Does CO₂ heat the troposphere ?

This graphic, seen on many websites, was not part of Vonk's essay, but added by Anthony to visually tag the topic

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.

  1. 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.

  1. 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.

  1. 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.

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Enneagram

I don’y know what anyone of you think about it, but from a common sense point of view, wherever you find cold around you find CO2, from baking soda to dry ice. BTW there is a “something” in it which produces and urgent need of peeing in betwetters 🙂

Scott

Moderators – typographical error in this sentence:

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

The word looses should be loses.
-Scot

Scott

And then I misspelled my own name. 🙁
-Scott

Tom

So why is this relevant to climate discussions? The contentious aspect (or at least a contentious aspect) of climate discussions is the question, what happens to the atmospheric temperature as the amount of CO2 in it increases? What you have presented has nothing to say about this – as you explicitly state the caveat:

if one changes the parameters of the system, for example composition… then the dynamics of the system change too.”

The question “Does CO2 heat the troposhpere?” then has two answers:
1. No – so long as the composition of the atmosphere remains constant.
2. Yes (or at least maybe, this discussion is irrelevant) – if the amount of CO2 is increasing.

londo

I’m sorry to ask but, is there actually a point beeing made in this post? I confess that lost interest through about half of it. If anybody could point out the interesting stuff, I would be happy to reread it.

Daniel H

It’s amazing that they can sink billions of dollars into ever more complex, highly sophisticated computer models, and yet the end result still looks suspiciously like a third-rate reproduction of a Georgia O’Keeffe painting. I’m not impressed.

kfg

Tom Vonk says: “– your bedroom is in LTE .”
If only.

steveta_uk

Tom, I think you must add this:
3. Yes, if the system is no longer in LTE, due for example to increased IR from outside the volume.
In which case, ff there is no global warming, then there is no global warming, QED
So what?

paulw

I think that most people will not read the text and will be confused by the (not relevant) graphic image. The text is so long that even scientists will avoid reading.
Who made the graphic? What is the reference of the data? Which model does it compare the measured temperatures (by whom?) with?
The post title is ‘Does CO2 heat the troposphere?’ in the same way that Glen Beck asks questions for the sake of insinuating.
This is weak arguments and need much more work to fly.

steveta_uk

As far as I can tell, the original questioned contained the phrase “and submitted to infrared radiation” but this is completely ignored.

Paul

I don’t understand why this discussion is important. I thought that the greenhouse effect was the facile delivery of high energy photon to the earths surface. They are absorbed and heat the surface. Then the surface radiates IR photons. But greenhouse gases absorb the IR radiation and then reradiate in all directions. Since some of the radiation is back toward earth it slows the energy transfer from the surface. Greenhouse gases act as insulators. The higher the concentration of greehouse gases the more insulation you have and the the easier it is to maintain a temperature differential.
If your whole point is to nitpick the language, yes greenhouse gases do not heat the troposphere but they surely insulate the earths surface and keep it warm.

The missing hotspot might also mean that there is no additional water vapor, modelled as a positive feedback, causing the runaway warming. Either the CO2 induced warming is negligible, or consumed by some negative feedbacks.

Robin Kool

This post seems to state that in a mixture of N₂, O₂, CO₂ and H₂O that is in equilibrium there is no energy tranceference, because, in fact, the mixture is in equilibrium.
And it seems that the term ‘to heat’ or ‘to heat up’ is used in its strict meaning of ‘to raise the temperature’, while in common usage it is also used to mean ‘to keep at a higher temperature’.
As in, when I have a kettle on the stove with the water boiling, strictly speaking, the stove is not ‘heating’ the kettle as in ‘raising its temperature’, because the kettle stays on 100˚C, however long I leave it on the stove.
Strictly speaking the stove was only heating the kettle until it reached 100˚C.
But in normal usage, we do say that the kettle is being heated by the stove while it sits on the stove.
‘Heating’ here means ‘keeping the kettle at a higher temperature than it would be without the stove’. And in fact, when I turn off the stove, the kettle cools down to kitchen temperature.

Louis Hissink

Anthony
Your two graphics at the head of this post – RH side – I think the conclusion is that it’s not the models that are wrong, but that the theory itself.

mircea

I think people do not realise (yet) that Misckolczi’s paper is a breakthrough. It empirically proves that the atmosphere transparency remains constant because when CO2 increases H2O vapours decreases. The mechanism is as follows:
When CO2 is added to the atmosphere the transparency of the atmosphere to infrared radiation (LW) decreases. As such more energy remains in the atmosphere instead of radiating to space. This extra energy will either heat the atmosphere or will be radiated downward toward the ground.
Thus, the temperature in the troposphere at height H will increase from Te to Te1 and at a higher altitude H1 the air will now have the initial temperature Te.
The ground temperature (Tg) remains constant for some time after the increase in CO2 because the land/sea surface has a huge thermal capacity compared with the atmosphere. The extra downward infrared energy will just be absorbed by sea/land and will not increase the surface temperature (we are talking here of the short time effect after the increase of CO2 percentage in the atmosphere).
The effect of all these is that the atmospheric rate lapse changes as if the air would become moister (i.e. it cools slower when height increases)
But, because the temperature increased in the upper part of troposphere the H2O water vapours will raise higher up to the height H1 where temperature is Te. This is because temperature determines the condensation point (in average).
With H2O vapours raising higher concentration of H2O vapours in the column of air will decrease (same quantity in higher volume). A decreased concentration of H2O will dry up the atmosphere. A drier atmosphere cools faster when height increases and therefore the atmosphere rate lapse will come back to the initial values. The temperature at height H will decrease back to Te.
The H2O vapours that are higher then H will be now at a temperature lower then Te and will condense (most probably will fall back to land/see surface as rain). Therefore the concentration of H2O will decrease and the concentration of GHG gases in the atmosphere will decrease to the initial value and as such the atmosphere transparency decreases to the initial value.
The system comes back to the initial state with only the H2O concentration smaller and CO2 concentration higher. H2O concentration is highly variable but on average around 1% and CO2 concentration is around 0.039% . Therefore the decrease in H2O vapours for a doubling of C02 is negligible.
Please note that all these are empirically (measured, no computer models) proven by Misckolczi in his 2010 paper.
The paper can be found here:
http://www.friendsofscience.org/assets/documents/E&E_21_4_2010_08-miskolczi.pdf
Regards!

@ Tom : 9.25 am
and yet CO2 continues to rise but temperature does not.

R Stevenson

Tom asks the question which is at the heart of AGW – does increasing CO2 in the atmosphere absorb more IR photons and thereby increase the air temperature. If not then does retention time of photons increase and if so does it matter.

Tom W.

Point 1.
The statement: “It means that there exists a mechanism transferring net (e.g non zero) energy unidirectionaly” is solipsistic. A “net” transfer is always unidirectional since it means “taking both directions into account”. Is this what you meant here?
Point 2.
The statement: ““X heats Y” is equivalent to the statement “Y cannot cool X”” In a closed system this is absolutely false. If you are transferring heat from X TO Y then Y is necessarily transferring heat AWAY from X. I grant however that in some open systems and if you mean heating and cooling to mean “raising or lowering the temperature of” then this statement could be true.
Point 3.
If you are arguing that energy must move in BOTH directions on the microscopic level then I don’t know who is disagreeing with you. It must move in both directions for LTE to be established. But you’re conclusion “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.” Is not correct. In the most general way:
System is in LTE
A small amount of energy is added (in this case as IR radiation)
LTE is restored
But this is a NEW LTE! At a higher temperature than the old. And all of the modes, those of CO2 and of N2 will have a higher average energy (by the Equipartition Theorem). So you have added energy to one mode – a CO2 vibrational mode – and eventually this extra energy gets spread out into the others – 3 of which are translational modes of N2.

Buffoon

Tom
I struggle to grasp this. Help me out.
You say the CO2* + N2 reaction is exactly reversible, but it can’t be.
If you have a full dish of apple pie, and you hand out three pieces, you get three excited children. They run away. In the meantime, the chef brings you three new pieces of apple pie, and you fill your dish. The apple pie tastes bad, and the three children bring it back. But your pie is full, because the cook filled it already. They must, unfortunately, stay excited.
You have a finite number of CO2 in any concentration scenario. But you always have two sources of CO2*. One is photons (from the sun or other CO2*.) The other is N2*. N2* + CO2 is not perfectly reversible because there is an outside influence driving your system to the right by creating CO2*. In the absence of an additional energy loss mechanism, you have net warming.

cal

I am sorry but I still cannot see how this helps.
Firstly you seem to imply that there is no such thing as temperature if there is no LTE whereas my understanding is that you cannot define the temperature if LTE does not exist. This is quite different. For example Heisenberg states that you cannot define the position of a particle if you define momentum exactly. This is not the same thing as saying that if you define its position it does not have a momentum.
In the real world radiation from the warmer earth is constantly being absorbed by water vapour and CO2 in the troposphere and this energy is either re radiated or lost to N2 and O2 molecules by collision. Since there is a nett radiation upwards no part of the system is ever in equilibrium. Since it is not in equilibrium any measurement of temperature will have a statistical error but a temperature can still be inferred from the energy radiated by molecules: indeed this is what the satellite sensors rely on.
In summary, I feel that you are using a circular argument. You assume LTE which, by defintion, has no nett heat transfer between molecules and then prove that there is no nett heat transfer between molecules. I cannot see where you prove that LTE exists.
Did I miss something?
You also make the statement:
“X heats Y” is equivalent to the statement “Y cannot cool X”.
If I were to replace X and Y with real examples I might say:
“The whisky warms the ice is equivalent to the statement that the ice cannot cool the whisky”!
I am going to get myself a drink!

Not an english prof.

@Scott
Moderators – grammatical error in this sentence:
In simple words – the CO₂ has never time to emit any IR photons because it looses vibrational energy by collisions instead.
‘has never’ should read ‘never has’…and ‘looses’ should be ‘loses’
[Fixed, thx]
Great article, Mr. Vonk.

peterhodges

all the theorizing is nice but the pictures say it all
model predictions wrong

WOJ

No Scott, you misspelt your own name!

mircea

Tom says: August 31, 2010 at 9:25 am
“So why is this relevant to climate discussions? The contentious aspect (or at least a contentious aspect) of climate discussions is the question, what happens to the atmospheric temperature as the amount of CO2 in it increases? ”
Answer: The H2O vapour concentration decreases. See my post at 10:32 for why.
The question “Does CO2 heat the troposhpere?” then has the answer:
No, because the more CO2 increases the more H2O vapour decrease. The transparency remains constant.

Ulric Lyons

So the majority of `Global` warming since 1985 has been at higher latitudes and more so in the Arctic, but NOT in summer months.
How is CO2 supposed to do that?

Tom got it the wrong way around, I think.
The statement X heats Y is equivalent to X cannot cool Y.
In this context, Cal’s whisky / ice example makes sense.
The whisky heats the ice is equivalent to saying the whisky cannot cool the ice.
Otherwise, I’m going to have a glass of whisky (no ice) to calm my melting brains… 🙂
[Note: You will understand if we delete free advertising. ~dbs, mod.]

jorgekafkazar

“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.
Better phrased in English would be: the proportion is given by the ratio of the sum of the dark blue and light blue areas to the total area under the 35°C curve.
http://www.yourdictionary.com/proportion
Proportion; n; 1. the comparative relation between parts, things, or elements with respect to size, amount, degree, etc.; ratio

George E. Smith

Interesting exposition Tom.
First off; your definition of Temperature; aka the average energy of the set of molecules. Because we are talking about a dynamic system here, in which collisions are constantly occurring, the energy of any one molecule is constantly changing. So arguably your Maxwell Boltzmann distribution can be considered to be a snapshot of an instantaneous situation. Every one of those molecules plotted on the graph, has a serial number; but we can never know what they are. So in an instant; all the molecules interchange energies and this has the effect of scrambling whatever order the serial numbers were in the previous instant. So the end result is the same energy distribution results; even though the position of any particular serial numbered molecule on your graph changes.
As a consequence, it is also true, that any single (serial numbered) molecule, will at some time or other occupy any possible position within your MB distribution.
So the MB distribution you have plotted is an equally valid time distribution of the energy of ANY SINGLE MOLECULE.
So the definition you have given of Temperature; the mean energy of the set of molecules; can also be stated as the time average energy of any single molecule in the LTE volume. The instantaneous average you cite, is of course what all the textbooks cite for a definition of Temperature; but I think you can see that it is just as valid to say the Temperature is the time average energy of any single molecule within the LTE volume.
And that is why (facetiously) I often state that Mother Gaia has a thermometer in every single molecule.
I have no argument with your point about the CO2 excited states having some threshold energy (of the N2) required to excite the CO2 modes. Of coruse the O2 and Ar components of the atmosphere will likewise have their own slightly different energy distributions because of the different molecular masses; well different velocities anyhow.
Your “X heats Y” but “Y does not cool X” is troubling. Surely X must have lost energy in heating Y, and in an LTE situation, X must have a finite heat capacity.
Well in any case, I appreciate your very thoughtful exposition. Now I’m going to lose a lot of sleep; either in coming to agreement with you; or maybe stumbling over some error; if there is one.
But in any case, I am now more confident; that Mother Gaia does always get the right answer; because she has all the thermometers anyone could ever need.
George
PS I’m trying to remember; in a purely elastic scattering collision between two equally massive particles (classically), the two particles simply swap energy ? Izzat so ?

mircea

R Stevenson says: August 31, 2010 at 10:41 am
“Does increasing CO2 in the atmosphere absorb more IR photons and thereby increase the air temperature.”
The answer is NO. Because the H2O vapours in the atmosphere decreases so the IR photons absorbed in the atmosphere remain constant.
The rule of thumb is: The more CO2 the less H2O and the less CO2 the more H2O.
If it wasn’t for the water cycle then the answer to your question will have been YES.

“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”.
I knew that.

kfg

paulw says: August 31, 2010 at 10:09 am
“The text is so long that even scientists will avoid reading.”
If scientists find this text so long that they won’t read it, they’re certainly not going to be able to make it through Sears & Salinger and ought to, for the benefit of us all, excuse themselves both from class and climatology.
Have scientists now reduced themselves to the level of Maxim “readers?”
WOJ says: “No Scott, you misspelt . . .”
I’m a celiac; I misspelt too.

Bryan

Tom are you absolutely sure of this
……” * 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 moment I am reading a thermodynamics text book by Richard Fitzpatrick
http://farside.ph.utexas.edu/teaching/sm1/statmech.pdf
He seems convinced that the kelvin temperature depends only on the translational kinetic energy!

Dave Dardinger

I agree with Tom W, Buffoon and especially Cal. You’ve had enough people point out your errors between the previous post an this one that you should be taking heed of them. As it is, it’s hard to work up energy to repeatedly point out the problems when you don’t really discuss them. Which I suppose means you’re not in LTE. In which case it’s impossible to take your temperature.

I’m still confused by this argument. Why is “local thermodynamic equilibrium” an appropriate assumption here? Sure, molecular collisions are fast and mean free paths are short, so the molecular degrees of freedom are nearly in local equilibrium at any time. The “local gas temperature” is well-defined for the CO2/N2 mixture. But what about the radiation? The local radiation environment is NOT a Planck spectrum!
The real situation is (at best) a steady-state one, not an equilibrium one. So the kind of detailed-balance arguments given here don’t apply at all. Or have I missed something?

The concept that CO2 does not cause significant “heating” in the troposphere is not new. Look at what is found in Dr. Walter Elsasser’s 1942, “Infrared Radiation Heat Transfer in the Atm” on page 23 is particularly interesting!
(It’s page 24 of the ScribeD posting). Dr. Elsasser says this about CO2 in the troposphere:
“It may be noted that since the flux of CO2 band is equal at any level to a definite fraction of the blackbody radiation of the CO2 corresponding to the temperature at that level, both in the upward and downward direction, the resultant flux of the CO2 radiation vanishes in the approximation of the chart.” (He refers to his general radiation chart, derived in this paper.)
http://www.scribd.com/doc/34962513/Elsasser1942
Thus for Day to Day heat up and cool down calcuations only P, T, and humidity measurements be taken by the radio balloons. No consideration of CO2 content is made after this point.
Interesting, CO2 made moot, in 1942?
Max

James Sexton

Thank you Mr. Vonk. Your article is interesting and thought provoking. Is there any chance of a more “ambitious” question later? I know the questions (and thus answers) become much more complex, but radiative and/or boundary questions would in my estimation be the next logical progression in the discussion.

bob

You have to start an argument with a decent premise, at least one that is true.
This one is neither
“Given a gas mixture of CO₂ and N₂ in Local Thermodynamic Equilibrium (LTE) and submitted to infrared radiation, does the CO₂ heat the N₂?”
A gas mixture heated by infared radiation is not in Local Thermodynamic Equilibrium.
This is also an oversimplification.
“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.””
Even though the mean time between collisions is much shorter, there is still a chance the CO2 molecule will emit a photon before it loses the energy to collisions.
The mixture still has a temperature and is still likely to radiate in the infared as decribed by Planck’s blackbody radiation.
And where exactly is the earth’s atmosphere at local thermodynamic equilibrium?

John Whitman

Tom Vonk,
“The CO₂ does not heat the troposphere”
Is my following restatement essentially correct?
All the CO₂, CO₂* , N₂ and N₂⁺ will knock on each other until the CO2* gives up the IR in equal amount to the IR it received to make it CO2* in the first place; LTE is achieved again. This leaves the N₂ at its original temperature before the IR was received by CO₂.
Please let me know if I got that restatement right. : )
Tom Vonk, appreciate you spending time with us. I enjoy your work.
John

Scott

So I read the article this morning and didn’t think it worked out. After thinking about it some more, I still haven’t changed my mind. People aren’t arguing that the local atmosphere is in pseudo-equilibrium. However, when CO2 absorbs radiation, that equilibrium shifts to a higher energy (i.e. warmer) state! As long as there is more energy absorbed (from more CO2 in this case), then the system will warm up more.
I would argue that CO2 wouldn’t warm the atmosphere if it wasn’t in equilibrium, because then it wouldn’t warm N2…but you argue (correctly) that it does reach the same temperature as N2…thus if more energy is added to the CO2, it distributes to the N2.
The use of statistical thermodynamics mostly just seems to confuse people. Presenting from traditional thermodynamics would be more accessible to the masses and would also make it clear that the argument doesn’t seem to work.
Please let me know if my interpretation is wrong.
-Scott (spelled correctly this time 🙂

JP

“I don’t understand why this discussion is important. I thought that the greenhouse effect was the facile delivery of high energy photon to the earths surface. They are absorbed and heat the surface…If your whole point is to nitpick the language, yes greenhouse gases do not heat the troposphere but they surely insulate the earths surface and keep it warm”
Paul, GHGs do not heat anything. That is, they do not create “extra heat energy”. And the theory of how GHGs lead to AGW is an important theory in that it can lead to quite a bit of bad to catastrophic public policy. One of the signatures of AGW (based on increased CO2 concentrations) is the Mid-tropespheric Tropical Hotspot. If you do not understand, you need to hit the books and read up on it. Despite increases in CO2 concentrations, the tropics have failed to warm -period. The mid level hotspot is missing. Which leads one to observe that either the IPCC theory is wrong; or, it is right, and the warming of the last 30 some years has not been induced by human activity.

“Tom asks the question which is at the heart of AGW – does increasing CO2 in the atmosphere absorb more IR photons and thereby increase the air temperature. ”
The CO2-based GW theory is different – it says that more of the outgoing IR is absorbed and re-radiated back to the surface, warming the surface and atmosphere gets warmer as secondary effect.

I didn’t bother to read the entire post, since it is obvious that Tom Vonk has not learned a thing from the previous discussion.
1) Unless the radiation is in thermal equilibrium with the gas you cannot have LTE. At most you can have a steady-state approximation to LTE, in which there is in general a net transfer of heat between constituents. This is crucial.
2) That thermodynamic temperatures cannot be defined in the absence of LTE is false; so long as enthalpy and entropy are calculable, so is a temperature. In astronomy it is common for multiple temperatures to co-exist in the same volume, from the 2.7K of the microwave background to million degree plasmas and even terakelvin cosmic rays.
3) “X heats Y” is not equivalent to “Y cannot cool X”. I cannot see any sense, however perverted, in which this would be correct. “X heats Y” is equivalent to “Y cools X”.
Tom’s conclusions yet again amount to saying that if you have equilibrium you can’t have any heating (which is a tautology) and that you can determine what radiation does to the thermodynamics by ignoring the thermodynamics of the radiation (which is nonsense).

Tenuc

thanks Tom for a good post which helps clarify the misunderstandings people have about atmospheric heating.
peterhodges says:
August 31, 2010 at 11:06 am
…model predictions wrong,”
It’s worse than that – CO2 causes global warming wrong!

Peter Miller

This is all far too complex for anyone at the IPCC, therefore it will be ignored like all other inconvenient facts in its next review.

Enneagram

No CO2, no photosynthesis, no photosynthesis no glucose, no glucose, no cotton, no cotton, no underwear and no diapers for global warming bedwetters.

George E. Smith says:
August 31, 2010 at 11:40 am
“PS I’m trying to remember; in a purely elastic scattering collision between two equally massive particles (classically), the two particles simply swap energy ? Izzat so ?”
Only if they hit head on. In a glancing collision only a fraction of the energy is swapped. Think billiards.

Greenhouse gases cool the atmosphere. Without greenhouse gases, the atmosphere would be over 120 °C.
http://mc-computing.com/qs/Global_Warming/EPA_Comments/TheGreenhouseEffect.doc
The proof is trivial. Since the gases are “cooling” the atmosphere, that heat has to go somewhere and a part of that is what heats the surface.

CO2 and TSI have a little (if anything) to do with large temperature oscillations.
There is only one indicator which in main correctly signals in advance those oscillations -N. Atlantic precursor.
http://www.vukcevic.talktalk.net/CETnd.htm
More importunately it suggests significant cooling in decade to come. Variable delay is well within parameters of the precursor.
Any serious researcher should scrutinise it in detail. 10 year period 1695-1705 is the odd one out, but this may possibly be something to do with data reliability from the period.
For those ignoring its significance today, may render their views worthless tomorrow.

Michael J. Dunn

Enneagram says:
August 31, 2010 at 9:14 am
I don’t know what anyone of you think about it, but from a common sense point of view, wherever you find cold around you find CO2, from baking soda to dry ice.
But that only depends on where you are looking. Any fire will produce copious CO2 at very high temperature (consider coal). Combustion of dicyanoacetylene (C4N2) produces the hottest combustion flame, at 5260 K, giving only CO2 and N2 as products. Even thee and me, as we exhale, produce CO2 at body temperature (and if we didn’t, we would be dead).

John Eggert

Anyone who disputes the greenhouse effect due to radiant heat absorption must explain why air temperature drops at a slower rate on humid evenings than on dry evenings.