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.
Don V says:
“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.”
— — —
Don, if I may make a comment here. I like most of what you are saying, you seem to see it clearly but one area. I don’t interpret Ira ever implying electron level actions, only thermal ro-vibration interaction. Any upward movement to higher frequencies could only be via the tails of the Maxwell-Boltzmann distribution.
IR active molecules mostly thermalize instead of emitting but there is always a certain fraction doing so. Once thermalized all of that energy just gets evenly distributed maintaining the equipartition principle between all of the degrees of freedom. But there are always some higher velocity molecules present also and this is where the re-excitation at higher frequencies occurs by the statistical chance and this all depends on the temperature. This gets deep in statistical mechanics and I could easily be wrong but that is the way I see it. But of course not phosphoresce, that is at a much higher temperature.
ref: http://wattsupwiththat.files.wordpress.com/2010/08/vonk_kinetic_curve.jpg
explained here: http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/
@ur momisugly Don V, March 4, 2011 at 10:38 am
1) The heat capacities of H2O vs CO2 is not really an issue — the H2O and CO2 do not hold on to the energy from the IR absorption. Either of these GHGs will gain energy if they absorb IR from the ground (or from GHGs in the atmosphere), but then when they collide with other molecules (almost always N2 or O2) they will give away that extra energy until they are at the same average energy as the rest of the air. This would slowly warm the entire atmosphere.
The GHG can also emit IR, which would remove energy from that molecule. These molecules would (on average again) take energy from the next molecule they hit (again most likely N2 or O2).
The “heat retaining ability” of the atmosphere would be from the combined heat capacities of all the gases. Since N2 & O2 are by far the most common, they will contribute the bulk of the heat capacity of the atmosphere.
The fact that there much more H2O than CO2 will allow the H2O to more effectively heat or cool the atmosphere as a whole. But this would be related primarily to their IR properties, not their heat capacity properties.
2) I second what wayne said about the energy for the 7 um photons coming from the MB distribution of molecules in the atmosphere, not from some sort of 2 photon interaction of 15 um photons within a single CO2 molecule.
“”””” Phil. says:
March 4, 2011 at 10:57 am
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 “””””
Thanks Phil. Yes you caught the essence of my thinking on BB emission from gases. And yes my assumption is that because a gas is rather rarified compared to solids, that any thin layer of gas, would necessarily have a spectral emissivity much less than one, reflecting the small number of atoms or molecules that are present in such a thin layer; which is why I asked the question about the graphene; does a thin enough solid layer also have a poor emissivity ?
Now Ray PH in his article did say that this aspect of coninua emission from gases was not well understood (so I guess I’m in some really great company), but he talked about the colliding molecules “possibly” forming some, I suppose you could call it a super molecule that had its own energy levels for the duration of the collision interraction. Maybe so, but maybe not necessary, because it seems to me that any single neutral molecule or atom in collision, is going to undergo a distortion of the geometry of the nucleus, and the electron charges, and that is enough to create a radiating dipole moment; which is why I thought that was a plausible notion, and one inspired by someone else who had hinted at it. The very short collision interraction time, would certainly explain why a process with quantized energy values can be Heisenberg smeared into an effectively continuous spectrum; the grand daddy of line broadenings.
It’s amazing how kmuch stuff we accept, thinking we understand it, and then find out we don’t understand much of anything.
Much of the misunderstanding I read here, dave you might perk up and find this interesting, has to do with the lack of understanding of what LTE (local thermodynamic equilibrium) is. It is NOT TE (thermodynamic equilibrium) you read so often in thermodynamics texts.
The simplest, but true, visualization of LTE is that it only exists when there is hugely more collisions between atoms or molecules than there is radiation interactions. Above about 45+ km the atmosphere becomes radiation ruled and there it is in NLTE (non-LTE). To be in LTE the local parcel does not imply radiation equilibrium or even perfect TE.
In the ionosphere at 1,000+K above 100km you would think you would be very warm but if floating in a spacesuit there you would freeze to death. Why? This is NLTE. Temperatures can only be defined and exist if that point being measured is in LTE. It would be better to just list the mean velocities of the particles when above some 45 km for there is really no temperature at all, it can not even be defined there.
Wolfram’s definition of LTE is way wrong, totally. They are speaking only of radiation and in true LTE, radiation has a minor back seat. Wonder who wrote that.
Don’t rely on Wikipedia to straighten you out either, they make it somewhat messy, but one statement made there is correct and simple:
“It is important to note that this local equilibrium may apply only to a certain subset of particles in the system. For example, LTE is usually applied only to massive particles. In a radiating gas, the photons being emitted and absorbed by the gas need not be in thermodynamic equilibrium with each other or with the massive particles of the gas in order for LTE to exist.”
and correctly on what creates the LTE in the first place,
“… while local thermodynamic equilibrium (LTE) means that those intensive parameters are varying in space and time, but are varying so slowly that for any point, one can assume thermodynamic equilibrium in some neighborhood about that point.”
and
“The process that leads to a thermodynamic equilibrium…” (think local here in LTE’s case) ”…is called thermalization. … By interacting, they will share energy/momentum among themselves and reach a state where the global statistics are unchanging in time” albeit slowly.
If radiation is flowing into, out of and across such of a parcel in LTE it *does* require that the flow to be smooth everywhere locally and constant or changing very slowly. You cannot have a laser pumping energy into such parcel and call it LTE even if it is dense for it is not smooth across all particles in that local volume. A small parcel up in the atmosphere being warmed by the Earth below or sun above is smooth and acceptably constant so LTE is maintained.
That might help some reading here to keep their views correct in physic definitions when thermalization is brought into the conversation and radiation. If ever in doubt of my accuracy, just read.
Ira Glickstein, PhD,”Take a quantity of pure N2 and place it into a sealed metal container. Heat the container to some high temperature…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…Anyone here expert enough to know for sure?”
You don’t need an expert, Ira. Think about what you are proposing. What percentage N2 is the atmosphere between the aperture of your IR camera and the object you are taking a photo of (even if it is just a warmer patch of N2)?
If N2 absorbed/emitted in the IR spectrum, every IR image taken on earth would look like a photograph taken in a thick fog. There’d be no way for the IR to get from object to aperture without hitting a significant number of N2 molecules. Since this doesn’t happen, we can take crisp IR photos.
Well, I was waiting for Steve to return, but I’ll continue anyway.
Regarding cavity radiation, Steve gave a good answer to what happens to an incoming photon.
But that’s only part of it.
He said nothing about outgoing radiation.
Take our hypothetical N2 cavity again.
Now put a tiny hole in it and put a detector there to look inside and measure the emittance.
What will that detector see, what wavelengths, for a given temperature of the N2 inside that cavity?
Will the N2 do what ‘Phil’ thinks it will do by reference to some textbook or some spectral source book, or other?
(Hint: I’ve experimented with cavity radiation for over 25 years. So, if anyone wishes to bet on this stuff, put their money where there mouth is, I will be more than happy to do so.)
Good questions Domenic. The second container is metal, at room temperature. If you want a different kind of second container, let just say so and it will be provided at no etra charge,
Yes, prior to connecting, the hose was flushed with pure N2 at room temperature.
The IR camera was at room temperature and is looking through a small hole in the second container. A slight positive pressure of pure N2 is maintained such that no outside air gets into the second container through that hole. Alternatively, there is rock salt window or some more modern material that is totally transparent to the entire IR range. your choice.
The camera is sensitive to the entire range from UV (0.1μ) through far IR (70μ), and the video is “color” coded such that the viewer can determine which band or bands the plume occupies. (E.g., the UV range is shown in purples, the visual in blues, near-IR in greens, mid-IR in oranges, and far-IR in reds. If desired, any given band or bands may be expanded to the full color range for display since the video camera records the image digitally, characterizing each pixel according to its wavelength.)
Domenic:
Yes! You do see my point. That is one proper view of reality, and they are sometimes so very hard to get, and I’m no exception. There are so many obscuring variables and you have to correctly remove all *known but unimportant* variables to finally get down to see what is really happening.
One of my simple pet peeve on this line is viewing the Earth as a flat disk by climatologists when calculating energies, temperatures and radiation fluxes. You might as well wad it all up and throw it away.
Five temperatures across the globe averaged (272+285+271+301+297)/5 = 282.2K so radiation to space ignoring emissivities is 5.67E-08•282.2²² = 359.59 Wm-2 right?
WRONG!
The mean radiation from these five regions is actually:
5.67E-08•(272²²+285²²+271²²+301²²+297²²)/5 = 379.37 Wm-2.
The radiation from varying temperatures will ALWAYS be more than that of a constant temperature at the mean.
So why do we average temperatures then assume radiation flows? You see it every day. It is climatologists doing this and it’s a crying shame. 🙁
They may be lying by omission but probably just plain ignorance, bad teachers and texts.
Same for “back radiation”… well, I won’t go into that right now.
“”””” Domenic says:
March 4, 2011 at 3:12 pm
Well, I was waiting for Steve to return, but I’ll continue anyway.
Regarding cavity radiation, Steve gave a good answer to what happens to an incoming photon. “””””
Domenic, presumably you want this cavity to be isothermal, so that it has a definite Temperature.
How then would one know at the peephole, whether the emerging radiation (BB) was emitted by the N2, or by whatever formed the enegy impermeable solid wall of the cavity. I’m inclined to believe that Kirchoff’s law does apply to this hypothetical situation. but it is not clear to me, that one can say that the N2 or whatever gas filling you might put in there must radiate (and absorb) the exact same spectrum of the cavity walls.
The citation I gave earlier, from the “Handbook of Optics ” (Walter Driscoll), the author of the section on BB radiation specifically said the cavity walls must be “Black” in the radiation total absorption sense. I don’t think that is inconsistent with some other material being in the cavity, and in thermal equilibrium with it, so long as that body radiates and absorbs whatever spectrum IT has identically (the strange material that is.
Something absorbing some part of the cavity radiation, but simultaneously radiating the exact same things is basically invisible to the cavity; which presumably can go on radiating its BB spectrum regardless of some included foreign material.
I presume that freezing Platinum or Copper, Laboratory black body cavities, are not evacuated, but are open to the atmosphere; well maybe you do have to use them in a vaccuum chamber; because you certainly don’t want any physical window in them.
Pochas and Tim Folkerts: Thanks for your comments, I did not actually do the experiment. However, here is my thought process:
1) The hot N2 has lots more heat energy in it that the room-temp N2.
2) Once in the presence of the room-temp N2 it will exchange heat with the room-temp N2 molecules by conduction (when a hot N2 collides with a room-temp N2) and also by radiation. I believe the camera will record the radiation part, and, possibly some radiation that results from the collision part.
3) After all the hot N2 is released into the second container, and some time has passed, all the N2 in the second container will become about the same temperature, and eventually go down to room temperature.
The only part that is in question is my step 2, where, I suppose, you could conclude that all the heat tranfer is via conduction, with absolutely no radiation and therefore a blank video recording.
OK, if that is a problem, repeat the experiment in the vacuum of Space. Take a container of pure N2 and heat it up, then open the valve and let the N2 escape into Space. Do you see a plume with the IR camera now? I do.
The hot N2, we all know, will cool down. It cannot release its heat energy to other molecules or atoms because it is in empty Space. So the choices area: a) to stay hot (in other words, since heat is something like the average velocity of atoms, the N2 could just go off in all directions at high velocity, until they collide with something), or b) to radiate.
I vote for “b) radiate”, but does any WUWT reader know for sure?
Domenic says: “What will that detector see, what wavelengths, for a given temperature of the N2 inside that cavity?”
So IF you could build a 0% transmissive cavity out of nothing but pure N2? I think this falls in the “If my aunt had balls she’d be my uncle” category.
In a 0% transmissive cavity you will detect a near continuous band of wavelengths in a range commensurate to the temperature of the cavity. That does not mean that the molecules in the walls of the cavity have gained the ability to absorb/emit in a continuous band of wavelengths – there is reflection.
By definition there is no wavelength within our test bandwidths that will end up transmitted through the walls, so every photon is either absorbed or reflected. What happens if a photon just outside the absorption spectrum strikes the wall? It won’t be transmitted, right? (If it is, your walls must be made of something like N2) So the photon is reflected. A minuscule fraction of the photon’s energy is lost. The photon is now an ever-so-slightly longer wavelength. Compared to it’s previous wavelength, practically a continuous bandwidth. The photon continuous and strikes another wall. If it still isn’t within the absorption spectrum of the walls, the process continues until it is.
So within your 0% transmissive cavity you can indeed detect a practically continuous bandwidth of photons. This has nothing to do with the molecules of the walls gaining an ability to absorb/emit new bandwidths.
Ira Glickstein, PhD says:
March 4, 2011 at 3:56 pm
The only part that is in question is my step 2, where, I suppose, you could conclude that all the heat tranfer is via conduction, with absolutely no radiation and therefore a blank video recording.
Yes.
OK, if that is a problem, repeat the experiment in the vacuum of Space. Take a container of pure N2 and heat it up, then open the valve and let the N2 escape into Space. Do you see a plume with the IR camera now? I do.
The hot N2, we all know, will cool down. It cannot release its heat energy to other molecules or atoms because it is in empty Space. So the choices area: a) to stay hot (in other words, since heat is something like the average velocity of atoms, the N2 could just goo off in all directions at high velocity, until they collide with something), or b) to radiate.
Generally fly around until they hit something, take a look at the Earth’s thermosphere there’s plenty of hot N2 up there doing exactly that. Possibly get some emissions due to forbidden transitions occurring because of the very long lifetimes of the states (such as Lyman-Birge-Hopfield).
Good to hear from you again DonV. Have a look at the graphs that appear after my comment time stamped: March 1, 2011 at 7:19 am
Notice that, “(a) 20 Km looking down” (from near the top of the atmosphere, approximately Space, down towards the Earth), there is quite a bit of ~15μ energy (which in the description of my graphic includes all far-IR from about 12μ up) emitted by the Atmosphere. Note that there is a distinct depression between 14μ and 16μ, but still a fair amount of energy there, but not as much as at around 13μ and 17μ, which approximate the blackbody curve.
Now, have a look at the same wavelength range in “(b) Surface looking up” (from Earth towards the Atmosphere and Space). There is a maximum amount of energy around ~15μ, but HOLY COW! while the range between 14μ and 16μ is right up there around the blackbody curve, the surrounding wavelengths are depressed.
OK, my conclusion. The Earth radiates 14μ through 16μ at full force, but, when that radiation emerges from near the top of the Atmosphere, some of the 14μ and 16μ energy has been lost. Why? Well, CO2 and H2O molecules absorbed it and re-emitted some at their characteristic wavelengths, but, they also transferrred some of it, via collision (conduction) to other molecules in the Atmosphere, all in random directions.
The energized molecules re-emit the radiation at a variety of wavelengths, including in the ~7μ and ~10μ portions, which is why, looking again at the entire spectrum of the graph, there is proportionately more energy in the ~7μ and -especially- ~10μ regions. Where did that energy come from? Well, the only place I can think of is from the ~15μ portion where it is missing.
If little Johnny has blueberry juice all over his mouth and chin, and there is a piece of freshly baked blueberry pie missing, that is pretty good circumstantial evidence of where it came from!
I hope that explains what is going on in the dark blue box in my graphic, where ~15μ goes in to the Atmosphere at the bottom, and a variety of radiation, including ~7μ, ~10μ, and ~15μ, comes out at both the top and bottom. The deeper we delve into this thing, the more it is starting to look like my original (bouncing balls) graphic, is it not?
“OK, if that is a problem, repeat the experiment in the vacuum of Space. Take a container of pure N2 and heat it up, then open the valve and let the N2 escape into Space. Do you see a plume with the IR camera now? I do.”
You won’t detect any IR bandwidths that don’t show up in the absorption/emission spectrum using an IR spectrometer. Molecules can cool by emitting microwaves, radio waves…
The temperature of the object determines which bandwidths within it’s emission spectrum will be emitted upon cooling. It doesn’t give the molecule a new emission spectrum (unless your temperature change causes a state change, which gives you a different molecular structure with a different emission spectrum).
Hi Ira,
The second container should have a matte, or black painted finish on the inside surface. If you left it shiny stainless steel, the thermal radiation would simply reflect throughout the vessel and mask any image.
The entry hose on top needs to be as far from the opposite surface as possible. It also needs to be far from the IR camera lens to prevent heating of the lens.
There should also be a second hose from the bottom of the second container to the first container to allow for fairly equal pressure to prevent cooling of the heated N2 gas as it enters the second container. Expanding gases cool themselves. Let it enter from slight natural convection, or low pressure rather than via high pressure. The hose should be insulated.
If you do that, with a wide band .5 to 50 micron or 70 micron window on a high quality IR camera with high sensitivity, I am pretty confident that you will see the N2 entering. (I mention the .5 to 50 micron window because it is fairly easy to find. It uses a K Br coating on silicon lens. Going out to 70 microns might be tough to find.) The image should look like water or a vapor plume entering the chamber from the first hose, with the image most clear right after exit from the hose.
I believe the highest sensitivity IR cameras are still those with liquid nitrogen cooled detectors. They have been around for years, and are extremely sensitive. I haven’t checked in a while, though.
IR cameras are used to check for gas leaks in industrial facilities. They see the air plumes due to either heating or cooling of the surrounding air by the escaping gas. Gas under pressure, but at ambient temp will chill the air molecules as it escapes. Heated gases depend on the temperature and pressure of the piping they are escaping from and the outside ambient air temp. The escaping gas plumes are pretty easy to see with most IR cameras using a wide spectrum lens to make sure it is sees all possible wavelengths.
Ira Glickstein, PhD says:
March 4, 2011 at 4:32 pm
Don V says:
March 4, 2011 at 10:38 am
Good to hear from you again DonV. Have a look at the graphs that appear after my comment time stamped: March 1, 2011 at 7:19 am
Notice that, “(a) 20 Km looking down” (from near the top of the atmosphere, approximately Space, down towards the Earth), there is quite a bit of ~15μ energy (which in the description of my graphic includes all far-IR from about 12μ up) emitted by the Atmosphere. Note that there is a distinct depression between 14μ and 16μ, but still a fair amount of energy there, but not as much as at around 13μ and 17μ, which approximate the blackbody curve.
OK, my conclusion. The Earth radiates 14μ through 16μ at full force, but, when that radiation emerges from near the top of the Atmosphere, some of the 14μ and 16μ energy has been lost. Why? Well, CO2 and H2O molecules absorbed it and re-emitted some at their characteristic wavelengths, but, they also transferrred some of it, via collision (conduction) to other molecules in the Atmosphere, all in random directions.
Around 15μm the atmosphere is ~opaque until the temperature of ~225K above which the CO2 emissions are able to make it up to 20km. Between 13μm and 10μm and 9.5μm and 8μm the atmosphere is transparent and you can see BB from the surface at ~268K
The energized molecules re-emit the radiation at a variety of wavelengths, including in the ~7μ and ~10μ portions, which is why, looking again at the entire spectrum of the graph, there is proportionately more energy in the ~7μ and -especially- ~10μ regions. Where did that energy come from? Well, the only place I can think of is from the ~15μ portion where it is missing.
No from the surface.
Good points, Steve, and you may be correct.
However, IR photos are generally taken fairly close to 10μ (FLIR video, looking for objects whose temperatures range around the blackbody temperature of the surface of the Earth to, for example, expose people walking across a field) where there is a window in the H2O absorption spectrum, or sometimes around 4μ (detection of vehicles with hot engines or tires) where there is another window. If you try to take an IR image in other parts of the spectrum the water vapor will make the image look like fog or nothing will be clearly seen.
The above is why I insisted that the second container be pure N2 with no H2O or CO2 to fog or block the image.
As I said, you may be right that hot N2, even when it is injected into a room-temp container of N2 will not emit photons as it cools off. On the other hand, I think the hot N2, which has lots of energy, when it collides with room-temp N2, will release radiation somewhere in the IR spectrum, perhaps around 4μ, and possibly, given a fairly large mass of hot and room-temp N2 colliding and mixing, over a wider number of wavelengths. The reason I think so is that collisions may distort the molecules and change their exact geometry, which could affect the wavelength of the ejected photons. I am no expert in this area, so I do not know for sure.
Thanks Domenic, you seem to confirm that plume I saw in my mind’s eye. And, not only that, but someone stole my idea for detecting escaping gas even before I came up with it :^). At what approximate wavelength(s) will the plume of hot N2 be observed?
Hi Steve,
Steve wrote:
“In a 0% transmissive cavity you will detect a near continuous band of wavelengths in a range commensurate to the temperature of the cavity. That does not mean that the molecules in the walls of the cavity have gained the ability to absorb/emit in a continuous band of wavelengths – there is reflection…..and so on.”
Yes, you used the reflections to account for decreasing energy states by increasing the wavelengths, to points where the photons would finally be absorbed by the N2 at its ‘supposed’ absorption spectra. Thus 100% absorption.
But that is not what I asked in this part of the question.
I am asking about what is being EMITTED only through the pinhole to the outside detector. From the inside to the outside. The inside is completely N2 cavity walls with no transmissive effects from the outside world.
Are you trying to use DECREASING energy states via reflections to account for what you will ‘see’ as a continuous spectrum? A continuous spectrum that includes wavelengths of shorter length (thus HIGHER energy states) than those possible by ‘supposed’ published N2 absorption spectra at a given temperature? Thus, the assumed equivalent emission spectra at a given temperature?
Because, if you isothermally heat that ‘all N2’ cavity up, above outside ambient temperature, the emitted radiation from the pinhole will follow near perfect BB radiation. It will not follow the ‘supposed’ N2 spectra for emission at all. And you can’t say that the emitted shorter wavelengths are from reflections due to entering photons from outside through the pinhole into the cavity, because the outside temperature is below the cavity temperature. Thus, there is NO source for the apparent shorter wavelengths except the N2.
And yet, strangely enough, that is the way the real world works with cavity radiation. It doesn’t violate the laws of thermodynamics, but it does shift the emitted energy to BB like wavelengths regardless of the source molecules.
I will leave you to ponder that mystery…
Hi Ira,
I don’t know the exact wavelengths, as I have never mearusred them. Nor have I seen anyone else measure them.
As I’ve stated many times here, N2 and O2 have never been tested rigorously over the continuous spectrum. It’s tedious and difficult to do so, and there has never been a good reason to do so before. But I am pretty confident there will be detectable emissions in the .5 to 50 micron range from playing around with a .5 to 50 micron hand held detector many, many years ago, aiming it at the sky, the sun, etc and comparing the output to a detector using much narrower bands.
I think it is a time to stop this discussion. Everything emits and absorbs in any frequency range. This is true but physically meaningless statement. The real question is how much. Good physics dictates to put numbers on magnitude of each effect, and reasonably sort things out in accord with their relative contribution to main effect.
Yes, pairs of N2-N2 and O2-O2 do present in air at normal pressure and temperature when all molecules collide in gaseous mess. Yes, this is called “collision-induced absorption”, and it does exist, has been measured several times, and is fully theoretically accounted for:
http://pubs.giss.nasa.gov/docs/2003/2003_Boissoles_etal.pdf
At atmospheric conditions, the spectrum is continuous, centered at 100um, with measurable tails down to 30um. It is even considered important in certain cases of dry stratosphere when looking along a very slant path. Unfortunately I could not find a straight answer in their articles about relative contribution of N2-O2 collision-induced continuous background to vertical path of air filled by CO2, water vapor, and water droplets.
I can only offer this simple estimate, that the entire relative energy of Earth IR emission is less than 0.02% above of 30um (see spectralcalc.com). Therefore, regardless of how strong N2-O2 absorb at its respective band, 30um-10000um, this effect cannot exceed 0.02%. Now you need to decide if this amount (about 0.05W/m2, or 50mW) is really important in explaining why the Earth surface is warmer than the planetary emission temperature is.
Regarding the “hot N2 jet experiment” and depending on how hot it really is, it is possible that a good cooled IR camera will see some emission in 70um end; the modern IR cams are fairly sensitive.
Domenic says:
March 4, 2011 at 8:03 pm
Hi Ira,
I don’t know the exact wavelengths, as I have never mearusred them. Nor have I seen anyone else measure them.
As I’ve stated many times here, N2 and O2 have never been tested rigorously over the continuous spectrum.
Which is of course not true!
It’s tedious and difficult to do so, and there has never been a good reason to do so before. But I am pretty confident there will be detectable emissions in the .5 to 50 micron range from playing around with a .5 to 50 micron hand held detector many, many years ago, aiming it at the sky, the sun, etc and comparing the output to a detector using much narrower bands.
It’s easy to do, in fact every time I ran my FTIR spectrometer I did so, but of course there’s no signal. The only wavelength where anything could possibly show up would be the incredibly weak lines at 4.3μm, which is of no practical significance. Your confidence is misplaced.
@Ken Coffman
I hate to rank on you, because I do believe you also are fair-minded and trying to understand (and hopefully you will believe the same of me). However, the water hose analogy that has been used is just that — an analogy, and it has several shortcomings. Just to be clear — in the analogy, spraying water molecules are used as representing the IR photons.
“There is no side channel or alternate path for the fire hose and the blue-water squirt gun. They must use the same path.”
Unlike water molecules, photons can and do easily cross without colliding with each other. They can use the “same path” without any (significant) chance of blocking each other.
“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. ”
The original analogy was flawed here, and this further exaggerates the flaw. For IR photons there are similar numbers going each way, as i will explain now
* The ground is a little warmer than the atmosphere, so that factor will mean some more photons going up than down (but since the back radiation is mostly from low layers, the atmosphere emitting the back radiation will not be that much cooler than the land so the effect from temperature will not be TOO great)
* The ground is close to a black body for IR (emissivity = 1 for all IR frequencies), but the atmosphere has bands where it does not emit or absorb well (emissivity ~0) and other bands where it does emit or absorb well (emissivity ~1). So this factor by itself might mean ~ half as much going down as up (depending on just how big the “windows” are. (BTW, I am combining the emissions from H2O with the CO2 here. But the CO2 by itself would also return photons in significant bands of wavelength.)
* Also, I believe the original graphic of the “windows” included water vapor (ie humidity) but not water droplets (ie clouds). Clouds cover much of the earth and would absorb and re-emit almost 100% of the IR photons.
Combine these three factors , and there are nearly as as many photons going down as up. (In fact, if you believe Trenberth, the ratio is 324:390 ~ 83%. Even if you don’t believe the numbers exactly, the back radiation just from the clouds should be close to 70% since clouds cover about 70% of the earth.)
The analogy should be more like a 4″ fire hose vs a 3″ fire hose if we are representing the relative numbers of IR photons.
There are other aspects of the analogy worth discussing, but as you said “once the sale is made, shut the hell up.” 🙂
Before everybody start jumping on me, I’d like to correct myself. Energy fraction under Planck curve above 30um at 255K is somewhat larger than 0.02%. I pull this number out of my mind meaning that the N2-O2 spectrum corresponds to a blackbody with 100um peak, which would correspond to a body at about 40K with total emission of 45mW/m2. More, the entire region above 15um is already filled with CO2 and water over the top, see Ira’s spectrum above, so there is no way to see any N2 or O2 effect in practice.
Ira,
I would say this a little differently:
OK, my conclusion. The Earth radiates 14μ through 16μ at full force, but, when that radiation emerges from near the top of the Atmosphere, some of the 14μ and 16μ energy has been lost. Why? Well, CO2 and H2O molecules absorbed it and re-emitted some at their characteristic wavelengths, but, they also transferrred some of it, via collision (conduction) to other molecules in the Atmosphere, all in random directions.
The Earth radiates 14μ through 16μ at full force, but 100% of those photons are absorbed by the GHGs fairly near the surface. None of that radiation actually emerges from the TOA.
GHGs near the TOA can and do radiate their own new 14μ through 16μ IR photons. However, since the atmosphere is much cooler there, they radiate much less energy in those bands — hence the much reduced intensity compared to the wavelengths that came all the way up though the “windows” where the atmosphere is nearly transparent.