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# Why the logarithm? (CO2 forcing)

Discussions of climate sensitivity tend to talk about "climate response to a doubling of greenhouse gases". More generally, the additional forcing is expressed in terms of a logarithm $$\Delta F = \alpha \ln (C/C_0)$$ where $\Delta F$ is forcing, $\alpha$ is a constant (the climate sensitivity), $C$ is the current GHG concentration and $C_0$ is a reference value.

What is the reason for the logarithm? Does it arise from some fundamental physical principle, or is it an empirical fit to modeling output? Any references to a derivation? The only one that I found is "Logarithmic Response and Climate Sensitivity of Atmospheric CO2", S.H. Lam, 2007 (see here), but there must be others.

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1.
edited December 2012

Thanks for the link. Since some time I've been having the intention to look into this and perhaps write a blog article about.

I thought the log-law follows from a calculation (though there's maybe some fit included because of the absorption spectrum) but why it's intuitively obvious is not clear (at least to me). I'll check your link.

Two years ago Nathan Urban gave a link here but that's already broken.

Comment Source:Thanks for the link. Since some time I've been having the intention to look into this and perhaps write a blog article about. I thought the log-law follows from a calculation (though there's maybe some fit included because of the absorption spectrum) but why it's intuitively obvious is not clear (at least to me). I'll check your link. Two years ago Nathan Urban gave a link [here](http://forum.azimuthproject.org/comments.php?DiscussionID=74&Focus=1474#Comment_1474) but that's already broken.
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2.

I uploaded a cached copy of the slides that you refer to.

Comment Source:I uploaded a [cached copy](https://dl.dropbox.com/u/101453219/radiative-transfer.pdf) of the slides that you refer to.
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3.

@Nathan, thanks!

Comment Source:@Nathan, thanks!
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4.

It may be simpler to understand the atmosphere as a whole. From observation, the pressure or density decreases approximately exponentially as you go up, and the temperature decreases approximately linearly, see lapse rate. So that gives a logarithmic relationship. You could then argue that CO2 behaves like the whole thing, when restricted to certain wavebands where CO2 is optically thick and the other gases are transparent.

The pressure curve is not hard to understand since pressure depends on the amount of air above. Why something as complicated as the atmosphere has a linear temperature gradient is a mystery to me.

Comment Source:It may be simpler to understand the atmosphere as a whole. From observation, the pressure or density decreases approximately exponentially as you go up, and the temperature decreases approximately linearly, see [lapse rate](http://en.wikipedia.org/wiki/Lapse_rate). So that gives a logarithmic relationship. You could then argue that CO2 behaves like the whole thing, when restricted to certain wavebands where CO2 is optically thick and the other gases are transparent. The pressure curve is not hard to understand since pressure depends on the amount of air above. Why something as complicated as the atmosphere has a linear temperature gradient is a mystery to me.
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5.
edited December 2012

@ Graham: to get the background temperature profile you can consider a small parcel of air, then use the ideal gas law, mechanical equilibrium and adiabatic /isentropic displacements (this is not really true, hence the 'approximately linear') -- these are three equations for three unknows -- and then you get the linear temperature gradient. This is only valid in the troposphere, in the stratosphere the temperature goes up because of ozone absorption. So the constant lapse rate does not take the greenhouse effect into account.

But I am not sure if your argument is applicable (perhaps I understand it incorrectly) because I thought that the temperature rise related to the doubling of carbon dioxide in the atmosphere is for the average temperature at the earth's surface. I don't see how the change in average surface temperature can be related to the linear decrease of the temperature with height in the troposphere, the former relates to the greenhouse effect, the latter to insentropic profiles for ideal gases.

Comment Source:@ Graham: to get the background temperature profile you can consider a small parcel of air, then use the ideal gas law, mechanical equilibrium and adiabatic /isentropic displacements (this is not really true, hence the 'approximately linear') -- these are three equations for three unknows -- and then you get the linear temperature gradient. This is only valid in the troposphere, in the stratosphere the temperature goes up because of ozone absorption. So the constant lapse rate does not take the greenhouse effect into account. But I am not sure if your argument is applicable (perhaps I understand it incorrectly) because I thought that the temperature rise related to the doubling of carbon dioxide in the atmosphere is for the average temperature at the earth's surface. I don't see how the change in average surface temperature can be related to the linear decrease of the temperature with height in the troposphere, the former relates to the greenhouse effect, the latter to insentropic profiles for ideal gases.
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edited December 2012

James Hansen explains the logithmic response as a physical phenomenon due to saturation. He plots response on a log-log scale against frequency. Lam says his derivation is analytic so that leaves me lost.

Comment Source:James Hansen [explains](http://www.columbia.edu/~jeh1/mailings/2011/20110415_EnergyImbalancePaper.pdf) the logithmic response as a physical phenomenon due to saturation. He plots response on a log-log scale against frequency. Lam says his derivation is analytic so that leaves me lost.
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7.
edited December 2012

I've read Lam, 2007 and it seems to me that the log-response basically follows from a linear Taylor expansion of "the logarithm of the coarse-grained-averaged absorption spectrum" in function of the wavelength, near the edges of the major band of the carbon dioxide absorption spectrum (centered around 15 micron).

The edges of the absorption spectrum matter because of the saturation effect: before the rise in carbon dioxide a certain (to be chosen appropriately) wavelength is still "optically thin", i.e. radiation passes through easily, and it then shifts to "optically thick", i.e. most radiation is blocked, when carbon dioxide levels rise (n.b. this thin/thick binary behavior is of course a simplified view). The middle of the absorption band doesn't matter because it was already optically thick from the beginning (i.e. the saturation). When carbon dioxide levels rise more and more, more and more channels become saturated and thus the temperature response becomes less sensitive.

The article has a funny sentence: "This picture is understood and is generally accepted."

So I would consider the log-response as the combination of a physical picture (the saturation) and general mathematics (Taylor first order), although perhaps I should additionally look up why the absorption spectrum looks exponential in wavelength near the edges of a band, that sounds physical too.

Comment Source:I've read Lam, 2007 and it seems to me that **the log-response basically follows from a linear Taylor expansion** of "the logarithm of the coarse-grained-averaged absorption spectrum" in function of the wavelength, near the edges of the major band of the carbon dioxide absorption spectrum (centered around 15 micron). **The edges of the absorption spectrum matter because of the saturation effect**: before the rise in carbon dioxide a certain (to be chosen appropriately) wavelength is still "optically thin", i.e. radiation passes through easily, and it then shifts to "optically thick", i.e. most radiation is blocked, when carbon dioxide levels rise (n.b. this thin/thick binary behavior is of course a simplified view). The middle of the absorption band doesn't matter because it was already optically thick from the beginning (i.e. the saturation). When carbon dioxide levels rise more and more, more and more channels become saturated and thus the temperature response becomes less sensitive. The article has a funny sentence: "This picture is understood and is generally accepted." So I would consider the log-response as the combination of a physical picture (the saturation) and general mathematics (Taylor first order), although perhaps I should additionally look up why the absorption spectrum looks exponential in wavelength near the edges of a band, that sounds physical too.
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8.

John Roe wrote:

What is the reason for the logarithm?

I don't understand it. My impression is that it's 1) only approximate, 2) not very easy to explain, 3) not understood by many people, 4) deserving of a blog post!

Can I coax you to write one? Or anyone else here?

Comment Source:John Roe wrote: > What is the reason for the logarithm? I don't understand it. My impression is that it's 1) only approximate, 2) not very easy to explain, 3) not understood by many people, 4) deserving of a blog post! Can I coax you to write one? Or anyone else here?
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9.
edited December 2012

(1) yes, (2) no (well, maybe I'm wrong), (3) yes (well, at least for scientists like us outside of radiation physics) (4) yes

I'm a candidate, it's winter and I need to collect material for my lost bet.

Comment Source:(1) yes, (2) no (well, maybe I'm wrong), (3) yes (well, at least for scientists like us outside of radiation physics) (4) yes I'm a candidate, it's winter and I need to collect material for my lost bet.
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10.

The edges of the absorption spectrum matter because of the saturation effect:

There were some moves to do something like flattening these edges which I found completely incomprehensible.

Comment Source:> The edges of the absorption spectrum matter because of the saturation effect: There were some moves to do something like flattening these edges which I found completely incomprehensible.
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11.

Frederik wrote:

I’m a candidate, it’s winter and I need to collect material for my lost bet.

GREAT! I don't know what winter has to do with it...

...but I would love to see a good explanation of this logarithmic effect on the Azimuth blog, so that would be a wonderful way to start paying off your bet concerning neutrinos.

Comment Source:Frederik wrote: > I’m a candidate, it’s winter and I need to collect material for my lost bet. GREAT! I don't know what winter has to do with it... ...but I would love to see a good explanation of this logarithmic effect on the Azimuth blog, so that would be a wonderful way to start paying off your [bet concerning neutrinos](http://johncarlosbaez.wordpress.com/2011/10/05/a-bet-concerning-neutrinos-part-2/). (Link provided for people who don't know what we're talking about.)
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12.
• Jim wrote:

There were some moves to do something like flattening these edges

Are you referring to the Lam or the Hansen paper? (I have only read half of the former and none of the latter yet, sorry for being lazy and asking before I read the Hansen paper fully, which I should do anyway)

• John wrote:

GREAT!

Ok, I'll start up a draft.

I don’t know what winter has to do with it…

In winter my mean free-time presence migrates indoors, so I'm closer to my laptop.

Comment Source:* Jim wrote: > There were some moves to do something like flattening these edges Are you referring to the Lam or the Hansen paper? (I have only read half of the former and none of the latter yet, sorry for being lazy and asking before I read the Hansen paper fully, which I should do anyway) * John wrote: > GREAT! Ok, I'll start up a draft. > I don’t know what winter has to do with it… In winter my mean free-time presence migrates indoors, so I'm closer to my laptop.
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13.
edited December 2012

Correction: from reading the slides uploaded by Nathan it seems that Graham's intuition is correct!

Before I replied to Graham:

I don’t see how the change in average surface temperature can be related to the linear decrease

but what I didn't capture was the "optically thick/thin" argument, even though Graham hinted at it.

I now believe it goes like this: the radiative equilibrium is for an effective temperature (let's call it like this) somewhere in the atmosphere, above which the atmosphere can be considered thin for the important wavelengths. When carbon dioxide doubles, this effective temperature level shifts upwards (with a log relation wrt the carbon dioxide concentration because that scales with pressure, which drops exponentially) towards a colder temperature (in non-equilibrium). In equilibrium this temperature at this higher level should become the same temperature as the previous (before the doubling) equilibrium temperature at the lower level and because the temperature in the atmosphere is linearly related to the temperature at the ground, we have our logarithmic response.

This sounds much more physical than my previous invoking of a Taylor expansion, and I discard my comment 8. Of course I'll have to read Lam more closely to see how this previous picture fits with his explanation (or if it doesn't fit, where's the error) but at least the saturation effect plays a role (determines the level where the atmosphere becomes thin).

Anyway, don't worry about apparent inconsistencies, it's work in progress and when I write the blog article I will have figured it out! But let's make comment 10, (2) yes ;-)

I think I should precede the log-article with an article about atmospheric concentration and lapse rate, to provide the reader with the necessary background.

Comment Source:Correction: from reading the slides uploaded by Nathan **it seems that Graham's intuition is correct!** Before I replied to Graham: > I don’t see how the change in average surface temperature can be related to the linear decrease but what I didn't capture was the "optically thick/thin" argument, even though Graham hinted at it. I now believe it goes like this: **the radiative equilibrium is for an effective temperature** (let's call it like this) somewhere in the atmosphere, **above which the atmosphere can be considered thin** for the important wavelengths. **When carbon dioxide doubles, this effective temperature level shifts** upwards (**with a log relation** wrt the carbon dioxide concentration because that scales with pressure, which drops exponentially) towards a colder temperature (in non-equilibrium). In equilibrium this temperature at this higher level should become the same temperature as the previous (before the doubling) equilibrium temperature at the lower level and because **the temperature in the atmosphere is linearly related to the temperature at the ground**, we have our logarithmic response. This sounds much more physical than my previous invoking of a Taylor expansion, and I discard my comment 8. Of course I'll have to read Lam more closely to see how this previous picture fits with his explanation (or if it doesn't fit, where's the error) but at least the saturation effect plays a role (determines the level where the atmosphere becomes thin). Anyway, don't worry about apparent inconsistencies, it's work in progress and when I write the blog article I will have figured it out! But let's make comment 10, (2) yes ;-) I think I should precede the log-article with an article about atmospheric concentration and lapse rate, to provide the reader with the necessary background.
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14.

Can these explanations of the approximate log-dependence for CO2 also explain the approximate square-root-dependence for methane (at the current very low concentrations ~2 ppm)? Or would they give an incorrect log dependence?

Comment Source:Can these explanations of the approximate log-dependence for CO2 also explain the approximate square-root-dependence for methane (at the current very low concentrations ~2 ppm)? Or would they give an incorrect log dependence?
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15.

Or would they give an incorrect log dependence?

I'd like to answer that, but my atmospheric chemistry and spectroscopic knowledge has many holes...

• Is methane well-mixed? This is needed to invoke the 'scales with pressure' argument.

• How about the optical thickness in methane's wavelengths?

I'll try to find an answer to this too. If not, perhaps it's a nice exercise to end the blog article with.

Comment Source:> Or would they give an incorrect log dependence? I'd like to answer that, but my atmospheric chemistry and spectroscopic knowledge has many holes... * Is methane well-mixed? This is needed to invoke the 'scales with pressure' argument. * How about the optical thickness in methane's wavelengths? I'll try to find an answer to this too. If not, perhaps it's a nice exercise to end the blog article with.
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16.

Methane is well-mixed. I think the square-root dependence arises from its low concentration; my impression is that the logarithmic law generically transitions to a square root law at low concentrations, but it could be something methane-specific.

Comment Source:Methane is well-mixed. I think the square-root dependence arises from its low concentration; my impression is that the logarithmic law generically transitions to a square root law at low concentrations, but it could be something methane-specific.
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17.
edited December 2012

Frederik boldly wrote:

I now believe it goes like this: the radiative equilibrium is for an effective temperature (let's call it like this) somewhere in the atmosphere, above which the atmosphere can be considered thin for the important wavelengths. When carbon dioxide doubles, this effective temperature level shifts upwards (with a log relation wrt the carbon dioxide concentration because that scales with pressure, which drops exponentially) towards a colder temperature (in non-equilibrium). In equilibrium this temperature at this higher level should become the same temperature as the previous (before the doubling) equilibrium temperature at the lower level and because the temperature in the atmosphere is linearly related to the temperature at the ground, we have our logarithmic response.

This is great! Nothing I'd read had made it sound so simple. Of course there's an 'Achilles heel' to this simplicity, pointed out by Graham: why should the temperature in the atmosphere vary about linearly with height? But that's okay. If this approximate linearity has no simple explanation, we've at least pushed back the problem one step. And if it has some simple explanation, that would be great!

I want to start learning more about this stuff. Raymond Pierrehumbert gave a really great talk at the AGU, which should eventually appear online - I'll post about it on the blog when it does. In it, he mentioned his book Principles of Planetary Climate, and he said someone described it as "death by theoretical physics". That could be good.

It has rave reviews on Amazon, and includes exercises in Python.

At the AGU someone told me a book that could serve as an introduction to climate science, which has the word "radiative" in the title. It could be

• Gary E. Thomas and Knut Stamnes, Radiative Transfer in the Atmosphere and Ocean, Cambridge U. Press, 2002.

I don't know! Can anyone guess?

One interesting thing Pierrehumbert pointed out is that the 'standard model' of the Earth's atmosphere is similar to an upside-down version of the standard model of the Sun's upper atmosphere, which gets hotter as you go down, and has a convective zone sitting on top of a nonconvective zone.

Comment Source:Frederik boldly wrote: > I now believe it goes like this: **the radiative equilibrium is for an effective temperature** (let's call it like this) somewhere in the atmosphere, **above which the atmosphere can be considered thin** for the important wavelengths. **When carbon dioxide doubles, this effective temperature level shifts** upwards (**with a log relation** wrt the carbon dioxide concentration because that scales with pressure, which drops exponentially) towards a colder temperature (in non-equilibrium). In equilibrium this temperature at this higher level should become the same temperature as the previous (before the doubling) equilibrium temperature at the lower level and because **the temperature in the atmosphere is linearly related to the temperature at the ground**, we have our logarithmic response. This is great! Nothing I'd read had made it sound so simple. Of course there's an 'Achilles heel' to this simplicity, pointed out by Graham: why should the temperature in the atmosphere vary about linearly with height? But that's okay. If this approximate linearity has no simple explanation, we've at least pushed back the problem one step. And if it has some simple explanation, that would be great! I want to start learning more about this stuff. Raymond Pierrehumbert gave a really great talk at the AGU, which should eventually appear online - I'll post about it on the blog when it does. In it, he mentioned his book _Principles of Planetary Climate_, and he said someone described it as "death by theoretical physics". That could be good. <img src = "http://math.ucr.edu/home/baez/emoticons/tongue2.gif" alt = ""/> It has [rave reviews on Amazon](http://www.amazon.com/Principles-Planetary-Climate-Raymond-Pierrehumbert/dp/0521865565/ref=sr_1_1?s=books&ie=UTF8&qid=1354918732&sr=1-1&keywords=pierrehumbert+planetary), and includes exercises in Python. At the AGU someone told me a book that could serve as an introduction to climate science, which has the word "radiative" in the title. It could be * Gary E. Thomas and Knut Stamnes, _Radiative Transfer in the Atmosphere and Ocean_, Cambridge U. Press, 2002. I don't know! Can anyone guess? One interesting thing Pierrehumbert pointed out is that the 'standard model' of the Earth's atmosphere is similar to an _upside-down_ version of the standard model of the Sun's upper atmosphere, which gets hotter as you go _down_, and has a convective zone sitting _on top_ of a nonconvective zone.
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18.
edited December 2012

why should the temperature in the atmosphere vary about linearly with height?

this is fine, I vaguely described above how one can understand it (mechanical equilibrium, ideal gas law, isentropic profile). In the blog post I'll work it out carefully.

told me a book that could serve as an introduction to climate science, which has the word “radiative” in the title

I don't know that book. Btw, most introductory books without the word radiative have an introductory chapter about radiation transfer too, but they don't consider the log-relation.

Of course I’ll have to read Lam more closely to see how this previous picture fits with his explanation

This is now fine too in principle (the shape of the absorption cross-section follows from the same principles that govern the log response) I only need to work it out. [Edit: if it sounded like your crackpot index, it's really coincidental!]

Comment Source:> why should the temperature in the atmosphere vary about linearly with height? this is fine, I vaguely described above how one can understand it (mechanical equilibrium, ideal gas law, isentropic profile). In the blog post I'll work it out carefully. > told me a book that could serve as an introduction to climate science, which has the word “radiative” in the title I don't know that book. Btw, most introductory books without the word radiative have an introductory chapter about radiation transfer too, but they don't consider the log-relation. About my later comment: > Of course I’ll have to read Lam more closely to see how this previous picture fits with his explanation This is now fine too in principle (the shape of the absorption cross-section follows from the same principles that govern the log response) I only need to work it out. [Edit: if it sounded like your crackpot index, it's really coincidental!]
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19.

Frederik, you seem to have made more sense of my vague intuition than I could! Thanks for the explanation about the temperature variation with height. I'm happy to accept that a calculation can be done for heat transfer by convection, but what about other mechanisms? I guess conduction is too small to worry about. I have little idea about radiation. The one that worries me is water, as vapour, droplets, ice crystals. It seems that might transfer a lot of heat a long way in a nonlinear kind of way.

On the approximate square-root-dependence for methane. For extremely low concentrations of a gas, the dependence is linear - at least that's been said in earlier discussions. My guess would be that methane is in betwen the linear regime and the logarithmic one, and a square root approximation happens to work well. I wonder if there's a simple formula like log(1+x) that works for all concentrations.

Comment Source:Frederik, you seem to have made more sense of my vague intuition than I could! Thanks for the explanation about the temperature variation with height. I'm happy to accept that a calculation can be done for heat transfer by convection, but what about other mechanisms? I guess conduction is too small to worry about. I have little idea about radiation. The one that worries me is water, as vapour, droplets, ice crystals. It seems that might transfer a lot of heat a long way in a nonlinear kind of way. On the approximate square-root-dependence for methane. For extremely low concentrations of a gas, the dependence is linear - at least that's been said in earlier discussions. My guess would be that methane is in betwen the linear regime and the logarithmic one, and a square root approximation happens to work well. I wonder if there's a simple formula like log(1+x) that works for all concentrations.
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I’m happy to accept that a calculation can be done

Are you referring to the calculation of the log-response? This is actually for radiation. The linear temperature background is a static background (no convection) and the other part of the calculation is about the radiative properties of a well-mixed trace gas in the atmosphere.

Heat conduction doesn't play a role in the atmosphere.

Convection and latent heat transfer can be large, but they only redistribute heat in the troposphere, the heat isn't lost to outer space. Of course you could argue that the temperature distribution in the troposphere is altered by these forms of heat transfer, but this is a local temperature alteration of the background temperature profile. I would guess that the final temperature in the stratosphere is not altered, because most evaporated water (and phase transitions) stays in the troposphere (and eventually precipitates) so the temperature profile in the stratosphere is not altered by convection and latent heat transfer. Convection and evaporation happen because of solar radiation on the surface (which absorbs radiation much better than air) so one would also expect that these phenomena stay largely confined to the surface (the boundary layer or at most the whole troposphere).

On the approximate square-root-dependence for methane [...]

Interesting suggestion. I will think about it.

Comment Source:> I’m happy to accept that a calculation can be done Are you referring to the calculation of the log-response? This is actually for radiation. The linear temperature background is a static background (no convection) and the other part of the calculation is about the radiative properties of a well-mixed trace gas in the atmosphere. Heat conduction doesn't play a role in the atmosphere. Convection and latent heat transfer can be large, but they only redistribute heat in the troposphere, the heat isn't lost to outer space. Of course you could argue that the temperature distribution in the troposphere is altered by these forms of heat transfer, but this is a local temperature alteration of the background temperature profile. I would guess that the final temperature in the stratosphere is not altered, because most evaporated water (and phase transitions) stays in the troposphere (and eventually precipitates) so the temperature profile in the stratosphere is not altered by convection and latent heat transfer. Convection and evaporation happen because of solar radiation on the surface (which absorbs radiation much better than air) so one would also expect that these phenomena stay largely confined to the surface (the boundary layer or at most the whole troposphere). > On the approximate square-root-dependence for methane [...] Interesting suggestion. I will think about it.
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21.

Are you referring to the calculation of the log-response?

No, it is just that

to get the background temperature profile you can consider a small parcel of air,

sounded like convection to me.

Comment Source:> Are you referring to the calculation of the log-response? No, it is just that > to get the background temperature profile you can consider a small parcel of air, sounded like convection to me.
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22.

I think you can consider it as "virtual convection" to determine the equilibrium profile.

Comment Source:I think you can consider it as "virtual convection" to determine the equilibrium profile.
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23.

Frederick, I am excited to see your exposition when it appears. Thanks for picking up on my question here.

Comment Source:Frederick, I am excited to see your exposition when it appears. Thanks for picking up on my question here.
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24.

I'll try to finish it before Christmas/New Year. That is, I'll write an "intuitive" exposition (one-wavelength discussion) first.

Later I'd like to extend that to all wavelengths (in a correct manner) for which I need more one additional piece, and I also want to take a closer look at some "misconceptions". But if I split those parts from the "intuitive" exposition I can already finish something in the next weeks.

Comment Source:I'll try to finish it before Christmas/New Year. That is, I'll write an "intuitive" exposition (one-wavelength discussion) first. Later I'd like to extend that to all wavelengths (in a correct manner) for which I need more one additional piece, and I also want to take a closer look at some "misconceptions". But if I split those parts from the "intuitive" exposition I can already finish something in the next weeks.
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25.

Frederik wrote:

I’ll try to finish it before Christmas/New Year.

Okay, so have you written it yet? (You didn't say which Christmas.)

I really want to get this piece finished. Maybe I'll have to finish it myself.

Comment Source:Frederik wrote: > I’ll try to finish it before Christmas/New Year. Okay, so have you written it yet? (You didn't say _which_ Christmas.) I really want to get this piece finished. Maybe I'll have to finish it myself.
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26.
edited January 2014

Last year I meant last year's Christmas, which was a complete failure.

[When I wrote that it was actually around Christmas 2012: that's already two calendar years ago...]

Actually I started editing again yesterday.

Maybe I’ll have to finish it myself.

Perhaps this is a good way to add some pressure on me. But if you think I'm going too slow and you really want to finish it faster, don't hesitate to ask me questions if you think I could help.

(maybe the best way to speed my writing up would be to ask me questions about what I have written, messy as it is)

Comment Source:Last year I meant last year's Christmas, which was a complete failure. [When I wrote that it was actually around Christmas 2012: that's already **two calendar years ago**...] Actually I started editing again yesterday. > Maybe I’ll have to finish it myself. Perhaps this is a good way to add some pressure on me. But if you think I'm going too slow and you really want to finish it faster, don't hesitate to ask me questions if you think I could help. (maybe the best way to speed my writing up would be to ask me questions about what I have written, messy as it is)
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27.

Basically what I should add to the notes of last year is that only those wavelengths which are not optically thick yet matter. Since carbon dioxide is already present in the atmosphere, this is only at the edges of the absorption spectrum (and hence the more CO2, the less it matters). For gases of low concentration the response should be stronger, e.g. methane, but why linear (as Graham wrote) or a square root (as Nathan wrote) I can't answer yet.

And I have to clean up all the mess.

Comment Source:Basically what I should add to the notes of last year is that only those wavelengths which are not optically thick yet matter. Since carbon dioxide is already present in the atmosphere, this is only at the edges of the absorption spectrum (and hence the more CO2, the less it matters). For gases of low concentration the response should be stronger, e.g. methane, but why linear (as Graham wrote) or a square root (as Nathan wrote) I can't answer yet. And I have to clean up all the mess.
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28.
edited January 2014

I will start by editing Blog - the log forcing. That will make me ask questions. I would like all 3 parts to be done by June, when my sabbatical ends! This stuff is very important and the world needs a good clear explanation.

I will probably bring Pierrehumbert's book on climate science with me to Germany. He discusses this stuff, so it will be helpful.

Comment Source:I will start by editing [[Blog - the log forcing]]. That will make me ask questions. I would like all 3 parts to be done by June, when my sabbatical ends! This stuff is very important and the world needs a good clear explanation. I will probably bring Pierrehumbert's book on climate science with me to Germany. He discusses this stuff, so it will be helpful.
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29.

If the concentration of some gas is optically thin enough at some wavelength w, and the rest of the atmosphere is transparent for wavelength w, then you can ignore the possibility of a photon wavelength w being emitted and re-absorbed by the gas. Then it seems obvious that the response is linear: double the concentration and you double the (very small) probability of it being absorbed once.

Comment Source:If the concentration of some gas is optically thin enough at some wavelength w, and the rest of the atmosphere is transparent for wavelength w, then you can ignore the possibility of a photon wavelength w being emitted and re-absorbed by the gas. Then it seems obvious that the response is linear: double the concentration and you double the (very small) probability of it being absorbed once.
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30.

Graham said:

Then it seems obvious that the response is linear

Yes, this is very reasonable (and physically much more meaninful than my own "close to the expansion point everything has a linear approximation") but I was also thinking how one could get a square root.

Comment Source:Graham said: > Then it seems obvious that the response is linear Yes, this is very reasonable (and physically much more meaninful than my own "close to the expansion point everything has a linear approximation") but I was also thinking how one could get a square root.