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.
To wayne regarding the Fresnel lens experiment at room temperature.
The critical factor you missed in your thought experiment is that the radiation from the wall is not all approaching the lens in parallel beams. The lens will only focus the heat radiation at the target that came from a spot on the all that is directly perpendicular to the surface of the lens. In order to focus the entire heat radiation from the wall onto the target, all of the radiation from the wall would need to be focused in parallel beams first, before it hits the lens. The best way to accomplish this would be to move the wall a long distance away, and heat it up so that the radiation from it reaching the plane of the lens is equal to the total radiation coming from the wall at room temperature. The water in the target would then heat up nicely, and the laws of thermodynamics are satisfied.
Joel Shore;
the rest of you should consider carefully reading what Dave, David Hoffer, Ira, myself and others are writing here and actually trying to understand the correct science.>>>
Joel, I am going to take that as a compliment, and my hat is off to you, you and Dave Springer have been doing yeoman’s work in this thread. Your explanations have been bang on and I have to admit I’ve gotten a bit ill tempered in this particular topic of late. Your patience and clarity is appreciated to no end by me as a skeptic. Pardon me, a RAGING skeptic. For those skeptics who want to be part of winning the debate with the warmists, I have this suggestion:
On the matter of “backradiation” and “greenhouse gases”, I recommend that you accept Joel Shore’s explanations at face value. CAGW is a farce of gigantic proportions, perpetrated by incompetant and/or fraudulent science and I have firm and specific reasons for saying that. Though a warmist he may be, I detect neither incompetance nor fraud in any of his explanations in this thread. And as I am certain Joel will tell you, if I did, he would hear about.
Joel is explaining physics that is used and verified by design engineers by the thousands every single day, all over the world. As are Ira and Dave Springer and others. Some of the arguments that I’ve seen presented in this thread amount to no more than saying “the earth is flat, just LOOK at it”. Yup, I looked, and yup, it looks flat, and NOPE it isn’t.
jae, please! It is better to sit quietly in the corner and let some people think you are stupid, rather than open you mouth and prove it!
Yes, you are missing something. Everyone but you knows that “3 +/- 1” means a range from “3 + 1 = 4” to “3 – 1 = 2”. In other words, the IPCC says the amplification is between 2 and 4.
Joel accepts this IPCC estimated range. Some of us, including me, believe the amplification is less than 2, which would make the IPCC-specified range wrong.
Now, please go back to the kiddie table or I’ll tell your mother and she’ll wash your mouth out with soap. :^)
Boris Gimbarzevsky,
Your reasoning from the lunar daytime temperature is wrong, that temperature has more to do with the heat capacity of the surface and the insulating properties of regolith. Even here on earth it gets hot enough on the hood of a dark colored automobile to fry eggs, even though the ambient temperature is much lower. See David Springer’s comment above for a better consideration of the lunar analogy.
http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655571
BTW, thanx again for your temperature readings. I think the ones at night will be the most interesting, especially if you can also report the humidity along with the ambient temperature.
wayne says: May 8, 2011 at 3:01 pm
“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.”
No, Wayne, you are misunderstanding the physics involved. I can’t off-hand remember the name of the theorem involved , but you can’t focus the spot any hotter than the surface creating the light. (Similarly, the spot focused by a magnifying glass aimed at the sun can never be hotter/more intense than the light at the surface of the sun.)
So no matter how thermal EM photons are focused, they can not make the surface they approach any warmer than the surface emitting the photons. The IR from the atmosphere could never (by itself) make the surface any warmer than the atmosphere. And the light from the sun could never make any part of the earth any warmer than 5700 K no matter how it was focused. Of course, the IR from the atmosphere PLUS some energy from the sun could warm the earth above the temperature of the atmosphere.
(PS I don’t have time to address the other question of εσ( 288K^4 – 280K^4 ) vs. εσ( 288K^4 ) – εσ( 280K^4 ) .. i”ll have to see about doing that some other time.
I’ll also have to see about addressing Jae’s concerns about Venus vs Earth vs Mars re the GH effect. Lapse rate is important, but I don’t believe it is the whole story — along the lines that some people have already mentioned.)
JAE says:
May 9, 2011 at 11:28 am
the lapse rate depends upon only Cp and g. ??
The Dry Adiabatic Lapse Rate (DALR) depends upon only Cp and g. The actual (environmental) lapse rate depends on other things and varies through out the day.
ferd berple says:
May 9, 2011 at 1:27 pm
My question is why we assume that N2 in the atmosphere doesn’t back radiate
Because it has been measured and N2 does not radiate at the temperatures normally found at the surface.
wayne, I think you are right, I misread what you said.
However, I stand by my comment that the temperature of the clouds has little effect on the surface temperature. On cloudy nights, the back radiation is almost the same as the radiation emitted from the surface. This is either because the clouds are the same temperature as the surface or because clouds simply scatter the surface radiation back toward the surface. Since clouds emit both up and down, they should be able to emit in each direction only half what they receive. In addition, since the water vapor mixing ratio generally decreases with altitude, if the surface temperature was the same as the cloud temperature, then there would be fog. Therefore, since the energy emitted toward the surface equals what is received, and since there is no fog, scattering makes more sense. Unfortunately, I have not found the measurements needed to be certain about this.
Thanks to the following posters who talked about Latent Heat
Dave in Delaware
Nullius in Verba
Bryan
Dave Springer
Martin Lewitt
They’ve set me at ease for I now know I’m not the only one to have noticed that radiative models are missing a large piece of the puzzle.
Even the climate keepers at Wikipedia have allowed it to give 23 % back to the atmosphere
http://en.wikipedia.org/wiki/Earth's_energy_budget
Sensible heat flux is mentioned too (convection) with 7 %.
None of this has anything to do with greenhouse warming so comparing the blackbody example to earth now as a gauge of greenhouse warming is a pointless exercise.
Joel Shore
You challenged me to discus the science.
I am here to do exactly that so you cant avoid the direct question posed below.
If you prefer to avoid the science and instead issue smears instead then I will not be surprised, as insulting others, seems to be your default mode.
Tell me which part of the query below is not science.
This item is directly addressed to Ira’s post and deserves an answer from you!
In your co-written Halpern et al comment paper you state that the Stephan Boltzmann Law can be used to work out the thermal energy exchange between atmospheric shells.
There is a whole series of statements on pages 8,9,and 10(0r 1316 to 1318 of pdf) referring to the atmosphere, Stephan Boltzmann Law and indeed including the SB formula(constant x T^4).
(The pdf below will allow readers to verify for themselves)
However there is a major problem for you!
A filtered spectrum does not vary with T^4.
The Stephan Boltzmann Law cannot be used for the filtered spectrum of atmospheric gases.
Do you now agree with me and withdraw your comments?
Or will you persist in this unphysical assumption.
Depending on your answer we can call on Ira’s spreadsheet to settle the matter.
That is, to prove that the reduced Planck function does not vary with T^4.
This has implications not just for your absurd paper but for ANY climate model with the T^4 assumption built in!
Comment on ‘Falsification Of the atmospheric CO2 greenhouse effects within the frame Of Physics’ by Joshua B. Halpern, Chistopher M. Colose, Chris Ho-Stuart, Joel D. Shore, Arthur P. Smith, Jorg Zimmermann.
http://scienceblogs.com/stoat/upload/2010/05/halpern_etal_2010.pdf
Charlie Foxtrot, I saw Dave Springer’s comment and it may be that the average temperature of the moon is -23 C, but what I’m interested in is the immediate surface temperature of a planet, not an extremely low pass filtered temperature. I’ve worked in the NWT in areas of permafrost and it was very nice to be able to just dig a hole in the ground to keep our food cool when the outside temperature was 90 F during the longest days of the year. The temperature a couple of feet below the ground is below freezing but what we’re interested in is the surface temperature which is where we live. During a hot summer day I have to open all of my vehicle doors for 5 minutes before I can get into the vehicle and touch the steering wheel without getting burned. What I’m trying to get at is that the extremes of temperature get smoothed out by the low pass filter of the atmosphere and that heat transfer between sections of the atmosphere is probably far more important than any minimal effects of greenhouse gasses. The frequency response of the filters is what we’re interested in — I don’t care about the very long time constant of below ground temperatures when a relatively short pulse of heat will melt plastic items in my vehicle.
As far as temperature of the chunk of air above my head goes, it seems to vary over a fairly small range. Just now stepped outside and found a minimum temperature of -12 F when the IR thermometer was vertical and humidity is given as 52% right now. Sky completely clear without a cloud to be seen and great night for stargazing but I have to work in a few hours:-(. When I’ve played around with this IR thermometer before I always seem to get a low value and it increases as one points the IR thermometer at angles of less than 90 degrees. If I was really obsessive about this I could mount the IR thermometer on a jig where I could precisely vary the angle and measure the temperature as a function of angle. I suspect this data could be used to compute a temperature profile of the atmosphere using similar math that one uses to generate an image with a CT scanner. The forecast is for rain tomorrow and will be interesting to see what kind of reading I get from the base of clouds.
Wayne says:
‘Robert, I come up with a close conclusion, not exact, but close. However, the 3600 m is more than 36 times that is given in other conversations I have read speaking of it’s absorption very near the surface. Do you have an average absorption coefficient you are using, or a mean optical thickness for CO2 from which that is calculated? Maybe Hottel specifies emissivity direct then what equation are you using to get the 3600 m extinction distance?’
Wayne,
I read on a website (Gary Novak – Global Warming – not caused by carbon dioxide) that CO2 in air absorbs to extinction at its 15 micron peak in about 10m (Heinz Zug). I tried to confirm this figure using Plank – Hottel and got 3600m . I found that CO2 was too low at 0.038%. I used an absorptivity of 0.1905 for a hemispherical gas mass of radius L=3600m. Hottel gives CO2 emissivity direct from charts based on direct measurements of total emission. The precise method of calculating the effective absorptivity or emissivity is quite complex and is given in -(1) Trans. Am. Inst. Chem Engrs., 31, 517-549 (1935) Hottel, H.C., and H.G. Mangelsdorf. (2) Trans ASME, 57, 463-470 (1935) Hottel, H. C., and V. C. Smith.
Wayne,
When I said CO2 too low at 0.038% I meant too low for an extinction distance of 10m or so. However using the same technique (Planck- Hottel) for water vapour which has a partial pressure or concentration 100 x that of CO2 the extinction distance is much less at 120m (absorptivity 0.5734).
Ferd berple says:
‘My question is why we assume that N2 in the atmosphere doesn’t back radiate. All objects radiate depending on their temperature. As the atmosphere is mostly N2, that must be the source of most of the back radiation’.
Ferd
The emissive power of N2 is small and insignificant compared with the emissive power of the entire spectrum of Earth radiation .
David says: May 10, 2011 at 12:54 am
…I now know I’m not the only one to have noticed that radiative models are missing a large piece of the puzzle [Latent Heat].
David, I’m not sure why this came as a surprise to you. When trying to understand radiative effects (like Ira is doing here), naturally people concentrate on EM radiation. However, more advanced models include many more effects in much more detail.
You note that “Even the climate keepers at Wikipedia have allowed it [latent heat] to give 23 % back to the atmosphere … Sensible heat flux is mentioned too (convection) with 7 %”. But, of course, those same terms were included in the Trenberth energy balance diagrams, so it is pretty clear that they have been considered important for a long time in all serious models.
http://en.wikipedia.org/wiki/Global_circulation_model gives a quick overview of how the more advanced models work.
Ira:
Funny that you replied only to the comment which I already realized and acknowledged (next comment) was off-base. How about the other comments? Still that irritating silence from the warmers and luke-warmers.
Robert says:
“The Dry Adiabatic Lapse Rate (DALR) depends upon only Cp and g. The actual (environmental) lapse rate depends on other things and varies through out the day.”
Yes. And your point is??
Robert Clemenzi says:
May 9, 2011 at 11:56 pm
“Because it has been measured and N2 does not radiate at the temperatures normally found at the surface.”
ALL matter with a temperature above absolute zero radiates. This is one of the most fundamental laws of physics. Either you are making things up out of thin air or the source you got it from doesn’t know his ass from elbow.
You can’t heat anything to a higher temperature than the temperature of the photons doing the heating. If you used a parabolic mirror aimed at a warm wall to heat water you could heat it to the temperature of the wall then the heating stops because the object being heated, now at the same temperature as the heat source, emits back as much radiation as it receives. It’s precisely because of back radiation that you can’t heat something hotter than the source of the heat.
Things are more complicated than that. Water vapor is a condensing greenhouse gas. In frigid conditions most of it gets frozen out of the air. Antarctica has by far the dryest air on the planet’s surface. A second complication is that the IR absorption response of CO2 (and water vapor too) isn’t linear except at low concentrations (see my apple orchard vs. number of pickers in the field analogy) which is one of the surprising things John Tyndall discovered 150 years ago. Complicating things yet more is the albedo of water changes drastically between liquid and solid phase. The ocean absorbs almost all incident light while a glacier reflects almost all of it.
So. Given the earth with an albedo similar to the moon (i.e. all rocks) it would have an average surface temperature well below freezing. But even your average rock is fairly dark and absorbs near 90% of incident light. But the earth is a water world so while the global ocean has an unfrozen surface it absorbs about 10% more light than rock. But what happens if the surface ever becomes almost completely frozen over? There would be hardly any water vapor in the atmosphere for a greenhouse effect and the snow & ice would reflect 90% of the sunlight. We would have a very frigid planet with no escape.
But there is an escape hatch. If the earth is covered in snow & ice there are can be no green plants taking CO2 out of the atmosphere nor will there be any chemical processes (formation of carbonate compounds) removing CO2 from the atmosphere, nor will it be mixing with the ocean because the ocean is frozen over. In the meantime volcanoes will continue to vent CO2 into the atmosphere. CO2 level will keep on rising until the greenhouse effect from it kicks in and it starts melting ice first at the lowest latitudes and progressively higher latitudes. Water vapor again enters the atmosphere, ocean albedo changes from 90% to 1%, and the melt accelerates in one big hurry. Green plants bloom like crazy from the CO2-rich air/water and dissolved nutrients in the ocean accumulated over eons of nothing removing it.
So basically CO2 greenhouse effect is what eventually stops a permanently frozen earth. Were we farther from the sun, like Mars, where it gets cold enough for CO2 to condense we wouldn’t have an escape hatch and the earth would be a cold dead rock much like Mars.
Because CO2’s ability to absorb IR increases linearly at low concentrations (under 100ppm) a minimal amount even absent most water vapor serves to keep the earth just warm enough to prevent a snowball earth episode most of the time. It is somewhat controversial but it is generally believe that earth has had a few snowball episodes in the past and the CO2 hypothesis above is the only reasonable explanation for how it ever manages to melt once the ice takes over. The sun’s output has been gradually increasing over the billions of years and is about 10% higher now than in the distant past which is probably why there haven’t been any recent snowball episodes.
My hypothesis taken from the above is that the first 100ppm of CO2 serves as a kind of kindling that fires up the water cycle i.e. it raises the average temperature just enough so that liquid water dominates frozen water. Evidently, at 280ppm, it’s just barely enough because the earth has been in an ice age for the past 3 million years with just some small cyclic variations in eccentricty & inclination being enough to transition between glacial periods and interglacial periods with the glacial ages lasting about 100,000 years and the interglacials about 15,000 years.
Clearly the earth’s climate is at a tipping point but that tipping point is tipping back to a glacial period. Because CO2 at 280ppm is in the exponential part of its IR absorption curve it’s likely we can’t possibly enrich the CO2 content enough to end the 3 million year-old ice age but it would be great for all life on the planet if it did including us.
In any case the indisputable testimony of the geologic column reveals that in the last 500 million years (since the Cambrian explosion when most of the modern phyla appeared and Ediacaran phyla disappeared) and animals started crawling out onto land the CO2 level in the atmosphere has, most of time, been 10 to 20 times greater concentration than today. Today’s atmosphere is CO2-depleted in historic terms and life is not nearly as abundant. The normal state of affairs for the earth is complete absence of any polar icecaps – it’s green from pole to pole. The living world today is a shadow of what it is during the good times with life struggling to survive in the nutrient-poor air and through the long cold winters. We should want to be moving away from the point where the climate tips back over into a glacial period. Evidently as a group we’re too stupid to know what’s good for us.
Actually it appears that Mars too was once a liquid water world but it didn’t last long. Mar’s problem is insufficient gravity. Gravity is what gives us our air pressure of 14.7psi at sea level. This in turn raises the boiling point of water 212F above the freezing point. Mar’s lower gravity lowers the boiling point of water at the surface a lot so its oceans literally boiled away. For a time the excessive water vapor in the atmosphere would have given it an uber-greenhouse effect but slowly but surely it was electrolized by energetic radiation and blown off by solar wind – never to return. Mars is smaller than the earth so its liquid iron core cooled off a lot faster. A liquid iron core generates a magnetic field which deflects high energy radiation. The tipping point for Mars was probably when the iron core cooled off enough to solidify and its radiation shield vanished paving the way for it to be stripped of its water.
>>Robert Clemenzi says: May 9, 2011 at 11:56 pm
>>“Because it has been measured and N2 does not radiate at the
>>temperatures normally found at the surface.”
>Dave Springer says: May 10, 2011 at 6:32 am
>ALL matter with a temperature above absolute zero radiates.
>This is one of the most fundamental laws of physics. Either you
>are making things up out of thin air or the source you got it from
>doesn’t know his ass from elbow.
I think the point is that N2 doesn’t radiate APPRECIABLE amounts of energy at typical atmospheric temperatures.
The amount of energy radiated depends on the emissivity. Highly polished metal has an emissivity ~ 0.03 and will not radiate much energy. Rough, oxidized, dark surfaces often have an emissivity above 0.9. So it is easy for one material to emit 30x more radiation than another at the same temperature.
It is easy to confirm that even many meters of N2 have little effect on IR – neither absorbing nor emitting appreciable amounts. The image at the top of the post — showing the IR spectrum looking up — shows quite clearly that for most wavelengths, there is almost no IR emitted by the atmosphere. The bulk of the IR that is seen is clearly attributed to H2O & CO2. Hence the entire column of N2 overhead emits no radiation (or close enough to zero for practical practical).
The original statement is (for all practical purposes in climate modelling that I can imagine) correct.
I presented evidence for my claim. Can you present evidence that shows IR emission from N2 is appreciable in any situations?
The toy global circulation models take latent heat flux into account so it isn’t missing although one might question the accuracy of the number or how it changes with temperature. The problem lies in how they model clouds. They believe that clouds have a net warming effect so as temperature rises from additional CO2 we get more clouds and more clouds cause even more warming. Clearly this is wrong because the end result is a runaway greenhouse and the earth has never in billions of years experienced a runaway greenhouse even when as recently as several million years ago CO2 levels were far higher than we could ever raise them through fossil fuel consumption. There just ain’t enough economically recoverable fossil fuel to get CO2 concentration into undiscovered territory. It’s questionable whether we can raise it high enough to cancel the effect of the Milankovich cycle which triggers glacial periods.
Prior to 3 million years ago Milankovich cycles were not enough to trigger ice ages. It’s thought that the arrangement of continents are the other contributing factor. Depending on how the continents are arranged the global ocean conveyor belt changes and having a land mass over a pole blocks warm water from getting at the ice to melt it. Currently we have one large continent directly over a pole and one pole with none. The water sequestered at the south pole is a large amount which reduces the surface area of the global ocean and because land is higher albedo than water it raises the planetary albedo and thus cools it. If we had a continent over both poles at the same time we’d probably get a snowball earth episode that would last until either CO2 built up in the atmosphere to melt it or the continents drifted off the poles or some combination of both.
CO2 extinction altitude can be seen in the spectrum looking down from 20 kilometers above the arctic ocean. IIRC the spectrum follows a 270K blackbody curve where the atmosphere is IR transparent i.e. it “sees” the temperature of the ocean surface but in the 15um region it drops down to follow a 250K blackbody curve. Going by dry adiabatic lapse rate (arctic air is fairly dry) of 1K per 100 meters the IR sensor is “seeing” the air temperature in the 15um range at a height of 2000 meters. I’m not an optics expert but I believe that altitude represents the optical depth of the atmosphere at 15um or in other words the extinction altitude. In any case there is no way in hell the extinction depth is in tens of meters.
One of the concerns about increasing CO2 even if it doesn’t raise surface temperature all that much is that as it increases the extinction point will be lowered and near-surface adiabatic lapse rate will change as a result potentially causing changes in the weather but again I must go back to the indisputable testimony of the fossil record which reveals that the earth is increasingly friendly to life as CO2 rises even when it rises to 20 times the current level (greater than four doublings) so one might reasonably conclude it doesn’t cause climate disaster.
Dave Springer:
“ALL matter with a temperature above absolute zero radiates. This is one of the most fundamental laws of physics. Either you are making things up out of thin air or the source you got it from doesn’t know his ass from elbow.”
Huh? I hope you are not saying that everything radiates in the IR frequencies. To repeat a comment I made above somewhere:
http://www.wpi.edu/Academics/Depts/Chemistry/Courses/General/infrared.html
The key paragraph:
The fact that a molecule vibrates does not in itself insure that the molecule will exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the vibrational motion associated with that mode must produce a change in the dipole moment of the molecule. HCl, for example, with a center of positive charge at the H atom and a center of negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further, stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that cause changes in the value of the dipole moment from 0 to some non-zero value. An example is CO2, which has no permanent dipole moment because the individual bond dipoles exactly cancel. However, when CO2 undergoes a bending vibration, its dipole moment changes from zero to some non-zero value. This vibration produces a change in dipole moment and is IR active.”