Guest post by Ira Glickstein
A real greenhouse has windows. So does the Atmospheric “greenhouse effect”. They are similar in that they allow Sunlight in and restrict the outward flow of thermal energy. However, they differ in the mechanism. A real greenhouse primarily restricts heat escape by preventing convection while the “greenhouse effect” heats the Earth because “greenhouse gases” (GHG) absorb outgoing radiative energy and re-emit some of it back towards Earth.
The base graphic is from Wikipedia, with my annotations. There are two main “windows” in the Atmospheric “greenhouse effect”. The first, the Visible Light Window, on the left side of the graphic, allows visible and near-visible light from the Sun to pass through with small losses, and the second, the Longwave Window, on the right, allows the central portion of the longwave radiation band from the Earth to pass through with small losses, while absorbing and re-emitting the left and right portions.
The Visible Light Window
To understand how these Atmospheric windows work, we need to review some basics of so-called “blackbody” radiation. As indicated by the red curve in the graphic, the surface of the Sun is, in effect, at a temperature of 5525ºK (about 9500ºF), and therefore emits radiation with a wavelenth centered around 1/2μ (half a micron which is half a millionth of a meter). Solar light ranges from about 0.1μ to 3μ, covering the ultraviolet (UV), the visible, and the near-infrared (near-IR) bands. Most Sunlight is in the visible band from 0.38μ (which we see as violet) to 0.76μ (which we see as red), which is why our eyes evolved to be sensitive in that range. Sunlight is called “shortwave” radiation because it ranges from fractional microns to a few microns.
As the graphic indicates with the solid red area, about 70 to 75% of the downgoing Solar radiation gets through the Atmosphere, because much of the UV, and some of the visible and near-IR are blocked. (The graphic does not account for the portion of Sunlight that gets through the Atmosphere, and is then reflected back to Space by clouds and other high-albedo surfaces such as ice and white roofs. I will discuss and account for that later in this posting.)
My annotations represent the light that passes through the Visible Light Window as an orange ball with the designation 1/2μ, but please interpret that to include all the visible and near-visible light in the shortwave band.
The Longwave Window
As indicated by the pink, blue, and black curves in the graphic, the Earth is, in effect, at a temperature that ranges between a high of about 310ºK (about 98ºF) and a low of about 210ºK (about -82ºF). The reason for the range is that the temperature varies by season, by day or night, and by latitude. The portion of the Earth at about 310ºK radiates energy towards the Atmosphere at slightly shorter wavelengths than that at about 210ºK, but nearly all Earth-emitted radiation is between 5μ to 30μ, and is centered at about 10μ.
As the graphic indicates with the solid blue area, only 15% to 30% of the upgoing thermal radiation is transmitted through the Atmosphere, because nearly all the radiation in the left portion of the longwave band (from about 5μ to 8μ) and the right portion (from about 13μ to 30μ) is totally absorbed and scattered by GHG, primarily H2O (water vapor) and CO2 (carbon dioxide). Only the radiation near the center (from about 8μ to 13μ) gets a nearly free pass through the Atmosphere.
My annotations represent the thermal radiation from the Earth as a pink pentagon with the designation 7μ for the left-hand portion, a blue diamond 10μ for the center portion, and a dark blue hexagon 15μ for the right-hand portion, but please interpret these symbols to include all the radiation in their respective portions of the longwave band.
Sunlight Energy In = Thermal Energy Out
The graphic is an animated depiction of the Atmospheric “greenhouse effect” process.
On the left side:
(1) Sunlight streams through the Atmosphere towards the surface of the Earth.
(2) A portion of the Sunlight is reflected by clouds and other high-albedo surfaces and heads back through the Atmosphere towards Space. The remainder is absorbed by the Surface of the Earth, warming it.
(3) The reflected portion is lost to Space.
On the right side:
(1) The warmed Earth emits longwave radiation towards the Atmosphere. According to the first graphic, above, this consists of thermal energy in all bands ~7μ, ~10μ, and ~15μ.
(2) The ~10μ portion passes through the Atmosphere with litttle loss. The ~7μ portion gets absorbed, primarily by H2O, and the 15μ portion gets absorbed, primarily by CO2 and H2O. The absorbed radiation heats the H2O and CO2 molecules and, at their higher energy states, they collide with the other molecules that make up the air, mostly nitrogen (N2), oxygen (O2), ozone (O3), and argon (A) and heat them by something like conduction. The molecules in the heated air emit radiation in random directions at all bands (~7μ, ~10μ, and ~15μ). The ~10μ photons pass, nearly unimpeded, in whatever direction they happen to be emitted, some going towards Space and some towards Earth. The ~7μ and ~15μ photons go off in all directions until they run into an H2O or CO2 molecule, and repeat the absorption and re-emittance process, or until they emerge from the Atmosphere or hit the surface of the Earth.
(3) The ~10μ photons that got a free-pass from the Earth through the Atmosphere emerge and their energy is lost to Space. The ~10μ photons generated by the heating of the air emerge from the top of the Atmosphere and their energy is lost to Space, or they impact the surface of the Earth and are re-absorbed. The ~7μ and ~15μ generated by the heating of the air also emerge from the top or bottom of the Atmosphere, but there are fewer of them because they keep getting absorbed and re-emitted, each time with some transfered to the central ~10μ portion of the longwave band.
The symbols 1/2μ, 7μ, 10μ, and 15μ represent quanties of photon energy, averaged over the day and night and the seasons. Of course, Sunlight is available for only half the day and less of it falls on each square meter of surface near the poles than near the equator. Thermal radiation emitted by the Earth also varies by day and night, season, local cloud cover that blocks Sunlight, local albedo, and other factors. The graphic is designed to provide some insight into the Atmospheric “greenhouse effect”.
Conclusions
Even though estimates of climate sensitivity to doubling of CO2 are most likely way over-estimated by the official climate Team, it is a scientific truth that GHGs, mainly H2O but also CO2 and others, play an important role in warming the Earth via the Atmospheric “greenhouse effect”.
This and my previous posting in this series address ONLY the radiative exchange of energy. Other aspects that control the temperature range at the surface of the Earth are at least as important and they include convection (winds, storms, etc.) and precipitation that transfer a great deal of energy from the surface to the higher levels of the Atmosphere.
I plan to do a subsequent posting that looks into the violet and blue boxes in the above graphic and provides insight into the process the photons and molecules go through.
I am sure WUWT readers will find issues with my Atmospheric Windows description and graphics. I encourage each of you to make comments, all of which I will read, and some to which I will respond, most likely learning a great deal from you in the process. However, please consider that the main point of this posting, like the previous one in this series, is to give insight to those WUWT readers, who, like Einstein (and me :^) need a graphic visual before they understand and really accept any mathematical abstraction.
Oliver Ramsey;
That the radiated energy from cold ice can heat a warmer object to a higher temperature than it was before…. now that’s problematic.
If you are convinced that that does occur I would be genuinely interested to see an explanation with some quantification>>
How about…read the original article and then work your way down the comments?
Or go straight to why we’re splitting hairs and it doesn’t matter:
http://knowledgedrift.wordpress.com/2011/03/03/why-the-co2-greenhouse-gas-debate-doesnt-matter/
“”””” kuhnkat says:
March 3, 2011 at 3:17 pm
George E. Smith,
” they have no recollection of their origin, nor any knowledge about where they are going.”
Please read the details of the 2 hole experiments as it relates to photons and electrons in quantum mechanics. Your statement is not correct. “””””
Well off hand, I don’t quite see the connection; BUT, I’ll take your word for it; and see if I can understand what you are saying. I’m here to learn.
Since the greenhouse, in general and for what ever purpose, is to create the optimal condition for what ever you wish to grow. One could say that with a greenhouse you want to retain the optimal summer conditions.
High humidity, high water levels and through put, high proper nutrient levels, higher concentration of nitrogen and and much higher levels of CO2.
Since most people sporting greenhouses aren’t growing pine trees heat actually have to be added sun and all. Depending on what greenery you grow heat is used to not place the plants in hibernation mode, some seeds wont even sprout if they’re not warm enough. Consider that most of land is in the northern most hemisphere and away from large bodies of warm water, so you need to add heat, and lots of it, by gas, coal, or nuclear, because god ol’ sodding Sol only does so much for as long as it’s rays reaches the ground, and H2O and CO2 what do they do but dissipate their retained “heat” the moment after they got it, until everything equalize with it’s surrounding, which happens before midnight if you don’t add heat.
On the one hand you can only grow bananas north of Anchorage as it is, and only in the boiler room so to speak, if you add heat but who’s so dumb enough to say it would be bad to be able to actually grow pineapples, coconuts and rice even north of Anchorage as well, preferably without adding expensive wind powered derived heat.
Point is, without the added heat the inside of a greenhouse gets to be the same temperature as the outside no matter what the levels and concentrations of humidity and CO2 you have inside. The only thing that makes a greenhouse hotter ‘an the outside in blazing sunlight is the enclosure, or rather the roof because if you open the roof but still keep your denser atmosphere heats go out to space anyhow.
>>
kuhnkat says:
March 3, 2011 at 3:17 pm
George E. Smith,
” they have no recollection of their origin, nor any knowledge about where they are going.”
Please read the details of the 2 hole experiments as it relates to photons and electrons in quantum mechanics. Your statement is not correct.
<<
If you mean the two-slit experiment where the particle-wave duality properties are tested, then you don’t understand what the experiment is doing. George is correct (as usual).
Jim
Yup, there is a need for new physics textbooks that are customized to any bias. We could call our books New Clear Physics. We could have a textbook version with pre-canned exams that would ask the exact same questions each year, but change the answers.
The lawyers tell about the town that had only one lawyer, and he was starving for business. Then another lawyer moved into town and they are both doing just fine with plenty of business for all. With different versions of our textbooks there would be conflict galore. We could each get a government grant to study the textbooks and conclude which wwer true and which false. Of course, our textbooks would be mostly true, but written with studied ambiguity. Each of our govenment-funded studies would come to different conclusions, so we’d get hired by industry and environmentalist think tanks to write reports to resolve the differences.
The problem with the above paragraph is that it is a not too exaggerated account of how government-supported science is actually done now. (/sarc)
Thanks Tim Folkerts for your similar physical analogy. For WUWT readers who might have missed my first entry in this series (with the bouncing balls) see here. More coming over the next couple weeks. Stay tuned.
to Tim Folkerts
Yes, of course, cavity radiation assumes no transmissivity through the walls.
The N2 thickness required is not known.
Now, with regards to N2 (and O2), the actual absorption properties must be tested at many, many wavelengths, and under the same conditions that N2 is found in the atmosphere. In the atmosphere, N2 is in temperature states from approx 300K to near 3K (from the surface of the earth to near space at night), and at the associated pressures depending on altitude. In addition, the source radiation needed to drive the signal through the sample must vary from the high temp solar equivalent to a very low temp 3K equivalent.
Those conditions are not easy to duplicate in a laboratory. So, N2 has never really been tested through the full range of its enviromental conditions for all wavelengths of absorption. Most of the transmission/absorption data for the atmosphere comes from military research. But the military just wanted the ‘atmospheric windows’ to see clearly through, narrow wavebands are fine, and could care less about the rest of the spectrum and conditions.
There was never any good reason to gather the actual real life N2 and O2 absorption data.
Perhaps now there is. With all the whining of their ‘certainty’ by the AGW crowd, it may be necessary.
I can almost guarantee you there will be surprises when, or if, the actual data is taken. I’ve seen it happen so many times in my professional experience.
I spent a number of years overseeing design, manufacture, calibration, application and sales of IR sensors in the US, Europe and Asia.
Ira,
I read your thoughts about the town with only one lawyer. In the town I grew up in there was a saying..when two farmers have a fight over a cow, two lawyers wind up with a side of beef each.
It just hit me how similar that is to climate scientists. Of course there seems to be one subtle difference. They wind up with a side of beef each, but the farmers didn’t even know there was a fight.
George E. Smith says:
March 3, 2011 at 12:07 pm
Thank you for your response. It may or may not be true, but I personally don’t question the statement that gases emit “blackbody-like” radiation–in part because “blackbody-like” is not a well defined term. What I have trouble with is applying Planck’s blackbody radiation law to matter that does not have a clearly defined surface area.
My answer to your specific question:
How about a graphene layer, that is a single atom thick solid; can it emit a black body like (carbon is black) thermal spectrum.
And if not; why not ?
I don’t know.
“Cavity radiation is an interesting phenomenon. In a completely enclosed isothermal cavity, it makes no difference what material the interior walls are composed of. The internal radiation in the cavity will conform precisely to blackbody radiation at the temperature the material is at….And if you inject any photons into this hypothetical N2 constructed radiation cavity, of any wavelength, it will all be absorbed by the N2 molecules within 100% absorption regardless of wavelength.”
Someone has duped you regarding cavity radiation. If you make a box out of clear glass and shine a light into the box, those photons are still going to shine through the glass box just as if your were shining your light through a window. Glass doesn’t gain magical absorption properties because you shape the glass into a box.
to Steve
You didn’t read my posts correctly. Cavity radiation assumes no transmissivity.
1. A thin layer of any of many substances on the outside of the glass can suffice, or
2. The glass itself must be thick enough to stop transmission. Glass, like water, does eventually stop all transmission given enough thickness.
Domenic says:
March 3, 2011 at 7:15 pm
to Tim Folkerts
Yes, of course, cavity radiation assumes no transmissivity through the walls.
The N2 thickness required is not known.
Although at least thicker than the atmosphere?
Now, with regards to N2 (and O2), the actual absorption properties must be tested at many, many wavelengths, and under the same conditions that N2 is found in the atmosphere. In the atmosphere, N2 is in temperature states from approx 300K to near 3K (from the surface of the earth to near space at night), and at the associated pressures depending on altitude.
You may want to reconsider the 3K since the coldest part of the atmosphere is the mesopause at ~180K. Testing at many wavelengths are what spectrometers are made for, really easy.
In addition, the source radiation needed to drive the signal through the sample must vary from the high temp solar equivalent to a very low temp 3K equivalent.
No, this is irrelevant, the wavelength is the only parameter of interest, the temperature of the source isn’t ‘remembered’ by the photon!
Those conditions are not easy to duplicate in a laboratory. So, N2 has never really been tested through the full range of its enviromental conditions for all wavelengths of absorption.
Really, care to justify that assertion?
Most of the transmission/absorption data for the atmosphere comes from military research. But the military just wanted the ‘atmospheric windows’ to see clearly through, narrow wavebands are fine, and could care less about the rest of the spectrum and conditions.
There was never any good reason to gather the actual real life N2 and O2 absorption data.
Oh you don’t think that it is the gas that usually fills the light paths in our spectrometers might be a factor? Or that N2 is frequently the diluent in the cell when measurements are made of other gases.
The reason why we don’t “gather the actual real life N2 and O2 absorption data” is that there isn’t any to be gathered!
“You didn’t read my posts correctly. Cavity radiation assumes no transmissivity.”
Considering your conclusion (that N2 can be forced to emit/absorb any wavelength if placed in a cavity), I’m thinking you didn’t read your post correctly.
“Cavity radiation assumes no transmissivity. 1. A thin layer of any of many substances on the outside of the glass can suffice.”
Ahhhh, so it isn’t the glass or N2 that is absorbing/emitting at new wavelengths, is it?
“or 2. The glass itself must be thick enough to stop transmission. Glass, like water, does eventually stop all transmission given enough thickness.”
Pure water? I hope you aren’t making this assumption because the ocean depths are dark. The ocean isn’t pure water – it is full of salts and organic debris. I’ll grant that given enough of anything, in the reality, you’ll run across enough impurities to stop all transmission (even an optic fiber). But then you really aren’t talking about the absorption/emission spectrum of your pure molecule, are you?
The cavity doesn’t create quantum magic, it ensures that all reflected wavelengths are eventually absorbed. The molecules of the cavity walls don’t gain an ability to absorb wavelengths that they never could before. With each reflection, the photon loses a tiny amount of energy, becoming a slightly lower frequency, until it eventually does fall within a wavelength of the wall’s absorption spectrum. But the photon must begin as a wavelength that interacts with that wall’s molecules (reflection or absorption). Hence, no transmission is allowed.
Saying,”If you can make a box of N2 that doesn’t transmit light that it normally does, you can force N2 to absorb light that it normally doesn’t.” is a tautology. Restating the case with different words doesn’t make it possible.
to Steve
Ahh…finally, a thinker here.
Just as perfect blackbody does not exist, nor does a perfect reflector. That’s why I posted earlier that the opposite of a blackbody is a perfect reflector, as I saw a poster struggling with the idea of a blackbody.
Thermal radiation comes into the atmosphere from the sun, is converted to different wavelengths, and leaves it again. N2 and O2 are indeed part of that process, shifting wavelengths through energy lost due to bouncing around. Kinetic energy for the molecules and wavelength shift for the photons.
Because there are no perfect reflectors.
There are no perfect reflectors.
(Also, there no perfect transmission mediums other than a vacuum.)
My whole point was that N2 and O2 are indeed part of the ‘greenhouse gases’.
And there are vast quantities of N2 and O2 in the atmosphere. Especially compared to CO2.
And yet, they are ignored.
No one accounts for them.
Phil doesn’t understand that. The AGW crowd doesn’t understand that.
I am happy to see questioners and thinkers like you here.
Take a quantity of pure N2 and place it into a sealed metal container. Heat the container to some high temperature. After a period of time, the N2 must assume a temperature approaching that of the heated container.
Now, take a short hose and connect it between the container of hot N2 and and a much larger container of pure N2 that is at room temperature and pressure. As the hot N2 puffs into the room temperature container, take a video using an IR camera. Do you see the plume of hot N2? I do, which means the N2 is radiating IR energy. The container has nothing but N2 in it, so the radiation must be coming from N2.
Based on previous comments and links in this thread, I think much of the radiation from the plume will be in the shorter IR bands, perhaps around 3-4μ, but I would not be surprised if some of it spreads out into the longer IR bands, up to 7μ to 10μ to 15μ or longer as the hot N2 mixes with the room temperature N2 .
Anyone here expert enough to know for sure?
Ira
I need a complete description of the experiment.
1. What kind of container was the second one?
2. Was the hose evacuated of all gases except N2?
3. Was the IR camera inside the second container or loooking through some sort of window?
4. What was the spectral range of the camera?
“”””” Phil. says:
March 4, 2011 at 7:40 am
Domenic says:
March 3, 2011 at 7:15 pm
to Tim Folkerts
Yes, of course, cavity radiation assumes no transmissivity through the walls.
The N2 thickness required is not known.
Although at least thicker than the atmosphere?
Most of the transmission/absorption data for the atmosphere comes from military research. But the military just wanted the ‘atmospheric windows’ to see clearly through, narrow wavebands are fine, and could care less about the rest of the spectrum and conditions.
There was never any good reason to gather the actual real life N2 and O2 absorption data.
Oh you don’t think that it is the gas that usually fills the light paths in our spectrometers might be a factor? Or that N2 is frequently the diluent in the cell when measurements are made of other gases.
The reason why we don’t “gather the actual real life N2 and O2 absorption data” is that there isn’t any to be gathered! “””””
Well I didn’t get any sleep all night thinking about this problem; maybe I’m having another CO2 Ice moment.
I thought Phil and I were pretty much on the same page with this; but maybe not.
I think we both agree that in the lower atmosphere (how about the two meter Owl box height) the generally 15 micron band radiation gets captured by CO2, but then is immediately thermalized through molecular collisions, long before the excited CO2 gets a chance to spontaneously decay, in a presumably allowed transition. So the captured energy gets shared with the regular gas molecules N2, O2, Ar. I’m comfortable with that and I think Phil and I agree on that.
But at this point we apparently seem to diverge.
I assumed that heated atmosphere being above absolute zero, then emits a “black body like” thermal continuum spectrum based on the Temperature. By the way, no mystery to “BB like” I use that term to mean a continuous spectrum that generally conforms to the Planck Radiation formula, and is bound by the Planck formula as a limiting envelope, but may have deviations here and there due to emissivity variations. But it is distinctly different from a line or band spectrum, which I take as characteristic of isolated atoms or molecules respectively. So then my atmosphere is emitting a spectrum that carries no signature of the species that grabbed the energy in the first place; or it even could have been solar radiation caused or simply conduction from the ground or somewhere else.
One of Phil’s recent comments indicated to me, that he believes, that the N2/O2/Ar do NOT radiate, as per my concept; BUT those gases are in thermal equilibrium with the trace CO2/H2O/O3 /whatever other GHG molecules that are IR active, so they match in Temperature, and then the IR active GHG molecules radiate according to whatever vibrational/rotational etc states those molecules have; so that the “Atmospheric LWIR radiation” then does carry the signature of the various GHG species, rather than being “BB like”. I think Phil’s position is somewhat along those lines. I know we both agree that in the rarified stratosphere where mean free paths are long, the GHG molecules can spontaneously radiate at the same frequencies they absorbed.
But back to the denser atmosphere where that isn’t possible.
Now Tom Vonk, some time back talked about this same situation in several posts, and he basically said the same thing. That the ordinary atmosphere gases were in local Thermal Equilibrium with the GHG (CO2), so neither one was heating the other; or perhaps they all are heating each other; and then in Tom’s view (as far as I recall), then it was the CO2 etc that was radiating whatever LWIR was coming from the atmosphere, not the N2/O2/Ar.
Now Peter Humbug says almost exactly the same thing in his “Physics Today” Featured Article. Now I assume, everyone knows that is my whimsical nickname for Raymond T. Pierrehumbert; the Louis Block Professor of Geophysical Sciences at the University of Chicago. And let me say here categorically, that is in no way derogatory; I’m not into name calling; it helps me remember his name, and it fits.
Well if Foster Brooks or whoever likes to call himself Tamino, the sappy hero of Mozart’s Magic Flute, that’s ok with me; or Eli Rabbett for example. So Prof RTPH, is Peter Humbug, and with no malice.
So he basically describes the same sort of scenario that Tom Vonk presented some time ago, and apparently is along the lines of what Phil has said here.
Well among all the stacks of books, that crowd me out of my cube, there’s not a single one that describes the specific origins of the black body radiation; nor is there one that states categorically that it does not apply to gases. Those that mention the subject simply say EVERY body that is above absolute zero, emits black body (like) thermal radiation, that is characteristic of the Temperature, and may be modified by a spectral emissivity.
Unfortunately most of those books are “handbooks”, not Physics Textbooks. I lost every single textbook I ever had or used in school 50 years ago, in a box that got shipped, and never delivered. I know we actually did the derivation of the Planck Formula and the Raleigh Jeans Formula, and there never was any mention of any “energy level structure”, just that the (average) energy assigned to each degree of freedom had to be quantized (in the Planck case). And It was my impression (at the time) that the radiation mechanism was simply the acceleration of electric charge; but I can’t say that at the time, I understood that to be as a consequence of the molecular or atomic interactions during collisions. That concept came to me only quite recently; although once again it was a post by somebody here at WUWT (can’t remember who) that led me to start thinking along those lines. And I think that is a very plausible mechanism; because I can see how during a collision, the electron cloud, and the nucleus, can get displaced from each other (CG wise) and create a dipole moment for the duration of the interaction. And my Quantum mechanics stops well short of being able to calculate that all out.
In any case, now I’m not sure what to believe, but it seems that Phil, and RTPH, and Tom Vonk all have a similar view, that is not what I had thought. The forgotten poster here, had said something to the effect that a thermal continuum was possible (in gases) due to perturbations of the quantum states during collision; well as best as I can recall.
Now I really don’t know anybody who is really handy that I know for sure would know the answer to this. Two names come to mind; but both are more solid state Physicists; both are recipients of the National (Presidential) Science Medal. One of them in my view should be a Nobel Laureate in Physics, but isn’t (yet). I’m going to try and contact him, if I can and ask him. The other one works (occasionally) in the building next to me, so I might be able to get to him; he’s a former student, of that other person.
In any case the atmospheric look up/look down spectra should solve the puzzle; as Phil says here; there ain’t any N2/O2 LWIR spectral data to be had. Even Mother Gaia evidently doesn’t have it.
So I may be having crow for lunch again today; but I do get so rejuvenated whenever that happens.
I still believe that H2O is firmly in control of the climate; not CO2, but I sure am curious about that gaseous BB radiation question.
And I think we are being pushed well off onto a back burner here; too much stuff up above.
[thanks G , I really do appreciate it]
Ira Glickstein, PhD says:
March 4, 2011 at 9:48 am
“I do, which means the N2 is radiating IR energy. The container has nothing but N2 in it, so the radiation must be coming from N2.”
??? Ira, N2 is not a greenhouse gas. It is transparent to both incoming shortwave and outgoing longwave radiation. It does not radiate IR energy at any wavelength relevant to greenhouse theory. In a closed oven it will of course come to the equilibrium temperature inside the oven, but this involves conduction and convection and not radiation as long as no other gases are present. Here is a more complete graphic than the one you used. Transmission spectra for all of the greenhouse gasses are shown. Notice that Nitrogen is not present, and if nitrogen is transparent to IR it is also non-radiating in the IR bands.
http://upload.wikimedia.org/wikipedia/commons/7/7c/Atmospheric_Transmission.png
Ira Glickstein, PhD says: March 4, 2011 at 9:48 am
Do you see the plume of hot N2? I do, which means the N2 is radiating IR energy.
That is an insightful thought experiment. Have you ACTUALLY seen the radiated IR from that plume? I would love to see an IR video (or even a still picture) of such a thing.
I suspect that it is not at all easy to see because I am pretty well convinced that N2 cannot radiate any appreciable amount of IR in the 5-50 um range associated with the applicable temperatures.
There are a couple things you might see.
1) If there was a small amount of H2O or CO2, then those molecules could radiate some IR. Using pure enough N2 or filtering for specific frequencies NOT related to CO2 & H2O should eliminate those possibilities
2) The change in temperature will change the index of refraction, which could lead to a wavering effect (like you see over hot roads of hot roofs). IR light shining thru from the far wall could get distorted, which would create some sort of blurry features on the IR detector.
The Fire Hose Analogy
The second law of thermodynamics refers to the NET transfer of heat energy that must be from warmer to cooler bodies.
But, as was noted earlier in this thread, if you are spraying me with a fire hose and I am simultaneously spraying you with a garden hose that has blue-tinted water, I will get a lot wetter, but you will get some blue water on your clothes.
– Dr. Ira Glickstein
– http://wattsupwiththat.com/2011/02/28/visualizing-the-greenhouse-effect-atmospheric-windows/
I hate to rank on Dr. Glickstein; he seems like a smart and fair-mined man, but this kind of argument is used to justify the human-caused global warming theme. The idea is: outgoing infrared radiation stimulates atmospheric CO2 which heats up and “back radiates” to the earth…making us hotter. Honestly, at this point, does anyone believe this is physically possible? This “greenhouse effect” and analogy is complete nonsense.
What can I do to help Dr. Glickstein? Let’s fix his analogy for him. There is no side channel or alternate path for the fire hose and the blue-water squirt gun. They must use the same path. So, I get the fire hose, you get the squirt gun. Open fire. How much of your blue water will travel into the nozzle of my fire hose? That’s right my friends. None!
However, this is still not right because your finger working vigorously on the squirt gun’s trigger is an independent source of energy. This is not analogous to the human-caused global warming theory. CO2 does not have an independent source of energy. The only energy it has comes from other places…like collisions with the other atoms and molecules of the atmosphere, from infrared radiation and conduction from the earth’s surface. CO2 is not a battery, it is not being combusted and it is not a nuclear reactor.
Let’s refine the analogy further. Instead of a squirt gun, you get a funnel with a blue die injector. A really teenie-tiny, small one to represent 390PPM of CO2 in the atmosphere. Bend the spout of the funnel back toward me…now you capture water from my fire hose and direct blue stuff back toward my nozzle.
How much of your redirected water travels into my nozzle and up my fire hose?
There’s a cardinal rule in the selling business…once the sale is made, shut the hell up.
This is me shutting up now.
Upgrade your paradigm. Buy the Dragon Slayer book. Thanks.
There is some excellent conversation going here.
Would someone comment on this please? As I understand it, for N, N2, O, O2 and Ar…if there is mass and a temperature, this stuff will radiate, but the wavelength is really, really tiny, hard to measure, and we don’t know much about it because it is not really useful to us.
Right? Not right?
Ira, I thank you for graciously for replying to my post. I know this is a late response, but I didn’t realize that you had replied.
First, let me respond that I am not saying the “green house” effect, or delayed radiative loss, or whatever you want to call it, doesn’t occur to some extent. I agree. It most certainly does. Whenever the atmosphere above the surface, especially clouds, have accumulated energy during a hot sunny day, portions of that energy are prevented from direct and rapid release of it to the cold of outer space by back radiation. The atmoshpere does have a certain heat retaining “capacity” which is entirely dependent on radiative loss to outer space. However, by far the greatest contributor to this effect is water vapor, not CO2. Water has a MUCH higher energy retaining capacity than CO2 and serves as the largest energy “sponge” in the atmosphere.
What I objected to was what appeared to be incorrect portrayal of this process in your animation. In the dark blue box all the way to the right you illustrate “15μ aborption” and you claim “7μ, 10μ, and 15μ” re-emission as a direct result of this “15μ absorption”. I agree that it is possible that water vapor might provide 7μ, and possibly 10μ, back radiation as a result of the numerous SHORTER wavelength NIR bands in which it receives energy. But that is not what the dark blue box is claiming. It strongly suggests 7μ re-emission that comes directly from 15μ illumination. How can that be? Your graphic is at the very least misleading. I know of only a few materials capable of down converting light in this manner, and they all have to be charged with much more energetic light to be able to “phosphoresce” in this manner (see: http://www.roithner-laser.at/cards.html ), and neither CO2, nor water vapor, nor CH4 are capable of this kind of behavior. Any energy that is taken up by these molecules at 15μ can ONLY re-emit at LONGER wavelengths – not shorter, higher energy wavelengths. If you are claiming that they are capable of re-emitting at a shorter wavelength after “excitation” at 15μ, I believe you are mistaken.
There are only two substantial absorption bands that enable CO2 participates in this absorption/re-emission activity, and they are quite narrow when compared to all of the many NIR and IR bands that water vapor absorbs at. Every illustration I have seen of the combined absorption spectrum associates the shoulder at 15μ as being CO2’s main contribution. However the extremely low concentration of CO2 in the atmosphere almost guarantees that nearly all of the energy absorbed by CO2 at either of it’s absorbtion bands is almost immediately given up to one of the other atmospheric gases through thermal collision. The net effect is that very little, if any, IR radiation is “re-emitted” at 15μ. If this were not the case, and if your graphic was correct, then think about it, the atmosphere would appear to be somewhat “transparent” in the 15μ band, you would see energy in the 15μ appearing in the top graphic . . . . but your graphic doesn’t show any 15μ IR at all, coming “through” to outer space. Are you saying that the 15μ IR only radiates back towards the warmer surface? I don’t think so. Your analogy of a firehose and a garden hose doesn’t hold water either. The source of blue water coming from the garden hose must originate from the fire hose. Since ALL of the water leaving the fire hose is either consumed (ie. absorbed at 15μ) in the fire or goes right thru the house (doesn’t even see CO2 at all or leaves at 10μ), there is NO water left to dye blue and squirt back at a much lower pressure at me.
Other’s have claimed that increasing CO2 has narrowed the 10μ window that allows the IR radiation caused by all of the shorter wavelength aborption bands of water and the one CO2 band to radiate to outer space. But this concept doesn’t hold validity either since the spectra from outer space showing the width of the IR band at this window has not shown any narrowing with increasing CO2 concentration! Every IR spectrum looking down looks the same. This suggests that the current concentration of CO2 in the atmosphere is high enough, thank you, to completely absorb all of the IR at 15μ, (and promptly give up that energy in thermal collisions) without re-emitting any of it, AND any change in concentration of CO2 will ONLY drop the altitude at which complete loss of IR radiation at 15μ occurs!
Maybe the new post by Anthony reviewing the new paper by Hermann Harde titled “How much CO2 really contributes to global warming? Spectroscopic
studies and modelling of the influence of H2O, CO2 and CH4 on our
climate,” will clear up some of the confusion on this subject. It predicts the sensitivity of the climate to changes in CO2 concentration at 1/7 what the IPCC claims.
What you think? I welcome any additional insights.
Domenic says: March 4, 2011 at 9:16 am
“My whole point was that N2 and O2 are indeed part of the ‘greenhouse gases’. ”
The quantum theory of molecular vibrations says there should be no (significant) contributions from diatomic N2 in the IR range. The same theory says there should be contributions from triatomic H2O, CO2 and O3.
The experimental data of downward IR presented earlier in the thread show clear contributions of H2O, CO2, and O3. In the regions where these three do not emit IR, there is practically no IR.
Theory and experiment both agree that N2 (or other monatomic or diatomic gases) is not a significant GHG. What about the the theory or data do you disagree with? What theory or data can you present that suggest N2 IS a significant GHG?
Now, N2 DOES play a role. The true GHGs will collide with the N2 and transfer energy to/from the N2 molecules. N2 acts as a “reservoir” of thermal energy simply by the N2 molecules gaining/losing speed when they collide with GHGs. In this sense they are “part” of the GHG since N2 surrounds and interacts with the GHGs.
Unless you can refute clearly either the theory or the evidence presented so far, you have not convinced me that they are themselves GHGs that absorb/emit IR.
George E. Smith says:
March 4, 2011 at 10:19 am
Those that mention the subject simply say EVERY body that is above absolute zero, emits black body (like) thermal radiation, that is characteristic of the Temperature, and may be modified by a spectral emissivity
Hi George, I’ve emphasized what may be the key to your understanding, for N2 the spectral emissivity is zero for the wavelength range of interest. That’s why they showed the dotted lines of BB at different T on the downwelling plot, the BB sets the limit which is reached in spectral regions where emissivity is 1. So you can tell what temperature the emitting gas was at. Regarding the origins of BB radiation as far as I’m aware it requires a dense medium where the molecules (atoms) are in close proximity whereas in a gas they are acting as individuals. I’d tend to think of it as a very dense gas, in that case I’d expect extreme broadening so that all the lines would completely merge and ultimately you could end up with BB.
Your point about collision induced effects is good, there are continua associated with such effects but they are extremely weak, there are also dimer effects (sticky collisions and such like). The proof of the pudding is in the eating and when we look at the measured atmospheric spectra there is nothing there from N2 or O2!
Ken Coffman says:
March 4, 2011 at 10:37 am
There is some excellent conversation going here.
Would someone comment on this please? As I understand it, for N, N2, O, O2 and Ar…if there is mass and a temperature, this stuff will radiate, but the wavelength is really, really tiny, hard to measure, and we don’t know much about it because it is not really useful to us.
Right? Not right?
Not right.