Guest post by Ira Glickstein
The Atmospheric “greenhouse effect” has been analogized to a blanket that insulates the Sun-warmed Earth and slows the rate of heat transmission, thus increasing mean temperatures above what they would be absent “greenhouse gases” (GHGs). Perhaps a better analogy would be an electric blanket that, in addition to its insulating properties, also emits thermal radiation both down and up. A real greenhouse primarily restricts heat escape by preventing convection while the “greenhouse effect” heats the Earth because GHGs absorb outgoing radiative energy and re-emit some of it back towards Earth.
Many thanks to Dave Springer and Jim Folkerts who, in comments to my previous posting Atmospheric Windows, provided links to emission graphs and a textbook “A First Course in Atmospheric Radiation” by Grant Petty, Sundog Publishing Company.
Description of graphic (from bottom to top):
Earth Surface: Warmed by shortwave (~1/2μ) radiation from the Sun, the surface emits upward radiation in the ~7μ, ~10μ, and ~15μ regions of the longwave band. This radiation approximates a smooth “blackbody” curve that peaks at the wavelength corresponding to the surface temperature.
Bottom of the Atmosphere: On its way out to Space, the radiation encounters the Atmosphere, in particular the GHGs, which absorb and re-emit radiation in the ~7μ and ~15μ regions in all directions. Most of the ~10μ radiation is allowed to pass through.
The lower violet/purple curve (adapted from figure 8.1 in Petty and based on measurements from the Tropical Pacific looking UP) indicates how the bottom of the Atmosphere re-emits selected portions back down towards the surface of the Earth. The dashed line represents a “blackbody” curve characteristic of 300ºK (equivalent to 27ºC or 80ºF). Note how the ~7μ and ~15μ regions approximate that curve, while much of the ~10μ region is not re-emitted downward.
“Greenhouse Gases”: The reason for the shape of the downwelling radiation curve is clear when we look at the absorption spectra for the most important GHGs: H2O, H2O, H2O, … H2O, and CO2. (I’ve included multiple H2O’s because water vapor, particularly in the tropical latitudes, is many times more prevalent than carbon dioxide.)
Note that H2O absorbs at up to 100% in the ~7μ region. H2O also absorbs strongly in the ~15μ region, particularly above 20μ, where it reaches 100%. CO2 absorbs at up to 100% in the ~15μ region.
Neither H2O nor CO2 absorb strongly in the ~10μ region.
Since gases tend to re-emit most strongly at the same wavelength region where they absorb, the ~7μ and ~15μ are well-represented, while the ~10μ region is weaker.
Top of the Atmosphere: The upper violet/purple curve (adapted from figure 6.6 in Petty and based on satellite measurements from the Tropical Pacific looking DOWN) indicates how the top of the Atmosphere passes certain portions of radiation from the surface of the Earth out to Space and re-emits selected portions up towards Space. The dashed line represents a “blackbody” curve characteristic of 300ºK. Note that much of the ~10μ region approximates a 295ºK curve while the ~7μ region approximates a cooler 260ºK curve. The ~15μ region is more complicated. Part of it, from about 17μ and up approximates a 260ºK or 270ºK curve, but the region from about 14μ to 17μ has had quite a big bite taken out of it. Note how this bite corresponds roughly with the CO2 absorption spectrum.
What Does This All Mean in Plain Language?
Well, if a piece of blueberry pie has gone missing, and little Johnny has blueberry juice dripping from his mouth and chin, and that is pretty good circumstantial evidence of who took it.
Clearly, the GHGs in the Atmosphere are responsible. H2O has taken its toll in the ~7μ and ~15μ regions, while CO2 has taken its bite in its special part of the ~15μ region. Radiation in the ~10μ region has taken a pretty-much free pass through the Atmosphere.
The top of the Atmosphere curve is mostly due to the lapse rate, where higher levels of the Atmosphere tend to be cooler. The ~10μ region is warmer because it is a view of the surface radiation of the Earth through an almost transparent window. The ~7μ and 15μ regions are cooler because they are radiated from closer to the top of the Atmosphere. The CO2 bite portion of the curve is still cooler because CO2 tends to be better represented at higher altitudes than H2O which is more prevalent towards the bottom.
That is a good explanation, as far as it goes. However, it seems there is something else going on. The ~7μ and ~15μ radiation emitted from the bottom of the Atmosphere is absorbed by the Earth, further warming it, and the Earth, approximating a “blackbody”, re-emits them at a variety of wavelengths, including ~10μ. This additional ~10μ radiation gets a nearly free pass through the Atmosphere and heads out towards Space, which explains why it is better represented in the top of the Atmosphere curve. In addition, some of the radiation due to collisions of energized H2O and CO2 molecules with each other and the N2 (nitrogen), O2 (oxygen) and trace gases, may produce radiation in the ~10μ region which similarly makes its way out to Space without being re-absorbed.
There is less ~15μ radiation emitted from the top of the Atmosphere than entered it from the bottom because some of the ~15μ radiation is transformed into ~10μ radiation during the process of absorption and re-emission by GHGs in the atmosphere and longwave radiation absorbed and re-emitted by the surface of the Earth.
Source Material
My graphic is adapted from two curves from Petty. For clearer presentation, I smoothed them and flipped them horizontally, so wavelength would increase from left to right, as in the diagrams in my previous topics in this series. (Physical Analogy and Atmospheric Windows.)
Here they are in their original form, where the inverse of wavelength (called “wavenumber”) increases from left to right.
Source for the upper section of my graphic.
Top of the Atmosphere from Satellite Over Tropical Pacific.
[Caption from Petty: Fig. 6.6: Example of an actual infrared emission spectrum observed by the Nimbus 4 satellite over a point in the tropical Pacific Ocean. Dashed curves represent blackbody radiances at the indicated temperatures in Kelvin. (IRIS data courtesy of the Goddard EOS Distributed Active Archive Center (DAAC) and instrument team leader Dr. Rudolf A. Hanel.)]
Source for the lower section of my graphic.
Bottom of the Atmosphere from Surface of Tropical Pacific (and, lower curve, from Alaska).
[Caption from Petty: Fig. 8.1 Two examples of measured atmospheric emission spectra as seen from ground level looking up. Planck function curves corresponding to the approximate surface temperature in each case are superimposed (dashed lines). (Data courtesy of Robert Knutson, Space Science and Engineering Center, University of Wisconsin-Madison.)]
The figures originally cited by Dave Springer and Tim Folkerts are based on measurements taken in the Arctic, where there is far less water vapor in the Atmosphere.
[Fig. 8.2 from Petty] (a) Top of the Atmosphere from 20km and (b) Bottom of the Atmosphere from surface in the Arctic. Note that this is similar to the Tropical Pacific, at temperatures that are about 30ºK to 40ºK cooler. The CO2 bite is more well-defined. Also, the bite in the 9.5μ to 10μ area is more apparent. That bite is due to O2 and O3 absorption spectra.
Concluding Comments
This and my previous two postings in this series Physical Analogy and Atmospheric Windows 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 (clouds, rain, snow, etc.) that transfer a great deal of energy from the surface to the higher levels of the Atmosphere.
For those who may have missed my previous posting, here is my Sunlight Energy In = Thermal Energy Out animated graphic that depicts the Atmospheric “greenhouse effect” process in a simlified form.
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 Emissions Spectra 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 ones 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.
Phil. says:
March 10, 2011 at 9:44 am
“But a warm object with a continuous heat source surrounded by a radiatively active atmosphere which is warmer than space will reach an equilibrium temperature which is warmer than it would otherwise be in the absence of such an atmosphere.”
——————–
I’m not aware of anywhere on this spinning, tilting, cloud-bedecked planet for which this description would be apt.
Furthermore, the fact that the atmosphere also interacts with the incoming radiation that is the heat source for the surface complicates speculation about what it it woud all be like “if..”.
“Since gases tend to re-emit most strongly at the same wavelength region where they absorb, the ~7μ and ~15μ are well-represented, while the ~10μ region is weaker.”
We went over this before. Kirchoff’s first of law radiation: solids, liquids, and dense gases emit continuous blackbody spectrums with the peak emission frequency determined by the temperature. The earth’s atmosphere below the thermosphere is dense in the context of Kirchoff’s law. When the gas molecules are densely packed they give up energy through collisions. Collisions don’t generate emission line spectra they generate continuous blackbody spectra. The “temperature” is a measure of the average speed of the molecules and hence the average energy involved in collisions between molecules.
Vince Causey
……..”Actually heat is collected from the ground. Thermal heat collectors based on heat pumps are able to extract that heat from the 2 metres below the surface and can be used to warm houses.”……..
This fully complies with the second law.
Clausius said that heat will not flow spontaneously from a lower to a higher temperature.
You can of course use external energy as work in a device to pump the heat.
Like a refrigerator or heat pump.
GaryP says:
March 10, 2011 at 10:47 am
Here is an actual experiment. I was using a hot filament in high vacuum to heat a sample to high temperature (~1000°C). I could not get one sample hot enough so I put a shiny metal cylinder around but not touching the filament or sample. The reflected heat increased my sample temperature over 100°C. The added radiation from the relatively cold cylinder increased the temperature of the white hot filament. The net flow of heat of course was from hot to cold. This is a pure radiation example and some heat did radiate from the cold reflector to the hot filament.
A slightly more sophisticated version of this is used with incandescent light bulbs to improve their efficiency. The glass envelope is coated with a coating which reflects IR and transmits visible, so the visible is emitted but the IR is reflected back towards the filament, the filament gets hotter and therefore brighter as well as shifting to more visible. Consequently the bulbs emit more visible light for a given electrical input than a conventional bulb thanks to ‘back radiation’.
John Marshall says:
March 10, 2011 at 2:04 am
Before you go about “understanding” the laws of thermodynamics in your head you should first learn about the results of experiments conducted circa 1859 by John Tyndall with regard to thermal absorption of gases. If the answer you get from the thermodynamic model in your head is not the same answer that comes from experiment then your mental model is flawed. Your understanding is flawed if you do not acknowledge “back radiation”. Tyndall was the first to actually measure it in a series of thousands of experiments with various gases, mixtures of gases, at various pressures, through varying column lengths, and with thermal energy sources at different temperatures. He gave countless lectures and demonstrations about it and published collections of papers in a few books which are available in their entirety free of charge from books.google.com. You might want to start with “Heat: A Mode of Motion” and skip up to the chapters on gases. It’s great reading especially the ingenuity of the experimental setup which is described in great detail along with anecdotes about he obtained various materials required for its construction.
I recently voted for this as being the best scientific blog but, in response to this article, we still get people who say that back radiation cannot happen or wish to divert the discussion to some strange theories about the temperature on Venus.
But there is a serious problem with Ira’s article!
As expressed before (6:10AM); his blackbody curves are showing a peak at about 18 micron when they should be peaking at about 10 micron for a 300K blackbody. So, the graphs don’t make sense
Now Joel Heinrich at 2.57 AM and Phil at 8.21AM, March 10, say this is because Ira has converted wave number to wavelength.
But this does NOT cut the mustard. This is a perfectly acceptable thing to do, albeit ending up with a non-linear scale rather than a linear scale. In fact, the
Source Data itself includes the wavelength scale, so it is nothing to do with Ira’s conversion.
On the ‘wave number’ scale the peak of the blackbody curves should be around a wave number of 1000 – which they are not! They peak around a wave number of 600 – which is WRONG!
So the whole article and its conclusions are WRONG!
Now, can any mathematician or physicist please explain to me why I am mistaken? What am I missing?
Come on Ira.
tallbloke on March 10, 2011 at 4:26 am says: “…it was reduced albedo 1979-1998 allowing more Solar energy to enter the oceans that caused the majority of the global warming at the end of the last millennium.” Rubbish. There was no such warming before 1998, it’s all fake. Satellite temperature measurements contradict this fantasy that appears on NASA, NOAA and Met Office charts. What satellites do see in this time slot is a temperature oscillation, up and down by half a degree for twenty years, but no warming until the super El Nino shows up. The oscillations belong to the alternation of warm El Nino and cool La Nina periods that are part of the ENSO system in the Pacific. Read “What Warming?” available on Amazon.com.
Since such articles are very comment-rich, I assume the science is far from settled, even the basic theory.
While the charts are ok, nobody yet calculated how much is this backward IR actually affecting the earth surface temperature. Remove oceans and 99% of atmosphere (oxygen and nitrogen) with their tremendous heat-keeping capacity and run the experiment with those GHGs again. Or look at the Mars with 6,000 ppm of CO2 and its temperature like blackbody, since oceans, nitrogen and oxygen are missing.
Phil. says:
March 10, 2011 at 10:53 am
If I said surface I misspoke as I know the air isn’t going to heat the surface. My point was and is that -33 C which is used as the starting point for GHG effect may not be correct and that 0 C (STP) is a more proper place to start with any effect of GHG’s.
Again with the angry. I never said I had all the answers about gas physics I said I was asking questions and there was a difference of opinion. But now I know that, “At any given temperature they will not emit more that given by S-B at that wavelength. ”
So does this say I can use Wein Displacement Law for a temperature find the wavelength and figure the energy being emitted?
CommieBob,
Is that flashlight cooler than your eyes? I never said, nor have I seen anyone else on this thread say, that radiation from a hot source would somehow “block” the radiation from a cooler source. I don’t see how this example refutes anything anyone is stating. If you shine a flashlight in your eyes, it will warm you, not the other way around. The flashlight will not warm the sun. The filament in the flashlight is not a hot as the photosphere of the sun. The photons that represent the information being exchanged between the sun and the filament will result in a net warming of the filament, not the sun.
I would expect my students to state that the atmosphere (or any insulation layer), changes the rate at which energy is transmitted.
You say that wearing a coat warms the body. I say putting on a coat slows the rate at which your body transfers thermal energy to the surroundings. (and yes, this is obviously conduction, not radiation) There is not really much difference mathematically. It is that statement that greenhouse gases HEAT the earth” by re-radiating IR that is not going to sit well with the thermodynamics vocabulary police.
Phil. says:
March 10, 2011 at 11:16 am
Yes I can read wikipedia also. http://en.wikipedia.org/wiki/Bond_albedo
from which you have clearly quoted verbatim.
The fact is that again you are either missing the point or you are deliberately taking things out of context.
Harry is talking about ALBEDO, you are talking about BOND ALBEDO. You have shifted the goal posts.
ALBEDO in context to Harry’s point is only relevant to visible light. Also from wikipedia, Quote: “When quoted unqualified, it usually refers to some appropriate average across the spectrum of visible light.”
And from the same wiki page: The term is derived from Latin albedo “whiteness”
http://en.wikipedia.org/wiki/Albedo
So as usual we are now unnecessarily arguing semantics, when it is perfectly clear to anyone what Harry actually means by albedo.
As for this
Phil. says:
March 10, 2011 at 11:04 am
I am no more interested in responding to waffle than I am arguing semantics.
So I refer you to my first post.
http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/#comment-617396
“”””” P. van der Meer says:
March 10, 2011 at 3:15 am
Ira Glickstein, why don’t you explain to your readers why the various curves for blackbody radiation in your article peak in the range of 17μm to 19μm when the http://spectralcalc.com/blackbody_calculator/blackbody.php site comes up with a peak wavelength of 9.659μm for 300K and 11.828μm for 245K. This is also confirmed by the Wikipedia graph (http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png), showing a peak at about 9μm for 310K.
I look forward to your explanation. “””””
“”””” Joel Heinrich says:
March 10, 2011 at 2:57 am
The graphics have one (unfortunately very common) error. They are plotted as a wavenumber distribution but labeled with wavelength. You CANNOT just transform the wavenumber into wavelength as they have different peaks. Much like the difference between wavelength and frequency: http://commons.wikimedia.org/wiki/File:PlanckDist_ny_lambda_en.png
The peak of a distribution in wavelength for 280K is at 10.5 µm. “””””
Well the explanations are quite simple:
For Joel Heinrich’s comment; you can in fact plot a graph of anything you like against anything else that you like, even if they aren’t in any way related.
But in this case they are. Wavelength and wave number also are related, and you can get from one to the other or from the other to one if you prefer.
And some of the Graphs Ira gives, have horizonatal axes, in both Wavelength and Wave number, as you can plainly see.
But as you can also plainly see the vertical scale has specific units. And ALL of the graphs have the wrong labels. The vertical axis is not RADIANCE. It IS “Spectral Radiance”, which is the contribution to total RADIANCE, per increment of either WAVE NUMBER as it is in these graphs, or WAVELENGTH which is also quite co0mmon. I actually am more used to seeing blackbody radiation plots using a Wavelength horizontal scale, or a logarithmic wavelength horizontal scale, and a per unit wavelength Spectral Radiance for the vertical scale; but chemists tend to use Wave Numbers and per wavenumber units, and the BB curves are different depending on which you use.
At 288 K , the BB Planck function has about a 10 micron peak spectral radiance; which is also 1000 cm^-1 BUT ONLY if the Spectral Radiance is specified per unit of WAVELENGTH increment. If the spectral radiance units are per wavenumber (cm^-1) then the peak is closer to 600 cm^-1 or around 18 microns.
Obviously the global warmists prefer the per wave number version, since the peak of the curve is right on the CO2 band, making CO2 seem more important; whereas the same data on a per micron wavelength plot peaks at 10 microns, or 1000 cm^-1 with the CO2 15 micron band now down lower on the long wavelength tail.
Although I am more used to the wavelength and the per micron spectral radiance scale, I do agree there is some merit in using frequency units, since the photon energies are proportional to frequency, and not to wavelength. But actually neither version is without problems.
On a per wavelength scale, you get high spectral radiance for solar UV; but remember that the range of wavelengths available for solar UV is just 2-3 hundred nanometres, versus the tens of microns available for the LWIR emissions from the surface.
Conversely, on a per wave number spectral radiance plot, the CO2 band gets a miserable 100 wave numbers, whereas that low amplitude water tail has many hundreds of wavenumbers bandwidth; so neither representation is without problems.
Some people graph black body radiation curves using numbers of photons, versus wavelength or wave number, and then you get a different peak again.
Now guffawingly, Peter Humbug in his now famous Physics Today special paper, emitted the per spectral units interval all together, giving a totally ridiculous plot, that implied an infinite amount of total energy; which interestingly he also called flux, rather than spectral radiance.
But we will cut him some slack, and put his error down to a simple typo; we know what you meant Peter. And incidently as he plotted it he did mean per wave number interval.
Perhaps if he had paid more attention to his subject and left out the political BS, about climate disruption, and 800,000 Kelvin earth Temperatures, then he might have caught his mistake. But as I said we will grant him some leeway. I personally never make typeos, because I review everything carefully.
Arno Arrak says:
March 10, 2011 at 12:10 pm
tallbloke on March 10, 2011 at 4:26 am says:
“…it was reduced albedo 1979-1998 allowing more Solar energy to enter the oceans that caused the majority of the global warming at the end of the last millennium.”
Rubbish. There was no such warming before 1998, it’s all fake. Satellite temperature measurements contradict this fantasy that appears on NASA, NOAA and Met Office charts. What satellites do see in this time slot is a temperature oscillation, up and down by half a degree for twenty years, but no warming until the super El Nino shows up. The oscillations belong to the alternation of warm El Nino and cool La Nina periods that are part of the ENSO system in the Pacific.
Yeah, that’s the point. The oceans had been saving up excess solar energy and hiding it away from surface measurements in the subsurface pacific warm pool for years. Then just like I said, “at the end of the last millenium” kapow! super el nino. Large amounts of solar derived energy released into the atmosphere and recycled round the ocean surface, and an upward step change in surface temperature post 2000 as the system bounced back from la nina.
@Ira
I don’t think it’s a good idea to go into quantum mechanics which is generally reserved for quantum scale phenomena. Wavelengths in the micron range in a dense gas means a single wavelength spans millions of molecules at once which is in the domain of classical thermodynamics. Temperature itself is not a quantum measure but rather an average speed of a great number of molecules. That said I’ll take a stab at a lay description of individual molecules and photons in the dense portion of the earth’s atmosphere.
I just finished reading an article in this month’s issue of Scientific American about a new technique for cooling matter down closer to absolute zero than anyone has obtained before and even better a way to do it with elements that previous techniques didn’t work with. Background information in the article talked about the speed of motion of room temperature gases which if I recall correctly ranges from about 2000 miles per hour to down near zero. The gadget they constructed used a pair of laser beams that act like Maxwell’s Demon allowing lower speed molecules to pass through the pair of beams in one direction only. The volume on one side is smaller than the other and when they trap all the atoms on the smaller side (which for some esoteric quantum reason I don’t really understand doesn’t raise their temperature by compression) they turn off the lasers and let the atoms expand back into the full volume and when they expand they cool down. By doing that over and over and over each time dropping the temperature a little bit they get to within a few millionth’s of a degree of absolute zero.
Preceding the laser stage they got most of their cooling by letting a gas of atoms out through a pinhole into a vacuum. This had the effect of both cooling the atoms and getting them all moving at close to the same speed. They aimed this beam of atoms at the receding edge of a fan blade moving at half the speed of the atoms which had the effect of slowing them down even more without disturbing their distribution or speed relative to each other. After having that done to them a few times they passed into the laser chamber for the final cooling cycles.
I just thought I’d throw that in because it’s way interesting, bleeding edge, and descriptive of how temperature is a measure of speed of motion.
Anyhow, certain gas molecules have a shape that is resonant with certain frequencies of thermal radiation. When photon of that frequency intersects the molecule it is absorbed. The absorbed energy raises the molecule to a higher excitation state. In a rarefied gas it would eventually emit a photon of the same frequency and fall back to the lower excitation state in a quantum jump. In a dense gas where it is rubbing shoulder to shoulder with other molecules however the higher excitation results in greater speed and means it’s going to collide with another molecule right away, before the average amount of time elapses for a quantum transition to occur. Don’t ask me how long it takes for a quantum jump to occur. All I know is that it isn’t a fixed amount of time but rather a probability of occuring where there is a small probability of being instantaneous, a small probability of taking forever, and a greater probability of some length of time in between those two. A quantum physicist should be able to give a more definitive answer. But the answer is moot because it has been experimentally demonstrated that in a dense gas collisions occur before any significant number of quantum emissions happen. Thus Kirchoff’s law predates quantum mechanics by over half a century.
Due to the absorption of a thermal photon at narrow resonant frequences in so-called GHGs the higher excitation means that molecule is going to have a more energetic collision than it would have otherwise. A collision will also cause a photon to be emitted but the photon frequency is determined by the collision energy not the energy added by the resonant photon. Since the molecules are all moving at anywhere from 0-2000mph to start with the collisions will cause photons to be emitted over continuously varying range of energies (or frequencies as photon energy and frequency are the same thing). So what we end up with is a continuous blackbody emission spectrum reflective of the individual collision energies in billions of atoms all moving at different speeds. A few of the collisions will be very low energy giving us lower frequency photons and some will be higher energy giving us higher energy photons. The closer we get to the average speed of the molecules the more collisions we get with that amount of energy in them. Thus we get a spectrum with a peak energy at a certain frequency that falls off in a smooth curve to either side of that frequency i.e. a continuous blackbody spectrum. Since temperature is a measure of the average speed of a great many individual molecules the peak emission frequency of the spectrum is a function of temperature.
Bomber_the_Cat says:
March 10, 2011 at 12:01 pm
The graphs presented reflect the change in temperature, and thus wavelength emitted, associated with the atmospheric lapse rate as measured from TOA or from the surface. Quite a different picture from a blackbody spectrum where all the emissions would be from the surface of the substance.
TB, how do you think the ocean is heated if not at the surface? Geothermal? And why does it matter how far IR penetrates the surface? At what penetration distance do you think it would make a difference? All we need is for the top micron to be warm in order to start things cooking, by whatever means.
Willis’ point was spot on–without atmospheric heating of the ocean it would freeze. In your defense you only contradict yourself, claiming long term equilibrium, which equilibrium could only be arrived at through atmospheric heating. And globally we would see that temp/time lags even out, or no equilibrium could be achieved. That is, being a good heat sink, half the time the ocean heats the air and the other half the sky heats the ocean. If the ocean were always a net emitter of energy it would freeze fast.
Your claim that H2O molecules rise because they’re lighter is worse than the notion that IR hits the ocean directly from on high. Do you think there is a layer of water vapor that floats to the top of the atmosphere? It takes months for gases to mix through the atmosphere, and it all happens through collisions. The only thing the light molecules have in their favor is their faster speed, which does accelerate dispersion through permeable membranes and through the air.
The top film of a warm ocean is hot and dense due to radiation and evaporation. But that superdensity contributes negliigibly to circulation–it cannot compete with wind and wave action or even conduction. But we know the sea heats from the top down–geothermal heat is negligible, and the coldest water is at the bottom.
Some confusion on wavelengths and wave numbers:
To go from wave number to wave length:
769 wave number, move the decimal 4 places to the left, or 0.0769 and hit the 1/x reciprocal key giving 13 µm.
To go from wave length (in µm) to wave number:
Take 13 µm, hit 1/x, gives you 0.0769, move the decimal 4 to the right to give 769/cm wave number. That simple. Maybe that will help.
The real confusion is these two type of charts are normally also flipped right to left on the x axis to each other. I agree, that makes it hard for everyone including myself to compare apples to oranges.
What is incredible for me it’s that everyone who talk about the outgoing and the back scattered spectra, always considers the vertical paths only.
Does anybody out of here have graphs of the back scattered spectrum at ground for different angles at the same place? That is the azimuthal (90°), the 60°, the 45° and the 30° for example.
What I’m guessing here is that the graphs could be very different.
But the most interesting should be the outgoing spectrum at the TOA for the different angles, because the following links shows how the so called “limb radiation” is almost complementary to the nadir one (see the topmost graph of figure 3).
http://www.atmos-chem-phys-discuss.net/6/4061/2006/acpd-6-4061-2006-print.pdf
(The research pertain a radiosonde at abt. 34km, but how you can see from the radiance emitted from the deep space at that height there is almost no more energy back scattered).
What I’m arguing is that the increased absorption pit at 15um due to the doubling of the CO2 seen at the satellite nadir view doesn’t mean that the radiation is backscattered or held by the atmosphere, that energy just exits the atmosphere under different angles not seen by the satellites “eye”.
We do observe back radiation from the atmosphere I think though that what comes back is much less than is emitted by the Earth. Long wave radiation has less energy than short wave radiation and we are told that the molecules in the atmosphere do not absorb shortwave radiation only long wave radiation at particular frequencies but the Earth absorbs shortwave radiation and emits long wave radiation there has to be a net transfer from the Earth to the atmosphere and to space because of this. I think that the presentation was more realistic than many that we have been given, it does attempt to give empirical data rather than analogy and thought experiments.
We are clearly talking at crossed purposes. One of the things I have expertise on is EM radiation. If the problem is vocabulary, so be it. If you are saying that a weak emitter can not radiate energy toward a stronger emitter, you are just plain, flat, dead, completely wrong.
What I can also tell you as an expert is this: If I build and energize a transmitter antenna, I can still measure the signals on it that are caused by adjacent transmitters. (The total voltage on the antenna will be the sum of the local transmitted signal and all the received signals.) In fact, a good example would be continuous wave radar. A signal leaves the transmitting antenna, strikes the sharp edge of an aircraft, re-radiates at the same frequency and, shifted by the doppler effect, returns to the sending antenna and is detected. http://www.tpub.com/content/fc/12404/css/12404_24.htm
For experimental evidence with gases and heat, Dave Springer has a post above where he describes the work of John Tyndall. http://en.wikipedia.org/wiki/John_Tyndall
A G Foster says:
March 10, 2011 at 1:23 pm
TB, how do you think the ocean is heated if not at the surface? Geothermal?
The ocean is heated by the solar shortwave radiation that penetrates tens of metres into it.
There are fuller replies to your other misconceptions on my blog where you posted the same comment.
http://tallbloke.wordpress.com/2011/03/03/tallbloke-back-radiation-oceans-and-energy-exchange/#comment-5571
Fred Souder says:
March 10, 2011 at 12:32 pm
“I would expect my students to state that the atmosphere (or any insulation layer), changes the rate at which energy is transmitted. You say that wearing a coat warms the body. I say putting on a coat slows the rate at which your body transfers thermal energy to the surroundings. (and yes, this is obviously conduction, not radiation)”
If I were in a physics class I’d say the coat slows down how fast your body loses heat. If I were at a soccer game on a cold day and my kid came off the field shivering I’d say put on this coat to warm yourself up.
“There is not really much difference mathematically. It is that statement that greenhouse gases HEAT the earth” by re-radiating IR that is not going to sit well with the thermodynamics vocabulary police.”
Agreed. But in the end the surface is warmer and it’s difficult to describe how the equilibrium temperature between the surface and the 3 Kelvin cosmic microwave background temperature rises such that the heat transfer rate through the additional insulation rises commensurately. Then you might have to explain that because it’s a dynamic system with a plethora of other things changing both more slowly and more rapidly than GHGs the equilibrium temperature is a moving target that is approached but never attained for very long but that when the system goes farther out of equilibrium the harder it tries to get back to equilibrium.
Arno Arrak says:
March 10, 2011 at 10:11 am
………….Miskolczi used this database to calculate how the global annual infrared optical thickness of the atmosphere varied throughout these years. And he found that the optical thickness of the atmosphere in the infrared where carbon dioxide absorbs remained constant for 61 years, with a value of 1.87. This tells us that the transparency of the atmosphere in the infrared did not change for 61 years despite constant addition of CO2 to the atmosphere through all this long stretch of time. Hence, the greenhouse absorption signature of the added carbon dioxide which we are told about simply isn’t there. This is an empirical finding, not something derived from theory, and it overrides any calculations from theory that do not agree with it. Theories that disagree must either be modified or discarded. I want to point out also that his work came out in 2009 and no one so far has attempted to present any peer-reviewed arguments against it. Miskolczi concludes: “It will be inferred that CO2 does not affect the climate through the greenhouse effect.”
On the one hand you have very educated people doing back of the envelope calculations using formulae that do not quite apply to the task with inputs to the formulae that are based on oversimplified assumptions, and willfully ignoring major aspects of the atmosphere.
On the other you have an empirical scientist going out and using actual atmospheric measures to invalidate the claims.
Stands to reason that the ‘back of the envelope incomplete formula disregarding major parts of the system’ will be the one people believe.
AGW looks more and more like the Phlogiston theory – that was supported by scientific consensus too.
commieBob says:
March 10, 2011 at 1:52 pm
“What I can also tell you as an expert is this: If I build and energize a transmitter antenna, I can still measure the signals on it that are caused bay adjacent transmitters. (The total voltage on the antenna will be the sum of the local transmitted signal and all the received signals.) In fact, a good example would be continuous wave radar. A signal leaves the transmitting antenna, strikes the sharp edge of an aircraft, re-radiates at the same frequency and, shifted by the doppler effect, returns to the sending antenna and is detected.”
I only worked on a conventional pulsed (weather) radar 35 years ago. CW Doppler can’t do ranging without adding in frequency modulation. What a nightmare sorting out the return signal would have been with old analog electronics!
That said, constructive and destructive interference happens in all situations and really muddies up the picture. As long as the frequencies are different you sort it out using analog filters or a DSP running FFTs on it. In fact new fangled cars have active noise reduction where road & engine noise in the passenger compartment are drastically reduced by destructive interference – there are listening devices that generate sound of equal amplitude but 180 degrees out of phase. The energy in the sound waves has to go somewhere so in the car I reckon that heats the air a little bit.
But this raises a puzzling question for me when it comes to electromagnetic waves in a vacuum. Suppose we have two emitters at exactly the same frequency but 180 degrees out of phase. The wavefronts meet and perfectly cancel out. Say they are separated by a two light-seconds and each starts emitting at exactly the same time. Theoretically the wavefronts will meet halfway and cancel each other out so the radiation from either will never reach the other. The puzzling part is where does the energy go?
I had an interesting email conversation with a couple of academics who’d published an article in SciAm about common misconceptions about the Big Bang. They were describing the expansion of the universe and how the 3000K temperature of the early universe (the point at which it became transparent to radiation) had fallen to 3K today due the universe expanding by a factor of 1000 since that time. They explained it in quantum terms using photons and how the individual photons lost energy over time. I objected and said that photons propogating through a vacuum are immortal and unchanging. If the photon lost energy, I asked, where does the lost energy go and what form does it take? I went on further to say the particle description was inappropropriate. They should have explained it using waves which is appropriate for EM propogating through a vacuum – the fabric of space simply expanded and stretched out the electromagnetic waves along with it. That satisfies the law of conservation of energy. They came back with some vague BS about gravitational energy and refused to answer any further questions. There may very well be a gravitational answer but unfortunately there’s currently no theory of quantum gravity. Why do people insist on making things so complicated when there are simple answers staring them in the face?
Well when I think about these issues, I ask myself; How can I explain this to those ten year old kids, on “Are You Smarter than a Fifth grader ?”
Which is not to suggest that visitors to WUWT are no smarter than fifth graders; simply that some visitors kn0w Physics, or Physical Chemistry, or Extra Terrestrial Biology, and some understand how to turn ordinary mud into China, or draw a bunch of chicken scratching line drawings, that illustrate some political fox pass of this week’s news cycle. So if you can explain anything to a ten year old, then probably most adults, regardless of their specific skill sets, can grasp what you are talking about, (about ANYTHING).
So does putting on the coat; say after you run in off the soccer field, warm you up, or doesn’t it ? Well of course it does. Assuming that you had breakfast before going out on the field, then your body is taking energy out of the chemical decomposition that is going on in your GI tract, so you are able to get out on that field and run after the guy with the ball. Or instead, you could perhaps climb up K2, instead of kicking the leather Bucky ball.
Now without the coat on, you are going to freeze your arse off; specially if you decide to climb K2. Because you are connected to the environment (atmosphere) by the largest organ in the human body; your skin; and it was specially designed for getting rid of EXCESS HEAT, both by way of conduction directly to the cold atmospheric molecules that touch your skin (which is at 98.6 deg F), and also by evaporation of H2O through the pores of your skin, removing something like 645 calories per gram of water lost (depending on the Temperature). So your skin is going to cool down, despite the fire that is raging within. Remember that “Calories” of food, are actually kilocalories, of energy (disguised so as to not shake up the weight conscious ladies.)
So climbing K2 in a speedo, is not too cool; you’ll freeze; same thing for coming off the soccer field in winter.
So you put on that jacket; maybe lined with Eider down, or some synthetic super insulator, or some breathable fabric, that lets air in and out, but provides a high thermal resistance to heat flow, either by conduction, or convection, and even by radiation (silver linings).
Now instead of your skin assuming -40C/F deg on K2, or just say 20 deg F off the soccer field, the internal heater (decaying hot dogs) continues to try to drive heat out through your flesh and skin, but now since you raised the external thermal resistance, the Temperature drop across your internal thermal reistance goes down, so your skin warms up. Most people when they get goose bumps, actually experience it on their skin, in the form of little protrusions, not unlike goose bumps, and it makes them shiver, which raises the thermal activity level to try and stop that goosebumpy skin from complaining it is cold out there. So it is your skin that tell you; hey it is bloody cold out here !
Do not expect your gall bladder to tell you when it is too cold; or for that matter when it is too hot; gall bladders just do not know hot and cold.
Guess how a lot of people who are dumb enough to try and climb K2, end up dying up there. No they don’t smash their limited brain capacity on a glacier 4,000 feet below them; nor do they freeze their arse off at -40C/F deg.
They COOK their insides (gall bladder included). That coat (which of course couldn’t possibly warm them), stops their skin from freezing or even goose bumping, so their skin cannot get rid of all the excess internal heat energy generated from the decay of hot dogs, and drinking too mcuh 5-hour energy.
So their internal organs, which are mostly lacking in nerve endings, and specially don’t have a goose bumper, or a sweat gland to tell them; hey its getting too damn hot in here, simply cook; the person dies from heat stress, because that coat that they need to stop their skin from freezing, also stops their body core, from dissipating the excess heat that is generated, specially from doing dumb things like climbing K2. (or playing soccer ).
So the key thing about the coat, warming, or a blanket on your bed for example, is that there is a heat (energy) soure INSIDE the thermal barrier.
So try putting an inflatable pin-up doll in your guest bed overnight along with a Thermometer to measure the Temperature.
Next night try putting a real pin-up doll in your guest bed overnight, along with the Thermometer to read the Temperature (record your results). For a real clincher test, join the pin up doll in your guest bed over night along with the Thermometer, to record the Tempertaure. NOTE Reread the above about how idiots who climb K2 die, before trying this last experiment.
Well of course it is the same with the atmosphere and GHG.
You see there IS a continuous source of HEAT inside that enveloping blanket of GHG’d atmosphere.
The atmosphere is to a large extent (but not completely) transparent to the incoming solar spectrum EM radiation that constantly arrives from the sun, and most of it goes right into the oceans to depths from a few cm down to hundreds of metres, and the rest gets absorbed in rocks, or plants, or Urban heat islands; and the bulk of that absorbed radiant energy is quickly converted into waste heat; pretty much the same as decaying hot dogs; and that waste heat source is now totally INSIDE the enveloping high thermal resistance atmospheric blanket, that surrounds the earth.
So no you can say the blanket is not cooking the earth; just like the down jacket does not cook the K2 climber; but the internal heat source most surely does; and it is the continuous input of solar spectrum EM radiation energy, that ultimately becomes extra heat to warm the planet; because the “blanket”, slowed down the rate of exit of that excess “HEAT.”
Now I’ll bet, that any ten year old kid can understand that.