# CO2 heats the atmosphere…a counter view

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)

The simplistic view of CO2 heat trapping

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 .

1. 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 .

## 343 thoughts on “CO2 heats the atmosphere…a counter view”

1. Patrick Davis says:

Fascinating read! I wonder how many “climate scientists” have read Tom’s posts before? Thankyou Tom and Anthony.

Thank you, thank you, thank you for such a clear and concise refutation of the big lie, for which you have just taken down the cult of CAGW’s main tenet. May this be the final nail in their coffin!

3. Michael Schaefer says:

Oopsie!

4. richard telford says:

Nothing can warm a gas at thermodynamic equilibrium. That’s what thermodynamic equilibrium means!

— > swish it goes over the head

6. John Marshall says:

Is this not like saying that the laws of thermodynamics must be obeyed? To comply with the 2nd law the heated CO2, which heats rapidly, will cool rapidly. Also adiabatic cooling of convecting air, containing CO2, will radiate heat to a cooler area, ie above, not to the warmer surface. So heat will dissipate to space. The 2nd law also means that the CO2 must loose heat because entropy must increase, so it cannot store heat-nothing can. The Trenberth paper, which the IPCC rely on for their energy transfer calculations and which describe the GHG actions, does not include night time or the fact that the earth is shaped like a ball. Its use of the Stefan Boltzman law is not appropriate because a black body cannot be found in nature, or anywhere else.

7. merrick says:

I’m sorry I’m really busy today and getting a late start. I suspect this is very well done, but I saw an error near the top.it is stated (with my minor edits for clarification:
“In addition this interaction (with infrared radiation) will take place only if the molecule presents a non zero dipolar [moment]”
That is incorrect. CO2, for instance, is a linear molecule symmetric along its axis in the ground vibrational state and therefore has no dipole moment. To make you statement correct it should read:
“In addition this interaction (with infreared radiation) will take place only if there is a change in the dipole moment of the molecule between the ground and excited states.”
Where excited state is defined as the higher state of energy after radiation is absorbed and the ground state is defined as the lower state of energy before the radiation is absorbed or after radiation is re-emitted (the same rule applies for emission of radiation as well).
I hope you don’t draw incorrect conclusions later on because of this misunderstanding or that it’s simply a mistake made while writing this up. But I just don’t have time to look any closer right now.

8. i>Now the most important question is “What are the rates of the → and the ← processes ?”
The question that precedes that is, what do you mean by rate? How do you quantify it? What number are you talking about?
In chemical kinetics, where the considerations are rather similar, the relevant number is the rate constant. In this context, it would be something like
d[N2+]/dt = k [CO2*][N2], where k is the constant.
Then your argument that the rates must be equal would have to take the concentrations (partial pressures) into account.
To put it another way, in any interaction between a CO2 and a N2, it is much more likely that the CO2 will be excited than the N2 (because of the disparity in numbers).

9. merrick says:

Ok. Spent just 5 more minutes and I really this you’ve missed the boat on this one. For starters, V->T processes occur at very low probability. In your overly simplified scenario where the N2 then interacts with the CO2 molecule to re-excite a vibration one must assume that the interaction occurs, for instance, before the excited N2 molecule has collided with another N2 molecule and that they both are now translationally excited, but both with less than the total energy required to excite the CO2 vibration upon subsequent interactions. Remember, those are quantized, so the collision which re-excites that mode must have exactly the right combination of both translational energy AND impact parameter.
You also seem to have left out rotations. Any collisions between relatively translationally excited N2 (or any other molecule) and CO2 are far more likely to transfer energy into excited rotational states and have exactly no effect on the vibrational state of CO2.
I really think the energy transfer between molecules is not handled correctly in this treatment.
Again, I hope that’s helpful, and let me know if you think I’m off the mark on my criticisms.

10. Jared says:

Thanks Tom. It may be above some, but most should be able to understand your “Conclusions and caveats” section.

11. BillD says:

The conclusions is ….(?) that the green house effect works as scientists have known for over a century, but some of the simplified terminology often found in textbooks is misleading (?)

12. merrick says:

Again, sorry. In your hohlraum thought experiment the enclosed volume likely would emit the same amount of 15 um radiation as it absorbed, though your setup wouldn’t find that detail. This is a blackbody at equilibrium. Now, add a source at greater than 15C (like a warm earth surface) and ad long as the rate of incoming 15 um radiation is greater than the 15 um radiation rate you already measured from your hohlraum there will be disequilibrium and the temperature of the hohlraum (not just the CO2 but all of the gas) will increase until the hohlraum is again emitting the same amount of 15 um radiation as is coming in. That’s heating in my book. And when the external source goes away the gas in the hohlraum will start cooling.
Certainly in the actual atmosphere there are other issues, like the lapse rate as you mention, but that doesn’t mean heating/cooling isn’t also occuring. It is.

13. PSU-EMS-Alum says:

“The simplistic view of CO2 heat trapping”

And by “simplistic”, you mean “wrong”, right?
What is trying to be demonstrated by the graphic is the long-wave radiation absorptivity of the atmosphere. What it is showing is that the atmosphere “reflects” such radiation back toward the ground, which is just bad science.
Would it have been so difficult to terminate the smaller atmospheric absorption arrow in the atmosphere itself and then have a separate set of arrows (both toward the surface and into space) showing the radiative energy from the atmosphere?
It still would have been “simplistic”, but also “correct”.

14. Jan says:

As I don’t understand this post, am I supposed to take it to prove that climate change doesn’t exist, as any argument will suit this purpose?
Anyway, it tries to prove something that flies in the face of the evidence that CO2 keeps us warm, and more CO2 heats up the planet (as can be seen in the nice curves that are not believed by a majority of bloggers in attendance, let alone the obvious increase in average temperatures over the last decades).
Tom Vonk did not win a Nobel price for his three papers on Differential cross sections for rotationally inelastic collisions (all he wrote as far as I can see on Scopus), and did not publish a peer-reviewed paper on CO2. But because he’s blogging a lot, he must know a lot, so I believe him.

15. Does this scepticism know no end? Doubt about apocalyptic alarm ends up with a challenge to the last bastion of the ‘the simple science’…struck down by simple science…of text books textbook science…known for 100 years.
So how does this placed the results, sometimes experimental results, of Peixoto 1992, Goody 1964, Howare 1955, Schmidt 1913, Rubens and Aschkinass 1898, and of course Tyndall. Would Lindzen agree? Are they seeing emission as reflection (back to earth) rather than dispersal of energy? I am sure the answer is implicit in what you say, but someone might elaborate for us non-scientists.

16. kwik says:

Thank you Tom for taking your time. Very nicely put, and very interesting.

17. Randy says:

Tom,
Thanks for your essay. I think I understand “most” of what you were saying. Could you please comment on the ideas presented in this video and explain where they violate the physics.
http://earthguide.ucsd.edu/earthguide/diagrams/greenhouse/
Thanks, I would appreciate it.

18. Roy Spencer says:

If local thermodynamic equilibrium exists in a certain volume of a gas, and you add more CO2 at the same temperature, it is true that the volume’s temperature will not change.
No one I know of would disagree with this.
But it’s when that volume is exposed to outside influences — like IR radiation from the solar-heated surface of the Earth passing through that volume — that a temperature change can occur as a result of adding more CO2 to the volume.

19. Ed Fix says:

It’s great to see someone clearing the air about the basic physics of the system. Too often we see sweeping, simple-to-understand analogies that are simply wrong.
like:
“CO2 traps heat in the atmosphere like a blanket”
“Infrared energy (which is heat)…”
Stuff like that is like fingernails on a blackboard to me, and I’m not even a physicist. I just paid attention during my freshman year.
Ed

20. John W. says:

II have some questions:
1. Who are you?
2. Do you have a Ph.D.? From what university?
3. How many peer reviewed papers have you published?
4. Do you now or have you ever taken any form of payment from a profit making organization?
Just wanted to get those out of the way. 8^)
Seriously, thanks for a well written and informative essay. I do have a serious request. Your essay addresses absorptivity and emissivity. If you can find the time, could you also address transmissivity?

21. Bernie Anderson says:

If the above is true then how do NDIR CO2 measuring instruments work. They detect CO2 either by the attenuation of an IR beam passing through the cell with the test gas or by detecting the temperature rise of the gas in the cell. Both of these would require absorption of IR by the CO2 and that energy has to go somewhere.

22. beng says:

In defense of the “Scienceofdoom” site, it seems like a detailed but standard explanation of basic CO2 science. It’s listed on WUWT’s site as a “Pro AGW View” next to SurrealClimate, but that doesn’t seem fair. I don’t think he comes up w/anything other than the standard ~1.2C temp rise for doubling CO2.
Undetermined feedbacks could raise that amount or OTOH drop it to near zero. We don’t know, and I don’t think he makes any claims beyond that.

23. HR says:

The theory of AGW says that extra CO2 causes a minor warming (less than 0.5degree) which then causes the atmosphere to absorb more water vapour. H2O content of the atmosphere is dependant on temperature – if you warm the air by 0.5C it can hold a little more H2O. This small increase in H2O then increases atmospheric temperature since H20 is a strong GHG which then allows more H2O to be absorbed etc (positive feedback loop).
The problem that the AGW’ers ignore is that when that first stage of extra H2O absorption happens, the H2O comes from evaporation from the oceans/surface water. Evaporation is a Endothermic process which causes a COOLING effect, so the first stage of H2O absorption MUST causes a slight cooling of the oceans which will offset the slight warming of the initial CO2. They have ignored this.
Not only that, but the slight cooling of the oceans from step1, results in the oceans being able to absorb more CO2, since CO2 absorption in the oceans is dependant on ocean temperature (when temperatures go up, oceans release CO2, when temps go down, oceans absorb CO2). So the slight cooling from the evaporated H2O will actually cause the oceans to absorb some of the CO2 back.
ie. Nature ballances itself out.

24. Gnomish says:

Thanks. That was eminently digestible and thoroughly nutritious.

25. DocWat says:

Help me here… A system in equilibrium quickly returns to equilibrium at a higher level when it absorbs an IR photon: CO2+N2CO2+N2 becomes CO2*+N2CO2+N2+ (pardon the limited special character skills). This looks like heating to me, and, the temperature is controlled by the variance in the rate of absorbed and emitted IR photons for any small volume.
What I really don’t understand is why water and CO2 are better, by a factor of 20, at this as N2 and O2

26. Jan K. Andersen says:

I am sorry to say that this was a rather disappointing article on this otherwise excellent blog.
The fact that CO2 absorbs infrared energy and heat the atmosphere is no theory,, it is a fact. The flaw in the article is that it does not take into account that the absorbed radiation is outgoing, but the emitted radiation go in all directions.

27. Richard111 says:

Many, many thanks for this.
I am going to be busy for a while. < 🙂

28. Aeronomer says:

Greenhouse gases delay the transport of heat to the top of the atmosphere. The net effect is to make surface temperatures higher than they would be without these gases. So it may be sensible to say to the layperson “greenhouse gases warm the atmosphere”. I don’t know that saying such things automatically destroys credibility.
Also, the use of “microscopical” vs. “microscopic” and “dipolar momentum” vs. “dipole moment” seemed weird to me.

29. DocWat says:

My above entry would have looked better if the system had not erased my left and right carrots from the equations.
[Note: “Caret.” And the angle brackets can be created so they will not disappear with the HTML command: &, lt [or gt], ; (Delete the commas & spaces but keep the semicolon.) Put together, the “lt” command will create a ‘less than’ angle bracket; “gt” will create a ‘greater than’ bracket: < > ~dbs]

30. James Sexton says:

Heh, the instinctual aversion to chemistry I had in my youth is still with me! I almost fell asleep reading the article just like I did when I was taking organic chemistry 20+ yrs ago! That being said, if I understand this basic chem lesson properly, given the multi-directional release of energy by the CO2 molecule. It releases equally. Now, I’m going to have to try an absorb the significances. Thanks Tom and A.
PS, any way we can put this in some algebraic equation?

31. Vince Causey says:

Interesting article Tom. However, there is one point I did not quite understand.
In equation (2) you wrote 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⁺ “.
But later you wrote: “As we have seen above that this relaxation cannot happen through collisions with N2 because no net energy transfer is permitted , only the process (1) is available .”
But doesn’t equation (2) say that CO2* collides with an N2 molecule and relaxes?

32. There is no question that greenhouse gas H2O warms the atmosphere. Why would CO2 be different?

33. Rich says:

In other words CO2 which absorbs strongly the 15µ IR , will emit strongly almost exactly as much 15 µ radiation as it absorbs .
Why isn’t it “exactly”?

34. Robert says:

Fantastic paper about what CO2 actually does. just one question, in this equation CO2* + N2 ↔ CO2 + N2⁺ (3) you say that there is no net energy transfer between the CO2 and N2. That makes sense, but doesn’t some energy get lost during the collisions? With this energy loss, would the atmosphere actually cool a little bit, or would some outside source of energy come in and add whatever energy was lost, thereby maintaining the temperature?

35. Thank you very much for this post, it’s something I’ve wanted to read for ages but haven’t taken enough time to hunt down. It’ll take a couple more readings to absorb.
I am confused by your central point:

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 .

If we take a handful of air with a 15µ IR flux passing through, is it really in LTE? What happens in the case where the handful is not illuminated and in LTE, then illuminated briefly, and then not. The handful will settle back into LTE, and hence has a temperature – will that temperature be higher? Your quote above requires that the illuminated air be in LTE, so whether it is or not is my central confusion.
Also, I’d appreciate any notes on the time scale for CO2* to relax vs the collision time between CO2 and N2 at different atmospheric levels (i.e. pressure & temperature).

36. I feel smarter for having read this.

37. Stephan says:

OT but {snip]
[reply] But me no buts, butter me no parsnips, off to Tips and Notes with you. 😉 RT-mod

38. John Prendergast says:

This is brilliant, and validates Lord Kelvin’s second law and many others. I have long beleived that CO2 did pass any trapped heat to other molecules until equilibrium was restored If it were not so, there would have been reports of airliners flying into patches of incadescent CO2 , and suffering engine stall while they melted! Even UFOs can’t do this.

39. trbixler says:

I know that this is the first part of a series but having covered the action inside little boxes or spheres where delta size goes to a molecular level, what happens as the density changes?

40. Nylo says:

Sorry but your demosntration is wrong. You have demonstrated that in LTE:
1) The ammount of energy going from CO2 to N2 is the same as the ammount going from N2 to CO2.
2) The ammount of radiation received by the CO2 is the same as the ammount it radiates back.
But you have not demonstrated that:
3) The percentage of molecules in excited state remains the same if you change the initial conditions regarding the percentage of CO2 in the gas.
And if the percentage of molecules in excited state changes, the gas changes its temperature. Perhaps the fraction of time when this happens cannot be called LTE conditions, I don’t know. But it happens anyway.
A 5% of the CO2, when this gas is a 0,04% of the atmosphere, means very few molecules excited. A 5% of CO2, if it was 50% of the atmosphere, would mean a huge ammount of molecules excited and a higher temperature. Even though they would emit the same IR energy that they would absorb.
By the way, CO2 doesn’t heat, that’s true, and that makes your small experiment work. What it does is slow down the cooling of the surface of the Earth. It reduces the net radiation at 15um. As a result, you get higher temperatures than you would have without CO2. But all the initial heat came from the Sun, of course.
This is the experiment you need: put a ball at “room temperature” in a room filled with CO2 and N2, but then make the walls of the room cool to nearly 0K. Then try to demonstrate that the the speed at which the ball cools down is independent of the ammount of CO2.

41. Amino Acids in Meteorites says:

Since there is no tropospheric warming that was predicted to happen in global warming seeing this explanation partly helps to clear up why.
Next I know there will be the argument of downward flux to the surface of energy from added co2 causes warming. But that isn’t happening either. That argument poorly incorporates the effects of H2O. There is found to be cooling at the surface when H2O is effected by that downward flux.
The global warming hypothesis is wrong.

42. Thank you.
I now must take some C13H18O2 after reading this excellent piece.
(Isn’t carbon (and hydrogen and oxygen) a wonderful thing!)
😉

43. Nice presentation, but an incorrect conclusion.

44. Jan says:

@Jared 4.23: I’m not a physicist, and it takes a physicist to shoot holes in stories like Tom’s. It also takes a degree in physics to be able to say if the conclusions and caveats are worthy of discussion. As far as I can see, most physicists in attendance are shooting big holes in the story…. But then again, maybe they’re wrong, not Tom. That’s why I like this better than theoretical physics.

45. Thanks Tom. It might not have been as elegant as your own post, but basically I think we were saying the same thing – and it is great to have my model validated at last!
Submission to science and technology committee of UK parliament
Noddy Science: CO2 Warms the Planet?
(24) The complex bond structure within CO2 means that it can readily absorb and emit radiation in the infra-red (IR) band where thermal radiation is given off by a blackbody9 at the temperature of the earth. Much of this IR is at wavelengths at which other atmospheric constituents do not interact, so if CO2 is exposed to a warmer surface like the earth, it will absorb radiation that would otherwise pass through into the cold of space AND likewise if CO2 is exposed to the cool of outer space it will emit vast quantities of IR at wavelengths which other gases cannot emit.
(25) When CO2 is present low in the atmosphere, it tends to block transmission of these wavelengths into space and reduce heat loss to space. When CO2 is present high in the atmosphere, it helps emit IR, so causing cooling of the atmosphere acting as a vector by which other gases can lose heat into space. Like triple glazing, the system is complicated by the movement of air. Air warmed at the surface naturally tends to rise above the majority of the (blocking) atmosphere and it cannot descend until it has cooled by the emission of IR into the cool of outer space. CO2 cooling is as natural as CO2 warming, the atmosphere being a highly dynamic and complex system: a natural cooling system taking heat from the surface of the earth up into space via convective currents.
(26) Simple physics could suggest CO2 is a cooling gas as easily as warming and “obvious” assertions must be validated against real evidence, not the preconceptions of “scientists”. CO2 could impact the atmosphere in other ways: changes in specific heat capacity, density, interaction with water droplets and cloud formation. Other gases like water vapour also have their effects. It would be wrong to say that increases in CO2 can not affect the climate, but it is equally absurd, in such a complex system, to say this or that effect must dominate in the absence of the normal rigorous testing required by science.
http://www.publications.parliament.uk/pa/cm200910/cmselect/cmsctech/387b/387we32.htm

46. Enneagram says:

This is really how it works, not to mention all the rest of CO2 related reactions in nature, all endothermic, energy consuming, to build living organisms, beginning with glucose and cellulose. As I said: No CO2=No underwear. 🙂
However, sadly, you won´t find a chemist-politician.

47. Jean Parisot says:

Thank you, that was helpful. One question: you mentioned the non-LTE areas of earth and space boundaries; would the dynamic mechanism of thunderstorms also fall into a non-LTE area and be a factor in some of the emerging work on storm related energy transfers?

48. Thanks Tom for that well-written explanation. Some writers swear blind that CO2 cannot lose vibrational energy by collision and some swear otherwise. Whilst it was obvious to me that an equation like your (3) must be possible (for interactions with energy equal to a quantum level difference in CO2 vibration), I have been puzzled why some otherwise expert physicists have claimed otherwise. I think your explanation in terms of LTE explains why. Thanks again.

49. Leon Brozyna says:

A fine starting off point in this post, especially with all the “yes, but … ” comments. Will have to come back to this later today to see what else has sprouted.

50. Gnomish says:

Rich says:
August 5, 2010 at 5:36 am
In other words CO2 which absorbs strongly the 15µ IR , will emit strongly almost exactly as much 15 µ radiation as it absorbs .
Why isn’t it “exactly”?
because some portion of what is absorbed via radiation is transferred via collision and other frequencies.
the velocity of these molecules is around 1000 mph, too.

51. Anthony, Tom makes several mistakes in this piece, equal time for a rebuttal?

52. I guess you overstretch the principle of local thermal equilibrium.
Suppose you put some water into a microwave oven. The water is in LTE. Then you switch on the microwave generator. For a while there is no LTE, there is more absorption by the water than emission. Eventually, there will be LTE again, but at quite a different temperature.
Replace the microwave oven by the black body radiation of earth’s surface. Without CO2 and other infrared active gases, and forgetting collision induced absorption of N2 or O2, there is no absorption in the atmosphere. Near room temperature you even may forget about N2 and O2 vibrations. Then, the temperature is determined by the kinetic energy of the molecules. Its value is governed by collisions of the gas moelcules with the earth’s surface, in fact by the temperature of the earth’s surface (plus a temperature gradient of roughly 1 Celsius per 100 m change of altitude on the basis of the so-called dry-adiabatic limit).
Add CO2 or H2O to the atmosphere, there is no LTE for a while, as parts of the black body radiation is absorbed by CO2 and H2O. Eventually, a new equilibrium is established.
The main difference to the old temperature gradient is that everywhere the new temperatures are somewhat higher than the old ones, just as in the microwave oven example. Now add further CO2 generated by burning of fossile fuels…

53. Edvin says:

It’s been a few years since i studied physical chemistry. But something here doesn’t seem to add up.
“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 .”
I believe this would only hold true if there was no net production or loss of either type of the two molecules CO2* and N2(+). This asumes there are no collisions like:
N2(+) + N2 = N2(0.5+) + N2 (0.5+)
Or similar collision of N2(+) with a non-exited CO2 molecule. A collision where the kinetic energy is split between two (or more) molecules, leading to a decay in number of molecules at N2(+) or near N2(+) kinetic energy levels. Which will eventually lead to the molecules reaching a Maxwell-Boltzmann kinetic enegy distribution.
Is there no such decay in kinetic energy levels for a system containing only CO2 and N2?
If there is such a decay then additional CO2* can be formed by IR absorption and feed the N2(+) which in turn decays into several N2(Y+) and CO2/y+) (where Y<1) and the distribution of these N2(y+) and CO2(Y+) defines the temperature.

54. steveta_uk says:

As far as I can tell, the points raised in “caveat 1” seems be saying that while this treatment may be valid for a volume of gases in LTE, that never happens in the real world.
Any volume of gases with incoming IR cannot be in LTE unless precisely matched with outgoing LTE, and since the the missing N2-N2 collisions (and O2 of course) will prevent much of the outgoing IR from existing, the gas in not in LTE.

55. James Sexton says:

Jeez guys, for those that are taking Tom to task about what he didn’t show, did you expect an entire advanced chem semester in one article? The way I took this lesson, is to bring people up to speed about how CO2 acts when excited by infrared heat. And then touched on the interaction between CO2 and N2. I don’t believe this was supposed to be an all encompassing comprehensive article explaining the entire radiative chemical processes of our atmosphere. Obviously, there’s more going on up there than this.
Tom, once again, thanks for putting this in terms easily expressed. I’d often thought the energy emissions of CO2 were equally outward as they were back to earth, but as I’ve alluded to earlier, chemistry never really held much of an interest to me during my college years.

56. ClimateWatcher says:

There is no question that greenhouse gas H2O warms the atmosphere. Why would CO2 be different?

CO2 and H2O cool the atmosphere (at most levels):
http://rtweb.aer.com/lblrtm_frame.html
The follow on is that while they cool the atmosphere, they warm the surface even more,
warming which is subsequently shared to the atmosphere via convection.
absorption bands meaning energy which left earth from the surface or troposphere,
(at a higher T) now would leave earth from the stratosphere
(at a lower T) thus, ‘All other things being equal’ less energy should leave earth.
‘All things’ usually not equal, and negative feedbacks (which IPCC seems to preclude)
seem to me as likely as positive feedbacks (which IPCC seems to ‘guarantee’).
But more likely still would be net zero feedbacks which would leave us with a small warming, which is what we’ve observed for the last third to half century.
That being said, the actual measurements of outgoing IR seem much more variable,
with no clear trend, making one wonder if CO2 is just not significant compared
to other factors governing energy loss to space:
http://www.climate4you.com/images/OLR%20Global%20NOAA.gif

57. Great presentation! Back to text book basics. I understand it because 60 years ago I used the rigid-rotator, harmonic-oscillater approximation to calculate classical thermodynamic functions, as a function of temperature. I doubt there are many post modern “climatetoligist” who have done that. My statistical analysis tends to confirm your conclusions. http://www.kidswincom.net/CO2OLR.pdf.

58. Jean Parisot says:

The main difference to the old temperature gradient is that everywhere the new temperatures are somewhat higher than the old ones, just as in the microwave oven example.
Is that microwave oven in a vast freezer? How does the potential to hold more energy work when it is bordered by an area with an unlimited potential to adsorb it.

59. Pamela Gray says:

I would suggest the author edit some of the sentence structures and grammar before posting on such a hot topic. It detracts from his attempt to state his case. Further, I am not at all convinced that the treatise presented here is a strong case against increased CO2 (from any source) adding to the heating of the Earth’s atmosphere.
As a skeptic, my central issue is that I am not convinced that our noisy planet, with its intrinsic short and long term temperature variability from natural sources, can be adequately measured in such a way to detect anthropogenic CO2’s warming affect.

60. Gail Combs says:

From the comments I have read so far I think everyone is missing the very important point you made in defining the initial conditions:
Essentially you are setting the initial conditions to equilibrium and THEN detail your proof FOR THAT CONDITION.
Initial conditions:
1. LTE requires only that the equilibrium exists in some neighborhood of every point…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 .
2. …a temperature cannot be defined for a volume which is not in LTE
3. …the energy distribution in a volume in LTE follows known laws and can be computed .”
I hope that clears up the misunderstandings.

61. Ed Fix says:

We’ve all seen the “proof” starting with a=1 and b=2, going through a series of seemingly valid algebraic manipulations, and ending with the conclusion that a=b. The trick, of course, is to hide a divide-by-zero operation somewhere in the proof.
Ton Vonk has made some similar errors.
First, throughout the article, he uses “temperature” when he is actually talking about “heat”. Except that sometimes, “temperature” really does mean temperature, so he can come to some seemingly valid conclusions that are completely wrong. I suspect that untangling “heat” and “temperature” throughout this essay would clearly show some of the places where he has gone off the rails.
Second, he treats the “energy equipartition law” as if it were an iron-clad law of physics that holds exactly at all scales of size and time. This is simply not true.
The equipartition law, like all such empirical laws, is a statistical approximation that becomes less and less accurate as the population of gas molecules becomes smaller. Yet this article treats it as if it must be obeyed exactly by any random pair of molecules. Also, this law can be temporarily violated by transient responses.
Let’s put one molecule of CO2 in a chamber with 78 molecules of N2 and 21 of O2 (to pull some random numbers out of thin air). Hit that lone CO2 with a 15 micro-meter photon of infrared light and let it absorb the photon. Instantly, the equipartition law is violated, because the CO2 molecule has gained more vibrational energy, which increases the total vibrational energy of the gas sample relative to the rotational and translational energy. Before long, the excessively vibrating CO2 molecule will bump into another molecule. Depending on the exact geometry of the collision, both could use that excess vibration to rebound with excess translational or rotational energy, or some combination. Both molecules then go on to collide with other molecules, and the energy increase eventually distributes throughout the volume, re-establishing local equilibrium, and re-establishing the energy partition. In the end, there is an infinitesimally higher heat energy content of our gas sample, and consequently an infinitesimally higher temperature.
So, yes, CO2 absorbing infrared light does indeed increase the temperature of the atmosphere.
The question that is usually un-asked, let alone satisfactorily answered, is “how much?”

62. Basil says:

DocWat says:
August 5, 2010 at 5:26 am
My above entry would have looked better if the system had not erased my left and right carrots from the equations.
I blame Bugs Bunny.

63. wolfwalker says:

To make my position clear: I come at this after many years of seeing creationists mis-use basic physics such as entropy and the laws of thermodynamics to “prove” that evolution can’t happen. And just as many years of arguing against those mis-uses. I don’t know much about climatology, but I know a LOT about how to spot bad arguments.
The opening paragraphs of this article give me the same tingling-down-the-back-of the-neck feeling as those creationist arguments about thermodynamics. It doesn’t feel right. The slightly patronizing writing style, the drastic simplifying of a very complex subject, the theme of ‘armchair genius uses basic facts to prove a major field of science wrong’ — all just like any pseudo-scientific creationist tract.
I have been a skeptic on AGW for several years and will likely remain so for years yet to come. But this article strikes me as bad science. Bad science on one side is no more justified than bad science on the other.

64. The whole of this unfortunately fallacious argument is based on a false premise: that we have LTE. We don’t.
Nowhere in the atmosphere is at equilibrium (except in the sense of a dynamic equilibrium with continuous energy flows).
What we have are ensembles in which the various components (sunlight, thermal radiation, oxygen, nitrogen, water, greenhouse gases, etc.) and their various energy modes (translational, rotational, vibrational, directional, polarisation, chemical, etc.) all have slightly different thermodynamic temperatures (except for the incoming, scattered and reflected sunlight, which has a radically different temperature from the rest). What we call “the” temperature of the gas is just an average.
We do not have equipartition. It is the departures from equipartition that drive the system.
There are many other minor mis-statements in the article (for example, infra-red does interect with the translational modes; conservation of momentum requires it), but the crucial error is in the assumption of equilibrium. At equilibrium, nothing can warm anything else, by definition. If you have warming, you don’t have equilibrium.

65. Deanster says:

This essay is all fine and good. But i have a question.
OK .. we know that CO2 can transfer no net energy to O2 or N2, but what about H2O?? Where does water fall within your essay??
We all know that Water Vapor is the most important Greenhouse Gas. Thus, while your essay addresses the main “gas” characters in the atmosphere, it does not address the role of water vapor.
Can you expand on this??

66. Bill Illis says:

Someone posted on Roy Spencer’s blog a nice point.
“You just can’t have a “hot” CO2 molecule beside a “cold” N2″ molecule for more than a microsecond.” [That is about the right number but it can even less time than this].
The greenhouse theory and the textbooks are written like the N2 and the O2 molecules play no role at all. Our thermometres tell us that they do.
The fact that there is lapse rate where temperatures decline from 288K at the surface to 220K at 10 kms high (and the temperature then stays stable at 220K for the next 10 kms) tells us what really happens.

67. “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 .
Of course there is a net energy transfer from CO2 to N2, as in every equilibrium reaction: if you add something at one side of the equation, that pushes the reaction to the other side. In this case, you add energy to a CO2 molecule as it captures an IR foton. That increases the total energy content of the (micro)system: energy can’t be destroyed (except for energy-mass transfers, which isn’t the case here)… The increased energy content either is re-emitted by the CO2 molecule or redistributed over all the existant molecules by collisions. In the first case, half of it is directed to the surface. In the second case, the temperature of the mass increases. In the atmosphere, the warmer gas lifts up, causing an increase of the surface temperature due to the lapse rate.
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 .
That is true if there is no net radiation energy coming in as in your example. But (as Tyndal proved over 100 years ago): put a beam of IR, including the right frequency, through a volume of pure N2 and nothing happens. Put some CO2 in the same volume, and part of the IR is absorbed, increasing the temperature of the volume of gas. This proves beyond doubt that your reasoning doesn’t hold for a dynamic process where one of the components changes (either the amount of IR or the amount of CO2…).

68. Griz says:

I agree that IR does not heat the atmosphere. The heating affect is from the redirection of the IR. CO2 absorbs some IR traveling from earth to space and reradiates in some random direction. Some significant percentage is aimed back at the ground which can be warmed by IR.

69. Robert Vertullo says:

So, a system at Local Thermodynamic Equilibrium doesn’t change temperature. Isn’t that the definition of Local Thermodynamic Equilibrium? I’m not sure I learned anything here.

70. John F. Hultquist says:

DocWat says:
August 5, 2010 at 5:23 am
“What I really don’t understand is why water and CO2 are better, by a factor of 20, at this as N2 and O2 ”
Here is a very rough analogy. Recall the small hand exercise thing shaped like a V with a spring. Squeeze the handles, let go, and the thing springs back to its original shape. Think of your squeezing as the absorption and spring-back part as the release of that energy. Think of this as a CO2 molecule. Now cut two 25 cm long pieces, one blue and one green, from wood broom handles. Blue can be N2 and green can be O2. There is nothing to squeeze together. No squeeze – no energy absorbed. So, think of the nitrogen and oxygen gases (major components of earth’s atmosphere) as having no role to play in this little game. Only molecules with particular characteristics (read the post for them), and CO2 is one such, can play in this game.

71. Julio says:

This is an amazing exercise in disinformation: how to thoroughly muddle the waters while pretending to clarify.
All the greenhouse physics is hidden in caveat 1 and the weasel-y reference to “much more complex and not well understood aspects.”
Here’s a better “thought experiment” for you. Shine light on a black rubber sphere, uniformly, from all sides, until it gets hot. How hot? If the incoming energy flux is Iin, then it will get hot enough to radiate out at a rate Iup = Iin, in equilibrium.
Suppose that most of the radiation that makes up the flux Iup is not visible, but infrared. Now put the sphere inside a larger glass sphere, made of a glass that is transparent in the visible but partly absorbing in the infrared. So Iin watts/m2 still comes through, but not all the Iup makes it out. A fraction is absorbed and–in equilibrium–reradiated in all directions: some out, some back in. As a result of this, the total Iout is not equal to Iup anymore. Let us say it is equal to 0.8*Iup. So what happens?
Well, since in equilibrium we need to balance things, we need the total radiation out to equal the radiation in,
Iout = Iin
but Iout is only 80% of the radiation emitted upwards by the sphere,
Iout = 0.8*Iup
so we end up with
Iup = Iin/0.8 = 1.25 Iin
which means the “earth” has to radiate more with the glass in place, which means it has to get hotter.
Now tell me exactly what part of “CO2 traps part of the infrared radiation in the atmosphere” you think is untrue.

72. Merrick says:

Sorry, just need to weigh in one more time.
When transferring translational energy, start by thinking about classical systems and let’s start with the simplest classical system – spheres. Imagine billiard balls. Imagine one billiard ball moving toward another and the vector that the moving billiard ball is travelling passes exactly through the center of the stationary billiard ball. When they collide some of the energy of the moving billiard ball will be transferred to the stationary billiard ball and both will be moving. In a completely frictionless environment with perfectly rigid balls the greatest energy possible will be transferred to the stationary ball and both balls will continue in the same direction as the original vector of originally moving ball. The amounf of energy transferred is also dependent upon the relative masses of the two balls.
Now, imagine that the center of the stationary ball is not exactly on the vector of the moving ball. the distance between the center of mass of the stationary object and the infinite line described by the direction of motion of the moving ball is called the impact parameter, usually denoted as b. The impace parameter can take on any value from 0 to infinity. The first case we considered was the b=0 case. The maximum interesting value of b is b r1+r2 then the balls don’t collide. With 0 < b < r1+r2 we have an interesting case where the amount of energy transferred is now determined by more than just the relative masses of the particles and the two objects also change direction after collision.
Now, if we replace one of the balls with two balls connected by a rigid stick, when the collision occurs the moving ball will put energy, in general, into BOTH translation AND rotational degrees of freedom of what I'll call the rigid rotor. If we're talking about macroscopic objects both the translational and rotational degrees of freedom and the energies that each of the modes can take on are continuous: i.e., the translational and rotational energies can take on any continuous value.
Now, replace the rigid stick with a spring. Our imagined collision between what I'll now call the vibrotor (which is not moving translationationally, rotationally, or vibrationally) and a moving sphere results in a vibrotor that IS moving, rotating, and vibrating. And, again, if the objects are macroscopic, all of the modes can take on any continuous value.
In the case of CO2 and N2 molecules, all of the translations are still rigorously classical – they can take on any real value. However, both the rotations and vibrations are classical and can only take on certain discrete values. For instance, as mention above, there are well known excitation modes in CO2 with an energetic value equivalent to 15 um radiation. If an object (for example, an N2 molecule) is going to collide with a CO2 molecule and excite it with to an excited state that N2 molecule has to have translational energy equivalent to at least 15 um radiation and the impact must occur with exactly the right impact parameter for the reaction to occur. Far more likely is for the transfer of translational (and rotational) energy from the N2 molecule to end up in translational and rotational degrees of freedom in the CO2 molecule.
I think Tom makes the common mistake that chemists see physicists make here, and that is assuming that microreversibility applies on the macroscopic scale. That's rarely true. As a physicist he's probably more used to seeing the term detailed balance, which is essentially the same thing, though in fact microscopic reversibility is technically a broader term. And, as misroscopic reversibility applies in laser cavities, as mentioned, we are still forced in almost all practical cases to find laser gain media with extremely long-lived states which are typically only available through non-radiative processes (i.e., the initially excited state does NOT re-emit!) in order to get a useful laser – but that's a digression (but Tom brought up the laser!).
Tom points out that re-excitation of CO2 vibrations occurs all of the time in an N2:CO2 laser. Quite right, but the effective internal temperature in the laser cavity is extremely higher than, for instance, 15C, so that there is in fact plenty of translational energy available at well above the vibrational excitation energy. At 15C this reverse step is in fact almost never going to occur due to the fact that all other energy transfer options are so much more likely and with much higher transition probabilities and that so few of the N2 molecules have energy significantly in excess of the equivalent of 15 um to overcome the impact parameter issues (i.e., most imapct parameters are going to be greater than 0).

73. Symon says:

There’s a throwaway remark in the article “Even Internet sites that are said to have a good scientific level like “Science of doom” publish statements similar to those quoted above .” I wonder if anyone could point me at some specific examples of this from “Science of doom”?
Thanks.

74. DR says:

I’d think the heat capacity of a CO2 molecule vs water vapor would be of importance. No?

75. Pascvaks says:

“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.”
__________________________
The fact that today’s “science” of AGW is quite murky and has a disgusting odor therefore suggests that scientists of all stripes are not policing their turf well at all, and are themselves therefore very complicit in the “political” shenanigans that have been going on for the past 30 years. If it ain’t the math it must be the money. Why else would so many do so little to stop so few from making a mockery of their profession? For the right price, you can get anyone to do or say anything you want; especially in today‘s world. Now we know how Lot felt.

76. Rob Vermeulen says:

It is a very strange statement!
All the non-equilibrium thermodynamics of continuous systems is based on the local equilibrium assumption but however clearly allow for heat transfer, mass transfer, non-equilibrium reactions etc.
The rates of your processes will be equal only at true equilibrium, i.e. when all the activities of the reactants and products will be such that the equilibrium constant is reached. If you prefer, the rates of the processes do not only depend on the cross-section but also on the relative amounts of everything – including the photon flux! This is why, for example, one can have non-equilibrium reactions AND equipartition of energy all together.

77. Gail Combs says:

Jan says:
August 5, 2010 at 4:33 am
As I don’t understand this post, am I supposed to take it to prove that climate change doesn’t exist, as any argument will suit this purpose?
__________________________________________________________
No that is the whole point of this blog. If you do not understand the post ASK QUESTIONS! or google the terms you do not understand.
This post is more or less a mathematical proof.

78. Merrick says:

Oh. I also should have mentioned – in the N2:CO2 laser the N2 is *vibrationally* excited, not translationally (though it certainly is probably travelling quite rapidly at these energies, thank you bery much!). It is this vibrationally excited N2 that transfers the vibrational energy to CO2 in the N2:CO2 laser. V->V reactions occur with significantly higher probablility than T->V reactions. To the points that in most chemical kinetics problems no one ever bothers to add the T->V channels because they occur with so little probability.
How did the N2 get vibrational excited in the laser cavity, you might ask – the electron impact as a result of the multikilovolt discharge across the cavity to initiate the laser build up.
Electronic and nuclear degrees of freedom the only to remaining modes that energy can be stored in chemical systems that we haven’t discussed so far. They really don’t contribute to the problem of atmospheric heat transfer.

79. 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”.

80. Tom in Florida says:

Isn’t the real question “How much does CO2 affect warming” rather than “Does CO2 affect warming”? I am in no position to enter into a discussion on the physics of CO2 in the atmosphere however everyone of us knows why deserts experience extreme cooling at night and places with high humidity, like south Florida, do not. I grew up in southern New England and during summer when humid air flowed from the south it stayed hot and humid at night but when the wind shifted to come out of the north bringing in drier air, things cooled off at night. Both of these scenarios have nothing to do with CO2. So again, isn’t the question really how much warming can be credited to CO2? If we can get a final answer on that we will have made real progress.

81. Enneagram says:

CO2, the gas we all EXHALE does not stay there just looking down to earth in the atmosphere, it enters into the living organisms chain, first to make GLUCOSE (Plants do breath CO2 ya know, and btw exhale O2- a gas we like to breath, ya know…), then cellulose, etc., etc. then we EAT those vegs in order to live, or cattle eats them to make the MEAT we eat, or chicken eat corn grains to make the EGGS YOU EAT scrambled every morning, ya know buddy….all these transformations USE the energy CO2 ATE first, the HEAT, ya know, which makes possible for YOU TO WALK AROUND , fool around, repeating, like a parrot, silly things like: “CO2 is a greenhouse gas”.
And, what you don’t know is that all this is BUSINESS it’s just an invention to take the money out from your work, from your pockets, for some guys up there, who you will never know, and who will enjoy it in some distant sunny beaches while you keep on working.
So WAKE UP!

82. Laws of Nature says:

Aeh well .. lets just say, that a gas in local equilibrium heats up under the influence of external IR-radiation just as much as water in local equilibrium under the influence of external microwave radiation, then we can skip “all that complicated math” and have an experiment everybody can test at home . .

83. Clyde Rhodes says:

It seems to me that it is a circular argument to assume equilibrium in order to prove equilibrium. The whole point is that when energy is added to the system, that shifts the equilibrium. I don’t quite understand how someone capable of such a detailed proof would not see this. Perhaps I’m the one who is missing something, but I just don’t see it.

84. Pamela Gray says:

Merrick, I appreciate your cogent and respectful critique. Your comments are also well written given that they are probably composed as you type. I love debate that rises to this level.

85. mkelly says:

Werner Weber says:
August 5, 2010 at 6:23 am
Better yet go to your micro wave and set it to 5 minutes and hit start. Then when the bell rings check the temperature. Nada. The CO2 did not heat up. The air did not heat up.
If I have three identical rooms each filled with a different gas, one O2, one N2 and one CO2 and have say electric resistance heaters in each. Set the themostats of each to 70 F and see which goes up fastest and if any go above the 70 F.
Shut off the heat and see which cools down fastest.
See which one draws the most electricity maintaning the 70 F.
If CO2 can add heat and make something hotter then the CO2 room should be above 70 F.
Also if the earth is at say 300 K and is emitting IR at 10 mirco and CO2 absorbs at 200 K at 15 micro just how much of the earths IR emissions can be absorbed? 200 K does not heat 300 K.

86. Clyde Rhodes says:
August 5, 2010 at 7:36 am
“It seems to me that it is a circular argument to assume equilibrium in order to prove equilibrium. The whole point is that when energy is added to the system, that shifts the equilibrium. ”
Exactly. And when energy is continuously being added to – and lost by – the system, you never reach thermodynamic equilibrium, even locally, only at best a new steady state. Like pouring water into a bucket with a hole in it.

87. Ian E says:

Seems like you have proved that microwave ovens dont work – I wonder if the manufacturers will take mine back?

88. Buffoon says:

Tom,
I’m with you until your conclusion on your time symmetrical equation. You conclude that, because
CO2* + N2 ↔ CO2 + N2⁺
is time invariant, there is no net transfer of energy from CO2 to N2.
You noted, of course, that CO2* -> CO2 + y
CO2 + +y CO2* + N2 CO2 + N2⁺ CO2 + -y
It is noted then, that +y are photons taken in by your gas system from whatever sources, and -y are photons emitted from your system. The case for “”heating” of the atmosphere” (or whatever scorned terminology) is thus the statement +y > -y, in terms of the system described. I don’t believe your time invariance of the middle relationship touches on the ability of either outer relationship to be asymmetrical, so your conclusion that conventional wisdom is truly incorrect on that basis seems to be a bit of a stretch.
Also,
CO2* + N2 CO2 + N2⁺,
you state that this relationship is time symmetric, however, the system more accurately described (for the purposes of this all brand new argument!) is
CO2 + +y CO2* + N2 CO2 + N2⁺ N2 + N2⁺,
where the last portion represents the probability asymmetry of N2⁺ decay modes. Given the markedly different partial pressures of the two gasses, it is much more likely that N2⁺ + N2 decays to N2 + N2⁺ than N2⁺ + CO2 decays into N2 + CO2*. Probability asymmetries suggest bulk movement of energy into the largest component of the medium.
Sorry for the verbosity, I look forward to constructive criticism of exactly where I’m wrong above.

89. Chris V says:

So, if CO2 can’t warm the atmosphere, how does the sun do it?
The atmosphere is mostly transparent to sunlight.
Sunlight hits the ground, warms it up, and then…?

90. I find it facinating just reading the comments. I can follow the physics and arguments but the disagreements are eye opening. We just don’t have all the answers do we?

91. Pamela Gray says:

There’s a hole in the bucket dear Liza, dear Liza, there’s a hole in the bucket, a hole.
Love that little ditty.

92. Buffoon says:

(((MOD: PLEASE DELETE MY PREVIOUS POST AND THIS NOTE. THANKS GREATLY.)))
[Note: might help if you chose a different screen name, and used a working email address. You email is fake, no more posts from you until you fix it. see the WUWT policy page ~mod]
Tom,
I’m with you until your conclusion on your time symmetrical equation. You conclude that, because
CO2* + N2 ↔ CO2 + N2⁺
is time invariant, there is no net transfer of energy from CO2 to N2.
You noted, of course, that CO2* » CO2 + y
CO2 + +y «» CO2* + N2 «» CO2 + N2⁺ «» CO2 + -y
It is noted then, that +y are photons taken in by your gas system from whatever sources, and -y are photons emitted from your system. The case for “”heating” of the atmosphere” (or whatever scorned terminology) is thus the statement +y > -y, in terms of the system described. I don’t believe your time invariance of the middle relationship touches on the ability of either outer relationship to be asymmetrical, so your conclusion that conventional wisdom is truly incorrect on that basis seems to be a bit of a stretch.
Also,
CO2* + N2 «» CO2 + N2⁺,
you state that this relationship is time symmetric, however, the system more accurately described (for the purposes of this all brand new argument!) is
CO2 + +y «» CO2* + N2 «» CO2 + N2⁺ «» N2 + N2⁺,
where the last portion represents the probability asymmetry of N2⁺ decay modes. Given the markedly different partial pressures of the two gasses, it is much more likely that N2⁺ + N2 decays to N2 + N2⁺ than N2⁺ + CO2 decays into N2 + CO2*. Probability asymmetries suggest bulk movement of energy into the largest component of the medium.
Sorry for the verbosity, I look forward to constructive criticism of exactly where I’m wrong above.

93. Chris V says:
August 5, 2010 at 8:05 am (Edit)
So, if CO2 can’t warm the atmosphere, how does the sun do it?
The atmosphere is mostly transparent to sunlight.
Sunlight hits the ground, warms it up, and then…?

The sun hits the sea more than the land, penetrates it up to a depth of tens of metres, gives up it’s heat energy to it, and then the ocean emit’s longwave IR into the atmosphere. It heats it mainly by convection, and the latent heat of evaporation and condensation. Then the atmosphere loses the heat to space, after bouncing it up and down to the surface a few times.

94. Bob Kutz says:

merrick says:
August 5, 2010 at 4:32 am
. . . This is a blackbody at equilibrium. Now, add a source at greater than 15C (like a warm earth surface) and ad long as the rate of incoming 15 um radiation is greater than the 15 um radiation rate you already measured from your hohlraum there will be disequilibrium and the temperature of the hohlraum (not just the CO2 but all of the gas) will increase until the hohlraum is again emitting the same amount of 15 um radiation as is coming in. . . .
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.
Thanks.

95. JDN says:

>”First is that a temperature cannot be defined for a volume which is not in LTE . ”
Completely untrue. The most famous example is the fluorescent light bulb. Its electron temperature is very high, which generates the UV light, whilst its ion temperature is low, which is why the bulb doesn’t melt. They clearly have different temperatures which can be defined. What you’re searching for is called temperature coupling. This is the LTE you’re looking for. Plasmas have uncoupled temperatures of multiple species, while, gasses near atmospheric pressure have well coupled temperatures.
It’s been a while since I’ve studied atmospheric plasmas. I don’t know whether any of the assumptions of LTE, equipartition, or Boltzmann statistics actually hold at the altitudes you are interested in. In fact, I don’t think you’ve specified altitude. I’m a little busy now, so, I’ll read the rest of your article later.

96. Rod Everson says:

Well, as a non-scientist, here’s my take from the comments thus far, right or wrong:
1. The people who know what they’re talking about are making technical comments that they claim rebut the original article.
2. The people who don’t know what they’re talking about are making comments along the line of: “Thanks, Tom, good to have that all clearly stated, great effort, puts it all to rest, I’ll have to set aside time to really absorb what you’ve explained so elegantly, etc.” (paraphrasing all, so as not to insult any one particular individual.)
Leaving me with an unchanged opinion that goes as follows: Apparently the human-caused increase in CO2 is heating the planet to at least a minor degree, but probably not to any degree worth getting exercised over. In the end, the next ice age will still prevail, no matter how much CO2 we pump out by burning fossil fuels, and that ice age will be caused by the same cyclical factors that caused the others. The trick is in the meantime to avoid impoverishing ourselves attempting to avoid a scare trumpeted by people who desire additional, even complete, control over our lives, helped along by their useful idiots. (I’ll leave it to the individual alarmists to decide whether they desire control, or are simply one of the useful, but those will eventually prove to have been the only two choices in my view of the matter.)
Note that I consider myself a skeptic even though I grant that we are probably subtly adding to the earth’s temperature by burning fossil fuels. My skepticism revolves around the politics/economics of the situation, and the corrupting of science as a means to reach ends that I find objectionable.

97. Richard Garnache says:

JohnW says:
Seriously, thanks for a well written and informative essay. I do have a serious request. Your essay addresses absorptivity and emissivity. If you can find the time, could you also address transmissivity?
Dr Heins Hug had an excellent article several years back. He measured the transmisibility of 15 micron radiation through standard atmosphere and dtermined that all the radiation is absobed in the first 10 meters.
http://www.john-daly.com/artifact.htm
You can’t pump 20 watts per square meter into the air and absorb it in the first ten meters without warming thing up a bit.

98. dp says:

For this to be true it would have to be accepted as the dinosaur killer of AGW’s foundation. Somebody would have noticed that by now. I fear something you are sure of is wrong.
I eagerly await Part II.

99. Jeff Id says:

Your article is quite interesting but your experiment is moot to AGW, and the result is misinterpreted.
If define an experiment this way:
A thin rectangular box having two very large sides A and B parallel and opposite to each other, the other 4 sides having negligible area. Side A is a 1000Watt 15um IR emitter, side B is a perfect IR absorber.
If we take N2 and fill the volume, the N2 will reach an equilibrium temperature based on the temperature of the box and whatever radiation is absorbed. Now if we instantly change the gas content to be 50/50 N2 and CO2 which has better absorption of the 15um IR wavelength, an IR detector at side B would detect an instantaneous drop in IR energy at the 15um wavelength. I’m sure we can agree on that.
What would then happen over time, is the gas inside the box would stabilize to an equilibrium temperature. As the system stabilized, an energy detector at the absorbing B side would eventually record the same amount of energy transfer from the A side as before – return to equalibrium.
However, the gas temperature will stabilize slightly higher than the N2 gas. An effect proven by the initial drop in IR at your detector and the time delay until the system re-stabilizes.
None of this violates your equations or assumptions.
Good try though.

100. Gail Combs says:

Roy Spencer says:
August 5, 2010 at 4:57 am
If local thermodynamic equilibrium exists in a certain volume of a gas, and you add more CO2 at the same temperature, it is true that the volume’s temperature will not change.
No one I know of would disagree with this.
But it’s when that volume is exposed to outside influences — like IR radiation from the solar-heated surface of the Earth passing through that volume — that a temperature change can occur as a result of adding more CO2 to the volume.
_______________________________________________________________
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.
As you said “No one I know of would disagree with this.” but most lay people do not understand that.

101. cal says:

Mike Haseler wrote:
(24) The complex bond structure within CO2 means that it can readily absorb and emit radiation in the infra-red (IR) band where thermal radiation is given off by a blackbody9 at the temperature of the earth. Much of this IR is at wavelengths at which other atmospheric constituents do not interact, so if CO2 is exposed to a warmer surface like the earth, it will absorb radiation that would otherwise pass through into the cold of space AND likewise if CO2 is exposed to the cool of outer space it will emit vast quantities of IR at wavelengths which other gases cannot emit.
(25) When CO2 is present low in the atmosphere, it tends to block transmission of these wavelengths into space and reduce heat loss to space. When CO2 is present high in the atmosphere, it helps emit IR, so causing cooling of the atmosphere acting as a vector by which other gases can lose heat into space. Like triple glazing, the system is complicated by the movement of air. Air warmed at the surface naturally tends to rise above the majority of the (blocking) atmosphere and it cannot descend until it has cooled by the emission of IR into the cool of outer space. CO2 cooling is as natural as CO2 warming, the atmosphere being a highly dynamic and complex system: a natural cooling system taking heat from the surface of the earth up into space via convective currents.
(26) Simple physics could suggest CO2 is a cooling gas as easily as warming and “obvious” assertions must be validated against real evidence, not the preconceptions of “scientists”. CO2 could impact the atmosphere in other ways: changes in specific heat capacity, density, interaction with water droplets and cloud formation. Other gases like water vapour also have their effects. It would be wrong to say that increases in CO2 can not affect the climate, but it is equally absurd, in such a complex system, to say this or that effect must dominate in the absence of the normal rigorous testing required by science.
I would recommend this simple summary to anyone who is trying to understand the effect that CO2 has or might have. I have tried to make the same points in the past but the above statements are more succinct.
Just to embelish a few of the points.
The evidence from recent ice ages suggests that periods with high CO2 concentrations correlate with periods of cooling (but could easily be coincidental). There is absolutely no correlation with warming periods.
At low altitude CO2 absorbs most 13 -18 micron radiation within a few feet and re radiates it in all directions. The energy radiated downwards warms the earth (and air) below whilst that radiated upwards gets absorbed by other CO2 molecules and reradiated… and so on. Just below the tropopause the density of CO2 is such that there is a high probability that upward directed radiation will be emitted to outer space. This radiation from just below the tropopause represents the nett loss from the earth at that wavelength. It represents 18% of all the energy radiated from the earth. Most of the remaining 82% is radiated by water vapour molecules from various heights in the atmosphere (depending on wavelength) and the surface of the earth (particularly at wavelengths around 10 micron where the atmosphere is almost transparent).The total energy radiated by all these three (plus a few other small contributors like methane) has to balance the sun’s energy absorbed by the earth during the day. If it does not the earth will either heat up or cool down until it does.
I therefore have two questions that I have yet to get answers to:
1)Has amount of energy radiated into space at 13-18 micron reduced over the past 30 years? This is what make AGW a theory because it can be disproved if there has been no decrease.
Since this cannot be easily measured until the next generation of satellites goes up this year an alternative question could be:
2) Has the CO2 layer radiating directly into space decreased in temperature? The increased number of radiating CO2 molecules will have increased the radiation into space so the temperature will have to decrease significantly to compensate for this increase and reduce radiation still further.
The climate4you website has a section on long wave radiation which is very informative. It also shows how difficult it is to answer the first question. The global average of long wave radiation measured by satellites fluctuates by +/- 20%. This fluctuation dwarfs the tiny change that we are trying to measure over the longer term.

102. Enneagram says:

tallbloke says:
August 5, 2010 at 8:13 am
That is almost right. Water volumetric heat capacity=4.186 Jcm-3 K-1, Soil=~2.0,
Air=0.001297 Jcm-3 K-1 (3227 times less than water)
To everyone: Try preparing your breakfast using a hair dryer 🙂

103. Gail Combs says:

DocWat says:
August 5, 2010 at 5:23 am
Help me here… A system in equilibrium quickly returns to equilibrium at a higher level when it absorbs an IR photon: CO2+N2CO2+N2 becomes CO2*+N2CO2+N2+ (pardon the limited special character skills). This looks like heating to me, and, the temperature is controlled by the variance in the rate of absorbed and emitted IR photons for any small volume.
What I really don’t understand is why water and CO2 are better, by a factor of 20, at this as N2 and O2
___________________________________________________________
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
Temperature: A measure of the amount of heat in a system
[K] or [◦ C]. More precisely, it is a measure of the average velocity
of the particles in matter (see §XIII.G).
from PHYS-2010: General Physics I: Course Lecture Notes: Section XIII
Hope that helps.

104. bushy,
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.

105. Gail Combs says:

Jan K. Andersen says:
August 5, 2010 at 5:23 am
I am sorry to say that this was a rather disappointing article on this otherwise excellent blog.
The fact that CO2 absorbs infrared energy and heat the atmosphere is no theory,, it is a fact. The flaw in the article is that it does not take into account that the absorbed radiation is outgoing, but the emitted radiation go in all directions.
_____________________________________________________________-
There IS no flaw. Tom Vonk carefully defined the case he was discussing. I again defined the simple case under discussion here: http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/#comment-448340

106. Chris V says:

tallbloke says:
“The sun hits the sea more than the land, penetrates it up to a depth of tens of metres, gives up it’s heat energy to it, and then the ocean emit’s longwave IR into the atmosphere. It heats it mainly by convection, and the latent heat of evaporation and condensation. Then the atmosphere loses the heat to space, after bouncing it up and down to the surface a few times.”
So anything that warms up the oceans (eg, sunlight, IR emitted downward by water vapor in the atmosphere, or IR emitted downward by CO2) will also warm the atmosphere.
Even without going into the atmospheric physics, that seems to kill the OP right there.

107. Dave Springer says:

A few questions for Tom Vonk:
Is it fair to say CO2 is an insulator that slows down how fast the ground cools at night?
Is it also fair to say that like any other insulator each fixed incremental addition of more insulation is less effective than the previous increment – so that like having on five layers of clothing in the arctic winter adding a sixth won’t help keep you warm nearly as much as going from no clothes to adding the first layer?
Does a 25ppm increase in CO2 from 1880 to 1950 have about the same net surface warming effect as the 50ppm increase from 1950 to 2000?
The big questions though are the ones you can’t answer. Is there a negative feedback to increasing CO2 (in the water cycle) that will negate its insulating effect? Latent heat of vaporization at the surface, especially over the oceans, carries a tremendous and not easily quantifiable amount of energy straight through the CO2 like it wasn’t there and releases it much higher up where the path out the door to space has much less resistance compared to ground level. Is the effect of higher concentration of CO2 i.e. the CO2 signal or fingerprint, hopelessly swamped by other variables, some cyclic and some chaotic, some probably unknown and possibly unknowable and greatly variable in just about any timeframe from minutes to millions of years?
I guess I’ll finish up with a biology and geology question. What is the global optimum average temperature for maximum productivity of the primary producers (green plants) in the food chain? What is the optimal concentration of CO2 for the same producers?
It seems clear enough from evidence of the geologic past that before the earth started ringing like a bell every 120K years from glacial to interglacial with the former dominating the other 10:1 in persistence, the Eocene optimum 50 million years ago the earth was ice-free, green from pole to pole, it was about 11F warmer overall, with the most dramatic warming in the highest latitudes (right where you’d want it if you could ask for it), and atmspheric CO2 was several times what it is today, which makes sense in light of much warmer global ocean not able to hold as much CO2. Happily enough, terrestrial also plants also require less water per unit of growth as CO2 rises because they don’t have to open their stomata as much to get the respiratory exchange of gases.
So what am I supposed to fear from rising CO2? A warmer, greener earth? I’m more looking forward to it!

108. Vince Causey says:

Wolfwalker
“The opening paragraphs of this article give me the same tingling-down-the-back-of the-neck feeling as those creationist arguments about thermodynamics. It doesn’t feel right. The slightly patronizing writing style, the drastic simplifying of a very complex subject, the theme of ‘armchair genius uses basic facts to prove a major field of science wrong’ — all just like any pseudo-scientific creationist tract. ”
No need to get your spidey senses tingling, Wolf. I think you’ll find that far from trying to prove a major field of science wrong, Tom’s article is just an explanation of how greenhouse gases work at the molecular level.

109. James Sexton says:

Julio says:
August 5, 2010 at 7:15 am
“This is an amazing exercise in disinformation: how to thoroughly muddle the waters while pretending to clarify………..
Well, since in equilibrium we need to balance things, we need the total radiation out to equal the radiation in,
Iout = Iin
but Iout is only 80% of the radiation emitted upwards by the sphere,
Iout = 0.8*Iup
so we end up with
Iup = Iin/0.8 = 1.25 Iin
which means the “earth” has to radiate more with the glass in place, which means it has to get hotter.
Now tell me exactly what part of “CO2 traps part of the infrared radiation in the atmosphere” you think is untrue.”
Julio, I think you’re taking the wrong message from the posting. Admittedly, I’m rather ignorant of chemistry, especially, this type. What I took from the posting, given the equal distribution of emission of the CO2 molecule, of course there would be a part that is emitted back downward towards the earth. However, I think you have the equation wrong. You’re representing the percentage back out as a constant. It isn’t. Now hang with me as I attempt to clarify while over simplifying.
I’m going to assign a value to the infrared coming in as 10. For no particular reason other than it is a easy number to work with. As it hits the CO2 part of our atmosphere, the CO2 absorbs and then emits. Some back out, some downward towards the earth. Right? So, 10 goes to CO2 and 5 goes out and 5 goes in. So, you’re correct as far as you’re assertion goes, only my constant is 0.5. But that’s not where the story ends. But so far, the equation looks like this;
10—>CO2= 5up and 5 down. But, 10 is the constant representing atmospheric infrared constantly bombarding the earth. So the 5 up is gone back out harangue some other planet. the 5 down is now bouncing back up to combine with the constant 10 so 10 +5 —> CO2 = 7.5 up and 7.5 down. This appears to validate most of the CO2 warming crowd, let’s continue. the 7.5 bounces back up, so now we have 10+7.5 –>CO2 = 8.75 up and 8.75 down. Still going, 10+8.75—>CO2= 9.375 up/down…..10+9.75—>CO2 = 9.875 up/down. Lets see what CO2 is adding to the earth. First is was 5, the 2.5, then 1.25, then 0.625, o.3125…………
Yes, it’s adding, at least that’s my take on the multi-directional emission properties of CO2, but also, do you see how much it is adding? And do you see where you’ll never get. My take in the simplest of terms. If I over simplified, I apologize, if someone thinks I took the wrong bit of information from the post, feel free to correct.
Best wishes.

110. Dave Dardinger says:

Folks, it would be fun to try grading all the responses. The fact is that Tom has made a fundamental error. Roy Spencer has pointed it out, as you might expect. Werner Weber and Edvin have added more detail on where he went wrong. There are probably others who have gotten it right but I didn’t notice as I read them.
OTOH, a lot of people have either taken Tom’s words as gospel or gone off on tangents. I’m interested in seeing if Anthony gets it right or not.
BTW, in case you didn’t get it, the basic error by Tom is demanding that there be LTE at all times, even when a photon of IR from outside the local area is absorbed. In fact, he’s also misleading when he claims that a violation of LTE means the temperature of the local area is undefined. I’d like to see it worked out quantitatively, but in any case, if we’re talking a reasonable size local area, the uncertainty of temperature by one or a few IR photons being absorbed is too small to worry about. It’s only when there’s a constant input of such outside photons being absorbed that the local area must warm to produce a new LTE eventually.
One more thing. when we’re talking warming the atmosphere by added GHGs, the initial action is that a local area will absorb a larger % of the (unchanged at this point) IR entering it. This will, despite what Tom says, warm the local area and cause it to, when a new LTE is reached, emit more IR in all directions. The added IR which reaches the surface will warm it (See Roy Spencer’s site if you’re one of those who don’t think a colder body [atmosphere] can warm a hotter body [the surface] via radiation) and this is the raw material of the climate change theory. The CAGW crowd then make a number of wrong assumptions at this point, but I think all the skeptics here are quite aware of that.

111. RW says:

[SNIP] Repeated use of the d-word made your posting effort a waste of time. Read the site policy. ~dbs, mod.]

112. I think too many folk get caught up in detail (e.g. considering individual molecule interaction vs homogeneous ‘energy’ interaction in the composite atmosphere); rather it would simplify things to consider energy flux flows in 3-D and apply the concepts of S-Parameters (Scattering matrix parameters) to the analysis problem.
A similar problem was manifest years ago when electrical engineers attempted to characterize device (transistor) performance at RF/microwave frequencies using the usual I and E (or e) quantities but encountered shortcomings using that approach until the discovery a revolutionary new way of ‘handling’ a myriad of interactions at very high frequencies … the S-Parameter method of device characterization was born.
http://www.microwaves101.com/encyclopedia/sparameters.cfm

S-parameters refer to the scattering matrix (“S” in S-parameters refers to scattering). The concept was first popularized around the time that Kaneyuke Kurokawa of Bell Labs wrote his 1965 IEEE article Power Waves and the Scattering Matrix.

http://en.wikipedia.org/wiki/Two-port_network#Scattering_parameters_.28S-parameters.29

S-parameters are … defined in terms of incident and reflected waves at ports. S-parameters are used primarily at UHF and microwave frequencies where it becomes difficult to measure voltages and currents directly.

Measuring S-parameters: The First 50 Years
At low frequencies, voltages and currents (and impedance) are used to represent the electrical properties of an electrical circuit at a certain instant in time. When the excitation frequency increases to the point where the physical size of the circuit has the same order of magnitude as the associated wavelength, wave propagation must be taken into account. Voltage and current can no longer be defined unambiguously.
…. [S-Parameter] form put[s] transmission, reflection and impedance into a single two-dimensional representation, which could be readily measured and easily visualized, thus revolutionizing high frequency measurement and design

.

113. Gail Combs says:
August 5, 2010 at 8:27 am
“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.”
No system that has net radiative energy flows can be in LTE. It may be quite close to LTE, but the departures from LTE cannot be neglected when considering radiative heating and cooling. LTE is not a valid or relevant assumption for discussing the greenhouse effect. It is important to distinguish LTE (local Thermodynamic Equilibrium) from LDE (Local Dynamic Equilibrium), which merely means the temperatures, pressures, composition, inputs, outputs, etc., aren’t varying (or varying only on timescales much greater than that of the molecular processes concerned).
Note that within Earth’s troposphere, the gross molecular flow of heat energy through any surface is of order 100MW/m2 (heat content times molecular speed). The net energy flows (radiation, convection, etc.) are however only ~300W/m2, a few millionths of the total. Being so comparatively small, they normally produce only very slight departures from LTE. Nevertheless, they are quite adequate to create and maintain strong vertical temperature gradients and major temperature differences across the globe.

114. Dave Springer says:

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”.

Love it! Expression in my culture is “the tail does not wag the dog”.
What’s the optical depth of 15um earth vs. Mars atmosphere?
I think the biggest effect the atmosphere has on climate is 14.7psi at the surface raises the vaporization point of water enough that can have a global ocean covering 70% of the surface to an average depth of 4000 meters.
The the big picture becomes “the sun heats the ocean, the ocean heats the atmosphere, and the atmosphere radiates it into the frigid depth of outer space.
Everything else is a minor detail in comparison. The ocean is the dog. The atmosphere is the tail.

115. Gail Combs says:

DocWat says:
August 5, 2010 at 5:23 am
Help me here…
What I really don’t understand is why water and CO2 are better, by a factor of 20, at this as N2 and O2
________________________________________________________
It has to do with the shape of the molecule. CO2 is
O-C-O and can bend a little in collisions so it is no longer symmetrical
Water, H2O, is not linear like CO2 but shaped in a V (hope the drawing works)
H…. H
\ O /
N2 is N =N (triple bond) and O2 is O=O (double bond)
O2 and O3 grabs much higher energy wavelengths from the sun
This gives a picture of what molecules grabs which wavelength of energy from where (sun or earth) and a ball park of how much.

116. Pamela Gray says:

Gail, you point out again that much of chemistry is engineering design and physics, not chemical properties. Which is why I prefer to teach chemistry from a visual building blocks perspective. I wish Legos would sell a chemistry set. That would be so cool.

117. frank says:

This post is an insult to honest climate skepticism. The author’s thesis is correct, but incomplete, and therefore totally irrelevant to climate. “A volume of gas in Local Thermodynamic Equilibrium (LTE) cannot be heated by CO2” UNLESS MORE PHOTONS ENTER THAT VOLUME THAN LEAVE. WHEN more photons enter than leave, then that gas will heat up until it radiates away as many photons as it receives. Of course, carbon dioxide can’t heat the atmosphere by itself – it needs an outside source of energy!
Radiant energy is constantly FLOWING from the sun to the earth’s surface and atmosphere and from the earth’s surface and atmosphere to space. Within the atmosphere energy flows by both radiation and convection. The temperature at various locations in the atmosphere and on the surface of the earth is determined by the net flux of energy at that location (and never reaches true equilibrium because the energy input from the sun changes with night/day and the seasons). We are now conducting a planet-wide experiment to see what happens when we double the concentration of the most important molecule mediating the flow of energy from the upper atmosphere to space. (There is little water vapor in the upper atmosphere and most energy leaves the lower atmosphere by convection to or radiation to the upper atmosphere.)
Tom Vonk is correct when he says that the following statements are over-simplifications and need corrections (in caps): “CO2 absorbs AND EMITS the outgoing infrared energy and warms the atmosphere TO A HIGHER TEMPERATURE THAN IT WOULD HAVE WITHOUT CO2” – or – “CO2 traps part of the infrared radiation between ground and the upper part of the atmosphere” AND IS THE MAJOR SOURCE OF INFRARED RADIATION FROM THE UPPER ATMOSPHERE TO SPACE. The credibility of the skeptical community is not helped when Tom makes similar over-simplifications and then buries the important qualifications in Caveat 1: “The concentrations of CO2 (and H2O) play a role in [the] dynamics [of the Earth-atmosphere system] but it is not the purpose of this post to examine these much more complex and not well understood aspects.” What is the purpose of Vonk’s post: To correct the over-simplifications of CAGW’s or to confuse readers into thinking that CO2 plays no role in the temperature of the atmosphere? The future climate of the planet depends on these “much more complex and not well understood aspects”.

118. Gary P says:

This article is misleading. If one assumes constant equilibrium conditions then indeed nothing changes because, well, you have assumed constant equilibrium conditions. The collisional frequency and excited decay times are important because the collisions are so frequent that local equilibrium exists in the bulk of the the atmosphere. All excited states of all the gasses are in equilibrium in the lower atmosphere. This is not true 100 km up. Adding CO2 does increase the adsorption of IR in a closed cell and the temperature must go up to until the IR emission equals the absorption at a new equilibrium.
Closed cells are not at all like the open atmosphere. Adding IR absorption will increase convection and how this affects the hydrodynamic cycle is the critical process. My pet theory says that the tropopause will rise slightly due to CO2 increases and dry out the stratosphere so that the net optical density remains constant per Miskolczi.

119. RW says:

[snip]

120. I’ve only skimmed this, and will take a closer look after I’ve had some grub, but if this is all standard text book stuff, and has been for 100years, why don’t Jones, Hansen, Mann et al know about it….Actually, thinking about it, thats probably a stupid question.

121. Gail Combs says:

Paul Birch says:
August 5, 2010 at 9:43 am
No system that has net radiative energy flows can be in LTE. It may be quite close to LTE, but the departures from LTE cannot be neglected when considering radiative heating and cooling. LTE is not a valid or relevant assumption for discussing the greenhouse effect. It is important to distinguish LTE (local Thermodynamic Equilibrium) from LDE (Local Dynamic Equilibrium), which merely means the temperatures, pressures, composition, inputs, outputs, etc., aren’t varying (or varying only on timescales much greater than that of the molecular processes concerned)…..
________________________________________________________
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.
I think both side of the discussion should be looking at Tom’s definition of the LDE (Local Dynamic Equilibrium) AS an Equilibrium. Jump on him if his definition of LDE (Local Dynamic Equilibrium) and the physics of that state is incorrect otherwise jump on him when he finishes defining his terms and goes on from there.
Many of us have little or no thermodynamics or even physics or chemistry background so I find this definition of terms highly useful in advancing my knowledge.
So please stick to the very small portion of Thermodynamics, LDE (Local Dynamic Equilibrium) Tom has described. Is Tom’s description of this equilibrium state wrong and if so where and why.

122. Dave Springer says:

stevengoddard says:
August 5, 2010 at 8:43 am
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.

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.

123. Liam says:

Seems like we’ve got a good peer review process going on here.

124. Andy says:

I guess the real science was settled after all.

125. Bryan says:

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.

126. Karel says:

Sorry guys, but I’m not nearly smart enough to understand this post. Anyone game for a layman’s translation ?

127. coaldust says:

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?

128. JDN says:

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?

129. Dave Springer says:

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.

130. cba says:

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.

131. DeWitt Payne says:

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.

132. anna v says:

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?

133. 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.

134. wayne says:

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.

135. Dear Dr. Tom:
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???

136. dp says:

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?

137. Tenuc says:

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!

138. Merrick says:

Bob Kutz,
***********************************************************************
[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]
**********************************************************************
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.

139. Merrick says:

Correction above:
…otherwise GHGs WOULDN’T warm the planet…

140. Merrick says:

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!

141. Dave Springer says:

Karel says:
August 5, 2010 at 10:52 am
Sorry guys, but I’m not nearly smart enough to understand this post. Anyone game for a layman’s translation ?

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.

142. Reed Coray says:

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.

143. Merrick says:

Jeff Id, thanks for your brilliant thought experiment. Wish I’d thought of it…

144. Merrick says:

Gail Combs:
*********************************************************************
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.
*******************************************************************
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:
********************************************************************
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 ?
********************************************************************
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.

145. Dave Springer says:

“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.

146. Merrick says:

Gail Combs:
**********************************************************
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
***********************************************************************
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.

147. RW says:

If there is a particular word in what I post that you don’t like, why not just delete that word, instead of junking paragraphs that I’ve put time and effort into writing?
—-
LTE is a poor assumption in much of the Earth’s atmosphere. In any case, the logic does not work. There is energy being absorbed by CO2 molecules which is not absorbed directly by N2 or O2 molecules, and it is transferred collisionally to these molecules. It’s as simple as that. Even if the rate of CO2->N2 energy transfer is exactly equal to the rate of N2->CO2 energy transfer, the fact that there is energy being put into the CO2 side means that N2 is heated up.
To claim otherwise is simply anti-science.
The supposed experiment does not work, because it is obviously in thermal contact with the rest of the atmosphere. Doubling the CO2 concentration inside a tiny packet of air will make a negligible difference to the temperature of the whole atmosphere.

148. Merrick says:

Pamela Gray says:
August 5, 2010 at 10:06 am
Gail, you point out again that much of chemistry is engineering design and physics, not chemical properties. Which is why I prefer to teach chemistry from a visual building blocks perspective. I wish Legos would sell a chemistry set. That would be so cool.
This is where I’m compelled to say that much of physics and engineering design is physical chemistry, the physicists and engineers just don’t know it.
It reminds me of the days when one of my post-doctoral prefessors (a physicist) used to call gaseous calcium a “molecule” because it has two valence electrons and therefore electron correlation. “That’s what defines a molecule, right?”

149. Jan K. Andersen says:

Well I consider myself as an AGW skeptic, but I’m not a skeptic to the existence of a natural greenhouse effect. Tom Vonk says in essence that no greenhouse effect can exist.
He defines the system to be in equilibrium and then proof that the system indeed is in equilibrium.
If you substitute CO2 with glass you may use the same arguments to proof that conventional greenhouses don’t work.

150. Reed Coray says:

Nylo says:
August 5, 2010 at 5:51 am
By the way, CO2 doesn’t heat, that’s true, and that makes your small experiment work. What it does is slow down the cooling of the surface of the Earth. It reduces the net radiation at 15um. As a result, you get higher temperatures than you would have without CO2. But all the initial heat came from the Sun, of course.

I am uncomfortable with the statement CO2 slows down the cooling of the surface of the Earth.
To me, a “slowing down” of the cooling of an object means less energy per unit time leaves the object. Assume a fixed rate of energy per unit time is being absorbed by and/or generated within an object; and at temperature T the object is in “heat transfer” equilbrium–i.e., the amount of outgoing energy per unit time is equal to the incoming radiation per unit time. If you “slow down the cooling”, aren’t you decreasing the rate at which energy leaves the object? But if you slow down the cooling without decreasing the rate of energy input, as a function of time the object must retain more and more heat. The object will stop retaining heat only when the outgoing rate of energy equals the input rate of energy. A rise in object temperature may be the mechanism that re-establishs input rate/output rate equality; but once reached, the outgoing rate of energy will be equal to the input rate of energy (which hasn’t changed) and hence cooling (or the rate of cooling) will also unchanged. As such, it may be correct to say that CO2 may temporarily affect the rate of cooling, but without a change in the rate of energy input, eventually the rate of cooling will return to its original value, at which point the “cooling” has not been “slowed down”. Bottom line, CO2 has been added but for a fixed rate of energy input, the “cooling or rate of cooling” is unchanged.

“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.”
you must add the qualifier that this is “in an isolated system at thermodynamic equilibrium”. In the Earth’s atmosphere, there is a very real external IR source .. the surface of the planet. IR radiation absorbed from this source very definitely WILL heat the atmosphere (if convection from the surface is ignored). Energy WILL be transferred to the nitrogen. The way the article’s conclusions are stated implies that this does not happen.
Then your conclusions say “The main point is that every time you hear or read … you may be sure that this source is not to be trusted … ”
THAT is way beyond misleading. (The caveats help, but don’t go far enough.) To compare a closed system LTE analysis with open system real world results is simply wrong!
A more complete analysis that includes convection changes the heat balance so that adding CO2 actually causes increased atmospheric cooling. (The proof is trivial.)

152. stephen richards says:

Merrick says:
August 5, 2010 at 7:15 am
Sorry, just need to weigh in one more time.
That’s better ! I’m a physicist but was having great difficulty with this essay.

153. wayne says:

Dr. Vonk, (and I should include anna v)
I include you anna because I am going to bring back up a subject that you and I went around and around months ago and I was unable to have you see exactly what I was trying to portray.
Dr.Vonk, you mention in your article that CO2 (and I in silently add any GHG from here out) absorbs radiation from the ground but due to the LTE cannot hold that energy and I totally agree. However this gets into the intersection of topics with ‘back radiation’. That back radiation is photons emitted toward the nadir hemisphere since the remittance direction is randomized.
My point to both of you is a huge point that is being missed here. Huge. Since the LW radiation from the ground is always in the azimuth hemisphere’s direction and ½ of the reemitted photons are in the nadir hemisphere, a radiation pressure exists here which will cause and infinitesimal but real expansion of the atmosphere which will exactly cause a temperature decrease that exactly counteracts any warming from the downward radiation. anna, I stated this in January or February and never could get you to see.
Get it? Please, please give this a moment of thought this time through. I will keep bring this up every month or two until 1) someone says they clearly see what I am talking about or 2) someone clearly explains how this radiation pressure created by these LW radiations, up from the ground, ½ down from GHGs does not cause a pressure which will expand the atmosphere and by thermodynamics fundamental equations will cool the atmosphere in exactly an equal amount. Please give me a moment.
Wish I was better with words, you might have to read some proper terms in to make it clear to you. I give you permission to read between my lines.
Dr. Spencer, if you are reading, this is the one point you and I disagree with the common acceptance that CO2 (or any LW interacting GHG) will always warm. I still don’t see it because of the above explanation of increased radiative pressure.
Thanks for the moment.
– wayne

154. stephen richards says:

Jan K. Andersen says:
August 5, 2010 at 12:28 pm
Well I consider myself as an AGW skeptic, but I’m not a skeptic to the existence of a natural greenhouse effect. Tom Vonk says in essence that no greenhouse effect can exist.
There will be no ‘greenhouse effect” if the IR is ‘given up’ in sufficiently small timescales. So what I think I’m trying to say is that when a mole absorps radiation in the IR in translate to the first excited level and falls back to the ground state within a fraction of a second.
That is a very poor description but I’m tired and fed up.

155. stephen richards says:

Merrick says:
August 5, 2010 at 12:28 pm
Pamela Gray says:
August 5, 2010 at 10:06 am
Gail, you point out again that much of chemistry is engineering design and physics, not chemical properties. Which is why I prefer to teach chemistry from a visual building blocks perspective. I wish Legos would sell a chemistry set. That would be so cool.
I used to think that physical chemistry was a subset of physics. Was I arrogant or simply wrong?

156. K~Bob says:

Karel,
Yeah, I’ll take a whack at it.
Someone made an attempt to describe something, and a bunch of intelligent people decided it was darned near fraudulent, or something.
Because arguing in good faith is evidently not scientific enough?
Hey, you ruffled feather types: just deal with the argument and stop assuming some fraud is being perpetrated. Make your case and let the guy respond.

157. DirkH says:

Thanks Tom. This provides an excellent explanation for my idea of a “CO2 fog” that just redistributes the infrared radiation. You stirred up a record number of concern trolls! (“Uh, this is such a disappointing article on this otherwise excellent blog”. Yeah right you people go back to RealClimate Hell where you belong.)
For the people who said: “But you disturb the LTE when you add more CO2” please consider the time scale! The re-radiation processes work in milliseconds; the rise in concentrations over years. It’s practically unrelated.

158. mkelly says:

Max Hugoson says:
August 5, 2010 at 11:36 am
Dear Dr. Tom:
Dr. Elsasser’s classic, “Infrared Heat Transfer in the Atmosphere” is available at ScribeD.
Thanks Max. I have read this several times and find it fasinating. I will save the link this time.

159. DirkH says:

A lot of people have stated that Tom Vonk’s explanations mean that there cannot be a greenhouse effect. This is false; it is NOT what Tom’s explanations result in.
Rather, it results in 50% of the LWIR radiation in the CO2 absorption band passing through to space and 50% being re-radiated back to Earth for a sufficiently dense CO2 fog. This is of course a greenhouse effect; without the LWIR-absorbing gas 100% would reach space.
Of course, this effect will not be affected by further CO2 concentration rises as the CO2 fog is already dense enough to catch every LWIR photon in its absorption band emitted by the surface.

160. JAE says:

LOL. I think y’all will find that Vonk really understands this physics. Can’t wait for his responses!
“stevengoddard says:
August 5, 2010 at 8:43 am
bushy,
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.”
And places like Phoenix also have relatively low amounts of GHGs, yet have very hot temperatures (relative to Atlanta, which is at virtually the same latitude and altitude, but which has over 3 times as much GHG). Are you forgetting about solar insolation?
Why do so many folks keep trying to explain atmospheric temperatures with only infrared radiation? What about heat storage, convection, conduction? No matter what you believe about the greenhouse effect, it is only ONE effect!

161. wayne says:
August 5, 2010 at 1:04 pm
“Please, please give this a moment of thought this time through. I will keep bring this up every month or two until 1) someone says they clearly see what I am talking about or 2) someone clearly explains how this radiation pressure created by these LW radiations, up from the ground, ½ down from GHGs does not cause a pressure which will expand the atmosphere and by thermodynamics fundamental equations will cool the atmosphere in exactly an equal amount. ”
Within any gas in which radiation is being absorbed and re-radiated anisotropically, there is indeed a differential radiation pressure on the gas. So the partial trapping of solar energy near the Earth’s surface by clouds and greenhouse gases does cause the atmosphere to fill a volume greater than it otherwise would at that temperature. However, the magnitude of the effect is very small ~1E-6 N/m2 (radiative flux/speed of light), compared to the atmospheric pressure ~1E5 N/m2. Furthermore, it does not cause net ongoing cooling to offset the greenhouse warming because once the atmosphere has made that very slight adjustment it ceases to expand.
Although this mechanism has negligible effect on planetary atmospheres it does become significant in stellar atmospheres, especially in stars of very high luminosity. Radiation pressure limits the luminosity of stars of mass M solar masses to ~33,000M solar luminosities (the Eddington Limit) . Above that limit the radiation pressure would blow the star apart.

162. JAE
The absolute humidity in Phoenix is orders of magnitude higher than Antarctica.
I have been in Phoenix when the dew point was over 70F. Even on a dry day, there is plenty of water vapour.

163. Dave Wendt says:

Dave Springer says:
August 5, 2010 at 10:18 am
stevengoddard says:
August 5, 2010 at 8:43 am
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.
“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 CO2 signature at the South Pole would seem to be greater than anywhere else on the planet. Spectral analysis of Downward Longwave Radiation done there suggests that CO2 provides a full third of the DLR signal virtually yearround.
http://journals.ametsoc.org/doi/abs/10.1175/JCLI3525.1
“Annual cycles of downwelling broadband infrared radiative flux and spectral downwelling infrared flux were determined using data collected at the South Pole during 2001. Clear-sky conditions are identified by comparing radiance ratios of observed and simulated spectra. Clear-sky fluxes are in the range of 110–125 W m−2 during summer (December–January) and 60–80 W m−2 during winter (April–September). The variability is due to day-to-day variations in temperature, strength of the surface-based temperature inversion, atmospheric humidity, and the presence of “diamond dust” (near-surface ice crystals). The persistent presence of diamond dust under clear skies during the winter is evident in monthly averages of clear-sky radiance.
About two-thirds of the clear-sky flux is due to water vapor, and one-third is due to CO2, both in summer and winter”
This percent of CO2’s contribution to DLR doesn’t appear to be exceeded anywhere else on the planet. To me this would suggest that if increasing CO2 concentrations in the atmosphere are the prime driver of temperature trends, it should be most evident at the South Pole, but it is fairly evident that temps there have been laying about like a dead dog for 50 years.

164. kwik says:

John W. says:
August 5, 2010 at 5:04 am
“II have some questions:….”
John W, why do you have these questions? Because you must rely on authority?

165. Steven mosher says:

Tom V;
Thanks for that. I think however that many people will miss the point.
can you explain for folks what you mean by this? and how this caveat related to AGW.
Also, Just for the record, do you accept the fundamental physics of RTE’s
Or simply, to a first order, what is the impact of having more C02 or H20 or any GHG in the atmosphere?
The Caveat that needs some explantion:
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 .”

166. Savant says:

Anthony,
Nice to have these guest posts, but it would greatly improve the science if you would peer review these first. This is a classic unhelpful post with much that is correct (and could help the debate) yet the conclusion is all wrong as a result of some simple errors (e.g. not closed system due to IR); Ferdinand Engelbeen post on CO2 is very helpful; well written and referenced. More of this.

167. George E. Smith says:

Well Tom, I understood most of what you wrote; but I am not sure I can agree with your conclusions.
CO2 consists of only one in about 2570 molecules of the atmosphere so CO2-CO2 collisions are highly unlikely, and CO2-N2 and CO2 O2 collisions ought to occur in about 4:1 ratio.
But N2-N2, and N2-O2, or O2-O2 collisions are far more likely.
So a CO2 molecule colliding with either a N2 or an O2 molecule is highly unlikely to encounter the same N2 molecule again and retrieve the exchanged energy.
Yes I know you said that LTE is assumed; but how real is that in an open atmosphere with convection currents as well as both a Temperature gradient, and a density gradient.
Which brings to mind another question.
A solid material; say ice for example can and does emit black body like thermal continuum radiation since it is above absolute zero. Heat it above zero deg C and it becomes a liquid and continues to emit a continuum thermal radiation with a roughly black body spectrum and Stefan Boltzmann output.
But perish the thought that you should heat the water to above 100 deg C whereupon it becomes a gas, and immediately stops radiating thermal radiation; because everybody knows; or seems to think that gases do not emit black body like thermal radiation.
Of course nobody explained this to the sun; or the argon or other gas in a gas filled incandescent light bulb. By some miracle, a very large fraction of the broadband radiation emitted from a gas filled lamp comes from that very gas filling, which of course can’t emit BB like continuum radiation.
A consequence of the model you have proposed would seem to be that the “back radiation” due to CO2 interception of surface emitted (from solid or liquid continuum thermal radiation can consist only of the specific wavelengths that the CO2 absorbed in the first place; since you say no net energy is exchanged between the CO2 and the Atmosphere.
I agree with your contention that symmetrical non polar molecules do not interract with the IR radiation; but in collisions, even symmetrical polyatomic molecules become polar, and therefore can absorb and radiate.
By the same token, in collisions even symmetrical molecules like N2 or O2 can exhibit an accelerating electric charge and therefore emit EM radiation in a continuum spectrum.
Do you have a rigorous QM proof that gases do not emit ordinary BB like thermal continuum spectra just like any other material above absolute zero does ?
No I am not asking for your proof here; just do you have or know of one ?
If I accept your premise that the trace GHG gases do not transport thermal energy to the ordinary atmospheric gases; then that of course would apply to H2O as well; so how come we are not that frozen ice ball.
In any case; the question seems to me to be moot, since there is general agreement that CO2 and H2O and other GHG molecules DO capture LWIR from the surface or other atmospheric layers; which must increase the net energy (and Temperature) of THAT layer.
So is it relevent that CO2 and N2 can’t unilaterally transport energy following your explanation; the atmospheric Temperature is nevertheless increased by the GHG captures.
I don’t see any assymmetry in your N2-CO2 exchange. On average the same applies to any two molecules so there isn’t any net transfer from one N2 molecule to another, or N2-O2 either.
But anyway’ I appreciate your explanation. Any insight is always better than none at all.
George

168. George E. Smith says:

As to the world’s dries deserts; I recently watched a PBS show on the Andes; and they stated quite specifically that a particular high desert on the eastern side of the Andes (with real sand) was the driest desert on earth.
In any case could people please distinguish bewteen Absolute humidity and Relative humidity when using the term so we know what they are referring to.
I am under the impression that the water content; by molecular abundance of the driest desert air anywhere on earth is always higher than the molecular abundance of CO2 in the same location at the same time.

169. Lazy Teenager says:

Nearly all of the details of the argument are correct but it fails because of a faulty assumption. The assumption is that all parts of the atmosphere are in perfect local thermal equilibrium. This is not the case.
In actual fact the atmosphere are in APPROXIMATE local thermal equilibrium.
There is a thermal gradient between the surface (25C) and outer space (-270C) and this cannot happen if there is LTE.

170. Deanster says:

To follow up on Juraj Vs post.
Without doubt, scientifically proven, and indisputable, density of the atmosphere is the primary driver of temperature of the atmosphere.
SO .. can one of you more informed folks than me speculate what impact an increase in CO2 would have on density??

171. wayne says:

Paul Birch says:
August 5, 2010 at 2:02 pm
… Furthermore, it does not cause net ongoing cooling to offset the greenhouse warming because once the atmosphere has made that very slight adjustment it ceases to expand.
___
But that is exactly what I was inferring. The energy flux through any volume is equal after added GHG. It is only that initial increase in warming that persists, the warming doesn’t keep growing. But the expansion also persists at that new higher temperature in exact like manner. The expansion only has to account for some 0.7K/288K of the initial rise. I can’t seem to accurately calculate to see if this equates (my cry for expertise).

172. Dr A Burns says:

This discussion reminds me of the “CO2 warming violates the 2nd Law” paper by Gerlich :
http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf
I like Ed Fix’s comment. While these articles may not prove that CO2 doesn’t lead to warming, what is great is that articles like these encourage critical thinking, rather than the common blind acceptance of political AGW dogma.

173. wayne says:

@ Paul Birch
First, thanks for the moment. Second… reset.
My response was wrong. Went out for a break going over my response to you in my mind and I realized it made no sense. I said ‘increased temperature’ where I should have said ‘increased radiation’.
Here is another way to look at my unexplained cycle, all happening in an one instant…
Increase a LW interacting GHG in the atmosphere… causes absorption / random radiation… the ½ down increases radiation pressure… atmosphere expands… thus it cools… BUT, radiation ½ down causes an increase in temperature… warming equals (??) the cooling? Neither is continuous but a static one time differential since all heat must still equate on both sides, ground absorption and LW at TOA.
See, that is the loop I cannot totally explain as I go over again and again. Spent months with this popping up in my thoughts now and then as I read the thousands of comments and papers I have absorbed in the last year.
There, that might explain it better.

174. Rob R says:

Steve Goddard
At the North Pole or at least inside the Arctic Circle the mean summer temp is about 2 or 3 deg C as you are well aware, but of course this is at sea level. At the South Pole one is not at sea level but at 3000 to 4000 m altitude (very approx). The atmospheric lapse rate effectively demands that the summer temp is very cold over most of Antartica. But so what? At sea level around the fringe of Antarctica the mean summer temps are, as one would expect, nowhere near as cold.
If you correct for the environmental lapse rate over the interior of Antarctica the mean summer temp is similar to that of summer within the Arctic Circle. I am not saying it is the same given the rather different geography (one polar region is isolated by a very cold Southern Ocean, while the other is ringed by land).
So why is it cold in summer both in the Arctic and the Antarctic. I would venture to suggest that despite high values for summer insolation, the primary reasons are:
1. The angle of incidence, which is low, aiding in the reflection of incoming light.
2. The surface albedo at both poles is such that much of the light that gets to the surface is reflected regardless of the angle of incidence.
So although low greenhouse gas (H2O) concentrations over the Arctic and Antarctic may contribute to low summer surface temperatures these gases are only part of the story.

175. Bryan says:

dp
You are talking to the converted.
I think that H2O provides most of the radiative effects even in deep midwinter Antartica.

176. Julio says:

Reed Coray says:
August 5, 2010 at 12:00 pm
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. The point being, once the temperature of the “Earth” is such that Iup = 1.25*Iin (in my example), then everything is self-consistent: 80% of that upward flux, which is exactly 0.8*1.25*Iin = Iin, passes through the glass, and goes away; since that exactly matches Iin, no further thermal energy accumulates inside the enclosure. Starting with a “cold” ball, the temperature increases only until that equilibrium temperature is reached, and no further.
The real world is, of course, more complicated; instead of a “glass” you have the greenhouse gases in the atmosphere, but the point is that as long as *some* of the energy they re-radiate goes down, and not up into space, the Earth will have to get warmer than it would be if they were not there, in order for the next flux into space to match the solar input Iin. The other essential point is the spectral asymmetry: namely, that the greenhouse gases are more transparent at visible wavelengths (which is where most of Iin lies) than at infrared, which is where Iup is concentrated.
That’s the big picture, and nothing in Tom’s post has any bearing on it.

177. Gnomish says:

DR says:
August 5, 2010 at 7:20 am
I’d think the heat capacity of a CO2 molecule vs water vapor would be of importance. No?
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
Totally. The heat capacity of water when it changes from liquid to vapor or vice versa is 50,000 times greater than that of the CO2 in a volume of atmosphere with 1% H20 and 500 ppm CO2. The temperature does not change. The BB does not change.

178. cal says:

In my previous post I did not comment on Tom’s paper because I thought it was an earnest and sincere attempt to explain some of the detailed mechanisms that occur in the atmosphere. However, as many have pointed out, his intitial analysis only applies to LTE which is not realistic, and his development to dynamic systems and his caveats are incomplete.
But I would like to come to his defence because this is the nature of physics. I have studied physics and applied it all my life but I have never been able to use it to directly and completely solve any real life problem. However, by changing the problem to one I could solve (by simplification and limitation) I have gained insights that have allowed me to make and test predictions and create models that are close enough to reality to make them useable. After numerous iterations of this process something close to “good enough” is achieved. This is all I have ever hoped for.
Using such complex but proven models, that have evolved over time, engineers (who in general do not understand the minutae) build the world we see around us. No one uses quantum mechanics to design an aircraft but material science based on QM has provided clues as to what alloys might be worth testing for such a purpose.
Tom has tried to clarify some of those underlying principles but they are only really of academic value. They pertain to systems as defined but are not consistent with the real world. However he has done nothing different from what the climate science community has done except that they have encapsulated these approximations into complex models. The problem is that these models have not been tested and as yet cannot be tested. Rather than attack Tom you should see him as an example of climate scientists of all persuasions who are trying to analyse the behaviour of complex chaotic systems by the application of simplistic relationships studied in a laboratory. The analysis is helpful but there is no certainty in their predictions.

179. wayne says:
August 5, 2010 at 3:36 pm
“Increase a LW interacting GHG in the atmosphere… causes absorption / random radiation… the ½ down increases radiation pressure… atmosphere expands… thus it cools… BUT, radiation ½ down causes an increase in temperature… warming equals (??) the cooling? Neither is continuous but a static one time differential since all heat must still equate on both sides, ground absorption and LW at TOA.”
OK, let me try to explain for you. “Increase GHG … atmosphere expands … thus it cools…” Fine so far, but note that the expansion due to radiation pressure is of order only 1 part in 1E11, so the adiabatic cooling is only of order 1 nanokelvin (and that much only if you go from optically thin all the way to optically thick). Really, really tiny. “… radiation down causes temperature increase … ” Yes. Other things equal, an increase in the absolute temperature of ~28% for an optical depth difference of one. So ten trillion times as big as the cooling! Now the radiation pressure in the cavity between surface and absorber has increased by a factor of 2.7, due to the higher temperature, but the additional expansion and cooling this causes is still utterly negligible (a further 1.7 parts in 1E11). The warming of the atmosphere in the cavity also causes thermal expansion (of 28%). Note that the temperature reached within the cavity is unaffected by the expansion of the atmosphere, or the (~10%) adiabatic cooling temporarily caused thereby; it just takes a little longer to heat up to its final value; the extra energy is stored in the greater gravitational potential energy of the expanded atmosphere. Once the system reaches a state of dynamic equilibrium the energy lost by the ground equals the energy reaching the ground.

180. Julio says:

Julio says:
August 5, 2010 at 4:07 pm
Sorry, a typo: I meant the “net flux,” not the “next flux.”
Also, about those balance equations, notice that they work also from the Earth’s point of view: it is radiating at a rate 1.25*Iin, but it is also receiving Iin (directly from the sun) plus the 20% of Iup that is sent back to it by the “greenhouse glass”, which is 0.2*1.25*Iin = 0.25*In. Total, 1.25*Iin.

181. Steven mosher says:

Jeffid,
Brilliant.

182. Have a watch going on “global warming” & have noticed the climate alarmists’ new catch phrase is “undeniable” & no longer “settled science.”

183. Rob R
I go up to 14,000 feet pretty regularly in the summer, but have yet to encounter temperatures of -40C . Normally they are about 40-55C warmer than at the same elevation in Antarctica, even though Antarctica receives more solar insolation.

184. Chris de Freitas says:

The focus is on the gas (CO2) rather than on the radiation budget. The greenhouse gases affect the allwave radiation budget of the Earth’s surface as well as the allwave radiation budget of the atmosphere. For the Earth as a whole and on average, the atmosphere is in radiation deficit, whereas there is a surplus at the surface. Surface-to-atmosphere sensible and latent heat fluxes bring about equilibrium. If there is an increase in CO2, say, the surface-atmosphere budget changes. We get a higher air temperature (i.e. by increased sensible heat flux) and increase in evapotranspiration (i.e. increased latent heat flux).

185. Spector says:

I believe what is missing from this article and perhaps much of the work of many in this field is an equal focus on the cooling effect of Earthshine emitting and absorbing (i.e. greenhouse) gases. For each ‘greenhouse’ gas there must be some altitude at which half the photons emitted going straight up escape to outer space. I am somewhat puzzled that this or an equivalent mean escape altitude is not a well known parameter for each of the Earth’s greenhouse gases.
Once energy from CO2 and H2O begins to leak into outer space, LTE is violated, temperatures *must* fall until a more global thermal equilibrium is established with incoming thermal radiation and convection. At the 15µ band, solar radiation has diminished to an insignificant trickle and this allows temperatures at the tropopause level to be extremely cold (average -56.6 deg C.) such that the normal adiabatic lapse rate makes life possible at the surface.
At about 90 km up, I note that the atmosphere of Venus cools down to very low temperatures in the vicinity of -112 deg C. I suspect this is the CO2 thermal radiation escape altitude there.

186. Kevin Kilty says:

Paul Birch
…Note that within Earth’s troposphere, the gross molecular flow of heat energy through any surface is of order 100MW/m2 (heat content times molecular speed). The net energy flows (radiation, convection, etc.) are however only ~300W/m2, a few millionths of the total. Being so comparatively small, they normally produce only very slight departures from LTE. Nevertheless, they are quite adequate to create and maintain strong vertical temperature gradients and major temperature differences across the globe…

I don’t follow, here. The net power flowing through a square meter per unit path via advection (is this what you mean by molecular flow?) is specific heat x density x velocity (dot) temperature gradient. I find values of maybe 10 kW per meter squared at a temperature gradient of a degree per hundred meters in a stiff breeze. What is this gross molecular flow of heat?

187. Caleb says:

Very interesting stuff. I’ve only had time to struggle through the article and read the first twenty comments, and can see I have a lot to learn. However I crave to have water vapor included in this sort of discussion. (If it occurs further down in the comments, I will relish it. But I will dare comment before reading all the comments.)
These other gases are rather boring, for they pretty much stay in the gaseous state. Where’s the fun in that? Water vapor, on the other hand, turns from gas to liquid to solid, in the rising tower of a thunder head, and with each change in state there is a vast release of latent energy.
Some of this latent energy is released in the middle of a cloud, even though the relative humidity is 100%, because cloud droplets grow in size.
However the real release of latent energy occurs where the warm air at 100% relative humidity comes in contact with freezing cold air. And where might that be? On the very skin of a cloud, especially the upper skin.
In big thunderstorms, and especially in big hurricanes, this upper skin of clouds, releasing vast amounts of latent energy, is way, way up there, twice as high as Mount Everest. At that altitude there is very little air of any sort, (and especially very little CO2,) between the vast amounts of latent energy being released, and the cold, merciless drain of empty outer space.
In my humble opinion the amount of heat released by water vapor changing state, in a force five hurricane, is so huge that it makes our interesting discussions about the heat involved in CO2, N2 and O2 look a bit like nit-picking.

188. maxwell says:

Tom,
as a physicist, you should know that molecular vibrations are not ‘quantified’. They are quantized. That’s why they call it ‘quantum mechanics’.
All in all, this argument fails for many of the reasons presented already. Most importantly, it’s observationally incorrect. Given that experimental data matters most in physics, I’d retract this whole post if I were you.

189. IMHO this topic does not belong on your excellent Blog. Too many assumptions and caveats are made without any rational explanation as to how they correspond with the actual atmosphere of the Earth.
For example, if I assume I have a perfectly smooth elephant with negligible mass lots of interesting circus tricks would follow. Sadly, Tom assumes LTE, energy equipartition, time symmetrical equations invariant under time reversal, and then caveats “the dynamics of the Earth-atmosphere system” as well as “relaxation time”.
Don’t get me wrong, I don’t buy into the AGW alarmist story that we are anywhere near a tipping point or even that human CO2 and land use activities are responsible for most of the warming we have experienced in the past 150 years. I think the actual warming is 30% to40% less than the 0.8C claimed by the climate research “team” because they have fudged the data and adjusted for UHI effects improperly. I accept that, after a certain point (about 300 ppmv) CO2 “greenhouse” effects tail off drastically due to saturation.
However, I think it is clear that human activities are responsible for perhaps 10% of the warming. I am also concerned about the long-term effects of the continuing rapid rise in CO2 levels, perhaps half of which is due to human activities. None of this justifies any kind of alarm or any action that would destroy the economy, but it cannot be totally dismissed by the “proof” Tom provides. Indeed, the only readers who will accept Tom’s argument are those who understand none of it but happen to like the conclusion they think he has “proven”.

190. wayne says:

Paul Birch
August 5, 2010 at 4:39 pm
Ok, don’t see where you gave where any of the numbers are coming from but it suffices you seem sure the radiation pressure from added GHG molecules (H2O or CO2), let’s say 10 per million is so infinitesimally small that the lift of the atmosphere in every cubic meter per one meter layer up to say 80 km is so insignificant it can be totally ignored. At one in 10 trillion, I would agree.
Do you mind me asking, how did you come up with the 1 or 1.7 in 10 trillion expansion figure? Very roughly is fine. Was it taking the 390 W/m2 of back radiation in common energy balance charts and calculating from that the upward pressure and therefore the volume expansion?

191. stevengoddard says:
August 5, 2010 at 6:04 am
Nice presentation, but an incorrect conclusion.

Would you be so kind as to tell me how is it that the conclusion is incorrect?
I have calculated the totabs and toemiss of carbon dioxide and other gases in the atmosphere through several well-known algorithms (T, σ and ρ) and found that the carbon dioxide cannot be a warmer of the atmosphere either the surface by any means.
Ira says:
August 5, 2010 at 6:53 pm
…I am also concerned about the long-term effects of the continuing rapid rise in CO2 levels, perhaps half of which is due to human activities… Indeed, the only readers who will accept Tom’s argument are those who understand none of it but happen to like the conclusion they think he has “proven”.
I am not concerned about the CO2 levels. CO2 is good for life. I understand Tom’s argument and accept it because, as I have told to Steven Goddard, the physics of thermal energy transfer supports the Tom’s argument.
Just to give an example, the mean free path length of photons in the CO2 at its current density in the atmosphere is 48 meters. Isn’t it too long? Compare it with the mean free path length in water vapor, nitrogen, oxygen and argon at their actual densities in the atmosphere and you’ll find the roots of AGW mistakes.

192. Gnomish says:

What I got out of the article was a detailed explanation of the distinction between temperature and heat, with special reference to the nature of radiation in the equation.
Temperature is not heat. I thought that was worth logical discrimination and enjoyed the article.

193. Caleb says:

PS to above post:
I should probably mention that the idea that most of the latent energy in a thermal is released in the “skin” of a cloud is not my own. It was the idea of a commenter in WUWT who was commenting on a comment I made, in the dregs of a posting, when most people likely had moved on to the next fascinating subject. I wish I could remember his name, for his was a really provocative idea.
The idea is that little latent energy can be released in the middle of a cloud, for relative humidity is already at 100%, and it is in some ways hard for water vapor to change to liquid state in such an environment. (Once fog has formed, how can it become more foggy than fog? The answer is “drizzle,” of course, however the latent energy released doesn’t seem enough to power a thermal up to the Stratosphere.)
Therefore the uplift of a thermal could largely be supplied by the “skin” of the thermal. Rather than the air within the thermal lifting the thermal like the air within a hot air balloon, the real uplift is supplied by the skin itself, as if the rubber skin of a hot air balloon was supplying more uplift than the air within the balloon.
This is a really neat idea. It makes you look at clouds in a new way. You entertain the idea a cloud is not “homogenized” uplift. Nor is the latent energy released “homogenized.” Instead the energy is largely released on the “skin,” where it is most likely to radiate out into outer space.
Whoever the fellow was, who shared this idea with me, he deserves, in my humble opinion, more stimulus money than any dweeb fiddling about with the temperatures from 1899, for he challenged me to look at clouds in a new and interesting way.
Of course, he is unlikely to see a cent.
However, in the unlikely event his idea “went viral,” you can bet your bippy Stimulus Money would be spent to shoot his idea down, (as any sort of suggestion that negative feedbacks exist are a threat to “Global Warming,” and therefore a threat to “Cap and Trade.”
But that is a really neat thing about WUWT. Most people don’t post here “for the money.” Rather they are interested in this thing called “Truth.”

194. Reed Coray says:

Julio says:
August 5, 2010 at 4:07 pm
Reed Coray says:
August 5, 2010 at 12:00 pm

My knowledge of the quantum aspects of radiation is insufficient to either agree or contend with what Tom said. In that light, this response may be off topic. However, your “glass shield” model still confuses me.
I don’t believe it makes a difference to the discussion, but I’m going to assume that the energy source for the enclosed object is radioactivity so deep within the enclosed object that no radioactive energy directly escapes the enclosed object, but rather is converted into heat within the enclosed object. Furthermore, I define the “system” to be comprised of two “subsystems”: (1) the enclosed object including its radioactive internal energy source, and (2) the glass shield. In this “system”, In is the rate of the energy generated by the enclosed object’s internal radioactivity. As I understand your model, the enclosed object thermally radiates energy at the rate 1.25*In, and 80% of this radiated energy passes through the glass shield. For this model, the “system’s” net rate of energy is zero–i.e., In is being generated by radioactivity and In is being radiated to space. However, although energy-rate equilibrium exists for the “system”, it doesn’t exist for both subsystems. For example, consider the “enclosed object subsystem”. The rate of energy input to this subsystem is In, while the rate of energy leaving this subsystem is 1.25*In. Where does the extra 0.25**In rate of energy come from? It doesn’t come from radioactivity. If it comes from the glass shield, then the energy input to the enclosed object is no long In, and I argue we must now “iterate” because we have changed the rate of energy input to the enclosed object. If it comes from thermal energy contained within the enclosed object, the temperature of the enclosed object will decrease.
Bottom line, I don’t see how you can say that without iteration “everything is self-consistent”. The “enclosed object subsystem’s” net rate of energy is not self-consistent.

195. Spector says:

RE: Caleb: (August 5, 2010 at 6:10 pm) “However I crave to have water vapor included in this sort of discussion.”
As the concentration of gaseous water (water ‘vapor’) in the atmosphere is on the order of 3o,000 to 40,000 ppm, this trace greenhouse gas really *is* the invisible 800 lb gorilla in the room even if we ignore the phase change effects.

196. jae says:

goddard sez:
“I go up to 14,000 feet pretty regularly in the summer, but have yet to encounter temperatures of -40C . Normally they are about 40-55C warmer than at the same elevation in Antarctica, even though Antarctica receives more solar insolation.
Please, provide some actual data to throw around in this food-fight. Are you talking about Long’s Peak or Mt. Shasta in the “mid-latitudes? Where? The average solar radiation at Barrow, AK FOR TWO AXIS FLAT TRACKING COLLECTORS at the peak of summer is 8.3 KWH/m-2. If you look at the “zero-tilt” figures, it’s only 4.9 KWH m-2. It’s over 11 KWH m-2 (more than twice the radiation) for most mid-latitude sites in the northern hemisphere. Here is the data: http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/27502.txt

197. Hi again Tom,
Thanks for this interesting article. I found it enlightening re how different competent people have been saying things that seem inconsistent, but now the question is, have you got it right?
I followed you until, roughly, this point:

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 .

The problem here is that the energy equipartition law is a statistical result, not a precise law such as “Action and reaction are equal and opposite.” An abundance of transfers one way or the other are expected and will happen, all they do is increase the probability that an increase in transfers the other way will restore the equilibrium sooner or later. But talking about a single interaction necessitating an interaction in the opposite direction? No.
Here is the complete interaction sequence, in which I label the three states:
CO2 + γ + N2 (state A) ↔ CO2* + N2 (state B) ↔ CO2 + N2⁺ (state C)
All three states are in equilibrium for some given composition of the components. Photons (state A) will be absorbed giving excited CO2 (state B) and the reverse, and excited CO2 will interact with N2 to give translationally more energetic N2 (state C) and the reverse. Accordingly, you tell us “For example only about 5 % of CO2 molecules are in a vibrationally excited state at room temperatures , 95 % are in the ground state.”
But that assumes the existing proportions of the ‘ingredients’. If more CO2 molecules were pumped into the atmosphere, then it might be, for example, that 6% would be excited and only 94% in the ground state. Similarly, if the photon flux were doubled, that would drive more CO2 into the excited state and in turn drive more energy into the N2 translational mode. The relative balances amongst states A, B, and C depends upon the concentrations of the components, in just the same way as with a normal chemical reaction: change something and the ratios of all the rest change too. For this reason, whilst I thank you for explaining something that has puzzled me, I don’t agree that your conclusion follows.

198. wayne says:
August 5, 2010 at 7:15 pm
Paul Birch
August 5, 2010 at 4:39 pm
Ok, don’t see where you gave where any of the numbers are coming from but it suffices you seem sure the radiation pressure from added GHG molecules (H2O or CO2), let’s say 10 per million is so infinitesimally small that the lift of the atmosphere in every cubic meter per one meter layer up to say 80 km is so insignificant it can be totally ignored…

Radiation Pressure doesn’t depend on the density of the substance, but on the incident power exerted by a photon stream on a given surface. A real life example:
The insolation today was 417 W*s/m^2. The surface absorbed ~191.25 W/m^2 and emitted 181.7 [(N * m/s) /m^2] to the atmosphere. The Prad of the surface photon stream on the atmosphere was 0.575 μPa. On the other hand, the Prad exerted by the power of the incident solar radiation upon the surface was 1.39 μPa, i.e. 2.4 times more intense than the Prad exerted by the surface radiation. Regarding the Prad of the atmosphere, at its current physical conditions, it was 0.221 μPa, that is, ~40 times weaker than the Prad exerted by the surface. According to the second law of thermodynamics, the surface cannot affect to the solar photon stream and the radiation pressure exerted by the atmosphere cannot affect to the surface. The solar Prad affects the surface and the surface Prad affects the atmosphere. The atmosphere cannot affect the surface by any means. It would be as if one tried to inflate a balloon by extracting the air from its interior.

199. anna v says:

wayne says:
August 5, 2010 at 1:04 pm
My point to both of you is a huge point that is being missed here. Huge. Since the LW radiation from the ground is always in the azimuth hemisphere’s direction and ½ of the reemitted photons are in the nadir hemisphere, a radiation pressure exists here which will cause and infinitesimal but real expansion of the atmosphere which will exactly cause a temperature decrease that exactly counteracts any warming from the downward radiation. anna, I stated this in January or February and never could get you to see.
The sun hits the earth, there is radiation pressure, but the forces are so tiny with respect to the bulk of the earth, nobody includes this in the calculation of the orbits.
Some of that momentum bounces back as infrared, part of it interacts and changes energy and momentum of the bulk of the atmosphere ( in complicated manner not yet well understood as Tom says) , but still tiny forces with respect to the bulk atmosphere .
Your error is assuming linearity in the response:
http://en.wikipedia.org/wiki/Statistical_mechanics
You can see in the article that there are no linearities : exponentials in probability functions are involved in the temperature changes so your “will exactly cause a temperature decrease that exactly counteracts ” is not self evident. Care to do the calculations necessary to see how much pressure changes affect temperature changes in this framework, i.e. the photon qm framework? http://star-www.st-and.ac.uk/~mmj/Fluids_notes/fluids_05.pdf
Actually, I had used radiation pressure as an argument that finally all radiation reflected has to vanish into space taking its momentum with it, otherwise from momentum conservation H2O and CO2 should end up in the stratosphere because their numbers are limited, but radiation is continuous and unlimited, and if they are being continuously hit-by/absorbing photons momentum conservation would push them to the stratosphere.

200. anna v says:

maxwell says:
August 5, 2010 at 6:36 pm

Tom,
as a physicist, you should know that molecular vibrations are not ‘quantified’. They are quantized. That’s why they call it ‘quantum mechanics’.

Play nice.
English is not Tom’s first language or the language he read physics in.

201. Nasif Nahle says:

Just to give an example, the mean free path length of photons in the CO2 at its current density in the atmosphere is 48 meters.

I checked your paper: http://www.biocab.org/Mean_Free_Path.pdf , and I got confused. Your calculations seem to show that the MFP is 4.8cm, not 48m. I must be missing something obvious, but I can’t spot my oversight. Your help will be appreciated.

202. Bulldust says:

Enneagram says:
August 5, 2010 at 6:11 am
*SNIP*
However, sadly, you won´t find a chemist-politician.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
How quickly we do forget:
“She read chemistry at Somerville College, Oxford and later trained as a barrister.”
http://en.wikipedia.org/wiki/Margaret_Thatcher

203. pochas says:

Tom,
You write “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 .”
I would urge you to visit the Wiki page on CO2 lasers wherein it is explained how energy is transferred from Nitrogen to CO2 and finally to Helium, in order to make the thing work.
http://en.wikipedia.org/wiki/Carbon_dioxide_laser

204. stevengoddard says:
August 5, 2010 at 2:13 pm

JAE
The absolute humidity in Phoenix is orders of magnitude higher than Antarctica.
I have been in Phoenix when the dew point was over 70F. Even on a dry day, there is plenty of water vapour.

Ah. If Phoenix could be just a little bit drier it would be as cold as Antarctica.
Got it. And the hottest days in Phoenix must have the highest dew point.

205. Barry Moore says:

I am still a bit puzzled about the CO2 to N2 energy transfer being equal and opposite the CO2 molecule is outnumbered 25 000 to 1 by the other molecules in the atmosphere N2 the largest then O2 and H2O so the chance of the CO2 molecule colliding with a different molecule is 25 000 to 1 the chance of another molecule colliding with a CO2 is 1 in 25 000 but the other molecules will collide with each other thus diluting the energy so I do not think the equal and opposite argument stands.

206. Bill Illis says:

Here is a nice chart showing the effective temperature that is being radiated across the IR Earth spectrum.
http://cimss.ssec.wisc.edu/goes/sndprf/spectra.gif
Essentially, in the CO2 bands, the effective emission temperature is 220K. In other words, the 10km to 20km layer (where CO2 relaxation/emission now has a 50% chance of being emitted directly to space rather than being absorbed by other near-by CO2 molecules or by other atmospheric molecules through collision). It should also reflect the Ozone layer temperatures where CO2 steals energy from Ozone and emits that directly back to space but the numbers are a little off here.
The H20 bands emit at 230K to 270K, or let’s say at the beginning of the troposphere at the top of the thunderstorm and cloud layers.
There is also an Ozone and Methane bands which are similar to H20.
After that, we have the “Atmospheric Windows” where emission is direct to space with no interception in the GHG bands and this emission comes from the hard surface and any other atmospheric molecule. It is, effectively, at the blackbody radiation temperature (and all molecules including N2 and O2 absorb and emit blackbody radiation – this seems to not be understood by many).
What is unusual about the Atmospheric Window emission temperatures is that they are emitting at a higher temperature than they really should be. The Windows are emitting at 296K (versus an average 288K). One can also review the Arctic atmospheric window emission temperatures and find that it is also higher than it really should be – the Arctic windows are emitting at 268K or -5C which is quite a bit higher than the average temperature).
http://i165.photobucket.com/albums/u43/gplracerx/PettyFig8-2.jpg
This means that the climate science community has been focused on the CO2 and GHG bands and forgot that solar energy is also being emitted by blackbody radiation. The fact that it is higher than the surface temperature means that some GHG band long-wave absorptions are being transferred to N2 and O2 and then being partially emitted by blackbody radiation in the atmospheric windows.
This is the kind of issue that the blinders of textbook greenhouse theory puts on those that fully accept every word coming out of pro-AGW scientists. It is not what really happens in the real atmosphere apparently.

207. kuhnkat says:

Tom,
Could you please explain whether nobel gasses emit black body or not??
If they do, could you also speak to whether the IR emitted from the molecular bond is the same as from the blackbody??
I have wondered whether the molecular bond radiation is being conflated with black body radiation which actually shows the TEMPERATURE of particles. This would cause a number of inccorrect assumptions as to where and what is emitting.
That is 15 micron band is also the band of a particular TEMPERATURE emission. How would someone with an IR detector tell whether CO2 was emitting or something at that temperature was emitting? All CO2 can’t be at that same temperature right??

208. kuhnkat says:

Pochas:
“I would urge you to visit the Wiki page on CO2 lasers wherein it is explained how energy is transferred from Nitrogen to CO2 and finally to Helium, in order to make the thing work.”
I would urge you not to conflate an environment where extremes of energy are being pumped with an environment of low temps and energy levels. The stuff happens differnt!!

209. Ron House says:
August 5, 2010 at 9:20 pm
Nasif Nahle says:
Just to give an example, the mean free path length of photons in the CO2 at its current density in the atmosphere is 48 meters.
I checked your paper: http://www.biocab.org/Mean_Free_Path.pdf , and I got confused. Your calculations seem to show that the MFP is 4.8cm, not 48m. I must be missing something obvious, but I can’t spot my oversight. Your help will be appreciated.

You’re correct, Ron. Thanks a lot! I made a mistake when writting the article. I apologize. Here the reviewed version:
http://www.biocab.org/Mean_Free_Path.pdf
Sorry and thanks again.

210. Ric Werme,
Antarctica is not a little drier. It is a lot drier. The first few ppm account for most of the greenhouse effect.
Antarctica receives more solar energy in midsummer than Phoenix does. Instead of being knee-jerk sarcastic, think about the implications.

211. Spector says:

[Accidently pre-posted in “Why the CO2 increase is man made…”]
I note in this discussion a reference to returning to a ‘ground state’ analogous to what an excited electron in an atom does. At the atomic level, the emission spectrum just corresponds to an allowed set of energy states. If this were the case with molecular vibration, I would think we would only see simple sharp spectral lines
However, when I look at the HITRAN data available online I see a broad array of many sharp spikes. This gives me the impression that molecular vibrations may be in a transition region between the quantum and the continuous worlds. I suspect that each band represents a vibration mode and perhaps each narrow spike represents one of the allowed modal vibration energy states. Imagine a bell that changes tone in steps as the vibration dies out.
As these molecules are continuously striking one another, each molecule should always be quivering in some agitated state. Perhaps any given photon emission or absorption just corresponds to a single quantum change in an allowed set of multi-level modal vibration energy states — food for thought.

212. @Barry Moore…
The number of moles, the mass and the number of molecules per gram are the properties that have a major influence on the mean free path length of photons through the gases comprising the atmosphere.

213. Nylo says:

Reed Coray,
Yes you are right, the CO2 will only temporarily slow down the cooling of the Earth, and later on the rate of cooling will again rise to match the energy input. But in the mean time, the Earth’s temperature rises. The Earth with more CO2 needs a higher temperature to equal the rate of cooling that it had with less CO2. However it is still wrong to say that CO2 warms the Earth. The only warming source here is the sun. CO2 only makes the cooling more difficult.

214. Bryan says:

This paper should be of great interest for discussion on this topic.
It 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.
There are some people who want to exaggerate the effect of CO2 and they will not find anything in this result to support them.
http://www.webpages.uidaho.edu/~vonw/pubs/TownEtAl_2005.pdf

215. K~Bob says:

stevengoddard wrote:
“Antarctica receives more solar energy in midsummer than Phoenix does.”
Phoenix is noticeably smaller than Antarctica, but is that what you mean?
Or do you mean that a square foot of ground in McMurdo receives more energy in the Antarctic midsummer than does a square foot of ground in Phoenix during an Arizona midsummer? Phoenix is only about five degrees or so North of the Tropic of Cancer.
Why doesn’t it’s more-nearly perpendicular exposure to the axis of radiation at midsummer provide Phoenix more energy per square unit than someplace further from the “tropic” parallels?
Or did I miss something?
[reply] The number of daylight hours maybe? RT-mod

216. Edward Bancroft says:

Enneagram says:
August 5, 2010 at 6:11 am
*SNIP*
However, sadly, you won´t find a chemist-politician.
As has been noted, Margaret Thatcher was such a chemist-politician.
The UK MP Graham Stringer PhD was an analytical chemist. Significantly, he is one of the few UK MP’s to publicly question AGW.
Ed

217. cal says:

Bill Illis says
Here is a nice chart showing the effective temperature that is being radiated across the IR Earth spectrum.
http://cimss.ssec.wisc.edu/goes/sndprf/spectra.gif
Essentially, in the CO2 bands, the effective emission temperature is 220K. In other words, the 10km to 20km layer (where CO2 relaxation/emission now has a 50% chance of being emitted directly to space rather than being absorbed by other near-by CO2 molecules or by other atmospheric molecules through collision). It should also reflect the Ozone layer temperatures where CO2 steals energy from Ozone and emits that directly back to space but the numbers are a little off here.
The H20 bands emit at 230K to 270K, or let’s say at the beginning of the troposphere at the top of the thunderstorm and cloud layers.
There is also an Ozone and Methane bands which are similar to H20.
After that, we have the “Atmospheric Windows” where emission is direct to space with no interception in the GHG bands and this emission comes from the hard surface and any other atmospheric molecule. It is, effectively, at the blackbody radiation temperature (and all molecules including N2 and O2 absorb and emit blackbody radiation – this seems to not be understood by many).
What is unusual about the Atmospheric Window emission temperatures is that they are emitting at a higher temperature than they really should be. The Windows are emitting at 296K (versus an average 288K). One can also review the Arctic atmospheric window emission temperatures and find that it is also higher than it really should be – the Arctic windows are emitting at 268K or -5C which is quite a bit higher than the average temperature).
Thanks Bill. I love this site, someone always comes up with the data I need!
These figures confirm the points I was making in my first post. Just to clarify a point for those why have not seen such a plot before. This effectively shows at what height in the atmosphere photons of various wavelengths are emitted. It does not show the energy radiated at each wavelenth. As can be seen the circa 15 micron photons characteristic of CO2 are emitted near the tropopause where the lowest temperatures in the atmosphere occur. Without repeating all of my previous post I would just ask how the radiation in this band can be reduced any further? To answer my own question it can only do so by reducing the temperature of the tropopause or by losing more energy to other molecules in the atmosphere at lower levels. Both are possible but I would want to see evidence before I would accept it was happening.
I have, without successs, tried to get the Hadley centre to relook at their explanation of the greenhouse effect. They say that increasing CO2 increases the height at which effective radiation takes place and this increased height means lower temperatures and therefore reduced radiation to space for which a warmer surface has to compensate to maintain the radiation balance. Sounds fair enough but then they go on to say that the current height is about 8km where the temperature is around 250K such that there is still room to cool further. When I showed them similar satellite data giving temperatures much lower than theirs and pretty well at the minimum already they stopped talking to me!

218. Ok here is my theory-of-everything, which seems to combine two different views: during the night, IR back radiation keeps the night surface a bit warmer, but this surface warmth must be trapped by bulk atmosphere, which is in physical contact with the surface. So the night air gets heated by convection or conduction from the warmer surface. Then this warmed bulk atmosphere warms our bodies. Otherwise like on Mars, you get plenty of IR flowing down and up, but in such a thin atmosphere it has not much effect.
During the day, direct sunlight does the surface warming job more than enough, as we can see in deserts with low humidity. In this case, IR active gases are rather responsible for cooling the surface by clouds and evaporation.
So the “atmospheric effect” is a result of both presence of IR active gases and bulk atmosphere. The ratio in which they cooperate on the final effect – relatively narrow range of survivable temperatures – is another question; it is for sure, that the most abundant IR active gas likes to create clouds, which effectively cool the surface, then it rains down and again effectively cools the surface and sometimes gets frozen to snow, which reflects sunlight and again effectively cools the surface. My bet is, that atmosphere with water cools the day and warms the night by different physical mechanisms at the same time.
Ouch, I almost forgot those 0.004% of CO2, but compared with hundreds times more water vapor, clouds and ice reflecting 30% of direct sunlight, it keeps me rather indifferent.

219. Julian Braggins says:

I would have thought that though interesting in itself this discussion is rather irrelevant in the greater scheme of climate change, ever since Ferenc Miscolski published his paper showing with the aid of 60 yrs of data and the resources of NASA behind him, that the transparency of the atmosphere to IR has not changed in those 60 years.
If heat can escape as easily through all that time, then surely the source of any increased heat on Earth must be Solar, whether it be visible light, IR or UV, and indirect effects thereof ?

220. Julian Braggins says:

Sorry, Miscolski did not limit the transparency to IR.

221. Patrick Davis says:

Strewth! It’s taken a while to read all the replies, but WOW! What a read.
Thanks again Anthony for posting the article.

222. TomVonk says:

Nick Stokes
The question that precedes that is, what do you mean by rate? How do you quantify it? What number are you talking about?
A rate is dN/dt
Merrick
You also seem to have left out rotations. Any collisions between relatively translationally excited N2 (or any other molecule) and CO2 are far more likely to transfer energy into excited rotational states and have exactly no effect on the vibrational state of CO2.
Yes I did and said so . As long as I the energy equipartition law is respected and the rotationnal quantum levels obey Maxwell Boltzmann statitics like vibrational do the argument stays unchanged . The focus of this post was the process (1) and the length of the post was limited . I could have added R/V and R/T processes for a much longer post . What you wrote is of course true .
Jan
As I don’t understand this post, am I supposed to take it to prove that climate change doesn’t exist, as any argument will suit this purpose?
No . I focus on N2-CO2 collisons and the question of where there can be or not be a net energy transfer .
Randy
Thanks for your essay. I think I understand “most” of what you were saying. Could you please comment on the ideas presented in this video and explain where they violate the physics.
http://earthguide.ucsd.edu/earthguide/diagrams/greenhouse/
Thanks, I would appreciate it.

There is nothing specially wrong with this diagram with the exception of the hand waving at the end . The only wrong phrase is “the absorbed energy is remitted towards the surface” which should read “PART of the absorbed energy is reemitted towards the surface .”
Bernie Anderson
Both of these would require absorption of IR by the CO2 and that energy has to go somewhere.
This is the process I called (1) . The absorbed energy is reemitted .
Jan K.Andersen
The fact that CO2 absorbs infrared energy and heat the atmosphere is no theory,, it is a fact.
CO2 absorbs (and emits) IR . That is a fact . It does not heat the atmosphere by transferring energy to N2 by collisions . This is another fact .
Vince Causey
But later you wrote: “As we have seen above that this relaxation cannot happen through collisions with N2 because no net energy transfer is permitted , only the process (1) is available .”
But doesn’t equation (2) say that CO2* collides with an N2 molecule and relaxes?

Yes you are right . The first phrase should have been “As we have seen above that no variation in the distribution of the CO2 vibration quantum states can happen through collisions with N2 because no net energy transfer is permitted , only the process (1) is available . Sorry for that .
Steven Goddard
There is no question that greenhouse gas H2O warms the atmosphere. Why would CO2 be different?
Unless you mean by that “H2O may heat the atmosphere by latent heat exchange” it is wrong . H2O may not heat the N2 by collisions by the same arguments exposed here .
Rich
In other words CO2 which absorbs strongly the 15µ IR , will emit strongly almost exactly as much 15 µ radiation as it absorbs .
Why isn’t it “exactly”?

Because I have taken here a simplified view where only 15µ is absorbed . In reality because of line broadening there will be many additional lines very near to 15 µ .
Ric Werme
If we take a handful of air with a 15µ IR flux passing through, is it really in LTE?
Yes . If a volume is not in LTE you cannot define a temperature .
Nylo
But you have not demonstrated that:
3) The percentage of molecules in excited state remains the same if you change the initial conditions regarding the percentage of CO2 in the gas.
And if the percentage of molecules in excited state changes, the gas changes its temperature. Perhaps the fraction of time when this happens cannot be called LTE conditions, I don’t know. But it happens anyway.

No I have not demonstrated that , Boltzmann did . It goes the other way round – for a given temperature this percentage is a constant in LTE . It will change if the temperature changes but if the temperature stays constant , the percentage is independent of the concentration of CO2 .
Jean Parisot
One question: you mentioned the non-LTE areas of earth and space boundaries; would the dynamic mechanism of thunderstorms also fall into a non-LTE area and be a factor in some of the emerging work on storm related energy transfers?
There is no special role of LTE in thunderstorms . LTE is just a very important condition to be able to talk about well defined temperatures .
Edvin
I believe this would only hold true if there was no net production or loss of either type of the two molecules CO2* and N2(+).
Yes exactly . There can be no net production or loss of CO2* because as mentioned , the distribution of the quantum states is constant for a given temperature . The same is true for N2(+) because the distribution of velocities is also constant (Maxwell Boltzmann) .
Pamela Gray
I would suggest the author edit some of the sentence structures and grammar before posting on such a hot topic.
Even if english is not my mother language , I generally make little to no grammar errors . As for the sentence structure it is hard for me to judge . Feel free to suggest better sentence structures , I would be happy to have a sentence structure review . Sorry if I can’t do better .
Ed Fix
The equipartition law, like all such empirical laws, is a statistical approximation that becomes less and less accurate as the population of gas molecules becomes smaller. Yet this article treats it as if it must be obeyed exactly by any random pair of molecules. Also, this law can be temporarily violated by transient responses.
That is partly right and partly wrong . I have already written that it is a statistical law valid in certain conditions (density and velocity) etc . However I have supposed nowhere that it should be obeyed exactly by any random pair of molecules . It actually has not to .
Wolf Walker
The opening paragraphs of this article give me the same tingling-down-the-back-of the-neck feeling as those creationist arguments about thermodynamics. It doesn’t feel right.
Sorry to take you as proxy for several posts of this kind . As physicist I have no satisfactory answer to tinglings in the back and feelings . I guess that’s how you feel but it will be hard for you to contribute something concrete if all you have are just feelings .
Paul Birch
The whole of this unfortunately fallacious argument is based on a false premise: that we have LTE. We don’t.
We do . Really . Look up about every textbook . When we don’t , you must do a full quantum mechanical treatment . But in the troposphere we do .
Merrick
I think Tom makes the common mistake that chemists see physicists make here, and that is assuming that microreversibility applies on the macroscopic scale.
I am not assuming that . I am saying that the statistics of a process (2) in a neighborhood of a point of a gas in LTE reflect the time symmetry of the process (2) . This is a classical assumption in mechanical statistics too .
V->V reactions occur with significantly higher probablility than T->V reactions.
Yes .
Unfortunately as I live in Europe , most reactions took place while I was sleeping and that makes the task of answering very hard . Thanks for so many reactions which allow me to see how many additional points/questions I should have addressed . But a science blog like WUWT is there for that – constructive discussion . I must stop here now but will try to come back and answer/comment other posts .

223. Kevin Kilty says:
August 5, 2010 at 5:52 pm
Paul Birch says:…Note that within Earth’s troposphere, the gross molecular flow of heat energy through any surface is of order 100MW/m2 (heat content times molecular speed). …
“I don’t follow, here. The net power flowing through a square meter per unit path via advection (is this what you mean by molecular flow?) is specific heat x density x velocity (dot) temperature gradient. I find values of maybe 10 kW per meter squared at a temperature gradient of a degree per hundred meters in a stiff breeze. What is this gross molecular flow of heat?”
It’s the statistical motion of molecules at the molecular scale. The number of molecules passing through unit area per unit time multiplied by the average energy carried per molecule. If we divide the disequilibriating fluxes by this figure we get an order of magnitude indication of how closely we can expect to approximate local equilibriation. The answer for terrestrial conditions is “very close”. But these small departures from LTE and equipartition still matter.

224. Mike Edwards says:

Tom Vonk’s caveat #1 says it all. His exposition does not deal with the real situation of the atmosphere, which is in effect a layer of gas dealing with an infrared energy flux emanating from the Earth’s surface and which passes out into the void of space.
A simple model for this is to consider first 2 flat plates, separated by some distance, with one plate at Earth’s mean surface temperature and the other at absolute zero (OK, real universe is at ~3K, but it won’t change the description by much). Then consider a layer of gas placed halfway between those 2 plates – and the only assumption you have to make is that the gas has some absorption/emission of infrared. The dynamics of the gas as discussed by Tom is not particularly relevant.
The gas layer will absorb some of the radiation coming from the “warm” plate – its temperature will rise until it is emitting as much radiation as it is receiving – but although all the incoming radiation is from the “warm” plate, the emitted radiation goes in all directions, which can be modelled as a flux towards the “cold” plate and one towards the”warm” plate. The warm plate thus experiences a radiation flux towards it that was not there without the gas layer.

225. TomVonk says:

A specific comment on several posts mentioning circularity .
In a sense yes , that is what I did and I think that people who mentioned that have understood the argument .
There is also “circularity” in the whole huge body of mathematics .
The fractions are really the “same” thing like the natural numbers because there is an isomorphism between N and Q .
So when you demonstrate a theorem about fractions it is just “circular” because you keep repeating that 1+1=2 with different words .
This kind of circularity is a trademark of the consistency .
Yet despite “circularity” nobody would contest that there is added value and new insights by defining Q on top of N .
The “circularity” in my post tries to show that if there is LTE , time symmetry and energy equipartition , then not anything goes in collisional processes .

226. wayne says:
August 5, 2010 at 7:15 pm
“Do you mind me asking, how did you come up with the 1 or 1.7 in 10 trillion expansion figure? Very roughly is fine. Was it taking the 390 W/m2 of back radiation in common energy balance charts and calculating from that the upward pressure and therefore the volume expansion?”
Sorry, the “ten trillion” was a slip (I originally wrote the sentence the other way round and referred to ten trillionths). I meant “a tenth of a trillion” (1E11). Radiation pressure equals power per unit area divided by the speed of light. So (300W/m2)/(3E8m/s) =1E-6 N/m2. Atmospheric pressure is 1E5 N/m2. So the ratio of radiation pressure to atmospheric pressure is 1E-11. Adiabatic expansion will reduce the absolute temperature by about a third of that (for dry air with Cp/Cv=1.4 it’s 0.4/1.4=.286); that is, ~300K/3 * 1E-11 = 1E-9 K. For an optical depth of unity, the radiative power in the cavity is increased by a factor of e (2.718). The fourth root of that gives the temperature ratio (since radiant power goes as temperature to the fourth power); that’s a factor of 1.284, or a rise of 0.284 in absolute temperature. Comparing this 0.284 with the earlier 0.286*1E-11, we have neat ratio of just about 1E-11, which is ten trillionths, ie, negligibly small. Because the effect is so tiny this crude calculation is good enough for present purposes. Note that if the optical depth is less than one (as it is for CO2 in Earth’s atmosphere) both effects are reduced more or less in proportion, so the ratio of the two effects is still the same (close enough).

227. cba says:

================
Spector says:
August 5, 2010 at 11:11 pm
I note in this discussion a reference to returning to a ‘ground state’ analogous to what an excited electron in an atom does. At the atomic level, the emission spectrum just corresponds to an allowed set of energy states. If this were the case with molecular vibration, I would think we would only see simple sharp spectral lines
However, when I look at the HITRAN data available online I see a broad array of many sharp spikes. This gives me the impression that molecular vibrations may be in a transition region between the quantum and the continuous worlds. I suspect that each band represents a vibration mode and perhaps each narrow spike represents one of the allowed modal vibration energy states. Imagine a bell that changes tone in steps as the vibration dies out.
======================
That’s because of several factors. First off, The HITRAN database itself consists of single frequency ( or wavelength) line descriptions. The bands like co2 15 um are nothing but a tremendous number of lines close together. To use HITRAN in a spectrum, one must create a line width based upon temperature, pressure, including partial pressures of the molecule of interest and the ‘other molecules present’, and the line data associated with width. For instance, there’s pressure broadening along with wavelength shift due to pressure corrections to be made.
By transisitional between quantum and continuous, I think you’re looking simply at the realm of readily accessible energy states of lower energy levels. With a hydrogen atom, the first transition from ground state gives a hard uV photon energy. If you excite that atom somewhat above that energy, only then can you get any emissions from it in the visible and IR. Burning hydrogen from a leak in a tank or pipe is quite dangerous as it emits no IR or visible light so you don’t see it or feel any warmth of it radiating. To see the pink hydrogen emission nebulae, the gas there is excited by extremely hot uV emitting type O and B stars which raise the hydrogen above the ground state so it can emit visible light.
A bell, like a drum head, has overtones which are not harmonics and those die out at different rates. They’re still quantized so to say and have particular modes of vibration, Chladni patterns. It’s the one dimensional things like strings and pipes that produce the quantized harmonic series of frequencies.

228. TomVonk says:
August 6, 2010 at 4:08 am
Paul Birch says: The whole of this unfortunately fallacious argument is based on a false premise: that we have LTE. We don’t.
“We do . Really . Look up about every textbook . When we don’t , you must do a full quantum mechanical treatment . But in the troposphere we do .”
We don’t. Really. We have an approximation to LTE. We have an approximation to equipartition. But it is the departures from LTE, caused by and causing superimposed radiative fluxes at different temperatures (as well as the disequilibria from convection, condensation, evaporation etc. ) that among other things produce greenhouse warming and drive the weather system. Your caveat 1, “… there are 2 interfaces – the void on one side and the surface of the Earth on the other side . Neither the former nor the latter is in LTE and the arguments we used are not valid…”, applies throughout the entire volume, because the whole of the volume “sees” (receives radiation from and radiates to) both of those non-LTE boundaries.

229. Julio says:

Reed Coray says:
August 5, 2010 at 7:38 pm
Reed, let me try to explain how it works.
Suppose you make money at the rate of 100 dollars a day. Your money bin is full, so you want to get rid of this annoying cash inflow. You consider giving it away to the federal government, but because of some bizarre tax rule they are obligated to give you back 20% of everything you give them. How do you ensure a net zero flow?
Answer: you give them 125 dollars a day, \$25 more than what you make from your other sources. They give you back 20% of 125, which is \$25, and they use the rest cheerfully for their own nefarious purposes. Your net cash flow is now zero.

230. Spector says:

Bill Illis says: (August 5, 2010 at 10:06 pm) “… all molecules including N2 and O2 absorb and emit blackbody radiation – this seems to not be understood by many).”
I believe N2 and O2 are largely transparent and thus have only a few allowed radiation emission and absorption bands just like CO2 except these do not interfere with earthshine or sunshine to any appreciable extent.
I believe the operative statement here is “only black bodies emit true blackbody radiation.” As I recall, this is based on the theory that the structure of the black body surface is so complex that it has a continuous spectrum of oscillatory mechanisms and can thus radiate or absorb at any frequency with equal probability. To a greater or lesser degree, this applies to the surface of any opaque object. Think of a black body as a potential ‘white noise’ optical radiator if it were white hot.
Also, I believe, ionized gases or plasma can become sufficiently complex and interactive to radiate and absorb over a broad range of frequencies as well.
Transmission of blackbody radiation through transparent materials is another issue.

231. TomVonk says:

Ok I have read now the remaining posts (while all of you sleep :)) .
Trying a summary :
1)
Posts confusing thermodynamic equilibrium and local thermodynamic equilibrium .
This is a common trap and that’s why I spent some time to define what LTE is and what it is not . A volume in LTE has not a constant temperature . For instance the troposphere has a variable temperature yet it is in LTE . So saying or implying that I have somehow supposed thermodynamic equilibrium of the whole atmosphere is not correct .
2)
Posts establishing relations between radiative transfer or green house effect and my post . There are none . My purpose is only and specifically to examine the question whether there can be a net energy transfer between CO2 and N2 by collisions supposing that we have LTE , time symmetry and energy equipartition . In other words whether CO2 can heat N2 . The answer is no .
3)
Posts that deny explicitely or implicitely that CO2 emits IR at frequencies corresponding to vibrationally excited states .
This is in contradiction with experience which shows that CO2 radiates strongly at those frequencies .
4)
What I miss and what I hoped because that would give , I think , the most interesting discussion would be an argument denying my conclusion AND showing how an existence of a net energy transfer by collision is consistent with the time symmetry of the process within a small volume in LTE or even in TE for that matter because such an argument if it existed would have a very general validity .

232. jae says:

stevengoddard says:
August 5, 2010 at 10:57 pm
“Ric Werme,
Antarctica receives more solar energy in midsummer than Phoenix does. Instead of being knee-jerk sarcastic, think about the implications.”
Just where do you get such an idea? I don’t have data for Antartica, but compare Barrow, AK with Phoenix, AZ here: http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/state.html

233. Clyde Rhodes says:

If you make the same assumptions and caveats with regard to a paper towel in a microwave, would you come to the conclusion that water molecules do not heat the paper? From the circularity, I’m still coming to the conclusion that that is what you’d conclude. In reality, a dry paper towel in the microwave is not (substantially) heated, but a wet paper towel is heated by the water which is excited by the microwaves. The presence of water in the paper towel is, in my mind, analogous to the presence of CO2 in N2 in that its presence increases the temperature of the entire system because of the presence of a body which absorbs the radiation. Perhaps I’m too stuck in what seems like common sense to see the theory or perhaps you’re too stuck in theory to do the check of theory vs. common sense. I’m really not sure which because you sound like you know what you’re talking about.

234. Bob Kutz says:

Merrick says:
August 5, 2010 at 11:49 am
Thanks for your very patient response. I had confused more radiation being absorbed due to the molecular nature of CO2 vs. N2 with just simply adding heat to the system, which didn’t seem fair to the argument.
On a separate but closely related point; I have heard the argument put forth that CO2 absorption of IR in the bandwidths that apply is 100% in the first several or 10’s of meters above the surface of the earth, so additional CO2 will not have any meaningful effect on surface T. Now, understanding that we’re really dealing with an equilibrium, so 100% absorption on one level only translates to 100% emission of that energy (in whatever bandwidth is applicable), does that argument (100% IR absorption is already present, so no additional warming can be achieved through higher concentrations of CO2) hold any water? I thought I’d ask since you did such a great job of replying to my prior questions.
Thanks again.

235. Spector says:

RE: TomVonk: (August 6, 2010 at 5:57 am) “My purpose is only and specifically to examine the question whether there can be a net energy transfer between CO2 and N2 by collisions supposing that we have LTE , time symmetry and energy equipartition . In other words whether CO2 can heat N2 . The answer is no.”
I believe, in local thermal equilibrium we must assume, by definition, that there can be no net transfer of energy between any of the components of the atmosphere. But as soon as we let energy exit or enter that local region, then we must look to the potential establishment of a new, more global, thermal equilibrium.
If CO2 and H2O molecules now are cooled below the previous equilibrium point by having their radiation allowed to escape to outer space, then I believe these molecules must then tend to absorb more energy than yield energy with each interaction with the other components of the atmosphere until that atmosphere as a whole reaches a new thermal equilibrium where the net radiation going out and the net radiation coming in (primarily from the sun and the surrounding atmosphere) is the same.

236. Julio says:

TomVonk says:
August 6, 2010 at 5:57 am
“saying or implying that I have somehow supposed thermodynamic equilibrium of the whole atmosphere is not correct .”
You have ignored all boundary conditions, so your LTE implicitly extends throughout the whole atmosphere.
“Posts establishing relations between radiative transfer or green house effect and my post . There are none . ”
Thanks for acknowledging that your post has nothing to do with the greenhouse effect.
“My purpose is only and specifically to examine the question whether there can be a net energy transfer between CO2 and N2 by collisions supposing that we have LTE , time symmetry and energy equipartition . In other words whether CO2 can heat N2 . The answer is no .”
You didn’t need such a long post to show that two substances at the same temperature will not heat each other up!

237. Richard Sharpe says:

Tom Vonk says:

Posts establishing relations between radiative transfer or green house effect and my post . There are none . My purpose is only and specifically to examine the question whether there can be a net energy transfer between CO2 and N2 by collisions supposing that we have LTE , time symmetry and energy equipartition . In other words whether CO2 can heat N2 . The answer is no .

Tom, it is my understanding that in a CO2 laser, excited nitrogen (N2*?) is used to transfer energy to CO2 to get it to lase. If this is so, why can’t transfers occur in the opposite direction, or have I misunderstood something?

238. Henry@Tom
I am afraid this does not explain what I observe.
Standing in the African sun, as the humidity in tha air increases the direct heat from the sun on my skin becomes less. How do you explain this?
http://wattsupwiththat.com/2010/08/03/new-carbon-dioxide-emissions-model/#comment-447343
How can we measure the radiation from the absorptive wavelenghts of CO2 as it bounced back from the moon to earth unless it had come from earth first? So the path was: sun-earth-moon-earth. How would you explain that?
What I see is happening is this : The photon hits on the molecule, but at its absorbative wavelengths some are absorbed until the molecule is filled. At that stage the molecule becomes like a “mirror” at that wavelength and therefore this light is bounced back.
Because of the random position of the molecule 50% goes back to space and that is why I observe cooling (on my skin) as the humidity in the air increases.
The same argument can be made for the greenhouse effect. In fact this was the definition that I picked up, (but it seems that someone changed it later) :
Quote from Wikipedia (on the interpretation of the greenhouse effect);
“The Earth’s surface and the clouds absorb visible and invisible radiation from the sun and re-emit much of the energy as infrared back to the atmosphere. Certain substances in the atmosphere, chiefly cloud droplets and water vapor, but also carbon dioxide, methane, nitrous oxide, sulfur hexafluoride, and chlorofluorocarbons, absorb this infrared, and re-radiate it in all directions including back to Earth.”
end quote.
My interpretation:
Water and carbon dioxide behave similarly when exposed to infra red radiation. Each molecule excepts a number of photons. Once this transaction is completed the molecule becomes sort of like a little mirror to infra red radiation and the molecules start reflecting the infra red. Because of the random position of the molecules we may assume that at least 50% of the infra red from earth is radiated back to earth. The process then repeats itself.
the point that I have been making is that many substances in the air e.g ozone, carbon dioxide, water vapor, methane, etc.
cause both cooling (by deflecting or re-radiating sunshine) and warming (by re-radiating earthshine) but nobody can tell me what the nett effect is.
We don’t need more stories and theories. We need actual tests and measurements.
At this stage, unless I see some actual measurments in the right SI dimensions, I can not be sure that carbon dioxide is a green house gas,
i.e. that the warming effect is > than the cooling effect….

239. Spector says:

RE: cba: (August 6, 2010 at 5:44 am)“To use HITRAN in a spectrum, one must create a line width based upon temperature, pressure, including partial pressures of the molecule of interest and the ‘other molecules present’, and the line data associated with width.”
In this case, it is the mere existence of the line series that I considered important. Instead of dealing with the allowed energy states of an electron, I visualize we now have energy states associated with multi-proton atoms moving with respect to each other with several degrees of freedom. All the lines around 15µ give me the impression that these may represent allowed quantum energy states of a classical vibration mode.
My expectation is that the continual collisions with other molecules should cause each molecule to be in a continuously excited state and that photon emissions or absorptions only cause the vibration amplitude to jump from one allowed energy step to another.
REF: HITRAN Spectrum Plotting Tool
http://savi.weber.edu/hi_plot/

240. TomVonk says:
August 6, 2010 at 5:57 am
“1)Posts confusing thermodynamic equilibrium and local thermodynamic equilibrium .
This is a common trap and that’s why I spent some time to define what LTE is and what it is not . A volume in LTE has not a constant temperature . For instance the troposphere has a variable temperature yet it is in LTE . So saying or implying that I have somehow supposed thermodynamic equilibrium of the whole atmosphere is not correct .”
You still do not seem to appreciate that if you have radiant flows at different temperatures you do not have LTE. You also do not seem to appreciate that even systems not in LTE have thermodynamic temperatures. In space, for instance, there are billions of temperatures all superimposed (one for every star and nebula that’s visible plus temperatures for neutral gases, excited gases, ionised gases, electrons, cosmic rays, magnetic fields, dust, microwave background, neutrinos, and probably lots of others I haven’t mentioned).
“2) Posts establishing relations between radiative transfer or green house effect and my post . There are none . My purpose is only and specifically to examine the question whether there can be a net energy transfer between CO2 and N2 by collisions supposing that we have LTE , time symmetry and energy equipartition . In other words whether CO2 can heat N2 . The answer is no .”
The title of your article says “CO2 heats the atmosphere – a counter view”. Not “Can CO2 heat N2 under conditions of LTE, time symmetry and energy equipartition?”, to which the obvious answer, by definition, is simply, “no” – at least if by “heat” you mean “cause an elevated temperature” as distinct from “transfer energy, but not necessarily net energy, to)”. You specifically related your article to conditions in the atmosphere and common explanations of the greenhouse effect within that atmosphere. And as such, it’s plain wrong.

241. TomVonk says:

Richard Sharpe
Tom, it is my understanding that in a CO2 laser, excited nitrogen (N2*?) is used to transfer energy to CO2 to get it to lase. If this is so, why can’t transfers occur in the opposite direction, or have I misunderstood something?
This is true . But there is no LTE in a CO2/N2 laser . Basically when you mix a very hot gas with a cool gas you will expect a transient period where equilibrium will be established . It is during this period that CO2 degrees of freedom get excited .
Clyde Rhodes
The presence of water in the paper towel is, in my mind, analogous to the presence of CO2 in N2 in that its presence increases the temperature of the entire system because of the presence of a body which absorbs the radiation.
I can see many differences but no analogy . Gases , liquids and solids do very different things . When a direct approach of a problem is possible one should avoid analogies especially if one is not sure that every process is effectively analogous .
Paul Birch
We don’t. Really. We have an approximation to LTE. We have an approximation to equipartition. But it is the departures from LTE, caused by and causing superimposed radiative fluxes at different temperatures etc
There are 2 methods to answer that .
One is to put a dozen of links showing that this is the working hypothesis especially in all radiative transfer models .
I prefer the second which is to ask you to specify what you mean by approximate LTE
You know , a system not in LTE can only be treated by QM .
A system in LTE can be treated semi clasically .
There is a rather sharp distinction between the 2 states . As I have written LTE is a statement about a neighborhood of a point . It doesn’t say how large it must be , it only says that it must exist . And questions of existence are not “approximate” , they have mostly binary yes or no answers .

242. TomVonk says:

Henry Pool
Water and carbon dioxide behave similarly when exposed to infra red radiation. Each molecule excepts a number of photons. Once this transaction is completed the molecule becomes sort of like a little mirror to infra red radiation and the molecules start reflecting the infra red. Because of the random position of the molecules we may assume that at least 50% of the infra red from earth is radiated back to earth. The process then repeats itself.
This is a non conventional and partly wrong view of what really happens .
You should not focus on individual molecules but on an ensemble with billions of them .
At every instant some of them will absorb IR and increase their vibrational energy . At the same instant others will emit IR and decrease their vibrationnal energy . And others still at the same instant will undergo collisions in which they will or will not change their quantum states . If you followed an individual molecule you would see an apparently chaotic and random behaviour in which it keeps jumping up and down among its quantum states . The order appears only when you make a statistic on a very large number of them .
The reason of 50% is not in the position of the molecules but in the isotropy of the emission .
While there is a non isotropic flow of IR (from ground up) as long as it is not absorbed , it is isotropically reemitted . So in a plane approximation you can say that statistically over a large number of reemissions , half will be up and half will be down .

243. Look over this graph at http://upload.wikimedia.org/wikipedia/commons/7/7c/Atmospheric_Transmission.png
There is an atrocious omission and flawed data there. Have you got it? The vertical axis says: “major” components (of the atmosphere); nevertheless, they omitted Argon. Argon mass fraction in the atmosphere is more than 20 times higher than CO2 mass fraction. Argon’s totabs and totemiss is 128% higher than CO2 totabs and totemiss. The absorption spectrum of Argon is wider than CO2 absorption spectrum. What’s the reason the author omitted such important component of the atmosphere, i.e. Argon, and included CO2, which mass fractions and totabs and totemiss are so insignificant?

244. Clyde Rhodes says:

Thanks to Jeff for finding the hole in this argument and debunking the seemingly illogical conclusion in a more recent blog entry. I guess it was Tom who got tied up in the theoretical intricacies and neglected the check the conclusions against common sense. I’m glad it wasn’t me, but next time it might just as well be me.

245. George E. Smith says:

“”” Chris de Freitas says:
August 5, 2010 at 5:49 pm
The focus is on the gas (CO2) rather than on the radiation budget. The greenhouse gases affect the allwave radiation budget of the Earth’s surface as well as the allwave radiation budget of the atmosphere. For the Earth as a whole and on average, the atmosphere is in radiation deficit, whereas there is a surplus at the surface. Surface-to-atmosphere sensible and latent heat fluxes bring about equilibrium. If there is an increase in CO2, say, the surface-atmosphere budget changes. We get a higher air temperature (i.e. by increased sensible heat flux) and increase in evapotranspiration (i.e. increased latent heat flux). “””
Nice to see your post here Dr de Freitas to give some of your insights. Particularly to remind us of the other transport mechanisms such as Latent Heat.
I’m always aware that evaporative and convective cooling near the surface is contributing to the heating of the atmosphere; but do tend to push it aside, to concentrate on what the purely EM radiative effects are.
Trenberth’s energy budget schematic appears to claim a quite assymmetrical atmospheric radiation distribution; since he gives an outgoing longwave flux of 235 W/m^2 of which 40 W/m^2 is actually a direct path from the surface; not an atmospheric radiation. But he has 324 W/m^2 “back radiation” from the atmosphere to the ground, which is hugely in excess of the 195 W/m^2 outgoing from the atmosphere.
It would seem that any layer of the atmosphere should be emitting isotropically, so that the downward and outgoing amounts are about equal. After that; my back of the envelope thinking is that the outgoing escape path should be favored over the back radiation downward path; because both the Temperature lapse rate and the density gradient should lead to a narrowing of the GHG absorption lines at increased altitude; but broadening of those lines for lower layers. So upper layers should allow more wavelenghts to escape the GHG traps, while the downward radiation has to run the gauntlet of an increasingly widening blockage, leading to a greater probablility of recapture going down than going up.
So I would expect more than 50% of the total atmospheric Radiation to escape, and less than 50% to return to the surface; which is totally at odds with Trenberth’s schematic.
If you have a simple answer to this quandrary, It would be nice to know where the truth lies. Trenberth’s 40 W/m^2 escape via the atmospheric window, also seems too low to me; implying almost complete absorption of the surface emission by the atmosphere whereas spectral plots would seem to allow capture of a much smaller fraction of the total energy.
George

246. George E. Smith says:

“”” TomVonk says:
August 6, 2010 at 9:42 am
Henry Pool
……………………………..
The reason of 50% is not in the position of the molecules but in the isotropy of the emission .
While there is a non isotropic flow of IR (from ground up) as long as it is not absorbed , it is isotropically reemitted . So in a plane approximation you can say that statistically over a large number of reemissions , half will be up and half will be down . “””
Tom, I agree with you that the atmospheric emission from any layer ought to be isotropic leading to a 50-50 up/down split.
However, I have argued elsewhere, that because of both temperature and density gradients, the escape path to space is favored over the return path to the surface; because of re-absorption in subsequent atmospheric layers.
A higher layer is both cooler and less dense, so both the Doppler, and Collision broadening of the GHG absorption lines are reduced fro the higher layer but increased for a lower denser and warmer layer.
So the atmospheric layer below some arbitrary layer is even more (spectrally) absorptive, while the layer above is less absorptive, and so allows more of the spectrum to escape. So I believe the escape route is favored over the return route; since every recapture (in the atmopshere) event leads to another 50-50 split of subsequent emissions; and less returns to the surface than escapes to space. My mathematics has gotten just a bit too long in the tooth to do the necessary computations of this multiple cascade of absorptions and emissions along with computing the Doppler and Collision broadening of the absorption lines to compute the degree of assymmetry; but I do believe there is one.

247. George E. Smith says:

“”” Nasif Nahle says:
August 6, 2010 at 9:46 am
Look over this graph at http://upload.wikimedia.org/wikipedia/commons/7/7c/Atmospheric_Transmission.png
There is an atrocious omission and flawed data there. Have you got it? The vertical axis says: “major” components (of the atmosphere); nevertheless, they omitted Argon. Argon mass fraction in the atmosphere is more than 20 times higher than CO2 mass fraction. Argon’s totabs and totemiss is 128% higher than CO2 totabs and totemiss. The absorption spectrum of Argon is wider than CO2 absorption spectrum. What’s the reason the author omitted such important component of the atmosphere, i.e. Argon, and included CO2, which mass fractions and totabs and totemiss are so insignificant? “””
So Nasif; it would seem you have the rest of us at a disadvantage; in that you appear to have data on the Infra-Red absorption spectrum of Argon. I wasn’t previously aware that Argon, even had an IR absorption spectrum or that any Noble gas could absorb at LWIR wavelengths.
Can you point us to a location where we can find this Argon LWIR spectrum ?

248. George E. Smith says:

“”” cba says:
August 6, 2010 at 5:44 am
================
Spector says:
August 5, 2010 at 11:11 pm
………………………………………………Izapped
======================
That’s because of several factors. First off, The HITRAN database itself consists of single frequency ( or wavelength) line descriptions. The bands like co2 15 um are nothing but a tremendous number of lines close together. To use HITRAN in a spectrum, one must create a line width based upon temperature, pressure, including partial pressures of the molecule of interest and the ‘other molecules present’, and the line data associated with width. For instance, there’s pressure broadening along with wavelength shift due to pressure corrections to be made. “””
So –cba– you hit a nerve somewhere or lit up my alarm board somehow.
I have used some other molecular spectra program that Phil graciously pointed me to; but I began to wonder just what they were calculating the spectrum of.
Obviously the IR or any spectrum of say CO2 or H2O in isolation will be way different from what happens in a real atmosphere. Phil has pointed out both CO2 and H2O spectra to me (and anyone else who wanted it) that look as similar as the warts on a wart hog are to the feathers on a bald eagle. Those spectra immediately had me screaming insanely for some information on just what the sample being modelled (computed) were; and how they were related to a real atmospheric sample.
So your comments on what needs to be done pre-HITRAN modelling are most appropriate.
I assume that HITRAN is one of those things that is available to institutionalized folks; but beyond the reach of us peons.
Thanks for bringing up the issue of a real world sample to model.
I also see very little discussion of the result of Doppler and Collision broadening on the up/down split of the atmospheric LWIR emission.
I have argued that escape is favored over return to surface, since higher layers are colder and lower density, and so must exhibit narrower GHG absorption lines; while lower layers must have broader absorption lines; so recapture is more likely going down than going up. Each recapture of course then leads to another 50-50 up/down split of subsequent emissions.

249. Sorry Tom. I don’t follow you.
The paper that confirmed to me that CO2 is (also) cooling the atmosphere by re-radiating sunshine is this one:
http://www.iop.org/EJ/article/0004-637X/644/1/551/64090.web.pdf?request-id=76e1a830-4451-4c80-aa58-4728c1d646ec
they measured this radiation as it bounced back to earth from the moon. So the direction of the radiation was:sun-earth-moon-earth. Follow the green line in fig. 6, bottom. Note that it already starts at 1.2 um, then one peak at 1.4 um, then various peaks at 1.6 um and 3 big peaks at 2 um.
Obviously this also happens as 4-5 um and at other absorbative wavelenghts of CO2 that were not included in this investigation. As to what is happening here: do you not see this as cooling?
And would the opposite – i.e. the trapping of (IR) radiation- not also cause warming?
So which is it?
Your theory does not explain my simple observation that humidity (water vapor) reduces heat – which is why coastal areas are always cooler than more inland (where it is drier)

250. anna v says:

George E. Smith says:
August 6, 2010 at 10:24 am
Hi George,
Are you willing to think along the lines of momentum conservation as far as isotropy claims go?
I say that isotropy exists in the center of mass system of the molecule, BUT, when the molecule absorbed a photon it also got a momentum impulse of h*nu/c and thus it is no longer in the center of mass system. This is true for the “average” molecule absorbing a photon. When it emits a photon, again on the average the real quantum mechanical solution should show the asymmetry that the conservation of momentum imposes on the gas.
Now if all this momentum does not escape as other photons with different energy distributions, the H2O and CO2 etc that are involved in this absorption business would end up in the stratosphere, imo. There is a finite number of molecules but a practically infinite number of infrared photons impinging continuously on them.

251. Reed Coray says:

Julio says:
August 6, 2010 at 5:56 am
Reed Coray says:
August 5, 2010 at 7:38 pm
Reed, let me try to explain how it works.
Suppose you make money at the rate of 100 dollars a day. Your money bin is full, so you want to get rid of this annoying cash inflow. You consider giving it away to the federal government, but because of some bizarre tax rule they are obligated to give you back 20% of everything you give them. How do you ensure a net zero flow?
Answer: you give them 125 dollars a day, \$25 more than what you make from your other sources. They give you back 20% of 125, which is \$25, and they use the rest cheerfully for their own nefarious purposes. Your net cash flow is now zero.

Again, thank you for taking the time to respond to my comments. Continuing the discussion, I believe your federal government tax analogy doesn’t apply to radiation from an object enclosed by a glass shield. Here’s why. In your federal government analogy, the federal government equates to the glass shield. As such, I believe you are saying that the 20% of the energy radiated from the enclosed object that does not pass through the glass is returned to the enclosed object. If any of this “unpassed” energy is not returned to the emitting object, then the emitting object (in your analogy, my “money bin”) loses energy (money) with time. E.g., (a) my employer deposits \$100 dollars per unit time into my “money bin”, (b) from my “money bin” I’m giving to the government \$125 per unit time, and (c) the government is returning to my “money bin” less than \$25 per unit time–say for the sake of argument, \$20 per unit time. Using these numbers my “money bin” is being depleted at a rate of \$5 per unit time.
I believe that only for special geometries and special “reflection conditions” can it be shown that all of the energy not passed by the glass is returned to the emitting object. Specifically, I believe all “un-passed” energy will be returned to the emitting object only for (a) a spherical glass shield, (b) specular reflection off the glass, (c) a spherical emitting object, and (d) the centers of the spherical emitting object and the spherical glass sphere are collocated.
Taking these one at a time.
(a) Spherical glass shield. If the glass shield is not spherical, with the exception of (i) a spherical enclosed emitting object co-centered with a surrounding glass sphere and (ii) specular reflection off the glass shield, I believe a point on the surface of the emitting object can always be found such that radiation emanating from that point that encounters the glass shield will not return to the emitting object; but rather will be directed towards a different part of the surface of the glass shield. Since the glass shield passes 80% of the energy incident on it, 80% of such energy will pass through the glass–never to be returned to the enclosed emitting object. As such, only a fraction of the “20% energy initially not passed by the glass shield” will return to the emitting object.
(b) Specular reflection off the glass shield. If the “20% energy initially not passed by the glass shield” is returned to the interior of the glass shell with directionality other than specular reflection–say either Lambertian (amount proportional to the cosine of the angle between the radiated direction and the normal to the surface of the glass) or isotropically in the interior half-plane (which I believe is the model the AGW community uses), then a portion of the energy emitted from the enclosed object when encountering the glass surface will be directed towards another portion of the glass surface. 80% of such energy will pass through the glass shield never to be returned to the enclosed radiating object.
(c) The enclosed object must be a sphere. If energy is spectrally reflected from a spherical glass shield, I believe the only emitting-object shape for which the reflected energy is guaranteed to strike the emitting object before striking the glass shield is a sphere, and the center of that sphere must be coincident with the center of the spherical glass shield. [Note: I’m almost positive that energy leaving the surface of a spherical object that is specularly reflected from the surface of a co-centered larger spherical object will be returned to the emitting object. I’m not positive other geometries don’t exist with these properties, but I doubt it. In any event, I can find geometries for which energy leaving a non-spherical object reflected from a spherical glass shell will not be directly returned to the emitting object.] For these cases, some of the “20% unpassed energy” will encounter the glass shield before encountering the emitting object, and 80% of that portion of the “20% unpassed energy” will never return to the emitting object.
(d) The center of the spherical emitting object must be coincident with the center of the spherical glass shield. If the two objects are not co-centered (say the enclosed sphere is offset where at one point the two surfaces almost touch, then energy from the “almost touching point” on the enclosed sphere radiated tangent (or nearly tangent) to the surface of the enclosed sphere when specularly reflected by the surface of the larger sphere will be reflected towards the larger sphere and not back towards the emitting sphere. As such, some of this reflected energy will pass through the glass shield never to be returned to the emitting objects.
Bottom line, for your “tax analogy” to apply to energy passing through a glass shield, I believe you must prove that the 20% of the energy that doesn’t initially pass through the glass shield is returned to the emitting object. With the exception of the special case mentioned above (and maybe a few other special cases, but definitely not the general case), I don’t believe this can be done.

252. George E. Smith says:

“”” cal says:
August 6, 2010 at 2:36 am
Bill Illis says
Here is a nice chart showing the effective temperature that is being radiated across the IR Earth spectrum.
http://cimss.ssec.wisc.edu/goes/sndprf/spectra.gif “””
A very interesting graph Bill; but I am having a hard time trying to understand what it really means.
For a start, I would ask, is this a calculated spectrum from some model or other of the atmosphere or is this actual measured earth emission observed from outer space.
I don’t understand your comment that some bands are emitting at a higher temperature than they should be; (296 versus 288). Clearly they are emitting at exactly the temperature they should be.
I have long argued that the earth is doing it’s most efficient cooling from the hottest desert surfaces during the heat of the day; where surface temeprature can get over +60 deg C or 333 K which is even higher than your 296. Because of the sigma. T^4 effect, the hotter surfaces radiate much more proportionately than do the colder surfaces so basing a radiation budget on some fictional global average temeprature of 288 K is clearly not real. And at the other end; at cold places like Vostok, the radiant cooling is more than an order of magnitude lower than from the hottest deserts.
But back to this brightness Temperatures business. What I am hearing, seems to be that it is assumed that ALL of the surface radiation is absorbed in the lowest atmosphere layers EXCEPT Trenberth’s ridiculously small 40 W/m^2 that escpaes immediately via “the amosheric window”. Then that lowest atmosphere layer emit and a 50-50 split sends it half up and half down; and the up ward is again absorbed by a higher and now cooler layer; which in turn emits but now at a lower temperature; until finally some much higher and much cooler layer gets to emit radiation that actually escapes to space and that radiating temperature is the one that must balance with the incoming TSI insolation rate.
And that is where I get hung up. It seems to me that any layer from the surface to the highest limits of the atmosphere is radiating some roughly blackbody looking spectrum corresponding to its own Temperature; and much of that spectrum exits directly to space (assuming cloudless skies for the moment) with a spectrum corresponding to the emission temperature of that surface; but now with holes in it from absorption by GHG molecules or the atmospheric gases themselves.
Each higher and cooler layer in turn emits thermal radiation corresponding to its temperature; and much of that also escapes directly to space around the absorption bands of the higher atmosphere layers; and so on; so that the total LWIR emission from the earth should then be a composite of roughly BB spectra but with source temepratures ranging ove the entire surface Temeprature range, as well as the range of atmospheric emitting Temperatures.
Observations of earth emission spectra seen outside the atmosphere from satellites should co0ntain components that are emitted from surface that are 333 K or even higher; and this is important since the Wien displacement Law, would shift these emission peaks even further away from the CO2 15 micron nand as the spectral peak moves from its nominal 10.1 microns at 288 K down to about 8.7 microns at 333 K.
I’m having a hard time keeping straight what people are citing as actual emissions from the earth apart from what some computer model is calculating that satellites should be able to see (but maybe aren’t).
The earth’s surface SHOULD be emitting at a higher effective Temperature than 288 K because the hotter surfaces far more than make up for the laziness of the colder surfaces.

253. TomVonk says:
August 6, 2010 at 9:26 am
Paul Birch says: We don’t. Really. We have an approximation to LTE. We have an approximation to equipartition. But it is the departures from LTE, caused by and causing superimposed radiative fluxes at different temperatures etc
Tom Vonk says: “There are 2 methods to answer that .
One is to put a dozen of links showing that this is the working hypothesis especially in all radiative transfer models .”
It is a useful approximation for some of the analysis. Not when taken to your absurd extreme.
“I prefer the second which is to ask you to specify what you mean by approximate LTE”
A state in which only a small proportion of molecules at any one time belong to a population with a significantly different temperature to the majority, or in which the temperature difference between populations is only slight.
“You know , a system not in LTE can only be treated by QM .”
Not true. One can for example treat radiative systems using classical absorption, scattering and transmission coefficients. One can consider mean lifetimes of the various populations against energy transfer to other populations. Quantum mechanics may ultimately lie behind the physics, but bulk macroscopic parameters are fine for most practical thermodynamic analysis.
“There is a rather sharp distinction between the 2 states . As I have written LTE is a statement about a neighborhood of a point . It doesn’t say how large it must be , it only says that it must exist . And questions of existence are not “approximate” , they have mostly binary yes or no answers .”
If you really want to get this pedantic, then the answer is that LTE does not exist in the real universe, and if it ever did exist it could not be observed. It is a mathematical fiction. A limiting case never actually achieved.

254. http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/#comment-449799
Sorry.
A theory must describe what we see happening. Otherwise it is no good.
If it were not for the ozone and the water vapor and the CO2 and the oxygen in the air then we would all fry….the extra 30% radiation or so would make us toast….
Unfortunately some of these gases also cause warming…. by re-radiating some earthshine at certain wavelengths. But what is the net effect of the cooling and warming of each the gases in the atmosphere (including the noble gases)???
Where is the research on that? If you don’t have anything on that, then what is the use discussing any of these so-called theories?
Cheers…
Henry

255. Julio says:

Reed Coray says:
August 6, 2010 at 11:49 am
Hi Reed,
Please don’t take this wrong, but I think you are tying yourself up in knots unnecessarily.
To go back to my silly example, it doesn’t matter the exact amount the government gives me back, as long as it is a fixed percentage of my donation. If it only gives me back \$20 a day for every \$125 I donate, that means the rate of return is only 16%, and my “equilibrium donation rate” then has to be 100/(1-0.16) = \$119.05; 16% of this is \$19.05, and I’m in business again (or not, of course, depending on your perspective).
Back to the Earth: it radiates upward, and without an atmosphere all that radiation would just go out into space. With greenhouse gases, some of it comes back to the Earth, some of this goes up again, etc. In steady state, you still have to have that the net amount going out into space equals the amount coming in, or else the Earth + atmosphere system would just keep getting hotter and hotter. So what you end up with is some amount of radiation “trapped”, kind of bouncing back and forth between the upper atmosphere and the earth, like the \$19.05 above: every day I give the government \$119.05, and it gives me back \$19.05; the next day I get a fresh \$100 in dividends from my bank, add the previous day’s \$19.05, and give it to the government; by the evening, the \$19.05 is back, and so on.
Back to the atmosphere: you’re saying that you don’t believe that the back-reflected flux plus the transmitted flux equals the total upward flux. Well, what else is the energy to do? Think of the glass example. The energy could go back; it could go through; or it could–what, continually build up inside the glass until the glass melts? That’s not what happens in real life. Check out any real greenhouse. The glass gets hot, up to a point, and then it reaches its own optimal equilibrium temperature where it radiates out (in both directions) as much as it is getting.
You just have to imagine a steady state in which energy just flows through without building up anywhere, although there may be large, constant pools of it in various places–like a lake in the course of a stream; that would be the equivalent of the “trapped energy.”

256. George E. Smith says:
August 6, 2010 at 10:30 am
“”” Nasif Nahle says:
August 6, 2010 at 9:46 am
Look over this graph at http://upload.wikimedia.org/wikipedia/commons/7/7c/Atmospheric_Transmission.png
There is an atrocious omission and flawed data there. Have you got it? The vertical axis says: “major” components (of the atmosphere); nevertheless, they omitted Argon. Argon mass fraction in the atmosphere is more than 20 times higher than CO2 mass fraction. Argon’s totabs and totemiss is 128% higher than CO2 totabs and totemiss. The absorption spectrum of Argon is wider than CO2 absorption spectrum. What’s the reason the author omitted such important component of the atmosphere, i.e. Argon, and included CO2, which mass fractions and totabs and totemiss are so insignificant? “””
So Nasif; it would seem you have the rest of us at a disadvantage; in that you appear to have data on the Infra-Red absorption spectrum of Argon. I wasn’t previously aware that Argon, even had an IR absorption spectrum or that any Noble gas could absorb at LWIR wavelengths.
Can you point us to a location where we can find this Argon LWIR spectrum ?

There cannot be any data on IR absorption by Argon because Argon is transparent to IR, more transparent than the CO2. Argon is a major component of the atmosphere which is in a higher concentration than CO2.

257. John Whitman says:

TomVonk,
Thank you for your piece of the puzzle on CO2 in the atmosphere. It was concise and lucid.
The more pieces of the puzzle that are offered and discussed the better we are to be able to recognize the picture/pattern of the total puzzle even before all the pieces of the puzzle are available and fitted in.
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.
Anthony and team, you guys sure know how to give us all great learning episodes : ) Thanks.
John

258. cba says:

Spector,
I’m not totally sure what you’re driving at with ‘quantized classical vibration’ but I did not see anything wrong with your statement. It is the effect of the total spectrum for transmission or absorption with contributions from all lines of all molecules involved.
The HITRAN database includes parameters for each line that include the starting and ending energy states for a line plus information about line width and the effects of ‘other’ molecules upon the particular line. You might want to continue reading below.
George E. Smith,
HITRAN is not a spectrum nor does it generate one as is. You have to generate your own – which is what I had to do. The papers they provide on their website include details on how to accomplish that effort – Rothman et al 1994, appendix. What I did was to create a program that reads an output file from their javahawks program (supplied) that permits selection of molecules (&isotopes) along with temperatures and pressures and frequency ranges (I use wavelength ranges). My program then takes the partial extracted database and corrects for the temperature and pressures and permits further selection of molecules. It then builds a set of transmission parameters in an array of wavelengths for the number of layers being used. A layer has concentrations for molecules, a temperature, and a pressure.
Once the output file is prepared, I import it into excel where I have a series of transmission arrays. The lines have widths and heights and the contribution for a wavelength range for each line is added together. Each array is for a 1cm thickness of the gas for that layer. Actual transmission and absorption is determined for a layer is the exponential of that value times the thickness of the layer in cm. 1-transmission fraction gives the absorption fraction.
depending on what I’m doing, I can do straight attenuation or do radiative transfer where there is both absorption and emission in a layer.
The absorption occurs when a BB like spectrum enters the layer and it is multiplied by the layer’s attenuation. Emission from the layer is the multiplication of the BB spectrum for a black body at the temperature of the layer with the absorption factors. The entire atmosphere is modeled by typically using over 50 layers.
My program system is limited to between about 0.1 um and 75um. The resolution is variable, permitting me to look with high resolution at very narrow ranges or lower resolution at much wider ranges. For validation, I compared my results to some medium and ultrahigh resolution measured spectrums in the visible spectrum (not associated with climatology). It works pretty good for its intended application (which is not climatology).
To get HITRAN, one must request it. As I recall, when I requested access, I was not associated with an institution nor had I been involved with one or with basic scientific research for decades. When I requested it, my purpose was primarily to play with atmospheric studies on my own. Only later did I start to apply my work to any actual (non climatological) research effort. Note- I don’t recommend this approach to anyone else. Going to this resolution doesn’t really offer improved accuracy as one is really dealing with over 50% cloudy skies and variations that are significant.
as for one molecule or all, I use all and subtract out a single molecule for determining the effects. What I’ve used most recently is rather low resolution and has the ability to adjust co2 and h2o levels. There isn’t a lot of actual power absorption overlap but there is several percent difference.

259. George E. Smith says:

“”” Nasif Nahle says:
August 6, 2010 at 2:18 pm
George E. Smith says:
August 6, 2010 at 10:30 am
“”” Nasif Nahle says:
August 6, 2010 at 9:46 am
Look over this graph at http://upload.wikimedia.org/wikipedia/commons/7/7c/Atmospheric_Transmission.png
There is an atrocious omission and flawed data there. Have you got it? The vertical axis says: “major” components (of the atmosphere); nevertheless, they omitted Argon. Argon mass fraction in the atmosphere is more than 20 times higher than CO2 mass fraction. Argon’s totabs and totemiss is 128% higher than CO2 totabs and totemiss. The absorption spectrum of Argon is wider than CO2 absorption spectrum. “””
Well there I have cut and pasted just what you said:- “”” The absorption spectrum of Argon is wider than CO2 absorption spectrum. “””
So is the argon LWIR absorption spectrum wider than that of CO2 or is it non-existent; I’m having a hard time getting your point.
If Argon is a totally raiatively inert component of the atmosphere then it is not unlike a mountain or a building that is also not part of the active atmosphere.
They have also omitted helium and xenon that also are parts of the atmosphere having no effect on climate.

260. cba says:

Tom Vonk,
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.
If there is an increase in the radiation amount (in the spectral area of the ghg molecules), then there will be additional absorption by the CO2 but the emission is temperature dependent. For energy balance, there must be an increase in the CO2 temperature if more power is being absorbed to permit more power to be emitted. During this transient time, the co2 molecules must heat up and become hotter than the n2 and o2. In order for LTE to be restored, the co2 must heat up the n2 and o2 to the new temperature. Once LTE is restored, then one returns to the condition that n2 and o2 no longer receive or give back net power to the co2. Also, the co2 must be in radiative equilibrium, giving off as much radiation energy as it receives.

261. cba says:

Nasif,
argon is a noble gas. It isn’t interested in forming any molecules and so requires high energy uV to excite it to higher levels before it is capable of absorbing and emitting in the visible and infrared. The only way you’re really going to get much IR is to have a molecule where you can have low energy states without having to raise the lower state well above the ground state.

262. sky says:

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.
We know empirically that backradiation has no diurnal cycle. This is a strong indicator that it doesn’t come from GHG emissions alone. The bulk constituents of the atmosphere do heat heat up as result of all the the heat-transfer processes, not just radiation. What the usual explanation of the “greenhouse effect” usually misses, moreover, is that the backradiation is simply part of a nearly null-net exchange between the surface and the base of the atmosphere. With the surface being more than 70% water, which evaporates in response to IR rather than raising its temperature, there is no radiation-only algebra that can determine temperatures. While HITRAN and similar codes do an excellent job of determining the transmitivity of signals through the atmosphere for remote sensing, they do not solve any thermodynamic problem that involves convective transfer of latent heat. Observations of the Bowen ratio show that the latter is the principal mode of heat transfer to the atmosphere. Ths science is not settled–it hasn’t even been done!

263. suricat says:

High five Tom! It’s a long time since we spoke on the, now defunct, CA Forum.
Unfortunately, I’m still unconvinced on the ‘microscopic’ ideology. I still only see this as an ‘insulation’ factor for ‘radiative theory’ and an ‘interpretation’ of GHE (greenhouse effect) WRT altitude.
Best regards, Ray Dart.

264. Reed Coray says:

Julio says:
August 6, 2010 at 1:55 pm
Reed Coray says:
August 6, 2010 at 11:49 am
Hi Reed,
Please don’t take this wrong, but I think you are tying yourself up in knots unnecessarily.

Julio, I don’t take your comment “wrong”; in fact, I kind of like it. I’ve tied myself in knots before and I may be doing so now. I accept the idea that for a system surrounded by a vacuum when radiation-rate-equilibrium is reached, the amount of energy per unit time leaving a system via radiation is equal to the amount of energy per unit time entering the system. In fact, this is the definition of radiation-rate-equilibrium. For a “system” to be in radiation-rate-equilibrium, I believe all subsystems of that system must also be in energy-rate-equilibrium–not radiation-rate-equilibrium because the subsystems may exchange energy via convection and conduction.
I also believe a body surrounding a second body that possesses an internal energy source will affect the temperature of the second body. For example, (a) Assume a black body exists internal to which is an energy source supplying a fixed amount of energy per unit time. (b) Assume the temperature of the surface of that black body is everywhere the same, T Kelvins. (c) In radiation-rate-equilibrium with a vacuum at 0 Kelvins, the temperature of the black body surface will be such that A*sigma*T^4 is equal to the rate of internal energy supplied to the black body (where A is the black body radiator’s surface area and sigma is the Stefan-Boltzmann constant). (d) If you surround this black body with another black (or grey) body which has no internal energy source, the presence of the surrounding black/grey body will likely alter the surface temperature of the emitting black body. So, I agree that the temperature of the emitting black body can be affected by a surrounding body.
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. Any physical material with a non-zero specific heat capacity “traps heat”. “Trapping heat” is not limited to “greenhouse materials.” In fact, adding material that “traps heat” may even lead to cooling. Before you say, “now he’s off his rocker”, consider the following. A black body, spherical, Earth has an internal energy source deep within its core that is supplying energy at a fixed rate, H. The radius of Earth is R. In radiation-rate-equilibrium with a vacuum at 0 Kelvins, the temperature, T of the Earth’s surface is given by the equation:
T = { H / (4*pi*R*R*sigma) }^(1/4)
where sigma is the Stefan-Boltzmann constant. If we increase the radius of the Earth to R0 by adding “black body material” that is devoid of an internal energy source, the added material will “trap heat”, but the temperature of the Earth surface will decrease. The latter happens because the formula for the Earth surface temperature has the same form as above, but the radius R is replaced with a larger radius, R0. Thus, we are “trapping heat” and the temperature of the body decreases.
In addition, isn’t it true that the CO2 surrounding the Earth will “trap heat” independent of the temperature of the surface of the Earth? If so and if “trapping heat” causes the temperature to rise, then the “trapping of heat” doesn’t stop when the black body’s temperature rises, and the black body’s temperature will continue to rise until the CO2 reaches a temperature where its molecular motion has sufficient speed to escape the gravitational pull of the Earth.
I believe a better way to state what is happening is the following. If a black body with a fixed-rate energy source is in radiation-rate-equilibrium with the vacuum of space at 0 Kelvins, placing additional material separate from but surrounding the black body will likely cause the temperature of the surface of the black body to change in such a way that energy-rate-equilibrium is re-established for the black body. Only now heat can leave the black body via conduction and convection as well as radiation, so I don’t want to say the black body reaches radiation-rate-equilibrium, rather it reaches energy-rate-equilibrium. Furthermore, once energy-rate-equilibrium is established, it holds for any subsystem. That is, it holds for the CO2 as a whole, any “block of atmosphere”, etc. In particular, I believe the temperature of the surface of the Earth would be altered by a pure Nitrogen or Argon or Helium or any atmosphere. Although greenhouse gases may affect the temperature of the surface of the Earth differently than non-greenhouse gases, I don’t agree with the claim that it’s the presence of greenhouse gases that causes the Earth’s temperature to be on the order of 33 Kelvins than it would be in the absence of greenhouse gases. (I even question the number 33, but that’s another story.) 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.
I’d appreciate it if you’d answer a few 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”?
(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?
(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?”
(4) If there is an Earth surface temperature at which CO2 ceases to “trap heat”, what roughly is that temperature?
You can see where I’m going with this [I’m tying bigger and more complicated knots :)] If the premises that (a) “CO2 trapping heat” causes the temperature to rise, and (b) CO2 traps heat independent of the temperature of the Earth’s surface are valid, then a CO2 atmosphere should cause the Earth surface temperature to rise without bound.
About five months ago I got interested in developing from first principles equations for radiative energy flow between two “grey bodies” in a vacuum where one grey-body completely surrounds the other grey body. For most geometries, the mathematics is too unwieldy (at least for me) to generate a closed-form equation for the rate of energy transfer. However, for the geometry of two spherical, co-centered, grey-body shells, the mathematics is tractable. I wrote a paper providing all the mathematics. If you’re interested, I’d be happy to send you the paper (microsoft word 2007) for your review and comment. I don’t know your E-mail address, but I’m sure we can get Anthony to either send you my E-mail address or have him send me your E-mail address.
Final note. I’ve enjoyed our exchange of comments. Thank you for your time.

265. Icba says:
August 6, 2010 at 5:26 pm
Nasif,
argon is a noble gas. It isn’t interested in forming any molecules and so requires high energy uV to excite it to higher levels before it is capable of absorbing and emitting in the visible and infrared. The only way you’re really going to get much IR is to have a molecule where you can have low energy states without having to raise the lower state well above the ground state.

Please, read my post at Nasif Nahle. August 6, 2010 at 2:18 pm

266. stevengoddard says:
August 5, 2010 at 10:57 pm
Ric Werme,

Antarctica is not a little drier. It is a lot drier. The first few ppm account for most of the greenhouse effect.
Antarctica receives more solar energy in midsummer than Phoenix does. Instead of being knee-jerk sarcastic, think about the implications.

Sorry Steven, you make it so easy to be knee jerk sarcastic. It would help if you were truly interested in sharing some stuff that you don’t. BTW, I did think about length of day – the areas just poleward of the arctic circles get a bit more daylight thanks to refraction of the atmosphere, though that’s not germane.
So, implications of slightly more insolation for a couple weeks on either side of midsummer at the SP and PHX:
1) Peak insolation at the SP is sin(23.44)N where N is solar flux in w/m^2 on a surface perpendicular to the Sun – about 40%, but good for all day. At PHX, latitude 33.43, at local noon the Sun will be at altitude 90 – 33.43 + 23.44 = 80°, sin(80.01) is 0.985, more than twice as much, but falling off quickly more than a few hours before and after.
Implication 1) Phoenix’s temperature likely has a bigger diurnal variation than the south pole.
2) The SP has greater insolation for only a few weeks. The seasonal lag in New Hampshire between peak insolation and peak daily temperatures is about five weeks.
Implication 2) The many weeks where the SP has less insolation than PHX has a significant effect on temperature differences between the two places.
3) The altitude of the SP is 2,835 m. The altitude of PHX is 340 m. That’s good for some 23°C in adiabatic lapse rate. Possibly more given the dense polar air.
4) The albedo at the SP is much higher than at PHX. I’m sorry, I can’t estimate the temperature differential that’s good for.
5) CO2 concentration in ppm should be about the same since rain doesn’t take much CO2 out of the air on its way to the SP, but H2O will be quite a bit less, thanks to the low dew point.
Implication 3+5) the air above PHX has quite a bit more CO2 and H2O thanks to the extra 777 m of air LWIR has to navigate.
So, given all that, I’m disappointed that you ascribed the temperature difference between the two sites as solely due to humidity without giving the merest mention to all the other variables.

267. Oops – stupid math error (actually, number error)
Implication 3+5) the air above PHX has quite a bit more CO2 and H2O thanks to the extra 2500 m of air LWIR has to navigate.

268. Jim D says:

On Antarctic receiving more solar radiation than Phoenix in mid-summer.
The math is a little difficult for Phoenix, but it can be shown it receives more than
the equator at the equinox (when the sun goes overhead there), so I would say it is plausible for Phoenix.
In Antarctic mid-summer the sun stays at 23 deg above the horizon for 24 hours. sine 23 deg = 0.4. Average daily flux = 0.4 times the overhead value.
At the equator equinox the sun goes up over ahead and down in 12 hours. 24-hour average value of sine (positive part only) is 1/pi. Average daily flux is 1/pi= 0.32 times the overhead flux.
So, strange but true. In a clear sky, the mean daily flux reaching a horizontal surface is greater for Antarctica than the equator at equinox.

269. Spector says:

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.

270. anna v says:

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?

271. Jim D says:
August 6, 2010 at 7:13 pm

On Antarctic receiving more solar radiation than Phoenix in mid-summer.
The math is a little difficult for Phoenix, but it can be shown it receives more than
the equator at the equinox (when the sun goes overhead there), so I would say it is plausible for Phoenix.

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. 🙂

272. K~Bob says:

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.

273. TomVonk says:

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 .

274. TomVonk says:

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 .

275. 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.

276. cba says:

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.
.
.

277. cba says:

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.

278. anna v says:

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.

279. Julio says:

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.)
“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…

280. Roger Clague says:

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

281. 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.

282. Spector says:

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?

283. Bryan says:

cba says:
…..”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

284. John Whitman says:

TomVonk says:
August 7, 2010 at 3:06 am

John Whitman
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 .
—————–
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

285. Spector says:

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.

286. Julio says:

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.

287. R Stevenson says:

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.

288. 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.

289. Kevin Kilty says:

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.

290. Kevin Kilty says:

Roger Clague says:
August 7, 2010 at 6:28 am
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

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.

291. Reed Coray says:

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?

292. 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.

293. Reed Coray says:

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?”

294. Julio says:

Reed Coray says:
August 7, 2010 at 11:57 am
Hi Reed,
(1) “Here’s what I think you mean when you use the term “trapping” as applied to greenhouse gases. Greenhouse gases “trap” radiative energy because they absorbed IR radiation from the Earth’s surface which then continually “bounces up and down”. Is that correct?”
Yes, that is what I mean.
(2) Nice example with the cube and the sphere. Clearly, I spoke too soon, but I should say in my defense that I have trouble visualizing your constant internal source of heat, physically. Blackbodies are traditionally heated from the outside, not the inside (although the sun is a good counterexample).
Anyway, you add a layer of material to the surface of your sphere, and then…? You have more material to heat from the same constant internal source. At first sight it would certainly seem that you are spreading the energy thinner. Of course, you could in principle postulate any kind of internal energy distribution that you wanted. Technically, only the surface needs to be a blackbody… It does not look like a well-posed problem to me.
Regarding the real world earth and the real world atmosphere–granted that the temperature measures different things for different objects, but in all cases there is a proportionality constant between the (change in) thermal energy content of an object and its (change in) temperature, and that is the heat capacity. So when you bring objects at different temperatures into contact, how much energy flows from one to the other depends on the heat capacities. The atmosphere has a very small heat capacity compared to the Earth, so it can only absorb a small fraction of the Earth’s thermal energy before reaching thermal equilibrium. That’s the main point I was trying to make.
(3) “Will trapping of heat by an object cause a temperature rise in the object?” If by heat you mean thermal energy, yes, of course, as long as the object has a positive heat capacity. If you put thermal energy into an object, its temperature will go up. What else could it do?
Is infrared radiation “heat”? Well, it is energy. Does it count as thermal energy or not? It depends on the context. Is the infrared radiation “trapped” in the Earth’s atmosphere part of the whole planet’s (Earth + atmosphere) “thermal energy”? Probably yes, depending on how you choose to keep your books. It’s hard to imagine what else it could count as. Does it keep the Earth’s surface warmer than it would be it it wasn’t there? Definitely yes.

295. sky says:

TomVonk says:
August 7, 2010 at 3:29 am
I’m not disparaging narrowly-based advances in physical understanding. As an applied scientist, I do look for their utility in the real-world setting. Sometimes, looking from the oppsite viewpoint of the particle-wave duality provides additional insights. I’m not sure that your formulation of energy transfer from CO2 to N2 molecules helps explain the extinction of 15 micron terrestrail radiation within ~100m of the surface. Please enlighten me.

296. Kevin Kilty says:
August 7, 2010 at 11:55 am
“The only temperature profile I have seen for Venus looks pretty linear in the region of measurement, and the linear extrapolation to the surface comes quite close to the surface temperature. I wonder though if the adiabatic lapse rate on Venus shouldn’t be a bit steeper at high elevation (below 300K) than near the surface simply because the specific heat of Co2 is larger at high temperature. I’d guess a 20% change, but I’m unsure of the effect of gasses like So2.”
I’m not sure if it’s significant, but the lapse rate going through the cloud layers ought to be some sort of “wet” rate for the sulphuric acid. Also, the lapse rate doesn’t depend on the specific heat but on the ratio of specific heats Cp/Cv; perhaps this is what you meant. Higher up, the composition of the atmosphere changes too – less CO2 and more diatomic species like N2, which have a higher Cp/Cv and therefore higher lapse rates.

297. cba says:

Spector says:
August 7, 2010 at 7:52 am
RE: cba: (August 7, 2010 at 4:00 am) “Spector, rather than 100/1, I’d suggest it’s closer to 2/1, excluding those that are simply not going very far. That’s based on there being a factor of two in absorption, roughly co2 total contribution 30W/m^2 versus h2o at 64w/m^2 in the approximately current clear sky atmosphere calculations.”
Is this, 64w/m^2, perhaps a measure of that portion of the blackbody radiation energy emitted from the earth (‘earthshine’) that is blocked by the saturated H2O absorption spectrum as opposed to the relative ability of any given parcel of air to capture or export heat via the H2O photon radiation path?
Spector,
the 64 w/m^2 is the difference between h2o and no h2o in my model atmosphere which incorporates radiative transfer. Starting with the surface, each layer attenuates the BB spectrum with the ghg lines and will also emit at the ghg spectral line wavelengths but with a lower power level if the layer temperature is less. If I included only what was emitted by the surface, it would be somewhat more than that – and it would be rather irrelevant because all those h2o molecules in the atmosphere are going to be involved in radiating their characteristic spectrum at their respective temperatures. As one goes higher in the atmosphere, the temperature will be cooler so the higher energy states and shorter wavelengths will not be radiated as much and the total energy radiated will be less. Also, at higher altitudes, the individual lines will become narrower due to lower pressure.

298. cba says:

Bryan says:
August 7, 2010 at 8:17 am
cba says:
…..”rather than 100/1, I’d suggest it’s closer to 2/1, excluding those that are simply not going very far. That’s based on there being a factor of two in absorption, roughly co2 total contribution 30W/m^2 versus h2o at 64w/m^2 in the approximately current clear sky atmosphere calculations”………..
I disagree, the paper below shows that that even for deep midwinter Antarctica with extremely low humidity, clear skies conditions, the radiative flux from H2O vapour was more than twice that for CO2.
Since CO2 is well mixed in the atmosphere its radiative contribution will not change much as we move to more average conditions.
However if we move to more average Earth condition with average levels of humidity introducing condensation phase change then the radiative contribution from H2O will increase dramatically.
On the other hand there does not seem to be any physical reason for CO2 to become dramatically more radiative at more average conditions.
————
Thanks for the reference Bryan,
I’ve only time right now to read the abstract. It doesn’t look that far off for summer. I’ve got a model with 107w/m^2 (out to only 75 um) which is a few % shy of the whole spectrum, out past 100um and into radio). that 107 is total absorption of outgoing. It’s going to be close (I think) to their back radiation measurements of 110-120w/m^2 summer. I’m using a 1976 std atmosphere model which is much closer to that of the summer than the winter. Essentially, I’ve got about 2/1 just like they have – with around 10 or 15 w/m^2 due to all but co2 and h2o. It looks like I’m definitely in their ballpark from their abstract.
I hope to have some more time tomorrow or later tonight to read the whole paper.

299. Reed Coray says:

Hi Julio,
At some point we’re going to have to “scale back” our interchange–if for no other reason than it consumes a lot of my time, and I’m sure yours as well. Since I’m retired, it’s not a serious issue with me; but if it is with you, just let me know and I’ll scale back.
I don’t claim to have all the answers–or for that matter, even a few of the answers. I am trying to understand if a logical and internally consistent argument for the claims made by the AGW community exists. Specifically the claims that (a) greenhouse gases rather than something else (e.g., gases in general, or friction resulting from atmospheric and liquid convection currents that arise because various regions of the Earth are at different temperatures and because gases and liquids in motion encounter the “Corlios Force” which exists because the Earth is rotating relative to inertial space, etc.) are primarily responsible for the Earth’s surface temperature being higher than it would be for a perfect-thermal-conducting black body Earth; (b) man’s current and future use of fossil fuel will cause a “significant” change to the Earth’s surface temperature, (c) such a change will be bad for mankind, and (d) the AGW community cure (drastically curtailing our use of fossil fuels to generate energy) won’t be worse than the disease. When I see discussions that employ terms that are not well defined and/or appear to be self-inconsistent, I ask questions–mostly to myself, but sometimes to the authors of those statements.
In your August 7, 2010 at 1:05 pm response you said: “Anyway, you add a layer of material to the surface of your sphere, and then…? You have more material to heat from the same constant internal source. At first sight it would certainly seem that you are spreading the energy thinner“. If by thinner you mean the ratio of (a) the “rate-of-internal-energy-generation” …to… (b) the “object’s volume” is decreasing, of course I agree with you. I’m just not sure this observation has anything to do with the surface temperature of the body.
In another regard, I agree with the idea that if body “A” completely surrounds body “B”, the presence of body “A” may and even “likely will” affect the temperature of body B. To give a specific example. Assume body “B” is a thin spherical shell (external radius, RI) with a radioactive source (make it an alpha particle emitter so that the energy of the radioactive particles are quickly converted to random molecular motion) uniformly (i.e., symmetrically) distributed throughout the shell wall. Further assume that the shell’s internal and external surfaces radiate as a black body. Place the shell in the vacuum of space where the temperature of “space” is 0 Kelvins. Because the energy source is symmetrically distributed throughout the shell wall, at some point the external surface of the shell will reach a temperature such that the rate of energy radiated outward by this surface will be equal to the rate at which radioactive decay energy is generated internal to the shell wall. Let that temperature be Tisolated.
Now surround body “B” with body “A”, which is also a thin spherical shell (external radius RE) with internal and external black body radiating surfaces, but contains no internal radioactive energy source. Let the internal radius of spherical shell “A” be greater than the external radius of spherical shell “B”, and don’t allow the two shell surfaces to come into contact. I don’t think it’s a requirement, but for simplicity assume the two shells are co-centered. Finally, assume the thermal conduction properties of the outer shell (shell “A”) are “almost perfect” so that the temperature of the outer shell’s internal surface and the temperature of the outer shell’s external surface are infinitesimally different. For these conditions, when radiation-rate-equilibrium is reached for the “two-shell system” (i.e., when the rate of energy being radiated outward by the outer shell equals the rate of energy being generated in the wall of the inner shell, I believe the presence of body “A” will affect the temperature of the external surface of the inner shell. Specifically, under these conditions I believe the temperature, Tdual of the external surface of shell “B” will be given by:
Tdual = Tisolation * [1 + (RI/RE)^2]^0.25
If valid, the above equation indicates that unless RE is infinite, the presence of the surrounding shell will, in radiation-rate-equilibrium, result in an increase in the temperature of the external surface of the inner shell. In this sense, provided energy is supplied to the Earth but not its atmosphere, I believe it is possible for an atmosphere to raise the temperature of the Earth’s surface. The physics that led to the above equation isn’t trivial (at least not to me); but compared to the physics of the Earth and its atmosphere, it’s less than trivial, it’s super trivial.
In part, this is why I am skeptical of anthropogenic global warming. The physics of atmospheric thermodynamics, especially with all the unknowns that exist, is I believe far too complex to use simple arguments like “radiation is trapped, and therefore the temperature should rise by X degrees”. I think it may even be too complex to even argue that the presence of greenhouse gases at the levels they are present on the Earth will cause the Earth’s temperature to rise any measurable amount. I think it likely, but I haven’t come across a conclusive argument yet. That, by the way, doesn’t mean (a) a conclusive argument doesn’t exist, or (b) one does exist that I’ve seen but I don’t have the wherewithall to understand it. What it does say, is that I am FAR from being sufficiently convinced of the argument to be willing to make the drastic changes in the way the western world generates energy advocated by a large section of the AGW community.
One last point. In response to my question: (3) “Will trapping of heat by an object cause a temperature rise in the object?”; you answered: If by heat you mean thermal energy, yes, of course, as long as the object has a positive heat capacity. If you put thermal energy into an object, its temperature will go up. What else could it do? Answer: Over the short term, the energy could go into a change of phase state–ice at 32 degrees Fahrenheit to water at 32 degrees Fahrenheit, which doesn’t involve a temperature rise. Eventually though, if you keep adding energy in the form of heat to a system, I believe the system temperature will rise. However, if (a) after the atmosphere’s temperature has risen to achieve radiation-rate-equilibrium the atmosphere is still “trapping heat”, and (b) “trapping heat” will cause temperature to rise, isn’t it correct to conclude that the atmosphere temperature will “rise some more”, and this rise will only stop when the atmosphere ceases to “trap heat?”
Thanks again,

300. Spector says:

RE: Paul Birch: (August 7, 2010 at 3:49 am) “All gases absorb and radiate at all wavelengths.”
RE: Paul Birch says: (August 7, 2010 at 11:10 am) “2) Collisions between molecules are not quantised; they produce some absorption at all wavelengths.
I do not believe this second statement to be true or the atmosphere of the Earth would not be optically transparent and perhaps there would be a visible glow in the sky at night due to the reciprocal photon emission process. I suspect that photon emission or absorption requires a quantum electromagnetic field change.
Perhaps such collisions can only cause kinetic energy exchanges and perhaps trigger or assist in the release or absorption of photons characteristic of the molecules involved in the collision. Both molecules must return to one their normal allowed quantum electromagnetic states after impact.

301. Spector says:

RE: cba: (August 7, 2010 at 5:17 pm) “Spector, the 64 w/m^2 is the difference between h2o and no h2o in my model atmosphere which incorporates radiative transfer.”
Thanks. My basic question is this: if one looked down from above the tropopause, how much energy is being exported out of tropopause region and below by H2O and how much by CO2 in those saturated bands where no radiative transport from the surface is possible?
I would expect the factor of 77 to 103 greater abundance of H2O relative to CO2 to be reflected in the role H2O serves in cooling of this vital region unless this advantage was compensated by a higher rate of absorption of that energy caused by the same high abundance of H2O.
I believe the numbers that you are quoting represent the ratio of the saturated absorption bandwidths of H2O and CO2 as seen from the surface.

302. cba says:

Nasif,
The number I’m using, 64 W/m^2, is not from trenberth nor is it the same thing. Rather it is the difference in radiative transmission between h2o being present and being totally removed as calculated at the tropopause over a 75 um range and assuming conditions of a 1976 std. atmosphere under clear sky conditions. If one were to take the model and plot a curve of emission versus wavelength rather than simply summing up the values (integrating over wavelength), it would show a spectral curve of a black body at 288.2k with the absorption lines of the atmosphere dipping down to the point where there is a spectral curve for a lower temperature at which there is emission going on in the wavelength bands associated with ghg absorption.

303. cba says:

Spector,
See above explanation to Nasif Nahle. I think there’s probably good correlation between what is seen at the surface coming down and my numbers but it is not at all the same thing. Mine is a calculation made at the tropopause. It is not made with only saturated lines or this or that, it is made up of the whole spectrum – as calculated. This includes the individual line contributions including their wings and including the overlaps of various line widths at the same wavelength. And, there are tens of thousands of lines from 39 different ghg contributor molecules. Saturated simply means that the lengths of the air column are much longer than the mean free path of a photon and that is an extremely complex function of wavelength. Some path lengths are in the vicinity of a few cm while right next to that wavelength, the pathlength is longer than the thickness of the atmosphere. I get the function by adding up all the contributing amounts from all the wavelengths and molecules and then calculate the path length information.

304. Spector says:
August 7, 2010 at 7:19 pm
RE: Paul Birch: (August 7, 2010 at 3:49 am) “All gases absorb and radiate at all wavelengths.”
RE: Paul Birch says: (August 7, 2010 at 11:10 am) “2) Collisions between molecules are not quantised; they produce some absorption at all wavelengths.
“I do not believe this second statement to be true or the atmosphere of the Earth would not be optically transparent and perhaps there would be a visible glow in the sky at night due to the reciprocal photon emission process. I suspect that photon emission or absorption requires a quantum electromagnetic field change.”
The atmosphere of the Earth is not optically transparent. It is nearly transparent. Your suspicion is not correct: continuum emission and absorption is well known.
“Perhaps such collisions can only cause kinetic energy exchanges and perhaps trigger or assist in the release or absorption of photons characteristic of the molecules involved in the collision. ”
This is not the case. The energy and momenta of the absorbed photons are transferred to the molecules. Since the speed, direction and impact parameters of the molecules are not quantised, nor is the absorption spectrum. An interesting feature of this mechanism is that the absorption per unit mass is proportional to the density (because so is the collision frequency), which means that the mechanism is significant in dense stellar or planetary atmospheres (like those of Venus or Jupiter), but quite hopeless in interstellar space. At high densities collisions between three or more molecules at a time also add to the absorption rate.
By the way, scattering due to Brownian density fluctuations in the gas also contributes to the opacity.

305. cba says:

Tom Vonk,
Hi Tom,
Thinking back, I may have applied an equilibrium concept to LTE that isn’t part of the usual interpretation concerning photons but it is a similar concept to the original. When I was saying that the photons were in LTE equilibrium, I was referring to the concept that the continuum spectrum coming in to the gas cloud was at the same temperature as the temperature of the gas cloud of your conceptual example. If that is the case and if the continuum is coming in from all directions, then there is no net radiative power transfer going on and in fact, one would not observe any absorption spectra (or emission spectra) at all. For the geometry we have, Earth at T = ~288k, space T = ~2.7k, it is not really possible to have that situation as each small layer of the atmosphere between surface and space will radiate outward and inward its characteristic spectrum based upon its temperature and the incoming radiation to that layer will be a combination of what is left of the surface continuum PLUS what makes it through to the layer from those layers lower down radiating outward and those layers above radiating downward. In the case where there is rather little change in the layer’s temperature and pressure and molecular constituency, there will be little net power flow happening at those wavelengths of significant interaction. For wavelengths other than those, there is little or no interaction going on. I think this is the reason why physical meteorology concepts are capable of describing convection using adiabatic assumptions with any success.

306. Spector says:

RE:Paul Birch: (August 8, 2010 at 4:39 am) “Collisions between molecules are not quantised; they produce some absorption at all wavelengths.
I note there is a paper in Nature that is titled “Collision-induced Absorption in the Earth’s Atmosphere” by C. B. Farmer & J. T. Houghton, March 26, 1966.
The abstract of this report says most of the absorption [and thus radiation] observed in the atmosphere is due to its trace components such as water-vapor, carbon dioxide and ozone. It goes on to say that the major components have no dipole moment so that their vibration and rotational energy states cannot be excited directly by absorption of radiation, but dipole moments are induced during the collision process. The results of these collisions, they say, can be seen in very long [very minor] absorbing paths in the atmosphere.
Their report shows that absorption of the solar spectrum near 4.2 microns is due to collision-induced absorption by nitrogen and they present a short discussion of the importance of the rotational collision-induced bands of oxygen and nitrogen in the radiation budget of the atmosphere.

307. Spector says:
August 8, 2010 at 10:21 am “…”
Yes, that’s just as I said.

308. TomVonk says:
August 6, 2010 at 4:08 am
Merrick
“You also seem to have left out rotations. Any collisions between relatively translationally excited N2 (or any other molecule) and CO2 are far more likely to transfer energy into excited rotational states and have exactly no effect on the vibrational state of CO2.”
Yes I did and said so . As long as I the energy equipartition law is respected and the rotationnal quantum levels obey Maxwell Boltzmann statitics like vibrational do the argument stays unchanged . The focus of this post was the process (1) and the length of the post was limited . I could have added R/V and R/T processes for a much longer post . What you wrote is of course true .
Jan
“As I don’t understand this post, am I supposed to take it to prove that climate change doesn’t exist, as any argument will suit this purpose?”
No . I focus on N2-CO2 collisons and the question of where there can be or not be a net energy transfer .

Which where you get it wrong. There is necessarily a net transfer of energy to the non-CO2 molecules approximately equal to the total energy absorbed by the CO2.
Your simplification of the process to a single step renders the analysis nonsense.
If we just reduce the absorption to a single line then we will have CO2 molecules with a particular state (v=1, r), the multiple collisions will deactivate this excited state to a variety of lower states (v=1 or 0, many r). The large numbers of collisions will distribute these states over a Boltzmann distribution appropriate to the steady state temperature reached. The time spent by an individual molecule in a particular state is extremely small at atmospheric conditions, orders of magnitude less than the mean radiative lifetime which is why emission is extremely unlikely, and most of the energy ends up thermalized. If the world worked the way Vonk thinks it does then my experiments on Laser Induced Fluorescence would be much simpler, even with much shorter lived states (~10ns) collisional quenching dominates the energy cascade and often only one molecule in a thousand will fluoresce! In order to perform the analysis you need to set up the kinetic equations between each state. I hope to follow the rest of this thread but I may not get back frequently as I’m away from internet access at present (relying on visits to coffee shops).

309. Gnomish says:

An experiment can illuminate a principle.
An analogy can give it perspective.
If one shines a really bright light on the Mississippi can he boil Lake Superior?

310. Remo says:

Tom dumbed his presentation down a little for you folks.
The trouble with Tom’s LTE analysis is that it ignores differential calculus. The temperature and emissions will be infinitesimally less immediately above the sample as compared with below it. The smaller the sample, the smaller the difference. But it will always be there and cumulatively: It will add up to the difference between the ground temperature and the temperature at the top of the troposphere where CO2 radiates to space.
Add CO2 and you will catch an infinitesimally more thermal radiation at each layer. The result is that the ground becomes warmer because there is more “thermal resistance” to pushing thermal energy from the earth to the top of the troposphere. Tom is correct that adding CO2 to a black box does not change the temperature inside the box. But if you have a thermal gradient, it does slow the transfer of heat via infrared radiation from one side of the gradient to the other.
Adding a differential equation to account for the slowing of the radiative flux would bring Tom’s analysis in line with the mainstream.

311. anna v says:

On the momentum conservation question I raised:
1) the effect of the asymmetry per molecule is small. I make it out for CO2 at a 15micron absorption , if we assume about 300 meters/second velocity of the molecule a 10^-5 in percent addition to the motion. It will add more to the part of the distribution that has smaller velocities depending of course on direction. A model is needed :). Nevertheless in principle when there is energy transfer there is momentum transfer and isotropy is broken.
2) in total CO2 in its collisions with other atmospheric molecules will transfer this upward impetus and it will become part of the “breathing” of the atmosphere, (limited by gravity as in the example of cbs with the ball) together with convection.
3) It should play a part in keeping CO2 aloft, as it is heavier than other molecules and would stratify lacking convection and maybe radiation pressure. I have not done the calculation, it probably again is a very small effect.

312. TomVonk says:

cba
Thinking back, I may have applied an equilibrium concept to LTE that isn’t part of the usual interpretation concerning photons but it is a similar concept to the original.
Yes I think that I have seen what you tried to say .
My point was that LTE (condition that I am using) is independent of the questions whether there is or is not radiative equilibrium . That’s why I did no assumption about the precise properties of the radiation (isotropic , continuous , black body etc) .
The purpose was to observe what happens in collisional V/T interactions and whether there is or is not energy transfer from N2 towards CO2 via a T->V process .
The fact that LTE is a property of material particles only allows to decouple this question from much more complex questions concerning radiative transfer over large distances (measured in km) about which there is enough papers anyway .
For this same reason if there is a conclusion about T/V interactions , it will be largely independent from detailed assumptions concerning the nature of the IR radiation .
The problem with many reactions is that they focus on something else altogether – namely radiative transfer and then seem to construct relations between radiative transfer and my post which are simply not there .
Phil
The time spent by an individual molecule in a particular state is extremely small at atmospheric conditions, orders of magnitude less than the mean radiative lifetime which is why emission is extremely unlikely, and most of the energy ends up thermalized.
This is just hand waving . How “extremely unlikely” is the emission ? What is the curve of the radiative life (probability vs time) ? What is “extremely small” ?
How much is “most” , more than 50% ? What is the probability of CO2 vibrational excitation , do you take it for 0 ?
Needless to add that you have adressed neither of the arguments I used .
If somebody used the same kind of hand waving arguments you do , you would probably qualify them as “nonsense” . I have more respect for people’s opinions so I won’t . What’s really your point ?

313. cba says:

Tom Vonk,
But Tom, doesn’t the requirement of LTE require a transfer of energy between the CO2 and N2? When you change the conditions of radiative energy coming in, the temperature of the CO2 will need to adjust in order to radiate it away. If there is no interaction with N2, energy transfer-wise, the N2 will remain at the old temperature and you’ve got the loss of the LTE condition.

314. TomVonk says:

cba
But Tom, doesn’t the requirement of LTE require a transfer of energy between the CO2 and N2? When you change the conditions of radiative energy coming in, the temperature of the CO2 will need to adjust in order to radiate it away. If there is no interaction with N2, energy transfer-wise, the N2 will remain at the old temperature and you’ve got the loss of the LTE condition.
Yes it does . Actually the whole point of LTE and energy equipartition law is to make sure that all degrees of freedom of material particles interact sufficiently so that all energy forms are “well mixed” .
It is when they are not well mixed that very difficult problems begin .
An example . One can define vibrational and translational “temperatures” by taking the average V and T energies . There is à priori no reason that they should be equal .
But when they are (what means that V and T degrees of freedom often and strongly interact) then we have LTE .
It is precisely this kind of graphics that shows that the troposphere is in LTE while the stratosphere is not . That’s also why radiative transfer models (see f.ex SAMM2 http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA427001&Location=U2&doc=GetTRDoc.pdf) must distinguish between LTE and non LTE conditions what is apparently beyond the understanding of people like Birch .
And yes . If you change the radiation properties then there will be a transient in which all the degrees of freedom will interact and adjust themselves .
Clearly if you begin to fire a 1 kw IR laser in a volume , the final state will not be the same as what it was without the laser 🙂
In that sense it is true , there is an unsaid assumption in my post that the radiation properties are not dramatically changing at the scales and times i consider .

315. TomVonk says:
August 9, 2010 at 4:27 am
Phil
“The time spent by an individual molecule in a particular state is extremely small at atmospheric conditions, orders of magnitude less than the mean radiative lifetime which is why emission is extremely unlikely, and most of the energy ends up thermalized.”
This is just hand waving

Really? That’s all you’ve done in your flawed analysis.
How “extremely unlikely” is the emission ?
Didn’t you notice ‘orders of magnitude’?
What is the curve of the radiative life (probability vs time) ?
Usually a Poisson distribution, in the case of CO2 in the 15 μm band the mean is order 10ms, compared with a collisional lifetime of order ns.
What is “extremely small” ?
ppm as you chose to ignore I pointed out that parts per thousand was an experimental result for a much shorter lived excited state, inconvenient fact?
How much is “most” , more than 50% ?
More than 99%.
What is the probability of CO2 vibrational excitation , do you take it for 0 ?
No, what is the relevance of this question?
Needless to add that you have adressed neither of the arguments I used .
If somebody used the same kind of hand waving arguments you do , you would probably qualify them as “nonsense” .

Actually I’ve addressed them both, your model of collisional deactivation I showed was wrong, and your ideas about excited state lifetimes wrt emission I also showed to be wrong.
And I’d have followed up the suggested subjects in the literature and found they were right, you apparently couldn’t be bothered to do that.
I have more respect for people’s opinions so I won’t . What’s really your point ?
Not enough respect to follow up the points but instead you dismiss them out of hand. The point is that your analysis is flawed, try looking up Collisional Quenching or the Stern-Volmer equation, I gave you some hints before.

316. cba says:

Tom Vonk,
Tom, that’s some reference. I saved it to go through later. I did notice that it stated while LTE conditions start to happen around 45-50km (above the stratosphere) but that co2 appeared to stay in LTE conditions past 100km. This 45-50km is the area that is at the bottom of the ionosphere – although usually the D layer doesn’t go below 60km.

317. Remo says:

Phil — you are right and wrong.
The time spent by an individual molecule in a particular state is extremely small at atmospheric conditions, orders of magnitude less than the mean radiative lifetime which is why emission is extremely unlikely, and most of the energy ends up thermalized.
Yes, almost always, unless you are in a vacuum, the energy absorbed by a CO2 molecule will thermalize. However, what happens if you are indeed in thermal equilibrium is that there will always be enough CO2 molecules with sufficient energy to radiate out at exactly the same rate as the flux in. Increase the flux in, the sample warms until it is in equilibrium and then there will be sufficient CO2 molecules with an energy state high enough to shoot out as many photons as are coming in. This is all Planck’s law of Black Body radiation.
You need to be sure not to look at the mean time a single CO2 has sufficient energy, but the total time based on the number of CO2 molecules bumped up.
As an example, I was running out some of calcs the other day and at the temperature I was using, only 1 in 800 CO2s had the energy necessary to shoot off a photon (and consequently, one N2 in 800 also had that energy), but that was exactly enough to balance the absorption I was dealing with. This is all your Maxwell – Boltzmann stuff.
As I pointed out in an earlier post, the weakness in Tom’s argument is you have to deal with a temperature gradient and the radiative flux gradient which is ignored in LTE model. It is this gradient that is the driver in exporting thermal energy from the earth’s surface to space.
Do you know any circuitry stuff? You can do a quick simple model of the total gas in the atmosphere as the thermal induction and the total CO2 as the thermal resistance. If you increase the resistance by increasing CO2 than the temperature (i.e., voltage) has to increase at the surface in order to maintain the same outflow of energy into space. (In actuality, there will also be a slight dip in the radiation into space in the CO2 band and an increase in some of the other bands. But the net flux into space has to remain a constant). Hope this helps.

318. TomVonk says:

Remo
However, what happens if you are indeed in thermal equilibrium is that there will always be enough CO2 molecules with sufficient energy to radiate out at exactly the same rate as the flux in. Increase the flux in, the sample warms until it is in equilibrium and then there will be sufficient CO2 molecules with an energy state high enough to shoot out as many photons as are coming in. This is all Planck’s law of Black Body radiation.
Absolute agreement . What Phil. doesn’t understand is that his thesis not only unexplicably violates the time symmetry of the processes which is at the basis of the statistical equilibrium but it would prevent any equilibrium to form . In order to have an equilibrium it is simply necessary that the rates of the V/T processes be equal . And yes this is text book stuff . Looking at the reference I linked for CBA helps too .
As I pointed out in an earlier post, the weakness in Tom’s argument is you have to deal with a temperature gradient and the radiative flux gradient which is ignored in LTE model. It is this gradient that is the driver in exporting thermal energy from the earth’s surface to space.
Yes and no . I actually don’t deal with radiative transfer , there are enough papers (see among others the reference above) that do that quite nicely .
I deal with small volumes in LTE and argue about what happens with the statistics of molecular collision processes WITHIN the volume . As already mentionned in a discussion with CBA above , LTE is independent of the radiation conditions and a property of material systems only so I do not need to make any specific assumptions about radiation . Sure implicitely I suppose that the IR intensity coming from outside of the small volume is constant so that equilibrium may be established . This is a reasonable assumption at the time scales we consider .
I observe that after 300 comments , not a single one contested either the time symmetry or the energy equipartition law . One commenter contests the existence of LTE (“it never exists”) and Phil. contests the existence of IR emission by CO2 . Both are inconsistent with observational evidence .

319. R Stevenson says:

Heat and mass balances are tools used by chemical engineers in heat and mass transfer analyses;
Energy in – energy out = accumulation
Material in – material out = accumulation
At equilibrium, accumulation is zero

320. TomVonk says:
August 9, 2010 at 4:27 am
“My point was that LTE (condition that I am using) is independent of the questions whether there is or is not radiative equilibrium .”
And that is where you are fundamentally wrong. You cannot have LTE without radiative equilibrium too. The most you can have is an approximation to LTE. The reason you cannot have full LTE is that the molecules interacting with the non-equilibrium radiation are pumped into non-equilibrium populations. Absorption of “hotter” radiation puts more molecules into excited states; re-radiation depletes the excited states. If the radiation is anything other than isotropic black body radiation at the temperature of the gas, the steady-state population of the excited states cannot be thermal. It must depart from LTE to some degree. This is how a laser works, and although the radiative pumping in Earth’s atmosphere is very very weak by comparison, the same processes are occurring there.

321. TomVonk says:
August 10, 2010 at 1:53 am
“I observe that after 300 comments , not a single one contested either the time symmetry or the energy equipartition law . One commenter contests the existence of LTE (“it never exists”) and Phil. contests the existence of IR emission by CO2 . Both are inconsistent with observational evidence .”
Perhaps if you read the comments more carefully your observations would be less inaccurate. Several commenters (including myself) have pointed out that time symmetry does not apply to conditions in the Earth’s atmosphere, or to any situation in which there is a macroscopic entropy change. Several commenters (including myself) have pointed out that equipartition is not an invariable law and is not valid under conditions of radiative pumping for example. Furthermore, I do not claim that there is no such thing as LTE, but rather that LTE is a limiting case never quite achieved but only approximated; it was you who denied the existence of approximate LTE and pedantically asserted that LTE is a binary either-or; for so pedantic a definition, then, one has to say that it does not exist. I would also invite you to think about how perfect LTE could possibly be observed if it did exist; any device you use to measure the thermal radiation or the distribution of velocities or the population of excited states must itself be at a different effective temperature from the gas in question, and must absorb energy from it, disturbing the very equilibrium you are trying to observe.

322. TomVonk says:
August 10, 2010 at 1:53 am
Absolute agreement . What Phil. doesn’t understand is that his thesis not only unexplicably violates the time symmetry of the processes which is at the basis of the statistical equilibrium but it would prevent any equilibrium to form . In order to have an equilibrium it is simply necessary that the rates of the V/T processes be equal . And yes this is text book stuff . Looking at the reference I linked for CBA helps too .
No I understand it very well but when you apply these laws to a process which doesn’t happen, such as Vonk’s fictitious V-T exchange, you get garbage out.
I observe that after 300 comments , not a single one contested either the time symmetry or the energy equipartition law . One commenter contests the existence of LTE (“it never exists”) and Phil. contests the existence of IR emission by CO2 . Both are inconsistent with observational evidence .
No I don’t, I apply the Stern-Volmer equation which indicates that, for vibronically excited CO2 (bending mode) at atmospheric conditions near the surface, emission will be a very unlikely mode of energy loss. This is in fact consistent with observation, emission from excited CO2 will increase with decreasing pressure (Stern-Volmer plot). It is Vonk who is so wedded to his beliefs that he is blind to the actual physics involved.

323. sky says:

I leave it to theoreticians and specialists in molecular and laser physics to fight it out amongst themselves on the question of what sort of energy transfers may or may not occcur on the microscopic level between absorbent and non-absorbent gas species in the atmosphere. From the macro perspective of geophysics, that question is largely mooted by the fact that radiative transfer does not operate as the sole means of thermal energy transfer from surface to space. In fact, scores of careful total energy-flux experiments repeatedly show the Bowen ratio (sensible to latent heat transfer) almost invariably well below unity in the marine environment and whenever surface moisture (dew) is present. Yes, it may rise to values 5 or more in the ultra dry environment of Antarctica and some deserts, but these constitute a minor fraction of the surface.
What we have as a consequence is a totally different system than the one usually depicted by AGW proponents. Instead of a paradigm that leads people to think that any impedance to OLR necessarily must raise surface temperatures to maintain TOA radiative balance, Earth has a parallel non-radiative path of thermal energy flow, with the endothermic process of evaporation playing a central role. Neither this nor the paradigm system can be realistically modeled by a circuit with the bulk of the atmosphere representing an inductor or capacitor and the GHGs representing a resistor. Resistance is a dissipative mechanism, which in any isentropic (energy budgetr) analysis is disallowed. In fact, the atmosphere does thermodynamic work in in sustaining macro-scale pressure gradients that drive the geostropic winds and in overcoming vapor pressure during evaporation. This component of enthalpy is what almost everyone neglects.
It may surprise many that Earth would still have a “greenhouse effect” even if the atmosphere were totally transparent to IR. The lowest levels of the atmosphere would still be warmed by (dry) convection, and backradiation from the atmosphere would allow the surface to retain more energy in storage than it would adjacent to the heat sink of space.

324. sky says:
August 10, 2010 at 4:54 pm
“It may surprise many that Earth would still have a “greenhouse effect” even if the atmosphere were totally transparent to IR. The lowest levels of the atmosphere would still be warmed by (dry) convection, and backradiation from the atmosphere would allow the surface to retain more energy in storage than it would adjacent to the heat sink of space.”
I would agree with you concerning the importance of convection as a heat transfer mechanism within the atmosphere. It does magnify the night-time greenhouse effect by warming the clouds or the higher levels of the atmosphere, thus increasing the amount of heat radiated back to the surface; though the overall effect is to reduce net planetary greenhouse warming by limiting the temperature gradient. However, in the quoted sentence you contradict yourself: if the atmosphere were “totally transparent to IR” there could be no backradiation at all.
Vertical convection would not arise in a transparent atmosphere under constant insolation. Convection due to a diurnal cycle can cool the surface during the day and warm it at night; that is, it permits a larger fraction of the atmosphere to buffer the diurnal temperature changes. Because of the fourth power law of radiation, the average temperature is then higher than in the absence of this buffering. (Fully buffered, with heat capacity -> infinity, T~(average insolation)**1/4; unbuffered, for a simple model in which the sun switches full on for 12 hours then off for the other 12 hours, T~0 at night and ~(2*average insolation)**1/4 during day, so Tav~(0+2**1/4)/2=0.59, much lower than the buffered case). Whether this should be called greenhouse warming is debatable. I’d say no. It’s a different phenomenon (though one that happens in greenhouses too!) and only helps to bring the temperature up to the black body temperature calculated on the basis of the total insolation; it doesn’t warm it above that level.

325. sky says:

Paul Birch says:
August 11, 2010 at 4:16 am
“you contradict yourself: if the atmosphere were “totally transparent to IR” there could be no backradiation at all.”
An IR transparent atmosphere means there’s no IR absorption by its constituents. It doesn’t mean no IR emissions from them. Every substance with a temperature above 0K emits in the EM spectrum regardless of the process by which it got thermalized. With conduction and convection being such processes at terrestrial temperatures, there is no requirement that the constituents be IR absorbers (i.e., GHGs) to obtain backradiation.
No doubt, with an IR transparent atmosphere and a dry planet, the lapse rate would stay close to the stable dry adiabatic. But the surface being fractal and composed of materials with different specific heats, it would be subject to differential heating and thus convection, at least in the boundary layer. The atmosphere would thus still be heated from below, but not as effectively as with deep moist convection. Its isotropic emissions would produce backradiation, which is the essence of the “greenhouse effect.”
Your simple model with T = 0 at night is an excercise in self-deception, which I will not take the time to dissect. Suffice it to say that backradiation operates steadily 24/7, without a diurnal cycle. This is a clear indication that that it does not come principally from GHGs, whose re-emissions should coherently follow the surface temperature, but from the non-absorbent bulk constituents thermalized by the chaotic process of convection.

326. sky says:
August 11, 2010 at 3:06 pm
Paul Birch says:“you contradict yourself: if the atmosphere were “totally transparent to IR” there could be no backradiation at all.”
“An IR transparent atmosphere means there’s no IR absorption by its constituents. It doesn’t mean no IR emissions from them. Every substance with a temperature above 0K emits in the EM spectrum regardless of the process by which it got thermalized. With conduction and convection being such processes at terrestrial temperatures, there is no requirement that the constituents be IR absorbers (i.e., GHGs) to obtain backradiation. ”
Sorry, but this is one thing Vonk has got right. Any molecular species or assemblage that can radiate IR can also absorb it. It’s a fundamentally symmetrical process. Yes, every substance does emit IR to some degree; but then it absorbs IR to the same degree. Under non-equilibrium conditions, the actual rates of absorption and emission can differ, but both processes must still occur together. If you have any backradiation at all the atmosphere cannot be perfectly transparent.
“No doubt, with an IR transparent atmosphere and a dry planet, the lapse rate would stay close to the stable dry adiabatic.”
No, it would be isothermal, so long as the insolation were constant. Otherwise it would be isothermal above a convective buffer layer. The depth of that layer would depend on the ratio of the insolation times the sol period (day length) to the volumetric heat capacity of the atmosphere.
“Your simple model with T = 0 at night is an excercise in self-deception, which I will not take the time to dissect. Suffice it to say that backradiation operates steadily 24/7, without a diurnal cycle. ”
The model as given was for a transparent atmosphere with no backradiation. Your discourteous rejection of it displays your own ignorance. The mechanism I described is quite well known in planetary science. The lack of buffering is why airless bodies and the skins of spacecraft display much greater temperature variations than the surface of planets like the Earth. Planets like Mars, with very low atmospheric pressures, are interesting intermediate cases. Even on Earth, higher altitude sites, with less atmosphere to buffer them, display wider diurnal temperature ranges. Obviously, night time temperatures do not in practice fall as far as the near zero temperature of deep space (~2.7K); heat is conducted into the ground during the day, and released again during the night, buffering the temperature extremes; the same basic mechanism, but employing the heat capacity of the regolith instead of the atmosphere. On the Moon, the effective buffering depth is ~1m, which for comparison gives about a tenth of the buffering capacity of the Earth’s atmosphere, while the sol is of course ~29 times as long, so the temperature swing of the lunar surface is much greater, reaching ~120C during the lunar day.

327. Spector says:

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.”
Update: My 100/1 ratio came a result of a seemingly general statement from a document at the DOE NETL website that implied the H2O concentration in the atmosphere was 3 to 4 percent and even went so far as to compare this with the concentration of CO2 expressed as a percentage. I now find that this probably applies only near the surface in very humid air. So far, I have not yet run across any good accepted numbers for tropopause-level, typical mono-molecular H2O concentrations.

328. sky says:

The isothermal atmosphere exists only in academic textbooks. I was making a practical and not a theoretical point. Instead of saying “totally transparent to IR,” my intent would have been clearer had I said “without water vapor or any of the GHGs identified as AGW culprits.” And my reference to the more realistic dry adiabatic in my comment to you was a nod to the usual meteorological criterion for convective stability.
You can justify T=0K at night perhaps in your mind, but not mine. The whole point of my original comment directed to no one in particular was to get lay people to think in more in terms of “thermal mass and inertia” and all the means of thermal transfer, rather than relying upon simplistic radiative vector algebra. Your accusation of “ignorance” on my part about the “buffering” effects of a thermalized air mass is as amusingly misplaced as your ideas about what convection does throughout the diurnal cycle.
I see no prospect of fruitful discussion. Good night!

329. Spector says:

RE: sky says: (August 10, 2010 at 4:54 pm) “From the macro perspective of geophysics, that question is largely mooted by the fact that radiative transfer does not operate as the sole means of thermal energy transfer from surface to space.”
From above the tropopause level, radiation is the *only* ticket out. I am not sure that we really understand the process by which infra-red emitting gases at top of the atmosphere exhaust the heat convected up to that level and contribute to what I call the tropopause ‘ice-locker’ effect.

330. sky says:
August 11, 2010 at 7:53 pm
“The isothermal atmosphere exists only in academic textbooks.”
For the simple reason that no atmosphere is totally transparent to IR. However, most atmospheres do have an approximately isothermal upper region. They can also have approximately isothermal (or at any rate, nonconvective) lower regions, when layers considerably more opaque to sunlight than to thermal radiation shield them from further insolation.
“I was making a practical and not a theoretical point. Instead of saying “totally transparent to IR,” my intent would have been clearer had I said “without water vapor or any of the GHGs identified as AGW culprits.” ”
OK, but that wasn’t what you said. You could easily have replied, “Sorry, I didn’t mean totally transparent, just one without any of the usual suspects, like CO2 or H2O”. Instead, you tried to defend the obvious contradiction of combining backradiation with total transparency, arguing that you can get emission from substances that don’t absorb. Which you can’t.
However, even your “practical point” isn’t really true. If there were none of the strong absorbers present, like H2O, CO2 and aerosols, the backradiation would be far too weak to notice. There would be no significant warming from this cause. It is the strong absorbers that are also the strong emitters; the backradiation comes overwhelmingly from those strong absorbers (which are in very close to local thermodynamic equilibrium with the bulk gas).
“And my reference to the more realistic dry adiabatic in my comment to you was a nod to the usual meteorological criterion for convective stability.”
Fine, but it still wasn’t correct. We only get any sort of continuous vertical lapse rate when energy is being continually lost from the top of the atmosphere, by radiation. Convection only stops the lapse rate significantly exceeding the adiabatic rate, it doesn’t stop it being less than that. If the surface can radiate happily into space through a transparent atmosphere, it will be less than that. This is what we see on Mars, for instance.
“You can justify T=0K at night perhaps in your mind, but not mine. The whole point of my original comment directed to no one in particular was to get lay people to think in more in terms of “thermal mass and inertia” and all the means of thermal transfer, rather than relying upon simplistic radiative vector algebra. Your accusation of “ignorance” on my part about the “buffering” effects of a thermalized air mass is as amusingly misplaced as your ideas about what convection does throughout the diurnal cycle.”
So again you would rather sneer than attempt to address the science. Do you deny that the temperature of the night sky into which planets radiate is close to absolute zero? Do you deny that a non-illuminated surface radiating into space would (in the absence of warming from above or below) cool to that very low temperature? Do you deny that the rate at which the temperature falls would depend upon the heat capacity of the cooling mass? Do you deny that increasing the mass, by coupling to sub-surface material by conduction, or a (transparent) atmosphere by convection, would reduce the rate of change of temperature, both during the cooling phase, and in the warming phase when the illumination is switched on again? Do you deny that this is buffering the temperature extremes? Do you deny that the daytime vertical convection processes I describe are real? (Hint, they’re called thermals).
Note that, at night, vertical convection in a (mostly) transparent atmosphere will cease, because the required temperature gradient is the wrong way round; the ground is cooling the atmosphere, no longer warming it; this relies upon conduction and radiation. So buffering the cooling then comes mainly from the heat capacity of the ground. Which is why continental climates have much colder nights than maritime climates (heat capacity of water versus soil, plus land-sea convection which allows the atmosphere to continue buffering during the night).

331. sky says:

Spector says:
August 11, 2010 at 8:45 pm
Of course radiation is the only energy ticket out to space. But it’s by no means the only ticket by which thermal energy is transfered from the surface to a level that is radiatively “seen” from space. IMO our understanding of the dynamics of moist convection and its consequences is even less than that of the “ice-locker” effect. In particular, our understanding of thermal capacitance and heat exchange in the realistic, cloud-filled troposphere is minimal.
The role of the bulk constituents in that heat exchange is usually minimized by trundling out backradiation spectra from the Antarctic or the Sahara and showing the prominent peak of 15 micron radiation over a narrow range of wavenumbers. But the thermal range of radiation extends from a fraction of a micron out to ~100microns, and even in those ultra dry environments the total (integrated) power associated with water-vapor is considerably more. The far infrared, where the bulk constituents (along with overlapping H2O) radiate in a continuum is almost never shown.
GHGs being strong IR radiators makes them excellent dispersers of thermal energy and very poor capacitors. The 24/7 nature of backradiation and the general absence of a diurnal cycle in-phase and coherent with the surface temperature (in marine environments) points to an atmospheric capacitance that GHGs suis generis cannot provide. Thus clouds and the bulk constituents are likely what provide the atmosphere the little heat retentivity that it posseses.

332. sky says:

Your clarification that you had T=0K in mind for the sink of space, rather than the surface, is helpful. It’s ironic, however, that I should be presented with a long list of elementary facts put up as strawmen for me to “deny” about the “buffer” effects of the atmosphere and the role of the diurnal cyle. Perhaps you’re unaware that it was I who first argued the absorption/capacitance/discharge viewpoint of the “greenhouse effect” here at WUWT. This ploy is what makes me think that a fruitful discussion is unlikely. I do have one question for you: do you deny that a poor absorber of IR can be heated mechanically (conduction, convection, eddy diffusion) to a temperature that is asymmetric with its IR absorption aloft?

333. sky says:
August 12, 2010 at 5:07 pm
“Your clarification that you had T=0K in mind for the sink of space, rather than the surface, is helpful. It’s ironic, however, that I should be presented with a long list of elementary facts put up as strawmen for me to “deny” about the “buffer” effects of the atmosphere and the role of the diurnal cyle. Perhaps you’re unaware that it was I who first argued the absorption/capacitance/discharge viewpoint of the “greenhouse effect” here at WUWT. This ploy is what makes me think that a fruitful discussion is unlikely. I do have one question for you: do you deny that a poor absorber of IR can be heated mechanically (conduction, convection, eddy diffusion) to a temperature that is asymmetric with its IR absorption aloft?”
First, you seem to have forgotten what I said at the beginning of my first comment to you:
“I would agree with you concerning the importance of convection as a heat transfer mechanism within the atmosphere.”
I have never disputed that.
Second, the T~0 (I never wrote T=0) was for the temperature of a completely unbuffered unilluminated surface, contrasted with the even black body temperature of a completely buffered surface. Those are the two extremes. There are real physical examples approaching both of them. Heavenly bodies such as the Earth and the Moon lie between those extremes. Different parts of the Earth are buffered more than others.
Third, I have put up no “strawmen”. I presented a list of simple statements for you to accept or deny because you had failed to make any rational argument to back up your sneers. You have still not answered those questions; you have still failed to put forward any rational argument concerning them. There is no “ploy” here on my part. I seriously do not know whether you accept this basic physics or are fostering some crank notion of your own. Your unwillingness to answer inclines me to the latter view. If this seems unfair, you can help me by answering the questions directly.
Fourth, I do not know what you mean by “absorption/capacitance/discharge viewpoint”. I would guess that it means that you don’t think CO2 can be a major player because there isn’t enough of it to hold much energy. If so, then I have already answered that point: “It is the strong absorbers that are also the strong emitters; the backradiation comes overwhelmingly from those strong absorbers (which are in very close to local thermodynamic equilibrium with the bulk gas)” (emphasis added). There is no requirement for the thermal energy and latent heat to be stored in the same molecular species that provide the IR absorption and emission. If this isn’t what you meant, please explain.
Fifth, in answer to your question, it is certainly the case that poor IR absorbers can be heated by other means. I don’t know what you mean by “a temperature that is asymmetric to its IR absorption aloft”. If its temperature is the same as the surface below, then it will radiate as much to that surface as it absorbs from that surface. If its temperature is less, then it will in general radiate less to the surface than it absorbs from the surface. Since it radiates both up and down, but is absorbing only from below (assuming no IR from above), then in the absence of additional heat transfer (by eg., conduction or convection), its absolute temperature would fall to ~0.84 Tsurface. This is true for both poor and good absorbers, so long as the optical depth is less than unity. When the optical depth is greater than unity, the topmost layer no longer clearly “sees” the surface, only the layer immediately below it, so the overall temperature ratio would be correspondingly greater. However, to maintain such a temperature difference, the pressure ratio would have to be 1.85:1 or greater (in a dry atmosphere). Unless the gas is a very poor radiator or is very sparse (ie, very transparent), conduction alone will not be sufficient and convection will occur in the lower levels. Note that this is ignoring the effect of any absorption of energy from external sources (eg., sunlight) in the upper atmosphere.

334. sky says:

Paul Birch says:
August 14, 2010 at 7:08 am
As someone who types very slowly and enjoys his weekends, I will not spend more time on a low-level discussion that grows increasingly more confused and OT relative to Tom Vonk’s clearly thought-provoking, high-level post.

335. sky says:
August 14, 2010 at 2:24 pm
“As someone who types very slowly and enjoys his weekends, I will not spend more time on a low-level discussion that grows increasingly more confused and OT relative to Tom Vonk’s clearly thought-provoking, high-level post.”
You have typed far more words wriggling out of answering my questions than you would have done simply answering them. The confusion is obviously on your side, not least in foolishly considering the debate off topic, but if there was anything you did not understand, you only had to ask. [snip]

336. Spector says:

RE: sky says: (August 12, 2010 at 4:24 pm )
“Of course radiation is the only energy ticket out to space. But it’s by no means the only ticket by which thermal energy is transfered from the surface to a level that is radiatively “seen” from space. IMO our understanding of the dynamics of moist convection and its consequences is even less than that of the “ice-locker” effect. In particular, our understanding of thermal capacitance and heat exchange in the realistic, cloud-filled troposphere is minimal.”
When I said “radiation is the only ticket out,” I was speaking *exclusively* of the tropopause altitude and above. This is the altitude where convection ceases to be an effective heat-transport factor. Also, there are very few clouds above this altitude. If the tropopause could not be cooled to a typical temperature of 220 deg K, the adiabatic lapse rate would translate the increase to equivalently warmer temperatures at the surface.
Heat transferred from the surface by radiation is already on that train unless it gets off at some intermediate stop.