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.
@Ira — sorry about all that extra work 🙂
I agree that adding more CO2 molecules will increase the chances of photons of a particular energy impinging upon another molecule with an energy band exactly matching that of the photon energy (another CO2 or H2O molecule) and so slowing the progress of that “packet” of energy towards space and freedom.
In short, it will improve the insulation properties of the atmosphere.
As you say, because of the statistical chance of hitting another molecule, as the concentration of molecules increases, so will the chance of such an encounter, but the effect is, indeed, logarithmic in nature (Log base e, not Log base 10).
Look ONLY at CO2 and you will find that a doubling does have some headroom to make a measurable difference. However, if you treat CO2 and H2O molecules as effectively the same, which they are in this context, then a doubling of CO2 amounts to a much, much smaller fractional increase when considering all H2O and CO2 molecules combined, and I would be exceedingly surprise if you could actually measure any difference.
Ira Glickstein
The famous experiment by R W Wood proved two things;
1. Glasshouses(greenhouses) heat up by stopping convection.
2. The heating radiative effect at atmospheric temperatures is so small as to be negligible for most practical purposes.
People tend to forget the second part.
Dave Springer says: March 1, 2011 at 7:53 am
Enough already with the argument that certain gases that don’t absorb infrared can’t emit it.
A black body absorbs and emits all frequencies of light perfectly. A non-black body will absorb and emit light less well at some (or all) frequencies. But the frequencies where it absorbs poorly are the same as the frequencies where it emits poorly. Thus a material (like the atmosphere) that absorbs some wavelength of light (say 10 um) poorly will also emit that wavelength poorly.
Since you agree that N2 absorbs 10 um light poorly, then the only conclusion is that it also emits 10 um light poorly.
In fact, the very image you reference at http://www.sundogpublishing.com/fig8-2.pdf shows that there is almost no radiation coming down at 8-12 um. If the N2 was indeed emitting significant amounts of thermal radiation at those wavelengths, where is it? I see the (nearly) black body radiation above and below those limits due to the H2O and CO2 (and other GHGs) which can (almost perfectly) absorb those wavelengths.
Bill Illis says: March 1, 2011 at 5:14 am
But the Sun emits mostly like a blackbody.
It is not being emitted in the Hydrogen frequencies only, it is emitting across the entire spectrum close to what the blackbody curve predicts.
Quite true. But the circumstances were quite different from what I was describing. The sun is not a single H atom, but a huge collection of ionized H+ and electrons. Ionized H+ cannot have a Balmer spectrum because it has no orbiting electron. (And as you point out, non-ionized H in the cooler outer layers can and does then absorb the Balmer lines, making them dimmer than other wavelengths). Certainly a large enough collection of ionized CO2 or N2 or any other gas could emit a significant continuous spectrum like the sun does. But that does not describe the situation in the earth’s atmosphere.
Peter Taylor;
It took me a long time to realise that the Radiative Forcing (RF) that is at the heart of the model calculations is centred on a notional point toward the top of the troposphere…defined by the temperature at which the earth will reradiate the balancing energy – which is -19 C and about 10km up>>>
This was one of my very first epiphanies in regard to the IPCC numbers. They keep quoting 3.7w/m2 and 1 degree, but a quick SB Law calculation versus temp at earth surface doesn’t support that. From your post it sounds like you couldn’t find the supporting docs in AR4, but I did. I’d have to dig like crazy because it was a long time ago, but from memory:
IPCC calculates sensitivity at the “theoretical black body” temperature of earth. I thought it was -15, but could very well have been -19
They specificaly state that surface temperature changes cannot be directly related to changes in RF. In fact, they even say it is possible that a change in RF results in climate change, but potentially NO surface temp change.
The implication that CO2 is linear was very carefully worded, but turned out to be a reference to a small change of some sort where treating it as linear was a reasonable approximation. BUT, if you go to the section on ECONOMIC scenarios, that make it VERY difficult to work the numbers backward to CO2 concentrations, but the log functions are very clear in those sections.
This subterfuge is very misleading, deliberately so, and I’ve remarked on it often. If someone wants to write a detailed expose, I’ll be happy to go scrounge up all the references even if it means reading the whole d*** thing again.
Thanks, Dave Springer for your very helpful expert comments and useful links in this and other topic threads on WUWT. I have learned a lot from reading your postings and I hope you continue to contribute your special knowledge to my topic threads.
My PhD is in System Science with my dissertation being in Hierarchy Theory (an application of Shannon’s Information Theory to Optimal Span – the “magical number 7 +/- 2” that may usefully be applied to Quantifying the Mythical Man-Month and Management Span of Control).
Fortunately, I got my PhD late in my career as a system engineer, after I had done useful work in advanced visionics and automation for military aircraft. Thus, being educated beyond my intelligence did not set in until about 15 years ago, when I was within five years of retirement from my real job. However, the PhD came in time for me to qualify as an Associate Professor and I am currently afflicting students with my knowledge of system engineering in an online graduate course.
Taking your comment to heart, I promise to eschew obsfuscation and modify all my convolutions into pointwise products of Fourier transforms, if you get my drift :^)
Slacko says:
March 1, 2011 at 6:19 am
Tim Folkerts says:
February 28, 2011 at 7:23 pm
“However, it is an observed fact (supported by theory) that monatomic gases (like argon) or symmetric diatomic gases (like N2 and O2) do not have vibration modes or rotation modes that would allow them to absorb (or emit) IR photons.
… IR is not absorbed or emitted by N2 & O2.”
A hot body that won’t radiate infrared?? That’s gotta be the best-kept secret in physics, if it’s true.
A gas is not a hot body and gases that don’t have a dipole, like N2, O2 and Ar, don’t emit IR, this has been known to physics and chemistry for quite some time. See Herzberg G., ‘Molecular spectra and molecular structure. I. Spectra of diatomic molecules’, New York: Van Nostrand Reinhold, 1950, the classic text on the subject.
mkelly says:
March 1, 2011 at 8:44 am
Dave Springer says:
March 1, 2011 at 7:53 am
“Nitrogen doesn’t absorb infrared radiation but it can certainly gain kinetic energy by excited molecules of CO2 and H2O bumping into nitrogen molecules. The kinetic energy gained by the nitrogen molecules will be lost in a continuous blackbody spectrum characteristic of the temperature of the gas.”
Agreed. My point to Cal earlier. If N2 and O2 could not shed energy (trapping it) then we would have a problem.
But that leads to a question: what difference is there between the spectrum of N2 and O2 both at the exact same temperature and radiating. I have not found anything on that.
As stated above absolutely wrong, the only way that N2 and O2 molecules can shed energy is by collisional exchange with CO2, H2O etc. molecules which can then radiate.
Here’s a low resolution spectrum showing N2, O2 and CO2, I’ve left out H2O for simplicity but it occupies the 5-10 micron region at a similar intensity to CO2 (10^-20). Note that the vertical scale is log base 10!
http://i302.photobucket.com/albums/nn107/Sprintstar400/CO2N2O2.png
mkelly says:
March 1, 2011 at 8:44 am
Dave Springer says:
March 1, 2011 at 7:53 am
Nitrogen doesn’t absorb infrared radiation but it can certainly gain kinetic energy by excited molecules of CO2 and H2O bumping into nitrogen molecules. The kinetic energy gained by the nitrogen molecules will be lost in a continuous blackbody spectrum characteristic of the temperature of the gas.
Agreed. My point to Cal earlier. If N2 and O2 could not shed energy (trapping it) then we would have a problem.
But that leads to a question: what difference is there between the spectrum of N2 and O2 both at the exact same temperature and radiating. I have not found anything on that.
OK so please describe in detail the mechanism by which N2 and O2 radiate this energy if it is not by the movement of electrons between different energy states (as a consequence of the vibrational modes of the molecule) which causes the characteristic absorption and emission lines.
I gave the explanation to your supposed paradox in my original comment. The N2 and O2 molecules gain energy and lose energy by collision with CO2, H2O and other greenhouse gases. There is a small amount of radiation as well but it is negligible in the context of the greenhouse gas emissions. Just look at the emission spectra of N2 and O2 and compare it with the IR spectrum that Ira gives. Further all bodies above absolute zero radiate but the emissivities vary dramatically. A black body is 100% efficient a white body (which cannot really exist) is 0% efficient. N2 and O2 are closer to being white bodies at these temperatures.
I thought it was still a matter of dispute whether a gas (the molecules in a gas) behaved/radiated according to BB “laws”. There seem to be very positive opinions being asserted above that hinge on that specific issue.
BTW Ira, I’ve just gone thru the first part of your novel, and I wonder if you still “believe” the scenarios and causalities for climate behavior and fluctuation you posit and “explain” in Chapter 1. The verbiage circles round and round, but boils down to domination by CO2 changes, with some drastic natural and not-so-natural interventions and intervening events forcing it back down to “normal” levels.
Which, IMO, is alla buncha hooya.
to Phil and Cal
Nope. The full characteristics of N2 and O2 in the long wavelength IR have never been actually tested.
I have been looking for long wavelength IR real test data for both N2 and O2 , and cannot find it.
The only ‘graphs’ out there are calculated from models, not from lots of actual test data. With 20+ years in IR measurement and quantification experience, I don’t trust the graphs out there.
There are no such things as ‘white bodies’ out there. The opposite of a ‘blackbody’ is a ‘perfect mirror’, not a ‘white body’.
The only materials that exhibit close to ‘perfect mirror’ thermal radiation properties are pure metals. Highly polished gold, aluminum, etc, or their atoms.
From my direct IR measurement experiences, materials with nitrides (N, N2, etc) and oxides (O, O2, etc) in their surface composition have very high emissivity, closer to blackbodies. Thus they have very small reflective components. Metal oxides are perfect examples. Bare metals are highly reflective. But let the surface oxidize heavily, and their IR reflection capability disappears…and their emittance approaches a blackbody.
There is something very fishy in the spectral graphs bandied about from calculations regarding the emittance and reflectance of N2 and O2. They are not metals.
Both N2 and O2 must be tested thoroughly in the long wavelengths and not guessed at.
Many thanks, Ira, for a clear explanation, and I understand that you have only attempted to describe the radiative effects in this post.
My view is that the term ‘Greenhouse Effect’ is dead right because it describes exactly how a greenhouse does not work. A real greenhouse works just as well with polythene, styrene or, classically, rock salt windows because, although the blocking of long wave IR is real it’s irrelevant compared with the greenhouse’s blocking of the convective currents which are the main mechanism for removal of hot air, directly and by evaporation.
Similarly, although I’m sure there is a radiative effect which would be active in a static atmosphere, the real atmosphere is anything but static, being thoroughly turbulent and full of convective cells. Radiation thus only comes into effect at the top of the troposphere, where the density of all gases is too low to do much blocking.
In my view the talk of radiative physics is a waste of time since it’s an extremely minor effect.
Domenic says:
March 1, 2011 at 11:43 am
The only materials that exhibit close to ‘perfect mirror’ thermal radiation properties are pure metals. Highly polished gold, aluminum, etc, or their atoms.
————————-
I have read that variations in emissivity between substances (except for metals) are largely due to the topography of the surfaces. I imagine that must also be true for metals, since the differences between polished and unpolished surfaces are so marked.
I’m suprised that our coarse rubbing with abrasives is able to produce such a strong effect, even though I have seen it with my own eyes. ???
Phil. says:
March 1, 2011 at 10:15 am
“As stated above absolutely wrong, the only way that N2 and O2 molecules can shed energy is by collisional exchange with CO2, H2O etc. molecules which can then radiate.”
Thanks for the information Phil. Your photo while nice does not answer the question of N2 and O2 at the same temperature. Since temperature drives the emittion frequency will they both be the same? I math (Weins Law) says yes, but I have found no info on the subject.
By the way by your statement above you are in the camp that says without GHG we would be a very hot world indeed. Not at -18 C. Once a N2 or O2 molecule was heated via conduction with the ground it would stay heated for a very long time with no H2O or CO2 to reduce its energy. Correct?
“”””” Scottish Sceptic says:
February 28, 2011 at 2:56 pm
“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 would suggest CO2 in the atmosphere is much like the “correct” expanation of the Crookes radiometer (the black and white rotating thing in a glass vacuum bulb). “””””
I believe that the original conception of the Crookes Radiometer, was to demonstrate “Radiation pressure” The Kinetic energy of Photons impinging on a surface.
The idea was that the blackened surface would absorb the photons, setting up a pressure on the black side of the paddles. The shiny side of the paddles however would not absorb the photons but would reflect them, creating twice the momentum change, so the paddles would rotate, with the shiny side pushing the black side before it.
Of course it rotates the other way instead; and the reason is that the black side is slightly hotter than the shiny side, and it heats the residual gas molecules close to the surface more than the shiny side does thereby creating a higher pressure on the black side. They couldn’t create a high enough vaccuum to get the radiation pressure to dominate the residual gas differential pressure.
Wow! There are soo many problems and issues being thrown around that it is nearly impossible to address everything. As someone with a degree in physical science (including physics and astronomy) I guarantee you cannot learn the basics from one page, let alone all of the complexities, e.g. spectroscopy.
I just want to cover four simple items and see if I can provide other insight later:
1) Steve at 141 is correct: The nearly absent atmosphere is the reason Mars is so cold. Other comparisons: the Moon vs the Earh (basically the same distance from the Sun but no atmosphere on the Moon) and Mercury vs Venus (Venus is further away but with an extemely dense, CO2 heavy atmosphere which causes Venus to be much hotter than Mercury).
2) On a large scale (planetary), the method of heat transfer is radiation. There may be convection and conduction within the atmosphere (up to about 10 Km as found by E. O. Hulburt in 1931), but between the Earth (including the atmosphere) and “space”, heat is transferred via radiation since space is a vacuum (you need matter for conduction and convection, also shown by Hulburt in 1931).
3) CO2 and H2O
Ugh, this comes up soo often it is getting tiresome. So let’s look at two pieces of information.
3a) Absorption by components
Please look at the following picture showing the absorption from CO2 and H2O, as well as other constituents. While water is a broad absorber of long wave IR radiation (greater than 10 microns), it is not complete. But look at CO2. This is very important. There is a major absorption feature between ~13-20 microns which is very good at absorbing LWIR radiation. This is important becasue H2O does not completely absorb the radiation in this wavelength region. So when CO2 levels increase, it further impedes LWIR radiation from escaping the atmosphere. This is even more important since according to Planck’s law, a blackbody at the Earth’s surface temperature will radiate in this wavelength region. While the Earth may not be a blackbody, it does radiate in this spectral region.
3b) Composition of atmosphere relative to CO2 and H2O
H2O is primarily limited to being located close to the surface of the Earth and is not well mixed into the entire atmospheric column, unlike CO2. However, when we look at things spectrally, it does not matter if the gas we are looking through is localized in a small area or spread out through a larger area/volume (ignoring clouds of course). It will have the same effect. When we measure gases using spectral instruments, we need to take into account the distance from the source to the sensor. We calculate the amount of the gas we see and divide by the distance and get an average over that distance. Same is true when looking down from a satellite towards the Earth. When we consider the entire column we find:
H2O: ~0.40%
CO2: ~0.04%
While H2O is higher than CO2, it is not orders of magnitude larger. The challenge is that the H2O amount is not changing by 100%. Remember from 3a, CO2 is absorbing outgoing radiation where H2O is not and the amount of CO2 is increasing.
4) Be careful with these plots and simple explanations. Total absorption and opaque are not quite the same, i.e. if 100% is already absorbed than more CO2 cannot be a problem. The graph at the top of the page suggests that the atmosphere is a single step from Earth to space. In reality, models use multiple levels to account for the differences in gas constituents at various heights.
Last lesson: Be very careful with LWIR. Most people are okay with near IR and short wave IR since this radiation is identical to visible sunlight, which we are accustomed to as humans since our eyes are optimized for that region. In this part of the spectrum, we talk about reflection and absorption. But when we head towards long wave IR, you have absorption/emission, transmission and reflection. It is a somewhat messier region that can be easily confused. Most people familiar with LWIR come from a lab setting using instruments in a controlled setting. But when you start looking at/using the atmosphere, it becomes a much more difficult problem.
Hopefully NASA’s Glory satellite may help answer more questions when it finally launches.
Thanks, Ira, for an interesting post.
I think there is an important point, however, on which your analysis is misleading. I agree with George E. Smith (February 28, 2011 at 3:23 pm) that the IR absorption occurs by exciting the vibrational modes of GHGs. The reason that N2 and O2 and H2 do not absorb IR is not that they have no dipole (since neither does CO2), but because, being diatomic, they have no asymmetric vibrational modes, which are the ones that absorb in the IR range. George is right in saying that the vast majority of the absorbed light rays are immediately reradiated and do not have time to affect the thermal distribution of the surrounding gases. The reradiating becomes even more preponderant the farther up in the atmosphere you get because of the decreasing density of gases with height. So Ira’s assertion that GHGs exert a large thermal effect on the atmosphere seems to me to be a bit unlikely, except possibly in the lower troposphere. The other thing to consider is that where thermal transfer does take place, any blackbody radiation that occurs will be at the new temperature rather than at the original temperature. Because of adiabatic lapse (think PV=nRT), the temperature of the atmosphere drops rather quickly as a function of height until you get up to the thermosphere, where there is a large rise in temperature, but so rarefied an atmosphere as not to have any significant impact on the atmospheric windows. So even assuming you could model radiation from a layer of atmosphere as if it were a black body (which it is not), the distribution is shifted toward yet longer wavelengths than are radiated by the earth.
My main point is that my substantial physics training leads me to believe that the main impact of GHGs is vibrational excitation followed by return to the ground state, emitting a photon of the same wavelength but in a random direction, rather than by any significant transfer of internal vibrational energy to thermal energy of surrounding molecules.
Hi Oliver,
I like these discussions because they have been encouraging me to remember my IR skill set.
Yes, it’s a surface condition effect. And it is quite dramatic.
But what I suddenly realized, from this discussion is something that I had not thought of before.
The stars. Outer space.
Do you realize that it impossible from earth to know whether all the ‘stars’ that are seen in the sky, or from space, are indeed really ‘stars’ and not ‘reflections’ from massive pure metallic bodies, or perfect reflectors, of nearby true stars?
It is literally impossible to know from a distance. Impossible.
Now that is wild.
I have never seen that idea looked at in astrophysics.
But from a radiational physics point of view, it is entirely possible.
There is confusion on this thread about the meaning of the term ‘greenhouse effect’. Some of the commenters seem to think that it is merely something that reduces the rate of heat loss (and they draw analogies with blankets and igloos which nobody disputes). But the greenhouse effect according to climate ‘science’ is not the insulation of a blanket or a reduction in the rate of cooling, but a positive addition of heat (33C or more) because of ‘backradiation’. If this theory were true it would mean that the Earth emits more energy than it receives. This is a contravention of the laws of thermodynamics. It is based on the fallacy that if the escape of heating is blocked, then the temperature will continue to rise until a ‘radiative equilibrium’ is reached and the heat bursts through the barrier. It is claimed that the laws of thermodynamics are not broken because the emission at the top of the atmosphere is equal to the incoming solar radiation. But, like the theory of the runaway greenhouse, it is a fallacy. If it were true we could generate huge amounts of energy from a small input simply by placing an infra-red barrier around a radiator – just tap some of the heat off at regular intervals – free energy.
“”””” _Jim says:
February 28, 2011 at 3:45 pm
George E. Smith February 28, 2011 at 3:23 pm :
…
So the atmosphere does radiate a black body like thermal spectrum, and the presence of the gHG molecules simply means that there will be bands of that continuum emission, that are also captured by the GHG molecules, as well as the emissions from the surface.
I’ll wait to see how this is adjudged.
(IR Spectroscopy explaining spectral response regarding gas molecule vibrational modes would seem to indicate otherwise.) “””””
Jim needs to take remedial English Reading courses. IR Spectroscopy indicates nothing of the sort.
The IR Spectroscopy item you cited, refers to ABSORPTION SPECTROSCOPY which as it properly indicates, relates to RESONANCE MODES which are indeed characteristic of specific molecules, and their excited state energy levels.
No where did I make any statement that is inconsistent with this; it is the reason we talk about the “15 micron” or 666 cm^-1 degenerate bending mode of the linear CO2 molecule. It is degenerate since there are two identical and indistinguishable modes at right angles to each other; each of which can gather its own photon.
But the question which you chose to challenge, is the quite different THERMAL SPECTRUM EMISSION; which has nothing whatsoever to do with atomic or molecular structure or energy levels. It is a classical physics consequence of the accelerations of electric charge; which according to Maxwell’s equations must (and do) result in the radiation of an EM wave (or photon if you wish; which has a continuous energy spectrum; not a quantized one. Ordinary atoms, once ionized by the removal of one or more electrons, also emit a continuum EM spectrum, which is not quantized, because a newly captured free electron can have any initial energy before being captured by the ion.
The accelerations of Electric charge in molecular collisions, are entirely unpredictable as the the redistribution of momentum and energy; although one can talk statistically about expected distributions.
One can learn a thing of two from the Particle Accelerator Physicists. The Stanford two mile long electron linear accelerator is one of if not the most powerful electron accelerators in the world. It is linear, because then the only acceleration the electrons see, is due to the accelerating electric (traveling) waves in the beam line. On the other hand circular track accelerators, have to steer the beam around corners continuously to get them to repass through the accelerator gaps, and gain more energy. To compensate for this increased energy, the bending magnet fields have to increase synchronously with the particle energy, to maintain them in the correct orbits. Since the charged particles are traveling in a circle, they are subject to a continuous acceleration, since acceleration is a change in velocity and velocity is a vector quantity, having a direction as well as a “speed”.
So particles in a circular accelerator continuously radiate, EM energy as the go around the track. Well pedantically, the path is a number of curved arcs, interleaved with a number of straight sections, and it is only in the arced sections that the particles radiate.
So the reason why proton beams can be accelerated in a circle, and electrons can’t (beyond certain energies, is that the proton is far more massive than the electron.
So consider two neutral atoms (or molecules) colliding. Each molecule or atom is electrically neutral, so the electrons will go more or less, where the protons in the nucleus goes (on average), but in the process of collisions and recoil of the particles, the more massive nuclei consisting of Protons, and Neutrons, undergo far lower accelerations, than do the much lighter electrons; so even though and single atom or molecular is electrically neutral, in collision, the electrons undergo much higher accelerations than do the Protons or nuclei; so just as in the particle accelerator, it is the electrons that radiate the EM wave energy, and the much smaller opposing field due to the proton accelerations does not compensate. So one can practically ignore the proton positive charges, and consider only the acceleration of the net electron charge to figure out the radiated EM fields.
The thermal radiation continuum depends only on the Temperature, and has a BB like spectrum, that does not depend on the atomic or molecular species; only on the Temperature.
Some of you people need to get your noses out of facebook, and wiki, and hit some Physics books, to learn what you should have learned around the 8th grade.
Nothing in the IR Spectroscopy paper you cited is inconsistent with that; they talk ONLY about absorption spectra; not emission.
And Kirchoff’s law applies only in equilibrium, and the atmosphere is never in equilibrium.
“”””” Tim Folkerts says:
March 1, 2011 at 9:37 am
Dave Springer says: March 1, 2011 at 7:53 am
Enough already with the argument that certain gases that don’t absorb infrared can’t emit it.
At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity.
http://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation “””””
“”””” At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity. “””””
“””””At thermal equilibrium, ……. “””””
Can people stop citing Kirchoff’s law, and then TOTALLY ignoring it (spectral) emmissivity equals (spectral) absoptance, ONLY in “THERMAL EQUILIBRIUM ”
Earth’s atmosphere is NEVER in thermal Equilibrium; it has a continuous energy input from somewhere else, and a continuous energy output to someplace else.
So Kirchoff’s Law NEVER applies to earth atmospheric situations; EMISSIVITY NEVER EQUALS ABSORPTANCE.
[[[ Ira Glickstein, PhD says:
March 1, 2011 at 6:01 am
…Your analogy of the black balls in the pool does not work because the mechanism is not at all like that of the “greenhouse effect” GHG in the atmosphere. There is more H2O (the main GHG) in the Atmosphere than CO2, but the percentage of H2O is also miniscule, yet, the tiny percentage of GHGs (mostly H2O but also CO2) are responsible for the Earth being at livable temperatures. ]]]
OK Ira,
I agree that Earth’s Atmosphere helps make our Earth livable. I think water in all three states, solid, liquid, and gas, clearly plays the major role. The alleged “CO2 effect” is critical that it be accurately quantified, as it is being considered a “problem” and will cost us all dearly (money) if current politicians impose their planned restrictions.
I want to shrink the “alleged CO2 effect” to a simple provable visual experiment that the average layperson can easily absorb. Some of the material presented in this wonderful blog is quite advanced and beyond the expertise of many lay readers.
How about this…..
1) Chill 1 bottle of distilled water and 1 bottle of CO2 injected mineral water in the refrigerator (to keep the CO2 in solution).
2) Find 2 identical glass or plastic bottles with a screw top that seals.
3) Place a thermometer upsidedown in each bottle so the bulb won’t be in the liquid.
4) Fill each bottle half way with identical volumes, one with distilled water, the other with the CO2 impregnated mineral water.
5) Secure the cap and place in full sun on the window sill.
6) Keep room temperature just slightly below the thermometer temperature as it changes.
7) If CO2 really does have a big influence in keeping Earth warmer, at some point in the experiment either during the day, or at sunset, or in the evening, or at night, the air above the mineral water should be warmer.
Both will have a mix of normal atmosphere and water vapor, but the mineral water will have released additional CO2 into its “atmosphere”, boosting its “greenhouse effect” that will keep retained/reflected heat in its sealed environment.
Remember to keep the room temperature just slightly below the bottle temperature to keep any heat transfer influence from the room air at a minimum.
If there is a “CO2 effect” it should show up here on the thermometers.
How about this one Ira?
@-mkelly says:
March 1, 2011 at 12:24 pm
“By the way by your statement above you are in the camp that says without GHG we would be a very hot world indeed. Not at -18 C. Once a N2 or O2 molecule was heated via conduction with the ground it would stay heated for a very long time with no H2O or CO2 to reduce its energy. Correct?”
If I may reply as someone who would agree with Phil’s spectralcalc picture of the emissivity of N2 and O2 as several orders of magnitude smaller than that of CO2 for present surface temperature. I understand that such values are calculated from quantum mechanical first principles, perhaps those that doubt them can find direct measurements that contradict that. But the research on radiative transfer carried out in connection with heat sensor/seeking systems for military purposes would seem to make it unlikely that any such major error has gone unnoticed.
Without GHG we would not be hotter because the N2 and O2 atmosphere has a lower emissivity. They could only warm by conduction with the surface to a temperature equal to the surface. Without GHG the surface would lose energy by IR radiation faster because there is no downwelling or back radiation from the atmosphere. Therefore it would be colder, and could not heat up the N2 and O2 molecules to any greater temperature than it had reached.
“”””” Mark Nodine says:
March 1, 2011 at 12:35 pm
Thanks, Ira, for an interesting post.
I think there is an important point, however, on which your analysis is misleading. I agree with George E. Smith (February 28, 2011 at 3:23 pm) that the IR absorption occurs by exciting the vibrational modes of GHGs. The reason that N2 and O2 and H2 do not absorb IR is not that they have no dipole (since neither does CO2), but because, being diatomic, they have no asymmetric vibrational modes, which are the ones that absorb in the IR range. George is right in saying that the vast majority of the absorbed light rays are immediately reradiated and do not have time to affect the thermal distribution of the surrounding gases. “””””
Well Mark, you were doing ok until you said this:- “”””” George is right in saying that the vast majority of the absorbed light rays are immediately reradiated and do not have time to affect the thermal distribution of the surrounding gases. “””””
Because absolutely never in my life have I ever said any such thing; the exact opposite in fact.
The CO2 (or other GHG molecule) certainly absorbs specific LWIR energies, according to the energy levels of its excited molecular states; ” BUT ” in the lower atmosphere it NEVER (hardly ever) has enough time to re-radiate that absorbed Photon, at the same energy and frequency (wavelength). Molecular collisions happen many thousand times faster, than spontaneous emission at specific (CO2) frequencies. The energy is lost in thermalization collisions, which warm the ordinary gases of the atmosphere. From that time on, the GHG gas molecule has no influence whatsoever on what happens. The (slightly heated) atmospehric gases now radiate a complete thermal spectrum continuum, depending ONLY on the atmospheric Temperature; AND THEN the CO2 or other GHG molecule can re-engage with THAT spectrum, and once again carve out its specific absorption band chunk.
The proof of this is in the satellite (and calculated) LWIR spectra of the earth. They are BLACK BODY LIKE SPECTRA with the GHG absorption band partly missing (attenuated) by the GHG molecules.
If the only LWIR emission from the atmopshere was from the GHG molecules, itwould be ONLY at th4e specific GHG resoanance absorption band frequencies; and it is not; it is a full BB like spectrum.
You have to note that the actual spectrum is the overlaying Temperature BB spectrum, MODIFIED by the spectral radiant emissivity of the radiating material.
Since a thin gas layer, has few molecules, it has a correspondingly low emissivity.
A 50 Angstom gold film, also has a correspondingly low emissivity; but of course higher than the gas layer, because it still has a greater atomic or molecular density.
There is nothing in the phase transition from solid ice, or liquid water, to H2O vapor by sublimation or evaporation; that suddenly turns off the ability of those molecules to emit a BB like thermal continuum radiation spectrum according to Planck formula and other applicable laws (don’t forget the emissivity).
But I have never said the the atmospheric LWIR spectrum consists of the GHG resonance absorption frequewncies only.
That condition is approached in the stratosphere, when molecular mean free paths, and intercollision times become long compared to the lifetimes of the excited states. Then the spontaneous emission can occur; but the molecular density is so low, that it cannot be a significant contributor to the total earth LWIR emission spectrum.
“”””” izen says:
March 1, 2011 at 1:31 pm
@-mkelly says:
March 1, 2011 at 12:24 pm
“By the way by your statement above you are in the camp that says without GHG we would be a very hot world indeed. Not at -18 C. Once a N2 or O2 molecule was heated via conduction with the ground it would stay heated for a very long time with no H2O or CO2 to reduce its energy. Correct?”
If I may reply as someone who would agree with Phil’s spectralcalc picture of the emissivity of N2 and O2 as several orders of magnitude smaller than that of CO2 for present surface temperature. I understand that such values are calculated from quantum mechanical first principles, perhaps those that doubt them can find direct measurements that contradict that. “””””
Those specracalc images that Phil posted, are ABSORPTION SPECTRA; NOT EMISSION SPECTRA.
to Izen
you wrote: “But the research on radiative transfer carried out in connection with heat sensor/seeking systems for military purposes would seem to make it unlikely that any such major error has gone unnoticed.”
Heat seeking systems are ridiculously simply compared to accurate quantification and complete summation of thermal radiation properties for specific gases.
Heat seeking sensor systems only need a ‘window’ or two, or three to locate a target.
Heat seeking sensor systems are qualitative in nature, not quantitative. The military doesn’t care much about the details. They just want to find and lock onto the target.
If you had any experience with quantifiable IR, you would know this.
And that is one of the biggest problems. Much research about atmospheric IR has been done by the military in order to locate and track heat signatures. But that data is really of minimal use when absolute quantification is necessary for establishing real world temperature effects of gases in the atmosphere regarding ‘greenhouse effects’.