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.
My last message should have been addressed to:-
sky says:
September 1, 2010 at 3:09 pm
“”” anna v says:
September 2, 2010 at 12:18 pm
George,
yes, it is hbar=c=1 that define useful units “””
Thanks Anna. I do recall that when we were doing that stuff probably in my third year of “Physics”; as distinct from “Radio-Physics” or “Mathematical Physics” which I also did; it just seemed like a dumb idea because you might have to explain it in an appendix to any paper you used it in.
But I don’t recall that I actually ever saw it in the form of E = m (c =1) but now that I do, my reaction is “Duh !! ” Now it makes all the sense in the world. But why hbar, if E = h.nu ; unless nu is in radians per sec ?
Well Shakespeare seemed awfully dumb when I was in school too .
“”” Merrick says:
September 2, 2010 at 11:46 am
George E. Smith – how pressure broadening effects absorption is a little complicated, but not terribly so. In what is commonly referred to as the Beer limit, it effects the spectrum only, and not the strength of the absorption. This is the limit in which the absorption is not near saturation and the integrated absorption is directly proportional to concentration: i.e., doubling the concentration results in a doubling of the absorption. “””
Merrick, that helps a whole lot. I can savvy most of that, but I can’t give a dissertation on it quite yet. I get the point that the broadening doesn’t change the absorption; just the spectral components that will get absorbed. Must admit that I had not thought about that issue before. If one is to scratch out explanations with a stick in the sand; then one doesn’t want to include things that aren’t too material.
Thanks a bunch; it’s of no earthly use in my work; but it sure jogs the “fancy that” excitement nodes; that are about all that keeps me alive these days. Very much appreciated.
George
seems that either I or Tom V are missing something basic. LTE demands that there be one temperature in the region and when that happens, then there is no net transfer of energy between co2 and N2, etc. and that the temperature of each is the same. However, the energy transfer mechanisms going on for co2 are IR absorption, IR emission, energy gain due to collisions, and energy loss due to collisions. The fact that collisions occur at a shorter mean time than does the mean time decay of an excited state by emission should not be inhibiting or preventing the emissions. To do so would cause a continued increase in T of the local area.
Would it be correct to say that using a ‘grey-body’ representation for calculating the radiation emitted from the atmosphere involves using the Planck’s law black-body formula with a wavelength or frequency dependent emissivity (degree of blackness) being equal to the fraction of the incoming radiation that would be absorbed (100% = 1.000) in the atmosphere at each wavelength or frequency and also using a wavelength or frequency dependent temperature that is the fractional rate-of-absorption weighted effective value averaged over the full atmospheric absorption path for that wavelength or frequency?
Spector,
I think the term greybody is still using a stefan’s law with less than 1 emissivity. I don’t think one uses the term ‘greybody’ or ‘greybody approximation’ to refer to what you are asking about. When one considers this sort of thing, one must deal with both absorption and emissions and it goes by the name ‘radiative transfer’ and usually is based on a plane situation and uses what is called the Eddington approximation.
Using a plancks law wavelength dependent (or freq.) eqn lets you do what you’re talking about EXCEPT that just using a single atmospheric layer is a far bigger problem than you might think. Both temperatures and pressures become involved and these change as you move through the atmosphere. The width of the lines is a function of your pressure. A 1-dimensional model is where you create layers of atmosphere so that lower down you have warmer temperatures and higher pressures and sometimes different amounts of various molecules.
essentially, you create a layer assumed to be a constant pressure and temperature and a constant molecular consistency. You can then determine the optical thickness by some means – such as a molecular database calculation or a measurement. Once you have the optical thickness, you can determine the amount of power absorbed in that thickness. You can use the planck formula at the temperature of the layer to determine the state distribution of energy as it will provide what a BB radiator emits if you integrate it over a hemisphere. You can then use the absorption information and planck function to calculate the emission function – both as functions of wavelength or frequency. Each layer will absorb going through and also emit outward a particular amount of power. If the temperature of the incoming power BB spectrum corresponds to the temperature of the layer, it will have no net absorption or emission from that layer. If the layer is colder, then there will be less emission than absorption and vice versa if hotter.
I did this with hitran over the last couple of years. I use a 50+ layer atmosphere approximation. It is good more or less for clear skies but once clouds are present, the whole idea pretty much falls apart and one has even more to worry about that become far more important.
RE: cba: (September 2, 2010 at 5:03 pm)
Subj: Radiation Emitted from the Atmosphere
Thanks, I had a feeling that this might be a layer by layer calculation.
BTW, I note the Savi-Weber HITRAN tool indicates that water, H2O, has strong absorption lines in the 40 to 120 micron range. Do the lower vibration energy levels of this range mean that water should expel heat more effectively than CO2 at the tropopause level (minus 55 degrees C), for same the reason the author of the main article indicates in his second figure, above, or does the higher energy of CO2 photons compensate for the fact that they are on the far right side of the excitation probability distribution curve.
Tom,
I’ll ask this again,
In you initial condition of LTE of a mixture of two gases, N2 and CO2.
By what means is the N2 remaining in equilibrium as it is losing heat due to radiation, what is keeping it warm?
If you can’t provide an answer to that, then your whole argument falls apart.
thanks
Maybe part of the confusion comes from the term Local Thermodynamic Equilibrium.
“Local” allows for changes of the local quantities as the locus is changed but “equilibrium” brings to mind balance and the thoughts go to “global”. “Local” allows for a T(x,y,z,t,…), temperature is a function of relevant variables. “Local” just ensures continuity, the way that in mathematics one studies continuous variables. It disallows going into small distances where variables are discontinuous, as in the QM framework.
That is why confusion and double counting may arise. Thermodynamic concepts require local continuity, QM concepts offer the reverse.
bob says:
September 2, 2010 at 8:24 pm
Tom,
I’ll ask this again,
In you initial condition of LTE of a mixture of two gases, N2 and CO2.
By what means is the N2 remaining in equilibrium as it is losing heat due to radiation, what is keeping it warm?
It is the thermodynamic equilibrium that is necessary, it does not mean the temperature remains constant, or univorm through a volume. It means it changes in coherence with thermodynamic equilibrium: all thermodynamic quantities change coherently in accord with the thermodynamic equations.
Merrick says:
September 2, 2010 at 11:46 am
“And for any one of those boxes you’ll find that the integrated absorption through that box is NOT saturated for CO2. This means two VERY impoortant things:
1) Pressure broadening isn’t impacting total integrated absorption within any cell.
2) Increasing CO2 concentration IS causing more energy to be absorbed at *lower elevations* on it’s way eventually back out into space.”
These two statements are not quite correct. (1) should read “Spectral broadening is impacting the total absorptivity within each cell only slightly”. (2) would be correct in an other-things-being-equal scenario, but not necessarily when there is complex feedback from water vapour, clouds and strong convection. The preamble is also somewhat misleading, since even if the difference in absorptivity in any one cell is small, it is the integrated change over the whole path length that counts.
Note that all forms of spectral broadening, for all the absorber species in the gas (not just CO2), contribute in a similar fashion (not just the pressure broadening, but also the doppler broadening, etc.), and further, that spectral broadening does change the total absorption whenever the path is not optically thin at all wavelengths (which is the case throughout the troposphere).
It looks like many of us have been assuming that LTE is general equilibrium in a local area. The Wikipedia article on Equilibrium seems to say that it applies when conditions around any one point are varying so slowly, that kinetic thermodynamic equilibrium can be assumed.
regarding “Spector says:
September 2, 2010 at 8:20 pm ”
Spector,
I’d have to say I think it’s not too relevent down there. I’m afraid I lopped off my calcs at around 65 um anyway. Here’s the reason why. Stefan’s law is the result of integrating over all wavelengths (or freqs.) over a hemisphere of emission angles as one would have with a BB surface so you can integrate (or sum up) the individual bin values of a planck function to get the same answer for bb radiation. For temperatures of 288k, you have an emission curve to 65 or 75 um that is around 387W/m^2 and a stefan’s law calc (to infinite wavelength) that is around 391 w/m^2 giving a difference between them of about 4w/m^2 power emissions beyond the 65 or 75 um total. Note, it’s early and i’m working from memory here so it might not be exactly correct but it should be fairly close.
Now, at 288k we’re talking very little power out in the range you mentioned – although we’re missing the shorter wavelength range and that will be more. Wein’s displacement law indicates that the wavelength of peak emission will be proportional to the absolute temperature. Also, stefan’s law indicates the amount of power emitted for a bb will increase with the 4th power of the temperature. What these mean is that the peak shifts location but there is never a decrease of power over any small band of wavelengths as temperature is increased because the whole graph is rising and shifting shape. If there’s only 4 w/m^2 in the spectrum tail beyond 65 um while at 288k, then there’s going to be 4 w/m^2 or less in this tail while at 230k. Also, Wein’s law indicates that for 288k, we have the peak at around 10um, and if we drop the T to 230k, it shifts the peak only to around 12.5 um and the total power emission over all the spectrum has dropped from 390 to only 159 w/m^2.
btw, when I did an energy balance estimate, it was early on and there were still potentially some accuracy problems but there were some interesting results. Lower down in the troposphere, substantial power was necessary, approaching 100w/m^2 deficit – meaning more power was necessary (convection/latent heat was needed). But this decreased to practically 0 by the tropopause with the noise in the calculations being a bit uncomfortable by that time and for above that altitude. This though is clear sky calculations and thunderheads poking through the tropopause are not clear sky conditions, LOL.
I hope this answers your questions
FYI
I don’t know if everyone here agrees with this definition but it is the one I’ve assumed all through this thread. I fail to see how it helps Vonk’s case. LTE is an approximation used in analysis of systems where there is no global thermodynamic equilibrium. A volume of high density mixed gases such as the troposphere can certainly be in LTE in any sufficiently small volumes but that doesn’t mean that each small volume cannot change in temperature due forces from outside the volume nor does it mean that a large volume can be considered in thermal equilibrium simply because it is composed of smaller volumes that each approximate thermal equilibrium. So it really means nothing to Vonk’s assertion that CO2 cannot thermalize N2. In fact it makes it possible for exactly that thing to happen so it supports exactly the opposite of what he would have us believe. In the rarified upper atmosphere where there is no LTE he would be correct but in the troposphere he is dead wrong.
http://wapedia.mobi/en/Non-equilibrium_thermodynamics#3.
Dave Springer says:
September 3, 2010 at 7:32 am
“I don’t know if everyone here agrees with this definition but it is the one I’ve assumed all through this thread.”
It’s a reasonable definition (for LTE of Matter), though by restricting it to matter – and not including the radiation or magnetic fields in approximate equilibrium with the matter – it does rather beg the question. Note that the text says, “These conditions are unfulfilled … in the boundary layers of a star, where radiation is passing energy to space”, which also describes the Earth’s atmosphere, to a degree. However, the crucial point is that it’s only an approximation, which allows one to do calculations wrt bulk properties like pressure, density, volume and temperature; this is the background to the interesting non-equilibrium stuff, like the absorption of sunlight from above and thermal radiation from below, for which it is the (locally small) departures from perfect equilibrium that drive the system, and which when integrated have strong macroscopic effects. Vonk’s misuse of LTE is like claiming that rocks can’t roll down a mountain because in a sufficiently small neighbourhood of any point the altitude above sea level is the same.
George E Smith says:
‘But if instead that captured energy simply raises the atmospheric temperature; it is easy to see that te resultant LWIR emission should be a thermal (BB like) spectrum that reflected the Temperature of the ordinary atmospheric gases; so that spectrum should be quite independent of the GHG species that caused it; whether CO2 or H2O or anything else.’
Could you point to a reference that confirms the above statement that ordinary atmospheric gases emit a black-body spectrum similar to that of a brick .
Dave Springer says:
September 3, 2010 at 7:32 am
A volume of high density mixed gases such as the troposphere can certainly be in LTE in any sufficiently small volumes but that doesn’t mean that each small volume cannot change in temperature due forces from outside the volume nor does it mean that a large volume can be considered in thermal equilibrium simply because it is composed of smaller volumes that each approximate thermal equilibrium.
I agree, if you mean that a function of ( x,y,z,t) will describe the temperature in this volume that is has LTE in all neighborhoods.
So it really means nothing to Vonk’s assertion that CO2 cannot thermalize N2. In fact it makes it possible for exactly that thing to happen so it supports exactly the opposite of what he would have us believe.
I find this a non sequitur.
CO2 and N2 both will be at the same temperature T(x,y,z,t,…) as defined in statistical mechanics above.
CO2 does not get “hotter” when it absorbs a photon. It gets a bit heavier but keeps the common to both temperature; then the photon energy cascades down the CO2 available soft rotational and vibrational transitions and reaches thermal energies that are transferred by collisions continuously to the whole gas, N2 and CO2 keeping the same temperature for both though it might be higher as a value than before the photon cascade.
@Paul Birch
I believe LTE pretty much applies for magnetic fields and radiation in the troposphere when keeping in mind we can define arbitrarily tiny volumes and short temporal spans if need be so long as our volume contains enough molecules to remain in the domain of statistical thermodynamics. So far as I know nothing really qualifies. The radiative inputs change slowly and in the troposphere of the earth I don’t think magnetic fields get strong enough to have any effect of practical significance in this context. When considering volumes of billions of atoms each volume approximates LTE very well even in what might be viewed as macroscopic extremes like an iceberg floating in the ocean or even a tornado. Maybe at the interface of troposphere and an erupting volcano is arguably not in LTE.
Do you know of any rapidly changing radiative or magnetic forces acting on tiny volumes of troposphere such that the volume would not approximate thermodynamic equilibrium?
anna v says:
September 3, 2010 at 9:59 am
“N2 and CO2 keeping the same temperature for both though it might be higher as a value than before the photon cascade.”
It *might* be higher? Under what conditions might it be higher and under what conditions might it not be higher? When I hear words like “may” and “might” in regard to statistical mechanics it doesn’t inspire exactly inspire confidence that first principles in statistical thermodynamics are being properly understood.
MartinGAtkins says:
September 2, 2010 at 12:44 pm
Let’s clear up the misunderstandings. Neither evaporation nor the mixing of vapor into the atmosphere is a system-wide forcing, in the proper physics sense of the term. They are, however, very important processes in transferring thermal energy from surface to atmosphere and increasing its thermal capacitance. Without them, atmospheric temperatures would be quite different.
Sure, if you know the ocean temperature, than you can determine its effective BB radiation. My point, however, was that, because of on-going phase changes, that temperature cannot be determined for any radiative forcing from the Stefan-Boltzman law, as with a true blackbody.
Hope this finally settles these matters, which, important as they may be to system operation, are quite tangential to the posted thread. Have a good weekend.
there’s a problem with presuming photons are in thermal equilibrium or LTE rather than just matter. There is a radiation pattern based upon surface temperature and modified by spectral absorption. For the photons , thermal equilibrium would mean the bb spectrum would have to be at the temperature of the gas parcel as I understand it.
Dave Springer says:
September 3, 2010 at 1:34 pm
anna v says:
September 3, 2010 at 9:59 am
“N2 and CO2 keeping the same temperature for both though it might be higher as a value than before the photon cascade.”
As english is not my mother tongue, let me rephrase the meaning of the sentence:
Even though the new temperature may be higher than the value before the cascade, it does not mean that CO2 provided that energy. That heat came from the original photon filtering down through the collective gas statistical properties.
In my opinion , it is people who do not understand the equipartition theorem, ( and how black body radiation with its continuum spectrum can come about, and are mixing up quantum mechanical concepts in a purely thermodynamic macroscopic problem,) who are the ones talking through their hats.
It is not that the quantum mechanical formulation is useless. It just confuses people who only know the rudiments of physics.
Classical thermodynamics reigns macroscopically. It only needs waves for electromagnetic observations and statistical mechanics. Quantum mechanics is useful when the scale becomes microscopic and the LTE fails because of the discontinuities. In between there exists quantum statistical mechanics, that is aware of discrete properties in the constituents of mass and treats it in bulk. This last is not what happens when people sprout energy levels and absorption emission lines. Those are interesting to measure in the lab and as underlying stratum in quantum statistical mechanics, but one needs to use the tools of large numbers developed for the purpose: ensembles and partition functions etc etc. when wanting specific predictions.
This is not what is happening in the disorienting “explanations” about CO2 and forcings and what not hand wavings of climate science.
Matter is radiating continuously a continuous spectrum, the black body formula. This may be modified in the gray body formula, and in formulae fitted for gases , but radiate and absorb according to a continuous spectrum, it does.
How does this continuum spectrum arise from the discrete microscopic behavior of molecules? Each molecule in collision is exchanging photons, that is what collisions are about. Some elastically, most inelastically, i.e. part of the kinetic energy leaves as a soft photon and the originating molecule brakes, contributing to the lowering of temperature. This is the continuum spectrum observed. It has little to do with the solutions of the potential well that each molecule is. “Little” because it does utilize higher order moments of rotational and vibrational states, if one goes into details, but statistically the continuum framework is adequate since the molecules have a continuous spectrum of velocities.
Extra lines in the potential wells are like the cherries on the top of the cake. Interesting, useful, but not much to do with energy and equipartition and the real description of the problems.
@ur momisugly anna v
Speaking as Tom Vonk’s Joe the Plumber of Physics, I have to say that I don’t understand.
Surely, that photon would have been long gone into the void if CO2 hadn’t absorbed it and therefore it’s reasonable to say that CO2 introduced that energy to the atmosphere.
I have to echo Dave Springer’s aversion to “might”. That does sound implausibly haphazard.
anna v says:
September 3, 2010 at 9:01 pm
It is not that the quantum mechanical formulation is useless. It just confuses people who only know the rudiments of physics.
Unless it is a communication problem you are the one confused about the physics of gases. Photons of appropriate energy excite rotational/vibrational transitions which are deactivated by collisions with surrounding gas molecules thereby heating the gas mixture, that’s all!