A guest post by Ken Coffman and Mikael Cronholm
In clicking around on the Internet, I found an outstanding paper called Thermodynamics of Furnace Tubes – Killing Popular Myths about Furnace Tube Temperature Measurement written by Mikael Cronholm. The paper was clever and wise…and made a lot of sense. Clearly Mikael knows a lot about infrared radiation and I’m a guy with questions. A match made in heaven?
We exchanged e-mails. I want to be clear about this…Mikael corrected some of my wrong ideas about IR. I’ll repeat that for the slow-witted. Some of my ideas about infrared radiation were wrong. I am considered a hard-headed, stubborn old guy and that’s completely true. However, I want to learn and I can be taught, but not by knuckleheads spewing nonsense and not by authoritarians who sit on thrones and toss out insults and edicts.
Ken Coffman (KLC) is the publisher of Stairway Press (www.stairwaypress.com) and the author of novels that include Hartz String Theory and Endangered Species.
Mikael Cronholm (MC) is an industry expert on infrared radiation, a licensed, level III Infrared Training Center Instructor and holds two Bachelor of Science degrees (Economics and Business Administration).
The following is a summary of our conversation.
KLC: Hello Mikael. I found your paper called Thermodynamics of Furnace Tubes and I found it very informative, practical and interesting. I hope you’ll bear with me while I ask a couple of dumb questions. I am an electrical engineer, so I have some knowledge about thermodynamics of conduction and convection, but not so much about IR radiation. In return for your time, I would be happy to make a donation to the charity of your choice.
If I take an inexpensive IR thermometer outside, point it at the sky and get a temperature reading of minus 25°C, what am I actually measuring? Is there anything valid about doing this?
MC: Just as a matter of curiosity, how did you find my paper? I checked your website and I guess this has to do with the Dragon, no? If you want to make a donation I would be happy to receive that book. If you can, my postal address is at the bottom. I don’t follow the debate more than casually, but I am a bit skeptical to all the research that is done on climate change…it seems that the models are continuously adjusted to fit the inputs, so that you get the wanted output…and they argue “so many scientists agree with this and that”…well, science is not a democracy…anyway…
About radiation, then. There is more to this than meets the eye. Literally!
Looking at the sky with an infrared radiometer you would read what is termed “apparent temperature” (if the instrument is set to emissivity 1 and the distance setting is zero, provided the instrument has any compensation). Your instrument is then receiving the same radiation as a blackbody would do if it had a temperature of -25°C, if that is what you measure. It is a quasi-temperature of sorts, because you don’t really measure on a particular object in any particular place, but a combination of radiation, where that from outer space is the lowest, close to absolute zero, and the immediate atmosphere closest to you is the warmest. (I have once measured -96°C on the sky at 0°C ground temperature.) What we have to realize though, is that temperature can never be directly measured. We measure the height of a liquid in a common thermometer, a voltage in a thermocouple, etc, and then it is calibrated using the zeroth law of thermodynamics and assuming equilibrium with the device and the reference.
KLC: Global warming (greenhouse gas) theory depends on atmospheric CO2 molecules absorbing IR radiation and “back radiating” this energy back toward the earth. If you look at the notorious Ternberth/Keihl energy balance schematic (as shown in Figure 1 of this paper: http://www.cgd.ucar.edu/cas/Trenberth/trenberth.papers/TFK_bams09.pdf ), you see the back radiation is determined to be very significant…more than 300W/m2. From your point of view as an IR expert, does this aspect of the global warming theory make any sense?
MC: The paper you sent me mentions Stefan-Boltzman’s law, but it does not talk about Planck’s law, which is necessary to understand what is happening spectrally. I suggest you read up on Planck and Stefan-Boltzman at Wikipedia or something. Wien’s law would be beneficial as well—they are all connected.
Planck’s law describes the distribution of radiated power from a blackbody over wavelength. You end up with a curve for each blackbody temperature. The sun is almost a blackbody, so it follows Planck quite well, and it has a peak at about 480nm, right in the middle of visual (Wien’s law determines that). The solar spectrum is slightly modified as it passes through the atmosphere, but still pretty close to Planckian. When the radiation hits the ground, the absorbed part heats it. The re-radiated power is going to have a different spectral distribution, with a peak around 10um (micrometer). Assuming blackbody radiation it would also follow Planck’s law.
S-B’s law is in principle the integral of Planck from zero to infinity wavelength. Instruments do not have equal response from zero to infinity, but they are calibrated against blackbodies, and whatever signal they output is considered to mean the temperature of the blackbody. And so on for a number of blackbodies until you have a calibration curve that can be fitted for conversion in the instrument.
That means that the instrument can only measure correctly on targets that are either blackbodies, or greybodies with a spectral distribution looking like a Planck curve, but at a known offset. That offset is emissivity, the epsilon in your S-B equation in that paper. It is defined as the ratio of the radiation from the greybody to that of the blackbody, both at the same temperature (and wave length, and angle…). Some targets will not be Planckian, but have a spectral distribution that is different. If you want to measure temperature of those, you need to measure the emissivity with the same instrument and at a temperature reasonably close to the one you will measure on the target later.
So, of course, the whole principle behind the greenhouse effect is that shorter wavelengths from the sun penetrates the atmosphere easily, whereas the re-radiated power—being at a longer wavelength—is reflected back at a higher degree. I have no dispute about that fact. It is reasonable. So I think the Figure 1 you refer to is correct in principle. My immediate question is raised regarding the numbers in there though. The remaining 0.9 W/m2 seems awfully close to what I would assume to be the inaccuracies in the numbers input to calculate it. You are balancing on a very thin knifes edge with such big numbers as inputs for reaching such a small one. An error of +/- 0.5% on each measurement would potentially throw it off quite a bit, in the worst case. But I don’t know what they use to measure this, only that all the instruments I use have much less accuracy than that. But with long integration times…well, maybe…but there may be an issue there.
KLC: I am interested in some rather expensive thermopile-based radiation detectors called pyrgeometers (an example is the KippZonen CGR 3 instrument http://www.kippzonen.com/?product/16132/CGR+3.aspx).
If a piece of equipment like this is pointed into the nighttime sky and reads something like 300W/m2 of downwelling IR radiation, what is it actually measuring? If I built a test rig from IR-emitting lightbulbs calibrated to emit 300W/m2 and placed this over the pyrgeometers, would I get the same reading?
MC: “What is it actually measuring?” Well, probably a voltage from those thermopiles…and that signal has to be calibrated to a bunch of blackbody reference sources to covert it either to temperature or blackbody equivalent radiation.
Your experiment will fail, though! If you want to do something like that, you have to look at a target emitting a blackbody equivalent spectrum, which is what the instrument should be calibrated to. IR light bulbs emitting 300W/m2 is simply impossible, because 300W/m2 corresponds to a very low temperature! Use S-B’s law and try it yourself. Like this: room temp, 20°C = 293K. The radiated power from that is 293K raised to the power of 4. Then multiply with sigma, the constant in S-B’s law, which is 5.67*10-8, and you get 419 W/m2 or something like that, it varies with how many decimals you use for absolute zero when you convert to Kelvin. For 300 W/m2 radiation I get -23.4°C at 300 W/m2 when I calculate it (yes, minus!). Pretty cool light bulb.
I don’t know what your point is with that experiment, but if it is to check their calibration you need a lot more sophisticated blackbody reference sources if you want to do it at that temperature. But you could do a test at room temperature though. Just build a spherical object with the inside painted with flat black paint, make a small hole in it, just big enough for your sensor, and measure the temperature inside that sphere with a thermocouple, on the surface. Keep it in a stable room temperature at a steady state as well as you can and convert the temperature to radiation using S-B’s law. You should get the same as the instrument. Any difference will be attributable to inaccuracy in the thermocouple you use and/or the tested instrument. Remember that raising to the power of 4 exaggerates errors in the input a lot!
I hope I have been able to clarify things a little bit, or at least caused some creative confusion. When I teach thermography I find that the more you learn the more confused you get, but on a higher level. Every question answered raises a few more, which grows the confusion exponentially. It makes the subject interesting, though.
Let me know if you need any more help with your project!
KLC: I found your paper because one of the FLIR divisions is local and I was searching their site for reference information about IR radiation. I know what a 100W IR lamp feels like because I have one in my bathroom. If someone tells me there is 300W/m2 of IR power coming from space, and I hold out my hand…I expect to feel it. What am I missing?
MC: Yeah, you put your hand in front of a 100W bulb, but how big is your hand…not a square meter, I’m sure. It is per area unit, that is one thing you are missing. The 100W of the bulb is the electrical power consumption, not the emitted power of the visual light from it. That’s why florescent energy-saving lamps as opposed to incandescent bulbs give much more visual light per electrical Watt, because they limit the radiation to the visual part of the spectrum and lose less in the IR, which we cannot see anyway. The body absorbs both IR and visual, but a little less visual.
And, here is the other clue. Your light bulb radiation in your bathroom is added to that of the room itself, which is 419 W/m2, if the room is 20°C. Your 300 W/m2 from space is only that. You will feel those 300 W/m2, sure. It will feel like -25°C radiating towards your hand. But you don’t feel that cold because your hand is in warmer air, receiving heat (or losing less) from there too.
Actually, we cannot really feel temperature—that is a misconception. Our bodies feel heat flow rate and adjust the temperature accordingly. It is only the hypothalamus inside the brain that really has constant temperature. If you are standing nude in your bathroom, your body will radiate approximately 648 W/m2 and the room 419 W/m2, so you lose 229 W/m2. That is what you feel as being cooled by the room, from radiation only. Conduction and convection should be added of course. The earth works the same way—lose some, gain some. It is that balance that is being argued in the whole global warming debate.
KLC: I still feel like I’m missing something. IR heat lamps are pretty efficient, maybe 90%? Let’s pick a distance of 1 meter and I want to create a one-square meter flooded with an additional 300W/m2. It must be additional irradiation, doesn’t it? That’s going to take a good bunch of lamps and I would feel this heat. However, I go outside and hold out my hand. It’s cold. There’s no equivalent of 300W/m2 heater in addition to whatever has heated the ambient air.
Perhaps I’m puzzled by something that is more like a flux…something that just is as a side-effect of a temperature difference and not really something that is capable of doing any work or as a vehicle for transporting heat energy.
It’s a canard of climate science that increasing atmospheric CO2 from 390PPM to 780PPM will raise the earth’s surface temperature by about 1°C (expanded to 3°C by positive feedbacks). From my way of thinking, the only thing CO2 can do is increase coupling to space…it certainly can’t store or trap energy or increase the earth’s peak or 24-hour average temperature.
Any comments are welcome.
MC: Efficiency of a lamp depends on what you want, if heat is what want then they are 100% efficient, because all electrical energy will be converted to heat, the visible light as well, when it is absorbed by the surrounding room. If visible light is required, a light bulb loses a lot of heat compared to an energy saving lamp. Energy cannot be created or destroyed—first law of thermodynamics.
When you say W/m2 you ARE in fact talking about a flux (heat flow is what will be in W). If you have two objects radiating towards each other, the heat flow direction will be from the hotter one, radiating (emitting) more and absorbing less, to the cooler one, which radiates less and absorbs more (second law of thermodynamics). The amount of radiation emitted from each of them depends on two things ONLY, the temperature of the object and its emissivity. So radiation is not a side effect to temperature, it is THE EFFECT. Anything with a temperature will radiate according to it, and emissivity. (If something is hotter than 500°C we get incandescence, emission of visible light.) Assuming an emissivity of unity, which is what everyone seems to do in this debate, the radiation (flux. integrated from zero to infinity) will be equal to what can be calculated by Stefan-Boltzmann’s law, which is temperature in Kelvin, raised to the fourth power, multiplied by that constant sigma. It’s that simple!
With regard to your thought experiment, it is always easier to calculate what an object emits than what it absorbs, because emission will be spreading diffusely from an object, so exactly where it ends up is difficult to predict. I am not sure where you are aiming with that idea, but it does not seem to be an easy experiment to do in real life, at least not with limited resources.
CO2 is a pretty powerful absorber of radiated energy, that fact is well known. Water vapor is an even stronger absorber. In the climate debate it is also considered a reflector, which probably also true, because that is universal. Everything absorbs and reflects to a degree. So I guess that the feedback you mention has to do with the fact that increasing temperature increases the amount of water vapor, which increases absorption, and so on. But my knowledge is pretty much limited to what happens down here on earth, because that is what matters when we measure temperature using infrared radiation. However, it is important to remember, again, that we talk about different spectral bands, the influx is concentrated around a peak in the visual band and the outgoing flux is around 10 micrometer in the infrared band, and the absorption may not be the same.
With so many scientists arguing about the effects of CO2 I am not the one to think I have the answers. I really don’t know what the truth is. And the problem that all these scientists have is that they will never be able to test if their theories are correct, because the time spans are too long. For a theory to be scientifically proven, it has to be stipulated and tested, and the test must be repeatable and give the same results in successive tests for the theory to be proven.
If not, it is not science, it is guessing.
More like a horoscope…
George Smith
dave Springer maintains that ghg’s – particularly c02 slows the rate of cooling of the atmosphere, and uses the analogy of a thermal flask (which limits conduction and convection.
i’m arguing that c02 doesn’t make much measurable difference to atmospheric heat content and cool rate
although, George, it depends what posts you’re referring to
steven mosher [February 13, 2011 at 3:39 pm] says:
“yup, but there is more to it than that. If you want to do IR design today ( of sensors, or counter measures or stealth technology) the physics you use is the same physics that many skeptics deny. good thing we don’t let them design the machines that our safety as a nation depends upon.“
Got out of bed the wrong side today Mosh??? Only an idiot would think that tarring all sceptics with the same brush is a sensible way to have a useful debate.
My own take on this is simple. Although lab experiments show CO2 could raise Earth’s overall temperature by a tiny amount, the IPCC’s predictions of catastrophic temperature increases produced by carbon dioxide have been challenged by many scientists and found to have no substance in the real world.
The importance of water vapour is frequently overlooked by environmental activists and by the media. The large temperature increases predicted by many computer models are non-physical and inconsistent with results obtained by basic measurements – much scepticism is warranted when considering computer-generated projections of global warming when these same models cannot even predict existing observations.
ok, I think i gleaned something re: the shoulders of co2. 15microns is the peak.
all I can fathom is that Dav Springer argues that the entire shoulders and peaks of c02 count, but that at the shoulders – or else the band either side around 15microns – then they become miniscule as heat absorbers. At the tropospheric level where the outer bands absorb, there are more oxygen and nitrogen competing for heat than c02
A C Osborn says:
February 14, 2011 at 9:40 am
Oliver Ramsay says:
February 14, 2011 at 8:51 am “In fact the warmer the surface becomes the more the energy will be radiated to space without bouncing around in the atmosphere.”
WUWT?
So the surface has a direct connection to space then?
—————
Yes
cal says:
February 14, 2011 at 8:27 am
Most of the energy the earth receives is visible light from the sun. The 5200K blackbody spectrum carries a significant but minor fraction of its total energy in the near infrared but no practical amount of the energy is in the far infrared where CO2 absorption begins. Conversely the earth (discounting reflected visible light during the day) emits essentially no energy in the near infrared and essentially all of it in the far infrared. The amount of the energy emitted in the far infrared necessarily over time is equal to the amount of shortwave energy absorbed by the ocean durin the day. Near infrared hitting the ocean surface is absorbed within the first few microns and doesn’t heat the ocean but is rather carried off in water vapor which rises by convection where the energy is released when the water vapor condenses into a cloud. Visible light however penetrates over 100 meters and is almost completely absorbed with very little of it reflected. Land surfaces to a small degree and clouds to a high degree reflect incoming visible light straight back out into space. Estimates of the earth’s average albedo are in the 35% range plus or minus a few percentage points depending on who you ask and what they need to stick into their climate models to better reproduce paleo-climate data. Albedo is used as a fudge factor to tune climate models for better fitting climate hind-casts.
So anyhow about 60% of the solar energy arriving at the top of the atmosphere works to heat the ocean and all that energy eventually radiates out at night at much lower frequency. It’s the difference in CO2 frequency response to short wave versus long wave radiation that allows it to act as an insulator. It’s transparent to short wave radiation and opaque at a few significant long wave frequencies so it almost all solar energy through to the ocean unimpeded during the day but impedes long wave energy emitted by the surface at night. The simplest and quite accurate way to conceptualize this is that CO2 is an insulator – it’s like a blanket over a dark rock where you remove the blanket during the day so the rock can heat up in the sun and you put the blanket back at night to retain more of the warmth from the sun during the night. The end result of blanket vs. no blanket is a warmer rock with the blanket than the rock would be without it.
Mikael Cronholm says:
February 13, 2011 at 9:56 pm
Ian W says:
February 13, 2011 at 12:35 pm
When water vapor in the atmosphere condenses into liquid water and then changes state again and becomes ice, it gives off latent heat for both state changes.
Does that latent heat release follow Stefan-Boltzmann’s radiative equation ?
No Ian, latent heat is not really related to S-B.
So if you consider all the clouds and it is estimated that 62% of earth’s surface is covered by clouds – the amount of energy released as latent heat is huge. Yet people persist in working out the amount of heat leaving the surface using Stefan-Boltzmann radiation equations. This is obviously incorrect.
izen says:
February 14, 2011 at 8:15 am
Some energy leaves the SURFACE by convection, but it does not get very far.
It is not clouds that block it… strange idea!
Its the adiabatic lapse rate.
Convection can only move air until the increased bouancy from the lower number of molecules per cubic metre is offset by the lower density with increasing altitude. It is the lower density of the atmosphere with altitude that blocks convection.
The low altitude of most clouds indicates that the temperature falls below freezing just a few Km above ground level and the atmosphere regains that latent heat of evaporation well below the tropopause. Above the main low cloud layer water vapor is much less important as an atmospheric absorber/emitter.
Well I don’t know about ‘not getting very far’ 60,000ft or more in the tropics is the level of the tropopause (the top of the convective atmosphere) and in the Inter-Tropical Convergence Zone (ITCZ) the cloud tops of towering storms are (by definition) up to that level. (A recent Northwest Airlines flight experienced liquid rain hitting the aircraft in cruise above 30,000ft in a storm between Hong Kong and Tokyo despite the ambient outside-air-temperature being well below zero. The rain froze into ice on hitting the airframe) The height of the cloud tops all depends on why the clouds are there.
Humid air is more buoyant than dry air, so just being more humid is enough for a volume of air to start convection. Drier air sucked in will then pass over the wet surface and Henry’s law applies and more water will evaporate into the dry air cooling the surface. No external heat source is required. As the humid air rises the air temperature will drop at the wet adiabatic lapse rate and water will start to condense around nucleation particles and form clouds. This type of cloud with a gentle wind will become the low stratocumulus over oceans and coastlines. And each and every cloud droplet has taken heat from the surface and radiated it as it becomes a droplet and again when it becomes ice.
With the application of tropical heat and humidity convective storms develop every day some towering ten miles high into the atmosphere. If the conditions are conducive these storms merge and the Coriolis force on the air being rapidly drawn in at the base of the storms starts the winds in the storm system to rotate. There is a lot of discussion of the ACE index here, but to put it in perspective, the amount of energy released by a hurricane can be calculated based on the latent heat emitted and / or the kinetic energy.
An average hurricane in a day transfers latent heat equivalent to 200 times the world-wide electrical generating capacity.
http://www.aoml.noaa.gov/hrd/tcfaq/D7.html
As confirmed in response to my question at the top of this thread, the temperature of the cloud has no bearing on the amount of latent heat being released. The heat being released is linked to the amount of water changing state and the latent heat of that state change.
The missing heat is almost certainly in the misunderstood and underestimated hydrologic cycle that is ignored in favor of a simple Stefan-Boltzmann radiation equation.
Slacko,
Following up on the reply already by cal, solar radiation contains a small fraction of IR, maybe only a few W/m2, so most of the warmth you feel is from absorbing the visible/UV rays.
The other part of Slacko’s question: yes, N2 can’t radiate IR, nor can O2 or argon. This is from physics.
Robert Clemenzi,
Yes, 324 W/m2 corresponds to 34 F. This is supposed to be a global and annual average value, which obviously varies greatly locally. I think your suggestion is that it implies reflection because the clear atmosphere is probably colder than this. Actually clouds behave as almost perfect black bodies, so the cloud bases are emitting at their temperature, which may account for the average being as high as 324 W/m2. In the tropics even clear skies can emit more than 324 W/m2 to the surface because of the high H2O contents.
The paragraph about Mikael Cronholm (MC) states … ‘and holds two Bachelor of Science degrees (Economics and Business Administration)’.
I think it likely that’s a typo, and should read ‘Bachelor of Arts’.
Of course I could be wrong. Are there really universities that issue Bachelor of Science (Economics) and Bachelor of Science(Business Administration) degrees?
“”””” at 15microns peak, that makes not an iota of difference to atmospheric temperatures. Besides, gases are not blackbodies, and are 3 dimensional. Radiative equations can’t be applied to them (air has a very poor conductor), and so is irrelevant to climatology. With ghg’s, *blocking* operates in all directions, so doesn’t make any difference to the atmospheric heat content. “””””
Either P.Wilosn or Dave Springer said the above; well it contains both of their names at the top. I’m inclined to discount Dave, as the source of that statement; but can’t tell where if anywhere Dave stops, and P. starts.
In any case, while it may be pedantically true that gases aren’t black bodies; we know that since absolutely nothing real is a black body.
But where in the Physics texts does it say that water or ice stops radiating thermal radiation, the instant that it changes from a liquid or solid, into a gas at the same Temperature. Has anybody ever recorded a motion picture video of that shut-off process actually taking place in any Physical system.
Can somebody cie some peer reviewed literature that forgives gases from radiating thermal radiation; that is Electromagnetic Radiation of a spectral nature that depends on on the Temperature of the gas. And where in that peer reviewed paper, did it say that the sun is further excused from obeying the non thermal EM radiation prohibition of gases ?
One thing we can be fairly sure of, is that the Raleigh scattering of short wavelength sunlight, which makes the entire daytime sky appear “Sky blue” when viewed in any direction (not counting near the sun); and that means it looks the same looking up as looking down; because the large angle scattering due to the RS process makes the atmosphere an isotropic source of sky blue light which originally came from the sun.
By the same token, the LWIR emissions from the atmosphere; whatever their source; are slso isotropic, since there is no preferred dirction of emission. So the infrared sky looking up towards space is pretty much the same as looking down from outer space; except (apparently) that according to Trenberth, 40 W/m^2 of LWIR radiation that is actually emitted from the earth’s surface, actually escapes unharmed to space, and the other 350 W/m^2 is absorbed by the atmosphere. Apparently per P.Wilson, it cannot be subsequently emitted from the atmosphere per the radiation laws governing BBs and other thermal radiation laws; which he says don’t apply to gases.
So now would somebody not as dense as I am, like to explain to me what is the source of quite thermal spectrum looking (grey body) radiation that is seen from outer space, when looking at the earth. Onl;y 40 W/m^2 of it can be coming from the ground per Trenberth and it does not have a narrow CO2 absorption band spectrum as one would expect, if direct emission of molecular resonance radiation spectra from CO2 was the source.
Several have noted that a lot of “heat” transport to the upper atmosphere is a result of convection of hotter surface gases into the upper regions. Somebody could explain for us how that eventually escapes to space, seeing as how gases can’t radiate thermal spectra according to the black body or S-B radiation laws.
Clearly the ten years, that I formally studied Physics in School, plus the subsequent 50 years of using it daily in practice to accomplish things; was not enough for me to come across that jewel of knowledge, that gases do not radiate according to the radiation laws.
Readers might also find it intersting to look at Fi 11; “Higher Members of the Balmer Series of the H Atom (in Emission) Starting from the Seventh Line, and showing the Continuum (Hertzberg41) ”
The reference to “hertzberg 41” is the G. Herzberg Ann. Physik (4) 84, 565 1927 citation
And that photo of an actual real measured, scientific observation of a (non teracomputer simulation of) Balmer spectrum including a “Continuum Spectrum” along twith the Balmer line specrum is in; “Atomic Spectra and Atomic Structure” by Gerhard Herzberg; Prentice Hall 1937. Herzberg mentions that in Emission the continuum corrsponds to a free electron of any energy being captued by a proton, adn going in to “the orbit” having the principal quantum number n=2 (this is of course in the Bohr Atom interpretation. Modern Quantum formulations may be different.)
And nowhere does temperature get mentioned here. This continuum radiation still corresponds to an energy level transition; in this case from an ionised state. Well of course you won’t find any ionised states on the sun I would imagine. that spectrum is commonly observed in stars. But The end of the Balmer series, is at about 3800 Angstroms; so that continuum is in the UV region; not in the LWIR region where the thermally originated LWIR emission spectrum occurs.
The extraterrestrial LWIR spectrum from the earth, does show a narrow spectral dip at the 9.6 micron Ozone band, and a wider one at the 15 micron CO2 band from about 13.5 to 16.5 microns. otherwise it looks pretty much like any other near bB-spectrum as limited by the Planck and S-B laws (Wien also)
The question posed early on this thread from KLC; ” . . . From your point of view as an IR expert, does this aspect of the global warming theory make any sense?”
RE: “. . . notorious Ternberth/Keihl energy balance schematic (as shown in Figure 1 of this paper: http://www.cgd.ucar.edu/cas/Trenberth/trenberth.papers/TFK_bams09.pdf ), you see the back radiation is determined to be very significant…”
Maybe I missed it but was there a summary or (for want of a better term) a consensus from the posters here that can answer this with some authority? MC replied he had issues with the numbers used. I was wondering what others thought re: reasonable numbers if those referred to (fig 1) are not reasonable.
MC also noted, “Every question answered raises a few more, which grows the confusion exponentially. ”
I was hoping to clear up some confusion on my part.
-Barn
“”””” P Wilson says:
February 14, 2011 at 9:44 am
although, George, it depends what posts you’re referring to “””””
I cut and pasted the entirety of the post headed by your name, and containing a reference to Dave Springer. That post, and only that post was being referred to by me.
i see. That was entirely my post. As a response to Dave Springer. I just cut and pasted his name and time to show what post I was replying to
Hope that clears things up
barnErubble,
MC was not concerned with the main number of > 300 W/m2 from IR as back radiation in those papers. He was concerned with the accuracy of the residual 0.9 W/m2, and how anyone could accurately state such residuals given the large canceling numbers involved. This is rather related to Trenberth’s “missing energy” issue, and is a valid concern.
George E. Smith says:
February 14, 2011 at 10:28 a
“Clearly the ten years, that I formally studied Physics in School, plus the subsequent 50 years of using it daily in practice to accomplish things; was not enough for me to come across that jewel of knowledge, that gases do not radiate according to the radiation laws.”
of course they do, as only you would know. I’m referring to the SB equation in particular as applied to c02, and the fanciful results on paper that it generates. I think you’ve implied before that except for a tiny fingerprint of black body radiation, most radiation evades c02 in its escape to space.
Before energy escapes into space through radiation, it doesn’t matter whether the energy is in contact with CO2, water vapor or nitrogen.
Jim D says:
“solar radiation contains a small fraction of IR, maybe only a few W/m2, so most of the warmth you feel is from absorbing the visible/UV rays.”
Actually, more than 50% of the solar energy at the top of the atmosphere is IR. At the surface, the percent is increased because some of the UV was absorbed at a higher level. On a cloudy day you can get a sunburn. However, since you will not feel a temperature difference between overcast light and full shade, the “heat” you feel on a sunny day will not be from the UV.
Steeptown mentioned:
…it is evident to me that the radiative efects of CO2 in the atmosphere are of 2nd or 3rd order compared to the radiative, convective and latent heat effects of H2O.
Well, when do 2nd order effects become important? The problem with increasing CO2 in the atmosphere is the ever increasing 2nd order effect. So a little bit of energy doesn’t escape. No big problem. But, if the effect is compounded, which appears to be the case for CO2, suddenly a little bit of energy becomes a lot of energy. When do we care?
Water in the atmosphere has two very different properties versus CO2. First, the amount of water in the atmosphere is not likely to double unless there is a major change in the atmosphere. So even if H2O is a first order effect, the much smaller changes in the amount of water is not going to be significant. However, CO2 is increasing fairly quick. And a doubling or possibly tripling the amount of CO2 is very real so the second order impact is likely to be much larger than the small deviations of the first order impact. Second, CO2 is throughout the entire atmosphere. Sure, it is only 0.04% of the atmosphere. But if we look at the entire atmosphere, not just the surface, water is only 0.4%. Doubling or tripling CO2 is now ~0.1 %. Suddenly CO2 is not that small after all.
Something else folks seem to be forgetting with CO2 is the bands it absorbs. In particular, it is absorbing in bands that water is not. At lower levels of CO2, a sufficient amount of LW radiation was escaping to provide an energy balance. With the increased amount of CO2, the balance has been upset and the Earth is trying to reach a new equilibrium.
Some folks are very taken with Mr. Cronholm’s one comment:
For a theory to be scientifically proven, it has to be stipulated and tested, and the test must be repeatable and give the same results in successive tests for the theory to be proven.
This may be true, but we only have one planet. We don’t have an alternate Earth that we can play with to see what happens. Also, tested does not necessarily mean run in a lab or a model. Consider evolution. We don’t “test” evolution, we look at the available data and the composite result from this evidence is the fundamental reasons for evolution. Climate change is similar: we look at temperature, CO2, extended growing seasons, northern and southern boundaries of migrating birds and insects, extreme weather events, receding glaciers, etc.
One comment that has been discussed time and again is that in the geological record, an increase in CO2 occurs after an increase in temperature. Fine, let’s assume that is true. The problem we have is the measurements we are making now show CO2 increasing and temperature following. What does it mean? Simple: a different mechanism is now responsible, not the natural processes we have been able to identify from past changes.
Of course, there could be another natural phenomenon we have not found/discovered. But as Dr. Alley from Penn State has pointed out: we built the satellites, we made the measurements, when someone disagreed we made new satellites and new measurements. We have spent billions of dollars and so far the overwhelming evidence is in support of AGW.
Think of it like Children’s Hospital. The doctors there have one job: to put themselves out of a job. If they could solve all the medical issues they see, they would no longer be needed. And that is actually their goal! They don’t want kids to suffer. Do people really think climate change scientists want AGW to be true? If they are right, the end result is really bad.
AC Osborne: re photon-photon collision
I am not a physicist but, back when I went to college and took physics, I was taught that photon-photon collisions are not possible. Being virtually massless and only possessing momentum, two photons can only indirectly interact with each other. Their interaction is called Delbrück scattering. When they “interact” they annihilate into a virtual electron-positron pair, which then annihilates back into two real photons again. They lose no momentum or energy in the process.
Don’t know if that is completely right, my memory is hazy, but my rule of thumb has always been that for first approximations photons are massles and that massless quanta’s of energy basically just go right thru each other.
Tenuc says:
February 14, 2011 at 9:46 am
steven mosher [February 13, 2011 at 3:39 pm] says:
“yup, but there is more to it than that. If you want to do IR design today ( of sensors, or counter measures or stealth technology) the physics you use is the same physics that many skeptics deny. good thing we don’t let them design the machines that our safety as a nation depends upon.“
Got out of bed the wrong side today Mosh??? Only an idiot would think that tarring all sceptics with the same brush is a sensible way to have a useful debate.
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I wrote: “the same physics that many skeptics deny”
You accused me of “tarring all sceptics”
There is a reason why I chose the word “many”
There is a reason why I did not use the term “all” or “Most”
The reason was to trap idiots who are not careful with words.
I found the explanations both fundamentally accurate to my understanding level and very nicely presented in what should be a very understandable way.
One major error though is the misconception that 0.9w/m^2 imbalance is measured. It is a model calculation based upon plenty of presumptions. The references to the CERES & ERBE data actually turn up to have accuracy problems of several watt’s per m^2 so despite all the discussions in the papers, they resort to modeling to get their value. In the earlier papers, such as hansen’s, the uncertainties in the modeling for such things as cloud formation versus T are discussed and decisions made that gives rise to their claim that cloud formation decreases with temperature which is not the case for all presumption options like cloud formation depending upon absolute humidity levels.
Don V :
February 14, 2011 at 1:47 am
Thank you Don, well thought out comment!
Found not a single thing within I would argue about.
You have a clear mind. Could we talk some physics here? I have one big question I need some help answering (or try to answer at least, and, it’s on a very simple logical level).
Barn,
the details isn’t so much radiative theory as it is to the system of Earth and its atmosphere. You can rather safely bet that the warmer crowd will minimize every number that limits AGW and exaggerate every number that supports it.
Primary amongst the real uncertainties is that of the albedo and cloud cover and the behavior of the cloud cover with conditions such as temperature. The whole premise being used that there is one surface temperature average for a given blockage of outgoing IR is totally flawed. While incoming solar average power is given great importance, the notion of what the albedo reflects away is often considered to be constant and not the actual variable it really is, driven by many other factors. If you consider the comments of the warmers, they give lip service to land use changes and loss of snow and ice cover as being the important factors with albedo. Funny how about 80% + of the albedo is due to a 60% + cloud cover and of that under 20% surface contribution, around 70% is oceans and that leaves next to nothing for contributions from the land surfaces, never mind ice and snow and a little rain forest or two.
K&T97 admit possible errors of up to 20% in their cartoon values. Amazing how close they got to that by underestimating cloud cover and overestimating land and surface contribution. It actually looks like they forgot to take cloud cover into account and gave a clear sky only value. Later they use the same values and apparently forgot to include their serious margin of error.
Ian W says:
February 14, 2011 at 10:06 am
“As confirmed in response to my question at the top of this thread, the temperature of the cloud has no bearing on the amount of latent heat being released. The heat being released is linked to the amount of water changing state and the latent heat of that state change.
The missing heat is almost certainly in the misunderstood and underestimated hydrologic cycle that is ignored in favor of a simple Stefan-Boltzmann radiation equation.”
The ‘Science of Doom’ site linked to from here (top right of page) has a good discussion of some of these issues.
The amount of energy transported by latent heat changes is complex, there is one simple way to quantify it however. The total global yearly rainfall is a direct measure of how much energy was moved by evaporation and condensation in the hydrological cycle.
But the first LoT kicks in. The energy is still around. The only way it can get of the planet is by radiating IR photons into space. All the hydrological cycle can do is move the location of that emission around. That evens out the temperature so that water does not boil at the equator and can melt at the poles, but the same amount of energy has to leave the planet. Otherwise it will warm until the S-B T^4 relationship increase the emissions enough to compensate.
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Smoking Frog says:
February 14, 2011 at 4:36 am
The average of the minimum and maximum overlaps gives exactly 62%:
((49 + 6 + 20) + 49) / 2 = 62
but I’m not sure that this is what we’d get with random overlap, even if, as my calculation assumes, there are no real-world constraints.
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Interesting. It may be what the authors were doing.
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The Inclusion-Exclusion Principle requires that we know the overlap(s) to begin with, so it can’t give us the answer.
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But it does give us an answer that is almost the same:
.49 + .06 + .20 – (.49)*(.06) – (.06)*(.20) – (.49)*(.20) + (.49)*(.06)*(.20) = 0.61648.
Jim
You accused me of “tarring all sceptics”
There is a reason why I chose the word “many”
There is a reason why I did not use the term “all” or “Most”
The reason was to trap idiots who are not careful with words.
Steve, That may well be the case but seeking to trap idiots is no more valid than your rivals statement. I believe also that your comment ‘that most skeptics deny’ is also and POSSIBLY incorrect. I know I feel somewhat agrieved because while I am a skeptic (and a physicist) I understand well enough the ‘science of CO²’ but I have never been convinced that the phrase that ‘increasing CO² causes global warming’ has ever been proven (show me where) or is in anyway true. Suggest you read those words very carefully. :))