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
February 13, 2011 12:26 pm

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

Ian W
February 13, 2011 12:35 pm

When water vapor in the atmosphere condenses into liquid water and then changes state again and becomes ice, it gives off latent heat for both state changes.
Does that latent heat release follow Stefan-Boltzmann’s radiative equation ?

Dennis Wingo
February 13, 2011 12:36 pm

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
February 13, 2011 12:36 pm

“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.

ferd berple
February 13, 2011 12:36 pm

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
February 13, 2011 12:42 pm

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

February 13, 2011 12:42 pm

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
February 13, 2011 12:57 pm

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
February 13, 2011 12:59 pm

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.

February 13, 2011 1:00 pm

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
February 13, 2011 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.

Charles Nelson
February 13, 2011 1:06 pm

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
February 13, 2011 1:25 pm

Interesting.

Domenic
February 13, 2011 1:30 pm

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
February 13, 2011 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.

Domenic
February 13, 2011 2:01 pm

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
February 13, 2011 2:28 pm

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
February 13, 2011 2:32 pm

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
February 13, 2011 2:35 pm

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
February 13, 2011 2:36 pm

>>
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

February 13, 2011 2:37 pm

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

Bruckner8
February 13, 2011 2:42 pm

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
February 13, 2011 2:43 pm

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
February 13, 2011 2:49 pm

>> 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
February 13, 2011 2:51 pm

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
February 13, 2011 2:54 pm

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.

February 13, 2011 3:39 pm

#
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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
February 13, 2011 3:40 pm

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
February 13, 2011 3:41 pm

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.

tallbloke
February 13, 2011 3:43 pm

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

February 13, 2011 3:56 pm

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
February 13, 2011 3:58 pm

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
February 13, 2011 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.

richard verney
February 13, 2011 4:11 pm

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
February 13, 2011 4:18 pm

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
February 13, 2011 4:55 pm

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.

February 13, 2011 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.

u.k.(us)
February 13, 2011 5:10 pm

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
February 13, 2011 5:15 pm

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
February 13, 2011 5:34 pm

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
February 13, 2011 5:35 pm

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
February 13, 2011 5:40 pm

“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…

Molon Labe
February 13, 2011 5:44 pm

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
February 13, 2011 5:44 pm

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
February 13, 2011 5:48 pm

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 )
February 13, 2011 5:49 pm

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
February 13, 2011 5:51 pm

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
February 13, 2011 5:57 pm

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
February 13, 2011 6:08 pm

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
February 13, 2011 6:12 pm

“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

DirkH
February 13, 2011 6:18 pm

Bill Illis says:
February 13, 2011 at 4:55 pm
“The IR emitted by the surface is not skipping Nitrogen and Oxygen molecules and preferentially seeking out CO2 and H20 only.”
I would say the IR photons *do* skip the N2 and O2 molecules as these molecules can’t absorb IR. The IR photon does not “see” these molecules but passes through them as if they weren’t there.
Of course, after the IR photon is absorbed by, say CO2, the energy is thermalized and can be transferred from, say CO2 to N2 – but also back from N2 to CO2, and the CO2 can re-emit an IR photon.

David Falkner
February 13, 2011 6:19 pm

steven mosher says:
yup, but there is more to it than that. If you want to do IR design today ( of sensors, or counter measures or stealth technology) the physics you use is the same physics that many skeptics deny. good thing we don’t let them design the machines that our
safety as a nation depends upon.

Wow, mosher. You must be bored to start taking cheap shots at strawmen. As far as I know, there aren’t many skeptics that deny the physics, just the conclusions that Earth is going to see a large temperature increase because of it.

eadler
February 13, 2011 6:28 pm

Charles Nelson says:
February 13, 2011 at 1:06 pm
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.

Brilliant. Focus on one particular phenomenon, radiation from the tops of clouds, as if that is all that is going on. No need to look at any other effect to determine the plausibility of the idea that GHG’s emission are going to warm the planet. No need to do any mathematical modeling. All you need to do is believe what you want to believe, and select those phenomena which confirm your beliefs and focus on it.
It is great propaganda and apologetics, but hardly science.
REPLY:
“Brilliant. Focus on one particular phenomenon, radiation from the tops of clouds, as if that is all that is going on.”
Well said, we should stop focusing on CO2, as if that is all that is going on. – Anthony

Dave Springer
February 13, 2011 6:35 pm

There’s no need to be confused. The “canard” the author refers is no canard but a manifestation of his ignorance. The infrared absorption characteristics of various gases was measured experimentally 150 years by John Tyndall. If you shine far infrared light through a column of IR-transparent gas the amount of radiation you get on the far side of the column is the same as you get going through a column of vacuum. If you replace the transparent gas (like nitrogen) with an IR-absorptive gas (like water vapor) there is less radiation emitted at the far end of the column. The gas absorbs the radiation from a directional source and re-emits it in all directions. In Tyndal’s apparatus there were only two exits for the radiation since the column was a brass tube polished to a mirror finish on the inside – the two exits were plates of rock salt at either end of the tube. The radiation entering the column remains the same no matter what gas (or vacuum) is in the column but the radiation exiting the column at the far end varies according the kind of gas in the column. Energy must be conserved so what’s missing at the far end of the tube has to be going somewhere – in fact it heats the gas which increases the amount of energy the gas is emitting and since the gas emits in all directions some of the energy goes through the rock salt entrance plate instead of the exit plate while some smaller portion will heat the brass tube through conduction and some even smaller amount will heat the tube through absorption because even brass polished to a mirror finish isn’t quite 100% reflective.

LazyTeenager
February 13, 2011 6:53 pm

For 300 W/m2 radiation I get -23.4°C at 300 W/m2 when I calculate it (yes, minus!). Pretty cool
———-
I get the impression that there is some miscomprehension here.
I would say that the flux of 300W/m2 being calculated here, is not the same as the 300W/m2 of down-radiated atmospheric IR.
Maybe the definition of terms and conditions in the SB formula needs to be verified.

Latitude
February 13, 2011 6:54 pm

Well said, we should stop focusing on CO2, as if that is all that is going on. – Anthony
===============================================
Now that really was brilliant! good one………

wayne
February 13, 2011 7:01 pm

Hans Erren says:
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.
———-
That’s right. Every added particle to the atmosphere is a small upward radiator also, adding more particles therefore adds more radiators. That’s in a nutshell is why there is no real “greenhouse effect”. The two effects always cancel. You say ‘more warming downward’ and I say ‘also equal cooling upward’.
You see, it’s a two way street in real physics, you just ignore the part that doesn’t fit your (incorrect) view.
That to me that is why Miskolczi came up with the results he did. Now why, No one has answered that. Is it that with the small increase of temperature we have seen and the small expansion of the atmosphere does cause a higher fraction to escape to space because we do live on a sphere and not an infinite plane and that is merely a geometric effect (ex-secant correction)? Could it be that there is one or more negative feedbacks? Is it a combination of both? I’m still searching.

Domenic
February 13, 2011 7:08 pm

To Steve Reynolds
1. There is not a great deal of difference, if any, between liquid phase H2O spectral transmission and water vapor spectral transmission. I used those two graphs because data from NIST tends to be more reliable than most that I have seen.
But here’s another graph for you http://en.wikipedia.org/wiki/File:Atmosfaerisk_spredning.gif
See the wide absorption band at 5.5 to 7.7 microns…that is entirely by water vapor, H2O. Compare it to the narrow bands absorbed by CO2 in the same graph. In the vapor state, it is still at least 10X greater for this wavelength spread.
There is a problem though, and that is that we have to speculate over the entire spectral output of the dominating forces especially the sun, which outputs much energy in shorter wavelengths to the earth. To my knowledge, that data (spectral absorption and transmission and reflection of CO2 and H2O) has never really been assembled as there had never seemed to be a need for it.
2. CO2 does indeed have a reflective component as does H2O. Only a black body has no reflective component. The reflective component of H2O is better known than that of CO2 because water is more common, and has been tested for IR more thoroughly. All real world materials are reflective, absorptive and emissive. In general, metals are highly reflective (think Al foil, etc) on one end of the scale, organics and others less so, water being one of the most absorptive, least reflective. And the reflective component is highly wavelength measured dependent, as well as angularly dependent (the angle with which the wavelength strikes the molecule).
But for magnitudes of the effect on the atmosphere in terms of greenhouse effect, the known data is sufficient to see the HUGE differentials between water vapor effects vs. CO2 effects.

eadler
February 13, 2011 7:11 pm

Dennis Wingo says:
February 13, 2011 at 12:36 pm
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.
There is no such thing as absolute proof of a theory in science. We don’t need that in order to take action on the basis of the knowledge that we have.
More detailed observations can tell us more precisely what is happening in the present, so that we don’t have to wait 100 years to find out the outcome of increasing CO2 concentrations. That is the purpose of the scientific research that is going on. Observations of radiation, behavior of clouds, and other phenomena that are involved, plus innovation in computers and models, provide the information needed to produce better predictions of climate change.
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.
No one is claiming that as far as I have seen.

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.

I think you are wrong about this. Models provide for different emissivities for snow, forests, grassland, rocks ocean etc.

Also, the models do not take into account the dramatic differences in resulting temperature based upon altitude, especially in desert regions of the globe.

This is clearly wrong. Models do take altitude into account in the prediction of climate change.
http://journals.ametsoc.org/doi/abs/10.1175/1520-0442%281997%29010%3C0288%3AEDOTSC%3E2.0.CO%3B2
Elevation Dependency of the Surface Climate Change Signal: A Model Study

A lot of physics modeling operates by making simplifying assumptions, but how many of these assumptions are testable and repeatable?

These assumptions are constantly being tested. One of the first assumptions made, by Arrhenius in 1896, in his modeling of CO2’s impact on global temperature was that average Relative Humidity would be constant and water vapor concentration in the atmosphere would increase, resulting in positive feedback for temperature increase. Satellite observations are being made which seem to confirm this assumption.
http://news.cisc.gmu.edu/doc/publications/Chung%20et%20al%202010.pdf
In spite of significant biases in tropospheric temperature
and humidity in climate models [John and Soden,
2007] and resultant compensating effects in simulating the
clear-sky OLR, our analysis finds broad consistency between
the observed and modeled rates of clear-sky OLR radiative
damping. This consistency is noted over a broad range of
observable sources of climate variations, suggesting that the
strong correlations between water vapor and temperature
necessary to generate such sensitivities are a robust feature
of both models and observations. This analysis offers
further evidence to support the ability of climate models
to depict the physical processes related to the combined
water vapor and temperature climate feedback.

This is the basis of a lot of my skepticism on the models involved.
It seems then that the basis of your skepticism is not real.

Dave Springer
February 13, 2011 7:17 pm

Domenic says:
February 13, 2011 at 1:30 pm
“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.)”
True enough. It varies between 1% (Sahara) and 5% (Amazon) except for Antarctic interior where it is close to 0%. But that’s at the surface. Adiabatic lapse rate squeezes out the water vapor with falling temperature while CO2 concentration remains constant with altitude.
Tyndall (circa 1850) however found that the concentration of the gas doesn’t matter but rather the total amount of the gas in the column is all that matters. He confirmed that by varying the pressure and length of the column. Tyndall performed literally thousands of experiments with varying, varying pressures, varying column lengths, and varying infrared frequencies.
To begin understanding the physics of greenhouse gases one must at least be familiar with what was experimentally demonstrated by physicists in the mid-19th century.
There’s plenty to complicate the situation in the real world beyond comprehension because the atmosphere is a dynamic system with varying gases, pressures, and many physical processes other than radiation going on that are moving heat from one place to another and radically different and rapidly variable rates.

Dennis Wingo
February 13, 2011 7:17 pm

the physics you use is the same physics that many skeptics deny.
This statement is an insult to the intelligence of people who actually work in the field. The parameters related to the absorption and re-emission of IR radiation by CO2 were worked out by the USAF (the early parts were classified), in what used to be called “upper atmospheric research”. Spectrometer technology was specifically improved to the point to where the individual quantum absorption and emission lines were discerned by the mid-late 1950’s.
These measurements were used as a validation of the gaussian to Lorentz transform that characterizes the increase in the absorption/emission lines of CO2 and other IR absorbers/emitters. This was used to design the IR sensors of ALL of our inventory of IR sensing missiles.
If you find the equations for this emission/absorption, you will find that the increase in the line widths is proportional to the increase in the minor gas against the ENTIRE atmosphere, not just the concentration of the gas relative to the arbitrary baseline of 280 ppm. The absorption/emission of CO2 has two dependent variables, temperature and pressure, neither of which are accurately characterized in the models used by the AGW community.
All of this is set forth in any Quantum Mechanics book on the theory of light, of which the relevant IR wavelengths absorbed/emitted by CO2 are a part. The relevant text I use is from Loudon, pages 82-89.
Why don’t you have the moral courage to look that up and derive the effect of the increase of CO2 yourself. You will be surprised at the result.

Policyguy
February 13, 2011 7:20 pm

Anthony,
This exchange was a treat, it amazes me sometimes what can be learned on the web.
This quote
“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…”
—–
Is intriguing as well. Gore and company stopped long ago trying to prove anything. They were more interested in having enough of their theories “accepted” to stay at the grant trough. In a way it is reminiscent of the times when it was “accepted” that the Sun orbited the Earth. But then it was more for religious reasons than scientific. Today Gore strives to maintain his religious acceptance of CAGW to prop up the price of his offset credits.

J. Bob
February 13, 2011 7:28 pm

W. Falicoff says: “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.”
HHmmmm. If a basic sensor has a mean error of 0.75 deg., and 3 sigma of say +/- 0.25 deg., your true error will be between 0.5 & 1.0 deg no matter how many readings you take. You might want to add a few qualifiers.

Dave Springer
February 13, 2011 7:31 pm

A safe statement is “increasing atmospheric CO2 will result in increasing surface temperature if nothing else changed“. What causes “confusion at a higher level” is that lots of other things DO change and the amount of warming directly attributable by CO2 infrared absorption and re-emission is small and can be utterly swamped (lost in the noise) by many other dynamic processes. Water in all its phases has the starring role in shaping our climate. CO2 plays a major role in surface temperature only during so-called “snowball earth” episodes when most of the water vapor has been frozen out of the atmosphere and there’s little to no liquid water presenting on the surface. In that case there are no working CO2 sinks so volcanoes, which keep on belching out CO2, gradually build up the amount of CO2 in the column until there’s enough greenhouse effect to begin melting the planet. If it weren’t for CO2 the earth would likely be covered in ice with no hope of ever melting. That’s the most commonly accepted hypothesis at any rate and I personally haven’t seen anything that seriously disputes it.

February 13, 2011 7:32 pm

Maybe those who doubt the ‘frequency selective’ nature of CO2 IR spectra should investigate Molecular Spectroscopy as it relates to the vibrational modes of CO2 (and the other important GH gas WV) on account of it’s molecular properties:

By examining the emission spectrum of the CO2 laser, we are able to understand much about the CO2 molecule and about the dynamics of diatomic and triatomic molecules in general. The CO2 laser is a molecular laser, meaning that it generates light from the vibrations and rotations of the CO2 molecules in the plasma rather than from electronic transitions between energy levels, as in a He-Ne laser.
Like a spring between two masses, the binding forces between the atoms of the CO2 molecule cause the atoms to move in one of three vibrational modes: the symmetric stretching mode, asymmetric stretching mode, and the bending mode.

Also: http://en.wikipedia.org/wiki/Infrared_spectroscopy
High-school level physics, boys …
.

Richard Sharpe
February 13, 2011 7:33 pm

wayne says on February 13, 2011 at 7:01 pm

Hans Erren says:
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.

That’s right. Every added particle to the atmosphere is a small upward radiator also, adding more particles therefore adds more radiators. That’s in a nutshell is why there is no real “greenhouse effect”. The two effects always cancel. You say ‘more warming downward’ and I say ‘also equal cooling upward’.

What you have said, Wayne, does not seem correct to me.
It seems to me that approximately half the energy absorbed by CO2 would be re-radiated upwards and half downwards.
They do not cancel.
The energy remains in the atmosphere for a little longer than it otherwise would, but I suspect that it does not matter as H2O transports much more energy out of the atmosphere and any that is “trapped” by CO2 quickly departs during the night.

G. Karst
February 13, 2011 7:35 pm

There is no reason for the warmist to exhibit glee to the admission of CO2 narrow back-scatter effect. All substances absorb and emit IR (yes, even salt and quartz windows). It is all about spectrum or frequency and a molecule’s representative target cross sectional area .
The debate is entirely about CO2 back-scatter’s significance on the actual climate.
Since theoretically, CO2 IR spectrum is already saturated, any additional molecules, simply cannot affect much. The same is not true concerning the planets biomass. It will expand exponentially to increased temps, increased CO2, and increased available moisture. This too will have a feedback. As stated many times, this is about climate sensitivity as per much debated feedbacks and forcings. GK

wayne
February 13, 2011 7:39 pm

Dennis Wingo says:
February 13, 2011 at 7:17 pm
… The absorption/emission of CO2 has two dependent variables, temperature and pressure, neither of which are accurately characterized in the models used by the AGW community. …
– – – – – – – –
Dennis, I see you great point but could you expand a bit on your statement above? I’m wondering if that could be temperature and density, for a slightly warmer atmosphere will expand but the mass remains constant, therefore, density drops but the pressure remains constant at sea level (average that is). Does it make sense?

kuhnkat
February 13, 2011 7:40 pm

Anthony,
doesn’t your camera “see” in the near infrared? The earth emits in the far infrared. The cut-off is at about 4um between the two I believe.
Some of that wonderful military hardware some people keep talking about is working in the near infra-red and has little applicability to the back radiation issue!!
The scatter, reflection, absorption bit I have been trying to clarify also. Reflection generally refers to the effect where a particle or wave encounters and leaves an object or field at the same angle. Scatter is where the direction is random. Absorption, of course, is where there is no exiting wave or particle. They are different.
Electromagnetic radiation in the atmosphere is rarely reflected by gasses, only particles like water droplets or aerosols. They are either scattered or absorbed.
Someone please correct me as I am obviously not the expert here.

Domenic
February 13, 2011 7:41 pm

To Dave Springer
Re the measurement of IR absorption.
The polished tube method is not exact. It’s an approximation. It does not account for angular effects of wavelengths and reflective effects of wavelengths striking the molecules accurately. In addition, it relies on imperfect ‘spectal window transmission coatings’ on the sensor window material, and other factors.
The science is nowhere as accurate as many believe it to be.

chico sajovic
February 13, 2011 7:45 pm

“Back Radiation” is such a horrible analogy, description or way to think about heat transfer in the atmosphere, that I wish it would just go away. What we are concerned with is the flow of heat from the surface of the earth to space. Its not that the surface warms the co2 in the air then the co2 “back radiates” re-warming the surface, its that when the co2 in the air is warmed the flow of heat from the surface to the warmed co2 is reduced, thus increasing the temperature of the surface until the flow of heat returns to equilibrium.
1) sun rays heats surface of earth
2) warmed surface radiates, heating greenhouse gasses
3) smaller temp delta between surface and greenhouse gasses reduces heat flow
4) reduced heat flow from surface increases surface temperature
5) equilibrium is restored with higher temp surface and greenhouse gas
My problem with the greenhouse effect is how do you separate radiation from the effects evaporation, condensation, conduction and convection. Anecdotally I feel that conduction is a far superior mode of heat transfer and evaporation is even better.
To cool my cup of hot chocolate faster is should:
a) hold it up to the sky on a low humidity night
or
b) blow on it
When I am hot I should:
a) stand outside naked on a low humidity night
or
b) get soaked in water and stand in front of a fan

Fitzy
February 13, 2011 7:46 pm

Brilliant conversation, thanks for sharing.
Considering the energy smart rattle being waved by local authorities, one wonders how derelict Urban Authority, IR emission equipment callibration, will be.
And i’ve learnt the hard way that the moment one puts Red on a map, some policy wonk will smell promotion and run with even the most tenuous data, and as this site so often points out Policy = Misspent funding. Usually when one says IR, the colour red shows up somewhere, and RED = Hot right and away we go.
One may point out a scale can be quite tight, with the banding representing a single degree, case in point, IR work done in winter. All of a sudden ones city glows bright red, the Policy wonk will scream ‘Look at all that energy wasted!’, until Johnny lab coat points out, ‘That sir is a mere 14 degree’s centigrade’. And you’d be amazed when you point out that water holds a lot of heat, their eyes cross, they don’t like the notion the very landscape is bleeding heat.
I’d be keen to know what tolerances and ranges equate to a fair IR reading within an urban environment, and with emissivity being so key, one can imagine Urban Authorities taxing by the W/m radiated.
Good luck with your endeavors.

kuhnkat
February 13, 2011 7:46 pm

Oh no, the Ham and radio guys are gonna get me.
In my statement above I left out the effects of IONIZED gasses which DO reflect electromagnetic radiation.

February 13, 2011 7:47 pm

steven mosher says:
February 13, 2011 at 3:39 pm
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.
And if you had any idea what that was, you would have told us instead of delivering that empty and useless trollbuttal. What physics are being denied? Put up or shut up!

Dave Springer
February 13, 2011 7:54 pm

Steve Reynolds says:
February 13, 2011 at 2:43 pm
“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.”
Actually that’s not quite right. At low CO2 concentrations it’s nearly linear and as concentration rises it becomes logarithmic. John Tyndall experimentally discovered that too by experiment over 150 years ago. It’s pretty much a case of the proverbial low-hanging fruit. The first few molecules get all the fruit they can handle but as the number of molecules increase there’s less and less fruit available per molecule. IIRC correctly the curve is logarithmic by the time concentration reaches 100ppm.

February 13, 2011 7:55 pm

Guys/mods – Is my comment above stuck in the spam filter?

Mikael Cronholm
February 13, 2011 8:02 pm

WOW! When Ken asked me if he could have our conversation offered for publication somewhere I had no idea it would create such an interest ans so many extremely initiated comments. I am glad I could contribute even if it is on a fairly basic level, and I apprecite everyones inputs.
I want to repeat again that I am not on any “side” in the global warming debate. And my opinion is that anyone who claims to be absolutely sure about it may be politically correct but never scientifically correct. There are no scientific proof, one way or the other. Science is not democratic, so counting the forces in your camp is merely silly.
I do stand corrected on the 300W/m2 calculation, it turns out to be about -3.3C. It does not change the point I was making though.
I also understand the objection when I say “reflected” regarding the radiation from that is returned to the earth from the atmosphere. Anything WILL reflect, unless it is a blackbody, but true, in IR where I am familiar only about 5% is reflected from water or ice. The rest is absorbed and re-radiated. But my expertise, as I pointed out mainly down on earth.
Water is an amazing substance and the more I study it the more it amazes me. It has several unique features, among them its incredible ability to store heat, especially latent heat. Just the simple fact that it is a dipole and therefore orders the molecules when it freezes in such a way that the density goes down (unique!) is an important fact that is overlooked. If not for that, we could forget about fish in most freshwater lakes up north, for example.
But the most important thing with water is its ability to moderate temperature here on earth. The greenhouse effect as such is not a threat – it is an absolutely necessary condition for life on earth. It creates the moderated thermal equilibrium that allows us to live on a planet that does not change its temperatures too much over the day and year. Whether or not we are tinkering too much with that equilibrium is what this whole debate is about, and no proof is yet presented, one way or the other.
I will look through the comments again one by one when I have a little more time to spend and see if I can make any additional contributions.
Thanks Ken for pulling these things out of me, and thanks everyone for scrutinizing and commenting on our conversation!

Keith Minto
February 13, 2011 8:05 pm

Key point by Mikael Cronholm about the uncertainties……

……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.

Mix in variable convection and possible gas density ‘layering’ and the difficulty in measurement is compounded.

kuhnkat
February 13, 2011 8:13 pm

_Jim,
Your comment claims elementary physics around the CO2 molecular bond and transitions. Would you mind explaining whether the transition energy or the bond energy is more important at atmospheric temperatures and why?

Dave Springer
February 13, 2011 8:14 pm

richard verney says:
February 13, 2011 at 4:11 pm
“CO2 delays IR finding its way out to space. However, what I do not understand is how this effectively heats up the Earth.”
It doesn’t heat the earth any more than insulation in your attic heats your house in the winter. It slows down how fast heat can escape. Since the heat is arriving at the same but leaving at a slower rate this causes the surface to become warmer which then increases the rate at which heat moves across the boundary. A new higher surface equlibrium temperature is thus established. In reality too many other things change too rapidly for equilibrium to ever be attained but the theoretical equilibrium point will rise and thus the target which the system seeks is that much higher.

u.k.(us)
February 13, 2011 8:17 pm

Slacko says:
February 13, 2011 at 7:47 pm
=======
So, other than your rant, have you anything to add?
I can’t wait.

chico sajovic
February 13, 2011 8:20 pm

Mikael Cronholm,
In your paper you say “Radiation is the strongest heat transfer mode”. Please elaborate. From my anecdotal experience conduction and evaporation are stronger modes of heat transfer: sweating is better for cooling down than not, blowing on a cup of hot chocolate is better than letting it cool by radiation.
What do you think of the concept of “back radiation” whereby “downwelling radiation” from colder air warms the warmer surface of the earth.

Dave Springer
February 13, 2011 8:27 pm

Domenic says:
February 13, 2011 at 7:41 pm
“The polished tube method is not exact. It’s an approximation.”
It might not be exact but when you have a vacuum in the tube and nitrogen in the tube the thermopile output is uneffected but when you have a greenhouse gas in the tube the thermopile output is reduced.
That proves beyond a shadow of a doubt that there’s something very different about greenhouse gases illuminated by infrared.
Feel free to explain the difference via some mechanism other than some gases being transparent to IR and some being opaque. Alternative explanations have two defining characteristics: they are always entertaining and never true.

AJB
February 13, 2011 8:36 pm

Mikael Cronholm says February 13, 2011 at 8:02 pm

Water is an amazing substance …

All your Joule are belong to us.

Patrick Davis
February 13, 2011 8:37 pm

Awesome article.
“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.”
Every AGW supporter I have spoken to cannot get their heads around this concept.

Dave Springer
February 13, 2011 8:44 pm

Mikael Cronholm says:
February 13, 2011 at 8:02 pm
“There are no scientific proof, one way or the other.”
Insofar as some gases absorb infrared radiation and some do not was scientifically proven in the mid-19th century. Your continued denial only means you know less about the physical properties of gases than mid-19th century physicists.

eadler
February 13, 2011 8:44 pm

J. Bob says:
February 13, 2011 at 7:28 pm
W. Falicoff says: “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.”
HHmmmm. If a basic sensor has a mean error of 0.75 deg., and 3 sigma of say +/- 0.25 deg., your true error will be between 0.5 & 1.0 deg no matter how many readings you take. You might want to add a few qualifiers.

If the mean error is know, the readings can be adjusted. If the mean error remains constant, than the temperature anomaly, which is what is being sought will not suffer in accuracy. This is the important point, that many who argue about the problems with temperature data seem to ingore, or not understand.

February 13, 2011 8:51 pm

Bill Illis says:
February 13, 2011 at 4:55 pm
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.
What!?? Where I live it can sometimes fall from +40C to +36C between sunset and sunrise. I think maybe you left the atmosphere out of that equation.

Tsk Tsk
February 13, 2011 9:02 pm

Dave Springer says:
February 13, 2011 at 7:54 pm
“Actually that’s not quite right. At low CO2 concentrations it’s nearly linear and as concentration rises it becomes logarithmic. John Tyndall experimentally discovered that too by experiment over 150 years ago. It’s pretty much a case of the proverbial low-hanging fruit. The first few molecules get all the fruit they can handle but as the number of molecules increase there’s less and less fruit available per molecule. IIRC correctly the curve is logarithmic by the time concentration reaches 100ppm.”
Not sure I like that. The form of the equation shouldn’t change for different numbers unless there’s a discontinuity. It has to be logarithmic over the entire range of positive real numbers and obviously isn’t physical for negative numbers. I think you mean that for low concentrations of Co2 the dependence diverges only slightly from a simple linear dependence and the divergence grows as the concentration grows.

davidmhoffer
February 13, 2011 9:05 pm

MC – excellent discussion, simple straight forward answers referencing the actual physics. We need more of this in the discussion, not more slogans, so thanks for stepping up!
Now – H2O vs CO2
For all the skeptics who have been trying to present the argument that water vapour is so much stronger an absorber than CO2 that CO2 doesn’t matter, sorry, but you are wrong. And I am a hard core skeptic!
Water vapour is in fact a much stronger absorber than CO2 and in the same approximate spectrum. If water vapour and CO2 were mixed evenly in the atmosphere, that would make CO2 insignificant. But they are NOT. CO2 is reasonably well mixed throughout the troposphere, but water vapour concentration falls rapidly with temperature. As a consequence water vapour declines with both altitude and latitude. What ever CO2 does or does not do, well over 90% of it exists in the atmosphere at temperatures low enough that water vapour is also low enough that CO2 becomes significant by comparison.
That said, I repeatedly ask the logical next question that I have yet to see a reasonable answer for from the warmists. If CO2 reflects/re-radiates/back-scatters what ever you want to call it, upwardly bound long wave, it makes sense that this would result in a temperature increase which in turn would increase water vapour. BUT, and I repeat BUT, does it not also follow that any increase in water vapour works both ways? That is, certainly the water vapour would re-radiate upward bound LW from earth surface, but it would ALSO re-radiate DOWNWARD bound LW too. The downward bound LW that would normaly have been absorbed at earth surface now has an increased % chance of being re-radiate back up. And we’re not talking about just the downward LW from increased CO2, we’re talking ALL the downward LW from ALL sources. In other words, increased water vapour may in fact have a net positive feedback, but it has a huge, built in, negative feedback too that likely renders the whole calculation near meaningless.
THERMAL QUENCHING – I saw some comments on this too. This was a big issue for Ernst Beck (may he rest in peace) who felt that this was an under estimated effect of CO2 LW absorption. His explanation was way, Way, WAY over my head. It was a seriously complex issue as there are so many factors that govern how an individual molecule can or cannot aborb or lose a photon when in collision with another dismilar molecule, at what temperature, and at what time period before re-emission would happen anyway. I don’t have any certainty that he was correct, but the explanations and rebuttals showed one thing pretty clearly – no one has a real good grip on the complexities, in the atmosphere in particular, and measuring thermal quenching is darn near impossible.

February 13, 2011 9:09 pm

For completeness and accuracy of the article it should be noted that the “inexpensive” household IR (non-contact single point) thermometers are designed to measure emission from distant solid objects. Therefore they use the “atmospheric window” where air has the least interference with IR and CO2 does not absorb nor emit, so they can get better distant readings of objects they are designed to measure. Hence the device uses an IR detector working in a narrow 8-12um range, right in the center of the window. When pointing to sky, only clouds/haze/smog can affect readings.
Regarding the Antony’s picture, he should be able to recalibrate the scale of his FLIR camera to colder side, and it is likely that he would see IR images of clouds behind his house in minus-40-50 range, where they fly.

wayne
February 13, 2011 9:10 pm

Richard Sharpe says:
February 13, 2011 at 7:33 pm
What you have said, Wayne, does not seem correct to me.
It seems to me that approximately half the energy absorbed by CO2 would be re-radiated upwards and half downwards.
——-
Two units of energy goes up, cooling the suface, surface -2, and is absorbed warming the atmosphere, atmosphere +2, your radiators means one unit always goes upward to space, atmosphere now +1, space +1, and one back down to the surface, surface -1, atmosphere is now zero. Plaease tell me what has just happened, for the surface is -1 unit of energy and space is now +1. Why is this so confusing to you?
And I do know this is ignoring energy that goes via the radiative window directly to space but that effect is exactly the same.

Ian L. McQueen
February 13, 2011 9:11 pm

The following is pretty elementary compared with what has been presented in the text and in the comments, but…..this example may help visualization of “back radiation”. I shower in the same stall daily. In the summer, I have to run the water a little cool to keep from overheating myself when the temperature is, say, 20°C. In the winter I have to run the water considerably warmer and keep the air temperature around 22° to keep comfortable. I figure that the difference is due to the difference in temperature of the walls and ceiling in the two seasons.
Related to this, friends heat their house by means of electrical heating panels that warm the drywall panels in the ceiling. Even though they are barely warm to the touch, they are adequate to keep people in the room comfortable. Radiant energy.
IanM

February 13, 2011 9:11 pm

u.k.(us) says:
February 13, 2011 at 8:17 pm
Slacko, — other than your rant, have you anything to add?
I can’t wait.

What rant? I repelled a rant and posed a question.
So while you’re waiting, maybe you can figure out “What physics do skeptics deny?” If you lot think you can get away with accusing me of your own folly, you’re wasting your time.

Dave Springer
February 13, 2011 9:12 pm

chico sajovic says:
February 13, 2011 at 8:20 pm
“Mikael Cronholm,
In your paper you say “Radiation is the strongest heat transfer mode”.”
Ultimately it’s the ONLY mode as convection, conduction, and mechanical transport (evaporation/condensation) ends where the atmosphere ends.
But that is certainly a valid point as we live and breathe and raise our crops on or very near the surface and these mechanisms can accelerate the transfer of heat from surface to above the cloud layer resulting in practically no change at the surface. The increase in temperature caused by increased CO2 doesn’t have to be at ground level. It could be at 10,000 ASL and not effect our surface activities one tiny bit. In fact the climate boffins expected to find the CO2 “signature” as a hotspot in the upper troposphere where the air is very dry and they were confounded when the temperature rise was found to be greatest at the surface where the air is very wet. This should have been their first clue that their climate models were fundamentally wrong. But they’d already decided by then that fossil fuel consumption was going to cause great harm to the planet so they had to continue blaming it. The mantra morphed from “global warming” first to “climate change” then when that didn’t resonate with the unwashed masses they changed it again to “global climate disruption”. That’s not science it’s a marketing campaign for an ecoloon religion. A floundering campaign, by the way, which is losing ground at an accelerating rate with every passing day.

Tsk Tsk
February 13, 2011 9:14 pm

From the article:
“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.”
What’s missing is that the lamps are already radiating based on their room temperature. The 300W/m2 from the lamps when they are turned on is, of course, additional, whereas the 300W/m2 from the sky is the total heat flux from the sky. The two are not comparing the same thing at all.
It was also nice to see all the talk about the importance of emissivity. It’s a subtle point that many get wrong.

Phil
February 13, 2011 9:17 pm

So far the only things that have been discussed are the thermodynamics of CO2 vs H2O vapor. What about ICE clouds (slide 23)?

Mikael Cronholm
February 13, 2011 9:18 pm

@chico sajovic, 8.20. I think you talk about the furnace paper then, and in a furnace radiation is much stronger than any other mode. In the radiation section at the bottom, where the flames are, radiation is completely dominant. Up in the convection section, the tubes draw out the remaining energy they can from the exhaust gases, but that is much less than what is added to the feedstock in the radiation section. The clue is that radiation increases with the temperature to the power of 4, according to Stefan-Boltzman’s law, while conduction and convection don’t. Conduction is linear, convection probably less strong than linear, with temperature (logic: just because a surface is 1000C it does not create a storm around it by convection).

W. Falicoff
February 13, 2011 9:19 pm

my comment above has a mistake. I meant to say “square root” not “square”. Let me provide an example. If we take a photo of an astronomical object such as a star or nebula using a CCD camera that has a Signal to Noise Ratio (SNR) for one image of say x, then if we take N images of that same object, the SNR will increase by the square of N. That is the new SNR will be (square root of N) times x.

Mikael Cronholm
February 13, 2011 9:30 pm

@ Dave Springer. I was not discussing properties of gases. I just say there are no scientific proof that increased CO2 emission causes climate change, or that it does not. And I am not on any side in the debate, for that very reason.

Stephen Rasey
February 13, 2011 9:31 pm

The solar day on Venus is about 582 days.
So it has long nights of 291 days.
It has a suface temp of 735 K, 460 C
Doesn’t that mean it should be emitting 36 times more W/m^2 than the
Earth at 300K? (735^4)/(300^4)
How is it, then, that Venus remains so hot, even on its night side?
It is closer to the Sun, but shouldn’t that mean it is only getting a little over 2.2 times the energy per unit area? Venus also has a high albeto — that ought to help keep it cooler.
Venus has almost 300,000 times more partial pressure of CO2 than does earth. It has almost no water. If the IR saturated spectra argument is valid, would 10% CO2 be just a bad as 99.5% CO2?
Surface pressure 93 bar (9.3 MPa)
Composition ~96.5% Carbon dioxide
~3.5% Nitrogen
0.015% Sulfur dioxide
0.007% Argon
0.002% Water vapor
0.001 7% Carbon monoxide
0.001 2% Helium
0.000 7% Neon
trace Carbonyl sulfide
trace Hydrogen chloride
trace Hydrogen fluorid
http://en.wikipedia.org/wiki/Venus
Is it something as simple as PV=nRT ?
with the high altitude, low pressure, low temperature stratosphere being the governor for heat loss rate?

W. Falicoff
February 13, 2011 9:40 pm

I realize that many of you are not fond of the website realclimate.org. However, there are useful posts on the above subject (CO2 radiation exchange in upper atmosphere) at
http://www.realclimate.org/index.php/archives/2007/06/a-saturated-gassy-argument/
and more importantly at
http://www.realclimate.org/index.php/archives/2007/06/a-saturated-gassy-argument-part-ii/
Also an important data source for the absorptivity of CO2 is given in the data found in HITRAB at http://www.cfa.harvard.edu/hitran//
This software uses data from HITEMP https://kb.osu.edu/dspace/handle/1811/13476
“The HITRAN database has been recognized for more than 20 years as the international standard compilation of spectroscopic absorption parameters for atmospheric gases.”
The following paper has some of the CO2 data (available on line):
http://faculty.uml.edu/robert_gamache/papers/Rothman_et_al_Preprint.pdf
From what I have read the emissivity of CO2 does vary as a function of both pressure and wavelength. But I am not an expert on this subject.

DeNihilist
February 13, 2011 9:43 pm

Mikael, I am sitting in front of my wood burning heat-o-lator. When the door is closed and the fan running, the room heats up quite quickly, but my body doesn’t. If i want a quick kick of warmth, I turn off the fan, open the door, re-arrange the wood so that the ember side is now facing out, and Whammo! instant warming of the body.
Also reference the campfire effect. Side facing campfire feels warmth, as the body is cooler then the flames/embers. side away from campfire feels cool/cold, as body on that side is radiating heat to a lower temp atmosphere.

February 13, 2011 9:45 pm

Jim D says:
February 13, 2011 at 5:48 pm
The measured IR from the sky is pretty much all emitted by CO2 and H2O molecules …
No Jim. If that were true then the warmth of the sun on my face would be coming in from all directions equally. But even a blind man …
Or do you mean at night?
And do you really mean to tell us that nitrogen can’t radiate IR?

Tsk Tsk
February 13, 2011 9:45 pm

Mikael Cronholm says:
February 13, 2011 at 9:18 pm
“@chico sajovic, 8.20. I think you talk about the furnace paper then, and in a furnace radiation is much stronger than any other mode. In the radiation section at the bottom, where the flames are, radiation is completely dominant. Up in the convection section, the tubes draw out the remaining energy they can from the exhaust gases, but that is much less than what is added to the feedstock in the radiation section. The clue is that radiation increases with the temperature to the power of 4, according to Stefan-Boltzman’s law, while conduction and convection don’t. Conduction is linear, convection probably less strong than linear, with temperature (logic: just because a surface is 1000C it does not create a storm around it by convection).”
Be careful. Just because radiation goes as the 4th power of temperature doesn’t mean it’s always the dominant form of heat transfer. It also has an extremely low coefficient compared to that of conduction or convection of most materials. Also, both conduction and convection are linear with temperature but their respective ratios are dependent on the particular fluid’s properties. Finally, that 1000C surface may not create a storm, but it most certainly will drive noticeable amounts of air or any other surrounding fluid flow. Of course, if you’re using your hand to measure that then you’re probably more worried about the burns you’re getting by holding your hand too close to a very hot object.

Mikael Cronholm
February 13, 2011 9:56 pm

Ian W says:
February 13, 2011 at 12:35 pm
When water vapor in the atmosphere condenses into liquid water and then changes state again and becomes ice, it gives off latent heat for both state changes.
Does that latent heat release follow Stefan-Boltzmann’s radiative equation ?
No Ian, latent heat is not really related to S-B. Latent heat means that when you add or remove heat, energy, from a substance it will usually change its temperature, except when there is a phase change. At a phase change, for example melting or evaporation, you will be able to add or remove heat without a change in temperature taking place, the change of heat in the substance causes the phase change to take place instead. In the case of freezing water for example, there is molecular kinetic energy stored in translational movement, i.e. the molecules move around relative to each other (they have a mass, they move=energy). When the liquid turns to a solid crystalline structure, that energy of movement must be removed before the molecules can stand still in relation to each other. That is the latent heat that is released from the water when it freezes. Fruit orchard are actually sprinkled with water to prevent the plants from getting damaged on cold clear night in the spring. The water freezes and give off heat to the plant, preventing it from getting too cold.
So, as far as S-B is concerned, this process will only influence the input T in that equation, insofar as the cooling or heating of the substance reaches a plateau when latent heat takes effect. The radiation will still always depend on the temperature and emissivity of the substance.

Domenic
February 13, 2011 10:32 pm

To Dave Springer
In actuality, to my knowledge, the true greenhouse effect of any gases (including Nitrogen) has never been measured properly. They are calculated based on a lot of basically untested assumptions. (To explain all the assumptions currently used would require a very detailed and long technical paper.)
To do it correctly, you should use a radiative source at room temp to represent the typical earth radiation temperatures instead of a high heat source used historically in gas detectors. Then the detector itself should be near absolute zero to simulate outer space at night time rather than the room temperature state it normally is in a gas detector.
In other words, this proposed setup should be an exact simulation of night time conditions. The gases you put inside the tube can represent any kind of atmosphere conditions, or gases, that you wish.
THEN you DIRECTLY measure the longwave radiation transmission, to a great extent, etc of the various gases in the atmosphere.
For those of you with IR guns or devices, you can see true greenhouse effects directly at night by aiming your IR device at the center of the sky on a cloudless night, and then comparing that reading to another night by aiming at the center of the sky when the sky is full of clouds. That is a TRUE greenhouse effect differential measurement.

February 13, 2011 10:33 pm

hotrod ( Larry L ) says:
February 13, 2011 at 5:49 pm
It is perhaps a subtle distinction for some folks, but the CO2 and water vapor “back radiation” does not “warm the ground”
Why not? Does it have to do with the 10 micron wavelength?
or the ocean surface,
Why not? Those photons have to go somewhere. Why does this look like a “missing heat” problem?
it only “reduces the rate at which it is cooling by IR radiation” to the apparent surface of the very very cold sky.
So “back radiation … only reduces … cooling.”
Huh??

Mikael Cronholm
February 13, 2011 10:56 pm

@ Tsk Tsk. No, but at high temperature it is definitely dominant, and my paper deals with furnaces where temperatures are very high.
Another point to make about that is that radiation and conduction are the most predictable and easy to calculate of the three, at least if conduction is in a solid. Convection is a nightmare!!!

Oliver Ramsay
February 13, 2011 10:58 pm

chico sajovic says:
February 13, 2011 at 7:45 pm
“Back Radiation” is such a horrible analogy, description or way to think about heat transfer in the atmosphere, that I wish it would just go away. What we are concerned with is the flow of heat from the surface of the earth to space. Its not that the surface warms the co2 in the air then the co2 “back radiates” re-warming the surface, its that when the co2 in the air is warmed the flow of heat from the surface to the warmed co2 is reduced, thus increasing the temperature of the surface until the flow of heat returns to equilibrium.
——————————————-
I don’t quite get the preoccupation with “the surface” when it’s actually the air temperature that we measure, not the ground.
GHG’s don’t make the ground warmer in the daytime than insolation is able to do and at night, the paltry amount of energy accumulated in the surface doesn’t provide much residual warmth to the air, especially since it’s only a fraction of outgoing LWR that is absorbed.
The heat capacity of the air is not enormous but radiative heat loss is not really rapid at these temperatures, either.
Something like that?

P Wilson
February 13, 2011 11:11 pm

in answer to the question up there, the SB equation doesn’t work with dimensional gases.
Secondly, it is a theoretical equation that doesn’t work with climate generally,
Practical demo: The basal human metabolic rate is around 58wm2. An average human is around 2m2, so the average energy a human generates is around 110wm2
This creates more heat than does the atmosphere or the ground at night under the view of spectroscopes.
It is therefore safe to assume that there is much less than even 50w m2 re-radiation.
true, there may be air currents rising, but these are invisible to c02 which absorbs (well, it doesn’t really absorb but delays by a billionth of a second) energy at around 15 microns, which corresponds to subzero temperatures.
It then quickly thermalises with nitrogen and oxygen at its most active region – which is quite high in th eatmsphere where freezing temperatures correspond to c02 absorbtion. However, at this height (around -28C in the troposphere) it also competes with the peaks of nitrogen and oxygen absortion (they absorb radiation too).
It is questionable that c02 is even a greenhouse gas in real observable terms.

Mikael Cronholm
February 13, 2011 11:12 pm

Slacko 10.33, perhaps I can explain. If you look at the beginning of Ken’s and my conversation I mention something we call “apparent temperature”. It is the blackbody equivalent temperature that something radiates. We use it to determine the reflected radiation in commercial IR measurements. That is also what would be measured in that IR image at the top if the camera was able to measure that low (and emissivity was set to 1 and distance to zero).
So consider the exchange of heat between earth and the sky. It depends on the balance between incoming and outgoing radiation, the net difference is the gain or loss. If you have a very clear sky, the sky will have a low apparent temperature, so the temperature difference between the earth and the sky is the greatest. (You will still have a big influence from the atmosphere, otherwise the apparent temperature of the sky would approach absolute zero. -273C.)
Clouds will have a higher apparent temperature, meaning that they will radiate more towards the earth than a clear sky would do, so the difference is smaller and hence the heat loss is also smaller, than with a clear sky.
So, yes, it is true that radiation from the clouds will prevent cooling of the earth, allowing the earth to keep its heat to a larger degree. But as the clouds will not be hotter than the earth, they will not reverse the heat flow from the earth to become a gain rather than a loss, they can just make it less of a loss. The sun will give the positive contribution.

Roger Carr
February 13, 2011 11:21 pm

Slacko says: (February 13, 2011 at 7:47 pm)
      …that empty and useless trollbuttal.
On the web: No definitions were found for trollbuttal.
Anthony, Slacko may have coined a new, very expressive, and very useful word here on WUWT.
I hope you will note it with appropriate applause.

P Wilson
February 13, 2011 11:23 pm

Mikael Cronholm says:
February 13, 2011 at 11:12 pm
heat leaves earth by convection. Clouds prevent convection. They don’t radiate energy

randomengineer
February 13, 2011 11:24 pm

Falicoff — 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.
Apples and oranges, and utter nonsense.
What you describe is simple measurement repeatability with an assumed constant source and then averaging out many multiple readings. All modern measurement devices derive their accuracy spec (traceable to NIST etc) in this manner. Of course, I know this because I made measurement equipment for many years… The measurements in question are those of thermometers looking at a different condition each time, not a fixed, constant source, and the result is that the error is the stated resolution of the device.
If an 1880’s era thermometer was good to +/- 1 degree, that’s your error range. You can’t average out 220x 1880’s era thermometers and derive accuracy tighter than +/- 1 degree.

Konrad
February 13, 2011 11:26 pm

richard verney says:
February 13, 2011 at 4:11 pm
I believe this is a valid question. The response –
Dave Springer says:
February 13, 2011 at 8:14 pm
fails to address the question in ignoring the diurnal cycle of solar energy input for a given location on the Earth’s surface. For CO2 to cause CUMULATIVE warming, the near surface temperature just before dawn for a single point on the Earth’s surface would have to be greater for a local air mass with greater CO2 concentration than for one with less. A simple empirical experiment could clarify this. I would be interested if anyone could point me to the results of such an experiment.

AusieDan
February 13, 2011 11:32 pm

richard verney
You asked what sort of time delay does CO2 impose on the re-emission of the sun’s energy back into space.
I do not know the answer, but suggest we consider what happens when the hot summer ends and chilly winter appears?
What happens when day follows night?
What happens to the measured temperature at individual locations for well over a century, when you are able to adjust for the rising level of UHI?
Not much – that’s my simple answer.

February 13, 2011 11:38 pm

John S says:
February 13, 2011 at 4:05 pm

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.

How can it be explained that ANY increase in temperature does not cause this “thermal runaway” ‘tipping point’? If the temperature as recently as the MWP was higher than today, which it certainly was in at least some areas and possibly all, why was this mythical tipping point not reached? As it was not, what makes you believe it will be during the present warm period? Is there any evidence for this? Models are NOT evidence, BTW, as they ONLY produce what they are programmed to produce.

P Wilson
February 13, 2011 11:38 pm

chico sajovic says:
February 13, 2011 at 7:45 pm
When I am hot I should:
a) stand outside naked on a low humidity night
or
b) get soaked in water and stand in front of a fan
from what we’re told. you should do c) go to a special room that has 280ppm c02, regardless of its temperature

davidmhoffer
February 13, 2011 11:55 pm

Slacko says:
February 13, 2011 at 10:33 pm
hotrod ( Larry L ) says:
February 13, 2011 at 5:49 pm
QUESTION
It is perhaps a subtle distinction for some folks, but the CO2 and water vapor “back radiation” does not “warm the ground”
Why not? Does it have to do with the 10 micron wavelength?
ANSWER: Because the downward re-reradiated LW is MUCH more likely be be absorbed by water vapour or other absorbers and be re-radiated back up than it is to ever hit the ground. Double CO2 instantly, wait for a new equilibrium to be established, and you have the EXACT same amount of SW going in, and the EXACT same amount of LW coming out. The EFFECTIVE black body temperature of the earth is what the IPCC claims is going to rise by 1 degree, and this is at a point high up in the atmosphere (about 14,000 feet if I recall) not at earth surface.
QUESTION
or the ocean surface,
Why not? Those photons have to go somewhere. Why does this look like a “missing heat” problem?>>
QUESTION
LW radiatiance cannot penetrate more than a micron or so of water before being absorbed. The result being that any longwave that does strike water is absorbed in a layer so thin that it immediately evaporates taking the extra energy from the LW, plus any energy that was already in that water with it into the atmosphere. Any temperature changes in the ocean have to be attributed as a consequence to other factors such as fluctuationm of SW from the Sun, rainfall, runoff and so on. As for “missing heat” that’s a measurement problem.
QUESTION
it only “reduces the rate at which it is cooling by IR radiation” to the apparent surface of the very very cold sky.
So “back radiation … only reduces … cooling.”
Huh??
ANSWER – that is an imperfect way of thinking about it, but at day’s end it is a fair description. A given photon might, in theory, be radiated upward and go straight out to space. Or it might hit one CO2 molecules be absorbed, and then re-radiated in a random direction. Up, sideways, down, what ever. For rough figuring, call it 2/3 up or sideways and 1/3 down. So no matter how much CO2 you have, the end results is always more ups than downs, and the photon eventually escapes to space, but not have hundred, to millions of absorptions and re-radiations. So adding CO2 does not add a single additional photon, not one, to the equation. All it does is increases the average number of zig zags through the atmosphere before eventually escapting to space. So yes, extra CO2 would add no extra heat at all, it would just increase the amount of time it takes any given photon to escape, and “slow down the escape” could roughly be equated to “slow down the cooling”.

cal
February 14, 2011 12:49 am

Slacko says:
February 13, 2011 at 9:45 pm
Jim D says:
February 13, 2011 at 5:48 pm
The measured IR from the sky is pretty much all emitted by CO2 and H2O molecules …
No Jim. If that were true then the warmth of the sun on my face would be coming in from all directions equally. But even a blind man …
Or do you mean at night?
And do you really mean to tell us that nitrogen can’t radiate IR?
Sorry Slacko but Jim D is right. The sun emits energy in the visible and UV part of the spectrum and a good job it does too. The atmosphere is almost transparent at these wavelengths so unless it gets reflected by clouds or the surface it will make it to the ground where it will be absorbed to warm the earth. The earth on the other hand emits at longer wavelengths with a peak at 10 micron (infra red starts at about 0.8 micron) and a lot of this does get absorbed by the CO2 and water vapour in the atmosphere before being finally radiated into space by the same molecules.
And no, nitrogen does not radiate (or absorb) in the infra red.
On a more general point. The picture at the top of this piece is a bit misleading. When you point an infra red camera at the sky at night you will only capture the infrared radiation from molecules (mainly CO2 and H2O ) in the atmosphere. There is no other significant source of infra red radiation in space to detect. Since these molecules are radiating in narrow bands and a lot of it in the far infrared that the camera may not detect the total energy received by the camera is quite small. As explained in the responses by MC the camera will assume it is looking at a grey or black body with energy distributed across the whole spectrum according to Planck’s law. So it will calculate what temperature a back or grey body would have to be to output that amount of energy. Because the energy in the narrow bands that are detected is then spread out over the whole spectrum that calculated temperature will appear much lower than it really is and the sky will be shown as black.

February 14, 2011 1:11 am

” 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…”
Love it!

February 14, 2011 1:15 am

I agree with Berényi Péter with respect to
“For 300 W/m2 radiation I get -23.4°C at 300 W/m2 when I calculate it”
My temperature conversion calculator correlates 300 W/m2 with -3.3°C (assuming an emissivity of one), and -23.4°C with either an emissivity of 73.4% or 220 W/m2. A few lines later
your body will radiate approximately 648 W/m2
implies that your skin temperature is at least 129°F (don’t think so).
To Jim D:
Yes, IR is reflected. The 324 W/m2 back radiation from Kiehl and Trenberth (1997) implies a temperature of 34°F if the emissivity is one, but 58°F (15°C) with an emissivity of 0.8 (-20% in my calculator). Since I don’t think that the atmosphere is able to be the same temperature as the surface (remember, the troposphere cools with increasing height), some of the energy must be reflected and not emitted. Some calculations indicate that even 80% is too high .. in the absence of clouds. On the other hand, since about 50% of the planet is covered with clouds, it sort of makes sense to assume that some of the energy is “reflected” by the cloud bottoms. In fact, on cloudy nights, the hand held radiometers indicate an apparent cloud temperature within 2°F of the surface temperature. In addition to the expected reflection by the droplets themselves, the heat from the surface evaporates droplets at the cloud bottom, which then causes the vapor rise a few inches, where they recondense. For both evaporation and condensation, the associated spectrum is nearly blackbody, with no spectral lines. Notice that this is similar to reflection since the energy emitted is not affected by the temperature of the cloud, but only by the temperature of the surface. (“Scattering” may be a better term.)
For Jim Masterson:
In Kiehl and Trenberth (1997), the downward flux of 324 W/m2 is specifically for cloudy days, it is 278 W/m2 for clear days.
To all:
At a single frequency, the change in absorption is logarithmic. However, when the entire IR band is considered, and over the range from 200 ppm to 500 ppm, the change in CO2 absorption is logarithmic with R^2=0.9988 and linear with R^2=0.9932. This is because as each frequency becomes more saturated, a new frequency starts to absorb.

Mikael Cronholm
February 14, 2011 1:47 am

P Wilson says:
February 13, 2011 at 11:23 pm
Mikael Cronholm says:
February 13, 2011 at 11:12 pm
heat leaves earth by convection. Clouds prevent convection. They don’t radiate energy
Reply: Sorry Mr Wilson, but I do not agree with what you say. (To other readers: I was not the one who made the statement above.)
Convection takes place in substances in the form of fluids and the atmosphere itself is a fluid, so convection can take place in it, including within the clouds themselves (they don’t “prevent” convection, they take part in it). It is true that heat transfers from the ground to the atmosphere, but it is still part of the heat that surrounds the planet. So assuming that we count the atmosphere as a part of our planet, as opposed to space around it, no heat will enter space by convection because there is no fluid that can circulate into space. Clouds do radiate, because everything with a temperature does.
Let us make this clear. The only way that Earth, and its atmosphere, can exchange heat with space is by radiation.

Don V
February 14, 2011 1:47 am

I have been fascinated by this posting – as well as all of the many cogent comments from the insightful experts that have been attempting to answer many of the questions that have been posed in the subsequent comments here.
If I may, I would like to add my two cents worth to clear up some things I have read here, and use this forum as a sounding board/perhaps clarify some misconceptions. Please correct me if I am wrong.
First, regarding reflection, vs absorption/re-radiation. Although, at a molecular level, it may seem they are the same, there is a difference. Corner cube reflection in a prism of glass or reflection that occurs inside a droplet of water generally 1)absorbs incoming light at the first material boundary, transmits through the second material, then reemits the incoming light back into the first material and maintains all spectral integrity and most of the intensity of the incoming light, (polarization may occur) 2)there is very little if any loss of energy EXCEPT at those wavelengths that the second material absorbs light and 3) reflection/refraction occurs where either a phase change, or significant density change is encountered. It isn’t widely understood but, gases can be seen to refract and change the apparent direction of incoming light and this could be considered “reflected” light, but it is nowhere near as dramatic an effect as what occurs in the reflection of nearly all visible light when it encounters many many tiny droplets of liquid water (clouds or rainbows). Two examples of air refracting/reflecting light are the Schlieren waves you see when you look at a distant scene across a hot parking lot or runway http://hiviz.org/hsi/ss/schlieren/index.htm
and the “mirage” you see of a distant city or the reflection of the sky to give the appearance of an oasis of water on a hot desert.
But that is NOT what happens when a molecule like water or CO2 ABSORBS light at their respective infrared bands. The definition of absorbance is A = 1 – R – T where R is the reflectance and T is the transmittance. Absorbance is the loss of light at specific wavelengths of light, and these unique wavelenghts for both CO2 and water correspond to electronic, vibrational, and rotational energy modes that are unique to each molecule. They happen at unique wavelengths because only quanta of energy that match the exact electronic, vibrational and rotational energies for each of these molecules can cause them to absorb and become excited at these quanta of energy. Generally, if a molecule is excited at a specific wavelength, but it immediately gives up that energy and decays back to an unexcited state and thus reemits that photon, it is no different than transmitted light. The resulting photon seems to pass right through the excited->unexcited molecule. Outside of absorbance bands this is usually what happens to both visible and infrared light when it interacts with gases like CO2 and water. However, at very short UV wavelengths light absorbance/scatter has the effect of making the absorbing gases appear to “glow” in the visible spectrum – Raleigh scattering and Mie scattering – which is why the sky appears sky blue, but at sunrise or sunset appears to have a rainbow of colors, and why the sun appears yellow.
At absorbance bands, however, scatter, transmittance and reflectance don’t often happen. The incoming photons with quanta of energy that match absorbance bands of any given gas molecule are absorbed by that gas molecule and it is excited up into a higher electronic or vibrational, or rotational energy state, and if or when it reemits that energy it necessarily must experience some LOSS – entropy gets it’s cut. Because the molecule has an “affinity” for quanta of energy that correspond to its natural “frequencies”, it is also very rare that light that is “absorbed” at a given wavelength is reemitted at that wavelength. If it were you wouldn’t see the loss of light – absorbance – at that wavelength. Instead, energy that is absorbed at one quanta is usually released only when 1) even more energy is pumped into the molecule at the same quanta (rare), or 2) it reemits that light at a longer wavelength (fluorescence – does CO2 fluoresce in the infrared? I’m not sure but I doubt it.), or 3) the molecule collides with another molecule that has less energy (the usual case). When collisions occur, (very frequently) the absorbed quanta of energy is transferred between the two molecules so that they balance out – generally both of them just pick up more translational energy – they move a little faster.
In general devices that “measure” the temperature of a gas or a liquid actually transfer (come into equilibrium with) the translational energy of that gas or liquid to a visual media – indicator liquid that expands in a fixed volume – or transfer the translational energy to an equivalent vibrational energy in a solid material which is used to derive an electronic (thermistor, thermocouple etc.) or visual signal (bimetal). So an observed “temperature” of a gas is an average of all of the molecule’s translational energies in a given volume that interact with the surface of the measuring device.
Now having said that, I would like to make the following ascertions:
1) As stated by several previous posters, water has many more, and much broader absorbance bands in the infrared than CO2,
2) Water vapor, and water droplets are much, much more abundant in the lower atmosphere than CO2,
3) Even at the now greater concentration of CO2 in the atmosphere the relatively small amount of IR energy it might absorb is quite quickly transferred during collisions to the much greater abundance of water liquid, water vapor, N2, or O2 molecules,
4) Water vapor, water liquid, and ice all have greater heat capacity than any of the other gas molecules in air. Liquid water’s heat capacity is significantly greater!
5) Because water experiences phase changes and significant density changes, it is redistributed throughout the atmosphere in ways that have a significantly greater impact on the transport of energy from the lower atmosphere and planet’s surface to outer space than any of the other gases that make up our air,
5) Since water vapor regularly experiences both gas to liquid, liquid to solid, and even gas to solid transitions in the atmosphere:
a) At any given instance the loss of incoming solar radiation energy (on the day sid of the planet) by direct reflection, refraction and back scatter, by clouds, snow, rain and the oceans are significantly greater in magnitude than all of the IR absorbed by CO2 combined.
b) At any given instance the total amount of incoming solar radiation energy that results in phase change of water liquid to water vapor, greatly and significantly exceeds all of the IR energy absorbed by CO2 combined.
c) At any given instance the total amount of energy contained in water vapor, water liquid droplets (clouds) and water ice particles (high clouds) in the planet’s atmosphere by a huge amount dwarfs the total amount of energy contained in IR excited CO2 gas molecules, so much so that I would think (I have no proof) that most of CO2’s translational energy content is created more by collisions with water molecules than IR photons by a very large ratio.
d) And finally and most imortantly, changes in the concentration of CO2 that have been observed have gradually been increasing over DECADES. But in any given DAY the concentration of water vapor, water liquid and ice particles in the atmosphere change by orders of magnitude greater concentration than the few hundred parts per million that CO2 has changed. These significant concentration changes even at the poles have a far greater effect on both immediate weather and of course long term climate. Alarmists would have us believe that water cycle does not now have the capacity to mitigate and regulate the small change in CO2 concentration that have occurred after centuries and centuries of significantly higher CO2 concentrations and significantly LOWER CO2 concentrations in the past.

Feet2theFire
February 14, 2011 1:56 am

@chemman Feb 13, 2011 at 12:57 pm:

…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.

I’ve always assumed they all deal with it as 3D re-radiation – including only the downward re-radiation in the heat flow system. Otherwise some of the skeptics would have certainly ripped into them about it.

Steeptown
February 14, 2011 2:08 am

This is a very useful discussion, particularly for those not of a physics bent. As someone who has worked for many years in the field of fluid flow and heat and mass transfer, it is evident to me that the radiative efects of CO2 in the atmosphere are of 2nd or 3rd order compared to the radiative, convective and latent heat effects of H2O. Our climate is dominated by the water cycle in all its forms.

Feet2theFire
February 14, 2011 2:12 am

I have some hands on industrial R&D experience with heat flow vs temperature, and also with the reality of emissivity.
I ran experiments using what are called “engineering plastics” in order to find the best plastic for a 420°F application. In one, we heated the plastic to temperature and then subjected it to weight loading to see what kind of strain we got. In the process, I handled up to 500°F pieces of plastic (a bit higher than the real application, for “cushion”). I did this barehanded. Even with a thermocouple telling me the plastic was 500°F, the plastic was almost not even warm in my hand. It is not the temperature that burns; it is the heat flow. See the next paragraph…
As to emissivity, we also had highly polished and chrome plated tool steel (H13) that was oil-heated to the 420°F temperature. The polishing and chrome plating made the emissivity extremely low (how I still don’t “get”). I could put my skin well within 1/16″ of the surface of the hot metal and not even feel warmth. (It is not easy to do without jittering, but I did manage to do it, bracing my hand on something cooler, just to see.) However, once the heat transfer method went from radiation to conduction – when I actually touched the metal – it was instant BURN!
Low emissivity is amazing.
Low heat flow is also amazing.
(Note: By far the best fix for the occasional burns was using freon from a spray can. I was SERIOUSLY [snip . . irritated?] when they banned CFCs to protect the ozone hole. And I completely believe that in time that theory will be completely refuted.)

Blade
February 14, 2011 2:16 am

steven mosher [February 13, 2011 at 3:39 pm] says:
“yup, but there is more to it than that. If you want to do IR design today ( of sensors, or counter measures or stealth technology) the physics you use is the same physics that many skeptics deny. good thing we don’t let them design the machines that our safety as a nation depends upon.

Ouch! That is seriously nasty.
ATTENTION: Thomas Fuller or CTM … would one of you please give Steve Mosher a call and let him know that one his laptops was stolen and that somewhere nearby in some dirty sleazy bar or seedy hotel with free Wi-Fi there is a nasty old miserable drunk posting on WUWT in his stead.
😉

Man BearPigg
February 14, 2011 2:28 am

“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” … W. Falicoff
“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.” … Mikael Cronholm
So all you have to do now W.F. is show your repeatable science and you will have done something that would change the entire scientology of AGW

P Wilson
February 14, 2011 3:47 am

Mikael Cronholm says:
February 14, 2011 at 1:47 am
Re convection:
It is known as convective inhibition when cloud cover at night = warmer than cloudless night.
Clouds form a barrier. From IR it block some solar radiation, and from earth it prevents convection currents of rising air from leaving. Most energy leaving is via convection. It is known as atmospheric convection

ThomasU
February 14, 2011 3:50 am

Very interesting post & posts! Great source of knowledge down to little known details (such as spraying water on fruit). Lot of food for thought!
I only have one question to add to this conversation: Is it at all possible to measure the energy balance (or radiation balance) of the earth. By energy (radiation) balance I mean the difference between energy input and output. Please excuse if this question is put forward in terms which fall short of physicists standarts – that´s because I am just a curious layman.

P Wilson
February 14, 2011 3:53 am

to clear the matter: Since IR radiation from earth is weak, it can’t move through the atmosphere and is absorbed by water vapour mainly, and c02.
Since warmer air is less dense than cold air convection occurs to take energy upwards, via what they call advection

wayne Job
February 14, 2011 3:56 am

We have a little blue planet that is stirred internally by a big yellow monster, this same monster bombards us with the full spectrum of radiation. Mitigating factors protect us from this monster or we would perish. Like wise if it failed to arise tomorrow we would perish very quickly. Our little blue planet works its fanny off trying to balance an out of balance heat input caused by all sorts of wobbles and perturbations. The occasional large volcanic eruption is compensated for and the world returns to normal. A flea on elephant comes to mind as to effect of CO2 on the planetary climate. The cyclic changes we see if we live long enough are the result of the rules of thermodynamics in an open heat pump trying its utmost to balance itself.
It would be pure folly and ego on our part to contemplate that we could change the climate. Looking backward to Newton and celestial mechanics using real geometry is the key to finding the reasons behind ice ages and interglacials. When these are understood then we can look at the minor perturbations with some understanding and a modicum of confidence. Some scientists are looking at this aspect and finding a surprising correlation to the suns recalcitrant attitude.

stephen richards
February 14, 2011 3:58 am

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.
If the CO² molecules has radiatively absorped and is therefore above ground state state it will likely ‘reflect’ further incoming IR.

P Wilson
February 14, 2011 4:06 am

ThomasU says:
February 14, 2011 at 3:50 am
This is indeed a big problem. Temperature measurements are used, which are inadequate, sparse and unreliable.
Atmospheric energy should be measured in joules per cubic metre. Not in temperature

Myrrh
February 14, 2011 4:06 am

Cal says Feb 14, 12:49 am, Slacko says, Jim says
Sort it out, will you.
http://www.wisegeek.com/how-does-infrared-heat-work.htm
“To observe how infrared heat works, the heating and cooling of the earth is probably the most helpful example. Apart from driving all weather events and patterns, as well as ocean currents, infrared heat from the sun is what the Earth during the day, and it is what the ground gives off at night after the sun sets. It is interesting to note that while the sun does emit a huge amount of infrared heat, only about half of the heat we feel on the ground is from direct infrared radiation from the sun. The other half comes from energy from visible light that is absorbed by objects on earth, and then emitted later as infrared heat.”
So which is it?

Myrrh
February 14, 2011 4:10 am

“infrared radiation is what warms the earth during the day”
Infrared is the heat you feel, but cannot see.

February 14, 2011 4:15 am

I’m not understanding this idea that a 1deg rise in temps caused by CO2 will cause a further 2deg rise through the extra water vapour that is created.
Surely if that were true it would be possible to surround a small lake with something like the Eden Project biodomes, pump it full of CO2, and then extract the heat that all the extra water vapour created. It would be the ultimate passive energy creator, no windmills, solar or thorium required. At the very least we could do it on a small scale to test the thesis that that is what would actually happen.

Bomber_the_Cat
February 14, 2011 4:18 am

Steve Reynolds says:, “I agree with everything MC said except… that IR is not reflected by CO2, it is absorbed and then re-emitted in all directions”
To which Latitude asks,
“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?”.
The gases in our atmosphere (N2,O2 and CO2) do not ‘reflect ‘ light. If they did it would be impossible for us to see coherent images because the light would be bouncing all over the place. We can only form clear images where light travels in straight lines.
The same is true in the infra-red.You no doubt have seen pictures from infra-red imaging cameras, e.g. police helicopters. These are only viable because atmospheric gases do not ‘reflect’ infra-red radiation but allow it to travel in straight lines.
However, CO2 does absorb infra-red within a narrow wavelength band. The important absorption occurs at wavelengths between 14 and 16 microns (other CO2 absorption bands are not significant In the context of greenhouse warming because they do not obstruct the radiation emitted from the Earth’s surface). Photons of infra-red radiation around 15 micron have just exactly the right amount of energy to raise a CO2 molecule from one vibrational state to another one. Being thus ‘tuned’, they may be captured by a CO2 molecule which then changes state. Almost instantaneously, within nanoseconds, the CO2 molecule re-emits the photon and reverts to its original state. Because the photon can be re-emitted in any direction (spherically random) about half of them will be re-directed back to Earth and will represent an additional heat flux if they ever get there.
If the ‘excited’ CO2 molecule collides with another molecule, say of Nitrogen. before it can re-emit its captured photon, then it becomes ‘thermalised’ (or as Mark Wagner says, ‘quenched’). Its extra energy is transformed into heat and it can no longer emit a photon. If this happens the atmosphere is warmed.

P Wilson
February 14, 2011 4:27 am

Its physically impossible for a 1C rise in temperature by c02.
What causes temperature changes are weather systems (via convection).
Storms, snow, cyclones, clouds – the whole gamut are caused by convection. When air convects, it cools sufficiently for it to fall below its dew point. Thats when clouds form.
I don’t know why c02 was brought into meteorology and climatology.
Maybe it was to paint a human face on climate.

P Wilson
February 14, 2011 4:32 am

Bomber_the_Cat says:
February 14, 2011 at 4:18 am
what you haven’t mentioned is the temperatures at which this 14.77 peak of c02 absorbs radiation. It corresponds to -28C which is up in the lower troposphere. When it leaves that temperature range it thermalises. Now: There is no physical mechanism that can cause subzero temperature ranges to penetrate back to earth and warm it up beyond the nominal 14-15C. That is tantamount to saying that putting cold water in the freezer will cause it to be lukewarm

stephen richards
February 14, 2011 4:36 am

Best post I’ve seen here for some time. It’s a clear, concise essay which encourages the reader to do their own research elsewhere. Great stuff.

Smoking Frog
February 14, 2011 4:36 am

Jim Masterson 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.”
The Inclusion-Exclusion Principle requires that we know the overlap(s) to begin with, so it can’t give us the answer. The average of the minimum and maximum overlaps gives exactly 62%:
((49 + 6 + 20) + 49) / 2 = 62
but I’m not sure that this is what we’d get with random overlap, even if, as my calculation assumes, there are no real-world constraints.

AJB
February 14, 2011 4:59 am

Don V says February 14, 2011 at 1:47 am
An excellent post, Sir. IMHO non radiative transfer of energy to the tropopause by water completely swamps any radiative effects of increasing CO2. What happens at and above this point is more interesting, however.

Bomber_the_Cat
February 14, 2011 5:59 am

P. Wilson says, “What you haven’t mentioned is the temperatures at which this 14.77 peak of C02 absorbs radiation”.
The CO2 absorption of photons at 15 micron is not dependent on temperature. A molecule of CO2 will absorb any photon with the ‘correct’ quanta of energy which impacts it, irrespective of temperature.
More interestingly, “There is no physical mechanism that can cause subzero temperature ranges to penetrate back to earth and warm it up beyond the nominal 14-15C”.
I assume that subzero here means below zero on the Celsius or Fahrenheit scales. Now ALL objects above absolute zero emit radiation; that includes you, me and an ice cube. The amount an object emits (Stephan’s Law) and the wavelengths of the radiation emitted (Plank’s Law) is determined by their temperature. So even cold objects, such as an ice cube, emit radiation. If that radiation impacts a warmer object then it is absorbed by the warmer object. The warmer object therefore receives energy that it wouldn’t receive if the cold object was not there – and so the warm object is kept warmer than it otherwise would be. Radiation is not somehow preferentially attracted only to colder objects. Thus the presence of cold objects can keep warm objects warmer!!! This does not infringe the 2nd law of thermodynamics, only a schoolboy misunderstanding of it.
As for making the Earth warmer, remember that the Earth is warmed by the Sun. The CO2 ‘blanket’ , by sending radiation back to the surface, simply acts as insulation. The incoming solar energy is what causes the Earth’s surface to rise.
At the top of the atmosphere of course, the amount of radiation leaving the Earth will always balance the radiation coming in – no matter how much CO2 there is in the atmosphere. But we don’t live at the top of the atmosphere, we live on the surface – and it is the surface temperature which is effected by the greenhouse gases.
P. Wilson says, “What you haven’t mentioned is the temperatures at which this 14.77 peak of C02 absorbs radiation”.
The CO2 absorption of photons at 15 micron is not dependent on temperature. It will absorb any photon with the ‘correct’ quanta of energy which impacts it, irrespective of temperature.
More interestingly, “There is no physical mechanism that can cause subzero temperature ranges to penetrate back to earth and warm it up beyond the nominal 14-15C”.
I assume that subzero here means below zero on the Celsius or Fahrenheit scales. Now ALL objects above absolute zero emit radiation; that includes you, me and an ice cube. The amount an object emits (Stephan’s Law) and the wavelengths of the radiation emitted (Plank’s Law) is determined by their temperature. So even cold objects, such as an ice cube, emit radiation. If that radiation impacts a warmer object then it is absorbed by the warmer object. The warmer object therefore receives energy that it wouldn’t receive if the cold object was not there – and so the warm object is kept warmer than it otherwise would be. Radiation is not somehow preferentially attracted only to colder objects. Thus the presence of cold objects can keep warm objects warmer!!! This does not infringe the 2nd law of thermodynamics, only a schoolboy misunderstanding of it.
As for making the Earth warmer, remember that the Earth is warmed by the Sun. The CO2 ‘blanket’ , by sending radiation back to the surface, simply acts as insulation. The incoming solar energy is what causes the Earth’s surface to rise.
At the top of the atmosphere of course, the amount of radiation leaving the Earth will always balance the radiation coming in – no matter how much CO2 there is in the atmosphere. But we don’t live at the top of the atmosphere, we live on the surface – and it is the surface temperature which is effected by the greenhouse gases.
P. Wilson says, “What you haven’t mentioned is the temperatures at which this 14.77 peak of C02 absorbs radiation”.
The CO2 absorption of photons at 15 micron is not dependent on temperature. It will absorb any photon with the ‘correct’ quanta of energy which impacts it, irrespective of temperature.
More interestingly, “There is no physical mechanism that can cause subzero temperature ranges to penetrate back to earth and warm it up beyond the nominal 14-15C”.
I assume that subzero here means below zero on the Celsius or Fahrenheit scales. Now ALL objects above absolute zero emit radiation; that includes you, me and an ice cube. The amount an object emits (Stephan’s Law) and the wavelengths of the radiation emitted (Plank’s Law) is determined by their temperature. So even cold objects, such as an ice cube, emit radiation. If that radiation impacts a warmer object then it is absorbed by the warmer object. The warmer object therefore receives energy that it wouldn’t receive if the cold object was not there – and so the warm object is kept warmer than it otherwise would be. Radiation is not somehow preferentially attracted only to colder objects. Thus the presence of cold objects can keep warm objects warmer!!! This does not infringe the 2nd law of thermodynamics, only a schoolboy misunderstanding of it.
As for making the Earth warmer, remember that the Earth is warmed by the Sun. The CO2 ‘blanket’ , by sending radiation back to the surface, simply acts as insulation. The incoming solar energy is what causes the Earth’s surface to rise.
At the top of the atmosphere of course, the amount of radiation leaving the Earth will always balance the radiation coming in – no matter how much CO2 there is in the atmosphere. But we don’t live at the top of the atmosphere, we live on the surface – and it is the surface temperature which is effected by the greenhouse gases.
P. Wilson says, “What you haven’t mentioned is the temperatures at which this 14.77 peak of C02 absorbs radiation”.
The CO2 absorption of photons at 15 micron is not dependent on temperature. It will absorb any photon with the ‘correct’ quanta of energy which impacts it, irrespective of temperature.
More interestingly, “There is no physical mechanism that can cause subzero temperature ranges to penetrate back to earth and warm it up beyond the nominal 14-15C”.
I assume that subzero here means below zero on the Celsius or Fahrenheit scales. Now ALL objects above absolute zero emit radiation; that includes you, me and an ice cube. The amount an object emits (Stephan’s Law) and the wavelengths of the radiation emitted (Plank’s Law) is determined by their temperature. So even cold objects, such as an ice cube, emit radiation. If that radiation impacts a warmer object then it is absorbed by the warmer object. The warmer object therefore receives energy that it wouldn’t receive if the cold object was not there – and so the warm object is kept warmer than it otherwise would be. Radiation is not somehow preferentially attracted only to colder objects. Thus the presence of cold objects can keep warm objects warmer!!! This does not infringe the 2nd law of thermodynamics, only a schoolboy misunderstanding of it.
As for making the Earth warmer, remember that the Earth is warmed by the Sun. The CO2 ‘blanket’ , by sending radiation back to the surface, simply acts as insulation. The incoming solar energy is what causes the Earth’s surface to rise.
At the top of the atmosphere of course, the amount of radiation leaving the Earth will always balance the radiation coming in – no matter how much CO2 there is in the atmosphere. But we don’t live at the top of the atmosphere, we live on the surface – and it is the surface temperature which is effected by the greenhouse gases.
P. Wilson says, “What you haven’t mentioned is the temperatures at which this 14.77 peak of C02 absorbs radiation”.
The CO2 absorption of photons at 15 micron is not dependent on temperature. It will absorb any photon with the ‘correct’ quanta of energy which impacts it, irrespective of temperature.
More interestingly, “There is no physical mechanism that can cause subzero temperature ranges to penetrate back to earth and warm it up beyond the nominal 14-15C”.
I assume that subzero here means below zero on the Celsius or Fahrenheit scales. Now ALL objects above absolute zero emit radiation; that includes you, me and an ice cube. The amount an object emits (Stephan’s Law) and the wavelengths of the radiation emitted (Plank’s Law) is determined by their temperature. So even cold objects, such as an ice cube, emit radiation. If that radiation impacts a warmer object then it is absorbed by the warmer object. The warmer object therefore receives energy that it wouldn’t receive if the cold object was not there – and so the warm object is kept warmer than it otherwise would be. Radiation is not somehow preferentially attracted only to colder objects. Thus the presence of cold objects can keep warm objects warmer!!! This does not infringe the 2nd law of thermodynamics, only a schoolboy misunderstanding of it.
As for making the Earth warmer, remember that the Earth is warmed by the Sun. The CO2 ‘blanket’ , by sending radiation back to the surface, simply acts as insulation. The incoming solar energy is what causes the Earth’s surface to rise.
At the top of the atmosphere of course, the amount of radiation leaving the Earth will always balance the radiation coming in – no matter how much CO2 there is in the atmosphere. But we don’t live at the top of the atmosphere, we live on the surface – and it is the surface temperature which is effected by the greenhouse gases.

kwik
February 14, 2011 6:32 am

Bomber_the_Cat says:
February 14, 2011 at 5:59 am
Well if I wasnt confused before, I am cerainly now, after reading your post. Was it an attempt on demonstrating backradiation with words? The same stuff coming again, and again, and again? Looked like a very unstable post to me.

David L
February 14, 2011 6:56 am

W. Falicoff says:
February 13, 2011 at 3:40 pm
“… 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. …”
Actually, repeated measurements cannot improve accuracy, it can only improve precision. If there is bias in the gauge then repeated measures will keep giving you the biased answer. Precision (e.g. the standard deviation) of the average value of the measurement can improve by the inverse square root of the sample size.
But there’s far more to it than increasing sample size. Check out MSA (Method System Analysis) or Gauge R&R (Repeatability and Reproducibility). In a manufacturing environment a Gauge (or measurement system) is not considered capable until the precision is at least 1/10th the specification range. So in terms of AGW, if the world is going to melt down at an additional 2C then the gauge should be able to repeatably and reproducibly measure to +/- 0.2C on the individuals….not multiple measurements to construct an average.

izen
February 14, 2011 6:58 am

@- davidmhoffer says:
February 13, 2011 at 11:55 pm
“LW radiatiance cannot penetrate more than a micron or so of water before being absorbed. The result being that any longwave that does strike water is absorbed in a layer so thin that it immediately evaporates taking the extra energy from the LW, plus any energy that was already in that water with it into the atmosphere. ”
The LW absorbed in the surface layer may not be enogh to liberate a water molecule from the bonds at the surface layer, it may just increase the random kinetic energy of the molecules in the surface layer that is transported to the deeper layers millimetres further down by surface turbulence. Water is rarely so still that this transport mechanism is insignificant.
Even if it does provide enough energy for a water molecule to break free of surface bonds and ‘evaporate’ that molecule has a ~50% chance of colliding with the other molecules and bouncing back into the water surface adding to the thermal kinetic energy of the surface layer.
There is no way that the water surface acts like a one-way street for all incident radient energy. It is not Maxwell’s demon!

Dave Springer
February 14, 2011 7:01 am

P Wilson says:
February 14, 2011 at 4:32 am
“what you haven’t mentioned is the temperatures at which this 14.77 peak of c02 absorbs radiation”
The temperature of the gas has little to do with IR absorption. The earth emits LWIR in a continuous blackbody spectrum with a peak emission at around 10um which corresponds to a blackbody at approximately 52F which is the average surface temperature of the ocean. CO2 absorbs narrow bands out of that continuous spectrum which excites the CO2 molecule and because the molecule is part of a cold dense mix of gases the excited molecule almost instantly bumps into a neighbor (most likely N2) which then thermalizes the N2. Re-emission up the upwelling narrow band energy is a continuous blackbody spectrum corresponding to the temperature of the gas at whatever altitude the re-emission occurs. The process starts from the ground up and proceeds to saturation. Looking down at the atmosphere from above with a spectrometer one sees a continuous blackbody spectrum with narrow absorption bands where the energy level falls off. The energy missing from those narrow bands is redistributed across the rest of the spectrum. The altitude/temperature to which you refer is the energy level at the top of the 15um band.
See figure 8.2 (spectrograph from 20km looking down on the north pole):
http://www.sundogpublishing.com/AtmosRadFigs.html
The blackbody emission curve is at about 265K which corresponds to the surface temperature at the pole at the time. You’ll notice a big hole in the spectrograph centered on 15um with the bottom of the hole following the 225K blackbody curve. The missing energy at 15um is what CO2 has absorbed beginning from the ground up and has been completely thermalized by an altitude where the air temperature is 225K which, applying the dry adiabatic lapse rate of 10K per 1000 meters is about 3000 meters or 10,000 feet (middle region of the troposphere).
The missing energy in the 15um, when thermalized, is re-emitted as continuous blackbody spectrum so the top of the curve at 265K (which is the surface temperature) is a bit higher than it would be otherwise. That bit higher surface temperature is the effect of CO2’s action as a greenhouse gas. It impedes the flow of 15um energy from surface to space which in effect acts like insulation making the surface temperature a little higher than it would be otherwise. The temperature of the cosmic void is about 3K and doesn’t change. The increased differential between surface and space raises the rate at which energy flows from surface to space re-establishing a new surface temperature equilibrium point between energy-in (short wave energy from the sun) and energy-out (long wave energy from the earth).
This is all undisputed except by cranks and and other assorted ignoramuses who refuse to accept the radiative absorption and emission characteristics of various gases according to physics theories that have been around for over 200 years and which were experimentally confirmed 150 years ago by John Tyndall and which have remained as well established as any theory in physics since then.
The controversy isn’t about the direct effect of increased CO2. That’s cut and dried number crunching of basic physics formulas. The bone of contention is in the feedbacks. The climate boffins on the “hockey team” insist there is a large positive feedback which will somehow cause a runaway greenhouse despite the fact that all paleo-climate evidence of every kind shows the earth has never in its history experienced a runaway greenhouse despite atmospheric CO2 levels far higher than could be obtained by burning every last drop, wisp, and crumb of recoverable fossil fuels. ALL the evidence says the feedback is negative which limits the maximum global surface temperature and where the temperature increase (we’re in a very COLD period of the earth’s history) is concentrated in the higher latitudes i.e. the tropics won’t get much warmer but the temperate and polar regions will. The usual state of affairs for the earth over the past billion years is warm and friendly for living things from pole to pole. The current terrestrial biosphere is a shrunken frozen shadow of itself much of the time over much of its extent compared to the warm (non-ice age) periods which is to say 90% of the time for uninterrupted periods lasting as long as hundreds of millions of years. The earth has been in a ice-age for the past 3 million years and ice ages are not the normal state of affairs – they are the exception to the rule of a planet lush, warm, and green from pole to pole with far higher atmospheric CO2 content and a far larger/faster carbon cycle driven by living things.

barnErubble
February 14, 2011 7:11 am

Could someone (obviously way smarter than I) post a brief summary of what I learned reading thru all this, comparatively to what I would’ve learned (had I understood any of it) reading the paper that contains the ‘notorious Ternberth/Keihl energy balance schematic’. Perhaps I just missed it but what was the definitive answer to, ” . . . does this aspect of the global warming theory make any sense?”
MC had issues with ‘the numbers’ used in fig1 etc. and I was hoping among the learned posters here that some discussion/debate over those numbers would lead to an understanding of what numbers were most likely to be considered real or reasonable.
-Looking for knowledge in all the right places . . .
-Barn

NoIdea
February 14, 2011 7:29 am

Hello Izen
Quote “There is no way that the water surface acts like a one-way street for all incident radient energy. It is not Maxwell’s demon!”
But CO2 is a demonic gas, with its wavenumber of 666.666 (15 µm)!
Is it not Maxwell’s demons casting back all the IR radiation?
The blackbody temperature to emit a peak IR of 15 µm is according to Wiens law 193.18456 Kelvin.
This equates to -79.97 Centigrade (Please do not miss the MINUS sign, yes nearly minus 80C!)
How hot can this 15 µm -80 Centigrade IR heat anything?
NoIdea

P Wilson
February 14, 2011 7:30 am

Bomber_the_Cat says:
February 14, 2011 at 5:59 am
Thanks for the lesson on radiative physics theory – i’m quite aware if it. (The theory, that is)
Which is all that it is. A theory.
In theory, I receive energy from my surroundings, but they never surpass my basal metabolic rate of 58wm2. Similarly I transmit to cool night air energy at this basal rate, although only to the immediate few inches at most, and this energy soon thermalises. (disappears as the form of heat. Heat is not a permanent)
There’s little point in invoking the rather absurd SB equation, which is a thought experiment that leads to rather absurd conclusions.
The c02 blanket is a non existent phonomenon. Its a theoretical phenomenon – since it paints a human face on a climate system, just like Big Bang paints a human face on the unknown origins of time and the universe.
It was established well before AGW ideology that c02 delays the transit of 8% of total IR radiation, regardless of its quantity. As you know, as 14.77 microns the saturation window closes. 14.77 microns does broadly correspond to -28C. 10 microns corresponds to around 15C, the average global temperature. So the last place that c02 absorbs radiation is at the surface, which on average is above subzero. C02 is invisible to these temperatures.
In climatology the fix is made by saying that something is going up in the upper levels of the lower troposphere where indeed these subzero (C) temperatures occur and where c02 captures radiation at its peaks. Only it really isn’t, and what radiation is absorbed at subzero temperatures is quickly released.
There is no radiative mechanism by which -28C can increase to 15C or upwards from a base point of -28C. Unless, of course you increase the heat source (The sun). But then c02 becomes invisible again to SW radiation (It absorbs/delays at subzero).
if you want to prove that an ice cube will heat a cup of coffee -a tepid one, to give you the benefit of the doubt (which is the gist of your argument – that used by alarmists to prove the c02 conjecture) then we’d all be grateful for your video test results)

P Wilson
February 14, 2011 7:59 am

Dave Springer says:
February 14, 2011 at 7:01 am
only the problem with this theory is that the radiation goes in all directions equally, which does not change the amount of heat in the atmosphere.

P Wilson
February 14, 2011 8:04 am

NoIdea says:
February 14, 2011 at 7:29 am
“The blackbody temperature to emit a peak IR of 15 µm is according to Wiens law 193.18456 Kelvin.
This equates to -79.97 Centigrade (Please do not miss the MINUS sign, yes nearly minus 80C!)
How hot can this 15 µm -80 Centigrade IR heat anything?”
reply
oops. I’m way out on the temperature range at which c02 captures energy.

Dave Springer
February 14, 2011 8:14 am

Bomber_the_Cat says:
February 14, 2011 at 5:59 am
Nice. It’s heartening to see some commenters who understand the physics. “School-boy” misunderstanding might be a bit harsh but I understand where you’re coming from. A good understanding of the four years of science taught in grades 9-12 covers almost everything you need to know about this subject in biology, chemistry, and physics from how the greenhouse gases work to what acid rain from sulfate emissions does to how these thing effect the biosphere. When PhDs get involved in the conversation they start such a pedantic bickering over small details in their fields of expertise that the big picture gets lost and laypersons then proverbially can’t see the forest because there are too many trees blocking the view.

izen
February 14, 2011 8:15 am

@-P Wilson says:
February 13, 2011 at 11:23 pm
“heat leaves earth by convection. Clouds prevent convection. They don’t radiate energy”
Some energy leaves the SURFACE by convection, but it does not get very far.
It is not clouds that block it… strange idea!
Its the adiabatic lapse rate.
Convection can only move air until the increased bouancy from the lower number of molecules per cubic metre is offset by the lower density with increasing altitude. It is the lower density of the atmosphere with altitude that blocks convection.
The low altitude of most clouds indicates that the temperature falls below freezing just a few Km above ground level and the atmosphere regains that latent heat of evaporation well below the tropopause. Above the main low cloud layer water vapor is much less important as an atmospheric absorber/emitter.

Dave Springer
February 14, 2011 8:21 am

P Wilson says:
February 14, 2011 at 7:30 am
” As you know, as 14.77 microns the saturation window closes. 14.77 microns does broadly correspond to -28C. 10 microns corresponds to around 15C, the average global temperature. So the last place that c02 absorbs radiation is at the surface, which on average is above subzero. C02 is invisible to these temperatures.”
This is laughably wrong! PEAK emission frequency corresponds to those temperatures. A blackbody emits a CONTINUOUS spectrum with broad shoulders that fall off slowly on either side of the peak frequency. There’s plenty of 15um energy coming from a blackbody with 10um peak frequency. You obviously have no idea what a continuous blackbody spectrum looks like. What you’re saying is like saying no colors of visible light come from sun except for yellow because that’s the peak emission frequency of a 5200K blackbody source. Ridiculous misunderstanding.

izen
February 14, 2011 8:26 am

Hello NoIdea
as usual you ask a deep question.
Any blackbody emitter above -80 Centigraqde will ALSO emit more energy in the 15 µm band, as well as more at shorter wavelengths.
But CO2 is NOT a blackbody emitter. Like the other atmospheric gases, including water vapor it is a very poor emitter over most of the spectrum. It just has strong bands of emission/absorption like water vapor at wavelengths related to its molecular vibrational modes.
Very hot CO2 will still have a peak emssion spectra in the 15 µm band becuase that is its surface ‘colour’, it does not have a blackbody emission/absorption spectra but one vastly biased to the energy level of the molecular vibrational modes.
In the Infra-red longwave spectra CO2 has a surface ‘colour’ that modifies its emission spectra; think of it as a ‘Deep Purple…..’
-grin-

cal
February 14, 2011 8:27 am

Myrrh says:
February 14, 2011 at 4:06 am
Cal says Feb 14, 12:49 am, Slacko says, Jim says
Sort it out, will you.
http://www.wisegeek.com/how-does-infrared-heat-work.htm
“To observe how infrared heat works, the heating and cooling of the earth is probably the most helpful example. Apart from driving all weather events and patterns, as well as ocean currents, infrared heat from the sun is what the Earth during the day, and it is what the ground gives off at night after the sun sets. It is interesting to note that while the sun does emit a huge amount of infrared heat, only about half of the heat we feel on the ground is from direct infrared radiation from the sun. The other half comes from energy from visible light that is absorbed by objects on earth, and then emitted later as infrared heat.”
So which is it?
I accept that I should have said that the sun radiates at all wavelengths from UV to far infrared, but that the peak is in the visible region. Moreover a lot of the incoming energy in the infra red region is absorbed by CO2 and H2O in the atmosphere and is then re-radiated downwards (and upwards ultimately to space) in exactly the same way as the energy radiated by the surface. UV is also absorbed (particularly by ozone) but the peak energy at the surface is still in the visible region of the spectrum. So at night all the infrared radiaton is from CO2 and H2O and during the day the majority is. However the main point I was making was about infra red radiation at night.

Phil.
February 14, 2011 8:34 am

Tsk Tsk says:
February 13, 2011 at 9:02 pm
Not sure I like that. The form of the equation shouldn’t change for different numbers unless there’s a discontinuity. It has to be logarithmic over the entire range of positive real numbers and obviously isn’t physical for negative numbers. I think you mean that for low concentrations of Co2 the dependence diverges only slightly from a simple linear dependence and the divergence grows as the concentration grows.

No need for a discontinuity, if you expand the terms in the equation for small [CO2] you get a linear dependence, for medium values you get √ln([CO2]) and for large values √[CO2]. For the range of values in the atmosphere ln[CO2] is a good fit.

P Wilson
February 14, 2011 8:35 am

Izen.
Thanks for the reply.
Clouds keep the earth warmer (or a given portion of the earth) by preventing heat from escaping. In fact, convectional currents can rise a long way into the atmosphere before they lose their buoyancy. Something like a hot air balloon does.
This uplift, and cooling of air is what causes most weather. Cumulonimbus clouds are often 10,000 metres in height from surface.
actually, watervapour above this level acts as an absorber of radiation. the so called greenhouse effect.

J. Bob
February 14, 2011 8:47 am

eadler says
“If the mean error is know, the readings can be adjusted. If the mean error remains constant, than the temperature anomaly, which is what is being sought will not suffer in accuracy.”
And those “if’s” can be significant. Much of the mean error “adjustment” ability depends on the sensor, electronics, environment, calibration protocols, and how well these protocols are ACCUALLY followed.
My first “outside” job, over 60 years ago, was taking the hi-lo temperatures, for a neighbor, on a old Taylor mechanical thermometer. It had mechanical slides that indicated the hi-lo points, and if I remember, 2 deg. graduations. The neighbor then sent the data, to some government agency. At $0.50 a week that was big money back then. A long way from 3/4 wire platinum RTD’s
Falicoff’s
comment about taking multiple readings may “enhance” the resolution, or quantization, but not necessarily the accuracy. This is also known as “resampling”, as denoted in p.334, “The Handbook of Astronomical Image Processing”, Berry & Burnell.

Dave Springer
February 14, 2011 8:48 am

P Wilson says:
February 14, 2011 at 8:04 am
“How hot can this 15 µm -80 Centigrade IR heat anything?”
Because it sits between a 15C ocean surface and the -253C of the cosmic void. It doesn’t heat anything. It slows down the rate of cooling by interposing something warmer than the black of space between the surface and the black of space. This isn’t rocket science. It’s about as difficult to understand as is understanding why sitting a cup of hot coffee on a block of dry ice versus a block of water ice. In both cases the coffee won’t get any hotter but it will cool a lot faster sitting on the dry ice versus sitting on the regular ice. Or even better hot coffee in an insulated thermos vs. a non-insulated thermos. In both cases the coffee is going to get cooler but the insulated thermos will slow down the rate of cooling. To get an even closer analogy consider an insulated versus and uninsulated hot water heater where the heating element is turned on once a day for a set period of time. The insulated water heater will have a higher temperature because the amount of energy added to each one is equal but the rate of escape of that added energy is lower for the insulated vessel. The end result is the water in the insulated vessel will have a higher maximum and minimum daily temperatures. In the case of the ocean it’s the sun doing the heating on a daily basis and greenhouse gases are the insulators which slow down how fast the ocean cools when the sun isn’t heating it. As Ernest Rutherford said “if you can’t explain a theory in physics such that a bartender can understand it then the theory is probably wrong”. A bartender can understand the difference between an insulated and uninsulated thermos. Why can’t you?

Oliver Ramsay
February 14, 2011 8:51 am

Bomber_the_Cat says:
February 14, 2011 at 5:59 am :
“If that radiation impacts a warmer object then it is absorbed by the warmer object. The warmer object therefore receives energy that it wouldn’t receive if the cold object was not there – and so the warm object is kept warmer than it otherwise would be. Radiation is not somehow preferentially attracted only to colder objects. Thus the presence of cold objects can keep warm objects warmer!!! This does not infringe the 2nd law of thermodynamics, only a schoolboy misunderstanding of it.”
———————-
Hoping to have eluded the Groundhog Day Effect that your comment was caught in, I’ll remark that the warmer object will now be inclined to radiate a lot more energetically than it otherwise would have.
At lower troposphere densities it will be even more inclined to pass on the energy, through collisions, to non-absorbing species which then convect upwards.
When your feet are cold, it takes a painfully long time for them to warm up by merely putting on wool socks. If you change those socks out repeatedly for a colder pair, your toes will turn blue.
It’s not hard to understand that GHG’s provide the air with warmth, but it seems the ground is not really significant in the back-radiation scenario. In fact the warmer the surface becomes the more the energy will be radiated to space without bouncing around in the atmosphere.

Ryan
February 14, 2011 8:52 am

“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”
Is that so? What if we had a month off for CO2 emissions as a globe, with CO2 output deliberately reduced by say 30%. This would create a point of inflexion in our CO2 output which in theory should then be followed by a point of inflexion in rising temperatures, thus proving AGW theory and demonstrating cause and effect. It would then allow us to go further and estimate the actual impact of given increases in CO2. Got to be worth doing just as a global experiment before making any solid commitments to changing our habits.
(Well in theory, in practice I guess it would just get lost in the noise of normal weather patterns which would only prove the futility of the whole AGW nonsense)

Dave Springer
February 14, 2011 8:54 am

P Wilson says:
February 14, 2011 at 8:35 am
“actually, watervapour above this level acts as an absorber of radiation. the so called greenhouse effect”
Water vapor is invisible. A cloud is composed of water droplets not water vapor. Water vapor coming out of a tea kettle is invisible until it has cooled enough to condense into water droplets. This is barely high-school level physical science. You need to go back to high school and relearn this stuff and that’s presuming you ever learned it in the first time through which at this point is a matter of grave doubt. You’d do miserably on the TV show “Are you smarter than a fifth grader?”

Oliver Ramsay
February 14, 2011 8:59 am

I didn’t mean to suggest that heating the surface more would cool the air!
Just that returning energy to the surface provides more of a direct avenue for escape than if it relied only on radiation from GHG’s.

P Wilson
February 14, 2011 9:10 am

Dave Springer says:
February 14, 2011 at 8:21 am
Largely irrelevant comments re c02 though correct ones regarding the sun. However, the argument is that c02 is largely irrelevant to the climate, since most heat escapes via convection and conduction, hence the quick cool down at night. this cooldown, and corresponding cool upper layers show just how quickly this takes place. Nothing to do with the SB equation. To do with observed phenomena
the main reason is that there is no heat being generated, or stored by greenhouse gases, and so back radiation doesn’t occur. Even if it did, the quantities would be almost infinitessimal.
In your later analogy however, it is convection that is being vastly reduced in a thermal flask, which prevents quick cooling.
this convectional trapping process does not occur with outgoing IR radiation through c02 or any other ghg.
however, even at ground-10,000 metres, what c02 does is so infinitessimally small that it might as well be factored out of climate projections.
at 15microns peak, that makes not an iota of difference to atmospheric temperatures. Besides, gases are not blackbodies, and are 3 dimensional. Radiative equations can’t be applied to them (air has a very poor conductor), and so is irrelevant to climatology. With ghg’s, *blocking* operates in all directions, so doesn’t make any difference to the atmospheric heat content.
if you’re talking about the shoulders of c02, then at this level there is more nitrogen and oxygen per CO2 molecule in this area. Dilution reduces the temperature increase per unit of energy.

P Wilson
February 14, 2011 9:14 am

Dave Springer says:
February 14, 2011 at 8:54 am
i said
“actually, watervapour above this level acts as an absorber of radiation. the so called greenhouse effect”
Water vapour is indeed invisible.
the above comment was a contradiction of Izens, who said that water vapour higher than clouds have no effect. He said “Above the main low cloud layer water vapor is much less important as an atmospheric absorber/emitter.”
I think you’ve got it muddled somewhere, although if you wish to trace a source of the confusion between clouds and water vapour, I suggest you consult Izens first.

George E. Smith
February 14, 2011 9:26 am

An interesting dialog; but perhaps KLC didn’t ask the right questions, so in turn MC didn’t really provide the needed answers; although what MC DID say is not incorrect.
One issue raised in the discussion was the 100 Watt light bulb, versus the 300 W/m^2 from the sky. the first one feels “hot”; while the second one doesn’t. MC says that the 300 W/m^2 corresponds to -23.4 deg C. That ties in well with Trenberth’s 390 W/m^2 corresponding to +15 deg C or 288 K.
But here’s what MC didn’t tell KLC; and we have all seen it with that 100 W light bulb “radiating” onto two samples of dry air; one containing more CO2 than the other. Well “The Science Guy” Bill Nye, has performed that quite fraudulent experiment in public.
Here’s what’s wrong with it and what MC forgot to tell KLC.
The 100 Watt light bulb, is a tungsten filament lamp (likely gas filled), and it will emit a roughly black body like spectrum (thermal spectrum); and MC explained how near BB and real BB sources can be related via an emissivity; and of course most practical sources, are not a singl4 Temperature source so they will be somewhat spectrally discombobulated as well.
But back to that 100 Watt lamp. Which is fairly typical of the so called “warm white”light source. It will radiate a spectrum that is something like a 28-2900 deg thermal spectrum; ie a somewhat BB like source of radiation with a color temperature of about 2900 K. THAT IS HALF THE TEMPERATURE OF THE SUN SURFACE.
The sun emits a thermal (BB like) spectrum at about 64 million Watts per square metre (at the sun’s surface). Our innocent 100 Watt lamp, at half the temperature is only emitting 4 million Watts per square metre, since that goes as T^4, per S-B.
More importantly, the 100 Watt “light bulb” is ten times the Temperature of the 288 K Trenberth average Temperature of the earth, and the presumed source of that 390 W/m^2.
So the incandescent light bulb is ten times the average earth Temperature, and is radiating 10,000 times as much energy per square metre as the average earth.
So now we kick in with the Wien Displacement Law that MC also mentioned.
At 288 K (+15 deg C or +59 deg F), the LWIR radiation spectrum, that is emitting a total 390 w/m^2, has a spectral peak wavelength at 10.1 microns. And assuming a single Temperature source, 98% of the LWIR radiation is contained between 1/2, and 8 times the peak wavelength so from 5.0 to 80 microns, is the spectral range of the average earth Temperature roughly BB spectrum. So colder places, will radiate even longer wavelengths, but being colder, they also are radiating even less, so they don’t really contribute much energy on the long wave end. For the highest surface desert Temperatures which can be above +60 deg C, the spectral peak could be as low as about 8.8 microns from the Wien Law, so that will radiate down to maybe 4.4 microns, and also the total S-B emission can be 1-8-2.0 times as high as Tenberth’s 390 W.m^2. So it is those hot dry desert regions that are the principle radiative coolers of the earth. The 8.8 micron peak is even further away from the 15 micron CO2 band, and the peak moved further into the “atmospheric window” where water vapor is somewhat benign. Well there’s little water vapor in those arid deserts anyway. But don’t forget that Ozone kicks in a dip at around 9.6 microns; but it is quite narrow, because of the height, and low density and Temperature of the ozone layer (less line broadening)
Now back at our phony heat lamp at 2880 K,; not only is it emitting 10,000 times the emittance of the average earth; but the spectral peak is not 10.1 microns either, but is now 1.0 microns.
Well wouldn’t you know it; water (H2O) has several absorption bands near there at 0.94 microns, and also at 1.1; and supposedly humans are 95% H2O.
So the peak radiant emission from a 100 Watt light bulb is at the correct wavelength to cause strong absorption in human flesh; no wonder it feels warm.
The 288 K mean earth surface on the other hand is radiating at 10.1 microns, and the water absorption coefficient there is about 1000 cm^-1; which gives a 1/e transmission depth of 10 microns; or a 99% absorption depth of 50 microns; 2/1000 inches.
It doesn’t even make it through the surface of your skin; and even if it did, it wouldn’t register against your body Temperature of 98.6 deg F or 37 deg C.
So no wonder the 100 Watt lamp feels warm to your skin, and no wonder it warms the air samples. CO2 has absorption bands at around 2.7 microns, and I think also at 4.0 microns. The 2880 K lamp spectrum is going to put 98% of its energy in the spectral range from 0.5 Microns (green) to 8.0 microns, so both of those CO2 bands are going to be activated. And even the 15 micron band will absorb much more energy from the 2880 K source than from the 288K source.
One should also note that although the S-B equation says the total emittance goes as the 4th power of temperature; the spectral peak emittance actually goes as the fifth power of the Temperature (T^5).
I’ll leave it to the math geeks out there to figure out how much more spectral emittance you get at 15 microns, from a 2880 K sourcves compared to a 288 K source (Watt’s per m^2 per micron wavelength.
So like I said the 100 Watt lamp demonstration is a total fraud. A much better source fo 300-390 W/m^2 LWIR radiation that is properly spectrally peaked at about 10 microns, would be an ordinary 16 ounce bottle of water. I don’t recommend the French Evian water; it’s too expensive and doesn’t radiate any more than any other water; use tap water it’s cheaper.

George E. Smith
February 14, 2011 9:37 am

“”””” P Wilson says:
February 14, 2011 at 9:10 am
Dave Springer says:
February 14, 2011 at 8:21 am
Largely irrelevant comments re c02 though correct ones regarding the sun. However, the argument is that c02 is largely irrelevant to the climate, since most heat escapes via convection and conduction, hence the quick cool down at night. this cooldown, and corresponding cool upper layers show just how quickly this takes place. Nothing to do with the SB equation. To do with observed phenomena
the main reason is that there is no heat being generated, or stored by greenhouse gases, and so back radiation doesn’t occur. Even if it did, the quantities would be almost infinitessimal.
In your later analogy however, it is convection that is being vastly reduced in a thermal flask, which prevents quick cooling.
this convectional trapping process does not occur with outgoing IR radiation through c02 or any other ghg.
however, even at ground-10,000 metres, what c02 does is so infinitessimally small that it might as well be factored out of climate projections.
at 15microns peak, that makes not an iota of difference to atmospheric temperatures. Besides, gases are not blackbodies, and are 3 dimensional. Radiative equations can’t be applied to them (air has a very poor conductor), and so is irrelevant to climatology. With ghg’s, *blocking* operates in all directions, so doesn’t make any difference to the atmospheric heat content.
if you’re talking about the shoulders of c02, then at this level there is more nitrogen and oxygen per CO2 molecule in this area. Dilution reduces the temperature increase per unit of energy. I don’t know about other readers; but I have absolutely no idea who said what, in this joint effort by P-Wilson, and Dave Springer.
Perhaps, P. Wilson, being the outermost poster, can find some simple methodology to separate what he is citing from Dave Springer (if anything) and what is the product of his own thoughts.
I often see people ascribe whole rafts of stuff to me; when I never said one word of what they posted.
Trying to italicize things doesn’t do a thing, since the browsers can make mincemeat out of font twiddling.
So I dunno P. Wilson; just what did you add if any to what Dave said if any ?

A C Osborn
February 14, 2011 9:38 am

Don V says:
February 14, 2011 at 1:47 am
Thanks.

A C Osborn
February 14, 2011 9:40 am

Oliver Ramsay says:
February 14, 2011 at 8:51 am “In fact the warmer the surface becomes the more the energy will be radiated to space without bouncing around in the atmosphere.”
WUWT?
So the surface has a direct connection to space then?

A C Osborn
February 14, 2011 9:41 am

Can I ask all the Physics guys a simple question?
What happens when 2 photons travelling in opposite directions collide?

P Wilson
February 14, 2011 9:43 am

George Smith
dave Springer maintains that ghg’s – particularly c02 slows the rate of cooling of the atmosphere, and uses the analogy of a thermal flask (which limits conduction and convection.
i’m arguing that c02 doesn’t make much measurable difference to atmospheric heat content and cool rate

P Wilson
February 14, 2011 9:44 am

although, George, it depends what posts you’re referring to

Tenuc
February 14, 2011 9:46 am

steven mosher [February 13, 2011 at 3:39 pm] says:
“yup, but there is more to it than that. If you want to do IR design today ( of sensors, or counter measures or stealth technology) the physics you use is the same physics that many skeptics deny. good thing we don’t let them design the machines that our safety as a nation depends upon.“
Got out of bed the wrong side today Mosh??? Only an idiot would think that tarring all sceptics with the same brush is a sensible way to have a useful debate.
My own take on this is simple. Although lab experiments show CO2 could raise Earth’s overall temperature by a tiny amount, the IPCC’s predictions of catastrophic temperature increases produced by carbon dioxide have been challenged by many scientists and found to have no substance in the real world.
The importance of water vapour is frequently overlooked by environmental activists and by the media. The large temperature increases predicted by many computer models are non-physical and inconsistent with results obtained by basic measurements – much scepticism is warranted when considering computer-generated projections of global warming when these same models cannot even predict existing observations.

P Wilson
February 14, 2011 9:54 am

ok, I think i gleaned something re: the shoulders of co2. 15microns is the peak.
all I can fathom is that Dav Springer argues that the entire shoulders and peaks of c02 count, but that at the shoulders – or else the band either side around 15microns – then they become miniscule as heat absorbers. At the tropospheric level where the outer bands absorb, there are more oxygen and nitrogen competing for heat than c02

Oliver Ramsay
February 14, 2011 9:58 am

A C Osborn says:
February 14, 2011 at 9:40 am
Oliver Ramsay says:
February 14, 2011 at 8:51 am “In fact the warmer the surface becomes the more the energy will be radiated to space without bouncing around in the atmosphere.”
WUWT?
So the surface has a direct connection to space then?
—————
Yes

Dave Springer
February 14, 2011 10:02 am

cal says:
February 14, 2011 at 8:27 am
Most of the energy the earth receives is visible light from the sun. The 5200K blackbody spectrum carries a significant but minor fraction of its total energy in the near infrared but no practical amount of the energy is in the far infrared where CO2 absorption begins. Conversely the earth (discounting reflected visible light during the day) emits essentially no energy in the near infrared and essentially all of it in the far infrared. The amount of the energy emitted in the far infrared necessarily over time is equal to the amount of shortwave energy absorbed by the ocean durin the day. Near infrared hitting the ocean surface is absorbed within the first few microns and doesn’t heat the ocean but is rather carried off in water vapor which rises by convection where the energy is released when the water vapor condenses into a cloud. Visible light however penetrates over 100 meters and is almost completely absorbed with very little of it reflected. Land surfaces to a small degree and clouds to a high degree reflect incoming visible light straight back out into space. Estimates of the earth’s average albedo are in the 35% range plus or minus a few percentage points depending on who you ask and what they need to stick into their climate models to better reproduce paleo-climate data. Albedo is used as a fudge factor to tune climate models for better fitting climate hind-casts.
So anyhow about 60% of the solar energy arriving at the top of the atmosphere works to heat the ocean and all that energy eventually radiates out at night at much lower frequency. It’s the difference in CO2 frequency response to short wave versus long wave radiation that allows it to act as an insulator. It’s transparent to short wave radiation and opaque at a few significant long wave frequencies so it almost all solar energy through to the ocean unimpeded during the day but impedes long wave energy emitted by the surface at night. The simplest and quite accurate way to conceptualize this is that CO2 is an insulator – it’s like a blanket over a dark rock where you remove the blanket during the day so the rock can heat up in the sun and you put the blanket back at night to retain more of the warmth from the sun during the night. The end result of blanket vs. no blanket is a warmer rock with the blanket than the rock would be without it.

Ian W
February 14, 2011 10:06 am


Mikael Cronholm says:
February 13, 2011 at 9:56 pm
Ian W says:
February 13, 2011 at 12:35 pm
When water vapor in the atmosphere condenses into liquid water and then changes state again and becomes ice, it gives off latent heat for both state changes.
Does that latent heat release follow Stefan-Boltzmann’s radiative equation ?
No Ian, latent heat is not really related to S-B.

So if you consider all the clouds and it is estimated that 62% of earth’s surface is covered by clouds – the amount of energy released as latent heat is huge. Yet people persist in working out the amount of heat leaving the surface using Stefan-Boltzmann radiation equations. This is obviously incorrect.

izen says:
February 14, 2011 at 8:15 am
Some energy leaves the SURFACE by convection, but it does not get very far.
It is not clouds that block it… strange idea!
Its the adiabatic lapse rate.
Convection can only move air until the increased bouancy from the lower number of molecules per cubic metre is offset by the lower density with increasing altitude. It is the lower density of the atmosphere with altitude that blocks convection.
The low altitude of most clouds indicates that the temperature falls below freezing just a few Km above ground level and the atmosphere regains that latent heat of evaporation well below the tropopause. Above the main low cloud layer water vapor is much less important as an atmospheric absorber/emitter.

Well I don’t know about ‘not getting very far’ 60,000ft or more in the tropics is the level of the tropopause (the top of the convective atmosphere) and in the Inter-Tropical Convergence Zone (ITCZ) the cloud tops of towering storms are (by definition) up to that level. (A recent Northwest Airlines flight experienced liquid rain hitting the aircraft in cruise above 30,000ft in a storm between Hong Kong and Tokyo despite the ambient outside-air-temperature being well below zero. The rain froze into ice on hitting the airframe) The height of the cloud tops all depends on why the clouds are there.
Humid air is more buoyant than dry air, so just being more humid is enough for a volume of air to start convection. Drier air sucked in will then pass over the wet surface and Henry’s law applies and more water will evaporate into the dry air cooling the surface. No external heat source is required. As the humid air rises the air temperature will drop at the wet adiabatic lapse rate and water will start to condense around nucleation particles and form clouds. This type of cloud with a gentle wind will become the low stratocumulus over oceans and coastlines. And each and every cloud droplet has taken heat from the surface and radiated it as it becomes a droplet and again when it becomes ice.
With the application of tropical heat and humidity convective storms develop every day some towering ten miles high into the atmosphere. If the conditions are conducive these storms merge and the Coriolis force on the air being rapidly drawn in at the base of the storms starts the winds in the storm system to rotate. There is a lot of discussion of the ACE index here, but to put it in perspective, the amount of energy released by a hurricane can be calculated based on the latent heat emitted and / or the kinetic energy.
An average hurricane in a day transfers latent heat equivalent to 200 times the world-wide electrical generating capacity.
http://www.aoml.noaa.gov/hrd/tcfaq/D7.html
As confirmed in response to my question at the top of this thread, the temperature of the cloud has no bearing on the amount of latent heat being released. The heat being released is linked to the amount of water changing state and the latent heat of that state change.
The missing heat is almost certainly in the misunderstood and underestimated hydrologic cycle that is ignored in favor of a simple Stefan-Boltzmann radiation equation.

Jim D
February 14, 2011 10:19 am

Slacko,
Following up on the reply already by cal, solar radiation contains a small fraction of IR, maybe only a few W/m2, so most of the warmth you feel is from absorbing the visible/UV rays.
The other part of Slacko’s question: yes, N2 can’t radiate IR, nor can O2 or argon. This is from physics.
Robert Clemenzi,
Yes, 324 W/m2 corresponds to 34 F. This is supposed to be a global and annual average value, which obviously varies greatly locally. I think your suggestion is that it implies reflection because the clear atmosphere is probably colder than this. Actually clouds behave as almost perfect black bodies, so the cloud bases are emitting at their temperature, which may account for the average being as high as 324 W/m2. In the tropics even clear skies can emit more than 324 W/m2 to the surface because of the high H2O contents.

John_in_Oz
February 14, 2011 10:27 am

The paragraph about Mikael Cronholm (MC) states … ‘and holds two Bachelor of Science degrees (Economics and Business Administration)’.
I think it likely that’s a typo, and should read ‘Bachelor of Arts’.
Of course I could be wrong. Are there really universities that issue Bachelor of Science (Economics) and Bachelor of Science(Business Administration) degrees?

George E. Smith
February 14, 2011 10:28 am

“”””” at 15microns peak, that makes not an iota of difference to atmospheric temperatures. Besides, gases are not blackbodies, and are 3 dimensional. Radiative equations can’t be applied to them (air has a very poor conductor), and so is irrelevant to climatology. With ghg’s, *blocking* operates in all directions, so doesn’t make any difference to the atmospheric heat content. “””””
Either P.Wilosn or Dave Springer said the above; well it contains both of their names at the top. I’m inclined to discount Dave, as the source of that statement; but can’t tell where if anywhere Dave stops, and P. starts.
In any case, while it may be pedantically true that gases aren’t black bodies; we know that since absolutely nothing real is a black body.
But where in the Physics texts does it say that water or ice stops radiating thermal radiation, the instant that it changes from a liquid or solid, into a gas at the same Temperature. Has anybody ever recorded a motion picture video of that shut-off process actually taking place in any Physical system.
Can somebody cie some peer reviewed literature that forgives gases from radiating thermal radiation; that is Electromagnetic Radiation of a spectral nature that depends on on the Temperature of the gas. And where in that peer reviewed paper, did it say that the sun is further excused from obeying the non thermal EM radiation prohibition of gases ?
One thing we can be fairly sure of, is that the Raleigh scattering of short wavelength sunlight, which makes the entire daytime sky appear “Sky blue” when viewed in any direction (not counting near the sun); and that means it looks the same looking up as looking down; because the large angle scattering due to the RS process makes the atmosphere an isotropic source of sky blue light which originally came from the sun.
By the same token, the LWIR emissions from the atmosphere; whatever their source; are slso isotropic, since there is no preferred dirction of emission. So the infrared sky looking up towards space is pretty much the same as looking down from outer space; except (apparently) that according to Trenberth, 40 W/m^2 of LWIR radiation that is actually emitted from the earth’s surface, actually escapes unharmed to space, and the other 350 W/m^2 is absorbed by the atmosphere. Apparently per P.Wilson, it cannot be subsequently emitted from the atmosphere per the radiation laws governing BBs and other thermal radiation laws; which he says don’t apply to gases.
So now would somebody not as dense as I am, like to explain to me what is the source of quite thermal spectrum looking (grey body) radiation that is seen from outer space, when looking at the earth. Onl;y 40 W/m^2 of it can be coming from the ground per Trenberth and it does not have a narrow CO2 absorption band spectrum as one would expect, if direct emission of molecular resonance radiation spectra from CO2 was the source.
Several have noted that a lot of “heat” transport to the upper atmosphere is a result of convection of hotter surface gases into the upper regions. Somebody could explain for us how that eventually escapes to space, seeing as how gases can’t radiate thermal spectra according to the black body or S-B radiation laws.
Clearly the ten years, that I formally studied Physics in School, plus the subsequent 50 years of using it daily in practice to accomplish things; was not enough for me to come across that jewel of knowledge, that gases do not radiate according to the radiation laws.
Readers might also find it intersting to look at Fi 11; “Higher Members of the Balmer Series of the H Atom (in Emission) Starting from the Seventh Line, and showing the Continuum (Hertzberg41) ”
The reference to “hertzberg 41” is the G. Herzberg Ann. Physik (4) 84, 565 1927 citation
And that photo of an actual real measured, scientific observation of a (non teracomputer simulation of) Balmer spectrum including a “Continuum Spectrum” along twith the Balmer line specrum is in; “Atomic Spectra and Atomic Structure” by Gerhard Herzberg; Prentice Hall 1937. Herzberg mentions that in Emission the continuum corrsponds to a free electron of any energy being captued by a proton, adn going in to “the orbit” having the principal quantum number n=2 (this is of course in the Bohr Atom interpretation. Modern Quantum formulations may be different.)
And nowhere does temperature get mentioned here. This continuum radiation still corresponds to an energy level transition; in this case from an ionised state. Well of course you won’t find any ionised states on the sun I would imagine. that spectrum is commonly observed in stars. But The end of the Balmer series, is at about 3800 Angstroms; so that continuum is in the UV region; not in the LWIR region where the thermally originated LWIR emission spectrum occurs.
The extraterrestrial LWIR spectrum from the earth, does show a narrow spectral dip at the 9.6 micron Ozone band, and a wider one at the 15 micron CO2 band from about 13.5 to 16.5 microns. otherwise it looks pretty much like any other near bB-spectrum as limited by the Planck and S-B laws (Wien also)

barn E. rubble
February 14, 2011 10:29 am

The question posed early on this thread from KLC; ” . . . From your point of view as an IR expert, does this aspect of the global warming theory make any sense?”
RE: “. . . notorious Ternberth/Keihl energy balance schematic (as shown in Figure 1 of this paper: http://www.cgd.ucar.edu/cas/Trenberth/trenberth.papers/TFK_bams09.pdf ), you see the back radiation is determined to be very significant…”
Maybe I missed it but was there a summary or (for want of a better term) a consensus from the posters here that can answer this with some authority? MC replied he had issues with the numbers used. I was wondering what others thought re: reasonable numbers if those referred to (fig 1) are not reasonable.
MC also noted, “Every question answered raises a few more, which grows the confusion exponentially. ”
I was hoping to clear up some confusion on my part.
-Barn

George E. Smith
February 14, 2011 10:33 am

“”””” P Wilson says:
February 14, 2011 at 9:44 am
although, George, it depends what posts you’re referring to “””””
I cut and pasted the entirety of the post headed by your name, and containing a reference to Dave Springer. That post, and only that post was being referred to by me.

P Wilson
February 14, 2011 11:01 am

i see. That was entirely my post. As a response to Dave Springer. I just cut and pasted his name and time to show what post I was replying to
Hope that clears things up

Jim D
February 14, 2011 11:04 am

barnErubble,
MC was not concerned with the main number of > 300 W/m2 from IR as back radiation in those papers. He was concerned with the accuracy of the residual 0.9 W/m2, and how anyone could accurately state such residuals given the large canceling numbers involved. This is rather related to Trenberth’s “missing energy” issue, and is a valid concern.

P Wilson
February 14, 2011 11:16 am

George E. Smith says:
February 14, 2011 at 10:28 a
“Clearly the ten years, that I formally studied Physics in School, plus the subsequent 50 years of using it daily in practice to accomplish things; was not enough for me to come across that jewel of knowledge, that gases do not radiate according to the radiation laws.”
of course they do, as only you would know. I’m referring to the SB equation in particular as applied to c02, and the fanciful results on paper that it generates. I think you’ve implied before that except for a tiny fingerprint of black body radiation, most radiation evades c02 in its escape to space.
Before energy escapes into space through radiation, it doesn’t matter whether the energy is in contact with CO2, water vapor or nitrogen.

February 14, 2011 11:23 am

Jim D says:
“solar radiation contains a small fraction of IR, maybe only a few W/m2, so most of the warmth you feel is from absorbing the visible/UV rays.”
Actually, more than 50% of the solar energy at the top of the atmosphere is IR. At the surface, the percent is increased because some of the UV was absorbed at a higher level. On a cloudy day you can get a sunburn. However, since you will not feel a temperature difference between overcast light and full shade, the “heat” you feel on a sunny day will not be from the UV.

Sensor operator
February 14, 2011 11:52 am

Steeptown mentioned:
…it is evident to me that the radiative efects of CO2 in the atmosphere are of 2nd or 3rd order compared to the radiative, convective and latent heat effects of H2O.
Well, when do 2nd order effects become important? The problem with increasing CO2 in the atmosphere is the ever increasing 2nd order effect. So a little bit of energy doesn’t escape. No big problem. But, if the effect is compounded, which appears to be the case for CO2, suddenly a little bit of energy becomes a lot of energy. When do we care?
Water in the atmosphere has two very different properties versus CO2. First, the amount of water in the atmosphere is not likely to double unless there is a major change in the atmosphere. So even if H2O is a first order effect, the much smaller changes in the amount of water is not going to be significant. However, CO2 is increasing fairly quick. And a doubling or possibly tripling the amount of CO2 is very real so the second order impact is likely to be much larger than the small deviations of the first order impact. Second, CO2 is throughout the entire atmosphere. Sure, it is only 0.04% of the atmosphere. But if we look at the entire atmosphere, not just the surface, water is only 0.4%. Doubling or tripling CO2 is now ~0.1 %. Suddenly CO2 is not that small after all.
Something else folks seem to be forgetting with CO2 is the bands it absorbs. In particular, it is absorbing in bands that water is not. At lower levels of CO2, a sufficient amount of LW radiation was escaping to provide an energy balance. With the increased amount of CO2, the balance has been upset and the Earth is trying to reach a new equilibrium.
Some folks are very taken with Mr. Cronholm’s one comment:
For a theory to be scientifically proven, it has to be stipulated and tested, and the test must be repeatable and give the same results in successive tests for the theory to be proven.
This may be true, but we only have one planet. We don’t have an alternate Earth that we can play with to see what happens. Also, tested does not necessarily mean run in a lab or a model. Consider evolution. We don’t “test” evolution, we look at the available data and the composite result from this evidence is the fundamental reasons for evolution. Climate change is similar: we look at temperature, CO2, extended growing seasons, northern and southern boundaries of migrating birds and insects, extreme weather events, receding glaciers, etc.
One comment that has been discussed time and again is that in the geological record, an increase in CO2 occurs after an increase in temperature. Fine, let’s assume that is true. The problem we have is the measurements we are making now show CO2 increasing and temperature following. What does it mean? Simple: a different mechanism is now responsible, not the natural processes we have been able to identify from past changes.
Of course, there could be another natural phenomenon we have not found/discovered. But as Dr. Alley from Penn State has pointed out: we built the satellites, we made the measurements, when someone disagreed we made new satellites and new measurements. We have spent billions of dollars and so far the overwhelming evidence is in support of AGW.
Think of it like Children’s Hospital. The doctors there have one job: to put themselves out of a job. If they could solve all the medical issues they see, they would no longer be needed. And that is actually their goal! They don’t want kids to suffer. Do people really think climate change scientists want AGW to be true? If they are right, the end result is really bad.

Don V
February 14, 2011 12:00 pm

AC Osborne: re photon-photon collision
I am not a physicist but, back when I went to college and took physics, I was taught that photon-photon collisions are not possible. Being virtually massless and only possessing momentum, two photons can only indirectly interact with each other. Their interaction is called Delbrück scattering. When they “interact” they annihilate into a virtual electron-positron pair, which then annihilates back into two real photons again. They lose no momentum or energy in the process.
Don’t know if that is completely right, my memory is hazy, but my rule of thumb has always been that for first approximations photons are massles and that massless quanta’s of energy basically just go right thru each other.

February 14, 2011 12:01 pm

Tenuc says:
February 14, 2011 at 9:46 am
steven mosher [February 13, 2011 at 3:39 pm] says:
“yup, but there is more to it than that. If you want to do IR design today ( of sensors, or counter measures or stealth technology) the physics you use is the same physics that many skeptics deny. good thing we don’t let them design the machines that our safety as a nation depends upon.“
Got out of bed the wrong side today Mosh??? Only an idiot would think that tarring all sceptics with the same brush is a sensible way to have a useful debate.
#####
I wrote: “the same physics that many skeptics deny”
You accused me of “tarring all sceptics”
There is a reason why I chose the word “many”
There is a reason why I did not use the term “all” or “Most”
The reason was to trap idiots who are not careful with words.

cba
February 14, 2011 12:03 pm

I found the explanations both fundamentally accurate to my understanding level and very nicely presented in what should be a very understandable way.
One major error though is the misconception that 0.9w/m^2 imbalance is measured. It is a model calculation based upon plenty of presumptions. The references to the CERES & ERBE data actually turn up to have accuracy problems of several watt’s per m^2 so despite all the discussions in the papers, they resort to modeling to get their value. In the earlier papers, such as hansen’s, the uncertainties in the modeling for such things as cloud formation versus T are discussed and decisions made that gives rise to their claim that cloud formation decreases with temperature which is not the case for all presumption options like cloud formation depending upon absolute humidity levels.

wayne
February 14, 2011 12:15 pm

Don V :
February 14, 2011 at 1:47 am
Thank you Don, well thought out comment!
Found not a single thing within I would argue about.
You have a clear mind. Could we talk some physics here? I have one big question I need some help answering (or try to answer at least, and, it’s on a very simple logical level).

cba
February 14, 2011 12:30 pm

Barn,
the details isn’t so much radiative theory as it is to the system of Earth and its atmosphere. You can rather safely bet that the warmer crowd will minimize every number that limits AGW and exaggerate every number that supports it.
Primary amongst the real uncertainties is that of the albedo and cloud cover and the behavior of the cloud cover with conditions such as temperature. The whole premise being used that there is one surface temperature average for a given blockage of outgoing IR is totally flawed. While incoming solar average power is given great importance, the notion of what the albedo reflects away is often considered to be constant and not the actual variable it really is, driven by many other factors. If you consider the comments of the warmers, they give lip service to land use changes and loss of snow and ice cover as being the important factors with albedo. Funny how about 80% + of the albedo is due to a 60% + cloud cover and of that under 20% surface contribution, around 70% is oceans and that leaves next to nothing for contributions from the land surfaces, never mind ice and snow and a little rain forest or two.
K&T97 admit possible errors of up to 20% in their cartoon values. Amazing how close they got to that by underestimating cloud cover and overestimating land and surface contribution. It actually looks like they forgot to take cloud cover into account and gave a clear sky only value. Later they use the same values and apparently forgot to include their serious margin of error.

izen
February 14, 2011 12:53 pm

Ian W says:
February 14, 2011 at 10:06 am
“As confirmed in response to my question at the top of this thread, the temperature of the cloud has no bearing on the amount of latent heat being released. The heat being released is linked to the amount of water changing state and the latent heat of that state change.
The missing heat is almost certainly in the misunderstood and underestimated hydrologic cycle that is ignored in favor of a simple Stefan-Boltzmann radiation equation.”
The ‘Science of Doom’ site linked to from here (top right of page) has a good discussion of some of these issues.
The amount of energy transported by latent heat changes is complex, there is one simple way to quantify it however. The total global yearly rainfall is a direct measure of how much energy was moved by evaporation and condensation in the hydrological cycle.
But the first LoT kicks in. The energy is still around. The only way it can get of the planet is by radiating IR photons into space. All the hydrological cycle can do is move the location of that emission around. That evens out the temperature so that water does not boil at the equator and can melt at the poles, but the same amount of energy has to leave the planet. Otherwise it will warm until the S-B T^4 relationship increase the emissions enough to compensate.

Jim Masterson
February 14, 2011 1:04 pm

>>
Smoking Frog says:
February 14, 2011 at 4:36 am
The average of the minimum and maximum overlaps gives exactly 62%:
((49 + 6 + 20) + 49) / 2 = 62
but I’m not sure that this is what we’d get with random overlap, even if, as my calculation assumes, there are no real-world constraints.
<<
Interesting. It may be what the authors were doing.
>>
The Inclusion-Exclusion Principle requires that we know the overlap(s) to begin with, so it can’t give us the answer.
<<
But it does give us an answer that is almost the same:
.49 + .06 + .20 – (.49)*(.06) – (.06)*(.20) – (.49)*(.20) + (.49)*(.06)*(.20) = 0.61648.
Jim

stephen richards
February 14, 2011 1:22 pm

You accused me of “tarring all sceptics”
There is a reason why I chose the word “many”
There is a reason why I did not use the term “all” or “Most”
The reason was to trap idiots who are not careful with words.
Steve, That may well be the case but seeking to trap idiots is no more valid than your rivals statement. I believe also that your comment ‘that most skeptics deny’ is also and POSSIBLY incorrect. I know I feel somewhat agrieved because while I am a skeptic (and a physicist) I understand well enough the ‘science of CO²’ but I have never been convinced that the phrase that ‘increasing CO² causes global warming’ has ever been proven (show me where) or is in anyway true. Suggest you read those words very carefully. :))

Oliver Ramsay
February 14, 2011 1:44 pm

izen says:
February 14, 2011 at 12:53 pm
“But the first LoT kicks in. The energy is still around. The only way it can get of the planet is by radiating IR photons into space. All the hydrological cycle can do is move the location of that emission around.
——————————–
You make it sound like water just shuffles energy around horizontally.
This is claiming the taxi delivers you to your holiday hotel door, not the airplane.

cba
February 14, 2011 1:47 pm

frog, jim,
the details show in kt97 that the assumptions are 100% optical thickness for the two main cloud types and 50 or 60% thickness for the 6% coverage contributor. The 62% result they come up with is assuming 100% thickness with random overlap. It does seem to agree well with what little is known of the actual coverage, which can vary substantially over time, something like over +/- 5% as I recall. That also leads to an albedo variation of around 5% and a peak to peak difference in reflected incoming light power of about 10 w/m^2, far more than a mere co2 doubling.

JAE
February 14, 2011 1:58 pm

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. Then why is the maximum temperature there NEVER as hot as it is almost every day from June-Sept. in Phoenix, AZ?

George E. Smith
February 14, 2011 2:21 pm

“”””” Ian W says:
February 13, 2011 at 12:35 pm
When water vapor in the atmosphere condenses into liquid water and then changes state again and becomes ice, it gives off latent heat for both state changes.
Does that latent heat release follow Stefan-Boltzmann’s radiative equation ? “””””
Latent heat and anything to do with heat, is a property of physical matter; it has nothing to do with Electromagnetic Radiation; so it has nothing to do with the Stefan Boltzmann equation.
EM radiation can go anywhere it wants to; heat can’t.

Matter
February 14, 2011 2:25 pm

Interesting article, and Al Tekhasski covers a good point too.
Detectors are tuned to work in a certain wavelength. We can make materials that have energy levels (or bands) that only allow transitions of certain energy. One common way of doing it is to create a ‘quantum well’ (look it up!) using sandwiched layers of different materials, but there are others.
The atmosphere is largely transparent in some wavelengths, it doesn’t absorb well here. This means that there aren’t many available transitions, so it can’t emit here well either. So if you point your scanner upwards and measure, and your scanner is tuned to this energy band then you won’t measure many photons coming down. This will mean a small electrical signal (the voltage output is related to the number of photons) and your device thinks ‘small voltage, so not many photons, so what I’m measuring is cold’.
This works because most objects have emissivity close to 1 at these ranges, whilst the atmosphere doesn’t. So you can measure the photons coming from far away objects and not the atmosphere; if the atmosphere had emissivity 1 for these wavelengths then it would absorb the light from the object before it could reach your scanner, and you would always measure the atmospheric temperature – not useful for a long range thermometer!
The strength of CO2 absorption can be tested empirically by spectral measurements from satellites (and changes can be measured from ground stations) for a variety of conditions. Look up some of Philipona’s papers or the Harries 2001 paper. They provide experimental confirmation of the physics.

Latitude
February 14, 2011 2:26 pm

physics schymisics,
real world
When CO2 levels were 4000 ppm – we had an ice age
when CO2 levels were 3000 ppm = we had an ice age
when CO2 levels were 2000 ppm – we had an ice age
and here we are wringing our hands and wetting the bed over 390 ppm……………….
real world says the only tipping point is when CO2 levels get in the thousands, we can have another ice age……………..

Jim D
February 14, 2011 2:57 pm

I will correct my answer to Slacko, taking into account Robert Clemenzi’s point.
Yes the IR at night is mostly due to H2O and CO2. During the day, the sun produces near-IR in the 1-2 micron range, while the atmosphere only emits at the wavelengths longer than about 4 microns. These two types of IR may be of comparable magnitude in the upper atmosphere, but much of the near-IR is absorbed before reaching the surface.

kuhnkat
February 14, 2011 4:07 pm

“Sensor operator says:
February 14, 2011 at 11:52 am
Think of it like Children’s Hospital. The doctors there have one job: to put themselves out of a job. If they could solve all the medical issues they see, they would no longer be needed. And that is actually their goal! They don’t want kids to suffer. Do people really think climate change scientists want AGW to be true? If they are right, the end result is really bad.”
Sensor, I don’t have to THINK that is what some climate scientists think. Phil Jones actually stated it in his e-mails.

Bill Illis
February 14, 2011 4:11 pm

How does Nitrogen and Oxygen warm up and cool down in the atmosphere.
The way it is decribed by some here, the radiation theory assumption is that both of these gases are at absolute zero.

Myrrh
February 14, 2011 4:12 pm

Infrared is Heat, or rather far and mid is heat. Near infrared is not felt as heat, think remote control. You’re not going to get warm by flicking your remote at yourself.
Visible light and UV do not feel hot, they are cool. UV doesn’t penetrate very far into the body, it tans, or burns, the skin – can do so even on cloudy days as Robert Clamenzi says (Feb 14, 11:23 am).
We receive IR here on earth from the Sun. Why is this excluded in AGW literature?

kuhnkat
February 14, 2011 4:28 pm

“Don V says:
February 14, 2011 at 12:00 pm
AC Osborne: re photon-photon collision”
Wikipedia has an interesting article on photon-photon collisions.
http://en.wikipedia.org/wiki/Two-photon_physics
Of course it points out that photons can’t collide since they don’t become photons, or quantized, until they interact with matter. This gets to the root of the issues with backradiation. Until the radiation from the CO2, or the earth for that matter, actually interacts with a particle they are best described through wave mechanics which was well mapped like here:
http://en.wikipedia.org/wiki/Phase_cancellation
http://www.mrelativity.net/Papers/4/Rykov.htm
Many people ask what would happen to the energy if 2 waves cancelled.
http://newsgroups.derkeiler.com/Archive/Rec/rec.radio.amateur.antenna/2005-12/msg00243.html
Sometimes the most simple concepts are missed. Actually it may not be simple reflection but scatter, although the scatter would seem to be when less than 100% cancellation occurs.

George E. Smith
February 14, 2011 4:51 pm

“”””” Bill Illis says:
February 14, 2011 at 4:11 pm
How does Nitrogen and Oxygen warm up and cool down in the atmosphere.
The way it is decribed by some here, the radiation theory assumption is that both of these gases are at absolute zero. “””””
Funny you should notice that too Bill. N2 and O2 at 288 or 300 K may not seem to be emitting any thermal radiation; but that is because we are used to feeling our “heat” at much shorter wavelengths, and much higher Temperatures.
Considering how difficult it is to observe and measure thermal radiation at 288 K; especially spectrally resolved; it is no wonder that ordinary humans are not even aware of its existence. We certainly can’t feel it on our skin.
But shift the wavelength down by ten, and raise the Temperature by ten, and the Total Radient emittance by 10,000 times, and the spectral peak emittance by a factor of 100,000 and humans finally can be made aware of it’s presence.
But I don’t see a whole lot of 100 Watt “heat lamps” turned up skywards, anywhere I’ve ever been !

John Whitman
February 14, 2011 4:51 pm

Mikael Cronholm,
You set up a wonderful discussion here. Thanks.
It is a main event.
John

David Ball
February 14, 2011 4:57 pm

Turns out I was right about you on the other thread, Mr. Mosher. You have shown your true colors. Why do you feel you have to trap anybody? You are not a nice man.
REPLY: David, you don’t know jack, I know Mosher personally, you don’t. You’re simply wrong on this point. – Anthony

Jim D
February 14, 2011 4:59 pm

Bill Illis, nitrogen and oxygen heat and cool by conduction and convection only, not by radiation.

George E. Smith
February 14, 2011 5:00 pm

“”””” JAE says:
February 14, 2011 at 1:58 pm
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.” “””””
So what is the magic of these particles that they know to only radiate downwards? Common sense would say that whatever in the atmosphere is emitting radiation of any kind, is doing so pretty much isotropically with no directional bias.
So whatever your particles are emitting downwards, they must be emitting a like amount upwards; whcih escapes to space. So whatever the originals ource of the energy that your particles are radaiting; be it direct incoming sunlight or surface emitted LWIR radiation, it seems that half of it si going to escape to space, and not reach the ground. Particularly when the sun is the source of that energy your particles are radiating downwards; that is an amount of sunlight that will never reach the ground; so it will get less hot than if your particles did not intercept that solar raiation.
No matter how you try to skin the cat; anything in the atmosphere that absorbs any incoming solar energy or even widely scatters it, such as the blue skylight due to Raleigh scattering, must result in less solar energy reaching the surface of the earth (ocean) and getting stored in earth’s thermal sink.

Ian W
February 14, 2011 5:19 pm

izen says:
February 14, 2011 at 12:53 pm
Ian W says:
February 14, 2011 at 10:06 am
“As confirmed in response to my question at the top of this thread, the temperature of the cloud has no bearing on the amount of latent heat being released. The heat being released is linked to the amount of water changing state and the latent heat of that state change.
The missing heat is almost certainly in the misunderstood and underestimated hydrologic cycle that is ignored in favor of a simple Stefan-Boltzmann radiation equation.”
The ‘Science of Doom’ site linked to from here (top right of page) has a good discussion of some of these issues.
The amount of energy transported by latent heat changes is complex, there is one simple way to quantify it however. The total global yearly rainfall is a direct measure of how much energy was moved by evaporation and condensation in the hydrological cycle.
But the first LoT kicks in. The energy is still around. The only way it can get of the planet is by radiating IR photons into space. All the hydrological cycle can do is move the location of that emission around. That evens out the temperature so that water does not boil at the equator and can melt at the poles, but the same amount of energy has to leave the planet. Otherwise it will warm until the S-B T^4 relationship increase the emissions enough to compensate.

So when the heat is released at 30,000 feet (*) in the atmosphere as liquid water turns to ice, well past the dense part of the atmosphere it does not radiate to space?
How does it know that it can only radiate down? 😉
It certainly appears that significant IR is being emitted from these clouds and weather systems independent of their temperature
http://www.ssd.noaa.gov/goes/east/natl/flash-rb.html
(*) The updrafts in the ITCZ as aircraft have found out to their cost can be in the order of 100KnH or more and liquid water can get that high before freezing – hence my quote of the Nortwest Airlines Airbus.

cba
February 14, 2011 5:22 pm

Bill Illis
N2 and O2 are at the same temperature in a small parcel of air, just like the co2. The co2 absorbs some energy and the average time it takes to radiate it away is more than enough time for it to be transferred away by collisions, most likely n2 followed by o2. just because a co2 molecule excited by a photon is likely to have its energy reduced by a collision, so too is the likelihood that a co2 molecule will be excited by a collision and capable of emitting a photon or capable of being ‘defused’ by yet another collision. net result is that a certain fraction of these molecules will emit a photon of a particular energy (wavelength) based upon the temperature and upon the proclivity of the co2 molecule to absorb or emit that energy. absorption doesn’t really depend much on temperature but the emission is highly dependent. The blackbody curve for a give temperature is actually a portrayal of the fraction of molecules at particular energies that are capable of emitting photons. The bb curve must be a solid or liquid – or an optically thick enough gas to be thick at all wavelengths. For the sun to essentially emit a 5800k BB curve from the photosphere, one has many heavier elements present that are ionized and so are capable of emitting a continuum rather than merely a spectrum.

Mikael Cronholm
February 14, 2011 5:24 pm

Myrrh, it may depend on what definition you use for heat, but the one I stick to is that heat is the total kinetic energy of the particles in a substance (mass times velocity squared over two). Then infrared is not heat, but it is caused by, and can cause, heat. Emission creates radiation energy by converting heat energy to electromagnetic waves (or photons, if you prefer) and absorption converts in the opposite way.
Thermal radiation, if you define it as radiation that can heat or cool a surface by exchange of radiation, will be wavelengths from somewhere in the UV throughout visible and IR. Most of the heat the earth receives from the sun is in fact in the visible band. In the shorter wavelengths, X-ray and gamma, most objects will transmit most of it, and hence no transfer of energy. Longer wavelengths, microwave, and radio wave, do not significantly heat things, I am not sure why, but if they did heat us we would be cooking with all the radio waves around. Microwaves can only heat by directly agitating water molecules, because they are dipoles, so it is a different process than we normally think of as heat transfer by radiation. But it is a matter of definition, largely.
Normally, the shorter the wavelength the better the penetration. If you consider human skin, UV, visible and IR penetrate in that order, UV the most, visible a little bit, and IR hardly at all.
The reason a remote control does not feel hot is merely a question of magnitude. A couple of AA batteries don’t have a lot of energy to give off. By the way, you can use a normal CCD camera to check if your remote works, a simple one in a cell phone will do fine. CCD’s work up to around 1 micrometer.

enough
February 14, 2011 5:28 pm

One question I am looking for a good answer.
At sea level, when an ir photon is emitted from earth in a upward direction, what is the distribution of its path length before it excites(is absorbed) by a CO2 molocule and is thermalized into the atmosphere. Believe I know the answer, but looking for indepenant confirmation.

Harold Pierce Jr
February 14, 2011 5:33 pm

davaidmhoffer says:
CO2 is reasonably well mixed throughout the troposphere,
————-
This is not true and is just flat out wrong. In real air, there is no unifrom distribution of mass of the atmosphere in space and time as shown by weather maps. High pressure cells have more regional mass and dry air than do low pressure cells with moist air. And these are constantly moving sometimes quite rapidly. Humidity lowers the density of dry air by a much as 5%.
Tropical air at ca 30 deg C and with 100% humidity has 80% of the mass per unit volume than does dry cold air at STP ( 0 deg C and 1 atm pressure).
Comprised of nitrogen, oxygen, the inert gases, which are the fixed gases, and CO2, purified dry air (PDA) at STP has presently 390 ml, 17.4 millimoles, 766 mg or 0.000766 kg of CO2 per cubic meter and has a density of 1.2929 kg per cu per meter. PDA does not occur in the earth’s atmosphere. The composition of PDA (i.e, rel amounts of the fixed gases and CO2) is fairly uniform thru out the earth’s atmosphere and is independent of site, temperature, pressure, humidity which includes water vapor and clouds except for minor local variation in particular with prespect to CO2.
If PDA is cooled to -53 deg, the amount of CO2 is 21.6 mmole cubic meter and concentration is 390 ppmv. If PDA is heated to 45 deg C, the is concentration is still 390 ppmv but there is only 12.5 mmoles of CO2 per cubic meter.
GCM calculations generally use the concentration of CO2 in ppmv which is the incorrect metric and thus are fatally flawed. The correct metric is mass (or millimoles) of CO2 per unit volume.
The water droplets of clouds contain CO2 which can be released if they dissipate or transport CO2 to the surface if they turn into raindrops. How much CO2 is sequestered in the clouds? Probably a lot.

cba
February 14, 2011 5:57 pm

enough says:
February 14, 2011 at 5:28 pm
One question I am looking for a good answer.
no real good answer I’m afraid. Depending on the wavelength of the photon, it might travel from surface to space without a hint of a capture OR it might not make it 2 centimeters without a sure thing capture. Most of the critical wavelength bands fall somewhere in between. at lower altitudes (higher pressures) the individual lines are spread out essentially forming bands. At higher altitudes and lower pressures, the lines become much sharper taller and narrower. Molecules here are less likely to absorb or emit a photon whose wavelength is further from the peak.

cba
February 14, 2011 6:01 pm

mikael,
that’s a very restricted definition of heat energy. energy can add velocity to a molecule or it can increase the internal energy state. IR is merely an electromagnetic form of energy. The same goes for radio waves, light, IR, uv, xrays and gamma rays.

eadler
February 14, 2011 6:03 pm

J. Bob says:
February 14, 2011 at 8:47 am
eadler says
“If the mean error is know, the readings can be adjusted. If the mean error remains constant, than the temperature anomaly, which is what is being sought will not suffer in accuracy.”
And those “if’s” can be significant. Much of the mean error “adjustment” ability depends on the sensor, electronics, environment, calibration protocols, and how well these protocols are ACCUALLY followed.

Calibration error doesn’t affect the temperature anomaly if the error remains constant. In the case of equipment changes, so that the temperature trend becomes discontinuous, the change if detected is corrected for by the use of adjacent location data that is consistent. These adjustments are normallyl done by using computer programs. Calibration of the equipment is not really necessary. The new data base, which is the subject of this thread will be corrected for equipment discontinuities in the same way as the previous thermometer data bases have done it.
This is getting off topic of this thread. There are other threads on this web site that deal with corrections of the temperature record, including the new data base being developed by Muller.

enough
February 14, 2011 6:37 pm

One question I am looking for a good answer.
no real good answer I’m afraid. Depending on the wavelength of the photon, it
Sorry, question was not to the point, If an IR photon in the co2 band is emitted at sea level, what is its path length before being absorbed by a co2 molecule.

Bill Illis
February 14, 2011 6:57 pm

Jim D says:
February 14, 2011 at 4:59 pm
Bill Illis, nitrogen and oxygen heat and cool by conduction and convection only, not by radiation.
—————
Accordingly to your theory, these are meaningless concepts. They play no part in the greenhouse effect. Why do you bring them up now?

Mikael Cronholm
February 14, 2011 7:23 pm

@ cba. Yes, I know it is a restricted definition, which I kind of pointed out, implicitly. I just though a simplified answer would be good enough, since the main point I was making was that radiation in itself is normally not defined as heat. But there are two definitions of heat, differing only in the semantics.

cba
February 14, 2011 7:54 pm

mikael,
heat is energy in transit. loosely, radiation readily fits the definition. One too easily tends to forget that there is more to energy states than kinetic energy of the motion of individual molecules.

jae
February 14, 2011 7:59 pm

cba says:
“Bill Illis
N2 and O2 are at the same temperature in a small parcel of air, just like the co2. The co2 absorbs some energy and the average time it takes to radiate it away is more than enough time for it to be transferred away by collisions, most likely n2 followed by o2. just because a co2 molecule excited by a photon is likely to have its energy reduced by a collision, so too is the likelihood that a co2 molecule will be excited by a collision and capable of emitting a photon or capable of being ‘defused’ by yet another collision. net result is that a certain fraction of these molecules will emit a photon of a particular energy (wavelength) based upon the temperature and upon the proclivity of the co2 molecule to absorb or emit that energy. absorption doesn’t really depend much on temperature but the emission is highly dependent. The blackbody curve for a give temperature is actually a portrayal of the fraction of molecules at particular energies that are capable of emitting photons. The bb curve must be a solid or liquid – or an optically thick enough gas to be thick at all wavelengths. For the sun to essentially emit a 5800k BB curve from the photosphere, one has many heavier elements present that are ionized and so are capable of emitting a continuum rather than merely a spectrum.”
I think this is all correct.
But, I would like to posit another “world” for the warmistas. Suppose the atmosphere of Earth consisted of ONLY N2 and O2. Now, these gases could not cool by IR emissions, but they would warm by conduction from the surface. And they could not radiate to space. So would they continually warm for millions of years?? Would they melt the planet? WTF, folks?

Jim D
February 14, 2011 8:00 pm

Bill Illis, you are saying conduction and convection are meaningless when they explain the temperature profile of the atmosphere and a large part of the heat flux from the ground. They are vital to any complete theory. Some people have trouble with the concepts of conduction, convection and radiation all being important at the same time, I guess.

Mikael Cronholm
February 14, 2011 8:25 pm

@ cba. Yes, that is the other definition. Both are equally valid, but when I teach heat transfer and IR temperature measurement to engineers, I find the definition that implies that heat is contained in objects to be easier to work with. I accept both definitions and the systems of nomenclature that they create, I just prefer one because of simplicity. And following this discussion it seems to me that the definition I prefer for practical reasons also seems to be the one that people use here.
If I define heat as being the thermal energy flow I run into problems explaining heat capacity for example. The term “heat transfer” becomes a bit funny if “heat” in itself is a transfer, a bit like saying”free gift”.
http://en.wikipedia.org/wiki/Heat#Semantic_misconceptions

marky48
February 14, 2011 8:28 pm

Great experts for this place. An electrical engineer and a business/economics major. And these musings disprove global warming and the properties of CO2? Lol. Check any temperature charts lately? Any Mars missions planned from your garage? How about no Watts in the bulb?

February 14, 2011 8:48 pm

marky48,
And what is your expertise?
No one here is trying to disprove global warming. It is the continuation of the natural warming cycle from the LIA. And no one is arguing the properties of CO2. It is a trace gas that delays the emission of radiation to space.
The question is whether a minor trace gas controls the climate. If you believe you have solid evidence that it does, provide it.

Mikael Cronholm
February 14, 2011 9:03 pm

@ marky48. Great contribution! Very informative! Naah, not really…
When you find something interesting to post that questions or discusses the facts presented instead of arguing ad hominem and attempting to ridicule me, you can try again.
In absence of degrees on the subject, I have worked with and studied IR thermography, as it is applied in industry and research, for over 20 years, and I have been teaching it for 15 years or so, and written books about it. This discussion is on the fringes of my expertise, and I don’t claim to be a climate scientist or expert at all, but I am quite enjoying learning about it here. And if you look a little carefully you will find that I am on neither side in the debate. You have apparently taken sides though, based on what, I don’t know. If you would attempt to enlighten me you will have to show a little more intelligence than you just did.

Oliver Ramsay
February 14, 2011 9:25 pm

jae says:
February 14, 2011 at 7:59 pm ………………..
Suppose the atmosphere of Earth consisted of ONLY N2 and O2. Now, these gases could not cool by IR emissions, but they would warm by conduction from the surface. And they could not radiate to space. So would they continually warm for millions of years?? Would they melt the planet? WTF, folks?
———————-
jae, there was a good thread on this several months ago. Although I’d like to track it down for you, I think somebody will recall it better than I do and save me the trouble.
Obviously, the surface could still radiate and accept heat from the atmosphere through collisions. Convection would arise.

marky48
February 14, 2011 9:28 pm

“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.”
Try http://en.wikipedia.org/wiki/John_Tyndall. And http://en.wikipedia.org/wiki/Svante_Arrhenius. Late to the party as usual, sport.

marky48
February 14, 2011 9:32 pm

Then try these nuggets on for size. Only fools and not even oil companies dispute man’s contribution to the greenhouse effect and the observed effects of global warming. If you are smarter than NASA you haven’t shown it here. Get out or lose your reputation. If you value the one you claim to have, that is.

marky48
February 14, 2011 9:38 pm

“The question is whether a minor trace gas controls the climate. If you believe you have solid evidence that it does, provide it.”
See the above smokestack. Ever heard of exponents? It sounds like you’ve had too much of the emissions already though like the rest of these toasties.

Oliver Ramsay
February 14, 2011 9:49 pm

@ cba
That CO2 imparts warmth to N2 and O2 seems straight-forward, but it also seems at odds with the notion that it can heat the terrestrial surface at the same time by radiation.
If CO2 happily re-radiates (I know some don’t like that expression) the energy it absorbs, why is the surface not simply tossing it right back out when it receives it?
Sure, the minerals of the surface will briefly become hotter but then they will radiate more vigourously and a percentage of that radiation will be at frequencies incompatible with CO2 absorption, so it will be gone, gone.
What does “back-radiation” have to do with anything? Isn’t it like radiation within an iron bar that I heat with an oxy-acetylene flame? Overwhelmed?

Oliver Ramsay
February 14, 2011 10:10 pm

@marky48
Including your IQ as part of your screen-name puts you at a bit of a disadvantage.
Very few will take your incredibly astute comments seriously and the planet will slip a little further into catastrophe.
Yes, it’s somewhat your fault but don’t beat yourself up too much!

David Ball
February 14, 2011 10:24 pm

marky48, right on que with insults and wiki references no less, ……….. here we go again.

kuhnkat
February 14, 2011 10:32 pm

marky48.
Arhenius having the wrong bandwidths I guess doesn’t really matter. Believers KNOW they are right so it is OK to adjust things to fit their incorrect models based on 100 year old incorrect science.
HAHAHAHAHAHAHAHAHAHA

February 14, 2011 10:38 pm

@All…
Imagine you have 450 boxes on ground and each box can contain only one cat. Imagine you have only one box buoying in the air and it only can contain one cat. Now, imagine the Sun is sending 900 cats towards the ground, i.e. towards the place where the 450 boxes are placed. Remember, only one cat per box. From the 900 cats, only 450 cats hit on the ground, where the boxes are placed. 45 cats are “bounced” towards the outer space and 405 cats go into their respective boxes. There will be 45 empty boxes. But… ~365 cats get scaping from their correspondent boxes and they want to go back to the source of cats. They find that one box is obstructing their passage to the glory and one cat is “trapped”, momentaneously, into the box floating in the air. Question:
How many cats can be “absorbed” by the empty box that is floating in the air?
Let’s continue:
The cat inside the floating box tries to scape to the glory and jumps out from the box. Question:
Would the cat go back to the ground looking for another empty box? For answering this question, remember that there are not empty boxes on the ground because the Sun is sending cats continuously, therefore, the empty boxes on ground are occupied immediately by another cat incoming from the Sun.
Let’s continue: The cat finds another empty box floating few meters far away from the first box from which it had scaped and try to jump toward it. It fails and feels a strong cats current from the ground pushes it towards the outer space; additionaly, it finds that there is a very strong pressure exerted from the surface against which it cannot struggle. Questions:
Will the cat be re-absorbed by any of the occupied boxes on the ground?
Could it be that other kinds of containers do exist in the air, at shorter distances than the prior box where it was “trapped”?
Could the cat prevail over that strong current of cats and the powerful pressure, which push it towards the outer space, if it doesn’t bump into another empty container?
Keys:
Cats = IR quantum/waves.
Cats from the Sun = short wavelength IR quantum/waves.
Cats from the ground = long wavelength IR quantum/waves.
Cats jumping from the boxes floating in the air = very low energy density long wavelength IR quantum/waves.
Boxes on ground = limited and limiting configurations the energy absorbed by the ground can adopt.
Boxes floating in the air = limited and limiting configurations the energy absorbed by CO2 molecules can adopt.
Other empty containers floating in the air = limited and limiting configurations that the energy can adopt without minimizing the free energy.
Cats’ current = short and long wavelength IR quantum/waves outgoing from the emitters.
Pressure = radiation pressure.
Conclusion: Through spontaneous processes of heat transfer, it is impossible for a colder system to do work on another warmer system. In other words, it is impossible for a colder system to transfer energy towards a warmer system.
😐

David Ball
February 14, 2011 10:40 pm

” Observed effects of global warming”. You mean like all the hurricanes ? Maybe you mean the cold and snow. Oh, and flooding, cause thats never happened before. It’s all covered by the theory as well. Global warming, it’s whats for dinner, ……

David Ball
February 14, 2011 10:49 pm

I can’t resist. Those were some mighty big nuggets you had there marky48.

Smoking Frog
February 14, 2011 11:13 pm

[Smoking Frog]The Inclusion-Exclusion Principle requires that we know the overlap(s) to begin with, so it can’t give us the answer.
[Jim Masterson] But it does give us an answer that is almost the same:
.49 + .06 + .20 – (.49)*(.06) – (.06)*(.20) – (.49)*(.20) + (.49)*(.06)*(.20) = 0.61648.

It can’t do it by itself, and you’re not using it by itself. You’re assuming random overlap without knowing it. (It didn’t occur to me that a random-overlap calculation could be that simple, so I thank you for the insight.)
Let the cloud layers be called A, B, C, corresponding to the 49%, 6%, 20%. Assume that the patches that make up each layer are infinitesimally small and randomly distributed in the spherical shell in which the layer resides. Then the most likely (A,B) overlap is (0.49)(0.06), the most likely (A,C) overlap is (0.49)(0.20), and so forth for (B,C) and (A,B,C).
Naturally, the patches are not infinitesimally small, but I doubt that this changes the result by much. Non-random distribution could change it by much, but it would take a meteorologist to deal with that problem.
Anyway, your answer is better than mine. I can see that, generally speaking, the two methods will give different answers, but I don’t see how to explain that the random-overlap answer is not the average of the two extremes.

Konrad
February 14, 2011 11:47 pm

I have read through this thread and have not yet seen a reasonable answer to the question raised in the comment –
richard verney says:
February 13, 2011 at 4:11 pm

Matt
February 15, 2011 12:26 am

What’s up with all the boxes of cats?
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 in visible light, and the atmosphere is shining on you in infrared. If you change the atmosphere so it shines more (e.g. adding more CO2 and other greenhouse gases) you will be warmer on the surface because there is more electromagnetic radiation coming down on you than otherwise. That’s it. More CO2 = more radiation on you.
All the complex logic might be important if you want to understand the details of radiative transfer (e.g. if you were grad student in physical oceanography or atmospheric science), but the previous description is good enough.
“I just say there are no scientific proof that increased CO2 emission causes climate change, or that it does not. And I am not on any side in the debate, for that very reason.”
Think for a bit. Is there scientific *proof* that Miami’s climate next winter is going to be warmer than Minneapolis’s?
Well, sure, because you can rely on basic physics. Because of the inclination of Earth’s axis, in the winter there is more electromagnetic radiation hitting Miami than Minneapolis, and this will be the case next winter too. You can handwave about advection and chaos and weather and unpredictability and yadda blah blah blah, but the overall number one physics controlling the temperature is the amount of EM radiation. After all, what else is there to make you above absolute zero? (only a tiny amount from radioactive decay of Uranium in the Earth’s center, and this can be measured and it is miniscule)
More E&M radiation = warmer climate.
And likewise, more CO2 in the atmosphere = more E&M radiation. This is not a theoretical prediction, this is a measured and observed *fact*.
With more CO2 in the atmosphere it is physically impossible for the climate *not* to change.
Next step is quantifying the amount, and then it gets complicated, but the people who have been doing this for 50 years are reasonably good at it, and you should believe them.

LazyTeenager
February 15, 2011 12:42 am

[snip]

LazyTeenager