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Discussion > That CO2 thing again..

It's not at all wrong for a long-term BH person to have different concerns to a recent visitor. But I've also said I agree with you. Why don't you give the subject a rest yourself and go back to some of the scientific points on which BH people dropped the ball on Sunday?

Aug 12, 2014 at 12:17 PM | Registered CommenterRichard Drake

"Rob also hasn't bothered to respond to SoD's excellent critique of his position - it's now over 48 hours of silence from Rob on that."

I assume I missed something, I don't look at this blog every day, so I can easily lose longish threads. I'll go and check.

Aug 12, 2014 at 2:11 PM | Unregistered CommenterRob Burton

I still don't really understand.

Under equilibrium temperature surely OLR is equal to Incoming Solar radiation (how do GHG's affect this balance?)

Aug 12, 2014 at 3:04 PM | Unregistered CommenterRob Burton

Rob,

GHGs don't affect that balance. The balance of the overall system is unchanged : energy in = energy out. if it wasn't we'd be slowly heating or freezing.

What changes is that adding GHGs makes it harder for outgoing LW to escape. The system is not able to lose the energy it must to stay at the balance, so the balance is upset - the temperature rises. As the temperature of the earth system rises, it releases more LW radiation (hotter things emit more energy).

Eventually the energy of LW escaping the system again equals the incoming SW and we are again at the balance.

But the earth is now at an increased temperature in order to supply all those extra LW photons to get around the 'blockage' that the extra GHG provides.

Aug 12, 2014 at 3:37 PM | Unregistered CommenterTheBigYinJames

And the warming to get it to the increased temperature was done by the incoming solar radiation - not by back radiation.

Aug 12, 2014 at 5:49 PM | Registered CommenterMartin A

And if two CO2 molecules are added, one on the day side and one on the night side the imbalance lasts for;

A nonosecond?
Ages and ages?

Aug 12, 2014 at 6:39 PM | Unregistered Commenterssat

And if the Earth is now at an increased temperature as TBYJ states, how does that fit with the sink temperature of space and S-B?

Aug 12, 2014 at 6:44 PM | Unregistered Commenterssat

ssat

The imbalance lasts for as long as it takes the earth system to heat up to a temperature where the incoming radiation is matched by outgoing radiation once again. Trying to reduce the problem to single molecules is a pointless exercise, since the temperature effect would be infinitesimal for such tiny numbers. The point is that the more absorbers you add, the more LW gets blocked* and the hotter the earth system has to become** to re-establish the balance.

* albeit the effect diminishes logarithmically.
** this is not some mysterious force, the sun is heating the earth, if we stop the earth cooling as fast, it gets hotter

How this fits with the sink temperature of space and SB is straightforward.

The sink temperature of space just means that we only need to worry about solar radiation, since deep space is so cold it contributes a negligible amount to the 'in' side of the equation. To all effects and purposes, it's only the sun and the earth.

SB only really applies to black-bodies, and the atmosphere isn't one, it's comprised of a small number gases, only some of which can radiate IR. (The ground is a pretty good black body, since it's made up of a mixture of elements and molecules covering the range of spectral emissions.)

This means when it gets hotter, the air can only cool to space by emitting more photons rather than shifting the energy of the photons, so the SB curve doesn't really apply, except in a limited way (3-atom molecules have more than one IR frequency available due to the various bond vibration modes discussed earlier) GHG molecules in the atmosphere release an IR photon, but are quickly re-excited again by kinetic collision in a hotter air, and can then re-emit more quickly than in a cooler gas.

Hope this helps.

Aug 12, 2014 at 7:07 PM | Unregistered CommenterTheBigYinJames

And if the Earth is now at an increased temperature as TBYJ states, how does that fit with the sink temperature of space and S-B?
Aug 12, 2014 at 6:44 PM | ssat

I originally made a wrong assumption about how the 2.7K comes into it.

SoD put us all right with

Globally annually averaged solar radiation absorbed by the earth: 240 W/m^2
Globally annually averaged deep space radiation absorbed by the earth: 0.000003 W/m^2

Making the assumption that the Earth's effective temperature in the absence of deep space radiation would be 252K due the Sun's radiation (and assuming SB applies) , I worked out what would be its temperature with the 2.7K background adding to the total energy received by the Earth. The increase in temperature was very small - something like a few millionths of a degree. [I don't have my working to hand so I might be an order of magnitude or two out but it was something like that]

So the background radiation makes no difference to anything of any importance.

Aug 12, 2014 at 9:02 PM | Registered CommenterMartin A

Rob,

"Under equilibrium temperature surely OLR is equal to Incoming Solar radiation (how do GHG's affect this balance?)"

To illuminate the problem I could dramatically over-simplify it to attempt to deal with the conceptual problem. This could be dismissed as a dramatic over-simplification, but maybe it will help.

Where is the "boundary" that is emitting energy to deep space? Is this at the same location as the "boundary" that is absorbing solar radiation (the surface)?

No. The boundary that is emitting energy to deep space is y km up in the atmosphere. The "effective radiating temperature" of this layer = 255K because it must emit 240 W/m^2 - otherwise the planet is warming or cooling.

The question then becomes, what is the temperature difference between the surface and this boundary? It is set by the lapse rate, L, so the boundary radiating to space has a temperature of 255K and the surface has a temperature, Ts = 255 + Ly

If we increase the GHG concentration we move the boundary up to y' km. So the new surface temperature Ts' = 255 + Ly'.

So the temperature increase, with a fixed lapse rate (ie no feedbacks), ΔTs = L(y'-y).

Over-simplification finished.

The key point is it isn't the surface that is radiating to space. A bit of surface radiation escapes but mostly a whole range of heights in the atmosphere. The atmosphere is mostly cooler than the surface.

Another way to think about it - more realistic but maybe won't snap that conceptual block..

if you do a calculation of current conditions (e.g. a range of typical atmospheric temperature profiles and GHG concentrations) with the radiative transfer equations and find OLR=240 W/m^2 - then you increase GHG concentration OLR drops.

So now the climate system must warm because incoming is greater than outgoing.

Don't know whether that helped.

Aug 12, 2014 at 9:43 PM | Unregistered CommenterScience of Doom

Thanks for the explanation SoD it makes sense. I just don't quite see how knowing the radiating boundary height tells you anything about the surface temperature past using the simplified explanation you explained very well. What do you think about something like Miscolski's model and observations of constant optical depth I believe??

Aug 12, 2014 at 10:17 PM | Unregistered CommenterRob Burton

Aug 12, 2014 at 3:37 PM | TheBigYinJames

I understand the simplistic explantion. So what is the Earth's temperature in the limiting case of being totally opaque to IR?? assuming surface pressure is the same.

Aug 12, 2014 at 10:30 PM | Unregistered CommenterRob Burton

Rob,

"..it makes sense.." That's awesome.

"..just don't quite see how knowing the radiating boundary height tells you anything about the surface temperature past using the simplified explanation you explained very well.."

In practice it's much more complicated. The point of the illustration is to help people understand why radiative balance doesn't mean the surface temperature stays constant.

"What do you think about something like Miscolski's model and observations of constant optical depth I believe??.."

I covered Miscolski at length. In brief, he has two parts. One is theory, one is experiment.

The theory part is a dog's breakfast, derived from invented laws of thermodynamics.

The experimental part is quite questionable. Radiosonde data going back into the 1950s has a lot of long term biases, some well known, others not clear - explained in detail in Water Vapor Trends.

Also to cite my conclusion from Part Six which I'm sure almost no one actually makes it to:

"If (and only if) water vapor has canceled out CO2 increases, then the increase in optical thickness due to these other gases (methane, nitrous oxide plus halocarbons) has destroyed the idea that optical thickness can be considered to be constant.

Of course, my calculations are rudimentary. My model is much less exact than the HARTCODE model used by Miskoczi and it would be interesting to see his results reproduced in full with the correct concentrations of all of the GHGs from 1948 – 2008.

As I commented earlier – this is one of the least important of the criticisms of Ferenc Miskolczi’s papers."

Aug 12, 2014 at 11:12 PM | Unregistered CommenterScience of Doom

Rob,

You asked: "So what is the Earth's temperature in the limiting case of being totally opaque to IR?? assuming surface pressure is the same."

The information in the question doesn't quite allow an answer.

Let me explain. Suppose our gas absorbs all the way through the spectrum of interest - say 4 μm - 100 μm and is constant with wavelength. This is often called the "semi-gray" approximation: "semi" because the gas is still transparent to solar radiation.

So there is an optical thickness, τ which equals the absorption coefficient x the number of molecules. Absorptivity = 1-exp(-τ).

Let's say optical thickness = 10, absorptivity = 1.0000. Which is "black" - a perfect absorber and a perfect emitted, in this part of the spectrum anyway. Just in case anyone wants to be picky, if we look a few more decimal places we will find it's actually less than 1. But within any kind of practical value for atmospheric physics, absorptivity = 1. (And for comparison if τ=5, absorptivity = 0.9933)

But, look back at where optical thickness comes from - it depends on the number of molecules, N, for a given absorption coefficient.

Are we saying τ =10 for a 10km thickness or a 1mm thickness? Of course, the number of molecules reduces as you go up, so you actually need to specify the path length and pressure (or just the number of molecules).

Because temperature changes with height, first reducing through the troposphere and then increasing through the stratosphere you would need to know the optical thickness as a function of N to be able to work out the resulting temperature.

It's an interesting question though and I might play with the Matlab RTE model I built to see what happens.

Aug 12, 2014 at 11:29 PM | Unregistered CommenterScience of Doom

TBYJ
Thanks for the detailed response. The 2 molecule example was theoretical and may have been better stated as n+2. The question was how long any warming effect (manifest as an increased temperature) would last. If I understand you correctly you are saying it would be as long as n+2 pertained and that n+4 would add to that (albiet diminishing logarithmically).

TBYJ & Marin A
Thanks again but I am at odds with you both here on the effect of the sink. Calculating the back radiation from the 2.7 degK as SoD did earlier in the thread... you can probably see why I have a problem with that? As I see it, at condition n+2 and radiative balance the temperature of Earth will be the same as at condition n. This is because the sink is infinite and it is the difference between the two temperatures that S-B seeks/demands to be maintained. The temperature of Earth is not set by the solar constant but by the sink temperature - if the sink temperature was 100 degK Earth's temperature would be higher. Invoking grey-body as a modifier negating S-B doesn't change the result, only the numbers.

SoD
I have absolutely no problem with your 'over-simplification' - increasing CO2 concentration will lift the height of the effective radiating temperature and which itself remains unchanged. I have a problem with the conclusion that, via the lapse rate, the surface temperature rises. My conclusion would have been that the bottom of atmosphere temperature rises. (On your site you state "If the place of emission of radiation – on average – moves upward for some reason then the intensity decreases. Why? Because it is cooler the higher up you go in the troposphere." I am not sure how that fits with your over-simplified description - perhaps I am misundersatanding 'intensity').

I'll jump ahead of any comments that may result from the above and state what I believe occurs. Increasing the concentration of CO2 raises the effective height of emission but its temperature remains the same as demanded by S-B, the BOA temperature increases as it is coupled by the lapse rate, the energy in the system remains constant, the actual surface cools to radiate less by an ammount that balances the increased radiating area of the atmosphere.

NiV, on an earlier derailed thread, urged me to forget the physics and contest the feedbacks. I think that may be premature even though the bulk of the alarm lies there. If it can be pointed out where I have gone wrong in my conclusion, I would be happy to take his advice.

Aug 13, 2014 at 8:42 AM | Unregistered Commenterssat

ssat:

NiV, on an earlier derailed thread, urged me to forget the physics and contest the feedbacks.

Do you mean this thread begun by Dung on 10th May? I'd be interested indeed if you could point us to the place, and precise words, where NiV urged you to "forget the physics". I feel I remember the thread well but I sure don't remember that.

Aug 13, 2014 at 8:59 AM | Registered CommenterRichard Drake

ssat,

Perhaps it's ok first to comment on your 2nd reply to others, because that seems to be the "biggest target", not in an attacking kind of way, but in a "try and point out what might be flaws in your thinking" kind of way..

"The temperature of Earth is not set by the solar constant but by the sink temperature - if the sink temperature was 100 degK Earth's temperature would be higher."

The temperature of "the climate system" - not necessarily the surface of the earth - is set by the total absorbed radiation.

In the case of 2.7K we saw that we had to add 0.000003 W/m^2 to 240 W/m^2 and obviously we don't know the solar value to anything like that kind of accuracy so we can ignore it.

In the case of say 50K the value will be 0.35 W/m^2 and in the case of 100K the value will be 5.67 W/m^2. It's now 2% of the solar constant (calculated as average absorbed over the surface of the earth) so it's the same as the solar value increasing by 2%. That's all.

Perhaps you are thinking of the Carnot cycle with the maximum hypothetical efficiency? I'm just stretching here.

But the first part of your statement cited above is not true, while the second part is true but only as just explained.

Aug 13, 2014 at 9:09 AM | Unregistered CommenterScience of Doom

TBYJ & Marin A
Thanks again but I am at odds with you both here on the effect of the sink. Calculating the back radiation from the 2.7 degK as SoD did earlier in the thread... you can probably see why I have a problem with that? As I see it, at condition n+2 and radiative balance the temperature of Earth will be the same as at condition n. This is because the sink is infinite and it is the difference between the two temperatures that S-B seeks/demands to be maintained. The temperature of Earth is not set by the solar constant but by the sink temperature - if the sink temperature was 100 degK Earth's temperature would be higher. Invoking grey-body as a modifier negating S-B doesn't change the result, only the numbers.

ssat -
First, I'd suggest not using the term 'back radiation' - it results in all sorts of confusion and cross purposes discussion.

"it is the difference between the two temperatures that S-B seeks/demands to be maintained"

Could you give a reference to that where it's explained? As I see it, the 'temperature of the Universe' detemines the power reaching the Earth which the Earth then has to re-radiate (along with the power it receives from the Sun).

The Earth knows nothing about the temperature of the universe - all it sees is incoming photons whose energy it has to get rid of (for it to remain at steady temperature).

In doing its radiating, the Earth uses the SB formula to calculate how to do its radiating. This tells it what its temperture has to be (in absolute units - not as a temperature difference between the Earth and something else).

The Earth then gets to work radiating the same power it receives. The photons it radiates are gone forever - the Earth has no information about what eventually happens to them so the temperature of where they finish up has no influence on anything.

Aug 13, 2014 at 9:13 AM | Registered CommenterMartin A

Rob

So what is the Earth's temperature in the limiting case of being totally opaque to IR?? assuming surface pressure is the same.

Very good question.. I imagine you would get a very hot turbulent atmosphere, the temperature limited by the number of GHG molecules at the TOA which can radiate out to open space, and the heat energy being transported up to them by convection. Venus, in other words. Calculating the actual temperature... unfortunately there is no pat equation which links IR opacity and temperature, if there was, there would be no AGW debate.

Aug 13, 2014 at 9:17 AM | Unregistered CommenterTheBigYinJames

ssat,

'you state "If the place of emission of radiation – on average – moves upward for some reason then the intensity decreases. Why? Because it is cooler the higher up you go in the troposphere." I am not sure how that fits with your over-simplified description - perhaps I am misunderstanding 'intensity').'

Intensity is emission of radiation. Just as an example, if you radiate from an atmosphere with a total emissivity of 0.7 at a temperature of 250K the flux = 155 W/m^2. If your emission location moves up 100m, with a typical lapse rate this will be a flux of 153 W/m^2. You have reduced the emission of radiation from the climate system.

'Increasing the concentration of CO2 raises the effective height of emission but its temperature remains the same as demanded by S-B, the BOA temperature increases as it is coupled by the lapse rate, the energy in the system remains constant, the actual surface cools to radiate less by an ammount that balances the increased radiating area of the atmosphere.'

The flaw is understanding how this happens. When you reduce the flux to space, which is what happens with increased GHGs (all other things remaining equal: lapse rate; absolute humidity), then the climate system stores energy.

So no, the energy in the system cannot remain constant. The mechanism by which the climate system restores balance is an increase in surface & atmospheric temperature. This is fundamental.

I give an example of not the climate system in A Challenge for Bryan with the one graph in the article showing the dynamics. This is the essence of how an imbalance in energy gets restored. It's different in detail for the climate system but the process is the same.

There is no "law" that says "its temperature remains the same as demanded by S-B". The law that applies is the first law of thermodynamics: energy in - energy out = energy gained. Energy gained becomes (mostly) temperature (sometimes ice melting and other phase changes).

Aug 13, 2014 at 9:25 AM | Unregistered CommenterScience of Doom

ssat

The question was how long any warming effect (manifest as an increased temperature) would last

Ah I see, you were asking if the temperature raising effect of extra GHG molecules is transient. The answer is no, as long as the extra GHG is in the atmosphere, the temperature has to stay at the elevated position in order to emit enough LW photons to keep the earth in radiative balance. I think you were hinting that the effect might be transient, but it's not... it's there as long as the extra GHG is there.

As I see it, at condition n+2 and radiative balance the temperature of Earth will be the same as at condition n. This is because the sink is infinite and it is the difference between the two temperatures that S-B seeks/demands to be maintained. The temperature of Earth is not set by the solar constant but by the sink temperature - if the sink temperature was 100 degK Earth's temperature would be higher. Invoking grey-body as a modifier negating S-B doesn't change the result, only the numbers.

This is where I find the 'sink' analogy breaks down, especially with people who have had training in electronics, (i.e. me!)

Yes, you are right, the eventual radiative balance on n and n+2 is exactly the same. Before we add the extra GHG, energy in = energy out. And at some point after the GHG molecules have been added, energy in = energy out.
There is no argument here. There is only energy coming from the sun. The 'sink' of empty space is just another way of saying 'no other energy input and no other things impeding the escape of energy output'

But you're dead wrong to state that the temperature of the earth is set by the sink temperature, not by the insolation. If that was the case, every body in the solar system (and indeed the universe) would be at the same temperature, since the sink of space is always the same temperature.

You state

This is because the sink is infinite and it is the difference between the two temperatures that S-B seeks/demands to be maintained

The sink is NOT infinite, it's 2.7 Kelvin. When you feed that into the SB equations, you get

case (n) : cooling rate proportional to T(n) minus 2.7K
case (n+2) : cooling rate proportional to T(n+2) minus 2.7K

So the cooling rate for (n+2) is slightly higher, because T(n+1) is slightly higher.

What you were doing was putting infinity into those equations instead of 2.7.

Aug 13, 2014 at 9:57 AM | Unregistered CommenterTheBigYinJames

Wow, this is becoming a good thread. The following is with a view to make it better.

At 2:11 PM yesterday Rob Burton quoted me (without mentioning it was me) as saying:

Rob also hasn't bothered to respond to SoD's excellent critique of his position - it's now over 48 hours of silence from Rob on that.

Sorry for the pejorative "bothered to", to which Rob understandably took exception. But it's clear from what he went on to say that he misunderstood what I referring to by "SoD's excellent critique of his position". Here's I wrote before the quoted sentence, which was also less than positive:

I felt that the response overall to SoD was slow at best, uninformed at worst - such as Rob Burton seeking to correct me and laugh off the situation with the geostrophic winds as a trivial case at 11:41 AM on 10th. Rob also hasn't bothered to respond to SoD's excellent critique of his position - it's now over 48 hours of silence from Rob on that.

The excellent critique I was referring to was this one at 12:08 PM, demonstrating the complexity of the arguments put forward to show that the earth's rotation causes geostrophic winds. I return to this not to get at Rob but because I think the question SoD originally asked Martin A, backed up by his subsequent posts, is well worth further consideration:

Can you prove that the earth's rotation is what causes the geostrophic winds? Not tests in a lab, not mathematical equations, actual direct evidence. Where is it? There isn't any. Maybe the winds all rotate in a certain way because of other reasons. The only direct evidence would be to stop the planet rotating and see what happens. Or find a similar planet that wasn't rotating and investigate.

For me a better question than the current one about the limiting case of the atmosphere being 'totally opaque to IR'. Rob didn't do justice to this less familiar case - but then none of us have. Still, as I say, a thread that is already a cut above many of the BH debates of yesteryear. Thanks all.

Aug 13, 2014 at 10:17 AM | Registered CommenterRichard Drake

Martin A

First, I'd suggest not using the term 'back radiation' - it results in all sorts of confusion and cross purposes discussion.
Absolutely agreed which is why I see no point in calculating it

Could you give a reference to that where it's explained? As I see it, the 'temperature of the Universe' detemines the power reaching the Earth which the Earth then has to re-radiate (along with the power it receives from the Sun).
I see it explained by the S-B equation using T1^4 - T2^4. As I see it, all energy flows from source to sink, sun to universe. Put a planet in the way and you get sun to planet, planet to universe.

The Earth knows nothing about the temperature of the universe - all it sees is incoming photons whose energy it has to get rid of (for it to remain at steady temperature).
The Earth gets rid of all its photons in a steady state whatever its temperature is. That temperature is coupled to the temperature of space via the T1^4 - T2^4 relationship. It does indeed know the temperature of space just as you know if you are sitting in a warm room or a cold room.

In doing its radiating, the Earth uses the SB formula to calculate how to do its radiating. This tells it what its temperture has to be
Agreed.
(in absolute units - not as a temperature difference between the Earth and something else).
Disagree for reasons above.

The Earth then gets to work radiating the same power it receives.
Agreed.
The photons it radiates are gone forever - the Earth has no information about what eventually happens to them so the temperature of where they finish up has no influence on anything.
Disagree, the photons it radiates are part of the process of the system attaining maximum entropy (if I remembered entropy correctly).

Thanks for your time answering, I now have to meet a work deadline and must leave for a while. Hence a little curt in my response.


SoD,
Thanks for that. I will look at the link and consider your comments - but for the moment I must get back to work!

Aug 13, 2014 at 10:46 AM | Unregistered Commenterssat

ssat

It does indeed know the temperature of space just as you know if you are sitting in a warm room or a cold room.

And here is where the crux lies. It does not "know" the same way you know sitting in a room.

A body in space radiates ONLY in proportion to its own temperature. It does not know, care or have the ability to know what the temperature of anything else is. It has no sensors or mind or knowledge to know what is around it. What it does do - the only thing it does - is radiate photons in proportion to its own temperature, and the body cools appropriately.

At the same time, photons from the environment fall on the body. These heat the body. When you arithmetically subtract this warming from the cooling that occurred due to temperature, you get a NET temperature change.

Now you may want to call these photons falling on the body "knowledge" and you may wish to interpret this as the body "seeing the remote temperature" and "adjusting its photon output accordingly" but although these may be useful interpretations, they do not describe what is actually happening. They provide a handy arithmetic rule only. Underneath, it's an exchange of energy, the net vector sum of which is in proportion to the difference in temperatures. This does not mean that the bodies know the temperatures concerned or change their behaviour based on it.

Aug 13, 2014 at 11:21 AM | Unregistered CommenterTheBigYinJames

TBYJ (11:21 AM): I'm with SoD, you and others that "a body in space radiates ONLY in proportion to [the fourth power of] its own temperature" but I'm eager to see the source for any theory of relativity for S-B!

ssat (10:46 AM): I'm sure you missed my suggestion that you had traduced NiV (at 8:42 AM) or you would have mentioned it. I'm mindful it's time-consuming to dig out exact quotes when you have other work to do. I suggest then that you retract the statement for now and later find the passage you were thinking of and quote it, verbatim, with link.

Aug 13, 2014 at 11:40 AM | Registered CommenterRichard Drake