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 .
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It is a very strange statement!
All the non-equilibrium thermodynamics of continuous systems is based on the local equilibrium assumption but however clearly allow for heat transfer, mass transfer, non-equilibrium reactions etc.
The rates of your processes will be equal only at true equilibrium, i.e. when all the activities of the reactants and products will be such that the equilibrium constant is reached. If you prefer, the rates of the processes do not only depend on the cross-section but also on the relative amounts of everything – including the photon flux! This is why, for example, one can have non-equilibrium reactions AND equipartition of energy all together.
Jan says:
August 5, 2010 at 4:33 am
As I don’t understand this post, am I supposed to take it to prove that climate change doesn’t exist, as any argument will suit this purpose?
__________________________________________________________
No that is the whole point of this blog. If you do not understand the post ASK QUESTIONS! or google the terms you do not understand.
This post is more or less a mathematical proof.
Oh. I also should have mentioned – in the N2:CO2 laser the N2 is *vibrationally* excited, not translationally (though it certainly is probably travelling quite rapidly at these energies, thank you bery much!). It is this vibrationally excited N2 that transfers the vibrational energy to CO2 in the N2:CO2 laser. V->V reactions occur with significantly higher probablility than T->V reactions. To the points that in most chemical kinetics problems no one ever bothers to add the T->V channels because they occur with so little probability.
How did the N2 get vibrational excited in the laser cavity, you might ask – the electron impact as a result of the multikilovolt discharge across the cavity to initiate the laser build up.
Electronic and nuclear degrees of freedom the only to remaining modes that energy can be stored in chemical systems that we haven’t discussed so far. They really don’t contribute to the problem of atmospheric heat transfer.
6,000 ppm of CO2 in the atmosphere of Mars does not create any measurable “greenhouse effect”. Its black body T = its actual T = 210K. Without digging into theory (and I am graduated analytical chemist specialized on spectral analytical methods), this is enough proof for me to believe, that IR active gas *alone* does nothing.
Mars: thin atmosphere (albeit composed of 95% CO2) –> no “greenhouse effect”
Earth: denser atmosphere –> some real “atmospheric effect”
Venus: 95x denser atmosphere than on Earth –> powerful “atmospheric effect”
No atmosphere – no “greenhouse effect”
We have such a joke: an ant and elephant are crossing the wooden bridge. Ant says: “What a clatter we do!”
For developing a “greenhouse effect”, you need the elephant – a bulk atmosphere. But then call it more properly “atmospheric effect”.
Isn’t the real question “How much does CO2 affect warming” rather than “Does CO2 affect warming”? I am in no position to enter into a discussion on the physics of CO2 in the atmosphere however everyone of us knows why deserts experience extreme cooling at night and places with high humidity, like south Florida, do not. I grew up in southern New England and during summer when humid air flowed from the south it stayed hot and humid at night but when the wind shifted to come out of the north bringing in drier air, things cooled off at night. Both of these scenarios have nothing to do with CO2. So again, isn’t the question really how much warming can be credited to CO2? If we can get a final answer on that we will have made real progress.
CO2, the gas we all EXHALE does not stay there just looking down to earth in the atmosphere, it enters into the living organisms chain, first to make GLUCOSE (Plants do breath CO2 ya know, and btw exhale O2- a gas we like to breath, ya know…), then cellulose, etc., etc. then we EAT those vegs in order to live, or cattle eats them to make the MEAT we eat, or chicken eat corn grains to make the EGGS YOU EAT scrambled every morning, ya know buddy….all these transformations USE the energy CO2 ATE first, the HEAT, ya know, which makes possible for YOU TO WALK AROUND , fool around, repeating, like a parrot, silly things like: “CO2 is a greenhouse gas”.
And, what you don’t know is that all this is BUSINESS it’s just an invention to take the money out from your work, from your pockets, for some guys up there, who you will never know, and who will enjoy it in some distant sunny beaches while you keep on working.
So WAKE UP!
Aeh well .. lets just say, that a gas in local equilibrium heats up under the influence of external IR-radiation just as much as water in local equilibrium under the influence of external microwave radiation, then we can skip “all that complicated math” and have an experiment everybody can test at home . .
Roger Pielke, Sr and Ben Herman on the Greenhouse Effect of CO2:
http://pielkeclimatesci.wordpress.com/2010/07/23/the-greenhouse-effect-by-ben-herman-and-roger-pielke-sr/
http://pielkeclimatesci.wordpress.com/2010/07/27/%e2%80%9cthe-greenhouse-effect-part-ii%e2%80%9d-by-ben-herman-and-roger-pielke-sr/
It seems to me that it is a circular argument to assume equilibrium in order to prove equilibrium. The whole point is that when energy is added to the system, that shifts the equilibrium. I don’t quite understand how someone capable of such a detailed proof would not see this. Perhaps I’m the one who is missing something, but I just don’t see it.
Merrick, I appreciate your cogent and respectful critique. Your comments are also well written given that they are probably composed as you type. I love debate that rises to this level.
Werner Weber says:
August 5, 2010 at 6:23 am
Better yet go to your micro wave and set it to 5 minutes and hit start. Then when the bell rings check the temperature. Nada. The CO2 did not heat up. The air did not heat up.
If I have three identical rooms each filled with a different gas, one O2, one N2 and one CO2 and have say electric resistance heaters in each. Set the themostats of each to 70 F and see which goes up fastest and if any go above the 70 F.
Shut off the heat and see which cools down fastest.
See which one draws the most electricity maintaning the 70 F.
If CO2 can add heat and make something hotter then the CO2 room should be above 70 F.
Also if the earth is at say 300 K and is emitting IR at 10 mirco and CO2 absorbs at 200 K at 15 micro just how much of the earths IR emissions can be absorbed? 200 K does not heat 300 K.
Clyde Rhodes says:
August 5, 2010 at 7:36 am
“It seems to me that it is a circular argument to assume equilibrium in order to prove equilibrium. The whole point is that when energy is added to the system, that shifts the equilibrium. ”
Exactly. And when energy is continuously being added to – and lost by – the system, you never reach thermodynamic equilibrium, even locally, only at best a new steady state. Like pouring water into a bucket with a hole in it.
Seems like you have proved that microwave ovens dont work – I wonder if the manufacturers will take mine back?
Tom,
I’m with you until your conclusion on your time symmetrical equation. You conclude that, because
CO2* + N2 ↔ CO2 + N2⁺
is time invariant, there is no net transfer of energy from CO2 to N2.
You noted, of course, that CO2* -> CO2 + y
Your system, more correctly, is
CO2 + +y CO2* + N2 CO2 + N2⁺ CO2 + -y
It is noted then, that +y are photons taken in by your gas system from whatever sources, and -y are photons emitted from your system. The case for “”heating” of the atmosphere” (or whatever scorned terminology) is thus the statement +y > -y, in terms of the system described. I don’t believe your time invariance of the middle relationship touches on the ability of either outer relationship to be asymmetrical, so your conclusion that conventional wisdom is truly incorrect on that basis seems to be a bit of a stretch.
Also,
CO2* + N2 CO2 + N2⁺,
you state that this relationship is time symmetric, however, the system more accurately described (for the purposes of this all brand new argument!) is
CO2 + +y CO2* + N2 CO2 + N2⁺ N2 + N2⁺,
where the last portion represents the probability asymmetry of N2⁺ decay modes. Given the markedly different partial pressures of the two gasses, it is much more likely that N2⁺ + N2 decays to N2 + N2⁺ than N2⁺ + CO2 decays into N2 + CO2*. Probability asymmetries suggest bulk movement of energy into the largest component of the medium.
Sorry for the verbosity, I look forward to constructive criticism of exactly where I’m wrong above.
So, if CO2 can’t warm the atmosphere, how does the sun do it?
The atmosphere is mostly transparent to sunlight.
Sunlight hits the ground, warms it up, and then…?
I find it facinating just reading the comments. I can follow the physics and arguments but the disagreements are eye opening. We just don’t have all the answers do we?
There’s a hole in the bucket dear Liza, dear Liza, there’s a hole in the bucket, a hole.
Love that little ditty.
(((MOD: PLEASE DELETE MY PREVIOUS POST AND THIS NOTE. THANKS GREATLY.)))
[Note: might help if you chose a different screen name, and used a working email address. You email is fake, no more posts from you until you fix it. see the WUWT policy page ~mod]
Tom,
I’m with you until your conclusion on your time symmetrical equation. You conclude that, because
CO2* + N2 ↔ CO2 + N2⁺
is time invariant, there is no net transfer of energy from CO2 to N2.
You noted, of course, that CO2* » CO2 + y
Your system, more correctly, is
CO2 + +y «» CO2* + N2 «» CO2 + N2⁺ «» CO2 + -y
It is noted then, that +y are photons taken in by your gas system from whatever sources, and -y are photons emitted from your system. The case for “”heating” of the atmosphere” (or whatever scorned terminology) is thus the statement +y > -y, in terms of the system described. I don’t believe your time invariance of the middle relationship touches on the ability of either outer relationship to be asymmetrical, so your conclusion that conventional wisdom is truly incorrect on that basis seems to be a bit of a stretch.
Also,
CO2* + N2 «» CO2 + N2⁺,
you state that this relationship is time symmetric, however, the system more accurately described (for the purposes of this all brand new argument!) is
CO2 + +y «» CO2* + N2 «» CO2 + N2⁺ «» N2 + N2⁺,
where the last portion represents the probability asymmetry of N2⁺ decay modes. Given the markedly different partial pressures of the two gasses, it is much more likely that N2⁺ + N2 decays to N2 + N2⁺ than N2⁺ + CO2 decays into N2 + CO2*. Probability asymmetries suggest bulk movement of energy into the largest component of the medium.
Sorry for the verbosity, I look forward to constructive criticism of exactly where I’m wrong above.
Chris V says:
August 5, 2010 at 8:05 am (Edit)
So, if CO2 can’t warm the atmosphere, how does the sun do it?
The atmosphere is mostly transparent to sunlight.
Sunlight hits the ground, warms it up, and then…?
The sun hits the sea more than the land, penetrates it up to a depth of tens of metres, gives up it’s heat energy to it, and then the ocean emit’s longwave IR into the atmosphere. It heats it mainly by convection, and the latent heat of evaporation and condensation. Then the atmosphere loses the heat to space, after bouncing it up and down to the surface a few times.
merrick says:
August 5, 2010 at 4:32 am
. . . This is a blackbody at equilibrium. Now, add a source at greater than 15C (like a warm earth surface) and ad long as the rate of incoming 15 um radiation is greater than the 15 um radiation rate you already measured from your hohlraum there will be disequilibrium and the temperature of the hohlraum (not just the CO2 but all of the gas) will increase until the hohlraum is again emitting the same amount of 15 um radiation as is coming in. . . .
Merrick, please help me on this (I think) last of your comments;
If you add additional heat the whole experiment does indeed fall apart. But I think that misses the point; if you’ve got a hollow sphere, filled with a mixed gas at equilibrium at a certain temperature, and then you replace some of the gas with CO2 (raise the concentration) does the sphere necessarily retain more heat? I thought he did a good job of explaining why (at least very locally) this is not really possible. You will change the level of equilibrium of the several forms of heat transfer, but that’s just the speed that the heat moves around. The temperature itself will not change. Or is that not correct?
Can you please explain, by way of increasing the concentration of CO2 in the sphere, rather then increasing the amount of heat being added, how he is incorrect in his analysis. The way I read your comment (applied to the real world) is that if the Earth’s surface would get warmer, so would the atmosphere. That is an argument I would probably tend to agree with, though many on the AGW side of this debate have denigrated the notion that the Sun’s variations aren’t really responsible for the recent increase in global surface T, so I’m not sure your argument is fruitful in that regard.
Thanks.
>”First is that a temperature cannot be defined for a volume which is not in LTE . ”
Completely untrue. The most famous example is the fluorescent light bulb. Its electron temperature is very high, which generates the UV light, whilst its ion temperature is low, which is why the bulb doesn’t melt. They clearly have different temperatures which can be defined. What you’re searching for is called temperature coupling. This is the LTE you’re looking for. Plasmas have uncoupled temperatures of multiple species, while, gasses near atmospheric pressure have well coupled temperatures.
It’s been a while since I’ve studied atmospheric plasmas. I don’t know whether any of the assumptions of LTE, equipartition, or Boltzmann statistics actually hold at the altitudes you are interested in. In fact, I don’t think you’ve specified altitude. I’m a little busy now, so, I’ll read the rest of your article later.
Well, as a non-scientist, here’s my take from the comments thus far, right or wrong:
1. The people who know what they’re talking about are making technical comments that they claim rebut the original article.
2. The people who don’t know what they’re talking about are making comments along the line of: “Thanks, Tom, good to have that all clearly stated, great effort, puts it all to rest, I’ll have to set aside time to really absorb what you’ve explained so elegantly, etc.” (paraphrasing all, so as not to insult any one particular individual.)
Leaving me with an unchanged opinion that goes as follows: Apparently the human-caused increase in CO2 is heating the planet to at least a minor degree, but probably not to any degree worth getting exercised over. In the end, the next ice age will still prevail, no matter how much CO2 we pump out by burning fossil fuels, and that ice age will be caused by the same cyclical factors that caused the others. The trick is in the meantime to avoid impoverishing ourselves attempting to avoid a scare trumpeted by people who desire additional, even complete, control over our lives, helped along by their useful idiots. (I’ll leave it to the individual alarmists to decide whether they desire control, or are simply one of the useful, but those will eventually prove to have been the only two choices in my view of the matter.)
Note that I consider myself a skeptic even though I grant that we are probably subtly adding to the earth’s temperature by burning fossil fuels. My skepticism revolves around the politics/economics of the situation, and the corrupting of science as a means to reach ends that I find objectionable.
JohnW says:
Seriously, thanks for a well written and informative essay. I do have a serious request. Your essay addresses absorptivity and emissivity. If you can find the time, could you also address transmissivity?
Dr Heins Hug had an excellent article several years back. He measured the transmisibility of 15 micron radiation through standard atmosphere and dtermined that all the radiation is absobed in the first 10 meters.
http://www.john-daly.com/artifact.htm
You can’t pump 20 watts per square meter into the air and absorb it in the first ten meters without warming thing up a bit.
For this to be true it would have to be accepted as the dinosaur killer of AGW’s foundation. Somebody would have noticed that by now. I fear something you are sure of is wrong.
I eagerly await Part II.
Your article is quite interesting but your experiment is moot to AGW, and the result is misinterpreted.
If define an experiment this way:
A thin rectangular box having two very large sides A and B parallel and opposite to each other, the other 4 sides having negligible area. Side A is a 1000Watt 15um IR emitter, side B is a perfect IR absorber.
If we take N2 and fill the volume, the N2 will reach an equilibrium temperature based on the temperature of the box and whatever radiation is absorbed. Now if we instantly change the gas content to be 50/50 N2 and CO2 which has better absorption of the 15um IR wavelength, an IR detector at side B would detect an instantaneous drop in IR energy at the 15um wavelength. I’m sure we can agree on that.
What would then happen over time, is the gas inside the box would stabilize to an equilibrium temperature. As the system stabilized, an energy detector at the absorbing B side would eventually record the same amount of energy transfer from the A side as before – return to equalibrium.
However, the gas temperature will stabilize slightly higher than the N2 gas. An effect proven by the initial drop in IR at your detector and the time delay until the system re-stabilizes.
None of this violates your equations or assumptions.
Good try though.