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.
The response to the article shows the wide level of interest in the mechanisms of the molecules making up the atmosphere.
How they respond to radiation and how they interact as the temperature changes.
There seems to be contradictory ideas about what goes on.
Perhaps Merrick can be persuaded to do an extended article covering broadly the same ground as the Tom Vonk post.
“”” Spector says:
September 1, 2010 at 4:26 am
RE: George E. Smith: (August 31, 2010 at 4:24 pm)
“And in the lower atmospehre the mean lifetime of the CO2 excited states is longer than the mean time between collisions; so that energy becomes thermalized before a spontaneous photon emission from the CO2 can occur. In the higher less dense regions in the stratospehre the time between collisins is great enough for spontaneous decay to occur.” “””
I should point out Spector, that “Phil” has made this point on several occasions; to the extent that it has even sunk into my thick skull.
When I started thinking about these climate/weather Physics issues some years ago; I was under the mistaken impression, that the 15 micron band photons that say CO2 captured; were subsequently re-emitted by the molecule; but in a now isotropic radiation pattern so at least half of it propagated upwards; but the escape was delayed by a multiple cascade of such photon absorption/emission events.
Once I realized that this could only occur in the rarified upper reaches of the atmosphere; where because of that rarified atmosphere the total energy involved was considerably reduced; it then started to make sense that the CO2 was acting almost as a catalyst but in a physical process, that conveyed radiant energy from the surface to the main gases of the atmosphere to warm them.
The distinction is important, because spontaneous emission from the CO2 would of course be in the form of a line spectrum, somewhere in the 15 micron band (and maybe other bands).
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
Some people still like to believe that gases do not emit Planck type thermal radiation; which flies in the face of the edict that any body above zero Kelvins should emit thermal radiation; and the sun certainly doesn’t mind doing that.
But you might be right, on your point that there should be an altitude at which the transition from primarily spontaneous decay takes over from thermalization and BB like spectral emission, and it maybe a region of atmospheric discontinuity.
We have to be mindful of the fact that the LWIR radiant emission from the atmospheric gases; whether N2, O2 or CO2 etc is pretty much the same as the LWIR radiant emission from an ordinary brick just sitting around at a Temperature of perhaps 288 K.
This radiation is not sensed by any human sensory organs; we do not detect it as “heat”; we don’t detect it at all, and in fact one has to jump through hoops to detect it with instrumentation designed for that purpose.
The lay public has visions of CO2 in the atmosphere creating a blow torch of heat that makes us all feel hot. Which is why I scoff at these laboratory or TV demonstrations of how an ordinary light bulb heats air containing increased CO2. They need to start using a cold brick for their LWIR radiation source.
Well that ordinary light bulb may have a color temperature of 2700-2800 K, and it is emitting a spectrum that is full of a lot of near one micron wavelength radiation that we sense as heat in ordinary sunlight; in fact the peak of the emission may be at about one micron for that temperature. Incidently the gas filling that ordinary light bulb is what is that big yellow glowing mass in there; a mass of gas, that of course cannot emit a thermal radiation spectrum like solids and liquids can. Well somehow the gas doesn’t know it is not supposed to do that.
Tom says:
“Merrick quite nicely resumed the purpose here :Yes, of course, a system in thermodynamic equilibrium (local or otherwise) cannot possibly have net energy flow – that violates the initial statement. When used in the physical sciences, equilibrium is defined as that state of a system in which an external force is required to change the state of the system. Posit BOTH equilibrium AND no external forcing (remember – no radiative coupling by Tom’s stipulation) and the argument is rock solid. By definition.”
Tom. I’m actually flattered that you selected my comment in this way. And as far as that statement you quoted there is absolutely NO doubt that we couldn’t possibly be in greater agreement. I think if you want that to be the total sum of your effort on this point and announce it clearly, perhaps in another (very short) article, that would be great and almost everyone would agree.
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.”
When the temperature of the earth’s surface and the atmosphere above it are not the same then the system is in disequilibrium. That condition occurs over the vast majority of the earth’s surface for nearly 24 hours of every day. That is absolutely true. Please look at a daily temperature profile for just about any place on the planet and I believe you will quickly be convinced.
THAT is the bone of contention that I have been belaboring and the contention I believe I understand in each of the various comments I’ve read.
Nylo says: August 31, 2010 at 7:33 pm (sorry for late answer, I didn’t see it until now)
“That theory is wrong. It has a wrong point. It says that “With H2O vapours raising higher concentration of H2O vapours in the column of air will decrease (same quantity in higher volume)”. No, they won’t. Because as they start to decrease near the surface, more moisture comes from the ocean. And you do not need any warming in the ocean to achieve this. ”
You have a good point, but if the concentration of H2O would increase back to the initial value then the temperature would increase and the water would expand and then, as per your logic, more water would get in the atmosphere which would heat more and so on until all the water would be in the atmosphere. A run off.
The concentration and distribution of the H2O vapours is a result of temperatures distribution and air pressures. Not the other way around.
The fact that H2O vapours decrease when CO2 increases it’s measured and one can see the trend from 1961 to present day.
Does this makes sense?
Best Regards!
Hum. Remind me of this one:
Climatologist: I have a system of undetermined complexity and undetermined composition, floating and spinning in space. It has a few internal but steady state and minor energy sources. An external energy source radiates 1365 watts per meter squared at it on a constant basis. What will happen?
Physicist: The system will arrive at a steady state temperature which radiates heat to space that equals the total of the energy inputs. Complexity of the system being unknown, and the body spinning in space versus the radiated energy source, there will be cyclic variations in temperature, but the long term average will not change.
Climatologist: Well what if I change the composition of the system?
Physicist: see above.
Climatologist: Perhaps you don’t understand my question. The system has an unknown quantity of CO2 in the atmosphere that absorbs energy in the same spectrum as the system is radiating. There are also quantities of carbon and oxygen that are combining to create more CO2 which absorbs more energy. Would this not raise the temperature of the system?
Physicist: there would be a temporary fluctuation in temperature caused by changes in how energy flows through the system, but for the long term average… see above.
Climatologist: But the CO2 would cause a small rise in temperature, which even if it was temporary would cause a huge rise in water vapour which would absorb even more of the energy being radiated by the system. This would have to raise the temperature of the system.
Physicist: there would be a temporary fluctuation in the temperature caused by changes in how energy flows through the system, but for the long term average… see above.
Climatologist: That can’t be true. I’ve been measuring temperature at thousands of points in the system and the average is rising.
Physicist: The temperature rise you observe can be due to one of two factors. It may be due to a cyclic variation that has not completed, or it could be due to the changes you alluded to earlier resulting in a redistribution of energy in the system that affects the measurement points more than the system as a whole. Unless the energy inputs have changed, the long term temperature average would be… see above.
Climatologist: AHA! All that burning of fossil fuel is releasing energy that was stored millions of years ago, you cannot deny that this would increase temperature.
Physicist: Is it more than 0.01% of what the energy source shining on the planet is?
Climatologist: Uhm… no.
Physicist: rounding error. For the long term temperature of the planet… see above.
Climatologist: Methane! Methane absorbs even more than CO2.
Physicist: see above.
Climatologist: Clouds! Clouds would retain more energy!
Physicist: see above.
Climatologist: Ice! If a fluctuation in temperature melted all the ice less energy would be reflected into space and would instead be absorbed into the system, raising the temperature. Ha!
Physicist: The ice you are pointing at is mostly at the poles where the inclination of the radiant energy source is so sharp that there isn’t much energy to absorb anyway. But what little there is would certainly go into the surface the ice used to cover, raising its temperature. That would reduce the temperature differential between equator and poles which would slow down convection processes that move energy from hot places to cold places. The result would be increased radiance from the planet that would exceed energy input until the planet cooled down enough to start forming ice again. As I said before, the change to the system that you propose could well result in redistribution of energy flows, and in short term temperature fluctuations, but as for the long term average temperature…. see above.
Climatologist: Blasphemer! Unbeliever! The temperature HAS to rise! I have reports! I have measurements! I have computer simulations! I have committees! United Nations committees! Grant money! Billions and billions and billions! I CAN’T be wrong, I will never explain it! Billions! and the carbon trading! Trillions in carbon trading!
Physicist: how much grant money?
Climatologist: Billions. Want some?
Physicist: Uhm…
Climatologist: BILLIONS
Climatologist: Hi. I used to be a physicist. When I started to understand the danger the world was in though, I decided to do the right thing and become a climatologist. Let me explain the greenhouse effect to you…
Humour found at http://knowledgedrift.wordpress.com, btw.
And then we see interesting statements from the supposedly “expert” literature that is a major underlying source. I admit to never having seen the book and I don’t know the authors from Adam – but to quote from above:
“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.”
The first excited state of CO2 results from an approximately 15 micron transition, as we have discussed repeatedly. The above quote is correct that, as homonuclear diatomics, N2 and O2 are IR inactive. But, unfortunately for the authors, the first excited state of O2 has a transition energy roughly eqiuvalent to 6 micron radiation (somebody please correct that if I’m wrong). That means that it takes more than 2 (i.e., 3) vibrational quanta of the CO2 fundamental to populate the O2 fundamental (3 CO2 quanta ~ 5 micron). This CANNOT happen via V/V transitions between CO2 and O2. N2 is both a lighter molecule and has a stronger bond, so that it’s vibrational fundamental is higher yet. I’m not suggesting that no O2 or N2 molecules get vibrationally excited at normal tropospheric conditions, and an excited O2 or N2 molecule colliding with CO2 would be energetically capable of exciting a CO2 vibration (possibly a non-radiative one!) but the reverse simply couldn’t happen without a lot of other preconditions. So, no, V/V collisions involving O2 and N2 are NOTimportant contributors to establishing thermal equilibrium in the atmosphere at normal tropospheric conditions. At those temperatures and given the atmospheric composition, process which work toward equilibrium are dominated by translational and rotational modes of N2 and O2,T/R/V modes of H2O and CO2, and T modes of Ar.
I’m not remotely interested inlaunching off on another topic that apparently requires teaching – I’m on vacation! – but among the things that determines the probably that a quanta will be transferred between two species when they collide is if the lower energy molecule has an available energetic state with a very similar transition energy. The next is that the change of the quantum number for both molecules is relatively small. So, for instance, a molecule typically has vibrational energy spacings approximately 100 times more energetic than rotational spacings. For a vibrationally excited CO2 molecule to transfer it’s quanta to rotational quanta in another CO2 molecule through collision the change in quantum number for one of the molecules is on the order of 100: that’s a low-probability transition. However, if that vibrationally excited CO2 molecule collides with an N2 molecules, with rotational energy spacings about an order of magnitude larger than for CO2, then the change in rotational quantum number for N2 would only be about 10: a much more likely occurence. Now, if that N2 molecule could immediately recollide with a non-excited CO2 molecule it would at least have the possibility of putting the energy “back.” But, since that N2 molecule is >99.9% more likely to collide with non-CO2 molecules at any given point in time, that energy (heat!) passed to the N2 molecule is going to be passed to all of the other N2 and O2 rotation and translations, etc., and more and more CO2 vibrational quanta funnelled into the system before the N2 population is sufficiently energetic to have either a significant fraction in highly rotationally excited states or vibrationally excited states to start significantly driving energy back into CO2 vibrations. The above quopted statement is not consistent with the facts and that’s a little more daylight in T/R/V dynamics.
@Vonk:
LTE = everything is the same temperature
Your argument: When everything is the same temperature in a closed system, nothing is heating anything else? Circular argumenting don’t you agree?
About the way the heat is transferred in LTE and the rest of the universe:
CO2 will always transfer heat via radiation and collision when above 0K. All matter does. No claiming it is only one or the other.
Interestingly, in LTE the CO2 would actually still heat the N, those being the CO2 molecules on the active state in your graph. There would be a similar amount of cooling happening to N in the other end of the CO2 activity spectre. Result: CO2 heats N in the LTE on a molecular level… and cools it too. Since this is all about heating I will make the claim that CO2 heats N indefinitely in LTE. An endless stream of energy from CO2 to N. (I hope everyone understands that there is also an endless stream of energy to the other direction…)
If you don’t disprove the above your argument is not valid so I’m sure that I’ll see a response when I get to work tomorrow, that being only 9hrs away now.. 🙁
Mircea,
The concentration of H2O in air capable of absorbing IR photons is 100 times greater than CO2 ( 380ppm) and the two are largely independent, CO2 is a permanent gas H2O is vapour. Increasing CO2 in air doesn’t decrease H2O. CO2 absorbs IR in three bands 2.36 to 3.02 microns, 4.01 to 4.8 microns and 12 to 16.5 microns; H2O also in three bands 2.24 to 3.27 microns, 4.8 to 8.5 microns and 12 to 25 microns. When the two appear together their emissivities are added (H2O is by far the larger) with a correction because both bands overlap. My question concerned traverse of black-body radiation from Earth through the atmosphere. Increasing CO2 would mean a shorter traverse to take out all the absorbable IR but no increase in total heat absorbed. However I could envisage that photons subsequently released would have a longer traverse before subsequently escaping ie a potentially longer retention time. My view is that convective mixing within the atmosphere and the dominance of water vapour would mean retention time is not significant.
Nylo says:August 31, 2010 at 7:33 pm (sorry for the delay, I didn’t see it until now)
“That theory is wrong. It has a wrong point. It says that “With H2O vapours raising higher concentration of H2O vapours in the column of air will decrease (same quantity in higher volume)”. No, they won’t. Because as they start to decrease near the surface, more moisture comes from the ocean. ”
But it doesn’t decrease near the surface of the ocean (or it is totally negligible compared with the normal variations of humidity near the ocean). The lapses are almost identical close to the sea/land surface. The variation in H2O vapour happens in the upper troposphere.
I have a plane to catch but I will write more about this here and at my blog tomorrow.
Merrick says:
September 1, 2010 at 12:51 pm
“… unfortunately for the authors, the first excited state of O2 has a transition energy roughly eqiuvalent to 6 micron radiation (somebody please correct that if I’m wrong). That means that it takes more than 2 (i.e., 3) vibrational quanta of the CO2 fundamental to populate the O2 fundamental (3 CO2 quanta ~ 5 micron). This CANNOT happen via V/V transitions between CO2 and O2. …
… among the things that determines the probably that a quanta will be transferred between two species when they collide is if the lower energy molecule has an available energetic state with a very similar transition energy.”
What I think you’re missing here is that these transitions are not simply quantised in a gas of finite density and temperature. They are broadened by the velocity distribution of the molecules; so they have a finite amplitude over the entire spectrum. A vibrationally excited CO2 molecule can quite easily vibrationally excite an O2 molecule, if it collides with a suitable velocity, which at atmospheric temperatures it often will. All of these exchanges occur with significant probability; they all contribute (in varying amounts) to the absorption, re-radiation and thermalisation processes.
“”” R Stevenson says:
September 1, 2010 at 1:29 pm
Mircea,
The concentration of H2O in air capable of absorbing IR photons is 100 times greater than CO2 ( 380ppm) and the two are largely independent, CO2 is a permanent gas H2O is vapour. Increasing CO2 in air doesn’t decrease H2O. CO2 absorbs IR in three bands 2.36 to 3.02 microns, 4.01 to 4.8 microns and 12 to 16.5 microns; H2O also in three bands 2.24 to 3.27 microns, 4.8 to 8.5 microns and 12 to 25 microns. When the two appear together their emissivities are added (H2O is by far the larger) with a correction because both bands overlap. “””
Well I don’t know who said what above since I can’t locate the originals; but I hate to point out that BOTH H2O and CO2 are PERMANENT gas components of the atmospehre. It is true that H2O is regarded as a VAPOR; but only when it is inequilibrium with either the liquid or solid phase of H2O. That is often true over the oceans; but over some deserts there can be plenty of H2O gas in the atmospehre (more than the CO2, with no liquid or solid water around for it to be in equilibrium with.
And if there were, it would only be in the vicinity of the phase interface; that we would properly regard H2O as a vapor. CO2 is also a vapor; just at a different Temperature range.
Bottom line is that an H2O molecule in the atmosphere at say 1000 metre’s from the surface; has no knowledge whatsoever regarding the presence or absence of any other H2O molecule 1km below; so there is no way that the H2O molecule would know to change hats from its gas hat to its vapor hat. And it behaves as an H2O molecule no matter which hat it is wearing.
True at some heights there could be clouds consisting of either liquid or solid H2O, and close to those, H2O gas could revert to its vapor behavior; which isn’t any different from its Ga behavior.
Why the AGW cloud keeps insisting that at times the atmosphere is devoid of H2O, whcih is therfore not a GHG is beyond me; In the part of the troposphere where people and things live and weather happens; there is always more H2O molecules that CO2 molecules; and it has pretty much always been that way for as long as it matters.
And some people might be surprised to learn that H2O starts to absorb EM radiation at least as short as 760 nm which is well inside the solar spectrum where much solar energy resides and is capable of being absorbed by H2O; but NOT CO2. H2O can absorb perhaps as much as 20% of the air mass one incoming solar spectrum at sea level. Only 1% of the total solar spectrum energy persists at longer than about 4 microns; and that 4 micron CO2 band has virtually no effect on the surface emitted LWIR spectrum, which peaks at about 10.1 microns (for a 288 K source Temeprature; and only 1% of that emission is at wavelengths as short as 5.05 microns; so the 4 micron CO2 band is rather inactive in any weather or climate issue on planet earth; although it might serve some function on Venus; but who cares ?
MartinGAtkins says:
September 1, 2010 at 9:04 am
“Evaporation neither adds or subtracts from the systems heat content.”
While this may be true for the planetary climate system, evaporation always subtracts (589 calories per gram at 288K) from the ocean heat content, which consequently doesn’t respond to radiative forcing as a blackbody.
Merrick says:
September 1, 2010 at 12:51 pm
The explanation that you provide is indeed consistent with the observed facts, including the fairly rapid extinction of 15 micron radiation with altitude. While there is widespread physical misunderstanding of how the climate system operates, Tom Vonk’s thesis is no narrowly constrained that it’s largely of pedagogical, rather than practical, interest.
tmtisfree: September 1, 2010 at 12:24 pm
LOL! Simply great!
Since Tom has responded and is in defense of the condition of Local Thermodynamic Equilibrium, I would like to ask a few questions to clarify my understanding of the initial conditions of this thought experiment.
Initial Conditions
““Given a gas mixture of CO₂ and N₂ in Local Thermodynamic Equilibrium (LTE) and submitted to infrared radiation, does the CO₂ heat the N₂?””
In this initial condition, what is keeping the N2 gas at temperature as it is radiating away its energy?
What happens when the mixture of the two gases comes under infared radiation?
The CO2 can absorb the radiation and thus its energy increases and thus its temperature increases.
The Nitrogen since it is opaque to the infared does not increase in temperature, and is decreasing in temperature as it is radiating in the infared.
You now have two gases in close contact and one is cooling and one is heating and what are we concluding is happening?
The nitrogen is not being heated by the CO2?
I’ll leave it to the reader as an excercise as to how this violates the first law of thermodynamics.
Increasing the concentration of CO2 in the atmosphere has to raise the temperature or violate the first law of thermodynamics, it’s as simple as that.
I found the following statement extraordinary
“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. ”
OK, I have a Ph.D. in physics. All I can say is posts like this degrade the
value of WUWT. One experiences relativity crackpots, and quantum
mechanics crackpots, and the whole global warming thing has now resulted
in more thermodynamics crackpots. Dealing with stuff like this is essentially
a waste of time. Energy transfer in two component gases is relatively
well known, and is of some importance in one of the more topical areas of
physics, namely cold-atom physics.
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 (and NOT on the scale of rotational spacings of small molecules, which is why we CAN measure rotationally resolved spectra of small molecules in the atmosphere).That is: on the order of a wavenumber or less. The CO2 vibrational energy is on the order of 650 wavenumbers. The O2 vibrational energy is almost exactly 1000 wavenumbers higher, or about 1650 wavenumbers. The N2 vibrational energy is even higher than that.
Please now explain to me how <1 wavenumber of spectral broadening bridges a 1000 wavenumber gap in V/V collisional transfer from CO2 to O2 (or larger to N2).
And, oh by the way, only the pressure broadening impacts the internal states of the system. The Doppler broadening only impacts your ability to *measure* the energy in the lab frame. So the true broadening effecting on the system in the way you suggest is even a smaller fraction of the already vague <1 wavenumber broadening we spectrally observe.
Real Physicist: OK, now I paint the surface of the planet black. What happens to the equilibrium temperature of the planet?
Armchair Physicist:It gets warmer.
Real Physicist: But it’s long been know that most black paints can only be relied upon to absorb strongly in the visible region and are often highly reflective in the IR.
Armchair Physicist: But it’s also well known that cars with black interiors get much hotter in direct sunlight than cars with lighter interiors. The point is that the black coloration absorbs light where it matters: in the visible peak of the solar radiation. So, we conclude, the planet will get warmer.
Real Physicist: Adding CO2 to the atmosphere darkens the atmosphere in *exactly* the portion of the relavent spectrum that matters: near the IR peak in terrestrial gray body radiation.
Armchair Physicist: ?
Jim says:
September 1, 2010 at 7:53 pm
…OK, I have a Ph.D. in physics. All I can say is posts like this degrade the
value of WUWT. One experiences relativity crackpots, and quantum
mechanics crackpots, and the whole global warming thing has now resulted
in more thermodynamics crackpots. Dealing with stuff like this is essentially
a waste of time.
——
Two possible solutions to that problem:
Either all the contributors and lurkers abandon the thread and find something else to do
or
you do
RE: George E. Smith says: (September 1, 2010 at 11:48 am)
“Once I realized that this could only occur in the rarified upper reaches of the atmosphere; where because of that rarified atmosphere the total energy involved was considerably reduced; it then started to make sense that the CO2 was acting almost as a catalyst but in a physical process, that conveyed radiant energy from the surface to the main gases of the atmosphere to warm them.”
My current view, so far uncontested, is there must be at altitude at which half the photons emitted by the ‘earthshine’ emitting/absorbing gases going straight up escape to outer space without any further interference in the thin atmosphere above. Going above this altitude, I would expect to see a rapidly widening cone defined by the vertical offset angle of the fifty percent escape path in the exponentially thinning atmosphere.
I presume the tropopause altitude is the best candidate for this point as this is the level where convection stops being an effective heat transfer mechanism and direct radiation to outer space is the only process that can expel the heat that has been wafted up to the coldest layer in the atmosphere.
A statement that you made caused me to wonder if the increasing mean time between collisions might be another factor causing an increased emission rate from the tropopause, but I have since been informed that this can only be a minor factor at the tropopause level.
As regards this article, I know that there are people, even AGW skeptics, who seem to have been convinced that CO2 and other ‘greenhouse’ gases are intrinsic sources of heat. If this is what the author is addressing when he says “The CO₂ does not heat the troposphere,” he is correct. For those of us who know this already, however, that point may seem so obvious that we automatically assume he is saying something far more revolutionary.
Jim
…..”Energy transfer in two component gases is relatively
well known, and is of some importance in one of the more topical areas of
physics, namely cold-atom physics.”……
I was the first poster to query Tom Vonk on his definition of Kelvin Temperature and he has yet to respond.
This area is where classical Kinetic Theory leaves off and Quantum Mechanical effects start to become important
I have a degree in Physics yet I find your familiarity with atmospheric gas mixture conditions as shown in quote above difficult to achieve.
When I google likely word combinations the sources revealed are mostly behind pay walls.
When I contrast what is known about say a CO2 laser with CO2 effects in the atmosphere this is what I find
CO2 laser; precise details of operating conditions, wavelengths,gas mixture percentages(CO2,N2 and others).
The variations used to produce desired output power and the background science is agreed.
CO2 in the atmosphere; a complete spectrum of contrasting opinion with very little agreed science.
Yet the IPCC propose to dislocate the economy of the world on the basis of very shaky science.
Tom Vonk has put himself in the limelight and leaves himself open for folk to pick holes in his view of this area.
However if folk like yourself and Merrick stay in the sidelines and snipe without offering a more comprehensive view we are unlikely to move forward.
bob says:
September 1, 2010 at 6:46 pm
The CO2 can absorb the radiation and thus its energy increases and thus its temperature increases.
This statement is wrong if one is speaking of spectral lines of the CO2 molecule, not rotational or vibrational. When an atom/molecule is raised from the ground state by absorbing a photon, two things happen:
1) its mass increases by h*nu
2) its momentum changes by h*nu/c , by momentum conservation.
On average there will be a change in the velocity of the distribution but one needs to solve (mass of CO2)*vnew=(mass of CO2)vold +h*nu/c . The maximum corrected momentum vnew=vold +(h*nu/c)/(mass of CO2)
This correction would be very small.
So no, CO2 does not heat up because its maxwell distribution changes very little.
It does have a heat capacity though, in the sense that this energy it absorbed as one lump photon, is later cascaded down to soft photons .
I think all this discussion of photons absorbed and radiated is really a red herring as far as the thermodynamics of the atmospheric gas system is concerned. Classical thermodynamics should be enough to describe gases at normal pressures with modified gray body distributions for radiation.
@anna v
“I think all this discussion of photons absorbed and radiated is really a red herring as far as the thermodynamics of the atmospheric gas system is concerned. Classical thermodynamics should be enough to describe gases at normal pressures with modified gray body distributions for radiation.”
Classical mechanics is not a should it’s a must when dealing with large numbers of molecules or even when dealing with electromagnetic radiation impinging on matter when the wavelength encompasses many molecules at once.
This is discussed in many places. Here’s a good place to start:
http://en.wikipedia.org/wiki/Correspondence_principle
The problem that happens over and over and over is that people attempt to describe macroscopic systems in quantum mechanical terms. It seldom works out. Not because QM doesn’t work though! It seldom works out because the macroscopic system descriptions are intractible in a quantum framework and due to the correspondence principle if the QM analysis doesn’t yield the same result as the classic statistical mechanics analysis then the QM analysis is buggered up not the other way around.
It was experimentally demonstrated by John Tyndal in 1859 that longwave radiation is absorbed by some gases in a mix while others are transparent and that the absorbing gases thermalize the others in the mix. One of his many publications “Heat a Mode of Motion” is great reading. A tour de force of experimental ingenuity leading to discovery and/or verification of basic thermodynamic principles.
You can read the whole book here:
http://books.google.com/books?id=m-TUAAAAMAAJ&pg=PR16&dq=heat+a+mode+of+motion&output=text
Vonk has nothing to say about the obvious problems put forth in several posts regarding:
– Circular argumenting (no temp difference, no heat transfer)
– X-Y heating logic being mutually exclusive
– The fact that CO2 actually DOES heat N by radiation AND collisions above zero K, in all LTE:s, all real life and in all hypothetical situations on a molecular level, as per Bohlzmann distribution.
– You manage to prove that mixing CO2 and N doesn’t cause an exothermic reaction. This is pretty much known…
So the argument put forth is false in every way.