Guest post By Tom Vonk (Tom is a physicist and long time poster at many climate blogs. Note also I’ll have another essay coming soon supporting the role of CO2 – For a another view on the CO2 issue, please see also this guest post by Ferdinand Engelbeen Anthony)

If you search for “greenhouse effect” in Google and get 1 cent for statements like…
“CO2 absorbs the outgoing infrared energy and warms the atmosphere” – or – “CO2 traps part of the infrared radiation between ground and the upper part of the atmosphere”
…you will be millionaire .
Even Internet sites that are said to have a good scientific level like “Science of doom” publish statements similar to those quoted above . These statements are all wrong yet happen so often that I submitted this guest post to Anthony to clear this issue once for all.
In the case that somebody asks why there is no peer reviewed paper about this issue , it is because everything what follows is textbook material . We will use results from statistical thermodynamics and quantum mechanics that have been known for some 100 years or more . More specifically the statement that we will prove is :
“A volume of gas in Local Thermodynamic Equilibrium (LTE) cannot be heated by CO2.”
There are 3 concepts that we will introduce below and that are necessary to the understanding .
- The Local Thermodynamic Equilibrium (LTE)
This concept plays a central part so some words of definition . First what LTE is not . LTE is not Thermodynamic Equilibrium (TE) , it is a much weaker assumption . LTE requires only that the equilibrium exists in some neighborhood of every point . For example the temperature may vary with time and space within a volume so that this volume is not in a Thermodynamic Equilibrium . However if there is an equilibrium within every small subvolume of this volume , we will have LTE .
Intuitively the notion of LTE is linked to the speed with which the particles move and to their density . If the particle stays long enough in a small volume to interact with other particles in this small volume , for example by collisions , then the particle will equilibrate with others . If it doesn’t stay long enough then it can’t equilibrate with others and there is no LTE .
There are 2 reasons why the importance of LTE is paramount .
First is that a temperature cannot be defined for a volume which is not in LTE . That is easy to understand . The temperature is an average energy of a small volume in equilibrium . Since there is no equilibrium in any small volume if we have not LTE , the temperature cannot be defined in this case.
Second is that the energy distribution in a volume in LTE follows known laws and can be computed .
The energy equipartition law
Kinetic energy is present in several forms . A monoatomic gas has only the translational kinetic energy , the well known ½.m.V² . A polyatomic gas can also vibrate and rotate and therefore has in addition to the translational kinetic energy also the vibrational and the rotational kinetic energy . When we want to specify the total kinetic energy of a molecule , we need to account for all 3 forms of it .
Thus the immediate question we ask is : “If we add energy to a molecule , what will it do ? Increase its velocity ? Increase its vibration ? Increase its rotation ? Some mixture of all 3 ?”
The answer is given by the energy equipartition law . It says : “In LTE the energy is shared equally among its different forms .”
As we have seen that the temperature is an average energy ,and that it is defined only under LTE conditions , it is possible to link the average kinetic energy <E> to the temperature . For instance in a monoatomic gas like Helium we have <E>= 3/2.k.T . The factor 3/2 comes because there are 3 translational degrees of freedom (3 space dimensions) and it can be reformulated by saying that the kinetic energy per translational degree of freedom is ½.k.T . From there can be derived ideal gas laws , specific heat capacities and much more . For polyatomic molecules exhibiting vibration and rotation the calculations are more complicated . The important point in this statistical law is that if we add some energy to a great number of molecules , this energy will be shared equally among their translational , rotational and vibrational degrees of freedom .
Quantum mechanical interactions of molecules with infrared radiation
Everything that happens in the interaction between a molecule and the infrared radiation is governed by quantum mechanics . Therefore the processes cannot be understood without at least the basics of the QM theory .
The most important point is that only the vibration and rotation modes of a molecule can interact with the infrared radiation . In addition this interaction will take place only if the molecule presents a non zero dipolar momentum . As a non zero dipolar momentum implies some asymmetry in the distribution of the electrical charges , it is specially important in non symmetric molecules . For instance the nitrogen N-N molecule is symmetrical and has no permanent dipolar momentum .
O=C=O is also symmetrical and has no permanent dipolar momentum . C=O is non symmetrical and has a permanent dipolar momentum . However to interact with IR it is not necessary that the dipolar momentum be permanent . While O=C=O has no permanent dipolar momentum , it has vibrational modes where an asymmetry appears and it is those modes that will absorb and emit IR . Also nitrogen N-N colliding with another molecule will be deformed and acquire a transient dipolar momentum which will allow it to absorb and emit IR .
In the picture left you see the 4 possible vibration modes of CO2 . The first one is symmetrical and therefore displays no dipolar momentum and doesn’t interact with IR . The second and the third look similar and have a dipolar momentum . It is these both that represent the famous 15µ band . The fourth is highly asymmetrical and also has a dipolar momentum .
What does interaction between a vibration mode and IR mean ?
The vibrational energies are quantified , that means that they can only take some discrete values . In the picture above is shown what happens when a molecule meets a photon whose energy (h.ν or ђ.ω) is exactly equal to the difference between 2 energy levels E2-E1 . The molecule absorbs the photon and “jumps up” from E1 to E2 . Of course the opposite process exists too – a molecule in the energy level E2 can “jump down” from E2 to E1 and emit a photon of energy E2-E1 .
But that is not everything that happens . What also happens are collisions and during collisions all following processes are possible .
- Translation-translation interaction . This is your usual billiard ball collision .
- Translation-vibration interaction . Here energy is exchanged between the vibration modes and the translation modes .
- Translation-rotation interaction . Here energy is is exchanged between the rotation modes and the translation modes .
- Rotation-vibration interaction … etc .
In the matter that concerns us here , namely a mixture of CO2 and N2 under infrared radiation only 2 processes are important : translation-translation and translation-vibration . We will therefore neglect all other processes without loosing generality .
The proof of our statement
The translation-translation process (sphere collision) has been well understood since more than 100 years . It can be studied by semi-classical statistical mechanics and the result is that the velocities of molecules (translational kinetic energy) within a volume of gas in equilibrium are distributed according to the Maxwell-Boltzmann distribution . As this distribution is invariant for a constant temperature , there are no net energy transfers and we do not need to further analyze this process .
The 2 processes of interest are the following :
CO2 + γ → CO2* (1)
This reads “a CO2 molecule absorbs an infrared photon γ and goes to a vibrationally excited state CO2*”
CO2* + N2 → CO2 + N2⁺ (2)
This reads “a vibrationally excited CO2 molecule CO2* collides with an N2 molecule and relaxes to a lower vibrational energy state CO2 while the N2 molecule increases its velocity to N2⁺ “. We use a different symbol * and ⁺ for the excited states to differentiate the energy modes – vibrational (*) for CO2 and translational (⁺) for N2 . In other words , there is transfer between vibrational and translational degrees of freedom in the process (2) . This process in non equilibrium conditions is sometimes called thermalization .
The microscopical process (2) is described by time symmetrical equations . All mechanical and electromagnetical interactions are governed by equations invariant under time reversal . This is not true for electroweak interactions but they play no role in the process (2) .
Again in simple words , it means that if the process (2) happens then the time symmetrical process , namely CO2 + N2⁺ → CO2* + N2 , happens too . Indeed this time reversed process where fast (e.g hot) N2 molecules slow down and excite vibrationally CO2 molecules is what makes an N2/CO2 laser work. Therefore the right way to write the process (2) is the following .
CO2* + N2 ↔ CO2 + N2⁺ (3)
Where the use of the double arrow ↔ instad of the simple arrow → is telling us that this process goes in both directions . Now the most important question is “What are the rates of the → and the ← processes ?”
The LTE conditions with the energy equipartition law give immediately the answer : “These rates are exactly equal .” This means that for every collision where a vibrationally excited CO2* transfers energy to N2 , there is a collision where N2⁺ transfers the same energy to CO2 and excites it vibrationally . There is no net energy transfer from CO2 to N2 through the vibration-translation interaction .
As we have seen that CO2 cannot transfer energy to N2 through the translation-translation process either , there is no net energy transfer (e.g “heating”) from CO2 to N2 what proves our statement .
This has an interesting corollary for the process (1) , IR absorption by CO2 molecules . We know that in equilibrium the distribution of the vibrational quantum states (e.g how many molecules are in a state with energy Ei) is invariant and depends only on temperature . For example only about 5 % of CO2 molecules are in a vibrationally excited state at room temperatures , 95 % are in the ground state .
Therefore in order to maintain the number of vibrationally excited molecules constant , every time a CO2 molecule absorbs an infrared photon and excites vibrationally , it is necessary that another CO2 molecule relaxes by going to a lower energy state . As we have seen above that this relaxation cannot happen through collisions with N2 because no net energy transfer is permitted , only the process (1) is available . Indeed the right way to write the process (1) is also :
CO2 + γ ↔ CO2* (1)
Where the use of the double arrow shows that the absorption process (→) happens at the same time as the emission process (←) . Because the number of excited molecules in a small volume in LTE must stay constant , follows that both processes emission/absorption must balance . In other words CO2 which absorbs strongly the 15µ IR , will emit strongly almost exactly as much 15 µ radiation as it absorbs . This is independent of the CO2 concentrations and of the intensity of IR radiation .
For those who prefer experimental proofs to theoretical arguments , here is a simple experiment demonstrating the above statements . Let us consider a hollow sphere at 15°C filled with air . You install an IR detector on the surface of the cavity . This is equivalent to the atmosphere during the night . The cavity will emit IR according to a black body law . Some frequencies of this BB radiation will be absorbed by the vibration modes of the CO2 molecules present in the air . What you will observe is :
- The detector shows that the cavity absorbs the same power on 15µ as it emits
- The temperature of the air stays at 15°C and more specifically the N2 and O2 do not heat
These observations demonstrate as expected that CO2 emits the same power as it absorbs and that there is no net energy transfer between the vibrational modes of CO2 and the translational modes of N2 and O2 . If you double the CO2 concentration or make the temperature vary , the observations stay identical showing that the conclusions we made are independent of temperatures and CO2 concentrations .
Conclusion and caveats
The main point is that every time you hear or read that “CO2 heats the atmosphere” , that “energy is trapped by CO2” , that “energy is stored by green house gases” and similar statements , you may be sure that this source is not to be trusted for information about radiation questions .
Caveat 1
The statement we proved cannot be interpreted as “CO2 has no impact on the dynamics of the Earth-atmosphere system” . What we have proven is that the CO2 cannot heat the atmosphere in the bulk but the whole system cannot be reduced to the bulk of the atmosphere . Indeed there are 2 interfaces – the void on one side and the surface of the Earth on the other side . Neither the former nor the latter is in LTE and the arguments we used are not valid . The dynamics of the system are governed by the lapse rate which is “anchored” to the ground and whose variations are dependent not only on convection , latent heat changes and conduction but also radiative transfer . The concentrations of CO2 (and H2O) play a role in this dynamics but it is not the purpose of this post to examine these much more complex and not well understood aspects .
Caveat 2
You will sometimes read or hear that “the CO2 has not the time to emit IR because the relaxation time is much longer than the mean time between collisions .” We know now that this conclusion is clearly wrong but looks like common sense if one accepts the premises which are true . Where is the problem ?
Well as the collisions are dominating , the CO2 will indeed often relax by a collision process . But with the same token it will also often excite by a collision process . And both processes will happen with an equal rate in LTE as we have seen . As for the emission , we are talking typically about 10ⁿ molecules with n of the order of 20 . Even if the average emission time is longer than the time between collisions , there is still a huge number of excited molecules who had not the opportunity to relax collisionally and who will emit . Not surprisingly this is also what experience shows .
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.


RE: Caleb: (August 5, 2010 at 6:10 pm) “However I crave to have water vapor included in this sort of discussion.”
As the concentration of gaseous water (water ‘vapor’) in the atmosphere is on the order of 3o,000 to 40,000 ppm, this trace greenhouse gas really *is* the invisible 800 lb gorilla in the room even if we ignore the phase change effects.
goddard sez:
“I go up to 14,000 feet pretty regularly in the summer, but have yet to encounter temperatures of -40C . Normally they are about 40-55C warmer than at the same elevation in Antarctica, even though Antarctica receives more solar insolation.
Please, provide some actual data to throw around in this food-fight. Are you talking about Long’s Peak or Mt. Shasta in the “mid-latitudes? Where? The average solar radiation at Barrow, AK FOR TWO AXIS FLAT TRACKING COLLECTORS at the peak of summer is 8.3 KWH/m-2. If you look at the “zero-tilt” figures, it’s only 4.9 KWH m-2. It’s over 11 KWH m-2 (more than twice the radiation) for most mid-latitude sites in the northern hemisphere. Here is the data: http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/27502.txt
Just what is your point?
Hi again Tom,
Thanks for this interesting article. I found it enlightening re how different competent people have been saying things that seem inconsistent, but now the question is, have you got it right?
I followed you until, roughly, this point:
The problem here is that the energy equipartition law is a statistical result, not a precise law such as “Action and reaction are equal and opposite.” An abundance of transfers one way or the other are expected and will happen, all they do is increase the probability that an increase in transfers the other way will restore the equilibrium sooner or later. But talking about a single interaction necessitating an interaction in the opposite direction? No.
Here is the complete interaction sequence, in which I label the three states:
CO2 + γ + N2 (state A) ↔ CO2* + N2 (state B) ↔ CO2 + N2⁺ (state C)
All three states are in equilibrium for some given composition of the components. Photons (state A) will be absorbed giving excited CO2 (state B) and the reverse, and excited CO2 will interact with N2 to give translationally more energetic N2 (state C) and the reverse. Accordingly, you tell us “For example only about 5 % of CO2 molecules are in a vibrationally excited state at room temperatures , 95 % are in the ground state.”
But that assumes the existing proportions of the ‘ingredients’. If more CO2 molecules were pumped into the atmosphere, then it might be, for example, that 6% would be excited and only 94% in the ground state. Similarly, if the photon flux were doubled, that would drive more CO2 into the excited state and in turn drive more energy into the N2 translational mode. The relative balances amongst states A, B, and C depends upon the concentrations of the components, in just the same way as with a normal chemical reaction: change something and the ratios of all the rest change too. For this reason, whilst I thank you for explaining something that has puzzled me, I don’t agree that your conclusion follows.
wayne says:
August 5, 2010 at 7:15 pm
Paul Birch
August 5, 2010 at 4:39 pm
Ok, don’t see where you gave where any of the numbers are coming from but it suffices you seem sure the radiation pressure from added GHG molecules (H2O or CO2), let’s say 10 per million is so infinitesimally small that the lift of the atmosphere in every cubic meter per one meter layer up to say 80 km is so insignificant it can be totally ignored…
Radiation Pressure doesn’t depend on the density of the substance, but on the incident power exerted by a photon stream on a given surface. A real life example:
The insolation today was 417 W*s/m^2. The surface absorbed ~191.25 W/m^2 and emitted 181.7 [(N * m/s) /m^2] to the atmosphere. The Prad of the surface photon stream on the atmosphere was 0.575 μPa. On the other hand, the Prad exerted by the power of the incident solar radiation upon the surface was 1.39 μPa, i.e. 2.4 times more intense than the Prad exerted by the surface radiation. Regarding the Prad of the atmosphere, at its current physical conditions, it was 0.221 μPa, that is, ~40 times weaker than the Prad exerted by the surface. According to the second law of thermodynamics, the surface cannot affect to the solar photon stream and the radiation pressure exerted by the atmosphere cannot affect to the surface. The solar Prad affects the surface and the surface Prad affects the atmosphere. The atmosphere cannot affect the surface by any means. It would be as if one tried to inflate a balloon by extracting the air from its interior.
wayne says:
August 5, 2010 at 1:04 pm
My point to both of you is a huge point that is being missed here. Huge. Since the LW radiation from the ground is always in the azimuth hemisphere’s direction and ½ of the reemitted photons are in the nadir hemisphere, a radiation pressure exists here which will cause and infinitesimal but real expansion of the atmosphere which will exactly cause a temperature decrease that exactly counteracts any warming from the downward radiation. anna, I stated this in January or February and never could get you to see.
The sun hits the earth, there is radiation pressure, but the forces are so tiny with respect to the bulk of the earth, nobody includes this in the calculation of the orbits.
Some of that momentum bounces back as infrared, part of it interacts and changes energy and momentum of the bulk of the atmosphere ( in complicated manner not yet well understood as Tom says) , but still tiny forces with respect to the bulk atmosphere .
Your error is assuming linearity in the response:
http://en.wikipedia.org/wiki/Statistical_mechanics
You can see in the article that there are no linearities : exponentials in probability functions are involved in the temperature changes so your “will exactly cause a temperature decrease that exactly counteracts ” is not self evident. Care to do the calculations necessary to see how much pressure changes affect temperature changes in this framework, i.e. the photon qm framework? http://star-www.st-and.ac.uk/~mmj/Fluids_notes/fluids_05.pdf
Actually, I had used radiation pressure as an argument that finally all radiation reflected has to vanish into space taking its momentum with it, otherwise from momentum conservation H2O and CO2 should end up in the stratosphere because their numbers are limited, but radiation is continuous and unlimited, and if they are being continuously hit-by/absorbing photons momentum conservation would push them to the stratosphere.
maxwell says:
August 5, 2010 at 6:36 pm
Tom,
as a physicist, you should know that molecular vibrations are not ‘quantified’. They are quantized. That’s why they call it ‘quantum mechanics’.
Play nice.
English is not Tom’s first language or the language he read physics in.
Nasif Nahle says:
I checked your paper: http://www.biocab.org/Mean_Free_Path.pdf , and I got confused. Your calculations seem to show that the MFP is 4.8cm, not 48m. I must be missing something obvious, but I can’t spot my oversight. Your help will be appreciated.
Enneagram says:
August 5, 2010 at 6:11 am
*SNIP*
However, sadly, you won´t find a chemist-politician.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
How quickly we do forget:
“She read chemistry at Somerville College, Oxford and later trained as a barrister.”
http://en.wikipedia.org/wiki/Margaret_Thatcher
Tom,
You write “As we have seen above that this relaxation cannot happen through collisions with N2 because no net energy transfer is permitted , only the process (1) is available .”
I would urge you to visit the Wiki page on CO2 lasers wherein it is explained how energy is transferred from Nitrogen to CO2 and finally to Helium, in order to make the thing work.
http://en.wikipedia.org/wiki/Carbon_dioxide_laser
stevengoddard says:
August 5, 2010 at 2:13 pm
Ah. If Phoenix could be just a little bit drier it would be as cold as Antarctica.
Got it. And the hottest days in Phoenix must have the highest dew point.
I am still a bit puzzled about the CO2 to N2 energy transfer being equal and opposite the CO2 molecule is outnumbered 25 000 to 1 by the other molecules in the atmosphere N2 the largest then O2 and H2O so the chance of the CO2 molecule colliding with a different molecule is 25 000 to 1 the chance of another molecule colliding with a CO2 is 1 in 25 000 but the other molecules will collide with each other thus diluting the energy so I do not think the equal and opposite argument stands.
Here is a nice chart showing the effective temperature that is being radiated across the IR Earth spectrum.
http://cimss.ssec.wisc.edu/goes/sndprf/spectra.gif
Essentially, in the CO2 bands, the effective emission temperature is 220K. In other words, the 10km to 20km layer (where CO2 relaxation/emission now has a 50% chance of being emitted directly to space rather than being absorbed by other near-by CO2 molecules or by other atmospheric molecules through collision). It should also reflect the Ozone layer temperatures where CO2 steals energy from Ozone and emits that directly back to space but the numbers are a little off here.
The H20 bands emit at 230K to 270K, or let’s say at the beginning of the troposphere at the top of the thunderstorm and cloud layers.
There is also an Ozone and Methane bands which are similar to H20.
After that, we have the “Atmospheric Windows” where emission is direct to space with no interception in the GHG bands and this emission comes from the hard surface and any other atmospheric molecule. It is, effectively, at the blackbody radiation temperature (and all molecules including N2 and O2 absorb and emit blackbody radiation – this seems to not be understood by many).
What is unusual about the Atmospheric Window emission temperatures is that they are emitting at a higher temperature than they really should be. The Windows are emitting at 296K (versus an average 288K). One can also review the Arctic atmospheric window emission temperatures and find that it is also higher than it really should be – the Arctic windows are emitting at 268K or -5C which is quite a bit higher than the average temperature).
http://i165.photobucket.com/albums/u43/gplracerx/PettyFig8-2.jpg
This means that the climate science community has been focused on the CO2 and GHG bands and forgot that solar energy is also being emitted by blackbody radiation. The fact that it is higher than the surface temperature means that some GHG band long-wave absorptions are being transferred to N2 and O2 and then being partially emitted by blackbody radiation in the atmospheric windows.
This is the kind of issue that the blinders of textbook greenhouse theory puts on those that fully accept every word coming out of pro-AGW scientists. It is not what really happens in the real atmosphere apparently.
Tom,
Could you please explain whether nobel gasses emit black body or not??
If they do, could you also speak to whether the IR emitted from the molecular bond is the same as from the blackbody??
I have wondered whether the molecular bond radiation is being conflated with black body radiation which actually shows the TEMPERATURE of particles. This would cause a number of inccorrect assumptions as to where and what is emitting.
That is 15 micron band is also the band of a particular TEMPERATURE emission. How would someone with an IR detector tell whether CO2 was emitting or something at that temperature was emitting? All CO2 can’t be at that same temperature right??
Pochas:
“I would urge you to visit the Wiki page on CO2 lasers wherein it is explained how energy is transferred from Nitrogen to CO2 and finally to Helium, in order to make the thing work.”
I would urge you not to conflate an environment where extremes of energy are being pumped with an environment of low temps and energy levels. The stuff happens differnt!!
Ron House says:
August 5, 2010 at 9:20 pm
Nasif Nahle says:
Just to give an example, the mean free path length of photons in the CO2 at its current density in the atmosphere is 48 meters.
I checked your paper: http://www.biocab.org/Mean_Free_Path.pdf , and I got confused. Your calculations seem to show that the MFP is 4.8cm, not 48m. I must be missing something obvious, but I can’t spot my oversight. Your help will be appreciated.
You’re correct, Ron. Thanks a lot! I made a mistake when writting the article. I apologize. Here the reviewed version:
http://www.biocab.org/Mean_Free_Path.pdf
Sorry and thanks again.
Ric Werme,
Antarctica is not a little drier. It is a lot drier. The first few ppm account for most of the greenhouse effect.
Antarctica receives more solar energy in midsummer than Phoenix does. Instead of being knee-jerk sarcastic, think about the implications.
[Accidently pre-posted in “Why the CO2 increase is man made…”]
I note in this discussion a reference to returning to a ‘ground state’ analogous to what an excited electron in an atom does. At the atomic level, the emission spectrum just corresponds to an allowed set of energy states. If this were the case with molecular vibration, I would think we would only see simple sharp spectral lines
However, when I look at the HITRAN data available online I see a broad array of many sharp spikes. This gives me the impression that molecular vibrations may be in a transition region between the quantum and the continuous worlds. I suspect that each band represents a vibration mode and perhaps each narrow spike represents one of the allowed modal vibration energy states. Imagine a bell that changes tone in steps as the vibration dies out.
As these molecules are continuously striking one another, each molecule should always be quivering in some agitated state. Perhaps any given photon emission or absorption just corresponds to a single quantum change in an allowed set of multi-level modal vibration energy states — food for thought.
@Barry Moore…
The number of moles, the mass and the number of molecules per gram are the properties that have a major influence on the mean free path length of photons through the gases comprising the atmosphere.
Reed Coray,
Yes you are right, the CO2 will only temporarily slow down the cooling of the Earth, and later on the rate of cooling will again rise to match the energy input. But in the mean time, the Earth’s temperature rises. The Earth with more CO2 needs a higher temperature to equal the rate of cooling that it had with less CO2. However it is still wrong to say that CO2 warms the Earth. The only warming source here is the sun. CO2 only makes the cooling more difficult.
This paper should be of great interest for discussion on this topic.
It shows that that even for deep midwinter Antarctica with extremely low humidity, clear skies conditions, the radiative flux from H2O vapour was more than twice that for CO2.
Since CO2 is well mixed in the atmosphere its radiative contribution will not change much as we move to more average conditions.
However if we move to more average Earth condition with average levels of humidity introducing condensation phase change then the radiative contribution from H2O will increase dramatically.
There are some people who want to exaggerate the effect of CO2 and they will not find anything in this result to support them.
http://www.webpages.uidaho.edu/~vonw/pubs/TownEtAl_2005.pdf
stevengoddard wrote:
“Antarctica receives more solar energy in midsummer than Phoenix does.”
Phoenix is noticeably smaller than Antarctica, but is that what you mean?
Or do you mean that a square foot of ground in McMurdo receives more energy in the Antarctic midsummer than does a square foot of ground in Phoenix during an Arizona midsummer? Phoenix is only about five degrees or so North of the Tropic of Cancer.
Why doesn’t it’s more-nearly perpendicular exposure to the axis of radiation at midsummer provide Phoenix more energy per square unit than someplace further from the “tropic” parallels?
Or did I miss something?
[reply] The number of daylight hours maybe? RT-mod
Enneagram says:
August 5, 2010 at 6:11 am
*SNIP*
However, sadly, you won´t find a chemist-politician.
As has been noted, Margaret Thatcher was such a chemist-politician.
The UK MP Graham Stringer PhD was an analytical chemist. Significantly, he is one of the few UK MP’s to publicly question AGW.
Ed
Bill Illis says
Here is a nice chart showing the effective temperature that is being radiated across the IR Earth spectrum.
http://cimss.ssec.wisc.edu/goes/sndprf/spectra.gif
Essentially, in the CO2 bands, the effective emission temperature is 220K. In other words, the 10km to 20km layer (where CO2 relaxation/emission now has a 50% chance of being emitted directly to space rather than being absorbed by other near-by CO2 molecules or by other atmospheric molecules through collision). It should also reflect the Ozone layer temperatures where CO2 steals energy from Ozone and emits that directly back to space but the numbers are a little off here.
The H20 bands emit at 230K to 270K, or let’s say at the beginning of the troposphere at the top of the thunderstorm and cloud layers.
There is also an Ozone and Methane bands which are similar to H20.
After that, we have the “Atmospheric Windows” where emission is direct to space with no interception in the GHG bands and this emission comes from the hard surface and any other atmospheric molecule. It is, effectively, at the blackbody radiation temperature (and all molecules including N2 and O2 absorb and emit blackbody radiation – this seems to not be understood by many).
What is unusual about the Atmospheric Window emission temperatures is that they are emitting at a higher temperature than they really should be. The Windows are emitting at 296K (versus an average 288K). One can also review the Arctic atmospheric window emission temperatures and find that it is also higher than it really should be – the Arctic windows are emitting at 268K or -5C which is quite a bit higher than the average temperature).
Thanks Bill. I love this site, someone always comes up with the data I need!
These figures confirm the points I was making in my first post. Just to clarify a point for those why have not seen such a plot before. This effectively shows at what height in the atmosphere photons of various wavelengths are emitted. It does not show the energy radiated at each wavelenth. As can be seen the circa 15 micron photons characteristic of CO2 are emitted near the tropopause where the lowest temperatures in the atmosphere occur. Without repeating all of my previous post I would just ask how the radiation in this band can be reduced any further? To answer my own question it can only do so by reducing the temperature of the tropopause or by losing more energy to other molecules in the atmosphere at lower levels. Both are possible but I would want to see evidence before I would accept it was happening.
I have, without successs, tried to get the Hadley centre to relook at their explanation of the greenhouse effect. They say that increasing CO2 increases the height at which effective radiation takes place and this increased height means lower temperatures and therefore reduced radiation to space for which a warmer surface has to compensate to maintain the radiation balance. Sounds fair enough but then they go on to say that the current height is about 8km where the temperature is around 250K such that there is still room to cool further. When I showed them similar satellite data giving temperatures much lower than theirs and pretty well at the minimum already they stopped talking to me!
Ok here is my theory-of-everything, which seems to combine two different views: during the night, IR back radiation keeps the night surface a bit warmer, but this surface warmth must be trapped by bulk atmosphere, which is in physical contact with the surface. So the night air gets heated by convection or conduction from the warmer surface. Then this warmed bulk atmosphere warms our bodies. Otherwise like on Mars, you get plenty of IR flowing down and up, but in such a thin atmosphere it has not much effect.
During the day, direct sunlight does the surface warming job more than enough, as we can see in deserts with low humidity. In this case, IR active gases are rather responsible for cooling the surface by clouds and evaporation.
So the “atmospheric effect” is a result of both presence of IR active gases and bulk atmosphere. The ratio in which they cooperate on the final effect – relatively narrow range of survivable temperatures – is another question; it is for sure, that the most abundant IR active gas likes to create clouds, which effectively cool the surface, then it rains down and again effectively cools the surface and sometimes gets frozen to snow, which reflects sunlight and again effectively cools the surface. My bet is, that atmosphere with water cools the day and warms the night by different physical mechanisms at the same time.
Ouch, I almost forgot those 0.004% of CO2, but compared with hundreds times more water vapor, clouds and ice reflecting 30% of direct sunlight, it keeps me rather indifferent.
I would have thought that though interesting in itself this discussion is rather irrelevant in the greater scheme of climate change, ever since Ferenc Miscolski published his paper showing with the aid of 60 yrs of data and the resources of NASA behind him, that the transparency of the atmosphere to IR has not changed in those 60 years.
If heat can escape as easily through all that time, then surely the source of any increased heat on Earth must be Solar, whether it be visible light, IR or UV, and indirect effects thereof ?