Guest Post by Ira Glickstein
Solar “light” radiation in = Earth “heat” radiation to Space out! That’s old news to those of us who understand all energy is fungible (may be converted to different forms of energy) and energy/mass is conserved (cannot be created nor destroyed).
My Visualizing series [Physical Analogy, Atmospheric Windows, Emission Spectra, and Molecules/Photons] has garnered almost 2000 comments, mostly positive. I’ve learned a lot from WUWT readers who know more than I do. However, some commenters seem to have been taken in by scientific-sounding objections to the basic science behind the Atmospheric “Greenhouse Effect”. Their objections seemed to add more heat than light to the discussion. This posting is designed to get back to basics and perhaps transform our heated arguments into more enlightened understanding :^)
As I’ve mentioned before, during my long career as a system engineer I’ve worked with many talented mathematical analysts who always provided precise results, mostly correct, but some precisely wrong, usually due to mistaken assumptions. I got into the habit of doing a “back of the envelope” calculation of my own as a “sanity check” on their results. If their results matched within reasonable limits, I accepted them. If not, I investigated further. In those days my analysis was really done using a slide rule and scrap paper, but I now use spreadsheets.
The graphic above is based on an excellent spreadsheet from http://serc.carleton.edu/files/introgeo/models/mathematical/examples/XLPlanck.xls. It uses Planck’s Law to calculate the black body radiation spectrum from the Sun, as observed at the top of the Earth’s Atmosphere. It also may be used to calculate the radiation spectrum from the Earth System (Atmosphere and Surface, see below for explanation) at any assumed temperature. (I will refer to this spreadsheet as “Carleton” in this posting.)
I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).
Since the mean temperature at the surface of the Earth is 288 Kelvins (+15ºC which is 59ºF), that leaves 33 Kelvins (33ºC which is 58ºF) to be accounted for. Guess how we acount for it?
The yellow curve (above left) shows that Solar radiation is in a tall, narrow “shortwave” range, from about 0.1μm (microns, or millionths of a meter) to about 4μm, which we call ultra-violet, visual, and near-infrared. The vertical axis is Intensity of the radiation, measured in Watts/m^2/μm, and the horizontal axis is Wavelength, measured in μm. If you divide the area under the yellow curve into vertical strips, and add up the total area, you get 240 Watts/m^2.
Since we humans sense the visual portion of this radiation as “light”, that is the name we give it, and that has led to the false assumption that it contains no “heat” (or “thermal”) energy.
The violet curve (above right) shows that, assuming a mean temperature of 255 K, Earth System radiation to Space is in a squat, wide “longwave” range, from about 5μm to beyond 40μm, which we call mid- and far-infrared. If you divide the area under the violet curve into vertical strips, and add up the total area, you get the same 240 Watts/m^2 as is under the yellow curve.
DETAILED EXPLANATION
The graph on the left shows the actual observed Solar radiation spectrum (in red) as measured at the top of the Atmosphere. It is superimposed on a black body model (in blue) showing very good correlation. Thus, while the Sun is not exactly a black body, it is OK to assume it is for this type of “sanity check” exercise.
If you calculate the area under the curve you get about 1366 Watts/m^2. That means that a square meter of perfect black body material, held perpendicular to the Sun, would absorb 1366 Watts.
However, the Earth is not a perfect black body, neither is it a flat surface perpendicular to the Sun! So, to plot the yellow curve at the top of this posting, I had to adjust that value accordingly. There are two adjustments:
- The Earth may be approximated as a sphere, with the Sun shining on only half of it at any given time. The adjustment factor for this correction is 0.25.
- The albedo (reflectiveness) of the Earth system, primarily clouds and light-colored areas on the Surface such as ice, causes some of the Solar radiation to be reflected back out to Space without contributing any energy to the Earth System. The adjustment factor for this correction is 0.7.
After applying these adjustments, the net Solar energy absorbed by the Earth System is 240 Watts/m^2.
The graph on the right shows the black body model for an Earth System at a mean temperature of 255 K, a temperature that results in the same 240 Watts/m^2 being emitted out to Space.
Of course, the Earth System is not a perfect black body, as shown by the graph in the upper panel of the illustration below, which plots actual observations from 20 km looking down. (Adapted from Grant Petty, A First Course in Atmospheric Radiation, Figure 8.2, http://www.sundogpublishing.com/AtmosRad/Excerpts/index.html.)
The actual measured radiation is the dark squiggly curve. Note that it jigs and jags up and down between the topmost dashed curve, which is the black body spectrum for a temperature of 270 K and a lower dashed curve which is the black body spectrum for 230 K. This data was taken over the Arctic, most likely during the daytime. The Petty book also has a graph looking down from over the Tropical Pacific which ranges from 300 K down to 210 K. Observations will vary by tens of degrees from day to night, summer to winter, and Tropical to Polar.
However, it is clear that my result, based on matching 240 Watts/m^2, is within a reasonable range of the true mean temperature of the Earth System as viewed from Space.
NOTE ABOUT THE ABOVE ILLUSTRATION
WUWT readers will notice some apparent inconsistencies in the graphs above. The top and bottom panels, from Petty, peak at 15μm to 20μm, while the purple, blue, and black curves in the middle panel, and the Earth System curves from the Carleton spreadsheet I used (see above) peak in the 9μm to 11μm range. Also, the Petty black body curves peak at a “Radiance” around 100 mW/m^2/sr cm^-1 while the black body curves from Carleton peak at an “Intensity” of around 14 W/m^2/μm. Furthermore, if you look closely at the Petty curves, the labels on the black body curves are mirror image! What is going on?
Well, I know some of the reasons, but not all. (I hope commenters who are more fluent in this than I am will confirm my explanations and provide more information about the differences between “Radiance” and “Intensity”.) I have Googled and Wikied the Internet and am still somewhat confused. Here is what I know:
- The horizontal axis in Petty’s plots are what he calls “Wavenumber”, increasing from left to right, which is the number of waves that fit into a cm (centimeter, one hundredth of a meter).
- This is proportional to the frequency of the radiation, and the frequency is the inverse of the wavelength. Thus, his plots are the mirror image of plots based on wavelength increasing from left to right.
- The spreadsheet I used, and my previous experience with visual, and near-, mid-, and far-IR as used in military systems, always uses wavelength increasing from left to right.
- So, when I constructed the above illustration, I reversed Petty’s curves, which explains why the labels on the black body curves are mirror image.
- Fortunately, Petty also included a wavelength legend, which I faithfully reproduced, in non-mirror image, at the top of each plot.
But, that still does not explain why the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet and other graphics on the Internet. I tried to reproduce Petty’s blackbody curves by multiplying the Carleton values by the wavelength (μm) and that did not move the peak to the right enough. So, I multiplied by the wavelength again (μm^2) and, voila, the peaks agreed! (I hope some WUWT reader will explain why the Petty graphs have this perverse effect. advTHANKSance!)
ANSWERING THE OBJECTIONS TO BASIC ATMOSPHERIC “GREENHOUSE EFFECT” SCIENCE
First of all, let me be clear where I am coming from. I’m a Lukewarmer-Skeptic who accepts that H2O, CO2 and other so-called “greenhouse gases” in the Atmosphere do cause the mean temperature of the Earth Surface and Atmosphere to be higher than they would be if everything was the same (Solar radiation, Earth System Albedo, …) but the Atmosphere was pure nitrogen. The main scientific question for me, is how much does the increase in human-caused CO2 and human-caused albedo reduction increase the mean temperature above what it would be with natural cycles and processes? My answer is “not much”, because perhaps 0.1ºC to 0.2ºC of the supposed 0.8ºC increase since 1880 is due to human activities. The rest is due to natural cycles and processes over which we humans have no control. The main public policy question for me, is how much should we (society) do about it? Again, my answer is “not much”, because the effect is small and a limited increase in temperatures and CO2 may turn out to have a net benefit.
So, my motivation for this Visualizing series is not to add to the Alarmist “the sky is falling” panic, but rather to help my fellow Skeptics avoid the natural temptation to fall into an “equal and opposite” falsehood, which some of those on my side, who I call “Disbelievers”, do when they fail to acknowledge the basic facts of the role of H2O and CO2 and other gases in helping to keep temperatures in a livable range.
Objection #1: Visual and near-visual radiation is merely “light” which lacks the “quality” or “oomph” to impart warmth to objects upon which it happens to fall.
Answer #1: A NASA webpage targeted at children is sometimes cited because they say the near-IR beam from a TV remote control is not warm to the touch. Of course, that is not because it is near-visual radiation, but rather because it is very low power. All energy is fungible, and can be changed from one form to another. Thus, the 240 Watts/m^2 of visible and near-visible Solar energy that reaches and is absorbed by the Earth System, has the effect of warming the Earth System exactly as much as an equal number of Watts/m^2 of “thermal” mid- and far-IR radiation.
Objection #2: The Atmosphere, which is cooler than the Earth Surface, cannot warm the Earth Surface.
Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.
Objection #3: The Atmospheric “Greenhouse Effect” is fictional. A glass greenhouse works mainly by preventing or reducing convection and the Atmosphere does not work that way at all.
Answer #3: I always try to put “scare quotes” around the word “greenhouse” unless referring to the glass variety because the term is misleading. Yes, a glass greenhouse works by restricting convection, and the fact that glass passes shortwave radiation and not longwave makes only a minor contribution. Thus, I agree it is unfortunate that the established term for the Atmospheric warming effect is a bit of a misnomer. However, we are stuck with it. But, enough of semantics. Notice that the Earth System mean temperature I had to use to provide 240 Watts/m^2 of radiation to Space to balance the input absorbed from by the Earth System from the Sun was 255 K. However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.
John Shade says:
May 11, 2011 at 3:07 am
“I am with the commenter who sees the atmosphere primarily as a coolant – taking heat out of the tropical surface, and helping it escape to the temperate/polar zones and to space.”
I try to make analogies to help people (myself included) understand things. In other commentary on WUWT I compared the poles to the radiator on an automobile. The equator is the engine. When we have a continent over a pole that’s like a blockage in the radiator limiting how well it can dump the heat (south pole today). When we have continents surrounding a pole with restricted openings that also, to a lesser degree, also limits the effectiveness of the radiator.
I think heat transport in the atmosphere is almost all vertical with evaporation, convection, and condensation being the major players there.
As far as heat capacity goes the ocean has roughly 1000 times as much as the atmosphere. So that relationship is akin to a dog and its tail. The tail doesn’t wag the dog of course. However if the dog is sleeping a tug on its tail can wake it up. The climate boffins I think are worried that anthropogenic CO2 is tugging the tail on a sleeping dog and believe it’s always best to let sleeping dogs lie. Or maybe it’s more like a tiger to them. 🙂
ferd berple says:
May 10, 2011 at 1:43 pm
Here is a thought experiment. If we hold everything else equal and double the amount of N2 in the atmosphere, because N2 does not participate in the back radiation, the surface equilibrium temperature should remain “unchanged” according to the radiative transfer model.
However, comparisons of earth and venus suggest that the temperature of the atmosphere will only remain unchanged at 1 atmosphere pressure. As we now have more atmosphere, the pressure at the surface will be increased as compared to present. Under the gravitational model, the surface temperature would be expected to increase as a result of doubling the N2 in the atmosphere.
Which of these two is correct? Adding N2 will not increase temperature, or adding N2 will increase temperature (holding all else equal)?
Adding N2 will not increase temperature since it will have no impact on the radiative emission to space.
The problem for me is the lapse rate. Doubling the N2 in the atmosphere will raise the height of the atmosphere. If the surface temperature does not increase, then the lapse rate dictates that since the atmosphere extends higher, that further up, past the present altitude it must get colder than it is at present. That seem nonsensical.
Your assumption of a continuation of the lapse rate is flawed, the tropopause is the region where the positive lapse rate of the troposphere changes to the negative lapse rate of the stratosphere (heated from above). I see no reason why addition of N2 would change that very much.
Someone hand Joel Shore a hanky.
Blogrule #1: Never demand an apology. It is pitiful and impotent.
Joel’s begging for an apology for his hurt feelings is simply his tactic to avoid answering the points I raised above. Here is the comment I made, which he still avoids answering:
What is untrue about that comment? Truth doesn’t require apologies. I regularly ask Joel Shore and others to produce verifiable evidence of global damage caused by the rise in CO2. But rather than address that point, Joel has repeatedly accused me of being ideological; I am not. He is deliberately avoiding the central question in the whole trumped-up “carbon” debate.
Joel Shore is wrong; I am only ‘biased’ against alarmist scientists feeding at the public trough, and who are trying to scare the public by spending a large part of their workdays wasting our tax money by writing long comments on blogs, instead of doing what they are paid to do.
I am sorry Mr Shore’s boss allows him to misappropriate public funds in this manner. Gavin Scmidt and James Hansen have set a terrible precedent in this regard. There is no doubt that if someone like John Christy began blogging throughout the workday like Joel Shore does, the screams of outrage from the hypocritical alarmists would be deafening.
If it is ‘hounding’ Joel Shore for any putative real world, testable evidence showing that rising CO2 – a tiny trace gas – causes measurable harm to the planet, he needs to stop prevaricating and produce it, or take being ‘hounded’ until he produces evidence, or admits that there is none. Calling others ideologically biased is just Joel Shore’s way of avoiding the question.
I agree with much of what Dave Springer says, but I can’t agree with some of his conclusions about optical depth.
Dave Springer says: May 10, 2011 at 8:55 am
> IIRC the spectrum follows a 270K blackbody curve where the atmosphere is
>IR transparent i.e. it “sees” the temperature of the ocean surface but in the
>15um region it drops down to follow a 250K blackbody curve.
From the graph at the top, the temperatures are closer to 265 and 225 K. That gives a difference of 40 K, not 20 K. Not a huge change, but significant.
> Going by dry adiabatic lapse rate (arctic air is fairly dry) of 1K per 100 meters
I would love a little more info here. The key is the relative humidity. Certainly cold arctic air has a very low absolute humidity, but is the relative humidity low?
The observed environmental lapse rate is closer to 0.65 K/100 m (speaking in general — I don’t know details specifically for the arctic), and the temperature difference is closer 40 which would put the altitude closer to 40K / (0.65 K/100m) = 6000 m. So we have a very large uncertainty in the altitude here.
>the IR sensor is “seeing” the air temperature in the 15um range at a height of
>2000 meters. I’m not an optics expert but I believe that altitude represents
>the optical depth of the atmosphere at 15um.
I would not define the optical depth the way you seem to be doing.
In other words, of a region of the atmosphere has an optical depth of 1, then 70% of the incoming light has been blocked. (Or equivalently, 70% of the light we see would come from that region). If most of the light can pass from 2000 m up out of the atmosphere (passing thru well over 1/2 the total mass of the atmosphere) then, then most of the light passing up from the surface could presumable ALSO pass thru well over half the atmosphere. I.e. the optical depth looking up would be well over 2000 m. However as the altitude gets higher, the optical depth gets smaller.
There is, however, another issue here. All we really know is that the photons come from a region where the temperature is about 220 K. First, I put the altitude closer to 6000 m, not 2000 m. But the temperature of 220 K is pretty close to the temperature of the tropopause, which typically extends for several km upward. The photons could come from pretty much anywhere over a range of several km of altitude and still appear to be ~ 220 K. The photons might be traveling a relatively short distance — for example, from space down to just 15 km altitude. So a very tiny part of the atmosphere could stop most of the photons — so a relatively short distance near the surface could stop the photons. We just can’t know very accurately using this method.
(If someone has more details about the temperature and altitude range of the arctic tropopause, I would love to hear it.)
>In any case there is no way in hell the extinction depth is in tens of meters.
After all that, the simper way to determine optical depth near the surface is to look UP. From the graph at the top, the “temperature” of the 15 um IR radiation when looking up is practically the same as the temperature of the surface (certainly no more than 5 K different). The photons must come from no more than 500 m up (using the lapse rate argument), so the optical depth near the surface is well below 500 m.
Whether the optical depth for 15 um photons near the surface is 3 m or 30 m or 300 m, I can’t tell from just this. But there is no way in hell it could be 2000 m.
Dave Springer says:
May 11, 2011 at 5:06 am
JAE says:
May 10, 2011 at 8:56 am
“Huh? I hope you are not saying that everything radiates in the IR frequencies. To repeat a comment I made above somewhere:”
Solids, liquids, and dense gases emit continous blackbody spectra characteristic of their temperature. We were talking about atmospheric nitrogen which is a dense gas and its temperature, at least in the lower and middle atmosphere, puts its blackbody emission spectrum in the far infrared.
As far as blackbody radiation is concerned N2 is not a dense gas in the earth’s atmosphere and doesn’t emit as a blackbody. If it were a blackbody it would emit in the far IR.
Heh. Come on, folks, I am not supporting the teacher’s different methods of calculatig the GHE; I only pointed it out. Don’t shoot the messenger. I have a strong hunch that NONE of these radiative calcs mean anything, anyway, because any of the effects are instantly overwhelmed by other phenomenon, such as convection and water evaporation/condensation.
In that vein, I’m still really interested why there is so much silence surrounding this (and other related expositions):
http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html
Martin Lewitt says:
May 10, 2011 at 8:28 pm
I’ll need to think about that some more. Shiva operates in the very near infrared (1um) which correlates to a temperature in the thousands of degrees. Moreover, it focuses 20 beams coming from different directions on its target and achieves heating to millions of degrees through compression (shock waves).
Perhaps a more familiar technology is the microwave oven. Exceedingly cold matter emits blackbody spectra in the microwave region of the spectrum. The cosmic microwave background radiation is 3K for instance. That then raises the question of how a microwave oven can heat up a cup of coffee. It isn’t accomplished by microwave photons per se but rather by phase variation which causes dialectric molecules (primarily water in this case) to constantly re-orient with the magnetic phase. Friction then does the actual heating because microwave radiation in and of itself can’t raise the temperature of anything that isn’t within a couple degrees of absolute zero.
jae says:
May 11, 2011 at 6:30 am
Heh. Come on, folks, I am not supporting the teacher’s different methods of calculatig the GHE; I only pointed it out. Don’t shoot the messenger.
You asked my reasons for saying that his method was wrong, I gave them, that’s all.
Phil. says:
May 11, 2011 at 6:23 am
“As far as blackbody radiation is concerned N2 is not a dense gas in the earth’s atmosphere and doesn’t emit as a blackbody. If it were a blackbody it would emit in the far IR.”
A gas at 1 bar isn’t dense? In general a gas is considered dense if the molecules are close enough to each other to conduct heat. Nitrogen at 1 bar will certainly conduct heat. Try again.
Joel quoted Richard Lindzen:
“The dominant role of cumulus convection in the tropics requires that temperature approximately follow what is called a moist adiabatic profile. This requires that warming in the tropical upper troposphere be 2-3 times greater than at the surface. Indeed, all models do show this, but the data doesn’t and this means that something is wrong with the data.”
I’m sorry Joel, but you need to get your sarcasm detector seen too. Lindzen’s tongue was so far in his cheek it should have been clearly visible to everyone from at least a mile away.
Smokey says:
No…It is you who are desperately trying to avoid the fact that you have spent months attacking me (including in this thread) for a statement that Richard Lindzen makes in even more extreme terms than I do. That you are unwilling to address this when called out on it shows a completely lack of ethics and personal accountability on your part.
As for what you ask: I have explained this stuff many times. It is also explained in many reports by reputable scientific organizations. I can’t convince someone who is unconvinceable.
Joel Shore
I said…… “No wonder an educated person like Professor Gerlach “blew his top” when he found your paper full of elementary mistakes.”……
You said …”While any actual physicists reading our paper may say we should have been a little more precise in our usage of the word “heat” in a few places”…
You have a particularly snide way of phrasing your replies.
Are you here implying that Professor Gerlach is not a physicist?
That a Professor from a German University is an impostor perhaps!
That’s very impertinent from someone who finds coping with reading a big problem.
For someone who is a bit “rusty” in the thermodynamics department yet was happy to add his name to a collection of elementary gaffs.
Its too late now to withdraw you name from the Halpern et al embarrassment.
I could recommend some useful textbooks to bring you back up to speed with Physics.
jae says:
Silence?
http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655676
http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655686
http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655880 (last paragraph)
David Springer, you need to slow down a little!
“A gas at 1 bar isn’t dense? In general a gas is considered dense if the molecules are close enough to each other to conduct heat. “
No. A gas is considered “optically dense” if it absorbs most of the photons at that wavelength.
N2 does not absorb much IR at any wavelength through earth’s entire atmosphere, so the entire depth of N2 in earth’s atmosphere is not considered “optically dense”. You have already been corrected once on this. You might consider doing a little research before being so sure of your own knowledge.
In another post David discusses microwaves.
“Perhaps a more familiar technology is the microwave oven. Exceedingly cold matter emits blackbody spectra in the microwave region of the spectrum. The cosmic microwave background radiation is 3K for instance. That then raises the question of how a microwave oven can heat up a cup of coffee. It isn’t accomplished by microwave photons per se ….”
Once again, this is not right. It IS true that THERMAL microwaves would come from objects at a few Kelvins and that they could not warm an object above that same temperature. But microwave ovens don’t produce their microwaves by heating the walls of the oven to a few K. They use devices called “klystrons” or “cavity magnetron”. These devises are quite capable of producing very intense beams of microwave photons — several hundred watts of microwaves from a few square inches. It is exactly those microwave photons that heat the food.
Let me, as a “thought experiment”, slow the radiation circuits down to a speed where we can see it happening (At least in “our mind’s eye”). And let’s also think of things as single units i.e. radiation = 1 ray, energy =1 unit (Watt) & CO2 = 1 molecule etc. Furthermore lets do what the scientists do on occasions like this; ignore everything other than radiation, i.e. adiabatic warming /cooling, Conduction, convection and advection etcetera. (after all the atmosphere is not needed for radiation.)
Short Wave Radiation (SWR) comes in from space and warms the surface which begins to radiate Long Wave Radiation (LWR) back towards space, say at a constant energy rate of 1 W/m². Having sent this 1 W/m² away the surface must cool down respectively. CO2 intercepts and absorbs this energy and now CO2 warms up respectively and therefore begins to increase it’s radiation, – half of which (estimates may vary) returns to the surface and is now near to the end of it’s first circuit.
Meanwhile a ray of SWR has been absorbed by the surface and the 1 W/m² that LWR took away earlier has been replenished, so now that one part, however small, of LWR returns for absorption it must mean that the surface will warm a little bit extra. (That is if this theory is correct)
However what happens on the dark side of the Earth (or what we call at night)?
Where the Sun does not shine and SWR therefore does not replenish the energy which LWR takes away, – there can be only cooling.
–
To find out if the “cooling capacity” is great enough to bring the temperature right back to what it was the previous morning (i.e. heating starts again at “The zero line”) or if a part of the extra heating remains in the system and the heating starts again at a slightly higher temperature than yesterday it seems logical to me to bring back the Atmosphere (especially the Troposphere) and all it’s air-movements, pressures and lapse rates etcetera. And we must not forget the “Atmospheric Window”
Do scientists have enough data to anything bar guessing?
Dave Springer says:
We are not talking about conduction but radiative absorption. For N2, radiative absorption does not occur for the isolated molecule and can only occur because of collisional processes between molecules. Here is a paper discussing the measurement of such absorption lines at pressures of 0 to 10 atmospheres: http://www.opticsinfobase.org/view_article.cfm?gotourl=http%3A%2F%2Fwww.opticsinfobase.org%2FDirectPDFAccess%2FBC73981D-EC8E-6BE9-5F66346A49A16C1A_60399.pdf%3Fda%3D1%26id%3D60399%26seq%3D0%26mobile%3Dno&org=Rochester%20Institute%20of%20Technology
Note that even for the strongest absorption line, the measurements rely on obtaining ultra-high purities of nitrogen because any small contamination by CO or CO2 overwhelms the measurement:
So, apparently only 1 part per million of CO2 was, along with the CO, enough to render the measurements useless. Imagine what 380 parts per million does! Furthermore, as this graph shows http://www.learner.org/courses/envsci/visual/img_med/electromagnetic_spectrum.jpg , the absorption line of N2 that they are talking about, which is at 4.3 um, would already place it quite far out in the wings of the terrestrial radiation spectrum.
Supplement to my previous comment:
Oh yes, – “The Atmospheric window” – last time I looked at how it was once presented by “Climate Scientists” it showed that out of the 390 W/m² of long wave Infra Red (IR) radiation emitted by the surface 40 W/m² was let through that window “scott free”.
Well, according to my rough calculations that is just around 10.25 – (or 10 %) – How can that be correct? – Yes, it may be correct in so far as they can say that; “around 10% of the wavebands emitted by IR radiation are made up of wave-lengths that cannot be absorbed by “Greenhouse Gases” (GHGs), but that cannot possibly mean that 0.04%, in the case of CO2 concentration but certainly less than 10% of the Atmosphere as a total has got what must be a “supernatural” ability to stop LWR. After all, according to scientific measurements the bulk of the Atmosphere, which is transparent to IR radiation, is made up from; Nitrogen (78%), Oxygen (21%) and Argon (0.9%) and the rest = 0.1% is made up from “trace gases” of which some are said to be “GHGs”. – Water Vapour (WV) does not mix evenly in with the other gases and atmospheric WV content varies from location to location but I believe it is estimated to be around 4 to 5%. – Of course the most dense WV concentration must be in the lower half of the Troposphere which may give the impression (by looking at Hydrometers) that is far more WV than a miserly 4-5%. However I doubt very much that the WV concentration near the surface amounts to more than say 40 – 50%. But of course I have got no official data as proof – and I could be wrong.
So, as far as I can see, the case must be that whatever percentage of LWR passes through the “Atmospheric window” must be added to whatever passes through the “Transparent” part of the Atmosphere.
Joel Shore:
You say: “Silence?”
Yeah, silence! Have you even read the article at the link? I don’t see how any of your links relate to it; all they do is repeat again the radiatitive GHE theory. What I want to know is why ALL planetoids with atmospheres, regardless of the gases (and solids) in those atmospheres, have the same temperature at 1000 bar, when distance to the sun is factored in. Are the articles wrong? If not, that simply cannot be due to any GHE. Just curious, man….
Bryan says:
That is very generous, but I think you had better look through the latest edition of such textbooks to make sure they pass through your “political correctness” filter. The last time that you quoted a physics textbook as a source, it turned out that the latest edition of that textbook had a whole page on the greenhouse effect and global warming (see http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/#comment-644823 ). To put it mildly, it did not provide any support whatsoever for your views on the subject!
Several people are having trouble understanding the adiabatic lapse rates.
Dry adiabatic lapse rate (DALR) means the change in temperature with a change in pressure (altitude) when the contained water vapor is NOT changing state.
The Moist adiabatic lapse rate (MALR) (aka the Saturated adiabatic lapse rate – SALR) means the same thing except that the water vapor is condensing to form clouds.
The difference has nothing to do with the amount of water in the atmosphere, but only with whether or not the water vapor is changing state.
Bryan says:
“You have a particularly snide way of phrasing your replies.”
I would put it: whiny.
Instead of misdirecting the conversation to Prof Lindzen, I’m still waiting for an answer based on the scientific method to the question: where is the putative global damage caused by CO2? In fact, if there was such evidence, it would have been trumpeted 24/7/365 by the CAGW crowd.
Mr Shore, please provide testable, measurable, reproducible and convincing evidence demonstrating verifiable global harm specifically due to the rise in CO2 – you know: per the scientific method. Claiming that “I have explained this stuff many times” is a copout. We need solid evidence, because $trillions more in wasted tax dollars are at risk, based not on empirical evidence, but on GIGO models invented by people who can’t even get the definition of heat right.
Alternatively, Joel Shore can be a man-up and admit that there is no such evidence, and then the whole baseless CAGW conjecture becomes clear to everyone: the purpose of the scare is to provide a free ride on the grant gravy train for these climate charlatans at taxpayer expense.
R Stevenson,
I would very much like a link to that spreadsheet. My attempts to locate Hottel’s method keep hitting paywalls.
Using 1.0% water vapor, my program computes 282.68 W/m2 .. close to your value.
For power absorbed, my program incorrectly assumes 1 km at constant temperature and pressure. However, it also plots the distance to absorb x% of the available radiation. As long as those values are under 100 meters, I feel that assuming constant temperature and pressure causes only minor errors.
At any rate, I would like to investigate WHY the programs are giving different values to absorb the radiation.
David Springer must won’t let go of this nonsense:
“Solids, liquids, and dense gases emit continous blackbody spectra characteristic of their temperature. We were talking about atmospheric nitrogen which is a dense gas and its temperature, at least in the lower and middle atmosphere, puts its blackbody emission spectrum in the far infrared.
What part of that don’t you understand?’
I would have ignored this, except for the snide remark.
Come on, David, I gave you a link to a very clear textbook exposition on the interaction of IR with gases, which states unequivocally that N2 is not “IR active,” and it provides specific detailed reasons why this is true (no dipole moment). Density probably has nothing to do with it, unless it is so dense it is a liquid (and that occurs at about 1600 psi at 25 C).
I can see how we can debate something as complicated as the existence or amount of the GHE effect, but we should not be debating such basic, easily demonstrable, and widely accepted facts like the lack of signficant emissions of IR from N2.
Smokey says:
“Instead of misdirecting the conversation to Prof Lindzen, I’m still waiting for an answer based on the scientific method to the question: where is the putative global damage caused by CO2? In fact, if there was such evidence, it would have been trumpeted 24/7/365 by the CAGW crowd.”
Come on, Smokey, you must not be listening: IPCC and the warmistas keep yelling to us that there are “multiple lines of evidence” for CAGW. That’s Orwell-speak for “we don’t have any direct clear evidence, but, considering all things that MIGHT happen…(whatever)…” When dealing with the “progressives,” you gotta learn a new vocabulary. Such phrases as “Kinetic Military Action” (war); “Overseas Contingency Operations” (war on terror); “man-caused disaster” (terrorist act), “climate disruption” (man destroying the world), “we must be civil” (the Conservative bastards must be civil). Ad nauseum.
Problem with the CAGWers is that they can’t seem to locate one single clear line of evidence. An even bigger problem for them is that the “multiple lines of evidence” are actually demonstrating rather clearly that they are wrong, I’m sure to their inner horror (it’s a travesty…). Nothing seems to be working out for them, since the sea level is leveling off, the glaciers are not really melting everywhere, the temperature is not going up for the last 10-15 years, no “hot spot,” as predicted by models, etc., etc.
On the other hand, I guess if you allow yourself to blame cold on warming, storms and any sort of malady on “climate change,” then you can allow yourself to say that, “multiple lines of evidence” show that we are killing ourselves through some sort of unclearly defined mechanism, which we will just call CAGW. LOL.
It all just Brings tears to my eyes.
But they are gonna fix the problem by setting up special forums and getting the liberal MSM to explain their (false) science better, so we morons can understand their brilliance. Or something like that.
O H Dahlsveen says: May 11, 2011 at 8:45 am
“but that cannot possibly mean that 0.04%, in the case of CO2 concentration but certainly less than 10% of the Atmosphere as a total has got what must be a “supernatural” ability to stop LWR. ”
hmmmm … lets see try an order of magnitude estimation …
A “typical” IR photon is ~ 10 um, or 2E-20 J
The ground emits ~ 400 W/m^2, which implies around 400/ (2e-22) = 2E22 photons per second being emitted.
Air applies 100,000 N/m^2 of pressure, so a square meter column of contains about 10,000 kg of air. CO2 is ~ 400 ppm by volume (and slight more than 400 ppm by mass), so there are about 10,000 kg x (400/1,000,000) = 4 kg =4000 g of CO2 in each square meter column of atmosphere.
At 44 g/mole, that would be 100 moles of CO2, or ~ 6E25 molecules of CO2 per square meter.
That makes 6E25 / 2E22 = more than 1,000 CO2 molecules for every IR photon passing by. CO2 only absorbs ~ 1/10 of the possible IR wavelengths, so that’s ~ 10,000 CO2 molecules per absorbable photon passing by each second. And the CO2 molecule only “holds” that energy for a faction of a second, so a given molecule could in principle absorb many photons every second.
Each CO2 molecule absorbing 1 out of 10,000 photons passing doesn’t seem “supernatural” to me ….
(I did this pretty quickly, so people are welcome to find flaws. I would not be surprised if my numbers are off by 10x or possibly even 100x, but that still doesn’t cahnge the fact that there are plenty of CO2 molecules around to absorb photons even at an “insignificant 0.04%” concentration.)