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.
The solar radiation hitting the earths surface is 184 W/m^2 per Trenberth 2009 or 198 W/m^2 per Trenberth 1997. The surface albedo must be applied to this to get the net gain. Back radiation adds another 333 or 324 W/m^2 respectively.
There seems to be a whole lack of common sense going around these days.
So, according to T, the earth receives 184 w from the sun, radiates some of this back outwards, where some of it is absorbed by the atmosphere and re-radiated back to earth, where it provides 333 w.
In other words, 184 w ends up as 666 w, because the back radiation is omni-directional at at most 50% reaches the earth.
Of course, it is like two mirrors pointing in towards each other. Start out with 184w and after it reflects back and forth a bunch of times, you end up with 333w.
Only one problem. If you start out with 184w from the sun, you can NEVER absorb more that that. Where is the excess energy coming from? The sun is the energy source.
The atmosphere cannot manufacture 333 w of back radiation for a 184 w power source. The back radiation from the atmosphere must be less than the original 184 w from the sun.
Otherwise we could put an ingenious array of one-way mirrors around a 100 watt light bulb, and generate millions of watts of power. What climate science has invented is a perpetual motion device, where the energy out is more than the energy in.
“If there were some mechanism that were taking volumes of air high in the atmosphere and transporting them lower in the atmosphere his thesis would have some merit but there is no such mechanism at work on Venus, Earth, or Mars.”
There is massive movement of air from high altitudes to low altitudes, as a result of an equally massive movement of air from low altitudes to high altitudes.
http://en.wikipedia.org/wiki/Hadley_cell
Here is an experiment to disprove the existence of back radiation.
CAUTION: Only perform with parental supervision.
CAUTION: Treat any large Fresnel lens as if it were an open flame!
CAUTION: Since this experiment is indoors, pay close attention to ANY light or infrared (IR) source, which could be focused by the lens. That includes sunlight through windows, hot kitchen stove filaments, etc.
The only thing great expense in performing this experiment with be the Fresnel lens itself. One square meter Fresnel lenses are available through science supply sites. The lens need to be a plano-convex radial lens to focus parallel rays onto a one square centimeter spot (or even down to one square millimeter). I have heard an F-stop near one is best. This gives a one ten-thousand multiplication though that will to be reduced by up to ~50% due to refraction, scattering and optical imperfections in the facets.
You will need a very small black metal cap to hold one cm³ (1 gram) of water or ice. This will be placed at the focal point of the lens. You will need an infrared (IR) heat source, a space heater, large wattage hair drier, something to add additional warmth to a wall later in the experiment. You will later need an ice cube.
Similar to:
http://www.modulatedlight.org/optical_comms/fresnel_lens_comparison.html
http://www.scientificsonline.com/large-fresnel-lens.html?&cm_mmc=Mercent-_-Google-_-NULL-_-3052833&mr:trackingCode=CC6E1735-DB81-DE11-8C0A-000423C27502&mr:referralID=NA
Now, ready? If back-radiation proponents are correct, we should be able to use this huge lens to focus back-radiation from any warm wall within your house, at the ambient room temperature of say 20 ºC (68 ºF), in some room to boil water in seconds.
Use a small flashlight to illuminate the wall and so the exact focal point can be located. At this point place the black cap with the one gram of water. Wait. If they are correct the water should evaporate quickly, even boil, for that 70 ºF wall is giving off 5.67e-8*(273K+20K)^4 or 418 Watts per square meter that is going to focus down to 400 W x 5,000 multiplication factor or 2,000,000 Watts per square centimeter and we only need 2250 Joules to vaporize that gram of water. Wait, maybe there are some loses within your setup so be patient. Wait. Don’t know about your experiment results but mine is not working. Half an hour later I gave up, something mystical is blocking all of those back radiation photons from entering that lens. Watts up with that?
Now replace the water and cap with an ice cube. it should develop a hole slowly at the focus. Here the temperature at the focus is less than the temperature of the wall and radiation flows.
But I want to see the water boil, so cover the lens and warm up that wall in front of the Fresnel lens to a much higher temperature, let’s say 104 ºF (40 ºC). The watt flux should now be something near 5.67e-8*(313K-293K)^4 = 0.009 W/m^2 or 50% efficiency of 0.009 W/m^2 x 5,0000x = 45 W/cm^2. Uncover the lens. Well, it probably won’t boil but you should now evaporate that gram of water in about 2250 J/g ÷ 45 J/s or 50 seconds if you have a perfect setup with no loses. Let’s just say in a couple of minutes.
Careful, but now fill the cap and point that heat source, directly into the lens from five or six feet away, this is why Fresnel lenses can be dangerous, the water should now boil and quickly in seconds.
Thermal radiation heat going backwards does not exist in reality, at all, if you are a scientist type and not merely a AGWChurch member, run the experiment, prove it in reality.
Sorry back-radiation lover’s, you love is just in your minds (and wrong application of Stefan-Boltzmann equation).
That is also why Fresnel lenses are not dangerous in a very warm dark room at the focus. That room is at radiative equilibrium due to everything being at the same temperature and there is, in reality, no radiation at all, or you could focus it.
Ferd Berple,
You are just looking at the downward radiation, the upward radiation balances it, there is no energy problem. The mirrors on two sides of the bulb has some merits as an analogy, you wouldn’t doubt that the total flux of photons going back and forth between the mirrors sums to more than the 100 watts being input, but at equilibrium, only 100 watts escapes. A completely enclosed bulb is not a good analogy, because the energy is allowed to accumulate, the filament’s resistance will increase with temperature, reducing the amount of energy input, but still the bulb components will eventually melt.
In the climate case, the system is very complex, because the increase in temperature in response to the back radiation, increases a number of other responses to increased temperature before equilibrium is reached.
The idea that the CO2 effect is insignificant, and adiabatic explains all, fails for the common observation of another greenhouse gas, H2O. If it isn’t back radiation, how do you explain the difference in heat loss, on a clear calm humid night vs a clear calm dry night?
DirkH;
David, this means that at -25 deg the upper atmosphere is no more colder than the ground; an equilibrium is reached. Of course, nothing happens from there; but your premise was to show that the people who say a cold object cannot warm a warm object are somehow wrong – your example shows that two equally cold objects are in equilibrium and does not contradict those people at all. In your example, the upper atmosphere is no more colder than the ground.>>>>
Read more carefully my friend. I said that at at 25,000 feet it is colder than that even at the equator. Two points:
1. According to AMSU-A http://discover.itsc.uah.edu/amsutemps/ satellite data the global average at that altitude is in the range of -38C. It would be warmer than that over the equator, and colder… much colder, over the arctic.
2. 25,000 feet isn’t “upper atmosphere”. In fact, it is only part way up the troposphere which is the lowest layer of the atmosphere. Troposphere ends at 11Km, then Stratosphere about 50Km, Mesosphere about 88Km and then the thermosphere about 500km and then the Exosphere.
Lot’s of other examples by the way. Build a house with no insulation and measure the temperature in the house and between the inner and outer walls with the furnace running at a constant rate. Then insulate the wall cavity. With the furnace running at the same rate the inside of the house will become warmer. The thermometer between the inner and outer walls will also read warmer, but still cooler than inside the house. The insulation absorbs outbound heat and reradiates it back into the house. Pile snow up around a house, and despite being colder than the inside of the house, the house gets warmer with the exact same heat sources inside, that’s how the early pioneers in NA insulated their sod shacks in the winter. That’s how the Inuit survived centuries in the far north by living the winter in igloos where the cold ice walls nonetheless kept them warmer than they would have been in the open…It’s even the same principle that a freezer works on as does an air conditioner. A kiln gets much hotter than an open fire, but put a thermometer inside the wall of the kiln and you will find that it is cooler than the temperature inside the kiln. Same with a forge. And no, these examples have nothing to do with convection, you can build a forge or a kiln with thin walls or thick walls and have the exact same air circulation, but the thick walled ones have a higher temperature inside, and the temperature of the walls is cooler.
I am a raging skeptic. An angry, they are misleading us with how they represent the results, producing results from tragically flawed and sometimes outright fraudulant “science” and are being used by lobbyists who are more interested in lining the pockets of their lobby than they are in what’s best for humanity, and I’m disgusted by the whole thing. But cold things radiate photons carrying energy and if they bump into something warmer, there are any number of factors that determine if they will be absorbed or not. But the relative temperature of the two bodies has zippo to do with it. The relative temperature of the two bodies only defines the magnitude and direction of the NET energy exchange, which can only be from warmer to cooler. But take the cooler body away, and the warmer body’s temperature must fall because it is no longer recieving any heat from the colder body.
Martin Lewitt says:
May 8, 2011 at 3:11 pm
The idea that the CO2 effect is insignificant, and adiabatic explains all, fails for the common observation of another greenhouse gas, H2O. If it isn’t back radiation, how do you explain the difference in heat loss, on a clear calm humid night vs a clear calm dry night?
——
Simply the difference in the temperatures of one layer to the other (clouds in this case). The air above the cloud tops will be cooler than normal shielded by the water mass in the clouds and the air between the surface and the cloud base warmer than normal. Please, pick up a calculator and calculate these layer’s flux, the overall flux from surface to TOA when all added remains the same, but, the inter-atmosphere profile of the temperatures causing the flux at each layer will be different in the two cases.
Wayne,
For how much of the infrared black body frequency range is your Fresnel lens able to focus at that one spot? Is your lens material transparent to infrared?
Joel Shore says:
…”Okay…So it is time to debunk the Postma nonsense again, even though we already did it in the last thread that Ira wrote.”……
Joel, in case any readers are unaware, is one of the authors of one of the most imaginative pieces of fiction ever to have made its way into a science publication.
Readers of Ira’s last thread will know that Joel postulated that;
1. Heat can move from a colder surface to a warmer surface spontaneously.
2. Invented statements and then proceeded to attack them.
3. Has the Stephan Boltzmann equation applying to gases.
Perhaps Joel is still a bit “rusty” about when and where the Stephan Boltzmann Equation applies.
It applies to black bodies issuing a continuous spectrum centred around a characteristic temperature.
It is the full integration of the Planck function.
It does not apply to the line spectra issued by H2O and CO2 making up less than 1% of the atmosphere.
Joels proof of the “greenhouse theory” involves getting rid of the Oceans and atmosphere.
Postma is absolutely correct in stating that the long wave radiation produced by the Earth system is released at an average height.
Some radiation from the ground leaves directly through the “window”.
Radiation also leaves from different heights above the surface.
The average height is about 5Km.
The lapse rate which is completely independent of the radiative effects then determines the surface temperature .
The greenhouse theory is like Joel’s paper, …… pure fiction.
Bare Rock Earth is Joel’s favorite topic and he rarely strays from it.
Charlie Foxtrot: exactly. Well-said.
Dr. Glickstein: I appreciate your efforts to draw analogies to explain the CO2-induced warming of the Earth. However, I’m suspect the effort is doomed from the start because one cannot apply a steady-state model to a dynamic process. The Earth does indeed receive energy from the Sun, and radiates energy into space. But, the system is not at steady state due to Earth’s rotation around its axis, slight eccentricity of its orbit around the sun, greater or lesser cloudiness over time, absorption of energy into oceans or loss of energy from the oceans, and likely a host of other factors not mentioned in this list.
I’ve done a fair bit of outdoor cooking, over a campfire with a chunk of meat rotating on a spit above the fire. One would have a tough time creating a valid model of how the meat cooks using a steady-state model. Same thing applies to the Earth’s energy balance.
1 Watt = 1 Joule/second
1 Watt/m2 = 1 Joule/second/m2
Now convert the whole concept into Joules and bring time into the equation – say a 24 hour period. It will look much different.
The problem with all the calculations is that the definition of “Watt” has a time dimension already built into it so we tend to forget that all this has to happen in time.
The Sun is beating down at 960 Joules/second/m2 at the height of the day but there is almost no energy coming at night. How come the average Earth temperature is only the equivalent of 418 Joules/second/m2 in the daytime and 364 Joules/second/m2 at night.
Yes, ferd berple, when a gas is compressed its temperature increases. However, once compressed, that increase in temperature will radiate away.
For example, when you pump up your bicycle tire, the pump and the air in the tire will be warmer than the ambient (uncompressed) air in the room. However, the air in the tire, although it is compressed to several times the normal ambient pressure, will not remain warmer for very long. The heat energy will radiate and convect and conduct away until the tire reaches equilibrium with the ambient temperature of its surroundings.
The same is true of the Atmosphere. If we transported an Atmosphere’s worth of air to a planet that had no Atmosphere, the force of gravity of that planet would indeed compress the Atmosphere and warm it. However, before long, the Atmosphere would cool down to the temperature of the planet. If we then placed that planet in an Earth-like orbit around the Sun, short-wave Solar radiation would be absorbed by the surface of the planet and the surface would re-emit long-wave radiation to its Atmosphere where some would pass through and out to Space, while some would be absorbed by the H2O and CO2 and other “greenhouse” gases in the Atmosphere. The heated gas molecules would bump into other air molecules and warm them, and like any material above absolute zero, the Atmosphere would emit radiation at a variety of long-wave wavelengths in random directions, some of which would be absorbed by the surface of the planet, warming it further. Just like our Earth.
Theo, please read before you comment. I wrote “… 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. ..”
So, I specifically said the Earth System is NOT a perfect black body and I presented a graph to prove how it differs from a black body! I said further: “… 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. …”
So, I specifically said how the Earth’s radiation to Space differs from a black body and how it jigs and jags between black body curves as high as 300 K and as low as 210 K. How could I have been more clear?
My use of a black body model at 255 K was only for what I said was a “sanity check” on the supposed experts who claim that 255 K is the temperature a black body Earth would have to be at to absorb and emit the same amount of Solar radiation the actual, non-black body Earth absorbs and emits. Got it?
Martin Lewitt, it’s an experiment for you to perform, look on the specifications when your lense arrives, or better, mid-IR lens are also available, specify your choice when ordering. Let us know your results. ☺
“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.”
Maybe the effect of compressing the atmosphere might cause an increase in of 33 degrees C.
I wish my fingers would learn that ‘lens’ is singular, ‘lenses’ are the plural. ☺
“Yes, ferd berple, when a gas is compressed its temperature increases. However, once compressed, that increase in temperature will radiate away.
For example, when you pump up your bicycle tire, the pump and the air in the tire will be warmer than the ambient (uncompressed) air in the room. However, the air in the tire, although it is compressed to several times the normal ambient pressure, will not remain warmer for very long. The heat energy will radiate and convect and conduct away until the tire reaches equilibrium with the ambient temperature of its surroundings.”
A tire is an enclosed space where the air pressure is the same in all locations. The atmosphere is not. The air circulates. As it warms, it rises and cools, and as it cools it falls and warms. Go ahead and bat away a fact of science and try and shove your less than fully thought out process down some other people’s throats.
And continuing.
I would like to see the back radiation proponents do the following.
Take a spherical black body object with a small heat source in its center and shine a light on it and read the temperature.
Take the same spherical object and enclose it with a black body enclosure that does not touch the surface except at some required points to center it around it. use the same lamp at the same distance, and measure the temperature at the surface of the original sphere. Now make another black body enclosure and repeat the procedure. Do this until you have 12 layers of black body. If your argument is correct, then the temperature should sky rocket to the point of starting a fire. According to back radiation proponents statement of how it works.
Ira Glickstein, PhD says:
May 8, 2011 at 4:26 pm
“The same is true of the Atmosphere. If we transported an Atmosphere’s worth of air to a planet that had no Atmosphere, the force of gravity of that planet would indeed compress the Atmosphere and warm it. However, before long, the Atmosphere would cool down to the temperature of the planet. If we then placed that planet in an Earth-like orbit around the Sun, short-wave Solar radiation would be absorbed by the surface of the planet”
You’re right so far.
” and the surface would re-emit long-wave radiation to its Atmosphere where some would pass through and out to Space, while some would be absorbed by the H2O and CO2 and other “greenhouse” gases in the Atmosphere.”
Here it starts going downhill. There are three modes of energy transport from the surface upward: LWIR Radiation, conduction and convection, LWIR radiation being the weakest. IOW, the oceans get warmed by visible and UV radiation and will give the energy to the atmosphere mostly by surface conduction; kinetically.
” The heated gas molecules would bump into other air molecules and warm them, and like any material above absolute zero, the Atmosphere would emit radiation at a variety of long-wave wavelengths in random directions, some of which would be absorbed by the surface of the planet, warming it further. Just like our Earth.”
The gas molecules who get excited by LWIR photons would reradiate some immediately; in some cases they would give the energy to neighbouring molecules, thermalizing the energy. But Kirchhoff’s Law states that for any number of thermalization events there must be an equal number of dethermalization events as long as the gas is in local thermal equilibrium, which it is in the lower atmosphere. So no heat is trapped there.
See
http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/
So, the consequence is that LWIR radiation is transporting energy to space. Some is re-emitted back to Earth from where it is re-emitted back upwards. Whether there is 0.03 % or 0.04 % of CO2 in the atmosphere only influences how often the photons get absorbed and re-radiated on their way to space – an increase in CO2 delays the process a little but does not change it fundamentally and *Does* *Not* *Trap* *Heat* any more than a sieve traps water.
I agree completely Dave Springer! Anthropogenic forcing (human-caused warming due to land use changes that reduces albedo and burnign of fossil fuels that raises CO2 levels, etc.) is most proabably below the lower IPCC estimate of 0.6 W/m^2.
In previous postings here at WUWT I have estimated the human contribution to net warming since 1880 at 0.2ºC, the natural cycles and processes contribution over which we humans have no control at 0.3ºC to 0.4ºC, with the remainder of the supposed warming of 0.8ºC due to data bias and cooking of the books by the official climate Team.
My main purpose in this Visualizing series was not to justify any particular number for the warming effect, but rather to correct those of our fellow Skeptics who have bought into the false notion that the Atmospheric “greenhouse effect” is fiction. In some sections of the media and public opinion, outspoken, non-scientific Disbelievers make us look foolish by association.
Ira,
Please detail exactly why you think Swedish climatologist Dr. Hans Jelbring and chemical engineer William Gilbert are “foolish” “non-scientific Disbelievers” and “irrational” in their explanation of the so-called “greenhouse effect” based simply upon the adiabatic lapse rate:
http://www.tech-know.eu/NISubmission/pdf/Politics_and_the_Greenhouse_Effect.pdf
as well as physicist Joe Postma who explains the same
http://www.globalwarmingskeptics.info/forums/attachment.php?aid=327
Bryan says:
Actually, what readers of that thread will know is that Bryan is a purveyor of pseudo-science. He has absolutely no serious desire to discuss the science and, like all purveyors of pseudo-science, tries to distract people by nitpicking words…even after the wording has been corrected to everyone’s satisfaction. (See here for more discussion specifically on the tactics that Bryan, Gerlich and Tscheuschner and other purveyors of such pseudo-science employ: http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/#comment-649274 ) The existence of people like Bryan within the skeptic movement, willing to actively engage in such deception, certainly helps to keep you guys marginalized in the scientific realm.
More nonsense. Simple toy models of the greenhouse effect illustrate the concept and give one a qualitatively understanding of what is going on. Line-by-line radiations codes do the full calculations for the actual absorption lines as they exist in the atmosphere in order to nail down the effects quantitatively. Yes, the simple models don’t have all the gory details in them…That is the whole point.
Fine…Everything you say is basically correct. The one thing that you and Postma are leaving out is what I said in my last post…that the average height of 5km is determined by the IR-absorptivity of the atmosphere, i.e., by the greenhouse gases in the atmosphere and clouds. (The technical statement is something like this: At any given wavelength, the height at which most of the radiation escapes to space is given by the height at which the “optical depth” [ http://en.wikipedia.org/wiki/Optical_depth ] of all of the atmosphere above that point to radiation of this wavelength is of order 1. More quantitative details can be found in books on atmospheric radiation, such as Ray Pierrehumbert’s book “Principles of Planetary Climate”.)
I haven’t had time to read all of the comment but I haven’t seen any comment on land-use/botany. Even in the oceans there are plants. Much of the earth is a GREEN body that is converting solar energy into some form of biomass. How many watts does it take to produce a bushel of wheat or a 1,000 board feet of pine lumber or produce a nice turf (or Veg. Garden) at the White House? It isn’t being factored in!
WOW Anton Eagle, that is quite a question! My short answer is “yes, that is the essence of my claim.” However there are some caveats:
1) The atmosphere would be composed of pure nitrogen and other gases that transparently pass both long-wave and short-wave radiation without any absorption at all.
2) The albedo of the Earth System would be the same as it is now, around 30%. This could be accomplished by covering a portion of the Surface with reflective materials such that only about 240 Watts/m^2 would be absorbed by the Earth Surface, the remainer being reflected back out to space through the transparent Atmosphere.
3) The effects of winds, storms, rain, snow, H2O phase change, ocean currents, and so on – including convection and conduction from the Surface to the Atmosphere and the reverse, would have to be simulated or corrected for.
Under the above conditions, I believe the mean Earth Surface temperature would be closer to 255 K than to the present 288 K.
I would appreciate hearing the answer to this question (and perhaps more caveats) by others who are more knowledgeable than I am.
Ira has explained it here: http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655793
I have explained it here: http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655686
The only way that adiabatic compression could possibly explain the earth’s surface temperature would be if the atmosphere (and / or the earth itself) were undergoing continual gravitational collapse, which we know is of course not the case. Otherwise, such a hypothesis does not even satisfy the First Law of Thermodynamics (basically, conservation of energy): Without substances in the atmosphere that absorb terrestrial radiation, the earth’s surface at its present temperature would be emitting back out into space way more energy than it receives from the sun and hence would rapidly cool down.
The temperature structure as a function of altitude, i.e. the lapse rate, in the troposphere is set by the considerations discussed regarding adiabatic expansion and compression (basically because a lapse rate higher than the appropriate adiabatic lapse rate for the given water vapor content is unstable and leads to convection until marginal stability is restored).
However, this alone does not determine the temperature at the surface. That is determined by consideration of the absorption of the atmosphere of terrestrial radiation (and radiation emitted by the atmosphere), which essentially ends up determining at what altitude the temperature has to be determined via radiative balance between the Earth system (earth + atmosphere) and the sun and space [which for the earth system with its current albedo is ~255 K]. Then, by using the lapse rate, you can get the temperature at the surface.
However, as one adds greenhouse gases, one raises this altitude and hence the required surface temperature. (In the simplest case, this is under the assumption that the lapse rate doesn’t change, whereas the more complicated calculations assume that there is somewhat of a decrease in lapse rate, by the well-understood physics of the moist adiabatic lapse rate in the tropics. This decrease in lapse rate is a negative feedback that is included in all of the climate models.)
IRA: It is getting more and more conspicuous that you are completely ignoring the alternative explanations for the Earth’s temperature. You have completely ignored me and many others who have suggested some other theories to this point, unless I missed something. You need to respond. With some facts. Otherwise, my advice is to shut up.