Guest Post by Tom Vonk
In a recent post I considered the question in the title. You may see it here : http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/
The post generated great deal of interest and many comments.
Even if most of the posters understood the argument and I answered the comments of those who did not, I have been asked to sum up the discussion.
Before starting, I will repeat the statement that I wished to examine.
“Given a gas mixture of CO₂ and N₂ in Local Thermodynamic Equilibrium (LTE) and submitted to infrared radiation, does the CO₂ heat the N₂?”
To begin, we must be really sure that we understood not only what is contained in the question but especially what is NOT contained in it.
- The question contains no assumption about the radiation. Most importantly there is no assumption whether a radiative equilibrium does or does not exist. Therefore the answer will be independent from assumptions concerning radiative equilibrium. Similarly all questions and developments concerning radiative transfer are irrelevant to the question.
- The question contains no assumption about the size or the geometry of the mixture. It may be a cube with a volume of 1 mm³ or a column of 10 km height. As long as the mixture is in LTE, any size and any geometry works.
- The question contains no assumption about boundary conditions. Such assumptions would indeed be necessary if we asked much more ambitious questions like what happens at boundaries where no LTE exists and which may be constituted of solids or liquids. However we do not ask such ambitious questions.
Also it is necessary to be perfectly clear about what “X heats Y” means.
It means that there exists a mechanism transferring net (e.g non zero) energy unidirectionaly
from X to Y .
Perhaps as importantly, and some posters did not understand this point, the statement
“X heats Y” is equivalent to the statement “Y cannot cool X”.
The critical posts – and here we exclude posts developing questions of radiative transfer which are irrelevant as explained in 1) above – were of 2 types.
Type 1
The argument says “LTE never exists or alternatively LTE does not apply to a mixture of CO₂ and N₂.”
The answer to the first variant is that LTE exists and I repeat the definition from the original post : “A volume of gas is in LTE if for every point of this volume there exists a neighborhood in which the gas is in thermodynamic equilibrium (TE)”
2 remarks to this definition:
- It is not said and it is not important how large this neighborhood of every point is. It may be a cube of 1 mm³ or a cube of 10 m³ . The important part is that this neighborhood exists (almost) everywhere.
- LTE is necessary to define local temperature. Saying that LTE never exists is equivalent to saying that local temperatures never exist.
The second variant admits that LTE exists but suggests that a mixture of CO₂ and N₂ cannot be in LTE.
The LTE conditions are given when energy at every point is efficiently spread out among all available degrees of freedom (translation, rotation, vibration).
The most efficient tool for energy spreading are molecular collisions.
Without going in a mathematical development (see statistical thermodynamics for those interested), it is obvious that LTE will exist when there are many molecular collisions per volume unit.
This depends mostly on density – high density gases will be often in LTE while very low density gases will not.
For those not yet convinced, hold out a thermometer in your bedroom and it is probable that it will show a well defined temperature everywhere – your bedroom is in LTE .
We deal here with a mixture of CO₂ and N₂ in conditions of the troposphere which are precisely conditions where LTE exists too.
Type 2
The argument says “The mean time between collisions is much shorter than the mean decay time (e.g time necessary to emit a photon) and therefore all infrared energy absorbed by the CO₂ molecules is immediately and unidirectionaly transferred to the N₂ molecules.”
In simple words – the CO₂ never has time to emit any IR photons because it loses vibrational energy by collisions instead.
This statement is indeed equivalent to the statement “CO₂ heats N₂”.
Now let us examine the above figure.
The good understanding of this figure will do much better than only answering the original question. It will also make clear to everybody what is really happening in our gas mixture in LTE.
The figure shows the distribution of the kinetic energy (Ox axis) among the N₂ molecules (Oy axis).
This typical curve is called the Maxwell Boltzmann distribution, has been known for more than 100 years and experimentally confirmed with high accuracy.
We know that the temperature is defined by <E>, the energy average.
Hence it is the curve shown in the figure that defines the temperature of a gas.
Another way to say the same thing is to say that the curve depends only on temperature. If we wanted to have the distribution for another gas than N₂ , f.ex CO₂ or O₂, it would be given by an identical curve.
The blue curve gives the distribution of kinetic energy at 25°C while the red curve gives the distribution at 35°C.
The minimal energy is small but non-zero and there is no maximal energy.
A very important point on the Ox axis is the energy of the first vibrationally excited state of a CO₂ molecule.
You notice that at 25°C the majority of N₂ molecules has insufficient kinetic energy to excite this vibrational state.
Only the proportion of them given by the dark blue surface has enough energy to excite the vibrational state by collision.
When the temperature increases to 35°C, you notice that the proportion of N₂ molecules able to excite the vibrational CO₂ state by collision has significantly increased .
This proportion is given by the sum of the dark blue and light blue surface.
You also notice that as there exists no maximal energy, there will be a proportion of N₂ molecules able to excite the vibrational CO₂ state at any temperature.
Trivial so far? Well it will not get much more complicated.
First 2 technical points which play no role in the argument but which I would like to mention for the sake of completness.
- The figure shows the translational kinetic energy. Even if in some (popular) literature the temperature is defined as being an average of the translational kinetic energy, this is not strictly true.
The temperature is really defined as an average of all energy modes. So what about the vibrational and rotational energy?
At the low tropospheric temperatures we are considering, the distribution of the vibrational energy is extremely simple : about 5% or less of the molecules are in the first excited state and 95% or more are in the ground state.
As for the rotational energy, it can be computed classically without quantum corrections and the result is that it also follows a Maxwell Boltzmann distribution.
Therefore if we wished to plot the total energy (Etranslational + Evibrational + Erotational) we would rescale the Ox axis and obtain exactly the same curve as the one that is shown.
However as we are interested in studying the T/V interactions, it is the curve of the translational kinetic energy that interests us.
- We find the omnipresence of LTE again. This curve has been derived and experimentally confirmed only, and only if, the gas is in TE. Therefore the following 2 statements are equivalent :
“The gas is in LTE” , “The energy distribution at every point is given by the Maxwell Boltzmann distribution” .
If you feel that these statements are not equivalent, reread carefully what is above.
Now we can demonstrate why the Type2 argument is wrong.
Imagine that you mix cold N₂ represented by the blue curve in the Figure with highly vibrationally excited CO₂. The mixture would then not be in LTE and a transient would take place.
In the molecular process (1) CO₂* + N₂ → CO₂ + N₂⁺ which says that a vibrationally excited CO₂ molecule (CO₂*) collides with an N₂ molecule , decays to the ground state (CO₂) and increases the translational kinetic energy of N₂ (N₂⁺) , there would be a net energy transfer from CO₂* to N₂ .
As a result of this transfer the temperature of N₂ would increase and the blue curve would move to the red one.
However doing that, the number of molecules able to excite CO₂ vibrationally would increase (see the blue surfaces in the figure).
That means that during the increase of the temperature of N₂ , the rate of the opposite molecular process (2) CO₂ + N₂⁺ → CO₂* + N₂ where N₂ molecules (those from the blue surface in the figure) vibrationally excite CO2 molecules, will increase too.
Of course the transient net energy transfer from CO₂ to N₂ will not continue forever because else the mixture would transform into superheated plasma.
A local equilibrium will be established at each point and in this equilibrium the rate of the process (1) will be exactly equal to the rate of the process (2).
The curve of energy distribution will stop moving and the Maxwell Boltzmann distribution will describe this distribution at every point.
This is exactly the definition of LTE.
The transient will stop when the mixture reaches LTE and its characteristic feature is that there is no local net energy transfer from CO₂ to N₂.
This result demonstrates both that the Type2 argument is wrong and that the answer on the question we asked at the beginning is “No”.
In very simple words, if you take a small volume (for example 1 m³) of the CO₂ and N₂ mixture in LTE around any point , then there cannot be any net energy transfer from CO₂ to N₂ within this volume.
To establish the last step we will take the following statements.
- The result obtained for the CO₂ and N₂ mixture in LTE is equally true for a mixture containing 78% of N₂ , 21% of O₂ , x% of CO₂ and 1-x % H₂O in LTE.
- The mixture defined above approximates well the troposphere and the troposphere is indeed in LTE
- From the 2 statements above and the demonstrated result follows :
“The CO₂ does not heat the troposphere” what is the answer on the question asked in the title.
Caveat1
I have said it both in the initial post and in this one.
Unfortunately, I know that it can’t be avoided and that some readers will still be confused about the result established here and start considering radiative transfers or radiative equilibriums.
That’s why I stress again that LTE and the result established here is totally independent of radiative equilibriums and radiative transfer properties.
However it does falsify one misconception concerning radiative properties of CO₂ that has also figured in the comments and that is that “CO₂ does not radiate at 15µ because it “heats” N₂ instead”.
It is also to be noted that we consider only the T/V process because it is only the vibrational modes that interact with IR radiation.
There are also rotational/translational and rotational/vibrational transfers.
The same argument used for T/V applies also for the R/T and R/V processes in LTE – e.g there is no net energy transfer between these modes in LTE even if for example the R/T process has a much higher probability than a T/V process.
For the sake of clarity we don’t mention specifically the R/T and R/V processes.
Caveat2
The result established here is a statistical thermodynamics bulk property.
This property is of course not sufficient to establish the whole dynamics of a system at all time and space scales.
If that was our ambition – and it is not – then we would have to consider boundary conditions and macroscopic mass, momentum and energy transfers, e.g convection, conduction, phase changes, lapse rates etc.
More specifically this result doesn’t contradict the trivial observation that if one changes the parameters of the system, for example composition, pressure, radiation intensity and spectrum, etc, then the dynamics of the system change too.
Yet it contradicts the notion that once these parameter are fixed there is a net transfer of energy from CO₂ to the troposphere. There is not.
Caveat3
It will probably appear obvious to most of you but it has also to be repeated.
This result says little about comparisons between the dynamics of 2 very different systems such as, for example, an Earth without oceans and atmosphere, and an Earth with oceans and atmosphere. Clearly the dynamics will be very different but it stays that in the case of the real Earth with an atmosphere in LTE, there will be no net energy transfer from the CO₂ to the atmosphere.
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.
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.
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
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.
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.
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 .
—————–
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
@Vonk
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.
@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.
@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.
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.)
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.
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.
@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.
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?
@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.
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.
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?
————————-
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
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?
Paul Birch said :
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.
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.
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.
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.
“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