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 .
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Reed Coray says:
August 7, 2010 at 11:57 am
Hi Reed,
(1) “Here’s what I think you mean when you use the term “trapping” as applied to greenhouse gases. Greenhouse gases “trap” radiative energy because they absorbed IR radiation from the Earth’s surface which then continually “bounces up and down”. Is that correct?”
Yes, that is what I mean.
(2) Nice example with the cube and the sphere. Clearly, I spoke too soon, but I should say in my defense that I have trouble visualizing your constant internal source of heat, physically. Blackbodies are traditionally heated from the outside, not the inside (although the sun is a good counterexample).
Anyway, you add a layer of material to the surface of your sphere, and then…? You have more material to heat from the same constant internal source. At first sight it would certainly seem that you are spreading the energy thinner. Of course, you could in principle postulate any kind of internal energy distribution that you wanted. Technically, only the surface needs to be a blackbody… It does not look like a well-posed problem to me.
Regarding the real world earth and the real world atmosphere–granted that the temperature measures different things for different objects, but in all cases there is a proportionality constant between the (change in) thermal energy content of an object and its (change in) temperature, and that is the heat capacity. So when you bring objects at different temperatures into contact, how much energy flows from one to the other depends on the heat capacities. The atmosphere has a very small heat capacity compared to the Earth, so it can only absorb a small fraction of the Earth’s thermal energy before reaching thermal equilibrium. That’s the main point I was trying to make.
(3) “Will trapping of heat by an object cause a temperature rise in the object?” If by heat you mean thermal energy, yes, of course, as long as the object has a positive heat capacity. If you put thermal energy into an object, its temperature will go up. What else could it do?
Is infrared radiation “heat”? Well, it is energy. Does it count as thermal energy or not? It depends on the context. Is the infrared radiation “trapped” in the Earth’s atmosphere part of the whole planet’s (Earth + atmosphere) “thermal energy”? Probably yes, depending on how you choose to keep your books. It’s hard to imagine what else it could count as. Does it keep the Earth’s surface warmer than it would be it it wasn’t there? Definitely yes.
TomVonk says:
August 7, 2010 at 3:29 am
I’m not disparaging narrowly-based advances in physical understanding. As an applied scientist, I do look for their utility in the real-world setting. Sometimes, looking from the oppsite viewpoint of the particle-wave duality provides additional insights. I’m not sure that your formulation of energy transfer from CO2 to N2 molecules helps explain the extinction of 15 micron terrestrail radiation within ~100m of the surface. Please enlighten me.
Kevin Kilty says:
August 7, 2010 at 11:55 am
“The only temperature profile I have seen for Venus looks pretty linear in the region of measurement, and the linear extrapolation to the surface comes quite close to the surface temperature. I wonder though if the adiabatic lapse rate on Venus shouldn’t be a bit steeper at high elevation (below 300K) than near the surface simply because the specific heat of Co2 is larger at high temperature. I’d guess a 20% change, but I’m unsure of the effect of gasses like So2.”
I’m not sure if it’s significant, but the lapse rate going through the cloud layers ought to be some sort of “wet” rate for the sulphuric acid. Also, the lapse rate doesn’t depend on the specific heat but on the ratio of specific heats Cp/Cv; perhaps this is what you meant. Higher up, the composition of the atmosphere changes too – less CO2 and more diatomic species like N2, which have a higher Cp/Cv and therefore higher lapse rates.
Spector says:
August 7, 2010 at 7:52 am
RE: cba: (August 7, 2010 at 4:00 am) “Spector, rather than 100/1, I’d suggest it’s closer to 2/1, excluding those that are simply not going very far. That’s based on there being a factor of two in absorption, roughly co2 total contribution 30W/m^2 versus h2o at 64w/m^2 in the approximately current clear sky atmosphere calculations.”
Is this, 64w/m^2, perhaps a measure of that portion of the blackbody radiation energy emitted from the earth (‘earthshine’) that is blocked by the saturated H2O absorption spectrum as opposed to the relative ability of any given parcel of air to capture or export heat via the H2O photon radiation path?
Spector,
the 64 w/m^2 is the difference between h2o and no h2o in my model atmosphere which incorporates radiative transfer. Starting with the surface, each layer attenuates the BB spectrum with the ghg lines and will also emit at the ghg spectral line wavelengths but with a lower power level if the layer temperature is less. If I included only what was emitted by the surface, it would be somewhat more than that – and it would be rather irrelevant because all those h2o molecules in the atmosphere are going to be involved in radiating their characteristic spectrum at their respective temperatures. As one goes higher in the atmosphere, the temperature will be cooler so the higher energy states and shorter wavelengths will not be radiated as much and the total energy radiated will be less. Also, at higher altitudes, the individual lines will become narrower due to lower pressure.
I hope that made sense.
Bryan says:
August 7, 2010 at 8:17 am
cba says:
On H2O/CO2 radiative flux ratio
…..”rather than 100/1, I’d suggest it’s closer to 2/1, excluding those that are simply not going very far. That’s based on there being a factor of two in absorption, roughly co2 total contribution 30W/m^2 versus h2o at 64w/m^2 in the approximately current clear sky atmosphere calculations”………..
I disagree, the paper below shows that that even for deep midwinter Antarctica with extremely low humidity, clear skies conditions, the radiative flux from H2O vapour was more than twice that for CO2.
Since CO2 is well mixed in the atmosphere its radiative contribution will not change much as we move to more average conditions.
However if we move to more average Earth condition with average levels of humidity introducing condensation phase change then the radiative contribution from H2O will increase dramatically.
On the other hand there does not seem to be any physical reason for CO2 to become dramatically more radiative at more average conditions.
————
Thanks for the reference Bryan,
I’ve only time right now to read the abstract. It doesn’t look that far off for summer. I’ve got a model with 107w/m^2 (out to only 75 um) which is a few % shy of the whole spectrum, out past 100um and into radio). that 107 is total absorption of outgoing. It’s going to be close (I think) to their back radiation measurements of 110-120w/m^2 summer. I’m using a 1976 std atmosphere model which is much closer to that of the summer than the winter. Essentially, I’ve got about 2/1 just like they have – with around 10 or 15 w/m^2 due to all but co2 and h2o. It looks like I’m definitely in their ballpark from their abstract.
I hope to have some more time tomorrow or later tonight to read the whole paper.
Hi Julio,
At some point we’re going to have to “scale back” our interchange–if for no other reason than it consumes a lot of my time, and I’m sure yours as well. Since I’m retired, it’s not a serious issue with me; but if it is with you, just let me know and I’ll scale back.
I don’t claim to have all the answers–or for that matter, even a few of the answers. I am trying to understand if a logical and internally consistent argument for the claims made by the AGW community exists. Specifically the claims that (a) greenhouse gases rather than something else (e.g., gases in general, or friction resulting from atmospheric and liquid convection currents that arise because various regions of the Earth are at different temperatures and because gases and liquids in motion encounter the “Corlios Force” which exists because the Earth is rotating relative to inertial space, etc.) are primarily responsible for the Earth’s surface temperature being higher than it would be for a perfect-thermal-conducting black body Earth; (b) man’s current and future use of fossil fuel will cause a “significant” change to the Earth’s surface temperature, (c) such a change will be bad for mankind, and (d) the AGW community cure (drastically curtailing our use of fossil fuels to generate energy) won’t be worse than the disease. When I see discussions that employ terms that are not well defined and/or appear to be self-inconsistent, I ask questions–mostly to myself, but sometimes to the authors of those statements.
In your August 7, 2010 at 1:05 pm response you said: “Anyway, you add a layer of material to the surface of your sphere, and then…? You have more material to heat from the same constant internal source. At first sight it would certainly seem that you are spreading the energy thinner“. If by thinner you mean the ratio of (a) the “rate-of-internal-energy-generation” …to… (b) the “object’s volume” is decreasing, of course I agree with you. I’m just not sure this observation has anything to do with the surface temperature of the body.
In another regard, I agree with the idea that if body “A” completely surrounds body “B”, the presence of body “A” may and even “likely will” affect the temperature of body B. To give a specific example. Assume body “B” is a thin spherical shell (external radius, RI) with a radioactive source (make it an alpha particle emitter so that the energy of the radioactive particles are quickly converted to random molecular motion) uniformly (i.e., symmetrically) distributed throughout the shell wall. Further assume that the shell’s internal and external surfaces radiate as a black body. Place the shell in the vacuum of space where the temperature of “space” is 0 Kelvins. Because the energy source is symmetrically distributed throughout the shell wall, at some point the external surface of the shell will reach a temperature such that the rate of energy radiated outward by this surface will be equal to the rate at which radioactive decay energy is generated internal to the shell wall. Let that temperature be Tisolated.
Now surround body “B” with body “A”, which is also a thin spherical shell (external radius RE) with internal and external black body radiating surfaces, but contains no internal radioactive energy source. Let the internal radius of spherical shell “A” be greater than the external radius of spherical shell “B”, and don’t allow the two shell surfaces to come into contact. I don’t think it’s a requirement, but for simplicity assume the two shells are co-centered. Finally, assume the thermal conduction properties of the outer shell (shell “A”) are “almost perfect” so that the temperature of the outer shell’s internal surface and the temperature of the outer shell’s external surface are infinitesimally different. For these conditions, when radiation-rate-equilibrium is reached for the “two-shell system” (i.e., when the rate of energy being radiated outward by the outer shell equals the rate of energy being generated in the wall of the inner shell, I believe the presence of body “A” will affect the temperature of the external surface of the inner shell. Specifically, under these conditions I believe the temperature, Tdual of the external surface of shell “B” will be given by:
Tdual = Tisolation * [1 + (RI/RE)^2]^0.25
If valid, the above equation indicates that unless RE is infinite, the presence of the surrounding shell will, in radiation-rate-equilibrium, result in an increase in the temperature of the external surface of the inner shell. In this sense, provided energy is supplied to the Earth but not its atmosphere, I believe it is possible for an atmosphere to raise the temperature of the Earth’s surface. The physics that led to the above equation isn’t trivial (at least not to me); but compared to the physics of the Earth and its atmosphere, it’s less than trivial, it’s super trivial.
In part, this is why I am skeptical of anthropogenic global warming. The physics of atmospheric thermodynamics, especially with all the unknowns that exist, is I believe far too complex to use simple arguments like “radiation is trapped, and therefore the temperature should rise by X degrees”. I think it may even be too complex to even argue that the presence of greenhouse gases at the levels they are present on the Earth will cause the Earth’s temperature to rise any measurable amount. I think it likely, but I haven’t come across a conclusive argument yet. That, by the way, doesn’t mean (a) a conclusive argument doesn’t exist, or (b) one does exist that I’ve seen but I don’t have the wherewithall to understand it. What it does say, is that I am FAR from being sufficiently convinced of the argument to be willing to make the drastic changes in the way the western world generates energy advocated by a large section of the AGW community.
One last point. In response to my question: (3) “Will trapping of heat by an object cause a temperature rise in the object?”; you answered: If by heat you mean thermal energy, yes, of course, as long as the object has a positive heat capacity. If you put thermal energy into an object, its temperature will go up. What else could it do? Answer: Over the short term, the energy could go into a change of phase state–ice at 32 degrees Fahrenheit to water at 32 degrees Fahrenheit, which doesn’t involve a temperature rise. Eventually though, if you keep adding energy in the form of heat to a system, I believe the system temperature will rise. However, if (a) after the atmosphere’s temperature has risen to achieve radiation-rate-equilibrium the atmosphere is still “trapping heat”, and (b) “trapping heat” will cause temperature to rise, isn’t it correct to conclude that the atmosphere temperature will “rise some more”, and this rise will only stop when the atmosphere ceases to “trap heat?”
Thanks again,
RE: Paul Birch: (August 7, 2010 at 3:49 am) “All gases absorb and radiate at all wavelengths.”
RE: Paul Birch says: (August 7, 2010 at 11:10 am) “2) Collisions between molecules are not quantised; they produce some absorption at all wavelengths.
I do not believe this second statement to be true or the atmosphere of the Earth would not be optically transparent and perhaps there would be a visible glow in the sky at night due to the reciprocal photon emission process. I suspect that photon emission or absorption requires a quantum electromagnetic field change.
Perhaps such collisions can only cause kinetic energy exchanges and perhaps trigger or assist in the release or absorption of photons characteristic of the molecules involved in the collision. Both molecules must return to one their normal allowed quantum electromagnetic states after impact.
RE: cba: (August 7, 2010 at 5:17 pm) “Spector, the 64 w/m^2 is the difference between h2o and no h2o in my model atmosphere which incorporates radiative transfer.”
Thanks. My basic question is this: if one looked down from above the tropopause, how much energy is being exported out of tropopause region and below by H2O and how much by CO2 in those saturated bands where no radiative transport from the surface is possible?
I would expect the factor of 77 to 103 greater abundance of H2O relative to CO2 to be reflected in the role H2O serves in cooling of this vital region unless this advantage was compensated by a higher rate of absorption of that energy caused by the same high abundance of H2O.
I believe the numbers that you are quoting represent the ratio of the saturated absorption bandwidths of H2O and CO2 as seen from the surface.
Nasif,
The number I’m using, 64 W/m^2, is not from trenberth nor is it the same thing. Rather it is the difference in radiative transmission between h2o being present and being totally removed as calculated at the tropopause over a 75 um range and assuming conditions of a 1976 std. atmosphere under clear sky conditions. If one were to take the model and plot a curve of emission versus wavelength rather than simply summing up the values (integrating over wavelength), it would show a spectral curve of a black body at 288.2k with the absorption lines of the atmosphere dipping down to the point where there is a spectral curve for a lower temperature at which there is emission going on in the wavelength bands associated with ghg absorption.
Spector,
See above explanation to Nasif Nahle. I think there’s probably good correlation between what is seen at the surface coming down and my numbers but it is not at all the same thing. Mine is a calculation made at the tropopause. It is not made with only saturated lines or this or that, it is made up of the whole spectrum – as calculated. This includes the individual line contributions including their wings and including the overlaps of various line widths at the same wavelength. And, there are tens of thousands of lines from 39 different ghg contributor molecules. Saturated simply means that the lengths of the air column are much longer than the mean free path of a photon and that is an extremely complex function of wavelength. Some path lengths are in the vicinity of a few cm while right next to that wavelength, the pathlength is longer than the thickness of the atmosphere. I get the function by adding up all the contributing amounts from all the wavelengths and molecules and then calculate the path length information.
Spector says:
August 7, 2010 at 7:19 pm
RE: Paul Birch: (August 7, 2010 at 3:49 am) “All gases absorb and radiate at all wavelengths.”
RE: Paul Birch says: (August 7, 2010 at 11:10 am) “2) Collisions between molecules are not quantised; they produce some absorption at all wavelengths.
“I do not believe this second statement to be true or the atmosphere of the Earth would not be optically transparent and perhaps there would be a visible glow in the sky at night due to the reciprocal photon emission process. I suspect that photon emission or absorption requires a quantum electromagnetic field change.”
The atmosphere of the Earth is not optically transparent. It is nearly transparent. Your suspicion is not correct: continuum emission and absorption is well known.
“Perhaps such collisions can only cause kinetic energy exchanges and perhaps trigger or assist in the release or absorption of photons characteristic of the molecules involved in the collision. ”
This is not the case. The energy and momenta of the absorbed photons are transferred to the molecules. Since the speed, direction and impact parameters of the molecules are not quantised, nor is the absorption spectrum. An interesting feature of this mechanism is that the absorption per unit mass is proportional to the density (because so is the collision frequency), which means that the mechanism is significant in dense stellar or planetary atmospheres (like those of Venus or Jupiter), but quite hopeless in interstellar space. At high densities collisions between three or more molecules at a time also add to the absorption rate.
By the way, scattering due to Brownian density fluctuations in the gas also contributes to the opacity.
Tom Vonk,
Hi Tom,
Thinking back, I may have applied an equilibrium concept to LTE that isn’t part of the usual interpretation concerning photons but it is a similar concept to the original. When I was saying that the photons were in LTE equilibrium, I was referring to the concept that the continuum spectrum coming in to the gas cloud was at the same temperature as the temperature of the gas cloud of your conceptual example. If that is the case and if the continuum is coming in from all directions, then there is no net radiative power transfer going on and in fact, one would not observe any absorption spectra (or emission spectra) at all. For the geometry we have, Earth at T = ~288k, space T = ~2.7k, it is not really possible to have that situation as each small layer of the atmosphere between surface and space will radiate outward and inward its characteristic spectrum based upon its temperature and the incoming radiation to that layer will be a combination of what is left of the surface continuum PLUS what makes it through to the layer from those layers lower down radiating outward and those layers above radiating downward. In the case where there is rather little change in the layer’s temperature and pressure and molecular constituency, there will be little net power flow happening at those wavelengths of significant interaction. For wavelengths other than those, there is little or no interaction going on. I think this is the reason why physical meteorology concepts are capable of describing convection using adiabatic assumptions with any success.
RE:Paul Birch: (August 8, 2010 at 4:39 am) “Collisions between molecules are not quantised; they produce some absorption at all wavelengths.
I note there is a paper in Nature that is titled “Collision-induced Absorption in the Earth’s Atmosphere” by C. B. Farmer & J. T. Houghton, March 26, 1966.
The abstract of this report says most of the absorption [and thus radiation] observed in the atmosphere is due to its trace components such as water-vapor, carbon dioxide and ozone. It goes on to say that the major components have no dipole moment so that their vibration and rotational energy states cannot be excited directly by absorption of radiation, but dipole moments are induced during the collision process. The results of these collisions, they say, can be seen in very long [very minor] absorbing paths in the atmosphere.
Their report shows that absorption of the solar spectrum near 4.2 microns is due to collision-induced absorption by nitrogen and they present a short discussion of the importance of the rotational collision-induced bands of oxygen and nitrogen in the radiation budget of the atmosphere.
Spector says:
August 8, 2010 at 10:21 am “…”
Yes, that’s just as I said.
TomVonk says:
August 6, 2010 at 4:08 am
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 .
Which where you get it wrong. There is necessarily a net transfer of energy to the non-CO2 molecules approximately equal to the total energy absorbed by the CO2.
Your simplification of the process to a single step renders the analysis nonsense.
If we just reduce the absorption to a single line then we will have CO2 molecules with a particular state (v=1, r), the multiple collisions will deactivate this excited state to a variety of lower states (v=1 or 0, many r). The large numbers of collisions will distribute these states over a Boltzmann distribution appropriate to the steady state temperature reached. The time spent by an individual molecule in a particular state is extremely small at atmospheric conditions, orders of magnitude less than the mean radiative lifetime which is why emission is extremely unlikely, and most of the energy ends up thermalized. If the world worked the way Vonk thinks it does then my experiments on Laser Induced Fluorescence would be much simpler, even with much shorter lived states (~10ns) collisional quenching dominates the energy cascade and often only one molecule in a thousand will fluoresce! In order to perform the analysis you need to set up the kinetic equations between each state. I hope to follow the rest of this thread but I may not get back frequently as I’m away from internet access at present (relying on visits to coffee shops).
An experiment can illuminate a principle.
An analogy can give it perspective.
If one shines a really bright light on the Mississippi can he boil Lake Superior?
Tom dumbed his presentation down a little for you folks.
The trouble with Tom’s LTE analysis is that it ignores differential calculus. The temperature and emissions will be infinitesimally less immediately above the sample as compared with below it. The smaller the sample, the smaller the difference. But it will always be there and cumulatively: It will add up to the difference between the ground temperature and the temperature at the top of the troposphere where CO2 radiates to space.
Add CO2 and you will catch an infinitesimally more thermal radiation at each layer. The result is that the ground becomes warmer because there is more “thermal resistance” to pushing thermal energy from the earth to the top of the troposphere. Tom is correct that adding CO2 to a black box does not change the temperature inside the box. But if you have a thermal gradient, it does slow the transfer of heat via infrared radiation from one side of the gradient to the other.
Adding a differential equation to account for the slowing of the radiative flux would bring Tom’s analysis in line with the mainstream.
On the momentum conservation question I raised:
1) the effect of the asymmetry per molecule is small. I make it out for CO2 at a 15micron absorption , if we assume about 300 meters/second velocity of the molecule a 10^-5 in percent addition to the motion. It will add more to the part of the distribution that has smaller velocities depending of course on direction. A model is needed :). Nevertheless in principle when there is energy transfer there is momentum transfer and isotropy is broken.
2) in total CO2 in its collisions with other atmospheric molecules will transfer this upward impetus and it will become part of the “breathing” of the atmosphere, (limited by gravity as in the example of cbs with the ball) together with convection.
3) It should play a part in keeping CO2 aloft, as it is heavier than other molecules and would stratify lacking convection and maybe radiation pressure. I have not done the calculation, it probably again is a very small effect.
cba
Thinking back, I may have applied an equilibrium concept to LTE that isn’t part of the usual interpretation concerning photons but it is a similar concept to the original.
Yes I think that I have seen what you tried to say .
My point was that LTE (condition that I am using) is independent of the questions whether there is or is not radiative equilibrium . That’s why I did no assumption about the precise properties of the radiation (isotropic , continuous , black body etc) .
The purpose was to observe what happens in collisional V/T interactions and whether there is or is not energy transfer from N2 towards CO2 via a T->V process .
The fact that LTE is a property of material particles only allows to decouple this question from much more complex questions concerning radiative transfer over large distances (measured in km) about which there is enough papers anyway .
For this same reason if there is a conclusion about T/V interactions , it will be largely independent from detailed assumptions concerning the nature of the IR radiation .
The problem with many reactions is that they focus on something else altogether – namely radiative transfer and then seem to construct relations between radiative transfer and my post which are simply not there .
Phil
The time spent by an individual molecule in a particular state is extremely small at atmospheric conditions, orders of magnitude less than the mean radiative lifetime which is why emission is extremely unlikely, and most of the energy ends up thermalized.
This is just hand waving . How “extremely unlikely” is the emission ? What is the curve of the radiative life (probability vs time) ? What is “extremely small” ?
How much is “most” , more than 50% ? What is the probability of CO2 vibrational excitation , do you take it for 0 ?
Needless to add that you have adressed neither of the arguments I used .
If somebody used the same kind of hand waving arguments you do , you would probably qualify them as “nonsense” . I have more respect for people’s opinions so I won’t . What’s really your point ?
Tom Vonk,
But Tom, doesn’t the requirement of LTE require a transfer of energy between the CO2 and N2? When you change the conditions of radiative energy coming in, the temperature of the CO2 will need to adjust in order to radiate it away. If there is no interaction with N2, energy transfer-wise, the N2 will remain at the old temperature and you’ve got the loss of the LTE condition.
cba
But Tom, doesn’t the requirement of LTE require a transfer of energy between the CO2 and N2? When you change the conditions of radiative energy coming in, the temperature of the CO2 will need to adjust in order to radiate it away. If there is no interaction with N2, energy transfer-wise, the N2 will remain at the old temperature and you’ve got the loss of the LTE condition.
Yes it does . Actually the whole point of LTE and energy equipartition law is to make sure that all degrees of freedom of material particles interact sufficiently so that all energy forms are “well mixed” .
It is when they are not well mixed that very difficult problems begin .
An example . One can define vibrational and translational “temperatures” by taking the average V and T energies . There is à priori no reason that they should be equal .
But when they are (what means that V and T degrees of freedom often and strongly interact) then we have LTE .
It is precisely this kind of graphics that shows that the troposphere is in LTE while the stratosphere is not . That’s also why radiative transfer models (see f.ex SAMM2 http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA427001&Location=U2&doc=GetTRDoc.pdf) must distinguish between LTE and non LTE conditions what is apparently beyond the understanding of people like Birch .
And yes . If you change the radiation properties then there will be a transient in which all the degrees of freedom will interact and adjust themselves .
Clearly if you begin to fire a 1 kw IR laser in a volume , the final state will not be the same as what it was without the laser 🙂
In that sense it is true , there is an unsaid assumption in my post that the radiation properties are not dramatically changing at the scales and times i consider .
TomVonk says:
August 9, 2010 at 4:27 am
Phil
“The time spent by an individual molecule in a particular state is extremely small at atmospheric conditions, orders of magnitude less than the mean radiative lifetime which is why emission is extremely unlikely, and most of the energy ends up thermalized.”
This is just hand waving
Really? That’s all you’ve done in your flawed analysis.
How “extremely unlikely” is the emission ?
Didn’t you notice ‘orders of magnitude’?
What is the curve of the radiative life (probability vs time) ?
Usually a Poisson distribution, in the case of CO2 in the 15 μm band the mean is order 10ms, compared with a collisional lifetime of order ns.
What is “extremely small” ?
ppm as you chose to ignore I pointed out that parts per thousand was an experimental result for a much shorter lived excited state, inconvenient fact?
How much is “most” , more than 50% ?
More than 99%.
What is the probability of CO2 vibrational excitation , do you take it for 0 ?
No, what is the relevance of this question?
Needless to add that you have adressed neither of the arguments I used .
If somebody used the same kind of hand waving arguments you do , you would probably qualify them as “nonsense” .
Actually I’ve addressed them both, your model of collisional deactivation I showed was wrong, and your ideas about excited state lifetimes wrt emission I also showed to be wrong.
And I’d have followed up the suggested subjects in the literature and found they were right, you apparently couldn’t be bothered to do that.
I have more respect for people’s opinions so I won’t . What’s really your point ?
Not enough respect to follow up the points but instead you dismiss them out of hand. The point is that your analysis is flawed, try looking up Collisional Quenching or the Stern-Volmer equation, I gave you some hints before.
Tom Vonk,
Tom, that’s some reference. I saved it to go through later. I did notice that it stated while LTE conditions start to happen around 45-50km (above the stratosphere) but that co2 appeared to stay in LTE conditions past 100km. This 45-50km is the area that is at the bottom of the ionosphere – although usually the D layer doesn’t go below 60km.
Phil — you are right and wrong.
The time spent by an individual molecule in a particular state is extremely small at atmospheric conditions, orders of magnitude less than the mean radiative lifetime which is why emission is extremely unlikely, and most of the energy ends up thermalized.
Yes, almost always, unless you are in a vacuum, the energy absorbed by a CO2 molecule will thermalize. However, what happens if you are indeed in thermal equilibrium is that there will always be enough CO2 molecules with sufficient energy to radiate out at exactly the same rate as the flux in. Increase the flux in, the sample warms until it is in equilibrium and then there will be sufficient CO2 molecules with an energy state high enough to shoot out as many photons as are coming in. This is all Planck’s law of Black Body radiation.
You need to be sure not to look at the mean time a single CO2 has sufficient energy, but the total time based on the number of CO2 molecules bumped up.
As an example, I was running out some of calcs the other day and at the temperature I was using, only 1 in 800 CO2s had the energy necessary to shoot off a photon (and consequently, one N2 in 800 also had that energy), but that was exactly enough to balance the absorption I was dealing with. This is all your Maxwell – Boltzmann stuff.
As I pointed out in an earlier post, the weakness in Tom’s argument is you have to deal with a temperature gradient and the radiative flux gradient which is ignored in LTE model. It is this gradient that is the driver in exporting thermal energy from the earth’s surface to space.
Do you know any circuitry stuff? You can do a quick simple model of the total gas in the atmosphere as the thermal induction and the total CO2 as the thermal resistance. If you increase the resistance by increasing CO2 than the temperature (i.e., voltage) has to increase at the surface in order to maintain the same outflow of energy into space. (In actuality, there will also be a slight dip in the radiation into space in the CO2 band and an increase in some of the other bands. But the net flux into space has to remain a constant). Hope this helps.
Remo
However, what happens if you are indeed in thermal equilibrium is that there will always be enough CO2 molecules with sufficient energy to radiate out at exactly the same rate as the flux in. Increase the flux in, the sample warms until it is in equilibrium and then there will be sufficient CO2 molecules with an energy state high enough to shoot out as many photons as are coming in. This is all Planck’s law of Black Body radiation.
Absolute agreement . What Phil. doesn’t understand is that his thesis not only unexplicably violates the time symmetry of the processes which is at the basis of the statistical equilibrium but it would prevent any equilibrium to form . In order to have an equilibrium it is simply necessary that the rates of the V/T processes be equal . And yes this is text book stuff . Looking at the reference I linked for CBA helps too .
As I pointed out in an earlier post, the weakness in Tom’s argument is you have to deal with a temperature gradient and the radiative flux gradient which is ignored in LTE model. It is this gradient that is the driver in exporting thermal energy from the earth’s surface to space.
Yes and no . I actually don’t deal with radiative transfer , there are enough papers (see among others the reference above) that do that quite nicely .
I deal with small volumes in LTE and argue about what happens with the statistics of molecular collision processes WITHIN the volume . As already mentionned in a discussion with CBA above , LTE is independent of the radiation conditions and a property of material systems only so I do not need to make any specific assumptions about radiation . Sure implicitely I suppose that the IR intensity coming from outside of the small volume is constant so that equilibrium may be established . This is a reasonable assumption at the time scales we consider .
I observe that after 300 comments , not a single one contested either the time symmetry or the energy equipartition law . One commenter contests the existence of LTE (“it never exists”) and Phil. contests the existence of IR emission by CO2 . Both are inconsistent with observational evidence .