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) | |||
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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.
kuhnkat @ur momisugly 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
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>)
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.
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.
Tim Folkerts @ur momisugly 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.
R. Gates @ur momisugly 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.
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%.
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.
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.
Tim Folkerts @ur momisugly 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 @ur momisugly October 30, at 11:59 am
Thanks for the link Wayne…… very interesting; must study.
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.
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!!
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.”
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
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..
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.
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.
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?
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??
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.
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.
Ask why is it so? says:
October 28, 2011 at 8:05 am
….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.
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The temperature depends on selective absorption. For example I have a livestock trailer, It is painted midnight blue (almost black) and white. When inside (painted uniform gray) I can tell by the difference in heat whether the paint on the outside is white or blue. At high noon the dark area is very warm to the touch while the white area is not.
That is one part (absorbed or reflected)
The second is the actual amount of energy in the photon. Depending on the wavelengths it might take two or three LW photons to equal the energy of an EUV photon.
Myrrh.. it hurts my brain to read your posts.
Visible can move electrons but doesn’t have the power to move whole molecules into vibration.
Visible has higher energy than infrared, so it most certainly does have “the power” to vibrate atoms.. visible light has too high of a frequency (and thus, the sign of the electric field switches too fast) for most atomic dipoles to react quickly enough, and this is what makes water more transparent to visible light. Electrons move at higher frequencies, so they can match with higher-frequency light than dipoles (atoms) can.
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.
Not true – 1 cm of water is about 99% transparent to deep-red visible light, and about 99.99% transparent to deep-blue visible light. Not purely transparent in either case. Absorption of visible light is mostly at the overtone frequencies of the stretch and/or bend frequencies… so the absorption is weaker, but definitely there.
Will you take wikipedia as a source? If not, I can provide papers from physicists.
http://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water
Observe, the absorption in the visible spectrum is not zero.
You cannot feel visible light, it is not hot.
Want to bet? Humans are reasonably opaque to visible light, which means it gets absorbed, which means, yes, you can feel it if it’s strong enough. Try it! Go stand in front of a 10,000-Watt set of concert LED lights, with a few panes of glass in between to block the infrared and any convection, and you’ll see for yourself.
We tend to feel more infrared because there’s more of it around (from low-efficiency lighting and background radiation), not because it’s more energetic or more absorbed.
As Robert mentioned, there are other ways for high-energy light to interact with matter, even “transparent” matter, via ionization, scattering, conversion to chemical energy, etc.
Wayne says: “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. ”
That is not quite how I would say it. The ~ 330 W/m^2 should be though of as the integrated affect of the atmosphere acting on a given square meter of the surface. It does not come from any specific cubic meter of space, or from some specific square meter of surface in the atmosphere somewhere.
I could set some sort of 1 m^2 empty picture frame on the ground. Presumably ~ 390 W/m^2 of IR would be heading up thru the frame from the ground. The downward radiation through the frame would be a combination of radiation from many parts of the atmosphere heading many directions. In the bands that CO2 or H2O gas emit IR, much of the IR comes from quite close by — within a few meters of the ground. Along the edges of the bands, the distance could be 100’s of meters. In the bands where these don’t emit, then the IR come be coming from several km away from clouds that emit a continuous spectrum of IR. The net value should on average be ~ 330 W/m^2
If you take that same frame to the “top of atmosphere” (perhaps 50 km up), you get a completely different situation. The IR in CO2 bands should come from near/just below the tropopause. Since there is little H2O that high, the IR from H2O gas would come from a little lower in the atmosphere (as I understand it). The IR “in the gaps” would come from the tops of the clouds far below.
There is no reason to expect this value to be ~ 330 W/m^2 since it is coming from very different places. You cannot simply double the DLR arriving at the surface and expect to get the ULR at any particular location.
Tim Folkerts,
good job side stepping the issue of the surface irradiating itself SLOWING THE COOLING and invalidating the correctness of SB!! 8>)
So, since it irradiates itself similar to the GHG’s, except it is a wider range of frequencies and higher amplitude I think the simplified models have more problems. Are the Big Boys dealing with this any better?