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.
To cba:
The total emissive power Eb of 391W/m^2 is the area under Planck’s law spectrum curve at 288 K. 79.8W/m^2 is total emissive power within the CO2 absorption bands (mainly 12.5 to 16.5 microns) and is the area under the curve and bounded by the band wavelengths. The relevancy is that this band energy is completely absorbed or ‘filtered out’ and reduced to zero after 4000m for 350ppm CO2, 2000m for 700ppm CO2 and 7000m for 150ppm CO2. I didn’t say that it remained absorbed and not released or radiated which of course it is; I am merely pointing out that increasing (or decreasing) CO2 in the atmosphere makes no difference to the radiant heat that is absorbed (or radiated) it remains the same fixed amount that is radiated from Earth in the wavelength bands.
Also a correction has to made for Earth’s emissivity say 0.9
To Bryan:
A correction has to be made to gas emissivity (and absorptivity) due to the spectral overlap of water vapour and CO2 – because each gas is somewhat opaque to the other. The total absorption or radiation due to both is less than the sum of the separately calculated effects. When concentrations are roughly equal and mean beam lengths are small the correction factor is not very large; but when the concentration difference is as large as in the atmosphere the absorption due to CO2 could be, as you point out, greatly reduced as it is only in the first 120m that they are both absorbing.
In an NG fired furnace N2 (and O2 from xs air) do not radiate; only H2O and CO2 radiate to the chamber walls and heat sink. N2 and O2 lose heat by convection to the walls and sink as do H2O and CO2. Intimate mixing also transfers heat from the nonradiating to the radiating gases.
Bryan says:
September 6, 2010 at 6:13 am
If we accept these figures as accurate one thing puzzles me.
The main wavelength said to be absorbed by CO2 is around 15um.
However H2O also absorbs this wavelength admittedly not as strongly as CO2.
However because on average there are about 30 H2O molecules for every CO2 then the absorption must be mainly by H2O in the 15um region
CO2 takes 3600m to completely absorb 15um
H2O completely absorbs 15um in 240m.
The absorption isn’t ‘monolithic’ as you portray it, the absorption is distributed over a range of wavelengths (see below). The absorption in the 15 μm band is mainly by CO2.
http://i302.photobucket.com/albums/nn107/Sprintstar400/H2OCO2.gif
R Stevenson says:
September 6, 2010 at 9:49 am
“… the nonradiating … gases.”
There is no such thing. All gases absorb, radiate and scatter throughout the entire spectrum, due to collisions. But the line emissions are much stronger (except when the gas is compressed to near solid densities), which is why in discussions of the greenhouse effect the continuum is often ignored.
Paul Birch says:
September 6, 2010 at 4:11 am
September 6, 2010 at 4:45 am
Thanks Paul. I understand what you were saying better now.
My comment back to you was getting so long I gave up and re-wrote to be short and to the point. I’m not sure you meant exactly what you wrote back in a couple of areas but I have changed much over the last few days due to some deep reading, even after my comment you were commenting on.
I’ve kept very close tie to science since college but thermodynamics has not been my forte or interest till a year ago or so, you know, the climate. The mere mention of many trillions of dollars of taxes jerked me knee deep here at WUWT. Time for me to go back to school, so to speak, and refresh my understanding of areas in science I have so ignored. My interests have been heavy in astrophysics, astronomy, math, etc, professional programmer, but not TD. Rarely got into planetary or solar atmospheres specifically very much.
I’ve come to realize that Dr. Vonk is basically correct. My little square meter example above seems now to be just a sub-set of what he is saying. Many here might view Tom’s post better if they just realized that all of the sun is in LTE but the upper atmosphere. That is where LTE is broken on the sun.
Here are some links that lead me to that conclusion, this is better that me just rambling on and on trying to tell you what is written in these books on radiation and LTE:
http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBIQFjAA&url=http%3A%2F%2Fads.harvard.edu%2Fbooks%2F1989fsa..book%2FAbookC15.pdf&rct=j&q=AbookC15&ei=RaaFTPu9IIWClAea-tS6Dw&usg=AFQjCNGK7bItMyUEKDweQhltLBzuLVC_rg
Breakdown of Local Thermodynamic Equilibrium – Harvard
http://www.scribd.com/doc/34962513/Elsasser1942 thanks to Max in comment above.
Others found by search on LTE, radiation, etc.
My view of the terms you highlighted:
A thermodynamic system is in thermodynamic equilibrium when it is in all of thermal equilibrium, mechanical equilibrium, radiative equilibrium, and chemical equilibrium, i.e., a state of balance. For every internal event within, of any of the types above, there is an opposite to maintain this balance, not at the quantum level or individual atom or molecule level, but as a whole.
My little example was in thermal equilibrium but LTE doesn’t require it. Form those links I now realize that basically LTE is a misnomer, that is, you tend to only see LTE being broken in the spectrum you look at very low pressures. A certain line should be a certain strength by quantum calculations at a specific temperature and it isn’t, equipartition is broken, black body curve is violated. Something else is then amiss causing this. You then have to go to the quantum level to explain these divergence outside of LTE.
Steady state is usually termed in chemistry or flow of mass but also generally mean all processes or events involved are reversible and the rates of any change are balanced.
Thanks for your time again.
wayne says:
September 6, 2010 at 9:59 pm
Paul Birch says:
September 6, 2010 at 4:11 am
September 6, 2010 at 4:45 am
“Thanks Paul. I understand what you were saying better now. ”
I’m afraid you don’t. If you did you would realise that I am right and Vonk is wrong. I note that you have totally failed to address the science as I explained it (as, of course, has Vonk). Trying to make the words mean something else doesn’t cut it. Thus, for example, steady state does not mean that all the processes are reversible. It means only that the description of the state doesn’t change with time. A waterfall can be steady state; it is certainly not reversible. Entropy is being produced continually throughout the atmosphere, as energy cascades down from 600oK sunlight to thermal radiation at 300K and below. This is not LTE and will never be LTE no matter how many people want to believe it is. LTE has to be isentropic. The atmosphere isn’t. Not even locally. Your example was not and could not be in thermal equilibrium, because it is in a continuous irreversible exchange with non-equilibrium radiation. This is also true throughout the Sun, by the way, most strongly in the core and photosphere, but also in the main body, where, as well as convection, there is irreversible radiative transfer upwards, at ever longer wavelengths; because the photon mean free path is so short, relative to the size of the sun, the departure from LTE is very small (there is only a very small entropy gain with each scattering); but you still can’t ignore it if you want to know what the radiation is doing.
Phil. says: (September 6, 2010 at 10:09 am )“The absorption isn’t ‘monolithic’ as you portray it, the absorption is distributed over a range of wavelengths (see below). The absorption in the 15 μm band is mainly by CO2.”
The Savi-Weber online HITRAN absorption spectrum plotting utility shows multiple isolated H2O spikes longer than 15 microns. The highest peaks are around 23.81, 28.45, 39.38, 49.34, 58.68, 66.45, 71.95, 75.38, 82.05, 100, and 108.07 microns as read from the graphs.
At a typical tropopause temperature of 220 deg K, my Planck’s Law calculations seem to indicate that the typical black body emission of lines from 15 to about 30 microns is equal or slightly greater than that of the CO2, 15 micron line.
… I REALLY don’t want to get reinvolved in this pointless debate. But since everyone has gone on over and over about how a helium balloon can’t be a perfect insulator, etc…
Is everyone just clueless about a basic weatherproofing step that everyone USED to understand? The concept of double-paned windows with a dead airspace? The whole point of that basic item used by virtually everyone who could possibly be posting here RELIES on the fact that T/T energy transfer is among the poorer mechanisms for energy transfer. So much so that purging the system of all bodies with available vibrational and rotational modes even makes a measureable difference. It’s measureable, but not large. So you still get scam artists trying to convince you to fill your window with argon (no vibrational or rotational modes to populate) to get the best insulating effect. It’s actually pretty substantial. On the order of a 10-15% increase in insulating efficiency over simple air, but the problem is that the argon leaks out within a few months long before you can see the cost benefit in reduced heating/cooling bills. However, since the major constituents of air are O2 and N2, and since water can’t easily move in and out of the window seals, and since O2 and N2 are further small molecules with relatively highly spaced rotational and very highly spaced vibrational states they act as very good insulators. Fill that window gap with a relatively large molecule with relatively low rotational and vibrational spacings and you’d see the insulative effect largely dissappear.
If that simple observation, something that everyone who owns a home SHOULD know, doesn’t convince people that rotations and vibrations are key to *rapid* randomization of energy in gasses then I for one give up trying.
Paul Birch quoted me and responded:
“And for any one of those boxes you’ll find that the integrated absorption through that box is NOT saturated for CO2. This means two VERY impoortant things:
1) Pressure broadening isn’t impacting total integrated absorption within any cell.
2) Increasing CO2 concentration IS causing more energy to be absorbed at *lower elevations* on it’s way eventually back out into space.”
These two statements are not quite correct. (1) should read “Spectral broadening is impacting the total absorptivity within each cell only slightly”. (2) would be correct in an other-things-being-equal scenario, but not necessarily when there is complex feedback from water vapour, clouds and strong convection. The preamble is also somewhat misleading, since even if the difference in absorptivity in any one cell is small, it is the integrated change over the whole path length that counts.
1) No. What I said is correct. Changing the word from absorption to absorptivity makes your statement correct, also, but that’s not the point I was making. Broadening does effect absorptivity – under ANY concentration scenario. Broadening does NOT effect integrated absorption within the Beer limit. The Beer limit is defined by how close one is to saturation within your absorption cell. If you’re talking about the whole atmosphere, then your statement is reasonable. However, since I was talking explicitly about absorption near the surface and its impact on lapse rates, my statement in the context I made it was correct. By the time one reaches more than a kilometer or so in the atmosphere any out-bound photons are as likely to have been re-radiated by intervening atmosphere as they are to have originated from the surface, so you have to be careful about what you’re calling your absorption path and how you’re defining saturation.
2) I never discounted or ignored Doppler. I pointed it out in my initial post, if you’d care to look back. I said nothing to exclude Doppler in any of the later posts on the subject. I was merely focusing on misstatements regarding pressure broadening. And there is absolutely nothing wrong with the preamble statement you quoted. But one has to understand how energy in the atmosphere (as opposed to that radiated from the ground directly into space) is moved to the upper atmosphere to understand why it’s important and actually can make a difference if a certain fraction of heat radiated from the surface is absorbed in the first 100 m as opposed to the first 200 m. It DOES make a difference regardless of how well saturated the total vertical absorption is or isn’t. That why my comments are relavent to the actual science regardless of whether or not you think it’s the total path length absorption that matters most.
Paul, I’m sorry, but you’re simply mistaken. You CANNOT transfer one quanta of vibrational energy from CO2 at 650 wavenumbers into a 1650 wavenumber vibrational mode of O@ur momisugly. It violates conservation of energy. You back this up by suggesting multiple modes of one molecule dumping into multiple modes of another molecule, basically rephrasing what I already stated the solution was – that MANY modes have to be populated before such a transfer was possible, but it’s does change your two initial mistatements and you’re still wrong:
1) “collisional” broadening is NOT something other than pressure broadening, but is in fact only a PART of the full umbrella of pressure broadening (and therefore has a smaller total effect than TOTAL pressure broadending effects which are still in total less than 1 wavenumber). Here’s a source: “http://en.wikipedia.org/wiki/Spectral_line#Spectral_line_broadening_and_shift” If you have an authoritative source which talks about total line broadening in atmospheric conditions on the order of hundreds or even thousands of wavenumbers I’d like to see it.
2) The original source I quoted was wrong and your defense of it was wrong. Yes, under the scenario I first laid out (MANY modes exicted before there was enough available energy to populate an O2 or N2 vibrational mode) it is finally possible to effect a transfer into an excited vibrational state of O2 or N2, but in such a state it’s as likely that the transfer would come from H2O or any of the other molecules with a lot of thermal energy. To single out the CO2 absorption is a bit misleading. It is true that once all modes are well populated that CO2 populated by an absorption is a likely source of energy transfer, but if one understands equipartition at ALL one understands that the energy is just as likely to end up in other modes and not the SPECIFIC CO2 -> O2/N2 that the original quote comes from. And, even then, it’s far more likely to be randomized into other O2/N2/H2O modes before it ends up back in a CO2 vibration (which I think was the original point).
Now, as I said up front, I haven’t read the source and they may put all of these qualifiers in proceeding and following paragraphs. But I simply commented on the quote provided. It’s wrong – or at least VERY misleading and incomplete.
To Paul Birch:
What you say is true but rather pedantic and of little practical use. For instance if black-body radiation passes through a gas mass containing say carbon dioxide, absorption occurs in certain regions of the infrared spectrum. Conversely, if the gas mass is heated, it radiates in those same wavelength regions. This infrared spectrum of gases has its origin in simultaneous quantum changes in the energy levels of rotation and of interatomic vibration of the molecules and, at the temperature levels reached in industrial furnaces, is of importance only in the case of heteropolar gases. Of the gases* encountered in heat transfer equipment, carbon monoxide, the hydrocarbons, water vapour, carbon dioxide, sulphur dioxide, ammonia, hydrogen chloride, and the alcohols are among those with emission bands of sufficient magnitude to merit consideration. Gases with symmretrical molecules, hydrogen, oxygen, nitogen, etc., have been found not to show absorption bands in those wavelength regions of importance in radiant heat transmission at temperatures met in industrial practice.
*The two most important gases are of course carbon dioxide and water vapour.
Spector says:
September 7, 2010 at 5:25 am
Phil. says: (September 6, 2010 at 10:09 am )“The absorption isn’t ‘monolithic’ as you portray it, the absorption is distributed over a range of wavelengths (see below). The absorption in the 15 μm band is mainly by CO2.”
The Savi-Weber online HITRAN absorption spectrum plotting utility shows multiple isolated H2O spikes longer than 15 microns. The highest peaks are around 23.81, 28.45, 39.38, 49.34, 58.68, 66.45, 71.95, 75.38, 82.05, 100, and 108.07 microns as read from the graphs.
At a typical tropopause temperature of 220 deg K, my Planck’s Law calculations seem to indicate that the typical black body emission of lines from 15 to about 30 microns is equal or slightly greater than that of the CO2, 15 micron line.
What about all the other CO2 lines in the 15 micron band?
Did you look at the comparative spectra I provided?
RE: Phil.: (September 7, 2010 at 9:35 am)
“What about all the other CO2 lines in the 15 micron band?
Did you look at the comparative spectra I provided?”
I was just listing some of the primary H2O absorption lines indicated Savi-Weber HITRAN online absorption data plotting tool. Unlike the tightly packed bundle of CO2 lines around 15 microns, the H2O lines appear to be few and far between. The Y-axis is labeled “Intensity (cm^-1/(molecule cm^-2))” and the selected values I picked had intensities ranging from about 3E-19 to 3E-18.
The peak CO2 line at 15 microns has an intensity of about 2.8E-19 with most of the others ranging from 3E-20 to 1.5 E-19.
Your charts appear to exclude frequency-proportional wave-numbers below 600 cm^-1, however, I believe the wavelengths I listed, including 15 microns, have a wave-number range from 91 to 667 cm^-1. This range may be most significant to all-important task of removing heat from the extremely cold upper atmosphere.
I hope this helps.
Paul Birch says:
September 7, 2010 at 3:23 am
Don’t you realize you are arguing with a proper physics definition? It’s awfully hard to argue with a definition.
I’m not going to pick at your words and I wish you wouldn’t do it to mine. You tossed the word “generally” in that simple statement above and then pounced on it. Enough said on that.
Paul, LTE is pretty deep, Wikipedia has a very short section on it and GTE but it can leave you still with the wrong view point. You really should read in some physics books in areas where LTE is a core concept with 10-20 pages to explain it. I gave you one link. You should then have learned that the three points you objected to in http://wattsupwiththat.com/2010/08/31/does-co%e2%82%82-heat-the-troposphere/#comment-471397 are in fact all correct. Physics can twist your mind sometimes and I’m human too and the #3 point, X heats Y…, especially looked suspicious to me at first, for two days in fact. You speak of Dr. Vonk needing to learn and it seems to turn out it is many here that could use some learning, not him. Dr. Vonk struggles with English I read so I always give him very wide berth.
Sorry, can’t go further.
From what I gather after looking at the Tom Vonk articles and replies from posters, there seems to be no clear settled view of the radiative effects of CO2.
If we could agree on CO2s methods of interaction with N2 , O2 and H2O, we could
then compare it to H2Os radiative contribution.
We then have to quantify the total radiative effect of heat transfer against the other three main methods.
I suspect it is far less important than convective and phase change effects.
Most posters use the results of equilibrium thermodynamics and elementary quantum mechanics to explain the atmosphere.
However the turbulent atmosphere is far from any equilibrium most of the time.
A full explanation would have to involve non equilibrium thermodynamics which throws up several presently insoluble partial differential equations.
This is the reason that the several attempts to “model” the atmosphere on a computer have all failed.
However the article and discussion has been most interesting and perhaps WUWT could host a more comprehensive feature on this topic.
The statement “X heats Y” is equivalent to the statement “Y cannot cool X”, when taken in isolation, with no reference to LTE, appears to be a direct contradiction of the fact that mixing one pound of boiling water and one pound of sugar at 25 deg C will cause, the water (X) to heat the sugar (Y) and the sugar to cool the water. Of course, one could argue that all this heating and cooling was actually caused by the entity (Z) performing the mixing operation.
The Wikipedia article on heat seems to indicate that the actual definition of heat has become confused as a result of the modern practice of defining heat in terms energy. This, the article says, differs from both the original historic scientific definition and the modern lay concept of heat. Friedrich Herrmann argues that the quantity “heat” as introduced by Joseph Black in the 18th century, is known today as entropy.
Merrick says:
September 7, 2010 at 8:22 am
“1) No. What I said is correct. Changing the word from absorption to absorptivity makes your statement correct, also, but that’s not the point I was making. Broadening does effect absorptivity – under ANY concentration scenario. Broadening does NOT effect integrated absorption within the Beer limit.”
What you said is not correct. Broadening affects both local absorptivity and integrated absorption over the path, unless that whole path is optically thin at all wavelengths (I assume that’s what you mean by the Beer limit). Even if there were no change in absorptivity within a particular cell, its absorption would still change if there were any change in absorption anywhere along the path, because absorption equals absorptivity times the amount of radiation passing through. Change the amount of radiation passing through and you change the absorption. What you seem to be thinking is that, although broadening reduces the absorptivity at a given frequency in the vicinity of a spectral line, it does not change the absorptivity integrated over the whole spectrum, merely moves it around a bit; however, this is true only when the absorptivity over the full path is very much less than unity, at every frequency, because otherwise the exponential has significant non-linear terms. This latter condition does not apply in the Earth’s lower atmosphere; there are always bands in which the optical depth is significant. It is not necessary that the band be fully saturated (optical depth very much greater than unity) for the non-linear terms to appear.
“It DOES make a difference regardless of how well saturated the total vertical absorption is or isn’t. That why my comments are relavent to the actual science regardless of whether or not you think it’s the total path length absorption that matters most.”
It makes a difference to the detailed profiles, though for calculating the overall greenhouse effect all you really need are the overall optical depths. But my point was that the microscopic and quantum mechanical details don’t actually matter, so long as you know the absorption, reflection and transmission coefficients (at each altitude and wavelength).
Merrick says:
September 7, 2010 at 8:48 am
“Paul, I’m sorry, but you’re simply mistaken. You CANNOT transfer one quanta of vibrational energy from CO2 at 650 wavenumbers into a 1650 wavenumber vibrational mode of O@ur momisugly. It violates conservation of energy.”
No, it doesn’t. The extra energy comes from the kinetic energy of the molecules, which is not quantised. This is an inelastic collision.
“1) “collisional” broadening is NOT something other than pressure broadening, but is in fact only a PART of the full umbrella of pressure broadening (and therefore has a smaller total effect than TOTAL pressure broadending effects which are still in total less than 1 wavenumber).”
You’re still not getting it. Broadening of spectral lines (involving the emission, absorption or scattering of photons) is quite distinct from the mechanical transfer of energy between molecules by physical collisions between them. Except in very thin gases, this latter process is highly effective at thermalising any energy obtained from incoming radiation. Not perfectly effective – it does take a finite time – but still highly effective. It is not limited by the sort of quantum mechanical restrictions you suggest.
“2) The original source I quoted was wrong and your defense of it was wrong. ”
No, they weren’t. What you quoted was this: “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″. Barring a few quibbles (eg that N2 and O2 are weakly active in the infra-red, not completely inactive), this is rather obviously true. Would you seriously claim that the bulk gas is unimportant in thermalising the energy of the vibrational levels … and thus can safely be ignored? Surely not.
Btw, I suspect that the V-V exchanges the author is mainly referring to are those in which collisions with N2 or O2 affect a change in CO2’s vibrational/rotational state, rather than exciting a vibrational state of the N2 or O2 (are you sure the first O2 vibration is at as low an energy as 6 micron? Offhand, I’d have expected a much higher energy), though the principle’s the same either way.
R Stevenson says:
September 7, 2010 at 9:33 am
“To Paul Birch: What you say is true but rather pedantic and of little practical use. ”
It is of more relevance than perhaps you realise. Continuum effects are of considerable importance in both stellar and planetary atmospheres. In the fully ionised plasma of the Sun’s interior they are predominant. Without them, the Sun would be a very different beast and the Earth as we know it would probably not exist. Even in the Earth’s atmosphere, they are far from insignificant; they are the reason why the sky is blue. Compton, Thompson and Rayleigh scattering are meteorologically and climatologically quite important; though, admittedly, scattering from aerosols is more important than from the bulk gas.
wayne says:
September 7, 2010 at 11:42 pm
“Don’t you realize you are arguing with a proper physics definition? It’s awfully hard to argue with a definition.”
I don’t know what definition you think I’m arguing with. I am using standard physics meanings.
“I’m not going to pick at your words and I wish you wouldn’t do it to mine. You tossed the word “generally” in that simple statement above and then pounced on it. Enough said on that.”
No idea what you mean here. I can’t find any statement to you in which I wrote “generally”. There is one in which I wrote “in general”, in which I noted that, in general, the temperatures of different sources are not the same (but occasionally, by coincidence, or special construction, they could be – the equality is not excluded). That’s what “generally” and “in general” generally mean!
“Paul, LTE is pretty deep, Wikipedia …”
I have no interest in hearing what Wikipedia says about it. Nor can I debate with a link. If you do not understand the basic physics well enough to be able to debate using your own words, and from first principles, you do not understand the topic, and will likely misinterpret any references you come up with. Stop wriggling, stop appealing to authority, and address the physics. If you look at it carefully enough, you will see that what I have said is essentially correct (though I won’t guarantee not to have made any slips or promulgated any ambiguous statements whatsoever). The basic rule is that you do not have LTE unless local entropy changes can be neglected, which in the absorption or scattering of non-equilibrium radiation they can’t be.
To Paul Birch:
A world treaty (as considered from Kyoto to Copenhagen) restricting human production of “greenhouse gases,”chiefly carbon dioxide, fearing that CO2 will result in “human caused global warming” with disastrous environmental consequences would be the result of bogus science. Bogus science as peddled by the IPCC and championed by Al Gore leading to treaty that would cost the US and all other developed economies trillions of dollars in expenditure that is not necessary and that would have crippling effects.
As a climate science layman I think that quantum mechanics should be excluded from the discussion of how N2 in the troposphere gains or loses heat, as nobody understands it. I’m sure it has little to do with whether a photon has been tricked into thinking it is a wave or a particle by someone spying on it as it passes through a slot. If Schrodinger were alive today I’m sure he’d be a great asset to the sceptic argument wondering whether his is cat was alive or dead.
To attract more support to the sceptic point of view the bogus science of AGW must be challenged more effectively in language that folks can understand.
RE: Paul Birch: (September 8, 2010 at 6:25 am)
“The basic rule is that you do not have LTE unless local entropy changes can be neglected, which in the absorption or scattering of non-equilibrium radiation they can’t be.”
In his post of September 1, 2010 at 6:39 am, the author of the main article maintained that LTE is a property of material particles, presumably their kinetic states at any given time, and he finds no need to assume the existence of any parallel state of radiative equilibrium.
He further challenges anyone to explain why the Maxwell Boltzmann distribution cannot be derived when radiative equilibrium is not established.
With regard to LTE, it seems to me that we may be getting lost in the weeds of moot (debating) point logic with little regard to physical reality. This has been a thought provoking exercise, however.
Spector says:
September 9, 2010 at 2:36 pm
RE: Paul Birch: (September 8, 2010 at 6:25 am) “The basic rule is that you do not have LTE unless local entropy changes can be neglected, which in the absorption or scattering of non-equilibrium radiation they can’t be.”
“In his post of September 1, 2010 at 6:39 am, the author of the main article maintained that LTE is a property of material particles, presumably their kinetic states at any given time, and he finds no need to assume the existence of any parallel state of radiative equilibrium.”
Yes, and he’s wrong. If you don’t have radiative equilibrium, the radiation doesn’t behave in equilibrium ways. Nor do the absorbers or radiators, or the things heated or cooled by those absorbers and radiators. The greenhouse effect is just such a non-equilibrium departure from LTE, so Vonk’s attempt to assume it away by construction is simply nonsense.
“He further challenges anyone to explain why the Maxwell Boltzmann distribution cannot be derived when radiative equilibrium is not established.”
I’ve answered this, several ways. First, there can be more than one such distribution (or approximation thereto) within the same volume. Superimposed populations do not have to be at the same temperature, and in general will not be, so long as the heating and cooling (or exciting and relaxing) mechanisms for them are different; then different types of thermometer will give different readings. For example, it is quite common, astronomically, to have the ionised and neutral media at radically different temperatures (eg, a millions degrees versus a hundred). Second, where there is any interaction with non-equilibrium radiation, the kinetic energy of the affected populations will not in general follow a Maxwell-Boltzmann distribution, because portions of the curve are being bumped up, and other portions drained down. If thermalisation is rapid, the differences will be slight, but not negligible. One has to be very careful how one uses these approximations, so as not to exclude the very effects one is trying to calculate. Vonk is unfortunately misusing them.
“With regard to LTE, it seems to me that we may be getting lost in the weeds of moot (debating) point logic with little regard to physical reality. This has been a thought provoking exercise, however.”
It is Vonk who is ignoring physical reality. LTE is an approximation, which it is sometimes appropriate to use and sometimes not. Here it is not, because it is specifically the behaviour of the non-equilibrium radiation that we are considering.
R Stevenson says:
September 9, 2010 at 12:40 pm
“To Paul Birch: …I think that quantum mechanics should be excluded from the discussion of how N2 in the troposphere gains or loses heat … ”
That’s why I said: “the microscopic and quantum mechanical details don’t actually matter, so long as you know the absorption, reflection and transmission coefficients (at each altitude and wavelength)”. Classical thermodynamics is then quite adequate for calculating temperature gradients, etc.. By concentrating on the quantum mechanics of vibrations, etc., people often lose sight of the importance of the continuum scattering, which they wouldn’t if they paid more attention to those macroscopic coefficients.
“To attract more support to the sceptic point of view the bogus science of AGW must be challenged more effectively in language that folks can understand.”
I agree with you to a point (which is that most people lack the background knowledge to understand the science, so if you put it simply enough for them you run the risk of over-simplifying). And, yes, the greenhouse effect can be explained much more simply: if it’s harder for heat to get out than to get in, it will get hotter. Water vapour, CO2 and other greenhouse gases make it harder for the heat to get out.
However, on this blog, which is not really intended for the average telly-watching
sheepvoter, but for those with a more active thirst for truth, it should be possible for us to go into greater scientific depth.