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 .
You don’t think it has anything to do with far lower insolation at the poles due to small angle of incidence compounded by an albedo of 0.85?
Not even a little bit? /sarc off
CO2 signature might be barely detectable at the south pole due to so little water vapor swamping the signal but the cold is due to the factors above.
The fact that the south pole is 3000 meters above sea level isn’t helping to keep it warm either. Dry adiabatic lapse rate is 9.8C per 1000 meters. North pole is near sea level. South pole should be 30C cooler than north because of altitude alone.
Seems like we’ve got a good peer review process going on here.
I guess the real science was settled after all.
stevengoddard says:
……”There isn’t any question that the greenhouse effect is real and that CO2 is a contributor. Places that have very little GHG (i.e. South Pole) also have very cold temperatures.”………..
CO2 is well mixed in the atmosphere.
Strangely enough even at the South Pole the H2O / CO2 backradiation ratio is at 2.6.
Sorry guys, but I’m not nearly smart enough to understand this post. Anyone game for a layman’s translation ?
The argument sounds to me like the following: A system (x) with CO2 and N2 under bombardment from IR will establish an equlibrium state with temperature Tx.
But this does not show the conclusion. In order to show the conclusion, two systems (a & b) under bombardment from the same amount of IR but with different CO2 concentrations must establish the same equilibrum state with temperature Ta = Tb.
So, Mr. Vonk, do they?
Bill Illis says:
August 5, 2010 at 9:07 am
Watch how fast Liquid Nitrogen absorbs infrared radiation/gains energy from the rest of the molecules in its space.
You’re kidding, right?
Physics in the macroscopic world gets real complicated real fast trying to analyze it at the quantum scale. I don’t think it’s worth trying and I’ve read a lot of equally credentialed physicists disagreeing about all sorts of stuff at the quantum level. One of my favorites keys off of time invariance mentioned in the OP. That leads to there being no such thing as time it’s merely an artifact of the law of entropy. It also leads to there being no way to destroy information such that if I burn every physics textbook in the world all the information in them is still in the universe recorded in completely time reversible quantum interactions. It implies that information obeys the same law of physics as energy – it can’t be created or destroyed but only change form and obey the laws of entropy while doing it. It also implies that maximum order in the universe was at the big bang and law of entropy has been taking its toll ever since. It also implies that the information content of the observable universe, if it’s a closed system, is invariant. Even more it tends to support a deterministic universe because every quantum change since the big bang is time invariant and so you can theoretically calculate every moment in the future and every moment in the past from a perfect snapshot of the quantum state of the universe at any point in time. Good luck with that though as there’s an estimated 10^80th particles in the universe and the math gets untractable when you have more than 3 particles in your closed system. Heck, even calculating precise gravitational interactions in a 3 or more body system becomes intractible and shortcuts have to be taken to get practical solutions for deep space exploration. There’s weird crap happening far out in the solar system on Voyager and Pioneer spacecraft not being at the position and velocity where theory says they should be and radiothermal power supplies not decaying at rates predicted upon what are axiomatically constant radioactive decay rate of the isotopes like it isn’t really a constant at all.
There are a great many unknowns in physics. People like to think we have a good handle on all of it and that’s simply not the case.
There are more things in heaven and earth, Horatio, than are dreamt of in the standard model.
Tom,
I think you’ve got almost all the basics right for a good explanation but there are some serious problems.
There’s also serious problems with most of the criticism on this thread.
The atmosphere will be in LTE very quickly with a change but not during the change. One description of LTE is the state that in a small area, all of the various molecule types are at the same temperature. With 10^11 typical interactions per second, it doesn’t take long for the energy to be distributed amongst both CO2 and N2. Without LTE, there is no temperature, rather one might have a separate temperature for CO2 and N2 and O2 etc – all in the same tiny area with interactions going on.
The primary problem is the presentation assumes a single temperature situation with the radiation. Yes, a batch of CO2 at a temperature will emit the same amount of IR it absorbs – IF the IR is coming from a source at that SAME temperature. The energy distribution of the IR is related to the temperature at which it was emitted. The absorption of the IR is only loosely related to the temperature at which it is absorbed.
If the IR is being emitted from the Earth’s surface as a BB continuum, there will be absorption occurring in a layer above that is based upon the amount of CO2 in that layer and upon the amount of IR present at each wavelength and the absorption at each wavelength will be a function of the the pressure and amount of CO2 present in that layer and to a far lesser extent, the temperature at that layer. Some wavelengths will be absorbed almost totally within a couple of cm while other wavelengths will have no absorption.
If the temperature of that layer is the same as the emitting surface, there will be no net absorption as it will radiate the same spectrum that it absorbs. Ultimately, that cannot be the case because what is being absorbed comes from below and what is emitted by this thin layer will be the same amount outward as is being absorbed PLUS that same amount downward. It’s a small conservation of energy problem. What’s happening in the atmosphere though is the layer is surrounded by layers above and below that are close in temperature (lapse rate) so there is some power coming up and some coming down – until one reaches the outer bounds and the downward effects dwindle. Note too that the conservation of energy is for all energy entering and leaving the layer, not just radiative energy.
If one had the case where the layer (or a clump of gas) was hotter than the radiating temperature of the black body behind it, the layer would radiate more power than it absorbed. There would be emission lines rather than absorption lines. Since the layer is pretty much guaranteed to be cooler than the black body surface, the distribution of energy states results in a distribution that would radiate a black body curve of lower peak frequency and lower total power.
The radiated spectrum can be constructed by superimposing the absorption spectrum (the likelihood of absorption in the layer for a given wavelength) on the black body spectrum for that lower temperature. The absorbed power in that layer is of course this same absorption spectrum times superimposed on the black body spectrum of the black body (Earth’s surface).
Note that these two conditions are only equal if we’re talking about one temperature for the black body surface and for the layer. It also assumes here we have no intervening layers involved.
Take a cell with transparent ends and perfectly insulated walls. The length doesn’t really matter. Now put a blackbody source at temperature T at one end and a spectrophotometer at the other. At equilibrium with cell contents and source at the same temperature, the spectrophotometer will see the same blackbody curve and total energy flux for T whether the cell is evacuated or filled with any gas or mixture of gases. SpectralCalc actually had a bug in their software for a while where this wasn’t true. If the cell contains an absorbing gas, you will see absorption dips and lower total flux at the detector if the temperature of the cell is maintained below T by continuously removing energy from the cell and emission peaks if the cell is maintained above T by adding energy. We see a greenhouse effect (higher surface temperature) at the Earth’s surface because the temperature of the gas (atmosphere) on average is less than the temperature of the source (surface). That means the temperature of the source has to increase to get the same flux at the detector (space) that would be present if the atmosphere were perfectly transparent.
Well, Tom.
Though I do not disagree with what you state, given your caveats, I would agree with those who say that LTE is not relevant when the temperature changes either up or down.
A ball of gas, whether composed of N2 or CO2, in space would radiate energy away until it reaches microwave temperatures or so, no? How does Local Thermodynamic Equilibrium enter this picture except in increments of dT/dt?
The earth is a ball of matter in space , part of it a gas envelope.
You are also catering to this absorption and decay line climate mania as if there is no continuum in quantum mechanical solutions. If there would be no continuum acceptable solutions how would a monatomic gas lose energy in vacuum?
Gail Combs says:
August 5, 2010 at 10:17 am
“I do not disagree. However you still have to define LTE (local Thermodynamic Equilibrium) and its physical characteristics before you can go on to describe LDE (Local Dynamic Equilibrium) were things ARE varying.”
In neither LTE nor LDE are conditions varying. The difference is that in LTE (which Tom relies upon in his analysis) there cannot be any net radiant fluxes. If there is any anisotropy in the radiation at any wavelength then one does not have LTE, one does not have complete equipartition, there is a net entropy increase and the system is not invariant under time conjugation, so Tom’s analysis is fundamentally invalid. Without realising it, he is trying to apply LTE rules to a system that is not in LTE but only LDE.
It’s not necessary to understand LTE or the complex absorption and heat transport modes to understand the greenhouse effect. In my view, it’s less confusing simply to look at the transmission of radiant energy through the atmosphere. If sunlight can get in more easily than thermal radiation can get out, then the surface will get hotter (and radiate more fiercely) until enough thermal radiation does escape to balance the incoming energy. The key parameters are the transmission and reflection coefficients for sunlight versus long wave thermal radiation (it’s rather like an electrical circuit). But we don’t need to get into the gory details to see that the greenhouse effect of CO2 in the Earth’s atmosphere is small, or that the effect of water vapour is large; just go outside on a clear and a cloudy night in winter and you can feel the difference for yourself.
Mr. Vonk,
Very good article. I had to wait until all of the trolls had thrown their mud. I must admit I saw a few places where a missing word or two, if inserted, would have prevented a lot of the contention in the comments above. But I am smart enough to add them as I read where other ‘intelligent’ commenters seem not able to do so.
You hit exactly on the principles I am becoming aware of when speaking of radiation transfers in the atmosphere. Wish you would have paralleled H2O molecules as you developed this article. For all statements you made also applies to H2O and we all know H2O has some additional properties which cause the two molecules to diverge in their actions. Hydrogen bonds, tri-states at normal temperatures, etc.
It is becoming apparent that the warming of clouds that is used as a point to prove ‘back-radiation’ from water vapor is not from water vapor (single water molecules) but instead is from the micro-droplets within the clouds which then act as gray bodies when radiating, spreading the energy into all IR bands. Moisture seems not to cause any additional warmth at the surface, clouds do. Does back-radiation warming occur? Yes, on a single molecule basis. No, on a macroscopic (thermodynamic) level.
Being a sailplane pilot I have experienced this first hand. At the very bottom of a cloud there is a warm super fine mist you always encounter before you ever really get high enough to say you are within the cloud. I have no measurement of how many molecules are contained within these super fine mist particles by I do know that once water has transformed from single molecules to a liquid state the radiation is no longer on a per band or per line basis but as a gray body curve devoid of radiation windows.
Seems the same 5% saturation (taken as roughly) that you mention about CO2 has a value also from H2O molecules. Putting the two together with water droplets will narrow the scope much closer to reality. I think in the end we will find small variances in concentrations has little to do with total energy flux through the atmosphere as a whole. Just my physics feeling.
It seems that somewhere within all of these factual properties the truth of our atmosphere lies. Wish I was better at thermodynamics and quantum mechanics to prove this myself, maybe you and other more knowledgeable commenters here can. It’s there, I can feel it, ever now and then I can almost see it, just can’t seem to put in the proper words within physics principles and laws.
Dear Dr. Tom:
Please use this reference:
Dr. Elsasser’s classic, “Infrared Heat Transfer in the Atmosphere” is available at ScribeD.
http://www.scribd.com/doc/34962513/Elsasser1942
Page 23 is rather interesting! It says: “It may be noted that since the flux in the carbon dioxide bad is equal at any level to a definite fraction of the black body radiation corresponding to the temperature at that level both in upward and downward directions, the RESULTANT (emphasis Elsasser’s) the of the carbon dioxide radiation vanishes in the approximation of the chart (Elsasser’s General Radiation Chart, used to calculate daily heat up and cool downs based on T,P and RH recorded by Radiosone balloons at multiple levels to 50,000 Ft.) This is a fair approximation to the truth in the lower atmosphere. For the upper atmosphere see section 12). (In section 12, CO2 is taken for “upper atmosphere” …or generally stratosphere, as an overall UPFLUX agent, or cooling agent. This has to do with the shape factor relation between the 4 Pi steradians of “view” and the percentage “seen” by “space” versus that “seen” by the Earth’s surface…)
I think that if the dear Dr. will take some time with this classic, he will be asking, as I do…”When did CO2 become net “downflux”? At what levels, and how intense. And what changed in the basic physics….which (let us not be “temporal provincialists” here) were well worked out by theory and observation by 1942???
Happy downloads.
Bryan says:
August 5, 2010 at 10:29 am
“stevengoddard says:
……”There isn’t any question that the greenhouse effect is real and that CO2 is a contributor. Places that have very little GHG (i.e. South Pole) also have very cold temperatures.”………..
CO2 is well mixed in the atmosphere.
Strangely enough even at the South Pole the H2O / CO2 backradiation ratio is at 2.6.”
CO2 represents very little GHG, even away from the poles. Water vapor is the majority GHG. What is the percentage of water vapor at the south pole?
Thanks Tom, for a good thought provoking article. One issue I have is that although I think that CO2 and water vapour cause a very small increase in temperature, the amount of extra energy retained is so small that the effect on climate/weather is imperceptible.
Jurai made some interesting points very eloquently, with which I am in complete agreement:-
“Juraj V. says: August 5, 2010 at 7:25 am
6,000 ppm of CO2 in the atmosphere of Mars does not create any measurable “greenhouse effect”. Its black body T = its actual T = 210K. Without digging into theory (and I am graduated analytical chemist specialized on spectral analytical methods), this is enough proof for me to believe, that IR active gas *alone* does nothing.
Mars: thin atmosphere (albeit composed of 95% CO2) –> no “greenhouse effect”
Earth: denser atmosphere –> some real “atmospheric effect”
Venus: 95x denser atmosphere than on Earth –> powerful “atmospheric effect”
No atmosphere – no “greenhouse effect”
We have such a joke: an ant and elephant are crossing the wooden bridge. Ant says: “What a clatter we do!”
For developing a “greenhouse effect”, you need the elephant – a bulk atmosphere. But then call it more properly “atmospheric effect”.”
Trying to isolate any bits of our weather/climate, which is a highly complex turbulent system, is fraught with difficulty. Only by investing in much, much better geographic instrument granularity and in the quality of data will the hard problem of understanding climate be solved – politics and carbon tax permitting!
Bob Kutz,
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[i]If you add additional heat the whole experiment does indeed fall apart. But I think that misses the point; if you’ve got a hollow sphere, filled with a mixed gas at equilibrium at a certain temperature, and then you replace some of the gas with CO2 (raise the concentration) does the sphere necessarily retain more heat? I thought he did a good job of explaining why (at least very locally) this is not really possible. You will change the level of equilibrium of the several forms of heat transfer, but that’s just the speed that the heat moves around. The temperature itself will not change. Or is that not correct?
Can you please explain, by way of increasing the concentration of CO2 in the sphere, rather then increasing the amount of heat being added, how he is incorrect in his analysis. The way I read your comment (applied to the real world) is that if the Earth’s surface would get warmer, so would the atmosphere. That is an argument I would probably tend to agree with, though many on the AGW side of this debate have denigrated the notion that the Sun’s variations aren’t really responsible for the recent increase in global surface T, so I’m not sure your argument is fruitful in that regard.[/i]
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Thanks for the thoughtful question. In a closed system I agree with you that simply adding CO2 (more specifically, increase the partial fraction of CO2 at the expense of the N2 partial fraction: I don’t want to add the complications of changes in heat content due to pressure that would occur if you simply added more material) in the very simplified hohlraum following from Tom’s original setup, I want to discuss two simple thought experiments.
1. For starters. Do you agree “before” and “after” in your thought experiment that radiation is entering the hohlraum at a higher rate than in the “after” case than in the “before” case. I beleive that is a given precondition of the question (I stipulated it in my original statement). If so, then the only way that equilibrium can be reestablished is for the hohlraum to emit more energy. By definition that can only happen if the temperature of the hohlraum increases to a higher equilibrium value.
2. Simply imagine the reverse. Instead of increasing the fraction of CO2 in the hohlraum decrease it to zero (and we’ll assume we now have 100% N2 at the original hohlraum pressure). The N2 can no longer absorb or emit 15 um radiation (to a first order approximation). Exactly the same amount of radiation passes through the hohlraum as in the “before” state from the first case, only it never actually spends any time in any internal energy states in the hohlraum. With CO2 in the hohlraum at least some of the molecules, everyone seems to agree, spend a fraction of there time with energy from the radiation field bound up in internal excitations of one or more forms. In the N2 only case the hohlraum simply doesn’t respond to the radiation. This, by the very thermodynamic definition of temperature, means that the temperature of the hohlraum is higher with some CO2 than with no CO2. Now, please take note, this second temperature elevation has taken place WITHOUT changing the amount of incident radiation.
Finally, to wrap it all up, although I think this proves the point you were interested in clarification on, it’s still important to point out that this is still too oversimplified an image of the problem. As others have pointed out the *effective* radiation field is increasing. It is true that, in the case we assume that the primary source of radiation, the Sun, remains constant, that it appears as though IN == OUT argument rules the day, but feedbacks are the key here (otherwise GHGs would warm the planet at all and we’d be in a perpetual ice age). The solar radiation coming in remains constant, but the atmosphere mostly doesn’t care because that light is mainly in the visible (that’s an oversimplification, but I’m not writing a thesis here). The warming of the ground results in the ~15 um radiation we are primarily concerned with. If there were absolutely no GHGs in the atmosphere then that 15 um would simply escape to space. That would establish some baseline temperature. However, because, as I think we all agree now, the atmosphere with GHGs does intercept and reradiate that light (and does so in all direction including back toward the ground) the total incoming radiation the ground sees is *slightly* higher than with no GHGs. So I refer you back to my case 1) above – the radiation field is increased and a new slightly higher equilibrium is established. And the argument itself is again based on case 2) above, that the simple fact that the atmosphere interacts with the radiation field means the atmosphere has to have gotten warmer.
This is still an oversimplification, but it has all of the *relevant* physics require to explain the effect fully covered. Like many here, I strongly disagree with the level of positive feedbacks in the system and therefore believe that AGW as it currently stands is radically overhyped, but that doesn’t mean that I believe GHGs don’t result in a warmer atmosphere and a warmer surface. They certainly do. I think most of the mods on board also agree with that statement (mods, feel free to disagree if I’m incorrect).
If you think I’ve got that wrong anywhere please let me know.
Correction above:
…otherwise GHGs WOULDN’T warm the planet…
Pamela, thanks for the comment. Actually, they’re all done on my blackberry while moving from meeting to meeting, etc. Which is why I often have more typos than I’m confortable with!
Yeah, CO2 is like insulation in the home attic. It slows down the rate at which energy can move across it from warmer to colder.
And it does a damn poor job of it because it only slows a small fraction of the total energy moving across the boundary and does nothing to slowing down a major way that energy is transported in the fluid turbulent atmosphere – convection – whereas the roof over your head pretty much eliminates all convection, all radiation, and forces the energy to move through it by conduction alone.
Given the state of confusion in my own mind, I can understand why others are confused. Julio (07:15, 2010) presents an argument that seems to (and maybe does) make sense. However, I believe Julio stopped his analysis in mid-stream so to speak. I now attempt to continue Julio’s logic. Some may call this a strawman. So be it. Julio concludes with the statement “which means the ‘earth’ has to radiate more with the glass in place, which means it has to get hotter.” OK, the earth gets hotter–how much is not said but let’s assume the temperature rise doesn’t appreciably change the spectral distribution of the energy radiated by the earth (the body enclosed by the glass)–that is, the temperature rise is small enough that the radiated energy is still predominately in the IR band. To be more specific, let T1 be the earth’s equilibrium temperature absent the glass shield, and let T2 be the earth’s temperature that corresponds to the outgoing radiation needed to compensate for the 20% of the original outgoing radiation that is “trapped” by the glass. Then applying Julio’s argument to the “higher temperature earth–i.e., to the earth at temperature T2”, won’t the T2 earth behave like the T1 earth in that won’t “Iout” for the T2 earth be equal to “0.8*Iup” of the T2 earth, with the result that the T2 earth’s temperature will rise approximately the same amount as the T1 earth’s temperature rose to equalize the 20% of the energy “trapped” by the glass? This leads to an earth temperature T3 equal approximately to T1 + 2*(T2 – T1) = 2*T2 – T1. This process repeats indefinitely with the earth getting hotter and hotter at each step.
I see only two ways out of this dilemma, although I admit there might be more. First, the temperature of the “earth” rises to the point that its radiation spectrum shifts to the “visible” and the glass lets the radiation pass. Second, the infinite process is bounded and eventually converges to a finite temperature TC higher than T1 but lower than the temperature at which the outgoing radiation’s spectral content peaks in the visible.
I don’t believe the former happens because I believe the temperature at which black body radiation peaks in the visible is much higher than the temperature of the earth (approximately 300 Kelvins). The latter may be valid, but I would like to see a computation that so demonstrates.
In their book, Third Edition, Part I, University Physics, Pages 396-397, Sears and Zemansky discuss the rate of energy transfer from the surface of a small object at temperature T (all temperatures in Kelvins) when surrounded by a large enclosure whose temperature is everywhere T0. The formula for the “net rate [from the enclosed object] of loss or gain of energy by radiation (or the heat transferred by radiation) is
H = A*e*sigma*(T^4 – T0^4)”
where “A” is the surface area of the enclosed object, “e” is the relative emittance of the enclosed object, and sigma is the Stefan-Boltzmann constant. The text states: “The heat transferred by radiation is proportional to the difference between the 4th powers of the temperatures of the body and of the enclosure, and to the relative emittance e of the surface of the body. It does not depend on the nature of the walls of the enclosure.” [my emphasis].
If Sears and Zemansky are correct, the spectral absorptivity properties of the glass shield in Julio’s example will affect the temperature of the enclosed body only if those spectral absorptivity properties affect the temperature of the glass shield. This is probably the case, but then the discussion of how the glass shield affects the temperature of the enclosed object should focus on the temperature behavior of the glass shield, not simply the amount of IR and visible radiation that passes or doesn’t pass through the glass.
Jeff Id, thanks for your brilliant thought experiment. Wish I’d thought of it…
Gail Combs:
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Tom Vonk is not talking about anything but what he has defined. Until the idea of “local thermodynamic equilibrium” and the physics that applies to it is understood you can not discuss anything else. I think this post comes under the heading of defining terms.
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If Tom had left it at defining terms I would generally agree with you, but for two things:
1) He definitely is wrong about how energy absorbed by CO2 in the LTE volume is released. He states pretty clearly that it results in no net temperature increase because it spends some small transient time in an N2 translation then comes right back out into CO2 vibrational excitation and re-emits right back out – that is by definition non-thermodynamic with CO2 and N2 in transient excited states. That is simply incorrect. CO2 begins re-emitting again after the average temperature of the LTE increases to the amount that CO2 in thermal equilibrium has a non-trivial excited state population in the relevant vibrational mode. Now there’s nots of excited CO2 to emit. re-emission after absorption simply doesn’t happen, and the reexcitation simply doesn’t happen non-thermodynamically (as modelled in the transient N2 translational excitation in the model given). Again, also, by the way, even if that model were correct it can’t be argued that the gas HASN’T been heated by the radiation because populations exist in the excited state by Tom’s own definition. This means, by thermodynamic definition, a higher average temperature.
But more directly in response to your point, in caveat 2 Tom clearly states:
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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 ?
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That statement is anything other than the simple definition of terms you suggest. I also beleive it is simply wrong as I and others have argued.
“What is the percentage of water vapor at the south pole?”
Not sure if the figures are totally accurate as it’s hard digging up average absolute humidities over different regions.
Absolute humidity 0.03% at South Pole.
The Sahara desert by contrast is 0.3% and temperate latitudes over land vary between low of 1% and high of 5%. Wet tropical climates and over the oceans is 5% or more on average. That’s near the surface however. At altitude it freezes and/or condenses out to very dry air then you get ice crystals that raise albedo, clouds that raise albedo, and cool rather than warm what’s underneath.
But keep in mind when water freezes high in the atmosphere there’s a fair amount of latent heat released when water at freezing point becomes ice at freezing point (lots of heat released with no change in temperature). Not as much latent heat as is released by condensation of water vapor into liquid water at the same temperature but still a lot.
Gail Combs:
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I am going to go out on a limb and drag up my physics from forty years ago (Eeek is was that long ago?!)
The absorption of a photon does not translate to heat it is the VELOCITY of the gas translates to heat. PV=nRT
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Gail. The V in that equation is volume, not velocity.
Technically, temperature is only rigorously defined for a system in thermodynamic equilibrium. In this case, enery is equipartitioned (almost, it’s slightly more complicated) which means there is an equal amount of energy in three translational degrees of freedom (for all species), two rotational degrees of freedom for linear molecules only and three rotational degrees of freedom in non-linear molecules, and 3n- (5 or 6) degrees of vibrational freedom for (linear or non-linear) molecules IF those rotational or vibrational degress of freedom are “accessible”. Small, light molecules tend to have very little energy stored, at equilibrium, in rotational/vibrational degrees of freedom while large, heavy molecules tend to have much energy stored in those modes.
So temperature IS proportional to translation energy given the further complications I’ve mentioned above. So when a molecule absorbs a photon (which, by definition results in an excited vibrational and/or rotational state) the very process of re-establishing equilibrium means that some of that energy winds up in translation. Hope that helps.