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.
First I apologise if I am rehashing something that has already been discussed, I have been out in the field and haven’t had a chance to catch up on all the posts in this thread as yet, however this one did.
The equation cited does not, in my opinion, imply backradiation. If it did there would be terms for the emissivity of both objects as well as the surface area of both objects. There aren’t. What the equation does account for is the net difference in temperatures between the two objects or one object and its surroundings.
The net difference in temperature is obviously one of the factors which determines the flow of energy from the hot body to the colder one. If this were water we would call it the hydraulic head, if it were electricity it would be the voltage. Furthermore, if the surroundings or other object did heat the hotter object through “backradiation” wouldn’t the equation not only require terms for the surface area and emissivity of the colder object, but also require integration of the effect of the backradiation? In other words, to consider just the colder object, as it emits energy to the warmer object it cools, but it also warms from the energy input of the warmer object which causes its temperature to rise as a function of the difference between its absortivity and its emissivity. The increased temperature would then cause it to backradiate more energy etc… While at the same time the hotter object would be behaving in a like fashion. So you would need to integrate the energy inputs to each from each as a function of not only emissivity but absorbtivity as the net temperature difference tended towards 0 ( assuming it could in fact tend towards 0)?
Martin Lewitt:
I’m not sure what you mean by this. Are you trying to say that you don’t believe that the estimates of the forcing due to increases in CO2 is not around the value that nearly all scientists, including Roy Spencer and Richard Lindzen, accept it to be? And, if so, what evidence do you have to make the assertion? It seems to me that you are taking one rather slender piece of evidence (the Wentz paper, which is interesting but not completely definitive) and then extrapolating it like crazy to suggest all sorts of things that they haven’t suggested…and which don’t really seem in line with the empirical data that we have.
Smokey says:
How am I supposed to convince you of something that involves seriously weighing of scientific evidence when I cannot even get you to admit you have made in error when it is right in front of your face for everyone to see? I might as well convince a rock to do the tango! You are so far from a reasonable person able to objectively weigh scientific evidence that it is a fool’s errand to try to convince you of anything.
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 additional N2 to doubling it’s contribution to total mass of the atmosphere will absolutely increase the temperature at the surface, quite a bit. Mainly water vapor and carbon dioxide will continue to be mixed and will absorb, thermalize and re-excite as normal, the lapse rate in a deeper atmosphere will do the rest.
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 view of the lapse rate is right on the spot. At the one bar level temperatures will be very close to what we experience today.
jae says:
No…I also explained why the general claims (by people like Postma) that the lapse rate alone can explain the surface temperature are completely wrong.
That article just shows this as being true for Venus and Earth (assuming the data shown is accurate), which could very well be a coincidence, especially considering that the factoring in of the distance from the sun without taking the very different albedos into account seems rather unjustified. Can you show me the data for the other planetoids? Then maybe it would be worth thinking about why this might be the case.
However, it would almost certainly have no bearing whatsoever on whether there is the GHE or not, since nobody has successfully told me how one can propose the earth and Venus have the surface temperature they do without invoking the greenhouse effect or abandoning conservation of energy.
Tim Folkerts
Perhaps you could shed some light on the CO2 photon capture aftermath.
I did a calculation some time ago (and from memory) at atmospheric temperatures about 4% of CO2 molecules have the vibrational mode active.
This means that 96% are available to absorb.
So in the Earth surface up there will be a reasonable flux of 15um photons which can be absorbed to increase the active population.
The relaxation time in this mode is relatively long compared with the chance of losing the energy by collision to probably N2, O2 molecules.
This causes local heating and a local temperature rise.
The fact that the re-emission of a 15um photon requires a CO2 molecule to reach a high and statistically unlikely translational speed means that the following seems more probable;
1. Local temperature rise.
2.15um Emissions are less than absorptions as a energy consequence of 1.
3. Radiative emissions (when they happen) will favour the longer wavelength, lower energy, more probable and readily available H2O bands.
Some people have trouble with 2 as Kirchhoff no longer holds.
This is perhaps a needless concern as;
1. LTE no longer holds – a condition for Kirchhoff
2. Photon energy has been diverted along more probable paths.
Thats my take – any comments?
Joel Shore says:
May 11, 2011 at 10:50 am
“…since nobody has successfully told me how one can propose the earth and Venus have the surface temperature they do without invoking the greenhouse effect or abandoning conservation of energy.”
Pressure, Joel, pressure. There I have successfully told you. Whether you accept is a different story.
JAE, regarding some other comments made here, corrections are in order.
Dr Richard S. Lindzen states: “If one assumes all warming over the past century is due to anthropogenic greenhouse forcing, then the derived sensitivity of the climate to a doubling of CO2 is less than 1C.” [my bold]
None but the most radical alarmists claim that all of the warming over the past century is anthropogenic. Natural variability is always at work. Lindzen also points out:
“There is ample evidence that the Earth’s temperature as measured at the equator has remained within ± one degree centigrade for more than the past billion years. Those temperatures have not changed over the past century.” [source]
As one approaches the poles the Earth’s temperature begins to vary more and more from one region to another, and from summer to winter, and day to night. This is normal, and has always been so. Greenland has very wide temperature swings, while Egypt is extremely uniform from year to year, from decade to decade, and from century to century.
By selectively picking a base year for comparison, anyone can show either warming or cooling. Pick a warm base year such as 1997, and you can show cooling. Pick a cool base year like 2001, and you can show a warming trend. But the Earth’s overall temperature is extremely constant, and a slight variation of a few tenths of a degree is routine and natural. Temperatures have fluctuated much more over the Holocene, sometimes by tens of degrees. Dr Lindzen points out how ridiculous the current arm-waving is, over the extremely *mild* natural variability of only 0.7°C over the past century and a half:
“Future generations will wonder in bemused amazement that the early 21st century’s developed world went into hysterical panic over a globally averaged temperature increase of a few tenths of a degree, and, on the basis of gross exaggerations of highly uncertain computer projections combined into implausible chains of inference, proceeded to contemplate a roll-back of the industrial age.”
Finally, regarding the question of the climate’s sensitivity to a doubling of CO2, the UN/IPCC preposterously claims it is 3°C or more. That is absurd. If sensitivity were that high, the temperature would closely track rises in CO2. It doesn’t.
More knowledgeable scientists than the IPCC’s political appointees, including its head railroad engineer/bodice-ripper author, and its WWF advisors, provide much lower sensitivity numbers:
Dr Chylek estimates the sensitivity at 1.4°C; less than one-half the IPCC’s lowest number. Dr Schwartz gives the sensitivity as 1.1°C. Prof Lindzen puts the sensitivity at under 1°C. Dr Spencer puts the sensitivity at 0.46°C, based on observations. Drs Idso, fils & pere, put the sensitivity at 0.37°C. Dr Miskolczi puts the climate’s sensitivity to a doubling of CO2 at 0.0°C.
And there is still zero evidence of any global harm as the result of the increase in that beneficial trace gas. Conclusion: CO2, at current and projected concentrations, is both harmless and beneficial. And that answers the key AGW question: Is the rise in CO2 a problem? In fact, the increase in CO2 is, on balance, a net benefit to the biosphere.
Bryan says: May 11, 2011 at 11:50 am
“The fact that the re-emission of a 15um photon requires a CO2 molecule to reach a high and statistically unlikely translational speed …
I agree with things before this line, but I disagree with this. The average KE of a gas molecule is 3/2 kT. At 250 K this works out to ~ 5e-21 J. This is the same approximate energy as photons with a wavelength of 15 um. Thus it seems that the KE due to thermal motion should be able to excite the modes needed to emit the IR photons. (it would take considerably more time to work out actual probabilities of having various specific energy.)
Without thinking about it too deeply, it seems that peak in BB radiation would correlate quite naturally with the peak in kinetic energies of the molecules emitting the radiation. The fact that the name “Boltzmann” is associated with both ideas adds to the supposition that the two are related.
>>Joel Shore says: May 11, 2011 at 10:50 am
>>“…since nobody has successfully told me how one can propose the earth
>>and Venus have the surface temperature they do without invoking the
>>greenhouse effect or abandoning conservation of energy.”
>mkelly says: May 11, 2011 at 12:12 pm
>Pressure, Joel, pressure. There I have successfully told you. Whether you accept is a >different story.
IMHO, pressure (and lapse rate) can indeed explain temperature differences for various layers within the atmosphere.
But, mkelly, can you explain how pressure can determine either the location or the temperature of the “top of atmosphere”? What calculation will predict that a pressure of 1 bar should be ~ 15 C? Or that the temperature at 5 km altitude should be ~ -20 C?
Until you can calculate such a temperature from first principles, you have not successfully told anything — you have simply made a bold, unsupported assertion.
Tim Folkerts says:
May 11, 2011 at 12:59 pm
IMHO, pressure (and lapse rate) can indeed explain temperature differences for various layers within the atmosphere.
Joel asked about the surface. I answered him. I believe your statement above would be valid for the layer that is at or immediately adjacent to the surface.
You ask some excellent questions and I was hoping someone else, more qualified than I in this thread would answer.
My short answer:
1) I do not attribute ALL the approximately 33ºC to the “back radiation” aspect of the Atmospheric “greenhouse effect”. However, if all else was equal except for the affect of so-caled “greenhouse gases” I am confident the mean temperature at the Surface would be about that much cooler.
2) I actually used the Trenberth model to estimate how the 1366 Watts/m^2 should be distributed into the Earth System, which I specifically said included BOTH the Atmosphere and the Surface. As I wrote, we must first multiply by 0.25 to account for the fact that half the Earth System is not illuminated by the Sun at any given time, and the hemisphere that is illuminated is not all at right angles to the Solar rays. The second correction factor is for the albedo of the Earth System, and for that I used the 77 W/m^2 refleted by clouds plus the additional 30 W/m^2 reflected by the Surface. Doing that, and rounding (as I always do since I am an engineer :^) I got 240 Watts/m^2. Trenberth gets 67 Watts/m^2 absorbed by the Atmosphere plus 168Watts/m^2 absorbed by the Surface, which is total of 235 Watts/m^2, very close to my number for the mean rate of Solar energy input.
3) Please keep in mind that the main purpose of this Visualizing series is to help those of our fellow Skeptics who (IMHO) have gone over the line to an equal and opposite error from the Alarmists, expressing total Disbelief in the very scientific basis of the ill-named “greenhouse effect” of the Atmosphere and in the total Earth System. I am completely open to those who do further analysis that casts doubt on Trenberth’s exact allocation and attribution of the Earth System energy budget.
4) Perhaps the 33ºC attributed to the effect of the absorption and emission Atmosphere is too low or too high, but, again IMHO, that value is close enough for purposes of a “sanity check”. I’ll leave the exact value to the experts to argue about.
5) Once we scientifically-oriented Skeptics accept the reality of the Atmospheric “greenhouse effect” we are, IMHO, better positioned to question the much larger issues which are: a) HOW MUCH does CO2 contribute to that effect, b) HOW MUCH does human burning of fossil fuels and land use changes that refuce albedo affect warming, and, perhaps most important, c) Does the resultant enhanced CO2 level and higher mean temperature actually have a net benefit for humankind?
5) Once we scientifically-oriented Skeptics accept the reality of the Atmospheric “greenhouse effect” we are, IMHO, better positioned to question the much larger issues which are: a) HOW MUCH does CO2 contribute to that effect, b) HOW MUCH does human burning of fossil fuels and land use changes that refuce albedo affect warming, and, perhaps most important, c) Does the resultant enhanced CO2 level and higher mean temperature actually have a net benefit for humankind?
How about using real heat transfer equations to explain this rather than back radiation. There is no input for back radiation in any heat transfer equation so taking as if it does something cannot be shown via standard equations.
mkelly says:
Sorry…Saying the world “pressure” does not get you around having to satisfy conservation of energy. As I have explained, those sorts of things explain the slope of the line (how temperature varies with height…or at least a limit on how steeply temperature can fall with height) but it does not determine the constant “b” in the equation “y = m*x + b”.
The only way you can get around the requirement that the steady-state temperature be such that the energy in from the sun equals the energy out from terrestrial radiation is to have some other significant source of energy. For a gaseous planet undergoing gravitational collapse, that source of energy could be the conversion of gravitational potential energy to other forms of energy…but that is not happening here on the earth. Energy flow from the interior of the earth is negligible.
O H Dahlsveen, I have noticed (and I think replied) to your comments in my other threads and I very much appreciate them because they evidence a sincere, interested, well-educated person, probably oriented towards engineering and technology but not necessarily science, per se, who has done quite a bit of reading about the subject of climate science.
However, when I read your statement I highlighted above, “I too have been led to conclude that light, and indeed that all kinds of other radiation contains no heat”, I get a disconnect!
As you must know, all energy is fungible, and may be transformed from one form to another. The potential energy of water in an elevated lake that is rushing down a river does not contain “heat” per se, but, if we dam the river and divert the water pressure to run a hydroelectric generator, we can transform the river energy into electrical energy and then use that to power an electric heater. Or, we could use that hydro-generated electricity to power a pump and pump that water up into a water tower, turning the electrical energy back into potential energy.
Thus, any form of energy contains “heat” and “potential energy” and “electrical energy” and so on. As I am sure you know, that water in the elevated lake got there because the Sun heated Surface water till it evaporated and rose high in the Atmosphere and came down as rain. You also know the energy in fossil fuels came about due to the Solar energy absorbed by long-ago living trees and sea life that got compressed and stored underground as coal and oil and natural gas. Thus, the Sun is the source of virtually all the energy on Earth (except for the relatively small amount we get from the hot core and atomic radiation).
It ALL came in as “light” and was transformed into “heat” when absorbed by the Surface, into living things when photosynthesis transformed it into living things, and into potential energy when raised into the air by evaporation, and into wind and waves, and so on.
That is what I mean when I say the Solar “light” contains “heat” or “thermal” energy. It is just a shorthand way of saying that energy is fungible and may be transformed into other forms.
mkelly says:
This is utter and complete hogwash. The term “back radiation” is just a label given to radiation emitted by the atmosphere that happens to travel toward the earth. All actually calculations of the greenhouse effect are done with the standard heat transfer equations. If you don’t like that label, you can get rid of it and it will make not one iota of difference in what you get out of the equations.
If the environmental lapse rate is larger than the dry adiabatic lapse rate, it has a superadiabatic lapse rate, the air is absolutely unstable — a parcel of air will gain buoyancy as it rises both below and above the lifting condensation level or convective condensation level.
what happens to the black.body spectrum of water vapor as it condenses, joel and ira and other radiation freaks? say it loud, if you dare.
mkelly says: May 11, 2011 at 1:55 pm
How about using real heat transfer equations to explain this rather than back radiation. There is no input for back radiation in any heat transfer equation so taking as if it does something cannot be shown via standard equations.
1) “back radiation” is radiation
2) radiation is included in standard heat transfer equations
Therefore “back-radiation” is included in standard heat transfer equations
If you want more detail, here are some “standard equations” that specifically deal with thermal radiation from one object of arbitrary size, shape, emissivity, and temperature to a second object of arbitrary size, shape, emissivity, and temperature:
Standard dQ/Dt (scroll about 2/3 of the way down).
“The radiation heat transfer from one surface to another is simply equal to the radiation entering the first surface from the other [added: in other words “back radiation”], minus the radiation leaving the first surface.”
http://en.wikipedia.org/wiki/Thermal_radiation
Standard form factors to go with the above equation.
http://www.engr.uky.edu/rtl/Catalog/tablecon.html
Do you contend these are not standard equations?
IF SO, what would you use instead?
WOW! I just read the linked webpage and it seems fairly convincing. I hope Joel Shore and Dave Springer and others comment and either set me straight or agree that I and virtually everybody else on the Internet who claims the Atmospheric “greenhouse effect” contributes about 33º to the average temperature of the Surface. If the correct number is closer to a third of that amount, established science has got a lot of ‘splainin to do. (I am pretty sure there is an error in the new claims, but I cannot figure it out on my own.)
In my “sanity check” I assumed that the Earth Surface emissivity was sufficiently high that, for all intents and purposes, I could use the black body model in the Carleton spreadsheet to estimate the temperature of the Surface absent the Atmospheric “greenhouse effect”. That is how I got an effective surface temperature of about 255 K which would emit about 240 Watts/m^2 and balance the mean input of Solar energy which is about 240 Watts/m^2 at the Surface.
The linked website points out that the albedo of the Earth System is about 30% (the same value I used). Therefore, they say, since absorption and emission tend to be about the same for many materials, the emissivity of the Surface is bound to be less than 1.0, and may be around 0.7, if I read them correctly.
Using the lower emissivity, they get 278.6 K (where I and almost all others get around 255 K).
I have thought about where they are going wrong and have some ideas, but I will appreciate confirmation and correction by those who know more than I and the linked site:
1) The albedo of the Earth System is indeed around 30%, but most of that is due to clouds. The albedo considering only the Surface is closer to 12% which would make emissivity (if it is equal to absorptivity) about 0.88 rather than 0.7. That change would decrease their 278.6 K by some amount, but not down to 255 K.
2) The albedo of the Earth Surface is 12% based on absorptivity of visual and near-visual Solar radiation (what we have been calling short-wave). The radiative emission from the Surface, based on its black body temperature, is in the mid- and far IR (long-wave), so it is not clear to me that that emissivity at long-wave would be the same as the absorptivity at short-wave. For example, common glass has an absorptivity of close to 0.0 at short-wave but about 1.0 at long-wave. So, perhaps they are comparing apples and oranges and getting a fruit salad rather than the correct result.
PLEASE EXPERTS: Comment on this very interesting link and argument and either confirm my conclusions or modify them or (Horrors!) accept the new argument as basically correct. advTHANKSance
[UPDATE 6:20PM: I’ve just noticed that Tim Folkerts says: May 10, 2011 at 8:14 pm has confirmed my objection (2) with the example of white paint albedo reflecting 90% of visible short-wave light but absorbing 90% of long-wave IR radiation. Thus the linked website is wrong to assume that the 30% albedo of the Earth System in the short-wave region equates to only about a 0.7 emissivity in the long-wave. Using Tim’s example, white paint would have an albedo of 0.9 in the short-wave but a an emissivity of 0.9 in the long-wave (rather than only 0.1 as the linked website would conclude. THANKS TIM and I appologize for not including you in my list of experts above. You certainly qualify, as do others here in this thread. – Ira]
Ira,
I have just a minor quibble with your wording here:
“Thus, any form of energy contains “heat” and “potential energy” and “electrical energy” and so on. As I am sure you know, that water in the elevated lake got there because the Sun heated Surface water”
I would change “contains heat” to “contains thermal energy”. In traditional thermodynamics terminology, “heat” is always energy being transferred, not contained in an object. An object doesn’t “contain heat” any more than it “contains work”. In many ways, it would be better if the language was more parallel. For example:
dU = δQ – δW
would be “the change in internal energy is equal to the ‘heat done to the system’ minus the work done by the system.” It sounds really awkward in English, but that is the traditional sense of the words in thermodynamics.
It seems nitpicky to many, I’m sure. But when talking about laws of thermodynamics in discussions like this, much of the confusion comes from imprecise wording, like using “to heat” to as a synonym for “to warm”. Or using “heat” when one means “U” (as you did above).
I do approve of your later use “the Sun heated Surface water”. There was indeed a transfer of energy from sun to water due to a temperature difference. 🙂
And I think you hit the nail on the head with:
“5) Once we scientifically-oriented Skeptics accept the reality of the Atmospheric “greenhouse effect” we are, IMHO, better positioned to question the much larger issues which are: a) HOW MUCH does CO2 contribute to that effect, b) HOW MUCH does human burning of fossil fuels and land use changes that reduce albedo affect warming, and, perhaps most important, c) Does the resultant enhanced CO2 level and higher mean temperature actually have a net benefit for humankind?”
Soft-science students of climate woul benefit greatly from performing a mass-weighted vertical integration of the standard atmospheric temperature profile. It nearly matches the theoretical expectation for the average graybody temperature of the Earth. This illustrates two vital points that are often overlooked: 1) the LWIR escaping into space comes NOT directly from the surface (except in the “window” wavelengths) but from a diaphanous mass of gases, and 2) there is NO additional thermal energy produced by the “greenhouse effect.”
Air temperature is a nonconservative, intensive variable whose local value depends not only upon the radiative fluxes driven by thermalization of insolation, but upon upon the atmoshperic pressure, in accordance with Boyle’s law. Indeed, if a parcel of air is moved adiabatically from the surfaceto aloft, the temparature drops accordingly. The ~33K increase in temps that we enjoy is available only because we live at the bottom of the atmosphere. It is NOT the product of any additional thermalization, for that would require TOA LWIR planetary emissions of ~390w/m^2 to maintain a steady state. The fundamental flaw of the “radiative greenhouse” paradigm is the illusion that thermalization is a recursive process. In fact, it can occur only once. Like income, it cannot be reused.
Ira Glickstein, PhD says:
May 11, 2011 at 3:08 pm
PLEASE EXPERTS: Comment on this very interesting link and argument and either confirm my conclusions or modify
them or (Horrors!) accept the new argument as basically correct. advTHANKSance
—
Since you have never responded to all of the real aspects I have supplied to you to think about, I don’t know why I am responding. Since you said please I guess. Seems you keep asking for an expert, which I will continue to maintain I am not, whether the surface of the Earth is a black body to long wave radiation peaking at 10 µm, the answer is no. It is very close to a gray body if the spectrum is taken merely meters from the surface. The is a paper by some Swiss scientists on the current radiation flux in the Alps that gives a great multi-altitude spectrum of the upwelling LW radiation. Search it out.
Have you read Miskolczi’s papers yet as I suggested you do, by an atmospheric physicist, which explains plainly why this is so?
Also, that is a rather cheap shot when you said if someone did not respond they will accept all you believe as laid out in this thread. That make you look so incredibly shallow.
Tim Folkerts said….”I agree with things before this line, but I disagree with this. The average KE of a gas molecule is 3/2 kT. At 250 K this works out to ~ 5e-21 J.”
I agree with this
Tim Folkerts said….” This is the same approximate energy as photons with a wavelength of 15 um.”
I dont agree with this.
Energy of 15um photon is almost three times bigger.
Redo your calculations and work out probabilities from Maxwell Boltzmann Statistics
Tim
I left a redundant top line of your post – ignore that part.
I am trying to figure out why a doubling of atmospheric N2 would necessarily cause a dramatic increase in temperatures at the surface, as I have seen some propose. N2 has little to no insulative properties, right?
By ideal gas laws, PV=nRT. N2 is roughly 75% of the mass of the atmosphere. I know that n is moles, not mass, so let’s assume that we just increase N2 by “a whole helluva lot”, so that n(2) = 1.75n(1), in terms of the entire atmosphere.
So we release a whole helluva lot of N2 gas into the atmosphere and then, much later, sample a cubic meter at sea level. What would we find is different about this volume V(2) as compared to V(1)? Is there any possibility that T(2) will have increased insignificantly compared to T(1)?
Assuming n is the only change on the right side of the equation, sure. I can think of a pretty simple solution. P could increase by a factor of 1.75, and that’s it – equation balanced. We end up with a thicker atmosphere (a factor of 1.75 more moles per cubic meter), but the same surface temperature.
I don’t think that is exactly what would happen, but I don’t see why increasing the mass of an atmosphere guarantees a dramatic increase in surface temperatures. It does guarantee an increase in surface pressures, but that just means a thicker atmosphere, not necessarily a hotter atmosphere.