CO2 heats the atmosphere…a counter view

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)

The simplistic view of CO2 heat trapping

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 .

  1. 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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Patrick Davis

Fascinating read! I wonder how many “climate scientists” have read Tom’s posts before? Thankyou Tom and Anthony.


Thank you, thank you, thank you for such a clear and concise refutation of the big lie, for which you have just taken down the cult of CAGW’s main tenet. May this be the final nail in their coffin!

Michael Schaefer

There goes “Cap’N’Trade”…

richard telford

Nothing can warm a gas at thermodynamic equilibrium. That’s what thermodynamic equilibrium means!


— > swish it goes over the head

John Marshall

Is this not like saying that the laws of thermodynamics must be obeyed? To comply with the 2nd law the heated CO2, which heats rapidly, will cool rapidly. Also adiabatic cooling of convecting air, containing CO2, will radiate heat to a cooler area, ie above, not to the warmer surface. So heat will dissipate to space. The 2nd law also means that the CO2 must loose heat because entropy must increase, so it cannot store heat-nothing can. The Trenberth paper, which the IPCC rely on for their energy transfer calculations and which describe the GHG actions, does not include night time or the fact that the earth is shaped like a ball. Its use of the Stefan Boltzman law is not appropriate because a black body cannot be found in nature, or anywhere else.


I’m sorry I’m really busy today and getting a late start. I suspect this is very well done, but I saw an error near the is stated (with my minor edits for clarification:
“In addition this interaction (with infrared radiation) will take place only if the molecule presents a non zero dipolar [moment]”
That is incorrect. CO2, for instance, is a linear molecule symmetric along its axis in the ground vibrational state and therefore has no dipole moment. To make you statement correct it should read:
“In addition this interaction (with infreared radiation) will take place only if there is a change in the dipole moment of the molecule between the ground and excited states.”
Where excited state is defined as the higher state of energy after radiation is absorbed and the ground state is defined as the lower state of energy before the radiation is absorbed or after radiation is re-emitted (the same rule applies for emission of radiation as well).
I hope you don’t draw incorrect conclusions later on because of this misunderstanding or that it’s simply a mistake made while writing this up. But I just don’t have time to look any closer right now.
I hope that’s helpful!

i>Now the most important question is “What are the rates of the → and the ← processes ?”
The question that precedes that is, what do you mean by rate? How do you quantify it? What number are you talking about?
In chemical kinetics, where the considerations are rather similar, the relevant number is the rate constant. In this context, it would be something like
d[N2+]/dt = k [CO2*][N2], where k is the constant.
Then your argument that the rates must be equal would have to take the concentrations (partial pressures) into account.
To put it another way, in any interaction between a CO2 and a N2, it is much more likely that the CO2 will be excited than the N2 (because of the disparity in numbers).


Ok. Spent just 5 more minutes and I really this you’ve missed the boat on this one. For starters, V->T processes occur at very low probability. In your overly simplified scenario where the N2 then interacts with the CO2 molecule to re-excite a vibration one must assume that the interaction occurs, for instance, before the excited N2 molecule has collided with another N2 molecule and that they both are now translationally excited, but both with less than the total energy required to excite the CO2 vibration upon subsequent interactions. Remember, those are quantized, so the collision which re-excites that mode must have exactly the right combination of both translational energy AND impact parameter.
You also seem to have left out rotations. Any collisions between relatively translationally excited N2 (or any other molecule) and CO2 are far more likely to transfer energy into excited rotational states and have exactly no effect on the vibrational state of CO2.
I really think the energy transfer between molecules is not handled correctly in this treatment.
Again, I hope that’s helpful, and let me know if you think I’m off the mark on my criticisms.


Thanks Tom. It may be above some, but most should be able to understand your “Conclusions and caveats” section.


The conclusions is ….(?) that the green house effect works as scientists have known for over a century, but some of the simplified terminology often found in textbooks is misleading (?)


Again, sorry. In your hohlraum thought experiment the enclosed volume likely would emit the same amount of 15 um radiation as it absorbed, though your setup wouldn’t find that detail. 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. That’s heating in my book. And when the external source goes away the gas in the hohlraum will start cooling.
Certainly in the actual atmosphere there are other issues, like the lapse rate as you mention, but that doesn’t mean heating/cooling isn’t also occuring. It is.


“The simplistic view of CO2 heat trapping”

And by “simplistic”, you mean “wrong”, right?
What is trying to be demonstrated by the graphic is the long-wave radiation absorptivity of the atmosphere. What it is showing is that the atmosphere “reflects” such radiation back toward the ground, which is just bad science.
Would it have been so difficult to terminate the smaller atmospheric absorption arrow in the atmosphere itself and then have a separate set of arrows (both toward the surface and into space) showing the radiative energy from the atmosphere?
It still would have been “simplistic”, but also “correct”.


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?
Anyway, it tries to prove something that flies in the face of the evidence that CO2 keeps us warm, and more CO2 heats up the planet (as can be seen in the nice curves that are not believed by a majority of bloggers in attendance, let alone the obvious increase in average temperatures over the last decades).
Tom Vonk did not win a Nobel price for his three papers on Differential cross sections for rotationally inelastic collisions (all he wrote as far as I can see on Scopus), and did not publish a peer-reviewed paper on CO2. But because he’s blogging a lot, he must know a lot, so I believe him.

Does this scepticism know no end? Doubt about apocalyptic alarm ends up with a challenge to the last bastion of the ‘the simple science’…struck down by simple science…of text books textbook science…known for 100 years.
So how does this placed the results, sometimes experimental results, of Peixoto 1992, Goody 1964, Howare 1955, Schmidt 1913, Rubens and Aschkinass 1898, and of course Tyndall. Would Lindzen agree? Are they seeing emission as reflection (back to earth) rather than dispersal of energy? I am sure the answer is implicit in what you say, but someone might elaborate for us non-scientists.


Thank you Tom for taking your time. Very nicely put, and very interesting.


Thanks for your essay. I think I understand “most” of what you were saying. Could you please comment on the ideas presented in this video and explain where they violate the physics.
Thanks, I would appreciate it.

Roy Spencer

If local thermodynamic equilibrium exists in a certain volume of a gas, and you add more CO2 at the same temperature, it is true that the volume’s temperature will not change.
No one I know of would disagree with this.
But it’s when that volume is exposed to outside influences — like IR radiation from the solar-heated surface of the Earth passing through that volume — that a temperature change can occur as a result of adding more CO2 to the volume.

Ed Fix

It’s great to see someone clearing the air about the basic physics of the system. Too often we see sweeping, simple-to-understand analogies that are simply wrong.
“CO2 traps heat in the atmosphere like a blanket”
“Infrared energy (which is heat)…”
Stuff like that is like fingernails on a blackboard to me, and I’m not even a physicist. I just paid attention during my freshman year.

John W.

II have some questions:
1. Who are you?
2. Do you have a Ph.D.? From what university?
3. How many peer reviewed papers have you published?
4. Do you now or have you ever taken any form of payment from a profit making organization?
Just wanted to get those out of the way. 8^)
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?

Bernie Anderson

If the above is true then how do NDIR CO2 measuring instruments work. They detect CO2 either by the attenuation of an IR beam passing through the cell with the test gas or by detecting the temperature rise of the gas in the cell. Both of these would require absorption of IR by the CO2 and that energy has to go somewhere.


In defense of the “Scienceofdoom” site, it seems like a detailed but standard explanation of basic CO2 science. It’s listed on WUWT’s site as a “Pro AGW View” next to SurrealClimate, but that doesn’t seem fair. I don’t think he comes up w/anything other than the standard ~1.2C temp rise for doubling CO2.
Undetermined feedbacks could raise that amount or OTOH drop it to near zero. We don’t know, and I don’t think he makes any claims beyond that.


I would like to add:
The theory of AGW says that extra CO2 causes a minor warming (less than 0.5degree) which then causes the atmosphere to absorb more water vapour. H2O content of the atmosphere is dependant on temperature – if you warm the air by 0.5C it can hold a little more H2O. This small increase in H2O then increases atmospheric temperature since H20 is a strong GHG which then allows more H2O to be absorbed etc (positive feedback loop).
The problem that the AGW’ers ignore is that when that first stage of extra H2O absorption happens, the H2O comes from evaporation from the oceans/surface water. Evaporation is a Endothermic process which causes a COOLING effect, so the first stage of H2O absorption MUST causes a slight cooling of the oceans which will offset the slight warming of the initial CO2. They have ignored this.
Not only that, but the slight cooling of the oceans from step1, results in the oceans being able to absorb more CO2, since CO2 absorption in the oceans is dependant on ocean temperature (when temperatures go up, oceans release CO2, when temps go down, oceans absorb CO2). So the slight cooling from the evaporated H2O will actually cause the oceans to absorb some of the CO2 back.
ie. Nature ballances itself out.


Thanks. That was eminently digestible and thoroughly nutritious.


Help me here… A system in equilibrium quickly returns to equilibrium at a higher level when it absorbs an IR photon: CO2+N2CO2+N2 becomes CO2*+N2CO2+N2+ (pardon the limited special character skills). This looks like heating to me, and, the temperature is controlled by the variance in the rate of absorbed and emitted IR photons for any small volume.
What I really don’t understand is why water and CO2 are better, by a factor of 20, at this as N2 and O2

Jan K. Andersen

I am sorry to say that this was a rather disappointing article on this otherwise excellent blog.
The fact that CO2 absorbs infrared energy and heat the atmosphere is no theory,, it is a fact. The flaw in the article is that it does not take into account that the absorbed radiation is outgoing, but the emitted radiation go in all directions.


Many, many thanks for this.
I am going to be busy for a while. < 🙂


Greenhouse gases delay the transport of heat to the top of the atmosphere. The net effect is to make surface temperatures higher than they would be without these gases. So it may be sensible to say to the layperson “greenhouse gases warm the atmosphere”. I don’t know that saying such things automatically destroys credibility.
Also, the use of “microscopical” vs. “microscopic” and “dipolar momentum” vs. “dipole moment” seemed weird to me.


My above entry would have looked better if the system had not erased my left and right carrots from the equations.
[Note: “Caret.” And the angle brackets can be created so they will not disappear with the HTML command: &, lt [or gt], ; (Delete the commas & spaces but keep the semicolon.) Put together, the “lt” command will create a ‘less than’ angle bracket; “gt” will create a ‘greater than’ bracket: < > ~dbs]

James Sexton

Heh, the instinctual aversion to chemistry I had in my youth is still with me! I almost fell asleep reading the article just like I did when I was taking organic chemistry 20+ yrs ago! That being said, if I understand this basic chem lesson properly, given the multi-directional release of energy by the CO2 molecule. It releases equally. Now, I’m going to have to try an absorb the significances. Thanks Tom and A.
PS, any way we can put this in some algebraic equation?

Vince Causey

Interesting article Tom. However, there is one point I did not quite understand.
In equation (2) you wrote 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⁺ “.
But later you wrote: “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 .”
But doesn’t equation (2) say that CO2* collides with an N2 molecule and relaxes?

There is no question that greenhouse gas H2O warms the atmosphere. Why would CO2 be different?


In other words CO2 which absorbs strongly the 15µ IR , will emit strongly almost exactly as much 15 µ radiation as it absorbs .
Why isn’t it “exactly”?


Fantastic paper about what CO2 actually does. just one question, in this equation CO2* + N2 ↔ CO2 + N2⁺ (3) you say that there is no net energy transfer between the CO2 and N2. That makes sense, but doesn’t some energy get lost during the collisions? With this energy loss, would the atmosphere actually cool a little bit, or would some outside source of energy come in and add whatever energy was lost, thereby maintaining the temperature?

Thank you very much for this post, it’s something I’ve wanted to read for ages but haven’t taken enough time to hunt down. It’ll take a couple more readings to absorb.
I am confused by your central point:

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 .

If we take a handful of air with a 15µ IR flux passing through, is it really in LTE? What happens in the case where the handful is not illuminated and in LTE, then illuminated briefly, and then not. The handful will settle back into LTE, and hence has a temperature – will that temperature be higher? Your quote above requires that the illuminated air be in LTE, so whether it is or not is my central confusion.
Also, I’d appreciate any notes on the time scale for CO2* to relax vs the collision time between CO2 and N2 at different atmospheric levels (i.e. pressure & temperature).

I feel smarter for having read this.


OT but {snip]
[reply] But me no buts, butter me no parsnips, off to Tips and Notes with you. 😉 RT-mod

John Prendergast

This is brilliant, and validates Lord Kelvin’s second law and many others. I have long beleived that CO2 did pass any trapped heat to other molecules until equilibrium was restored If it were not so, there would have been reports of airliners flying into patches of incadescent CO2 , and suffering engine stall while they melted! Even UFOs can’t do this.


I know that this is the first part of a series but having covered the action inside little boxes or spheres where delta size goes to a molecular level, what happens as the density changes?


Sorry but your demosntration is wrong. You have demonstrated that in LTE:
1) The ammount of energy going from CO2 to N2 is the same as the ammount going from N2 to CO2.
2) The ammount of radiation received by the CO2 is the same as the ammount it radiates back.
But you have not demonstrated that:
3) The percentage of molecules in excited state remains the same if you change the initial conditions regarding the percentage of CO2 in the gas.
And if the percentage of molecules in excited state changes, the gas changes its temperature. Perhaps the fraction of time when this happens cannot be called LTE conditions, I don’t know. But it happens anyway.
A 5% of the CO2, when this gas is a 0,04% of the atmosphere, means very few molecules excited. A 5% of CO2, if it was 50% of the atmosphere, would mean a huge ammount of molecules excited and a higher temperature. Even though they would emit the same IR energy that they would absorb.
By the way, CO2 doesn’t heat, that’s true, and that makes your small experiment work. What it does is slow down the cooling of the surface of the Earth. It reduces the net radiation at 15um. As a result, you get higher temperatures than you would have without CO2. But all the initial heat came from the Sun, of course.
This is the experiment you need: put a ball at “room temperature” in a room filled with CO2 and N2, but then make the walls of the room cool to nearly 0K. Then try to demonstrate that the the speed at which the ball cools down is independent of the ammount of CO2.

Amino Acids in Meteorites

Since there is no tropospheric warming that was predicted to happen in global warming seeing this explanation partly helps to clear up why.
Next I know there will be the argument of downward flux to the surface of energy from added co2 causes warming. But that isn’t happening either. That argument poorly incorporates the effects of H2O. There is found to be cooling at the surface when H2O is effected by that downward flux.
The global warming hypothesis is wrong.

Thank you.
I now must take some C13H18O2 after reading this excellent piece.
(Isn’t carbon (and hydrogen and oxygen) a wonderful thing!)

Nice presentation, but an incorrect conclusion.


@Jared 4.23: I’m not a physicist, and it takes a physicist to shoot holes in stories like Tom’s. It also takes a degree in physics to be able to say if the conclusions and caveats are worthy of discussion. As far as I can see, most physicists in attendance are shooting big holes in the story…. But then again, maybe they’re wrong, not Tom. That’s why I like this better than theoretical physics.

Thanks Tom. It might not have been as elegant as your own post, but basically I think we were saying the same thing – and it is great to have my model validated at last!
Submission to science and technology committee of UK parliament
Noddy Science: CO2 Warms the Planet?
(24) The complex bond structure within CO2 means that it can readily absorb and emit radiation in the infra-red (IR) band where thermal radiation is given off by a blackbody9 at the temperature of the earth. Much of this IR is at wavelengths at which other atmospheric constituents do not interact, so if CO2 is exposed to a warmer surface like the earth, it will absorb radiation that would otherwise pass through into the cold of space AND likewise if CO2 is exposed to the cool of outer space it will emit vast quantities of IR at wavelengths which other gases cannot emit.
(25) When CO2 is present low in the atmosphere, it tends to block transmission of these wavelengths into space and reduce heat loss to space. When CO2 is present high in the atmosphere, it helps emit IR, so causing cooling of the atmosphere acting as a vector by which other gases can lose heat into space. Like triple glazing, the system is complicated by the movement of air. Air warmed at the surface naturally tends to rise above the majority of the (blocking) atmosphere and it cannot descend until it has cooled by the emission of IR into the cool of outer space. CO2 cooling is as natural as CO2 warming, the atmosphere being a highly dynamic and complex system: a natural cooling system taking heat from the surface of the earth up into space via convective currents.
(26) Simple physics could suggest CO2 is a cooling gas as easily as warming and “obvious” assertions must be validated against real evidence, not the preconceptions of “scientists”. CO2 could impact the atmosphere in other ways: changes in specific heat capacity, density, interaction with water droplets and cloud formation. Other gases like water vapour also have their effects. It would be wrong to say that increases in CO2 can not affect the climate, but it is equally absurd, in such a complex system, to say this or that effect must dominate in the absence of the normal rigorous testing required by science.


This is really how it works, not to mention all the rest of CO2 related reactions in nature, all endothermic, energy consuming, to build living organisms, beginning with glucose and cellulose. As I said: No CO2=No underwear. 🙂
However, sadly, you won´t find a chemist-politician.


Thank you, that was helpful. One question: you mentioned the non-LTE areas of earth and space boundaries; would the dynamic mechanism of thunderstorms also fall into a non-LTE area and be a factor in some of the emerging work on storm related energy transfers?

Thanks Tom for that well-written explanation. Some writers swear blind that CO2 cannot lose vibrational energy by collision and some swear otherwise. Whilst it was obvious to me that an equation like your (3) must be possible (for interactions with energy equal to a quantum level difference in CO2 vibration), I have been puzzled why some otherwise expert physicists have claimed otherwise. I think your explanation in terms of LTE explains why. Thanks again.

Leon Brozyna

A fine starting off point in this post, especially with all the “yes, but … ” comments. Will have to come back to this later today to see what else has sprouted.


Rich says:
August 5, 2010 at 5:36 am
In other words CO2 which absorbs strongly the 15µ IR , will emit strongly almost exactly as much 15 µ radiation as it absorbs .
Why isn’t it “exactly”?
because some portion of what is absorbed via radiation is transferred via collision and other frequencies.
the velocity of these molecules is around 1000 mph, too.