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.
“Anyone who supports Trenberth’s diagram and considers that the back radiation theory is correct should be asked to explain why we are not utilising this fanastic energy source. After all, if it existed it should be the main focus of green energy projects.”
“- If back radiation is so powerful and it is a heat source, why don’t we use it?”
Good questions. Although what about this free energy oven. Does it qualify?
http://www.slayingtheskydragon.com/en/blog/111-a-pictures-worth-a-1000-words#comments
Joel Shore says:
You clearly do not know the basis for the Second Law, which is statistical physics.
The basis for the second law (and the first) is careful and accurate measurement, mostly paid for by wicked 19th century capitalists, who wanted to know how to get more from their steam engines, and also how to avoid being sold snake oil.
There is no need for a statistical ‘fudge’ unless you wish to introduce special pleading for “IPCC Approved” photons.
If I were to dump a billion microscopic ice particles into a tank of hot water (di-hydrogen monoxide), your statistical approach would insist that a few of them may have warmed the water (hydrogen hydroxide), especially using a computer model.
I use spreadsheets to calculate Earth IR emissions and absorption by CO2. Using mean beam lengths and CO2 partial pressures, I use PcL charts by Hoyt C hottel to obtain the emissivity of CO2 in the atmosphere. For current CO2 partial pressure or concentration I calculate a distance of 3600 m of traverse through the earth’s atmosphere by ‘IR radiation’ before the absorbable IR is ‘filtered out’ and reduced to zero. I then calculate that by doubling the atmospheric CO2 to 700ppm or so the ‘absorption distance’ is reduced to 2000 m but with no additional IR absorption by CO2. Therefore I conclude no global warming from increasing atmospheric CO2.
Ira,
“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”
Maybe I could help, or perhaps just confuse you further.
If you plot the radiant energy against wavelength for blackbody around 287K, then you get a peak intensity at about 10 micron. The graphs reproduced from Petty show the peak intensity at about 18 micron. This apparent anomaly arises because the vertical axes of the 2 graphs use different units.
The Plank distributions plotted against wavelength (the way I like to see them) have a vertical scale of ‘Watts per Square metre PER MICRON’. The graphs shown by Petty are plotted against wavenumber (not wavelength) and the vertical axes have units of ‘Watts per Square Metre per Steradian PER WAVENUMBER’.Note: graphs plotted in these different ways are NOT simply mirror images of each other. [Forget the Steradian bit, this is just a multiplying factor which doesn’t effect the peak or shape]
Say that the peak radiation on the wavelength plot occurs at 10.5 micron. Because the energy is plotted per micron it represents the total energy between 10 micron and 11 micron. In terms of wavenumber, 10 to 11 micron is equivalent to a range of 909 to 1000 per cm. So the energy for the 10.5 micron peak has to be dispersed over 91 different wavenumbers on the wavenumber plot. If we consider another point on the wavelength plot, say between 18 and 19 microns, this is represented by a wavenumber range from 526 to 555. So the energy around 18 microns has only to be shared between 29 wavenumbers whilst the energy around 10 microns has to be shared over 91 wavenumbers. As a consequence, the peak intensity no longer occurs at 10 micron, it now occurs at 18 micron. That is why the two methods of plotting give different peak intensities.
Did that help, or just confuse everyone further?
Boris Gimbarzevsky,
Thanks for sharing your thermometer readings It would be interesting to know if on clear nights, sky readings taken at the same ambient temperature but different absolute humidities varied due to increased water vapor greenhouse effect.
Dishman says:
That may be your point, but nothing you say backs it up. You say “the thermal energy will be reduced by 1/5th of the energy required to lift molecules by that delta-h.”
A quick calculation shows that the average molecular speed, assuming constant g, is sufficient to lift a molecule about 11 km. The mean free path of a molecule in air is about 65 nanometres. Dividing these, we see the energy needed to lift the molecule by your delta-h is about 1.7*10^-11 of its total energy. In other words, diddly squat. Gravitational effects make no difference to the result that could possibly be measured.
Charles Nelson said: Gotcha Henry! He thinks CO2 is like a toxin in the atmosphere! An amazing toxin which is NOT TOXIC at 380ppm but suddenly BECOMES TOXIC at 400ppm. There in a single post is the fuzzy thinking of the Warmists! Fantastic. Textbook example of ‘Carbonaphobia’!
I don’t read him that way. I think he’s making the point that the tiny size of the fraction, by itself, doesn’t tell us anything. If he’s making the point that you think he’s making, he’s doing a poor job of it.
Ira,
you do not consider conduction and convection, nor non-GHG absorbers/reflectors of radiation nor spatial issues. Simply saying that the difference in temperature between a perfect blackbody and the incoming radiation is due to “greenhouse” effects is disingenious. You can convince yourself by a thought experiment: would the atmosphere really be at zero degree Kelvin were it not for GHGs? I’d say obviously not and hence your simple difference doesn’t apply.
best regards,
Martin Mason said – If back radiation is so powerful and it is a heat source, why don’t we use it?
Because the earth’s surface is already at the temperature to which the back-radiation has raised it.
Allan M says:
May 9, 2011 at 2:04 am
“If I were to dump a billion microscopic ice particles into a tank of hot water (di-hydrogen monoxide), your statistical approach would insist that a few of them may have warmed the water (hydrogen hydroxide), especially using a computer model.”
Some actually will warm the water. Things get more than passing strange when you leave to domain of bulk matter and look at what’s going on with individual atoms and molecules. Some of the water molecules in your tank of hot water are colder than ice and some of the water molecules in the ice are more than boiling hot. Those molecules are few and far between so for most practical considerations they are ignored. Temperature is an artifact of motion. In bulk matter where the molecules are all bumping into each other chaotically some small fraction will happen to be moving very fast and some very slow. A good analogy would be what happens to a set of billiard balls after the break. Initially all the motion is in the cue ball but when it hits the rack the energy of its motion is distributed amongst all the balls but not equally. The cue ball could stop dead in its track only to get whacked back into motion by one of the other balls a moment later or if it’s a glancing blow the cue ball might remain the fastest mover for a few moments. But because energy can be neither created nor destroyed the total kinetic energy in all the balls on the table will always equal the initial energy in the cue ball. Temperature is analogous to the total kinetic energy of all the billiard balls.
Robert Stevenson says:
May 9, 2011 at 2:16 am
“Therefore I conclude no global warming from increasing atmospheric CO2.”
This was disproven by expermental physicist John Tyndall over 150 years ago. Back radiation from gases that absorb IR is quite real. Suggest you read the original work here (it’s free):
http://www.archive.org/details/heatconsideredas00tyndrich
We do use it all the time. One example is the smudge pots used to limit freeze damage in orange and avocado groves. The heat of the smoke from the smudge pots is inconsequential. The effect comes from the black carbon particles in the smoke which absorb upwelling IR from the ground and send a portion of it downwards and sideways. It works as an insulator. This is the same mechanism by which CO2 molecules insulate. Anyone who contests this is simply wrong and needs to take an introductory class in physics.
“The water will rise above the ‘V’ just enough so that the rate of input is equal to the rate of overflow. Let us now represent the back radiation by blocking some of the ‘V’ which represents the reduced outflow of radiation from the earths surface. ”
The problem is that Mr “T” says the energy from the sun is 184 watts and the amount being back radiated is 333 w. In other words, you are blocking more water than the maximum flow of the outlet, which is physically impossible in you analogy.
You’d like more back radiation. You’re getting quite a bit of it now. The temperature of the cosmos is 3K and if it weren’t for back radation when you point your IR thermometer at the sky on a clear night you’d get a reading of -454F.
Back radiation doesn’t come from just CO2. When a CO2 molecule absorbs upwelling IR it thermalizes surrounding non-GHG molecules via kinetic transfer. A common fallacy is that CO2 molecules emit a photon of the same frequency as that absorbed. That’s true in a very thin gas but in a cold dense gas the excited CO2 molecule almost instantly bumps into a neighboring molecule (likely a nitrogen molecule in the case of the earth’s atmosphere) and transfers some of its newly acquired energy to the neighbor.
Ira Glickstein, PhD says:
May 8, 2011 at 4:26 pm
“If we then placed that planet in an Earth-like orbit around the Sun”
Thanks Ira, what I read you to be saying is that if the atmosphere was pure N2O2, without any H2O, that it would have no effect upon the surface temperature of the earth. That since N2O2 do not absorb longwave radiation, then they cannot keep the earth warm.
I’m saying that this is simply wrong. I’m saying that an atmosphere of pure N2O2 would change the surface temperature of the earth significantly as compared to an earth without an atmosphere.
This is the calculation that must be undertaken first. How much of a difference would we see in the surface temperature of the earth due to N2O2.
The next calculation is the effect of phase change of H2O in the atmosphere. Not the GHG effect of H2O, but the effects of evaporation and condensation in transporting heat vertically.
Then, when you have completed those two calculations, what you have left is a candidate for the GHG effect.
From what I’ve seen, when you apply these calculations to real models, such as venus, earth and mars, then the first calculation, the effects of the atmosphere without regard to the composition provides nearly the same amount of heating on venus, earth and mars.
These are real observations, not theoretical models. And, observation trumps models, no matter how correct we think our models are. I don’t see this calculation in your work. I don’t see where you calculate the effects of a N2O2 atmosphere that does have any H2O or CO2.
As to why the atmosphere of venus, earth and mars show the same amount of heating regardless of composition, that is a matter of theory. Put 10 experts in a room and you will have 20 opinions. None of these opinions will change the facts.
The atmosphere of venus, earth and mars show a significant heating effect that is independent of the composition of GHG and this is not accounted for in your model.
As to why this is, my theory is that it is due to compression. That the molecules i the atmosphere all have roughly the same amount of energy due to convection, but since there are many more of them per cubic meter at lower altitudes, there is more energy per cubic meter, which we register as increased temperature.
For example, we could have 1 air molecule per cubic meter. When we went to measure the temperature of this meter of air, we would average out the energy of this one molecule over the volume and end up with a low temperature.
Now, if we were to have a whole lot more molecules per cubic meter, all will the same energy as our first example, we would see this as an increase in temperature. However, in both examples the molecules are at the same energy level, yet the samples taken from lower down in the atmosphere will appear to have a higher temperature.
However, this may not be the reason for what we observe. And it doesn’t really matter. What does matter is that FROM OBSERVATION the atmospheres of venus, earth and mars show similar warming at similar pressure, independent of their levels of CO2.
Ira,
Here’s why your bicycle tire analogy fails: you instead need a leaky tire that has to be inflated continuously. Would the air in that situation remain hot? Yes
The open atmosphere essentially acts as a continuous ‘air conditioner’ or pump. Cold air continuously descends to the surface and is heated by the surface as well as from compression. That air then ascends and expands and cools until the process repeats. Thus, adiabatic compression is able to maintain a higher surface temperature on a continuous basis, the so-called GHE.
We seem to have an unusually large number of comments from the anti-science brigade today who seem to have invented a whole new branch of previously unknown physics.
Imagine an ice cube, at zero degrees Celsius, alone in the vacuum of space (which is at, say, absolute zero). What happens to that ice cube? Well, it radiates away heat (as everything above absolute zero does) and eventually cools to absolute zero itself, asymptotically. Now imagine an another ice cube, this time at -10 degrees Celsius, being placed near to the original ice cube. This colder ice cube also radiates heat; some of it will impact the first ice cube and ‘warm’ it. The original ice cube still continues to cool, but not as fast as it was doing before because it now has a heat input which was not there in the first case. The colder ice cube therefore helps to keep the original ice cube warmer. Of course, it never gets above its starting value of zero degrees Celsius, but that is how the presence of cold objects can keep hot objects warmer i.e. they are still warmer than what would be in their place if they were not there (in this case the cold vacuum of space at absolute zero).
The 2nd Law of Thermodynamics? – Yes – it fully supports this – no contradictions. .
martin mason says:
In any model of the greenhouse effect, the net heat flows are from the warmer earth to the colder atmosphere. The atmosphere is not heating the earth itself…what it is doing is slowing down the cooling of the earth (for a given surface temperature). Since the steady-state temperature of the earth is determined by the balance between what it receives from the sun and what it emits back out into space, an IR-absorbing atmosphere will in fact cause the earth’s steady-state temperature to be higher than it would be if the atmosphere did not absorb IR. In a colloquial sense, one might call this “heating” the earth, although it is best to avoid such language since it leads to confusion and to instead state clearly what the effect of the atmosphere is, as I have explained above.
No…Such a hypothesis does not even satisfy energy conservation! What can be explained without recourse to GHGs is the lapse rate in the troposphere. What can’t be explained is what the temperature at any altitude is. It is as if you told me the slope m of a line of the form y = mx + b and claimed that I could now compute y for any x. In fact, I would need the value of b to do so. The value of b is in essence what is determined by the absorption of IR radiation by the atmosphere.
This is a strawman argument. (Almost) noone is claiming that there will be “runaway warming”. [Hansen is talking about the possibility under certain circumstances and has specific arguments as to why the current case may be different from what the earth has experienced in the past, so let’s leave him out of this.]
What we are talking about is a climate that is quite sensitive, but not unstable, to perturbations. See, for example, here for a discussion of what the paleoclimate record tells us: http://www.sciencemag.org/content/306/5697/821.summary
Really? You think the temperature on Venus is very pleasant? Did the runaway go on forever? No…An object that is linearly unstable will tend to find a point where stability is restored. However, a runaway did occur to produce Venus’s current very hot climate. Fortunately for earth, we are not as close to the sun and thus not subject to the runaway scenario that occurred on Venus.
How are you proposing that we harness it? As has already been noted by Smoking Frog, the back-radiation is already heating the earth. (Because most objects tend to be almost perfect blackbodies in the mid- and far-IR, almost all of the back-radiation that is received by the earth is absorbed.) Also, the back-radiation that we receive is diffuse (coming from all different angles) and so it cannot be focused in the same way that solar radiation (which is approximately from a point source) can. Furthermore, since the energy of an individual photon is proportional to its frequency, the back-radiation…unlike solar radiation…cannot be harnessed by photovoltaic devices, which rely on having photons of sufficient energy to cause electronic transitions (in particular, energies much larger than thermal energy of ~kT where K is Boltzmann’s constant). Finally, although the back-radiation we receive from the atmosphere is, in the global average (e.g., day and night at all latitudes), larger than the radiation we receive directly from the sun, the amount of radiation that we receive from the sun when it is actually shining and close to overhead is a fair bit larger.
Ira – regarding your summary comment 4) at May 8, 2011 at 7:51 pm
my comment – NO, the atmosphere does NOT emit LWIR across a distribution of wavelengths like a blackbody , see my earlier comment at Dave in Delaware says: May 8, 2011 at 7:00 am
Ira Glickstein, PhD says:
“4) As I understand it, the ~15μm radiation from the Surface to the Atmosphere is absorbed by H2O and CO2 molecules which, when excited, bump into nitrogen and oxygen and other air molecules, and heat the air. The result is emission of long-wave radiation at a range of wavelengths (not just in the ~15μm region, but including the ~7μm and ~10μm regions) in random directions, some of which finds its way back down to the Surface and is re-absorbed. That conversion of a portion of the ~15μm radiation into other wavelengths may explain why there is a “bite” out of the ~15μm radiation that seems to be missing from the radiation out to Space. ”
——————————————
This is the part of your comment I disagree with –
“The result is emission of long-wave radiation at a range of wavelengths (not just in the ~15μm region, but including the ~7μm and ~10μm regions) in random directions, …”
My reply – If CO2 absorbs at 15μm, it emits at 15μm. The Nitrogen and Oxygen in the atmosphere neither absorb nor emit in the LWIR wavelengths of interest. That is why there is an Atmospheric Window – the atmosphere neither absorbs nor emits in that 10μm region.
Think about the satellite-looking-down readings – the 10μm atmospheric window corresponds to the surface temperature. Over the Sahara, the 10μm readings show a surface temperature around 320 K. Over the Mediterranean, the 10μm readings show a surface temperature around 280 K. Over the Antarctic, the 10μm readings show a surface temp just below 220 K. If the atmosphere (mostly N2 and O2) emitted LWIR like a blackbody, the atmospheric window readings would look like some middle atmosphere temperature – not like the surface. There is an Ozone band in the middle of the atmospheric window, I expect from ozone in the upper atmosphere; if N2 was emitting blackbody LWIR there should be other N2 bands in the middle of the window, and there are not.
———————————-
Ira comment – “That conversion of a portion of the ~15μm radiation into other wavelengths may explain why there is a “bite” out of the ~15μm radiation that seems to be missing from the radiation out to Space. ”
My reply – Recall that the satellite-looking-down readings are Emissions; their temperature tells us something about where those emissions originated. Looking at the CO2 band satellite 15μm readings, the temperatures are around 220 K and nearly the same for the Sahara, the Mediterranean and the Antarctic. That looks to me like the Tropopause. It does not necessarily tell me how the energy got to the Tropopause, so I am not sure I could estimate a ‘bite’ value. That emitted energy could have arrived from lower atmosphere CO2 emissions, or from atmospheric lapse type convection, or possibly from the tops of thunderstorms.
My comments are referenced to these satellite IR spectra at
http://mensch.org/5223/IRspectra.pdf
Slide (c) Western Tropical Pacific is particularly interesting because it compares ‘clear’ with ‘Thunderstorm Anvil’. Note that the CO2 band temperature is identical in the clear and thunderstorm curves. Also note that the atmospheric window 10μm region temperatures are MUCH DIFFERENT for clear vs thunderstorm anvil. This suggests to me that while the atmosphere does not emit like a black body, the thunderstorm (liquid water) does.
Best regards to Ira – I hope my comments are constructive and instructive, and will help your ‘mental models’ and descriptions.
Allan M says:
Experiment did provide the evidence for the Second Law. However, our modern (20th century) understanding of where the law comes from is based on statistical physics. To dismiss an entire field of physics that has successfully explained a huge wealth of empirical observations as “a statistical ‘fudge'” because one implication of it does not line up with your ideology is bizarre.
This is essentially the project that Claes Johnson, one of the “Slaying the Sky Dragon” authors, has embarked upon. He wants to base things on the bizarre proposition that an artifact of numerical calculations of the differential equations actually governs the universe instead. The irony is that although he claims to have gotten rid of “back radiation” by coming up with a different interpretation of the terms in the equation governing radiative transfer between two bodies, he has in fact done nothing to change any numerical result using those equations…including all of the numerical results that support the existence of the greenhouse effect! What he has done is snookered a few people who want to believe the nonsense that he is peddling.
It depends what you mean by “microscopic”. It doesn’t take a very large particle size before the statistics make it vastly improbable for such a thing to occur. But, yes, if you particles were truly only a few molecules large, then you could detect behavior contrary to the 2nd Law. (Of course, the distinction between “ice” and “water” begins to lose meaning at such a small length scale since phases of matter are themselves a macroscopic concept.)
Circa 1850 Tyndall experiments with greenhouse gases in a nutshell.
Radiation flow:
heat_lamp -> -> adjustable shade -> salt crystal window -> mirrored tube filled with test gas -> salt crystal window -> adjustable shade -> galvanometer
Tyndal ran literally thousands of experiments with different gases and mixtures at different pressures, different infrared light sources, and different lengths of tubes. His setup took up a lot of space but I marvel at the cleverness he used to increase accuracy and precision using the technology of the day. His analog gear was so sensitive he read the reading from the galvinometer through a telescope because getting a warm body anywhere near the experiment mucked it up.
His best most stable lamp was a black painted face of a vessel filled with boiling water. His galvanometer response is not linear so the primary purpose of the adjustable shades was keep the radiative energy input to the face of the galvanometer in a range where you got maximum needle deflection from minimal change in radiation.
In lectures he used to take his galvanometer to the podium in the hall. He’d point it at a blank wall on the other side of the hall and ask someone from the audience to walk over to that wall. The galanometer needle would move dramatically when a person entered the scene.
Anyhow Tyndall found a great many gasses, most famously water vapor, that would dramatically lower the galvanmeter reading when present in the tube versus the same mixture without it. He also used some rather ingenious ways of processing his gasses prior to filling the tube with them including means of completely drying a sample of the atmosphere.
If one wants to dispute the concept of back radiation one must first explain why Tyndall’s galvanometer would read progressively lower as the absolute humidity of the atmospheric sample in the tube rose. No one will. Every attempt is fatally flawed in some way. Tyndall didn’t discover anything that wasn’t already predicted by the theoretical physics of the day. He confirmed the theory via experiment or in other words he was doing science the way it’s supposed to be done.
Tyndall did discover some things that no one had worked through on paper at the time. One notable thing he discovered was the non-linear absorptive response as the partial pressure of an IR-absorbing gas went up. He found it to be linear at the lowest partial pressures then progressing to exponential or in other words when it comes to IR absorbing gasses increasing the amount of it is a case of diminishing returns after a certain amount.
This is why the climate boffins talk about CO2 doublings and temperature rise. At the current atmospheric CO2 partial pressure its absorptive response is in the exponential range. The linear response range is all below the first 100ppm. It’s sort of like people picking low hanging fruit in an apple orchard. They can all pick as fast as they can for a while but as the number of people picking increases at some point they’ll be competing with each other for the same low hanging fruit. So putting more pickers in the orchard will always increase the rate at which fruit is picked but at some point it becomes a case of diminishing returns.
Dave Springer
You did not read what I had written – no additional heat absorbed- no temperature increase- no global warming. 150 years ago Tyndall new nothing about CO2 absorption bands
Charles Higley says:
May 7, 2011 at 9:33 pm
“It should also be pointed out that, as CO2 partitions 50 to 1 into the oceans, we would have to emit 50 times more CO2 that required to simply double atmospheric CO2. There is not enough carbon available to do this. If we really tried, we might be able to do 20%.”
It could also be pointed out that CO2 partitions 100,000 to 1 between the rocks and the air, so we would have to add 100,000 times more CO2 to just double atmospherics CO2…
This is wrong, as the ocean comparison is wrong because of time-scales. It takes the oceans centuries if not millenia to absorb additional CO2 so that partitioning will take effect eventually, for rcok/air the timescales are more like millions of years…
One little point about assumptions this is all based upon.
Incoming energy is spread over HALF the earth’s surface.
Outgoing energy is spread over the FULL surface of the earth, with an uneven distribution, probably (?) higher rate of outgoing radiation on the night side of the planet.
So its entirely incorrect to say that incoming = 240W/m^2 and outgoing = 240W/m^2.
Get the energy distributions right, then try again.
Here is a reference showing that Jupiter has 60K higher temperature than expected.
33K higher than expected on earth. 60K higher than expected on Jupiter. Co-incidence?
http://burro.astr.cwru.edu/stu/advanced/jupiter.html
•As seen in the table below, the average temperature of Jupiter is approximately 160 K. However, due to the equation for thermal equilibrium (below), it should only be about 100 K.