**Guest Post by Tom Vonk**

In a recent post I considered the question in the title. You may see it here : http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/

The post generated great deal of interest and many comments.

Even if most of the posters understood the argument and I answered the comments of those who did not, I have been asked to sum up the discussion.

Before starting, I will repeat the statement that I wished to examine.

“Given a gas mixture of CO₂ and N₂ in Local Thermodynamic Equilibrium (

LTE) and submitted to infrared radiation, does the CO₂ heat the N₂?”

To begin, we must be really sure that we understood not only what is contained in the question but especially what is **NOT** contained in it.

- The question contains no assumption about the radiation. Most importantly there is no assumption whether a radiative equilibrium does or does not exist. Therefore the answer will be independent from assumptions concerning radiative equilibrium. Similarly all questions and developments concerning radiative transfer are irrelevant to the question.

- The question contains no assumption about the size or the geometry of the mixture. It may be a cube with a volume of 1 mm³ or a column of 10 km height. As long as the mixture is in LTE, any size and any geometry works.

- The question contains no assumption about boundary conditions. Such assumptions would indeed be necessary if we asked much more ambitious questions like what happens at boundaries where no LTE exists and which may be constituted of solids or liquids. However we do not ask such ambitious questions.

Also it is necessary to be perfectly clear about what “X heats Y” means.

It means that there exists a mechanism transferring **net** (e.g non zero) energy unidirectionaly

**from** X **to** Y .

Perhaps as importantly, and some posters did not understand this point, the statement

“X heats Y” is equivalent to the statement “Y cannot cool X”.

The critical posts – and here we exclude posts developing questions of radiative transfer which are irrelevant as explained in 1) above – were of 2 types.

**Type 1**

The argument says “LTE never exists or alternatively LTE does not apply to a mixture of CO₂ and N₂.”

The answer to the first variant is that LTE exists and I repeat the definition from the original post : “* A volume of gas is in LTE if for every point of this volume there exists a neighborhood in which the gas is in thermodynamic equilibrium (TE*)”

2 remarks to this definition:

- It is not said and it is not important how large this neighborhood of every point is. It may be a cube of 1 mm³ or a cube of 10 m³ . The important part is that this neighborhood exists (almost) everywhere.

- LTE is necessary to define local temperature. Saying that LTE never exists is equivalent to saying that local temperatures never exist.

The second variant admits that LTE exists but suggests that a mixture of CO₂ and N₂ cannot be in LTE.

The LTE conditions are given when energy at every point is efficiently spread out among all available degrees of freedom (translation, rotation, vibration).

The most efficient tool for energy spreading are molecular collisions.

Without going in a mathematical development (see statistical thermodynamics for those interested), it is obvious that LTE will exist when there are many molecular collisions per volume unit.

This depends mostly on density – high density gases will be often in LTE while very low density gases will not.

For those not yet convinced, hold out a thermometer in your bedroom and it is probable that it will show a well defined temperature everywhere – your bedroom is in LTE .

We deal here with a mixture of CO₂ and N₂ in conditions of the troposphere which are precisely conditions where LTE exists too.

**Type 2**

The argument says “The mean time between collisions is much shorter than the mean decay time (e.g time necessary to emit a photon) and therefore all infrared energy absorbed by the CO₂ molecules is **immediately** and **unidirectionaly** transferred to the N₂ molecules.”

In simple words – the CO₂ never has time to emit any IR photons because it loses vibrational energy by collisions instead.

This statement is indeed equivalent to the statement “**CO₂ heats N₂**”.

Now let us examine the above figure.

The good understanding of this figure will do much better than only answering the original question. It will also make clear to everybody what is really happening in our gas mixture in LTE.

The figure shows the distribution of the kinetic energy (Ox axis) among the N₂ molecules (Oy axis).

This typical curve is called the Maxwell Boltzmann distribution, has been known for more than 100 years and experimentally confirmed with high accuracy.

We know that the temperature is defined by <E>, the energy average.

Hence it is the curve shown in the figure that defines the temperature of a gas.

Another way to say the same thing is to say that the curve depends only on temperature. If we wanted to have the distribution for another gas than N₂ , f.ex CO₂ or O₂, it would be given by an identical curve.

The blue curve gives the distribution of kinetic energy at 25°C while the red curve gives the distribution at 35°C.

The minimal energy is small but non-zero and there is no maximal energy.

A very important point on the Ox axis is the energy of the first vibrationally excited state of a CO₂ molecule.

You notice that at 25°C the majority of N₂ molecules has insufficient kinetic energy to excite this vibrational state.

Only the proportion of them given by the dark blue surface has enough energy to excite the vibrational state by collision.

When the temperature increases to 35°C, you notice that the proportion of N₂ molecules able to excite the vibrational CO₂ state by collision has significantly increased .

This proportion is given by the sum of the dark blue and light blue surface.

You also notice that as there exists no maximal energy, there will be a proportion of N₂ molecules able to excite the vibrational CO₂ state at **any** temperature.

Trivial so far? Well it will not get much more complicated.

First 2 technical points which play no role in the argument but which I would like to mention for the sake of completness.

- The figure shows the translational kinetic energy. Even if in some (popular) literature the temperature is defined as being an average of the translational kinetic energy, this is not strictly true.

The temperature is really defined as an average of **all** energy modes. So what about the vibrational and rotational energy?

At the low tropospheric temperatures we are considering, the distribution of the vibrational energy is extremely simple : about 5% or less of the molecules are in the first excited state and 95% or more are in the ground state.

As for the rotational energy, it can be computed classically without quantum corrections and the result is that it also follows a Maxwell Boltzmann distribution.

Therefore if we wished to plot the total energy (Etranslational + Evibrational + Erotational) we would rescale the Ox axis and obtain exactly the same curve as the one that is shown.

However as we are interested in studying the T/V interactions, it is the curve of the translational kinetic energy that interests us.

- We find the omnipresence of LTE again. This curve has been derived and experimentally confirmed only, and only if, the gas is in TE. Therefore the following 2 statements are equivalent :

“*The gas is in LTE*” , “*The energy distribution at every point is given by the Maxwell Boltzmann distribution*” .

If you feel that these statements are not equivalent, reread carefully what is above.

Now we can demonstrate why the Type2 argument is wrong.

Imagine that you mix cold N₂ represented by the blue curve in the Figure with highly vibrationally excited CO₂. The mixture would then not be in LTE and a transient would take place.

In the molecular process **(1)** CO₂* + N₂ → CO₂ + N₂⁺ which says that a vibrationally excited CO₂ molecule (CO₂*) collides with an N₂ molecule , decays to the ground state (CO₂) and increases the translational kinetic energy of N₂ (N₂⁺) , there would be a net energy transfer from CO₂* to N₂ .

As a result of this transfer the temperature of N₂ would increase and the blue curve would move to the red one.

However doing that, the number of molecules able to excite CO₂ vibrationally would increase (see the blue surfaces in the figure).

That means that during the increase of the temperature of N₂ , the rate of the opposite molecular process **(2)** CO₂ + N₂⁺ → CO₂* + N₂ where N₂ molecules (those from the blue surface in the figure) vibrationally excite CO2 molecules, will increase too.

Of course the transient net energy transfer from CO₂ to N₂ will not continue forever because else the mixture would transform into superheated plasma.

A local equilibrium will be established at each point and in this equilibrium the rate of the process **(1)** will be **exactly** equal to the rate of the process **(2). **

The curve of energy distribution will stop moving and the Maxwell Boltzmann distribution will describe this distribution at every point.

This is exactly the definition of LTE.

The transient will stop when the mixture reaches LTE and its characteristic feature is that there is **no local net energy transfer from CO₂ to N₂**.

This result demonstrates both that the Type2 argument is wrong and that the answer on the question we asked at the beginning is “No”.

In very simple words, if you take a small volume (for example 1 m³) of the CO₂ and N₂ mixture in LTE around **any** point , then there cannot be any net energy transfer from CO₂ to N₂ within this volume.

To establish the last step we will take the following statements.

- The result obtained for the CO₂ and N₂ mixture in LTE is equally true for a mixture containing 78% of N₂ , 21% of O₂ , x% of CO₂ and 1-x % H₂O in LTE.

- The mixture defined above approximates well the troposphere and the troposphere is indeed in LTE

- From the 2 statements above and the demonstrated result follows :

“*The CO₂ does not heat the troposphere*” what is the answer on the question asked in the title.

**Caveat1**

I have said it both in the initial post and in this one.

Unfortunately, I know that it can’t be avoided and that some readers will still be confused about the result established here and start considering radiative transfers or radiative equilibriums.

That’s why I stress again that LTE and the result established here is totally independent of radiative equilibriums and radiative transfer properties.

However it does falsify one misconception concerning radiative properties of CO₂ that has also figured in the comments and that is that “*CO₂ does not radiate at 15µ because it “heats” N₂ instead*”.

It is also to be noted that we consider only the T/V process because it is only the vibrational modes that interact with IR radiation.

There are also rotational/translational and rotational/vibrational transfers.

The same argument used for T/V applies also for the R/T and R/V processes in LTE – e.g there is no net energy transfer between these modes in LTE even if for example the R/T process has a much higher probability than a T/V process.

For the sake of clarity we don’t mention specifically the R/T and R/V processes.

**Caveat2**

The result established here is a statistical thermodynamics bulk property.

This property is of course not sufficient to establish the **whole** dynamics of a system at all time and space scales.

If that was our ambition – and it is not – then we would have to consider boundary conditions and macroscopic mass, momentum and energy transfers, e.g convection, conduction, phase changes, lapse rates etc.

More specifically this result doesn’t contradict the trivial observation that if one changes the parameters of the system, for example composition, pressure, radiation intensity and spectrum, etc, then the dynamics of the system change too.

Yet it contradicts the notion that once these parameter are fixed there is a net transfer of energy from CO₂ to the troposphere. There is not.

**Caveat3**

It will probably appear obvious to most of you but it has also to be repeated.

This result says little about comparisons between the dynamics of 2 very different systems such as, for example, an Earth without oceans and atmosphere, and an Earth with oceans and atmosphere. Clearly the dynamics will be very different but it stays that in the case of the real Earth with an atmosphere in LTE, there will be no net energy transfer from the CO₂ to the atmosphere.