Guest post by Ira Glickstein
The Atmospheric “greenhouse effect” has been analogized to a blanket that insulates the Sun-warmed Earth and slows the rate of heat transmission, thus increasing mean temperatures above what they would be absent “greenhouse gases” (GHGs). Perhaps a better analogy would be an electric blanket that, in addition to its insulating properties, also emits thermal radiation both down and up. A real greenhouse primarily restricts heat escape by preventing convection while the “greenhouse effect” heats the Earth because GHGs absorb outgoing radiative energy and re-emit some of it back towards Earth.
Many thanks to Dave Springer and Jim Folkerts who, in comments to my previous posting Atmospheric Windows, provided links to emission graphs and a textbook “A First Course in Atmospheric Radiation” by Grant Petty, Sundog Publishing Company.
Description of graphic (from bottom to top):
Earth Surface: Warmed by shortwave (~1/2μ) radiation from the Sun, the surface emits upward radiation in the ~7μ, ~10μ, and ~15μ regions of the longwave band. This radiation approximates a smooth “blackbody” curve that peaks at the wavelength corresponding to the surface temperature.
Bottom of the Atmosphere: On its way out to Space, the radiation encounters the Atmosphere, in particular the GHGs, which absorb and re-emit radiation in the ~7μ and ~15μ regions in all directions. Most of the ~10μ radiation is allowed to pass through.
The lower violet/purple curve (adapted from figure 8.1 in Petty and based on measurements from the Tropical Pacific looking UP) indicates how the bottom of the Atmosphere re-emits selected portions back down towards the surface of the Earth. The dashed line represents a “blackbody” curve characteristic of 300ºK (equivalent to 27ºC or 80ºF). Note how the ~7μ and ~15μ regions approximate that curve, while much of the ~10μ region is not re-emitted downward.
“Greenhouse Gases”: The reason for the shape of the downwelling radiation curve is clear when we look at the absorption spectra for the most important GHGs: H2O, H2O, H2O, … H2O, and CO2. (I’ve included multiple H2O’s because water vapor, particularly in the tropical latitudes, is many times more prevalent than carbon dioxide.)
Note that H2O absorbs at up to 100% in the ~7μ region. H2O also absorbs strongly in the ~15μ region, particularly above 20μ, where it reaches 100%. CO2 absorbs at up to 100% in the ~15μ region.
Neither H2O nor CO2 absorb strongly in the ~10μ region.
Since gases tend to re-emit most strongly at the same wavelength region where they absorb, the ~7μ and ~15μ are well-represented, while the ~10μ region is weaker.
Top of the Atmosphere: The upper violet/purple curve (adapted from figure 6.6 in Petty and based on satellite measurements from the Tropical Pacific looking DOWN) indicates how the top of the Atmosphere passes certain portions of radiation from the surface of the Earth out to Space and re-emits selected portions up towards Space. The dashed line represents a “blackbody” curve characteristic of 300ºK. Note that much of the ~10μ region approximates a 295ºK curve while the ~7μ region approximates a cooler 260ºK curve. The ~15μ region is more complicated. Part of it, from about 17μ and up approximates a 260ºK or 270ºK curve, but the region from about 14μ to 17μ has had quite a big bite taken out of it. Note how this bite corresponds roughly with the CO2 absorption spectrum.
What Does This All Mean in Plain Language?
Well, if a piece of blueberry pie has gone missing, and little Johnny has blueberry juice dripping from his mouth and chin, and that is pretty good circumstantial evidence of who took it.
Clearly, the GHGs in the Atmosphere are responsible. H2O has taken its toll in the ~7μ and ~15μ regions, while CO2 has taken its bite in its special part of the ~15μ region. Radiation in the ~10μ region has taken a pretty-much free pass through the Atmosphere.
The top of the Atmosphere curve is mostly due to the lapse rate, where higher levels of the Atmosphere tend to be cooler. The ~10μ region is warmer because it is a view of the surface radiation of the Earth through an almost transparent window. The ~7μ and 15μ regions are cooler because they are radiated from closer to the top of the Atmosphere. The CO2 bite portion of the curve is still cooler because CO2 tends to be better represented at higher altitudes than H2O which is more prevalent towards the bottom.
That is a good explanation, as far as it goes. However, it seems there is something else going on. The ~7μ and ~15μ radiation emitted from the bottom of the Atmosphere is absorbed by the Earth, further warming it, and the Earth, approximating a “blackbody”, re-emits them at a variety of wavelengths, including ~10μ. This additional ~10μ radiation gets a nearly free pass through the Atmosphere and heads out towards Space, which explains why it is better represented in the top of the Atmosphere curve. In addition, some of the radiation due to collisions of energized H2O and CO2 molecules with each other and the N2 (nitrogen), O2 (oxygen) and trace gases, may produce radiation in the ~10μ region which similarly makes its way out to Space without being re-absorbed.
There is less ~15μ radiation emitted from the top of the Atmosphere than entered it from the bottom because some of the ~15μ radiation is transformed into ~10μ radiation during the process of absorption and re-emission by GHGs in the atmosphere and longwave radiation absorbed and re-emitted by the surface of the Earth.
Source Material
My graphic is adapted from two curves from Petty. For clearer presentation, I smoothed them and flipped them horizontally, so wavelength would increase from left to right, as in the diagrams in my previous topics in this series. (Physical Analogy and Atmospheric Windows.)
Here they are in their original form, where the inverse of wavelength (called “wavenumber”) increases from left to right.
Source for the upper section of my graphic.
Top of the Atmosphere from Satellite Over Tropical Pacific.
[Caption from Petty: Fig. 6.6: Example of an actual infrared emission spectrum observed by the Nimbus 4 satellite over a point in the tropical Pacific Ocean. Dashed curves represent blackbody radiances at the indicated temperatures in Kelvin. (IRIS data courtesy of the Goddard EOS Distributed Active Archive Center (DAAC) and instrument team leader Dr. Rudolf A. Hanel.)]
Source for the lower section of my graphic.
Bottom of the Atmosphere from Surface of Tropical Pacific (and, lower curve, from Alaska).
[Caption from Petty: Fig. 8.1 Two examples of measured atmospheric emission spectra as seen from ground level looking up. Planck function curves corresponding to the approximate surface temperature in each case are superimposed (dashed lines). (Data courtesy of Robert Knutson, Space Science and Engineering Center, University of Wisconsin-Madison.)]
The figures originally cited by Dave Springer and Tim Folkerts are based on measurements taken in the Arctic, where there is far less water vapor in the Atmosphere.
[Fig. 8.2 from Petty] (a) Top of the Atmosphere from 20km and (b) Bottom of the Atmosphere from surface in the Arctic. Note that this is similar to the Tropical Pacific, at temperatures that are about 30ºK to 40ºK cooler. The CO2 bite is more well-defined. Also, the bite in the 9.5μ to 10μ area is more apparent. That bite is due to O2 and O3 absorption spectra.
Concluding Comments
This and my previous two postings in this series Physical Analogy and Atmospheric Windows address ONLY the radiative exchange of energy. Other aspects that control the temperature range at the surface of the Earth are at least as important and they include convection (winds, storms, etc.) and precipitation (clouds, rain, snow, etc.) that transfer a great deal of energy from the surface to the higher levels of the Atmosphere.
For those who may have missed my previous posting, here is my Sunlight Energy In = Thermal Energy Out animated graphic that depicts the Atmospheric “greenhouse effect” process in a simlified form.
I plan to do a subsequent posting that looks into the violet and blue boxes in the above graphic and provides insight into the process the photons and molecules go through.
I am sure WUWT readers will find issues with my Emissions Spectra description and graphics. I encourage each of you to make comments, all of which I will read, and some to which I will respond, most likely learning a great deal from you in the process. However, please consider that the main point of this posting, like the previous ones in this series, is to give insight to those WUWT readers, who, like Einstein (and me :^) need a graphic visual before they understand and really accept any mathematical abstraction.
UncertaintyRunAmok says on March 15, 2011 at 11:51 am:
“I’m just curious about something. Do any of the experts here happen to know the average global (direct plus diffuse) clear-sky solat irradiance measured at the surface, or the average clear-sky measured solar reflected? I only ask because it seems to have a direct bearing on the “global energy budget” diagram.”
The short answer to your question is; no, – it is all guess-work. The so-called “Solar Constant” is simply divided by 4 as ¼ is all that is left when you take the formula for working out the area of a circular disc and subtract it from the formula for the area of a sphere.
O H Dahlsveen says:
March 15, 2011 at 3:33 pm
The answer to your questions is contained in the answer to my previous question. However, I doubt that any of the experts posting on this thread can, or will, answer the question I posed. If it has not been answered by tomorrow, perhaps I will have time to provide the answer myself.
Sorry, O H, our replies must have crossed in moderation. However, you must not have understood the question, the values I am referring to ARE known.
UncertaintyRunAmok says on March 15, 2011 at 11:51 am:
“I’m just curious about something. Do any of the experts here happen to know the average global (direct plus diffuse) clear-sky solat irradiance measured at the surface, or the average clear-sky measured solar reflected? I only ask because it seems to have a direct bearing on the “global energy budget” diagram.”>>>
You can’t calculate a global average unless you have complete global data. Which we don’t. Polar regions in particular.
Mr Hoffman,
You are, of course, mistaken. I have the 1991 edition of the CRC Handbook of Chemistry and Physics, and it contains a table of total global clear-sky solar irradiance at the surface, and I believe it was first published at about the same time as the 1976 US Standard Atmosphere. It gives an estimated uncertainty of plus or minus ~12.5W, but the average is only 6W different from current measurements. Clouds are the thing which most interferes with the measurements today (satellites, you know – they can’t “see” through them). I didn’t ask the question for my own benefit. But I can’t continue tonight, I have to work very early.
Later.
A little update to my comment here http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/#comment-621292 :
The simple model that I described is actually called a random-walk with absorbing boundary conditions. Here is a link to one of the only basic discussions that I have found on this (as most modern work has moved on to more complicated scenarios): http://banach.millersville.edu/~bob/book/Brownian/main.pdf The question that I basically addressed is posed on p. 20; on p. 34, the theorem there shows that for the simple model that I have described implementing in that post, the probability of returning to 0 without having gone beyond N is (1 – 1/N) and the probability of getting to N without having returned to 0 is 1/N.
UncertaintyRunAmok
If the values you are referring to are known then please let me know what they are as all I can find are “estimates” and as there are also estimates around which recons only a few (less than 10) Watts per square meter can mean the difference between ice ages and interglacial eras I would be pleased to know what the real values are
George,
Not that is really matters here, but you got your relativity wrong when you said “Remember that the energy of a real particle travelling at a velocity (v) has an energy of mv^2 /2 ; NOT mv^2 .”
The energy of a particle is E = (gamma)mc^2
Doing a binomial expansion on this gives the infinite series
E = mc^2 + 1/2 mv^2 + 3/8 m v^4/c^2 + ….
The first term is the rest mass energy
The second term is the classical kinetic energy
The higher terms are relativistic corrections to the kinetic energy.
Joel Shore says:
March 12, 2011 at 6:28 am
“So, for example, if you model the atmospheric greenhouse effect by two shells ………. In fact, such a simplified model is easy to solve exactly and will have a temperature at the earth’s surface that is higher than the temperature in absence of the shells by a factor of the 4th root of 3. [In general, the model with N blackbody shells has a surface temperature that is higher by a factor of the 4th root of (N+1) from the no-greenhouse case of 0 shells.]”
Well, you are mixing things up. Yes, in this case the radiant flux = (sigma * A (T1^4 – Tn^4) )/n-1,
But the surface temperature will not change because of the shells. The outer blackbody shell n is the lowest in temperature and placing extra shells anywhere in between will neither change the temperature of the first nor the last shell, but just reduce Q.
T1 is fixed by the solar input and Tn by 3K space. All that the shells do is equilibrate with the radiation energy density. And with no shells at all the surface would still be T1.
Now lets apply above to the real thing; the radiating surface of the Sun the first surface, Earth second and then GHG shells. If your vision would be right, then greenhouse gasses would make the sun warmer and this would also go on to the core of the sun.
O H Dahlsveen says: March 15, 2011 at 3:33 pm
However the questions that bug me still remain:
1) Where do the 324 W/m² of “back radiation come from? – Or in your case
350 W/m²?
2) Why does radiation from GHGs (in the Energy Flow Chart) only flow in one direction. (towards the Earth)?
1) The molecules in the atmosphere are warm. All warm things emit EM radiation. It is pretty much that simple. Calculating the exact value or determining how much comes from CO2 vs H2O vapor vs H2O liquid (clouds) would not be trivial, but the principle of thermal radiation is undeniable.
2) That is more subtle, but the answer is that it really does emit all directions. The bottom layer of the atmosphere (which is close to the surface temperature) emits ~ 325 W/m^2 downward (which gets absorbed by the ground). This bottom layer also emits > 325 W/m^2 upward, almost all of which is absorbed by a slightly higher layer of the atmosphere.
This slightly higher layer is cooler than the bottom layer, so it emits less in each direction – perhaps 300 W/m^2 up and 300 W/m^2 down.
The layer above that is even cooler — emitting perhaps 275 W/m^2 up and 275 W/m^2 down.
Eventually the “top layer” emits ~ 200 W/m^2 upward (which escapes to outer space) and ~200 W/m^2 downward (which gets absorbed by the penultimate layer).
The intermediate layers are not shown in the simplified energy transfer diagrams – only the top layer (~200 W/m^2 escaping the atmosphere upward) and bottom layer (~325 W/m^2 escaping the atmosphere downward).
In reality there are not distinct “layers” of course, but the idea works. As with calculus, you could divide the “layers” as thin as you wanted to get a better approximation to the real situation.
UncertaintyRunAmok says:
March 15, 2011 at 6:07 pm
Mr Hoffman,
You are, of course, mistaken. I have the 1991 edition of the CRC Handbook of Chemistry and Physics, and it contains a table of total global clear-sky solar irradiance at the surface, and I believe it was first published at about the same time as the 1976 US Standard Atmosphere. It gives an estimated uncertainty of plus or minus ~12.5W>>>
Uhm… its Hoffer.
And uhm…you’ve got a book that “estimates” an error of +/- 12.5 W
Think, think, think…estimates the error. ESTIMATES the error. Sorry, been doing a lot of ranting tonight, don’t mean to tick you off.
They don’t give a standard deviation, do they?! They’ve estimated the error because a lot of those values are extrapolated. Further, of what value are measurments of +/- 12.5 degrees in a climate analysis where we are trying to correlate a fraction of a degree of warming to calculated CO2 effects of less than 2 watts?
Yes, honey, that’s six inches. Plus or minus a yard. errr meter. What ever.
I’ve been away from this thread for a while (grading Mid-Term Exams for my online grad course in System Engineering at U. Maryland) but it seems the discussion has really taken off. THANKS ALL!
I think the example of the incandescent light bulb is instructive. I think all agree that:
1) At a given input voltage, but with a clear bulb, the filament would be at temperature T1.
2) At a the same input voltage, but with a bulb that has an IR-reflective coating on the inside, the filament would be at temperature T2.
3) T2 is greater than T1.
4) If the electrical current in case (1) is equal to (or less than in) case (2), the input energy to the filament is the same in both cases (or less in case (2)).
5) The increase in temperature of the filament is therefore entirely due to the IR-reflective coating on the inside of the bulb. The coating on the inside of the bulb is at a temperature that is less than that of the filament (T1 or T2). Therefore, it is possible for radiation from a cooler surface to raise the temperature of a warmer surface.
6) The above process involves reflection because the energy is in the shortwave region (visible and near-IR*). It is therefore different from the absorption and re-emission of longwave energy from the Earth to the GHGs near the bottom of the Atmosphere and vice-versa. However, the energy aspects are similar. Just as the IR-reflective coating does not create any additional energy, neither do the GHGs, Nevertheless, just as the filament would be cooler absent the IR-reflective coating, the surface of the Earth would be cooler absent the GHGs.
If anyone objects to the above, please cite the statement number and detail your objection.
Discusssion: You could say that the IR-reflective coating “slows” the escape of energy from the filament, thus raising its temperature. But, it seems to me, while it is OK as a shorthand explanation, it is far from complete. My reason is that the filament produces radiation in a broad spectrum, from visual to near-IR (and probably some UV). The visual gets a nearly free pass throught the IR-reflective coating. The near-IR is mostly reflected back to the filament where it is absorbed, and re-emitted in a broad spectrum. Thus, in that transaction, some of the near-IR radiation energy is transformed into visual radiation energy.
Imagine an ideal case where the IR-reflective coating was perfect, allowing 100% of visual out and reflecting 100% of the near-IR. The reflected IR energy would bounce back and forth between the coating and the filament until all of it was transformed to visual.
Of course, in the case of the incandescent light bulb, all the energy comes from the electrical voltage and current input. No one claims that the IR-reflective coating contributed any energy to the system at all. All the IR-reflective coating does is reflect near-IR energy back to the filament.
In the case of the Atmospheric “greenhouse” effect, all the energy comes from the Sun in the form of shortwave light energy. No one claims that the GHGs contribute any energy to the system at all. All the GHGs do is absorb and re-emit longwave energy back to the Earth surface.
*When I write “near-IR” I mean ~1μ to ~4μ. “Shortwave” is ~0.1μ to ~4μ. “Longwave” is ~4μ to ~50μ.
Hans says: March 15, 2011 at 7:25 pm to Joel Shore
Well, you are mixing things up.
Actually, it is you, Hans, who are mixing things up. Joel was talking about CONCENTRIC shells around the earth with specific properties – no any old shells positioned any old place.
If you completely change the situation, then OF COURSE his conclusion will not apply! (PS even with your situation, I don’t agree with your conclusions!)
Joel Shore;
But, I had no bone to pick with you in this thread until you turned around and attacked me in regards to the comment that I made about Ray’s book. I was simply defending myself from this attack. Why you felt it necessary to attack me in this way, I don’t know.>>>
That wasn’t an attack. That was kicking dirt in your face. You’ll be very clear about it when I attack that there is a difference. The why is because your comments have a singular thread that runs through them. You understand stuff, you’ve got a PhD, you teach these subjects at the university level… but actually explain something yourself? Never happens. You refer to a book, or a study, or explain that you teach this stuff, so you know what you’re talking about. Stop telling us how smart you are and what books we should read. George E Smith caught me on an issues, I was wrong, admitted it, and submitted my reasoning as to why I didn’t think it was significant. You start participating like that, and I’ll stop kicking dirt in your face.
Joel Shore;
If you are asking about where your “physicist” is wrong in the thread that I linked to, then I would say he is wrong in arguing that the surface temperature of the earth does not depend on the atmospheric composition. >>>
Boy. You REALLY didn’t get the joke. And no, I didn’t ask, I know why the physicist is wrong. He’s not a very good physicist. That’s why its so easy to get him to sell out and become a climatologist. That reminds me, last time I kicked dirt in your face you said something about getting out of physics and into climate research. How’s that going for you?
@ur momisugly Ira
I can’t find a way to agree with 3)
The radiation from the filament is determined by its temperature.
Reduce those emissions by reducing the temperature.
Temperature can be reduced by conduction/convection.
Or, reduction of input energy.
Temperature can be increased by increasing input energy.
If temperature is increased, radiation is increased exponentially. Stefan-Boltzmann.
If the much cooler filter is capable of heating the filament at all, then it is capable of heating it infinitely unless there is some negative feedback mechanism.
TE at the filament’s temperature won’t even stop it because, if a cold thing can heat a hot thing and a hot thing can heat a hot thing then surely a same temperature thing can heat a hot thing.
As far as I know, these bulbs are made for “cool” light or special effects. In a cursory look around lighting sites I didn’t see them in the “energy saving” category.
Joel Shore
What you appear to miss out in your layer model is the effects of thermalisation.
Lets start with CO2 the villain of the IPCC.
At atmospheric temperatures only around 4% are in active ready to emit 15um mode while the other 96% are ready to absorb.(Using MB statistics)
The plentiful 15um surface up IR is readily absorbed.
However the relaxation time length indicates that the chances of re-emitting are unlikely compared to loss by collision with N2 and O2 (99% of atmosphere).
This causes local heating (thermalisation)
Kirchoff’s Law does not hold strictly for this situation as significant quantities of thermal energy are passed on to non emitters.
Some of this energy can come back to CO2 by collision but emission of 15um as we get to higher altitudes becomes increasingly unlikely due to temperature drop.
A more likely radiative outlet path would be H2O which has several wavelengths >15um available.
This accounts for the large “bite” missing around 15um as shown in the Ian’s satellite “looking down” graphs above.
So the net result is slightly increased troposphere temperature and shifting the radiating spectrum to longer wavelengths.
>>
O H Dahlsveen says:
March 15, 2011 at 3:33 pm
I cannot see where else it can be subtracted from but from the 66 W/m² surface radiation as that is the only IR long-wave radiation that can possibly contain wave-lengths which GHGs cannot absorb. The term “atmospheric window” means simply “passage through GHGs for IR wave-lengths which cannot be stopped by GHGs” – That must therefore mean that the circuit set up by 324 W/m² from GHG back- radiation is closed in and cannot escape from the earth system.
<<
The 66 W/m² is not the surface radiation. The surface radiation is 390 W/m². The 66 W/m² is a result of subtracting 324 W/m² from 390 W/m².
>>
However the questions that bug me still remain:
1) Where do the 324 W/m² of “back radiation come from? – Or in your case
350 W/m²?
2) Why does radiation from GHGs (in the Energy Flow Chart) only flow in one direction. (towards the Earth)?
<<
I don’t presume to know what your training is, but an electrical engineer would know where the extra energy comes from–it’s stored in the feedback loop. This isn’t creating energy, it’s just reusing energy. Cut off the source (solar input), and this oscillator would quickly decay (exponentially) to zero. Electrical circuits do this feedback thing all the time. (Without bi-stable multivibrators or flip-flops, our modern-day computers wouldn’t be possible. Flip-flops are bi-stable oscillators, and they store energy in the feedback loops.) I could run you through a few cycles of the KT 97 model (that would be a long and boring post), and you would see where the energy builds up in the atmosphere during each cycle.
The total energy in the atmosphere is 519 W/m² (according to KT 97). It’s not a one-way flow to the surface. 324 W/m² radiates toward the surface, but 195 W/m² radiates out to space. If you model the atmosphere response (using multiple layers), then the lower layers radiate more energy than the upper layers do. This skews the resultant flow towards the surface. It’s nothing Earth-shattering. There’s no violation of the Second Law or the First Law.
Jim
Ira Glickstein, PhD says:
March 15, 2011 at 8:12 pm
5) The increase in temperature of the filament is therefore entirely due to the IR-reflective coating on the inside of the bulb. The coating on the inside of the bulb is at a temperature that is less than that of the filament (T1 or T2). Therefore, it is possible for radiation from a cooler surface to raise the temperature of a warmer surface.
Exactly Ira. George’s mechanism of reduced cooling can’t happen because the dichroic doesn’t effect the ‘cooling’ of the filament, the same photons leave in the presence of the dichroic as in its absence. What is critical is that the dichroic directs the IR photons back towards the filament where they are absorbed thereby increasing the filament temperature.
Bryan says:
March 16, 2011 at 2:17 am
Kirchoff’s Law does not hold strictly for this situation as significant quantities of thermal energy are passed on to non emitters.
Kirchoff’s Law certainly does apply it’s just that you don’t know what it says!
Emissivity=absorptivity still holds true when collisional deactivation occurs.
Some of this energy can come back to CO2 by collision but emission of 15um as we get to higher altitudes becomes increasingly unlikely due to temperature drop.
Wrong it becomes increasingly likely due to the lower collision rate.
A more likely radiative outlet path would be H2O which has several wavelengths >15um available.
This accounts for the large “bite” missing around 15um as shown in the Ian’s satellite “looking down” graphs above.
No that is classic CO2 spectrum, go to MODTRAN and you can reproduce it exactly but not using H2O.
Hans says:
I am not speculating on what the blackbody shell model would show. I am solving the model exactly (which is quite trivial to do). You are just speculating and it is easy to see that you speculation is incorrect because in radiation balance the system has to re-emit all the power it receives back into space and since only radiation from the outer shell goes back out into space, the temperature of this shell is what is set by the amount of radiation that is absorbed from the sun.
Bryan says:
No…It does not ignore such effects. Each shell radiates according to its temperature. What it does ignore is the spectral behavior of the radiation…because it is a blackbody (or, in the more general case, greybody) model. But, that is of course the whole point of simple models: To get a model that is simple enough to easily solve analytically and to give you good intuition of what is going on, you give up on including some of the details.
By contrast, when you want to actually make quantitative calculations to determine, for example, what the radiative forcing is from doubling CO2, then you have to solve the full-blown radiative-convective problem. That gives you actual quantative results…but at a price in terms of understanding and ease of solution.
That is why it is good to have a whole hierarchy of models…But it is meaningless to object that the simple model is too simple when the evidence from the more complicated models show that in fact the simple model is fine for showing the basic qualitative behavior. Sure, it’s simple…That’s the point.
Phil. says:
“What is critical is that the dichroic directs the IR photons back towards the filament where they are absorbed thereby increasing the filament temperature.”
——————————
Then it does it again and gets even hotter, then again and again.
It does it at 100 degrees, at 500 degrees, at 1,000 degrees. What stops the process?
davidmhoffer says:
(1) In this thread, I have in fact been explaining things quite a bit. I have talked about what a simple shell model of the greenhouse effect illustrates. I have explained patiently to O H Dahlsveen where he is confused and even presented a simple random walk model that he could try simulating (even with a coin if he doesn’t want to do it on the computer). Yes, sometimes when people just make a lot of incorrect statements about the state of the science, I suggest that they might want to read a book that discusses the state of the science before making these statements; Is that a crime?
(2) I think it is ironic that you peg my posts as “telling us how smart you are”. Really, which is the more humble and which is the more arrogant approach? Approach 1: Don’t bother to particularly familiarize yourself with a scientific field but argue that you know more than the scientists who have been working in the field for years, that they are all wrong, actively deceiving people for whatever motivation, or some combination of the two. Approach 2: Read the scientific literature and textbooks to try to familiarize oneself with the science and accept the fact that one can learn a lot from the scientists who have actually dedicated their careers to studying these issues. And, point out to others who seem to think that they are so smart that they can just figure everything out off the top of their head that they could also learn a lot by familiarizing themselves with the science.
You are spinning your joke so bizarrely that I don’t really know what to say. So, you want the readers to realize that the climatologist is correct and the physicist is wrong in his arguments? Do you really think that you have written this in a way that makes readers want to think that?
davidmhoffer: At the introduction of your “joke” column, you say:
Do you mean this part to be a joke too, because it is not correct? (Admittedly, whether some parts of what you say are correct depends on whether you consider the response implied by the Stefan-Boltzmann to be a feedback or the 0th-order effect and what you mean by the long term. But, your statement is quite clear at the level that you make the claim that climate models are actually making an unphysical assumption…and if you have any evidence to back that up, I certainly haven’t seen it.)
I like most of what Jim said @ur momisugly March 16, 2011 at 3:00 am, except
“The total energy in the atmosphere is 519 W/m² ”
519 W/m^2 is the RATE at which energy is ENTERING the atmosphere. More specifically, 519 W is the average rate at which energy is entering an average column of air 1 m^2 on a side stretching from the ground to the top of the atmosphere. (It is also the rate at which energy is leaving that column — with any long-term imbalance leading to a long-term warming or cooling of the air).
The total energy in the atmosphere would millions of joules per 1 m^2 column
Ira Glickstein, PhD on March 15, 2011 at 8:12 pm you say: “I think the example of the incandescent light bulb is instructive. I think all agree that:” —And then you make 6 statements (numbered 1 to 6).
I understand what you are saying but I feel I need a bit more ‘info’ before I can seriously comply with: “If anyone objects to the above, please cite the statement number and detail your objection.”
1) What do you mean by; “At a given input voltage”? – As far as I know the voltage in The US is 110 V, while in Europe it is 220 V. You of course are the electrical guy and I am the dumb mechanical one but I would have thought wattage would have been more appropriate.
2) Same voltage problem as above.
3) T2 is greater than T1. – Do you make that statement because T2 and T1 have been measured, or because you see it as a logical development? – In either case, at what point do T2 or/and T1 stop rising or is there no limit to ‘absoluteT’? Furthermore I feel it would help me to further understand if I get to know if the bulbs are gas filled or vacuous. If they are gas filled equilibrium will continuously be sought, due to conduction between the filament, the gas and the inside of the bulb whether IR- coated or not.