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