Visualizing the "Greenhouse Effect" – Emission Spectra

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.

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John Marshall
March 10, 2011 2:04 am

A very complicated and flawed article.
What are we actually measuring when we measure temperature? It is the measure of Kinetic energy of the atom. In the laboratory CO2 will adsorb IR radiation and increase in temperature. In other words the atoms increase their vibrations. In the atmosphere, where there are other gasses, this CO2 will transfer these vibrations to the other gasses by collision. This is heat conduction.
When gas gets warmer it expands, density falls and it rises, convection. In the troposphere heat is lost through convection not radiation. Radiation is not possible because as this air rises its temperature falls due to adiabatic expansion. Its temperature will fall below that of the surface, and the 2nd law of thermodynamics forbids heat flow from cold to hot ( this is heat flow by any means available) so this rising warm air, relative to the surrounding air though colder than the surface, cannot warm the surface.
We can measure what we might think is the LW IR as back radiation but we probably forget that as energy flows through the atmosphere it has an effect on that atmosphere which will reduce that incoming energy which is revealed as a frequency change ( its speed remains constant) so that measured LW IR is altered solar energy not re-radiated IR from CO2.
At least this is how I see it bearing in mind those old Laws of Thermodynamics.
It is also flawed to use the black body formula for the earth since it is not in equilibrium at any time due to changes to cloud cover.

March 10, 2011 2:10 am

I’m glad you included the original Petty plots, because I think they are the most informative. The Arctic ones are very helpful. It’s a good idea to include lots of black body curves, so you can see where the radiation is coming from.
You said “some of the ~15μ radiation is transformed into ~10μ radiation during the process of absorption and re-emission by GHGs in the atmosphere”. Well, yes, but not much. Kirchhoff’s Law comes into play. Just as 10μ is easily transmitted, very little is emitted. That’s why the 10μ looks like it is coming from the ground, rather than higher up.

Morris Minor
March 10, 2011 2:21 am

“…. Perhaps a better analogy would be an electric blanket that, in addition to its insulating properties, also emits thermal radiation both down and up…..”
I am not sure about this ‘radiation down’ or so called ‘back-radiation’.
According to Trenberth et al. this amounts to about 330 W/m2, nearly twice that of the energy reaching the Earths surface from the sun (184 W/m2). If this is true why isn’t this energy collected and used as an energy source (better than solar energy as no need for storage – this can be collected 24 hours /day).
The reason I suspect we can’t use this energy is because it doesn’t exist – heat will not flow from the cold atmosphere to the warm surface of the Earth.
I think the emission spectra shows the scatter of infra-red from the atmospheric gases that wont reach the collimated collector of the sensors due to its direction of travel?
Thoughts please!

richard verney
March 10, 2011 2:27 am

Ira
I will enjoy reading and considering your post. I have only quickly glanced at the first few paragraphs and my immediate brief observations are:
I consider your better analogy (ie., that of an electric blanket) to be a worse analogy. An electric blanket has a power source and consumes energy in order to provide heat. Where is this seperate power source in your model Earth?
It would also be useful if you would briefly explain how the Earth emits a smooth blackbody curve given that temperatures vary between about 325K (desserts) and 220 to 235K (Artic/Antartic). What are the emission frequencies at these varying temperatures.

rbateman
March 10, 2011 2:50 am

Morris Minor says:
March 10, 2011 at 2:21 am
Don’t forget that light has 2 behaviors:
As a particle and as a wave.
As a wave it is contructive going down (in sympathy with incoming solar radiation) but in going up it is destructive and helps to add to Earth’s albedo.
Gets complicated in a hurry, doesn’t it?
Take away the water vapor and the C02 in Earth’s atmosphere would probably freeze out at the poles (most notably the Antarctic).
Remind you of another planet?

Joel Heinrich
March 10, 2011 2:57 am

The graphics have one (unfortunately very common) error. They are plotted as a wavenumber distribution but labeled with wavelength. You CANNOT just transform the wavenumber into wavelength as they have different peaks. Much like the difference between wavelength and frequency: http://commons.wikimedia.org/wiki/File:PlanckDist_ny_lambda_en.png
The peak of a distribution in wavelength for 280K is at 10.5 µm.
“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.”

March 10, 2011 3:11 am

Better …
.

P. van der Meer
March 10, 2011 3:15 am

Ira Glickstein, why don’t you explain to your readers why the various curves for blackbody radiation in your article peak in the range of 17μm to 19μm when the http://spectralcalc.com/blackbody_calculator/blackbody.php site comes up with a peak wavelength of 9.659μm for 300K and 11.828μm for 245K. This is also confirmed by the Wikipedia graph (http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png), showing a peak at about 9μm for 310K.
I look forward to your explanation.

peter_ga
March 10, 2011 3:21 am

If one defines “greenhouse” as the systems capacity to absorb shortwave radiation being different to and greater than its capacity to emit long-wave radiation, such that a net increase in temperature over a pure black-body temperature is needed for thermal equilibrium, then an important aspect is the ability of the oceans to absorb shortwave to a much greater depth than the depth that emits longwave.
This means that the temperature of the oceans would generally be greater than blackbody temperature, that they will tend to heat the atmosphere above them through various mechanisms, and evaporate water into the atmosphere as well. Earth is warmer because of its oceans, through an indirect greenhouse mechanism, that has nothing to do with co2.

Vince Causey
March 10, 2011 3:25 am

John Marshall,
“Its temperature will fall below that of the surface, and the 2nd law of thermodynamics forbids heat flow from cold to hot ( this is heat flow by any means available) so this rising warm air, relative to the surrounding air though colder than the surface, cannot warm the surface.”
The second law says no such thing. The misconception that the second law says that heat cannot flow from a cooler to a warmer body is the old nineteenth century understanding which dealt with conducting bodies. It was then rewritten in terms of entropy – entropy must always increase in a closed system.
This is what the second law actually says:
“The second law of thermodynamics is an expression of the tendency that over time, differences in temperature, pressure, and chemical potential equilibrate in an isolated physical system (Wikipedia).”
The key point to note is that over time, differences in temperature will equilibriate. The idea of greenhouse gases radiating energy back to the Earth’s surface does not contradict this description as long as the temperatures will equilibriate over time.
The second law allows energy from a cooler body to radiate to a warmer body (in fact, allow is the wrong word – all bodies above absolute zero must radiate energy) because it knows that the warmer body must be radiating even more energy to the cooler body and that their temperatures would most definately equilibriate over time.

Scarlet Pumpernickel
March 10, 2011 3:30 am

So what concentration of CO2 saturates these wavelengths?
There is already proof from Venus that the Greenhouse effect is not exponential http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html

James
March 10, 2011 3:31 am

The reason I suspect we can’t use this energy is because it doesn’t exist – heat will not flow from the cold atmosphere to the warm surface of the Earth.

Yes it will. When the energy is emitted it is in a random direction and has no knowledge of it is heading towards a warmer or colder body.
Heat will flow from a cold to hot body but at the same time more energy is going the other way. The flow energy from the atmosphere to the surface does exist.

Bryan
March 10, 2011 3:32 am

Ira Glickstein
You seem to think that an ordinary blanket is unlike an electric blanket in effect to radiation, perhaps you think it does not radiate?
For the record an ordinary blanket is a better radiator than for example the atmosphere.
At night the passive atmosphere is only capable of insulating the hotter Earth surface to some extent.
That is it reduces the heat loss compared to no atmosphere.
During daylight the DLR radiation is mainly redirected and degraded radiation of solar origin that has not reached the surface directly.
There will of course be some DLR which has an energy sourced by the warmer Earth surface.
I think you would do well to read some Physics books that deal with thermodynamics and heat transfer before making elementary errors.

richard verney
March 10, 2011 3:35 am

Ira
Further to my last comment, your article is interesting and I am still pondering upon its implications. Whilst ultimately at the very top of the atmospher all heat is radiated into space, I consider that below that boundary other forms of heat transfer are far more significant that radiation.
As I have commented numerous times, I do not consider the Earth to be a blackbody and the assumption that it behaves in this simplistic way distorts what is truely happening. The Earth is never at equilibrium. There are constant changes in albedo due to changes in cloud cover, ice, snow, vegetation., seasons etc These variations change the emission properties of the Earth. The oceans act as huge heat sinks such that diurnal changes over oceans is very different to changes in diurnal temperatures over land. There are peaks and valleys and all of this means that the Earth does not simplistically behave as a uniform blackbody.
You state: “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.” I accept that the curves set out in the “Bottom of the Atmosphere from Surface of Tropical Pacific (and, lower curve, from Alaska) tend to follow the 300K and 245k blackbody curve save for dips in the 8 to 14μ wavelengths, the effects of which you are seeking to highlight.
However, the “Top of the Atmosphere from Satellite Over Tropical Pacific” data is very different. This does not follow the 300k or 290k curve. Up to 7μ it seems to follow a 210/220k curve Between 7 and 8μ it takes a path crossing the 220 to 250K curve. Between 8 and 9μ it jumps up to about the 300K curve. Between 18 to 25μ it falls through the 270 down to the 250K curve. My view of this plot, ignoring the dips around the 9.5μ and the 13 to 18μ range is that it does not exhibit a smooth blackbody curve at any part through the 6 to 25μ wavelengths.
What is the explanation and how in the light of that explanation can it vbe asserted that the Earth (even in that snapshot in time) is exhibiting a smooth blackbody emission spectra?
I look forward to hearing your commenst so that I may better understand your article.

James
March 10, 2011 3:36 am

rbateman says:
As a wave it is contructive going down (in sympathy with incoming solar radiation) but in going up it is destructive and helps to add to Earth’s albedo.

The waves need to be in phase to interfere. The radiation from the surface and the atmosphere is at a completely different frequency and wavelength so doesn’t interfere with the solar radiation.

Steve in SC
March 10, 2011 3:45 am

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.

Your other modes of heat transfer are SIGNIFICANT and can not be dismissed.
At altitude radiation is indeed the mode of heat leaving the planetary system.
As altitude gets lower radiation becomes less important and transports less and less of the total heat.

cal
March 10, 2011 3:57 am

Thanks Ira. At last a description that I can pretty well totally agree with. My only caveat is the one Nick Stokes mentioned; that very little energy can be converted into “10 micron ” radiation in the atmosphere as I, and others, pointed out before on your previous thread.
As I also commented then, the facts with good explanations beat a good analogy. The graphs are excellent. It won’t stop some trying to deny them or interpret them in bizarre ways ( we have had a couple on this thread already) but the majority of WUWT readers will understand this contribution.
I also welcome the fact that you have taken the time to do this and persevere in trying to come to a “consensus”. I can’t believe I have just used that word!
I think that the AGWs have exploited the poor science displayed by some posters here to argue that the whole of the sceptic position is ill founded. The fact that the greenhouse gas effect exists is irrefuteable, as your graphs clearly show, and arguing to the contrary plays into the hands of those who wish to exploit the implications of increased CO2. While the effect of CO2 is clear the effect of increasing CO2 is far from obvious.
From the graphs it is clear that CO2 radiates from space from the top of the troposphere and re-radiates back to earth from relatively low altitudes. Paradoxically this is due to the fact that CO2 is indeed a VERY powerful greenhouse gas in the 14-18 micron window so that it will trap energy very close to the earth’s surface and will not radiate, unencumbered, into space until the air density is very low. Due to convection in the lower troposphere, high levels of humidity and clouds it is not at all obvious how increased CO2 and the subsequent reduced altitude of first absorption will effect downward radiation. Nor is it clear what will happen to radiation to space given that this is already taking place near the tropopause where the temperature no longer decreases as one goes higher.
These are the issues that need to be debated not the fundamental science. As far as I know there have been no actual measurements that confirm the signature of greenhouse forcing that would explain current warming on the basis of CO2 changes and this is a travesty!

March 10, 2011 4:02 am

Water vapour is the main so called “greenhouse gas” yet it does not produce a “greenhouse effect” as in a warming effect because it is a fluid gas. Though it may absorb IR, having done so it obviously expands and convects up through the atmosphere taking the IR up to cloud level. It does so because it is a highly mobile fluid gas.
CO2 does exactly the same thing by absorbing heat at the surface and transferring up to cloud level where it is emitted as IR. The cloud level, which begins at around 5000 meters is the beginning if the emission hight of the atmosphere. We know that CO2 begins emit the same level as water vapour because rain water is acidic. The acidity is caused by the dilution of CO2 into the water vapour as it form into clouds on emission, becoming carbonic acid.
Water vapour has a cooling effect on the surface. In exactly the same way so too does CO2.
How is this a “greenhouse effect” ???
The only place that the “greenhouse effect” is actually observed is in the computer models, not in the real world. The only reason that computer models predict a “greenhouse effect” is because they use a parameter known as “convective parameterisation”. In a computer model, if you over estimate convection you will not see a “greenhouse effect”. But if you underestimate this parameter, it will obviously and naturally result in a “greenhouse effect”.
Greenhouses are designed predominantly to inhibit convection as every body knows.
So in the models we have a “greenhouse effect” but in the atmosphere we do not.

Sjoerd
March 10, 2011 4:07 am

Ira,
Please leave out the “degree” when talking about Kelvin. It’s “degree Fahrenheit” and “degree Celsius”, but it’s “Kelvin” (without the “degree”). Same when abbreviated: It’s 270K, not 270ºK.
In your first article, you also abbreviated Argon as A, while it is Ar.
And it’s μm, not μ. The graphics you copied from other sources use it correctly, the graphics you made yourself are labelled wrongly.
Small mistakes like this distract from the article and undermine your credibility: If one can’t get simple details like proper usage of units correct, …

Gilbert K. Arnold
March 10, 2011 4:09 am

Bryan says:
March 10, 2011 at 3:32 am
Bryan: Did you even look at Ira’s brief CV at the bottom of his article. Does the phrase: “BS in Electrical Engineering” mean anything to you? Every EE program that I know of, includes Physics, Dynamics, and Thermodynamics as basic course requirements for graduation. Your insinuation that Dr. Gluckstein read some books that deal with heat transfer and Thermo is a bit of snark that is not needed.

RJ
March 10, 2011 4:17 am

Would a human being cook if he or she was enclosed in a container of CO2
In theory a percentage of the heat given off would return and increase the body temperature if the GHG theory is correct. Something surely is seriously wrong with the GHG theory.

tallbloke
March 10, 2011 4:26 am

Ira Glickstein
“the “greenhouse effect” heats the Earth because GHGs absorb outgoing radiative energy and re-emit some of it back towards Earth.”

No, No and a thousand times No.
Re-emitted radiation does not and cannot heat the Earth significantly, because downwelling IR does not and cannot penetrate the surface of the ocean beyond its own wavelength. The amount of energy from back radiation mixed into the ocean by wind and wave action is negligible and extra co2 therefore cannot account for the additional warming of the ocean bulk in the late C20th.
The greenhouse effect works by *SLOWING DOWN THE RATE THE EARTH COOLS AT*, by raising the altitude at which the atmosphere radiates to space . There is more than a semantic difference. Understanding it this way enables you to understand that it was reduced albedo 1979-1998 allowing more Solar energy to enter the oceans that caused the majority of the global warming at the end of the last millenium.
http://tallbloke.wordpress.com/2011/03/03/tallbloke-back-radiation-oceans-and-energy-exchange/

Bryan
March 10, 2011 4:38 am

Vince Causey
….”The second law says no such thing. The misconception that the second law says that heat cannot flow from a cooler to a warmer body is the old nineteenth century understanding which dealt with conducting bodies. It was then rewritten in terms of entropy – entropy must always”…….
Clausius was well aware of radiation when he formulated his second law.
He also invented the entropy concept.
He conducted tests with mirrors and lenses to confirm that for radiation as for conduction and convection;
…..”Heat flows from from a higher temperature surface to a lower temperature surface never the reverse.” one expression of the famous second law.
Vince Causey perhaps is confusing radiation with heat.
A colder surface will radiate to a warmer surface.
A colder surface cannot heat the warmer surface.
Heat carries the thermodynamic capacity to do work in the given situation.
Thus work can be done hot to cold.
Work cannot be done cold to hot.
Another way to think about it is the cold surface cannot increase the temperature of the hotter surface all it does is reduce somewhat the heat loss from the hotter surface.

tallbloke
March 10, 2011 4:46 am

Ira Glickstein
Perhaps a better analogy would be an electric blanket that, in addition to its insulating properties, also emits thermal radiation both down and up.

Yeah? and where is this ‘electric blanket’ getting its power from? Directly plugged into Trenberth’s mysterious reservoir of “missing heat” perhaps? Lol.
http://tallbloke.wordpress.com/2010/12/20/working-out-where-the-energy-goes-part-2-peter-berenyi/

Joe Lalonde
March 10, 2011 4:49 am

Ira,
As you have seen many people do not agree with your findings. Nor do I.
First, without planetary rotation, there is no convection as the planets own energy is the centrifugal force it generates at 1669.8Km/hr due to the vacuum of space. The atmosphere bends a great deal of light and solar energy with the suspended molecules in the atmosphere with the tilting of the planet to the sun. The hottest point of the sun is it’s equator that our planet drifts through due to proximity and size of the suns equator. Next very little consideration for the absorption and storage of heat that is then released at night.

DocMartyn
March 10, 2011 4:53 am

can you state how you generated the radiance curves for each temperature? Typically these type of figures have a fudge factor to correct for the black/white body fraction. I have never believed that one is allowed to do this from first principles.

Alberta Slim
March 10, 2011 5:00 am

Ira,
That is a terrific presentation.
Most interested people lose interest if an article gets too scientific and complicated.
You are fast becoming our new Isaac Asimov. [especially for GHG theory]
Thanks to you and Anthony for this.

Joe Born
March 10, 2011 5:06 am

Dr. Glickstein, I greatly appreciate your efforts, as well as those commenters who have made constructive criticisms.
In what I hope is one of the latter, I, too, will cast my vote against the electric-blanket analogy.

wayne
March 10, 2011 5:09 am

If someone could just toss up a link to a spectrum at 800 mbar in the tropics pointing sideways so we could just blow everyone’s mind quickly and hopefully be able to rebuild what’s left outside of this insidious AGWphysics this time. ☺
Fourier’s law of heat diffusion reigns when outside the windows frequencies in the bulk of the lower atmosphere, on both sides of the spectrum. You should not have brought in “back radiation” there unless you are only speaking of the lower 100 meters or so. The term back radiation has it’s place but it is only meaningful within the window wavelengths and mainly in relation to clouds, their BB radiation near 10 µm beams down the window to warm the ground on cloudy nights (and days, but not apparent). Within the thick of the atmosphere, all radiation outside the window can be properly viewed a fast, long-range conduction, that simple. Why? The atmosphere is all but totally black to these frequencies and most radiation only travels mere meters before re-absorption. That’s my current take, and still evolving though quickly now.
Ira, E for effort. I had high hopes as you began this whole series that you were going to be more open and all here could move a notch closer in truly understanding radiation’s relations within the atmosphere. You have many points exactly correct, and your animations can be deemed correct too, but only with the correct limitations and factors applied in the surrounding words. But beyond that, the whole story-line gets off of the track, IMO.

1DandyTroll
March 10, 2011 5:09 am

“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.”
This is why climate scientist ought not dabble in police work I think, for what if Charles was the one who nicked the piece and in passing Johnny smeared him blue with it?
But essentially the dissipation is 360˚ but mostly, naturally, concentrated at the nap of the earth, just like the stove which will burn if ones hand is too close but the farther away the hand gets the colder it gets, to the point of room temperature, the in between GHGs doing absolutely next to nothing to transfer the heat to that further away location due to it’s own dissipation rate even though the GHGs are pre-heated by being inside the house as well. But of course, eventually, the stove’ll raise the temperature in the kitchen, but only when the floor, ceiling, and the walls are more radiant, but only to a certain point (unless the stove doesn’t burst into flames the paint on the farthest wall ain’t gonna start cooking any time soon.) So whom is the construct of the thief that steals that heat? :p

Cementafriend
March 10, 2011 5:15 am

Ira, read the following http://climategate.nl/wp-content/uploads/2011/02/CO2_and_climate_v7.pdf which is the basis of a presentation by Dr (Ir) Noor Van Andel to KNMI in February and based on real measurements. Then, think again. -Incoming radiation to the sea surface from the sun controlled by cloud cover, evaporation at the sea surface, convection and pressure changes causing air movements, ToA radiation to space.
Dr Van Andel is a chemical engineer who has developed efficient heat exchangers for use with air see here http://www.xs4all.nl/~fiwihex/english/ Look at some of the links if you want some information on heat transfer theory.

March 10, 2011 5:16 am

tallbloke says:
March 10, 2011 at 4:26 am
“the hot-water bottle effect”
But what does heat the bottle, in the first place?
It is not “green-house effect”.
http://es.scribd.com/doc/28018819/Greenhouse-Niels-Bohr
Joe Lalonde says:
March 10, 2011 at 4:49 am
First, without planetary rotation, there is no convection
But….what does the Earth rotate…..an homopolar motor?
Then….heat by resistance?

william gray
March 10, 2011 5:16 am

How do we measure re-emittance of IR? I know absorbtion can be measured.

March 10, 2011 5:39 am

We live in an upside down world.
Water vapor makes up anywhere from 1-4% of the atmosphere. I can’t find the average because the percentage is so variable. CO2 is .04% of atmosphere. So water vapor is about 100x more plentiful than CO2 in atmosphere. Do anyone know the difference between the strength of CO2 vs WV as a greenhouse gas. Also, we know CO2 gets saturated in its’ greenhouse gas ability. Does this happen to WV? Then we know that man only creates 3% of the atmosphere CO2, well we don’t really know that, it’s just another big guess/estimate. How could we know that when we have forests densities completely changing throughout our world. Here in the US our forests are growing at a very fast rate, still recovering from the clearcuts of >100 yrs ago.
Then we have a possible feedback between CO2 and WV. Is it positive or is it negative? IPCC and modern “science” says CO2 always causes a positive increase in WV, causing amplification of greenhouse effect. But measurement of humidity and water vapor apparently doesn’t show this to be reality.
http://www.climate4you.com/GreenhouseGasses.htm, in fact WV is decreasing the last few years.
Then of course we have the oceans and their cylces, the sun and planetary alignment, undersea volcanoes etc etc etc. But THEY know everything. The SCIENCE is settled.
Yet after 11 years of increasing CO2 and amplification models, their data shows we are back where we started from in temperature with no trend whatsoever.
http://www.woodfortrees.org/plot/hadcrut3vgl/from:1999
We don’t know nothing about climate!

commieBob
March 10, 2011 5:41 am

1 – Ira has presented us with curves showing the measured values of upward and downward radiation.
2 – One of the graphs shows the energy leaving the planet. The only way energy can leave the planet is by radiation.
3 – The other graph shows radiant energy arriving at the surface in the far infra-red region of the spectrum. Most of this energy is radiated by the atmosphere. It is explained by the heating of the atmosphere by the sun and by heating of the atmosphere caused when the atmosphere is heated by radiation, conduction, evaporation, etc. from heat leaving the surface of the planet.
4 – Under some conditions, upward radiation probably explains most of the heat leaving the surface. The graph comparing Barrow with Nauru is pretty dramatic. There isn’t a lot of heat from the atmosphere beaming down on Barrow. 😉 There also isn’t a lot of heat moving up through the atmosphere in the form of thunder storms in Barrow in November (when the graph was created). Most of the heat leaves Barrow as radiation and little of it heats the atmosphere and returns as back radiation in the range between 13 um and 8 um. (It looks like much/most of it comes back around 15 um.)
On the other hand, the effect of thunderstorms in tropical regions is huge. Evaporation takes heat from the surface and deposits it higher in the atmosphere. From there it can be carried elsewhere or be radiated.
5 – For those who doubt that back radiation exists, consider this: Infra-red radiation is electromagnetic radiation. It is the same as radio waves and light. Here are two examples of radiation from a weaker source going toward a stronger one: a) If I stand with my back toward the sun, I will be able to see a flashlight being shone at me. Nothing about the sun’s radiation will prevent the radiation from the flashlight from reaching me. b) If I stand near a strong radio transmitter, I can still tune in weaker transmitters. Nothing about the stronger transmitter’s signal prevents the weaker signal from getting to me. The net energy flux will still be from the stronger source toward the weaker one. It is a net flux though, in other words, the net flux is the difference between the two signals.
6 – The discussion totally ignores heat removed from the tropics by ocean currents and moved toward the poles. The effect of that heat is also huge. Compare, if you will, the climate of Regina Saskatchewan and London England. In the winter, Regina is a lot colder than London in spite of the fact that London is somewhat further north. The average low January temperature in London is 2.4 deg. C. In Regina, the same figure is -21.6 deg. The difference is entirely explained by heat transported by the ocean.
7 – Ira’s article made stark the difference between a tropical humid atmosphere and an arctic dry one.
8 – Many commenters might spend more time carefully reading Ira’s article and less automatically gainsaying it on the basis of any preconceived notions they may have. It is what it is. It explains one aspect of climate science and doesn’t pretend to be encyclopedic. My only quibble is that it leaves the impression that most of the heat in the atmosphere is caused by radiation.

wayne
March 10, 2011 5:42 am

william gray says:
March 10, 2011 at 5:16 am
How do we measure re-emittance of IR? I know absorbtion can be measured.
——-
You should be able to get pretty close by taking a spectrum at say 800 mbar in the tropics aimed sideways at night. ☺ That’s the only way I can think of doing it. Wouldn’t that be close?

Domenic
March 10, 2011 5:46 am

Ira
I applaud your efforts, but you should start off with the simplest case first.
The Tropical western Pacific spectrum is not fixed. You have presented a simple snapshot of of it, I believe. Or is it an average? I don’t see any definition here of what that spectrum really means.
The tropics are EXTREMELY dynamic from a radiational heat transfer point of view. The absorption, transmission, and reflections, are constantly shifting due to water vapor. Everything changes from moment to moment, and it can be drastic. Just imagine lying on a beach on a sunny day, and a thick cloud passes overhead. The temperature you feel on your skin immediately plunges. That’s an immediate change in radiational heat transfer. Those curves you present are actually dancing around dramatically.
Even Barrow, Alaska is a rather complex situation.
I suggest trying to model and understand the interior Antarctic Polar region (Amundsen-Scott, Vostok, etc) first from a radiational heat transfer point of view. There is virtually no water in the atmosphere there. It is dominated by radiational heat transfer. And there are reliable surface temperature data, as well as documented of increased CO2 from 1957 to data.
You have to learn how to simply walk before you can dance the tango with its infinite complexities and beauty.

Fred Souder
March 10, 2011 5:53 am

Ira,
You should stop saying the greenhouse gases in the atmosphere warm the earth to this crowd. Too may engineers running around here. The atmosphere slows the rate at which the earth loses energy to deep space. Thus, the earth in the sun-earth-space system has a higher equilibrium temp. The “old” rules of thermodynamics still apply: a warm object cannot gain net thermal energy from a cold object.
Thank you for posting the good data though, especially of the IR toward and away from earth at the arctic. I am curious what the ground temperature was on the days that this data was collected, since at the arctic you often have an inverted lapse rate and the atmosphere could indeed “heat” the ground. I am curious if the ground temps are cold enough at the poles, and the lapse rate is positive instead of negative, so that the atmosphere could indeed “heat” the Earth, at least in these specific conditions and locations.

Alberta Slim
March 10, 2011 5:57 am

After reading all the comments I am, again, uncertain.
There are so many contradictory statements about frequency and wavelength, heat transfers, conduction, radiation etc., etc., …….
Has anyone presented a theory based on photons?
I know……… stupid question

Jeremy
March 10, 2011 5:59 am

Sorry but this article is filled with misinformation.
“Perhaps a better analogy would be an electric blanket that, in addition to its insulating properties, also emits thermal radiation both down and up.”
no a much better analogy is a blanket. There is no additional heating coming from the atmosphere.
They seem to give out degrees and doctorates in corn flakes boxes these days.

Alan McIntire
March 10, 2011 6:08 am

I too have been enjoying your articles. From prior posters, I thought of an additional analogy. Consider traffic flow over a section of road. The number of vehicles in the section being monitored represents joules. Joules are constantly entering the section at one end- radiation from the sun, and leaving at the other end-outgoing raiation cooling off the earth. Now, thanks to greenhouse gases, suppose there’s a traffic accident or a road construction crew closing one of the outgoing lanes The amount of traffic coming into the intersection-radiation from the sun, continues at the same rate, but the outflow due to fewer exit lanes results in a buildup of traffic- joules, in the system.
Note the NET effect is not a warming from the sky, which tends to confuse some posters here, but a constriction of outgoing radiation and a buildup of heat from the sun.

Larry Barnes
March 10, 2011 6:08 am

As in virtually all such articles focusing on longwave emissions/apsorption/etc, no account is given for the effect of the oceans which are the main determinant of global temperatures on this planet. The oceans absorb the vast majority of solar radiation on this planet and they do so to a considerable depth. Water is a relatively slow conductor of heat and therefore tends to hold it for a good length of time. Ocean currents and temperature layering cause the heat to move from the equator to the poles. Atmospheric temperatures in the tropics are nearly what one would expect from a black body analysis. The so called “green house effect” is really only visible at higher latitudes. This is due almost entirely to the redistribution of heat by the oceans. Atmospheric effects are quite small in comparison. This is not a new idea. until a few years ago oceanic scientists such as Doctor Robert Stevenson could have described all of this in considerable detail. Unfortunately, “climate scientists” with a poor understanding of the influence of the oceans have completely ignored it.

Bomber_the_Cat
March 10, 2011 6:10 am

Ira, there’s a problem here. as P. van der Meer says at 3:15 am. The blackbody curves are showing a peak at about 18 micron when they should be peaking at about 10 micron for a 300K blackbody.
In fact, if you refer to your previous post ‘Visualising the Greenhouse Effect – Atmospheric Windows’, you have the peak correct at 10 microns.
So, the current graphs don’t make sense – unless I am missing something? I have looked at the source material and that appears to be wrong also.
Has anyone got an explanation for this?

richard verney
March 10, 2011 6:29 am

tallbloke says: March 10, 2011 at 4:26 am
“….The greenhouse effect works by *SLOWING DOWN THE RATE THE EARTH COOLS AT*, by raising the altitude at which the atmosphere radiates to space . There is more than a semantic difference. Understanding it this way enables you to understand that it was reduced albedo 1979-1998 allowing more Solar energy to enter the oceans that caused the majority of the global warming at the end of the last millenium.
http://tallbloke.wordpress.com/2011/03/03/tallbloke-back-radiation-oceans-and-energy-exchange/
////////////////////////////////////////////////////////
Thanks for the link to your article. I was one of those who was arguing similar points with Willis and I have not seen your article before today. It is an interesting read.
I too respect Willis’ views but he was unable to even begin to explain the physics involved in how heat could be entrained by the oceans given the wavelength of DLR and its penatrative depth and thus become well mixed.
The only point he came up with (which did not answer the question) was that but for GHGs, the oceans would freeze and he referred to a link on scienceofdoom which suggested that without GHGs, the oceans would freeze within about 4 years. The underlying data and codes were not attached to the scienceofdoom article so that that assessment could not be verified. However, as I tried to point out to Willis, it is too simplistic looking at average temperatures and average conditions. The oceans are extremely complex and act as both a huge storage reservoir and a huge heat pump. For example, if one looks at the Baltic, in late summer, the sea temp is 16 to 18C and yet within about 4 months, it freezes over notwithsanding GHGs. There are many parts of the oceans (and inland lakes/seas) that freeze within months and this will tend to give the impression that when viewed on an average basis the seas would freeze within years. However, of course, there are great swathes of the Pacific, Indian Ocean, Atlantic etc receiving immense amounts of solar energy which energy is then pumped around by currents etc. It is almost certainly the case that it is this input and distribution that stops the majority of oceans from freezing over within seasons. Further, one may enquire rhetorically as to what causes the ice to melt/recede on these frozen seas/lakes? It is not an increase in GHCs but rather an increase in solar energy either directly and/or indirectly (via currents/circulation patterns).
I consider it probable that the vast majority of recent warming is due to natural variations and one of the key contenders for this being changes in cloud cover and changes in albedo allowing more solar energy to have penatrated the oceans.

March 10, 2011 6:34 am

I’ve been working with a combination of the reanalysis data and CO2 data to quantify the relative effects of atmospheric water and CO2 on OLR from the top of the atmosphere (click on my name). I think that a better approach would be to consider the average optical thickness of that blanket of atmosphere. First estimate the temperature at the top of the atmosphere using the S-B law. Then using skin surface temperature (SST) and lapse rates to calculate average optical thickness. (wet in the tropics and dry near the poles). Regress the calculated thickness on precipitation rate, precitatable water, and CO2. This method measures the effects of non-radiative processes of energy transfer as well as the “greenhouse effect”. (the formation of clouds, rain, and snow). You will find that any minor effect of CO2 is lost in the variability in the combination of effects of atmospheric water (vapor, clouds, rain, and snow).

wayne
March 10, 2011 7:02 am

Scarlet Pumpernickel says:
March 10, 2011 at 3:30 am
So what concentration of CO2 saturates these wavelengths?
There is already proof from Venus that the Greenhouse effect is not exponential http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html
——–
Thanks Scarlet, one impressive analysis. Also since NASA lists Venus’s *average* temperature at 464 C and if the surface was as high as 505 C then the natural dry lapse rate matches the graviation acceleration almost exactly, as it should. (778K-339K)/49.5km = ~8.87 °C/km (g=8.87 m/s2). Neat, 96.5% CO2 and no greenhouse effect at all. Now that’s some pure simple logic!

James
March 10, 2011 7:06 am

Re-emitted radiation does not and cannot heat the Earth significantly, because downwelling IR does not and cannot penetrate the surface of the ocean beyond its own wavelength.

But the radiation from the Sun has an even smaller wavelength. How come solar radiation can heat the Earth but re-emitted radiation?

nighttime
March 10, 2011 7:12 am

funny all this blanket stuff, we know that without greenhouse gases the surface can get very hot during the daytime – the moon, though at night leads to rapid cooling.
Seems to me that greenhouse gases lead to cooling during the day, and at night leads to a slow cooling , or in the desert, rapid due to less moisture.
After all if the sun didn’t come up next day how quick before the planet froze.
Such a simple elegant explanation.

syphax
March 10, 2011 7:13 am

The commentary for this article is pure gold. Mr. Watts, please continue a series of posts on basic planetary science processes; I’m very eager to see what your readership can come up with. I’d love some insight on, for example, how the Coriolis effect really works.

ShrNfr
March 10, 2011 7:21 am

I suggest we rid the atmosphere of o2 also. I mean it has this fat absorption/emission band at 50-70 GHz. One cannot be too careful you know.

March 10, 2011 7:25 am

A proper comparison of the atmospheric temperatures of Venus and Earth proves (to any competent, and honest, physical scientist) that there is no greenhouse effect such as is “visualized” here and promulgated by the IPCC scientists.
Venus: No Greenhouse Effect
Therefore, clearly something is fundamentally wrong with your understanding (and the IPCC’s).
From many comments I have seen on the internet, it is clear to me that many think that a “greenhouse effect” is proven merely by the ability of CO2 and other gases to absorb infrared radiation (Judith Curry clearly believes this, for example). And invariably, those who try to lecture on the greenhouse effect, focus on infrared emitted by the surface, and then absorbed by the “greenhouse gases”, and then supposedly radiated back to the surface again. All the visualizations of this supposed process I have seen, from the by-now-infamous Trenberth and Kiehl “Energy Budget” onwards, simply accept a huge loop of energy, larger than that coming from the Sun, between the surface and the atmosphere (and which is largely derived from wrongly assuming that the Earth’s surface is a blackbody). I know this is believed as the sacred scientific truth in many quarters (based on a belief in the present use of radiative transfer theory in climate science), so I won’t argue about it; I will simply tell you, as directly and honestly as I know how, that it is an incompetent belief, that violates the conservation of energy (no matter what believers such as ScienceofDoom, or Judith Curry, or Roy Spencer, etc. tell you). The critical evidence that decides the issue is the Venus/Earth comparison of actual temperatures I have done, and it is decisive against that incompetent belief.
To start with, everyone needs to consider that the Sun radiates a continuous spectrum that is about half infrared, most of which (except for the well-known atmospheric windows) is directly absorbed by the atmosphere (even the Trenberth diagram shows 20% of the incident radiation being so absorbed, and the actual fraction, in my present view, is around 40%). Absorbed going down, not coming back up from the surface. The Venus/Earth comparison directly indicates this, because the ratio of temperatures, Venus/Earth, is essentially a constant (1.17) which is just that due to the distances of the two planets from the Sun, NOTHING ELSE. It does not depend upon CO2 concentration (Earth has 0.04% CO2, and Venus a whopping 96.5%); and it does not depend upon albedo, either at cloud tops or planetary surfaces (Earth’s surface is 70% deep ocean, while Venus is solid crust) — These great differences in the two atmospheres and surfaces mean nothing, introduce no complicating effects. The Venus/Earth comparison shows they have no overall effect, because the solar distance explains the entire difference, over a broad range of atmospheric pressures. So both planets MUST be warmed, overall, by direct atmospheric absorption of the same (infrared) portion of the Sun’s incident radiation, not by the more complicated process of first warming of the surface. This is a revolutionary finding, but my Venus analysis is easily verified by any competent scientist, and should have been confronted and generally accepted nearly 20 years ago, when the 1991 Magellan mission returned the detailed Venus data.
So you have a complicated radiative transfer theory — obviously incompetently implemented when you consider the Earth’s surface a blackbody, or ignore the fact that you are showing thermal measurements, not directed radiation measurements — supposedly backing up a complicated “greenhouse effect”, versus a simple, planet-sized experiment (a competent Venus/Earth comparison) that definitively denies the “greenhouse effect” concocted by James Hansen et al..
Obviously, in the present heated intellectual atmosphere, it is going to take time for enough good scientists of dispassionate manner to confront and accept my clear and simple contribution. I submitted it to “Physics Today”, for a necessary open airing before the entire scientific community and the public, but of course have gotten no response. So I await signs of an incipient competency among climate scientists on this subject, or an overthrow of the current climate “consensus” by scientists in other fields (I am a physicist). The truth is quite different, and remarkably simpler, than anyone now seems to realize.

ferd berple
March 10, 2011 7:31 am

“I am not sure about this ‘radiation down’ or so called ‘back-radiation’.
According to Trenberth et al. this amounts to about 330 W/m2, nearly twice that of the energy reaching the Earths surface from the sun (184 W/m2). If this is true why isn’t this energy collected and used as an energy source (better than solar energy as no need for storage – this can be collected 24 hours /day).”
This simple question points a very large problem with the back radiation model. The simple fact that we cannot harness the power of the back radiation points to a problem with the back radiation model.
If Trenberth’s et al model is correct, then standing naked outside at midnight in still air should feel almost as warm as standing outside naked at noon on a sunny day.
We should also be able to harness the energy of the IR photons from the back radiation using solar panels sensitive in the IR spectrum to generate a large fraction of the power we generate in sunlight.
The simple fact that we cannot harness the energy of the back radiation suggests that it exists more as a function of our current theories in physics than as anything else. If the back radiation can’t do work, then something is missing in the explanation.

March 10, 2011 7:34 am

I’ve been working with a combination of the reanalysis data set and CO2 data in an effort to quantifiy the relative effects of atmospheric water and CO2 on OLR from the top of the atmosphere.(click on my name). I think a better approach is to estimate the “average optical thickness” of the atmospheric blanket and determine what effects water and CO2 have on it. First estimate the temperature at the top of the atmosphere using the S-B law and OLR at TOA. Then estimate “average optical thickness” using the difference between SST (skin surface temperature) and temperature at TOA with different lapse rates (wet for tropics and dry near the poles). Regress these values on precipitation rate, precipitable water, and CO2 to determine the magnitude of their effects. This technique captures the additional effects of non-radiative processes (formation of clouds, rain, and snow). I expect you will find that any minor effect of CO2 is statistically lost in the error variability of the atmospheric water factors.

tallbloke
March 10, 2011 7:34 am

richard verney says:
March 10, 2011 at 6:29 am (Edit)
tallbloke says: March 10, 2011 at 4:26 am
“….The greenhouse effect works by *SLOWING DOWN THE RATE THE EARTH COOLS AT*, by raising the altitude at which the atmosphere radiates to space .
http://tallbloke.wordpress.com/2011/03/03/tallbloke-back-radiation-oceans-and-energy-exchange/
Thanks for the link to your article. I was one of those who was arguing similar points with Willis and I have not seen your article before today. It is an interesting read.

Hi Richard,
Thanks for that. In fact, it was your exchange with Willis on the folie a deux part deux thread which prompted me to write the article and invite Willis to respond.
His position and Ira Glicksteins seem quite close, but niether of them seem willing to engage with the issue of the inadequacy of the mixing down of the back radiation warmed ocean surface to explain the rise in ocean heat content in the ’90’s.
The ocean freezing argument misses the real point. If the ocean is re-emitting whatever back radiation flirts with it’s surface in short order, an increase in co2 is not going to affect ocean heat content much, because it’ll just cause a bit more evaporation/convection, which cools the ocean surface.
Cheers

mkelly
March 10, 2011 7:42 am

Two identical plates at identical temperatures exactly parallel in a vacuum radiating at each other.
heat transfer between the two is:
q/a = (emissivity * sigma * T1^4) – (emissivity * sigma * T2^4) since e and sigma are the same
q/a = emissivity*sigma* (T1^4-T2^4) since they are both the same T no heat transfer.
There is no temperature increase at both plates because the other is radiating at it, no slow down of radiation, nothing. Even if you now put a gas (CO2) in between the plates you still end with the same thing no heat transfer between plates. The gas will not impede the plates from radiating. The gas will not make the plates hotter.
For people who say a cold body will transfer heat to a warmer body please present a formulaic argument for that.
For Phil who ask about why PV=nRT does not heat Titan: The gases on Titan are near critical temperature and therefore the formula cannot be used. However on Earth the critical temperature of nitrogen (-173 C or so) is so far away from the air temperature of the Earth that PV=nRT can be used at one atmosphere with a less than 1% error. Ergo my statement that we over state the effects of any GHG effect by at least 18 C.
And we have yet to have a definitive answer as to whether N2 and O2 follow the noraml law that “all matter will radiate according its temperature” or that N2 and O2 are immune form this.

Gary Swift
March 10, 2011 7:52 am

To John Marshall and Vince Causey:
In regard to the second law discussion, I would add this way of looking at it, which is really just a simpler way of saying what Vince said. When you have energy being transfered between the ground and the air it will have a rate of flow based on the difference in temperature between them. If you make the air warmer and the ground stays the same temperature, then the rate of energy flow will slow down. It’s not a matter of whether energy flows in both directions or not. The question is relative rate of flow. Changing the temperature of either medium relative to the other will change the rate of flow in the form of conduction.
To Ira Glickstein:
Are there any sources that show what those bottom/top of atmosphere graphs would look like with double CO2? I wonder how much it would change and how?

nighttime
March 10, 2011 8:06 am

this one must be for Roy Spencer and his Vacuum theory,
For people who say a cold body will transfer heat to a warmer body please present a formulaic argument for that.

March 10, 2011 8:09 am

Next up: Could you address for our reading audience LOWTRAN, MODTRAN and HITRAN and what that means for LWIR transmission and how it relates to CAGW?
Moshpit (S. Mosher) can advise where to obtain if required … so can cba I think.
.

Bryan
March 10, 2011 8:12 am

#
#
Gilbert K. Arnold
……” Did you even look at Ira’s brief CV at the bottom of his article. “……
Perhaps he has forgotten his thermo.
Other posters have pointed out mistakes with Kelvin temperature units and so on.
I was being helpful when I said he should revise his thermodynamics.

Phil.
March 10, 2011 8:21 am

Bomber_the_Cat says:
March 10, 2011 at 6:10 am
Ira, there’s a problem here. as P. van der Meer says at 3:15 am. The blackbody curves are showing a peak at about 18 micron when they should be peaking at about 10 micron for a 300K blackbody.
In fact, if you refer to your previous post ‘Visualising the Greenhouse Effect – Atmospheric Windows’, you have the peak correct at 10 microns.
So, the current graphs don’t make sense – unless I am missing something? I have looked at the source material and that appears to be wrong also.
Has anyone got an explanation for this?

Yes, the Petty data is spectral radiance (mW/m^2.sr.cm-1) plotted vs. cm-1 and is correct, the transformation to a plot in terms of wavelength is non-linear (and is why it’s misleading for Ira to have reversed the axis on his plots, should leave it in wavenumbers).

tallbloke
March 10, 2011 8:24 am

James says:
March 10, 2011 at 7:06 am
Re-emitted radiation does not and cannot heat the Earth significantly, because downwelling IR does not and cannot penetrate the surface of the ocean beyond its own wavelength.
But the radiation from the Sun has an even smaller wavelength. How come solar radiation can heat the Earth but re-emitted radiation [can’t]?

It’s because it has a smaller wavelength that it can penetrate the ocean. Most of the incoming solar energy is in the visible wavelengths, and it penetrates the ocean to around 300feet at most. The U.V. penetrates even further, to 1500feet or more, but carries a lot less energy than the visible. A big percentage is absorbed in the top 30feet, but that’s ok, because that’s well onto the zone which can get mixed further down by wind and wave action, tidal action and subsurface currents.

Phil.
March 10, 2011 8:24 am

Gary Swift says:
March 10, 2011 at 7:52 am
Are there any sources that show what those bottom/top of atmosphere graphs would look like with double CO2? I wonder how much it would change and how?

Go and use Modtran, you can play with it to your heart’s content, very instructive.
http://geoflop.uchicago.edu/forecast/docs/Projects/modtran.orig.html

March 10, 2011 8:27 am

Harry Dale Huffman says:
March 10, 2011 at 7:25 am
Thank you Harry.
And
mkelly says:
March 10, 2011 at 7:42 am
“And we have yet to have a definitive answer as to whether N2 and O2 follow the noraml law that “all matter will radiate according its temperature” or that N2 and O2 are immune form this.”
Ditto.
I have been saying exactly this for years. If N2 and O2 do not radiate according to their temperature then they must be the only two substances in the Universe which do not.
How convenient would it be, if that were the case (which of course it is not), that they happen to make up 99% of the atmosphere?

Domenic
March 10, 2011 8:30 am

to Harry Dale Huffman
Good post. Great analysis. Far more valid than the nonsense out there.
On earth, the data from the Antarctic interior regions, confirms exactly what you concluded. CO2 levels have no silly feedback effects.
It’s as if the climate scientists, ignorant of the basics of radiational heat transfer, have built an elaborate farce. They chase around bits and pieces, thousands of localized effects. Meanwhile they construct a farcical ‘greenhouse effect’, include in it what they think belongs there, blow them way out of proportion, and exclude or ignore what they ‘think’ doesn’t belong there.
Then they frighten themselves with their own farce of a nightmare. And they try to get others to join them in their self-created nightmare.
Incompetence abounds, not only as scientists, but as people.

March 10, 2011 8:32 am

“Morris Minor says:
March 10, 2011 at 2:21 am
“…. Perhaps a better analogy would be an electric blanket that, in addition to its insulating properties, also emits thermal radiation both down and up…..”
I am not sure about this ‘radiation down’ or so called ‘back-radiation’.
According to Trenberth et al. this amounts to about 330 W/m2, nearly twice that of the energy reaching the Earths surface from the sun (184 W/m2). If this is true why isn’t this energy collected and used as an energy source (better than solar energy as no need for storage – this can be collected 24 hours /day).
The reason I suspect we can’t use this energy is because it doesn’t exist – heat will not flow from the cold atmosphere to the warm surface of the Earth.
I think the emission spectra shows the scatter of infra-red from the atmospheric gases that wont reach the collimated collector of the sensors due to its direction of travel?
Thoughts please!”
Absolutely 100% right Morris Minor- Cooler air CANNOT heat the warm ground, this article is nothing more than lukewarmer nonsense-
( and by the way great choice of car, I own three! One Traveller, one convertible and a 4-door saloon project. )
regards
John

tallbloke
March 10, 2011 8:35 am

Gilbert K. Arnold says:
March 10, 2011 at 4:09 am
Did you even look at Ira’s brief CV at the bottom of his article.

I’ll reserve judgement on the value of the CV until Ira has cogently debated the points raised in opposition to his characterisation of the greenhouse effect.

Phil.
March 10, 2011 8:36 am

mkelly says:
March 10, 2011 at 7:42 am
For Phil who ask about why PV=nRT does not heat Titan: The gases on Titan are near critical temperature and therefore the formula cannot be used. However on Earth the critical temperature of nitrogen (-173 C or so) is so far away from the air temperature of the Earth that PV=nRT can be used at one atmosphere with a less than 1% error. Ergo my statement that we over state the effects of any GHG effect by at least 18 C.

As you’ve been told before you can’t use the gas laws this way, it’s utter nonsense to do so. I didn’t ask about Titan but Triton, where the atmosphere (N2) is far from the critical point.
And we have yet to have a definitive answer as to whether N2 and O2 follow the noraml law that “all matter will radiate according its temperature” or that N2 and O2 are immune form this.
Yes you have but it doesn’t fit with your preconceptions so you don’t believe it.

kwinterkorn
March 10, 2011 9:11 am

On the 2nd Thermodynamics Law, some above are confusing Net Flow of Heat, which must always be from hotter to colder, with Rate of Heat Flow, which can be influenced by placing a resistor in the system (eg the vacuum part of a thermos bottle, for example). The hot toddy in the thermos bottle still cools as heat is transferred out, but more slowly.
A blanket is a resistor to heat flow. The atmosphere is a resistor to heat flow.
The atmosphere resists heat flow more in the infra-red than in the visible light specrum. So heat from the Sun, mostly in the visible range, gets in with less resistance than heat radiated from the Earth, mostly in the infra-red. The Earth must rise in temperature until the imbalances in resistance are balanced by increased radiation from the warmer Earth. This higher equilibrium point, about which the Earth in reality fluctuates, is empirically confirmed by comparison with the average temps on the airless moon.
The issue for AGW (the CO2-part, not the urban heat island (UHI) or land use part) is whether elevating CO2 levels change the net resistance to flow in the assumed linear fashion (ie, doubling CO2 doubles CO2’s contribution to the air’s resistance to heat flow outgoing from the Earth). Logical application of saturation effects suggest that as CO2 rises, its added contribution diminishes.
This leaves aside the issue of feedbacks, positive or negative, related to clouds, storms, surface albedo, and so on.
In summary, one reaches a flat-earther or the-moon-shots-were-fakes level of denial when one tries to deny that a “greenhouse effect” exists because of the atmosphere’s resistance to heat flow. The reason that CAGW (catastrophic anthropogenic global warming) is in doubt is because:
1. The temp records poorly correlate with CO2 levels.
2. “Saturation” effects may limit the change induced by rising CO2.
3. The net feedbacks are most likely negative (characteristic of a stable system, as the Earth has been for eons), not positive.

Gil Dewart
March 10, 2011 9:19 am

Obvious take-aways: the “greenhouse” has a lot of broken windows and the “blanket” has a lot of holes.

mkelly
March 10, 2011 9:21 am

Phil. says:
March 10, 2011 at 8:36 am
“As you’ve been told before you can’t use the gas laws this way, it’s utter nonsense to do so. I didn’t ask about Titan but Triton, where the atmosphere (N2) is far from the critical point.”
Sorry for the mistake between the two moons. But my comment stands as I went back to my themodynamics book and what I said is a virtual quote about critical temperature and being able to use it about air within a 1% error here on earth. If you disgree then you disagree with not only me but my old text book.
Again with the preconceptions. There was a disagreement on the last thread about this and just because you post a link to somewhere does not make it definitive. But let’s say you are correct that N2 and O2 do not radiate according to thier temperature. Then why Phil: “Since 1979, NOAA satellites have been carrying instruments which measure the natural microwave thermal emissions from oxygen in the atmosphere.” Is NOAA wrong? The quote is from an earlier story here on WUWT.

Phil.
March 10, 2011 9:31 am

Domenic says:
March 10, 2011 at 8:30 am
to Harry Dale Huffman
Good post. Great analysis. Far more valid than the nonsense out there.

Except it’s wrong!
For example, the following:
The Venus/Earth comparison directly indicates this, because the ratio of temperatures, Venus/Earth, is essentially a constant (1.17) which is just that due to the distances of the two planets from the Sun, NOTHING ELSE. It does not depend upon CO2 concentration (Earth has 0.04% CO2, and Venus a whopping 96.5%); and it does not depend upon albedo, either at cloud tops or planetary surfaces
Venus has a bond albedo of 0.75 due to the sulfuric acid clouds whereas the earth has a bond albedo of 0.29, you bet the temperature depends on the albedo. Venus absorbs 25% of the light incident on it and Earth 71% but the ratio of temperatures still only depends on the distance from the sun? Even a physicist should see the problem with that.

Michael J
March 10, 2011 9:32 am

I’m an engineer, not a physicist, but I think I may be able to clarify the misunderstandings with respect to heat transfer and the second law of thermodynamics.
Any matter that is not at zero Kelvin will emit energy.
Where a cold surface meets a hot surface, both emit energy but the hot surface will emit more than the cold.
Although there will be energy transfer in both directions, the net transfer must be from the hot body to the cold one, causing the hot body to cool and the cold body to warm.
So when the second law forbids the transfer of heat from the cold body to the warm, I think it refers to net heat transfer.
Disclaimer: my thermodynamics are pretty rusty so that might be all wrong.

Gaylon
March 10, 2011 9:37 am

“Harry Dale Huffman says:
March 10, 2011 at 7:25 am
A proper comparison of the atmospheric temperatures of Venus and Earth proves (to any competent, and honest, physical scientist) that there is no greenhouse effect such as is “visualized” here and promulgated by the IPCC scientists.”
Excellent post, went to your site and read in entirety, very succinct and straightforward. Thanks for joining us here. I find it disheartening that not more of the posters here (only one other I’ve seen so far: Domenic says:
March 10, 2011 at 8:30 am and I agree with his comments also) have picked up on the import of your analysis.
My personal feeling is that too much time is spent on coddling over personalities, psychologies, and what you touched on: this preponderance over comples calculations concerning a dead hypothesis. I mentioned on another thread that the truth is ill-served by trying to “build bridges” or find “common ground” for discussion with scoundrels. Congratulations on your adept use of Occam’s Razor and thanks again.
I have bookemarked your site for frequent referencing.

March 10, 2011 9:37 am

mkelly says:
March 10, 2011 at 9:21 am
You are correct, all substances radiate according to their respective temperatures, there are no exceptions to that fact.

Gaylon
March 10, 2011 9:43 am

Gaylon says:
Your comment is awaiting moderation.
March 10, 2011 at 9:37 am
Hey all, in my previous post I am not refering to ALL AGW’s believers as scoundrels. I am refering to the scoundrels (Trenberth, Mann, Jones, et al). I realize that this science (climate science) is still in it’s infancy and that many climb on-board through a sincere desire to explore, explain, and to learn. Others, named, not so much.

Phil.
March 10, 2011 9:44 am

Fred Souder says:
March 10, 2011 at 5:53 am
Ira,
You should stop saying the greenhouse gases in the atmosphere warm the earth to this crowd. Too may engineers running around here. The atmosphere slows the rate at which the earth loses energy to deep space. Thus, the earth in the sun-earth-space system has a higher equilibrium temp. The “old” rules of thermodynamics still apply: a warm object cannot gain net thermal energy from a cold object.

But a warm object with a continuous heat source surrounded by a radiatively active atmosphere which is warmer than space will reach an equilibrium temperature which is warmer than it would otherwise be in the absence of such an atmosphere. Some of the BB emitted from the surface will be recycled to the surface (and be measured) ask some of the ChemEs what the effect of 50% recycle is on throughput, is that contrary to conservation of mass? Ask the MechEs why when they measure the exhaust from a gas turbine with a thermocouple they get a certain temperature but when they put a thin radiation shield around it the temperature goes up, is that contrary to the laws of thermodynamics?

Gaylon
March 10, 2011 9:55 am

“”Phil. says:
March 10, 2011 at 9:31 am
Domenic says:
March 10, 2011 at 8:30 am
to Harry Dale Huffman
Good post. Great analysis. Far more valid than the nonsense out there.
Except it’s wrong!””
________________
Phil, go back and actually read the analysis here:
http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html
The calculations are relatively simple, are applied at the 1000mb level, and are congruent with earth at 1000mb. It’s straight, simple, to the point and beyond refutation. Maybe Harry will come back and answer you directly, but I think when he writes, “The Venus/Earth comparison directly indicates this, because the ratio of temperatures, Venus/Earth, is essentially a constant (1.17) which is just that due to the distances of the two planets from the Sun, NOTHING ELSE. It does not depend upon CO2 concentration (Earth has 0.04% CO2, and Venus a whopping 96.5%); and it does not depend upon albedo, either at cloud tops or planetary surfaces (Earth’s surface is 70% deep ocean, while Venus is solid crust) — These great differences in the two atmospheres and surfaces mean nothing, introduce no complicating effects. The Venus/Earth comparison shows they have no overall effect, because the solar distance explains the entire difference, over a broad range of atmospheric pressures.”, his comments are substantiated by the math. Very simple.

March 10, 2011 9:58 am

Phil. says:
March 10, 2011 at 9:31 am
“Venus has a bond albedo of 0.75 due to the sulfuric acid clouds whereas the earth has a bond albedo of 0.29, you bet the temperature depends on the albedo. Venus absorbs 25% of the light incident on it and Earth 71% but the ratio of temperatures still only depends on the distance from the sun? Even a physicist should see the problem with that.”
I think Phil has either missed the point here or he hasn’t read Harry’s article. http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html
He clearly states that Earth and Venus are heated top down by incoming IR.
Quote: “This in fact indicates that the Venusian atmosphere is heated mainly by incident infrared radiation from the Sun, which is not reflected but absorbed by Venus’s clouds, rather than by warming first of the planetary surface. (It also indicates that the Earth atmosphere is substantially warmed the same way, during daylight hours, by direct solar infrared irradiation, and that the temperature profile, or lapse rate, for any planetary atmosphere is relatively oblivious to how the atmosphere is heated, whether from above or below.)”
Albedo applies to visible electromagnetic radiation not invisible IR.

March 10, 2011 10:11 am

You have gone to great lengths to explain the spectral absorption features of the greenhouse gases. These gases are in the atmosphere and that is what they do. But it is unhelpful to know what the gases that are sitting there do when the atmosphere is in a stationary state. What is important to know is how does the addition of more carbon dioxide to the atmosphere change this picture. We are told that if you do that the atmospheric absorption will simply increase at the wavelengths that the added gas absorbs and in proportion to the amount added. But is this really true? There are no direct instrumental measurements of this and we simply have to believe theory about it. But what if there was a way to actually observe how the total atmospheric absorption changes as we increase the amount of CO2 we add to the atmosphere? It turns out that there is as Ferenc Miskolczi has pointed out. NOAA has been keeping a database of weather balloon observations since 1948 and these can be used to determine the relevant atmospheric absorption parameters. Miskolczi used this database to calculate how the global annual infrared optical thickness of the atmosphere varied throughout these years. And he found that the optical thickness of the atmosphere in the infrared where carbon dioxide absorbs remained constant for 61 years, with a value of 1.87. This tells us that the transparency of the atmosphere in the infrared did not change for 61 years despite constant addition of CO2 to the atmosphere through all this long stretch of time. Hence, the greenhouse absorption signature of the added carbon dioxide which we are told about simply isn’t there. This is an empirical finding, not something derived from theory, and it overrides any calculations from theory that do not agree with it. Theories that disagree must either be modified or discarded. I want to point out also that his work came out in 2009 and no one so far has attempted to present any peer-reviewed arguments against it. Miskolczi concludes: “It will be inferred that CO2 does not affect the climate through the greenhouse effect.”

richard verney
March 10, 2011 10:11 am

I am enjoying the various views expressed in the various comments.
As regards the various arguments as to whether heat can flow from a cooler body to a warmer body, net heat flow, the rate of net heat flow etc, given the importance of this issue, it amazes me that there appears to be no experimental data on this. Whilst this is not my field, surely, it cannot be that difficult to devise a suitable experiment. Possibly along the lines:
1. A very larger insulated air chamber with a well mixed air temperature at say 290K.
2. Suspend within the chamber a modest size blackbody sphere at say 340K and measure the time taken to cool to 320K.
3. Repeat 2 above but this time additionally suspend within the chamber a significantly smaller blackbody sphere (say perhaps 1/10th surface area) at 300K say 2 metres away from the larger 340K blackbody. Measure the time taken for the 340K blackbody to cool to 320K. Measure the heat radiated on both sides of that blackbody to see whether there is a difference in the amount of heat being radiated on the side which is adjacent to the smaller cooler blackbody. Measure the heat being radiated by the smaller blackbody to see whether there is a difference in the amount of heat being radiated on the side adjacent to the warmer blackbody.
4. Repeat 3 above but with the smaller blackbody at 280K.
5. Repeat the experiment with different sizes of blackbodies, and different temperatures for each blackbody and different distances between the blackbodies.
6. Repeat the experiement with the chamber of air having 500ppm, 600ppm, 700 ppm and 800ppm of Co2.
7. Repeat the experiment but with a slow running fan placed under and sime distance away from the blackbody spheres.
8. Carry out a number of experimental runs as appropriate.
9. Collect data and analyse.
I am not suggesting that the above experiment but some experiment along those lines ought to establish what really happens in the real world.

Fred Souder
March 10, 2011 10:16 am

Phil,
In fact, I was one of those Chem E’s! Thank you for your compliment!
I agree with what you say. I am just telling Ira that his use of language is contrary to what we would teach in a thermodynamics classroom. An atmosphere won’t transfer net heat to a warmer body. It may slow the rate at which EMR is emitted from the body, or change the equilibrium temp of the body, but the “Heat Transfer” will always go from the Hot source to the Cold source. You can’t make statements like -“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” – to a group of engineers.

tallbloke
March 10, 2011 10:33 am

Phil. says:
March 10, 2011 at 9:44 am
a warm object with a continuous heat source surrounded by a radiatively active atmosphere which is warmer than space will reach an equilibrium temperature which is warmer than it would otherwise be in the absence of such an atmosphere.

And a warm object which has lower albedo for 25 years allowing more energy from the continuous heat source onto its surface will too.

March 10, 2011 10:45 am

Unfortunately we have low clouds in the middle of the optical path.
If the emission is a function of surface temperature.
Nobody knows the correct temperature.
The interface temperature (SSTint)
At the exact air-sea interface a hypothetical temperature called the interface temperature (SSTint) is defined although this is of no practical use because it cannot be measured using current technology.
The skin sea surface temperarature (SSTskin)
The skin temperature (SSTskin) is defined as the temperature measured by an infrared radiometer typically operating at wavelengths 3.7-12 µm (chosen for consistency with the majority of infrared satellite measurements) that represents the temperature within the conductive diffusion-dominated sub-layer at a depth of ~10-20 µm. SSTskin measurements are subject to a large potential diurnal cycle including cool skin layer effects (especially at night under clear skies and low wind speed conditions) and warm layer effects in the daytime.
http://www.ghrsst.org/SST-Definitions.html
On the other hand, consider a cloud that is made up of 10 µm-diameter water droplets: it behaves like a homogeneous medium with respect to the long wavelengths of microwave radiation even though it is quite inhomogeneous with respect to visible and infrared radiation.
in,
A First Course in Atmospheric Radiation page 72.
Unfortunately we have low clouds in the middle of the optical path.
In those circumstances it seems reasonable that the input is different from the output. Without necessarily indicate an energy imbalance (caused by CO2).

GaryP
March 10, 2011 10:47 am

I keep seeing the argument that heat cannot flow from a cold body to a hot body. If I follow this argument then a emergency reflective blanket sold in camping stores will not keep you warmer. The second law says there cannot be any NET flow from cold to hot but says nothing about an absorbing gas reducing the rate of heat flow from the hot body.
Here is an actual experiment. I was using a hot filament in high vacuum to heat a sample to high temperature (~1000°C). I could not get one sample hot enough so I put a shiny metal cylinder around but not touching the filament or sample. The reflected heat increased my sample temperature over 100°C. The added radiation from the relatively cold cylinder increased the temperature of the white hot filament. The net flow of heat of course was from hot to cold. This is a pure radiation example and some heat did radiate from the cold reflector to the hot filament.

Dave Springer
March 10, 2011 10:52 am

An electric blanket has an internal heat source and thus isn’t an appropriate comparison. An electric blanket can warm another object above the object’s starting temperature. A regular blanket can’t do that. A regular blanket can only slow down the rate of heat loss from an object it covers. Stick to a pair of black rocks that are exposed to the sun during the day and one of them gets a blanket over it at night. That’s about as close to reality as you’re going to get using ordinary objects in a way that almost everyone can understand from personal experience. Most people also understand that each additional blanket is less effective at retaining warmth than the previous blanket i.e. the first blanket makes a huge difference while a tenth blanket won’t help nearly as much as the first.
After that the argument is just over the exact amount of warming (not much for CO2 & CO2 equivalents alone which isn’t a matter of great dispute) and whether or not there is any positive feedback from water vapor. The water vapor feedback is the matter of great dispute and appears to be nothing more than a wholesale fabrication without a wisp of empirical evidence in support of it and posited for the sole purpose of turning a little bit of welcome, beneficial warming into something worth worrying about.

Phil.
March 10, 2011 10:53 am

mkelly says:
March 10, 2011 at 9:21 am
Phil. says:
March 10, 2011 at 8:36 am
“As you’ve been told before you can’t use the gas laws this way, it’s utter nonsense to do so. I didn’t ask about Titan but Triton, where the atmosphere (N2) is far from the critical point.”
Sorry for the mistake between the two moons. But my comment stands as I went back to my themodynamics book and what I said is a virtual quote about critical temperature and being able to use it about air within a 1% error here on earth. If you disgree then you disagree with not only me but my old text book.

Your old text book correctly tells you the accuracy with which the gas laws apply to the Earth’s atmosphere, they don’t tell you that PV=nRT heats the Earth which is what you said!
At the surface P is effectively constant so the gas laws tell us that the density of the atmosphere is a function of T, i.e. n/V=P/RT. The surface temperature is determined by radiational exchange with the sun and the surface heats the atmosphere, the hotter it gets the lower the density, Temperature does not depend on PV=nRT!
Again with the preconceptions. There was a disagreement on the last thread about this and just because you post a link to somewhere does not make it definitive. But let’s say you are correct that N2 and O2 do not radiate according to thier temperature. Then why Phil: “Since 1979, NOAA satellites have been carrying instruments which measure the natural microwave thermal emissions from oxygen in the atmosphere.” Is NOAA wrong? The quote is from an earlier story here on WUWT.
N2 and O2 do not emit BB radiation according to their temperature, they can only emit where there are transitions between energy levels which are determined by molecular structure. At any given temperature they will not emit more that given by S-B at that wavelength. O2 has pure rotationa transitions in the microwave spectrum around 60μm, the fraction of BB emissions from the Earth at that wavelength is extremely small and has no significant contribution (line intensities ~10^-26, BB spectral radiance ~1%) compared with H2O and CO2 (many more lines, line intensities ~10^-19, right in the heart of the BB radiance spectrum). Those microwave emissions are useful diagnostically because they’re in a region which is sparsely populated by other spectra (a little H2O), they don’t contribute to the energy balance of the atmosphere.
Learn something about the physics of gases rather than pick up snippets that you don’t understand and think it shows that those who’ve researched the subject and taught it at the graduate level missed something.

Ken Finney
March 10, 2011 11:01 am

@sjoerd: “”Please leave out the “degree” when talking about Kelvin. It’s “degree Fahrenheit” and “degree Celsius”, but it’s “Kelvin” (without the “degree”). Same when abbreviated: It’s 270K, not 270ºK.””
Only to kids. We old farts grew up with “degrees Kelvin”, before it was changed by a convention in the early 70’s. Just like many of us grew up with “degrees Fahrenheit”, before many countries started changing to “degrees Celsius”. We also used to know it by “Centigrade”, not “Celsius”. How about you wait till we’re all dead before waxing on about your superiority, huh?

Phil.
March 10, 2011 11:04 am

Will says:
March 10, 2011 at 9:37 am
mkelly says:
March 10, 2011 at 9:21 am
You are correct, all substances radiate according to their respective temperatures, there are no exceptions to that fact.

There are many exceptions to that ‘fact’ because it’s simply not true, if it were there would be no need to go to the difficulty of constructing a BB cavity for calibration purposes. Gases in particular emit line spectra not BB continua.

Bryan
March 10, 2011 11:07 am

Michael J says:
March 10, 2011 at 9:32 am
…..”Any matter that is not at zero Kelvin will emit energy.”….
Yes but some molecules radiate quite a lot at certain frequencies and some hardly any.
For O2 and N2 we can ignore any radiation in the IR for all practical purposes.
……”So when the second law forbids the transfer of heat from the cold body to the warm, I think it refers to net heat transfer.”……
Radiation goes from cold to hot but not heat.
Being an engineer you can think of lots of ways to get work done by a machine taking heat from a high temperature then e.g. operating a piston and rejecting unused heat to a lower temperature.
However you cannot think of one device that takes heat from a lower temperature does some work then rejects unused heat to a higher temperature.

Stephen Richards
March 10, 2011 11:09 am

John Marshall says:
March 10, 2011 at 2:04 am
May I respectfully suggest you go read a junior level physics text book.

Phil.
March 10, 2011 11:11 am

Arno Arrak says:
March 10, 2011 at 10:11 am
You have gone to great lengths to explain the spectral absorption features of the greenhouse gases. These gases are in the atmosphere and that is what they do. But it is unhelpful to know what the gases that are sitting there do when the atmosphere is in a stationary state. What is important to know is how does the addition of more carbon dioxide to the atmosphere change this picture. We are told that if you do that the atmospheric absorption will simply increase at the wavelengths that the added gas absorbs and in proportion to the amount added.

Who’s telling you that? It depends on the gas, for CO2 it’s approximately logarithmic at present atmospheric conditions.
But is this really true? There are no direct instrumental measurements of this and we simply have to believe theory about it.
There are plenty of instrumental measurements of this, the broadening of spectral lines is well understood.

Vince Causey
March 10, 2011 11:11 am

Bryan,
“Vince Causey perhaps is confusing radiation with heat.”
Yes, I was using the two terms interchangably. If we take heat flow to mean something getting warmer, then it is certainly impossible for a cooler body to make a warmer body still warmer. What I was trying to say was that energy will flow from the cooler body but heat will not – ie, the warmer body will not heat up, but will still cool down and equilibriate with the cooler body.
I was attempting to respond to a poster who asserts that the GHG effect violates the 2nd law of thermodynamics which forbids heat flowing from a cooler body to a warmer body. A better worded response would be to say that GHG does not depend on ‘heat’ flowing in that direction, merely the flow of energy, and that energy flow will have the effect of slowing down heat loss from the warmer body.

Stephen Richards
March 10, 2011 11:13 am

Fred Souder says:
March 10, 2011 at 10:16 am
2nd Law. Disorder to order. Low entropy to High Entropy. And everyone still appear to be mixing classical and quantum reactions. Kinetic and convective transfer are classical effects. Radiative is a quantum effect. What ‘see’ will be different for both models.

Stephen Richards
March 10, 2011 11:14 am

GaryP says:
March 10, 2011 at 10:47 am
Nail, head. Radiative not kinetic.

Phil.
March 10, 2011 11:16 am

Will says:
March 10, 2011 at 9:58 am
Phil. says:
March 10, 2011 at 9:31 am
“Venus has a bond albedo of 0.75 due to the sulfuric acid clouds whereas the earth has a bond albedo of 0.29, you bet the temperature depends on the albedo. Venus absorbs 25% of the light incident on it and Earth 71% but the ratio of temperatures still only depends on the distance from the sun? Even a physicist should see the problem with that.”
………..
Albedo applies to visible electromagnetic radiation not invisible IR.

No: “The Bond albedo is the fraction of power in the total electromagnetic radiation incident on an astronomical body that is scattered back out into space. It takes into account all wavelengths at all phase angles.
It is an important quantity for characterizing a planetary body’s energy balance.
For objects in the solar system, the relevant weighting of each wavelength is proportional to the solar power spectrum. Visible light is a major contribution because over 40% of solar output is in this range.
Like most albedos, the Bond albedo is a value between 0 and 1.”

Vince Causey
March 10, 2011 11:18 am

Morris Minor,
“If this is true why isn’t this energy collected and used as an energy source (better than solar energy as no need for storage – this can be collected 24 hours /day).”
Actually heat is collected from the ground. Thermal heat collectors based on heat pumps are able to extract that heat from the 2 metres below the surface and can be used to warm houses.

Stephen Richards
March 10, 2011 11:23 am

tallbloke says:
March 10, 2011 at 8:24 am
Be careful about wavelength v penetration in water. Submarines use very longwave radiation for communication and detection because higher frequencies and therefore higher energy waves are rapidly attenuated in water. I have not data at all beyond these details because I designed adaptive filters for one of the first sideways viewing ‘radars’ back in the late ’70s’.

Bryan
March 10, 2011 11:24 am

GaryP says:
……”Here is an actual experiment. I was using a hot filament in high vacuum to heat a sample to high temperature (~1000°C). I could not get one sample hot enough so I put a shiny metal cylinder around but not touching the filament or sample. The reflected heat increased my sample temperature over 100°C. The added radiation from the relatively cold cylinder increased the temperature of the white hot filament. The net flow of heat of course was from hot to cold. This is a pure radiation example and some heat did radiate from the cold reflector to the hot filament.”…..
All your reflector did was to insulate the hot filament.
When you put on clothes you reduce heat loss from your body.
The clothes also radiate IR as well as reducing heat loss by conduction and convection.
You would not saythat you wear your own greenhouse effect clothing would you?

commieBob
March 10, 2011 11:30 am

Fred Souder says:
March 10, 2011 at 10:16 am
… but the “Heat Transfer” will always go from the Hot source to the Cold source. You can’t make statements like -”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” – to a group of engineers.

Net heat will go from the warmer object toward the cooler. You are absolutely right about that. You are, however, wrong about radiation.
A weaker source will radiate toward a stronger source. The radiation from the stronger source does not magically drive back that of the weaker source.
I give a couple of examples in a longer post above which I will repeat here.

a) If I stand with my back toward the sun, I will be able to see a flashlight being shone at me. Nothing about the sun’s radiation will prevent the radiation from the flashlight from reaching me. b) If I stand near a strong radio transmitter, I can still tune in weaker transmitters. Nothing about the stronger transmitter’s signal prevents the weaker signal from getting to me. The net energy flux will still be from the stronger source toward the weaker one. It is a net flux though, in other words, the net flux is the difference between the two signals.

Infra-red is electromagnetic radiation. It behaves like any other kind of EM radiation.
Ira’s statement is just fine.

Stephen Wilde
March 10, 2011 11:36 am

peter_ga says:
March 10, 2011 at 3:21 am
“Earth is warmer because of its oceans, through an indirect greenhouse mechanism, that has nothing to do with co2.”
Bingo!
see here:
http://climaterealists.com/index.php?id=1487&linkbox=true&position=4
“The Hot Water Bottle Effect.”
and while I’m about it the one most important thing I have realised from the advice of solar expert Leif Svalgaard is that it is not all about radiative physics (sorry Ira).
If we are going to explain movements of the Earth’s air circulation systems to fit observations then radiative physics just does not work.
We see a cooling stratosphere/mesosphere when the sun is active and a warming stratosphere/mesosphere when the sun is inactive. That is the opposite of conventional climatology.
Thus the air circulation systems move poleward when the sun is active and equatorward when it is inactive. That affects global albedo and energy input to the oceans.
What we are left with is atmospheric chemistry involving ozone overriding radiative processes to shift the air circulations, affect global cloudiness and albedo and thereby switch the oceans from net energy gain to net energy loss. The tropospheric air temperatures then follow in due course.
This is my latest effort on that issue:
http://www.irishweatheronline.com/irishweather/how-the-sun-could-control-earths-temperature.html
“How The Sun Could Control Earth’s Temperature”
Alternative suggestions are welcome but they need to fit real world observations at least as well as do my proposals.

Oliver Ramsay
March 10, 2011 11:40 am

Phil. says:
March 10, 2011 at 9:44 am
“But a warm object with a continuous heat source surrounded by a radiatively active atmosphere which is warmer than space will reach an equilibrium temperature which is warmer than it would otherwise be in the absence of such an atmosphere.”
——————–
I’m not aware of anywhere on this spinning, tilting, cloud-bedecked planet for which this description would be apt.
Furthermore, the fact that the atmosphere also interacts with the incoming radiation that is the heat source for the surface complicates speculation about what it it woud all be like “if..”.

Dave Springer
March 10, 2011 11:41 am

“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.”
We went over this before. Kirchoff’s first of law radiation: solids, liquids, and dense gases emit continuous blackbody spectrums with the peak emission frequency determined by the temperature. The earth’s atmosphere below the thermosphere is dense in the context of Kirchoff’s law. When the gas molecules are densely packed they give up energy through collisions. Collisions don’t generate emission line spectra they generate continuous blackbody spectra. The “temperature” is a measure of the average speed of the molecules and hence the average energy involved in collisions between molecules.

Bryan
March 10, 2011 11:43 am

Vince Causey
……..”Actually heat is collected from the ground. Thermal heat collectors based on heat pumps are able to extract that heat from the 2 metres below the surface and can be used to warm houses.”……..
This fully complies with the second law.
Clausius said that heat will not flow spontaneously from a lower to a higher temperature.
You can of course use external energy as work in a device to pump the heat.
Like a refrigerator or heat pump.

Phil.
March 10, 2011 11:48 am

GaryP says:
March 10, 2011 at 10:47 am
Here is an actual experiment. I was using a hot filament in high vacuum to heat a sample to high temperature (~1000°C). I could not get one sample hot enough so I put a shiny metal cylinder around but not touching the filament or sample. The reflected heat increased my sample temperature over 100°C. The added radiation from the relatively cold cylinder increased the temperature of the white hot filament. The net flow of heat of course was from hot to cold. This is a pure radiation example and some heat did radiate from the cold reflector to the hot filament.

A slightly more sophisticated version of this is used with incandescent light bulbs to improve their efficiency. The glass envelope is coated with a coating which reflects IR and transmits visible, so the visible is emitted but the IR is reflected back towards the filament, the filament gets hotter and therefore brighter as well as shifting to more visible. Consequently the bulbs emit more visible light for a given electrical input than a conventional bulb thanks to ‘back radiation’.

Dave Springer
March 10, 2011 11:57 am

John Marshall says:
March 10, 2011 at 2:04 am
Before you go about “understanding” the laws of thermodynamics in your head you should first learn about the results of experiments conducted circa 1859 by John Tyndall with regard to thermal absorption of gases. If the answer you get from the thermodynamic model in your head is not the same answer that comes from experiment then your mental model is flawed. Your understanding is flawed if you do not acknowledge “back radiation”. Tyndall was the first to actually measure it in a series of thousands of experiments with various gases, mixtures of gases, at various pressures, through varying column lengths, and with thermal energy sources at different temperatures. He gave countless lectures and demonstrations about it and published collections of papers in a few books which are available in their entirety free of charge from books.google.com. You might want to start with “Heat: A Mode of Motion” and skip up to the chapters on gases. It’s great reading especially the ingenuity of the experimental setup which is described in great detail along with anecdotes about he obtained various materials required for its construction.

Bomber_the_Cat
March 10, 2011 12:01 pm

I recently voted for this as being the best scientific blog but, in response to this article, we still get people who say that back radiation cannot happen or wish to divert the discussion to some strange theories about the temperature on Venus.
But there is a serious problem with Ira’s article!
As expressed before (6:10AM); his blackbody curves are showing a peak at about 18 micron when they should be peaking at about 10 micron for a 300K blackbody. So, the graphs don’t make sense
Now Joel Heinrich at 2.57 AM and Phil at 8.21AM, March 10, say this is because Ira has converted wave number to wavelength.
But this does NOT cut the mustard. This is a perfectly acceptable thing to do, albeit ending up with a non-linear scale rather than a linear scale. In fact, the
Source Data itself includes the wavelength scale, so it is nothing to do with Ira’s conversion.
On the ‘wave number’ scale the peak of the blackbody curves should be around a wave number of 1000 – which they are not! They peak around a wave number of 600 – which is WRONG!
So the whole article and its conclusions are WRONG!
Now, can any mathematician or physicist please explain to me why I am mistaken? What am I missing?
Come on Ira.

March 10, 2011 12:10 pm

tallbloke on March 10, 2011 at 4:26 am says: “…it was reduced albedo 1979-1998 allowing more Solar energy to enter the oceans that caused the majority of the global warming at the end of the last millennium.” Rubbish. There was no such warming before 1998, it’s all fake. Satellite temperature measurements contradict this fantasy that appears on NASA, NOAA and Met Office charts. What satellites do see in this time slot is a temperature oscillation, up and down by half a degree for twenty years, but no warming until the super El Nino shows up. The oscillations belong to the alternation of warm El Nino and cool La Nina periods that are part of the ENSO system in the Pacific. Read “What Warming?” available on Amazon.com.

March 10, 2011 12:11 pm

Since such articles are very comment-rich, I assume the science is far from settled, even the basic theory.
While the charts are ok, nobody yet calculated how much is this backward IR actually affecting the earth surface temperature. Remove oceans and 99% of atmosphere (oxygen and nitrogen) with their tremendous heat-keeping capacity and run the experiment with those GHGs again. Or look at the Mars with 6,000 ppm of CO2 and its temperature like blackbody, since oceans, nitrogen and oxygen are missing.

mkelly
March 10, 2011 12:24 pm

Phil. says:
March 10, 2011 at 10:53 am
If I said surface I misspoke as I know the air isn’t going to heat the surface. My point was and is that -33 C which is used as the starting point for GHG effect may not be correct and that 0 C (STP) is a more proper place to start with any effect of GHG’s.
Again with the angry. I never said I had all the answers about gas physics I said I was asking questions and there was a difference of opinion. But now I know that, “At any given temperature they will not emit more that given by S-B at that wavelength. ”
So does this say I can use Wein Displacement Law for a temperature find the wavelength and figure the energy being emitted?

Fred Souder
March 10, 2011 12:32 pm

CommieBob,
Is that flashlight cooler than your eyes? I never said, nor have I seen anyone else on this thread say, that radiation from a hot source would somehow “block” the radiation from a cooler source. I don’t see how this example refutes anything anyone is stating. If you shine a flashlight in your eyes, it will warm you, not the other way around. The flashlight will not warm the sun. The filament in the flashlight is not a hot as the photosphere of the sun. The photons that represent the information being exchanged between the sun and the filament will result in a net warming of the filament, not the sun.
I would expect my students to state that the atmosphere (or any insulation layer), changes the rate at which energy is transmitted.
You say that wearing a coat warms the body. I say putting on a coat slows the rate at which your body transfers thermal energy to the surroundings. (and yes, this is obviously conduction, not radiation) There is not really much difference mathematically. It is that statement that greenhouse gases HEAT the earth” by re-radiating IR that is not going to sit well with the thermodynamics vocabulary police.

March 10, 2011 12:52 pm

Phil. says:
March 10, 2011 at 11:16 am
Yes I can read wikipedia also. http://en.wikipedia.org/wiki/Bond_albedo
from which you have clearly quoted verbatim.
The fact is that again you are either missing the point or you are deliberately taking things out of context.
Harry is talking about ALBEDO, you are talking about BOND ALBEDO. You have shifted the goal posts.
ALBEDO in context to Harry’s point is only relevant to visible light. Also from wikipedia, Quote: “When quoted unqualified, it usually refers to some appropriate average across the spectrum of visible light.”
And from the same wiki page: The term is derived from Latin albedo “whiteness”
http://en.wikipedia.org/wiki/Albedo
So as usual we are now unnecessarily arguing semantics, when it is perfectly clear to anyone what Harry actually means by albedo.
As for this
Phil. says:
March 10, 2011 at 11:04 am
I am no more interested in responding to waffle than I am arguing semantics.
So I refer you to my first post.
http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/#comment-617396

George E. Smith
March 10, 2011 12:56 pm

“”””” P. van der Meer says:
March 10, 2011 at 3:15 am
Ira Glickstein, why don’t you explain to your readers why the various curves for blackbody radiation in your article peak in the range of 17μm to 19μm when the http://spectralcalc.com/blackbody_calculator/blackbody.php site comes up with a peak wavelength of 9.659μm for 300K and 11.828μm for 245K. This is also confirmed by the Wikipedia graph (http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png), showing a peak at about 9μm for 310K.
I look forward to your explanation. “””””
“”””” Joel Heinrich says:
March 10, 2011 at 2:57 am
The graphics have one (unfortunately very common) error. They are plotted as a wavenumber distribution but labeled with wavelength. You CANNOT just transform the wavenumber into wavelength as they have different peaks. Much like the difference between wavelength and frequency: http://commons.wikimedia.org/wiki/File:PlanckDist_ny_lambda_en.png
The peak of a distribution in wavelength for 280K is at 10.5 µm. “””””
Well the explanations are quite simple:
For Joel Heinrich’s comment; you can in fact plot a graph of anything you like against anything else that you like, even if they aren’t in any way related.
But in this case they are. Wavelength and wave number also are related, and you can get from one to the other or from the other to one if you prefer.
And some of the Graphs Ira gives, have horizonatal axes, in both Wavelength and Wave number, as you can plainly see.
But as you can also plainly see the vertical scale has specific units. And ALL of the graphs have the wrong labels. The vertical axis is not RADIANCE. It IS “Spectral Radiance”, which is the contribution to total RADIANCE, per increment of either WAVE NUMBER as it is in these graphs, or WAVELENGTH which is also quite co0mmon. I actually am more used to seeing blackbody radiation plots using a Wavelength horizontal scale, or a logarithmic wavelength horizontal scale, and a per unit wavelength Spectral Radiance for the vertical scale; but chemists tend to use Wave Numbers and per wavenumber units, and the BB curves are different depending on which you use.
At 288 K , the BB Planck function has about a 10 micron peak spectral radiance; which is also 1000 cm^-1 BUT ONLY if the Spectral Radiance is specified per unit of WAVELENGTH increment. If the spectral radiance units are per wavenumber (cm^-1) then the peak is closer to 600 cm^-1 or around 18 microns.
Obviously the global warmists prefer the per wave number version, since the peak of the curve is right on the CO2 band, making CO2 seem more important; whereas the same data on a per micron wavelength plot peaks at 10 microns, or 1000 cm^-1 with the CO2 15 micron band now down lower on the long wavelength tail.
Although I am more used to the wavelength and the per micron spectral radiance scale, I do agree there is some merit in using frequency units, since the photon energies are proportional to frequency, and not to wavelength. But actually neither version is without problems.
On a per wavelength scale, you get high spectral radiance for solar UV; but remember that the range of wavelengths available for solar UV is just 2-3 hundred nanometres, versus the tens of microns available for the LWIR emissions from the surface.
Conversely, on a per wave number spectral radiance plot, the CO2 band gets a miserable 100 wave numbers, whereas that low amplitude water tail has many hundreds of wavenumbers bandwidth; so neither representation is without problems.
Some people graph black body radiation curves using numbers of photons, versus wavelength or wave number, and then you get a different peak again.
Now guffawingly, Peter Humbug in his now famous Physics Today special paper, emitted the per spectral units interval all together, giving a totally ridiculous plot, that implied an infinite amount of total energy; which interestingly he also called flux, rather than spectral radiance.
But we will cut him some slack, and put his error down to a simple typo; we know what you meant Peter. And incidently as he plotted it he did mean per wave number interval.
Perhaps if he had paid more attention to his subject and left out the political BS, about climate disruption, and 800,000 Kelvin earth Temperatures, then he might have caught his mistake. But as I said we will grant him some leeway. I personally never make typeos, because I review everything carefully.

tallbloke
March 10, 2011 12:57 pm

Arno Arrak says:
March 10, 2011 at 12:10 pm
tallbloke on March 10, 2011 at 4:26 am says:
“…it was reduced albedo 1979-1998 allowing more Solar energy to enter the oceans that caused the majority of the global warming at the end of the last millennium.”
Rubbish. There was no such warming before 1998, it’s all fake. Satellite temperature measurements contradict this fantasy that appears on NASA, NOAA and Met Office charts. What satellites do see in this time slot is a temperature oscillation, up and down by half a degree for twenty years, but no warming until the super El Nino shows up. The oscillations belong to the alternation of warm El Nino and cool La Nina periods that are part of the ENSO system in the Pacific.

Yeah, that’s the point. The oceans had been saving up excess solar energy and hiding it away from surface measurements in the subsurface pacific warm pool for years. Then just like I said, “at the end of the last millenium” kapow! super el nino. Large amounts of solar derived energy released into the atmosphere and recycled round the ocean surface, and an upward step change in surface temperature post 2000 as the system bounced back from la nina.

Dave Springer
March 10, 2011 1:12 pm

@Ira
I don’t think it’s a good idea to go into quantum mechanics which is generally reserved for quantum scale phenomena. Wavelengths in the micron range in a dense gas means a single wavelength spans millions of molecules at once which is in the domain of classical thermodynamics. Temperature itself is not a quantum measure but rather an average speed of a great number of molecules. That said I’ll take a stab at a lay description of individual molecules and photons in the dense portion of the earth’s atmosphere.
I just finished reading an article in this month’s issue of Scientific American about a new technique for cooling matter down closer to absolute zero than anyone has obtained before and even better a way to do it with elements that previous techniques didn’t work with. Background information in the article talked about the speed of motion of room temperature gases which if I recall correctly ranges from about 2000 miles per hour to down near zero. The gadget they constructed used a pair of laser beams that act like Maxwell’s Demon allowing lower speed molecules to pass through the pair of beams in one direction only. The volume on one side is smaller than the other and when they trap all the atoms on the smaller side (which for some esoteric quantum reason I don’t really understand doesn’t raise their temperature by compression) they turn off the lasers and let the atoms expand back into the full volume and when they expand they cool down. By doing that over and over and over each time dropping the temperature a little bit they get to within a few millionth’s of a degree of absolute zero.
Preceding the laser stage they got most of their cooling by letting a gas of atoms out through a pinhole into a vacuum. This had the effect of both cooling the atoms and getting them all moving at close to the same speed. They aimed this beam of atoms at the receding edge of a fan blade moving at half the speed of the atoms which had the effect of slowing them down even more without disturbing their distribution or speed relative to each other. After having that done to them a few times they passed into the laser chamber for the final cooling cycles.
I just thought I’d throw that in because it’s way interesting, bleeding edge, and descriptive of how temperature is a measure of speed of motion.
Anyhow, certain gas molecules have a shape that is resonant with certain frequencies of thermal radiation. When photon of that frequency intersects the molecule it is absorbed. The absorbed energy raises the molecule to a higher excitation state. In a rarefied gas it would eventually emit a photon of the same frequency and fall back to the lower excitation state in a quantum jump. In a dense gas where it is rubbing shoulder to shoulder with other molecules however the higher excitation results in greater speed and means it’s going to collide with another molecule right away, before the average amount of time elapses for a quantum transition to occur. Don’t ask me how long it takes for a quantum jump to occur. All I know is that it isn’t a fixed amount of time but rather a probability of occuring where there is a small probability of being instantaneous, a small probability of taking forever, and a greater probability of some length of time in between those two. A quantum physicist should be able to give a more definitive answer. But the answer is moot because it has been experimentally demonstrated that in a dense gas collisions occur before any significant number of quantum emissions happen. Thus Kirchoff’s law predates quantum mechanics by over half a century.
Due to the absorption of a thermal photon at narrow resonant frequences in so-called GHGs the higher excitation means that molecule is going to have a more energetic collision than it would have otherwise. A collision will also cause a photon to be emitted but the photon frequency is determined by the collision energy not the energy added by the resonant photon. Since the molecules are all moving at anywhere from 0-2000mph to start with the collisions will cause photons to be emitted over continuously varying range of energies (or frequencies as photon energy and frequency are the same thing). So what we end up with is a continuous blackbody emission spectrum reflective of the individual collision energies in billions of atoms all moving at different speeds. A few of the collisions will be very low energy giving us lower frequency photons and some will be higher energy giving us higher energy photons. The closer we get to the average speed of the molecules the more collisions we get with that amount of energy in them. Thus we get a spectrum with a peak energy at a certain frequency that falls off in a smooth curve to either side of that frequency i.e. a continuous blackbody spectrum. Since temperature is a measure of the average speed of a great many individual molecules the peak emission frequency of the spectrum is a function of temperature.

rAr
March 10, 2011 1:13 pm

Bomber_the_Cat says:
March 10, 2011 at 12:01 pm
The graphs presented reflect the change in temperature, and thus wavelength emitted, associated with the atmospheric lapse rate as measured from TOA or from the surface. Quite a different picture from a blackbody spectrum where all the emissions would be from the surface of the substance.

A G Foster
March 10, 2011 1:23 pm

TB, how do you think the ocean is heated if not at the surface? Geothermal? And why does it matter how far IR penetrates the surface? At what penetration distance do you think it would make a difference? All we need is for the top micron to be warm in order to start things cooking, by whatever means.
Willis’ point was spot on–without atmospheric heating of the ocean it would freeze. In your defense you only contradict yourself, claiming long term equilibrium, which equilibrium could only be arrived at through atmospheric heating. And globally we would see that temp/time lags even out, or no equilibrium could be achieved. That is, being a good heat sink, half the time the ocean heats the air and the other half the sky heats the ocean. If the ocean were always a net emitter of energy it would freeze fast.
Your claim that H2O molecules rise because they’re lighter is worse than the notion that IR hits the ocean directly from on high. Do you think there is a layer of water vapor that floats to the top of the atmosphere? It takes months for gases to mix through the atmosphere, and it all happens through collisions. The only thing the light molecules have in their favor is their faster speed, which does accelerate dispersion through permeable membranes and through the air.
The top film of a warm ocean is hot and dense due to radiation and evaporation. But that superdensity contributes negliigibly to circulation–it cannot compete with wind and wave action or even conduction. But we know the sea heats from the top down–geothermal heat is negligible, and the coldest water is at the bottom.

wayne
March 10, 2011 1:23 pm

Some confusion on wavelengths and wave numbers:
To go from wave number to wave length:
769 wave number, move the decimal 4 places to the left, or 0.0769 and hit the 1/x reciprocal key giving 13 µm.
To go from wave length (in µm) to wave number:
Take 13 µm, hit 1/x, gives you 0.0769, move the decimal 4 to the right to give 769/cm wave number. That simple. Maybe that will help.
The real confusion is these two type of charts are normally also flipped right to left on the x axis to each other. I agree, that makes it hard for everyone including myself to compare apples to oranges.

Massimo PORZIO
March 10, 2011 1:24 pm

What is incredible for me it’s that everyone who talk about the outgoing and the back scattered spectra, always considers the vertical paths only.
Does anybody out of here have graphs of the back scattered spectrum at ground for different angles at the same place? That is the azimuthal (90°), the 60°, the 45° and the 30° for example.
What I’m guessing here is that the graphs could be very different.
But the most interesting should be the outgoing spectrum at the TOA for the different angles, because the following links shows how the so called “limb radiation” is almost complementary to the nadir one (see the topmost graph of figure 3).
http://www.atmos-chem-phys-discuss.net/6/4061/2006/acpd-6-4061-2006-print.pdf
(The research pertain a radiosonde at abt. 34km, but how you can see from the radiance emitted from the deep space at that height there is almost no more energy back scattered).
What I’m arguing is that the increased absorption pit at 15um due to the doubling of the CO2 seen at the satellite nadir view doesn’t mean that the radiation is backscattered or held by the atmosphere, that energy just exits the atmosphere under different angles not seen by the satellites “eye”.

don penman
March 10, 2011 1:37 pm

We do observe back radiation from the atmosphere I think though that what comes back is much less than is emitted by the Earth. Long wave radiation has less energy than short wave radiation and we are told that the molecules in the atmosphere do not absorb shortwave radiation only long wave radiation at particular frequencies but the Earth absorbs shortwave radiation and emits long wave radiation there has to be a net transfer from the Earth to the atmosphere and to space because of this. I think that the presentation was more realistic than many that we have been given, it does attempt to give empirical data rather than analogy and thought experiments.

commieBob
March 10, 2011 1:52 pm

Fred Souder says:
March 10, 2011 at 12:32 pm
CommieBob,
… by re-radiating IR that is not going to sit well with the thermodynamics vocabulary police.

We are clearly talking at crossed purposes. One of the things I have expertise on is EM radiation. If the problem is vocabulary, so be it. If you are saying that a weak emitter can not radiate energy toward a stronger emitter, you are just plain, flat, dead, completely wrong.
What I can also tell you as an expert is this: If I build and energize a transmitter antenna, I can still measure the signals on it that are caused by adjacent transmitters. (The total voltage on the antenna will be the sum of the local transmitted signal and all the received signals.) In fact, a good example would be continuous wave radar. A signal leaves the transmitting antenna, strikes the sharp edge of an aircraft, re-radiates at the same frequency and, shifted by the doppler effect, returns to the sending antenna and is detected. http://www.tpub.com/content/fc/12404/css/12404_24.htm
For experimental evidence with gases and heat, Dave Springer has a post above where he describes the work of John Tyndall. http://en.wikipedia.org/wiki/John_Tyndall

tallbloke
March 10, 2011 1:54 pm

A G Foster says:
March 10, 2011 at 1:23 pm
TB, how do you think the ocean is heated if not at the surface? Geothermal?

The ocean is heated by the solar shortwave radiation that penetrates tens of metres into it.
There are fuller replies to your other misconceptions on my blog where you posted the same comment.
http://tallbloke.wordpress.com/2011/03/03/tallbloke-back-radiation-oceans-and-energy-exchange/#comment-5571

Dave Springer
March 10, 2011 2:21 pm

Fred Souder says:
March 10, 2011 at 12:32 pm
“I would expect my students to state that the atmosphere (or any insulation layer), changes the rate at which energy is transmitted. You say that wearing a coat warms the body. I say putting on a coat slows the rate at which your body transfers thermal energy to the surroundings. (and yes, this is obviously conduction, not radiation)”
If I were in a physics class I’d say the coat slows down how fast your body loses heat. If I were at a soccer game on a cold day and my kid came off the field shivering I’d say put on this coat to warm yourself up.
“There is not really much difference mathematically. It is that statement that greenhouse gases HEAT the earth” by re-radiating IR that is not going to sit well with the thermodynamics vocabulary police.”
Agreed. But in the end the surface is warmer and it’s difficult to describe how the equilibrium temperature between the surface and the 3 Kelvin cosmic microwave background temperature rises such that the heat transfer rate through the additional insulation rises commensurately. Then you might have to explain that because it’s a dynamic system with a plethora of other things changing both more slowly and more rapidly than GHGs the equilibrium temperature is a moving target that is approached but never attained for very long but that when the system goes farther out of equilibrium the harder it tries to get back to equilibrium.

Ian W
March 10, 2011 2:54 pm

Arno Arrak says:
March 10, 2011 at 10:11 am
………….Miskolczi used this database to calculate how the global annual infrared optical thickness of the atmosphere varied throughout these years. And he found that the optical thickness of the atmosphere in the infrared where carbon dioxide absorbs remained constant for 61 years, with a value of 1.87. This tells us that the transparency of the atmosphere in the infrared did not change for 61 years despite constant addition of CO2 to the atmosphere through all this long stretch of time. Hence, the greenhouse absorption signature of the added carbon dioxide which we are told about simply isn’t there. This is an empirical finding, not something derived from theory, and it overrides any calculations from theory that do not agree with it. Theories that disagree must either be modified or discarded. I want to point out also that his work came out in 2009 and no one so far has attempted to present any peer-reviewed arguments against it. Miskolczi concludes: “It will be inferred that CO2 does not affect the climate through the greenhouse effect.”

On the one hand you have very educated people doing back of the envelope calculations using formulae that do not quite apply to the task with inputs to the formulae that are based on oversimplified assumptions, and willfully ignoring major aspects of the atmosphere.
On the other you have an empirical scientist going out and using actual atmospheric measures to invalidate the claims.
Stands to reason that the ‘back of the envelope incomplete formula disregarding major parts of the system’ will be the one people believe.
AGW looks more and more like the Phlogiston theory – that was supported by scientific consensus too.

Dave Springer
March 10, 2011 3:24 pm

commieBob says:
March 10, 2011 at 1:52 pm
“What I can also tell you as an expert is this: If I build and energize a transmitter antenna, I can still measure the signals on it that are caused bay adjacent transmitters. (The total voltage on the antenna will be the sum of the local transmitted signal and all the received signals.) In fact, a good example would be continuous wave radar. A signal leaves the transmitting antenna, strikes the sharp edge of an aircraft, re-radiates at the same frequency and, shifted by the doppler effect, returns to the sending antenna and is detected.”
I only worked on a conventional pulsed (weather) radar 35 years ago. CW Doppler can’t do ranging without adding in frequency modulation. What a nightmare sorting out the return signal would have been with old analog electronics!
That said, constructive and destructive interference happens in all situations and really muddies up the picture. As long as the frequencies are different you sort it out using analog filters or a DSP running FFTs on it. In fact new fangled cars have active noise reduction where road & engine noise in the passenger compartment are drastically reduced by destructive interference – there are listening devices that generate sound of equal amplitude but 180 degrees out of phase. The energy in the sound waves has to go somewhere so in the car I reckon that heats the air a little bit.
But this raises a puzzling question for me when it comes to electromagnetic waves in a vacuum. Suppose we have two emitters at exactly the same frequency but 180 degrees out of phase. The wavefronts meet and perfectly cancel out. Say they are separated by a two light-seconds and each starts emitting at exactly the same time. Theoretically the wavefronts will meet halfway and cancel each other out so the radiation from either will never reach the other. The puzzling part is where does the energy go?
I had an interesting email conversation with a couple of academics who’d published an article in SciAm about common misconceptions about the Big Bang. They were describing the expansion of the universe and how the 3000K temperature of the early universe (the point at which it became transparent to radiation) had fallen to 3K today due the universe expanding by a factor of 1000 since that time. They explained it in quantum terms using photons and how the individual photons lost energy over time. I objected and said that photons propogating through a vacuum are immortal and unchanging. If the photon lost energy, I asked, where does the lost energy go and what form does it take? I went on further to say the particle description was inappropropriate. They should have explained it using waves which is appropriate for EM propogating through a vacuum – the fabric of space simply expanded and stretched out the electromagnetic waves along with it. That satisfies the law of conservation of energy. They came back with some vague BS about gravitational energy and refused to answer any further questions. There may very well be a gravitational answer but unfortunately there’s currently no theory of quantum gravity. Why do people insist on making things so complicated when there are simple answers staring them in the face?

George E. Smith
March 10, 2011 3:46 pm

Well when I think about these issues, I ask myself; How can I explain this to those ten year old kids, on “Are You Smarter than a Fifth grader ?”
Which is not to suggest that visitors to WUWT are no smarter than fifth graders; simply that some visitors kn0w Physics, or Physical Chemistry, or Extra Terrestrial Biology, and some understand how to turn ordinary mud into China, or draw a bunch of chicken scratching line drawings, that illustrate some political fox pass of this week’s news cycle. So if you can explain anything to a ten year old, then probably most adults, regardless of their specific skill sets, can grasp what you are talking about, (about ANYTHING).
So does putting on the coat; say after you run in off the soccer field, warm you up, or doesn’t it ? Well of course it does. Assuming that you had breakfast before going out on the field, then your body is taking energy out of the chemical decomposition that is going on in your GI tract, so you are able to get out on that field and run after the guy with the ball. Or instead, you could perhaps climb up K2, instead of kicking the leather Bucky ball.
Now without the coat on, you are going to freeze your arse off; specially if you decide to climb K2. Because you are connected to the environment (atmosphere) by the largest organ in the human body; your skin; and it was specially designed for getting rid of EXCESS HEAT, both by way of conduction directly to the cold atmospheric molecules that touch your skin (which is at 98.6 deg F), and also by evaporation of H2O through the pores of your skin, removing something like 645 calories per gram of water lost (depending on the Temperature). So your skin is going to cool down, despite the fire that is raging within. Remember that “Calories” of food, are actually kilocalories, of energy (disguised so as to not shake up the weight conscious ladies.)
So climbing K2 in a speedo, is not too cool; you’ll freeze; same thing for coming off the soccer field in winter.
So you put on that jacket; maybe lined with Eider down, or some synthetic super insulator, or some breathable fabric, that lets air in and out, but provides a high thermal resistance to heat flow, either by conduction, or convection, and even by radiation (silver linings).
Now instead of your skin assuming -40C/F deg on K2, or just say 20 deg F off the soccer field, the internal heater (decaying hot dogs) continues to try to drive heat out through your flesh and skin, but now since you raised the external thermal resistance, the Temperature drop across your internal thermal reistance goes down, so your skin warms up. Most people when they get goose bumps, actually experience it on their skin, in the form of little protrusions, not unlike goose bumps, and it makes them shiver, which raises the thermal activity level to try and stop that goosebumpy skin from complaining it is cold out there. So it is your skin that tell you; hey it is bloody cold out here !
Do not expect your gall bladder to tell you when it is too cold; or for that matter when it is too hot; gall bladders just do not know hot and cold.
Guess how a lot of people who are dumb enough to try and climb K2, end up dying up there. No they don’t smash their limited brain capacity on a glacier 4,000 feet below them; nor do they freeze their arse off at -40C/F deg.
They COOK their insides (gall bladder included). That coat (which of course couldn’t possibly warm them), stops their skin from freezing or even goose bumping, so their skin cannot get rid of all the excess internal heat energy generated from the decay of hot dogs, and drinking too mcuh 5-hour energy.
So their internal organs, which are mostly lacking in nerve endings, and specially don’t have a goose bumper, or a sweat gland to tell them; hey its getting too damn hot in here, simply cook; the person dies from heat stress, because that coat that they need to stop their skin from freezing, also stops their body core, from dissipating the excess heat that is generated, specially from doing dumb things like climbing K2. (or playing soccer ).
So the key thing about the coat, warming, or a blanket on your bed for example, is that there is a heat (energy) soure INSIDE the thermal barrier.
So try putting an inflatable pin-up doll in your guest bed overnight along with a Thermometer to measure the Temperature.
Next night try putting a real pin-up doll in your guest bed overnight, along with the Thermometer to read the Temperature (record your results). For a real clincher test, join the pin up doll in your guest bed over night along with the Thermometer, to record the Tempertaure. NOTE Reread the above about how idiots who climb K2 die, before trying this last experiment.
Well of course it is the same with the atmosphere and GHG.
You see there IS a continuous source of HEAT inside that enveloping blanket of GHG’d atmosphere.
The atmosphere is to a large extent (but not completely) transparent to the incoming solar spectrum EM radiation that constantly arrives from the sun, and most of it goes right into the oceans to depths from a few cm down to hundreds of metres, and the rest gets absorbed in rocks, or plants, or Urban heat islands; and the bulk of that absorbed radiant energy is quickly converted into waste heat; pretty much the same as decaying hot dogs; and that waste heat source is now totally INSIDE the enveloping high thermal resistance atmospheric blanket, that surrounds the earth.
So no you can say the blanket is not cooking the earth; just like the down jacket does not cook the K2 climber; but the internal heat source most surely does; and it is the continuous input of solar spectrum EM radiation energy, that ultimately becomes extra heat to warm the planet; because the “blanket”, slowed down the rate of exit of that excess “HEAT.”
Now I’ll bet, that any ten year old kid can understand that.

Phil.
March 10, 2011 3:54 pm

Will says:
March 10, 2011 at 12:52 pm
Phil. says:
March 10, 2011 at 11:16 am
Yes I can read wikipedia also. http://en.wikipedia.org/wiki/Bond_albedo
from which you have clearly quoted verbatim.
The fact is that again you are either missing the point or you are deliberately taking things out of context.
Harry is talking about ALBEDO, you are talking about BOND ALBEDO. You have shifted the goal posts.
ALBEDO in context to Harry’s point is only relevant to visible light. Also from wikipedia, Quote: “When quoted unqualified, it usually refers to some appropriate average across the spectrum of visible light.”

But I referred to Bond albedo and you said in reference to that that albedo only referred to visible which is not relevant to my correct statement.
How can the ratio of the temperatures of Earth and Venus only depend on their distance from the sun when the albedo (bond albedo used correctly as I did before) is so different? Harry’s original statement was wrong.

Phil.
March 10, 2011 4:17 pm

mkelly says:
March 10, 2011 at 12:24 pm
Phil. says:
March 10, 2011 at 10:53 am
If I said surface I misspoke as I know the air isn’t going to heat the surface. My point was and is that -33 C which is used as the starting point for GHG effect may not be correct and that 0 C (STP) is a more proper place to start with any effect of GHG’s.

As I told you before STP is an arbitrary point with no physical significance at all.
Again with the angry. I never said I had all the answers about gas physics I said I was asking questions and there was a difference of opinion.
And I told you that facts are not something you can have opinions about, perhaps you should start listening to the answers to those questions.
But now I know that, “At any given temperature they will not emit more that given by S-B at that wavelength. ”
So does this say I can use Wein Displacement Law for a temperature find the wavelength and figure the energy being emitted?

At any temperature a distribution of wavelengths will be emitted (S-B distribution), Wien’s Law will tell you the the wavelength of the peak of the distribution. It will give a different value if applied to the spectrum in the frequency domain, a cause of confusion above.

sky
March 10, 2011 4:24 pm

Glickstein’s analogy of the “greenhouse effect” to an ELECTRIC blanket is totally incorrect. Electric blankets operate on an INDEPENDENT source of energy to produce additional heat. In stark contrast, GHGs can only operate on the energy thermalized mainly at the surface, producing none on their own. It’s sad to see such basic physical confusion being perpetuated here on WUWT.

Dave Springer
March 10, 2011 4:30 pm

Massimo PORZIO says:
March 10, 2011 at 1:24 pm
“What I’m arguing is that the increased absorption pit at 15um due to the doubling of the CO2 seen at the satellite nadir view doesn’t mean that the radiation is backscattered or held by the atmosphere, that energy just exits the atmosphere under different angles not seen by the satellites “eye”.”
Over the arctic looking down from 20km the bottom of the pit conforms perfectly to a 225K blackbody curve while outside the pit the curve fits perfectly to a 265K curve. Clear arctic air is very dry so water vapor has a minimal effect. 265K is -8C which is the surface temperature of the ice. Yet in the 15um window the ice appears to be -48C. What we are seeing in the 15um window is the so-called emission altitude. Dry adiabatic lapse rate is 1 Kelvin per 100 meters. The emission altitude is 4000 meters. This is the height at which the CO2 in the atmosphere has absorbed all the 15um radiation available upwelling from the ground and what remains is thermalized radiation from the atmosphere at that height.
Were we to reduce the 265K curve to a temperature such that the total reduction in volume underneath the lowered curve is the same volume as the hole then we would have the equilibrium surface temperature absent ALL atmospheric CO2. I believe that works out to about 15C. The thing of it is that the first 100ppm or so of CO2 does the lion’s share of the work so by the time we get to the surface equilibrium temperature difference between 280ppm and 560ppm CO2 (a doubling) the increase is only about 1.0C.
This takes us round to my hypothesis that the MOST IMPORTANT climate function of CO2 is in raising the average surface temperature of the earth from -23C (no greenhouse gases – the average surface temperature of the moon) to -8C. At -23C the earth is a snowball and positive feedback from the exceedingly high albedo of ice keeps it that way. Once CO2 is in the atmosphere to help melt some snow and ice this lowers the albedo dramatically where it matters the most (at low latitudes where the sun is the strongest) and additionally starts pumping water vapor into the atmosphere from the liquid water surface which accelerates the melt and we are back to a water planet instead of an ice planet. Once we have a water planet the low albedo of the global ocean and the higher absolute humidity keeps it a water planet (barely, with a few excursions to a snowball once every few hundred million years).
So CO2 serves as “kindling” to light the fire which melts the ice and turns what would otherwise be a frozen world into a liquid water planet. The water cycle, once activated takes over the show. And water limits the maximum temperature through negative feedbacks of evaporation and cloud formation.
During most of the earth’s history CO2 has been up around 2000ppm. The current 380ppm is dangerously low which probably has a lot to do with why we’ve been in an ice age for the past 3 million years.
Despite even 2000ppm CO2 the earth has still entered ice ages once in while and a few times severe enough to freeze the entire planet. It’s a bit of a puzzle what could possibly kick the earth out of a snowball episode but after about 10 million years it does indeed melt. The most popular hypothesis for the melt is that when the earth is frozen over all the normal CO2 sinks are gone. The global ocean doesn’t absorb any CO2 and there are no forests that grow on glaciers taking in CO2 and no chemistry between CO2 and ice forming carbonate compounds. But volcanoes don’t stop belching out CO2 so over the course of millions of years atmospheric CO2 rises and rises and rises until its greenhouse effect starts driving back the ice at an accelerating rate and viola – a water world is born again and the water cycle caps the maximum temperature at a point nice and comfy for living things from pole to pole.

Phil.
March 10, 2011 4:32 pm

Bomber_the_Cat says:
March 10, 2011 at 12:01 pm
I recently voted for this as being the best scientific blog but, in response to this article, we still get people who say that back radiation cannot happen or wish to divert the discussion to some strange theories about the temperature on Venus.
But there is a serious problem with Ira’s article!
As expressed before (6:10AM); his blackbody curves are showing a peak at about 18 micron when they should be peaking at about 10 micron for a 300K blackbody. So, the graphs don’t make sense
Now Joel Heinrich at 2.57 AM and Phil at 8.21AM, March 10, say this is because Ira has converted wave number to wavelength.
But this does NOT cut the mustard. This is a perfectly acceptable thing to do, albeit ending up with a non-linear scale rather than a linear scale. In fact, the
Source Data itself includes the wavelength scale, so it is nothing to do with Ira’s conversion.
On the ‘wave number’ scale the peak of the blackbody curves should be around a wave number of 1000 – which they are not! They peak around a wave number of 600 – which is WRONG!
So the whole article and its conclusions are WRONG!
Now, can any mathematician or physicist please explain to me why I am mistaken? What am I missing?

What you have in your head is the spectral radiance wrt wavelength whereas what’s plotted in Petty’s paper is spectral radiance wrt wavenumber, as I said above you can’t linearly transform them and they don’t have the same shape nor a maximum in the same place. Both are correct they’re just showing the data in a different form, as George has explained above chemists tend to prefer one form, physicists another.
If you use Wien’s Law to determine the maximum there are two forms, a frequency form and a wavelength form, the peak frequency does not correspond to the peak wavelength using c=λν.

Stephen
March 10, 2011 5:10 pm

Please let me know if I am mistaken, but it seems to me that the blackbody radiation-spectrum gives an upper limit on downward emmissions from the atmosphere. I understand that if the atmosphere gave such a spectrum, that would would imply that it actually absorbed everything and re-emitted isotropically.
It looks like where CO2’s absorbtion and emmission are significant, the blackbody-spectrum has already just about been reached. Does this imply the small difference between current emmissions and those of a blackbody gives a low upper limit on the potential effect of additional CO2 on the temperature?

Tim Folkerts
March 10, 2011 5:15 pm

If only there was a FAQ for some of these issues! Then we could refer people to previous discussion rather than rehashing all the discussions like:
* What is “heat”? (confusion abounds about how “heat” relates to energy flow, energy content, temperature, of “thermal IR”. As long as different people use different definitions, people will continue to talk past each other.)
* Can energy flow from cool objects to warm objects? (yes!)
* Can NET energy (heat, as defined in thermodynamics) flow from cool objects to warm objects? (no!)
* Is “the greenhouse effect” the same as how a real greenhouse works? (not at all, so any arguments based on such analogies are pretty much useless from the start.)
* Do GHGs “heat” the ground? (no, by the definition of “heat” above) Do they “radiate IR energy” to the ground? (yes) Do they “increase they temperature” of the ground? (definitely, compared to an atmosphere with no GHGs) Do they “warm” the ground? (as I would use the word, my answer is yes)
What else would need to go in that FAQ?

sky
March 10, 2011 5:27 pm

Dave Springer says:
March 10, 2011 at 4:30 pm
“So CO2 serves as “kindling” to light the fire which melts the ice and turns what would otherwise be a frozen world into a liquid water planet.”
Don’t have time for a lengthy discussion, but the thermal mass of CO2 is way too small for it to play the “kindling” role that you suggest. A CO2-free Earth might be slightly cooler, but would not be frozen, because convection would nevertheless heat the base of the atmosphere to similar temperatures, thereby sharply reducing the radiative loss from the surface. It’s the PRESENCE of a dense atmosphere, rather than its chemical composition, that matters most thermodynamically. This is entirely lost in treatments that look only at radiative heating of the atmosphere.

Oliver Ramsay
March 10, 2011 5:36 pm

when it is too hot; gall bladders just do not know hot and cold.
George E. Smith says:
March 10, 2011 at 3:46 pm
“Guess how a lot of people who are dumb enough to try and climb K2, end up dying up there. No they don’t smash their limited brain capacity on a glacier 4,000 feet below them; nor do they freeze their arse off at -40C/F deg.
They COOK their insides (gall bladder included). That coat (which of course couldn’t possibly warm them), stops their skin from freezing or even goose bumping, so their skin cannot get rid of all the excess internal heat energy generated from the decay of hot dogs, and drinking too mcuh 5-hour energy.”
——————————–
This must be why the Canadian woods are filled with people running around harvesting bear gallbladders from the poor creatures that got cooked from the inside.

March 10, 2011 6:03 pm

John Marshall says:
March 10, 2011 at 2:04 am

Its temperature will fall below that of the surface, and the 2nd law of thermodynamics forbids heat flow from cold to hot ( this is heat flow by any means available) so this rising warm air, relative to the surrounding air though colder than the surface, cannot warm the surface.

I keep hearing this. It gets boring.
Cool Body radiates x in all directions, including towards Warm Body. Warm Body radiates 2x in all directions, including towards Cool Body.
Warm Body cools more slowly than if Cold Body were not present
Overall heat transfer is not from Cool Body to Warm Body, so that 2nd law has not been broken, and nobody (pun intended) goes to Thermodynamic Jail.

Fred Souder
March 10, 2011 6:06 pm

CommieBob,
I am not asserting that a cold object cannot radiate EMR to a warm body. Only that the net flow of energy will only go from a hot to cold.
Since you are an EMR expert (no sarc) , I do have a question that I have been wrestling around with, but have been unable to find an answer.
Any other EMR experts please feel free to chime in. None of my texts or classwork deal with this (at least through grad school), as it is a combination of thermo, general relativity, and quantum mechanics.
When an object drops from an excited state and emits a photon, that photon is then absorbed somewhere else in the universe by another object. To that photon, all distances are zero, time does not exist, and the photon transfers its information instantly. The photon doesn’t exist between between points (GR). Is it possible for that individual photon to transfer information from a less excited molecule (or electron or whatever) to one more excited? The answer isn’t the obvious yes that I thought initially. In fact, blackbody radiation would still work form a cold source to a warm source because of all the random motion and complete statistical range of excitation levels of the particles. I know there are some super smart people trying to tie in thermo with QM, QED, QLG, so it would seem if it ties in, then the photon would only be able to transmit information from an excited state to one less excited.
In other words, since this is all happening instantly from the perspective of the photon, can a photon “tell” an excited object to become more excited as it “tells” the less excited object to become even less excited. If it is all based on probabilities of states (QM), then the answer should be yes, even though this goes against the grain of the Thermo underpinnings.
Thanks!

March 10, 2011 6:13 pm

RJ says:
March 10, 2011 at 4:17 am

Would a human being cook if he or she was enclosed in a container of CO2
In theory a percentage of the heat given off would return and increase the body temperature if the GHG theory is correct. Something surely is seriously wrong with the GHG theory.

NO. They would just cool down more slowly, so the overall temperature of the human would increase slightly until equilibrium (energy gained = energy lost) is reached. This assumes they are living (which is not a given in a container of CO2, BTW), and is generating a constant amount of energy from that living process.
I really fail to see why something so obvious is so hard to get. When you add a blanket at night, do you ‘cook’? No. Your temperature just increases until equilibrium (energy gained = energy lost) is reached. Try it!

March 10, 2011 6:17 pm

Jer0me,
OK then, let’s see you falsify this.

March 10, 2011 6:21 pm

Joe Lalonde says:
March 10, 2011 at 4:49 am

First, without planetary rotation, there is no convection

I think a lot of convection is due to differing temperatures. This is why you get a lot of wind when clouds come over on a sunny day. Also why the wind (assuming few clouds) goes toward the sea (which is warmer) in the first part of the day, and way in the latter part. Simple temp.

as the planets own energy is the centrifugal force it generates at 1669.8Km/hr due to the vacuum of space.

No. 1. there is no such thing as centrifugal force. It is simple inertia. Centrifugal force, does exist, however, and that is what stops everything flying off. In this case, gravity. The rest of the concept you put forward makes no sense to me.

The atmosphere bends a great deal of light and solar energy with the suspended molecules in the atmosphere with the tilting of the planet to the sun. The hottest point of the sun is it’s equator that our planet drifts through due to proximity and size of the suns equator.

Well, between the tropics, anyway. It differs through the year.

Next very little consideration for the absorption and storage of heat that is then released at night.

That is a false assumption. Just because sunlight is shown does not mean the rest of it stops when the sun goes down.

Anton Eagle
March 10, 2011 6:27 pm

This might be too simplistic… but what I would like to see are some specta showing the measured (not calculated or theorized) downward radiation from the sun as measured at the TOP of the atmosphere. I find it amazing that the sun emits so little in the IR wavelengths.
Anyway, if there are good measured downward spectra at the top of the atmosphere, and good measured downward spectra at the bottom of the atmosphere, then the difference between the two should clearly show what’s being re-emitted (downward) by the atmosphere… shouldn’t it?
So… what do the data show?

pochas
March 10, 2011 6:33 pm

Back radiation is like what a diver sees when he is on a wreck 100 feet down with visibility of only 50 feet. He can’t see the surface but he can certainly see. Its like a blue glow that comes from all around. If we could see infrared and not visible, thats what we would see.

commieBob
March 10, 2011 6:43 pm

Dave Springer says:
March 10, 2011 at 3:24 pm
But this raises a puzzling question for me when it comes to electromagnetic waves in a vacuum. Suppose we have two emitters at exactly the same frequency but 180 degrees out of phase. The wavefronts meet and perfectly cancel out. Say they are separated by a two light-seconds and each starts emitting at exactly the same time. Theoretically the wavefronts will meet halfway and cancel each other out so the radiation from either will never reach the other. The puzzling part is where does the energy go?

The energy is still there and keeps on traveling. The stationary null that you describe is matched by a stationary peak somewhere else.
What you describe is similar to a standing wave in a cable. You get peaks and nulls. As you travel along the cable, the RF voltage varies in a predictable manner. The two waves continue to exist and travel along the cable and get to their respective ends where they reflect or are absorbed. http://en.wikipedia.org/wiki/Standing_wave
In three dimensional space, you get a predictable pattern. If I arrange two antennas wrt space and phase, I will get a particular pattern in the resulting field. No energy is created or destroyed. It is, however, redistributed. In the case of two antennas the pattern is basically based on hyperbolas. In other words, if you plot the nulls (or peaks, but nulls are easier to measure) you get hyperbolas.

George E. Smith
March 10, 2011 6:44 pm

“”””” Ira Glickstein, PhD says:
March 10, 2011 at 5:21 pm
Phil. says:
March 10, 2011 at 8:21 am
… the Petty data is spectral radiance (mW/m^2.sr.cm-1) plotted vs. cm-1 and is correct, the transformation to a plot in terms of wavelength is non-linear (and is why it’s misleading for Ira to have reversed the axis on his plots, should leave it in wavenumbers).
Thanks for your comment, Phil. The Petty plots have both wavenumber and wavelength, and I preserved his wavelengths in my simplified curves when I reversed the plots horizontally. The reason I think in wavelength is that my experience with IR is based on near-IR 1.06μ laser rangefinders and far-IR ~10μ Forward-Looking Infrared video sensors. However, the conversion between the two measures is simple:
Wavenumber = 10,000/Wavelength
The reason for the difference in peaks has nothing to do with the use of Wavenumber or Wavelength.
As Phil noted, Petty plots Radiance.
My earlier posting plots Spectral Intensity. (It is from http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png.
From Wikipedia:
The units of the spectral intensity are energy/time/area/solid angle/frequency. In MKS units this would be W·m-2·sr-1·Hz-1 (watts per square-metre-steradian-hertz). “”””
Notwithstanding ANYthing that you may find on wikipiddle, the units of “spectral Intensity” ARE NOT energy/time/area/solid angle/frequency.
“INTENSITY” (fancy word for candlepower) has NO MEANING for anything but a POINT SOURCE; so with /area/ in there the intensity (spectral or not) would be infinite.
The /area/ for a source of non zero area, converts Intensity (spectral or ot) into RADIANCE ore STEARANCE which some snobs like to use.
)
As to the per frequency for the “spectral Intensity” or “spectral Radiance”, it could be either per frequency (or wavenumber, or per wavelength; so long as it is stated; either /cm^-1 or /micron.
I like to use per wavelength; because I much prefer to use a normalized Black Body Radiation plot, which is totally universal; and acknowledges that the Planck Radiation Function is a function of the single independent variable lambda.T which immediately points to the Wien Displacement Law.
So a normalized plot plots on a horizontal scale (usually logarithmic) whose variable is (lambda.T)/|lambda.T|max; and the Y-axis, is spectral radiance/ |spectral radiance|max and usually is plotted on both a linear or a logarithmic vertical scale.
So the BB curve typically goes from say (0.1) t0 maybe (100) in X and (0) to (1) in Y or 1.0 down to 10^-5 on the logarithmic plot.
An excellent graph can be found in “Modern Optical Engineering” by the late Warren J Smith, who was for years with Infra-Red Industries in Santa Barbara California.
Now yes you can reformulate all of this in terms of frequency or wavenumbers; and then Wien’s Displacement law becomes T/f is constant.
And yes a frequency based plot acknowledges the Einstein relationship E = h.nu; but as I have already noted, neither a wavelength or wave number graph is very intuitive (on the spectral interval) since one form favors the UV, and the other favors the far IR, as far as viually indicating relative importance. There are simply huge wave number ranges associated with those miserable looking spectral radiance values in the H2O bands; but the very prominent looking CO2 band (in the wave number pictorial) has a measly 100 cm^-1 for spectral bandwidth.
Note that 25% (almost exactly) of BB radiation occurs at shorter than the spectral peak Wavelength on a per micron spectral radiance plot; I have no idea how much lies below the peak on a per wave number plot; but you can easily work it out.
As I said some people like their BB plots in numbers of Photons, not in direct energy or power units.
But nyet on the spectral Intensity; the earth surface or atmosphere is not a point source.
For the pedantic, the actual IRRADIANCE on a surface due to a finite area source (on axis) differs from that due to a perfect point source by less than 1% for a source to surface distance that is ten times the source diameter.
Also there is no such thing as a point source of anything (physical). For EM radiation the boundary conditions for Maxwell’s Equations cannot be satisfied for a point source; and for the Quantum Physicists, a point source by virtue of the Heisenberg uncertainty principle must have an infinite spectral bandwidth; so the energy or power in any finite wavelength or frequency range would be zero so it would be undetectable at any wavelength or frequency.
I don’t have a problem with Ira’s use of both the wavenumber and wave length horizontal scale, as Ira said, that makes no difference, except as Phil pointed out, the wavelength conversion scale from wave number would be non-linear. But just keep the per wave number or per micron units straight in the sepctral Radiance units, and then everything is kosher.
Peter Humbug solved that dilemma by simply leaving out the per frequency altogether; but I’ll give him a typo yellow card on that.

Phil.
March 10, 2011 6:52 pm

Fred Souder says:
March 10, 2011 at 6:06 pm
Is it possible for that individual photon to transfer information from a less excited molecule (or electron or whatever) to one more excited?

Yes provided that hν for the photon matched the energy difference between the two higher energy levels (in a Q branch for example).

richard verney
March 10, 2011 7:00 pm

Ira
Thanks for your further comments. I now understand why you suggested an electric blanket as analogy. However, I remain unconvinced that it is in fact a superior analogy. In fact in some ways, I liked the comment by Alan McIntire March 10, 2011 at 6:08 am comparing the position with traffic flow over a section of road with accidents, road works whathaveyou. My response to that is that yes at peak times, there are snarl ups and the journey time is increased, however, late in the evenning and at night, there is less and less traffic indeed during the night no traffic such that all the cars eventually exit the highway and the road is completely free come morning. All the traffic has disippated akin to all the energy received by the Earth finally finding its way out to space so that there has been no effective build up of energy.
I digressed, reverting to your electric blanket. Lets for the sake of argument assume that you are right with respect to the power source (although I have concerns regarding the lack of independent power source), one of the problems I have is how this all works in the real world (as opposed to in isolation in a laboratory), by which I mean that other atmospheric processes are at play which may overwhelm the effective influence of effective heat transfer from the sky to the Earth. In particular, convection. Further, does the energy absorbed by say CO2 thermalise before it has had an opportunity to radiate.
Say that you normally make your bed by creating a sandwich of a sheet, electric blanket and ordinary blanket. When the electric blanket is plugged in, this keeps you nice and snug. Now consider that you have a 4 poster bed but without any curtains. Instead of your electric blanket being placed on top of the sheet, it is suspended at the top of the 4 poster frame say about 3 feet above the matress. You now only have the sheet and the blanket directly over you. When the electric blanket is switched on, the down radiated heat is not sufficient to warm you, nor does the down radiated heat sufficiently prevent heat loss being radiated from your body. With this set up, you are cold at night. Even in the stillness of your bedroom, convection overcomes you feeling any real benefit from the down radiated heat. I am not saying that the electric blanket has no effect, only that it is overcome by other processes.
The fact is that radiation is weak compared to convection. I have gas central heating running radiators. When the system is on, even if I place my hand within 1/2 inch of the face of the radiator (ie., place my hand parallel with the radiator close to the surface of the panel), I struggle to feel the heat. Yet, if I place my hand several feet above the radiator, I can still feel the heat. Of course, if I touch the radiators they are very hot to touch and unpleasant to keep your hand there for any length of time.
I must admit that I struggle to see that the amount of energy that the Earth is said to receive from back radiation is more than the energy it receives from the sun in circumstances where the source of that back radiation has ultimately all come from the sun (if one ignores geothermal energy being produced from the centre of the Earth). If that were true, on a sunny day, there would not be a substantial difference in temperature between the temperature in the sun and under a cloud. At night, our bodies could be warmed and bask in the radiant energy from the sky which would be safer than sun bathing, and the photo electric cell would not be tuned to capture sun light but rather research would be directed at tuning it to work on back radiation which is available 24 hours a day and whether it is sunny or not. It appears to me that the heat transfer from the Earth from the ground to lower atmosphere is dominated by other processes (particularly the water cycle) which completely overwhelm the effects of back radiation.
I am not saying that there is no ‘greenhouse’ effect (how I dislike that expression) but rather in the real world, the effect of back radiation may be considerably less than is speculated (and I use this word since there is no body of empirical observational data backing up the claims) and I consider that the effect of any change in CO2 concentratioons above present levels is likely to be extremely modest. Indeed, Miskolczi could well be right on how it all works.
PS. Whilst I do not necessarily agree with various points you make, I liked your post and it certainly has generated many interesting comments and views.
PPS. I think you should consider carefully whether you wish to run with the eklectric blanket since this in my opinion in many ways confues the issue. Would not a better analogy be a one way mirror which is losing some of its mirrored surface such that it allows most of the incoming sunlight to pass through but only allows some of the LWR emitted from the Earth to pass back through towards space and reflects some part of the LWR back down towards the Earth.

March 10, 2011 7:38 pm

I really like Dr. Glickstein, but I think he’s way off base with his ideas about a cold, rarefied gas heating the earth’s surface. Trapping radiation? Really? You might delay it’s escape to space by a few microseconds, but storing heat energy? Back radiating? Unicorns?
Think about this. You have a heated bowling ball and a passive bowling ball. Suppose you could flip a switch and the make the passive ball instantly colder. What happens to the radiation between the balls?
A) Stays the same.
B) Increases.
C) Decreases.

Phil.
March 10, 2011 7:52 pm

richard verney says:
March 10, 2011 at 7:00 pm
Would not a better analogy be a one way mirror which is losing some of its mirrored surface such that it allows most of the incoming sunlight to pass through but only allows some of the LWR emitted from the Earth to pass back through towards space and reflects some part of the LWR back down towards the Earth.

I’ve used this as an example to show how the recycled IR can exceed the initial solar radiation.
Consider a blackbody illuminated by 300W/m^2 solar radiation via a dichroic mirror which passes 100% of the solar wavelengths but reflects 50% the IR (BB) incident on it. At equilibrium the temperature of the BB has increased so as to radiate 600W/m^2 of IR, 300W/m^2 of which is returned to the surface. If the reflectivity of the mirror is increased to 66.6% then the BB radiates 900W/m^2 of IR, 600W/m^2 of which is returned to the surface.

Brian H
March 10, 2011 7:57 pm

What’s the electric blanket plugged into, again? What is this external source of energy pumping the system up? I got lost somewhere in the bouncing frequency stuff.
What is also misrepresented is the minusculeness of the bounce-back. I refer you to Noor van Adler: http://climategate.nl/wp-content/uploads/2011/02/CO2_and_climate_v7.pdf

The trends of the temperature in the high atmosphere in the last half century are very negative, on and above this height where the deep convection reaches. Cloud tops radiate much more intensely
than the thin air on this height. This is the cause behind the cooling, as much as the CO2 increase is.
This cooling trend increases the effective environmental lapse rate and so reinforces the strength of deep convection. This means that in this respect, more CO2 has a cooling effect rather than a warming effect.

March 10, 2011 8:00 pm

All these posts about backradiation towards quantum mechanics made me go back to the basics, looking at it using the old thermodynamics and I ended with a result where there is backradiation but without any resulting effect. I dissected the first basic steps where radiation hits the surface and this seems enough to get answers.
In the radiative solution, the non-GHG surface temperature is found by using E = sigma T^4 (SB) where E comes only from the sun. If then T could go up because of GHG, this would mean more energy input would be needed. So where does this come from, are GHG capable of generating energy? Many people found a justification for this, but think of the following?
If we look at the CO2 idea as a plate with T2 above earth, you have something like the
classic radiation between two plates here, where the sun would heat the lower one which is T1.
So sure radiation will go both ways, there is backradiation and I go with that.
But read that in an end state (equation 19.2) where both plates would be T1, Qnet will be zero. Surface1 will never get any warmer by the (back)radiation from above, despite the accumulated (heat)energy in plate 2 and radiation going everywhere. Radiation is energy but is never heat.
So you can do this calculation with multiple radiation layers (like CO2 layers) and the
result will be the same. T1 will not get any warmer with Qnet getting zero, and all the
layers acting as insulation (like some people think CO2 does) and all plates actually
permanently getting radiation from both sides.
(Earth atmospheer would also be forced to become T1 everywhere form the surface up if it was’nt for gravity)
So what are the tricks with radiation that many people don’t see?
For starters I want to say that the photons of EM radiation can be regarde as cold, only interaction with matter gives any thermal energy (to get a temperature) which is taken away (partly) when a(nother) photon leaves.
Ideas where the delay of escaping photons by GHG causes heat accumulation have no basis, heat comes from a surface (matter) and cold energy went with the photons that took off. It does’nt matter how long it takes before the photon gets out to space (in the sun it takes up to 170.000 years)
Most greenhouse effect statements/sites claim that the radiative balance of earth must be because of the First law, the greenhouse theory makes fame by using energy in-out balances etc. convincing 99.99 percent of the people. But this is all dead wrong, the Second law rules here and this means simple energy conservation is not the whole story.
Earth could conserve it’s energy by accumulating it until it would be as hot as the sun,
nothing against the First law this way.
Is’nt it strange to say at first: well, because of energy conservation earth must
radiate away what it receives and that way we can calculate what the temperature must be by using SB and so we know for sure it must be -18C (not mentioning atmosfeer, and what is actualy measured).
Then secondly: listen, we also have GHG and this makes earth +15C because bla bla. So in fact saying: it’s wrong, earth does not simply radiate away what it receives and SB does not apply, but we can calculate the greenhouse effect using it.
Or: GHG introduces new physics, if one places multiple layers between two radiative
surfaces the old laws of physics don’t work anymore. Extra energy will turn into heat and raise the original surface temperature (think of the plates, it won’t happen).
So then in the GHG physics T of earth surface is supposed to go up with still the same
amount of energy coming from the sun. Now we know the Second law wants to establish a dynamic radiative equilibrium at the surface with a certain temperature but also T = dE/dS, so this means that for T to be able to go up by backradiation up entropy must decrease?? Strictly looking at this step.
Looking closer it must look like this:
Why does the sun heat the earth to T? Because HQ radiation from the sun can leave earth as LQ IR radiation (to the cold space) in the process gaining entropy by leaving WASTE HEAT on the surface (and they don’t call this irreversible for nothing, the downgraded (photon)energy won’t be able to do the same trick at this place again).
Some LQ IR returns (backradiation), but can never get rid of more heat. Because in that case it would have to leave as even lower quality IR, and hey …… this can only happen at a LOWER surface temperature (think of the BB-spectra of earth next to the sun, all the energy flows from the high spectrum to the low one).
See the contradiction, that’s the mighty second law of thermodynamics in action that says: it’s all wrong with this heating by backradiation philosophy.
So backradiation will simply leave in a reversible proces (like reflection does) with NO
waste heat, and with the same frequency as it had and bounce whatever way it wants with no effect on the surface. It’s like in the Qnet = zero situation between the two plates (lotsa radiation, no more heating by the cold photons).
So what is ‘the heat’ that can’t go from low to high temperature, and warming the earth surface and what’s all the confusion?
Sadi Carnot who laid the foundations of the Second law meant it to be like this: heat =
entropy (read here), but Clausius later changed the definition.
Maxwell’s classic Theory of Heat states: heat is something which may be transferred from to another, according to the Second law of thermodynamics.
In any case, heat needs matter which has a temperature (as kinetic energy) and other
matter at lower T. Radiation has no temperature and can never ever represent heat. And so the photons of radiation can be regarded as cold.
Heat is the waste from the irreversible radiation phenomena happening due to what the Second law dictates and is equivalent to the entropy gained and what is called
dissipation.
Heat is released when a photon interacts with matter. Almost everybody thinks ‘the
heat’ has to do with all the energy (E= h x v) related to the cold travelling photon. So
we have seen that this is not the case, this is imaginary heat (which is not heat but
energy) that is never released as long as matter is’nt involved. Energy is only released
when the photon disappears in the matter and after that heat from this matter may flow if other colder matter is there, but this heat is not equivalent with the photon energy but is a function of the temperature difference between surface temperature and colder matter above it (Q = k*A*dT). If there is no colder matter, there is no heat (dT=0).
If earth had no atmosfeer, the photons would warm earth surface to SB temperature but no heat would ever emerge or be exchanged. Then if there was a layer of CO2 around it at 10 meters or 10 km, this would create backradiation, but the vacuum inbetween would be as cold as in space, and the backradiation would not make the surface warmer, and there would be no heat despite all the radiation.
Now because heat is flowing from earth surface we know we have an irreversible process (waste heat, making entropy rise) then we also know for sure the new lower energy photon can never release waste heat at this temperature where it came from ever again, so to think backradiation can go beyond that and even create a higher temperature is impossible by the laws of thermodynamics.
This IR photon can release energy to matter at a higher colder level, and this energy
could flow as heat to matter at an even higher colder level, but that than is in fact
where this heat is. At a higher colder level, and this is the only direction heat can go.
So the heat coming from the surface is the part of the solar radiation that did’nt get
radiated away (directly) as IR (and the total energy of this IR is thus lower than the LWR that hits the surface). Heat is the part that has nothing to do with radiation, and the IR that left has nothing to do with heat.
And so the balance becomes: IR = LWR – HEAT. So the IR that leaves the surface does’nt only have a lower frequency, but the total energy of it is also lower than the incoming LWR. This is the second blow for the backradiation.
It is this HEAT that warms the lower atmosfeer and gives the temperature’s measured (in air, so does not even have to be representing the surface), and it can only take place by conduction from surface to air(surface), (or the process in the top ocean layer) it must be transported from matter at T1 to matter at T2 and it finds its way up to space dissected from the photons that gave up their energy.
As I Googled on the photons in the sun, the radiation from core to surface passed by on many pages and since these are not infected with the greenhouse virus the discriptions go along the lines of my view. Photons jiggle in the ‘random walk’ for tenthousands of years from layer to layer, and they do not heat the core. And also found; this degradation of high quality X-ray photons to relatively low quality optical photons, is only to be expected from the Second law of thermodynamics.
And look at this: a photon can only travel a tiny distance before running into another
hydrogen nucleus. It gets absorbed by that nucleus and the re-emitted in a random
direction. If that direction is back towards the center of the Sun, the photon has LOST
GROUND! It will get re-absorbed, and then re-emitted, over and over, trillions of times. The path it follows is called a "random walk"
— Sir Arthur Stanley Eddington:
"The law that entropy always increases holds, I think, the supreme position among the laws of Nature. If someone points out to you that your pet theory of the universe is in disagreement with Maxwell’s equations — then so much the worse for Maxwell’s equations. If it is found to be contradicted by observation — well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation."

Brian W
March 10, 2011 8:45 pm

Commie Bob (March 10, 2011@5:41am)
Your point no. 5 is completely wrong. Lets dissect. You say “For those who doubt that back radiation exists, consider this: Infra-red radiation is electromagnetic radiation. It is the same as radio waves and light. While infra-red, radio and light may propagate similarly and be subject to the same law (inverse square) you have forgotten about wavelength. The obscure rays from an incandescent solid are vastly superior in their ability to heat than the luminous rays. You go on “Here are two examples of radiation from a weaker source going toward a stronger one:” Radiation commieBob is not heat. Radiation produces heat, and heat produces radiation but the radiation in and of itself is not heat, it is a byproduct. Radiation propagates without difficulty in space where there is no effective temperature(forget the 3k). I can even generate and emit emr without heat as the cause simply by plugging a transmitter into the wall and hooking up an antenna. Simply put the 2nd law does not apply to radiation at all. If it did all electromagnetic radiation emitted from the antennas of radio, tv, etc. would automatically head for colder climes like the interior of antarctica. The coldest would win(net flow). I won’t bother with your flashlight as it’s only vague misdirection. Just like agw. Backradiation as claimed is an energy absorption(conversion to heat)/reemittance towards the surface from some fictitious emitting altitude and raising surface temperature through fake amplification is clearly Bogus.
5b is vacous nonsense. “If I stand near a strong radio transmitter, I can still tune in weaker transmitters. Nothing about the stronger transmitter’s signal prevents the weaker signal from getting to me. Of course not! A strong 30 meter wave doesn’t care what a weak 20m wave does! They don’t interact because their wavelength’s or frequencies are different so the stronger signal has no possible way to prevent the weaker signal from getting to you. “The net energy flux will still be from the stronger source toward the weaker one.” Here you are implying some form of energy transfer between a 30m wave and a 20m wave or vice versa. Do you have a magic wand?
“It is a net flux though, in other words, the net flux is the difference between the two signals.” Oh, please commieBob do not emit any more pseudoscience as the net result is I can only laugh out loud.

Brian H
March 10, 2011 10:04 pm

I advise all here to read through Hans’ post above, a few times. The English is weak, but the point about the nature of entropy and ‘waste heat’ is not to be missed.

Fred Souder
March 10, 2011 10:11 pm

Phil.
you say :Yes provided that hν for the photon matched the energy difference between the two higher energy levels (in a Q branch for example).
Aha! Eureka moment, maybe? This makes perfect sense! The photon does not exist between points, but instantaneously communicates information (in its reference frame) between the point where it is emitted and the point where it is absorbed. Of course the photon energy must match perfectly where it is absorbed (and emitted). This also explains why photons are made up of the same “stuff” as matter. They merely communicate the state of the matter, which is governed by expectation probabilities that are predicted by quantum mechanics. When one “decay” or energy level drop happens that emits a photon, the other state shift elsewhere is necessary and already mapped by the photon. That would explain the “spooky” experiments we learn about in quantum mechanics that involve photons with equal energies and opposite momentum. By observing the polarization of one, we “influence” the polarization of the other instantaneously at any distance. Of course, the photons must “know” instantly the properties of the other because they are the vehicles of communication relaying the change of states, and the photons already “know” of their destination, as they are not under the influence of time. The properties of the photon, the energy and momentum, are the communications between states of matter separated by a distance. This feels like a logical connection between spec. relativity and quantum mechanics.
Now, how does gravity fit in with all this?…
Need more help, Phil.

Anton Eagle
March 10, 2011 10:55 pm

Folks, re-radiative shells (excuse my made-up lingo) and their ability to indirectly raise the temperature of the radiating object (via reduction of the rate of heat loss) is real. NASA has been using this for decades to keep their space vehicles from freezing. Its easy to look this up… it’s not a mystery. Stop arguing that it isn’t real. You’re missing the point. It’s definitely real… but the far better question is “how much of an effect would it be in an atmosphere?” When NASA uses it on their satellites, they maintain a vacuum between the readiative layers (to eliminate conduction). The atmosphere would behave much differently.
I think all this discussion of thermodynamic, quantum theory, etc. is way off base. The question can be answered much more simply. Do what scientist used to do back when they were doing science. Measure.
Go out 1 hour after sunset. Point a spectrometer straight up and measure the spectrum. Whatever you measure is being emitted by the atmosphere (mostly). Its either being emitted because of its inherent temperature, or from absorption and reemission from the upwelling IR of the planet.
You could obtain measurements every hour, and watch the change. 1 hour after sunset may not be long enough… depending on the amount of solar radiation that scatters sideways in the atmosphere from the sun just over the horizon.
Regardless… what you measure will be reality. Clearly, we can probably all agree, you will definitely measure some amount of IR. To make the analysis even more precise, obtain temperature readings of the atmosphere at different altitudes, and subtract out how much inherent IR that temperature of air would naturally radiate, etc.
You may even be able to calculate or measure the upwelling IR from the ground. By carefully combining all this data, it shouldn’t be that hard to determine… not theorize… how much radiation is backscattered to the planet from the atmosphere.
I suspect the answer will not be zero (I believe there is at least some GHG re-radiation), but I also suspect that the answer will not be enough… not nearly enough… to account for the alarmists claims.
Anyways, all this discussion isn’t going to convince anyone… there are too many variables… to many unknowns. What fraction of heat transfer is re-radiation and what fraction is convection? No one knows. Does re-radiated IR from excited CO2 molecules make it back to earth, or does it somehow get taken up by the atmosphere along the way? No one knows. How much downwelling IR is from re-radiation of upwelling IR, and how much is due to scattered (and absorbed-reradiated) downwelling radiation from the sun? No one knows.
But if we stop theorizing and start measuring, we can determine how much IR is emitted from the atmosphere towards the earth (even if we don’t know the exact mechanism). This will be the only real way to get a handle on this issue.

Brian W
March 10, 2011 11:02 pm

Gary P. (March 10, 2011)
Here’s a better explanation for your “experiment”. The radiation being emitted by your filament where previously was allowed free escape is now being reflected by the interior surface of your metal cylinder. Reflection is not absorption/reemission. Energy reflected is not energy absorbed. To invoke backradiation your filament needs to heat the cylinder hot enough to radiate sensibly back to the object. This is clearly not happening. Your cylinder was “relatively cold”. In fact by placing a round reflector around a centered filament you have created a rudimentary cavity whereby radiation formerly lost is now harnessed and focused centrally. The metal of the cylinder did emit no sensible radiation of its own. There is technically no hot to cold flow at all since you have removed air. Oh, and your emergency reflective blanket does reflect the small amount of heat a body produces but it does a far more important thing to keep one warm. Normal blankets are porous and allow easy convection and consequent air exchange. A reflective blanket is far less porous and much better at trapping the air mass around the body reducing the level of air exchange. This allows your body to warm the water vapor in your cocoon. No backradiation needed thanks!

cal
March 10, 2011 11:53 pm

tallbloke says:
March 10, 2011 at 4:26 am
Ira Glickstein
“the “greenhouse effect” heats the Earth because GHGs absorb outgoing radiative energy and re-emit some of it back towards Earth.”
No, No and a thousand times No.
Re-emitted radiation does not and cannot heat the Earth significantly, because downwelling IR does not and cannot penetrate the surface of the ocean beyond its own wavelength. The amount of energy from back radiation mixed into the ocean by wind and wave action is negligible and extra co2 therefore cannot account for the additional warming of the ocean bulk in the late C20th.
The greenhouse effect works by *SLOWING DOWN THE RATE THE EARTH COOLS AT*, by raising the altitude at which the atmosphere radiates to space . There is more than a semantic difference. Understanding it this way enables you to understand that it was reduced albedo 1979-1998 allowing more Solar energy to enter the oceans that caused the majority of the global warming at the end of the last millenium.
————————————————————————————–
I find it hard to follow the logic of this post. If “downwelling IR does not and cannot penetrate the surface of the ocean beyond its own wavelength” (which I do not believe is true anyway) it is because it is being absorbed. That means the energy is retained by the oceans. The alternative is that it is being reflected in which case the sea would be red. If you look at a swimming pool the deep end always looks bluer. That is because the light reflected off the bottom has had the red light removed by absorption. Infrared is even more strongly absorbed by water molecules a fact that is exploited by microwave ovens.
You also write “he greenhouse effect works by *SLOWING DOWN THE RATE THE EARTH COOLS AT*, by raising the altitude at which the atmosphere radiates to space” . This I sort of agree with but would prefer to use the statement “changes the radiation balance” Although it is only a different way of coming to the same conclusion you cannot reduce the energy going into space in the 14 to 18 micron band without saying where the rest goes. The fact is that the remainder is back radiated to earth in exactly the way Ira describes. These descriptions are not in conflict with one another.
For all those others who come up with spurious reasons why there is no downward radiation, what does it take to convince you when you see graphs showing actual measurements of the energy being received. Where is this radiation coming from if it is not from the atmosphere? It cannot be the sun because the sun’s spectrum is insignificant beyond 3 micron.

March 11, 2011 12:15 am

Ira,
I follow your logic, but the electric blanket analogy was still a bad idea.
On the other hand I’m amused that half the criticism comes from either complaints about nomenclature, or the “cold things can’t warm up warm things” argument.
For the cold things can’t heat up warm things crowd, please stop with the complicated explanations and torchered arguments. Just ask yourself, do igloos work or don’t they? You can quote as many definitions of laws of thermodynamics as you want, construct as many analogies as you wish, with plates, bowling balls, flashlights, coats….but the fact is that igloos work. They’re made of cold stuff like ice and snow, and a person can survive very severe cold weather inside of one because “back radiation” keeps them warm. Someone dumb enough to sleep outside the igloo instead of inside is just a cryogenic experiment.

nighttime
March 11, 2011 12:33 am

do the inuit build fires in their igloos.

March 11, 2011 12:47 am

nighttime says:
March 11, 2011 at 12:33 am
do the inuit build fires in their igloos.>>>
sigh. i suppose they just might sometimes which has what to do with anything?
If you insist I exclude the possibility of fire keeping them warm….
An igloo, constructed of only snow and ice, containing no additional heat sources of any sort other than what is generated by the human being(s) within them, works.
The dope sleeping outside next to a fire on the other hand will still be a cryogenic experiment with a cooked spot opposite the fire.

Bryan
March 11, 2011 1:10 am

Brian H says:
I advise all here to read through Hans’ post above, a few times. The English is weak, but the point about the nature of entropy and ‘waste heat’ is not to be missed.
I would like to second this comment.
Hans also highlights the” quality” factor of the radiation or of energy generally.
A number of people think that once the quantity of energy is known that all there is to it.
The “quality” of the energy or radiation is perhaps even more important.
What does “quality” mean.
It means how easy it is to convert that energy into another type or for it to do work.
The classic example is high quality Solar radiation(short wavelength) arriving on Earth and an approximately equal quantity of degraded Long wavelength radiation leaving.
Now the first law of thermodynamics would not be violated if the whole process were reversed.
Its the second law of thermodynamics that states that this is impossible to convert low quality radiation into an equal quantity of higher quality radiation.

Massimo PORZIO
March 11, 2011 1:33 am

@ Dave Springer says:
March 10, 2011 at 4:30 pm
Hi Dave,
About your reply to my post, I agree with your explanation of what is seen at the arctic.
I would like to know your opinion about the small emission peak in the middle of the CO2 15um absorption pit.
It is my opinion that that could be the only real backscattered photon emission of CO2, and that the other emissions could be due just to temperature of the IR active gases in the atmosphere.
Maybe I’m wrong with that, but I believe that because the “peak” is always positive at ground and at the TOA, and looking down at the tropical atmosphere that peak appears only over the 19-20km where the CO2 became the predominant IR active gas.
I repeat, I could have wrote a very silly thing, I’m not an atmospheric physicist just an electronic engineer.

RJ
March 11, 2011 2:30 am

Thanks to Hans for this information (and other that counter what to me seems like fiction but I’m never sure).
Radiation leaving the surface then returning to heat the earth seems like nonsense. Even for reason of logic
http://slayingtheskydragon.com/images/stories/freeoven.jpg
but it helps that posters like Hans and other explain why in this way.

March 11, 2011 2:39 am

davidmhoffer says:
March 11, 2011 at 12:15 am
David, this is an interesting point but you are wrong about how igloos work. They do not work by back-radiation.
Igloos maintain a higher temperature than the outside environment in the same way that all enclosures, including greenhouses, do.
This is quite simple and holds for all internal 3 dimensional environments enclosed by two dimensional barriers.
It is the result of the reduced probability of all three modes of energy transfer from a three dimensional gaseous environment towards and absorption through the two dimensional solid barrier (wall, glass or in the case of igloos, ice). The further from the two dimensional surface, the lower the probability is that the energy will be able to leave the enclosed atmospheric environment.
Historically, traditional dwellings tended to be constructed around a central fireplace for this very reason.
I live in a house high on the Sussex Weald which dates back more than 200 years. The are 4 fireplaces on the ground floor built into one central brick stack in the centre of the house. The external walls are merely 25 mm weather boarded with a 50 mm cavity and the internal skin is just 15 mm plaster board. There is no cavity wall insulation whatsoever and no double glazing. Thats just 2″ inches of wood and plasterboard between me and the elements 600 feet above sea level high on the Sussex Weald.
In the winter it is evident that the above holds true in that the closer you are to the external walls on a cold night, the colder it feels. The same is true for greenhouses. Ask any horticulturalist which plants are his smallest and lowest yielding and he will tell you it is the plants nearest the glass. Ask any Eskimo which part of the igloo he prefers to sleep in and I guarantee he will tell you, “the middle”.
This is what the “greenhouse effect” is. The greenhouse effect is simply the reduced probability of energy loss from all three modes of transfer, towards and through an external two dimensional surface from a three dimensional gas.
This fact holds true for all internal environments from glass houses to timber clad houses and even to igloos.
It does not hold true for the open atmosphere and therefore there is no “greenhouse effect” in the atmosphere.
The two dimensional barrier, be it timber, brick, glass or even ice, simply inhibits all three modes of energy transfer. This is the so called “greenhouse effect”.
That is why humans have evolved to construct and reside in dwellings.
There is no “greenhouse effect” in the atmosphere.

RJ
March 11, 2011 3:00 am

Can someone please comment on the post by DavidM re igloo’s
“They’re made of cold stuff like ice and snow, and a person can survive very severe cold weather inside of one because “back radiation” keeps them warm.”
If David is serious isn’t this fiction (I’m not an expert). Don’t igloo’s keep people warm because the air is warmer due to conduction. It has nothing at all to do with back radiation.
For a start wouldn’t most of the heat leave the body by conduction not radiation. And the point that Hans raised above.

March 11, 2011 3:01 am

RJ;
but it helps that posters like Hans and other explain why in this way.>>>
I am not just a skeptic, but a raging skeptic. There is so much hooey in IPCC AR4 that they should replace the 4 with an S. Which is why it drives me crazy when people go to great lengths to explain why there is no backradiation or it defies the laws of thermodynamics, or it creates energy… simple things like an igloo show that there IS backradiation, cold things CAN radiate heat to warm things, and NONE of it breaks any laws of physics.
Here’s my oversimplified explanation on a photon by photon basis. Frankly it p**sed me off to no end to figure out that the surface does in fact get warmer (thought by darn little and maybe even cooler when all interelated effects are included) but as for the concept of backradiation, there is nothing wrong with the basic ghg explanation other than presumed order of magnitude:
http://knowledgedrift.wordpress.com/2011/02/27/co2-exactly-how-does-it-warm-the-planet/

wayne
March 11, 2011 3:17 am

Hans:
Brian H:
Brian:
Yes, I totally agree with Hans explanation, it is correct. There are others here saying many correct statements. I have tried many times, in somewhat words, different analogies, trying to get others to see reality. But fellows, don’t get your hopes up. There are some here are firmly stuck in the AGW-IPCC-Trenberth mentality, as this post highlights, and since they have yet to see the reality, they probably never will. If I could say two thing I see keeping them confused, is that they consistently either speak instantaneously and not averaged over time and they fail to separate sun-earth-to-space system level energy flows with the inter-Earth system energy flows. And, heat does always flows from warmer to cooler, never backwards.
davidmhoffer:
I don’t mean to cross you, I mean that, I respect many of things you have said but can I try one last time for you to critique your thoughts on this “back radiation” and warming?
All of this talk of igloos, quinzees, coats, blankets, layers of GHGs being insulators and increasing warmth by “back radiation” has one big flaw as my physics sees it. In the case of your body, the heat source is within your body and any of these insulators are wrapped around the energy source itself. In the sun-earth-to-space system’s case, when wrapping the Earth with blankets or layers of GHG’s, you do not have the energy source within, but instead, outside that system. Physics is a bitch, every time you try to fool it with ‘trapped heat’ you will find there is one big factor left out that in the end violates you whole concept.
GHG layers will keep the same amount out of the Earth system as it can ever keep within, since the ultimate energy source is outside that system (ignoring radioactive decay in the soil). That’s it. Don’t have papers in my hand proving that albedo/reflection will increase *equally* with any insulation wrapped about the Earth but I do know it is true from what I have learned over the years. As I always try, keep an open mind and critique your own thoughts, you should see something in what I just said.

richard verney
March 11, 2011 3:55 am

Phil and Ira
Phil, I found your comments (March 10, 2011 at 7:52 pm) very interesting. Is this based upon an actual experiment or only a theoretical calculation?
What I am getting at is whether this reflected (back) radiation has genuine power to do work.
Consider the attached power plant:. http://www.solar-green-wind.com/wp-content/uploads/2009/12/largest_solar_power_station-1528.jpg
Consider the Earth’s energy budget:
http://www.klimaatfraude.info/images/EnergyBudgetNew.jpg
The power station is using mirrors tuned to collect and focus 184 w/sqm (i.e., 161 + 23) of solar bandwidth energy. Of course, during a sunny day, these mirrors are also receiving some (or all) of the 333 m/sqm back radiation. However, they are not focused to collect that nor are they tuned to the wavelength of downwelling long wave radiation. They are of course, tracking the elevation of the sun and therefore are effectively tuned to collect and focus the 184 w/sqm solar bandwidth energy. I presume that this power station produces no power at night and presumably only modest amounts on cloudy days and may be none on heavy dark rain cloudy days. It can effectively only work in a sunny climate such as California, Spain, Affrica (although of course, the illustrated station appears to be in Germany but Germany are beginning to realise that the country is not well suited to solar power generation).
My question is: If there was real power in the 333m/sqm longwave back radiation, why isn’t the power station designed to collect and focus this longwave radiation? There is nearly twice as much energy (viz 333 cf 184 m/sqm) and this longwave energy is available 24 hours a day come rain or shine. This eliminated the storage issue which has always beset green energy projects.
If there was real power (ie, the ability to do work) in this backradiation it is inconceivable that someone would not be collecting it, or that there would not be mainstream research designed at collecting this. For the vast majority of countries solar radiation is a non starter (due to cloudiness and latitude) but these countries have plenty of backradiation.
I guess I am raising a question on the point made by Hans (Hans says:
March 10, 2011 at 8:00 pm) in that the radiation may be there but it does not have any energy/ability to do work.
Phil and Ira your further views/comments would be appreciated.

P. van der Meer
March 11, 2011 4:12 am

If, as so many people seem to claim, the atmosphere radiates like a blackbody than how come we can see landmasses sharply defined in the infrared satellite pictures that get beamed down every 6 hours. Would the land/sea boundary in local places not get fuzzy and streaky by either an onshore (cooler) wind or an offshore (warmer) wind. But no, the coastline remains sharp everywhere and at all times. These pictures are taken in the 8.3μm – 14.4μm band which is exactly the open infrared window WITH NO ATMOSPHERIC RADIATION.
Check it out for yourself at http://www.sat.dundee.ac.uk/geobrowse/geobrowse.php. You will have to register, but it costs nothing.

Dave Springer
March 11, 2011 4:18 am

Phil. says:
March 10, 2011 at 4:32 pm
“If you use Wien’s Law to determine the maximum there are two forms, a frequency form and a wavelength form, the peak frequency does not correspond to the peak wavelength using c=λν.”
The superimposed dashed line blackbody curves don’t correspond to the frequency/wavenumber on the horizontal axis. If you plug in the peak frequency indicated on the horizontal axis to Wien’s Law formula you get a very different temperature in Kelvin from that labeled on the dashed curves. Interestingly the labeled values correspond to degrees Rankine!
You can find it with calculator below which gives results in F, C, K, and R.
http://www.ajdesigner.com/phpwien/wien_equation_t.php
Something is definitely whacked in those plots but I can’t quite figure out what it is.

Dave Springer
March 11, 2011 4:30 am

wayne says:
March 11, 2011 at 3:17 am
“GHG layers will keep the same amount out of the Earth system as it can ever keep within”
No. The energy enters the system in short wavelengths (visible light) which is not absorbed by GHGs. The shortwave energy is absorbed by the ocean. The energy leaving the system, radiated from the ocean surface, is longwave which is absorbed by greenhouse gases. The insulation effect of the GHGs works only in one direction. They do not insulate the earth from the sun but rather insulate the earth from the frigid cold of the cosmic void (which is 3 Kelvins and is called the cosmic microwave background radiation).
The net effect is the GHGs don’t slow the warming of the ocean from the sun they slow the cooling of the ocean into the cosmic background. This causes the surface temperature to rise higher than it would be otherwise. The greater temperature differential between the surface and the cosmic background causes heat to move through the insulation faster which reestablishes equilibrium (energy in equals energy out). The insulating effect of the GHG raises the surface equilibrium temperature.

RJ
March 11, 2011 4:33 am

davidm
I don’t think people have said there is no back radiation. What I think they have said is this back radiation does not result in a colder body warming a warmer body from this back radiation.
Cold things can radiate to hot things. But it will not warm the hotter thing when the radiation arrives is how I understand this process.
Its like a brilliant teacher passing knowledge to a student. The student can pass knowledge back to the teacher but this will not improve or add to the teachers knowledge (heat). But the student can improve the knowledge of younger child. The same with the sun earth CO2 relationship.
And I see you as a luke warmer not a true sceptic. You are sceptical of the overstated alarmists predictions not the GHG science itself.
Even though the GHG science could lead to the chicken in the oven situation. Or a person in a CO2 container cooking themselves due to back-radiation.

Dave Springer
March 11, 2011 4:39 am

@wayne (con’t)
The one-way insulating effect of GHGs is why, in my blanket example, I use two black rocks which are both exposed to the sun during the day so they can heat up (equally) then just one rock has a blanket thrown over it at night. The next morning they are uncovered and the temperature of the blanketed rock will be higher than the unblanketed rock. The next day, since one rock is warmer in the morning than the other yet both will be exposed to the sun again getting the same amount of daytime heating, the rock that is blanketed at night will reach a higher daytime maximum temperature. This temperature increase will continue until a new (higher) daytime equilibrium temperature is reached. The hotter the rock is when the blanket is thrown over it at night the quicker it will lose heat through the blanket. Heat flow rate through the insulating barrier is proportional to the temperature difference between the warm and cold sides. As the difference increases heat is lost faster. It is that which prevents the blanketed rock from just getting hotter and hotter until it melts.

Dave Springer
March 11, 2011 4:49 am

RJ says:
March 11, 2011 at 4:33 am
“I don’t think people have said there is no back radiation. What I think they have said is this back radiation does not result in a colder body warming a warmer body from this back radiation”
Correct. But it causes the warmer body to cool at a slower rate.
Radiative transfer between two objects of different temperature is a two way street.
Say the warmer body is radiating at 2 w/m^2 and the cooler body is radiating at 1 w/m^2. The net flow is, of course from warmer to colder, at exactly 1 w/m^2.
The warmer body is adding energy to the cooler body but the cooler body is also adding energy to the warmer body. The cooler body is simply adding less energy to the warmer body than the warmer body is adding to the cooler body. This will always be the case. In most situations we only talk about the net flow of energy which always goes from warmer to colder but in reality there is flow in both directions and one flow is larger than the other. This smaller flow is “back radiation” and it is real beyond dispute.

RJ
March 11, 2011 5:39 am

Dave S
“or perhaps recycled back down to earth where the result is a warmer earth surface that generates photons at a higher rate to compensate”.
If two heat sources are in a vacuum. One is 20 degrees and one 10 degrees.
Option 1. Both heat sources have a power source to maintain their heat at 20 and 10 degrees. I assume the 10 degree source would rise slightly but the 20 degree source would not change.
Option 2. The power source is switched off. Would the 10 degree source slow the cooling rate of the 20 degree source
With conduction I assume it would as the air would be slightly warmer. But what about in a vacuum when there is only radiation energy loss.

Phil.
March 11, 2011 5:49 am

richard verney says:
March 11, 2011 at 3:55 am
Phil and Ira
Phil, I found your comments (March 10, 2011 at 7:52 pm) very interesting. Is this based upon an actual experiment or only a theoretical calculation?
What I am getting at is whether this reflected (back) radiation has genuine power to do work.

This was a theoretical example but for it not to work as outlined you have to overturn basic physics. In fact the approach is used in practical light bulbs, a dichroic coating is applied to the bulb envelope to reflect the IR back to the filament and thus heat it so you get more output from the same current (or the same output in the visible from a lower current).
http://www.bulbs.com/blogs/light_source/post/2010/12/03/Cooler-Than-Ever-Halogen-IR-Bulbs.aspx

March 11, 2011 5:55 am

The only way I could ever even partially accept the blanket analogy would be if the blanket in question were something like a water filled rubber duvet and you were attempting to get a good nights sleep, perched on an overhanging precipice at the top of Mount Everest.
Give it a go, I guarantee you will be stone cold dead by the morning.
Blankets, electric or otherwise, its pure pseudo-science. The atmosphere bares no resemblance to a blanket. It has many properties and they all must be considered.
Yet in the “greenhouse” hypothesis most of the atmospheres properties are ignored. This is a fact that we all know to be true.
See my first post as an example of exactly that: http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/#comment-617396

Bryan
March 11, 2011 6:12 am

Dave Springer
……”In most situations we only talk about the net flow of energy which always goes from warmer to colder but in reality there is flow in both directions and one flow is larger than the other. This smaller flow is “back radiation” and it is real beyond dispute.”……
There is radiation both ways but the radiation from the colder surface has fewer photons of every wavelength than is leaving the hot surface.
Also the QUALITY of the radiation is less as Hans pointed out and is restricted to the temperature profile of the colder surface.
This means that the colder surface radiation from say 300K can only effect a cancellation of that part of the output profile of say a 400K warmer surface.
That why it’s a misuse of language to say that the colder surface “heats” or “warms” the hotter surface.
To say that the cold surface insulates the warmer surface is a more accurate description of the process.
It also agrees better with your two rocks illustration

Alan McIntire
March 11, 2011 6:12 am

In this thread arguing about greenhouse gases and how they lead to a warmer earth, it occurs to me that clouds have the same effect, but more so. If the earth has a roughly 60% cloud cover, and those clouds act roughly like the surface and absorb most of the short wave radiation and reradiate at longer wavelengths, we get 60% of incoming sunlight absorbed by clouds, 30% radiated back to space, 30% to the earth’s surface.
40% of incoming sunlight directly reaches the surface, for a total of 70%, or roughly 240 watts out of 342.
The ground radiates energy back to space, 40% gets through, and 60% is absorbed by clouds, half radiated to space, half radiated back to earth. That would result in
a net warming to 10/7 *240 = 342 watts. You’ve got a significant fraction of the warming already without even considering greenhouse gases.
I realize there is plenty of reflected light in the daytime so the actual figures would be reduced somewhat, but obviously zero reflected light at night.

commieBob
March 11, 2011 6:31 am

Hi Fred Souder,
Sorry for the late reply, I missed your post. Your question about photons is interesting. I have no clue about the general case but I can answer for the case where I am confident that I know what I’m talking about. 😉
When a photon strikes a surface or a molecule, a variety of things can happen:
1 – It can promote something to a higher energy level. http://en.wikipedia.org/wiki/Infrared_spectroscopy
2 – It can have an elastic collision. (that’s how we see light reflecting from a surface) 3 – It can impart kinetic energy.
We are taught, and can practically confirm, that a photon can not cause something to jump to a higher quantum level unless it has enough energy to make it do so. In other words, if I want a gas to emit green light, I have to excite it with blue light. Red light won’t work.
On the other hand, low frequency (long wavelength) photons can cause a surface to heat without having to promote molecules to higher quantum levels.
I think the answer to your question (as I understand your question) is this: Promoting molecules to higher quantum levels is not the only way a photon can impart energy.
I have found a good lecture note on the subject: http://galileo.phys.virginia.edu/classes/252/black_body_radiation.html

RJ
March 11, 2011 6:49 am

Ira
RJ does not know re back radiation. I am just trying to understand this issue nothing more. Is the GHG theory flawed or not.
Re evidence of radiation. What are your views on this
http://slayingtheskydragon.com/Latest-News/climate-follies-encore.html
Lord: “Back radiation can be simply demonstrated by pointing a simple infrared detector at the underside of a cloud. Try it.”
Chorus: “My IR detector only cost $60! Simple! Agreed! Agreed!”
Slayer: “Clouds do not absorb and re-radiate heat back to Earth. Clouds add THERMAL MASS which takes longer to heat and cool. Warmists ‘support’ this false hypothesis with IR thermometer readings, but the IR readings of a hot Barbie is the same from any distance; ENERGY is not. Your $60 REMOTE thermometer is not measuring the radiant energy you are receiving, it is measuring the resonance of the Barbie.”
We know that this was the final act of the follies for there has been no reply in 36 hours from the Lord or the Luke Chorus.
At this point from what I have read I do not believe that a colder body can heat a warmer body. Otherwise we get a ridiculous situation where a chicken above zero could be cooked by nothing more than back radiation.
My views at this point are
That most of the energy leaves earths surface by conduction not radiation
The amount of CO2 is far to small to heat earth even if back radiation does occur and does heat the earth. (Which I doubt)
Colder bodies can not heat a warmer one. So CO2 might slow the rate of cooling but can not warm earth.

wayne
March 11, 2011 6:49 am

Dave Springer says:
“The insulation effect of the GHGs works only in one direction.”

The no belongs to you.
I have never come across a “one-directional insulator”. Have you?
You are imaginating out of what in physics has been proven over and over again.
There are spectral bands in CO2 where the solar specrum crosses and those will keep the same outside the Earth system. But, if you can ever invent a one-directional insulator, you will be a rich man. Go for it, you seem tho think CO2 is it.

tallbloke
March 11, 2011 6:56 am

cal says:
March 10, 2011 at 11:53 pm
tallbloke says:
March 10, 2011 at 4:26 am

I find it hard to follow the logic of this post. If “downwelling IR does not and cannot penetrate the surface of the ocean beyond its own wavelength” (which I do not believe is true anyway) it is because it is being absorbed. That means the energy is retained by the oceans. The alternative is that it is being reflected in which case the sea would be red. If you look at a swimming pool the deep end always looks bluer. That is because the light reflected off the bottom has had the red light removed by absorption. Infrared is even more strongly absorbed by water molecules a fact that is exploited by microwave ovens.
You also write “he greenhouse effect works by *SLOWING DOWN THE RATE THE EARTH COOLS AT*, by raising the altitude at which the atmosphere radiates to space” . This I sort of agree with but would prefer to use the statement “changes the radiation balance” Although it is only a different way of coming to the same conclusion you cannot reduce the energy going into space in the 14 to 18 micron band without saying where the rest goes. The fact is that the remainder is back radiated to earth in exactly the way Ira describes. These descriptions are not in conflict with one another.
For all those others who come up with spurious reasons why there is no downward radiation, what does it take to convince you when you see graphs showing actual measurements of the energy being received. Where is this radiation coming from if it is not from the atmosphere? It cannot be the sun because the sun’s spectrum is insignificant beyond 3 micron.

Hi Cal, and thanks for your reply.
I didn’t say there was no downward radiation, I said it doesn’t heat the bulk of the ocean. Now some say it must or the oceans would freeze, because of the amount of radiation measured as upwelling from the ocean surface, compared to the amount of solar shortwave entering it. This misses the point. Downwelling IR at the wavelengths found in nature doesn’t penetrate the ocean beyond it’s own wavelength. Microwave ovens are tuned to used wavelengths that cause water molecules to vibrate, and so heat up. Different wavelengths. The small amount of IR generated heat which is mixed into the near surface is not enough to account for the increase in ocean heat content, and the extra height of the space radiative atmosphere has not been elevated sufficiently by extra co2 or any water vapour feedback to account for slowing down the cooling of the ocean enough to cause the OHC increase either.
Therefore the most likely cause is the reduction in albedo over the tropics measured by the ISCCP cloud data from 1980-1998 allowing more insolation at the surface. Sunlight penetrates into the ocean many tens of meters. This is why you can see all the way to the bottom in clear tropical waters. It looks blue because the blue end of the spectrum penetrates the deepest of all, so it is blueish light which is being reflected off the white sand on the seabed or back out of the deep.
The downwelling IR is not just coming down vertically. over 95% of downwelling IR is emanating from molecules which are less than a kilometer above the Earth’s surface and the photons are travelling in all directions from straight down to horizontal. This means a significant portion get reflected by the sea surface anyway. This means a proportion of the radiation measured as coming off the sea surface is reflected radiation not absorbed and re-emitted radiation. Additionally, as Stephen Wilde says on my blog:
“What happens in practice is that increased DLR heats the ocean skin and thereby increases the rate of evaporation. The enthalpy of vapourisation (at current atmospheric pressure) dictates that for every unit of energy that provokes an evaporative event four more units oif energy are taken from the surrounding environment.
So, the increased evaporation from more downwelling IR is self limiting. One fifth of the DLR provokes extra evaporation but when the evaporation occurs it soaks up the other four fifths of the extra DLR and once the extra DLR is used up it cannot provoke any more evaporation.”
So it looks like the ocean maintains the atmosphere-ocean balance despite increased co2 and this is bourne out by the fact that satellite measurements of surface temperature show that the lower troposphere stayed at a pretty even temperature from 1980-1998 until the ocean belched out a big wad of energy in the super el nino. This then caused an upwards step change in global temperature. See Bob Tisdale’s posts on this
The key point is that this was sequestered solar energy from a more active than average sun shining through less than average tropical cloud amounts, penetrating deep into the oceans, where longwave radiation from clouds and co2 doesn’t reach.

Stephen Wilde
March 11, 2011 6:58 am

Dave Springer said:
“The net effect is the GHGs don’t slow the warming of the ocean from the sun they slow the cooling of the ocean into the cosmic background. This causes the surface temperature to rise higher than it would be otherwise. The greater temperature differential between the surface and the cosmic background causes heat to move through the insulation faster which reestablishes equilibrium (energy in equals energy out). The insulating effect of the GHG raises the surface equilibrium temperature.”
I think that needs slight refinement.
I don’t think that GHGs slow the cooling of the OCEAN into the cosmic background because all they do to the ocean is increase the rate of evaporation and radiation from the ocean skin with, I think, a zero effect on the energy flow from ocean bulk to air for reasons that I have set out elsewhere.
However they do slow the cooling of the AIR into the cosmic background so the rest of the comment is correct but then one has to put the effect into proportion.
Due to the ocean bulk temperature not being affected the necessary adjustments all have to be made in the air alone and that is achieved by shifting the speed of the hydro cycle and the air pressure distribution. The latitudinal position of the jetstreams is a good proxy for that.
However the scale of natural changes in the hydro cycle and the air pressure distribution is already huge over periods of say 500 years such as the time between MWP and LIA or LIA and the present. The jets appear to have shifted by up 1000 miles latitudinally between such peaks and troughs.
I have difficulty envisioning the effect of more CO2 contributing more than a mile to such shifts and possibly less than that when the full negative effects of a faster hydro cycle are factored in.
In other words, wholly inconsequential.

March 11, 2011 7:07 am

Dave Springer says:
March 11, 2011 at 4:30 am
Dave Springer,
Your reply to wayne regarding the oneway insulating effect of GHG’s would only be valid if the Incoming EMR were entirely or even mostly SW.
But this is not the case as it is mostly LW. Some say 50-50 but that is not the whole picture.
Your point only has validity if incoming EMR is mostly SW. It is not and therefore you are as far from reality as it is possible to be.
The SW EMR (light) emitted by the sun is a by product of the suns extreme temperature, not the other way round. Which is why there is a 600 km high bulge in the atmosphere under the solar point covering 25% of the atmospheres surface area, tracking the solar point around the Earth 24/7 called the Diurnal Atmospheric Bulge.
http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1966SAOSR.207…..J&db_key=AST&page_ind=0&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES

Bryan
March 11, 2011 7:16 am

What happens when radiation from cold surface reaches hotter surface?
One hurdle that the “cold” radiation suffers from.
It flows “up hill.”
The respective blackbody spectra of the hot and cold objects will be centred around different maxima characteristic of their temperature.
I will invent magnitudes rather than calculate accurate ones to save time.
Lets say that initially there is only the colder object present.
Lets say that some way out from the cold object placed in a vacuum, 100 Joules of IR centred radiation of wavelength 15um pass through one metre square area in one second.
Now place a hotter object radiating at 250 Joule/s of 5um centred radiation at that squaremetre.
So what happens to the radiation from the colder object?
Three viable solutions.
1. Classical wave theory approach.
A single Pointing vector of magnitude 150W/m2 with direction hot to cold.
2. Subtraction of photon streams will result in heat flow of resultant 150W/m2
(The insulation option)
3. Absorption of 100Joules/s of 15um centred radiation and emission of 100 J/s of 5um centred emission. (The “heating up” option)
I consider that options 1 or 2 give the correct solution
Whats wrong with option 3?
The consequence for that square metre of having the hotter object there is the up-conversion of 100J of 15um centred radiation into 100J of 5um centred radiation.
This also raises the possibility of increasing the temperature of the hotter object.
If so this would also contradict the Stephan Boltzmann Equation as shown below.
The increase in the “quality” of the radiation appears to violate the second law.
Option 2 gives the correct answer using Stephan Boltzmann Equation without speculating about quantum mechanical effects of photon absorption.
What do the textbooks say.
University Physics Young and Freedman Pg 484.
“If a body of absolute temperature T is radiating and its surroundings (at temperature Ts )is also radiating and the body absorbs some of this radiation.
If it is in thermal equilibrium with its surroundings then
T = Ts .
For this to be true the rate of absorption must be
= AεσTs^4 ”
If however T > Ts
Flow of heat from body = AεσT ^4 – AεσTs^4
H = Aεσ(T ^4 – Ts^4 )
“In this equation a positive value for H means heat flow out of the body.
The equation shows that for radiation as for conduction and convection the Heat current depends on the temperature difference between the bodies.”
So it looks like option 2, mirrors the equation above.
.

March 11, 2011 7:36 am

” wayne says:
March 11, 2011 at 3:17 am
Hans:
Brian H:
Brian:
Yes, I totally agree with Hans explanation”
Thanks Wayne, Brian, and RJ. I hoped there would be at least some readers here making it to the end of my bad English analysis.
“Ira Glickstein, PhD says:
March 11, 2011 at 6:23 am
RJ and others seem to accept the existence of “back radiation” but doubt that it can warm the surface of the Earth because it comes from a cooler source, the bottom of the Atmosphere. Thanks for setting them straight by reiterating that radiation goes both ways, with the Earth heating the bottom of the Atmosphere and the bottom of the Atmosphere heating the Earth, with the net flow of energy going from the Earth to the Atmosphere.”
Ira,
radiation goes both ways but radiation is not heat. Radiation is cold and is able to create heat and that heat is not related to the photon energy. It happens only if there is matter, and there must be matter at Thot and at Tcold. Only then heat emerges as the energy flow from hot towards cold. This heat is a function of the surface temperature not of the photons as I pointed out here.

Phil.
March 11, 2011 7:41 am

Bryan says:
March 11, 2011 at 7:16 am
What happens when radiation from cold surface reaches hotter surface?
One hurdle that the “cold” radiation suffers from.
It flows “up hill.”

A fundamental error in the first line, radiation doesn’t involve gradients, no such thing as ‘up hill’.

Bryan
March 11, 2011 7:43 am

Small correction
The consequence for that square metre of having the hotter object there is the up-conversion of 100J of 15um centred radiation into 100J of 5um centred radiation.
Should read
The consequence for that square metre of having the hotter object there is the up-conversion of 100J of 15um centred radiation into 100J of 5um centred radiation to add to the effect of the hotter surface.

Phil.
March 11, 2011 7:51 am

Dave Springer says:
March 11, 2011 at 4:18 am
Phil. says:
March 10, 2011 at 4:32 pm
“If you use Wien’s Law to determine the maximum there are two forms, a frequency form and a wavelength form, the peak frequency does not correspond to the peak wavelength using c=λν.”
The superimposed dashed line blackbody curves don’t correspond to the frequency/wavenumber on the horizontal axis. If you plug in the peak frequency indicated on the horizontal axis to Wien’s Law formula you get a very different temperature in Kelvin from that labeled on the dashed curves. Interestingly the labeled values correspond to degrees Rankine!
You can find it with calculator below which gives results in F, C, K, and R.
http://www.ajdesigner.com/phpwien/wien_equation_t.php
Something is definitely whacked in those plots but I can’t quite figure out what it is.

Nothing wrong with the plots, you’re using the wrong formula for Wien’s Law, you can’t use the wavelength formula and convert to wavenumber. You copied that statement from me and then ignored it and tried to do just that!
Try looking here:
http://www.wolframalpha.com/input/?i=Wien%27s+displacement+law&a=*FS-_**WiensDisplacementLaw.lambda-.*WiensDisplacementLaw.T–&f2=5780+K&f=WiensDisplacementLaw.T_5780+K&a=*FVarOpt-_**WiensDisplacementLaw.nu–
For a temp of 270K the peak frequency is 15.9 THz corresponds to 18.9 microns.

Phil.
March 11, 2011 7:55 am

Bryan says:
March 11, 2011 at 7:16 am
The increase in the “quality” of the radiation appears to violate the second law.

What on earth is the ‘quality’ of radiation? What are its SI units?

Phil.
March 11, 2011 8:44 am

Bryan says:
March 11, 2011 at 7:16 am
What happens when radiation from cold surface reaches hotter surface?
Lets say that initially there is only the colder object present.
Lets say that some way out from the cold object placed in a vacuum, 100 Joules of IR centred radiation of wavelength 15um pass through one metre square area in one second.
Now place a hotter object radiating at 250 Joule/s of 5um centred radiation at that squaremetre.
So what happens to the radiation from the colder object?
Three viable solutions.
1. Classical wave theory approach.
A single Pointing vector of magnitude 150W/m2 with direction hot to cold.
2. Subtraction of photon streams will result in heat flow of resultant 150W/m2
(The insulation option)
3. Absorption of 100Joules/s of 15um centred radiation and emission of 100 J/s of 5um centred emission. (The “heating up” option)
I consider that options 1 or 2 give the correct solution
Whats wrong with option 3?

None of the above, the cooler object will warm up until it is in equilibrium with the incoming radiation from the hotter object resulting in more radiating reaching the hotter object which will also increase in temperature until it is in equilibrium too.

Richard E Smith
March 11, 2011 8:45 am

Ira Glickstein said:
“Thanks Dave Springer for clearly explaining how “back radiation” from the Atmosphere to the surface of the Earth reduces the net flow of energy from the Earth to Space. This flow reduction causes the Earth to warm until, at the resultant higher temperature, the surface emits more longwave radiation such that energy in = energy out, and average temperatures stabilize.”
So now we have a new, sceptic, version of greenhouse warming. Backradiation does not directly add heat to the Earth but by reducing the rate of radiative cooling “it causes the Earth to warm”. Forgive me for not being able to distinguish between these two fallacies.
A heated body’s molecules are vibrating in accordance with the energy it is absorbing. A cooler body emitting towards the hotter body cannot heat it up any more because the emissions (backradiation) are not at the energy level to make the molecules vibrate any faster. Likewise the radiative cooling version of greenhouse theory cannot make the hotter body’s molecules vibrate any faster. Of course there is such a thing as radiative cooling, but in a constant irradiance model (such as Kiehl & Ternberth’s) the Earth is not cooling down so cutting off the flow of radiant energy cannot add any heat as it cannot make the incoming energy vibrate the molecules any more. Reduction in radiative cooling may have an effect at night (and if so the Earth will be less cold than it otherwise would be) but this is not the basis of the ‘greenhouse effect’ which is about a build up of additional heat below a radiation barrier until it is breached and a fictitious radiative equilibrium is arrived at. Further, the net flow argument which is used to demonstrate compliance with the laws of thermodynamics in this theory is a fraud. On this argument you could have any amount of backradiation – a million watts for example – but as the flow out would still be greater than the flow in (1 million plus the 66 watts of solar) then there would be compliance.

Bryan
March 11, 2011 8:50 am

Phil. says:
March 11, 2011 at 7:41 am
A fundamental error in the first line, radiation doesn’t involve gradients, no such thing as ‘up hill’.
Well its a reasonable analogy
Shorter wavelengths are characteristic of higher temperature sources .
Shorter wavelength photons have higher energy than longer wavelength ones

Phil.
March 11, 2011 9:07 am

Bryan says:
March 11, 2011 at 8:50 am
Phil. says:
March 11, 2011 at 7:41 am
“A fundamental error in the first line, radiation doesn’t involve gradients, no such thing as ‘up hill’.”
Well its a reasonable analogy
Shorter wavelengths are characteristic of higher temperature sources .
Shorter wavelength photons have higher energy than longer wavelength ones

Not reasonable at all because it has no relevance to the transport of radiation.

Domenic
March 11, 2011 9:09 am

It amazes me that so many are still chasing around the canard tossed out by the AGW people.
Just like “Hide the decline!”
“Hide the nitrogen, oxygen and water!”
They tell you to disregard the N2, O2 and H2O in the atmosphere.
They tell you to disregard the HUGE heat capacity and HUGE thermal mass represented by N2, O2 and H2O in the atmosphere.
Disregard the 99.05% of the atmosphere. It has no effect.
Focus only on the 0.05% CO2 that we tell you to.
Because CO2 is a magical molecule and it controls the atmosphere.
And like sheep, many do so.
This is a far more accurate picture of what is truly going on:
http://slayingtheskydragon.com/Latest-News/omgmaximum-co2-will-warm-earth-for-20-milliseconds.html
“Thermodynamics will provide exact, repeatable quantities for energy flows or resultant temperature based on equations with three main variables. These variables are the difference in temperature, the mass of the bodies and the specific heat of the bodies. The greater the temperature difference, the greater the energy movement, which is the ‘delta T’ component.
A pound of water is easier to heat than ten pounds of water, which is the ‘mass’ component. A pound of Aluminum is easier to heat than a pound of Lead, which is the specific heat component. Carbon Dioxide has a specific heat of 0.8J/Ks K, which means it gains or losses heat faster than standard air. This coefficient actually reduces the mass values we will now discuss, but we will assume this to be 1.0 J/Ks K to simplify discussion.
In determining the heat flow of the planet we must place the ‘human produced’ atmospheric component in perspective with the total system. Humans produce 28 Giga-tons of atmospheric carbon annually. For comparison, 28 G-tons of ocean would be 5.93 cubic miles or 28 G-tons of Earth would be 2.96 cubic miles.
What the Warmists are trying to tell you is that YOUR 6 cubic mile portion of the ocean controls the temperature of the entire 310,000 million cubic miles of ocean. Or alternatively, YOUR 3 cubic miles of dust controls the temperature of the 259 trillion cubic miles of Earth. As absurd as this assertion is, the actual particle physics is an equal distortion.
It’s a Small World After All
Carbon Dioxide is a benign molecule that is required for life and is currently less than 390 parts per million (PPM) of the atmosphere. Prolonged exposure to concentrations of up to 80,000 PPM, have shown no adverse side effects. All federal registries listed CO2 as ‘non-toxic’ until the recent EPA reclassification. Calling a substance a ‘toxin’ does not make that substance a toxin, except in the toxic mind of bureaucrats.
All substances absorb and emit electromagnetic energy in discrete spectrum bands. The Earths outgoing Infrared energy is in a narrow band and can be absorbed by CO2 only in the 5 and 15 micron wavelength range. There is a finite amount of this IR energy, so the absorption is not directly connected with the amount of CO2. The term ‘absorption’ is misleading also, for the amount of time that this IR flow is ‘interrupted’, called the lapse rate is a fraction of a millisecond.
The majority of the space around an atom is void. Most IR energy passes through the CO2 molecules with no impact. The further you go above the surface of the Earth, the less air and the less CO2 you and outgoing IR waves will impact. Professor Nasif Nahle of the UA de Nuevo Leon has done the ‘mean free travel’ calculations on the IR escape rate. Outgoing IR energy is delayed by at most 22 milliseconds.
That is the total extent of ‘global warming’. All of the Planets CO2, the 97% from natural and the 3% from man delay temperature change by an immeasurably small amount of time. And this ‘delayed’ heat transfer is NOT radiated back to Earth. It is leaving a ‘hot’ Earth at the speed of light for a ‘cool’ outer space and is only delayed momentarily.” end quote
CO2. In reality, in the overall scheme of things, an obscure bit player. Hardly worth looking at. A nothing. Just like the major AGW players: Keeling, Mann, Jones, Gore, Schmidt, Hansen….in reality they are nothing but bit players. Nothings. Not even worth a glance.
Soon, many more people will come to realize that.

Bryan
March 11, 2011 9:23 am

Phil. says:
What on earth is the ‘quality’ of radiation?
The ‘quality’ of radiation or energy generally is its ability to do work or in other words turn into other energy forms.
If you google the appropriate words you will find plenty of hits particularly from solar energy capture systems.
In the meantime here are two sources
http://www.scribd.com/doc/31017446/The-Second-Law-of-Thermodynamics
http://www.ijoticat.com/index.php/IJoT/article/viewFile/185/171

Bryan
March 11, 2011 9:29 am

Phil says of my 3 options above
…..”None of the above, the cooler object will warm up until it is in equilibrium with the incoming radiation from the hotter object resulting in more radiating reaching the hotter object which will also increase in temperature until it is in equilibrium too.”…..
So it appears that Phils answer to all questions on thermodynamics is…..
The heat death of the Universe

Phil.
March 11, 2011 9:57 am

Bryan says:
March 11, 2011 at 9:23 am
Phil. says:
“What on earth is the ‘quality’ of radiation?”
The ‘quality’ of radiation or energy generally is its ability to do work or in other words turn into other energy forms.
If you google the appropriate words you will find plenty of hits particularly from solar energy capture systems.

This is about science, the question you didn’t try to answer is what’s its units?
A quantum of light has a wavelength and frequency, its energy is given by the product of h and frequency. What’s the equation for quality?

Phil.
March 11, 2011 10:29 am

Bryan says:
March 11, 2011 at 9:29 am
Phil says of my 3 options above
…..”None of the above, the cooler object will warm up until it is in equilibrium with the incoming radiation from the hotter object resulting in more radiating reaching the hotter object which will also increase in temperature until it is in equilibrium too.”…..
So it appears that Phils answer to all questions on thermodynamics is…..
The heat death of the Universe

Questions on matters of thermodynamics I answer with thermodynamics, what a concept!

Bryan
March 11, 2011 10:30 am

Phil. says:
“What on earth is the ‘quality’ of radiation?”
So after informing you of something you knew nothing about wheres the;
Thank you Bryan?
……..” question you didn’t try to answer is what’s its units?”……..
I thought when you read the sources you could work that one out for yourself!
However sceptics are ever helpful!
“Quality” is a comparison.
Electrical Energy has a higher “quality” than an equal quantity of Heat Energy.
Black Body radiation centred around 2um has a higher “quality” than an equal quantity of Black body radiation centred around 20um

Domenic
March 11, 2011 10:30 am

to Bryan
((So it appears that Phils answer to all questions on thermodynamics is…..
The heat death of the Universe))
The laws of thermodynamics are simply a handy way of examining any closed system. They are the laws science uses to construct machines, toys, etc. That’s all. Period.
There is no evidence that the universe is a closed system. On the contrary, the evidence is overwhelmingly in favor of a open system, not closed.
What are the most powerful, energetic objects in the universe?
Quasars.
They are pouring additional energy into the universe.
But almost nothing about them is understood.

wayne
March 11, 2011 10:31 am

Phil. says:
March 11, 2011 at 7:51 am
For a temp of 270K the peak frequency is 15.9 THz corresponds to 18.9 microns.
——–
Phil, respectably, it seems you missed a factor in this calculation as explained by this paper: http://www.journal.lapen.org.mx/sep09/12_LAJPE_303_Lianxi.pdf
Best to just use equation λmax (µm) = 2897.77/T without needing to using the c/1.760 correction in subsequntly converting frequency back to wavelength with Wien’s frequency form of the equation.
For a temp of 270K the peak wavelength is 2897.77 µm·K/270K = 10.73 µm.
Dave Springer:
Look elsewhere than Wikipedia when possible for clarity, you are probably now totally confused, and yes, the top charts are incorrect.

Dave Springer
March 11, 2011 11:27 am

wayne says:
March 11, 2011 at 10:31 am

Phil, respectably, it seems you missed a factor in this calculation as explained by this paper: http://www.journal.lapen.org.mx/sep09/12_LAJPE_303_Lianxi.pdf
Best to just use equation λmax (µm) = 2897.77/T without needing to using the c/1.760 correction in subsequntly converting frequency back to wavelength with Wien’s frequency form of the equation.
For a temp of 270K the peak wavelength is 2897.77 µm·K/270K = 10.73 µm.
Dave Springer:
Look elsewhere than Wikipedia when possible for clarity, you are probably now totally confused, and yes, the top charts are incorrect.

Hey Wayne, before you go dissing wikipedia you ought to read the list of references for the papers you present in your comments.
The paper above (Lianxi) which you say explains a missed factor in Wien’s Law has a reference [5] which is to the Wikipedia article on Planck’s Law.
Physician, heal thyself. 🙂

Dave Springer
March 11, 2011 11:58 am

Phil. says:
March 11, 2011 at 9:57 am
“What’s the equation for quality?”
I think you’re being facetious but for the benefit of everyone else…
It depends on the application. Carnot was probably the first to quantify it for heat engines:
μC = (Ti – To) / Ti
where
μC = efficiency of the Carnot cycle
Ti = temperature at the engine inlet (K)
To = temperature at engine exhaust (K)
As a general rule of thumb energy quality is like pornography. It’s difficult to define but you know it when you see it.
For instance a solar pond with 1 million pounds of water that is 1 degree F above ambient air temperature has 1 million BTUs of energy available to do useful work but it’s exceedingly poor quality. A boiler with 1000 pounds of water 1000 degrees F above ambient air temperature also has 1 million BTUs of energy available to do useful work and it’s high quality energy.
Generally speaking water that is below the boiling point at STP, say 99C, is considered poor quality energy as it has a low Carnot efficiency. But if your application is supplying hot water for home use (clothes washing, hot shower, etc.) instead of running a heat engine then it’s very high quality energy.
Another example, a cloud might have an electric charge of 1 megawatt/hour but it’s low quality energy. A bank of Lithium/Ion batteries with an electric charge of 1 mw/h is high quality energy if you have an electric vehicle but it’s pretty low quality if you have a horse & buggy. 🙂

Dave Springer
March 11, 2011 12:39 pm

Richard E Smith says:
March 11, 2011 at 8:45 am

Ira Glickstein said:
“Thanks Dave Springer for clearly explaining how “back radiation” from the Atmosphere to the surface of the Earth reduces the net flow of energy from the Earth to Space. This flow reduction causes the Earth to warm until, at the resultant higher temperature, the surface emits more longwave radiation such that energy in = energy out, and average temperatures stabilize.”
So now we have a new, sceptic, version of greenhouse warming. Backradiation does not directly add heat to the Earth but by reducing the rate of radiative cooling “it causes the Earth to warm”. Forgive me for not being able to distinguish between these two fallacies.

Since I’m getting credit for it in this thread I get to name it. I’m calling it the “engineer’s version” because it’s the theory of operation behind millions of IR CO2 sensors around the world used to control ventilation systems in commercial structures.
http://v2010.raesystems.com/~raedocs/App_Tech_Notes/Tech_Notes/TN-169_NDIR_CO2_Theory.pdf
I’m not much on theory that has no demonstrated its usefulness in practical applications. In this case the absorption of longwave infrared by CO2 and thermalization of the other non-absorptive gases in the atmosphere is well established fact employed in practical applications around the world.

sky
March 11, 2011 1:27 pm

Hans says:
March 10, 2011 at 8:00 pm
Glad to see someone who deeply understands physics–entropy, in particular–and has the patience to explain it all to those weaned on soft science miscomprehensions upon which the “radiative greenhouse” paradigm rests. Danke schoen, Hans, very well done!

March 11, 2011 1:45 pm

“Phil. says:
What on earth is the ‘quality’ of radiation?”
That would be something like the ability do to work at the radiative equilibrium temperature of the matter that this radiation came from. So with the high frequency radiation of the sun at 5777 K versus the 255 K radiation of earth, the efficiency according to the Second law for the sun radiating to an object in space at say 100 K would be 1-(100/5777) = 98% and for earth doing this 1-(100/255) = 61%.
And for backradiation on earth from 255K towards 288 K it would be 1-(288/255) = – 13%, that’s a nasty negative heat engine.

March 11, 2011 3:16 pm

RJ, Will, wayne, whoever said this:
And I see you as a luke warmer not a true sceptic. You are sceptical of the overstated alarmists predictions not the GHG science itself.>>>>
I’m insulted. I am a hardcore skeptic. But that doesn’t change the fact that some elements of the AGW position are correct. To be an EFFECTIVE skeptic, set the record straight and help people get proper perspective, one needs to recognize those elements of the AGW argument that are correct, and attack them where they are dead wrong and also where they highly misrepresent the facts to create a false impression.
Stephan-Boltzman Law provides the math formula for calculating exactly how much energy flux a surface will emitt at a given temperature. You can do the calculation at an ambient temperature of -100C, 0, +100C and you will get the same number every time. You can do the calculation with another surface in proximity at -1000, +1,000C or 1,000,000 C, and you will STILL get the exact same answer. And those answers have been verified thousands upon thousands of times by experimentation.
Unless you can disprove SB, there’s very little logical thought to follow. Since the energy flux leaving the surface is unaffected by the temperature of objects around it be they colder or hotter, the photons leave at that specific rate. If a colder surface is in proximity, it is subject to the exact same laws. The colder surface emitts the exact same energy flux it would have not matter what environment it is in or what it is proximate to. The photons the colder surface emitts travel away, and if they strike a surface capable of absorbing them, they will be absorbed, warming that surface up by the amount of energy they carried, not matter what the temperature of the surface is.
At days end, the GHG explanation of CO2 IN ISOLATION OF ALL OTHER FACTORS is correct. It is however completely misrepresented by the AGW crown. The magnitude is tiny, it is modelled ats w/m2 at the top of the atmosphere which is ridiculous, feedbacks are ignored, exagerated, or even calculate with the wrong sine, and on and on and on.
There is SO MUCH to take the IPCC to task for where they are dead wrong or obviously manipulative. Take for example their calculation that doubling CO2 will produce a direct 3.7w/m2 at TOA which in term results in a 1 degree temperature increase. Here’s a brief rundown of the stuff they document in the fine print, but you’d never know if from the big bolded explanations:
1. They calculate +1 degree against the “effective blackbody” temperature of earth. That’s about -19C and occurs at about 15,000 feet altitude. Use their same exact physics to calculate the surface temperature change and you get only 0.6 degrees.
2. They then say that changes in RF from GHG are usefull for calculating changes in the effective black body temperature of earth, but that these may not translate into surface temp changes in linear fashion, and they EVEN say that there may be NO surface change in temperature, but that the climate will still change dramaticaly. Say what huh?
3. At no point do they explain their +1 degree in context other than average temperature. More hooey, there is no such thing. Assume worst case of +1 at effective black body, SB Law yields 0.6 at surface (it is is in fact linear which I doubt) average of +15. Now let’s use SB Law to calculate not the average increase, but the increase under various conditions. Like at +30 at the equator warms up to +30.1 while -50 in the arctic warms up to -46. See the issue here? The hottest parts of the planet, on the hottest days, change very little, most of the change comes in the coldest parts, during winter, at night time lows.
4. Then the vaunted IPCC makes what I consider to be the most egregious misrepresentation of all. They admit via their own math that CO2 is subject to the laws of diminishing return, that it is logarythmic. Then the merrily move forward with scenario after scenario showing what would happen under various circumstances, conveniently leaving out that the amount of fossil fuel we would need to consume to achieve those levels is orders of magnitude beyond what we could actually burn even if all we did with it was burn it!
I could go on…and on… and on… the CO2 backradiation argument is a rat hole that consumes everyone’s time, the basic physics of GHG is correct, but the magnitude, exagerated as far as the IPCC can possibly exagerate it through the information they bury in the fine print, is so small that it is at best splitting hairs.
So igloos work, if you’ve done any winter camping you know that they have a vent at the top to vent carbon dioxide laden air from exhaled breath, that they have several air changes per night as a result, that the warmest air leaves first and is replaced by the coldest air from entrance, and STILL the person inside is warmer than if they were outside the igloo, and the extra warmth is from backradiation from the colder surfaces of the igloo.
Stop fighting reality, start fighting the real battle.
CO2 is logarythmic, we cannot possibly produce enough more to make a significant difference.
The IPCC calculations are at a height in the atmosphere that does not translate to the same temp changes on the surface, and they even admit that there may be NO surface temp change.
The IPCC glosses over the fact that any warming will be insignifanct in areas that are already warm, and most pronounced in cold places where it makes the least possible difference.
With ammunition like that we’re fighting about backradiation why?
If someone says that a g

Brian H
March 11, 2011 5:45 pm

Good commentary and summary. The IPCC is notably more honest in the fine print that the political editors haven’t thoroughly cleansed.
Except — CO2 ain’t logging any r(h)ythms. But its effect declines logarithmically.
😉
:pPp

Joel Shore
March 11, 2011 5:47 pm

wayne says:

GHG layers will keep the same amount out of the Earth system as it can ever keep within, since the ultimate energy source is outside that system (ignoring radioactive decay in the soil). That’s it. Don’t have papers in my hand proving that albedo/reflection will increase *equally* with any insulation wrapped about the Earth but I do know it is true from what I have learned over the years. As I always try, keep an open mind and critique your own thoughts, you should see something in what I just said.

Sorry, but as a physicist, I can say that what you have just said is basically complete and utter nonsense. Okay…There is a germ of truth in your idea of the relation between absorption and emission of radiation (which is not exactly what you say, but seems to be what you are sort of getting at if we are generous in our interpretation of what you have written). In fact, by Kirkhoff’s Law says they are equal; HOWEVER, that equality only holds wavelength by wavelength. Thus, it does not hold for the total wavelength-integrated emissions and absorptions except in the case when two objects are at the same temperature (and thus have the same emission spectrum).
The sun and the earth are at very different temperatures and thus emit radiation with very different spectra…In fact, the temperatures are so different that their spectra are not only very different but even have very little overlap. As a result, there is no rule whatsoever that an increase in “insulation” needs to be accompanied by a compensating decrease in the amount of solar radiation that makes it through the atmosphere to reach the earth.
I’m not going to keep an open mind about well-understood and settled basic physics just because someone does not understand it (and presents no empirical evidence whatsoever to back up his views and overturn the last couple centuries of our understanding of thermodynamics and radiative physics). That’s openmindedness to the point of letting one’s brains fall out.

March 11, 2011 6:06 pm

Ira Glickstein, PhD says:
March 11, 2011 at 4:04 pm
“So, using Bryan’s misinterpretation of “quality”, the Sun sends Earth a quantity X Joules of Energy in 2μ photons and the Earth’s Atmosphere sends an exactly equal amount, X Joules of Energy in 20μ photons out to Space. The number of photons from the Sun to Earth is different than the number from the Atmosphere to Space, but the quantity of Energy, on average, is equal.”
Concerning Bryan’s quality, he is correct because that would be the ability do to work at the radiative equilibrium temperature of the matter that this radiation came from. So with the high frequency radiation of the sun at 5777 K versus the 255 K radiation of earth, the efficiency according to the Second law for the sun radiating to an object in space at say 100 K would be 1-(100/5777) = 98% and for earth doing this 1-(100/255) = 61%.
For backradiation on earth from 255K towards 288 K this would be 1-(288/255) = – 13%, that’s would be a negative heat engine.
Concerning quantity of Energy being equal, yes. But there are three flows in play at the surface: LWR in, IR out and HEAT out. So the IR coming from the surface is lower in quantity and lower in quality than the LWR hitting the surface. So IR = LWR – HEAT.
This heat has its own ways (thermal, evaporation) getting to the next higher layer at a lower temperature or the heat would not be there (no flow), and has no relation with the IR photons but with the delta T.
Backradiation from this next layer going down can occur if there is interaction with matter at this lower temperature, but it misses the HEAT component that does not go down but further up.
So backradiation hits the surface again and is believed to create more heat so we would get IR new = IR (old) – new HEAT.
And this cannot be, because IRnew would thus have to leave from a surface with a LOWER temperature. Or perhaps backradiation cools the surface?

Joel Shore
March 11, 2011 7:23 pm

Domenic:

This is a far more accurate picture of what is truly going on:
http://slayingtheskydragon.com/Latest-News/omgmaximum-co2-will-warm-earth-for-20-milliseconds.html

Wow…That was written by one of the co-authors of “Slaying the Sky Dragon”?!? That’s really embarrassing. If I were an AGW skeptic, I would be running as fast as possible to disassociate myself from that book and anyone remotely involved with it! The fact that people like you are actually endorsing such scientific nonsense guarantees that you will continue to be regarded as “flat-earthers” by any serious scientists.

CO2. In reality, in the overall scheme of things, an obscure bit player. Hardly worth looking at. A nothing. Just like the major AGW players: Keeling, Mann, Jones, Gore, Schmidt, Hansen….in reality they are nothing but bit players. Nothings. Not even worth a glance.
Soon, many more people will come to realize that.

And, you demonstrate that by posting a bunch of pseudoscientific nonsense?!? Let me give you a hint…If you actually want to convert serious scientists to your point-of-view, you would be much better off being a little more discerning in what arguments you choose to embrace. Stick with defending arguments for low climate sensitivity; they also may be pretty poor (e.g., as has recently been demonstrated regarding some of Roy Spencer’s work: http://arthur.shumwaysmith.com/life/content/roy_spencers_six_trillion_degree_warming ) but at least they aren’t patently ridiculous!

wayne
March 11, 2011 9:39 pm

Hans says:
March 11, 2011 at 6:06 pm
… So backradiation hits the surface again and is believed to create more heat so we would get IR new = IR (old) – new HEAT.
And this cannot be, because IRnew would thus have to leave from a surface with a LOWER temperature. Or perhaps backradiation cools the surface?

Like your fresh style Hans. Same thoughts, different way to say it.
That brings to mind the one case where I have always considered back-radiation literally, and not merely figuratively, real and proper; that is the case when low clouds are actually at a higher temperature that the surface, warmed from condensation. It does happen, but may be somewhat rare.
In that case your last equation would show positive.
Speaking in terms of individual photons as is common in these discussions, this is the very case where there would be more photons beaming down back to the surface than leaving from the surface traveling upward, literally heat moving downward and warming it. True back-radiation. Still don’t like that term, but since that what it is called, I will continue to use it. This is really just normal radiative warming from warm clouds.
Trying to stay parallel, can you it put that way?

Phil.
March 11, 2011 10:03 pm

wayne says:
March 11, 2011 at 10:31 am
Phil. says:
March 11, 2011 at 7:51 am
For a temp of 270K the peak frequency is 15.9 THz corresponds to 18.9 microns.
——–
Phil, respectably, it seems you missed a factor in this calculation as explained by this paper: http://www.journal.lapen.org.mx/sep09/12_LAJPE_303_Lianxi.pdf
Best to just use equation λmax (µm) = 2897.77/T without needing to using the c/1.760 correction in subsequntly converting frequency back to wavelength with Wien’s frequency form of the equation.
For a temp of 270K the peak wavelength is 2897.77 µm·K/270K = 10.73 µm.

Nope I guess I’ll have to explain it again.
When you plot spectral radiance as a function of frequency you get a maximum at a frequency of 15.9 THz corresponds to 18.9 microns as correctly shown in the data above from Petty.
When you plot spectral radiance as a function of wavelength it is a different functional form and the maximum is at 10.73 µm. Both are correct you’re just looking at different representations of the curves.
Try here as a source for the various equations:
http://www.spectralcalc.com/blackbody/blackbody.html
Dave Springer:
Look elsewhere than Wikipedia when possible for clarity, you are probably now totally confused, and yes, the top charts are incorrect.

No they are correct, perhaps it’s some of the commenters on WUWT that are confused?

Phil.
March 11, 2011 10:22 pm

Hans says:
March 11, 2011 at 1:45 pm
“Phil. says:
What on earth is the ‘quality’ of radiation?”
That would be something like the ability do to work at the radiative equilibrium temperature of the matter that this radiation came from. So with the high frequency radiation of the sun at 5777 K versus the 255 K radiation of earth, the efficiency according to the Second law for the sun radiating to an object in space at say 100 K would be 1-(100/5777) = 98% and for earth doing this 1-(100/255) = 61%.

Which is nothing to do with the radiation, you can’t tell me what the ‘quality’ of 5μm radiation is for example.
Dave Springer says:
March 11, 2011 at 11:58 am
Phil. says:
March 11, 2011 at 9:57 am
“What’s the equation for quality?”
I think you’re being facetious but for the benefit of everyone else…

Not being facetious at all, so far no-one has answered the question, bear in mind what I asked was “What on earth is the ‘quality’ of radiation?”, the simple answer is that there is no such thing!

Phil.
March 11, 2011 10:49 pm

Domenic says:
March 11, 2011 at 9:09 am
It amazes me that so many are still chasing around the canard tossed out by the AGW people.
Just like “Hide the decline!”
“Hide the nitrogen, oxygen and water!”
They tell you to disregard the N2, O2 and H2O in the atmosphere.
They tell you to disregard the HUGE heat capacity and HUGE thermal mass represented by N2, O2 and H2O in the atmosphere.
Disregard the 99.05% of the atmosphere. It has no effect.
Focus only on the 0.05% CO2 that we tell you to.

The canard appears to be of your making, who among the ‘AGW people’ tells you that nonsense, no one tells you to ignore the thermal mass of N2 and O2, that’s the sink that absorbs the energy absorbed by CO2. You should however be aware that N2 and O2 aren’t capable of absorbing IR radiation.

wayne
March 11, 2011 11:16 pm

Dave Springer says:
March 11, 2011 at 4:39 am
@wayne (con’t)
The one-way insulating effect of GHGs is why, in my blanket example, I use two black rocks which are both exposed to the sun during the day so they can heat up (equally) then just one rock has a blanket thrown over it at night. The next morning they are uncovered and the temperature of the blanketed rock will be higher than the unblanketed rock. The next day, since one rock is warmer in the morning than the other yet both will be exposed to the sun again getting the same amount of daytime heating, the rock that is blanketed at night will reach a higher daytime maximum temperature. This temperature increase will continue until a new (higher) daytime equilibrium temperature is reached. The hotter the rock is when the blanket is thrown over it at night the quicker it will lose heat through the blanket. Heat flow rate through the insulating barrier is proportional to the temperature difference between the warm and cold sides. As the difference increases heat is lost faster. It is that which prevents the blanketed rock from just getting hotter and hotter until it melts.

Sorry Dave, didn’t notice you had commented back, tsunami and all, thank the Lord more people were not lost.
But your example is a bit unreal in this aspect, you are physically removing the insulation from one rock only during the daytime, putting it back on during the night. Can’t do that for the atmosphere. Try running it again in your mind with the blanket left on it all the time, even during the day. One with a blanket, one without. What would the rocks show in their temperature profiles then?
To me they would be different temperature profiles for sure but it seems, off of the top of my head, that there would be two times every day that their temperatures would be identical, given enough days to equalize. The blanketed one would have less diurnal range. The emissivity and reflectivity of the blanket itself would be the controlling factor.
What would you guess?
See, I don’t believe one-way insulators exist in gases. To me, that very factor is why Venus with an atmosphere of 96.5% CO2 is approaching a mirror surface to solar radiation. The CO2 blanket that you seem to see cuts equally both ways, and you end up back at with no change at all. Physics is good at twisting you mind like that, forgotten factors. See Scarlet Pumpernickel’s link above concerning Venus. I found that very interesting. I have spent countless hours on Venus’s atmosphere, it’s pressure gradient, density gradient, lapse rate, temperature profile, trying to understand a union of all planetary atmospheres, but I was working the surface up. He worked top down. I give him a pat on the back for his method, even if it ends up having flaws. I learned a new approach there.

Oliver Ramsay
March 11, 2011 11:28 pm

@ davidmhoffer
I’ve always found winter camping very uncomfortable and I don’t do it if I can avoid it, however, now that I’ve gleaned some insight into the wonders of igloos I’m ready to give it another shot.
It takes quite a number of hours to build even a small one but I’m sure it’s well worth the effort since I will not have to pack in a tent or a down-filled sleeping-bag.
As I understand it, once in the snugness of my owner-built snow structure, I can safely remove my mukluks and caribou skin parka. These I can lay on the icy floor as a mat to reduce the dreaded conduction. Perhaps radiation from the floor would keep me warm but I’m not too sure of it.
I know that, inside my clothing, my skin temperature is about 22C and I know from experience that my body’s internal organs are able to maintain homeostasis when my skin is that temperature. It’s got to do with dilation of blood vessels, although I’ve read elsewhere that it’s quite different on mountain-tops.
So, I can confidently lie naked upon my folded clothing and feel the temperature of my skin rise to 23 degrees as the walls of the igloo benevolently shine their rays upon me. As this happens, the capillaries close to the surface of my skin will happily dilate so as to rid myself of excess heat and thus avoid the horrible fate of cooking from the inside.
I do have some apprehension about the snow’s ability to modulate its radiative re-heating of my person, but I assume that if the ice is heating me, it is perhaps doing it at the expense of its own temperature; that is to say, it’s getting colder.
There again, if cold ice can heat me so satisfactorily, what might even colder ice do?
David, you invoked, very convincingly, Stefan’s Law, and I know you don’t think much of the Second Law, but the First Law is starting to falter here, too.
Quality of energy has won itself a bad reputation in these parts but it’s the quantity I’m concerned about. How long can a person stay in an igloo before they spontaneously combust? Is there any way to control this relentless heating of oneself when surrounded by ice? If I put my parka back on will the ice stop with the heating?
Can I really have faith in you as my arctic adventure mentor? How do you know it’s radiation at work and not conduction and convection. After all, you’re a little off on how many air changes you think a person would need and you suggest that the vent is sited so as to shed heat as well as exhaust gases.
So we’ve got hot things heating cooler things, we’ve got cold things heating hotter things.
I suppose if two things are the same temperature, they just keep heating one another equally forever.
Why are you so keen to convince us that you’ve got it all figured out?
Is it lonely out there?

March 12, 2011 1:17 am

Oliver Ramsey;
You have got to be kidding. That’s a rebuttal? Ridiculous conjectures drawn from thin air and attributed to me? Followed up with reading my mind announcing that you know what I think, and then making fun of that? I musta some ticked you off at some point. I shall try and be civil in order to set the record straight if nothing else.
OR:It takes quite a number of hours to build even a small one (igloo)>>>
You can build a quinzee in a fraction of the time and they are a better shelter for one person.
OR:As I understand it, once in the snugness of my owner-built snow structure, I can safely remove my mukluks and caribou skin parka>>>
No, I never said that, and your sarcasm does nothing to advance the discussion. If you insist I be specific, it takes time for the warmth to build up inside, it might never get warm enough to take off anything, though in most cases you’ll wind up unzipping your coat a bit to cool off. But you’ll not freeze to death like you would outside the structure.
OR:So, I can confidently lie naked upon my folded clothing and feel the temperature of my skin rise to 23 degrees as the walls of the igloo benevolently shine their rays >>>
Again, I said no such thing. You want to lie naked in the hopes of warming up go ahead. You want to construe that I said the igloo would shine that strongly then fine, lest you put words in my mouth, I’ll answer that one too. If you stopped to think through the physics for a moment you might ask yourself how much benevolent sunshine there is shining out of your…body in the first place. The igloo doesn’t create more heat, it only passes a small portion of the heat radiated by your body back to you. So unless you yourself are hot enough to shine benificently upon those around you in the first place, you are just being ridiculous.
OR: I assume that if the ice is heating me, it is perhaps doing it at the expense of its own temperature; that is to say, it’s getting colder. >>>
Bingo! Everything radiates energy commensurate with its own temperature as defined by Stefan’s law. So yes, it is getting colder. But it isn’t “heating you” at its own expense. Its radiating energy at exactly the same rate it would if you were there or not. Since you are there some of it runs smack dab into you. Not enough to keep you from freezing to death since you were dumb enough to take your clothes off, but some.
OR: There again, if cold ice can heat me so satisfactorily, what might even colder ice do?>>>
Uhm…. heat you less satisfactorily? C’mon, no where did I say that the colder the ice was the more it could warm you.
OR: I know you don’t think much of the Second Law>>>
You clearly do not know what I think, you don’t even understand what I wrote. The Second Law is just as real as Stefan’s Law.
OR: How long can a person stay in an igloo before they spontaneously combust?>>>
Forever.
OR:Is there any way to control this relentless heating of oneself when surrounded by ice? If I put my parka back on will the ice stop with the heating?>>>
Per my previous comment, you only get a portion of what you yourself radiated out in the first place back. If you put your parka back on before you stupidly freeze to death, then YES! the ice suddenly be getting less energy radiated at it by you, and consequently radiate less back. For a while. Eventually your body heat will heat up the parka, and it will start radiating heat, a portion of which the ice will radiate back at you.
OR:How do you know it’s radiation at work and not conduction and convection. After all, you’re a little off on how many air changes you think a person would need and you suggest that the vent is sited so as to shed heat as well as exhaust gases.>>>
Because when you radiate heat, you lose energy at the rate defined by your temperature and Stefan’s Law. If you are dumb enough to to sit your naked but on ice, you will definitely have conduction, the rate at which you will lose energy to the ice via conduction will be massively higher than what you radiate, and far in excess of what your body can replace by burning calories. Best keep your pants and parka on, they are made of stuff that doesn’t conduct well. As for the vent, yes, it is sited at the top to exhaust gases, most notably the carbon dioxide from your breathing. Since the warmest air in the igloo is most likely your own breath, it will rise via convection, taking both the CO2 and the heat you breathed out with it.
In other words conduction is massively negative, and convection is also negative, that’s how I know that neither of them are what keep you warm inside an igloo.
OR: I suppose if two things are the same temperature, they just keep heating one another equally forever.>>>
Oddly, I know you meant that to be sarcastic, but you finally got one right.
OR: Why are you so keen to convince us that you’ve got it all figured out?
Is it lonely out there?>>>
Why are you so keen to misrepresent what I said, exagerate to the ridiculous claims that I never made, pretend to read my mind and be critical of the fictitious opinions you attribute to me, and end with the most devastating, sarcastic, cutting logic that you can summon up. “Is it lonely out there?”
You’re the one lying naked in the middle of an igloo freezing his appendages off and screaming see! I told you I wouldn’t spontaneously combust! So let me answer your question seriously.
The subject of climate change is WAY off the rails. The claims of the AGW crowd are preposterous, the warning of impending doom fictitious nightmares, and the solutions proposed at best a medicine for worse than the disease. AGW has become nothing more than a tool in the hands of those who grasp for power and wealth for themselves at the expense of those who are doing honest work, and at the expense of the poor and down trodden that they proclaim to be protecting.
I no more have it all figured out than do the climate “scientists” who spout doom and gloom while deleting anything that might contradict them. But there are certain aspects of how thermodynamics work that are known, they are the foundation upon which the AGW alarmism is built, and assailing a firm foundation with distorted criticisms of a ligitimate explanation wastes your time, my time, and discredits us as skeptics in front of the AGW scientists who happen to have taken the one piece of science they got right and built a house of cards upon it.
I’m keen to set the record straight with warmists and skeptics alike so that we can move on to discuss order of magnitude, feedback, distribution of change by lattidude and altitude, and all the other issues that make a mockery of the AGW claims. If you want to level a well thought out, valid criticism of my explanation, I’ll either answer with a better explanation or an hmmm never thought of that. But if all your going to do is throw up wild claims about I said as arguments, then I’ll just say this:
Those AGW dudes are planning on picking your pockets, and I’m keen to get their bullarky discredited before they get around to picking mine.

wayne
March 12, 2011 4:14 am

Joel Shore,
Thanks for catching that. After reading my own words back I should have been sure to mention the concept of shells, really any thickness. If a layer above “insulates” radiation going upward, then once given energy is above that same shell and is absorbed and radiated downwards that same shell will “insolate” to equally keep it outside the system. Same shell, same insolating property. Does that make it any clearer?

Joel Shore
March 12, 2011 5:08 am

Domenic:

This is a far more accurate picture of what is truly going on:
http://slayingtheskydragon.com/Latest-News/omgmaximum-co2-will-warm-earth-for-20-milliseconds.html

Wow…That was written by one of the co-authors of “Slaying the Sky Dragon”?!? That’s really embarrassing. If I were an AGW skeptic, I would be running as fast as possible to disassociate myself from that book and anyone remotely involved with it! The fact that people like you are actually endorsing such scientific nonsense guarantees that you will continue not to be taken seriously by any serious scientists.

CO2. In reality, in the overall scheme of things, an obscure bit player. Hardly worth looking at. A nothing. Just like the major AGW players: Keeling, Mann, Jones, Gore, Schmidt, Hansen….in reality they are nothing but bit players. Nothings. Not even worth a glance.
Soon, many more people will come to realize that.

And, you demonstrate that by posting a bunch of pseudoscientific nonsense?!? Let me give you a hint…If you actually want to convert serious scientists to your point-of-view, you would be much better off being a little more discerning in what arguments you choose to embrace. Stick with defending arguments for low climate sensitivity; they also may be pretty poor (e.g., as has recently been demonstrated regarding some of Roy Spencer’s work: http://arthur.shumwaysmith.com/life/content/roy_spencers_six_trillion_degree_warming ) but at least they aren’t patently ridiculous!

Joel Shore
March 12, 2011 5:41 am

For Oliver Ramsay and others who are saying that there is violation of the Second Law of Thermodynamics in the atmospheric greenhouse effect: Show us specifically where said violation occurs using any precise correct statement of the 2nd Law and anything from the full-blown line-by-line radiation calculations to any sort of simple quantitative model of the greenhouse effect (such as the simple shell model that Willis Eschenbach posted about once here on WUWT).
Hint: You won’t be able to do it because in fact none of these descriptions violate the Second Law.
When I teach my students about the Second Law in a few weeks, I will tell them what I always tells them: that the Second Law is a double-edged sword. On the one hand, it is a powerful tool that can easily cut through pseudoscientific nonsense (such as claims of perpetual motion machines). On the other hand, like any powerful tool, it can also be abused to actually spread pseudoscientific nonsense as when incorrect claims are made by creationists that evolution violates the 2nd Law or incorrect claims are made by AGW skeptics that the greenhouse effect violates the 2nd Law.
davidmhoffer says:

I’m keen to set the record straight with warmists and skeptics alike so that we can move on to discuss order of magnitude, feedback, distribution of change by lattidude and altitude, and all the other issues that make a mockery of the AGW claims.

Even though David Hoffer and I disagree strongly on the seriousness of AGW, I agree with him that if we are going to debate scientific issues, one might as well debate issues that are still open to real scientific debate (such as the magnitude of feedback effects and hence of the climate sensitivity) than to waste time with claims that are scientifically-ridiculous, such as claims that the greenhouse effect violates the 2nd Law or that humans are not responsible for the rise in atmospheric CO2 levels over the last couple hundred years. Arguing these things may seem fun and may be able to convince some members of the general public but it is only going to further marginalize you in the scientific community … and that is where the scientific debate ultimately matters.