Guest post by Bob Fernley-Jones by Bob Fernley-Jones AKA Bob_FJ
CAUTION: This is written in Anglo-Oz English.
Here is the diagram as extracted from their 2009 paper, it being an update of that in the IPCC report of 2007 (& also 2001):
The unusual aspect of this diagram is that instead of directly showing radiative Heat Transfer from the surface, it gives their depiction of the greenhouse effect in terms of radiation flux or Electro-Magnetic Radiation, (AKA; EMR and a number of other descriptions of conflict between applied scientists and physicists). EMR is a form of energy that is sometimes confused with HEAT. It will be explained later, that the 396 W/m^2 surface radiation depicted above has very different behaviour to HEAT. Furthermore, temperature change in matter can only take place when there is a HEAT transfer, regardless of how much EMR is whizzing around in the atmosphere.
A more popular schematic from various divisions around NASA and Wikipedia etc, is next, and it avoids the issue above:
- Figure 2 NASA
Returning to the Trenberth et al paper, (link is in line 1 above), they give that the 396 W/m2 of EMR emitted from the surface in Fig.1 is calculated primarily by using the Stefan–Boltzmann law, and global year average conditions. Putting aside a few lesser but rather significant issues therein, it is useful to know that:
1) The Stefan-Boltzmann law (S-B) describes the total emission from a flat surface that is equally radiated in all directions, (is isotropic/hemispherical). Stefan found this via experimental measurement, and later his student Boltzmann derived it mathematically.
2) The validity of equally distributed hemispherical EMR is demonstrated quite well by observing the Sun. (with eye protection). It appears to be a flat disc of uniform brightness, but of course it is a sphere, and at its outer edge, the radiation towards Earth is tangential from its apparent surface, not vertical. It is not a perfect demonstration because of a phenomenon called limb darkening, due to the Sun not having a definable surface, but actually plasma with opacity effects. However, it is generally not apparent to the eye and the normally observed (shielded) eyeball observation is arguably adequate for purpose here.
3) Whilst reportedly the original Stefan lab test was for a small flat body radiating into a hemisphere, its conclusions can be extended to larger areas by simple addition of many small flat bodies of collectively flat configuration, because of the ability of EMR waves to pass through each other. This can be demonstrated by car driving at night, when approaching headlights do not change in brightness as a consequence of your own headlights opposing them. (not to be confused with any dazzling effects and fringe illumination)
4) My sketch below demonstrates how radiation is at its greatest concentration in the lateral directions. It applies to both the initial S-B hemispherical surface radiation and to subsequent spherical radiation from the atmosphere itself.
5) Expanding on the text in Figure 3: Air temperature decreases with altitude, (with lapse rate), but if we take any thin layer of air over a small region, and time interval, and with little turbulence, the temperature in the layer can be treated as constant. Yet, the most concentrated radiation within the layer is horizontal in all directions, but with a net heat transfer of zero. Where the radiation is not perfectly horizontal, adjacent layers will provide interception of it.
A more concise way of looking at it is with vectors, which put simply is a mathematical method for analysing parameters that 
possess directional information. Figure 4, takes a random ray of EMR (C) at a modestly shallow angle, and analyses its vertical and horizontal vector components. The length of each vector is proportional to the power of the ray, in that direction, such that A + B = C. Of course this figure is only in 2D, and there are countless multi-directional rays in 3D, with the majority approaching the horizontal, through 360 planar degrees, where the vertical components also approach zero.
6) Trenberth’s figure 1 gives that 65% of the HEAT loss from the surface is via thermals and evapo-transpiration. What is not elaborated is that as a consequence of this upward HEAT transfer, additional infrared radiation takes place in the air column by virtue of it being warmed. This initially starts as spherical emission and absorption, but as the air progressively thins upwards, absorption slows, and that radiation ultimately escapes directly to space. Thus, the infrared radiation observable from space has complex sources from various altitudes, but has no labels to say where it came from, making some of the attributions “difficult”.
DISCUSSION; So what to make of this?
The initial isotropic S-B surface emission, (Trenberth’s global 396 W/m2), would largely be absorbed by the greenhouse gases instantaneously near the surface. (ignoring some escaping directly to space through the so-called “atmospheric window”). However, a large proportion of the initial S-B 396 surface emission would be continuously lateral, at the Trenberth imposed constant conditions, without any heat transfer, and its horizontal vectors CANNOT be part of the alleged 396 vertical flux, because they are outside of the vertical field of view.
After the initial atmospheric absorptions, the S-B law, which applied initially to the surface, no longer applies to the air above. (although some clouds are sometimes considered to be not far-off from a black body). Most of the air’s initial absorption/emission is close to the surface, but the vertical distribution range is large, because of considerable variation in the photon free path lengths. These vary with many factors, a big one being the regional and more powerful GHG water vapour level range which varies globally between around ~0 to ~4%. (compared with CO2 at a somewhat constant ~0.04%). The total complexities in attempting to model/calculate what may be happening are huge and beyond the scope of this here, but the point is that every layer of air at ascending altitudes continuously possesses a great deal of lateral radiation that is partly driven by the S-B hemispherical 396, but cannot therefore be part of the vertical 396 claimed in Figure 1.
CONCLUSIONS:
The vertical radiative flux portrayed by Trenberth et al of 396 W/m^2 ascending from the surface to a high cloud level is not supported by first principle considerations. The S-B 396 W/m^2 is by definition isotropic as also is its ascending progeny, with always prevailing horizontal vector components that are not in the field of view of the vertical. The remaining vertical components of EMR from that source are thus less than 396 W/m^2.
It is apparent that HEAT loss from the surface via convective/evaporative processes must add to the real vertical EMR loss from the surface, and as observed from space. It may be that there is a resultant of similar order to 396 W/m^2, but that is NOT the S-B radiative process described by Trenberth.
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ADDENDUM FOR AFICIONADOS
I Seek your advice
In figure 5 below, note that the NIMBUS 4 satellite data on the left must be for ALL sources of radiation as seen from space, in this case, at some point over the tropical Pacific. The total emissions, amount to the integrated area under the curve, which unfortunately is not given. However, for comparison purposes, a MODTRAN calculator, looking down from 100 Km gives some interesting information for the figure, which is further elaborated in the tables below. Unfortunately the calculator does not give global data or average cloud/sky conditions, so we have apples and pears to compare, not only with Nimbus, but also with Trenberth. However, they all seem to be of somewhat similar order, and see the additional tabulations.
| Compare MODTRAN & “Trenberth”, looking down from 2 altitudes, plus Surface Temperature | ||||
| Location | Kelvin | 10 metres | 100 Km. | (Centigrade) |
| Tropical Atmosphere | 300K | 419 W/m^2 | 288 W/m^2 | (27C) |
| Mid-latitude Summer | 294K | 391 W/m^2 | 280 W/m^2 | (21C) |
| Mid-latitude Winter | 272K | 291 W/m^2 | 228 W/m^2 | (-1C) |
| Sub-Arctic Winter | 257K | 235 W/m^2 | 196 W/m^2 | (-16C) |
| Trenberth Global | 288K ? | 396 W/m^2 | 239 W/m^2 | (15C ?) |
| Compare MODTRAN & “Trenberth”, looking UP from 4 altitudes: W/m^2 | ||||
| Location | From 10 m | From 2 Km | From 4Km | From 6Km |
| Tropical Atmosphere | 348 | 252 | 181 | 125 |
| Mid-latitude Summer | 310 | 232 | 168 | 118 |
| Mid-latitude Winter | 206 | 161 | 115 | 75 |
| Sub-Arctic Winter | 162 | 132 | 94 | 58 |
| Trenberth Global | 333 Shown as coming from high cloud area (= BS according to MODTRAN) | |||
Nasif Nahle says:
December 2, 2011 at 8:29 am
The molecule of carbon dioxide emits quanta/waves with wavelengths pertaining to radio, such quanta/waves are not thermal radiation; consequently, they cannot be transferred as heat.
Nasif Nahle says:
December 2, 2011 at 3:27 pm
@Tim O'Donovan Folkers…
Tim: “Whether or not it is “thermal radiation” depends on the source, not the wavelength. The cosmic microwave background is thermal radiation. Microwave ovens are not thermal radiation. Visible light from an incandescent bulb is thermal; visible light from a fluorescent bulb in not thermal.”
Thermal Radiation is the fraction of the electromagnetic spectrum between wavelengths 0.1 μm and 100 μm.
Thermal Radiation includes visible spectrum, almost the whole ultraviolet spectrum (Vacuum UV, Far UV, C-UV, Middle UV, B-UV, Near UV, and A-UV. Low UV is excluded), and almost the whole infrared (IR) spectrum, except the portion of the spectrum corresponding to wavelengths of Far-IR, from 101 μm to 1000 μm:
Pitts, Donald and Sissom, Leighton. Heat Transfer. 1998. McGraw-Hill. Pp. 289-311.
Please, can we sort this out once and for all?
Visible light is not thermal, it is Light not Heat.
What applies to: “wavelengths pertaining to radio, such quanta/waves are not thermal radiation; consequently, they cannot be transferred as heat”, applies to visible light. Visible light is not thermal, it cannot be transferred as heat.
Near infrared is not thermal, it is not hot, it does not transfer as heat. It cannot heat things up any more than visible light can heat things up. BECAUSE they are not thermal radiation.
>… such quanta/waves are not thermal radiation
Tim: “Whether or not it is “thermal radiation” depends on the source, not the wavelength. The cosmic microwave background is thermal radiation. Microwave ovens are not thermal radiation. Visible light from an incandescent bulb is thermal; visible light from a fluorescent bulb in not thermal.”
Thermal depends on the the wavelength, not the source. HEAT, is thermal energy, Heat in transfer from the Sun is thermal energy in transfer by radiation, thermal energy in transfer by radiation is thermal infrared, therefore, thermal infrared is heat. Visible light is not heat. It is not thermal energy in transfer from the Sun nor from an incandescent bulb. It cannot heat stuff up.
It cannot heat stuff up because it cannot heat stuff up. It takes energy capable of moving atoms and molecules into vibrational states to heat stuff up, because that is what heat is. That is what thermal energy of the Sun is, and thermal energy from an incandescent light bulb. 95% of the energy radiated out from an incandescent light bulb is thermal infrared, heat, only 5% is light. Thermal energy, thermal infrared, heat, is what you can still feel from the light bulb after you have switched it off.
It appears to me, that you have all become horribly, and nonsensically, confused by such concepts as ‘background radiation is thermal’ and ‘everything above absolute zero emits thermal radiation’. You really need to get your concepts sorted. You’re all talking absolute gobbledegook.
Visible light as claimed in the AGWSF (AGWScienceFiction) energy budget which y’all think is logical, is ABSOLUTE bloody nonsense. Visible light cannot heat land and oceans. It cannot heat land and oceans any more than radiowaves can heat land and oceans. It does not suddenly become ‘thermal’ because it comes from the Sun, it is not thermal from the Sun any more than it is thermal from a fluorescent bulb. Fluorescence is light, not heat. It takes heat to transfer as heat.
Do try and get your heads around the difference between Light and Heat, as delineated in traditional physics. Light, such as visible and uv, do not move molecules and atoms into vibrational states, they work on the electronic transition levels, not atom/molecule vibrational levels.
For example, visible light which is not heat from a fluorescent bulb and which therefore is not heat from the Sun, is scattered by the electrons of the molecules of nitrogen and oxygen in the atmosphere. Reflected/scattered is one of effects because of how visible and uv interact with matter, because one of the four things these wavelengths have as an effect. When an electron absorbs visible light and briefly energises the electron it then comes back to norm state sending the light back out back out the way it came, this is why we have a blue sky. It’s like a pinball machine, visible light hits the electrons and gets scattered. The energy is being utilised in this, not in heating the molecules of oxygen and nitrogen. Electronic transitions do not heat matter, their energies are used in non-thermal encounters, for example, chemical energy of photosynthesis, in creating sugars, this is not creating heat it is creating sugar. Water is transparent to visible light, this means that visible light waves do not even have electronic transitions with water molecules, but are transmitted through without being absorbed. The AGWSF energy budget which you all have bought into as if real physics and not the science fiction world it is, says that visible light heats the oceans. Prove it.
You don’t even know how to prove it, or understand what I’m asking, because you either think thermal means from which source or you can’t understand that visible is not heat any more than radiowaves are heat.
Until y’all can prove that the Visible light in the KT97 and ilk claim, which is now ubiquitous in education from this junk science fiction of AGW, can actually heat land and oceans as claimed you will not understand just how ridiculous your claims and your arguments with each other are, even more garbled nonsense because you don’t understand the difference between heat and light in first principles.
Nasif Nahle:
Hi, I asked you a number of questions in http://wattsupwiththat.com/2011/10/26/does-the-trenberth-et-al-%e2%80%9cearth%e2%80%99s-energy-budget-diagram%e2%80%9d-contain-a-paradox/#comment-816466 because I find your comment incomplete and the conclusions to be against what I think is accepted science based on experiments, analysis, etc, over the years by many folks (and with proven results). If your conclusions do contradict what is widely believed and you challenge the accepted views, your conclusions have to follow from your “proof”. That “proof” is rather incomplete.
Why do I mention this? Because you continue to make those unorthodox statements apparently without support. For example:
>> The carbon dioxide molecule emits quantum/waves with energy E = 5.4 x 10^(-26) J, which corresponds to wavelength (λ) = 3.75 m. those are radio waves, not thermal radiation. An excited molecule of carbon dioxide has not available microstates, so it cannot absorb more energy; consequently, it only is able to emit quanta/waves with longer wavelength and lower frequency: Serway, Raymond A., Moses, Clement J., Moyer, Curt A. Modern Physics-3rd Edition. Brooks Cole. 2005.
Also, for the benefit of the many of us who don’t have access to that text, can you quote the part of that “Modern Physics” text you believe supports your view that CO2 is only able to emit in the radio range or that the region where CO2 absorbs does not overlap the region where CO2 emits?
>> The values found at 300 K, p of 1 atm, 1000 m and density 0.00069 kg/m^3.
OK.
I think this is what I assumed you were doing. Not having the density handy, I could not verify numbers, so I assumed you were looking at a column of atmosphere many kilometers up. So let me ask this.
Why would you only look at 1000m?
Also, as I noted in my other reply, can you please fill in the steps that take us from this observation/calculation into your conclusion that CO2 cannot lead to any greenhouse effect, in particular (to pick an example), that doubling CO2 concentrations could not possibly result in any earth surface warming (global average)?
Bob,
“So why does the 396 not make it up there?
…..because we have an absorbent atmosphere, the critical wavelength photons from the surface are annihilated in quite short distances. When they are annihilated, they can no longer be observed.”
Yes, I agree. The ‘annihilated’ or absorbed radiative power from the surface is transferred to the atmosphere via collisions, which heat the gases of the atmosphere (mostly N2 and O2). The heated atmosphere then acts as grey body, isotropically radiating in the LW IR according to its temperature like any other heated object.
“See particularly my item 4 including Fig 3 in the article. Do you disagree that most radiation is sideways? Can you not see that the critical wavelengths are annihilated before escape to space is possible? Then please note that the horizontal vector components are persistently there as part of the isotropic emissions. (They do not go away).”
Yes, I agree that most of the emission has a sideways component to its direction and is not directly perpedicular to the surface. I also agree that all the radiative emission in the atmosphere is isotropic (equal in all directions).
“Are you hypothesising that when those spawning photons from the surface are absorbed that somehow their progeny after random collisional thermalization with N2 and O2 and whatnot are rotated through 90 degrees to the vertical?”
No, not at all.
Myrrh, I think you are using your body’s biological characteristics to intuitively yet wrongly claim that visible light in not “thermal” in the sense that it can’t impart kinetic energy onto matter.
Can you find a single text book or experiment that would support your claim?
Let me ask you something. How do you think your eyes can sense visible radiation?
Something in the visible light has to communicate with your eyes. There is an energy transfer. Your eyes are tuned in (via evolutionary changes) to accept visible light photons while rejecting most other types of photons.
Myrrh, sorry. I didn’t finish reading all of your comment.
So you think visible light photons can cause electron transitions in molecules but don’t impart any kinetic energy? What makes the visible light photons different than other types of photons?
CO2 has limited ability to accept photons, but a body full of molecules interacting with each other can exhibit a very different and wide range of abilities to absorb photons.
So, specifically, why do you think the earth and oceans (and most solids, especially in large amounts) cannot absorb visible light? Where do you think is the limit of what the earth can absorb.. and why do you think that?
Myrhh, http://en.wikipedia.org/wiki/Attenuation explains that light (eg, visible light) can lose its intensity as it passes through some material. See also http://en.wikipedia.org/wiki/Beer-Lambert_law
So if you have less light intensity at the end, what happened to that energy that was lost?
If that energy excited electrons, then do those electrons stay excited forever? If not, what happens to this energy?
Also, what do you think of this: http://en.wikipedia.org/wiki/Black_body
When you say people here are talking “gobbledegook”, do you realize that physicists have been talking this same gobbledegook for many years?
@Jose_X…
I am sorry, José, but I will not respond questions of this tone:
“Why do I mention this? Because you continue to make those unorthodox statements apparently without support. For example:
“text you believe supports your view that CO2 is only able to emit in the radio range or that the region where CO2 absorbs does not overlap the region where CO2 emits?”
It is not my fault that you have not enough knowledge on quantum thermodynamics as to understand that those are not my assertions or beliefs, but well demonstrated science.
I will only answer the following question:
can you please fill in the steps that take us from this observation/calculation into your conclusion that CO2 cannot lead to any greenhouse effect, in particular (to pick an example), that doubling CO2 concentrations could not possibly result in any earth surface warming (global average)?
And here my answer:
ΔT = (5.35 W/m^2 (LN (([CO_2]_f * 2) / [CO_2]_∞)) / (4 * σ * (T_st)^3) = 0.65 K.
That would be the change of temperature of only the fraction of carbon dioxide that absorbed that load of thermal radiation (5.35 W/m^2).
It is evident that 5.35 W/m^2 is not constant because it depends on the intensity of thermal radiation emitted by the source; additionally, the absorptivity of the absorbent system, carbon dioxide in this case, depends on its temperature and the change of temperature that such system exhibits before and after further absorptions. For example, carbon dioxide absorptivity collapses at partial pressures below 0.1 atm m at temperatures below 500 R and above 1500 R.
Myrhh, where do you draw the boundary between electromagnetic radiation that can impart kinetic energy and that which cannot? Why do you draw the boundary there?
And do you think it is impossible that a molecule excited/energized by radiation (photons of some wavelength) can then pass on some of that energy to a different molecule?
If you do think an energized molecule can impart some of that energy to another molecule, might that not apply to any molecule excited? and wouldn’t this mean that a visible light photon exciting some material can then later have that energy turn into kinetic energy gains on nearby molecules? So then wouldn’t this just be “thermal energy” being imparted thanks to the visible light?
[Bob Fernley-Jones]>> Are you hypothesising that when those spawning photons from the surface are absorbed that somehow their progeny after random collisional thermalization with N2 and O2 and whatnot are rotated through 90 degrees to the vertical?
Why do you think that the Trenberth diagram shows 396 at 90 degrees vertical rather than 396 above the horizontal plane (ie, into the upper hemi-space)? If you wanted to suggest “upper hemi-space” why would you absolutely not use an arrow pointing up? In other words, why does arrow pointing up have to mean 90 degrees up only [or some other reduced range around 90 degrees]?
Nasif Nahle:
>> ΔT = … = 0.65 K.
>> That would be the change of temperature of only the fraction of carbon dioxide that absorbed that load of thermal radiation (5.35 W/m^2).
So you are saying that a part of the atmosphere is hotter than everything else and remains so constantly?
Nasif Nahle:
[Jose_X]>> [why do you believe that] CO2 is only able to emit in the radio range or that the region where CO2 absorbs does not overlap the region where CO2 emits
[Nasif Nahle]>> It is not my fault that you have not enough knowledge on quantum thermodynamics as to understand that those are not my assertions or beliefs, but well demonstrated science.
Nasif Nahle, so you are saying that CO2 absorbs within one range of photon wavelengths and emits within another?
Considering the structure of molecules, in simple terms a reader of this blog might be likely to understand, would you do us a favor and communicate what electron and nucleus behavior is going on to justify that absorption and emission happening at different frequencies?
Now, I did some googling, and I came upon a website apparently maintained by Harvard University (USA). Look at what it says in part:
https://www.cfa.harvard.edu/~jbattat/a35/cont_abs_em.html
“So a cloud of cool gas that absorbs certain colors of light will also emit those same colors. If we look at the cool gas without the blackbody in the line of sight, we will see an emission line spectrum, and the colors of the lines we see are exactly the same colors that were missing from the absorption line spectrum.”
It seems to me they are roughly saying that absorption and emission happens in the same range of frequencies for a given gas.
From University of California: Riverside, http://physics.ucr.edu/~wudka/Physics7/Notes_www/node107.html
“When heated every element gives off light. When this light is decomposed using a prism it is found to be made up of a series of “lines”, that is, the output from the prism is not a smooth spectrum of colors, but only a few of them show up. This set of colors is unique to each element and provides a unique fingerprint: if you know the color lines which make up a beam of light (and you find this out using a prism), you can determine which elements were heated up in order to produce this light.
“Similarly, when you shine white light through a cold gas of a given element, the gas blocks some colors; when the “filtered” light is decomposed using a prism the spectrum is not full but shows a series of black lines (corresponding to the colors blocked by the gas); see Fig. 8.3…”
Then it clarifies:
“…For a given element the colors blocked when cold are exactly the same as the ones emitted when hot.”
I am sure I can find many more universities to make such statements.
Nasif says: “It is unphysical that the carbon dioxide could exhibits a pressure of 266 atm in each one of the centimeters of a column of air.”
You are mis-interpreting the units. “266 Atm-cm” means “the equivalent amount of gas that would fill a container 1 cm thick to a pressure of 266 Amt”.
OR a container 266 cm thick to a pressure of 1 atm.
OR a container 266 cm/ 0.00038 thick to a pressure of 0.00038 atm.
So a container with a CO2 partial pressure of 0.00038 atm would have to be 7000 m tall to absorb this well. (Actually, it would have to be a bit taller, since the pressure will drop considerably as you get above sea level. Of you could set the column horizontally to avoid this problem.)
There is nothing unphysical about 380 ppm CO2 spread thru a column 7000+ m tall in the atmosphere — in fact, it sounds very much like the true physical conditions of the atmosphere. Other than a few details like pressure broadening of the absorption/emission lines, all three columns of air have the same amount of CO2 and would absorb IR the same amount.
Aggh, Jose_X, I lost my post to you in trying to get a link to a previous post..
Jose_X says:
December 3, 2011 at 7:28 am
Myrrh, I think you are using your body’s biological characteristics to intuitively yet wrongly claim that visible light in not “thermal” in the sense that it can’t impart kinetic energy onto matter.
Can you find a single text book or experiment that would support your claim?
Let me ask you something. How do you think your eyes can sense visible radiation?
Something in the visible light has to communicate with your eyes. There is an energy transfer. Your eyes are tuned in (via evolutionary changes) to accept visible light photons while rejecting most other types of photons.
Jose_X says:
December 3, 2011 at 7:44 am
Myrrh, sorry. I didn’t finish reading all of your comment.
So you think visible light photons can cause electron transitions in molecules but don’t impart any kinetic energy? What makes the visible light photons different than other types of photons?
CO2 has limited ability to accept photons, but a body full of molecules interacting with each other can exhibit a very different and wide range of abilities to absorb photons.
So, specifically, why do you think the earth and oceans (and most solids, especially in large amounts) cannot absorb visible light? Where do you think is the limit of what the earth can absorb.. and why do you think that?
Jose_X says:
December 3, 2011 at 8:12 am
Myrhh, http://en.wikipedia.org/wiki/Attenuation explains that light (eg, visible light) can lose its intensity as it passes through some material. See also http://en.wikipedia.org/wiki/Beer-Lambert_law
So if you have less light intensity at the end, what happened to that energy that was lost?
If that energy excited electrons, then do those electrons stay excited forever? If not, what happens to this energy?
Also, what do you think of this: http://en.wikipedia.org/wiki/Black_body
When you say people here are talking “gobbledegook”, do you realize that physicists have been talking this same gobbledegook for many years?
==========================================
That’s the problem. That’s why you have GIGO. AGWSF has changed the basic properties and given to visible, Light, what is the true characteristic of thermal infrared, Heat.
I have already gone through this in above posts, please read them first and if you still have any of these questions at the end of that, post again.
What I am giving you is traditional science:
Real traditional physics as still taught. All the heat we feel from the Sun is thermal Infrared.All that thermal infrared is what is heating the Earth’s lands and oceans and atmosphere.
Thermal infrared direct from the Sun is powerful, without the Water Cycle the Earth would be 67°C, but water is the great absorber of thermal infrared and takes this heat away from the Earth through evaporation, water vapour rises with it even faster, it’s lighter than air anyway, and in the colder regions it gives up its heat and condenses into rain. (And, by the way, taking carbon dioxide out with it, all pure rain is carbonic acid because carbon dioxide spontaneously combines with water).
And, we feel the heat because it warms us up inside. That’s how thermal infrared saunas work. When we get too heated up from the thermal infrared from the Sun our bodies work to get rid of some of it, we sweat. Water is a great absorber of Heat, thermal infrared radiation, and its high heat capacity means that it holds on to the heat longer – our bodies are around 20% carbon and the rest mainly water, we are great absorbers of thermal infrared.
Follow the link in my posts above to the wiki page on http://en.wikipedia.org/wiki/Transparency_and_translucency
Note the differences between electronic transmissions which is how Light energies act, visible and uv, and thermal infrared heat energy – it takes movement of the whole molecule into vibrational states to get kinetic energy, visible light isn’t capable of this, it works on electronic transmission levels. You’ll need to get some perspective here, the scale of these energies, visible is tiny and highly strung, it gets knocked about the sky by the electrons of the molecules of oxygen and nitrogen, it doesn’t heat them, it doesn’t even get to the electrons of water molecules, water is a transparent medium and visible is passed through, transmitted.
@Jose_X
Read from the sources I provided and make calculations. You cannot rely exclusively on Internet information.
The transfer of energy from a quantum/wave to a particle of matter is an irreversible process:
http://www.brighthub.com/engineering/mechanical/articles/4616.aspx
And laws of classical thermodynamics apply also in quantum thermodynamics:
http://en.wikipedia.org/wiki/Second_law_of_thermodynamics
Many things you cannot explain and test through classical thermodynamics is well explained and can be tested through quantum thermodynamics.
What could you do with the following formula? Just a question, not pretentious:
λ = (1.986 x (0^(-25) J.m) / E
Notice the term E, which implies the energy emitted by a particle of matter is not the same as the absorbed energy by such particle.
And the following formula lets you know how much of the energy absorbed is emitted by the absorbent system:
E = (hc)/λ
ΔE = E_f – E_i
If ΔE is negative the shift would be towards blue; if ΔE is positive, the shift would be towards red. The effect is measurable and it has been measured.
Black bodies do not exist in the known universe:
http://en.wikipedia.org/wiki/Black_body
http://www.pnas.org/content/106/15/6044.full
http://www.iki.rssi.ru/asp/pub_sha1/Sharch06.pdf
And what you are describing as a molecule that absorbs the whole energy transported by a quantum/wave and emits the same amount of absorbed energy is a black body, i.e. something hypothesized (an entity) that does not exist in nature.
@Tim O'Donovan Folkerts…
“You are mis-interpreting the units. “266 Atm-cm” means “the equivalent amount of gas that would fill a container 1 cm thick to a pressure of 266 Amt”. OR a container 266 cm thick to a pressure of 1 atm. OR a container 266 cm/ 0.00038 thick to a pressure of 0.00038 atm.”
No, you are misinterpreting what partial pressure of a gas is. Again:
You are saying that carbon dioxide, if it was alone in the atmosphere, would exert a pressure 266 times higher than the whole column of air on the surface. Consequently, it is unphysical.
Anyway, I did such calculations taking into account the numbers you give and the results are the same, i.e. total emissivity of carbon dioxide is 0.002 at its current proportion in the atmosphere.
Please, do the calculations by yourself so speculations stop here. The formula is as follows… go on:
Ɛcd = [1 – (((a-1 * 1 –PE)/(a + b – (1 + PE)) * e^(-c (Log10 ((paL)m / paL)^2))] * (Ɛcd)0
I don’t know if I am coming across as impolite. I don’t know what to do.
Scientists supposedly publish things as a team. You have various work together to recheck the path they are taking because individuals have many lapses and weak spots. They take a bunch of time. After a paper appears done, it goes by editors and others. In the end, the idea is that if a paper passes, it is a team effort. A whole team was fooled if there are problems. Effort was made to avoid errors, but if the error fooled many, then it was not an easy error to avoid, generally.
I have thought about not commenting, but I feel like (a) there is a learning opportunity (for me and/or for others), and (b) I think I see attacks on a whole industry of people and livelihoods and using unsound arguments (obviously if I thought the argument was ok I would have little to complain about).
I would like to be part of a team here, but this seems like a very adversarial environment. I’m not sure what to do.
As to the use of the word “thermal radiation”, this is mostly matter of semantics.
I have no problem saying that “most thermal radiation” or “typical thermal radiation” is between 0.1 – 100 um, but I would not use this as a definition. I would use a more physical definition and say any photons emitted to to the temperature of an object make up the “thermal radiation”
* Lasers operate in the range 0.1 um – 100 um, but are not thermal radiation.
* Hot stars (40,000 K) emit about 1/2 of their “thermal radiation” below 0.1 um.
* the microwave background is thermal radiation that is mostly above 1000 um.
(If Myrrh wants to give a different definition of “thermal radiation” he is welcome to do that too. )
RECAP:
* Not all radiation between 0.1-100 um is thermal radiation.
* Not all thermal radiation is between 0.1-100 um.
@Jose_X…
I hope this message to you does not bother the administrators of this blog.
As you are questioning scientific knowledge, I think you are already part of WUWT team.
The only part I don’t go with concerns to personal references which cast doubts into the integrity of professionals posting here.
For example, if one made a mistake on writting a statement, references to his mental capabilities or knowledge are plainly unpleasant.
@Administrators… I sincerely apologize if I got out of bounds on this answer.
>You are saying that carbon dioxide, if it was alone in the atmosphere,
>would exert a pressure 266 times higher than the whole column of air
>on the surface. Consequently, it is unphysical.
NO! I am not saying that at all! Read what I am saying.
I am saying that (pressure)*(length) determines the absorption by gases. The same amount of gas will fill a column 1 cm deep @ur momisugly 266 atm or fill a column 266 cm deep @ur momisugly 1 atm or fill a column 7000 m deep @ur momisugly 0.00038 atm. (assuming of course, constant temperature and the same area for all the columns).
All of these have the same value of P*L = 266 atm*cm.
All of these have the same amount of gas in them.
All of these have the same absorption of a beam of IR light by that gas.
@Tim O'Donovan Folkerts…
It is not semantics, but delimitation.
Going simpler:
Thermal radiation is thermal energy in transit due to a spatial difference of internal thermal energy density. It can be transferred as heat or work.
I think that Myrrh refers to the capability of visible light to cause changes of temperature on thermodynamic systems.
I have to leave this dialogue for awhile because I am a bit busy on an experiment. Please, understand my position.
Jose_X says – there isn’t really a team here, but individuals sharing their points of view. There is a general background that AGW is junk science, because we’ve had just so much evidence on so many different fronts it’s an unavoidable conclusion, but within that – I’m arguing here against both pro and anti AGW, because I’m saying both have got their basic physics wrong in the energy budget. I don’t have a team, there may well be many who agree with me.., I don’t know. It’s a great place for learning, just don’t expect ‘consensus’ opinion, just work out what you think and who you agree with and don’t agree with and why and and argue the points and you’ll do fine.
@Tim O'Donovan Folkerts…
This will be my last message on this issue.
I recommend you to make the calculations using your numbers and find that the result I obtained is correct.
A couple years ago I made the calculations in several ways and found the same results. There are other formulas to obtain the total emissivity of atmospheric gases, although a bit more complicated; for example, the following formula which takes into account the time a photon takes on colliding with molecules of carbon dioxide and mean free path length of quantum/waves in an atmosphere composed exclusively by carbon dioxide at its current concentration:
ε = (1-e^(t * (-1/s)) / √π
The results are the same, i.e. total emissivity of carbon dioxide is 0.002.
Nasif Nahle, I haven’t read the details of your last couple of comments on this issue, but of CO2 emissivity, but without a reference to the model used to derive these formulas, you and anyone else can make a mistake and others would not be able to correct them. Can you quote a paper or something for these equations. Surely you don’t believe you are flawless in understanding the details of these formulas or that whoever came up with it is also flawless.
I also want to note that Tim was using a table from a published paper that specifically included those values he is quoting which are much closer to 0.2 for very large sections of the atmosphere. I think he provided a link (he did earlier) and the table is somewhere between page 10 and 30 I think. If you are interested in looking at it, I’ll look up the specifics again.
Myrhh:
>> Aggh, Jose_X, I lost my post to you in trying to get a link to a previous post..
That is a horrible feeling. I wish this blog page could be refreshed without trashing what one was writing. I try to use a text editor all the time, but sometimes I forget and then have an accident. Aggh is right.
>> NASA: Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared.
This page is designed for young students. You should not expect them to use the same meaning of “heat” and “thermal” as you might use in a more advanced physics course. These words are probably used in their more casual meanings than their more technical counterparts in thermodynamics.
From http://www.thefreedictionary.com/heat , note the difference in the physics meaning and the biology meaning:
heat [hiːt]
n
1. (Physics / General Physics)
a. the energy transferred as a result of a difference in temperature
b. the random kinetic energy of the atoms, molecules, or ions in a substance or body Related adjs thermal, calorific
2. (Life Sciences & Allied Applications / Physiology) the sensation caused in the body by heat energy; warmth
Anyway, you appear to trust NASA. I found another page http://helios.gsfc.nasa.gov/qa_gp_ht.html and a different quote which I believe contradicts your view of visible light:
> Yes, it is possible (and relatively easy) to change the wavelength of electromagnetic waves. As you have pointed out, visible light from the Sun can heat matter and that matter then radiates as infrared waves, which our eyes cannot see but our bodies can sense as “heat”.
So NASA stated that visible light from the Sun can heat matter.
Lastly, if you go back to the last of the comments I made that you quoted, I provide a link to wikipedia page on attenuation. I then asked how you would explain what happened to that missing energy? If molecules get excited with visible light and presumably through that mechanism remove energy from the light shining through it, then what happens to that energy in those excited molecules in that solid? It’s not to hard to suspect that some of that energy turns into vibrational energy that heats up the matter.
Anyway, I think the NASA quote above supports the view that visible light can help heat the planet.