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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Am I correct in assuming that for every external photon, coming or going, that contributes to the heating or cooling a local region of the atmosphere by interacting with CO2 molecules, there may be 60 to 100 other photons having that same effect by interacting with gaseous water, free H2O molecules, due to the relative abundance of these two trace gases in the atmosphere?
Perhaps this ratio might be moderated by the fact that CO2 radiation may begin to escape to outer space at lower levels in the atmosphere.
cba :
August 6, 2010 at 5:20 pm
I agree with you that LTE is good in defining the T in the state function of the gas, but changes in the state require a further analysis. What Tom is saying is that the temperature of all the components in the gas mixture that is the atmosphere is the same.
Now think of the wave of outgoing photons from the heated surface of the earth: photons have often been described as an ideal gas . These two states intermix obeying the basic laws of physics in the micro scale,i.e. conservation of energy momentum and angular momentum. Every body is concentrating on the conservation of energy.
People should also think of the conservation of momentum in this situation. This photon gas is not at the center of mass of the atmosphere,or the atmosphere levels,if one is computing in levels. It has definite momentum given by h*nu/c , and this has to be conserved.
Conservation of the photon gas momentum in its interaction with the gases of the atmosphere means that the famous 50% up 50% down isotropy in radiation of the bulk is wrong. To conserve momentum there must be more going up than down , and in fact, all this excess momentum must eventually be carried by photons to outer space because otherwise the “trapping ” molecules would end in the stratosphere.
Can you pick a hole in this argument?
Jim D says:
August 6, 2010 at 7:13 pm
One fairly minor item Steven and I both left out is that summer starts in PHX close to aphelion, the Earth’s far point from from the Sun. Summer starts at the SP close to perihelion. The ratio P/A is 94.5/91.4 (distances are Megamiles) or 1.0339. Squaring that gives the ratio of insolation at the two times, 1.069, so just by virtue of the shape of Earth’s orbit, the SP has a starting insolation 7% greater than PHX.
I thought it was only 4%, had I remembered it was 7%, I would have included it, so where I wrote sin(23.44)N above for my SP insolation, it should be 1.069 * sin(23.44) * N. Instead of 0.40, we have 0.43.
One other thing I should clarify – I mentioned in my initial comment “If Phoenix could be just a little bit drier.” That was a bit of a jab. At a 70F dewpoint (21C), the water vapor pressure is 2.5 kPa. Dry PHX air in June probably has a dewpoint of 40F (4.4C), and that has a vapor pressure of 0.84 – only 1/3 that of the pressure at 70F. So when I suggested “just a little bit drier” I had in mind that remaining 1/3, which would leave the air with no water at all.
Of course, PHX wouldn’t be as cold as the SP despite the dry air for all those other reasons. 🙂
Well I learned something new about energy hitting Antarctica. It feels a bit like a parlor trick, of course, but measurements and basic arithmetic are full of such things.
Thanks, everyone for this thread!
Paul Birch
LTE does not exist in the real universe, and if it ever did exist it could not be observed.
http://en.wikipedia.org/wiki/Thermodynamic_equilibrium
Begin there and continue to work towards statistical thermodynamics . What you wrote sofar doesn’t make much sense .
John Wittmann
Question: Given your conclusion with its caveats, and that they basically apply to other gases in the atmosphere beside N2 then please address the possibility of another case during the period of the detla CO2 conc/dt (time) or the time of transition from one CO2 conc to another. Does the rate of change of CO2 conc affect the situation? Of course I am assuming there is a situation where there is not a fundamentally an LTE situation.
If you suppose LTE then no . If you suppose non LTE then one basic assumption of my argument fails and then it becomes incorrect .
cba
Tom, I think there may be a problem in the area of LTE or not LTE with molecules and with radiation. It looks like you are dealing with the condition of LTE assumption for both molecules and with radiation. This is clearly not the case overall as there is radiation coming from the Earth’s surface at a higher BB emission temperature than one finds in most of the atmosphere.
The atmosphere essentially has LTE with molecules. The LTE example region is going to have all of the molecule types at the same temperature after a small time has expired from the last change in energy transport. This condition will have the energy transferred to and from the n2 and o2 molecules at the same rate and will not have a net transfer of energy. This also assumes no convection or conduction occurring in or out of the LTE example region. If there is total LTE, molecules and radiation included, then there will be the same amount of power radiated as absorbed with none going to the other molecules.
Thanks cba . I think that you perfectly understood the argument and I agree with almost everything you wrote .
Indeed I consider no mass transport in or out of the LTE region because the time scale of the collisionnal and radiative processes is vastly smaller than the mass transport time scale . I also do not consider space scales of kms and the radiative transfer within .
The point which is not correct is that the LTE assumption mixes up matter and radiation . LTE is a property of mass particles only . I should have probably written that explicitely in the post in order to avoid off topic discussions about radiative transfer or radiative equilibrium . The molecules (if they absorb/emit) do not need to be in equilibrium with photons in order to have LTE .
Spector
Am I correct in assuming that for every external photon, coming or going, that contributes to the heating or cooling a local region of the atmosphere by interacting with CO2 molecules, there may be 60 to 100 other photons having that same effect by interacting with gaseous water, free H2O molecules, due to the relative abundance of these two trace gases in the atmosphere?
Yes . Basically you are saying that most IR/matter interaction happens with H20 . This is true and Nasif Nahle has written a nice post about extinction length , mean free paths and absorption coefficients of CO2 and H20 . You might read it .
Annav
Conservation of the photon gas momentum in its interaction with the gases of the atmosphere means that the famous 50% up 50% down isotropy in radiation of the bulk is wrong. To conserve momentum there must be more going up than down , and in fact, all this excess momentum must eventually be carried by photons to outer space because otherwise the “trapping ” molecules would end in the stratosphere.
Can you pick a hole in this argument?
No , I think that it is correct . The non isotropy is probably small especially when you deal with thin layers . This because you may observe the non isotropy of the photons in the first 10 m or so as long as they have not been absorbed then this (tiny?) anisotropy will be conserved as you travel farther up .
Sky
Tom Vonk’s QM analysis may be correct, in a narrow pre-defined technical sense, but its relevance to the real-world problem of dynamic temperature levels seems very limited.
You are of course right in the first half of the sentence .
But only a handful of physicists work on a Theory of Everything .
The crushing majority of 99,99 % or more of scientific papers deal with narrow technically predefined matters .
The whole of the science as we know it , is the sum of such narrow but correct papers .
And be very sure that they are all relevant to the real world because they deal with properties of the real world .
If you read a paper “Collisionally induced emission coefficients of homonuclear linear molecules” you may consider it an extremely narrow and technical issue . Yet it is still very relevant to the real world .
One final (?) point. All gases are greenhouse gases. All gases absorb and radiate at all wavelengths. Some do it a lot better than others, but no gas is completely transparent. H20 is a pretty good absorber of IR, because it is a bent molecule. CO2 is poorer, because it is straight. N2 and O2 are poor because they are only diatomic; and monatomic gases like He and Ar are even worse. However, mixtures of gases (even poor absorbers) are better, not only because of different absorption spectra, but also because of the inter-species collisions (which to the radiation look like asymmetrical molecules). There would be greenhouse warming and cooling even if there were no H20, CO2, CH4 or anything but N2 and O2 in the atmosphere – just not very much of it.
Spector,
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.
.
.
Spector,
Anna,V
I’m going to ignore the angular momentum because there is conversion between angular and linear momentum when collisions occur – so just call it momentum. At the qm level, rotation is mostly for radio frequency absorption and emission.
The linear momentum, or just momentum must be conserved as you state. First off, the emissions of the slab of gas are going to cancel out , 50% up, 50% down. That leaves only the long wave IR from outside the slab of gas. Were the gas to be surrounded by other slabs of gas above and below, as is mostly the case, if those slabs are close in temperature they’re going to emit similar amounts upward and downward into the slab – giving equal amounts of momentum and canceling out – BUT there’s no radiative transfer in this condition because there’s thermal equilibrium. Rather, we have to look at the nature of pressure.
Given the case of a cylinder with an air tight ‘lid’ resting on a parcel of air inside, what is the situation for the lid? think bicycle pump with a weight pushing the handle down a bit and the outlet valve is stuck shut, totally sealing the cylinder. The air inside is at a higher pressure. What is generating that pressure is the momentum of the gas molecules colliding with the surface and being bounced back away from the walls. In some cases, there will be momentum conveyed to the wall and with twice the resulting momentum of the original particle of gas as it bounces back with equal and opposite momentum. Other collisions will result in a randomization of direction where initial motion was in an X direction and a non head on collision resulted in momentum in both the X and a new Y direction and all sorts of other activities are going on. Note that the cylinder lid is NOT being pushed upward, reducing the pressure back down to atmospheric. Note too, when I toss a tennis ball straight up, it has upward momentum. When it reaches its maximum height, it has zero velocity, (and for any CAGW fanatics who might read this – it is undergoing acceleration even at the max. height despite having zero velocity for a short period of time). At this point, the momentum is 0 and it wasn’t because all of the momentum was transferred to the air. After that, the momentum will reverse direction and by the time it reaches the height where it was released initially, it will have the same momentum in the negative direction, making that momentum change 2 * p from the original. Was there conservation of momentum? Of course! Was there conservation of momentum of only the ball? Of course not! It cannot be taken in isolation.
cba says:
August 7, 2010 at 4:48 am
Spector,
Anna,V
I’m going to ignore the angular momentum because there is conversion between angular and linear momentum when collisions occur – so just call it momentum. At the qm level, rotation is mostly for radio frequency absorption and emission.
The linear momentum, or just momentum must be conserved as you state. First off, the emissions of the slab of gas are going to cancel out , 50% up, 50% down.
The slab can only give 50% up 50% down if it is in its center of mass system. Before IR impinges on it , yes. Once an ensemble with a directional momentum is assimilated by the slab, it cannot keep the ratio. It will have to emit preferentially in the direction of momentum if it is not to be incrementally pushed to the stratosphere.
That leaves only the long wave IR from outside the slab of gas.
Were the gas to be surrounded by other slabs of gas above and below, as is mostly the case, if those slabs are close in temperature they’re going to emit similar amounts upward and downward into the slab – giving equal amounts of momentum and canceling out – BUT there’s no radiative transfer in this condition because there’s thermal equilibrium. Rather, we have to look at the nature of pressure.
Yes , it is a matter of pressure, but not of the example you give of a closed container that can balance the momentum. The only way the momentum can balance in gases is by collisions with other gases, and the vector sum of all these collisions will have to represent the original momentum of the impinging wave.
The ball example is also not relevant, the momentum is absorbed by the gravity creating body, as happens with the momentum of the photons that impinge on the ground and oceans. If though you continually bombarded the ball with a high pressure hose, ( the analogue of the continuous impinging radiation) it would go up and up and up until the force applied balanced the gravity.
Tom Vogt,
I know the effect is small, but it is continuous, day in day out, night in night out, there is blackbody radiation and absorption thereof by greenhouse gases. For the millions of years that the earth exists under the sun. I do not have the tools to do the calculations, but I expect if I could, it would show that those 26 petawatts per year integrated over so long a time have enough power to put H2O and CO2 into orbit.
Reed Coray says:
August 6, 2010 at 6:26 pm
“However, I’m uncomfortable with descriptions that say heat is “trapped”, and it is the “trapping of heat” that produces the rise in temperature of the emitting black body. ”
It is true that the “trapping” terminology is a bit ambiguous. You’ll notice I didn’t use it myself until my very last post, and by that point I think my meaning should have been quite clear.
Your example of adding mass to the Earth while keeping the same heat source is cute. Of course the temperature goes down, because you are spreading the same thermal energy over a larger volume. Put differently, you have to heat the extra material too. In the real world, the Earth does lose some amount of heat to the atmosphere in this way, by warming it. However, the heat capacity of the atmosphere is so small compared to, say, the oceans, that this should not have a large impact on the planet’s average temperature (see below).
“I think any atmosphere will alter the Earth’s surface temperature. How much is altered by greenhouse gases versus non-greenhouse gases, I don’t know. It’s a problem I would like someone more knowledgeable than I am to answer.”
An atmosphere will certainly help redistribute the heat around a planet, but without greenhouse gases the net effect on the average temperature will be negligible. Think densities: the density of air is one thousandth that of water. The entire troposphere (say, 20 km) can then only hold (at the same temperature) as much heat as the top 20 m layer of the ocean. (Much less in real life, of course, since it gets colder and thinner as you go up.) Suppose you add a 20 m layer to your 6,400,000 m-radius sphere. Plug this in your radiation formula and see the change in temperature you get!
(Incidentally, this was another reason why we didn’t need Tom’s post. The density of CO2 in the atmosphere is thousands of times smaller than the density of air (N2 and O2). To assume that CO2 could warm up the atmosphere by direct “contact” heating to any significant degree is ludicrous to begin with.
Radiation is a different matter, however. The scattering cross-section of a molecule or atom on resonance is of the order of the square of the wavelength. This means that, to a 15-micron photon, each individual molecule of CO2 appears to present a “target” about 15-micron wide. That is huge, by microscopic standards.)
OK, on to your questions:
“Assume the Earth is devoid of water, the Earth has an internal energy source providing energy at a constant rate, and the Earth’s surface temperature without any atmosphere is everywhere T.
(1) When a CO2 atmosphere is added, does the CO2 “trap heat”?”
A: not to any significant degree in the sense of “heat capacity”, as I have discussed above; yes in the sense of keeping some amount of infrared radiation bouncing up and down.
“(2) If yes, does this “trapping of heat” cause the temperature of the Earth to rise–i.e., can it be said that the “trapping of heat in the atmosphere” is the cause of the Earth’s surface temperature rise?”
A: Yes–the back-reflected radiation will warm the Earth. As a result, the Earth will radiate faster. However, the process is self-limiting. Go back to the money example. My bank gives my $100, I don’t want it, give it to the government, the government gives me back $20, that day I have made $20. The next day the bank gives me again $100, I give the whole $120 to the feds, this time they give me back $24 (always 20% of what I give them!). The next day the bank gives me $100, I give $124 away, get back $24.8. Can you see the series converging? Give $124.8, get back $24.96… give $124.96, get back $24.992, or rather $25, because they don’t make fractions of a penny. Now we have reached steady state.
“(3) If yes, is the CO2′s ability to “trap heat” a function of the Earth’s surface temperature? That is, will the CO2 trap heat for all Earth surface temperatures, or is there an Earth surface temperature at which CO2 ceases to “trap heat?”
A: The ability of CO2 to bounce radiation back will depend on the wavelength of the radiation, which in turn depends on the earth’s temperature, but that is not the self-limiting mechanism (see above).
“(4) If there is an Earth surface temperature at which CO2 ceases to “trap heat”, what roughly is that temperature?”
I don’t know! CO2 seems pretty transparent at visible wavelengths, so you could try the sun’s surface temperature, about 5000 K…
Paul Birch says ‘all gases are greenhouse gases’. I agree. However we should say all gases contribute to the atmospheric heating effect. Calling it a greenhouse effect assumes the cause is the geometry of the air molecules. It is not.
The atmosphere effect is caused by the density of the whole atmosphere, which is mostly N2 and O2.
This is confirmed by
The fact that Venus has a higher density and proportionally higher temperature. Similarly Mars has a less dense atmosphere and proportionally lower atmosphere effect.
The presence of linear adiabatic lapse rates in all 3 planets
TomVonk says:
August 7, 2010 at 3:06 am
Paul Birch: LTE does not exist in the real universe, and if it ever did exist it could not be observed.
“http://en.wikipedia.org/wiki/Thermodynamic_equilibrium
Begin there and continue to work towards statistical thermodynamics . What you wrote sofar doesn’t make much sense .”
Are you seriously appealing to wiki as an authority? You Fail! Marks 0/10.
If you understood the fundamentals of the subject – as distinct from having learned a few tricks, equations and undergraduate simplifications – you would have understood what I said, and why it makes perfect sense. Now, if you would like to query any particular statement you have failed to understand, and say why you consider it erroneous or confusing, I will be glad to explain it to you further. But there’s no point in my attempting to clarify my post without knowing where your difficulty lies.
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?
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.
http://www.webpages.uidaho.edu/~vonw/pubs/TownEtAl_2005.pdf
TomVonk says:
August 7, 2010 at 3:06 am
If you suppose LTE then no . If you suppose non LTE then one basic assumption of my argument fails and then it becomes incorrect .
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Tom Vonk,
Thank you, I appreciate your response.
A Modified Re-Question: CO2 conc changes in time. LTE at time 1 includes a CO2 conc. LTE at time 2 includes a different CO2 conc. I see from your post that at these two different times your argument is valid. My question is what is the proper approach for the transition period from one CO2 conc to another conc? Is LTE the approach still good at mini discrete times between time 1 and time 2 during the transition? Or does another concept besides LTE apply for analysis when the gas concentrations are changing?
John
RE: Paul Birch: (August 7, 2010 at 3:49 am) “One final (?) point. All gases are greenhouse gases. All gases absorb and radiate at all wavelengths.”
This is not true, except perhaps in the plasma state, gases have a limited number of natural mechanical vibration frequencies or wavelengths due to their particular physical structure. This is illustrated in the yellow figure at the top of this page. As each vibration mode has a quantized set of allowed amplitudes, these gases can only emit or absorb photons at those particular frequencies or wavelengths that correspond to the differences between the allowed energy steps of these vibrations.
Julio says:
August 7, 2010 at 6:28 am
Whoops, I was wrong, WAY wrong. I wrote:
“The scattering cross-section of a molecule or atom on resonance is of the order of the square of the wavelength. This means that, to a 15-micron photon, each individual molecule of CO2 appears to present a “target” about 15-micron wide. ”
WAY wrong!! That particular scattering (or absorption) cross-section turns out to be only of the order of 5*10^-22 m^2, so I was off by many, many orders of magnitude. It looks like electronic transitions in atoms (which is what I’m familiar with) scale very differently from vibrational transitions in molecules…
Anyway, even that small absorption cross section and the very low density of CO2 are enough to “catch” just about all the 15-micron photons the Earth radiates, apparently.
Sceptics do not deny that infra red is absorbed in the 2.8, 4.5 and 15 micron wavebands by CO2 in air thereby adding heat to the lower atmosphere. We deny that additional CO2 will absorb any more infra red or heat. Also at 15 C the peak emission wavelength is 10 microns at which the combination CO2/H2O is transparent. The peak emission wavelength is 15 microns when the temperature is -80 C (see Wien’s law)! If Planck and Wien were alive today they would both be confirmed sceptics.
Spector says:
August 7, 2010 at 8:39 am
RE: Paul Birch: (August 7, 2010 at 3:49 am) “One final (?) point. All gases are greenhouse gases. All gases absorb and radiate at all wavelengths.”
“This is not true, except perhaps in the plasma state, gases have a limited number of natural mechanical vibration frequencies or wavelengths due to their particular physical structure. This is illustrated in the yellow figure at the top of this page. As each vibration mode has a quantized set of allowed amplitudes, these gases can only emit or absorb photons at those particular frequencies or wavelengths that correspond to the differences between the allowed energy steps of these vibrations.”
1) All absorption lines are broadened by the velocity of the molecules, turning a quantised transition into a band.
2) Collisions between molecules are not quantised; they produce some absorption at all wavelengths.
Kevin Kilty says:
August 5, 2010 at 5:52 pm
I meant to say one degree per meter temperature gradient, not one per one hundred meters as I said.
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.
Hi Julio,
Three points.
(1) You said: “It is true that the “trapping” terminology is a bit ambiguous.” I whole heartedly agree with this statement. You then added: “You’ll notice I didn’t use it myself until my very last post, and by that point I think my meaning should have been quite clear.” 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?
(2) You said: “ Of course the temperature goes down, because you are spreading the same thermal energy over a larger volume. .” I believe this statement is incorrect. The temperature goes down because the black body surface area increases, not because the black body volume increases. The rate of energy radiated by a block body at temperature T Kelvins is proportional to the fourth power of T and the surface area A of the black body. The volume of the black body enters into the radiated rate-of-energy only to the degree the volume affects the surface temperature distribution. For example, consider a cube and a sphere of equal volume. The cube will have a slightly larger surface area than the sphere (for equal volume objects, the ratio of cube surface area to sphere surface area is I believe approximately 1.2407). If (a) the surfaces of both objects behave like a black body, (b) the surface temperature of each body is everywhere the same, and (c) the internal energy sources are equal (i.e., their rates-of-internal-energy-generation are the same), at radiation-rate-equilibrium the surface temperature of the cube will be lower than the surface temperature of the sphere by the ratio of the fourth root of 1.2407 or 1.0554. In this example, equal internal energy rates are being spread over equal volumes, but the surface temperatures are different. Furthermore, although it may take some time to raise the temperature of the added material, radiation-rate-equilibrium will eventually be reached. Thus, the facts that (a) the added material must be heated and (b) the same internal energy is being spread over a larger volume aren’t the reasons the temperature of the enlarged black body sphere drops at radiation-rate-equilibrium. I used the example of adding volume to a black body sphere because (1) I wanted to add material that would “trap radiation” (and black body material is the perfect “radiation trapper”), and (2) I wanted to show that “trapping” radiation (or trapping heat in any form) doesn’t necessarily imply a temperature increase.
You then said: “ Put differently, you have to heat the extra material too. In the real world, the Earth does lose some amount of heat to the atmosphere in this way, by warming it. However, the heat capacity of the atmosphere is so small compared to, say, the oceans, that this should not have a large impact on the planet’s average temperature (see below). .” I believe you are confusing “heat” and “temperature.” Temperature is not a measure of heat (energy) content. A thimble at 500 Kelvins is at a higher temperature than a large asteroid at 490 Kelvins, but if both objects are brought into thermal contact with a reservoir at 100 Kelvins, more energy will flow from the asteroid than from the thimble. Temperature doesn’t quantify heat content. I inferred from your statement that since the heat capacity of the atmosphere is small compared to “say, the oceans” that (a) it [the atmosphere] won’t store much energy, and (b) as such won’t change the total “heat content” of the Earth, and (c) as such won’t change the Earth’s temperature. I believe the perception that temperature is somehow a measure of the “heat content” of an object is incorrect. I believe temperature is used to specify the direction of heat (energy) flow between objects brought into thermal contact. If two objects are brought into thermal contact, heat will flow from the object at the higher temperature to the object at the lower temperature. If when three objects (A,B,C) are brought into thermal contact (a) heat flows from object A to object B, and heat flows from object B to object C, then object A is at a higher temperature than object B, and object B is at a higher temperature than object C, independent of the size, mass, heat capacity, etc. of the objects. Various numerical scales have been devised to quantitatively express temperature; but at its core, temperature is a physical parameter that specifies the direction of energy flow when objects are brought into thermal contact, not a physical parameter that quantifies the stored thermal energy of an object.
(3) Your response to my question: “When a CO2 atmosphere is added, does the CO2 “trap heat”?” was: ““Not to any significant degree in the sense of “heat capacity”, as I have discussed above; yes in the sense of keeping some amount of infrared radiation bouncing up and down.” I take this as a “yes.”
I then asked: “(2) If yes, does this “trapping of heat” cause the temperature of the Earth to rise–i.e., can it be said that the “trapping of heat in the atmosphere” is the cause of the Earth’s surface temperature rise?” I apologize. I asked the wrong question. What I meant to ask was: “Will trapping of heat by an object cause a temperature rise in the object?” The subtle difference between these questions is that if the answer to the second question is “yes”, then whenever heat is being trapped by an object, the object’s temperature will rise; whereas a “yes” answer to the first question only implies that the “trapping of heat” may cause a temperature rise, not that it must. So, if you don’t mind I’d appreciate your answer to the question:
Will trapping of heat by an object cause a temperature rise in the object?
Thank you for your time.
Julio says:
August 7, 2010 at 8:53 am
Julio says:
August 7, 2010 at 6:28 am
Anyway, even that small absorption cross section and the very low density of CO2 are enough to “catch” just about all the 15-micron photons the Earth radiates, apparently.
Not too much. Before any molecules of CO2 “catch” any 15 μm photon, a molecule of water vapor has done it. The overlapping section at all wavelengths between water vapor and carbon dioxide is quite small, 0.002. The latter magnitude must to be subtracted from the sum of both water vapor and carbon dioxide absorption power.
For example, considering the pressure and temperature, the water vapor’s total absorptivity is 0.4, while the total absorptivity of carbon dioxide is merely 0.0017. In consequence, we cannot say that 100% of 15 μm photons are absorbed by the carbon dioxide, but only 0.17%. In contrast, the water vapor interferes to 40% of those photons, i.e. 236 times more frequently and efficiently than the carbon dioxide. As the water vapor emits one photon, there will be another molecule of water vapor that interferes with it.
On the other hand, by taking into account the mean free path length of photons through the troposphere and the mass fraction of each absorbent gas, water vapor interferes with photons 84% more frequently than carbon dioxide, at any waveband you choose.
One cannot say the carbon dioxide is a good interferer of photons from an artificial deduction. Many scientists have worked on the natural process and derived many realistic algorithms through which one can calculate with high precision the thermal physical properties of substances.
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?
Yes, it seems this 64 W/m^2 is the flux of thermal energy emitted by the surface, which was taken from Trenberth’s scheme on Earth’s energy budget and after modified ad arbitrium by many people. However, it is not a magnitude of thermal energy transferred by radiation, but by conduction-convection at the boundary layer surface-atmosphere. The total load of thermal energy transferred from the surface to the air by conduction-convection at the boundary layer is ~68.7 W*s. Therefore, I don’t find any “strong” absorptivity of carbon dioxide at 15 μm. The “strong” aborptivity of CO2 at 15 μm is quite “weak”. It suffices a quick calculation of the absorptivity power of the carbon dioxide, whether by taking its density in the atmosphere or by taking the mass fraction of the carbon dioxide and the temperature of the surroundings, and the “solid” argument vanishes.
Julio, I missed one other point. In your August 6, 2010 at 6:26 pm response, you described a process leading to a constant outflow of energy. Specifically you wrote:
“A: Yes–the back-reflected radiation will warm the Earth. As a result, the Earth will radiate faster. However, the process is self-limiting. Go back to the money example. My bank gives my $100, I don’t want it, give it to the government, the government gives me back $20, that day I have made $20. The next day the bank gives me again $100, I give the whole $120 to the feds, this time they give me back $24 (always 20% of what I give them!). The next day the bank gives me $100, I give $124 away, get back $24.8. Can you see the series converging? Give $124.8, get back $24.96… give $124.96, get back $24.992, or rather $25, because they don’t make fractions of a penny. Now we have reached steady state.”
Previous to that comment (August 5, 2010 at 4:07 pm) you wrote:
“Reed, the balance equations are just as I wrote them, and you derive the necessary equilibrium temperature from them; there is no need to iterate things.”
Doesn’t your August 6, 2010 at 6:26 pm description involve an “iteration?”