A Conversation with an Infrared Radiation Expert

Visible and IR photo in Anthony's neighborhood - note that my IR camera doesn't go any lower than -20°C Photos by A. Watts

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…

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Murray Duffin

Interesting. Now I am confused at a higher level. Murray

Ian W

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 ?

Dennis Wingo

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.
Now that is a wise statement. As someone who designs temperature measurement systems (principally for spacecraft) I have always been amused that the AGW community can make the statement that they can use an instrument with 0.5 degree accuracy and get 0.001 degree temperature variation out of it.
In designing thermal control systems for spacecraft you have to be incredibly sensitive to the absorptivity/emissivity (a/e ratio). If you get this even slightly wrong in spacecraft design, you either run the equilibrium temperature too high and it will fail, or too low and it will fail.
MC makes the observation that the radiated temperature is 100% dependent upon emissivity, which is correct, but I have never seen this really integrated into AGW models, they simply use a blackbody approximation, which is a terrible reference in that this varies wildly around the world.
Also, the models do not take into account the dramatic differences in resulting temperature based upon altitude, especially in desert regions of the globe.
A lot of physics modeling operates by making simplifying assumptions, but how many of these assumptions are testable and repeatable?
This is the basis of a lot of my skepticism on the models involved.

David Ball

“The only thing Co2 can do is increase the coupling to space”. Thereby having a cooling effect. Good stuff. I would love to hear what Dr. Lindzen has to say about this. If I am not mistaken, this is what he has been saying.

ferdberple

Great 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. If not, it is not science, it is guessing.”
Why is it that leading climate scientists have tried to hide their data and methods? In what way are such tests repeatable? In what way have they stood the test of time?

Calvi36

Excellent dialogue Anthony. I found it quite gripping as it explained quite a lot for me. Many Thanks.

MC: For 300 W/m2 radiation I get -23.4°C at 300 W/m2 when I calculate it (yes, minus!). Pretty cool light bulb.
300 W/m² is more like -3.3°C (26°F). It is still cold though. At -23.4°C (-10°F) black body radiation is only 220 W/m².

chemman

Good post, thanks. The one area I struggle with is the reflector part. Do the “green house gas” molecules absorb and re-radiate energy? Yes. Is it like a reflector? Questionable. Since molecules are 3-dimensional and constantly in motion “green house gases” will re-radiate IR energy in 3 dimensions not just back at the earth like a reflector. Again it just may be the weakness of the terms used that is throwing me.

dp

The confusion of temperature vs energy is an interesting one. I’ve attempted to explain how it works by the following. Place a sheet of paper above a surface. In that sheet is a 6″ hole through which passes sunlight. That light lands on a thermometer which records the temperature. It is intuitively obvious the temperature will remain the same if we move the paper and thermometer around the immediate area so we determine there is a fixed relationship between the sun and the thermometer.
Now we place a 6″ lens over the paper and by good fortune the focus point is exactly on the thermometer. We know the same amount of energy is entering the lens as entered the hole in the paper but the temperature is likely to cause the thermometer to explode. The energy is concentrated.
This is clearly not a tale told to people of science with any expectation of creating revelation but it helps the grand kids to understand the relationship between heat and energy.

It seems that MC knows the subject matter extremely well. It also seems that he says you cannot be certain what is really going on with IR and CO2 in the atmosphere, despite those who seem to be absolutely certain.
Food for thought indeed!

Jim D

MC puts KLC right on a few of his misconceptions. I only fault MC’s terminology where he uses the word ‘reflected’ for IR from the sky, when in fact it is emitted. This is probably just loose wording on his part, but there is a scientific distinction.

Charles Nelson

Absolutely brilliant…thank you. I like the term ‘increase earth’s coupling with space’ I have been trying to explain this to the numpties for ever.
One question I ask CO2 Warmists is, ‘have you ever noticed the outside temperature when flying on a plane at cruising altitude…-30 to -60 degrees C. right? Have you ever looked down and wondered what is happening at the top of the cloud layer…heat loss right? So if the atmosphere was to get warmer wouldn’t these clouds simply rise a little higher because of convection and lose their energy through radiation into the vast reservoir of coldness above?
Interestingly I find that when I use words like radiation and convection they tend to glaze over and lose interest. In fact I don’t think that ‘science’ matters that much to believers…
Keep up the good work guys.

pat

Interesting.

Domenic

I am an IR expert, 20+ years in the field.
CO2 is an IR absorber of only narrow wavelength bands of IR.
Water vapor is a much more stronger IR absorber because it absorbs very large wavelength bands of IR.
In addition, water vapor is approx 3.5% of the atmosphere or 35000 ppm. (That’s a global average. At the poles the air is drier. In the tropics, the air is much wetter.)
CO2 is only about 390 ppm.
So, water vapor from a ppm point of view is probably 100X greater in effect as a greenhouse gas from an atmospheric percentage only compared to CO2. (I rounded up slightly because the tropics have more effect than the polar regions, having more water vapor to absorb and store solar energy.)
In addition, a molecule of H2O is also quite a few multiples greater in absorption of IR compared to a molecule of CO2.
Take a look at absorption spectra for H2O: http://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&Units=SI&Type=IR-SPEC&Index=1#IR-SPEC
For CO2: http://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Units=SI&Type=IR-SPEC&Index=1#IR-SPEC
Just eyeballing these two NIST absorption spectra curves for H2O and CO2, it appears H2O may be at least 10X greater at absorption per molecule than CO2 is.
Thus, the comparative effects on ‘global’ warming, changes in H2O composition in the atmostphere are probably 1000X greater than CO2 changes in terms of contribution to ‘greenhouse effect’.
give or take a bit for some error =>>>>> approximately 1000 TIMES!!!!
CO2 is a non-issue compared to water in the atmosphere.

Schadow

In the 1950’s, I was working on the Sidewinder air-to-air missile at the Naval Weapons Center at China Lake, CA. Sidewinder’s guidance system is built around an infrared sensor. I remember all the “gee whiz” stuff that was observed during the sensor’s testing, e.g., it would lock on to a quarter moon in the middle of a hot summer day and even Venus at night. The sensor has undergone much improvement since those days and advanced versions of the missile are still seen under-wing on today’s combat aircraft.
Not much to do with this topic but just an interesting side-note on the practical uses of infrared technology.

Domenic

As long as we’re on the subject, most people don’t realize that the most accurate CO2 measuring devices also are IR based. They utilize those narrow bandwidths of CO2 IR absorption to measure the amount of CO2 in an atmospheric air sample.
A while ago, I took a look at the history of the supposed pristine CO2 measurements at Mauna Loa. I pulled up two papers (Keeling 1960, and Thoning 1989) describing the methods, calibration protocol followed, etc.
They have been following very good protocol. However, they have to constantly calculate out the effects of the nearby Mauna Loa volcanic activity. Mostly during night times, due to the prevailing winds, the CO2 measuring devices do jump showing dramatic increases in CO2 from the volcanoes. They supposedly developed algorithms to eliminate those errors. I haven’t looked at those yet.
HOWEVER, that activity from Mauna Loa volcanoe, has another effect that I HAVE NOT SEEN THEM TAKE INTO CONSIDERATION.
If Mauna Loa volcanoe is potent enough to send their CO2 measurements skyward, that means that WARM AIR from the volcano at night, is also affecting their temperature data. It’s the same warm air that contains increased CO2. And I would bet it also has biased their night time temperature data. BUT THERE IS NO MENTION THAT THEY HAVE FACTORED THAT WARM AIR EFFECT FROM MAUNA LOA VOLCANOE OUT OF THEIR TEMPERATURE RECORDS!
In my opinion, the temperature data from Mauna Loa station is greatly suspect from what I can see.

DirkH

Berényi Péter says:
February 13, 2011 at 12:42 pm
“300 W/m² is more like -3.3°C (26°F). It is still cold though. At -23.4°C (-10°F) black body radiation is only 220 W/m².”
Gives “warmist” a whole new meaning. Co2 increases might even increase the temp to -2.3 deg C or so.

Steve Reynolds

Ken, I’m glad you are listening to someone that knows what he is talking about. I’ll claim to be somewhat of an expert here as well (I design the IR sensors that go into the kinds of instruments you are talking about).
I agree with everything MC said except (as Jim D noted) that IR is not reflected by CO2, it is absorbed and then re-emitted in all directions.

Mark Wagner

Since molecules are 3-dimensional and constantly in motion “green house gases” will re-radiate IR energy in 3 dimensions not just back at the earth
I doubt that Co2 re-radiates much at all. As soon as the CO2 molecule absorbs any radiation, 90% are immediately quenched by contact with other molecules in air, primarily nitrogen. The other molecules don’t “radiate” in the sense that CO2 does, with bending molecular bonds. They just convect and carry the heat up and away.
I’ve never seen any science that directly addresses quenching of “warmed” CO2 to other atmospheric gasses prior to “re-radiation.”

Jim Masterson

>>
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.
<<
This is a major problem with Trenberth, Fasullo, and Kiehl 2009. They use the same atmospheric window value of 40 W/m^2 as they do in Kiehl and Trenberth 1997.
There is a minor problem with cloud cover. The cloud cover is supposedly 62%. KT 1997 combines three cloud layers (49%, 6%, & 20%) to get that figure. They call it “random overlap,” whatever that means. It looks like an application of the Inclusion-Exclusion principle. I get 61.6% which rounds to 62%. Their famous energy diagram (fig. 7 in KT 1997 and fig. 1 in TFK 2009) should state: “62% cloud cover assumed.”
After calculating 62% for the global cloud cover, the term “cloudy” is ambiguous throughout the rest of KT 1997. Every time you see “cloudy,” does KT 1997 mean 100% cloudy, 62% cloudy, or something else?
My favorite computation is the value for the atmospheric window, and I quote from KT 1997:
“The estimate of the amount leaving via the atmospheric window is somewhat ad hoc. In the clear sky case, the radiation in the window amounts to 99 W/m^2, while in the cloudy case the amount decreases to 80 W/m^2, showing that there is considerable absorption and re-emission at wavelengths in the so-called window by clouds. The value assigned in Fig. 7 of 40 W/m^2 is simply 38% of the clear sky case, corresponding to the observed cloudiness of about 62%. This emphasizes that very little radiation is actually transmitted directly to space as though the atmosphere were transparent.”
This is really sloppy math. The term “cloudy” is again ambiguous. If KT 1997’s cloudy term means 62%, then the correct window value is 80 W/m^2. If they mean 80 W/m^2 is a 100% cloudy value then they should interpolate between 99 W/m^2 and 80 W/m^2 and get something like 87 W/m^2. Apparently the 80 W/m^2 cloudy value is thrown in as a detractor, because they interpolate between 99 W/m^2 and 0 W/m^2. Apparently, they obtain 37.62 W/m^2 and round up to 40 W/m^2.
That’s a slop of at least 2.38 W/m^2 (ignoring the other larger values). The 0.9 W/m^2 seems a little nonsensical to me.
Jim

I suggest that Dr Roy Spencer is asked to comment on these ideas as they seem to impinge on his area of work.

Bruckner8

The narrative is so clear, practical, non-confrontational, unemotional and straight-forward that I came away thinking “I *almost* understood what he said.” He brought up the measurement accuracy again too, which I posted about in my first post on WUWT a couple years ago, and another commenter has chimed in with his professional experience.
Are there any matter-of-fact scientific narrative examples similar to this on the AGW side?

Steve Reynolds

Domenic, you have a major error in your calculation: you used IR absorption data for _liquid water_, not gas phase water vapor!
Even when you correct that, you need to take into account that absorption reaches near 100% in narrow spectral lines, so adding more does not increase absorption linearly. That is why the effect of CO2 only increases approximately as the log of concentration.

Tom_R

>> Jim D says:
February 13, 2011 at 1:04 pm
MC puts KLC right on a few of his misconceptions. I only fault MC’s terminology where he uses the word ‘reflected’ for IR from the sky, when in fact it is emitted. This is probably just loose wording on his part, but there is a scientific distinction. <<
His statement was that there's always reflection. In a sense, that's true here. Whenever you have a change in the index of refraction you get a reflection at the boundary. I'm not sure how that applies to the continuous change as the atmosphere thins with altitude, but there might be some reflection from that.

G. Karst

Jim D:
“I only fault MC’s terminology where he uses the word ‘reflected’ for IR from the sky, when in fact it is emitted.”
I believe the reflection, re-emission, back-scattering terms are describing the same phenomenon, depending on the discipline. In the field of optics, it is considered reflection. From quantum physics we have no such phenomenon, as photons are absorbed and re-emitted. Even in nuclear reactors, old terminology, referred to the “moderator” as a “reflector”. It is a problem during inter-disciplinary discussions, especially when dealing with photons which can be regarded as both particle and wave. Personally, I prefer the term back-scattering as most descriptive of the term/process. GK

Steve Reynolds

Mark Wagner: “As soon as the CO2 molecule absorbs any radiation, 90% are immediately quenched by contact with other molecules in air, primarily nitrogen. …I’ve never seen any science that directly addresses quenching of “warmed” CO2 to other atmospheric gasses prior to “re-radiation.””
The effect you describe is included in the science. Re-radiation is assumed to occur at about the same temperature as the other gases at that altitude. That is why the measurements described in the article give -50 degrees C or thereabouts.

#
#
Schadow says:
February 13, 2011 at 1:30 pm
In the 1950′s, I was working on the Sidewinder air-to-air missile at the Naval Weapons Center at China Lake, CA. Sidewinder’s guidance system is built around an infrared sensor. I remember all the “gee whiz” stuff that was observed during the sensor’s testing, e.g., it would lock on to a quarter moon in the middle of a hot summer day and even Venus at night. The sensor has undergone much improvement since those days and advanced versions of the missile are still seen under-wing on today’s combat aircraft.
Not much to do with this topic but just an interesting side-note on the practical uses of infrared technology.
############
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.

W. Falicoff

Domenic- the CO2 measurements by Scripps Institute are taken in several locations around the world including the locations near the North and South pole. See http://scrippsco2.ucsd.edu/research/atmospheric_co2.html These measurements support the findings of the measurements at Mauna Loa. They also show there is a lag of CO2 from the Northern to Southern Hemisphere (the levels in Antarctica lag those in the Northern hemisphere), as expected given the gains in CO2 are primarily from anthropogenic sources.
The models used by climate modelers show that the heat balance in the upper atmosphere from increased concentrations of CO2 is what matters, not its effect at sea level where water vapor dominates .
Dennis Wingo- With regard to your statement that climate modelers assume Co2 emits as a black body (emissivity of 1) I believe is not true (the emissivity of CO2 in the IR range is quite low). Further, your statement that is not possible to achieve a measurement less than the accuracy of an instrument is also not correct, as the accuracy can be increased by taking many measurements (up to a threshold). The resultant accuracy is approximately proportional to the square of the number of readings. This holds true in several fields. For example one can determine the position of a smear image of an object on a digital sensor to the resolution of 1/5 the size of a pixel (so-called super-resolution techniques). The recent findings of earth-like planets around other stars uses the change in apparent brightness of the star when a planet or plants are in different positions on their orbit. The sensitivity of the systems used in these measurements is extraordinary, well beyond what scientists thought was possible with state-of-art equipment. By the way where did you see that AGW scientists are measuring temperature to an accuracy of one thousandth of a degree (K?)?

David Ball

Thank you for posting this Anthony. I am in awe of your sense of fair play. Takes a big man to have the courage to post something that does not necessarily fall in line with your own perspective. WUWT? sets the standard once again.

“what can be calculated by Stefan-Boltzmann’s law,”
The Stefan-Boltzmann Law. Two guys, one law.

Ken,
You referenced Figure 1 in the Trenberth article. This is very similar to the figure for the energy balance of the earth in IPCC AR4 FAQ. What strikes me about the energy balance is that about 160 w/m2 is absorbed by the earth’s surface, of which 17 ascends to the upper atmosphere in themals and 80 is used to evaporate water. That leaves a net of about 60 w/m2 to be radiated. Yet, the diagram shows well over 300 w/m2 being emitted to the CO2 “cloud”. Where does all that energy come from, and what ever happened to the First Law of Thermodynamics that energy can neither be created nor destroyed? There seems to me to be no need of considering the resulting temperature when the energy simply does not add up.
By the way, it matters little that the energy is circulated. That additional 300 w/m2 has to come from somewhere. It this diagram is correct, then we are getting free energy, in which case we do not need any more power plants at all.

Latitude

Steve Reynolds says:
February 13, 2011 at 2:32 pm
I agree with everything MC said except (as Jim D noted) that IR is not reflected by CO2, it is absorbed and then re-emitted in all directions
=================================================
Steve, Since it’s constantly being exposed to IR, when it’s saturated, does it just reflect because it can’t absorb any more?
Or is it in a constant state of absorbing and re-emitting? neutralizing itself?
Since it’s constantly being exposed to IR, does [?? ]

John S

Domenic Wrote:
“Thus, the comparative effects on ‘global’ warming, changes in H2O composition in the atmostphere are probably 1000X greater than CO2 changes in terms of contribution to ‘greenhouse effect’.”
This fact has been acknowledged by the AGW community, but as their argument goes, the relatively small added heat from CO2 creates a positive feedback loop that includes adding more water vapor into the atmosphere until there comes a ‘tipping point’ where the global temperatures go into thermal runaway.
CO2 is the “trigger” of of this positive feedback AGW, not necessarily the source of all the extra heat.
As a “warmist,” I cannot deny that CO2 does absorb and re-radiate heat energy. It is the existence of positive feedback vs. negative feedback mechanisms in the atmosphere that is at the heart of the global warming debate.

richard verney

What one is measuring is interesting. But one of the problems in this debate is the assumptions made by the AGW proponents and the fact that everything is treated as if it were an average. Averaging can be a useful tool. However, it is almost invariably the case that when one examines any given individual scenario, one is looking at something other than the average. In other words, the average is rarely encountered in the real world experience. This obviously has a bearing when one considers what they wish to see when making a measurement. Ken Coffman ponders upon an experiment where he is hoping to measure back radiation of 300w per sqm, but this is the so called average back radiation and this raises the question of precisely where on planet Earth would you see that amount of back radiation? It is highly unlikely to be seen at the location where he was conducting the experiment. That being the case, the results of the measuring experiment may tell one little about the effects of CO2.
I can understand that CO2 absorbs IR and then re-radiates this in all directions, some up, some sideways and some downwards etc. This scattering is a continuing process such that I can understand that as a consequence of that, CO2 delays IR finding its way out to space. However, what I do not understand is how this effectively heats up the Earth.
The average global temperature of the Earth will only heat up if the amount of energy received from the sun is not fully re-emitted back to space (by way of all forms of energy dispersion). But CO2 merely delays the escape of IR radiation, it does not provide an impenetrable shield preventing the IR from escaping back to space.
This then begs the question whether the encumbrance/delay caused by CO2 has any significant impact. The answer to this appears to me to be whether the delay is of such magnitude that all the IR does not have time to escape to space during the time when the Earth is not receiving energy from the sun. Whilst this process is happening 24 hours a day, simplistically the question is whether during the period of night there is sufficient time for all the energy received during the day to escape back into space. If there is sufficient time during night for this to happen, then there is no effective energy entrapment and the Earth will not be heating up.
IR travels at the speed of light. Approximately 300,000 km per second. If a photon of energy was to take a direct route to space this would involve a distance of about 120km. If because of the presence of CO2 in the atmosphere, the photon does not take a direct route but instead bounces around millions of times taking a zigzag course into space (bearing in mind that the density of CO2 decreases with height), we are delaying the IR escape into space by minutes. It is clear that all the energy received during the day, can (subject to water vapour and clouds etc) escape into space within less than an hour of sunset even if the concentration of CO2 were to double. So how does CO2 cause warming in the real world in which we live in?

John Cooper

I had to dig deeply through my old space program files to find this one:

I fully realize that I have not succeeded in answering all your questions. Indeed, I feel I have not answered any of them completely. The answers I have found only serve to raise a whole new set of questions, which only lead to more problems, some of which we weren’t even aware were problems. To sum it all up, in some ways I feel we are more confused than ever, but I believe we are confused on a higher level, and about more important things.

Bill Illis

Really nice article.
I would like to see much more discussion of the actual physics involved here because this is all happening at the quantum level – where physics is king – not in climate models where 20 km-square boxes and just 21 layers of the atmosphere is king.
The numbers is this debate are staggeringly huge as well as staggeringly small.
– the energy represented by a solar photon spends an average 43 hours in the Earth system before it is lost to space. Some spend just a millisecond while a very, very tiny percentage might get absorbed in the deep ocean and spend a thousand years on Earth or longer. In essence, the Earth has accumulated 1.9 days worth of solar energy. If the Sun did not come up tomorrow, it would take around 86 hours for at least the land temperature to fall below -200C.
– the energy represented by a solar photon spends time in 5 billion individual molecules on Earth before it escapes to space. That means it is bouncing around from molecule to molecule to molecule almost continuously. The IR emitted by the surface is not skipping Nitrogen and Oxygen molecules and preferentially seeking out CO2 and H20 only. Every molecule on Earth and in the atmosphere is participating in this process and does so continuously. Maybe CO2 or H20 provides the initial absorption, but that energy is shared amongst the rest of the atmospheric molecules almost immediately. What happens to it then?
– The surface accumulates almost none of the solar energy which hits the surface during the height of the day. 960.000 joules/m2/second is coming in and 959.083 joules/m2/second is moving up and away from the surface. At night, virtually no energy is coming in and only 0.001 joules/m2/second is flowing up and out to space. That is not consistent at all with the greenhouse theory and the back-radiation theory. It is more consistent with energy flowing from hot to cold continously (like the second law of thermodynamics) and it flows faster the more there is a differential between that hot and cold.
– We need a time perspective on radiation physics because it is happening at the speed of light and at the miniscule amount of time that a molecule absorbs that energy before passing it on through emission or collisional exchange. CO2 holds onto an absorbed IR photons for an average 0.000005 seconds before it is emitted or passed onto another molecule, Every atmospheric molecule hits another atmospheric molecule every 0.00000000015 seconds, an emitted IR photon from the surface could escape the atmosphere in just 0.000016 seconds at the speed of light – yet it actually takes 40 hours to make the journey. In the Sun, the average photon takes 200,000 years to make it out.
Is there a climate model that can simulate that accurately? It would be far too complicated for any computer even 1000 years from now to be able to simulate accurately. We have to empirically measure what is really happening in such a complicated system and base models on that instead. What is really happening is that the theory is off by at least half to date.

Adding water vapour in the tropics gives the tropics a greenhouse effect, I remember very well when I left the airconditioned hotel in Dhaka Bangladesh, a wall of humid heat hit me and I had to wipe my glasses.
Every added particle to the atmosphere is a small downward radiator, adding more particles therefore adds more radiators. That’s in a nutshell the greenhouse effect.

u.k.(us)

steven mosher says:
February 13, 2011 at 3:39 pm
“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.”
======
WOW, did you just say skeptics are too stupid to defend themselves??
Sounds like you helped protect us, thanks for that!!
I’ll say it again, thanks.
Now, it is time to protect ourselves, from ourselves.

wayne

Mark Wagner says:
February 13, 2011 at 2:35 pm

I’ve never seen any science that directly addresses quenching of “warmed” CO2 to other atmospheric gasses prior to “re-radiation.”
======================
Mark, this has been discussed right here a couple of times before. See:
http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/
http://wattsupwiththat.com/2010/08/31/does-co%e2%82%82-heat-the-troposphere/
Just if anyone else wants something to consider:
This is a two way street and the energy moving from CO2 to N2 and O2 can go backwards to both H2O and CO2 eventually, for all energy does leave the earth only by radiation. And since there are some 100x more H2O molecules up to the tropopause, mostly H2O accepts this energy and radiates it to space.
[ And yes, a bit less than half is directed downward, but “back radiation” is a null effect when speaking of surface to space transfer, it can’t actually re-warm the surface in bulk (thermodynamically), only if you speak of singular photons and ignore the other effect (cooling) that cancels. ]
If you are speaking of published scientific papers I have not seen any either.
All of Mikael Cronholm’s explanations are well worded and even the mention of reflection can be quite correct if placed in the right frame of reference. All reflection is an absorption and re-emission at the quantum level for Richard Feynman made that so perfectly clear in his explanation of his quantum electrodynamics.
http://vega.org.uk/video/subseries/8
If you listen to his lectures, just replace the glass with the atmosphere and the light with infrared and ask: Can the IR cancel just as he describes black reflection bands as his mental counter goes Zzzzzzzzz….. (on the complex plane)?

Richard Bell

Wonderful Stuff ……… I would like to confirm that CLOUDS ( shouting loudly ! ) are the winners in my book, I have just spent 5 days on a very remote beach on the west coast of Baja, no people, no lights, nothing ……. just the big blue watery thing called the PACIFIC, the big yellow thing called the SUN and on one cold windy day some white stuff called CLOUDS ………. they had a stunning effect for that day, then were gone.
In the isolation that I was experiencing the big three just made sense………Thanks for the very interesting IR info and discussion.

DirkH

John S says:
February 13, 2011 at 4:05 pm
“[…] This fact has been acknowledged by the AGW community, but as their argument goes, the relatively small added heat from CO2 creates a positive feedback loop that includes adding more water vapor into the atmosphere until there comes a ‘tipping point’ where the global temperatures go into thermal runaway.”
The tipping point, if it exists, should be observable first in hot, moist regions, then. So places like the Amazon should “tip” before, say, sub-zero wintery Siberia. Once a hot, moist place on Earth “tips over”, the generated increased moisture should spread, making neighbouring regions “tip over” into the “hothouse Earth” mode.
In fact, the threshold difference between the Amazon rain forest and Siberia should be so big that a much smaller CO2 concentration should suffice to make the rainforest tip over initially than Siberia.
So, if we want to find proof for the existence of the tipping point, we should watch the rainforest. And maybe one could do an experiment. CO2 is heavier than air. Erect a big circular “fence” of thin plastic, maybe lifted up with a number of balloons, and increase the CO2 concentration within the fence. That should bring us much closer to the tipping point. As surface LWIR is absorbed and re-emitted by CO2 after a few meters, the effect should be measurable even if the fence is only say a 100 m high.
The fence would also ensure that the “tipping” can’t spread to pristine wilderness. Any takers?

Horace the Grump

“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…”
Completely nailed AGW to the mast….
Or he could have said “More like a religion…”
Either is probably acceptable…

rokshox

steven mosher says @ February 13, 2011 at 3:39 pm:
“the physics you use is the same physics that many skeptics deny”
“Many skeptics,” indeed. Well, “some say” they are sick and tired of strawman characterizations of their skeptical positions.

Ian W

John S says:
February 13, 2011 at 4:05 pm
Domenic Wrote:
“Thus, the comparative effects on ‘global’ warming, changes in H2O composition in the atmostphere are probably 1000X greater than CO2 changes in terms of contribution to ‘greenhouse effect’.”
This fact has been acknowledged by the AGW community, but as their argument goes, the relatively small added heat from CO2 creates a positive feedback loop that includes adding more water vapor into the atmosphere until there comes a ‘tipping point’ where the global temperatures go into thermal runaway.
CO2 is the “trigger” of of this positive feedback AGW, not necessarily the source of all the extra heat.
As a “warmist,” I cannot deny that CO2 does absorb and re-radiate heat energy. It is the existence of positive feedback vs. negative feedback mechanisms in the atmosphere that is at the heart of the global warming debate.

It is more the insistence that there is one linear positive feedback – the rate of water evaporation into the atmosphere and then water vapor acting solely as a ‘green house gas’ that is at the heart of the debate. For example, if the water vapor feedback is initially positive but then becomes strongly negative due to albedo effects of clouds and release of latent heat at height above the optically thick CO2/H2O layers of the atmosphere (this appears to be shown to be the case by satellite metrics) then the AGW hypothesis fails.

Jim D

I will clarify my earlier statement regarding IR emission versus reflection. Reflection of IR photons is negligible if you remember (e.g. from “why is the sky blue?” arguments) that scattering is greater for shorter wavelengths (scattering being reflection), and IR has a long wavelength relative to visible light. The measured IR from the sky is pretty much all emitted by CO2 and H2O molecules (and the other greenhouse gases present). You don’t see the same photons coming back down as the ones going up from the earth.

hotrod ( Larry L )

Good discussion that adds to the recent posts I made about experiments using an IR thermometer to get a sense of the “effective” temperature of the sky as seen from the ground.
The important thing that this experiment shows, is that unless the ground or ocean surface is receiving direct sunlight its entire view to the sky is in effect a very very cold surface. As you mention it is frequently off scale low on my IR thermometer which reads down to -76 deg F (-60 deg C) at emissivity 0.95 setting.
As a result a surface like the ground, a roof, or the ocean surface must always be losing heat to the sky by IR if it is not directly illuminated by the sun (or scattered sunlight from clouds etc.) The only other source of heat energy to maintain the temperature of that surface must come from heat transferred from the air, which by comparison is very small unless there is air motion.
It is perhaps a subtle distinction for some folks, but the CO2 and water vapor “back radiation” does not “warm the ground” or the ocean surface, it only “reduces the rate at which it is cooling by IR radiation” to the apparent surface of the very very cold sky.
As I showed in the measurements I made in the IR, due to this constant IR exchange between surfaces and the cold sky, there is very little relationship between the air temperature and nearby surfaces. At night, concrete is a very efficient radiator in the IR being like asphalt a nearly perfect black body radiator with an emissivity in the high 90% range (different sources give different values but most cluster between 0.95-0.98 for both).
What is more important, in a built up area, is a given piece of ground has a significantly reduced view of the sky due to buildings. As a result it cannot cool as rapidly through IR radiation to the sky at night because most of its hemispherical view is no longer of the open sky but rather building walls that are only a few degrees cooler than the air temperature rather than deep into the subzero IR temperatures of the open sky.
The urban heat island effect is as much to do with changes in visibility of the sky, as it is about waste heat lost from buildings. Even if you built an empty town that used no energy, ground measurements would still be warmer at night due to this change in the IR view of the sky due to structures and modifications of the topography that limit the hemisphere of the sky that is viewable from any given surface.
An analogy could be made if you replaced IR radiation with visible light. The sky would be a deep black but all nearby surfaces are very bright . The illumination of the ground surface would be completely dominated by the bright surfaces in a built up area, where in the open prairie or desert the illumination of a flat surface would be completely dominated by the black sky, with only a trace of scattered light from the nearby ground.
No matter how much you insulate buildings or reduce their energy consumptions, you will not change this IR view issue to the sky. Surfaces will always be warmer in a built up area than they would be in the open unobstructed environment of open ground regardless of energy consumption.
Larry

Richard Sharpe

Hans Erren says on February 13, 2011 at 4:57 pm

Adding water vapour in the tropics gives the tropics a greenhouse effect, I remember very well when I left the airconditioned hotel in Dhaka Bangladesh, a wall of humid heat hit me and I had to wipe my glasses.
Every added particle to the atmosphere is a small downward radiator, adding more particles therefore adds more radiators. That’s in a nutshell the greenhouse effect.

Hmmm, surely, it is every second particle?
Secondly, which particles? H2O, CO2, O2, N2?
Which has the larger effect?

Steve Reynolds

Latitude: “Since it’s constantly being exposed to IR, when it’s saturated, does it just reflect because it can’t absorb any more?”
There is no reflection from a gas. Also, under normal conditions, CO2 in the atmosphere does not become ‘saturated’ if I understand how the term is being used.
Latitude: “Or is it in a constant state of absorbing and re-emitting? neutralizing itself?”
Yes, molecules are absorbing, equilibrating temperature with surrounding molecules, and emitting at the equilibrated temperature in random directions.

Latitude

Steve Reynolds says:
February 13, 2011 at 5:57 pm
Yes, molecules are absorbing, equilibrating temperature with surrounding molecules, and emitting at the equilibrated temperature in random directions.
======================================================
So it’s just absorbing and scattering the IR, same in and same out.
Wasn’t clear on that, I slept through class…….
thanks Steve

Dave Springer

“bulbs emitting 300W/m2 is simply impossible, because 300W/m2 corresponds to a very low temperature!”
It wouldn’t be a conventional “bulb” first of all because those are nowhere near blackbody spectrum and second of all because they’re near infrared not far infrared.
Your “light bulb” would be a piece of black metal with an area of 1 square meter maintained at a temperature at of -23.4C. In most cases you’d have to be cooling it below ambient temperature unless you were operating it outdoors the other night in Oklahoma when it hit -35C then you’d need to run a bit of electricity through it to heat it up some.
This is essentially what the great experimental physicist John Tyndall did 150 years ago when he experimentally verified infrared opacity in various gases. Tyndall’s main “light bulb” was a copper vessel painted with lamp-black filled with boiling water which emits a blackbody spectrum in the far infrared and maintains a pretty constant temperature without much hassle since water won’t heat above its boiling point without pressurizing the vessel. It was aimed at a brass tube polished to a mirror finish on the inside and capped on each end with a plate of rock salt (rock salt is practically transparent to infrared) with valves and a vaccuum pump to evacuate the tube and introduce the gases he wanted to test. On the far end had a thermopile and galvanometer. The galvanometer response was non-linear and so to keep the thermopile outputting a voltage in the linear high-sensitivity range of the device he put another “infrared source on the backside of the thermopile with an adjustable shield to regulate how much radiation the thermopile received from the backside. The setup was quite sensitive and just a warm body in the same room with it would compromise the experiments. To demonstrate its sensitivity Tyndall would bring the thermopile/galvnometer into halls where he was giving a lecture, aim it at a wall on the other side of the lecture hall then have a member of the audience step into its view and the galvanometer would peg at the max reading. So a warm body moving around the lab during an expermental runs didn’t compromise the results he read the galvanometer from a distance with a telescope. It was quite the ingenious experimental setup given the times (1850).
The thermopile was quite sensitive and just a warm body in the same room with it threw it off so