Guest post By Tom Vonk (Tom is a physicist and long time poster at many climate blogs. Note also I’ll have another essay coming soon supporting the role of CO2 – For a another view on the CO2 issue, please see also this guest post by Ferdinand Engelbeen Anthony)
If you search for “greenhouse effect” in Google and get 1 cent for statements like…
“CO2 absorbs the outgoing infrared energy and warms the atmosphere” – or – “CO2 traps part of the infrared radiation between ground and the upper part of the atmosphere”
…you will be millionaire .
Even Internet sites that are said to have a good scientific level like “Science of doom” publish statements similar to those quoted above . These statements are all wrong yet happen so often that I submitted this guest post to Anthony to clear this issue once for all.
In the case that somebody asks why there is no peer reviewed paper about this issue , it is because everything what follows is textbook material . We will use results from statistical thermodynamics and quantum mechanics that have been known for some 100 years or more . More specifically the statement that we will prove is :
“A volume of gas in Local Thermodynamic Equilibrium (LTE) cannot be heated by CO2.”
There are 3 concepts that we will introduce below and that are necessary to the understanding .
- The Local Thermodynamic Equilibrium (LTE)
This concept plays a central part so some words of definition . First what LTE is not . LTE is not Thermodynamic Equilibrium (TE) , it is a much weaker assumption . LTE requires only that the equilibrium exists in some neighborhood of every point . For example the temperature may vary with time and space within a volume so that this volume is not in a Thermodynamic Equilibrium . However if there is an equilibrium within every small subvolume of this volume , we will have LTE .
Intuitively the notion of LTE is linked to the speed with which the particles move and to their density . If the particle stays long enough in a small volume to interact with other particles in this small volume , for example by collisions , then the particle will equilibrate with others . If it doesn’t stay long enough then it can’t equilibrate with others and there is no LTE .
There are 2 reasons why the importance of LTE is paramount .
First is that a temperature cannot be defined for a volume which is not in LTE . That is easy to understand . The temperature is an average energy of a small volume in equilibrium . Since there is no equilibrium in any small volume if we have not LTE , the temperature cannot be defined in this case.
Second is that the energy distribution in a volume in LTE follows known laws and can be computed .
The energy equipartition law
Kinetic energy is present in several forms . A monoatomic gas has only the translational kinetic energy , the well known ½.m.V² . A polyatomic gas can also vibrate and rotate and therefore has in addition to the translational kinetic energy also the vibrational and the rotational kinetic energy . When we want to specify the total kinetic energy of a molecule , we need to account for all 3 forms of it .
Thus the immediate question we ask is : “If we add energy to a molecule , what will it do ? Increase its velocity ? Increase its vibration ? Increase its rotation ? Some mixture of all 3 ?”
The answer is given by the energy equipartition law . It says : “In LTE the energy is shared equally among its different forms .”
As we have seen that the temperature is an average energy ,and that it is defined only under LTE conditions , it is possible to link the average kinetic energy <E> to the temperature . For instance in a monoatomic gas like Helium we have <E>= 3/2.k.T . The factor 3/2 comes because there are 3 translational degrees of freedom (3 space dimensions) and it can be reformulated by saying that the kinetic energy per translational degree of freedom is ½.k.T . From there can be derived ideal gas laws , specific heat capacities and much more . For polyatomic molecules exhibiting vibration and rotation the calculations are more complicated . The important point in this statistical law is that if we add some energy to a great number of molecules , this energy will be shared equally among their translational , rotational and vibrational degrees of freedom .
Quantum mechanical interactions of molecules with infrared radiation
Everything that happens in the interaction between a molecule and the infrared radiation is governed by quantum mechanics . Therefore the processes cannot be understood without at least the basics of the QM theory .
The most important point is that only the vibration and rotation modes of a molecule can interact with the infrared radiation . In addition this interaction will take place only if the molecule presents a non zero dipolar momentum . As a non zero dipolar momentum implies some asymmetry in the distribution of the electrical charges , it is specially important in non symmetric molecules . For instance the nitrogen N-N molecule is symmetrical and has no permanent dipolar momentum .
O=C=O is also symmetrical and has no permanent dipolar momentum . C=O is non symmetrical and has a permanent dipolar momentum . However to interact with IR it is not necessary that the dipolar momentum be permanent . While O=C=O has no permanent dipolar momentum , it has vibrational modes where an asymmetry appears and it is those modes that will absorb and emit IR . Also nitrogen N-N colliding with another molecule will be deformed and acquire a transient dipolar momentum which will allow it to absorb and emit IR .
In the picture left you see the 4 possible vibration modes of CO2 . The first one is symmetrical and therefore displays no dipolar momentum and doesn’t interact with IR . The second and the third look similar and have a dipolar momentum . It is these both that represent the famous 15µ band . The fourth is highly asymmetrical and also has a dipolar momentum .
What does interaction between a vibration mode and IR mean ?
The vibrational energies are quantified , that means that they can only take some discrete values . In the picture above is shown what happens when a molecule meets a photon whose energy (h.ν or ђ.ω) is exactly equal to the difference between 2 energy levels E2-E1 . The molecule absorbs the photon and “jumps up” from E1 to E2 . Of course the opposite process exists too – a molecule in the energy level E2 can “jump down” from E2 to E1 and emit a photon of energy E2-E1 .
But that is not everything that happens . What also happens are collisions and during collisions all following processes are possible .
- Translation-translation interaction . This is your usual billiard ball collision .
- Translation-vibration interaction . Here energy is exchanged between the vibration modes and the translation modes .
- Translation-rotation interaction . Here energy is is exchanged between the rotation modes and the translation modes .
- Rotation-vibration interaction … etc .
In the matter that concerns us here , namely a mixture of CO2 and N2 under infrared radiation only 2 processes are important : translation-translation and translation-vibration . We will therefore neglect all other processes without loosing generality .
The proof of our statement
The translation-translation process (sphere collision) has been well understood since more than 100 years . It can be studied by semi-classical statistical mechanics and the result is that the velocities of molecules (translational kinetic energy) within a volume of gas in equilibrium are distributed according to the Maxwell-Boltzmann distribution . As this distribution is invariant for a constant temperature , there are no net energy transfers and we do not need to further analyze this process .
The 2 processes of interest are the following :
CO2 + γ → CO2* (1)
This reads “a CO2 molecule absorbs an infrared photon γ and goes to a vibrationally excited state CO2*”
CO2* + N2 → CO2 + N2⁺ (2)
This reads “a vibrationally excited CO2 molecule CO2* collides with an N2 molecule and relaxes to a lower vibrational energy state CO2 while the N2 molecule increases its velocity to N2⁺ “. We use a different symbol * and ⁺ for the excited states to differentiate the energy modes – vibrational (*) for CO2 and translational (⁺) for N2 . In other words , there is transfer between vibrational and translational degrees of freedom in the process (2) . This process in non equilibrium conditions is sometimes called thermalization .
The microscopical process (2) is described by time symmetrical equations . All mechanical and electromagnetical interactions are governed by equations invariant under time reversal . This is not true for electroweak interactions but they play no role in the process (2) .
Again in simple words , it means that if the process (2) happens then the time symmetrical process , namely CO2 + N2⁺ → CO2* + N2 , happens too . Indeed this time reversed process where fast (e.g hot) N2 molecules slow down and excite vibrationally CO2 molecules is what makes an N2/CO2 laser work. Therefore the right way to write the process (2) is the following .
CO2* + N2 ↔ CO2 + N2⁺ (3)
Where the use of the double arrow ↔ instad of the simple arrow → is telling us that this process goes in both directions . Now the most important question is “What are the rates of the → and the ← processes ?”
The LTE conditions with the energy equipartition law give immediately the answer : “These rates are exactly equal .” This means that for every collision where a vibrationally excited CO2* transfers energy to N2 , there is a collision where N2⁺ transfers the same energy to CO2 and excites it vibrationally . There is no net energy transfer from CO2 to N2 through the vibration-translation interaction .
As we have seen that CO2 cannot transfer energy to N2 through the translation-translation process either , there is no net energy transfer (e.g “heating”) from CO2 to N2 what proves our statement .
This has an interesting corollary for the process (1) , IR absorption by CO2 molecules . We know that in equilibrium the distribution of the vibrational quantum states (e.g how many molecules are in a state with energy Ei) is invariant and depends only on temperature . For example only about 5 % of CO2 molecules are in a vibrationally excited state at room temperatures , 95 % are in the ground state .
Therefore in order to maintain the number of vibrationally excited molecules constant , every time a CO2 molecule absorbs an infrared photon and excites vibrationally , it is necessary that another CO2 molecule relaxes by going to a lower energy state . As we have seen above that this relaxation cannot happen through collisions with N2 because no net energy transfer is permitted , only the process (1) is available . Indeed the right way to write the process (1) is also :
CO2 + γ ↔ CO2* (1)
Where the use of the double arrow shows that the absorption process (→) happens at the same time as the emission process (←) . Because the number of excited molecules in a small volume in LTE must stay constant , follows that both processes emission/absorption must balance . In other words CO2 which absorbs strongly the 15µ IR , will emit strongly almost exactly as much 15 µ radiation as it absorbs . This is independent of the CO2 concentrations and of the intensity of IR radiation .
For those who prefer experimental proofs to theoretical arguments , here is a simple experiment demonstrating the above statements . Let us consider a hollow sphere at 15°C filled with air . You install an IR detector on the surface of the cavity . This is equivalent to the atmosphere during the night . The cavity will emit IR according to a black body law . Some frequencies of this BB radiation will be absorbed by the vibration modes of the CO2 molecules present in the air . What you will observe is :
- The detector shows that the cavity absorbs the same power on 15µ as it emits
- The temperature of the air stays at 15°C and more specifically the N2 and O2 do not heat
These observations demonstrate as expected that CO2 emits the same power as it absorbs and that there is no net energy transfer between the vibrational modes of CO2 and the translational modes of N2 and O2 . If you double the CO2 concentration or make the temperature vary , the observations stay identical showing that the conclusions we made are independent of temperatures and CO2 concentrations .
Conclusion and caveats
The main point is that every time you hear or read that “CO2 heats the atmosphere” , that “energy is trapped by CO2” , that “energy is stored by green house gases” and similar statements , you may be sure that this source is not to be trusted for information about radiation questions .
Caveat 1
The statement we proved cannot be interpreted as “CO2 has no impact on the dynamics of the Earth-atmosphere system” . What we have proven is that the CO2 cannot heat the atmosphere in the bulk but the whole system cannot be reduced to the bulk of the atmosphere . Indeed there are 2 interfaces – the void on one side and the surface of the Earth on the other side . Neither the former nor the latter is in LTE and the arguments we used are not valid . The dynamics of the system are governed by the lapse rate which is “anchored” to the ground and whose variations are dependent not only on convection , latent heat changes and conduction but also radiative transfer . The concentrations of CO2 (and H2O) play a role in this dynamics but it is not the purpose of this post to examine these much more complex and not well understood aspects .
Caveat 2
You will sometimes read or hear that “the CO2 has not the time to emit IR because the relaxation time is much longer than the mean time between collisions .” We know now that this conclusion is clearly wrong but looks like common sense if one accepts the premises which are true . Where is the problem ?
Well as the collisions are dominating , the CO2 will indeed often relax by a collision process . But with the same token it will also often excite by a collision process . And both processes will happen with an equal rate in LTE as we have seen . As for the emission , we are talking typically about 10ⁿ molecules with n of the order of 20 . Even if the average emission time is longer than the time between collisions , there is still a huge number of excited molecules who had not the opportunity to relax collisionally and who will emit . Not surprisingly this is also what experience shows .
Sorry, Miscolski did not limit the transparency to IR.
Strewth! It’s taken a while to read all the replies, but WOW! What a read.
Thanks again Anthony for posting the article.
Nick Stokes
The question that precedes that is, what do you mean by rate? How do you quantify it? What number are you talking about?
A rate is dN/dt
Merrick
You also seem to have left out rotations. Any collisions between relatively translationally excited N2 (or any other molecule) and CO2 are far more likely to transfer energy into excited rotational states and have exactly no effect on the vibrational state of CO2.
Yes I did and said so . As long as I the energy equipartition law is respected and the rotationnal quantum levels obey Maxwell Boltzmann statitics like vibrational do the argument stays unchanged . The focus of this post was the process (1) and the length of the post was limited . I could have added R/V and R/T processes for a much longer post . What you wrote is of course true .
Jan
As I don’t understand this post, am I supposed to take it to prove that climate change doesn’t exist, as any argument will suit this purpose?
No . I focus on N2-CO2 collisons and the question of where there can be or not be a net energy transfer .
Randy
Thanks for your essay. I think I understand “most” of what you were saying. Could you please comment on the ideas presented in this video and explain where they violate the physics.
http://earthguide.ucsd.edu/earthguide/diagrams/greenhouse/
Thanks, I would appreciate it.
There is nothing specially wrong with this diagram with the exception of the hand waving at the end . The only wrong phrase is “the absorbed energy is remitted towards the surface” which should read “PART of the absorbed energy is reemitted towards the surface .”
Bernie Anderson
Both of these would require absorption of IR by the CO2 and that energy has to go somewhere.
This is the process I called (1) . The absorbed energy is reemitted .
Jan K.Andersen
The fact that CO2 absorbs infrared energy and heat the atmosphere is no theory,, it is a fact.
CO2 absorbs (and emits) IR . That is a fact . It does not heat the atmosphere by transferring energy to N2 by collisions . This is another fact .
Vince Causey
But later you wrote: “As we have seen above that this relaxation cannot happen through collisions with N2 because no net energy transfer is permitted , only the process (1) is available .”
But doesn’t equation (2) say that CO2* collides with an N2 molecule and relaxes?
Yes you are right . The first phrase should have been “As we have seen above that no variation in the distribution of the CO2 vibration quantum states can happen through collisions with N2 because no net energy transfer is permitted , only the process (1) is available . Sorry for that .
Steven Goddard
There is no question that greenhouse gas H2O warms the atmosphere. Why would CO2 be different?
Unless you mean by that “H2O may heat the atmosphere by latent heat exchange” it is wrong . H2O may not heat the N2 by collisions by the same arguments exposed here .
Rich
In other words CO2 which absorbs strongly the 15µ IR , will emit strongly almost exactly as much 15 µ radiation as it absorbs .
Why isn’t it “exactly”?
Because I have taken here a simplified view where only 15µ is absorbed . In reality because of line broadening there will be many additional lines very near to 15 µ .
Ric Werme
If we take a handful of air with a 15µ IR flux passing through, is it really in LTE?
Yes . If a volume is not in LTE you cannot define a temperature .
Nylo
But you have not demonstrated that:
3) The percentage of molecules in excited state remains the same if you change the initial conditions regarding the percentage of CO2 in the gas.
And if the percentage of molecules in excited state changes, the gas changes its temperature. Perhaps the fraction of time when this happens cannot be called LTE conditions, I don’t know. But it happens anyway.
No I have not demonstrated that , Boltzmann did . It goes the other way round – for a given temperature this percentage is a constant in LTE . It will change if the temperature changes but if the temperature stays constant , the percentage is independent of the concentration of CO2 .
Jean Parisot
One question: you mentioned the non-LTE areas of earth and space boundaries; would the dynamic mechanism of thunderstorms also fall into a non-LTE area and be a factor in some of the emerging work on storm related energy transfers?
There is no special role of LTE in thunderstorms . LTE is just a very important condition to be able to talk about well defined temperatures .
Edvin
I believe this would only hold true if there was no net production or loss of either type of the two molecules CO2* and N2(+).
Yes exactly . There can be no net production or loss of CO2* because as mentioned , the distribution of the quantum states is constant for a given temperature . The same is true for N2(+) because the distribution of velocities is also constant (Maxwell Boltzmann) .
Pamela Gray
I would suggest the author edit some of the sentence structures and grammar before posting on such a hot topic.
Even if english is not my mother language , I generally make little to no grammar errors . As for the sentence structure it is hard for me to judge . Feel free to suggest better sentence structures , I would be happy to have a sentence structure review . Sorry if I can’t do better .
Ed Fix
The equipartition law, like all such empirical laws, is a statistical approximation that becomes less and less accurate as the population of gas molecules becomes smaller. Yet this article treats it as if it must be obeyed exactly by any random pair of molecules. Also, this law can be temporarily violated by transient responses.
That is partly right and partly wrong . I have already written that it is a statistical law valid in certain conditions (density and velocity) etc . However I have supposed nowhere that it should be obeyed exactly by any random pair of molecules . It actually has not to .
Wolf Walker
The opening paragraphs of this article give me the same tingling-down-the-back-of the-neck feeling as those creationist arguments about thermodynamics. It doesn’t feel right.
Sorry to take you as proxy for several posts of this kind . As physicist I have no satisfactory answer to tinglings in the back and feelings . I guess that’s how you feel but it will be hard for you to contribute something concrete if all you have are just feelings .
Paul Birch
The whole of this unfortunately fallacious argument is based on a false premise: that we have LTE. We don’t.
We do . Really . Look up about every textbook . When we don’t , you must do a full quantum mechanical treatment . But in the troposphere we do .
Merrick
I think Tom makes the common mistake that chemists see physicists make here, and that is assuming that microreversibility applies on the macroscopic scale.
I am not assuming that . I am saying that the statistics of a process (2) in a neighborhood of a point of a gas in LTE reflect the time symmetry of the process (2) . This is a classical assumption in mechanical statistics too .
V->V reactions occur with significantly higher probablility than T->V reactions.
Yes .
Unfortunately as I live in Europe , most reactions took place while I was sleeping and that makes the task of answering very hard . Thanks for so many reactions which allow me to see how many additional points/questions I should have addressed . But a science blog like WUWT is there for that – constructive discussion . I must stop here now but will try to come back and answer/comment other posts .
Kevin Kilty says:
August 5, 2010 at 5:52 pm
Paul Birch says:…Note that within Earth’s troposphere, the gross molecular flow of heat energy through any surface is of order 100MW/m2 (heat content times molecular speed). …
“I don’t follow, here. The net power flowing through a square meter per unit path via advection (is this what you mean by molecular flow?) is specific heat x density x velocity (dot) temperature gradient. I find values of maybe 10 kW per meter squared at a temperature gradient of a degree per hundred meters in a stiff breeze. What is this gross molecular flow of heat?”
It’s the statistical motion of molecules at the molecular scale. The number of molecules passing through unit area per unit time multiplied by the average energy carried per molecule. If we divide the disequilibriating fluxes by this figure we get an order of magnitude indication of how closely we can expect to approximate local equilibriation. The answer for terrestrial conditions is “very close”. But these small departures from LTE and equipartition still matter.
Tom Vonk’s caveat #1 says it all. His exposition does not deal with the real situation of the atmosphere, which is in effect a layer of gas dealing with an infrared energy flux emanating from the Earth’s surface and which passes out into the void of space.
A simple model for this is to consider first 2 flat plates, separated by some distance, with one plate at Earth’s mean surface temperature and the other at absolute zero (OK, real universe is at ~3K, but it won’t change the description by much). Then consider a layer of gas placed halfway between those 2 plates – and the only assumption you have to make is that the gas has some absorption/emission of infrared. The dynamics of the gas as discussed by Tom is not particularly relevant.
The gas layer will absorb some of the radiation coming from the “warm” plate – its temperature will rise until it is emitting as much radiation as it is receiving – but although all the incoming radiation is from the “warm” plate, the emitted radiation goes in all directions, which can be modelled as a flux towards the “cold” plate and one towards the”warm” plate. The warm plate thus experiences a radiation flux towards it that was not there without the gas layer.
A specific comment on several posts mentioning circularity .
In a sense yes , that is what I did and I think that people who mentioned that have understood the argument .
There is also “circularity” in the whole huge body of mathematics .
The fractions are really the “same” thing like the natural numbers because there is an isomorphism between N and Q .
So when you demonstrate a theorem about fractions it is just “circular” because you keep repeating that 1+1=2 with different words .
This kind of circularity is a trademark of the consistency .
Yet despite “circularity” nobody would contest that there is added value and new insights by defining Q on top of N .
The “circularity” in my post tries to show that if there is LTE , time symmetry and energy equipartition , then not anything goes in collisional processes .
wayne says:
August 5, 2010 at 7:15 pm
“Do you mind me asking, how did you come up with the 1 or 1.7 in 10 trillion expansion figure? Very roughly is fine. Was it taking the 390 W/m2 of back radiation in common energy balance charts and calculating from that the upward pressure and therefore the volume expansion?”
Sorry, the “ten trillion” was a slip (I originally wrote the sentence the other way round and referred to ten trillionths). I meant “a tenth of a trillion” (1E11). Radiation pressure equals power per unit area divided by the speed of light. So (300W/m2)/(3E8m/s) =1E-6 N/m2. Atmospheric pressure is 1E5 N/m2. So the ratio of radiation pressure to atmospheric pressure is 1E-11. Adiabatic expansion will reduce the absolute temperature by about a third of that (for dry air with Cp/Cv=1.4 it’s 0.4/1.4=.286); that is, ~300K/3 * 1E-11 = 1E-9 K. For an optical depth of unity, the radiative power in the cavity is increased by a factor of e (2.718). The fourth root of that gives the temperature ratio (since radiant power goes as temperature to the fourth power); that’s a factor of 1.284, or a rise of 0.284 in absolute temperature. Comparing this 0.284 with the earlier 0.286*1E-11, we have neat ratio of just about 1E-11, which is ten trillionths, ie, negligibly small. Because the effect is so tiny this crude calculation is good enough for present purposes. Note that if the optical depth is less than one (as it is for CO2 in Earth’s atmosphere) both effects are reduced more or less in proportion, so the ratio of the two effects is still the same (close enough).
================
Spector says:
August 5, 2010 at 11:11 pm
I note in this discussion a reference to returning to a ‘ground state’ analogous to what an excited electron in an atom does. At the atomic level, the emission spectrum just corresponds to an allowed set of energy states. If this were the case with molecular vibration, I would think we would only see simple sharp spectral lines
However, when I look at the HITRAN data available online I see a broad array of many sharp spikes. This gives me the impression that molecular vibrations may be in a transition region between the quantum and the continuous worlds. I suspect that each band represents a vibration mode and perhaps each narrow spike represents one of the allowed modal vibration energy states. Imagine a bell that changes tone in steps as the vibration dies out.
======================
That’s because of several factors. First off, The HITRAN database itself consists of single frequency ( or wavelength) line descriptions. The bands like co2 15 um are nothing but a tremendous number of lines close together. To use HITRAN in a spectrum, one must create a line width based upon temperature, pressure, including partial pressures of the molecule of interest and the ‘other molecules present’, and the line data associated with width. For instance, there’s pressure broadening along with wavelength shift due to pressure corrections to be made.
By transisitional between quantum and continuous, I think you’re looking simply at the realm of readily accessible energy states of lower energy levels. With a hydrogen atom, the first transition from ground state gives a hard uV photon energy. If you excite that atom somewhat above that energy, only then can you get any emissions from it in the visible and IR. Burning hydrogen from a leak in a tank or pipe is quite dangerous as it emits no IR or visible light so you don’t see it or feel any warmth of it radiating. To see the pink hydrogen emission nebulae, the gas there is excited by extremely hot uV emitting type O and B stars which raise the hydrogen above the ground state so it can emit visible light.
A bell, like a drum head, has overtones which are not harmonics and those die out at different rates. They’re still quantized so to say and have particular modes of vibration, Chladni patterns. It’s the one dimensional things like strings and pipes that produce the quantized harmonic series of frequencies.
TomVonk says:
August 6, 2010 at 4:08 am
Paul Birch says: The whole of this unfortunately fallacious argument is based on a false premise: that we have LTE. We don’t.
“We do . Really . Look up about every textbook . When we don’t , you must do a full quantum mechanical treatment . But in the troposphere we do .”
We don’t. Really. We have an approximation to LTE. We have an approximation to equipartition. But it is the departures from LTE, caused by and causing superimposed radiative fluxes at different temperatures (as well as the disequilibria from convection, condensation, evaporation etc. ) that among other things produce greenhouse warming and drive the weather system. Your caveat 1, “… there are 2 interfaces – the void on one side and the surface of the Earth on the other side . Neither the former nor the latter is in LTE and the arguments we used are not valid…”, applies throughout the entire volume, because the whole of the volume “sees” (receives radiation from and radiates to) both of those non-LTE boundaries.
Reed Coray says:
August 5, 2010 at 7:38 pm
Reed, let me try to explain how it works.
Suppose you make money at the rate of 100 dollars a day. Your money bin is full, so you want to get rid of this annoying cash inflow. You consider giving it away to the federal government, but because of some bizarre tax rule they are obligated to give you back 20% of everything you give them. How do you ensure a net zero flow?
Answer: you give them 125 dollars a day, $25 more than what you make from your other sources. They give you back 20% of 125, which is $25, and they use the rest cheerfully for their own nefarious purposes. Your net cash flow is now zero.
Bill Illis says: (August 5, 2010 at 10:06 pm) “… all molecules including N2 and O2 absorb and emit blackbody radiation – this seems to not be understood by many).”
I believe N2 and O2 are largely transparent and thus have only a few allowed radiation emission and absorption bands just like CO2 except these do not interfere with earthshine or sunshine to any appreciable extent.
I believe the operative statement here is “only black bodies emit true blackbody radiation.” As I recall, this is based on the theory that the structure of the black body surface is so complex that it has a continuous spectrum of oscillatory mechanisms and can thus radiate or absorb at any frequency with equal probability. To a greater or lesser degree, this applies to the surface of any opaque object. Think of a black body as a potential ‘white noise’ optical radiator if it were white hot.
Also, I believe, ionized gases or plasma can become sufficiently complex and interactive to radiate and absorb over a broad range of frequencies as well.
Transmission of blackbody radiation through transparent materials is another issue.
Ok I have read now the remaining posts (while all of you sleep :)) .
They actually are mostly either off topic or adress points that I have already adressed in my answers above .
Trying a summary :
1)
Posts confusing thermodynamic equilibrium and local thermodynamic equilibrium .
This is a common trap and that’s why I spent some time to define what LTE is and what it is not . A volume in LTE has not a constant temperature . For instance the troposphere has a variable temperature yet it is in LTE . So saying or implying that I have somehow supposed thermodynamic equilibrium of the whole atmosphere is not correct .
2)
Posts establishing relations between radiative transfer or green house effect and my post . There are none . My purpose is only and specifically to examine the question whether there can be a net energy transfer between CO2 and N2 by collisions supposing that we have LTE , time symmetry and energy equipartition . In other words whether CO2 can heat N2 . The answer is no .
3)
Posts that deny explicitely or implicitely that CO2 emits IR at frequencies corresponding to vibrationally excited states .
This is in contradiction with experience which shows that CO2 radiates strongly at those frequencies .
4)
What I miss and what I hoped because that would give , I think , the most interesting discussion would be an argument denying my conclusion AND showing how an existence of a net energy transfer by collision is consistent with the time symmetry of the process within a small volume in LTE or even in TE for that matter because such an argument if it existed would have a very general validity .
K~Bob
http://earthobservatory.nasa.gov/Features/EnergyBalance/images/annual_solar_insolation.png
W/day/m^2 in summer is greater in Antarctica than any place else on earth. The sun is closest to the earth in December, and they get 24 hours of sunshine.
If it weren’t for the high albedo and lack of GHG, Antarctica would be quite warm in the summer.
stevengoddard says:
August 5, 2010 at 10:57 pm
“Ric Werme,
Antarctica receives more solar energy in midsummer than Phoenix does. Instead of being knee-jerk sarcastic, think about the implications.”
Just where do you get such an idea? I don’t have data for Antartica, but compare Barrow, AK with Phoenix, AZ here: http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/state.html
If you make the same assumptions and caveats with regard to a paper towel in a microwave, would you come to the conclusion that water molecules do not heat the paper? From the circularity, I’m still coming to the conclusion that that is what you’d conclude. In reality, a dry paper towel in the microwave is not (substantially) heated, but a wet paper towel is heated by the water which is excited by the microwaves. The presence of water in the paper towel is, in my mind, analogous to the presence of CO2 in N2 in that its presence increases the temperature of the entire system because of the presence of a body which absorbs the radiation. Perhaps I’m too stuck in what seems like common sense to see the theory or perhaps you’re too stuck in theory to do the check of theory vs. common sense. I’m really not sure which because you sound like you know what you’re talking about.
Merrick says:
August 5, 2010 at 11:49 am
Thanks for your very patient response. I had confused more radiation being absorbed due to the molecular nature of CO2 vs. N2 with just simply adding heat to the system, which didn’t seem fair to the argument.
On a separate but closely related point; I have heard the argument put forth that CO2 absorption of IR in the bandwidths that apply is 100% in the first several or 10’s of meters above the surface of the earth, so additional CO2 will not have any meaningful effect on surface T. Now, understanding that we’re really dealing with an equilibrium, so 100% absorption on one level only translates to 100% emission of that energy (in whatever bandwidth is applicable), does that argument (100% IR absorption is already present, so no additional warming can be achieved through higher concentrations of CO2) hold any water? I thought I’d ask since you did such a great job of replying to my prior questions.
Thanks again.
RE: TomVonk: (August 6, 2010 at 5:57 am) “My purpose is only and specifically to examine the question whether there can be a net energy transfer between CO2 and N2 by collisions supposing that we have LTE , time symmetry and energy equipartition . In other words whether CO2 can heat N2 . The answer is no.”
I believe, in local thermal equilibrium we must assume, by definition, that there can be no net transfer of energy between any of the components of the atmosphere. But as soon as we let energy exit or enter that local region, then we must look to the potential establishment of a new, more global, thermal equilibrium.
If CO2 and H2O molecules now are cooled below the previous equilibrium point by having their radiation allowed to escape to outer space, then I believe these molecules must then tend to absorb more energy than yield energy with each interaction with the other components of the atmosphere until that atmosphere as a whole reaches a new thermal equilibrium where the net radiation going out and the net radiation coming in (primarily from the sun and the surrounding atmosphere) is the same.
TomVonk says:
August 6, 2010 at 5:57 am
“saying or implying that I have somehow supposed thermodynamic equilibrium of the whole atmosphere is not correct .”
You have ignored all boundary conditions, so your LTE implicitly extends throughout the whole atmosphere.
“Posts establishing relations between radiative transfer or green house effect and my post . There are none . ”
Thanks for acknowledging that your post has nothing to do with the greenhouse effect.
“My purpose is only and specifically to examine the question whether there can be a net energy transfer between CO2 and N2 by collisions supposing that we have LTE , time symmetry and energy equipartition . In other words whether CO2 can heat N2 . The answer is no .”
You didn’t need such a long post to show that two substances at the same temperature will not heat each other up!
Tom Vonk says:
Tom, it is my understanding that in a CO2 laser, excited nitrogen (N2*?) is used to transfer energy to CO2 to get it to lase. If this is so, why can’t transfers occur in the opposite direction, or have I misunderstood something?
Henry@Tom
I am afraid this does not explain what I observe.
Standing in the African sun, as the humidity in tha air increases the direct heat from the sun on my skin becomes less. How do you explain this?
Read this:
http://wattsupwiththat.com/2010/08/03/new-carbon-dioxide-emissions-model/#comment-447343
How can we measure the radiation from the absorptive wavelenghts of CO2 as it bounced back from the moon to earth unless it had come from earth first? So the path was: sun-earth-moon-earth. How would you explain that?
What I see is happening is this : The photon hits on the molecule, but at its absorbative wavelengths some are absorbed until the molecule is filled. At that stage the molecule becomes like a “mirror” at that wavelength and therefore this light is bounced back.
Because of the random position of the molecule 50% goes back to space and that is why I observe cooling (on my skin) as the humidity in the air increases.
The same argument can be made for the greenhouse effect. In fact this was the definition that I picked up, (but it seems that someone changed it later) :
Quote from Wikipedia (on the interpretation of the greenhouse effect);
“The Earth’s surface and the clouds absorb visible and invisible radiation from the sun and re-emit much of the energy as infrared back to the atmosphere. Certain substances in the atmosphere, chiefly cloud droplets and water vapor, but also carbon dioxide, methane, nitrous oxide, sulfur hexafluoride, and chlorofluorocarbons, absorb this infrared, and re-radiate it in all directions including back to Earth.”
end quote.
My interpretation:
Water and carbon dioxide behave similarly when exposed to infra red radiation. Each molecule excepts a number of photons. Once this transaction is completed the molecule becomes sort of like a little mirror to infra red radiation and the molecules start reflecting the infra red. Because of the random position of the molecules we may assume that at least 50% of the infra red from earth is radiated back to earth. The process then repeats itself.
the point that I have been making is that many substances in the air e.g ozone, carbon dioxide, water vapor, methane, etc.
cause both cooling (by deflecting or re-radiating sunshine) and warming (by re-radiating earthshine) but nobody can tell me what the nett effect is.
We don’t need more stories and theories. We need actual tests and measurements.
At this stage, unless I see some actual measurments in the right SI dimensions, I can not be sure that carbon dioxide is a green house gas,
i.e. that the warming effect is > than the cooling effect….
RE: cba: (August 6, 2010 at 5:44 am)“To use HITRAN in a spectrum, one must create a line width based upon temperature, pressure, including partial pressures of the molecule of interest and the ‘other molecules present’, and the line data associated with width.”
In this case, it is the mere existence of the line series that I considered important. Instead of dealing with the allowed energy states of an electron, I visualize we now have energy states associated with multi-proton atoms moving with respect to each other with several degrees of freedom. All the lines around 15µ give me the impression that these may represent allowed quantum energy states of a classical vibration mode.
My expectation is that the continual collisions with other molecules should cause each molecule to be in a continuously excited state and that photon emissions or absorptions only cause the vibration amplitude to jump from one allowed energy step to another.
REF: HITRAN Spectrum Plotting Tool
http://savi.weber.edu/hi_plot/
TomVonk says:
August 6, 2010 at 5:57 am
“1)Posts confusing thermodynamic equilibrium and local thermodynamic equilibrium .
This is a common trap and that’s why I spent some time to define what LTE is and what it is not . A volume in LTE has not a constant temperature . For instance the troposphere has a variable temperature yet it is in LTE . So saying or implying that I have somehow supposed thermodynamic equilibrium of the whole atmosphere is not correct .”
You still do not seem to appreciate that if you have radiant flows at different temperatures you do not have LTE. You also do not seem to appreciate that even systems not in LTE have thermodynamic temperatures. In space, for instance, there are billions of temperatures all superimposed (one for every star and nebula that’s visible plus temperatures for neutral gases, excited gases, ionised gases, electrons, cosmic rays, magnetic fields, dust, microwave background, neutrinos, and probably lots of others I haven’t mentioned).
“2) Posts establishing relations between radiative transfer or green house effect and my post . There are none . My purpose is only and specifically to examine the question whether there can be a net energy transfer between CO2 and N2 by collisions supposing that we have LTE , time symmetry and energy equipartition . In other words whether CO2 can heat N2 . The answer is no .”
The title of your article says “CO2 heats the atmosphere – a counter view”. Not “Can CO2 heat N2 under conditions of LTE, time symmetry and energy equipartition?”, to which the obvious answer, by definition, is simply, “no” – at least if by “heat” you mean “cause an elevated temperature” as distinct from “transfer energy, but not necessarily net energy, to)”. You specifically related your article to conditions in the atmosphere and common explanations of the greenhouse effect within that atmosphere. And as such, it’s plain wrong.
Richard Sharpe
Tom, it is my understanding that in a CO2 laser, excited nitrogen (N2*?) is used to transfer energy to CO2 to get it to lase. If this is so, why can’t transfers occur in the opposite direction, or have I misunderstood something?
This is true . But there is no LTE in a CO2/N2 laser . Basically when you mix a very hot gas with a cool gas you will expect a transient period where equilibrium will be established . It is during this period that CO2 degrees of freedom get excited .
Clyde Rhodes
The presence of water in the paper towel is, in my mind, analogous to the presence of CO2 in N2 in that its presence increases the temperature of the entire system because of the presence of a body which absorbs the radiation.
I can see many differences but no analogy . Gases , liquids and solids do very different things . When a direct approach of a problem is possible one should avoid analogies especially if one is not sure that every process is effectively analogous .
Paul Birch
We don’t. Really. We have an approximation to LTE. We have an approximation to equipartition. But it is the departures from LTE, caused by and causing superimposed radiative fluxes at different temperatures etc
There are 2 methods to answer that .
One is to put a dozen of links showing that this is the working hypothesis especially in all radiative transfer models .
I prefer the second which is to ask you to specify what you mean by approximate LTE
You know , a system not in LTE can only be treated by QM .
A system in LTE can be treated semi clasically .
There is a rather sharp distinction between the 2 states . As I have written LTE is a statement about a neighborhood of a point . It doesn’t say how large it must be , it only says that it must exist . And questions of existence are not “approximate” , they have mostly binary yes or no answers .
Henry Pool
Water and carbon dioxide behave similarly when exposed to infra red radiation. Each molecule excepts a number of photons. Once this transaction is completed the molecule becomes sort of like a little mirror to infra red radiation and the molecules start reflecting the infra red. Because of the random position of the molecules we may assume that at least 50% of the infra red from earth is radiated back to earth. The process then repeats itself.
This is a non conventional and partly wrong view of what really happens .
You should not focus on individual molecules but on an ensemble with billions of them .
At every instant some of them will absorb IR and increase their vibrational energy . At the same instant others will emit IR and decrease their vibrationnal energy . And others still at the same instant will undergo collisions in which they will or will not change their quantum states . If you followed an individual molecule you would see an apparently chaotic and random behaviour in which it keeps jumping up and down among its quantum states . The order appears only when you make a statistic on a very large number of them .
The reason of 50% is not in the position of the molecules but in the isotropy of the emission .
While there is a non isotropic flow of IR (from ground up) as long as it is not absorbed , it is isotropically reemitted . So in a plane approximation you can say that statistically over a large number of reemissions , half will be up and half will be down .
Look over this graph at http://upload.wikimedia.org/wikipedia/commons/7/7c/Atmospheric_Transmission.png
There is an atrocious omission and flawed data there. Have you got it? The vertical axis says: “major” components (of the atmosphere); nevertheless, they omitted Argon. Argon mass fraction in the atmosphere is more than 20 times higher than CO2 mass fraction. Argon’s totabs and totemiss is 128% higher than CO2 totabs and totemiss. The absorption spectrum of Argon is wider than CO2 absorption spectrum. What’s the reason the author omitted such important component of the atmosphere, i.e. Argon, and included CO2, which mass fractions and totabs and totemiss are so insignificant?