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
August 31, 2010 9:14 am

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
August 31, 2010 9:19 am

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
August 31, 2010 9:19 am

And then I misspelled my own name. 🙁
-Scott

Tom
August 31, 2010 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? 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
August 31, 2010 9:53 am

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
August 31, 2010 10:01 am

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
August 31, 2010 10:08 am

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

steveta_uk
August 31, 2010 10:09 am

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
August 31, 2010 10:09 am

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
August 31, 2010 10:11 am

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

Paul
August 31, 2010 10:12 am

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.

August 31, 2010 10:12 am

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
August 31, 2010 10:28 am

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
August 31, 2010 10:29 am

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
August 31, 2010 10:32 am

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!

August 31, 2010 10:34 am

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

R Stevenson
August 31, 2010 10:41 am

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.
August 31, 2010 10:44 am

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
August 31, 2010 10:47 am

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
August 31, 2010 10:51 am

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.
August 31, 2010 11:05 am

@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
August 31, 2010 11:06 am

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

WOJ
August 31, 2010 11:12 am

No Scott, you misspelt your own name!

mircea
August 31, 2010 11:23 am

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
August 31, 2010 11:28 am

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?

August 31, 2010 11:29 am

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
August 31, 2010 11:36 am

“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
August 31, 2010 11:40 am

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
August 31, 2010 11:41 am

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.

August 31, 2010 11:44 am

“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
August 31, 2010 11:47 am

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
August 31, 2010 11:48 am

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
August 31, 2010 11:49 am

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.

August 31, 2010 11:50 am

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?

August 31, 2010 11:51 am

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
August 31, 2010 11:53 am

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
August 31, 2010 12:12 pm

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
August 31, 2010 12:27 pm

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
August 31, 2010 12:28 pm

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
August 31, 2010 12:30 pm

“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.

August 31, 2010 12:34 pm

“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.

August 31, 2010 12:36 pm

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
August 31, 2010 12:38 pm

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
August 31, 2010 12:39 pm

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
August 31, 2010 12:40 pm

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

August 31, 2010 12:49 pm

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.

August 31, 2010 12:52 pm

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.

August 31, 2010 12:58 pm

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
August 31, 2010 12:58 pm

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
August 31, 2010 1:07 pm

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.

kfg
August 31, 2010 1:20 pm

Robert C says: “Without greenhouse gases, the atmosphere would be over 120 °C.”
There’s a dark & a troubled side of life
There’s a bright, there’s a sunny side, too
Keep on the sunny side, always on the sunny side,

Oliver Ramsay
August 31, 2010 1:30 pm

“Given a gas mixture of CO₂ and N₂ in Local Thermodynamic Equilibrium (LTE) and submitted to infrared radiation, does the CO₂ heat the N₂?”
Surely, it’s a safe assumption that “submitted to infrared radiation” doesn’t mean a five-second burst and then nothing. It must mean a continuous bombardment for the duration of the examined period.
Mention was made that this is all to be considered in the context of the real planet Earth, where there is known to be a heat-sink for the atmosphere.
If the conclusion is that CO2 does not heat N2, that must mean that CO2 does not act as a portal for heat to be transferred from the Earth to the atmosphere through radiation.
This wouldn’t surprise me if I had already accepted that radiation from CO2 is busy heating the Earth ( or keeping it warm, etc.).
After all, we’ve all tried spending the same dollar in two different places and we know how that works out.
It would follow that, if CO2 wasn’t heating the atmosphere at density x, then it wouldn’t do it at density 2x. Likewise, for variations in radiative intensity.
If this is the case, then the surface must be heating the atmosphere by collisions, condensation and convection.
It also suggests that you can’t heat a gas radiatively, which sounds pretty unlikely to me.
If I was hoping for good counter-argument, I’m disappointed.
We’ve got some confusion of solipsism and tautology, accompanied by a declaration of “absolute” falsity that is then followed by a retraction.
There are excited children with slices of yucky pie. Beyond reading that they are obliged to stay excited we don’t learn what they do with the pie in the sky.
This is what we want to know!

Jan K Andersen
August 31, 2010 1:31 pm

Robin Kool says
“‘Heating’ here means ‘keeping the kettle at a higher temperature than it would be without the stove’”
What a good allegory. This is to the point where the logic in the article fails.

George E. Smith
August 31, 2010 1:31 pm

“”” Paul Birch says:
August 31, 2010 at 12:49 pm
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. “””
I’ll buy that restriction. Now what if the particles are differnet masses ?
I assume that everybody has at one time done that experiment where you hold a ping pong ball on top of a basketball, and then drop the pair onto a hard floor, trying to not let the BB rotate when you let it go. You have to do it to believe it.

John Whitman
August 31, 2010 1:33 pm

James Sexton says:
August 31, 2010 at 11:53 am
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.

——————-
Tom Vonk,
I wholeheartedly second James Sexton’s suggestion that you please post on the next step of analyzing the boundary conditions between atmosphere and the ocean/earth. Also, it would be helpful to see a post treating the radiative transfer mechanisms at play.
Thank you for your current post.
John

Spector
August 31, 2010 1:34 pm

RE: Main Article “X heats Y” is equivalent to the statement “Y cannot cool X”.
I believe this might be more properly worded as: “X is heating Y” is equivalent to the statement “X cannot be cooling Y.” Imagine X as a heat-pump and Y as a home. Of course, even this might not be true if X happened to be cooling the basement floor.
I also assume that CO2 is very unlikely to re-emit the energy of a photon it has absorbed before that energy is lost while colliding with other molecules, but I also believe these same collisions should cause a continuous flux of emitted photons. The fraction of collisions that cause those emissions might be very small, but I suspect that the number of collisions per cubic meter per second is a very large number, even at the tropopause altitude.
LTE could only be maintained if these photons were being reflected back into the local region or if there was a continuous, incoming flux of replacement photons or their equivalent energy.

Scott
August 31, 2010 1:35 pm

John Eggert says:
August 31, 2010 at 1:07 pm

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.

I don’t dispute the greenhouse effect, but your observation can be explained without it too. Water vapor’s heat capacity, ~1.9 J/(g*K), is higher than dry air’s, ~1.0 J/(g*K). Thus, with the same energy output, the humid air will drop temperature more slowly. Also contributing to the slower temperature change is that some of the energy may go into condensing H2O onto surfaces (the opposite of evaporative cooling) once the dew point is reached.
A better analogy would be to compare temperature drop rates at night with identical humidities/temperatures, but one in a cloudy situation and the other with clear skies. Even that isn’t quite a fair argument, since some of the cloud contribution is from radiation scatter and not absorption, but it’s closer to being fair.
-Scott

August 31, 2010 1:43 pm

If it is still not sufficiently clear that a gas under non-equilibrium irradiation is not in equilibrium, and that its constituent species will in general not all be equally excited (ie, will not be at the same temperature) but will be heating and cooling each other through collisions, then consider how convection fits in.
Parts of the atmosphere are continually being warmed by radiation, parts are being cooled by radiation. As a parcel of gas is warmed, it starts to rise; as it is cooled, it falls; then as it rises it cools; as it falls, it warms. The parcel of gas is not in equilibrium. So, instead of sticking with the same gas, let’s stick with the same volume, in which it is possible that the measured temperature may sometimes stay the same; even then, gas is entering the volume at one temperature, and leaving at a higher or lower one. Again, no equilibrium. So there’s no reason heat can’t be transferred from radiation to the bulk gas via H20 or CO2, or vice versa.

George E. Smith
August 31, 2010 1:46 pm

As to the question of whether the Thermodynamic Temperature is just the translational energy; and does not include rotational or vibrational; doesn’t that conflict with the equi-partition law, that the energy must be distributed equally among all degrees of freedom. The total energy at a particular Temperature might change with the nature of the molecule depending on the number of degrees of freedom it has; but I am not sure that the definition changes.
I like my definition of temperature as the time averaged molecular energy for even a single molecule; but you don’t get something for nothing, so it takes time to ascertain the Temperature. So there’s a sort of Heisenbergy uncertainty equivalency, in that you either have to examine a whole bunch of molecules at once; or watch one for some time to narrow down the error in the Temperature. And not to confuse with the quantum nature of uncertainty. My PhD cohorts say of course that applies at all scales; unless you believe there is some scale at which delta x. delta p < h/2pi. But I agree it may be irrelevent at macro scales.

Edward Palmer
August 31, 2010 1:53 pm

Are there no air movements at the Troposphere? Because if there are, the LTE assumption fails.

Bomber_the_Cat
August 31, 2010 1:54 pm

Tom,
I would like to say that this is a very good post but then that would imply that I fully understand it, which I don’t.
I assume, from what I do understand, that you are not actually saying that a greenhouse effect doesn’t exist (or are you saying that?).
Certainly, CO2 absorbs energy in the 15 micron band, we all agree on that. The absorbed energy has to have some effect. It either heats the local atmosphere or it is re-emitted as 15 micron radiation in some random direction. Do we both agree on that? If it is re-emitted as radiation, some of this will come back to the surface and produce extra warming. Is that right?
Has anyone actually measured the intensity and frequency spectrum of radiation coming back to Earth at night, for example, when there is no incoming solar radiation? Is the incoming spectrum consistent with a blackbody distribution (showing that it is from a warm atmosphere) or is it concentrated around a wavelength of 15 microns, showing that it being emitted by CO2. It seems very hard to find this information. Has no climate scientist performed this simple experiment? Even computer models require some input data.
Sorry, more questions than answers.

David A. Evans
August 31, 2010 1:55 pm

John Eggert says:
August 31, 2010 at 1:07 pm
Try the fact that the energy content of humid air is significantly higher than dry air.
DaveE.

kfg
August 31, 2010 2:02 pm

Oliver Ramsay says: “we’ve all tried spending the same dollar in two different places and we know how that works out . . .”
. . .quite well if done by two different carriers. If it is spent in a third place we may even receive our own dollar back again after having spent it. How cool is that?

August 31, 2010 2:03 pm

“actual” temperatures measured by balloons.
There is no such thing. There are instruments on balloons that record values. Those values are taken to be an estimate of “temperature” The instruments have error, the sampling method has changed, the instruments have changed. What we have in the end is a model of the observations. A model with error. On the GCM side we also have models with errors. In addition to the ballons, we have satillite “observations” These also are not data. They are values processed through an alogorithm that estimates the temperature. A comparison the these three models, the model of temperature from balloons, the model of temperatures from satillites, and the model of temperature from GCM are not in perfect agreement. Which model is wrong? Well, they are all wrong. (It;s interesting to note that the satillite “data” is being readjusted. readjusted upward, because the alogorithm that processed the ‘data’ –bits from a sensor– was biased. so the data was ‘wrong’ or rather the processing algorithm was ‘wrong.’
The spread of the models 3 models currently doesnt overlap very well. We can conclude from this that one or more of the models needs modification. We cannot logically conclude which needs the adjustment. each or all could require modification. You are not comparing “data’ with model. you are comparing model with model. models of data with models of physics.

Tom
August 31, 2010 2:10 pm

My comment (at 9:25) was taking Tom V’s argument at face value – on its own terms, is it relevant? My thermodynamics has been unused for too many years to easily tell if his argument is actually correct. So:
@steveta_uk – No, that is confusing radiation equilibrium with LTE. Of course increased radiation will increase temperature, by direct radiation onto both gases. But the rates of transfer between CO2 and N2 will remain constant. CO2 has no extra warming effect in this case.
@mircea – Maybe. I completely lack the expertise to assess the CO2 / H2O cycle. But it’s not relevant to either my point or Tom V’s.
In summary: Tom V has made the fundamental error, either deliberately or naively, of making equilibrium statements about a system that is in constant transient. “CO2 does not heat a mixture in LTE” may well be true, but the composition of the atmosphere is changing, hence there is no LTE, as he himself states. It looks suspiciously like a headline that is designed to push buttons – effectively “CO2 cannot cause global warming, so how much we produce is irrelevant,” which, in the text is qualified with, “so long as we don’t produce any.” Looking at the headline, alarmists see something to get upset about and skeptics see something that “proves” their ideas. When you dig to the detail, there is nothing for an alarmist to get upset about and nothing for a skeptic to care about. It fundamentally fails to address the question, “Do increased CO2 emissions increase global temperatures?” and, if you read the text carefully, it admits as much.
Well done to Anthony for providing a forum where these ideas can be discussed. Perhaps a response along the outline above is in order, though?

Terry
August 31, 2010 2:16 pm

Tom
Tom W is right
at http://wattsupwiththat.com/2010/08/31/does-co%e2%82%82-heat-the-troposphere/#comment-471298
When you add hv energy at a given LTE a new LTE is established due to the increased TE imparted to the interts and the lowering of the population of excited state CO2. While you are correct in stating that there is no NET transfer, that is ONLY true if there is no NET input to the system. In the atmosphere there is a net input from radiation and that is dependent on the IR flux and [CO2]. Since IR flux is relatively constant, the NET input is dependent on [CO2]

MartinGAtkins
August 31, 2010 2:16 pm

R Stevenson says:
August 31, 2010 at 10:41 am
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.
Not in our lower troposphere because the density of the CO2 layer is sufficient enough to absorb all the LW radiation available to it.
If not then does retention time of photons increase and if so does it matter.
The retention time doesn’t increase. The photon received is almost in an instant re-emitted. The work done by CO2 is a product of its absorption wavelength 15 µm which would not be caught by the other gases and so would radiate directly into space and be lost.

Oliver Ramsay
August 31, 2010 2:19 pm

Paul Birch says:
‘As a parcel of gas is warmed, it starts to rise; as it is cooled, it falls; then as it rises it cools; as it falls, it warms.’
————–
What makes it a parcel?

Roger Clague
August 31, 2010 2:27 pm

Tom Vonk says
“As a result of this transfer the temperature of N2 would increase
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”
Paul Birch says
“Tom’s conclusions yet again amount to saying that if you have equilibrium you can’t have any heating”
I agree that the reasoning is circular. He states his definition of LTE as a proof.
Energy is transferred so we do not have LTE.
The temperature of N2 increases. So energy has been transferred to the N2 from CO2
Energy of all types is transferred between all types of molecule in the atmosphere. The average temperature depends on what happens at the outer edge

August 31, 2010 2:49 pm

Tom,
Do you think that ozone, by absorbing UV, heats the stratosphere?

John Whitman
August 31, 2010 2:50 pm

Oliver Ramsay says:
August 31, 2010 at 1:30 pm

“Given a gas mixture of CO₂ and N₂ in Local Thermodynamic Equilibrium (LTE) and submitted to infrared radiation, does the CO₂ heat the N₂?”

[I edited out some lead in paragraphs]
If the conclusion is that CO2 does not heat N2, that must mean that CO2 does not act as a portal for heat to be transferred from the Earth to the atmosphere through radiation.
[I edited out a paragraph here]
It would follow that, if CO2 wasn’t heating the atmosphere at density x, then it wouldn’t do it at density 2x. Likewise, for variations in radiative intensity.
If this is the case, then the surface must be heating the atmosphere by collisions, condensation and convection.
[I edited out some paragraphs here]

—————–
Oliver Ramsay,
I excerpted from your post some of the paragraphs that I understood to be Tom Vonks point in his post. NOTE: I know you did not say you agreed with him, but I thought your statement of Vonk’s position seemed clear. Thanks for that what I take as a clear statement of Vonk’s position.
Hope Tom Vonk shows up to discuss soon.
John

Enneagram
August 31, 2010 2:54 pm

The next winter remember CO2, next to a warm fireplace tell your wife and kids how bad it is…..

Dave Springer
August 31, 2010 3:17 pm

@Vonk
How can you have local thermal equilibrium on a column of air many kilometers deep where one side of the column alternates between 5000K and 3K every 24 hours and the other side is relatively constant at around 280K.
The answer is you cannot. This article is worth far less than the cost to read it.

Milwaukee Bob
August 31, 2010 3:17 pm

Steven Mosher said at 2:03 pm
“actual” temperatures measured by balloons.
In addition to the ballons, we have satillite “observations” These also are not data.

And unlike the balloon “values” that are point in time and space specific (and thereby limited) each satellite “value” is an average electrical signal reading over a 750 Sq km area (as I recall from a previous post) that are then summed and averaged. A balloon reading taken somewhere within one of those areas at the EXACT same point in time and space would almost always be different (even if dead on accurate) than the average for that area. No way to sync, therefore.
“X heats Y” is equivalent to the statement “Y cannot cool X” is verbally wrong because we “say” Y cools X but scientifically it’s correct. There is no such thing as “cool”. (Unless your talking about CTM) 🙂 Cool doesn’t transfer. Only heat transfers and in all cases makes its source less warm. The whiskey transfers its heat/warmth to the ice. The ice can not transfer cool, because it does not have what doesn’t exist…. except in jazz players….

August 31, 2010 3:30 pm

Steven Mosher says:
August 31, 2010 at 2:03 pm
“actual” temperatures measured by balloons.
There is no such thing. There are instruments on balloons that record values. Those values are taken to be an estimate of “temperature” The instruments have error, the sampling method has changed, the instruments have changed. What we have in the end is a model of the observations. A model with error. On the GCM side we also have models with errors. In addition to the ballons, we have satillite “observations” These also are not data. They are values processed through an alogorithm that estimates the temperature. A comparison the these three models, the model of temperature from balloons, the model of temperatures from satillites, and the model of temperature from GCM are not in perfect agreement. Which model is wrong? Well, they are all wrong. (It;s interesting to note that the satillite “data” is being readjusted. readjusted upward, because the alogorithm that processed the ‘data’ –bits from a sensor– was biased. so the data was ‘wrong’ or rather the processing algorithm was ‘wrong.’

One of the things that always amused me Mosh was that when the UAH-MSU data first came out and showed the absence of warming it was justified by agreement with balloon data. After errors were corrected and a positive trend was established we were told that this data also agreed with balloon data!

Greg F
August 31, 2010 3:44 pm

Reasonable logic leads to bizarre result,
If true you should be able to setup some very surprising experiments and earn a nobel. Remember in real physics, awards are given when the experimental physicists prove what the theoretical physicists propose.
I can’t imagine experiments proving the claims of this article. As such it should be tossed as quickly as it can be.
fyi: Simple experiment – fire a high energy laser vertically through the atmosphere that is absorbed by CO2, but only by CO2. There must be a laser frequency that would do that.
Then measure the air temp with the laser on and off.
Experiments don’t get much simpler, and as I said if the air doesn’t heat, Vonk has won himself a Nobel.

August 31, 2010 4:14 pm

Tom,
I don’t want to take the time to address this again. I’m still frustrated that you wouldn’t listen before and my time is more valuable than this. This is incorrect:
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.
There is radiative input into the CO2 at the bottom of the column, it is emitted at the top of the column. Without the CO2’s presence the atmosphere would be cooler. The NET energy transfer happened when the CO2 was initially added, after that it is a constant but you should reword this.
As more CO2 is added, there is another net energy transfer initiated as you now admit in your post. The CO2 is capturing a portion of energy- global warming.
Clarity of phrasing would help your post a lot.
You’re whole long worded revised post says, ‘no change in CO2, no new energy delay, no new warming”. Where you go wrong, is to claim no energy transfer by CO2. The energy transfer occurs when the CO2 is added.
Correct your phrasing to say, at a static point without adding CO2 to the system, no new energy is trapped in the system, and I’m happy.
Of course that’s a ridiculously trivial statement but it is the correct one from this post.

George E. Smith
August 31, 2010 4:14 pm

“”” Nick Stokes says:
August 31, 2010 at 2:49 pm
Tom,
Do you think that ozone, by absorbing UV, heats the stratosphere? “””
I see no difference in the argument. Same goes for H2O vapor absorbing incoming solar spectrum radiation in the 750 nm to maybe 4.0 micron range. The absorption is real, the energy loss to the surface is real; evidently the heating of the atmosphere is what is not real.

John Eggert
August 31, 2010 4:22 pm

Scott Said:
I don’t dispute the greenhouse effect, but your observation can be explained without it too. Water vapor’s heat capacity, ~1.9 J/(g*K), is higher than dry air’s, ~1.0 J/(g*K). Thus, with the same energy output, the humid air will drop temperature more slowly. Also contributing to the slower temperature change is that some of the energy may go into condensing H2O onto surfaces (the opposite of evaporative cooling) once the dew point is reached.
Scott:
The heat capacity of moist air is ha + xhm, where ha is 1.006 for dry air and hm is your 1.84 (not 1.9) and x is the humidity ratio. For FULLY SATURATED (100% RH) air a 30C, x is 0.027. At 10C, it drops to 0.0076. Thus the change in heat capacity from 0% RH air to 100%RH air at 30 C is from 1.006 to 1.055. A 4% change. At 10C, the change is to 1.018. A 2% change. You can’t come close to fully explaining slowed cooling at night by the difference in heat capacity of completely dry versus saturated air. On a relatively dry day, the RH is certainly way above 0%, hence the difference in heat capacity of air on a relatively dry day versus a humid day is very small indeed.
Condensation is a heating process (or heat releasing if you prefer). On hot humid summer days, I don’t generally see much condensation until the temperature drops, hence I would argue that there is little or no condensation effect either.
The difference between the model and the reality, if the graphic that Anthony showed does indeed show such a disconnect can be explained by another means. My opinion is that the logarithmic change in heat absorption with increasing CO2 is only correct for lower levels of CO2. As levels increase, the rate of increase of heat absorption with increase in CO2 concentration drops off to near zero as CO2 concentration gets high. This levelling is seen in other areas where one considers radiant heat transfer in the atmosphere (blast furnace calcs, hot house calcs, etc.)
JE

George E. Smith
August 31, 2010 4:24 pm

“”” MartinGAtkins says:
August 31, 2010 at 2:16 pm
R Stevenson says:
August 31, 2010 at 10:41 am
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.
Not in our lower troposphere because the density of the CO2 layer is sufficient enough to absorb all the LW radiation available to it.
If not then does retention time of photons increase and if so does it matter.
The retention time doesn’t increase. The photon received is almost in an instant re-emitted. The work done by CO2 is a product of its absorption wavelength 15 µm which would not be caught by the other gases and so would radiate directly into space and be lost. “””
Well CO2 most certainly does not absorb all of the LW radiation available to it. The surface emission spectrum; at least from ana average Temeprature surface can be expected to contain significant energy between 5.0 microns and 80.0 microns (about 98% of it.
CO2 can absorb much of the range from about 13.5 to 16.5 Microns. If the Spectrum is roughly black body like; then 25% of the energy would reside below the peak wavelength which is 10.1 microns. So there is a good chunk of the spectrum that CO2 doesn’t touch.
And in the lower atmospehre the mean lifetime of the CO2 excited states is longer than the mean time between collisions; so that energy becomes thermalized before a spontaneous photon emission from the CO2 can occur. In the higher less dense regions in the stratospehre the time between collisins is great enough for spontaneous decay to occur.

DougB
August 31, 2010 4:28 pm

Tom W. above has it correct in my view, crystal clear:
“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.”
To me this is obvious.

INGSOC
August 31, 2010 4:34 pm

George E. Smith says:
August 31, 2010 at 11:40 am
“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.”
I knew someone was going to bring up Quantum Mechanics… My head is going to explode!
🙂

mircea
August 31, 2010 4:39 pm

Tom says: August 31, 2010 at 2:10 pm
“but the composition of the atmosphere is changing, hence there is no LTE”
– The composition of the atmosphere is changing but while CO2 increases H2O vapours decrease and as such there is LTE. Of course here we talk about an idealised atmosphere.
– More details about CO2/H2O are here: http://global-warming-explained.blogspot.com/2010/08/more-co2-means-less-h2o-dr.html

899
August 31, 2010 4:44 pm

“Of course the transient net energy transfer from CO₂ to N₂ will not continue forever because else the mixture would transform into superheated plasma.”
Well, no, it would not.
In order for that to take place more energy yet would have to be introduced into the system.
Rather, what would happen —in the entirely theoretical sense— is perpetual motion, which presumes that all of the energy trapped in the system remains within the system and is able to do work infinitely.
But otherwise …
🙂

Dave Springer
August 31, 2010 4:50 pm

Oliver Ramsay says: “we’ve all tried spending the same dollar in two different places and we know how that works out . . .”
kfg answers: ” . .quite well if done by two different carriers. If it is spent in a third place we may even receive our own dollar back again after having spent it. How cool is that?”
Evidently cool enough for the US government to do it a trillion times in one year and call it “stimulus spending”.

Truth and lies are faced alike; their port, taste, and proceedings are the same, and we look upon them with the same eye. I find that we are not only remiss in defending ourselves from deceit, but that we seek and offer ourselves to be gulled; we love to entangle ourselves in vanity, as a thing conformable to our being.
Michel de Montaigne
The Father of Modern Skepticism
1533-1592

kfg
August 31, 2010 4:51 pm

INGSOC says: “I knew someone was going to bring up Quantum Mechanics…”
Nah! He’s jes sayin’ that no two billiard balls on the table can occupy the same at the same time, but over a series of games any particular billiard ball will eventually have occupied all spaces.
Pure Newtonian objects in random motion.

John Eggert
August 31, 2010 5:05 pm

Mircea:
Your assertions about increasing CO2 leading to decreasing H2O is not supported by the paper you reference (see http://www.friendsofscience.org/assets/documents/E&E_21_4_2010_08-miskolczi.pdf). Miskolczi only states that he has observed no change in the amount of IR from the earth, even as CO2 increases. Regarding water, he asserts that Hansen et. al. have overestimated the water feed back. This is not the same as saying that increasing CO2 automatically leads to decreasing H2O.
JE

Steve Fitzpatrick
August 31, 2010 5:12 pm

After reading this post carefully, I must conclude that it is more obscure and meaningless than the last.
Really, there are only two alternatives that are important. Either 1) increasing CO2 in the atmosphere is expected to cause the Earth’s surface (and troposphere) to warm, or 2) increasing CO2 in the atmosphere is not expected to warm the Earth’s surface (and troposphere). Discussions of LTE are totally irrelevant, and only confuse and distract. Any analysis based on LTE tells us exactly nothing about how the Earth’s surface temperature should respond to increasing CO2 in the atmosphere. I find the application of an equilibrium analysis to a non-equilibrium system a pointless waste of time.
If Tom Vonk believes that a LTE analysis provides insight about the expected effect of increasing CO2 in the atmosphere on surface temperatures, then I sure which he would say what that insight is.

MikeA
August 31, 2010 5:30 pm

Anthony can we perhaps have a presentation on the Adiabatic Lapse Rate to go with the diagram. It obviously needs some sort of theory to falsify it?

Dave Springer
August 31, 2010 5:36 pm

John Eggert says:
August 31, 2010 at 4:22 pm
The heat capacity of moist air is ha + xhm, where ha is 1.006 for dry air and hm is your 1.84 (not 1.9) and x is the humidity ratio. For FULLY SATURATED (100% RH) air a 30C, x is 0.027. At 10C, it drops to 0.0076. Thus the change in heat capacity from 0% RH air to 100%RH air at 30 C is from 1.006 to 1.055. A 4% change. At 10C, the change is to 1.018. A 2% change. You can’t come close to fully explaining slowed cooling at night by the difference in heat capacity of completely dry versus saturated air. On a relatively dry day, the RH is certainly way above 0%, hence the difference in heat capacity of air on a relatively dry day versus a humid day is very small indeed.

Sorry, but no.
Air @ 30C, sea level, 17% RH = 42 kJ/kG
Air @ 30C, sea level, 93% RH = 95 kJ/kG
The wetter air contains over twice as much heat.
See psychometric calculator here: http://www.uigi.com/WebPsycH.html
Your figures were for sensible heat alone. You have to take latent heat into account for enthalpy.

Dave Springer
August 31, 2010 5:40 pm

Testing:
Two quite different comments of mine in a row were held up in the “special” moderation queue. What’s up with that?

kfg
August 31, 2010 5:52 pm

Dave Springer says: “. . . cool enough for the US government to do it a trillion times in one year . . .”
Ah, no. I didn’t say nuttin’ ’bout no havin’ one real dollar and one imaginary dollar and tryin’ to spend ’em both. That thar’s the road to daaaaaaamnation; even if you try to spend ’em both in the same place.
(No, actually, I haven’t got a clue why I’m talking like this. Thanks for asking and I’ll have to look into it)

Oliver Ramsay
August 31, 2010 6:00 pm

I was inspired by Anthony’s Urinal post to do my own research.
I set up an observation post with a full view of the entrance and exit of a public convenience structure for gentlemen.
The initial state of affairs was a steady stream of men entering and exiting the facility at the rate of one per minute. I noted that there was a group of ten individuals performing ablutions at any time that I stuck my head inside. Those entering and exiting kept perfect time; one per minute.
After 3 hours a coffee vendor set up shop nearby and, very soon, the rate at which the punters were entering the toilets increased to one every 45 seconds and, quite astonishingly, the rate at which they were leaving increased to exactly match the entry rate. Observation of the interior revealed that, at any time, there were ten people washing their hands; just as there had been before.
For a while I mused upon the variable elasticity of the human bladder and the diuretic effects of caffeine.
When I repeated the experiment the following day, everything unfolded exactly as it had on the first occasion, with the exception that, after the coffee-induced rate increase there were always 12 people milling around the sinks instead of ten.
I have to say that I never actually saw anybody go out by the In door, but that’s another story.
Clearly, everybody accelerated their micturition, their ablutions and their exit uniformly. Why did the number of lingerers increase on day 2? Were they in some way responsible for the haste of the egressors?
Or, was it just a dream?

Dave Springer
August 31, 2010 6:16 pm

kfg
If you spend the same dollar in two places one of them has to be imaginary. That’s how stimulus spending works. Uncle Sam trades an imaginary dollar to Chairman Mao for a real dollar. Sam then gives the real dollar to John Q. Public who uses it to buy a coffee pot from Mao to replace a broken one. Sam collects a tax on the transaction, Mao collects a tax on the transaction, and John can make his morning java. Everyone’s happy for the moment but a lot of us suspect someone is going to get screwed in the end and it probably won’t be Sam or Mao.

sky
August 31, 2010 6:20 pm

The thesis that the troposphere is everywhere in LTE, thus all collisional energy transfers between molecules are symmetrically bilateral leaves me perplexed. While I’m always reminding everyone that GHGs produce no energy on their own, it seems to go further by saying, in effect, that air cannot be warmed by radiative means.
This leaves many unanswered questions. Granted that moist convection plays a central role in heating the bulk constituents of the troposhere, what happens to terrestrial radiation at, say, 15 microns selectively abosrbed by CO2 (and overlapping water vapor). The claim that “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)” seems to preclude extinction via scattering to other wavenumbers through molecular collisions with N2. Yet 15 micron extinction is what is observed when the Earth is viewed from space. Surely, that energy does not simply disappear from the universe.
An explanation of this contradiction from Tom Vonk would be much appreciated.

August 31, 2010 6:27 pm

George E. Smith says: August 31, 2010 at 4:14 pm
” Nick Stokes says:August 31, 2010 at 2:49 pm
Tom, Do you think that ozone, by absorbing UV, heats the stratosphere? “
“I see no difference in the argument.”

Neither do I, and that’s my point. Ozone absorption of UV clearly does warm the stratosphere. Temperature increases with altitude. The absorbed heat is transmitted down to the tropopause, where GHG emission outwards is sufficient to reverse the gradient.

Dave Springer
August 31, 2010 6:29 pm

kfg:
P.S. The imaginary dollar that Sam spent to buy the real dollar comes from John’s increased future earnings because he’ll be more productive when he has a cup of Joe in the morning.
The biggest Ponzi Scheme evah!

August 31, 2010 6:34 pm

@sky – “Yet 15 micron extinction is what is observed when the Earth is viewed from space.”
No, it is NOT what is observed. The atmosphere is totally opaque at 15 microns. What is observed from space is the energy emitted from the stratopause, and above, at 15 microns.

Bill Illis
August 31, 2010 6:36 pm

“MikeA says:
August 31, 2010 at 5:30 pm
Anthony can we perhaps have a presentation on the Adiabatic Lapse Rate to go with the diagram. It obviously needs some sort of theory to falsify it?”
This is what is really happening in the atmosphere. It is reality versus theory.
The temperature (or let’s say the energy in the atmosphere) declines by 0.65C each 100 metres above the ground. It continues declining at this rate until about 10 kms high where it becomes a constant rate and the surface energy is then no longer being slowed down and more than 50% of it is just being emitted toward space (it might bounce around and continue to take a random walk in the upper atmosphere after that but it is not coming back to the ground).
Everything else is theory.
It is like 10 kms of insulation above us (and N2 and O2 obviously play a role in that in addition to the greenhouse gases).
Nobody has actually measured a photon’s path in the atmosphere, how far it travels, which molecules it interacts with, how long it spends in each molecule and where it goes after that.
But we can estimate some of those numbers. On average, without additional energy being added by the Sun each day, 4.5 W/m2 is emitted from the Earth each hour. Given the average collison rate and relaxation timeline for an energized molecule, each photon random walks through …
… 62 billion individual molecules in the atmosphere before it is emitted to space at 10 kms high;
… spends an average 0.000005 seconds in each molecule; and,
… travels an average 0.0001 millimetres before it is absorbed again.
The average photon takes 200,000 years to be emitted from the Sun’s surface after being generated in the core and it takes 86 hours for the average photon from the Sun to enter the atmosphere before it is emitted back to space (or after 86 hours they would all be gone give or take the oceans holding onto them a little longer).
That is a lot of really big and really small numbers. We need actual measurements to be able tell what is really going on rather than theory.

Dave Springer
August 31, 2010 6:59 pm

Oliver Ramsay says:
August 31, 2010 at 6:00 pm
Clearly, everybody accelerated their micturition, their ablutions and their exit uniformly. Why did the number of lingerers increase on day 2? Were they in some way responsible for the haste of the egressors?
Or, was it just a dream?

Actually there’s a good analogy to GHG warming there and I think yo meant to make it.
If you timed how long each individual person stayed inside the commode it would have increased. They were still entering and exiting at the same rate but there were 10% more of them on the inside. That’s what increased CO2 does. The same amount of heat is entering and exiting the ocean/atmosphere as before but there’s more heat trapped inside the system at any one time with more CO2.

jae
August 31, 2010 7:05 pm

Nick:
“Neither do I, and that’s my point. Ozone absorption of UV clearly does warm the stratosphere. Temperature increases with altitude. The absorbed heat is transmitted down to the tropopause, where GHG emission outwards is sufficient to reverse the gradient.”
Well, there is no LTE up there. As you know.

jae
August 31, 2010 7:10 pm

It seems to me that the question is this: At constant level of IR, can the presence of additional OCO move the temperature from the 25 C curve to the 35 C curve? If you energize more OCO molecules, do you increase the temperature of all the air?

Dave Springer
August 31, 2010 7:18 pm

Nick Stokes says:
August 31, 2010 at 6:27 pm
Neither do I, and that’s my point. Ozone absorption of UV clearly does warm the stratosphere. Temperature increases with altitude. The absorbed heat is transmitted down to the tropopause, where GHG emission outwards is sufficient to reverse the gradient.

I don’t think the emphasized portion is true. The path of least resistance for longwave radiation to move out of the stratosphere is back out into space. The denser longwave greenhouse gases below the ozone layer will still function as insulators only in this case they’re insulating the warmer stratosphere from the cooler layer below.
It works the same as the ground layer of CO2 only in reverse. The downwelling UV energy in the ozone absorption band is extinguished high in the atmosphere and re-emitted in all directions (some directly back out into space as longwave energy). The downwelling longwave is then absorbed (primarily by water vapor and CO2) and re-emitted in all directions (some directly back out into space).

George E. Smith
August 31, 2010 7:27 pm

“”” INGSOC says:
August 31, 2010 at 4:34 pm
George E. Smith says:
August 31, 2010 at 11:40 am
“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.”
I knew someone was going to bring up Quantum Mechanics… My head is going to explode!
🙂 “””
No! not to worry, no quantum mechanics; simply statistical mechanics. Tom’s MB distribution graph shows what fraction of all molecules have what kinetic energy at some gioven Temperature (which is presumed fixed for any one graph) Ergo, some molecule(s) have any one of the available energies represented by that graph; and nothing bars any particular molecule from at some point haveing any such available energy, and over time any single molecule will assume that whole range of energies with the same probability distribution. Hence the time averaged value of energy assumed by any one molecule is the same as the average of the whole set on a per molecule basis.
So it is just as valid to define thermodynamic Temperature as the time averaged energy per molecule for even a single molecule; rather than the average energy per molecule for the whole set of molecules.
So Mother Gaia knows the Temperature of each and every molecule; but only after looking at the thermometer for a while. But in the time scale of molecular collision frequencies; that is nearly instantaneous compared with the twice per day min/max readings of out pitiful global weather reporting stations. So she really does know what the true global mean temperature (say surface) of the earth is, averaged over a year; we have really no idea what it is; but likely can confine it somewhere betwen about +/-90 deg C; and maybe between +/-60 deg C. (well it is most unlikely to be near either extreme; but who knows in the 1/f noise spectrum; after all the big bang was mighty big; but so far as we know, it only happened once. Once is enough to cause all this havoc.

Dave Springer
August 31, 2010 7:29 pm

sky says:
August 31, 2010 at 6:20 pm
While I’m always reminding everyone that GHGs produce no energy on their own, it seems to go further by saying, in effect, that air cannot be warmed by radiative means.

It goes a lot more than further. It goes from true to false. The air certainly can be warmed by radiative means.

wayne
August 31, 2010 7:30 pm

Tom V,
I, for one, gained very much from your post as usual. I can bound the example to isolate your points. Keep posting such articles if only for me.
Would like you to post a very similar scenario with water molecules instead of carbon dioxide, being that water molecules possess an active dipole and linear carbon dioxide molecule do not. For I gather, water molecules do not have to immediately release excited energy as carbon dioxide molecules which do. Molecules without dipoles cannot hold on to vibrational energy, or so many universities teach. Could you be of any help on this? This type of information is hard to come by and even when you find it the question of correctness still lingers.
Thanks for the post. Keep it up.
Do you have a site where more long-term discussions on this subject occur?
-wayne

Nylo
August 31, 2010 7:33 pm

@mircea:
That theory is wrong. It has a wrong point. It says that “With H2O vapours raising higher concentration of H2O vapours in the column of air will decrease (same quantity in higher volume)”. No, they won’t. Because as they start to decrease near the surface, more moisture comes from the ocean. And you do not need any warming in the ocean to achieve this. At equilibrium for a given temperature there will be a given moisture next to the water, and if you take it away, more water will evaporate to mantain the equilibrium. Therefore no, H2O vapours raising do not cause it to stay at lower concentrations below. It causes more water to evaporate below to mantain the same moisture that it had before.

Oliver Ramsay
August 31, 2010 7:56 pm

@ Dave Springer,
Of course, I was drawing the analogy, but not necessarily subscribing to it, since I also suggested in the Day1 scenario that a non-heating stream of radiation that pauses briefly in the absorbing molecules is what is posited by Tom Vonk and others.
I don’t know which of the two is correct, but I wish the proponents of each side would present their views in a way that is clear to me.
For example, there seems to be the contention that CO2 both radiates IR back to Earth while at the same time thermalizing the entire atmosphere. Is there anything that molecule can’t do?
The opposite view is that gases catch and toss energy without getting hot.
Others say that it is not collisions between CO2 and N2 that warms the air but collisions between N2 and SiOx and friends at the surface that have been further heated by IR from CO2.
What’s the official position?

kfg
August 31, 2010 7:57 pm

Dave Springer says: “If you spend the same dollar in two places one of them has to be imaginary.”
Ah no. I didn’t say nuttin’, er, excuse me; I didn’t say anything about ME spending the same dollar in two places – at least not until round four. I specified two carriers. I spend a dollar, someone else spends the same dollar in a second place. Round three the dollar can come back to me when the third carrier spends it. All the same dollar. NOW I can, indeed, spend the SAME dollar in a second place. Mr. Smith would be so proud.
” Sam then gives the real dollar to John Q. Public . . .”
Now you’re just being silly; Sam hasn’t given John Q. Public a real dollar since 1964. One of those will still buy you 4 real gallons of imaginary Global Warming Carbon Footprint ™; (But WAIT! Don’t order yet. You’ll ALSO get someone else to pump it for you AND give you a free drinking glass to thank you for the privilege! I liked to hold out for the giant Sinclair “Dino” bar of Bronto soap though).
“The biggest Ponzi Scheme evah!”
See above. Don’t worry, I’ve been onto the game since about – 1964 (maybe a little earlier; being the inquisitive sort I asked my Grandfather what that $100 Gold Certificate under glass on the dresser was all about). If only I’d had enough real dollars to put aside back then.

mircea
August 31, 2010 8:06 pm

John Eggert says: August 31, 2010 at 5:05 pm
“Mircea:
Your assertions about increasing CO2 leading to decreasing H2O is not supported by the paper you reference (see http://www.friendsofscience.org/assets/documents/E&E_21_4_2010_08-miskolczi.pdf). Miskolczi only states that he has observed no change in the amount of IR from the earth, even as CO2 increases.”
See pag 260:
“As a final conclusion of this perturbation study, we can safely state that the dynamic
stability of the stationary value of the true greenhouse-gas optical thickness of the
atmosphere is mediated mainly by the amount and distribution of the water vapor in
the atmosphere, and by the surface and atmospheric temperatures.”
Also see fig. 9 page 257 and also pag 258: “Regarding the relative importance of the CO2 and H2O it was found that 1 ppmv increase in CO2 concentration (equivalent with 0.8 atm-cmSTP increase in column amount) can be compensated by 0.3 atm-cmSTP (2.4×10-4 prcm column amount) decrease in H2O. In other words, CO2 doubling would virtually, with no feedback, increase the optical thickness by 0.0246. Calculations here show that an equivalent amount of increase can be caused by 2.77 per cent increase in H2O.”
I have an open mind. Can you think at other reasons why the transparency remains constant?
Mircea

Nylo
August 31, 2010 8:32 pm

@Tom Wonk:
Sorry, but if you wanted to make a summary of the discussion, you did dead wrong. Sensible critics said that the fact of changing one gas to the other raises the temperature. Funnily enough, you seem to admit that yourself while explaining Type 2. Even more funnily, you don’t seem to notice that you did, as you use your explanation of Type 2 to somehow try to “prove” your point.
The key is here, I quote you:
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. […] As a result of this transfer the temperature of N₂ would increase […] Of course the transient net energy transfer from CO₂ to N₂ will not continue forever because else the mixture would transform into superheated plasma […] 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.
The “no local net energy transfer from CO2 to N2” is a characteristic feature of the new LTE. It is completely wrong if referred to the transient.
As Bill Clinton would say: “It is the transient, stupid!“.
Changing the composition of the atmosphere heats it up DURING THE TRANSIENT. If you substitue a gas which, alone, stays at a lower temperature for the given radiation of energy from Earth, for another gas with a higher temperature, the temperature of the new mix is higher. Even in the case that CO2 could not transmit its energy to N2 at all, even if it was absolutely imposible, the mere fact that there is more CO2 in the atmospheric composition and it is hotter than the other gases, that alone would raise the average temperature of the atmosphere. So CO2 to N2 transfer is to some point irrelevant.

Nylo
August 31, 2010 8:49 pm

The easiest way to see it is the opposite reasoning. Take a bottle of CO2+N2, put it next to a source of 15 microns radiation. It gets some temperature, and when the energy comming in equals energy getting out it is in LTE. Now, chemically or mechanically, whatever, take out the CO2 content alone, replace it for N2 at the same temperature that the mix had before. After the transient, the temperature of the new gas will be lower, because again energy out = energy in, and suddenly there is no energy in, so there must be no energy out, so the temperature must be 0 K.
If removing CO2 causes the temperature to drop, how the h*ll can you claim that adding CO2 cannot cause the temperature to raise? Increasing CO2 increases the gas’ capability of absorbing energy, therefore increases the incomming flux of energy, therefore increases the temperature that the gas must have to make energy in = energy out. It increases the needed temperature for reaching a new LTE.

Dave Springer
August 31, 2010 8:55 pm

I’m a bit stunned that some physicists don’t know how things work in the real world. Lots of jargon and mad math skillz but with such poor understanding of the real world they don’t know which equations to run the numbers through and can’t recognize the bogus results generated therefrom. Vonk is so far removed from reality in trying to conflate LTE to a deep column of air with wildly varying daily radiative input at the top of the column it’s comical. This is what gives skeptics a bad name.

Ian H
August 31, 2010 9:06 pm

This whole argument looks very circular to me. Isn’t this just a longwinded way of saying essentially “Things in thermal equilibrium don’t change temperature.”?
To which the only appropriate response is has got to be … … well duh!
Sorry but I am not convinced. In fact I’m not exactly sure what you were trying to convince me of.

Dave Springer
August 31, 2010 9:13 pm

kfg
You can’t eat gold. Well, you can, but I don’t think it’s very nourishing. If the monetary system collapses under its own weight no one is going to trade food, ammunition, medicine, fuel, and things of that nature for gold. Buy things that are practical barter goods and store them in a place you can control & defend.

Nylo
August 31, 2010 9:34 pm

@Tom Wonk:
A reasoning step by step, please tell me if any of the steps is wrong.
1) CO2 has a greater capability for absorbing 15 microns radiation than the remaining gasses in the atmosphere.
2) Earth is a source of 15 microns radiation (among other wavelengths).
3) Changing some O2 for some CO2, therefore increasing the CO2 concentration, increases the capability of the atmosphere to absorb 15 microns radiation (i.e. less 15um photons from the Earth’s surface escape unabsorbed to space).
4) If more photons are trapped by the atmosphere, there is more energy in.
5) At LTE, energy in equals energy out.
6) If energy in increases and energy out does not, there is a transient state where the atmosphere is not in LTE.
7) The transient finishes when we reach a new LTE.
8) The new LTE still gets an increased energy in, however now there is an increased energy out and balance is restored.
9) Given that the radiation of energy depends on the temperature, if there is an increased energy out, it is because the temperature is higher than in the previous LTE.
Is any of these points wrong? In case not, doesn’t 3 (increasing CO2 content of the atmosphere) inequivocally lead to 9 (temperature gets higher)?

kfg
August 31, 2010 9:48 pm

Dave Springer says: “You can’t eat gold. Well, you can, but I don’t think it’s very nourishing.”
A few years ago my mother asked me, “Is there a time when gold could become essentially worthless?”
I said, “Yeah, about 10 minutes after the food runs out.”
She put her money into tinned sardines. She never takes my advice; I had suggested tuna.
I’m still ahead of you, but I suppose I could use a bit more ammo. The ax handle will keep going as long as I can.
By the way, you didn’t ask me the critical question that has made this whole exchange on topic: “How could you get your own dollar back for “free”?”
Well, I couldn’t of course. I had to put energy into the system before it could reach that new equilibrium. Some physicists leave the tower and go to work as engineers.

kfg
August 31, 2010 10:01 pm

@Nylo: Now you are barking up Lindzen’s tree.

Oliver Ramsay
August 31, 2010 10:04 pm

@kfg
I suspect that if things got that dire, the value of lead (to a much lesser extent, axe handles) would far outstrip that of gold long before the food was all gone.

Dave Springer
August 31, 2010 10:31 pm

Oliver Ramsay says:
August 31, 2010 at 7:56 pm
@ Dave Springer,
Of course, I was drawing the analogy, but not necessarily subscribing to it, since I also suggested in the Day1 scenario that a non-heating stream of radiation that pauses briefly in the absorbing molecules is what is posited by Tom Vonk and others.
I don’t know which of the two is correct, but I wish the proponents of each side would present their views in a way that is clear to me.
For example, there seems to be the contention that CO2 both radiates IR back to Earth while at the same time thermalizing the entire atmosphere. Is there anything that molecule can’t do?
The opposite view is that gases catch and toss energy without getting hot.
Others say that it is not collisions between CO2 and N2 that warms the air but collisions between N2 and SiOx and friends at the surface that have been further heated by IR from CO2.
What’s the official position?

The official position that gases can be sensibly heated by longwave radiation was established experimentally in 1859.
“Heat a mode of motion
By John Tyndall
http://books.google.com/books?id=3DUJAAAAIAAJ&pg=PA341&dq=%22Heat+a+Mode+of+Motion%22&output=text
The chapter on radiative absorption by gases begins on page 321. The link above takes you to the discussion of results and skips past the details of the experimental setup.
Tyndall’s apparatus wasn’t sensitive enough to detect any absorption in perfectly dry air, which illustrates how little CO2 absorbs compared to water vapor, but he had little problem with humid air and concentrated gases of various kinds.
That’s what makes this so damn frustrating. This is knowledge from experimental physics 150 friggin’ years ago!

kfg
August 31, 2010 10:37 pm

@Oliver Ramsay:
Well, I was engaging in a bit of exaggeration for rhetorical purposes in that talk with my mother. In point of fact, under those conditions the value of lead and gold are about even ( I seem to recall a recent thread on this). Works great on the end of an ax handle too. The hammer reached its highest state of technological development about 10 minutes after metal plate armor did. Coincidence?
Let’s just say it’s a great way of transferring energy to the brain (much more effective than a cell phone); albeit at somewhat limited range and rate of “fire.”

dr.bill
August 31, 2010 10:38 pm

Dave Springer: August 31, 2010 at 8:55 pm
I’m a bit stunned that some physicists don’t know how things work in the real world. Lots of jargon and mad math skillz but with such poor understanding of the real world they don’t know which equations to run the numbers through and can’t recognize the bogus results generated therefrom. Vonk is so far removed from reality in trying to conflate LTE to a deep column of air with wildly varying daily radiative input at the top of the column it’s comical. This is what gives skeptics a bad name.

Tom Vonk is a physicist ???
/dr.bill (physicist, refraining from piling on)

Scott
August 31, 2010 10:43 pm

Regarding:

John Eggert says:
August 31, 2010 at 4:22 pmM

You asked why a more humid day cooled more slowly, I gave one. However, it’s pretty well established that the dominating effect is the “greenhouse” one, I just came up with an additional reason, if insignificant. As for as condensation, I agree it usually also plays a small role, but I did notice that your original question asked about “evenings” and your response to mine said “hot summer days”…big difference if you ask me. But again, just look at the first sentence of my response, I agree with you!…just wanted to illustrate that the question could be asked in a better way.
Also:

The difference between the model and the reality, if the graphic that Anthony showed does indeed show such a disconnect can be explained by another means. My opinion is that the logarithmic change in heat absorption with increasing CO2 is only correct for lower levels of CO2. As levels increase, the rate of increase of heat absorption with increase in CO2 concentration drops off to near zero as CO2 concentration gets high. This levelling is seen in other areas where one considers radiant heat transfer in the atmosphere (blast furnace calcs, hot house calcs, etc.)

Entirely agreed here, and that’s why I refuse to put the C in front of AGW (of which I believe is happening to a small degree). Beer’s law makes it very clear that higher concentrations of CO2 (or longer radiative pathlengths) lead to saturation. The only mechanism for absorbing more radiation is to have some absorb in regions of lower extinction coefficients/molar absorptivity. The problem is, the extinction coefficients rapidly drop off and/or go into regions already being absorbed by other molecules, thus leading to worse than logarithmic performance. I am, however, willing to listen to actual data/numbers telling me I’m wrong.
-Scott

August 31, 2010 10:46 pm

Phil.
“One of the things that always amused me Mosh was that when the UAH-MSU data first came out and showed the absence of warming it was justified by agreement with balloon data. After errors were corrected and a positive trend was established we were told that this data also agreed with balloon data!”
Don’t you know that data is always to be trusted.

Oliver Ramsay
August 31, 2010 10:46 pm

@ Nylo
I am not attempting to speak for either Tom Wonk or Tom Vonk, but I’m wondering why, at line 6) , you assume that the energy out doesn’t increase in line with the energy in, but then at line 8) you acknowledge that it does.
Does Kirchhoff suddenly wake up at line 7)?
If emission goes as the fourth power of temperature, then it will be considerably greater at the putative warmer temperature than at the previous cooler one.

Dave Springer
August 31, 2010 11:16 pm

kfg
“Some physicists leave the tower and go to work as engineers.”
An undergraduate degree in physics is something of a general purpose kind of thing. It’ll get your foot in the door at entry level a lot of different engineering specialties. A graduate degree in physics won’t help much. As a general rule years of experience and years in graduate school are equitable. An established salary history in the desired occupation trumps everything else though and you don’t get that salary history as a full time student.

Dave Springer
August 31, 2010 11:32 pm

@scott
Condensation plays a huge role when the sensible temperature reaches the dewpoint.

Oliver Ramsay
August 31, 2010 11:39 pm

Dave Springer says:
“That’s what makes this so damn frustrating. This is knowledge from experimental physics 150 friggin’ years ago!”
———–
It’s astonishing to me that people with serious degrees in Physics can be arguing about this, too.
However, the discussion is carried on in such truncated fashion that it’s not often clear what exactly is being disputed.
It’s hard to reconcile statements that emission is irrelevant at 1 bar with the assertion that IR is radiated back to the surface. Then , the twice-spent dollar crops up again when a photon that has gone translational is also able to speed back to Earth, warm up the ground, then zip back for a logarithmically reduced second shot.
Being mathematically naive, I am limited to simple arithmetic and a lot of what is said doesn’t add up.
Your link to Tyndall’s book was good enough to hint at the contents but I didn’t see much of the actual content. I’ll try Amazon or some such.

Dave Springer
August 31, 2010 11:39 pm

dr.bill says:
August 31, 2010 at 10:38 pm
Tom Vonk is a physicist ???
/dr.bill (physicist, refraining from piling on)

According to Anthony:

CO2 heats the atmosphere…a counter view
Posted on August 5, 2010 by Anthony Watts
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)

It’s more than a bit hard to believe at this point.

wayne
August 31, 2010 11:44 pm

Nylo says:
August 31, 2010 at 8:49 pm
The easiest way to see it is the opposite reasoning. Take a bottle of CO2+N2, put it next to a source of 15 microns radiation. It gets some temperature, and when the energy comming in equals energy getting out it is in LTE. Now, chemically or mechanically, whatever, take out the CO2 content alone, replace it for N2 at the same temperature that the mix had before. After the transient, the temperature of the new gas will be lower, because again energy out = energy in, and suddenly there is no energy in, so there must be no energy out, so the temperature must be 0 K.

I thought it was accepted that N2 does not radiate or absorb at normal temperatures so how is your example every to cool since you put it in without the CO2 at the same temperature? Does not absorbability = emissivity? (Or are you allowing conduction through the jars walls which makes little sense in your pure example?)
To me the N2’s temperature would never change, not absorbing or emitting.

Dave Springer
September 1, 2010 12:00 am

“Is any of these points wrong?”
No.
“In case not, doesn’t 3 (increasing CO2 content of the atmosphere) inequivocally lead to 9 (temperature gets higher)?”
Only if nothing else changes. The problem is other things change at the same time and these other things (all related to water in all its phases – albedo of snow, clouds, and liquid water – and phase changes which shuffle around enormous amounts of latent heat) and ) can easily drown (pun intended) the CO2 contribution. For instance when increased CO2 tries to throw the system further from equilibrium temperature it causes surface water to evaporate faster. Water vapor is lighter than air so it rises and condenses into a cloud and in the process carries an enormous amount of latent heat of vaporization straight through the densest layer of CO2 and releases it high in the sky where the underlying greenhouse gases then serve to insulate the warmer cloud from the cooler air below it. Latent heat of vaporization and to a lesser extent latent heat of fusion is the fatal flaw in the highly exagerated “climate sensitivity”. If there’s no positive feedback from CO2 heating that amplifies its effect by a factor of three, and there clearly ain’t no such thing indicated by any empirical observation, then catastrophic global warming becomes beneficial global warming.

Dave Springer
September 1, 2010 12:03 am

Nylo says:
August 31, 2010 at 9:34 pm
“Is any of these points wrong?”
No.
“In case not, doesn’t 3 (increasing CO2 content of the atmosphere) inequivocally lead to 9 (temperature gets higher)?”
Only if nothing else changes. The problem is other things change at the same time and these other things (all related to water in all its phases – albedo of snow, clouds, and liquid water – and phase changes which shuffle around enormous amounts of latent heat) can easily drown (pun intended) the CO2 contribution. For instance when increased CO2 tries to throw the system further from equilibrium temperature it causes surface water to evaporate faster. Water vapor is lighter than air so it rises and condenses into a cloud and in the process carries an enormous amount of latent heat of vaporization straight through the densest layer of CO2 and releases it high in the sky where the underlying greenhouse gases then serve to insulate the warmer cloud from the cooler air below it. Latent heat of vaporization and to a lesser extent latent heat of fusion is the fatal flaw in the highly exagerated “climate sensitivity”. If there’s no positive feedback from CO2 heating that amplifies its effect by a factor of three, and there clearly ain’t no such thing indicated by any empirical observation, then catastrophic global warming becomes beneficial global warming.

Dave Springer
September 1, 2010 12:18 am

Nylo says:
August 31, 2010 at 8:49 pm
I thought it was accepted that N2 does not radiate or absorb at normal temperatures

Everything radiates at any temperature above absolute zero.
Everything absorbs unless its surroundings are at absolute zero.
http://en.wikipedia.org/wiki/Absolute_zero

Absolute zero is the theoretical temperature at which entropy would reach its minimum value. The laws of thermodynamics state that absolute zero cannot be reached because this would require a thermodynamic system to be fully removed from the rest of the universe.

anna v
September 1, 2010 12:24 am

OK, I will chime in though my thermodynamics course is fifty years behind me.
What Tom is saying is that, at normal pressures ( not vacuum) once in local thermodynamic equilibrium, the type of constituent molecule is irrelevant to the temperature. All types should be at the same temperature by the way temperature comes out in statistical mechanics. Since they are at the same temperature, one type cannot heat the other by tautology. Radiation,incoming or outgoing is within this LTE again by construction. In this thermodynamic formalism the modifying effect of a new constituent molecule appears in changes in the heat capacity of the whole volume of gas under observation.
H2O is very good at that, as Dave has shown, doubling the heat capacity of air for high humidity values.
Is it then correct to say that H2O heats N2 an O2? I would say it delays the radiative cooling of the total gas volume, which is a different concept than heating, though connected to heat transfer. I would also accept that it changes the gray body constants of the gas volume. Again this cannot be called heating.
Again I will repeat that in my opinion the introduction of absorption and radiation lines is an unnecessary double counting when using the formalism of classical thermodynamics and statistical mechanics.

Dave Springer
September 1, 2010 12:26 am

Oliver Ramsay says:
August 31, 2010 at 11:39 pm
Your link to Tyndall’s book was good enough to hint at the contents but I didn’t see much of the actual content. I’ll try Amazon or some such.

The whole book is there. You can thumb back and forth a page at a time at the top left. The link I gave is the optical character recognition version so it has an occasional character that is wrong. You can also see the original pages as they were scanned and download the whole book in a single PDF (it’s a big file).

Dave Springer
September 1, 2010 12:37 am

@Oliver Ramsay
The link below takes you to the table of contents for Tyndal’s book with hotlinks to the chapters and sections. Google doesn’t make it easy to find the ToC.
http://books.google.com/books?id=m-TUAAAAMAAJ&pg=PR16&dq=heat+a+mode+of+motion&output=text

MartinGAtkins
September 1, 2010 12:50 am

George E. Smith says:
August 31, 2010 at 4:24 pm
Well CO2 most certainly does not absorb all of the LW radiation available to it. The surface emission spectrum; at least from ana average Temeprature surface can be expected to contain significant energy between 5.0 microns and 80.0 microns (about 98% of it.
“available to it” Has a specific meaning. It does not imply that CO2 absorbs all the LW radiation emitted from the black body. It means that CO2 absorbs all the energy in the wavebands that it can absorb, the ones outside those wavelengths are not available to it because of the atomic structure of the molecule.
And in the lower atmospehre the mean lifetime of the CO2 excited states is longer than the mean time between collisions; so that energy becomes thermalized before a spontaneous photon emission from the CO2 can occur. In the higher less dense regions in the stratospehre the time between collisins is great enough for spontaneous decay to occur.
I don’t dispute what you are saying but that is not what
R Stevenson asked.

If not then does retention time of photons increase and if so does it matter.

The answer I gave is correct retention time of the photon doesn’t increase with the density of the gas.

Volt Aire
September 1, 2010 12:51 am

Co2 does not heat the troposphere – correct in the sense that co2 is not a source of energy like the sun.
Co2 is however defacto heating the troposphere (in some amount) in the sense that it transforms energy that otherwise would pass through the troposphere and converts it to kinetic energy in the molecular level. I don’t suppose anyone seriously contends that?
There are a bunch of problems in this post that need to be addressed – being a circular argument, the best remedy might be just removing it as I can’t see how you could fix it if the basic logic is faulty. I do however appreciate the post, good read and made me think.

John Marshall
September 1, 2010 2:22 am

Some thermodynamisists state that the atmosphere can never be in LTE so temperature measurement will produce a meaningless number. The above argument also ignores the fact the warm air will convect and cool adiabatically a fact that the theory of GHG’s also ignores. I cannot see why an energetic CO2 molecule does not transfer this vibrational energy to other gasses. To say that this energy is transferred back only states that equilibrium will be established when all the gas in that system is at the same temperature, ignoring boundary exchanges. Unfortunately the atmospheric system does have boundaries and there are many energy exchanges.
The argument that temperature measurement in a room will produce a constant temperature shows that this gentlemen has not tried this experiment. Place the thermometer on the floor and the temperature will be fairly low and take the temperature at the ceiling and it will be higher. I have done this and the data shows this to be true so my room was not in LTE.
It is true that the troposphere does get warm and cools so must be by energy transfer from solar radiation and gas/gas interactions, convection and adiabatic cooling and warming due to vertical movement of gasses through the atmospheric column. What the alarmists state about backradiation and energy increasing does violate the laws of thermodynamics.

don penman
September 1, 2010 2:30 am

I think the point is that there is a lot more nitrogen in the atmosphere then there is co2 in the atmosphere ,it is going to be very difficult for co2 to raise the average energy in the troposhere by absorbing ir radiation but it would be very easy for nitrogen to raise the average energy in the troposphere by absorbing ir radiation.

Roger Clague
September 1, 2010 3:34 am

It is good that we are now discussing the thermodynamics ( a well established body of knowledge) of the gas N2 that makes up nealy 80% of the atmosphere.
The idea that CO2 traps heat is a return of the phlogiston theory that heat is a substance. We now know heat is the result of motion of every substance so cannot be trapped.

Spector
September 1, 2010 3:45 am

I believe one the problem with the exposition of the main article is that Local Thermodynamic Equilibrium is a condition that can only exist if the given local region is absolutely *not* being heated or cooled by any other external region. In practice, I think LTE never really applies in our atmosphere, even taken as a whole, because it is always interacting with the universe at large.
What I believe we do have is a dynamic Energy Exchange Equilibrium, where a given local region rises to a temperature that allows it to expel the same average amount of energy that it is receiving from the surrounding external environment.
I believe the logarithmic effect of added CO2 is the most important reason for discounting the AGW alarm. According to the MODTRAN online radiation calculator utility, the CO2 increase over the past 10 years should only have had a raw effect of about 0.06 degrees C.

barry
September 1, 2010 3:54 am

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.

The so-called ‘hotspot’ refers to the tropical troposphere, and increased heating there is not an artefact of GHG heating, but of any heating. If the sun gets hotter, or albedo somehow reduced – the anticipated signature of global warming is a hotspot in the tropical troposphere. It has no direct relation to CO2.
Therefore, if there is a missing ‘hotspot’, then there is a problem with the understanding of heat transport in the atmosphere, not the theory of global warming from increasing GHGs.
However, it appears that the hotspot may previously been obscured by poor data.

Merrick
September 1, 2010 4:08 am

Wow. I have been in the mountains for days and got a look at this for the first time when I was getting ready to go black again. I typed up some quick but long responses expecting to have to make extensive revisions to my comments to respond to what I expected to be further additions of (what I believe to be) incorrect statements about atmospheric heating in the follow-on comments, but am exceptionally pleased (and frankly shouldn’t be!) At the extremely sober comments on Tom’s article. The vast majority of Anthony’s visitors really get the basic issue with this contention:
Yes, of course, a system in thermodynamic equilibrium (local or otherwise) cannot possibly have net energy flow – that violates the initial statement. When used in the physical sciences, equilibrium is defined as that state of a system in which an external force is required to change the state of the system. Posit BOTH equilibrium AND no external forcing (remember – no radiative coupling by Tom’s stipulation) and the argument is rock solid. By definition.
The problem is, Tom isn’t saying that CO2 can’t heat N2 when the system is in LTE. If he stopped there he’d be on solid ground. The problem is that once having established that he then claims that therefore CO2 cannot heat the atmosphere at all and backs it up with very watery caveats about how the atmosphere is almost in LTE, etc. This is simply untrue.
I am so pleased that WUWT folks have so clearly understood this that I promise not to post my extremely lengthy replies unless I’m refuted again and am forced to.
And – by the way – contrary to what Tom says, all that V/R, R/T, and V/T transfer that Tom claims to be mere details IS the whole warp and weft of radiative heating in the atmosphere. Those modes, by and large, CANNOT get by energy coupling directly to the electromagnetic field (more details later if required) so the only way for energy to get into those modes is via the indirect path of collisional energy transfer after the electromagnetic field has energized the CO2 vibration. As long as the electromagnetic field is out of equilibrium with the atmosphere there is an external forcing. Period. If there is more capacity in the EM field then energy flows from the EM field into the CO2 vibration and energy moves from the CO2 vibration into all of the vibrations, rotations, and translations in all the atmospheric constituents until all of those constituents are in equilibrium with the EM field. This is called heating. The CO2 was the key constituent in the radiative heating process (water contributes in a similar fashion, but we’re discussing CO2 here). The higher the concentration of CO2, the more rapidly CO2 can couple EM energy into the atmosphere*. Period. So, yes Virginia, there is a “Greenhouse Effect” – even if it’s inappropriately named, as has been pointed out on many occasions. And, yes, there is ALSO a logarithmic effect, so at some point additional CO2 has a much smaller effect than the CO2 already there. We are well into that logarithmic rollover and IPCC CO2 doubling nightmare scenarios are simply rubbish.
*from above as marked, when the capacity of the EM field is less than that of the atmosphere (like when the earth cools after sunset) the CO2 also acts as an efficient mechanism for couplng the energy back out of the atmosphere. More photons go out of CO2 vibrations than are repopulated by the external forcing of the EM field, and energy flows from the V/R/T modes of the relatively warm atmospheric constitutents into CO2 vibrations to reestablish equilibrium and both of these energy flows continue until equilibrium is reestablished or the balance reverses and a warming earth under the risen sun starts pumping energy through CO2 (and water) back into the atmosphere.
Tom – just a suggestion: pull up a plot (perhaps accuweather.com?) of the hourly temperature records for yesterday for any place you’d like. Then tell me where it is so I can look at the same data and you tell me which parts of yesterday’s time record, according to the temperature record, were a good example of the atmosphere experiencing LTE conditions.

Spector
September 1, 2010 4:26 am

RE: George E. Smith: (August 31, 2010 at 4:24 pm)
“And in the lower atmospehre the mean lifetime of the CO2 excited states is longer than the mean time between collisions; so that energy becomes thermalized before a spontaneous photon emission from the CO2 can occur. In the higher less dense regions in the stratospehre the time between collisins is great enough for spontaneous decay to occur.”

This is an interesting comment. I have not considered that something like this might also be a factor in establishing the level of the tropopause before. I wonder at what altitude the mean time between collisions is equal to the typical CO2 or H2O excited state lifetime.

LazyTeenager
September 1, 2010 4:58 am

This article sounds like Xeno’s Paradox and Tom Volk has just proved that arrows can’t possibly fly.

LazyTeenager
September 1, 2010 5:01 am

Oops Zeno not Xeno
[Reply: that’s OK, you’re just a teenager.]

DocMartyn
September 1, 2010 5:05 am

Your Maxwell Boltzmann distribution is mis-labeled. You figure shows velocity, not energy. Energy is velocity squared, the population distribution against E is typically a plot of logE.
Moreover, you are applying classical equilibrium thermodynamics, local thermal equilibrium, to a steady state system. This is a no-no. Imagine applying equilibrium thermodynamics to a juggler, it is not possible.

LazyTeenager
September 1, 2010 5:17 am

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.
————
Yes he is absolutely sure of it and he is also absolutely wrong.

LazyTeenager
September 1, 2010 5:19 am

This claim is also wrong!
—————-
Of course the transient net energy transfer from CO₂ to N₂ will not continue forever because else the mixture would transform into superheated plasma.

Steve Fitzpatrick
September 1, 2010 5:25 am

Spector,
“I wonder at what altitude the mean time between collisions is equal to the typical CO2 or H2O excited state lifetime.”
Very high in the stratosphere. Collision rates are very high in the troposphere; much faster than the radiative half life of a vibrationally excited CO2 molecule. Of course, collisions can also bump a CO2 molecule into an excited vibrational state which can emit a photon (at 15 microns), so at any temperature there is always a modest population of CO2 molecules which are vibrationally excited, and 15 micron emission takes place continuously in all directions at a rate which depends mainly on the local temperature and CO2 concentration. Any emitted photon is likely going to be (surprise) absorbed by another (non-excited) CO2 molecule, unless the emission takes place high in the atmosphere, where it can escape to space unimpeded.
Emission and absorption of 15 micron photons by CO2 in the troposphere is taking place continuously, so the troposphere is essentially opaque to 15 micron radiation. That opacity falls with altitude, with the mean photon path length (following an emission from an excited CO2 molecule) going as the inverse of the pressure.

LazyTeenager
September 1, 2010 5:26 am

The basic problem with this article is that LTE is an approximation that is not relevant to the atmosphere and the way CO2 absorbs IR.
There is a temperature gradient between the ground (15C) and the top of the troposphere (-60C). This means there is also a temperature gradient across any cube of gas in the atmosphere. Thus means that any cube is not at LTE.

LazyTeenager
September 1, 2010 5:35 am

To put it another way.
The lower atmosphere has to heated somehow. At ground level this is a mixture of conduction (small), IR radiation (medium) and, evaporated water(large).
If we substitute CO2 with H2O in the above argument it is also proved that H2O does not heat the air either. Hence nothing can heat the air.
Reductio ad absurdum

LazyTeenager
September 1, 2010 5:57 am

George E Smith says:
As to the question of whether the Thermodynamic Temperature is just the translational energy; and does not include rotational or vibrational; doesn’t that conflict with the equi-partition law, that the energy must be distributed equally among all degrees of freedom.
—————-
No. The availability of additional degrees of freedom in more complex molecules means that the substance can absorb more heat for a given temperature rise as commonly represented by the substances heat capacity.
I am getting the impression of an assumption that temperature and heat are the same. They are not. Heat is the total energy per mol, temperature is the translational kinetic per mol.

wayne
September 1, 2010 6:36 am

Seems there are many knowledgeable commenters here.
There is one area that keeps eluding me and I would love to get a final answer. I have read both sides of this story. Do all gases radiate infrared at normal temperatures due to collisions or are all collisions between IR inactive gases completely elastic?
Take two identical huge cubical rooms to minimize the wall to volume ratio. One is filled with pure nitrogen and the other pure carbon dioxide. It’s zero degrees outside and both are fifty deg. C inside. The “nitrogen cannot radiate” side says the carbon dioxide room will cool fast and reach five deg. C much faster than the nitrogen room for the nitrogen had to cool the slow conduction way. Is this correct? Do you know of some truthful references?
If so, why do the accelerated electrons in collisions not cause radiation?
The other side says all matter always radiates above zero K.

TomVonk
September 1, 2010 6:39 am

Sorry for a late intervention again due to the fact that most comments took place during US time zone when I am sleeping .
First I will ignore the agressive ad homs . I have been looking hard in such posts and there is a couple , yet didn’t find a single argument dealing with physics .
Second are comments on the form . English is not my mother language . It is neither my second nor third . Even if people I meet qualify it as “fluent” , I am well aware that it is not and will never be equivalent to a native speaker . This is the case of almost all scientists who must publish in english while being more comfortable in other languages . However in this case my draft has been reviewed by a native english speaker .
So the form should be correct (even if apparently one typo passed through) .
Amusingly while my draft was more compact , I have been advised to separate the paragraphs more largely because the readers prefer that .
And of course there are some posts that say that the post should be more compact .
Well one can’t please everybody 🙂
Most of the discussion can be resumed by this question :
So why is this relevant to climate discussions?
I have answered this in my original post . It is relevant because there are many blogs and climate sites that explicitely or implicitely state what I called the Type 2 argument .
E.g “The CO2 emits no infrared radiation because it heats the N2 (atmosphere) by collisions isntead” .
This statement is simply wrong and if one does physics , one should not accept wrong statements .
Let us repeat it again : this statement is wrong .
The first and the second post were motivated by the wish to prove that this statement was wrong and should be removed from all blogs and sites where it appears .
To be perfectly clear that that was the purpose , I have introduced the Caveats to make sure that there would be no red herrings and no misinterpretations .
To my satisfaction most of the posters have understood the argument but a few are still repeating that LTE doesn’t exist . I am afraid that I cannot give a satisfactory answer to people who consider that temperature doesn’t exist .
Merrick quite nicely resumed the purpose here :
Yes, of course, a system in thermodynamic equilibrium (local or otherwise) cannot possibly have net energy flow – that violates the initial statement. When used in the physical sciences, equilibrium is defined as that state of a system in which an external force is required to change the state of the system. Posit BOTH equilibrium AND no external forcing (remember – no radiative coupling by Tom’s stipulation) and the argument is rock solid. By definition.
It is exactly that . Not more and not less . So why are people still writing nonsense in the form of Type2 argument ? Be my guest , propose an explanation other than the one that they understand nothing about physics .
Let’s just mention 2 red herrings that have already figured in the comments in the first post .
Things happen when parameters change
Sure . See caveats . But you are not free to redefine the question I asked and which is extremely clear and precise .
There may be many other questions you may want to ask and that are probably interesting too . But they are not the question I asked . You are just polluting the thread by talking about questions that are not being asked . I may have an opinion about these questions but this thread is not the place I will give it .
Most notably the question what happens when the composition of the mixture changes is a different question . There are hundreds of people and dozens of computer models that deal with this question . I do not think that anybody has accurately described the dynamics of such a system with variable composition . I certainly don’t .
Of course it follows from my argument that the equilibrium temperature of the mixture would increase if the CO2 concentration increased provided that nothing else changed . Just look at the Maxwell Boltzmann figure and ask yourself how would change the molecular processes (1) and (2) .
The second red herring is radiative equilibrium .
It is not necessary to have radiative equilibrium in order to have LTE . The LTE is a property or material particles , not one of radiation .
That is why , I do NOT assume radiative equilibrium and do not need to do so .
Anybody wanting to say that radiative equilibrium is necessary for LTE ?
If yes , then I want to hear it and especially the argument explaining why the Maxwell Boltzmann distribution can’t be derived if the radiative equilibrium is not established .
To understand the difference between LTE and non lTE conditions I strongly suggest reading : http://www.isbnlib.com/preview/9810245661/Non-LTE-Radiative-Transfer-in-the-Atmosphere-Series-on-Atmospheric-Ocean-and-Pla .
A big part of the book is available on line at the link .
One quote for those who still drone that LTE doesn’t exist and is unimportant anyway :
In general, LTE applies if the populations of the energy levels within a molecule are the same, or nearly the same, as they would be under true thermodynamic equilibrium conditions. LTE occurs when collisions are so frequent that the energy level populations depend predominantly on the local kinetic temperature, as defined by Maxwellian statistics of molecular motion, rendering negligible various other processes which may be going on at the same time. Obviously, it is necessary for the atmospheric scientist to understand when LTE applies and when it does not. ….. On the Earth this (LTE) would correspond to the altitude range below about 30 km .
This is exactly the same formulation like mine and that’s why I have written in the first post that the fact that CO2 doesn’t heat N2 in LTE is nothing new but text book knowledge .

John Whitman
September 1, 2010 6:41 am

R Stevenson says:
August 31, 2010 at 10:41 am
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. . . [edit]. . .

—————–
R Stevenson,
Yes, I think Tom Vonk is asking the right questions. More importantly he is discussing the possible answer with scientific concepts that are well known.
Keep on stimulating us Tom Vonk.
John

Dave Springer
September 1, 2010 7:49 am

@Vonk

E.g “The CO2 emits no infrared radiation because it heats the N2 (atmosphere) by collisions isntead” .
This statement is simply wrong and if one does physics , one should not accept wrong statements .

Everything with a temperature above absolute zero emits radiation. Even at temperatures up to that of the surface of the sun (around 10,000F) there is substantial longwave radiation in the emission.
This is high school physics, Tom. I’m certainly glad you can spot the errors that a bright high school student would spot. You seem to lose the plot shortly beyond that.

Dave Springer
September 1, 2010 8:02 am

@Tom Vonk
“One quote for those who still drone that LTE doesn’t exist and is unimportant anyway :”
LTE exists only as an idealized state and doesn’t exist in the actual universe unless you can perfectly isolate the system in question from the rest of the universe. This is also “textbook” physics – a basic law of thermodynamics. Your sophomoric knowledge is exposed when you start talking about it as being relevant to the earth’s atmosphere which is certainly NOT anywhere near isolated from the rest of the universe.
Your use of words like “drone” encourages hostility towards you, by the way. If anyone is droning it is you.

Dave Springer
September 1, 2010 8:36 am

@Vonk
“Obviously, it is necessary for the atmospheric scientist to understand when LTE applies and when it does not.”
Obviously. Your quote is lifted (out of context) from page 1 of
Non-LTE radiative transfer in the atmosphere
By Manuel López-Puertas, F. W. Taylor
It’s considered bad form to not cite your source when making a direct quotation, by the way.
If we skip ahead to page 162 in the same book we find:
“Very abundant molecules like N2 and O2, which are not themselves infrared active, but which are very important in thermalizing the energy of the vibrational levels and in exchanging quanta with CO2 through V-V collisions, must also be included, of course, since otherwise the model will calculate the wrong populations for CO2.”
Hoist by your own petard, Tom.

Michael J. Dunn
September 1, 2010 8:36 am

Nick Stokes says:
August 31, 2010 at 2:49 pm
Tom,
Do you think that ozone, by absorbing UV, heats the stratosphere?
Actually, ozone heats the stratosphere by energetically decomposing into oxygen. Oxygen is dissociated by UV photons, and recombines into ozone. Thus the cycle continues. But it doesn’t heat the lower atmosphere. The stratosphere below the ozone layer is colder. (The temperature profile of the atmosphere is worth pondering. It does funny things.)

Oliver Ramsay
September 1, 2010 8:37 am

You guys trash-talking aside, it doesn’t come over as droning; more like sniping.
Just don’t forget that there are thousands of people reading and hoping for some enlightenment.
Not that a little animus doesn’t provide some entertainment.

MartinGAtkins
September 1, 2010 9:04 am

LazyTeenager says:
September 1, 2010 at 5:35 am
The lower atmosphere has to heated somehow. At ground level this is a mixture of conduction (small),
Yes but time is notionally infinite.
IR radiation (medium)
Really? Would you like to quantify medium.
and, evaporated water(large).
Evaporation neither adds or subtracts from the systems heat content.
If we substitute CO2 with H2O in the above argument it is also proved that H2O does not heat the air either. Hence nothing can heat the air.
I have absolutely no idea how you can construct such a ridiculous conclusion from even your simplistic and at times wrong observations.
Reductio ad absurdum
Indeed.

Dave Springer
September 1, 2010 9:05 am

@Vonk
re:
Non-LTE radiative transfer in the atmosphere
By Manuel López-Puertas, F. W. Taylor
Since this book about non-LTE radiative transfer the discussion of LTE in the troposphere is in part explaining how satellites can measure the temperature at various levels in the troposphere through radiative emission. The satellite models must of course presume some semblance of LTE in the column of air they are looking at. Some semblance of LTE obviously is required to get an instantaneous measure of sensible heat. But anyone with a lick of common sense knows that the troposphere is a very busy layer of the atmosphere with all kinds of monkey wrenches keeping it out of thermal equilibrium even though at any one instant in time in any one local volume LTE is approximated well enough to record a temperature. The problem is that these local volumes are not quite in LTE since LTE is an idealized that doesn’t exist in nature and when you start adding up the small differences between small local volumes we find that the larger volumes fall farther and farther from LTE such that the temperature at the surface might be 30C at on point and a mile away where there’s a thunderstorm the temperature is 20C or where the surface temperature is 30C and the temperature at 1 kilometer (in clear air only) would be anywhere from 20C (dry adiabatic lapse rate) to 25C (saturated adiabatic lapse rate). At altitudes still well within the troposphere of 5 kilometers we’d be seeing sensible temperatures far below freezing even though it’s a balmy 30C on the surface.
The problem is that anyone physicist who can call a volume air with a temperature difference inside that volume of 50C in “LTE” is being utterly ridiculous.

September 1, 2010 9:09 am

This is exactly the same formulation like mine and that’s why I have written in the first post that the fact that CO2 doesn’t heat N2 in LTE is nothing new but text book knowledge .
You keep repeating the same meaningless stuff over and over again. No one’s interested in in a system in LTE with no energy flux through it. A mixture of N2 and CO2 with a flux of 15micron radiation through it and in LTE is at that equilibrium temperature because of the presence of the CO2, replace the CO2 and the LTE temperature will be lower. Increase the concentration of the CO2 and the LTE temperature will increase. If we consider a sub-nanosecond pulse of 15micron radiation added to a mix of N2 and CO2 in LTE we would see that the concentration of CO2* would increase above the M-B value, the system is now out of LTE. The concentration will rapidly achieve a new LTE at a slightly higher temperature and M-B will be satisfied again. In a real system there will be some form of heat loss and the original LTE will be reached. With a continuous energy flux the second LTE temperature will be maintained. That’s the real world, can we stop the constant rehashing of the meaningless LTE abstraction?

Dave Springer
September 1, 2010 9:14 am

@Vonk
Sorry for accusing you of not citing your source. I see that you did. I easily found it without the link you provided by googling a short phrase in the quote which was what I was focused on to the exclusion of the snippet above it which contained the reference.
Once again, you have my apology for that.

September 1, 2010 9:55 am

TomVonk says:
September 1, 2010 at 6:39 am
“The second red herring is radiative equilibrium .
It is not necessary to have radiative equilibrium in order to have LTE . The LTE is a property or material particles , not one of radiation . That is why , I do NOT assume radiative equilibrium and do not need to do so .
Anybody wanting to say that radiative equilibrium is necessary for LTE ?”
Radiative equilibrium is necessary for LTE. You’ve had this explained to you many times over, yet you keep ignoring the basic physics. First, LTE is not, in fact, merely “a property of material particles”, but of any and all mutually interacting degrees of freedom. Second, “material particles” in a radiation field are being excited by it and radiating to it; they will not have the same temperatures they would if the field were different. Third, if the system contains any degree of freedom that interacts differently from any other degree of freedom with any component of the radiation field, then the energy distribution between those degrees of freedom will depart to a greater or lesser extent from equipartition; they will have different degrees of excitation, different thermodynamic temperatures. They will not be mutually in LTE.
“If yes , then I want to hear it and especially the argument explaining why the Maxwell Boltzmann distribution can’t be derived if the radiative equilibrium is not established .”
I don’t recall anyone saying this. I have pointed out repeatedly that one can have multiple different “Maxwell Boltzmann distribution[s]” (different thermodynamic temperatures) within the same volume. In general, the different molecular species in a gas (and their vibrational and rotational modes) will all have slightly different (or, occasionally, very different) temperatures, and will therefore transfer heat between them. Some will heat the others, some will cool them.
“To understand the difference between LTE and non lTE conditions I strongly suggest reading …”
I strongly suggest that you stop pulling out of context chunks from works you don’t understand, and think about the underlying physical mechanisms instead. You might also try paying attention to what the physicists here are telling you.
From your quote: “Obviously, it is necessary for the atmospheric scientist to understand when LTE applies and when it does not” (my emphasis). LTE applies pretty well for calculations of bulk mechanical properties throughout most of the lower atmosphere. LTE does not apply to calculations of radiative heating and cooling, or other non-equilibrium phenomena, because it is precisely the departures from LTE that are significant there.

kwik
September 1, 2010 9:57 am

John Whitman says:
September 1, 2010 at 6:41 am
“Keep on stimulating us Tom Vonk.”
That was my response, too. Where are all the trolls?

John Whitman
September 1, 2010 10:07 am

Dave Springer says:
September 1, 2010 at 8:36 am
@Vonk
If we skip ahead to page 162 in the same book we find:
“Very abundant molecules like N2 and O2, which are not themselves infrared active, but which are very important in thermalizing the energy of the vibrational levels and in exchanging quanta with CO2 through V-V collisions, must also be included, of course, since otherwise the model will calculate the wrong populations for CO2.”
Hoist by your own petard, Tom.

————————-
Dave Springer,
Your above quote was from http://www.isbnlib.com/preview/9810245661/Non-LTE-Radiative-Transfer-in-the-Atmosphere-Series-on-Atmospheric-Ocean-and-Pla “Chapter 6 – Non-LTE Modeling of the Earth’s Atmosphere I: CO2”. The section is about Non-LTE Modeling. Your quoted paragraph comes from “Section 6.3.1 – Adoption of a reference atmosphere”. Your quote was excerpted from a discussion about the necessity of picking an appropriate reference atmosphere for a Non-LTE model in order for model to give the sought after info.
Please explain how then your claim of any “petard hoisting” has merit.
John

Djozar
September 1, 2010 10:14 am

Tom Vonk,
I’ve read this three times now and I think I understand, thanks to all the bloggers that have responded.
My only additional comment is that when apparently professional scientists like yourself and others on this blog can’t come to a consensus, how can the IPCC calim consensus on the larger picture?

NoAstronomer
September 1, 2010 10:21 am

Paul Birch said :

“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).”

Thanks for writing that so I didn’t have to. This whole article is an embarrassment. Does WUWT have a peer review process for articles? If not then it should.

Spector
September 1, 2010 10:39 am

RE: TomVonk: (September 1, 2010 at 6:39 am)
“This is exactly the same formulation like mine and that’s why I have written in the first post that the fact that CO2 doesn’t heat N2 in LTE is nothing new but text book knowledge.”
If this statement is meant to say that CO2 cannot manufacture or, on its own, be a net source of heat, then it is obviously true. It is a simple consequence of the law of conservation of energy.
This principle cannot be used, however, to imply that CO2 molecules are not able to receive externally generated radiant energy and transfer that energy as heat to local nitrogen or oxygen molecules. If LTE precludes the continual occurrence such external radiative transfers then LTE is not obtained.

Frank
September 1, 2010 10:41 am

Tom wrote: “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.”
Why exclude critical posts discussing radiative transfer? If you don’t consider radiative transfer, you aren’t discussing physics that is relevant to climate! Of course, CO2 doesn’t warm air in LTE when you don’t allow energy to enter or leave the system! That is just another statement of the law of conservation of energy. The real world in not in equilibrium – or anything close to equilibrium.
Infrared photons are absorbed and emitted by any parcel of air containing GHGs. Those photons come from places that are usually warmer or cooler, so absorption and emission are not in equilibrium. There must be a net upward flow of energy averaging 235 W/m2 through the atmosphere to compensate for the energy the sun delivers to the surface of the earth and lower atmosphere. The only way that energy leaves the earth for space is via infrared radiation. Different forms of energy are flowing in opposite directions because the earth is not in equilibrium.

Buffoon
September 1, 2010 10:57 am

Tom Vonk:
Your “statement to analyze” is:
“Given a gas mixture of CO₂and N₂in Local Thermodynamic Equilibrium (LTE) and submitted to infrared radiation, does the CO₂ heat the N₂?”
However, I think perhaps, this question is also acceptable:
Given a gas mixture of CO₂and N₂submitted to infrared radiation, does the CO₂ heat the N₂?
So.. Given a mixture of said gases at temperature X, exposed to high energy infrared radiation at 15um, can the mixture reach a temperature Y where Y > X?
Simple experiment. Mixture and source.
I say yes.
Disprove please.

Buffoon
September 1, 2010 11:10 am

“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.”
You mean, by suspecting that your whole argument should yield to the law of conservation of energy?
Let me try this another way:
deltaT = 0 iff Photonsin = Photonsout

Bryan
September 1, 2010 11:18 am

The response to the article shows the wide level of interest in the mechanisms of the molecules making up the atmosphere.
How they respond to radiation and how they interact as the temperature changes.
There seems to be contradictory ideas about what goes on.
Perhaps Merrick can be persuaded to do an extended article covering broadly the same ground as the Tom Vonk post.

George E. Smith
September 1, 2010 11:48 am

“”” Spector says:
September 1, 2010 at 4:26 am
RE: George E. Smith: (August 31, 2010 at 4:24 pm)
“And in the lower atmospehre the mean lifetime of the CO2 excited states is longer than the mean time between collisions; so that energy becomes thermalized before a spontaneous photon emission from the CO2 can occur. In the higher less dense regions in the stratospehre the time between collisins is great enough for spontaneous decay to occur.” “””
I should point out Spector, that “Phil” has made this point on several occasions; to the extent that it has even sunk into my thick skull.
When I started thinking about these climate/weather Physics issues some years ago; I was under the mistaken impression, that the 15 micron band photons that say CO2 captured; were subsequently re-emitted by the molecule; but in a now isotropic radiation pattern so at least half of it propagated upwards; but the escape was delayed by a multiple cascade of such photon absorption/emission events.
Once I realized that this could only occur in the rarified upper reaches of the atmosphere; where because of that rarified atmosphere the total energy involved was considerably reduced; it then started to make sense that the CO2 was acting almost as a catalyst but in a physical process, that conveyed radiant energy from the surface to the main gases of the atmosphere to warm them.
The distinction is important, because spontaneous emission from the CO2 would of course be in the form of a line spectrum, somewhere in the 15 micron band (and maybe other bands).
But if instead that captured energy simply raises the atmospheric temperature; it is easy to see that te resultant LWIR emission should be a thermal (BB like) spectrum that reflected the Temperature of the ordinary atmospheric gases; so that spectrum should be quite independent of the GHG species that caused it; whether CO2 or H2O or anything else
Some people still like to believe that gases do not emit Planck type thermal radiation; which flies in the face of the edict that any body above zero Kelvins should emit thermal radiation; and the sun certainly doesn’t mind doing that.
But you might be right, on your point that there should be an altitude at which the transition from primarily spontaneous decay takes over from thermalization and BB like spectral emission, and it maybe a region of atmospheric discontinuity.
We have to be mindful of the fact that the LWIR radiant emission from the atmospheric gases; whether N2, O2 or CO2 etc is pretty much the same as the LWIR radiant emission from an ordinary brick just sitting around at a Temperature of perhaps 288 K.
This radiation is not sensed by any human sensory organs; we do not detect it as “heat”; we don’t detect it at all, and in fact one has to jump through hoops to detect it with instrumentation designed for that purpose.
The lay public has visions of CO2 in the atmosphere creating a blow torch of heat that makes us all feel hot. Which is why I scoff at these laboratory or TV demonstrations of how an ordinary light bulb heats air containing increased CO2. They need to start using a cold brick for their LWIR radiation source.
Well that ordinary light bulb may have a color temperature of 2700-2800 K, and it is emitting a spectrum that is full of a lot of near one micron wavelength radiation that we sense as heat in ordinary sunlight; in fact the peak of the emission may be at about one micron for that temperature. Incidently the gas filling that ordinary light bulb is what is that big yellow glowing mass in there; a mass of gas, that of course cannot emit a thermal radiation spectrum like solids and liquids can. Well somehow the gas doesn’t know it is not supposed to do that.

Merrick
September 1, 2010 11:56 am

Tom says:
“Merrick quite nicely resumed the purpose here :Yes, of course, a system in thermodynamic equilibrium (local or otherwise) cannot possibly have net energy flow – that violates the initial statement. When used in the physical sciences, equilibrium is defined as that state of a system in which an external force is required to change the state of the system. Posit BOTH equilibrium AND no external forcing (remember – no radiative coupling by Tom’s stipulation) and the argument is rock solid. By definition.”
Tom. I’m actually flattered that you selected my comment in this way. And as far as that statement you quoted there is absolutely NO doubt that we couldn’t possibly be in greater agreement. I think if you want that to be the total sum of your effort on this point and announce it clearly, perhaps in another (very short) article, that would be great and almost everyone would agree.
My disagreement, and I’ll repeat it here one more time for clarity, is that one cannot extrapolate from that statement (one that is as true for earth’s atmosphere as it is for the vacuum of space and the core of the earth because it is a literal truism) regarding a system in equilibrium to a system which clearly is not in equilibrium. You make first this leap (or miss the point that the atmosphere is seldom in equilibrium with the IR radiation of earth’s surface) then take a yet further leap to get to what I believe to be a remarkable non sequitor: “CO2 *cannot* heat the atmosphere.”
When the temperature of the earth’s surface and the atmosphere above it are not the same then the system is in disequilibrium. That condition occurs over the vast majority of the earth’s surface for nearly 24 hours of every day. That is absolutely true. Please look at a daily temperature profile for just about any place on the planet and I believe you will quickly be convinced.
THAT is the bone of contention that I have been belaboring and the contention I believe I understand in each of the various comments I’ve read.

mircea
September 1, 2010 11:57 am

Nylo says: August 31, 2010 at 7:33 pm (sorry for late answer, I didn’t see it until now)
“That theory is wrong. It has a wrong point. It says that “With H2O vapours raising higher concentration of H2O vapours in the column of air will decrease (same quantity in higher volume)”. No, they won’t. Because as they start to decrease near the surface, more moisture comes from the ocean. And you do not need any warming in the ocean to achieve this. ”
You have a good point, but if the concentration of H2O would increase back to the initial value then the temperature would increase and the water would expand and then, as per your logic, more water would get in the atmosphere which would heat more and so on until all the water would be in the atmosphere. A run off.
The concentration and distribution of the H2O vapours is a result of temperatures distribution and air pressures. Not the other way around.
The fact that H2O vapours decrease when CO2 increases it’s measured and one can see the trend from 1961 to present day.
Does this makes sense?
Best Regards!

tmtisfree
September 1, 2010 12:24 pm

Hum. Remind me of this one:
Climatologist: I have a system of undetermined complexity and undetermined composition, floating and spinning in space. It has a few internal but steady state and minor energy sources. An external energy source radiates 1365 watts per meter squared at it on a constant basis. What will happen?
Physicist: The system will arrive at a steady state temperature which radiates heat to space that equals the total of the energy inputs. Complexity of the system being unknown, and the body spinning in space versus the radiated energy source, there will be cyclic variations in temperature, but the long term average will not change.
Climatologist: Well what if I change the composition of the system?
Physicist: see above.
Climatologist: Perhaps you don’t understand my question. The system has an unknown quantity of CO2 in the atmosphere that absorbs energy in the same spectrum as the system is radiating. There are also quantities of carbon and oxygen that are combining to create more CO2 which absorbs more energy. Would this not raise the temperature of the system?
Physicist: there would be a temporary fluctuation in temperature caused by changes in how energy flows through the system, but for the long term average… see above.
Climatologist: But the CO2 would cause a small rise in temperature, which even if it was temporary would cause a huge rise in water vapour which would absorb even more of the energy being radiated by the system. This would have to raise the temperature of the system.
Physicist: there would be a temporary fluctuation in the temperature caused by changes in how energy flows through the system, but for the long term average… see above.
Climatologist: That can’t be true. I’ve been measuring temperature at thousands of points in the system and the average is rising.
Physicist: The temperature rise you observe can be due to one of two factors. It may be due to a cyclic variation that has not completed, or it could be due to the changes you alluded to earlier resulting in a redistribution of energy in the system that affects the measurement points more than the system as a whole. Unless the energy inputs have changed, the long term temperature average would be… see above.
Climatologist: AHA! All that burning of fossil fuel is releasing energy that was stored millions of years ago, you cannot deny that this would increase temperature.
Physicist: Is it more than 0.01% of what the energy source shining on the planet is?
Climatologist: Uhm… no.
Physicist: rounding error. For the long term temperature of the planet… see above.
Climatologist: Methane! Methane absorbs even more than CO2.
Physicist: see above.
Climatologist: Clouds! Clouds would retain more energy!
Physicist: see above.
Climatologist: Ice! If a fluctuation in temperature melted all the ice less energy would be reflected into space and would instead be absorbed into the system, raising the temperature. Ha!
Physicist: The ice you are pointing at is mostly at the poles where the inclination of the radiant energy source is so sharp that there isn’t much energy to absorb anyway. But what little there is would certainly go into the surface the ice used to cover, raising its temperature. That would reduce the temperature differential between equator and poles which would slow down convection processes that move energy from hot places to cold places. The result would be increased radiance from the planet that would exceed energy input until the planet cooled down enough to start forming ice again. As I said before, the change to the system that you propose could well result in redistribution of energy flows, and in short term temperature fluctuations, but as for the long term average temperature…. see above.
Climatologist: Blasphemer! Unbeliever! The temperature HAS to rise! I have reports! I have measurements! I have computer simulations! I have committees! United Nations committees! Grant money! Billions and billions and billions! I CAN’T be wrong, I will never explain it! Billions! and the carbon trading! Trillions in carbon trading!
Physicist: how much grant money?
Climatologist: Billions. Want some?
Physicist: Uhm…
Climatologist: BILLIONS
Climatologist: Hi. I used to be a physicist. When I started to understand the danger the world was in though, I decided to do the right thing and become a climatologist. Let me explain the greenhouse effect to you…

tmtisfree
September 1, 2010 12:33 pm

Humour found at http://knowledgedrift.wordpress.com, btw.

Merrick
September 1, 2010 12:51 pm

And then we see interesting statements from the supposedly “expert” literature that is a major underlying source. I admit to never having seen the book and I don’t know the authors from Adam – but to quote from above:
“Very abundant molecules like N2 and O2, which are not themselves infrared active, but which are very important in thermalizing the energy of the vibrational levels and in exchanging quanta with CO2 through V-V collisions, must also be included, of course, since otherwise the model will calculate the wrong populations for CO2.”
The first excited state of CO2 results from an approximately 15 micron transition, as we have discussed repeatedly. The above quote is correct that, as homonuclear diatomics, N2 and O2 are IR inactive. But, unfortunately for the authors, the first excited state of O2 has a transition energy roughly eqiuvalent to 6 micron radiation (somebody please correct that if I’m wrong). That means that it takes more than 2 (i.e., 3) vibrational quanta of the CO2 fundamental to populate the O2 fundamental (3 CO2 quanta ~ 5 micron). This CANNOT happen via V/V transitions between CO2 and O2. N2 is both a lighter molecule and has a stronger bond, so that it’s vibrational fundamental is higher yet. I’m not suggesting that no O2 or N2 molecules get vibrationally excited at normal tropospheric conditions, and an excited O2 or N2 molecule colliding with CO2 would be energetically capable of exciting a CO2 vibration (possibly a non-radiative one!) but the reverse simply couldn’t happen without a lot of other preconditions. So, no, V/V collisions involving O2 and N2 are NOTimportant contributors to establishing thermal equilibrium in the atmosphere at normal tropospheric conditions. At those temperatures and given the atmospheric composition, process which work toward equilibrium are dominated by translational and rotational modes of N2 and O2,T/R/V modes of H2O and CO2, and T modes of Ar.
I’m not remotely interested inlaunching off on another topic that apparently requires teaching – I’m on vacation! – but among the things that determines the probably that a quanta will be transferred between two species when they collide is if the lower energy molecule has an available energetic state with a very similar transition energy. The next is that the change of the quantum number for both molecules is relatively small. So, for instance, a molecule typically has vibrational energy spacings approximately 100 times more energetic than rotational spacings. For a vibrationally excited CO2 molecule to transfer it’s quanta to rotational quanta in another CO2 molecule through collision the change in quantum number for one of the molecules is on the order of 100: that’s a low-probability transition. However, if that vibrationally excited CO2 molecule collides with an N2 molecules, with rotational energy spacings about an order of magnitude larger than for CO2, then the change in rotational quantum number for N2 would only be about 10: a much more likely occurence. Now, if that N2 molecule could immediately recollide with a non-excited CO2 molecule it would at least have the possibility of putting the energy “back.” But, since that N2 molecule is >99.9% more likely to collide with non-CO2 molecules at any given point in time, that energy (heat!) passed to the N2 molecule is going to be passed to all of the other N2 and O2 rotation and translations, etc., and more and more CO2 vibrational quanta funnelled into the system before the N2 population is sufficiently energetic to have either a significant fraction in highly rotationally excited states or vibrationally excited states to start significantly driving energy back into CO2 vibrations. The above quopted statement is not consistent with the facts and that’s a little more daylight in T/R/V dynamics.

Volt Aire
September 1, 2010 1:08 pm

@Vonk:
LTE = everything is the same temperature
Your argument: When everything is the same temperature in a closed system, nothing is heating anything else? Circular argumenting don’t you agree?
About the way the heat is transferred in LTE and the rest of the universe:
CO2 will always transfer heat via radiation and collision when above 0K. All matter does. No claiming it is only one or the other.
Interestingly, in LTE the CO2 would actually still heat the N, those being the CO2 molecules on the active state in your graph. There would be a similar amount of cooling happening to N in the other end of the CO2 activity spectre. Result: CO2 heats N in the LTE on a molecular level… and cools it too. Since this is all about heating I will make the claim that CO2 heats N indefinitely in LTE. An endless stream of energy from CO2 to N. (I hope everyone understands that there is also an endless stream of energy to the other direction…)
If you don’t disprove the above your argument is not valid so I’m sure that I’ll see a response when I get to work tomorrow, that being only 9hrs away now.. 🙁

R Stevenson
September 1, 2010 1:29 pm

Mircea,
The concentration of H2O in air capable of absorbing IR photons is 100 times greater than CO2 ( 380ppm) and the two are largely independent, CO2 is a permanent gas H2O is vapour. Increasing CO2 in air doesn’t decrease H2O. CO2 absorbs IR in three bands 2.36 to 3.02 microns, 4.01 to 4.8 microns and 12 to 16.5 microns; H2O also in three bands 2.24 to 3.27 microns, 4.8 to 8.5 microns and 12 to 25 microns. When the two appear together their emissivities are added (H2O is by far the larger) with a correction because both bands overlap. My question concerned traverse of black-body radiation from Earth through the atmosphere. Increasing CO2 would mean a shorter traverse to take out all the absorbable IR but no increase in total heat absorbed. However I could envisage that photons subsequently released would have a longer traverse before subsequently escaping ie a potentially longer retention time. My view is that convective mixing within the atmosphere and the dominance of water vapour would mean retention time is not significant.

mircea
September 1, 2010 1:44 pm

Nylo says:August 31, 2010 at 7:33 pm (sorry for the delay, I didn’t see it until now)
“That theory is wrong. It has a wrong point. It says that “With H2O vapours raising higher concentration of H2O vapours in the column of air will decrease (same quantity in higher volume)”. No, they won’t. Because as they start to decrease near the surface, more moisture comes from the ocean. ”
But it doesn’t decrease near the surface of the ocean (or it is totally negligible compared with the normal variations of humidity near the ocean). The lapses are almost identical close to the sea/land surface. The variation in H2O vapour happens in the upper troposphere.
I have a plane to catch but I will write more about this here and at my blog tomorrow.

September 1, 2010 2:35 pm

Merrick says:
September 1, 2010 at 12:51 pm
“… unfortunately for the authors, the first excited state of O2 has a transition energy roughly eqiuvalent to 6 micron radiation (somebody please correct that if I’m wrong). That means that it takes more than 2 (i.e., 3) vibrational quanta of the CO2 fundamental to populate the O2 fundamental (3 CO2 quanta ~ 5 micron). This CANNOT happen via V/V transitions between CO2 and O2. …
… among the things that determines the probably that a quanta will be transferred between two species when they collide is if the lower energy molecule has an available energetic state with a very similar transition energy.”
What I think you’re missing here is that these transitions are not simply quantised in a gas of finite density and temperature. They are broadened by the velocity distribution of the molecules; so they have a finite amplitude over the entire spectrum. A vibrationally excited CO2 molecule can quite easily vibrationally excite an O2 molecule, if it collides with a suitable velocity, which at atmospheric temperatures it often will. All of these exchanges occur with significant probability; they all contribute (in varying amounts) to the absorption, re-radiation and thermalisation processes.

George E. Smith
September 1, 2010 2:57 pm

“”” R Stevenson says:
September 1, 2010 at 1:29 pm
Mircea,
The concentration of H2O in air capable of absorbing IR photons is 100 times greater than CO2 ( 380ppm) and the two are largely independent, CO2 is a permanent gas H2O is vapour. Increasing CO2 in air doesn’t decrease H2O. CO2 absorbs IR in three bands 2.36 to 3.02 microns, 4.01 to 4.8 microns and 12 to 16.5 microns; H2O also in three bands 2.24 to 3.27 microns, 4.8 to 8.5 microns and 12 to 25 microns. When the two appear together their emissivities are added (H2O is by far the larger) with a correction because both bands overlap. “””
Well I don’t know who said what above since I can’t locate the originals; but I hate to point out that BOTH H2O and CO2 are PERMANENT gas components of the atmospehre. It is true that H2O is regarded as a VAPOR; but only when it is inequilibrium with either the liquid or solid phase of H2O. That is often true over the oceans; but over some deserts there can be plenty of H2O gas in the atmospehre (more than the CO2, with no liquid or solid water around for it to be in equilibrium with.
And if there were, it would only be in the vicinity of the phase interface; that we would properly regard H2O as a vapor. CO2 is also a vapor; just at a different Temperature range.
Bottom line is that an H2O molecule in the atmosphere at say 1000 metre’s from the surface; has no knowledge whatsoever regarding the presence or absence of any other H2O molecule 1km below; so there is no way that the H2O molecule would know to change hats from its gas hat to its vapor hat. And it behaves as an H2O molecule no matter which hat it is wearing.
True at some heights there could be clouds consisting of either liquid or solid H2O, and close to those, H2O gas could revert to its vapor behavior; which isn’t any different from its Ga behavior.
Why the AGW cloud keeps insisting that at times the atmosphere is devoid of H2O, whcih is therfore not a GHG is beyond me; In the part of the troposphere where people and things live and weather happens; there is always more H2O molecules that CO2 molecules; and it has pretty much always been that way for as long as it matters.
And some people might be surprised to learn that H2O starts to absorb EM radiation at least as short as 760 nm which is well inside the solar spectrum where much solar energy resides and is capable of being absorbed by H2O; but NOT CO2. H2O can absorb perhaps as much as 20% of the air mass one incoming solar spectrum at sea level. Only 1% of the total solar spectrum energy persists at longer than about 4 microns; and that 4 micron CO2 band has virtually no effect on the surface emitted LWIR spectrum, which peaks at about 10.1 microns (for a 288 K source Temeprature; and only 1% of that emission is at wavelengths as short as 5.05 microns; so the 4 micron CO2 band is rather inactive in any weather or climate issue on planet earth; although it might serve some function on Venus; but who cares ?

sky
September 1, 2010 3:09 pm

MartinGAtkins says:
September 1, 2010 at 9:04 am
“Evaporation neither adds or subtracts from the systems heat content.”
While this may be true for the planetary climate system, evaporation always subtracts (589 calories per gram at 288K) from the ocean heat content, which consequently doesn’t respond to radiative forcing as a blackbody.

sky
September 1, 2010 3:19 pm

Merrick says:
September 1, 2010 at 12:51 pm
The explanation that you provide is indeed consistent with the observed facts, including the fairly rapid extinction of 15 micron radiation with altitude. While there is widespread physical misunderstanding of how the climate system operates, Tom Vonk’s thesis is no narrowly constrained that it’s largely of pedagogical, rather than practical, interest.

wayne
September 1, 2010 3:21 pm

tmtisfree: September 1, 2010 at 12:24 pm
LOL! Simply great!

bob
September 1, 2010 6:46 pm

Since Tom has responded and is in defense of the condition of Local Thermodynamic Equilibrium, I would like to ask a few questions to clarify my understanding of the initial conditions of this thought experiment.
Initial Conditions
““Given a gas mixture of CO₂ and N₂ in Local Thermodynamic Equilibrium (LTE) and submitted to infrared radiation, does the CO₂ heat the N₂?””
In this initial condition, what is keeping the N2 gas at temperature as it is radiating away its energy?
What happens when the mixture of the two gases comes under infared radiation?
The CO2 can absorb the radiation and thus its energy increases and thus its temperature increases.
The Nitrogen since it is opaque to the infared does not increase in temperature, and is decreasing in temperature as it is radiating in the infared.
You now have two gases in close contact and one is cooling and one is heating and what are we concluding is happening?
The nitrogen is not being heated by the CO2?
I’ll leave it to the reader as an excercise as to how this violates the first law of thermodynamics.
Increasing the concentration of CO2 in the atmosphere has to raise the temperature or violate the first law of thermodynamics, it’s as simple as that.

Jim
September 1, 2010 7:53 pm

I found the following statement extraordinary
“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. ”
OK, I have a Ph.D. in physics. All I can say is posts like this degrade the
value of WUWT. One experiences relativity crackpots, and quantum
mechanics crackpots, and the whole global warming thing has now resulted
in more thermodynamics crackpots. Dealing with stuff like this is essentially
a waste of time. Energy transfer in two component gases is relatively
well known, and is of some importance in one of the more topical areas of
physics, namely cold-atom physics.

Merrick
September 1, 2010 7:56 pm

Paul Birch, I’m sorry but that is silly. Pressure and temperature broadening at typical tropospheric conditions is on the scale of rotational energy spacings for moderately large molecules (and NOT on the scale of rotational spacings of small molecules, which is why we CAN measure rotationally resolved spectra of small molecules in the atmosphere).That is: on the order of a wavenumber or less. The CO2 vibrational energy is on the order of 650 wavenumbers. The O2 vibrational energy is almost exactly 1000 wavenumbers higher, or about 1650 wavenumbers. The N2 vibrational energy is even higher than that.
Please now explain to me how <1 wavenumber of spectral broadening bridges a 1000 wavenumber gap in V/V collisional transfer from CO2 to O2 (or larger to N2).
And, oh by the way, only the pressure broadening impacts the internal states of the system. The Doppler broadening only impacts your ability to *measure* the energy in the lab frame. So the true broadening effecting on the system in the way you suggest is even a smaller fraction of the already vague <1 wavenumber broadening we spectrally observe.

Merrick
September 1, 2010 8:08 pm

Real Physicist: OK, now I paint the surface of the planet black. What happens to the equilibrium temperature of the planet?
Armchair Physicist:It gets warmer.
Real Physicist: But it’s long been know that most black paints can only be relied upon to absorb strongly in the visible region and are often highly reflective in the IR.
Armchair Physicist: But it’s also well known that cars with black interiors get much hotter in direct sunlight than cars with lighter interiors. The point is that the black coloration absorbs light where it matters: in the visible peak of the solar radiation. So, we conclude, the planet will get warmer.
Real Physicist: Adding CO2 to the atmosphere darkens the atmosphere in *exactly* the portion of the relavent spectrum that matters: near the IR peak in terrestrial gray body radiation.
Armchair Physicist: ?

Oliver Ramsay
September 1, 2010 8:41 pm

Jim says:
September 1, 2010 at 7:53 pm
…OK, I have a Ph.D. in physics. All I can say is posts like this degrade the
value of WUWT. One experiences relativity crackpots, and quantum
mechanics crackpots, and the whole global warming thing has now resulted
in more thermodynamics crackpots. Dealing with stuff like this is essentially
a waste of time.
——
Two possible solutions to that problem:
Either all the contributors and lurkers abandon the thread and find something else to do
or
you do

Spector
September 1, 2010 10:26 pm

RE: George E. Smith says: (September 1, 2010 at 11:48 am)
“Once I realized that this could only occur in the rarified upper reaches of the atmosphere; where because of that rarified atmosphere the total energy involved was considerably reduced; it then started to make sense that the CO2 was acting almost as a catalyst but in a physical process, that conveyed radiant energy from the surface to the main gases of the atmosphere to warm them.”
My current view, so far uncontested, is there must be at altitude at which half the photons emitted by the ‘earthshine’ emitting/absorbing gases going straight up escape to outer space without any further interference in the thin atmosphere above. Going above this altitude, I would expect to see a rapidly widening cone defined by the vertical offset angle of the fifty percent escape path in the exponentially thinning atmosphere.
I presume the tropopause altitude is the best candidate for this point as this is the level where convection stops being an effective heat transfer mechanism and direct radiation to outer space is the only process that can expel the heat that has been wafted up to the coldest layer in the atmosphere.
A statement that you made caused me to wonder if the increasing mean time between collisions might be another factor causing an increased emission rate from the tropopause, but I have since been informed that this can only be a minor factor at the tropopause level.
As regards this article, I know that there are people, even AGW skeptics, who seem to have been convinced that CO2 and other ‘greenhouse’ gases are intrinsic sources of heat. If this is what the author is addressing when he says “The CO₂ does not heat the troposphere,” he is correct. For those of us who know this already, however, that point may seem so obvious that we automatically assume he is saying something far more revolutionary.

Bryan
September 2, 2010 12:51 am

Jim
…..”Energy transfer in two component gases is relatively
well known, and is of some importance in one of the more topical areas of
physics, namely cold-atom physics.”……
I was the first poster to query Tom Vonk on his definition of Kelvin Temperature and he has yet to respond.
This area is where classical Kinetic Theory leaves off and Quantum Mechanical effects start to become important
I have a degree in Physics yet I find your familiarity with atmospheric gas mixture conditions as shown in quote above difficult to achieve.
When I google likely word combinations the sources revealed are mostly behind pay walls.
When I contrast what is known about say a CO2 laser with CO2 effects in the atmosphere this is what I find
CO2 laser; precise details of operating conditions, wavelengths,gas mixture percentages(CO2,N2 and others).
The variations used to produce desired output power and the background science is agreed.
CO2 in the atmosphere; a complete spectrum of contrasting opinion with very little agreed science.
Yet the IPCC propose to dislocate the economy of the world on the basis of very shaky science.
Tom Vonk has put himself in the limelight and leaves himself open for folk to pick holes in his view of this area.
However if folk like yourself and Merrick stay in the sidelines and snipe without offering a more comprehensive view we are unlikely to move forward.

anna v
September 2, 2010 1:13 am

bob says:
September 1, 2010 at 6:46 pm
The CO2 can absorb the radiation and thus its energy increases and thus its temperature increases.
This statement is wrong if one is speaking of spectral lines of the CO2 molecule, not rotational or vibrational. When an atom/molecule is raised from the ground state by absorbing a photon, two things happen:
1) its mass increases by h*nu
2) its momentum changes by h*nu/c , by momentum conservation.
On average there will be a change in the velocity of the distribution but one needs to solve (mass of CO2)*vnew=(mass of CO2)vold +h*nu/c . The maximum corrected momentum vnew=vold +(h*nu/c)/(mass of CO2)
This correction would be very small.
So no, CO2 does not heat up because its maxwell distribution changes very little.
It does have a heat capacity though, in the sense that this energy it absorbed as one lump photon, is later cascaded down to soft photons .
I think all this discussion of photons absorbed and radiated is really a red herring as far as the thermodynamics of the atmospheric gas system is concerned. Classical thermodynamics should be enough to describe gases at normal pressures with modified gray body distributions for radiation.

Dave Springer
September 2, 2010 1:39 am

@anna v
“I think all this discussion of photons absorbed and radiated is really a red herring as far as the thermodynamics of the atmospheric gas system is concerned. Classical thermodynamics should be enough to describe gases at normal pressures with modified gray body distributions for radiation.”
Classical mechanics is not a should it’s a must when dealing with large numbers of molecules or even when dealing with electromagnetic radiation impinging on matter when the wavelength encompasses many molecules at once.
This is discussed in many places. Here’s a good place to start:
http://en.wikipedia.org/wiki/Correspondence_principle

The rules of quantum mechanics are highly successful in describing microscopic objects, atoms and elementary particles. But macroscopic systems like springs and capacitors are accurately described by classical theories like classical mechanics and classical electrodynamics. If quantum mechanics should be applicable to macroscopic objects there must be some limit in which quantum mechanics reduces to classical mechanics. Bohr’s correspondence principle demands that classical physics and quantum physics give the same answer when the systems become large.

The problem that happens over and over and over is that people attempt to describe macroscopic systems in quantum mechanical terms. It seldom works out. Not because QM doesn’t work though! It seldom works out because the macroscopic system descriptions are intractible in a quantum framework and due to the correspondence principle if the QM analysis doesn’t yield the same result as the classic statistical mechanics analysis then the QM analysis is buggered up not the other way around.
It was experimentally demonstrated by John Tyndal in 1859 that longwave radiation is absorbed by some gases in a mix while others are transparent and that the absorbing gases thermalize the others in the mix. One of his many publications “Heat a Mode of Motion” is great reading. A tour de force of experimental ingenuity leading to discovery and/or verification of basic thermodynamic principles.
You can read the whole book here:
http://books.google.com/books?id=m-TUAAAAMAAJ&pg=PR16&dq=heat+a+mode+of+motion&output=text

Volt Aire
September 2, 2010 1:40 am

Vonk has nothing to say about the obvious problems put forth in several posts regarding:
– Circular argumenting (no temp difference, no heat transfer)
– X-Y heating logic being mutually exclusive
– The fact that CO2 actually DOES heat N by radiation AND collisions above zero K, in all LTE:s, all real life and in all hypothetical situations on a molecular level, as per Bohlzmann distribution.
– You manage to prove that mixing CO2 and N doesn’t cause an exothermic reaction. This is pretty much known…
So the argument put forth is false in every way.

September 2, 2010 2:17 am

Merrick says:
September 1, 2010 at 7:56 pm
“Paul Birch, I’m sorry but that is silly. Pressure and temperature broadening at typical tropospheric conditions is on the scale of rotational energy spacings for moderately large molecules …”
You are confusing pressure and temperature broadening of spectral lines (which under typical atmospheric conditions are indeed comparatively weak) with the kinetic broadening of collisional transfers of energy between molecules (which is much more powerful, the kinetic energy of the molecules being of the same order of magnitude as the energies of the various vibrational modes).

Dave Springer
September 2, 2010 2:56 am

Spector says:
September 1, 2010 at 10:39 am
This principle cannot be used, however, to imply that CO2 molecules are not able to receive externally generated radiant energy and transfer that energy as heat to local nitrogen or oxygen molecules. If LTE precludes the continual occurrence such external radiative transfers then LTE is not obtained.

It really doesn’t matter if the quantum interactions move energy from CO2 to other species. The net result that matters is that when infrared impinges on a mix of gases where some gas species absorb infrared the total energy in the mixture increases and it will raise the temperature of a glass bulb containing mercury or colored alcohol (a thermometer) or the temperature of a thermocouple, or whatever other instrument you’re using to measure the bulk sensible heat in the gas mixture. This heating of the mixture due to infrared-absorbing gas species was experimentally demonstrated 150 years ago. That’s the bottom line. If anyone argues that isn’t the case through contorted jargon filled quantum mechanical descriptions then they are simply and demonstrably WRONG.
I have no idea at this point what Vonk is trying to argue but if he’s arguing that exposing standard atmosphere to infrared radiation won’t raise its sensible temperature then he is simply and demonstrably WRONG and needs to review basic thermodynamic principles that were discovered experimentally 150 years ago.

September 2, 2010 2:56 am

anna v says:
September 2, 2010 at 1:13 am
“So no, CO2 does not heat up because its maxwell distribution changes very little.”
The net increase in the kinetic energy of the absorbing molecules is very small (of order (v/c)^2) ~1e-12) so the immediate temperature increase (before any thermalisation) is also very small (~3e-10K), though not actually zero. However, where photons are absorbed by pairs of molecules as they collide, which from the point of view of an infra-red photon they are doing much of the time, the immediate temperature rise is much larger (of the same order as the final rise), because much of the energy goes directly into the translational motion of the molecules.
I agree with you that this is mostly a red herring, though. It is better to work with macroscopic opacities and leave the QM on one side.

Spector
September 2, 2010 3:30 am

RE: anna v: (September 2, 2010 at 1:13 am )
“This statement is wrong if one is speaking of spectral lines of the CO2 molecule, not rotational or vibrational.”
By ‘not rotational or vibrational’ are you referring to spectral lines that are collision induced artifacts? I assume the family of lines around 15 um is associated with the basic CO2 bending vibration, perhaps with various superimposed rotational states.

R Stevenson
September 2, 2010 4:40 am

20.4% of black-body radiation from Earth at 288 K is absorbed as LWIR by CO2 most of it (97%) in the 12.5 to 16.5 micron waveband. 63.3% is absorbed again as LWIR by H2O the majority or it (78%) in the 12 to 25 micron band. Simple addition of the two is slightly too large because the absorption bands overlap and a correction may be applied (Hottel, Heat Transmission by McAdams). Absorption is complete after a relatively short transit through the atmosphere particularly in the case of H2O. Doubling CO2 to 700 ppm or increasing H2O for that matter would not absorb more heat it would merely shorten the length of transit of absorbable radiation through the atmosphere.

anna v
September 2, 2010 5:15 am

Spector says:
September 2, 2010 at 3:30 am
RE: anna v: (September 2, 2010 at 1:13 am )
“This statement is wrong if one is speaking of spectral lines of the CO2 molecule, not rotational or vibrational.”
By ‘not rotational or vibrational’ are you referring to spectral lines that are collision induced artifacts? I assume the family of lines around 15 um is associated with the basic CO2 bending vibration, perhaps with various superimposed rotational states.

I am trying to separate degrees of freedom that might enter in the equipartition budget from the basic potential well QM solution.
It might not be necessary to do this.
I am trying to point out that quantum mechanically E=mc^2 is the rule of how energy counts and is absorbed, and not billiard ball scattering transferring kinetic energy and therefore raising temperatures of the molecule (CO2 or H2O)selectively.

anna v
September 2, 2010 5:17 am

p.s.
I am saying that an excited CO2 is at practically the same temperature as before absorbing the photon, it is just a bit more massive.

Bryan
September 2, 2010 6:39 am

Its an odd little area of Physics because it overlaps areas where the Kinetic Theory gives a good account and yet quantum mechanical effects cannot be ignored.
For the rotational and vibrational modes Hooke’s Law derivation gets you most of the way plus a little QM to complete the picture.
Kirchoff’s Law is not a big help because CO2 readily absorbs 15um, yet post thermalisation (statistically using Maxwell Boltzmann) gives about 5 emissions to 100 absorptions.
My guess (though I would like a confirmation from someone who works in this area ) is that the radiative route results in more low energy emissions >20um .
All the above is what I have gathered bit by bit and perhaps a lot of it is not accurate.
Undergraduate textbooks in physics(of which I have a number) stop short of of going into detail on this topic.
These fairy weak photons accounts for a large part of the so called “backradiation” .
The measured magnitude of this backradiation is quite large yet it seems to have very little effect!

Merrick
September 2, 2010 6:58 am

No, Paul, I’m sorry, but you are confused. Those collisions ARE the pressure broadening. You really need to study this problem because you have some misconceptions.

Spector
September 2, 2010 7:09 am

RE: anna v: (September 2, 2010 at 5:17 am) “I am saying that an excited CO2 is at practically the same temperature as before absorbing the photon, it is just a bit more massive.”
I assume by this you mean that the CO2 molecule will not directly acquire any kinetic or translational energy as a result of absorbing a photon. I note that there has been a debate as to whether temperature should or should not include the energy of internal molecular rotations and vibrations.

Merrick
September 2, 2010 7:18 am

Paul says:
“…the kinetic energy of the molecules being of the same order of magnitude as the energies of the various vibrational modes).”
That, Paul, as you have stated it would be called T/V energy transfer. NOT V/V transfer. And, no, the vibrational state for the case you’re describing isn’t broadened to the 1000 wavenumbers required to get the 650 wavenumber mode of CO2 and 1650 wavenumber mode of O2 to overlap. The TOTAL energy of the molecules are simply raised to the point that individual molecules colliding with O2 (any molecule, not just vibrationally excited CO2) have a small probability of putting that energy into an O2 vibration.
That requires a lot of randomization of energy over a lot of modes in a lot of species before an average molecule has that much energy. But this is most definitely NOT a transfer of a 650 wavenumber vibrational quantum from CO2 into a 1650 wavenumber vibrational quantum in O2 enabled by collisional broadening of the states. That statement, in and of itself, violates conservation of energy. Even if the broadening were that dramatic, the quantum could only populate the O2 vibration transiently as a virtual excitation during the collision and could not remain in the O2 vibration as the two collision partners were seperating but return back to the CO2 vibration. Quantum effects can cause what appear to be “violations” of assumptions based on classical mechanics, but it doesn’t allow the violation of fundamental physical laws like conservation of energy.

anna v
September 2, 2010 7:21 am

Spector says:
September 2, 2010 at 7:09 am
Yes.
A molecule rotating about itself is different than the two constituent atoms rotating against each other. A molecule vibrating like a bell is also different than the constituent atoms vibrating against each other. The first , bell and total rotation, should count in the equipartition. The internal degrees of freedom should be described by the solutions of the QM equations for the molecule and excited states there will absorb the energy in-elastically.

September 2, 2010 7:23 am

anna v says:
September 2, 2010 at 1:13 am
When an atom/molecule is raised from the ground state by absorbing a photon, two things happen:
1) its mass increases by h*nu

No its energy increases by h*nu
2) its momentum changes by h*nu/c , by momentum conservation.

September 2, 2010 8:06 am

Merrick says:
September 2, 2010 at 6:58 am
“No, Paul, I’m sorry, but you are confused. Those collisions ARE the pressure broadening. You really need to study this problem because you have some misconceptions.”
You are the one that’s confused. There is more than one effect associated with the thermal motions of the molecules. First, the doppler shift as molecules move towards and away from the source of radiation; this causes a doppler broadening of the absorption and emission lines proportional to the square root of the temperature. Under terrestrial conditions this is generally small (~1e-6). Second, there are various forms of collisional pressure and density broadening, due to the interaction of two or more molecules with the emitted or absorbed photon; this can either broaden an existing line or enable continuum emission to occur. The photon energy is no longer restricted to a single value, since there are now three bodies to share the energy and momentum. All gases, even monotonic ones like helium, absorb and radiate across the entire spectrum by this process (albeit only weakly except at high densities). Third, quite apart from any radiation, molecules in collision exchange not only translational energy, but also rotational and vibrational energy. Again, because the kinetic energy is not quantised, rotational and vibrational energy is easily transferred between molecules without having to match quanta between donor and recipient species.

September 2, 2010 8:33 am

Merrick says:
September 2, 2010 at 7:18 am
Paul says: “…the kinetic energy of the molecules being of the same order of magnitude as the energies of the various vibrational modes).”
“That, Paul, as you have stated it would be called T/V energy transfer. NOT V/V transfer.”
No, a T/V transfer is one in which a vibrational mode is excited purely by translational energy. A V/V transfer is one in which some vibrational energy is transferred from one molecule to another. Perhaps, to be pedantic, one should call the former a TT/TTV transfer, and the latter a TTV/TTV transfer, since the translational motion of the molecules is a mediating factor in both processes (there is no way that the vibrational energy can get from one gas molecule to another without either a collision, or the emission and absorption of a photon). At the temperatures and energies we are talking about, these are not insignificantly low-probability exchanges; CO2 could thermalise with the O2 vibration (if it’s at the 6 micron you mentioned) in a fraction of a millisecond.

anna v
September 2, 2010 9:25 am

Phil. says:
September 2, 2010 at 7:23 am
Believe me, Phil: E=m*c^2 works in QM. I am used to units where c=1, particle physics being my specialty. If the energy changes, the mass changes.

September 2, 2010 9:32 am

anna v says:
September 2, 2010 at 9:25 am
Phil. says:
September 2, 2010 at 7:23 am
Believe me, Phil: E=m*c^2 works in QM. I am used to units where c=1, particle physics being my specialty. If the energy changes, the mass changes.

Indeed it does but energy = hν
mass=hν/c^2

George E. Smith
September 2, 2010 9:42 am

Say Anna, what is the rest of that weirdo Particle Physics set of units. I remember the c = 1 part; isn’t e also 1 and some others. I only got it in school, never worked with it, is it h or h/2pi that is also 1.
Well I’ll stop guessing; but your E = m.c^2 immediately makes the point.
George

TomVonk
September 2, 2010 9:50 am

Merrick
So you don’t know Prof. Manuel Lopez Puertas from IAA nor Prof. F.W.Taylor from Oxford .
Fine . Did you really expect to know everything ? As for me , I am sure that you have books in your library that I have never read and that there are american scientists that I have never heard about . OK so you have never been to the department of physics in Oxford . I don’t seea problem with it ?
I will tell you one thing . After my studies , I have begun working in plasma physics (Tokamak process) .
At one point I realized that I didn’t want to finish my life writing papers that interested a handful of people and that an even smaller handful of people understood .
I also realized that I was unwilling to deploy the amount of arrogance and agressivity that is a characteristic of the life in Academia .
I went to industry and kept only a link to Academia through my collaboration in the research in non-linear dynamics . Just because it interests me but I don’t depend on it to earn my living .
From the technical point of view I can’t say that I fundamentaly disagree with anything you wrote . Actually you even seem to show an uncharacteristic lack of agressivity which makes the discussion enjoyable .
I could always nitpick on this or that formulation for the sake of nitpicking but we are not writing a paper here and I don’t need to show off that I know something that you don’t .
There where you have a problem though , like 99% of people who are active scientists what I suspect you may be , is the inability to communicate efficiently .
This what the thousands of laypersons interested by science call the “ivory tower syndrom”
The first rule of communication is to know who you are talking to and what are their needs .
I didn’t write this post or you and I didn’t intend to publish a paper either .
I wrote it for the thousands of WUWT readers who are interested by the science but who have not necessarily a degree in molecular dynamics .
Most of them won’t even comment .
The point may be trivial for you but I can assure you that it is not trivial for very many readers as some e mails I received show .
So just to repeat – as those readers have often met a statement that “CO2 transfers energy to N2 by collisions and therefore only absorbs IR radiation and doesn’t emit” as well as many variants thereof , my intent was to show in a simple way that :
a) How molecular processes and specifically T/V interactions worked
b) The statement was wrong if the gas was in LTE
Then you say
My disagreement, and I’ll repeat it here one more time for clarity, is that one cannot extrapolate from that statement (one that is as true for earth’s atmosphere as it is for the vacuum of space and the core of the earth because it is a literal truism) regarding a system in equilibrium to a system which clearly is not in equilibrium. You make first this leap (or miss the point that the atmosphere is seldom in equilibrium with the IR radiation of earth’s surface) then take a yet further leap to get to what I believe to be a remarkable non sequitor: “CO2 *cannot* heat the atmosphere.”
Just a little nitpicking : as I deal explicitely with gases and their properties , there is obviously no way to extrapolate to vacuums or earth cores which are clearly not in LTE . But let’s chalk that up to your artistic creativity 🙂
But beyond the nitpicking .
20 minutes ago I was sitting on my terrace well protected from macroscopic air movements . The temperature measured 1 m from ground during 10 minutes was 27.6 (+/- 0.03) °C . This gas volume was clearly not in radiative equilibrium with the ground which was under sun at around 34°C . Yet it was approximately in TE what shouldn’t surprise because everybody should know by now that the troposphere is in LTE . If I did the measure 100 m higher I would observe something similar but the temperature would be slightly lower . Again not surprising because of LTE .
Of course this experiment could have been done with a higher accuracy in a more controlled environment .
What I am saying is that in this situation and beyond that at any t and and any (x,y,z) provided that we have LTE , there is no net energy transfer between the CO2 (which absorbs the IR) and the N2/O2 (or atmosphere if you wish) .
And the reason for that is that the processes (1) and (2) described in my post are an equilibrium . Or an approximate equilibrium if one wants to nitpick .
I could do the same experiment during the night with the same result . Of course the temperature would be lower because the ground would be at a lower temperature but there would still be no radiative quilibrium and no run away heating of the N2/O2 by collisions with CO2 .
That’s all I wanted to say in my post .
And if you want me to add that the value of the local equilibrium temperature which is guaranteed to exist at every moment , will change (increase or decrease) if the defining parameters like density , pressure , IR intensity etc change , then despite the fact that I have already said so , I can repeat it .

George E. Smith
September 2, 2010 10:06 am

The discussion between Merick and Paul Birch vis a vis the Doppler (Temperature) and pressure (collision) broadening of the atmospheric molecular absorption bands, raises some interesting points.
I’ve been aware of those processes of course; but never really spent any time figuring out the magnitude of those effects.
I have argued that both of those processes result in the absorption coefficients diminishing with height by reason of lower temperatures and densities. this would result in the upward escape route being favored over the downward return to earth direction in multiple cascades of absorption and eventual thermal emission.
But I must confess having not figured out the magnitude of that effect. So it could be one of those butterfly wing phenomena; real, but not significant.
If either of you has references to that sort of study that are readable; I would appreicate any leads. Phil has given references to spectra for CO2 and H2O presumably in atmospheric conditions; and the two spectra are so strikingly different ( other than simply different frequencies) that it is hard to believe they are plots of the same phenomenon in two different materials.

Merrick
September 2, 2010 11:46 am

George E. Smith – how pressure broadening effects absorption is a little complicated, but not terribly so. In what is commonly referred to as the Beer limit, it effects the spectrum only, and not the strength of the absorption. This is the limit in which the absorption is not near saturation and the integrated absorption is directly proportional to concentration: i.e., doubling the concentration results in a doubling of the absorption.
In the Beer limit, if the peak is broadened by one or more of the homogeneous or inhomogeneous broadening effects the peak in any given absorption feature will decrease as the width of the feature increases such that the integral under the peak is constant. So a broadband power meter will measure no difference in the fraction of light passing through the sample. One needs a high resolution spectrometer in this case to even determine that any difference has occurred. Repeat: in the Beer limit line broadening cannot change the fraction of light absorbed. A more technically correct statement is that the integrated absorption strength (the integrated area under all the rotational structure of the full vibrational feature) is constant.
It should be noted that there are certain conditions, such as the Dicke effect, that actually causes line narrowing, but most effect cause broadening.
In a highly saturated absorption situation, however, since portions of the spectrum near the peak are almost fully absorbed anyways, a little line broadening can move absorption strength from the peak to the wings where total absorption is not occuring. This can cause a net increase in the total amount of light absorbed.
I’ve already given you a feel for scale on this, but it bear a little more elucidation here.
The molecular properties which determine vibrational and rotational line spacings are indirectly related. In most cases the rotational line spacings are quite close to 1/100th of the vibrational spacings. Total homogeneous and inhomogeneous line broadening at room temeprature seldom approaches 1 wavenumber.
In CO2 the first vibrationally excited state is at about 650 wavenumbers. The rotational spacing is on the order of a few wavenumbers. This spacing is not constant. For the first few rotationally excited states the spacing is much larger than a wavenumber (perhaps 10, I don’t have any references with me on the hike!). But this spacing decreases and at higher rotational states (r > 25 ?) the spacing of the states starts to get close to the inhomogeneous broadening and the states merge spectrally.
In the atmospheric case the lower rotational lines are well saturated, so “filling in the gaps” between those states can have a real impact. But don’t get too excited about that as a panacea, because that’s a positive feedback. Whether any of us does or doesn’t believe that increased CO2 on whole is causing a catastrophic warming event we still have to be honest with the science. At higher rotational number, the absorption is not so strongly saturated so that so broadening is unlikely to have much effect there in any case.
This is where the topic honestly has to broaden and I hate to do it, but there it is. We often say these transitions are saturated and pull out plot of atmopheric absorption integrated through the atmosphere. But slice it any way you want, if you want to slice, let’s say, one narrow, vertical column of air into segments which are in rough equilibrium internally (let’s leave the radiation field out for now) then you will get a large number of boxes. And for any one of those boxes you’ll find that the integrated absorption through that box is NOT saturated for CO2. This means two VERY impoortant things:
1) Pressure broadening isn’t impacting total integrated absorption within any cell.
2) Increasing CO2 concentration IS causing more energy to be absorbed at *lower elevations* on it’s way eventually back out into space.
Now, with most folks here, I believe that there are a whole host of other processes and influences going on that far outweigh this effect. But that is the best, narrowest, full answer I can give you to what was a *relatively* narrow question about line broadening. I hope that helps.

Spector
September 2, 2010 11:58 am

RE: Dave Springer says: (September 2, 2010 at 2:56 am)
“It really doesn’t matter if the quantum interactions move energy from CO2 to other species. The net result that matters is that when infrared impinges on a mix of gases where some gas species absorb infrared the total energy in the mixture increases.”
I limited my comment to CO2 energy reception because that was the only apparent assertion made by the author. As I have stated earlier, I now suspect that he was only using a thermodynamic statement to dispel the belief some now seem to have that CO2 is a self-sufficient heat generating agent.

anna v
September 2, 2010 12:18 pm

George,
yes, it is hbar=c=1 that define useful units
this says it better than I would
diazona
May13-09, 11:54 PM
Basically, the way to come up with a system like that is to pick a unit of time, multiply by the speed of light to get a unit of length, and then just pick a unit of mass such that \hbar is equal to the length unit squared times the unit of mass divided by the unit of time. These units of length, mass, and time replace the meter, kilogram, and second from SI.

http://www.physicsforums.com/archive/index.php/t-313965.html

Merrick
September 2, 2010 12:34 pm

Tom says:
“Just a little nitpicking : as I deal explicitely with gases and their properties , there is obviously no way to extrapolate to vacuums or earth cores which are clearly not in LTE . But let’s chalk that up to your artistic creativity :)But beyond the nitpicking .20 minutes ago I was sitting on my terrace well protected from macroscopic air movements . The temperature measured 1 m from ground during 10 minutes was 27.6 (+/- 0.03) °C . This gas volume was clearly not in radiative equilibrium with the ground which was under sun at around 34°C . Yet it was approximately in TE what shouldn’t surprise because everybody should know by now that the troposphere is in LTE . If I did the measure 100 m higher I would observe something similar but the temperature would be slightly lower . Again not surprising because of LTE .Of course this experiment could have been done with a higher accuracy in a more controlled environment .What I am saying is that in this situation and beyond that at any t and and any (x,y,z) provided that we have LTE , there is no net energy transfer between the CO2 (which absorbs the IR) and the N2/O2 (or atmosphere if you wish) .And the reason for that is that the processes (1) and (2) described in my post are an equilibrium . Or an approximate equilibrium if one wants to nitpick .I could do the same experiment during the night with the same result . Of course the temperature would be lower because the ground would be at a lower temperature but there would still be no radiative quilibrium and no run away heating of the N2/O2 by collisions with CO2 .That’s all I wanted to say in my post .And if you want me to add that the value of the local equilibrium temperature which is guaranteed to exist at every moment , will change (increase or decrease) if the defining parameters like density , pressure , IR intensity etc change , then despite the fact that I have already said so , I can repeat it ”
First of all, I said equilibrium, and not LTE. And anything can exist in a state of equilibrium. I hope you understand that. There certainly are vacuums in equilibrium (and those that are not) and regions of the core that are in a state of equilibrium (and those that are not).
Second, my apologies for using “atmosphere” instead of “troposphere.” That was the word *you* chose for your title, as opposed to N2/O2 which you seem to think is at least important enough to “nit pick” here. (You shouldn’t be nit picking about word usage like that after complaining about others commenting on your English. Personally, I’m doing all of this with a blackberry while on vacation.)
Third, if you have a body of air at 27.6 C standing 1 meter over a surface at 34 C you absolutely do NOT have a system in equilibrium, local or otherwise. I’m sorry that you seem to think that, but if you do you are wrong. You might reply that the air 1 meter over the ground isn’t increasing in temperature, or isn’t doing so very rapidly, but that’s only because you have a moving airmass that is continuously bringing in more air. I know this because even if your contention about CO2 not being able to heat the troposphere is true, and it’s not, then the air would be heated simply due to contact with the ground (or do you now have an LTE argument that disproves convection?).
If you were in a more controlled environment with 27.6 C air standing over a 34 C surface the air would rapidly climb. That *IS* how a greenhouse works, by trapping air over a surface so that it cannot rise, expand, cool, and be replaced by air that hasn’t been heated yet by the surface (as opposed to the trapping of IR radiation so often stated). And the top of that rising, expanding, cooling column IS in thermodynamic contact with the bottom of that column while absolutely *not* being in equilibrium with the bottom of that column. That is the main mechanism by which heat is moved from the surface to the upper atmosphere. How heat gets into the atmosphere is by both convection and radiative coupling with the warm surface (and surrounding atmosphere) through both CO2 and H2O IR coupling.
So, now you’ve made an instrumental measurement and reported it. Take the next step: visit your favorite meteoroligical site where you can download local data and review the temperature profile for today. What hour of the day were good examples of equilibrium conditions? The hour when the temperature changed from 21 C to 22 C, or the next hour when the temperature changed from 22 C to 23 C? Do you see the point I’m getting at here when I suggest that your fine points about LTE in circumstances in which temperature is clearly changing are a bit incongruous?
I might also suggest that if just about anybody else who visits this board who has a science background was told that two locations in a system one meter apart had a temperature delta of about 6.5 C and that the system was also in a state of equilibrium would take great pause. Any comments from anyone on that alone? Moderators? Anthony?

MartinGAtkins
September 2, 2010 12:44 pm

Evaporation neither adds or subtracts from the systems heat content.”

While this may be true for the planetary climate system, evaporation always subtracts (589 calories per gram at 288K) from the ocean heat content, which consequently doesn’t respond to radiative forcing as a blackbody.

Why did you think I used the word system? The poster was positing that evaporation itself was a forcing. It is not. The resulting humidity may add to the mix of the atmosphere and become a forcing but to say the act of evaporation adds heat to the system is wrong.
The oceans do act as a blackbody radiator at depth. They contain more energy than the gaseous atmosphere expressed as joules for that very reason. The cooling of the surface is countered by the subsurface radiation and fluid convection.

MartinGAtkins
September 2, 2010 1:28 pm

My last message should have been addressed to:-
sky says:
September 1, 2010 at 3:09 pm

George E. Smith
September 2, 2010 2:54 pm

“”” anna v says:
September 2, 2010 at 12:18 pm
George,
yes, it is hbar=c=1 that define useful units “””
Thanks Anna. I do recall that when we were doing that stuff probably in my third year of “Physics”; as distinct from “Radio-Physics” or “Mathematical Physics” which I also did; it just seemed like a dumb idea because you might have to explain it in an appendix to any paper you used it in.
But I don’t recall that I actually ever saw it in the form of E = m (c =1) but now that I do, my reaction is “Duh !! ” Now it makes all the sense in the world. But why hbar, if E = h.nu ; unless nu is in radians per sec ?
Well Shakespeare seemed awfully dumb when I was in school too .

George E. Smith
September 2, 2010 3:08 pm

“”” Merrick says:
September 2, 2010 at 11:46 am
George E. Smith – how pressure broadening effects absorption is a little complicated, but not terribly so. In what is commonly referred to as the Beer limit, it effects the spectrum only, and not the strength of the absorption. This is the limit in which the absorption is not near saturation and the integrated absorption is directly proportional to concentration: i.e., doubling the concentration results in a doubling of the absorption. “””
Merrick, that helps a whole lot. I can savvy most of that, but I can’t give a dissertation on it quite yet. I get the point that the broadening doesn’t change the absorption; just the spectral components that will get absorbed. Must admit that I had not thought about that issue before. If one is to scratch out explanations with a stick in the sand; then one doesn’t want to include things that aren’t too material.
Thanks a bunch; it’s of no earthly use in my work; but it sure jogs the “fancy that” excitement nodes; that are about all that keeps me alive these days. Very much appreciated.
George

cba
September 2, 2010 3:39 pm

seems that either I or Tom V are missing something basic. LTE demands that there be one temperature in the region and when that happens, then there is no net transfer of energy between co2 and N2, etc. and that the temperature of each is the same. However, the energy transfer mechanisms going on for co2 are IR absorption, IR emission, energy gain due to collisions, and energy loss due to collisions. The fact that collisions occur at a shorter mean time than does the mean time decay of an excited state by emission should not be inhibiting or preventing the emissions. To do so would cause a continued increase in T of the local area.

Spector
September 2, 2010 4:08 pm

Would it be correct to say that using a ‘grey-body’ representation for calculating the radiation emitted from the atmosphere involves using the Planck’s law black-body formula with a wavelength or frequency dependent emissivity (degree of blackness) being equal to the fraction of the incoming radiation that would be absorbed (100% = 1.000) in the atmosphere at each wavelength or frequency and also using a wavelength or frequency dependent temperature that is the fractional rate-of-absorption weighted effective value averaged over the full atmospheric absorption path for that wavelength or frequency?

cba
September 2, 2010 5:03 pm

Spector,
I think the term greybody is still using a stefan’s law with less than 1 emissivity. I don’t think one uses the term ‘greybody’ or ‘greybody approximation’ to refer to what you are asking about. When one considers this sort of thing, one must deal with both absorption and emissions and it goes by the name ‘radiative transfer’ and usually is based on a plane situation and uses what is called the Eddington approximation.
Using a plancks law wavelength dependent (or freq.) eqn lets you do what you’re talking about EXCEPT that just using a single atmospheric layer is a far bigger problem than you might think. Both temperatures and pressures become involved and these change as you move through the atmosphere. The width of the lines is a function of your pressure. A 1-dimensional model is where you create layers of atmosphere so that lower down you have warmer temperatures and higher pressures and sometimes different amounts of various molecules.
essentially, you create a layer assumed to be a constant pressure and temperature and a constant molecular consistency. You can then determine the optical thickness by some means – such as a molecular database calculation or a measurement. Once you have the optical thickness, you can determine the amount of power absorbed in that thickness. You can use the planck formula at the temperature of the layer to determine the state distribution of energy as it will provide what a BB radiator emits if you integrate it over a hemisphere. You can then use the absorption information and planck function to calculate the emission function – both as functions of wavelength or frequency. Each layer will absorb going through and also emit outward a particular amount of power. If the temperature of the incoming power BB spectrum corresponds to the temperature of the layer, it will have no net absorption or emission from that layer. If the layer is colder, then there will be less emission than absorption and vice versa if hotter.
I did this with hitran over the last couple of years. I use a 50+ layer atmosphere approximation. It is good more or less for clear skies but once clouds are present, the whole idea pretty much falls apart and one has even more to worry about that become far more important.

Spector
September 2, 2010 8:20 pm

RE: cba: (September 2, 2010 at 5:03 pm)
Subj: Radiation Emitted from the Atmosphere
Thanks, I had a feeling that this might be a layer by layer calculation.
BTW, I note the Savi-Weber HITRAN tool indicates that water, H2O, has strong absorption lines in the 40 to 120 micron range. Do the lower vibration energy levels of this range mean that water should expel heat more effectively than CO2 at the tropopause level (minus 55 degrees C), for same the reason the author of the main article indicates in his second figure, above, or does the higher energy of CO2 photons compensate for the fact that they are on the far right side of the excitation probability distribution curve.

bob
September 2, 2010 8:24 pm

Tom,
I’ll ask this again,
In you initial condition of LTE of a mixture of two gases, N2 and CO2.
By what means is the N2 remaining in equilibrium as it is losing heat due to radiation, what is keeping it warm?
If you can’t provide an answer to that, then your whole argument falls apart.
thanks

anna v
September 2, 2010 8:29 pm

Maybe part of the confusion comes from the term Local Thermodynamic Equilibrium.
“Local” allows for changes of the local quantities as the locus is changed but “equilibrium” brings to mind balance and the thoughts go to “global”. “Local” allows for a T(x,y,z,t,…), temperature is a function of relevant variables. “Local” just ensures continuity, the way that in mathematics one studies continuous variables. It disallows going into small distances where variables are discontinuous, as in the QM framework.
That is why confusion and double counting may arise. Thermodynamic concepts require local continuity, QM concepts offer the reverse.

anna v
September 2, 2010 11:52 pm

bob says:
September 2, 2010 at 8:24 pm
Tom,
I’ll ask this again,
In you initial condition of LTE of a mixture of two gases, N2 and CO2.
By what means is the N2 remaining in equilibrium as it is losing heat due to radiation, what is keeping it warm?

It is the thermodynamic equilibrium that is necessary, it does not mean the temperature remains constant, or univorm through a volume. It means it changes in coherence with thermodynamic equilibrium: all thermodynamic quantities change coherently in accord with the thermodynamic equations.

September 3, 2010 3:01 am

Merrick says:
September 2, 2010 at 11:46 am
“And for any one of those boxes you’ll find that the integrated absorption through that box is NOT saturated for CO2. This means two VERY impoortant things:
1) Pressure broadening isn’t impacting total integrated absorption within any cell.
2) Increasing CO2 concentration IS causing more energy to be absorbed at *lower elevations* on it’s way eventually back out into space.”
These two statements are not quite correct. (1) should read “Spectral broadening is impacting the total absorptivity within each cell only slightly”. (2) would be correct in an other-things-being-equal scenario, but not necessarily when there is complex feedback from water vapour, clouds and strong convection. The preamble is also somewhat misleading, since even if the difference in absorptivity in any one cell is small, it is the integrated change over the whole path length that counts.
Note that all forms of spectral broadening, for all the absorber species in the gas (not just CO2), contribute in a similar fashion (not just the pressure broadening, but also the doppler broadening, etc.), and further, that spectral broadening does change the total absorption whenever the path is not optically thin at all wavelengths (which is the case throughout the troposphere).

Spector
September 3, 2010 3:28 am

It looks like many of us have been assuming that LTE is general equilibrium in a local area. The Wikipedia article on Equilibrium seems to say that it applies when conditions around any one point are varying so slowly, that kinetic thermodynamic equilibrium can be assumed.

cba
September 3, 2010 5:11 am

regarding “Spector says:
September 2, 2010 at 8:20 pm ”
Spector,
I’d have to say I think it’s not too relevent down there. I’m afraid I lopped off my calcs at around 65 um anyway. Here’s the reason why. Stefan’s law is the result of integrating over all wavelengths (or freqs.) over a hemisphere of emission angles as one would have with a BB surface so you can integrate (or sum up) the individual bin values of a planck function to get the same answer for bb radiation. For temperatures of 288k, you have an emission curve to 65 or 75 um that is around 387W/m^2 and a stefan’s law calc (to infinite wavelength) that is around 391 w/m^2 giving a difference between them of about 4w/m^2 power emissions beyond the 65 or 75 um total. Note, it’s early and i’m working from memory here so it might not be exactly correct but it should be fairly close.
Now, at 288k we’re talking very little power out in the range you mentioned – although we’re missing the shorter wavelength range and that will be more. Wein’s displacement law indicates that the wavelength of peak emission will be proportional to the absolute temperature. Also, stefan’s law indicates the amount of power emitted for a bb will increase with the 4th power of the temperature. What these mean is that the peak shifts location but there is never a decrease of power over any small band of wavelengths as temperature is increased because the whole graph is rising and shifting shape. If there’s only 4 w/m^2 in the spectrum tail beyond 65 um while at 288k, then there’s going to be 4 w/m^2 or less in this tail while at 230k. Also, Wein’s law indicates that for 288k, we have the peak at around 10um, and if we drop the T to 230k, it shifts the peak only to around 12.5 um and the total power emission over all the spectrum has dropped from 390 to only 159 w/m^2.
btw, when I did an energy balance estimate, it was early on and there were still potentially some accuracy problems but there were some interesting results. Lower down in the troposphere, substantial power was necessary, approaching 100w/m^2 deficit – meaning more power was necessary (convection/latent heat was needed). But this decreased to practically 0 by the tropopause with the noise in the calculations being a bit uncomfortable by that time and for above that altitude. This though is clear sky calculations and thunderheads poking through the tropopause are not clear sky conditions, LOL.
I hope this answers your questions

Dave Springer
September 3, 2010 7:32 am

FYI
I don’t know if everyone here agrees with this definition but it is the one I’ve assumed all through this thread. I fail to see how it helps Vonk’s case. LTE is an approximation used in analysis of systems where there is no global thermodynamic equilibrium. A volume of high density mixed gases such as the troposphere can certainly be in LTE in any sufficiently small volumes but that doesn’t mean that each small volume cannot change in temperature due forces from outside the volume nor does it mean that a large volume can be considered in thermal equilibrium simply because it is composed of smaller volumes that each approximate thermal equilibrium. So it really means nothing to Vonk’s assertion that CO2 cannot thermalize N2. In fact it makes it possible for exactly that thing to happen so it supports exactly the opposite of what he would have us believe. In the rarified upper atmosphere where there is no LTE he would be correct but in the troposphere he is dead wrong.
http://wapedia.mobi/en/Non-equilibrium_thermodynamics#3.

Local thermodynamic equilibrium of matter means that conceptually, for study and analysis, the system can be spatially and temporally divided into ‘cells’ of small (infinitesimal) size, in which classical thermodynamical equilibrium conditions for matter are fulfilled to good approximation. These conditions are unfulfilled, for example, in very rarefied gases, in which molecular collisions are infrequent; and in the boundary layers of a star, where radiation is passing energy to space; and for interacting fermions at very low temperature, where dissipative processes become ineffective. When these ‘cells’ are defined, one admits that matter and energy may pass freely between contiguous ‘cells’, slowly enough to leave the ‘cells’ in local thermodynamic equilibrium.

September 3, 2010 8:11 am

Dave Springer says:
September 3, 2010 at 7:32 am
“I don’t know if everyone here agrees with this definition but it is the one I’ve assumed all through this thread.”
It’s a reasonable definition (for LTE of Matter), though by restricting it to matter – and not including the radiation or magnetic fields in approximate equilibrium with the matter – it does rather beg the question. Note that the text says, “These conditions are unfulfilled … in the boundary layers of a star, where radiation is passing energy to space”, which also describes the Earth’s atmosphere, to a degree. However, the crucial point is that it’s only an approximation, which allows one to do calculations wrt bulk properties like pressure, density, volume and temperature; this is the background to the interesting non-equilibrium stuff, like the absorption of sunlight from above and thermal radiation from below, for which it is the (locally small) departures from perfect equilibrium that drive the system, and which when integrated have strong macroscopic effects. Vonk’s misuse of LTE is like claiming that rocks can’t roll down a mountain because in a sufficiently small neighbourhood of any point the altitude above sea level is the same.

Robert Stevenson
September 3, 2010 8:25 am

George E Smith says:
‘But if instead that captured energy simply raises the atmospheric temperature; it is easy to see that te resultant LWIR emission should be a thermal (BB like) spectrum that reflected the Temperature of the ordinary atmospheric gases; so that spectrum should be quite independent of the GHG species that caused it; whether CO2 or H2O or anything else.’
Could you point to a reference that confirms the above statement that ordinary atmospheric gases emit a black-body spectrum similar to that of a brick .

anna v
September 3, 2010 9:59 am

Dave Springer says:
September 3, 2010 at 7:32 am
A volume of high density mixed gases such as the troposphere can certainly be in LTE in any sufficiently small volumes but that doesn’t mean that each small volume cannot change in temperature due forces from outside the volume nor does it mean that a large volume can be considered in thermal equilibrium simply because it is composed of smaller volumes that each approximate thermal equilibrium.
I agree, if you mean that a function of ( x,y,z,t) will describe the temperature in this volume that is has LTE in all neighborhoods.
So it really means nothing to Vonk’s assertion that CO2 cannot thermalize N2. In fact it makes it possible for exactly that thing to happen so it supports exactly the opposite of what he would have us believe.
I find this a non sequitur.
CO2 and N2 both will be at the same temperature T(x,y,z,t,…) as defined in statistical mechanics above.
CO2 does not get “hotter” when it absorbs a photon. It gets a bit heavier but keeps the common to both temperature; then the photon energy cascades down the CO2 available soft rotational and vibrational transitions and reaches thermal energies that are transferred by collisions continuously to the whole gas, N2 and CO2 keeping the same temperature for both though it might be higher as a value than before the photon cascade.

Dave Springer
September 3, 2010 1:22 pm

@Paul Birch
I believe LTE pretty much applies for magnetic fields and radiation in the troposphere when keeping in mind we can define arbitrarily tiny volumes and short temporal spans if need be so long as our volume contains enough molecules to remain in the domain of statistical thermodynamics. So far as I know nothing really qualifies. The radiative inputs change slowly and in the troposphere of the earth I don’t think magnetic fields get strong enough to have any effect of practical significance in this context. When considering volumes of billions of atoms each volume approximates LTE very well even in what might be viewed as macroscopic extremes like an iceberg floating in the ocean or even a tornado. Maybe at the interface of troposphere and an erupting volcano is arguably not in LTE.
Do you know of any rapidly changing radiative or magnetic forces acting on tiny volumes of troposphere such that the volume would not approximate thermodynamic equilibrium?

Dave Springer
September 3, 2010 1:34 pm

anna v says:
September 3, 2010 at 9:59 am
“N2 and CO2 keeping the same temperature for both though it might be higher as a value than before the photon cascade.”
It *might* be higher? Under what conditions might it be higher and under what conditions might it not be higher? When I hear words like “may” and “might” in regard to statistical mechanics it doesn’t inspire exactly inspire confidence that first principles in statistical thermodynamics are being properly understood.

sky
September 3, 2010 5:07 pm

MartinGAtkins says:
September 2, 2010 at 12:44 pm
Let’s clear up the misunderstandings. Neither evaporation nor the mixing of vapor into the atmosphere is a system-wide forcing, in the proper physics sense of the term. They are, however, very important processes in transferring thermal energy from surface to atmosphere and increasing its thermal capacitance. Without them, atmospheric temperatures would be quite different.
Sure, if you know the ocean temperature, than you can determine its effective BB radiation. My point, however, was that, because of on-going phase changes, that temperature cannot be determined for any radiative forcing from the Stefan-Boltzman law, as with a true blackbody.
Hope this finally settles these matters, which, important as they may be to system operation, are quite tangential to the posted thread. Have a good weekend.

cba
September 3, 2010 6:25 pm

there’s a problem with presuming photons are in thermal equilibrium or LTE rather than just matter. There is a radiation pattern based upon surface temperature and modified by spectral absorption. For the photons , thermal equilibrium would mean the bb spectrum would have to be at the temperature of the gas parcel as I understand it.

anna v
September 3, 2010 8:35 pm

Dave Springer says:
September 3, 2010 at 1:34 pm
anna v says:
September 3, 2010 at 9:59 am
“N2 and CO2 keeping the same temperature for both though it might be higher as a value than before the photon cascade.”

As english is not my mother tongue, let me rephrase the meaning of the sentence:
Even though the new temperature may be higher than the value before the cascade, it does not mean that CO2 provided that energy. That heat came from the original photon filtering down through the collective gas statistical properties.
In my opinion , it is people who do not understand the equipartition theorem, ( and how black body radiation with its continuum spectrum can come about, and are mixing up quantum mechanical concepts in a purely thermodynamic macroscopic problem,) who are the ones talking through their hats.

anna v
September 3, 2010 9:01 pm

It is not that the quantum mechanical formulation is useless. It just confuses people who only know the rudiments of physics.
Classical thermodynamics reigns macroscopically. It only needs waves for electromagnetic observations and statistical mechanics. Quantum mechanics is useful when the scale becomes microscopic and the LTE fails because of the discontinuities. In between there exists quantum statistical mechanics, that is aware of discrete properties in the constituents of mass and treats it in bulk. This last is not what happens when people sprout energy levels and absorption emission lines. Those are interesting to measure in the lab and as underlying stratum in quantum statistical mechanics, but one needs to use the tools of large numbers developed for the purpose: ensembles and partition functions etc etc. when wanting specific predictions.
This is not what is happening in the disorienting “explanations” about CO2 and forcings and what not hand wavings of climate science.
Matter is radiating continuously a continuous spectrum, the black body formula. This may be modified in the gray body formula, and in formulae fitted for gases , but radiate and absorb according to a continuous spectrum, it does.
How does this continuum spectrum arise from the discrete microscopic behavior of molecules? Each molecule in collision is exchanging photons, that is what collisions are about. Some elastically, most inelastically, i.e. part of the kinetic energy leaves as a soft photon and the originating molecule brakes, contributing to the lowering of temperature. This is the continuum spectrum observed. It has little to do with the solutions of the potential well that each molecule is. “Little” because it does utilize higher order moments of rotational and vibrational states, if one goes into details, but statistically the continuum framework is adequate since the molecules have a continuous spectrum of velocities.
Extra lines in the potential wells are like the cherries on the top of the cake. Interesting, useful, but not much to do with energy and equipartition and the real description of the problems.

Oliver Ramsay
September 3, 2010 9:11 pm

@ anna v
Speaking as Tom Vonk’s Joe the Plumber of Physics, I have to say that I don’t understand.
Surely, that photon would have been long gone into the void if CO2 hadn’t absorbed it and therefore it’s reasonable to say that CO2 introduced that energy to the atmosphere.
I have to echo Dave Springer’s aversion to “might”. That does sound implausibly haphazard.

September 3, 2010 9:27 pm

anna v says:
September 3, 2010 at 9:01 pm
It is not that the quantum mechanical formulation is useless. It just confuses people who only know the rudiments of physics.

Unless it is a communication problem you are the one confused about the physics of gases. Photons of appropriate energy excite rotational/vibrational transitions which are deactivated by collisions with surrounding gas molecules thereby heating the gas mixture, that’s all!

Spector
September 4, 2010 12:34 am

It seems that LTE has been defined such that these events are excluded and the process of heating causes continual equilibration to a new LTE states. Some of this seems like asking if a window heats your house when it lets the sunlight come through.

anna v
September 4, 2010 1:24 am

Phil. says:
September 3, 2010 at 9:27 pm
The existence of a black body type radiation from gases too shows there is a continuum statistically of photon energies. It is not necessary to think of “appropriate energy levels” because the continuous motion and change of direction of molecules integrates over all phase space and creates a continuum. Individual molecules need appropriate energy to get excited and deexcited (except when at the “continuum” of their potential well spectrum, hi l and n,..) . Collectively, because of the motion of each molecule, which is of commensurate energy, the effect is a continuum and it is confusing people to keep talking as if there is no continuum of photons, i.e. electromagnetic waves, appropriate for the classical thermodynamic formulation.
I remind you that a monatomic gas also has black body type radiation which comes from the continuous collisions which are exchanged soft photons.

anna v
September 4, 2010 1:29 am

Oliver Ramsay says:
September 3, 2010 at 9:11 pm
Surely, that photon would have been long gone into the void if CO2 hadn’t absorbed it and therefore it’s reasonable to say that CO2 introduced that energy to the atmosphere.
Put a brick next to a fireplace. When hot enough wrap it and place it in the bed. Is the heat coming from the brick, or is it just that it has a large heat capacity and releases the heat slower than if it were made of iron?
CO2 changes the heat capacity of the atmosphere.
I have to echo Dave Springer’s aversion to “might”. That does sound implausibly haphazard.

I hope my second post clarif