enough to let radiation escape to space.
This level is at the cloud top, of course, and most of the radiation there comes from ice crystals. Emissivity of ice (and liquid water) in the thermal infrared is very close to one, that is, it behaves like a black body. This broad band radiation can hardly be intercepted any more by narrow absorption lines of dilute greenhouse gases at low pressure, which are not even subject to pressure broadening up there.
If the atmosphere gets a bit more opaque in the thermal infrared (due to GHGs), this intermediate water cycle simply gets more vigorous, making the overall average upward heat flux nearly constant.
Anyway, radiative transfer can’t give a serious contribution to heat flux between atmospheric layers as long as various forms of water are not involved, otherwise thermal conductivity of carbon dioxide could not be smaller than that of dry air (which, unlike carbon dioxide, neither emits nor absorbs em radiation in the thermal infrared band).

Leg
October 27, 2011 3:44 pm

Myrrh @5:15 am
As noted earlier by Gail, there are a number of processes that can happen to a photon as it interacts with matter(adsorber). Partial interaction of the photon with an atom will lessen the energy of the photon thereby changing its frequency/wavelength and the photon goes on to interact the adsorber at some distant spot. With complete adsorbtion, the atom will gain all of the photon’s energy. Interaction with a particular atom depends on the frequency/wavelength of the incomng photon and the chances that the photon will interact with an electron or the nucleus of the adsorber. The chances of interacting depend on: how many adsorber atoms there are; how big is the nucleus of each atom; and how many electrons associated with the atom? If there is enough of the absorber (lots of atoms), all photons will be absorbed.
Water does not adsorb light photons very well, but it does adsorb some and it will adsorb all of it if there is enough water. Therefore all of the energy from the sun’s light spectrum (and the rest of the sun’s electromagnetic spectrum photons such as microwave and IR) will get transferred to water just because there is so much water in the oceans.
When a photon, such as a light photon, is partially or completely absorbed by an atom, the atom’s electrons gain extra energy. The atom is now excited and its vibrational state increases. This is not the state that the electrons want to be in, so eventually that extra energy is re-radiated away from the atom at a different wavelength or the energy may be transferred to another atom. For example in water, the excess energy gets shared among the three atoms, H-O-H, and the water molecule is now in an excited state, which we call a “heat” state. I’m using the term heat very loosely here. When we experience hot water, we are experiencing the release of excess energy from the water in the form of photons that transfer energy to our atoms or as a direct transfer of energy to our body’s atoms that are in direct contact with the water molecules.
Anything will act as a photon adsorber, but some things are better adsorbers for specific wavelengths/frequencies than other adsorbers. A black piece of paper will stop most light photons, but not all. This is easily demonstrated using low intensity (few photons) light and a high intensity light (lots of photons). Some light will go through the paper with the high intensity. However, if you put enough pieces of paper together, the high intensity light will get stopped because the black (carbon) molecules are quite good at adsorbing the wavelength/frequencies of the light spectrum. If it was an X-ray photon, at its different wavelength/frequency, the black paper is basically transparent. The same basics apply to water as an adsorber no matter what the frequency/wavelength/frequency.
What is happening to the energy being deposited in the paper? It is exciting the paper molecules and some of it will expressed as heat, much of which gets radiated/transferred to the surrounds. If the energy deposition is too great, the paper molecules may shift and break down (burning).

George E. Smith;
October 27, 2011 4:04 pm

“”””” Kelvin Vaughan says:
October 27, 2011 at 1:16 am
Just a thought! anti phase sine waves of the same frequency cancel out. Does this happen at infra red wavelengths? “””””
Kevin, the question EXACTLY as you have phrased it has a very simple answer; YES.; well actually with one little detail you left out; those sine waves also have to be of the exact same amplitude, to completely cancel. And I bet you knew that too, if you had just thought about it.
So let’s just say that your sine waves are coming from some electronic oscillator circuit. You can of course switch the power to that circuit off and on; whenever you like.
Just what do you suppose is the likelihood, that if you switched the power off, and then on again, that the oscillator will come up in EXACTLY the same phase, that it might have had if you DIDN’T switch the power off; well not bloody likely as they say.
So ingeneral any oscillator goes on and off from time to time, either deliberately or accidently, so it may not remain PHASE CONTINUOUS all the time.
Well most light sources, and most Infra-red sources, are like that off and on sort of sources. They don’t remain phase continuous FOR VERY LONG; ACTUALLY THEY REMAIN CONTINUOUS FOR A VERY DAMN SHORT TIME.
The infra red radiation, may only travel two inches, and suddenly the phase jumps discontinuously to some other place in the cycle. We call it the COHERENCE LENGTH.
Ordinary light sources or IR sources, don’t normally cancel out, because they never remain exactly out of phase, because of theat coherence length.
The most obvious exception is the LASER. A laser, is not unlike an ordinary oscillator, with its phase remaining coherent for quite alarge distance (or time if you will).
As anybody knows, if you shine one of those little red laser pointers at a wall or something, you see this bright SPECKLE PATTERN apparently dancing in front of the wall, and as you move your eye the speckle patter seems to follow you around.
That speckle pattern is nothing more that the laser beam travelling towards the wall, INTERFERING with the beam reflected off the wall, in the space above the wall, where they overlap, and the speckle is a pattern of dark spots, where the two beams cancel, and the bright spots are where they add together to make an even brighter spot.
So the short answer to you r question is yes they can cancel, but only for the time duration when they really are out of phase.
Non laser ordinary light or infrared sources, usually do not cancel each other because the coherence length is too short..
Now Michelson, measured the length of the standard Metre bar against some lines in the spectrum of Cadmium, which happen to be very narrow bandwidth atomic spectral lines, and in that case they have a much longer coherence length than other sources. The Cadmium Red line as measured by Michelson came out at 6438.4696 Angstrom units, or aqbout 643.84696 nano metres. He had to have his coherence length at least 10 cm long to do that measurment, otherwise he would have seen no interference fringes.
There is a green line in the Mercury 198 spectrum, which is even sharper, than the Cadmium Red line, and makes even nicer interference patterns. Trouble is Mercury 198 is not a natural isotope, or is extrremely rare; so you have to make it in a nuclear reactor out of gold; so 198Hg is far more valuable than gold. There are some even better ones, before you have to go to lasers.
Ordinary sources; specially thermal ones, are very noisy so have low coherence lengths, maybe not even mm in extent.

Bob Fernley-Jones
October 27, 2011 4:22 pm

Tim Folkerts @ October 27, at 12:05 pm
The problem with your argument is that Trenberth gives:
Sunlight absorbed by surface (161) = energy leaving surface + the controversial heat retained of 0.9.
161 = 17 + 80 + 63 + 1
The radiative heat transfer from the surface of 63 can also be obtained from the net of the alleged up-down radiation: 396 – 333 = 63
Thus the 396 & 333 is at the surface, but you argue that it is constant up to the high cloud level, including clear skies.
You earlier seem to admit that this can only be the case for an ideal transparent atmosphere, but clear skies are closer to this condition than are cloudy skies

R. Gates
October 27, 2011 4:38 pm

davidmhoffer says:
October 27, 2011 at 1:26 pm
Tonyb;
Congrats on getting a response from R. Gates… well, sort of.
Me he continues to ignore entirely because the only answer he has left is to admit that he was totaly wrong and welched on his bet with me.
_____
Tony simply is far more polite than you and seems to have a better grasp of the the issues.

Bob Fernley-Jones
October 27, 2011 4:45 pm

Tim Folkerts @ October 27, at 3:22 pm & Bart @ October 27 1:43 pm
Tim said: Bob_FJ seems to be saying that only the ‘perpendicular component of the photon energy” gets deposited.
Please see this simple analysis of Lambert’s law (or cosine law in optics, Lambertian or black body reflection)
http://escience.anu.edu.au/lecture/cg/Illumination/lambertCosineLaw.en.html
In the first figure, see the effect of incoming light angle variations. There are more detailed descriptions around, but you can see the effect of vectors in the simple diagrams.

Leg
October 27, 2011 5:05 pm

In my previous post I neglected to mention the reflective property of water to light photons, which is relatively high compared to other matter. This is the result of partial interaction of the incoming photon without much energy deposition and is related to the unique structure of the water molecule, the frequency/wavelength of the incoming photon, and the incidence angle. Someone with a better theoretical physics background than I may be able to explain this phenomenon in laymen terms better than I. Obviously with reflected light photons, which do not enter into the water mass, this energy does not get adsorbed by the water. However, most of this light energy will be adsorbed by other matter, though a very small portion may escape into space.

October 27, 2011 5:10 pm

Berényi Péter said “Anyway, radiative transfer can’t give a serious contribution to heat flux between atmospheric layers as long as various forms of water are not involved, otherwise thermal conductivity of carbon dioxide could not be smaller than that of dry air (which, unlike carbon dioxide, neither emits nor absorbs em radiation in the thermal infrared band).”
I don’t understand this paragraph, CO2’s thermal conductivity peaks at -20C and the Kinematic viscosity of air at -20 is very high relative 15C. Antarctica has a conductive/convective relationship that appears to negate most of the radiant impact of a change in CO2 forcing. The reason it is not warming as advertised.

Bart
October 27, 2011 5:17 pm

Tim Folkerts says:
October 27, 2011 at 3:22 pm
“Whatever angle a photon hits a solar array, if that photon gets absorbed, it deposits all of its energy. “
Nobody is interested in individual photons. The question is, what percentage of photons in an incident plane wave are being absorbed. That is very much dependent on the angle of incidence.

R. Gates
October 27, 2011 5:18 pm

climatereason says:
October 27, 2011 at 1:29 pm
R gates
Thanks for your answer. So the observed 400 year warming trend only became a warming trend influneced by humans in the last 20 years or so? So for 380 years it was natural and then it switched to being man made. Is that correct?
tonyb
____
There was some recovery from the lower temps of the LIA, but it was hardly as strong as the temperature increases we’ve seen since 1980. Temperatures were wavering in the 19th century…up a little then back down before starting a pretty constant rise in the 20th century, becoming really pronounced after about 1980. Of course warming (or cooling) doesn’t switch from one cause to another, and I know you’re being facetious here, but climate is always a combination of factors, some reinforcing and some competing with each other. Human forcings on climate are a mixed bag, with some causing cooling and others causing warming. Over the several centuries since the industrial revolution, humans have been putting increasingly large amounts of carbon into the atmosphere and oceans. As this contribution has grown, and as the accumulated effect begins to show, you would expect the human forcing signal to begin to dominate over the natural background forcings at some point. Those other forcings are still there, but the anthropogenic signal would begin to dominate, with the signal getting easier and easier to see, even though the other forcings are still there. The next 10-30 years or so will be very interesting to watch, as if we have a quiet sun and cooler oceans, we will get to test first hand the relative strength of the human contribution. If you think this was similar to the Maunder Minimum period, we certainly know what happened then, so we can compare today’s climate response directly.

davidmhoffer
October 27, 2011 5:19 pm

R. Gates;
Tony simply is far more polite than you and seems to have a better grasp of the the issues.>>>
Then by all means, answer his follow up question, it was politely worded, was it not?
As for the notion that you don’t answer me because I’m not polite…in one of my comments I suggested you were a girl and you instantly jumped all over me. Seems to me when I’m totaly rude you respond right away. But to factual questions you either go silent or, as in your response above, come up with some inane excuse that is patently and obviously not true.
I’ve asked if you had some sort of relationship with Trenberth on many occassions. You finaly answered by claiming you had no relationship other than you lived in the same state. I repeat my follow up question:
Did you or did you not arrange for Anthony Watts and 20 guests from the WUWT forum to meet with Kevin Trenberth? You claimed very proudly in another thread that you did. How did you accomplish this without having any contact with Kevin Trenberth other than living in the same state as him?
As for our bet, you welched. You can fail to respond to me, or come up with excuses, but the record is there for all to see. You accepted the wager, and asked me how much I would like to bet. Then you tried to redefine the experiment. Then you tried to claim that Anthony’s results showed only that Anchor Hawking glass absorbs IR. That wasn’t the bet. The bet was that if the experiment were done as illustrated, it would not show the results illustrated. You said they would, and they didn’t.
Of course when it became totaly obvious that you’d lost the bet, you turned from defending Al Gore’s “illustration” to throwing the guy under the bus as if we’d forget what you said in his defense only a few weeks before.
Frankly R. Gates, I’m being perfectly factual, and while you might accuse me of being blunt, there is nothing impolite about anything I said or the questions I asked. If you are uncomfortable with answering the questions, might I suggest that your discomfort originates from another cause?
Ooops. I just did. How impolite of me.

Bart
October 27, 2011 5:23 pm

I do agree that every photon incident on the Earth will get processed in some way. But, the total percentage is 1/2 of the area of incidence (pi*r^2, being the projected area, rather than 2*pi*r^2, being the total area facing the Sun).
If what I am saying is not germane to the argument, nevermind. I haven’t read through the article carefully yet, I was just commenting on what appeared to be an erroneous argument which caught my eye.

Bill Illis
October 27, 2011 5:47 pm

Modtran results looking up from the surface in the tropics (looking at the back-radiation, the red line) when it is Clear.
http://img171.imageshack.us/img171/4308/rad12081141.gif
Modtran results looking up from the surface in the tropics when there is low cloud cover. A perfect black-body radiating at 20C (and it is cloudy 65% of the time).
http://img171.imageshack.us/img171/7268/tropicalsurfacelookingu.gif
Where is this described in Trenberth’s diagram?
It is an illustration which gives everyone a false impression of how the real atmosphere works, and how it really works over a 24 hour period.
Another good example, in the middle of the day in clear conditions, the solar irradiance coming in can be as high as 1200.000 joules/m2/second. But the surface energy level in the same conditions is only increasing by 0.008 joules/m2/second. It is flying out as fast as it is coming in. Where is that shown is Trenberth’s diagram?

KevinK
October 27, 2011 6:12 pm

Jim Masterson wrote;
“These diagrams are “steady-state.” That means the transients have had time to damp out or stabilize.”
Thank you, I fully understand the concept of “steady-state”.
Unfortunately, the complex climate system of the Earth is NEVER in a “steady-state” condition.
My point is if you model/analyze a complex chaotic system as if it is a simple “steady-state” problem you are VERY LIKELY to get the wrong answer, no matter how well you analyze the “steady-state” condition.
For example, an airplane flying towards the surface of the Earth at 500 mph is in a “steady-state” condition. Once the plane and the Earth’s surface intersect the “steady-state” becomes chaotic very quickly.
Cheers, Kevin.

October 27, 2011 6:12 pm

George E. Smith; says:
October 26, 2011 at 11:11 pm
“”””” CRISP says:
You CANNOT transfer heat from a colder body (the upper atmosphere) to a hotter body (the lower atmosphere and Earth surface) without doing work. The 2nd Law of Thermodymanics avbsolutely forbids it. “””””
EM radiation knows absolutely nothing about either “heat” or “Temperature.”, and it can go wherever it darn well pleases .
—————————————————–
George I hope you don’t want to imply that the 2nd Law doesn’t hold for radiation, Planck and Einstein surely build on it when they dug into the BB-radiation problem.
Anyways, I notice that many people think the entropy in k*ln*W has to do with the probability of jiggling atoms. It does not. It has to do with possible energy microstates, entropy is about the dispersal of energy itself. W stands for “Ways of energy distribution”, so in the case of atoms the energy distribution among all possible translation, rotation and vibrational states.
Why doesn’t the earth surface heat up infinitely by radiation? Because after absorption from a SWR photon the energy can leave dispersed as 20 diffuse LWR photons.
Backradiation may be real but it is clear that it cannot warm up the surface further because it can’t be dispersed any further than it was when it left the surface in the first place, it can only do this if it would be radiated away at an even LOWER surface temperature.
The ‘heat from hot to cold’ phrase that is often used, should be: Energy tends to flow from being concentrated in one place to becoming diffused and spread out and therefore less concentrated. This more concrete formulation with respect to the 2nd Law is well known in chemistry.
So EM radiation cannot transfer (heat)energy from a colder to a hotter body because the photon energy cannot be dispersed by the hotter body. Energy is still more concentrated here.
For an excellent website about the 2nd Law look here.

October 27, 2011 6:18 pm

Bob_FJ says: “Thus the 396 & 333 is at the surface, but you argue that it is constant up to the high cloud level, including clear skies.”
That was not what I said (or at the very least, not what I meant). This seems to be a combination of two different ideas.
1) The 333 W/m^2 downward would be the net result (on average, of course) of all the atmosphere. The higher up you go, the less the you would see looking up. This is because 1) there is less atmosphere above you and 2) the atmosphere is getting cooler.
The value must depend on the GHGs and clouds, since in an IR transparent atmosphere, it would be 0.00 W/m^2 of IR from the 3 K background of space.
2) Conversely, the 396 W/m^2 upward IR from the earth’s surface would continue upward in a transparent atmosphere. THIS is the number I claim would be constant at any altitude. (The curvature of the earth would make a small correction. By the time you are 64 km high, you are 1% of the earth’s radius up, and there would be a 2% correction to the IR — it would decrease by ~ 8 W/m^2 to 388 W/m^2.)
In a real atmosphere, this value will decrease. Some of photons from the warm surface will get absorbed. They will partially be replaced by the photons from the cooler GHGs/clouds higher up. This is exactly the “bites” taken out of the observed and calculated spectra you posted.

davidmhoffer
October 27, 2011 6:22 pm

Hans;
George I hope you don’t want to imply that the 2nd Law doesn’t hold for radiation, Planck and Einstein surely build on it when they dug into the BB-radiation problem.>>>
That’s not what George said. He said that all bodies radiate energy as per Stefan-Boltzmann and they do so in a spectrum as defined by Planck. The photons emitted by a body at any temperature don’t care WHAT the temperature of what they are aimed at is. They either get absorbed or they don’t depending on the properties of whatever they run into. SB, Planck and Einstein all require this to be the case. The 2nd Law holds up just fine if you define the NET exchange of radiation. So, a cold body does in fact send energy to a warm body, but not as much as it GETS from the warm body. That said, if you insert a cold body in between a warm body and a SUPER cold body (outer space for example) the warm body still cools, just not as fast as it would without the cold body in between.

October 27, 2011 6:37 pm

Bob Fernley-Jones says:
“Please see this simple analysis of Lambert’s law …
I don’t think that really applies here. There is no “surface” that is diffusely scattering the light. Mei scattering or Rayleigh scattering would be more applicable.
Even more applicable, however, would be isotropic emission of IR from the various sources in the atmosphere. The molecules of CO2 and H2O, along with droplets of water, would emit IR uniformly in all directions.
I think we are still not communicating clearly the key issues to each other. Hopefully various outer comments will also get at these central issues.

jae
October 27, 2011 7:17 pm

Again, again, again, only crickets when the facts appear. Nobody wants to address the real issues. Just argue about photons…
The Atmospheric GHG theory has been proven to be toast in several ways, yet that is ignored and the silly debates about radiation go on, as if there were no elephant in the room at all.
Rather deserving of a comical play, no?
Very worrisome, I think.
But, even of more concern, the same problem applies to many other totally irrational things going on now, because of a stupid, lazy, uneducated, rich society. Like substituting bumper sticker slogans for policy. “Hope and Change,” as a “policy,” e.g. Like calling a war a “kinetic military action” and Muslim terrorism “man-caused disasters.” Like the climate scientists and Obamabots saying that “our ideas are really right, we just need to communicate them better, and the (stupid fools) will learn to appreciate us. Like the absolute stupidity of the concepts of “green energy,” “renewable energy, and “sustainability.”
1984 is coming true in 2011, unless more people wake up!

October 27, 2011 7:49 pm

So enlighten us, JAE.
What do you consider to be “the Atmospheric GHG theory” and in what specific ways is it “proven to be toast” ?
How does that support or refute what Bob_FJ has written?

Bob Fernley-Jones
October 27, 2011 7:52 pm

Tim Folkerts @ October 27, at 6:37 pm
My comment related to Bart’s comment pertaining to the surface alignment of solar panels relative to sunlight. They are fairly devoid of specular reflection or if you like are close to Lambertian reflectors/emitters. It is a tad off topic, but had relevance particularly with it giving another demonstration of vector analysis.
We could also get into semantics again, in that there are different terminologies again between optics, and your view of physics/elementary quantum theory.

Bob Fernley-Jones
October 27, 2011 8:12 pm

Tim Folkerts @ October 27, at 6:18 pm

”Bob_FJ says: “Thus the 396 & 333 is at the surface, but you [Tim Folkerts] argue that it is constant up to the high cloud level, including clear skies.”
That was not what I said (or at the very least, not what I meant)…”

OK, maybe I misunderstood you. Are you now saying that although Trenberth shows the 396 & 333 arrowed up at the high cloud level, (and adjacent clear skies), that this is the net result at the surface from all of the above? In other words, the arrows are in the wrong place, and the columns depicting the EMR transmission should really taper-off towards the higher levels?
Not all that hard to draw it that way is it?

R. Gates
October 27, 2011 9:02 pm

davidmhoffer says: (to R. Gates)
October 27, 2011 at 5:19 pm
“Did you or did you not arrange for Anthony Watts and 20 guests from the WUWT forum to meet with Kevin Trenberth? You claimed very proudly in another thread that you did. How did you accomplish this without having any contact with Kevin Trenberth other than living in the same state as him?
___
100% true. I emailed him, and told him what we wanted to do…including who it was that would like to do it (i.e. a small group that I would bring and Anthony Watts and a group he would invite). He politely emailed me back his schedule of when he’d be in Boulder, and told me his assistant would be in contact. Within a day or so she contacted me and we saw that Nov. 10th was open. I then communicated directly with Anthony to set up the details. Very simple, very straight forward. Though I have been to NCAR several times, I’ve never met Dr.Trenberth in person, and have only exchanged a few emails (no more than 2 or 3) over the past few years. He’s always been quite generous and responded within a few days.

October 27, 2011 9:06 pm

So Gates is part of the conspiracy to backstab Anthony.
Figures.

R. Gates
October 27, 2011 9:18 pm

Smokey says:
October 27, 2011 at 9:06 pm
So Gates is part of the conspiracy to backstab Anthony.
Figures.
_____
Smokey, I would expect an apology from a reasonable person, but that might be asking too much from you. I would also like to see Anthony make some remark here, as he knows the full truth of the situation, and there was never any attempt to back-stab Anthony or anyone else. Though Anthony might have become more hesitant to go to NCAR after the BEST issue happened, the two were never connected in any manner, and based on things I heard, I think Dr. Trenberth and Anthony probably would have found some common ground in being critical of the way the BEST results were handled. it was a missed opportunity, but I respect Anthony’s decision not to go, but to insinuate that their was a “conspiracy to backstab Anthony” is…well, beyond the normal level of paranoia even for you.

REPLY: You wanted a response, OK here it is.
I made the decision not to go, because every time I’ve ever tried to outreach to climate science by invitation, I end up getting burned. It happened with NCDC after my visit in 2008, and it happened with BEST in the spring of this year. On both occasions, people that I confided in then went to great lengths to put me and the work I’ve done down in public and in very unprofessional ways. Karl et al used preliminary data against my objections, data that was put up briefly for the benefit of survey volunteers, and never part of the final product, and then Dr. Thomas Peterson had the gall to write a ghost authored “talking points memo” about it. He circulated it to every NOAA climate division and middle manager, naming me and my work and telling everyone how it was flawed, but didn’t have the integrity to put his own name to the internal paper. Then later, Karl, Menne, and Vose offer me an option to put my name on their upcoming paper…a week before it gets submitted, but I’m not allowed to offer any criticisms, revisions, or changes to it. Name only. I ask them for a letter of invitation, they refuse to even put it in writing. The idea was to get me to endorse their paper by putting my name on it. Sheesh.
I don’t know that Dr. Trenberth had any designs, he probably didn’t, but I do know this: I asked him personally not to address people as “deniers” at the AMS conference and he declined, only rewriting his speech to change the inaccuracies we called him out on. He didn’t even have the courtesy to respond. After my two previous experiences with “professional” climate scientists and the resulting outcome, I couldn’t see spending my own money, effort, and time to stand next to a man who has so little respect for me and others that he could not refrain from calling people like myself who question the issues on a scientific basis “deniers”.
That, and none of the other 8 scientists I invited wanted to attend. It seems they couldn’t justify making the effort for visiting Dr. Trenberth. So all in all it became simply R. Gates was the only one who really wanted this to happen. In better times, perhaps it would have worked out. But Dr. Muller has put the kiss of death on any further cooperation with his media blitzkrieg. And even now, he’s complaining that “I broke the embargo” because I published my essay on the BEST paper an hour before they got their website up…he seems to think that the Economist publishing a story hours before doesn’t count, and that I was bound by an embargo that somehow wasn’t broken by the Economist, but I did. I found myself being deluged with emails and requests for interviews even before BEST got their stuff out. Judith Curry published 30 minutes before BEST got their stuff on their website, but I’m the bad guy.
There’s no scruples it seems in climate science where the treatment of skeptics is involved. And after what I’ve been through, I won’t trust any of these people again.
Leaning from that, Dr. Trenberth can go fly a kite. – Anthony

Jim Masterson
October 27, 2011 9:20 pm

>>
KevinK says:
October 27, 2011 at 6:12 pm
Thank you, I fully understand the concept of “steady-state”.
<<
There’s no insult intended. Some here may not know what steady-state means.
>>
Unfortunately, the complex climate system of the Earth is NEVER in a “steady-state” condition.
<<
True, but I never said the climate system was. I said these models were in a steady-state mode.
>>
My point is if you model/analyze a complex chaotic system as if it is a simple “steady-state” problem you are VERY LIKELY to get the wrong answer, no matter how well you analyze the “steady-state” condition.
For example, an airplane flying towards the surface of the Earth at 500 mph is in a “steady-state” condition. Once the plane and the Earth’s surface intersect the “steady-state” becomes chaotic very quickly.
<<
Neither can these models balance your bank statement, fly you around the moon, or simulate supernovae. They are simply, back-of-the-envelope, global average energy flow models. If you want to model the climate (or a supernova) you’ll need more horsepower.
I have modeled these simple, steady-state models with a spreadsheet, and they require a macro to get around the self-referencing problem. That macro just copies the contents of one cell to another cell. After I make a small change, the macro needs to cycle several times. I’ve programmed a loop and it usually needs about fifty iterations before that small change finally stabilizes.
Jim

D. Patterson
October 27, 2011 9:21 pm

wayne says:
October 26, 2011 at 6:44 pm
Tim Folkerts says:
October 26, 2011 at 6:07 pm
Well, in a word: YES!
In a transparent atmosphere and ignoring the curvature of the earth, the average 396 W/m^2 would be constant all the way up. […]
Wrong.
It decrease with temperature with an increase in altitude.

Speaking as a former meteorologist who observed and reported innumerable surface and air temperatures with surface observations and rawindsonde probes during routine and special operations, Increasing altitude certainly does not result in a decrease in air temperature in all atmospheric layers or in all geographic locations and their time intervals. It is quite common for air temperatures to increase with increases in altitude in some atmospheric layers, in many geographic areas, at certain times, and in many common atmospheric conditions. There are a multitude of ways in which inversion layers develop as warm air masses become positioned above cooler air masses. The adiabatic lapse rates which assume decreasing temperatures with increasing altitude are based upon Tropospheric conditions and ideal assumptions about dry air conditions adjusted for dew point measurements in the upper air soundings. Variations in atmospheric temperature and density at different altitudes also change the quantities of electromagnetic energy being modulated by the air volumes. The presence of water vapor, water, and water ice changes the electromagnetic environment and thermal environment in complex ways. Models which fail to accurately account for these widespread and common exceptions are inherently lacking in reality.

davidmhoffer
October 27, 2011 9:25 pm

R. Gates;
100% true. I emailed him, and told him what we wanted to do…including who it was that would like to do it (i.e. a small group that I would bring and Anthony Watts and a group he would invite).>>>
Was that so hard? Of course you are now on record with the answer which I shall accept at face value unless evidence to the contrary emerges. In the meantime, thankyou for your clarification.
BTW, who is “we”?
How about the answer to Tonyb’s follow up question?
As for the bet…you’re still avoiding it. If you think Tonyb has a better grasp of the issues than I do, well, I think you are probably correct. But your over confidence in your own grasp of the issues is what led you to accept a wager that was in fact a sucker bet. There was no possible way the results as illustrated could be derived from the experiment as illustrated. Further, the experiment as illustrated could not possibly demonstrate the greenhouse effect because it used LW as the external heating source rather than SW. Even funnier, you, who claim to have a grasp of the issues, suggested changing the experiment by removing the globes because they were superfluous. Might I ask how one can demonstrate the greenhouse effect when there is nothing to absorb the SW and re-radiate it as LW? And you question MY grasp of the issues?

October 27, 2011 10:08 pm

Since no one has really picked up on how wrong the K&T cartoon is, think about this. If CO2 doubled and that caused water vapor forcing to double twice the CO2, there would be a total of 3 times 3.7 or 11.1 Wm-2 increase in that 333Wm-2 down welling radiation. 11.1/333 = 0.033 or 3 percent increase in forcing. If that 333 is related to all of the 33C greenhouse effect, a 3 percent increase would be 1.1 degrees of warming due to a doubling of CO2 forcing and water vapor forcing. Not just a no feed back forcing, the whole enchilata. That is what Richard Lindzen has said, K. Kimoto has said, and Monckton has said. They are right, if you use that cartoon.
Isn’t it ironic how some people defend the indefensable even when it is shooting themselves in the Foot? 🙂
Unfortunately, it is not as simple as discrediting another flawed icon of global warming past. The reality is a little more complicated.

Bart
October 27, 2011 11:36 pm

“Leaning from that, Dr. Trenberth can go fly a kite. – Anthony”
Wow. Thanks for the inside info. Things truly are much, much worse than I thought.

Editor
October 28, 2011 12:31 am

R Gates
Thank you very much for your reply in which you said;
“There was some recovery from the lower temps of the LIA, but it was hardly as strong as the temperature increases we’ve seen since 1980. Temperatures were wavering in the 19th century…up a little then back down before starting a pretty constant rise in the 20th century, becoming really pronounced after about 1980”
With respecf, that isn’t really so. I wrote an entire article on 19th Century temperatures viewed through the Life of Dickens
“Dickens life demonstrates the extraordinary variability of the British winters during that era, when the coldest and warmest winters in the CET records can be juxtaposed. Generally there are few examples of constant cold winters year after year-the LIA was becoming much more sporadic than it had been several centuries earlier, when bitter cold weather appears to have been the norm. To put this era into perspective mature English people might be surprised to learn they lived through a much colder winter than Dickens ever experienced. 1962/3 at -0.33C was the third coldest in the entire CET record compared to Dickens coldest year 1814 at 0.43c, the fourth coldest in the record. (1962/3 was a bit of a one off-Dickens experienced a greater number of relatively cold winters)
HH Lamb, (in ‘Climate, History and the Modern World’), says: “Indeed, the descriptions of ‘old-fashioned’ winters for which Charles Dickens became famous in his books may owe something to the fact – exceptional for London – that of the first nine Christmases of his life, between 1812 and 1820, six were white with either frost or snow.”
http://wattsupwiththat.com/2010/12/05/has-charles-dickens-shaped-our-perception-of-climate-change/
Within my article are also studies of Europe and the US.
The low point was the first decade of the 17th Century. The subsequent steady rise saw increases as great as the 1980’s. I am currently recostructing the temperature back to 1550 which is taking a great deal of research (and includes visits to the Met offive archives) . Good as part of Dr Manns experimental work was, it wasn’t as accurate as Hubert Lambs, although more recent evidence seems to suggest his adapted 1991 IPCC chart wasn’t wholly accurate. How Hubert Lamb would have loved the internet and the opportunities it gives someone like me to stand on the shoulders of giants
with best regards
tonyb

Bob Fernley-Jones
October 28, 2011 12:47 am

Dallas @ October 27, at 2:55 pm and more recently:
Sorry for a slow response, and I note that you have made a bunch of interesting comments, including, if I can paraphrase:
I think you inferred something like: Up or down radiant energy flux is OK until reaching the TOA, when radiation can then also escape freely to space sideways
Yes, indeed, and I was wondering at one stage if I should include an observation in the article that if the TOA could be treated as a surface, (which is a tad conceptual), then the idea of surface integration at “that level” might be applied. This is a concept that has been put forward by Tim Folkerts relative to the Earth’s surface, IF we had a transparent atmosphere. However, I thought it might get into a quagmire of semantics and whatnot, and I wanted to keep it simple and towards lay language.

Michel
October 28, 2011 1:08 am

Bob Fernley-Jones @ October 27, 2011 at 3:13 pm
My understanding of the 396 W m-2 is that it is an averaged net upwards oriented energy flux resulting from all phenomena taking place on a non-even emitting surface. The Trenberth and the NASA diagrams are just energy balance MODELS, not a fundamental explanation of all the physics going on on a lukewarm plate.
Even for lateral going radiation, a high enough concentration of IR absorbing gases (so called GHG) would be needed to absorb this energy. CO2 at 391 ppm and H2O at 0 – 40000 ppm cannot play that role on a very short distance (may be some methane at high concentration in cows’ flatulence could do the job). The major air components O2, N2, and Ar don’t absorb in the IR range.
To debate if [or not] a 15°C surface emits [or not] energy in its perpendicular [or not] direction is a “Nebenkriegschauplatz”, a battle not worth conducting.
What is more important is the unbalance shown in Trenberth, 333 W m-2 going only downward, and to understand what is wrong there (or if it’s right, to expain it).
And even more important is to know if a 3.7 W m-2 effect (GHG forcing for doubling CO2) can bring to a tipping point a system with huge shock absorbers like water evaporation and cloud albedo.

Bob Fernley-Jones
October 28, 2011 1:22 am

Re Smokey @ October 27, at 9:06 pm, and R.Gates.
ANTHONY, in your reply thereto; you have to be admired for having initially contemplated intercourse with Trenberth. But whilst you would like to make peace with such alarmists, I think it is a stretch to hope that such leopards could change their spots. For example, Trenberth’s and the IPCC’s treatment of Chris Landsea was disgusting. (and we don’t see any acknowledgement that Chris has been right so far!)

Bob Fernley-Jones
October 28, 2011 1:50 am

BTW everyone, may I suggest that regardless of what his relationship with Trenberth might be we should listen to R. Gates, because it seems to me that he does produce some arguments that we should consider as worthwhile thinking-on.

Jim Masterson
October 28, 2011 2:37 am

>>
MostlyHarmless says:
October 27, 2011 at 3:18 am
Secondly, a correction – K&T 2009 does not compute surface radiation for a “surface temperature of 16 °C and an emissivity of 1.0″.
In K&T 2009:
“To compute these effects more exactly, we have taken the surface skin temperature from the NRA at T62 resolution and sampling and computed the correct global mean surface radiation from (SB) as 396.4 W/m².
. . .”
I see plenty to argue about in the detail of K&T 2009, but let’s get the facts straight.
<<
In the first sentence (that you quote) they explicitly state that they use the S-B law to compute that 396.4 W/m². Using an emissivity of 1.0, the S-B constant, and absolute zero at -273.15 °C, we get exactly 16.00 °C to four significant digits. The rest is all hand waving.
Jim

Robert Stevenson
October 28, 2011 2:44 am

“There have been decades, such as 2000 – 2009, when the observed globally averaged surface-temperature time series shows little increase or even a slightly negative trend (a hiatus period). However, the observed energy imbalance at the TOA for this recent decade indicates that the net heat flux into the climate system of about 1Wm^-2 should be producing warming somewhere in the system “. This quote from the warmist publication ‘nature climate change’ shows a measure of desperation in their cause along with ‘climategate’ when hockey predictions of CAGW do not materialise. They resort to bogus claims of global energy imbalances by Trenberth (‘where has all the heat gone’) and Hansen; B F-J’s post showing the paradox in trenberth’s claims is excellent.

Kelvin Vaughan
October 28, 2011 3:14 am

George E. Smith; says:
October 27, 2011 at 4:04 pm
“”””” Kelvin Vaughan says:
October 27, 2011 at 1:16 am
Thanks for the lesson George, I appreciate it
Kelvin

October 28, 2011 3:14 am

Bob Fernley-Jones says:
“BTW everyone, may I suggest that regardless of what his relationship with Trenberth might be we should listen to R. Gates…”
Since R. Gates was actively trying to manipulate Anthony into a public meeting, I suspect a setup. Gates was doing what he was told, and I have no doubt that some kind of dog and pony show was planned to make Anthony look bad. Trenberth and Muller are two peas in a pod. Neither one can be trusted, and Gates is their water boy, even offering here to personally bus people from the airport to the meeting.
As I’ve often said, there should be a series of debates conducted on the subject of AGW. BUT they must be held in a neutral venue, with agreed-upon rules, and a debate moderator chosen by mutual agreement. After losing all debates the AGW crowd refuses to publicly debate any more, preferring instead to have proxies like Abraham take pot shots from the safety of his ivory tower. Alarmists fear debating with skeptics. So now Anthony has been invited to come to their home turf and be subject to their unstated rules. Based on their prior unethical treatment of a straight shooting, stand-up guy, IMHO Anthony was wise to decline their suspicious invitation.

Bob Fernley-Jones
October 28, 2011 4:17 am

Smokey @ October 28, at 3:14 am
Yep, I sort-of agree with your rant, but nevertheless, I think that R. Gates does come-up with some interesting thoughts. Call it playing devil’s advocate if you like, but worth considering, whereas for instance in comparison, the views of Myrrh are a waste of time and space when he cannot even accept that IR shielding glass allows high energy visible sunlight to pass through with observable consequences, but not IR which he thinks is the only source of heat.

Myrrh
October 28, 2011 5:36 am

Leif Svalgaard says:
October 27, 2011 at 8:24 am
Myrrh says:
October 27, 2011 at 5:15 am
For example, visible LIGHT cannot heat water because it doesn’t have the mechanism to do so, water is a transparent medium for visible LIGHT and so visible light is transmitted through without being absorbed, visible light cannot move the molecules of water into the vibrational state which is heat.
Yet it does.
No it bloody well doesn’t. Shine blue light on water and let me know when it’s hot enough for you to make a cup of coffee..

http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/watabs.html
“Transparency of Water in the Visible Range
“It doesn’t absorb in the wavelength range of visible light, roughly 400-700 nm, because there is no physical mechanism which produces transitions in that region – it is too energetic for the vibrations of the water molecule and below the energies needed to cause electronic transitions.”
If there is some other factor at play to alter that at some visible wavelengths it is still so insignificant that for all practical purposes it is zero. Get a grip on scale.
This Visible light direct from the Sun to Earth is what is being touted as the PRIMARY heating mechanism of the whole Earth! Claiming it is directly heating land and oceans. This is pure, unadulterated, unscientific, uneducated, stupidity. It’s high time this was acknowledged by everyone discussing this fictional AGW KT energy budget.
100 meters of water absorbs all the visible light falling on it, so heats it.
No it doesn’t. Water is a transparent medium for visible light. What is the problem here? How can you not know what transparent means? You’re happy enough to use transparent when talking about visible light passing through it to reach the surface, to heat the surface.
Transparent is used in error in the ‘AGW greenhouse scenario’ when it claims that the atmosphere is transparent to visible light. It isn’t, reflection and scattering, blue sky, are the result of ABSORPTION in the TECHNICAL physical sense, of being actually absorbed by whole or part of the molecule.
How hot is the sky from this? Why isn’t the figure in the KT?
Not all absorption creates heat, photosynthesis is chemical energy not heat energy, the creation of sugars, for example. This one dimension fictional science thinking that equates ‘all absorption of energy creates heat’, is quite frankly, pathetic coming from ‘educated scientists’. However, Water really is tranparent to visible light. “Absorption” here can not refer to the technical physical sense of the molecule or part thereof physically absorbing its energy. See Georgia link above.
Visible light does not affect the molecule of water, not on an electron scale nor on the atom/molecule scale, it cannot move the molecule into vibration, it can’t because it doesn’t have the mechanism to do this.
Visible light move water molecules into overtones of symmetric and anti-symmetric vibrations.
Codswallop, utterly stupid science fiction. ..I think you may have misread this, this is a description of how a molecule of water moves, but to get that moving into vibrational heat energy production takes thermal infrared, heat, visible isn’t capable of it.
Visible is transmitted through without being absorbed. Visible isn’t capable of moving the whole molecule into vibration.
Therefore, whatever reason visible travels through water to greater or lesser depths depending on its wavelength, colour, it isn’t because it is being absorbed by the molecules of water. The use here of “absorbed” can only mean in the general sense, that it disappears..
Visible is transmitted through without being absorbed. Visible isn’t capable of moving the whole molecule into vibration.
Water is a transparent medium for visible light, this means that it doesn’t even get in to play with the electrons. Visible light does not have the means to moves molecules of water, its effect on matter is on the electronic transition level, electron level.
In the atmosphere, visible light is absorbed by the electrons of the whole molecules of oxygen and nitrogen. What happens then? The electron briefly becomes a bit more agitated and then comes back to norm spitting the blue light out again, called reflection/scattering. Number two on this list of what electronic transitions can do in the post I set out the differences, here: http://wattsupwiththat.com/2011/10/18/replicating-al-gores-climate-101-video-experiment-shows-that-his-high-school-physics-could-never-work-as-advertised/#comment-778960
First: There is a real physical difference in the way Light and Heat energies from the Sun work:
“Mechanisms of selective light wave absorption include:
Electronic: Transitions in electron energy levels within the atom (e.g., pigments). These transitions are typically in the ultraviolet (UV) and/or visible portions of the spectrum.
Vibrational: Resonance in atomic/molecular vibrational modes. These transitions are typically in the infrared portion of the spectrum.”
Got that?
Now, in the list of what is possible that UV/Visible can and will do:

“UV-Vis: Electronic transitions
In electronic absorption, the frequency of the incoming light wave is at or near the energy levels of the electrons within the atoms which compose the substance. In this case, the electrons will absorb the energy of the light wave and increase their energy state, often moving outward from the nucleus of the atom into an outer shell or orbital.
The atoms that bind together to make the molecules of any particular substance contain a number of electrons (given by the atomic number Z in the periodic chart). Recall that all light waves are electromagnetic in origin. Thus they are affected strongly when coming into contact with negatively charged electrons in matter. When photons (individual packets of light energy) come in contact with the valence electrons of atom, one of several things can and will occur:
*An electron absorbs all of the energy of the photon and re-emits it with different color. This gives rise to luminescence, fluorescence and phosphorescence.
*An electron absorbs the energy of the photon and sends it back out the way it came in. This results in reflection or scattering.
*An electron cannot absorb the energy of the photon and the photon continues on its path. This results in transmission (provided no other absorption mechanisms are active).
*An electron selectively absorbs a portion of the photon, and the remaining frequencies are transmitted in the form of spectral color.”

The second is what happens to visible light in the atmosphere – it is being absorbed by the electrons of the molecules of oxygen and nitrogen and reflected/scattered back out.
The third is what happens to visible in water – it is transmitted through without being absorbed because water is a transparent medium for visible light.
VISIBLE LIGHT IS NOT ABSORBED BY WATER. THEREFORE, IT CANNOT HEAT WATER.
The AGW energy budget cartoon is just that, a cartoon. How anyone here can be discussing it as if it is serious physics is getting to be a bad joke, you’re wasting your time, better spent deconstructing it its core premises, which are here seen to be fictional memes. That visible light isn’t capable of heating land and oceans is basic, bog standard traditional physics, which knows the difference between Heat and Light. And it doesn’t need to be more complicated than these well understood category differences.

Myrrh
October 28, 2011 5:40 am

Bob Fernley-Jones says:
October 28, 2011 at 4:17 am
Smokey @ October 28, at 3:14 am
Yep, I sort-of agree with your rant, but nevertheless, I think that R. Gates does come-up with some interesting thoughts. Call it playing devil’s advocate if you like, but worth considering, whereas for instance in comparison, the views of Myrrh are a waste of time and space when he cannot even accept that IR shielding glass allows high energy visible sunlight to pass through with observable consequences, but not IR which he thinks is the only source of heat.
Yeah right, that’s why there’s a whole industry selling products to stop thermal infrared passing through window, but letting in as much light as possible. According to you, generic, that’s not stopping heat getting in because it’s visible that creates it…
You are of course fully entitled to take yourselves seriously.

October 28, 2011 6:05 am

Bob Fernley-Jones says: October 27, 2011 at 8:12 pm
” Are you now saying that although Trenberth shows the 396 & 333 arrowed up at the high cloud level, (and adjacent clear skies), that this is the net result at the surface from all of the above? In other words, the arrows are in the wrong place, and the columns depicting the EMR transmission should really taper-off towards the higher levels?
Not all that hard to draw it that way is it?”

Yes, that is pretty much what I am saying. This diagram is the simplest possible diagram to try to get across the points he is making. The atmosphere is being treated as a single object, and any of the arrows anywhere in or out of the atmosphere simply mean the energy is entering or leaving the atmosphere as a whole. It is not intended to mean that the back-radiation only occurs from the middle of the atmosphere or that it only occurs on the “right side” of the atmosphere.
But actually, it would be considerably more difficult to draw “correctly”. He could have drawn an arrow down from high (altitude) clouds and from low clouds and from middle clouds and from high O3 and low H2O and middle CO2. But soon the diagram would have 100+ arrows distinguishing flows to and from land and ocean and clouds and CO2 and sun and ….
The cartoon is what it is – a vast simplification to highlight major overall energy flows. People who want more detail need to actually read the papers in the field.

October 28, 2011 6:39 am

Myrrh says:
October 28, 2011 at 5:40 am
You are of course fully entitled to take yourselves seriously.
http://www.leif.org/EOS/unskilledandunaware.pdf

Bill Illis
October 28, 2011 7:10 am

Michel says:
October 28, 2011 at 1:08 am
The major air components O2, N2, and Ar don’t absorb in the IR range.
———————–
Sure they do.
About 0.00000000015 seconds after a CO2, H2O, or CH4 molecule absorbs a photon in the IR range.

October 28, 2011 7:12 am

Myrrh says:
October 28, 2011 at 5:36 am
http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/watabs.html
Your own link shows that the absorption of visible light by water is a million times weaker than that of infrared, but that only means that you need a layer of water a million times as thick, i.e 100 meters instead of a fraction of a millimeter. Since the oceans are more than 100 meters deep they will absorb all the visible light falling on them. The mechanism involves overtones of vibrational stretching as you have been shown many times.

Scottish Sceptic
October 28, 2011 7:29 am

Surely the real con in the diagram is that it treats clouds as static and not a variable and fails to show that the Latent heat rising causes cloud formation which then modulate the amount of solar radiation reaching earth and back radiation.
Or to translate from Sceptish: when the sun shines, the ground dries and this moisture rises up causing clouds. These clouds hide the sun during the day making it cooler and keep us warm at night.

Lars P.
October 28, 2011 7:35 am

Both the atmosphere and the oceans delay the incoming heat from being radiated out to space. We are living on the surface of the earth and probably this is why we think atmosphere first, but is it not in reality ruled by the oceans?
As Leif explained a couple of times the oceans are heated by visible light up to 200 m and more. The warming is distributed as water is almost 100% transparent to visible light.
As on the other side water is completely opaque to IR radiation, the deeper levels will not lose heat by radiation. Furthermore any IR from the atmosphere will hit the only the surface and not have any warming effect on the deeper levels.
As Stephen WIlde explains the oceans get themselves in equilibrium through evaporation & radiation at the surface.
“Importantly both the atmosphere AND the oceans delay the incoming solar heat from being radiated out to space.”
http://climaterealists.com/index.php?id=1487&linkbox=true&position=5

Ask why is it so?
October 28, 2011 8:05 am

Myrrh, thank you for your reply however I think you seem confused. I was talking about radiation not whether it is visible or invisible. The 2nd law of thermodynamics is often brought up to prove that CO2 cannot cause warming because heat does not travel from cold to hot, and the surface of the earth being hotter would not receive the heat CO2 apparently sends back to the surface to cause global warming. If it were heat, this would be true, however, radiation unlike heat can travel in any direction including back down to the surface. I believe it is important for it to be understood that heat is controlled by the Laws of Thermodynamics, radiation isn’t. I’m sorry if my explanation was too simple. I often get lectured about that but I believe the KISS theory much more effective than the complicated gobbledygook scientists use.
I’m still trying to work out how long wave radiation can produce a higher temperature than short wave radiation.

October 28, 2011 9:03 am

Bob_FJ
I linked to this image earlier in the thread and it might be worth repeating now: http://homepage.mac.com/williseschenbach/trenberth_mine_latest_big.jpg
It is a slightly more detailed version of the Trenberth diagram. For people who understand the original diagram, this one helps show a bit more detail by splitting the atmosphere into an upper layer and a lower layer. For those who don’t understand the original, it only makes things more confusing.

October 28, 2011 9:34 am

Bob FJ
“Yes, indeed, and I was wondering at one stage if I should include an observation in the article that if the TOA could be treated as a surface, (which is a tad conceptual), then the idea of surface integration at “that level” might be applied.”
To quote the silly movie, Night Shift, “I am a Big Picture, kinda guy.”
I don’t do complicated though, so I keep it as simple as possible, which is a challenge in atmospheric physics.
This article, http://redneckphysics.blogspot.com/2011/10/330-watt-weirdness.html could use a little better grasp of the Queen’s English. You may find it humorous and enlightening once you go back to the basics then work toward the true relationships. Besides the dig that Lindzen and K.Kimoto seems to be playing on K&T, the Arrhenius relationship based on perfect blackbody is illuminating.
Feel free to edit as you wish.
I am working on some Tropopause posts that explain the impact of the inversion more clearly. Fun stuff.

George E. Smith;
October 28, 2011 10:28 am

“”””” Hans says:
October 27, 2011 at 6:12 pm
George E. Smith; says:
October 26, 2011 at 11:11 pm
“”””” CRISP says:
You CANNOT transfer heat from a colder body (the upper atmosphere) to a hotter body (the lower atmosphere and Earth surface) without doing work. The 2nd Law of Thermodymanics avbsolutely forbids it. “””””
EM radiation knows absolutely nothing about either “heat” or “Temperature.”, and it can go wherever it darn well pleases .
—————————————————–
George I hope you don’t want to imply that the 2nd Law doesn’t hold for radiation, Planck and Einstein surely build on it when they dug into the BB-radiation problem. “””””
Well Hans, I just had a lengthy answer blown away, by an inadvertent key push.
so a shorter response.
1/ “Heat” as a noun or “heat” or “heating” as a verb has NO MEANING in the absence of real physical materials; atoms/molecules, because it is the kinetic energies (mechanical energy) of those atoms/molecules that defines what “heat” (noun) is. Absent, materials, NO HEAT !
Likewise “Temperature” has NO MEANING in the absence of atoms/molecules, since it is the mean kinetic energy per particle which defines Temperature.
2/ Radiation requires NO MATERIAL; it exists, and propagates, in the complete absence of atoms/molecules; which is why it is able to get to earth from the sun. “Heat” on the other hand CANNOT travel from the sun to the earth, because there is no material (atoms/molecules) to transport it.
3/ Radiation understands NOTHING about Temperature, so radiation has no way of knowing whether some destination (sink) has a higher or lower Temperature, than where it came from (source). So I’ll leave it up to you to explain, just how “Radiation” could possibly comply with the second law of Thermodynamics.
I would recommed to you the second law as phrased by Gustav Clausius; and probably preferrably in the original German.
“Thermal radiation” has nothing much to do with “heat” or “heating” , EXCEPT that the ORIGIN of thermal radiation, is the acceleration of electric charge; AS A DIRECT CONSEQUENCE of the interparticle collisions of a collection of particles that have a Temperature higher than zero Kelvins. All materials above absolute zero emit “Thermal radiation” or “Planckian ” radiation if you like, with a limiting (envelope) (continuum) spectrum given by the Planck Radiation formula.
4/ Note that the second law says nothing about ENERGY; just “heat” and remember “words have meaning”.
Readersa need to stop confusing the resonance lines/bands of molecular spectra, which are a consequence of, and characteristic of the particular molecule structure; with the quite separate continuum spectrum that arises (at the atomic and molecular level) solely due to the Temperature of those materials.
The 15 micron bending mode emissions of the atmospheric CO2 molecule don’t have anything to do with the Temperature of the CO2, other than the molecular collisions that arise as a consequence of that Temperature will alter the width of those spectral lines, due to Doppler and other effects.
The reason that SOLIDS and LIQUIDS appear to emit “Thermal Radiation” and not GASES, is simply that the atomic/molecular density numbers are such, that many more molecules are emitting in a given volume.
The reason that solids can absorb strongly and appear “black” at least in some spectral range, is there simply are more molecules to do the absorbing. A big enough volume of gas, eventually would look quite black too. So we don’t usually talk about gases as being good sources of near black body radiation, but each and every single gas molecule is emitting, and absorbing its fair share, according to the Planck formula.
For a particle Physicist, the cause is acceleration of electric charge (over some non zero distance or time interval) whereas the Radio-Physicist would simply say you need a varying electric current in some non zero length antenna. Well the physical antenna is superfluous, so long as the current can flow somehow (stream of charged particles). For our molecular sources; we would simply say, we need a non-zero electric dipole moment; and then we can emit or absorb to our heart’s content.
We get ENERGY from the sun (via radiation); we get NO HEAT from the sun; and we don’t need to because we can make all the heat we need right here on earth by simply wasting most of the energy we get from the sun. And remember ; WORDS HAVE MEANING, so use the right words.

October 28, 2011 11:51 am

On SW absorption in water. With a ;long enough path, it is absorbed. The more interesting part is suspended particulates at varing depths greatly impact the depth and degree of absorption. Woods Hole has quite a few interesting links. As a fisherman, this is something nice to know, in case you wonder why fish stocks and climate seem to correlate.

R. Gates
October 28, 2011 11:57 am

Smokey says:
October 28, 2011 at 3:14 am
Bob Fernley-Jones says:
“BTW everyone, may I suggest that regardless of what his relationship with Trenberth might be we should listen to R. Gates…”
Since R. Gates was actively trying to manipulate Anthony into a public meeting, I suspect a setup. Gates was doing what he was told, and I have no doubt that some kind of dog and pony show was planned to make Anthony look bad. Trenberth and Muller are two peas in a pod. Neither one can be trusted, and Gates is their water boy, even offering here to personally bus people from the airport to the meeting.
_____
This is so laughable as to be hardly worthy of a rebuttal, suffice to say that it indicates why perhaps you should consider a career as fiction writer…perhaps novels involving complex conspiracies. Rather than ‘manipulate” Anthony, I was trying to accomodate and act as a mediator or go-between to try and get these two men together. As far a “dog and pony” show being planned, Dr. Trenberth was asking exactly what we wanted him to talk about. Really Smokey, there are honest people in the world…get rid of your paranoia.

REPLY:
I’ll have to back up R. Gates here. I saw no evidence of anything like that in his dealings with me. – Anthony

jae
October 28, 2011 12:07 pm

Tim Folkerts:
“So enlighten us, JAE.
What do you consider to be “the Atmospheric GHG theory” and in what specific ways is it “proven to be toast” ?”
Multiple lines of evidence. I’ve linked to this twice in this thread alone:
http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html
If it’s wrong, why do you keep ignoring it?

R. Gates
October 28, 2011 12:25 pm

tonyb,
Do you agree or disagree with the trends and uncertainty shown in this chart:
http://www.appinsys.com/GlobalWarming/GW_Part1_HistoricalRecord_files/image010.gif
And if you agree, then how can you dispute the accelerated warming of the 20th century versus what we saw from 1600 to 1800? Seems the curve over the past century is far more steep than the previous 300 year average. It seems that temperatures starting warming in general about 1690-1700 (end of the Maunder), and continued a gradual warm-up, with a pause of course for the Dalton Minimum dip in the early 1800’s, which of course was the reason for the cold winters that influenced Dickens so greatly. At least to my eye, the warming accelerated in the later half of the 20th century.
In general of course (with some notable exceptions) , global temps have pretty much followed this curve of solar activity:
http://en.wikipedia.org/wiki/File:Sunspot_Numbers.png
With exceptions of course being during periods of strong volcanic activity, and also since about 1980, when global temps and solar activity begin to diverge greatly. I would welcome any evidence that shows conclusively that the divergence of global temperatures from solar activity since 1980 as not been caused, at least in part, by anthropogenic greenhouse gases.

October 28, 2011 12:26 pm

Gates,
I stand by what I wrote. No one else but you tried to push Anthony and his supporters into a meeting controlled by Trenberth, with no agenda or rules. If it was naive innocence on your part I understand. But those of us who are wise to the ways of the world and human nature would naturally have deep reservations when someone with reason to be hostile invites them onto their home territory with no pre-agreed ground rules. Anthony was smart to decline under the circumstances.

October 28, 2011 12:56 pm

Sorry, but I disagree strongly with much of what has been said about heat lately.
In classical thermodynamics, “heat” refers to the net flow of energy from one object to another due to a temperature difference. The 2nd law states that the net flow of energy is always from a warmer object to a cooler object. This applies to conduction and convection and radiation.
It is valuable to recognize that the modern definition is stated in terms of statistics and probabilities. There is a chance during any given time interval that the net flow is actually from cool to warm. But the larger the objects, the longer the time and the greater the temperature differences are, then the lower the probability is of heat going from cool to warm. If you look at objects with only a few atoms, or look for nanoseconds, or look at objects within microkelvins of each other, you might actually have a good chance of seeing a net transfer of energy from the warmer object to the cooler object
For typical objects, the probabilities are so vanishingly small of net energy flow from cool to warm, that we can conclude it never happens = 2nd Law.
This does NOT require that zero energy moves from the the cooler object to the warmer object. For example, if a 350 K coffee cup is placed in a 300 K room, the net flow of energy is going to be from warm coffee to cool air. However, individual air molecules could be traveling much faster than average, and when they hit the coffee, they could easily transfer energy FROM the cool air TO the warm coffee. There will simply be many more collisions each second that transfer energy TO the cool air FROM the warm coffee.
Same for photons. Photons from a 270 K atmosphere can and do transfer energy to 300 K ground. But the 2nd Law requires that MORE energy be transferred from the ground to the atmosphere, so there are more photons with more average energy moving from the warm ground to the cool air than vise versa. Yes — the 2nd law DOES apply to heating via radiation.
Even though the net flow of energy must be from warm ground to cool air, the cool air can still help keep the ground warm. The reason for this is simple. The warm ground would still radiate its ~ 395 W/m^2. Without an atmosphere radiating ~ 330 W/m^2 toward he ground, the ground would instead be receiving ~ 0 W/m^2 from the 3 K background radiation of outer space. Even though there is a net loss of energy to the atmosphere, there is a much SMALLER net loss than there would be for an IR transparent atmosphere. If we magically made the atmosphere transparent to IR, the surface would suddenly start cooling dramatically because it would be receiving ~ 330 W/m^2 less energy!

Robert Clemenzi
October 28, 2011 1:01 pm

Ask why is it so? asks:
October 28, 2011 at 8:05 am
I’m still trying to work out how long wave radiation can produce a higher temperature than short wave radiation.
Try this – the long wave energy is absorbed in a few millimeters of water, but the short wave energy is absorbed over 100 meters. Assuming that each group of frequencies carries the same amount of energy, the energy absorbed in the smaller volume will produce the greater temperature rise. If this case several million to one.
In the case of a solid surface – the absorptivity is frequency based. As a result, many more long wave photons are absorbed than shortwave. This is one of the main failings of the KT diagram – it does not indicate that the reflected radiation has a different spectrum than the absorbed radiation.

Robert Clemenzi
October 28, 2011 1:09 pm

Tim Folkerts says:
October 28, 2011 at 9:03 am
[This] is a slightly more detailed version of the Trenberth diagram.
http://homepage.mac.com/williseschenbach/trenberth_mine_latest_big.jpg
Please provide a reference. Was this created by Willis? That images significantly disagrees with how I understand the atmosphere. In particular, what does “Lowest Stratosphere” mean. In some sources (including the IPCC documents), that actually means the tropopause, in the others, it is the layer above the tropopause.
At any rate, that image makes no sense at all.

mkelly
October 28, 2011 1:28 pm

Tim Folkerts says:
October 28, 2011 at 12:56 pm
Sorry, but I disagree strongly with much of what has been said about heat lately.
Same for photons. Photons from a 270 K atmosphere can and do transfer energy to 300 K ground.
Mr. Folkerts please show me where in the this formula I can do what you say can or does happen.
W/m^2 = sigma*SB*(T1^4-T2^4)
If T1=T2 then the W/m^2 equal ZERO. If you can show me where ZERO equals some other number or amount please do.

Bob Fernley-Jones
October 28, 2011 2:44 pm

Robert Clemenzi @ October 28, at 1:09 pm, & Tim Folkerts @ 9:03 am
I recollect that Willis launched his version over at Steve McIntyre’s site several years ago, based on K&T97.
I’m a bit surprised that he has not joined in, in this here.

Old PI
October 28, 2011 3:06 pm

I’m not a scientist, just an old Air Force photo interpreter with 20+ years’ experience. I’ve worked with about every kind of photography imaginable, from B&W, Color, false-color IR, regular IR, Radar, and a few others I can’t talk about. I’ve physically SEEN that the atmosphere isn’t equally transparent to all wavelengths of EMR. LOTS of different things can degrade reflectivity (photography being the recording of reflected energy, regardless of the type). You can sometimes see quite well through very thin clouds, other times those same types of clouds are virtually opaque. The resolution of IR imagery is directly affected by the ambient-environment temperature in which it’s being recorded – the greater the difference in temperature, the better the resolution. RADAR has no-return zones, either caused by blocking the radar energy from a given area (building “shadow”, etc.), or in the case of relatively calm water, the majority of the reflected energy isn’t returned to the recording site (the angle of incidence is equal to the angle of reflection. Visible light has a greater variety of angles of incidence than other types of imagery). Dust and smog can also affect any imagery negatively.
The atmosphere works the same way for solar radiation. Some of the radiation is allowed through, some of it’s blocked, some of it is absorbed or reflected away in odd angles. Since the atmosphere is a constantly-changing soup of elements, chemicals, and aerosols, some of which exist in more than one phase, I can see no way to develop a direct measurement of its’ overall transparency for any given location, much less for the earth as a whole. The best we can do is come up with what we in the business call a SWAG – a ‘Scientific’ Wild-A$$ Guess”. Even that would require taking a constant, highly-accurate, 24/7/365 measurement of the transparency (both ways – receiving and emitting) from multiple points on the earth to develop that. Until then, all of this is just an exercise in futility, proving nothing.

Windchaser
October 28, 2011 3:10 pm

Ask why it is so? says:
I’m still trying to work out how long wave radiation can produce a higher temperature than short wave radiation.
Well, the short-wave radiation isn’t really changing, and the long-wave is increasing. This is based on the fact that different types of radiation are more readily trapped than others (in systems like Earth’s), at least until the system equilibriates.
The sun is radiating in shortwave. A general recipe for creating a greenhouse effect in something (in a car, a planet, whatever) is as follows:
Find a material that’s transparent at the incoming radiation frequency, but opaque to the outgoing radiation (the outgoing radiation being the blackbody radiation). When you add a layer of this material to your system, the incoming short-wave radiation will still get through, but now the outgoing longwave radiation will be absorbed by the covering layer, and some of it will be back-scattered/back-emitted. The heat will take longer to escape, so the average temperature will increase. In essence, you’re not changing how your system absorbs heat, but you’re slowing down how quickly it loses heat.. so it will heat up.
For Earth, water vapor is also an excellent greenhouse gas, and in the lower atmopshere its concentration is so much higher than that of CO2 that adding a little CO2 doesn’t change the absorption of long-wave radiation much. But at higher altitudes, the water vapor condenses out, and other greenhouse gases dominate the back-emission of long-wave radiation. And by slowing down the loss of heat through the stratosphere, the troposphere will also become warmer.. and thus the overall global warming effect.
Of course, we’re not breaking any laws of thermodynamics.. just moving our average temperature a bit closer to that of our heat source (the sun) and a bit further from that of our heat sink (the rest of the universe).

October 28, 2011 3:28 pm

Mkelly requests:
Mr. Folkerts please show me where in the this formula I can do what you say can or does happen.
W/m^2 = sigma*SB*(T1^4-T2^4)

I’m not sure how it could be more obvious in the equation.
(Net Power/Area) = (Rate of heat from 1 to 2) / Area
= sigma*SB*(T1^4-T2^4)
= sigma*SB*(T1^4) – sigma*SB*(T2^4)
= (power/Area leaving 1) – (power/ Area arriving from 2)
The first half of the equation is the “radiative flux” leaving object 1. The second half is the “radiative flux” arriving from object 2. These are measures of the photons heading in and out.
If T1 > T2, more energy leaves Object 1 than arrives from Object 2
** ie heat flows from Object 1 to Object 2.
**ie more photons with higher energy flow from 1 to 2 (but some still flow from 2 to 1)
If T1 = T2, the same energy leaves Object 1 as arrives from Object 2
** ie no heat flows
If T1 < T2, less energy leaves Object 1 than arrives from Object 2
** ie heat flows from Object 2 to Object 1.
(This version of the equation requires both objects to have emissivity = 1 (which you seem to call “SB”). Also, either Object 1 and Object 2 should be large and flat, or Object 2 should be surrounding Object 1 and “A” would be the area of Object 1. For a more general version, see http://en.wikipedia.org/wiki/Thermal_radiation)

Bob Fernley-Jones
October 28, 2011 3:30 pm

kuhnkat @ October 26, at 8:30 pm, you wrote:
The surface is a sphere. The adjoining points would NOT cancel. Surface irregularities will confuse this so that the surface is irradiating itself just as the surface of the water does. One of the requirements for S-B to apply is that the surface geometry does NOT allow it to irradiate itself. OOOPS!!!!
I doubt if this is a significant issue; for instance water that is wave affected only has a fairly momentary oopsy, because what was concave alternates with being convex, at which time, it will radiate to below the horizontal. (= radiate more than S-B). Also, on land, (even on collectively flat ground), vegetation might at first seem to be an issue, but broadly speaking it’s all in equilibrium. (Well, according to Trenberth anyway). Thus that vegetation will simultaneously radiate to below the horizontal in more or less isotropic fashion.

R. Gates
October 28, 2011 4:12 pm

Smokey says:
October 28, 2011 at 12:26 pm
Gates,
I stand by what I wrote. No one else but you tried to push Anthony and his supporters into a meeting controlled by Trenberth, with no agenda or rules.
_____
Smokey, read Anthony’s last reply when I said you’d make a good fiction writer. You are off base on this in a serious way, so let it go. It was a meeting that was completely above board and it simply didn’t work out. No one tried to push anyone into anything, and the Dr. Trenberth was willing to talk about any climate issue we wanted him to, and answer Q & A. After BEST, Anthony (understandably) became a bit gun-shy. Can’t blame him, but that’s all there was to it. End of story.

October 28, 2011 4:39 pm

Robert Clemenzi,
“David Socrates asks:
1. Does back radiation to the Earth’s surface occur at all or is it “unphysical”?
Yes it does. It is easy to measure it with an IR thermometer. It can be seen in lapse rate plots. Back radiation is what causes the morning temperature inversion over land.
2. If it does, what proportion of that back radiation is due to CO2?
That is the important question. No one knows. My analysis indicates that it is close to zero at the surface and about 100% in the stratosphere.
To all, the fact that the amount of energy toward space is less than the amount toward the surface is proof that the atmosphere is IR opaque over a significant part of the spectrum

In case someone else hasn’t answered this,
You are measuring the BRIGHTNESS of the atmosphere with your IR thermometer. It is most likely calibrated for an emissivity of greater than .95 which is too large for the atmosphere as it was NOT designed for atmospheric work. IF it has adjustable emissivity you should do some research and set it correctly.
Next it reads this brightness from about 7-14 microns. If you will look at the numerous charts of the atmospheric emissions you will find that this is primarily WINDOW where only water vapor has a small absorptivity and emissivity.
Additionally S-B is BUILT INTO the instrument. That is, the emissivity is used along with the temperature to do an S-B relationship to give you a reading. There is absolutely no direct reading of the actual IR emitted by CO2. So, when you claim you can tell how much DLR there is, you should understand that it just may not be quite as accurate as you are assuming. It is based on assumptions that your conditions may not meet.

October 28, 2011 4:56 pm

Tim Folkerts.
“And following up on that point, the calculations you have are all for clear skies. If you model a cumulus cloud base, the numbers are much higher. For example, the first line for looking up from 10 m in a clear tropical sky was 348 W/m^2 in your table, but 418 W/m^2 when cloudy. Arctic winter goes from 163 W/m^2 to 243 W/m^2.”
As most IR from the ground is absorbed within the first 10m, I would also think that any IR from above, if emitted from over 10m up would have little chance of getting to the ground!!! As this area is where much of the thermalization of the IR occurs I keep wondering what the real numbers are at sea level and the surface of the ocean. At higher altitudes we are cooler as there is more radiation directly to space due to less atmosphere.
Anyone got some actual ground measurements? The gradient at the surface is apparently a lot higher due to the density of the atmosphere and humidity!!!

Robert Stevenson
Reply to  kuhnkat
October 29, 2011 2:44 am

When absorbable photons (with the right frequency or in the correct waveband) are absorbed to extinction in the lower or near ground atmosphere (relatively) by CO2 and H20 molecules what happens to them? Do they become more ‘energetic’ ie with higher velocities and therefore higher KE which can be measured as an increase in temperature? The answer logically must be yes. The temperature reached must be less than the ground emitter to obtain a net flow of photons into this system. An increase in CO2 and H2O would reduce the depth of the absorption layer but not the number of first generation photons absorbed ie Q in the equation Q=M*Cp*dT would remain unchanged. Any change in dT would depend on the depth and composition of the absorption layer (M & Cp); but whatever the change, any increase in CO2 would have virtually no effect because it is an insignificant absorber when compared with H20. There is !00 times more H2O present and H2O’s absorption bands are much wider.

Bob Fernley-Jones
October 28, 2011 5:00 pm

Tim Folkerts @ October 28, at 6:05 am, you wrote:
But actually, it would be considerably more difficult [for Trenberth] to draw “correctly”. He could have drawn an arrow down from high (altitude) clouds and from low clouds and from middle clouds and from high O3 and low H2O and middle CO2. But soon the diagram would have 100+ arrows distinguishing flows to and from land and ocean and clouds and CO2 and sun and ….
Well yes, but very easy to draw it partly right, rather than really daft. See my quick sketch. (ignoring the values)
http://bobfjones.wordpress.com/2011/10/29/quick-sketch-for-trenberth-cartoon/

October 28, 2011 5:05 pm

Tim Folkerts,
“No. The IR comes from all directions, not just from one direction. There is no “night side” where IR from the atmosphere does not shine.”
You use a common dodge. How much thermal storage does a GHG molecule have?? How long can it emit without absorbing more IR??
Without the earth emitting IR as it cools and the non-GHG’s to collide with to gain energy the GHG’s would stop emitting within a second after sundown. Trying to explain a portion of an energy flow is misleading.

Bob Fernley-Jones
October 28, 2011 5:12 pm

Further to mine just above, does anyone think it a bit odd that Trenberth shows 40 escaping directly to space passing through a window in high clouds. It would be very easy to draw so as to avoid the depicted clouds.

October 28, 2011 5:12 pm

jae,
for a treatment of Venus that covers much of the misapprehensions and the actual DATA from the space missions John Ackerman’s paper is probably the best. The explanation of how it got so hot in the first place I would skip till later as it is speculation as everyone else’s ides. The rest on what is happening NOW probably comes closest to the reality of the planet.
http://www.firmament-chaos.com/papers/fvenuspaper.pdf

Jim Masterson
October 28, 2011 6:03 pm

>>
Bob Fernley-Jones says:
October 28, 2011 at 5:12 pm
Further to mine just above, does anyone think it a bit odd that Trenberth shows 40 escaping directly to space passing through a window in high clouds. It would be very easy to draw so as to avoid the depicted clouds.
<<
It’s drawn the same in KT 97. I question the computation of that 40 W/m² window. But I’m the only one who seems bothered by it.
Jim

October 28, 2011 6:05 pm

kuhnkat says: October 28, 2011 at 4:56 pm
“As most IR from the ground is absorbed within the first 10m, I would also think that any IR from above, if emitted from over 10m up would have little chance of getting to the ground!!!
IR within bands that are emitted & absorbed by CO2 and H2O may have traveled only short distances, but IR from outside those bands (for example, from the nearly BB radiation from clouds) can travel quite far. So at least SOME IR from high above the ground can reach the ground.
kuhnkat also says:
“You use a common dodge. How much thermal storage does a GHG molecule have?? How long can it emit without absorbing more IR??”
I’m really not sure what your concern is here. As you say in the next paragraph, the GHG’s can ‘recharge’ by either absorbing IR from the ground or by colliding with the other gases in the atmosphere. We agree on that point. Based on this, the IR from GHGs can continue to radiate all night long, which is all I was claiming.

October 28, 2011 6:19 pm

Bob FJ
Your diagram with sloping arrows has some good points.
My concern is that – by trying to indicate a depth to the atmosphere – you now open yourself to all sorts of questions about how quickly the arrows should taper of and how high the clouds are and how energy is transferred within the atmosphere and how high convection goes and ….
Trenberth apparently chose to avoid those details in his diagram. You are welcome to include more details (like Willis did), but I don’t think that necessarily makes one “better” than the other — just different.

October 28, 2011 6:20 pm

Bob Fj
“Further to mine just above, does anyone think it a bit odd that Trenberth shows 40 escaping directly to space passing through a window in high clouds. It would be very easy to draw so as to avoid the depicted clouds.”
I was reading a thesis on minimum local emissivity variances by David Taylor of UWM. There was 1RU ~ 20K unaccounted for in the Arctic that appeared to be absorbed by the atmosphere and not a calibration issue, i.e. missing in action. The Antacrtic flux readings are even more bizzare, +/- *0 Wm-2 in some areas. I believe the 40 is a don’t know indication. NASA shows the absorption and indicates atmospheric window radiation from the clouds as it should be, water and ice have a different spectrum.
About the only thing that can be learned from K&T is that they aren’t very sure what’s going on other than models are much more accurate determining 0.8Wm-2 net warming +/-0.18 CI 🙂

October 28, 2011 6:46 pm

Windchaser,
“And by slowing down the loss of heat through the stratosphere, the troposphere will also become warmer.. and thus the overall global warming effect.”
And just how much slowing do you think extra CO2 causes, 1 second 2???? If it isn’t at least 24 hours…

October 28, 2011 8:32 pm

Bob FJ,
“for instance water that is wave affected only has a fairly momentary oopsy, because what was concave alternates with being convex, at which time, it will radiate to below the horizontal. (= radiate more than S-B). Also, on land, (even on collectively flat ground), vegetation might at first seem to be an issue, but broadly speaking it’s all in equilibrium. (Well, according to Trenberth anyway). Thus that vegetation will simultaneously radiate to below the horizontal in more or less isotropic fashion”
For water there are always waves from small to large. The SB would be at worst there. Very small irregularities like cracks would cause small irregularities in SB. I haven’t seen any work on how much would cause a significant problem, but, my wag would be that Mountainous areas, water that isn’t very calm, and vegetation would not provide surface areas that would allow reasonable SB computations. The crude experiments where Solar Cookers are used to freeze water at night shows why this could be an issue. When surface areas radiate themselves there is a much larger energy flux than when just the atmosphere is at work. The water won’t freeze. When surface irregularities are blocked from irradiating the cookers view of a clear sky, water can be frozen at an air temp of 40F. Clouds also can interfere, but, not as much as trees, buildings…
We are told that DLR slows cooling of the surface. These crude experiments show that surface roughness can slow the cooling of the surface!!!!! As you point out there would appear to be more radiation in a horizontal mode. This would be an important issue here.
Do you know if there are computations where the Planck Equations have been used to estimate the actual emissions from the atmosphere or their absorption?? SB simply doesn’t apply and the Spectra simply doesn’t provide data as to where radiation is actually coming an going. Only that there are “holes” at the top and the bottom. It doesn’t tell us if that radiation has been frequency shifted by collision and emission.
The horizon issues are in reference to your geometry picture mostly. As anything but straight up is almost certainly to be reabsorbed the IR has a second (or 3,4,5,6…) chance to be transferred thru collision or emitted to the ground. I have issues as to whether much ever gets thru the dense surface layer to hit the ground, but, that is for another day.

October 28, 2011 8:44 pm

Tim Folkerts,
“IR within bands that are emitted & absorbed by CO2 and H2O may have traveled only short distances, but IR from outside those bands (for example, from the nearly BB radiation from clouds) can travel quite far. So at least SOME IR from high above the ground can reach the ground.”
SOME IR is NOT ~320w/m2.
“As you say in the next paragraph, the GHG’s can ‘recharge’ by either absorbing IR from the ground or by colliding with the other gases in the atmosphere. We agree on that point. Based on this, the IR from GHGs can continue to radiate all night long, which is all I was claiming.”
Thank you. Many people seem to think that GHG’s somehow BLOCK IR and RETAIN it in the atmosphere or HEAT the earth with it. I simply wish to make sure the flow is understood and that it never stops. A continuous COOLING except when an energy supply is present.
I apologize for the snarky comment.

Robert Clemenzi
October 28, 2011 10:08 pm

kuhnkat says:
October 28, 2011 at 4:39 pm
You are measuring the BRIGHTNESS
That is a good point. I interpret the readings as “number of photons”, more or less.That way I don’t really care about the emissivity. On a clear day, it might read -20F, but on a cloudy day, about 40F. This clearly shows that more photons come from clouds than from a clear sky. When pointed at the Earth, the “temperature” can also be interpreted as a number of photons.
The point is that a simple $10 tool can demonstrate the reality of back radiation. But kuhnkat is correct – it is wrong to interpret the reading as a “true temperature”, but I think is is still a good representation of an “effective temperature” – the temperature of a blackbody emitting the same number of photons that the atmosphere is currently emitting.
it reads this brightness from about 7-14 microns
Unfortunately, I have no way to determine the tool’s frequency range. I agree that this is one area that could (should) cause major problems. Thanks for pointing it out.

Bob Fernley-Jones
October 28, 2011 11:13 pm

In my October 28, at 2:44 pm, I wrote:
Robert Clemenzi @ October 28, at 1:09 pm, & Tim Folkerts @ 9:03 am
I recollect that Willis launched his version over at Steve McIntyre’s site several years ago, based on K&T97.
I’m a bit surprised that he has not joined in, in this here.

Well here it is; the Willis-wisdom from January 2008: Energy Balance at the Tropopause, at CA:
http://climateaudit.org/2008/01/10/energy-balance-at-the-tropopause/
If you have the energy to go through all the comments, you may recognise some commenters that have also appeared here.
In my previous life, under a “nom de blog” of Black Wallaby, (indicating Oz heritage), I also appeared there, but not as Bob_FJ.
Back in those days Willis Eschenbach, when he constructed his “improved” Trenberth thingy, he was obviously still learning on some of the physics. For instance, how about this absolute gem of his:

PS – for a most curious form of energy transport, consider that it is not always heat that flows in natural systems, sometimes it is cold that flows in natural systems. Not only that, but cold it can flow in either direction (warmer to colder, or colder to warmer). Go figure … nature is full of surprises.

Honest; me not joking; that is what he actually wrote…… (and with some repetition), read through the 2008 thread comments if you don’t believe me.

Bob Fernley-Jones
October 28, 2011 11:58 pm

kuhnkat @ October 28, at 8:32 pm
Thank you for your comments, but I need time to think on some of them, and will get back to you later

Michel
October 29, 2011 1:43 am

RE: Bill Illis October 28, 2011 at 7:10 am
Michel says:
October 28, 2011 at 1:08 am
The major air components O2, N2, and Ar don’t absorb in the IR range.
———————–
Sure they do.
About 0.00000000015 seconds after a CO2, H2O, or CH4 molecule absorbs a photon in the IR range.
_______________________________
Bill,
Sorry, you can’t be right, or you would have to show an absorption spectrum in the IR range for those molecules (O2, N2, Ar).
But you are also right: the consequence of absorbing IR radiations is a temperature increase, (energy absorbed divided by the heat capacity = delta T).
This heat is dissipated within the whole air mass by conduction (and later convection).
This is the mechanism explaining “forcing”.
And this is approx. 3.7 W m-2 for each doubling of CO2 concentration (it’s a logarithmic scale):
– from 280 to 560 ppm: 3.7 W m-2
– we are now at 391 ppm: therefore the forcing is at about 1.8 W m-2 as compared to the alleged historic value of 280 ppm
– from 391 to 782 ppm: another 3.7 W m-2
So, let’s not panic for these small and slow, nonetheless existing, effects.

wayne
October 29, 2011 3:49 am

Tim Faulkner says:
“Trenberth apparently chose to avoid those details in his diagram. You are welcome to include more details (like Willis did), but I don’t think that necessarily makes one “better” than the other — just different.”
I have to disagree. It is not just another way to portray the energy budget. Since radiation in a gas can emit in any direction in a full spherical manner, Trenberth, you I assume and others like to view it as ½ traveling upward, 396 Wm-2 from a 16 °C average surface and then there’s the 333 Wm-2 radiation raining down from the sky blue. But, under those conditions there must be 666 Wm-2 “up there” radiating in all directions ½, or 333 Wm-2, returning to keep the surface warm at 16 °C as people who believe in back-radiation see it.
Didn’t know before tonight that you have a PhD in physics so maybe you can clarify for us here just where this 666 Wm-2 (kind of prophetic) is located vertically in the sky and why there is not the other half 333 Wm-2 radiating upward to space plus the 396 Wm-2 from the surface. (seems it is you that said radiation “knows nothing of it’s surrounding matter”).
You talk like you know physics, so, don’t you see why so many rightfully have problems with Trenberth’s presentation of the IR portion? And that is just one aspect. Bob hit another one on the head in his article.

Lars P.
October 29, 2011 7:21 am

kuhnkat says:
October 28, 2011 at 4:39 pm
“2. If it does, what proportion of that back radiation is due to CO2?
That is the important question. No one knows. My analysis indicates that it is close to zero at the surface and about 100% in the stratosphere.”
kuhnkat if I correctly understand gases do radiate only in the same bands that they receive, only if the temperature is high enough. This would mean the CO2 from the stratosphere will radiate in the lower CO2 bandwidths, not the higher ones, and the radiation will be intercepted by CO2 about 10 meters below and above it. All this radiation in both directions is still net heat transfer from warm to cold.

Matt G
October 29, 2011 8:18 am

Michel says:
October 29, 2011 at 1:43 am
N2, O2 etc do absorb a little infrared, if you have ever used infrared spectroscopy before, the background absorbance reading is around 10 percent when none of the main greenhouse gases cover this band. (H2O, CO2, CH4, or O3 etc) This is down to the main gases in the air, nitrogen and oxygen. (only small yes, but still there.

Robert Clemenzi
October 29, 2011 8:47 am

Bob Fernley-Jones says:
October 28, 2011 at 11:13 pm
Well here it is; the Willis-wisdom from January 2008: Energy Balance at the Tropopause, at CA:
http://climateaudit.org/2008/01/10/energy-balance-at-the-tropopause/
Thanks, but I don’t see it there. Perhaps you are referring to greenhouse.bmp, which is not found. The earliest example I have found is
http://wattsupwiththat.com/2010/03/16/another-look-at-climate-sensitivity/
At any rate, it appears that Willis invented that image himself. In my opinion, it is wrong at many levels. First, the water in the troposphere emits a significant amount of energy directly to space. The proof is the increasing emissions from 400 to 600 cm-1 in figure 5 above. Next, the peak in the CO2 emissions to space (also in figure 5) comes from above the stratopause. In addition, there are no significant atmosphere to atmosphere photons that cross the tropopause. In addition, there is a significant energy transfer from the stratosphere it the tropopause.

October 29, 2011 12:24 pm

I don’t see any allowance in the charts for the nuclear furnace beneath our feet. Something other than the sun is heating mines of 1,000 feet to 120F degrees and more. Magma isn’t being kept fluid by the sun or gravity. Seems it should be at least a few percent, and should show up in measurements and models if we think they are accurate to 10ths of a percent.

RW
October 29, 2011 1:00 pm

There seem to be lots of flaws in Trenberth’s diagram. The most obvious to me is designating all the downward LW radiation received at the surface as ‘back radiation’. This is highly misleading, as downward LW has three potential sources and only a fraction of it is ‘back radiation’ (that which last originated surface emitted LW). The diagram makes it look like of the 396 W/m^2 emitted from the surface, 333 W/m^2 is coming back from the atmosphere, which is incorrect.
He also designates 78 W/m^2 of the post albedo is being ‘absorbed by the atmosphere’ and then brings this to the surface as part of the 333 W/m^2 designated as ‘back radiation’. This portion isn’t not ‘back radiation’ but ‘forward radiation’ from the Sun yet to reach the surface (key distinction).
Also, has anyone else noticed that he returns all the non-radiative flux from the surface to the atmosphere (latent heat and thermals) in the form of downward LW lumped in as part of the 333 W/m^2 designated as ‘back radiation’? What then is the source of the energy in the temperature component of precipitation? It’s not there.

Bob Fernley-Jones
October 29, 2011 2:18 pm

Robert Clemenzi @ October 29, at 8:47 am
Sorry, I did not check the links, but I can assure you that the Willis version was there, back in 2008. He remains proud of it, because he cited it again last August in his article here; “Radiating the Oceans”:
http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/#more-45114
Somewhere in the over 900 comments. He sure does generate debate!

Bob Fernley-Jones
October 29, 2011 2:43 pm

HEAT: Semantic misconceptions
Here is an extract from Wikipedia under that heading:

”There is some debate in the scientific community regarding exactly how the term heat should be used.[5] In current scientific usage, the language surrounding the term can be conflicting and even misleading. One study showed that several popular textbooks used language that implied several meanings of the term, that heat is the process of transferring energy, that it is the transferred energy (i.e., as if it were a substance), and that is an entity contained within a system, among other similar descriptions…”

http://en.wikipedia.org/wiki/Heat
My recollection is that physicists, not long ago, were quite happy to define HEAT at the quantum level as the state of kinetic energy within matter. Nowadays it seems that they prefer to consider it as a transfer of thermal energy from A to B.
The latter terminology is in conflict with engineering, where there is a whole field on “Heat Transfer”. In engineering, the heat content of matter can be calculated for any constant temperature; it is not necessarily a transient condition.
I think the sensible solution to the semantics issue is to consider the context of the use of the word.

October 29, 2011 3:19 pm

Ari Tai says: October 29, 2011 at 12:24 pm
I don’t see any allowance in the charts for the nuclear furnace beneath our feet.
Estimates I have seen are considerably less than 1 W/m^2 for geothermal energy flow. This is pretty small compared to most other energy flows.

Bob Fernley-Jones
October 29, 2011 7:02 pm

kuhnkat @ October 28, at 8:32 pm
I can’t really answer your questions, but would think that it is a case of “swings and roundabouts” and that the real world has an equivalence that is near enough to a flat surface for it to not matter within the scale of other uncertainties. Whatever, Trenberth also seems to assume that, and I would say it is the least of the problems in his diagram. (e.g. his emissivity = 1)
One of the things that have for a long time irritated me is the CAGW talk of significant albedo positive feedback when sea-ice melts and exposes “highly absorbent” water. (NO, I’m not going off-topic). Yet, it is well known that when the sun is low in the sky, as in the polar-regions, that specular reflection from water is high, and about the same as old snow, atop sea-ice. This led me, in connection with your issue, to wonder if there is specular reflection of IR at shallow angles on water. However, it seems to be a hard question, and I suspect that water would give close to Lambertian/ black body reflection/emission. Whatever the answer is, a lot of the self-absorbing issue that you raise would be minimised by the generally shallow angles involved.
Here is a simple description of Lambertian surfaces etc, from the perspective of optics. You can see the effect of shallow angles (and vectors) at the far right of figure 1. Caution: if you search around on the topic, you’ll find that some physicists seem to get over-excited by it, if you know what I mean.
http://escience.anu.edu.au/lecture/cg/Illumination/lambertCosineLaw.en.html

October 29, 2011 8:39 pm

Robert Clemenzi,
your simple $10 tool is not measuring what you think it is. It is not measuring the potential energy to be transferred. It isn’t even measuring the number of photons as you claim. The magnitude it measures is adjusted based on preset assumptions of emissivity of the object which you apparently haven’t checked.
I am glad that you have at least been awakened to the issue of the frequencies. There are a number of good manuals by manufacturers and other info on the net that explains how the instruments work and their limitations.
As mentioned above, NOTHING is a perfect Black Body with an emissivity of 1. The oceans and ice can be the closest with, I think, about .99 depending on their condition. The surface ranges from about .91-98. The atmosphere will also depend on the actual composition, but, I believe it is closer to .8 than .9. That may be something for you to research. I think I remember CO2 as about .16 in our atmosphere, but, like the surface, all the gases and aerosols contribute.
Basically, if you have the $10 unit it is designed to gather IR from up to about 1o ft from surfaces with an emissivity of about .98. I haven’t read what issues the distance would cause so can’t suggest anything there. The incorrect emmissivity will cause a reading higher than it should be.
The expensive instruments used by the big boys have adjustable emissivity. I would hope they have correctly computed the emissivity when they take their readings, except, the emissivity will depend on the humidity, clouds and aerosols and who has all that data?? They would need a massive instrumentation, satellite msu and/or sondes to even get in the ballpark.
Yeah, I don’t trust anybody!!!! 8>)

October 29, 2011 9:26 pm

Robert Stevenson,
Pekka Parilla over at Climate Etc. tells us that the time to collision is shorter than time to emission for GHG’s in the lower atmosphere. That and the multidirection emission when it does happen seems to be telling us that less IR would be going to the ground than the energy chart indicates. A smaller portion will get to the ground based on geometry. More energy is transferred by collision than gets emitted. The collisional energy will typically be convected. I don’t see how they can come up with those numbers. Emissions from higher up have a low probability of getting through the dense surface layer just like ground emissions are unlikely to get directly to the upper trop or TOA, yet, somehow we are to believe that there is almost as much radiation being absorbed by the ground as the atmosphere even though it would appear to be substantially biased up.

October 29, 2011 10:02 pm

Bob F-J,
thank you for the link.
I tend to think you may be right with the reflection issue in water unless the swells are large, which happens reasonably often. I had forgotten to consider the low angles and reflectivity. With the ground there is less reflectivity with most surfaces. We also have a lot more vertical roughness.
Of course, with the ocean, it would seem that pretty much all the DLR goes back up with evaporation anyway.

Bob Fernley-Jones
October 29, 2011 11:06 pm

Tim Folkerts @ October 28, at 6:19 pm, you wrote:
Bob FJ, Your diagram with sloping arrows has some good points.
My concern is that – by trying to indicate a depth to the atmosphere – you now open yourself to all sorts of questions about how quickly the arrows should taper [off] and how high the clouds are and how energy is transferred within the atmosphere and how high convection goes and ….

NO, definitely NO, it is not my problem, but a problem for Trenberth et al and his IPCC collusionists.

Bob Fernley-Jones
October 29, 2011 11:30 pm

R. Gates @ October 26, at 5:46 pm, you early-on wrote in part, concerning my article:
”There is lot’s to chew on here…”
I appreciate that since then you have probably been distracted by some rather hostile insinuations against your character/personality, but I would like you to put that aside and offer your views on my article.
Please.

October 30, 2011 10:47 am

Bob says:
“One of the things that have for a long time irritated me is the CAGW talk of significant albedo positive feedback when sea-ice melts and exposes “highly absorbent” water. (NO, I’m not going off-topic). Yet, it is well known that when the sun is low in the sky, as in the polar-regions, that specular reflection from water is high, and about the same as old snow, atop sea-ice. “
I don;t think this is as much of a problem as you might think for two reasons.
1) Snow typically reflects at least 60 % of light (http://en.wikipedia.org/wiki/File:Albedo-e_hg.svg). To get that sort of value for water, the light would have to hit at 85+ degrees from the normal (5 degrees from the horizontal) (http://en.wikipedia.org/wiki/File:Water_reflectivity.jpg). Much of the arctic ocean for much of the summer has the sun considerably higher than 5 degrees.
2) If the water is not perfectly smooth, the average angle if incidence will decrease (ie the water will hit more directly) leading to a lower reflection. The “upslope” of the wave facing the sun will block the light from the “downslope” on the other side of the wave. If the sun is 5 degrees above the horizon, but the wave slopes up at 10 degrees, the light hits at a 15 degree angle on that part of the wave. This will lower the reflectivity for that sunlight from 70% to 20%.

October 30, 2011 11:19 am

kuhnkat says: October 29, 2011 at 10:02 pm
Of course, with the ocean, it would seem that pretty much all the DLR goes back up with evaporation anyway.
I disagree. For starters, there is ~ 330 W/m^2 average DLR, but only ~ 80 W/m^2 goes up as evaporation. So no more than ~ 25 % of the DLR could go up as evaporation. The vast majority of the DLR must go up as “ULR”.
And of course, at least SOME of the evaporation is driven directly by sunlight, so less than 80 W/m^2 of the evaporation is due to DLR.

wayne
October 30, 2011 11:59 am

Bob says:
“… Yet, it is well known that when the sun is low in the sky, as in the polar-regions, that specular reflection from water is high, and about the same as old snow, atop sea-ice. “
Bob, see of MODIS and CERES. more precise than Tim’s logic:
NASA: Sea Ice and Snow Change, but Reflection Remains the Same
http://earthobservatory.nasa.gov/Features/ArcticReflector/arctic_reflector4.php
If anyone is worried of decreases in albedo with sea-ice melting at the poles, they need not, so far this effect cannot be found by the satellites of any significant amount, clouds compensate.

Bob Fernley-Jones
October 30, 2011 2:22 pm

Tim Folkerts @ October 30, at 10:47 am
I did not want to go off topic, so I’ll be brief.
1) If you look around, you should find a variety of albedo’s and incidence angles.
2) If the wave directions are “pointing at the sun”, part of them will absorb more, and part less. (another case of swings and roundabouts)
Wayne @ October 30, at 11:59 am
Thanks for the link Wayne…… very interesting; must study.

Bob Fernley-Jones
October 30, 2011 9:18 pm

Dear Moderator,
I’m puzzled and disappointed as to why Willis Eschenbach, has had nothing to say on this thread, since it touches something close to his heart; his own version of the Trenberth diagram. It might be something to do with this:
http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/#comment-727406
But whatever, please contact him and make sure that he is aware of this thread. I would appreciate his input.
BTW, a gerbil, as he described me, I discovered is a mouse-like creature not native to my homeland of Oz.

October 30, 2011 11:03 pm

Tim Folkerts,
No problem Tim, just my being very imprecise. I was trying to refer to the issue that little if any warming is from top down as it all goes back up one way or another.
My main thrust was that the objects on the surface irradiate themselves. Our host was pointing out that the reflection would prevent most of that in the case of the ocean and waves!!
What is your opinion of the effects of a rough surface causing self irradiation as another source of reduction of cooling?? Isn’t this why SB isn’t as accurate with a self irradiating geometry?? It would seem to be with solar cookers freezing water. When exposed to the sky only they can freeze water at higher air temps. When there is a wall, tree, or other object in their “view” it doesn’t work as well indicating significant radiation from solid objects!!
So, how is this included in the energy balance cartoons?? It isn’t. Only the DLR from GHG’s are which makes me think the effect is at least a little overestimated!!

Myrrh
October 31, 2011 6:45 am

Leif Svalgaard says:
October 28, 2011 at 7:12 am
Myrrh says:
October 28, 2011 at 5:36 am
http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/watabs.html
Your own link shows that the absorption of visible light by water is a million times weaker than that of infrared, but that only means that you need a layer of water a million times as thick, i.e 100 meters instead of a fraction of a millimeter.

?! And I thought my maths was bad, but the logic fail here.. 🙂 Water is a transparent medium for visible light, that is bog standard physics, well known knowledge used in countless industries, that chart shows it. It doesn’t absorb visible, not when there’s a millimetre of water or when there’s ten thousand feet of it. If there’s something happening at some points of zilch significance in the different visible wavelengths it doesn’t alter the basic mechanism which makes water a transparent medium for visible light, transparent means it is not absorbed, it cannot therefore be all absorbed the greater depth of water. Not absorbed means that is can’t heat water.
The AGWSF claim is that visible light, shortwaves, ‘is the energy direct from the Sun which is heating the Earth’s land and oceans’. This graphic shows just how utterly stupid the claim. And I’m supposed to be impressed by your reading of the information and the reading of other scientists like you priding themselves on their great scientific credentials? When you don’t even know what transparent means?? Transparent means the energy doesn’t have the mechanism to heat the medium, it is transmitted through without being absorbed.
From the link:
“Transparency of Water in the Visible Range
Water is strongly absorbing at most of the wavelengths in the electromagnetic spectrum, but it has a narrow window of transparency which includes the visible spectrum. The span of the absorption spectrum shown is from wavelengths on the order of a kilometer down to about the size of a proton, about 10-15 meters. It doesn’t absorb in the wavelength range of visible light, roughly 400-700 nm, because there is no physical mechanism which produces transitions in that region – it is too energetic for the vibrations of the water molecule and below the energies needed to cause electronic transitions.”
The water molecule keeps the tiddly visible out. Which words in the analysis of the information given by Georgia uni are you having a problem understanding in those I’ve bolded?
but? it? has? a? narrow? window? of? transparency? which? includes? the? visible? spectrum?
Since the oceans are more than 100 meters deep they will absorb all the visible light falling on them. The mechanism involves overtones of vibrational stretching as you have been shown many times.
So what changes the water to being not transparent the deeper it is..?
It? doesn’t? absorb? in? the? wavelength? range? of? visible? light?, roughly 400-700 nm?, because? there? is? no? physical? mechanism? which? produces? transitions? in? that? region?
What I’ve shown you many times is that vibrational stretching is not something that electronic transmissions can effect since visible works on the electronic transition levels of electrons.
Reflection/scattering in the atmosphere is the molecules of nitrogen and oxygen getting rid of piddling visible energy.
It takes the oomph of real heat energy, the invisible thermal infrared energy direct from the Sun, to heat land and oceans, to vibrate the whole molecules. Water is greatly absorbing in the thermal infrared because thermal can move whole atoms and molecules into vibration which is what it takes to heat matter.
Thermal infrared, heat, moves molecules into vibration, this is heat.
Visible can move electrons but doesn’t have the power to move whole molecules into vibration.
Visible is absorbed by the electrons of the molecules of oxygen and nitrogen in our atmosphere, and gets reflected back out by this, blue is more easily scattered so we have a blue sky.
Visible is being bounced all over the place like a ball in a pin ball machine, it doesn’t have the oomph to move the whole molecule, instead the molecule kicks it around all over the sky just with its electrons.
The AGWSF claim that the atmosphere is transparent to visible is another false, science fiction, meme introduced to confuse real physics as if real fact. How much is it heating the fluid gaseous air atmosphere above us since you claim that all absorption is directly creating heat and reflection/scattering is by nitrogen and oxygen electrons absorbing it?? Where is this in your ‘energy budget’? Logic fail. Science incompetance.
Water is really transparent to visible, (not like your fake greenhouse cartoon claims for visible in the atmosphere). Water is really transparent to visible Because there is no physical mechanism which produces transitions in that region.
Whatever you think you’re saying about vibrational stretching, it’s gobbledegook. Vibrational stretching is the movement of whole molecules which infrared can effect, visible’s limitations to electronic transitions can’t do this.
Either you have misunderstood it or you’re deliberately muddying the waters here. I don’t much care which, either way, you show yourself promoting a science fiction world where visible light has been given the properties of thermal infrared which is the only real heat energy direct from the Sun, the Sun’s real invisible thermal energy direct to us which heat we feel; we cannot feel visible light, it is not thermal. I have tended to give you credit for your mangling of the real facts as being deliberate because a great scientist and all that, but perhaps it’s just because you have no idea what you’re talking about..
… perhaps you’re simply out of your depth…
Again, the different mechanisms between Visible and Thermal Infrared, visible doesn’t do molecular vibration.
http://en.wikipedia.org/wiki/Transparency_and_translucency
“Mechanisms of selective light wave absorption include:
Electronic: Transitions in electron energy levels within the atom (e.g., pigments). These transitions are typically in the ultraviolet (UV) and/or visible portions of the spectrum.
Vibrational: Resonance in atomic/molecular vibrational modes. These transitions are typically in the infrared portion of the spectrum.”

Myrrh
October 31, 2011 7:53 am

Ask why is it so? says:
October 28, 2011 at 8:05 am
Myrrh, thank you for your reply however I think you seem confused. I was talking about radiation not whether it is visible or invisible. The 2nd law of thermodynamics is often brought up to prove that CO2 cannot cause warming because heat does not travel from cold to hot, and the surface of the earth being hotter would not receive the heat CO2 apparently sends back to the surface to cause global warming. If it were heat, this would be true, however, radiation unlike heat can travel in any direction including back down to the surface. I believe it is important for it to be understood that heat is controlled by the Laws of Thermodynamics, radiation isn’t. I’m sorry if my explanation was too simple. I often get lectured about that but I believe the KISS theory much more effective than the complicated gobbledygook scientists use.
I’m still trying to work out how long wave radiation can produce a higher temperature than short wave radiation.

I don’t have the time to go into this further with you here, sorry, I’ll just take your last sentence.
Longwave radiation is thermal, it is heat on the move, it is invisible. It is the heat you feel from the Sun. You cannot feel visible light, it is not hot. It is the heat in the Sun, the thermal energy of the Sun, that creates the visible. The visible light from the Sun is the product of the Sun’s great thermal energy, visible is not that energy. That great thermal energy of the Sun is what travels to us at the speed of light and reaches us direct on the surface of the Earth; we feel its heat, it is invisible. It heats us from the inside because water is the very great absorber of thermal energy, and we are mostly water, (and around 20% carbon). Visible light cannot heat water, water is really transparent to visible light, which means that the water molecule does not absorb visible ligh; visible light passes through water without being absorbed, this is called transmission.
See the link in my last post (to Leif), there are basic differences in how visible and thermal infrared, which is heat, work on meeting matter. Visible’s electronic transitions is interaction with the electrons, the visible light is too small to do even much here, as I explained, in the atmosphere the molecules of oxygen and nitrogen absorb visible in their electrons and bounce it back out again. This is the second possible way that electrons affect and are affected by matter in electronic transitions as described in this section on that page, the third possible is transmission through a transparent medium, when the energy is not absorbed.
I have extracted the information about these differences in another discussion here in this post: http://wattsupwiththat.com/2011/10/18/replicating-al-gores-climate-101-video-experiment-shows-that-his-high-school-physics-could-never-work-as-advertised/#comment-778960

The claim that visible light heats matter is falsified by real world physics, it is a scientific fraud to keep promoting it as if it is real world physics. The fraud is that visible light has been given the properties of the invisible heat energy of the Sun, which is thermal infrared.

Visible light cannot heat land and oceans, the invisible thermal energy of the Sun we receive directly here, heat from the Sun, can and does heat matter. Not only does water greatly absorb this heat direct from the Sun, but it has a very great capacity to store it, which is its heat capacity. Land also absorbs heat direct from the Sun, thermal infrared, but has a lower heat capacity than water, so it warms up more quicky but releases it quickly. This is how our weather comes to be on Earth, from the different temperature gradients created by this, and so we have winds, which is volumes of the fluid gas atmosphere on the move, as a heated volume of air rises and is displaced by the volume of cold air above it. The smaller energy of visible light is used in many other ways; we see the world by it, all the colours and so forms and it creates sugars, not heat, in photosynthesis, and so on.
Confusing by deliberately taking out the scale and property differences between these energies by saying they are all the same energy is giving the false impression that ‘highly energetic’ means greater power, it just means that the more highly energetic the smaller, gamma billions of times smaller than radio..
All these wavelengths are distinctly different from each other, they have their own properties distinct from each other, affect matter in different ways. There are also category distinctions, Heat and Light are the basic ones in the differences between ‘shortwave visible and invisible from the Sun around the Visible wavelength’, i.e. Solar, which includes near infrared which is not thermal, and the category Heat, which is the thermal energy from the Sun on the move, the invisible longer waves of thermal infrared. Light is not Heat, Light is not thermal, we do not feel it as heat.
It is utterly disgraceful that this deliberate change of giving visible light the properties of the thermal invisible infrared has been introduced into education to support the fake AGW money grab scheme. That all these discussions and arguments between those believing in AGW and those not are using this faked property in their posts is sure indication that we are seeing those who have walked through the looking glass with Alice, where you can believe all kinds of impossible things, create all kinds of fictional worlds.
Visible light heating land and oceans is a science fiction world, it is not reality here, where I am, where real world traditional physics contradicts it..

Robert Stevenson
October 31, 2011 8:20 am

The equation for the interchange of radiant heat between air at temp Tg containing CO2 at partial pressure Pc and the ground surface at temperature Ts through a distance L, per unit of surface, is:
Q = sigma x (aTs^4 – eTg^4)
where ‘e’ denotes gas or air emissivity and ‘a’ the gas or air absorptivity for black body radiation
from the surface at Ts.
Although the absorptivity equals its emissivity when Ts = Tg, a correction is made when ‘a’ for CO2 is evaluated as emissivity at Ts by multiplying the result by (Tg/Ts)^0.65.
As well as temperature ‘e’ for CO2 depends on product term PcL and the total pressure Pt (for the lower atmosphere this can be taken as 1atm).
Equating the absorption by CO2 of land IR ( Sigma x aTs^4) to the energy absorbable by the CO2 wavebands, gives a value of L (for Pc 0.0004 atm) of not less than 2000 m.
Repeating the calc for atmospheric water vapour glves a value for L of 120 m.

Robert Clemenzi
October 31, 2011 9:14 am

Myrrh, if water is transparent, then why can’t we see through clouds? Why is the sky clearer on a cold night than on a warm night?
I assume that you realize that sea water has many salts dissolved in it. Those salts provide the additional absorption. As a result, as every diver knows, the oceans do absorb visible light. Whether it is the water, or the stuff dissolved in it, does not matter – the light is still absorbed.
On the other hand, you claim that “Visible light cannot heat land and oceans” has some merit because most of the surface albedo is in the visible part of the spectrum. Thus, the reason we do not feel visible light as heat is simply because our bodies reflect it. This is why most of the heat we feel is from the infrared. However, when a pigment (or the ocean) absorbs a visible photon, that energy is eventually converted into heat. In the case of leaves, the light is first converted into sugar, but even that will eventually oxidize and produce heat at sometime in the future.
In the case of UV in the stratosphere, the photons are not directly converted to heat. Instead, they break molecular bonds and create free radicals. When the free radicals take part in various reactions, the stored energy is converted to heat. As a result, the temperature of the stratosphere increases with height.

mkelly
October 31, 2011 9:16 am

Robert Stevenson says:
October 31, 2011 at 8:20 am
Nice post Robert. Emissivity of an H2O-CO2 combination is always an important question. Hottel from memory had CO2 as very low on emissivity and always lowered emissivity of H2O when in combination.
Where is you get this, (Tg/Ts)^0.65, from?

October 31, 2011 11:13 am

Robert Stevenson,
please excuse my ignorance, but, wouldn’t you need to combine these two calculations in some way to give a net height for absorption as water vapor and CO2 absorptive wave lengths overlap?? Or am I completely misunderstanding what you are stating??

Myrrh
October 31, 2011 3:59 pm

Robert Clemenzi says:
October 31, 2011 at 9:14 am
Myrrh, if water is transparent, then why can’t we see through clouds? Why is the sky clearer on a cold night than on a warm night? etc.
Good grief. I’m arguing about the stupid fictional science of the ‘energy budget’ you’re all working to, where shortwave is falsely credited the heating mechanism for all land and oceans, having been given the properties of thermal infrared which is the real heat from the Sun and which we can all feel as the thermal energy of the Sun, which real heat energy you have excluded from your ‘energy budget’ saying it doesn’t even reach us even though we can feel it.., and, make the effort to give you real world physics on the difference in scale between these two energies of the sun, which shows that on visible scale which works to electronic transitions it is not possible to heat the gazzillions of gallons of water on our planet, while thermal infrared which is the real thermal energy from the Sun can and does, because electronic transitions cannot move even one molecule of water into vibration and is not absorbed which is what makes water transparent, while thermal infrared heats exactly by this doing this, its absorbed energy moving the molecule into vibration, and you widdle off about why you can’t see through clouds so how is it transparent..?
I’m talking here about the fact that the very basic physics premises have been changed in support of AGW, that non thermal shortwave Solar energies have been given the properties of heat, which is actually the invisible thermal infrared. This is a complete, and obviously, efficient scrambling of basic physics, leaving all you and your arguments in a totally ludicrous fictional world created by this sleight of hand. You’ve been had. Light is not at thermal energy, it is not Heat, it cannot and does not heat matter. And your claim is that it heats all land and oceans!
Water is transparent to visible light because it is transparent, transparent in physics, real physics here, means that visible energy cannot, I repeat, cannot be absorbed, the volumes of the molecules of water keep visible light out. Water does not allow visible light to play with its electrons and the tiddly visible isn’t big and strong enough to move a molecule into vibration, so it is passed through. This is called transmission.
It’s what you generic claim in your fictional energy budget that the atmosphere is for visible light, but, in the real physics in the real world the volume of the fluid gaseous atmosphere is not transparent to visible, because visible on the electronic transition level appropriate to its size is absorbed by the electrons of the molecules of the nitrogen and oxygen, before being kicked out, this is what creates the reflection/scattering of visible light. Since you claim that all ‘absorption creates heat’ in your one dimensional reality, how much is that blue in the sky heating your atmosphere? I don’t see it mentioned on your energy budget cartoons.
Either take it out of your budget and put back thermal infrared direct from the Sun to Earth, or admit you’re all garbling nonsense arguments at each other about a science fictional world because you’ve changed the basic properties and processes of Heat and Light energies from the Sun.

October 31, 2011 4:35 pm

Kuhnkat asks: “What is your opinion of the effects of a rough surface causing self irradiation as another source of reduction of cooling?? “
I can imagine several affects of surface roughness.
First of all, the area that should be used would be the area as seen from a point of interest. Or perhaps more precisely, it is related to the solid angle subtended by the various surfaces involved. So within a valley (or between large waves) the energy from the sky would be less and the energy from the ground (or water) would be larger. Hence frost not forming on the sides of cars facing a heated wall. However, once you are high enough that the horizon looks flat, then the surface topology should make less difference.
On another front, rough surfaces generally make the emissivity larger. See http://www.ib.cnea.gov.ar/~experim2/Cosas/omega/emisivity.htm for example. The extreme case would be a deep pits, which becomes an extremely effective black body (http://www.imagesfromhere.com/archives/Grate%20Work%206794.jpg). This might suggest that rough water should have a higher emissivity than smooth water. So it would be better at both emitting and absorbing light. There could a slight complication that water will form “whitecaps” when it gets rough. This could affect the emissivity.
“My main thrust was that the objects on the surface irradiate themselves.
True, but that also means that there is more surface area. If there are waves irradiating themselves, that means the waves have more surface area than the flat ocean.
Overall, I suspect the surface topography would have little affect on overall absorption or emission from he ocean.

Gail Combs