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.
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,
“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!
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.
“”” 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.
——————-
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
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.
John Eggert says:
August 31, 2010 at 1:07 pm
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
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.
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.
Are there no air movements at the Troposphere? Because if there are, the LTE assumption fails.
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.
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.
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?
“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.
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?
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]
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.
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?
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
Tom,
Do you think that ozone, by absorbing UV, heats the stratosphere?
—————–
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
The next winter remember CO2, next to a warm fireplace tell your wife and kids how bad it is…..
@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.
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….
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!