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…
“marky48 says:
February 15, 2011 at 9:08 am
Right, there is no Tyndall and Co. because they bios are on Wiki. Check. From the can’t- see-it-from-my-house-school-of -myopia. Check.
http://geosci.uchicago.edu/~rtp1/papers/PhysTodayRT2011.pdf
Learn.”
And because the natural variability of the solar system has decided to start cooling us off Tyndall and Co. matters how??
Teach.
My post seems to have gone missing!
Lets try again.
Mikael,
Did you see Bill Illis’ excellent question, and do you know the answer?
Bill Illis says:
February 15, 2011 at 4:38 am
Also Mikael you state at:
Mikael Cronholm says:
February 15, 2011 at 6:35 am
Quote: “Around the type of IR I work with, roughly 2-24 micron with the exception of the atmospheric absorption band at 6-8um”
Is 6-8 µm the absorption band of O2 and/or N2? Because it isn’t CO2 is it? CO2 absorption is 15 µm as we all know.
So when you refer to “the atmospheric absorption band at 6-8 µm” which part of the atmosphere is absorbing at 6-8 µm please?
Thanks
Will
To Will
Water. H2O absorbs nearly 100% in the 6 to 8 micron band. If you are an IR device operator, and you wish to estimate temperature measurement with any degree of accuracy from more than a few inches away, you had better not be using a filter that includes the 6-8 micron band.
CO2 is practically a non-issue in IR thermography (using IR cameras). CO2 is irrelevant.
That is one of the points that I keep making. Most physicists, scientists, who keep trying to point to CO2 as a major contributor to the greenhouse effect, have never spent time in the REAL WORLD working directly with IR and trying to quantify measurements with it. If they did, they would quickly see the foolishness of their assumptions regarding CO2, and would be amazed at the magnitude that H2O truly plays.
Forget CO2. It’s so minor that it is nearly irrelevant.
After wrestling with this question of CO2 heating the atmosphere, my take is this: CO2 absorbs at several different wavelengths. 15 microns is the one where water vapor isn’t already absorbing all of Outgoing Longwave Radiation. 15 microns wavelenght is in the far infrared, corresponding to -4 degrees F. So, when the ground is at -4 F, COLDER CO2 in the vicinity can absorb radiation and be heated. If the atmosphere is warmer than the ground, all bets are off. Now, since the atmosphere radiates too, any CO2 ABOVE air at -4 F, no matter how high up over the ground, can also absorb radiation and be heated. Of course, when CO2 absorbs 15 micron radiation, the molecule of CO2 is excited, and the atoms vibrate. Due to Brownian motion, this vibration is immediately absorbed as extra kinetic energy (= heat) by surrounding molecules. So, if the ground is at -4 F, the 4-parts-in-10,000 CO2 can absorb and transfer a little heat to the atmosphere without violating the Second Law.
It doesn’t seem enough to get all excited about. Water can freeze at 59 F ambient, outdoors with quiescent amposphere and a very clear night, it is in my Thermo textbook.
“”””” Matter says:
February 15, 2011 at 8:19 am
@Blade;
………………………….
The emission ability of a substance is related to its absorption (if it has energy levels allowing emission, it has the same levels allowing absorption). This value depends on wavelength for most materials – you can be transparent to visible light but absorb a lot in the infrared for example.
………………………
The emissivity/absorptivity of the atmosphere in the wavelengths sent by the Sun is relatively small (with notable exceptions like ozone absorbing UV). CO2/H2O doesn’t absorb here so adding it doesn’t reduce the amount of energy hitting Earth.
But it does absorb in the lower wavelengths. Since it absorbs there, it must be able to emit there too. The atoms move quickly and interact with each other, which allows CO2 to dump heat into the O2/N2 and vice versa.
Adding CO2 doesn’t prevent sunlight hitting Earth, but it does absorb light going up from Earth. Hence the warming. “””””
Well I can’t tell what @Blade said from what Matter said; but there’s enough misinformation to go around here.
“”””” The emissivity/absorptivity of the atmosphere in the wavelengths sent by the Sun is relatively small (with notable exceptions like ozone absorbing UV). CO2/H2O doesn’t absorb here so adding it doesn’t reduce the amount of energy hitting Earth. “””””
Take that for example; demonstrably false; and not just a little bit false. For a start, the solar spectrum received at earth contains 98% of its energy essentially in the wavelength range from 0.25 microns in the UV to 4.0 microns in the IR, with only 1% left over at each end. Ozone of course cuts off the short end at about 0.3 microns. About 2.5% of solar radiation lies below 300 nm. Ozone also takes asmall bite out of sunlight at around 0.6 microns; but not much.
CO2 on the other hand has several significant absorption bands in the solar spectrum IR region; first at 2.0 microns; where normal atmospheric CO2 absorbs about35% of that band, then there is a strong 2.7 micorn CO2 band where essentially 100 % is absorbed, and finally there is an equally strong CO2 band at 4.0 microns. The main GHG CO2 band is of course in the region of 13.5 to 16.5 microns.
For the 4.0 micron CO2 band; the solar spectrum has only 1% left in that area, so a fairly small but NOT zero CO2 absorption of incoming solar energy. At 2.7 microns, the solar spectrum still has 3% of its energy left; or 41 W/m^2, so CO2 can capture a good part of that (which therefore will NOT reach the ground as incoming solar energy.
Then there is that pesky non GHG feedback H2O. It has a strong absorption band starting at about 4.0 microns, and going up to about 6.5 microns, where the “Atmospheric Window” starts, so it overlaps the long edge of the CO2 4 micron band so doesn’t have much impact on that 1% remaining solar energy. BUT H2O has a strong band that starts at about 2.3 microns, and goes to about 3.3 microns completely enveloping the CO2 2.7 micron band. (but note that the high resolution lines are likely to be separated; but a good chunk of that solar energy will fall to water or CO2.
But water really gets going at shorter than the 2.3 to 3.3 micron band. there are bands at 1.6-1.8, 1.2-1.4, 0.94, 0.85, and 0.75 microns.
47% of the solar spectrum energy lies at longer than 0.75 microns wavelengths, and water plus CO2 could easily account for about half of that or say 20% of the total solar incoming energy can be absorbed by H2O and CO2 in the atmosphere, and not ever reach the surface as solar spectrum energy, where much of it would be deposited deep in the oceans.
So it simply is NOT true, to say that “”””” CO2/H2O doesn’t absorb here so adding it doesn’t reduce the amount of energy hitting Earth. “””””
Every single additional molecule of H2O or cO2 that is added to the atmosphere WILL reduce the amount of incoming solar energy that reaches the surface of the planet to warm it.
Yes that energy will warm the atmosphere as a result; and that warmed atmosphere; being above absolute zero, will radiate a thermal LWIR spectrum, and only half of that will be directed down towards the gound, so the rest will get lost to space.
Thermal radiation depends on Temperature; not on energy levels either atomic or molecular; it is a result of the thermal motion (and acceleration) of electric charge. Just think about the acceleration of charge that occurs when a warming GHG molecule collides with an ordinary gas molecules or atom (N2, O2, Ar). So every collision that results from the non zero Temperature of the gases, results in charge acceleration, and consequent Electromagnetic Radiation according to the Planck and Stefan-Boltzmann equations; not that you necessarily get a complete black body spectrum; but what you DO get, has that Planck BB spectrum as its outer envelope (can’t exceed).
People keep coming here to WUWT and insisting that gases do NOT emit thermal radiation. They are about the only people who believe that. Physicists don’t believe that; because they have never not observed such non-emission. Hot things radiate thermal spectra; they don’t suddenly stop doing so when the melt and vaporize, or simply sublime.
Some here may find this is interesting …
CALIPSO Spies Polar Stratospheric Clouds
“”””” izen says:
February 15, 2011 at 7:47 am
@- Myrrh says:
February 15, 2011 at 4:39 am
“The last is important, heat always travel from hot to cold.
…It’s been mentioned many times in this argument, that it is impossible for a cold body to warm up a warmer one, because heat travels from warm to cold.”
Heat or energy in the form of photons or radiation travels in ALL directions including from cold to hot, otherwise it would be impossible to see yourself in a mirror which is colder than you are.
It is just that MORE heat travels from the warm region to the cold than travels from the cool to the warm so the NET flow of heat is from warm to cold. But the amount flowing back from cold to warm modifies the NET flow “””””
“Heat” is NOT energy in the form of photons. Heat is purely MECHANICAL KINETIC ENERGY in the form of molecular or atomic vibrations due to collisions between particles (of matter) Sans matter; there is NO “HEAT”. Heat is characterized by TEMPERATURE.
Photon energy is Electromagnetic Radiation that obeys Maxwells equations; from down to; but not including DC; and extends all the way beyond the gamma ray spectrum.
EM radiation and photons don’t know ANYTHING about Temperature; and they can go where they damn well please; hot of cold.
The very same earthshine radiation that leves arth for the dark side of the moon will if it misses the moon proceed on and could reach the sun which has the same angular size as the moon, so the amount of earthshine that could hit the moon can also hit the sun.
Leif Svalgaard, has not reported ANY instances, of earthshine photons being refused landing permission on the sun; despite its 6,000 K Temperature.
And the second law of Thermodynamics as stated by Clausius, refers to cyclic machines.
“No cyclic machine can have no other effect, than to transport “HEAT” from a source at one Temperature, to a sink at a higher Temperature.” doesn’t say anything about EM radiation; just “HEAT”.
We wouldn’t have this problem if people just accpeted that “HEAT” is a verb, NOT a noun.
A C Osborn says:
February 15, 2011 at 10:03 am
Wow, I didn’t realise there were any Mountains that high.
The Earth’s Surface is in DIRECT contact with space?
Not going through any atmosphere for the photons to “bounce around” in then?
————————
Well, ACOsborn, it’s too bad your humour is commensurate with your knowledge and not with your truculence.
Since kuhnkat has given you a pointer to the physics, I’ll address your language usage and comprehension.
You asked about “direct connection” and then got it all confused with “contact”.
I know it’s difficult because words have literal and figurative and contextual meanings, but with patience and perseverance you will get the hang of it.
Wow, this discussion is interesting and lively. I really appreciate the facts and theories presentet here, and enjoy the challenge to try and follow the posts.
I had asked (Feb. 14, 3:50 am) if it was possible to actually measure the energy balance of our planet. P. Wilson – thanks for the reply – pointed to the fact, that temperature measurements are no good for this purpose.
What I had in mind was a satellite which should be able to actually measure the radiation coming to earth from space and also measure the radiation going from earth to space. Is it at all possible to do this? Has it already been done? Spetcral analyses of this kind should be possible I guess, I´d almost expect they were alredy made. If so: Is there anybody around who can comment the results?
It seems to me that a great deal of the theories which are claimed to substantiate the AGW hypothesis suffer from a frightful lack of evidence. I see the so called “glass-house effect” as a radiation delay. The gases in the atmosphere delay the radiation to space. If there was no delay, nighttime cooling would be similiar to that on the moon, I guess. The main source of energy on earth is – the sun. If its otput gets less – for whatever reasons – the earth is going to receive less and get cooler. And vice versa.
Well this discussion being as long as it is, I could say a lot more, but I want to come back to my question: Has the radiation (or energy) balance of earth ever been measured from space? Is it at all possible?
George E Smith, I made a humiliating series of errors there and apologise for my last post, it should be deleted.
My CO2/H2O energy approximation was a mistake from a quick misreading of the waveband units, absolutely stupid of me.
My thermal radiation mistake was from assuming a constant temperature and looking at changes in emission, which filters out the non-emissivity related features so I associated them with internal level energy transfers.
I took my conclusion from the results of line-by-line radiative transfer models along with measurements of net heat flow as a function of wavelength, which shows that a greenhouse effect does exist and that CO2 strengthens it. My understanding of the inbetween bits was wrong. Thanks for pointing me in the right direction George; I’ll take a bit more time out to check through the full spectral results now. 🙂
RE: Domenic says:
February 15, 2011 at 11:46 am
“That is one of the points that I keep making. Most physicists, scientists, who keep trying to point to CO2 as a major contributor to the greenhouse effect, have never spent time in the REAL WORLD working directly with IR and trying to quantify measurements with it.”
That statement I find most interesting. Are there any arguments to the contrary? (I mean other than the feeble flotsam drifting over from markyMark’s sinking theories.)
So if I understand correctly, the debate or questions about feedbacks (+/- &/or whatever) caused by increased CO2 levels (man made or otherwise) isn’t really moot but trivial?
I’m having a hard time here imagining I’ve come to the end of my journey – but thanks.
-barn
ThomasU says:
February 15, 2011 at 1:30 pm
The short answer is that it is possible to measure the in going and out going radiation from Earth. A satellite was built to do this named Triana later renamed DSCOVR. The satellite would have orbited earth in the L1 position, giving it a whole hemisphere view. There has been much debate about why this satellite was not launched with both sides of US politics claiming that the other side were afraid of the results. I believe the satellite, which was delayed by problems with the space shuttle, has been approved for reconditioning by the present administration for launch “sometime” in the future.
http://en.wikipedia.org/wiki/Deep_Space_Climate_Observatory
At present the work by Dr. Spencer utilizing existing lower orbiting satellites, has indicated that there is little radiation imbalance, and that any water vapour feed back to increases in temperature are negative. The DSCOVR satellite would however provided far better data to analyze.
Domenic says:
February 15, 2011 at 11:46 am
Sorry Domenic no offence but my question is to Mikael not you. You have missed the point with your H2O response. There is nothing in Mikael’s statement that indicates he is refering to water. He states clearly
“atmospheric absorption band at 6-8um”
This is clearly not a reference to H2O.
The atmosphere does not consist of H2O which incidently absorbs strongly in many regions not just the 6-8 µm.
So when you butt in and answer questions that are not directed at you, you miss the point and ultimately muddy the water. Whether this is deliberate or simple bungling interference, the end result is the same. You have diluted a pertinent question with irrelevant distractions.
So I ask once more, and this question is to:-
Mikael (and not any one else!)
Did you see Bill Illis’ excellent question, and do you know the answer?
Bill Illis says:
February 15, 2011 at 4:38 am
Also Mikael you state at:
Mikael Cronholm says:
February 15, 2011 at 6:35 am
Quote: “Around the type of IR I work with, roughly 2-24 micron with the exception of the atmospheric absorption band at 6-8um”
Is 6-8 µm the absorption band of O2 and/or N2? Because it isn’t CO2 is it? CO2 absorption is 15 µm as we all know.
So when you refer to “the atmospheric absorption band at 6-8 µm” which part of the atmosphere is absorbing at 6-8 µm please?
To barn E. rubble
Yes. CO2 is trivial.
To Will
I noticed Mikael has a day job, and would not be able to answer your question for a while. So, I gave you the answer to keep the thread going. It is interesting to me to see what kind of understanding of IR physics exists in those concerned about AGW and the greenhouse effect.
[snip. This isn’t “wingerville.” ~dbs, mod.]
[snip]
@ur momisugly Will. Your comment was not lost, but I was sleeping. I live in Thailand so I am on an almost opposite time zone compared to the US.
Anyhow, the absorption rates and wavelength bands of N2 and O2 are unknown to me, they are accounted for in the models we use together with all the rest in the air. Domenic correctly answered your other question about the absorption band that separates the two wavebands we use for thermal imaging, the dominant absorber there is water vapor.
I will explain a little deeper about the atmospheric compensation that is done when we measure temperature in IR. Two spectral bands are used, 2-5 and 8-14 um, approx. These days the longer one dominates the industry completely, for practical reasons and cost, because those detectors are un-cooled. (In furnaces we use a narrow band at 3.9 um to avoid H2o and CO2 generated by the flames from the burners. That is an extreme atmosphere.)
Remember, “atmospheric” does not mean we deal with higher altitudes. We assume sea level, I guess. The calculation uses the LOWTRAN atmospheric model, which is an empirically developed model. The camera has three inputs for the compensation; distance, air temperature and relative humidity. Air temperature is used for two things, to re-calculate the relative humidity to an absolute value, and to determine the emission from the atmosphere. The atmospheric emission, once calculated, is removed from the total signal. Then there is the absorption, the atmosphere gives and takes, so the calculated absorbed radiation is put back. Simple as that!
Notice that the model is empirical. That tells me that the scientists that figured all this out at some point, did not consider it worthwhile to try and calculate the influence of all the gases involved.
Notice also that although we are not in the H2O absorption band, humidity – water vapor – is still the only gas that is individually accounted for in the model, because it is the only one that changes significantly from time to time. That means the other gases are just a standard soup with an assumed recipe. So that is where O2 and N2 are accounted for, they are included in the empirical LOWTRAN model and their individual contributions are unknown to me, because I don’t need to know to do my job.
Now, if I would be manipulating these numbers to see what errors are involved, the one that throws the measurement off the most is distance. So the compounded effect of all gases together seems large H2O itself. Air temperature is the second most important, but seldom gives any large errors. Relative humidity is almost insignificant, it can usually be left at 50% and forgotten. I put it at 25% or 75% if I think I am closer to those values, but I never make a humidity measurement to be exact. It changes only on the decimals, if at all. This is valid in the distance ranges we normally work, up to about 20 m.
What we have to remember about absorption bands is that they do not abruptly cut on and off, they vary up and down with wavelength. I don’t think there is any gas that is COMPLETELY transparent in ANY part of the IR spectrum at least. And as a GENERAL rule I always assume that the shorter the wavelength the better the penetration. The band variations that exist I think have to do with optical effect on a molecular level due to the geometry and composition of the molecules, but I am not an expert on that, so don’t take my word for it. If anyone has initiated input I would appreciate it.
And, for the record, I meant to say I work in the 2-14um band, minus the absorption band, not 2-24um, that was a typo, sorry. I noticed my error in Will’s quote.
@ur momisugly Domenic. Yes, I have a day job, but halfway around the world from where most of you probably are, so, as I mentioned to Will, I was sleeping. 🙂
Matt says:
February 15 2011 at 12:26am
The simplest way I know to understand the greenhouse effect is to realize that when you’re standing on the ground in daytime the Sun is shining on you visible light and the atmosphere is shining on you in infrared.
Fifty percent of the energy radiating from the Sun is in Infrared, greater by the time it gets here, some 80% I read somewhere. This is what heats you, warms you up, which you feel as heat. When a cloud passes in front of the Sun you will feel cooler practically instantly even though visible light all around you hasn’t changed, this is because the Sun’s infrared isn’t reaching you for that moment. (The cloud isn’t as far away from you as the sun, doesn’t take minutes to feel lack of IR.)
……………..
Mikael Cronholm says:
February 15, 2011 at 6:35 am
If you roast a chicken, does it heat up evenly all the way through at once? Of course not! It will absorb the radiation on the surface and then it will conduct (no convection, no blood flow, dead chicken, hehe) throughout the meat. A microwave oven heats from the “inside” though, because it excites water molecules directly, as they are dipoles.
Nope. IR is deeply penetrating in an organic body, and the term for this is “conversion”, not “conduction” as in your dead chicken. Without IR, if you waited for your chicken to conduct heat from the skin to its insides you would probably have a thick layer of ash surrounding an empty bit in the centre, your stuffing uncooked. I’m not a scientist, I may be exaggerating in my thought experiment..
http://www.springerlink.com/content/eq0pu7155011728r/fulltext.html
“Abstract
This paper describes the synthesis of new upconverting luminescent nanoparticles that consist of YF etc. functionalized with poly(acrylic acid) (PAA).
Unlike the upconverting nanocrystals previously reported in the literature that emit visible (blude-green-red) upconversion fluorescence, these as-prepared nanoparticles emit strong near-infrared (NIR, 831 nm) upconverstin luminescence under 980 nm excitation. …………………….The new PAA-coated luminescent nanoparticles have the potential to be used in a variety of bioanalytical and medical assays involving luninescence detection and fluorescence imagin, especially in vivo flurorescence imaging, due to the deep penetration of NIR radiation.”
Odd that, other scientists using IR in practical applications understand it very differently from you.
A penetration depth of UV of 1mm is something I would consider deep penetration in skin.
Hardly through the epidermis, the first layer of skin, which is around 0.5mm eyelids to around 1.5mm soles of feet, palms.
IR is deep body penetration.
If you ask Max Planck he will tell you that the energy of a photon increases with shorter wavelengths.
Did he understand IR?
I don’t know what kind of sauna you use, but it sounds more like a toaster. My sauna is a wet sauna, with a wood fired stove with rocks on it where I throw water to get steam. The heat transfer to my body is..
You don’t know how peculiar that sound to me as you say you are an expert on IR…
An IR sauna doesn’t have that immediate great heat feeling from the water holding heat, the log fire using up your oxygen.., it’s comparatively cool, there’s no great sensation of heat as in the atmosphere of traditional saunas, which many find difficult to tolerate. IR heats the body, not the air. From what I can find, traditional saunas operate at around 70-120°C, IR saunas 35-55°C , so, comforably warm.
I used to have an IR panel heater which I used in a downstairs cloakroom of a limestone house; it never felt cold, never as warm as it would have done with a radiator, just comfortable because the IR first heated the walls which in turn were not drawing away one’s own heat on entry, and then heating one’s own body during the brief visits.
Common sense does not always make scientific sense.
Too true. However, here I’m making a judgement about explanations, and in that common sense is extremely valuable. Your “sounds like a toaster” doesn’t inspire me confidence that you understand IR at all.
Heat or energy in the form of photons or radiation travels in ALL directions including from cold to hot, otherwise it would be impossible to see yourself in a mirror which is colder than you are.
?? Yer what?
Energy travels in all directions, heat doesn’t.
It is just that MORE heat travels from the warm region to the cold than travels from the cool to the warm so the NET flow of heat is from warm to cold. But the amount flowing back from cold to warm modifies the NET flow.
Ah, this is what the warmistas say. Are you pretending you’re neutral? Energy may well travel in all directions, but definition of heat, that which always flows from hotter to cooler, or rather, the colder grabbing the heat from the hotter as I’ve seen it explained.
I don’t know Mikael, I thought one reply you gave in response to one poster’s statement that heat travels from the earth by convection, a very AGW type response.
Brushing aside that this was obviously referring to heat transfer in our atmosphere, you said something like ‘there’s no way for it to travel from the atmosphere into space but by radiation’.
Straw man. Why didn’t you engage with what he said?
Going back to something you said earlier, February 15, 2011 at 3:12 am:
Another proof for that is that objects with different colors will absorb differently in the visual, white the least, black the most, and other colors in between. In the IR, the absorptivity (or emissivity) will NOT depend on color. And this is definitely my area of expertise, I assure you!
I didn’t actually understand what you were talking about here, as this was added to your injunction that I read Herschel, which simply, is the discovery of IR outside the visible, found by measuring its greater heat than the other colours of the visible when he moved the thermometer to the side of the visible. That’s why some colours are described as cool, some hot.
Still, I have found this page on further exploration: http://windowoutdoors.com/WindowOutdoors/Color%20and%20Thermal%20Regulation.html
Which is interesting and explains what you mean by colours not affecting IR, as they all are black to it.
“Particularly, most paints and dyes are black in the infrared range. The exception is metallic (aluminum) paints and space blankets, which have low IR emissivity.”
The Herschel experiment had nothing to do with “absorptivity or emissivity” as such. It was strictly about the temperature of the different colours from which he found that there was a colour much hotter outside of the visible. That he might have been measuring one or the other or both, isn’t actually relevant, istm, and I have no idea what you mean by it being a proof.
You then confuse me further by your February 15, 2011 at 6:51 am post:
Good, you found old Herschel! Now, did you notice what that radiation went through before it hit his thermometers?
Yes. That’s how he got the visible light separated out enough to measure the differences in temps. Er, now let me think, it was one of the famous scientists who came up with a prism and found light wasn’t “white”, but split into different colours, wasn’t it? Now, what was his name?
A glass prism. Spectral things will happen to white light as it goes through glass.
Even a flat pane of glass?
And I can guarantee you that above approximately 2.3 micron there would be no more radiation to heat his bulbs.
? I’m at quite a loss to understand why I would need such a guarantee. Whatever that means. What does it mean?
If it relates to something I’ve said, please be explicit.
Now go and find a Planck curve for the solar spectrum and see all that energy right there in the visual part of the spectrum. What do you think happens to that ENERGY when it hits a surface? Three things could happen, absorption, transmission, reflection. And no, you are right the earth and us humans are not painted black, BUT WE ARE NOT SHINY POLISHED MIRRORS EITHER. Sorry, don’t know how to emphasize that in a better way. So all objects will be absorbing visual light, more or less, but for sure never zero. Not even close to zero, unless it is very shiny metal.
? I’m sure this means something to you, but as yet I’m not sure what. I meant by this that different colours affect absorption. As in plant photosynthesis, which takes in blue and red light via its pigment chlorophyll, which enables it to convert Carbon Dioxide into glucose. It is a poor absorber of green portions of spectrum, its green colour reflects green light.
Hmm, that’s interesting, I’ve just taken a look at the wiki page on chlorophyll, and it says:
“Chlorophyll molecules absorb in the blue and red bands, but not the green and infrared bands. Chlorophyll content meters measure the amount of absorption at the red band to estimate the amount of chlorophyll present in the leaf. To compensate for varying leaf thickness Chlorophyll Meters also measure absorption at the infrared band which is not significantly affected by chlorophyll.” Not sure what it means. Is it saying that green limits the absorption of IR?
But anyway, I still can’t see how anything you’re saying actually addresses my point, that colours have different degrees of “heat”. It is the Far Infrared which is tangibly felt as heat, hence, it is known as Heat Energy.
Most energy that reaches earth from the sun is in the visual spectrum. It will be absorbed and be converted to heat.
Not as I know it. Most of the energy from the Sun reaching earth is in the Infrared, some 80%.
OK, you try and find me some proof of what you claim and I’ll see what I can find to back up mine. And anyone else interested, either view.
But, you seem to be ignoring the references I made in my posts to pukkha medical understanding of light absorption, visible light does not penetrate the body to any significant extent, UV hardly at all, it is IR which penetrates deeply into organic material. What we feel as heat from the Sun is IR, Heat Energy. This is what warms the earth and us on it. Which is being released by the rocks in your sauna.
Put out the fire once the rocks have heated, add more rocks if needed, forget the water, and you’ve got yourself an IR sauna. Or join some Native Americans in a Sweat Lodge… Before the idea of heating the air in the room by convection, the heating systems across europe were mainly IR from ceramic stoves. Big corner stove with a small fire radiating heat energy into the walls, and into whoever was in the room. Like the Roman hypercaust system, (sp?), and back today as ‘underfloor heating’.
@ur momisugly George E. Smith, February 15, 2011 at 11:54 am
I just want to say I appreciate your input in this comment a lot. I have that solar spectrum curve in front of me if I close my eyes. And when I remind myself that the darn thing is in a logarithmic scale, all those numbers you give make a lot of sense. Thanks for that!
My only slight objection is here:
““Heat” is NOT energy in the form of photons. Heat is purely MECHANICAL KINETIC ENERGY in the form of molecular or atomic vibrations due to collisions between particles (of matter) Sans matter; there is NO “HEAT”. Heat is characterized by TEMPERATURE.”
Agreed to the most part. It is the last sentence I am a little bothered with. Heat and temperature are connected in such a way that they co-vary depending on heat capacity, BUT with the exception of latent heat during phase change. If you observe a temperature change you can safely assume that the amount of heat it the object has also changed, but if you have a change in heat it may not necessarily cause a change in temperature, if there is phase change going on.
Hi Mikael
Thailand? Good gracious!
One of the few places left in the world that I do wish to visit.
I was going to ask you what you were measuring out to 24 microns, as that is unusual for industrial/commercial work…but thanks for the correction.
I must admit I am impressed with your patience in answering the many questions here. The misconceptions about IR and radiational heat transfer are so vast in the scientific community…
To Myrhh
IR cannot penetrate into the body. It is absorbed by the skin to a minor depth. Most of skin composition is H2O. That is the main reason why.
H2O is a powerful absorber and emitter of long wavelengths. When you look at an IR image of a person’s head, you can see some arteries and other features. But it is not because the IR looks within the body, but rather because there are certain arteries that are close to the skin surface. And there are parts of the surface of the human body where there are lots of tiny arteries just below the skin, high blood perfusion, that show up as large hot spots.
This may help some begin to understand what radiational heat transfer really is. Take a look at this graphic
http://en.wikipedia.org/wiki/File:EM_Spectrum_Properties_edit.svg
Now, I really like this graphic because it uses a nice handy old fashioned thermometer figure at the bottom. So you have a nice easy way to look at the idea.
If you have a perfect ‘blackbody’ atom, at any temperature over absolute zero it emits radiation (heat energy). That perfect blackbody atom near absolute zero will first emit very very long radio waves. As you heat that blackbody atom up, the hotter it becomes, the heat energy shifts to shorter and shorter wavelengths.
Just follow the thermometer and the wavelengths.
That graphic is perfect in illustrating what basically happens in radiation in an ideal situation.
Everything emits energy above absolute zero. Everything. In effect there is no such thing as COLD. There is only less heat and more heat. Heat flux, or radiational heat transfer, is simply the NET flow of heat (or energy) from a hotter material to a less hot material. The forms that ‘heat radiation’ are classified as, is exactly as shown in the graphic.
But, in effect, all matter in the universe is in a form of constant sort of ‘communication’ due to electromagnetic radiation, but with net energy moving from hotter to less hot.
However, real world atoms (hence molecules, etc) are not perfect blackbodies, so they will GENERALLY follow this graphic, subject to unique quirks of each different type of atom or molecule. Those quirks are the tricky part. And science has not tested various atoms and molecules anywhere near enough to fully understand those quirks. Even N2 which makes up most of the atmosphere has hardly been tested at various temperatures, thus wavelengths to fully understand the nature of it’s true radiational properties.
@ur momisugly Myrrh.
“IR is deep body penetration.” Too generic a statement. IR is a very broad spectrum. Some IR may penetrate deeply, you may want to inform me exactly which wavelengths, if you know, I am truly interested (not sarcastic, I promise). But certainly NOT the kind of IR we look at with thermal imagers. From Hollywood people “know” that we can see through walls and stuff with IR. That is probably where that BS comes from originally. It is BS though, no less.
Saunas: Go ahead, use the one you prefer. I love mine! 🙂
If you don’t believe condensing steam will give off huge amounts of latent heat, put the kettle on and when it boils, stick your finger right in the steam that comes from it and you will find out.
““Particularly, most paints and dyes are black in the infrared range. The exception is metallic (aluminum) paints and space blankets, which have low IR emissivity.””
Wrong, total rubbish! They are approximately “grey” in IR images, if we refer to “black” as being blackbody absorbing. Paint will be about 0.90-0.95 emissivity in a long wave camera. I could go on and teach you for half a day or more about emissivity in thermography applications… You seem to just google for a while until you find something that supports your preconception, without bothering to understand. I don’t know where that statement came from, but I would not believe much from that source.
The guy with the other prism experiment that you don’t remember was probably Newton. He got in trouble with the religious guys for that if I am not remembering wrongly. Religious people sometime have a hard time changing their mind when they are “sure”, like the earth is flat, and so on…
And then we get to this:
“A glass prism. Spectral things will happen to white light as it goes through glass.
Even a flat pane of glass?
And I can guarantee you that above approximately 2.3 micron there would be no more radiation to heat his bulbs.
? I’m at quite a loss to understand why I would need such a guarantee. Whatever that means. What does it mean?
If it relates to something I’ve said, please be explicit.”
I understand that you don’t understand, because you don’t want to understand if it challenges your beliefs.
Glass is what is referred to as a selective radiator. It means that its emissivity, reflectivity, and transmissivity all change with wavelength. You seem to think that just because visible light passes through it easily, there is no change in the composition of he spectrum on the other side. There is. Not so much in visual, but in IR there are big things going on! We have to forgive Herschel, he had no spectrometers at hand, but in this day and age we do. Glass will start to lose transmissivity around 2 um and nosedive completely at 2.3um. Over 3um there is hardly any transmissivity AT ALL!!!!!! How can I know for sure? Because glass is totally opaque to any IR camera I use, in addition to all the spectral charts that are available. Can’t see through it at all. Had Herschel moved his thermometers a little further from NIR, into where IR should have been, they would have dropped completely. His result can only be seen as qualitative (“there is other light than visible”), rather than quantitative. The sun is still giving off more heat to the earth in the visible. Basta! Read George E. Smith, February 15, 2011 at 11:54 am for an initiated comment about that. Don’t argue with me about that any more, please, I will not oblige you further with any reply on it if you do. Search the net if you want more info.
FYI, in IR cameras we use lenses made of either silicon (3-5um) or germanium (8-14um). Glass lenses would definitely not work.
So, if you are able to accept that, we can go one step further into your claim that “IR is heat, visual is not”. Knowing that no IR over 2.3um will go through a normal window, will you accept that the heat you feel when you stand behind a window where the sun shines in comes from visual light? If not, I cannot help you any more, you have to remain ignorant about that.
The rest of your comment seems to use some kind of shotgun approach, firing off in all sorts of directions, chlorophyll, and so on. I don’t find that necessary to reply to, because you try to prove the same thing with all of that and I have just proven you wrong. At least one quote there was not even mine, so I will not defend it. So, “you seem to be ignoring the references I made in my posts to pukkha medical understanding of light absorption”. Yes I did. It is too limited. What these people are using may be some narrow band of NIR where deep penetration exists. I will not argue one way or the other and I don’t care to look into it – to prove my point I don’t need to.
And “warm” or “cool” colors may exist in psychology, but not science. If you think red is “warmer” than blue for example, that concept would be backwards, since blue incandescent light would need a much hotter object to dominate the spectrum. An object at 7245K would peak at 0.4um, for example. Much hotter than the sun. Consider arc welding. You have probably noticed how blue the arc looks. Very hot, that’s why!
I hope that was helpful to you.
@ur momisugly Domenic. It sounds like you are a colleague. You would be most welcome to visit Thailand. If you want to e-mail me, take my name and write both first and second together in lower case, and add “at” and gmail dot com. (Just avoiding spambots…)
@ur momisugly George E. Smith. I noticed afterward that all those nice numbers were in a previous post, not the one I referred to. Very useful nonetheless. Thanks!