A reply to Vonk: Radiative Physics Simplified II

Radiative Physics Simplified II

A guest post by Jeff Id

Radiative physics of CO2 is a contentious issue at WUWT’s crowd but to someone like myself, this is not where the argument against AGW exists.  I’m going to take a crack at making the issue so simple, that I can actually convince someone in blogland.  This post is in reply to Tom Vonk’s recent post at WUWT which concluded that the radiative warming effect of CO2, doesn’t exist.  We already know that I won’t succeed with everyone but when skeptics of extremist warming get this wrong, it undermines the credibility of their otherwise good arguments.

My statement is – CO2 does create a warming effect in the lower atmosphere.

Before that makes you scream at the monitor, I’ve not said anything about the magnitude or danger or even measurability of the effect. I only assert that the effect is real, is provable, it’s basic physics and it does exist.

From Tom Vonk’s recent post, we have this image:

Figure 1

Short wavelength light energy from the sun comes in, is absorbed, and is re-emitted at far longer wavelengths.  Basic physics as determined by Planck, a very long time ago.  No argument here right!

Figure 2 below has several absorption curves.  On the vertical axis, 100 is high absorption.  The gas curves are verified from dozens of other links and the Planck curves are verified by my calcs here.  There shouldn’t be any disagreement here either – I hope.

Figure 2 – Absorption curves of various molecules in the atmosphere and Planck curve overlay.

What is nice about this plot though is that the unknown author has overlaid the Planck spectrums of both incoming and outgoing radiation on top of the absorption curves.  You can see by looking at the graph (or the sun) that most of the incoming curve passes through the atmosphere with little impediment.  The outgoing curve however is blocked – mostly by moisture in the air – with a little tiny sliver of CO2 (green curve) effective at absorption at about 15 micrometers wavelength (the black arrow tip on the right side is at about 15um wavelength).  From this figure we can see that CO2 has almost no absorption for incoming radiation (left curve), yet absorbs some outgoing radiation (right curve).  No disagreement with that either – I hope.   Tom Vonk’s recent post agrees with what I’ve written here.

Energy in from the Sun equals energy out from the Earth’s perspective — at least over extended time periods and without considering the relatively small amount of energy projecting from the earth’s core.  If you add CO2 to our air, this simple fact of equilibrium over extended time periods does not change.

So what causes the atmospheric warming?

Air temperature is a measure of the energy stored as kinetic velocity in the atoms and molecules of the atmosphere.  It’s the movement of the air!  Nothing fancy, just a lot of little tiny electrically charged balls bouncing off each other and against the various forces which hold them together.

Air temperature is an expression of the kinetic energy stored in the air.  Wiki has a couple of good videos at this link.

“Warming” is an increase in that kinetic energy.

So, to prove that CO2 causes warming for those who are unconvinced so far, I attempted a thought experiment yesterday morning on Tom Vonk’s thread.   Unfortunately, it didn’t gain much attention.  DeWitt Payne came up with a better example anyway which he left at tAV in the comments.  I’ve modified it for this post.

Figure 3- Experimental setup. A – gas can of air with all CO2 removed at ambient temp and standard pressure. B – gas can of air diluted by 50 percent CO2, also at ambient temp and standard pressure. C ultra insulated laser chamber with perfectly transparent end window and a tiny input window on the back to allow light in from the laser. Heat exit’s the single large window and cannot exit the sides of the chamber.

Figure 4 is a depiction of what happens when  C contains a vacuum.

Figure 4 – Laser passes straight through the chamber unimpeded and a full 1000 Watt beam exits our perfect window.

The example in Figure 5 is filling tank C with air from tank A air (zero CO2) at the equilibrium state.

Figure 5 – Equilibrium of hypothetical system filled with zero CO2 air from canister A.

Minor absorption of the main beam causes infrared absorption and re-emission from the gas reducing the main beam from the laser. This small amount of energy is re-emitted from the gas through the end window and scattered over a full 180 degree hemisphere.

What happens when we instantly replace the no-CO2 air in chamber C with the 50% CO2 air mixture in B?

Figure 6 – Air in C is replaced instantly with gas from reservoir B

From the perspective of 15 micrometer wavelength infrared laser, the CO2 filled air is black stuff.  The laser cannot penetrate it.  At the moment the gas is switched, the laser beam stops penetrating and the 1000 watts (or energy per time) is added to the gas.  At the moment of the switch, the gas still emits the same random energy as is shown in Figure 5 based on its ambient temperature, but the gas is now absorbing 1000 watts of laser light.

Since the beam cannot pass through, the CO2 gains vibrational energy which is then turned into translational energy and is passed back and forth between the other air molecules building greater and greater translational and vibrational velocities.  —- It heats up.

As it heats, emissions from the window increase in energy according to Planck’s blackbody equation.  Eventually the system reaches a new equilibrium temperature where the output from our window is exactly equal to the input from our laser – 1000 watts. Equilibrium! – (Figure 7)

Figure 7 – Equilibrium reached when gas inside chamber C heats up to a temperature sufficient to balance incoming light energy..

The delay time between the instant the air in C is switched from A type air to B air to the time when C warms to equilibrium temperature is sometimes stated as a trapping of energy in the atmosphere.

“CO2 traps part of the infrared radiation between ground and the upper part of the atmosphere”

So from a few simple concepts, two gasses at the same temp, one transparent the other black (at infrared wavelengths), we’ve demonstrated that different absorption gasses heat differently when exposed to an energy source.

How does that apply to AGW?

The difference between this result and Tom Vonk’s recent post, is that he confuses equilibrium with zero energy flow.  In his examples and equations, he has a net energy flow through the system of zero, which is fine. Where he goes wrong is equating that assumption to AGW.

What we have on Earth, is a source of 15micrometer radiation (the ground) projecting energy upward through the atmosphere, exiting through a perfect window (space) – sound familiar?   Incoming solar energy passes through the atmosphere so we can ignore it when considering the most basic concepts of CO2 based warming (this post), but it is also an energy flow.  In our planet, the upwelling light at IR wavelengths is a unidirectional net IR energy flow (figure 2 – outgoing radiation), like the laser in the example here.

Of course adding CO2 to our atmosphere causes some of the outgoing energy to be absorbed rather than transmitted uninterrupted to space (as shown in the example), this absorption is converted into vibrational and translational modes (heating). Yes, Tom is right, these conversions go in both directions.  The energy moves in and out of CO2 and other molecules, but as shown in cavity C above, the gas takes finite measurable time to warm up and reach equilibrium with space (the window), creating a warming effect in the atmosphere.

None of the statements in this post violate any of Tom’s equations; the difference between this post and his, is only in the assumption of energy flow from the Sun to Earth and from Earth back to space.  His post confused equilibrium with zero flow and his conclusions were based on the assumed zero energy flow.   The math and physics were fine, but his conclusion that insulating an energy flow doesn’t cause warming is non-physical and absolutely incorrect.

Oddly enough, if you’ve ever seen an infrared CO2 laser cut steel, you have seen the same effect on an extreme scale.

————-

So finally, as a formal skeptic of AGW extremism, NONE of this should create any alarm.  Sure CO2 can cause warming (a little) but warmer air holds more moisture, which changes clouds, which will cause feedbacks to the temperature.   If the feedback is low or negative (as Roy Spencer recently demonstrated), none of the IPCC predictions come true, and none of the certainly exaggerated damage occurs. The CO2 then, can be considered nothing but plant food, and we can keep our tax money and take our good sweet time building the currently non-existent cleaner energy sources the enviro’s will demand anyway.  If feedback is high and positive as the models predict, then the temperature measurements have some catching up to do.

Even a slight change in the amount of measured warming would send the IPCC back to the drawing board, which is what makes true and high quality results from Anthony’s surfacestations project so critically important.

This is where the AGW discussion is unsettled.

====================================

My thanks to Jeff for offering this guest post – Anthony

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Alex
August 6, 2010 10:01 am

A much better post than that of Tom Vonk.
1. It would be crazy (just endlessly stupid) to deny greenhouse effect as such.
2. The direct effect of CO2 is easy to calculate. It gives less than 1 degree C per doubling of CO2 concentration.
3. The central question is the feedback. Is it positive or negative and how large is it. If it is negative (it must be negative!), we are fine. However, according to IPCC it is positive and uncertain. The uncertainty may result in some warming between 2 degree C and infinity (sic!).

August 6, 2010 10:02 am

A clear dissertation which appears to be eminently rational. If others with more math than I posess (which wouldn’t take much!) agree with this, I can go along with it.
If my high school math and science teachers had been as clear and succinct as many of those who post essays on WUWT, I may have stayed at school a bit longer. In the early 1950s in New Zealand, with full employment and a booming post-war economy,(as one of our better-known poets put it) ‘many high school teachers couldn’t get a job in the Post Office’. This had a serious negative effect on the education attainment of my generation (less than 3% of school leavers went on to higher education) which we are paying for now in terms of too many of us believing for too long the alarmist nonsense cranked out by the Greens, the closet Marxists and the extreme Left. A good education is a wonderful insurance against spin-doctors and snake-oil salesmen!

Steve Keohane
August 6, 2010 10:02 am

Nice clear piece Jeff, thanks. I have to agree CO2 has a warming or insulating effect, but the effect is small and limited, also the water vapor/cloud system appears to be a huge negative feedback.

Bird Stewart Lightfoot
August 6, 2010 10:04 am

Well done, Jeff.

Steven mosher
August 6, 2010 10:06 am

Thank you jeff.
Like you I believe in moving the discussion FORWARD. There are a couple places where the skeptical “movement” is retarded, err held back. Understanding and accepting the way the greenhouse effects works is one of those areas. The continual refusal to accept the basic physics, makes Skeptics look like the defenders of mann. Otherwise smart people, steadfastly refusing to accept a well proven physics. A well proven physics that works. The refusal on skeptics part to accept this physics is bewildering.
1. The design of working devices depends on this physics.
2. This physics is correct, But it has LITTLE BEARING on the alarmist claims.
3. Denying this physics, Weakens your position when you try to discuss the REAL UNCERTAINITIES.
4. When you deny the obvious and well proven physics, people will not take you seriously when you make valid objections to more questionable aspects of AGW.
5. When you accept this physics, you confuse the hell out of AGW folks, because THEY THINK you are anti-science.
6. When you accept this physics you have a better chance of being heard on the REAL issues.
And just for good measure we will throw in some appeal to authority:
Spenser, lindzen, Monkton,Christy,Willis,Anthony.. All accept the basic physics.
That is why they are not so easily dismissed. The real issue is Feedbacks, sensitivities, damages, and solutions.

Pamela Gray
August 6, 2010 10:07 am

No argument here. I live in NE Oregon. I lurvs me some greenhouse gas.

Enneagram
August 6, 2010 10:09 am

All physicists do not include LIFE in their equations and CO2 reactions are intimately related to life on earth. Life needs, in special we, warm blooded animals, an excess of energy to keep negentropy, life itself. This is why exists the UHI effect. Just calculate how much glucose, we the 7 billion inhabitants of the earth, need: We got the energy transforming championship.
Ask yourself how much, how many kilos of those nasty fat growing carbohidrates diets we consume or even how many kilos of jeans, t-shirts and underwear made from that polymer of glucose called cottom we use.
Then WE SHOULD BE REFUNDED , of course in CASH, not in carbon credits.

Enneagram
August 6, 2010 10:15 am
August 6, 2010 10:20 am

Nice post. Just a somewhat urgent off-topic comment.
In The Guardian, John Cook thinks that “A new iPhone climate change sceptics’ app [Our Climate] inadvertently reveals the strategies of disinformation and denial they employ”:
http://www.guardian.co.uk/environment/blog/2010/aug/06/iphone-climate-denial-app
Do you agree with him? Please create an account at the Guardian and freely express your opinion.

PJP
August 6, 2010 10:22 am

Other questions, to which the answer almost certainly exists, but I have no idea where to look:
The absorption spectrum of CO2 shown indicates some maximum absorption (attenuation of the 15 micrometer radiation). How does this change as the concentration of CO2 increases?
I think this is vitally important to answer. At some concentration the gas will become completely opaque to 15 micrometer radiation. Adding more will make no difference.
Are we already at that point? What is that point?

Scott
August 6, 2010 10:24 am

Can some back-of-the-envelope numbers using Beer’s Law and molar absorptivities/extinction coefficients be provided in this analysis? If not, where can I find something like that? That is where part of my skepticism lies, as anyone who’s run IR in organic chemistry can tell you that it doesn’t take a long pathlength before the CO2 absorption at 2350 wavenumbers (cm-1, around 4.25 microns) becomes saturated. Another fun thing I just realized is that CO2’s other main absorption band maxes out right around 666 cm-1…interesting.
Anyway, the main point of my comment is that I want someone to direct me to where I can get molar absorptivities for CO2 and hopefully water too. If no one has run the actual numbers with respect to saturation, I can start working on that (lots of numerical integrations, I know).
-Scott

Tom in Florida
August 6, 2010 10:25 am

On behalf of the Peanut Gallery, thank you for an example that was easy to understand.
One of the problems in this debate is that scientists sometimes forget that most of the people in the world are not science oriented, (the Peanut Gallery). The ability to communicate positively with these people is very important yet often forgotten. The AGW crowd, specifically Al Gore, has been winning this part of the war for years. Unless that changes, nothing else can change. This post is a very good start.

Edward Bancroft
August 6, 2010 10:28 am

OK. That’s a good thought experiment and I have no issues with the conclusions. However, you missed out the other half of the experiment.
Tak two cans, one with dry air, the other with 50% CO2, heat them to the same temperature well above the ambient. Now see how long each of the cans take to reach, say, half way to the ambient. The CO2 can will emit IR more than the dry air, and therefore cool the can quickest.
How does this apply to AGW? CO2 heating in the day is balanced(?) by CO2 cooling at night.
On another topic, why do the global heat flow diagrams only ever depict the situation in the daytime and not at night?
Ed

Layne Blanchard
August 6, 2010 10:29 am

Spencer also did a nice explanation on this at his site. It makes perfect sense that any barrier impeding the free flow of energy from a source will insulate that source and cause a higher temperature than a system without such insulation. Just as a cloud cover in winter holds warmth overnight, and a clear winter night becomes much colder. But our roiling convective atmosphere, heating during day, cooling overnight, facilitates the transfer of kinetic energy from CO2 to water vapor and then to relinquish that energy to space when water condenses. We live in a giant evaporative cooler.
I just traveled internationally a few days ago. I like to monitor parameters of flight on the seatback display. At the surface, temperature here was +63F. At 8000 ft it was near zero, and at 20,000 ft it was -50F. Yes, energy is being retained, but hardly what I would call a “hot spot”.

Stephen Wilde
August 6, 2010 10:30 am

Another opportunity to promote my description from a while ago in case some find it easier to follow:
http://climaterealists.com/index.php?id=1562&linkbox=true&position=4
“Greenhouse Confusion Resolved”

coaldust
August 6, 2010 10:30 am

An excellent illustration of the error in Mr. Vonks conclusion. Thank you.

Gail Combs
August 6, 2010 10:32 am

Actually I think we need both posts for a full explanation.
Tom Vonk’s post explains the physics fairly well for a system in local EQUILIBRIUM without any external input. (It could use a bit of rewrite) This post takes the base laid by tom’s post to the next step, a NON-EQUILIBRIUM situation. If you can understand what Tom was saying about the special case of a system at equilibrium, especially how photons get translated into velocity (heat) then the next step, adding energy in the form of photons that is then translated into heat becomes easier to understand. Also the transport of that energy to space and the time lag become understandable.
And yes I realize Tom was not trying to prove what he actually did prove (physics of a system at equilibrium), the physics and explanation were still OK.

Ed Caryl
August 6, 2010 10:37 am

Between Jeff, Tom, and Willis, the situation is getting clearer and clearer. I’m sitting here watching the thunderheads building over the Sacramento Mountains (NM), as they do nearly every day during the summer, pumping heat from the desert into the stratosphere. I’m looking forward to a rain shower this afternoon, which will bring the temperature down here from the mid-80’s (F) to the mid-60’s. The earth’s thermostat works just fine here. When it gets hotter, the rain is heavier and lasts longer.

CodeTech
August 6, 2010 10:37 am

See? Before I got to the diagram with the 1000-watt output reestablished at a different equilibrium, I was ready to throw rocks at you… 🙂
Yes, while reading through the comments here sometimes I wince at both “sides”. However, like many arguments, the real question usually turns out to be the definition of the words rather than the underlying reality (ie, we use the phrase “greenhouse” even though it’s not the same as a greenhouse).
Enneagram points out another obvious issue: the presence of life on this planet alters simple physical processes in a major way. You HAVE to account for:
1) the current influence of life on atmospheric and oceanic processes, and
2) the fact that life WILL change to adapt to conditions, and by doing so will change the conditions.
Are there any purely physical theories to account for 21% O2 in an atmosphere, that do not rely on the presence of life?

william Gray
August 6, 2010 10:42 am

Can someone post from Co2science Please? They have excellent papers on this subject.

joshv
August 6, 2010 10:44 am

“From the perspective of 15 micrometer wavelength infrared laser, the CO2 filled air is black stuff.”
I believe this is incorrect. Chemical bounds within CO2 absorb the energy of specific photons, and at equilibrium emit them at the same rate, though not necessarily in the same direction, as they are whizzing around, smacking into each other. It’s not black stuff, it’s “white” stuff, like a cloud. Does a 15 micrometer detector pointed at the earth from space see a black ball?

Bomber_the_Cat
August 6, 2010 10:45 am

Anthony, it is not clear from what you write whether this is a real experiment or a thought experiment (describing what you expect to happen).
If it is a real experiment, there are other possible explanations for what is said to happen. For example, I could say that the infra-red beam at 15 micron is effectively ‘scattered’ by the CO2 (via absorption and re-emittance) and impinges on the sides of the insulated chamber. It thus warms the insides of the chamber which in turn warm the gas by conduction and convection (which is how the atmosphere gets heated anyway?) Any radiation you detect with a ‘blackbody’ spectrum characteristic of the chamber temperature is probably coming from the warmed material of the chamber itself, and not necessarily from the gas. How do you eliminate that possibility? I think you need to repeat the experiment with the whole chamber made of what you call “perfectly transparent” window material. The radiation will not then heat the chamber itself, but simply pass through it.
By the way, isn’t the emissivity of the gas so low that it would be hard to detect its emissions anyway? Surely it’s coming from the interior walls of the chamber?

Rhys Jaggar
August 6, 2010 10:46 am

I guess you also must add to the equilibrium issue how more seeohtwo affects growth if temp also goes up a bit. Is that a feedback loop comparable to temp issue, smaller or bigger?
Seems to me that the ‘dummies’ posit that seeohtwo PLUS the solar output affect temperatures is arguable from this. Key spectral freqs for solar warm up??
As a moderate skeptic, what’s your thoughts about how radically or weakly seeohtwo warms the earth??

Richard Garnache
August 6, 2010 10:48 am

Jeff;
You are absolutely right, denying that CO2 contributes to the earth’s temperature weakens our position. Excellent post.

Warren in Minnesota
August 6, 2010 10:49 am

I somehow think or remember that each CO2 molecule can absorb no more than a finite quantity of the electromagnetic (em) radiation at 15 mm. With finite absorption, the temperature will never reach the 1000 watts of input energy as the excess laser em radiation will pass through the canister.

Honest ABE
August 6, 2010 10:49 am

I never got this far into physics and perhaps someone could enlighten me.
When CO2 absorbs infrared radiation and then emits it, does that increase the wavelength? If so, how many times would that radiation need to be absorbed and re-emitted by CO2/Earth until it was at a wavelength that CO2 is transparent to?

Gail Combs
August 6, 2010 10:51 am

Alexander K says:
August 6, 2010 at 10:02 am
… we are paying for now in terms of too many of us believing for too long the alarmist nonsense cranked out by the Greens, the closet Marxists and the extreme Left. A good education is a wonderful insurance against spin-doctors and snake-oil salesmen!
_______________________________________________
Don’t you believe that nonsense Alexander. Plain common sense and a grounding in reality are what are really needed.
The first place the Marxists targeted was education. In 1958 my brother attended his first year at a college in the frozen north of New York state near the Canadian border. He came home in November, less than three months later, a flaming Marxist. This is a guy who was an electrical engineering student with an IQ of over 200.
Dumbing Down America: http://www.ordination.org/dumbing_down.htm
This article shows what happen to our education system.

Ken Hall
August 6, 2010 10:52 am

Spot on article Jeff. This is what I have been trying to explain to many AGW alarmists for a loooooooong time.
When they incorrectly assert that I do not believe in CO2 having any warming property and they try to lecture me about Planck and long time established basic physics I have to tell them that the atmosphere is not a flask. I ask them to show me a flask with approximately 32 million trillion gallons of liquid water in it and 5140 trillion tonnes of water vapour filled air and then they will have something with a closer resemblance to earth.
Of course CO2 has an absorption band, but an increase of CO2 with a logarithmic absorption from 380ppm even to 600ppm will not and cannot cause >2 degrees c of warming when the increase from 200ppm to 380ppm caused only part of the alleged 0.6 degrees witnessed during the 20th century.
The “sophisticated” computer models of a chaotic, non-linear system are incomplete and therefore wrong.

latitude
August 6, 2010 10:53 am

“gas can of air diluted by 50 percent CO2”
But it’s less than 0.04%

dearieme
August 6, 2010 10:53 am

The layman might find it helpful to know that long before the AGW fuss, combustion engineers routinely included the effects of CO2 and water vapour in their calculations of radiative heat transfer in furnaces. This is not contentious stuff, it’s bog standard physics.

Richard111
August 6, 2010 10:55 am

In my simplistic non-scientific way I worked out that the CO2 in the atmosphere provides
a small but constant warming. Doubling CO2 quantity does not change the warming.
http://www.globalwarmingskeptics.info/forums/thread-249-post-3039.html#pid3039

Enneagram
August 6, 2010 10:55 am

Scott says:
August 6, 2010 at 10:24 am Another fun thing I just realized is that CO2′s other main absorption band maxes out right around 666 cm-1…interesting.

Which is the “Perfect Fifth” 2/3 :: 3/2, a complete “quanta” (not the Planck’s incomplete constant=0.66252)
http://upload.wikimedia.org/wikipedia/en/2/20/Perfect_fifth_on_C.mid

August 6, 2010 10:59 am

Why is it that when anyone talks about the “greenhouse effect”, they totally ignore the fact that the atmosphere doesn’t just sit there.
As everyone knows HOT AIR RISES.
So, considering only the radiation bands blocked by CO2, the CO2 laden air will absorb IR close to the ground, it will then heat up, causing the gas to expand and so rise, rise rise, rise, until it finds a way to emit that extra energy … and how is that done? Well it’s full of CO2 which is a superb cooling gas (CO2 provides a convenient extra pathway through which gases can LOSE HEAT by infrared emission. And once CO2 HAS CAUSED THE ATMOSPHERE TO COOL it starts to descend, descend, descend, until it is close enough to the ground to pick up the IR emitted from that blackbody radiator we call the ground.
So, what is the net effect of adding CO2 to the atmosphere? It is first to absorb more radiation at ground level, then to emit more radiation at higher levels where the relatively thin atmosphere creates a window into space. (It’s slightly more complicated because it doesn’t go all the way to the edge of space — but the argument still holds).
So, the net effect of adding CO2 is to heat up and cool the atmosphere more! It has two equal and opposite effects, and the big con of the global warming scaremongerers is to ignore its cooling function in order to pretend it only warms the atmosphere.
Next week, join me for a demonstration whereby a put a plug in the orifice out of which most global warmer communicate to prove to a global warmer that if you only consider what goes in and don’t consider what goes out … it’s pretty painful!

Joss
August 6, 2010 11:00 am

Jeff wrote ” In our planet, the upwelling light at IR wavelengths is a unidirectional net IR energy flow (figure 2 – outgoing radiation), like the laser in the example here.”
Sorry. I do not agree.
On earth, every point source emits isotropically. The emitted field is NOT a vector field. No poynting vector here.
Idem for the CO2 molecule and air volumes in the atmosphere. Therefore , you may not add, substract , calculate energy flows , energy conservation laws etc. as if they were vector fields.

Enneagram
August 6, 2010 11:07 am

Ken Hall says:
August 6, 2010 at 10:52 am
The “sophisticated” computer models of a chaotic, non-linear system are incomplete and therefore wrong
That’s because chaos only exists only in the mind of the beholders. Those who taught chaos and uncertainty were the rule, were lying on purpose. Their agenda was to derail the train of humanity.

Robinson
August 6, 2010 11:14 am

An excellent explanation, although not entirely relevant as the positive and negative feedbacks are where battle is usually joined.

A C Osborn
August 6, 2010 11:15 am

One thing that I find find very confusing about the Wavelength Diagram is how little the H2O has in the spectrum that affects the Incoming radiation and yet as everyone knows when it is cloudy the Temperature changes bu a very Significant amount. How does such a small bandwith have such a large effect?

CRS, Dr.P.H.
August 6, 2010 11:16 am

Thanks for posting, Jeff! Right on the money, denying the physics of carbon dioxide in the Earth’s atmosphere completely is a losing game.
Where the CAGW crowd collapses is in the “catastrophic” aspect, particularly Hansen’s beloved runaway “Venus” effect. Please see p. 22 of his slides at this download:
http://www.columbia.edu/~jeh1/2008/AGUBjerknes_20081217.pdf
Until Earth gets clouds composed of sulfuric acid, moves several million miles closer to the sun, and gets an atmosphere generating a surface pressure of 90 atm, I’m not terribly worried about taking lessons in Venusian.

Enneagram
August 6, 2010 11:18 am

Mike Haseler says:
August 6, 2010 at 10:59 am

Bravo!, chemically pure common sense.
That is why balloons fly when burning propane, which produces CO2+ Water. It goes up, up and away to give its heat back to space.
BTW, water vapor, on my head as low altitude clouds is a negative feedback. I feel colder under a cloud, not warmer.
Think we better prepare some stakes to send CO2 fanatics to explore space. Their heat will surely warm the cold Moon up 🙂

Julio
August 6, 2010 11:18 am

Jeff,
As a physicist, I appreciate your effort here, but I still think one needs to stress that the important point is that some of the energy reradiated by the CO2 in the atmosphere is sent downwards, that is, back to Earth. If all that happened was that the CO2 gets hot, while reradiating all the extra energy out to space (“forward”, like your figure 7 suggests), that would indeed have a negligible impact on the Earth’s temperature.

Jeff
August 6, 2010 11:27 am

I took his original point to be that CO2 can’t “create” new energy in the system, i.e. it can’t make it get hotter than the original energy input into the system would … that was the “heating” I thought he was talking about …
in a perfect black body one millimeter off the surface the temperature will be X depending on the amount of solar energy hitting said surface … X is the maximum value for air temperature … 20 millimeters above the surface it cannot and will not exceed X … 100% CO2 or 0% CO2 doesn’t change this …
how fast the heat transports is effected by the various gas or vapor concentrations …
so CO2 effects the speed of heat/energy transportation … it can’t “Heat” up anything … slow the loss, yes … so on a cronilogical basis you could say that the measured temp is hotter than it would have been with less CO2 …

August 6, 2010 11:27 am

In order to have a sane discussion about the theory of “global warming,” it is important for all camps to accept the fundamentals of science – which Jeff presents here well.
Neither side holds a monopoly on ignorance.

Jeremy
August 6, 2010 11:28 am

Thanks goodness a return to some refreshing sanity – before all credibility on WUWT is lost. I was appalled when I saw the Tom Vonk posts. I understand however that this is a blog and the beauty of blogs is that some have the cojones to stand to be corrected!!!
I gave up on The Economist – modern news medias are too arrogant and never publish corrections even when they publish bogus science and downright nonsense. (The Economist was once, many years ago, a half decent rag with a reputation that was worth defending – the modern version is so BS prone that you may be better served by the National Enquirer at your local supermarket checkout – at least they make no pretense about being “authoritative”)

Tufty
August 6, 2010 11:32 am

Joss is right and in any event, the ‘basic physics’ of CO2 warming the atmosphere is not basic physics at all. The CO2 claim is not falsifiable by experiment because we can’t do the necessary experiments on the atmosphere. That is to say we obviously can’t control and monitor the effect of raising and lowering CO2 in the real atmosphere while keeping other variables constant. We don’t even know what those variables would be. In science (including basic physics) not falsifiable means not true. Or false for that matter. CO2 may or may not warm the atmosphere. We don’t know.
As for the implied suggestion that credible AGW skeptics must accept that CO2 warms the atmosphere to avoid accusations of extremism – well that isn’t science either. That’s politics.

Gail Combs
August 6, 2010 11:34 am

PJP says:
August 6, 2010 at 10:22 am
Other questions, to which the answer almost certainly exists, but I have no idea where to look:
The absorption spectrum of CO2 shown indicates some maximum absorption (attenuation of the 15 micrometer radiation). How does this change as the concentration of CO2 increases?
I think this is vitally important to answer. At some concentration the gas will become completely opaque to 15 micrometer radiation. Adding more will make no difference.
Are we already at that point? What is that point?
________________________________________________________
Yes we are essentially at that point just look at the IR fingerprints for CO2 in the article or here: http://upload.wikimedia.org/wikipedia/commons/7/7c/Atmospheric_Transmission.png
CO2 is Logarithmic Explained: http://knowledgedrift.wordpress.com/2010/05/11/co2-is-logarithmic-explained/
Here is another paper you might want to read: http://www.friendsofscience.org/assets/documents/The_Saturated_Greenhouse_Effect.htm
Another very interesting site: http://www.co2web.info/

August 6, 2010 11:36 am

CRS, Dr.P.H.
Thanks for the Hansen link.

Charles Higley
August 6, 2010 11:37 am

There is no doubt that CO2 absorbing IR does convert a small fraction of the energy to heat energy. However, this factor is small and too small to do anything about.
The IPCC in their great unfounded wisdom has multiplied this factor by 12 and then considered water vapor as a 30-fold positive forcing factor. This is the problem – they have artificially magnified the effect, which creates a false, much overblown warming in models which are full of critical flaws.

jorgekafkazar
August 6, 2010 11:41 am

I thought Tom Vonk’s post was intended to cover a very narrow, simple topic, one that sets aside certain significant factors (as mentioned in his caveats) in order to discuss the theoretical physics of molecular absorption of light. I did notice a few areas where he glossed over definitions, and it now seems that his narrow case doesn’t apply to the larger system of the real world. Still, I’d like to see Tom’s response to Jeff’s post, indicating where the two are in agreement, and what differing assumptions were made that resulted in different results. .

CRS, Dr.P.H.
August 6, 2010 11:43 am

stevengoddard says:
August 6, 2010 at 11:36 am
CRS, Dr.P.H.
Thanks for the Hansen link.

——
Reply: Any time, Steve!
Thanks for your excellent posts! It will take a lot of folks, ranging from “everyman” to professionals, to shoot this stuff down, and your contribution is appreciated.

August 6, 2010 11:52 am

Julio says: “As a physicist, I appreciate your effort here, but I still think one needs to stress that the important point is that some of the energy reradiated by the CO2 in the atmosphere is sent downwards, that is, back to Earth. ”
That is why you can’t ignore the convective transport of heat through the atmosphere!
Hot air rises because it is hotter than the equilibrium temperature for such a gas at that level in the atmosphere. It therefore follows that rising air masses will emit radiation, it also follows that descending air masses tend (on balance) to absorb radiation.
In effect the whole atmosphere is just one great big heat engine. The heat source is the surface of the earth, and the heat sink is space. The heating causes the air to rise (now comes the fun bit) and because of the spin of the earth it begins to rotate IT IS LITERALLY LIKE AN ENGINE!!
Once the rising air has radiated heat into space by IR (or perhaps lost it by conduction), then it cools.
Warming: Now comes a bit that will confuse any global warmer
Remember those weather maps of low pressure areas with wind going into them? Well where does that wind go if all the air is heading inwards? You and I know that low pressures aren’t mysterious black holes, so we know it goes upward into the atmosphere, that upward movement is powered by the heat of the sun, and that heat is then lost to outer space – otherwise the fundamental thermodynamics of the heat engine are broken.
So, for each low pressure system at ground level, we can draw an equal an opposite outward going air system at highlevels. Likewise that air coming out of high pressure areas are part of the heat cycle of the atmosphere and similarly there are equal and opposite airflows going into high pressure areas.
HIGH PRESSURE = DESCENDING COOLED AIR
LOW PRESSURE = RISING WARMED AIR.
The whole system of atmospheric circulation will break down if you don’t have IR absorbing gases in the lower atmosphere allowing the natural cooling system of the planet to take that heat energy away from the surface where at higher levels, the same absorbing gases now act as equal an opposite cooling gases.
It’s all one big cooling cycle: the coolant (CO2) cools the surface of the earth, by absorbing IR, and then later on it cools itself by equal and opposite emission of IR at very similar wavelengths.
And, if you want to see how it works, go make a cup of black coffee, in a cold cup, then pour in milk DON’T STIR. If you’ve done it right, you’ll see cells of rising and descending liquid bounded by clearly defined lines just like our weather system … except they don’t rotate until you stir the coffee … add whisky and then drink.

ShrNfr
August 6, 2010 11:52 am

I don’t think anyone in their right mind disputes that CO2 and that horrible green house gas H2O do not cause an increase in the mean global temperature. Its the question of the first and second derivative. The last time I saw a first derivative of temperature with respect to CO2 concentration from a source I could respect, it was of the order of 0.00125 to 0.00250 degrees C per ppm of CO2. The second derivative was negative. Of course, this being the complex system that it is with convection and all even static first derivatives are at best a wag.
“CO2 traps part of the infrared radiation between ground and the upper part of the atmosphere” totally ignores the fact that convective activity takes the CO2 heated air upwards into the atmosphere and replaces it with cooler air from higher alititudes. The earth is not a greenhouse. There are no glass plates on the roof. Sometimes it takes that other greenhouse gas up with it and makes these funny things called clouds which make your “window” pretty darn reflecting instead of transparent. (Hint, that is why deserts are hot by and large).

Gail Combs
August 6, 2010 11:53 am

Bomber_the_Cat says:
August 6, 2010 at 10:45 am
…If it is a real experiment, there are other possible explanations for what is said to happen….. I think you need to repeat the experiment with the whole chamber made of what you call “perfectly transparent” window material. The radiation will not then heat the chamber itself, but simply pass through it.
________________________________________________________
A “perfectly transparent” window material is a plate of salt (NaCl) it is used in chemistry to hold the test sample in Infrared Spectrophotometers.
It was used by Woods in 1909 to show the true” greenhouse effect” found in greenhouses was from the normal glass allowing the sun’s energy (high wavelengths) to pass into the greenhouse but trapping the infrared energy (lower wavelengths) so it could not be re-radiated out. A greenhouse built with salt glass did not become warm.
An explanation of the experiment is here: http://hockeyschtick.blogspot.com/2010/06/greenhouse-theory-disproven-in-1909.html

UK John
August 6, 2010 11:54 am

Weather is cold, cloudy, and miserable in UK, a typical summer!
We could do with a bit of CO2 warming feedback! but it looks like it isn’t going to happen.
I am off to Spain! see you all soon!

Spector
August 6, 2010 11:54 am

RE: Joss: (August 6, 2010 at 11:00 am) “On earth, every point source emits isotropically.”
That is quite true, however, in the upper atmosphere there is a conical segment about straight up in which radiation may have a good chance of escaping to outer space.
For each earthshine absorbing/emitting (greenhouse) gas it might be useful to define an altitude where perhaps all radiation emitted within 45 degrees of straight up has a 50 percent chance of escaping to outer space.

Crazydung
August 6, 2010 11:59 am

Thank you for the post which helped me alot. However I see from the comments that I need to read a lot more.
As a none scientific sceptic I can comment on your point about us saying CO2 does not cause warming.
I believe what most of us have meant is that although we might be persuaded that CO2 has a warming effect, there are times when CO2 rises and the planet cools. To a none scientist it seems reasonable to conclude that during those times, CO2 has no warming effect ^.^

Kevin
August 6, 2010 12:16 pm

What is the saturation point of atmospheric CO2 and at what point does the increase in CO have a negligible effect?
As there has been some warming, but none matching any of the models I think this is where the battleground truly exists.

1DandyTroll
August 6, 2010 12:22 pm

The low down of radiative physics:
Crap gets radiated, crap has to go somewhere, crap hits some things some of the time, but most of the time it just keeps going, and going, and going into out space, and beyond. And that’s how all that outer space crap reach us in the first place.
Cheers

Roger Clague
August 6, 2010 12:27 pm

Steve Goddard says
In order to have a sane discussion about the theory of “global warming,” it is important for all camps to accept the fundamentals of science – which Jeff presents here well.
Who gave him the right to decide what is ‘fundamental science’. Science is process. Science is the sane discussion. That is science is the use of logic and numbers.
This complex thought experiment from Jeff Id does not at all resemble the earth’s atmosphere. So it can not show that CO2 causes it to heat up.
http://realplanet.eu/atmoseffect.htm
Here the author explains the effect using a thought experiment about the earth’s atmosphere. He does need to consider the chemistry of atmosphere

PSU-EMS-Alum
August 6, 2010 12:27 pm

Not this “re-radiate” stuff again. Objects “radiate”, they don’t “re-radiate”. When you eat a steak, are you “re-digesting” the grass that the cow ate?
Just drop the “re-” part and all your statements will not only still make sense, but they will be accurate, too.

Jan K. Andersen
August 6, 2010 12:32 pm

Thank you Jeff, An excellent explanation of the natural greenhouse effect.

Ronaldo
August 6, 2010 12:33 pm

Absorption of 15 micron radiation by CO2 raises the internal – vibrational – energy of the molecule, this is a quantized reaction and the CO2 rapidly relaxes to its ground state by emitting 15 micron radiation. CO2 molecules in the atmosphere will thus absorb outgoing thermal radiation and re-radiate it in a random direction. Some of this scattered radiation will reach the earth and provide a little extra warming. It is however important to recognise that, because of the quantum nature of this process, the CO2 molecules involved do not gain kinetic energy and are therefore incapable of “warming” (ie. adding kinetic energy)to the atmosphere unless there exist receptors such as dust, or vapours which absorb the 15 micron radiation kinetically.
It is thus correct to state that CO2 absorbs IR and re-radiates some of it back to earth, the so-called Greenhouse Effect. However, in my opinion ,FWIW, the impact on Global temperature pales into insignificance when faced with the complexity of natural feedbacks as Dr Spencer and others have ably demonstrate.

stephen richards
August 6, 2010 12:35 pm

Scott says:
August 6, 2010 at 10:24 am
Can some back-of-the-envelope numbers using Beer’s Law and molar absorptivities/extinction coefficients be provided in this analysis?
Google it. There are plenty of responses

Pamela Gray
August 6, 2010 12:42 pm

Gail, the Dewey denigrating article you linked to is a bit out of date, or if currently being spread, is somewhat out of touch. While I understand that historically, whole language become the preferred/only (and very unfortunate) method of reading instruction, it is no longer the case. Auditory phonemic awareness (sounds in spoken words) and phonological letter-sound association awareness is a key explicitly taught skill element in reading instruction in kgt through grade 2. Children are tested frequently throughout the year regarding decoding and recoding ability, with targeted phonics-based intervention applied when these skills are not being developed.
My concern is that commercial reading intervention products are not very strong in oral practice of phonics and rely too heavily on teacher talk and worksheets, or worse, try to do too much by including comprehension in the mix. Many also introduce spoken-sound practice and letter-sound practice in less than preferred ways in my opinion.
I have my favorites (one is out of print but I use it anyway and the other has just been re-issued at more than quadruple the price) and my less than favorites (which consists of most of the new programs), and then there is my list of bad, bad, bad interventions (also consisting of new programs) like “Read Naturally”. It is often touted as a computer based reading fluency intervention but is at the top of my bad list. It is a whole language holdover that needs to be entirely removed from our schools, burned till nothing is left but ashes, and the authors tarred and feathered.

Ronaldo
August 6, 2010 12:42 pm

Drat, that should be demonstrated.

Roy Clark
August 6, 2010 12:47 pm

There should be no doubt that the downward LWIR flux from both CO2 and H2O in the atmosphere exchanges LWIR flux with the surface and helps keep the surface warmer than it otherwise would be. Most (>90%) of this flux originates in the first km layer above the surface. However, the emission is determined by the species concentrations and the temperature profile (lapse rate) in the lower troposphere. The lapse rate is set by the bulk thermodynamics of the convection. As soon as the sun begins to heat the surface and sets up a temperature difference, the air is warmed at the surface and rises though the atmosphere. Under summer sun conditions with a dry surface, the surface temperature (the one under your bare feet) can easily reach 50 C. The air temperature will be about 25 C because the convection mixes cooler air down to the surface. Just using Stefan’s law, (sigma.T^4), the upper limit to the net LWIR flux radiated from the surface is about 200 W.m-2. The incoming solar flux is ~1000 W.m-2, so the balance is convection. ~80% of the thermal energy from a dry ground surface is transfered to the atmopshere as convection. If the surface is moist, evaporation will reduce the surface temperature and some of the energy will be tranported as latent heat.
The fundamental error in the AGW argument is equating the small increase in downard LWIR flux from 100 ppm CO2 (1.7 W.m-2) with an increase in ‘EQUILIBRIUM SURFACE TEMPERATURE’ using Stefans Law, with ‘WATER VAPOR FEEDBACK’ to fix the discrepancy in the presumed numbers. (These numbers come from the hockey stick calibration, but that is a separate fraud). In reality it is impossible to measure the effect of a 100 ppm increase in CO2 concentration/LWIR flux on the surface temperature. +1.7 in a flux that changes pseudorandomly from +1000 to -100 W.m-2 makes no difference to the surface/subsurface temperature of a cubic meter of soil. Changes in humidity alone will alter the downward LWIR flux by 50 W.m-2. Forget 100 ppm of CO2 – or 200 or even 2000 ppm! Alternatively the whole daily dose of 1.7 W.m-2 100 ppm CO2 LWIR flux is equivalent to less than 3 minutes of sunshine at 1000 W.m-2, or the evaporation of a layer of water 65 microns thick over an area of 1 m^2.
Furthermore, convection is mass transport, which means that the air mass as it rises through the atmosphere has to do work against gravity. That is undelying reason why the air cools as it rises. This is where the excess surface energy goes.
At night there is much less convection, so a stable air layer near the surface slowly cools by LWIR emission up through the atmosphere. The heat capacity of a 1 m^2 x 1 km column of air at the surface is about 1 MJ, so the radiative cooling rate is around 0.1 C/hr. The thermal gradient is reset each day by the surface convection.
The downward LWIR flux from CO2 helps to keep the surface warmer than it would otherwise be, but it does not control the surface temerpature. It is just a relatively stable component in a highly variable surface energy flux. To understand the greenhouse effect we need to look at the short term variations in the entire energy flux, not long term averages of small parts of the radiative flux.

Michael J. Dunn
August 6, 2010 12:48 pm

Misunderstandings are so unfortunate.
What you need to understand about Tom Vonk’s post is that he was explaining that the atmosphere does not warm by any radiative absorption process. (And it doesn’t. If you look at the temperature profile of the atmosphere with altitude, once it gets above the terrestrial boundary layer, there is only a weak relationship with surface temperature.) This is not identical with saying that a greenhouse process does not exist (which he wasn’t arguing).
In a nutshell, the picture is this: Sunlight heats the earth. The earth maintains the temperature of the atmosphere through convection. If certain gases are in the atmosphere, they function effectively as a spectrally-sensitive beam-splitter, scattering 50% of some outbound infrared spectra back to the earth, which can raise the temperature of the earth (greenhouse effect). The temperature of the atmosphere is unaffected by this beam-splitting process per se, because it is an equilibrium where input = output (no heat accumulation).
For those who are distracted by the “time lag,” it is negligible. Radiative equilibrium in an absorbing-emitting gas proceeds at the speed of light between molecules…very nearly the speed of light through open space, excepting for the random directions of the emitted photons. From a radiative heat transfer standpoint, the absorptive gases are like an opaque body with an internal heat conductivity that is thousands of times more rapid than normal conductivity. So, temperature equilibirum is almost instantly established, and the outbound radiation is unimpeded. (The substantial atmosphere is about 10 km deep. At 1/10 the speed of light, which is about a million times faster than molecular thermal velocities, the transit time is about 3/10 of a millisecond. Radiative equilibrium is established throughout the atmosphere faster than the air can gain temperature.)
And the air is worthless as a heat sink. We normally use it as an insulator. When nightfall hits the desert, it chills down rapidly. The air does not keep it warm. (The heat capacity of air is dominated by the condensibility of water vapor.)
My professional expertise includes performing detailed analysis of the radiative balance of high-energy laser beams engaging military targets. We have a definite interest in what happens to IR photons zipping through the air and how they affect solid materials. Tom is on firm ground. Try to understand it.
Jeff, your gedankenexperiment was okay except for two lapses of proper analogy. The first lapse is that the basic model should have been of a 15-micron laser hitting a black-body endwall (earth surface) and reflecting back out. But that doesn’t affect the experiment. The second lapse, critical to your argument—and which I have addressed above—is that the transition from Fig. 6 to Fig. 7 is essentially instantaneous. There is no heat accumulation. Also, there is no blackbody radiation! When we are speaking of molecular resonance absorption and emission, we are out of the realm of continuum radiation altogether. These are quantum mechanical processes for which (you guessed it) temperature cannot be defined. The earth emits as a black-body, but the atmospheric gases do not.
Experimental proof, anyone? The predicted tropospheric “hot spot” does not exist. As Tom has pointed out, it cannot exist. This is a key falsification of the “global warming” theory. (Don’t worry about stratospheric heating; that is chemical recombination of atomic oxygen, a much different process.) Gotta go. I know there is at least one regular out there who doesn’t like lectures….

Brego
August 6, 2010 12:48 pm

Re: Scott says:
August 6, 2010 at 10:24 am
[Anyway, the main point of my comment is that I want someone to direct me to where I can get molar absorptivities for CO2 and hopefully water too. If no one has run the actual numbers with respect to saturation, I can start working on that (lots of numerical integrations, I know).],
Scott, you can learn a lot about the optical properties of water here:
http://omlc.ogi.edu/spectra/water/abs/
The optical properties of CO2 can be found here:
http://www.spectralcalc.com/info/about.php

Scott
August 6, 2010 12:50 pm

A C Osborn says:
August 6, 2010 at 11:15 am

One thing that I find find very confusing about the Wavelength Diagram is how little the H2O has in the spectrum that affects the Incoming radiation and yet as everyone knows when it is cloudy the Temperature changes bu a very Significant amount. How does such a small bandwith have such a large effect?

Hi AC Osborn – the reason these don’t match up is because the diagram is an absorbance diagram. Clouds reduce the incoming radiation intensity (mostly) via a scattering mechanism. This is also the reason that aerosols in the atmosphere have a net cooling effect. Finally, it’s also the reason why skeptics complain that changes in cloud cover aren’t included in climate models. IIRC, according to Roy Spencer, all of the supposed warming in recent decades can be accounted for with only a 2% change in cloud cover.
-Scott

Peter
August 6, 2010 12:59 pm

Joss, the keyword is net IR energy flow.

donald penman
August 6, 2010 1:01 pm

So the idea that co2 does not warm the atmosphere is to be dismissed with a thought experiment,I am open to the idea that co2 does warm the atmosphere but have not seen any proof yet.If as suggested by tom vonk that the atmosphere does not warm up with outgoing infrared radiation from the earth then the atmosphere could still warm when h20 is evaporated from the ground and condenses in the atmosphere ,taking heat from the ground and adding it to the atmosphere,co2 does not condense in the atmosphere however it gets into the atmosphere .There could be infrared radiation going back to the earth from this as well as by backradiation caused by the absorption of h20 and co2.

Sandy
August 6, 2010 1:04 pm

Those two ‘black body radiation’ humps are very wrong.
The profile of a lower temp. body’s frequency spectrum is always lower than that of a higher temperature body. So the earth receives more IR in sunlight than it emits at night. Thus the blanket effect that warms the earth at night would by definition shade the earth from incoming sun IR which must be a bigger effect because of the extra IR.
Google ‘black body radiation’ and hit images to see what I mean.

Robert of Ottawa
August 6, 2010 1:04 pm

Good article. This also doesn’t address how the doubling of CO2 doesn’t double the amount of IR absorbed by the CO2.
But for me always, the arguments pro/con AGW is historical. If the atmosphere was unstable, as the AGWers maintain (positive feedbacks ARE unstable) then the temperature would have run away billions of years ago.

Gail Combs
August 6, 2010 1:05 pm

thegoodlocust says:
August 6, 2010 at 10:49 am
I never got this far into physics and perhaps someone could enlighten me.
When CO2 absorbs infrared radiation and then emits it, does that increase the wavelength? If so, how many times would that radiation need to be absorbed and re-emitted by CO2/Earth until it was at a wavelength that CO2 is transparent to?
______________________________________________________
Physicists please correct me if I am wrong:
The photon absorbed and the photon emitted should be exactly the same energy value. The only thing I can think of that would change that is “spectral line broadening”. If I recall “spectral line broadening” is caused by pressure effects: http://www.rsc.org/ebooks/archive/free/BK9780854045754/BK9780854045754-00001.pdf
There is one way the energy of the photon “packet” could be changed. It is when the infrared energized CO2 molecule collides with another gas (N2, O2, H2O…) and the photon energy is translated into velocity. Since velocity ca be of any value subsequent collisions can change the velocity (energy) by dividing it between the colliding partners.
The amount of “heat” (velocity) caused by the infrared energy absorbed by the CO2 depends on how much is translated into velocity of other air molecules and how much is re-radiated downward back to the earth’s surface. Do not forget that this is ONLY the energy from the specific IR bands shown in Figure 2, third line (green) on the right side under the bell shaped black line representing the gray body emissions of earth. Of course H2O (blue line) is going to be competing for some of that energy too. This can be seen in the shape of the black line at the top.

Robert of Ottawa
August 6, 2010 1:06 pm

As a thought experiment of the positive feedbacks that the AGWers claim, let’s model this feedback with some electric heaters in the above gas chamber. We will measure the temperature in the gas chamber and icnrease the electric heating proportionally to the increase in temperature. This is positive feedback.

Merrick
August 6, 2010 1:12 pm

I wouldn’t assume that Anthony would be particularly interested, and given my current load in professional and personal life I certainly can’t imagine finding the time, but when I read articles like this I’m glad that much of the basic physics required to understand the big picture is in place, but frustrated as usual that much of the important “physics” is left out.
As I alluded to in several posts under Tom’s article there’s a lot more to this than just the 15 um CO2 vibrational mode and translational energy. And, specifically, air temperature is NOT an expression of the average kinetic energy in air – despite what Wikipedia says – though it is related to the average kinetic energy. That definition completely disregards molecular internal energy, which is a little critical here, since light can only be absorbed into internal modes! One of the assumptions of Tom’s article, and tacitly in this one, seems to be that the way the 15 um CO2 vibrational excitation is relaxed non-radiatively is through V -> T (vibrational to translational) transfer from CO2 to N2. I haven’t taken the time to do a thorough analysis of this, but I suspect that’s a low-probability event. In general, V -> V and V -> R (rotational) energy are microscopically more likely. For a given type of transfer to take place it has to be energetically allowed and geometrically favored (through the impact parameter). The only impact parameters for strong V -> T coupling to N2 are those impact parameters near zero and along the axis of the N2 molecule. Given that there are a lot more impact parameters greater than zero than there are near zero and that there are a lot more approach angles off the N2 axis than near the axis this tends to result in it being much more likely for this particular energy transfer to be V -> R than V -> T.
Now, we’re talking about a lot of translational energy. The 15 um photon that is absorbed has an energy of a little less than 0.1 eV. At room temperature the available free energy is about 0.025 eV (k T). This means that the translationally excited N2 molecule (traveling at about 0.08 eV) running into any other molecule immediately after the CO2 collision which excited it, is generally going to run into a molecule with a lot less translational energy (about 0.025 eV). And therefore lose some translational energy.
We’re now back to the issue of energy transfer and the CO2 molecule. Very few N2 (or any other) molecules are going have sufficient translational energy to pass back to the CO2 molecule to excite a vibrational excitation when a subsequent collision occurs. And the very few molecules that still do have the required translational energy to transfer back must have alignment and impact parameter with the CO2 molecule such that essentially all of the energy is available to excite the vibration. And even then it’s still far more likely to excite a rotation or even just transfer the energy over as translation. All other alignments favor T -> R.
Now, in fairness, back-transfer DOES happen. One can get the order of magnitude of the fraction of excited state molecules in any *thermal* system from a simple Boltzmann calculation:
Ni / N = exp[-Ei / (k T)], where Ei is the excited state energy, T is the temperature, k is Boltzmann’s constant, and Ni and N are the excited state and ground state populations. [This is an oversimplification! – but of the right order of magnitude – there are partition function and degeneracy terms to account for]
For the 15 um CO2 vibrational state at 15C you get about 4% excitated state population of the CO2. So, as Tom suggested in his original article, there are definitely excited state CO2 molecules available to radiate 15 um radiation out in all directions. But it is the fact that the CO2 vibrational excitation, in general, cannot re-radiate before collisional cooling (as Tom originally stated in his article) and that it’s really hard to back-populate the CO2 vibrational excitation that makes you realize how much energy has to be placed into translational and internal molecular modes of all the other molecules before the back transfer starts to be likely to occur that gives you the appreciation for how CO2 is causing the atmosphere to heat up.
So, again, nothing in the article keeps the big-picture physics from being correctly understood, but the fine details leave out much of the more important chemistry that’s required for a more in depth discussion of the actual thermodynamics occurring.

Crazydung
August 6, 2010 1:14 pm

Mind blowing post by Michael J Dunn!
I hope he comes back ^.^

Gail Combs
August 6, 2010 1:14 pm

#
#
Mike Haseler says:
August 6, 2010 at 10:59 am
Why is it that when anyone talks about the “greenhouse effect”, they totally ignore the fact that the atmosphere doesn’t just sit there.
As everyone knows HOT AIR RISES.
So, considering only the radiation bands blocked by CO2, the CO2 laden air will absorb IR close to the ground, it will then heat up, causing the gas to expand and so rise, rise rise, rise, until it finds a way to emit that extra energy … and how is that done?…..
____________________________________________________________
NOPE.
Absorption of IR causes an electron to go from the ground state to the excited state. VELOCITY of the molecule is what we call heat. You missed a step.
Reread tom’s post closely he explains the difference. http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/

August 6, 2010 1:14 pm

Well, it went better than I thought so far. For those who are discussing convective heat transfer, of course you are right that this happens but it is related to the magnitude of the warming effect rather than whether the effect exists. As far as I’m concerned, I’ve got no idea about the magnitude of the warming.
Michael J. Dunn
“The second lapse, critical to your argument—and which I have addressed above—is that the transition from Fig. 6 to Fig. 7 is essentially instantaneous. There is no heat accumulation. Also, there is no blackbody radiation! When we are speaking of molecular resonance absorption and emission, we are out of the realm of continuum radiation altogether.”
I understand what you are trying to say, but if you are right, your laser cavity won’t need that water cooling jacket or gas circulation pump anymore. Go ahead and blast a thousand watts at an absorbing medium the size of a coffee cup, I think you’ll find it might get warm. 😀 The blackbody radiation comes into effect when the vibrating molecular bonds collide with other particles and it turns into translational energy.

August 6, 2010 1:16 pm

Gail Combs says ‘don’t you believe that stuff (about a good education), Alexander.
Sorry Gail, but I meant a non-dumbed down education mal-fashioned by Greens, Marxists and others of devious and malign intent. My idea of a ‘good education’ comes pretty much from the teachings of Aristotle, not some wet-behind-the-ears Marxist camped in an undeserved university tenured professorship. And I believe that many of the essayists who post here at WUWT are giving us a wonderful education.

Pamela Gray
August 6, 2010 1:17 pm

Mike, you have interesting points. CO2 is a pretty heavy gas. What equations do you have for its ability to rise once it has absorbed IR in this flow of rising warmed air? Or does it take a ride on water vapor and dust? Does it rise all by itself? Is water vapor the better mechanism for rising warmed air (it also absorbs IR)? How significant is CO2 compared to the other gasses in this process of rising air in order to cool the surface?

kuhnkat
August 6, 2010 1:27 pm

Jeff Id,
You ignore the FACT that the incoming IR from 1-4 microns is much larger than the outgoing through the earth’s complete power range among other things. It is really hard to compute RTE’s when you ignore large segments of the flux and concentrate on one area.
The graph you show has the sun’s output scaled to under 10 -6. Even Science of doom only used a graph at 10 -6 for the sun to get it on the same graph. Naughty naughty.

Steve in SC
August 6, 2010 1:37 pm

Your thought experiment is just so wonderful.
Trouble is, it can not be performed.
The world is fairly lossy, and will give you loads of trouble every time.
There is no perfectly transparent window and no material that will perfectly insulate a mass of gas.
As you add energy to the mass of gas the pressure will increase as it heats up.
That is where your new equilibrium will be established.
The other trouble is that at standard temperature radiation is not the primary mode of heat transfer and can not be neglected. The converse is probably more true.
I don’t think many people are disputing the basic physics. Its just that pesky everything else gets in the way in the real world. Nice try though.

Gail Combs
August 6, 2010 1:44 pm

A C Osborn says:
August 6, 2010 at 11:15 am
One thing that I find find very confusing about the Wavelength Diagram is how little the H2O has in the spectrum that affects the Incoming radiation and yet as everyone knows when it is cloudy the Temperature changes bu a very Significant amount. How does such a small bandwidth have such a large effect?
__________________________________________________________
Energy can be absorbed by a molecule, that is cause an electron to move to a higher state, or energy can be reflected.
A change in cloud cover causes a change in the albedo. The water in the clouds is reflecting the sun’s energy like a mirror does before it can even hit the ground. That is why skeptics have hissy fits about climate models that keep cloud cover constant.
Albedo: http://www.answers.com/topic/albedo
Changes in cloud cover:
Inter-annual variations in Earth’s reflectance 1999-2007.
http://bbso.njit.edu/Research/EarthShine/literature/Palle_etal_2008_JGR.pdf
Research Article: Automated Observations of the Earthshine
http://www.hindawi.com/journals/aa/2010/963650.html
“For a decade, we have been measuring the Earth’s reflectance by observing the earthshine, which is sunlight reflected from the Earth to the Moon and retroflected to the nighttime Earth. …..The earthshine observations reveal a large decadal variability in the Earth’s reflectance [7], which is yet not fully understood, but which is in line with other satellite and ground-based global radiation data…”

August 6, 2010 1:47 pm

Kuhn,
It’s the shape of the graph that matters, if you follow the verification link you’ll see that I demonstrated it both ways.
Not so naughty.
Also, it matters exactly zero what’s going on in 1-4 microns, that’s a magnitude of the warming issue, not a basic physics issue.

Spector
August 6, 2010 1:50 pm

RE: Gail Combs says: (August 6, 2010 at 1:05 pm) “The photon absorbed and the photon emitted should be exactly the same energy value.”
That only applies if there is absolutely no change in the energy of the entity that absorbed the photon in the first place. If you get $20 and it is stolen, then you cannot spend it. If the robber only takes $10, then you might be able to spend the other $10.
In the molecular world, the robbery may take place as a collision between two molecules. Such a collision may also result in the donation of energy.

Gnomish
August 6, 2010 2:05 pm

Very elegant experiment and explanation, thanks!.
Now get out your 1000W sunlamp and a fan and a full swimming pool. We need to evaporate all that and get to the bottom of it.

Latimer Alder
August 6, 2010 2:12 pm

@crazydung
‘I believe what most of us have meant is that although we might be persuaded that CO2 has a warming effect, there are times when CO2 rises and the planet cools. To a none scientist it seems reasonable to conclude that during those times, CO2 has no warming effect ^.^’
Perhaps a better statement would be ‘History shows that there have been times when the warming produced by an increase in CO2 has been outweighed by other, as yet unclear, cooling mechanisms, leading to an overall drop in temperature’
In other words, its a bit more complicated than just CO2.
(Shine on You Crazy Dungmand……)

Gail Combs
August 6, 2010 2:19 pm

Mike Haseler says:
August 6, 2010 at 11:52 am
Julio says: “As a physicist, I appreciate your effort here, but I still think one needs to stress that the important point is that some of the energy reradiated by the CO2 in the atmosphere is sent downwards, that is, back to Earth. ”
That is why you can’t ignore the convective transport of heat through the atmosphere!…
_________________________________________
Actually I like the idea of posts that take tiny bites out of the whole complex subject.
Tom Vonk’s recent post, even though it had problems gave me a very clear picture of the actual mechanism of infrared energy absorption by molecules and how the energy moves around and becomes “heat” (increased velocity)
This post takes the next step. A non-equilibrium state. Hopefully we will see other posts that build on the knowledge gained from these two posts until we have a decent picture of what is happening in a very complex system.

Tommy
August 6, 2010 2:25 pm

The problem I have with Tom’s figure is that it doesn’t say where the reflection happens. Does some of it happen above the surface (such as by clouds)? I certainly think so. And is this reflection just sunlight? What about GHGs above the clouds, do they emit energy? Don’t the clouds reflect this energy too? It seems reasonable to believe so.
So it seems to me that wherever there are clouds, the majority of high-altitude GHG emission goes into space.
During the day I doubt it matters. Why? Because while their emissions are getting lost in space, they are receiving sunlight both directly as well as reflected by the clouds.
During the night I think it matters in 2 ways.
1. There is no sunlight, and yet the clouds keep on reflecting the GHG emissions. So it seems to me that wherever there are clouds, high altitude GHGs cool down more quickly at night.
2. Seems to me the same clouds would reflect GHG emissions from underneath too, but that prevents them from getting lost in space, so it can be re-absorbed by these low altitude GHGs. So it seems to me that wherever there are clouds, low altitude GHGs have trouble cooling down. Now, if high altitude air is cooling down at a faster rate than low altitude, what happens? Well, you get cold air above warm air. Then what happens? Convection.
With convection happening, you can now imagine warmer GHGs ascending above the clouds, where they can cool more quickly (point 1 above). But while this is happening, there are cooler GHGs descending below the clouds. Well, since they are GHGs, they can absorb the glowing warmth of the planet down here, collecting it. Since they are below the clouds, they have trouble getting it back out to space by emission. But this makes convection ever the more eager to raise them above the clouds again.
So let me summarize what I expect from reflection of clouds + GHGs:
Day:
– warming above
– shading below
– upper warm air opposes convection, trapping heat above
Night:
– cooling above
– retaining warmth below
– upper cool air aids convection, transporting heat from below
How much does this matter? I don’t know. So then my question to the scientists is: how well do clouds reflect the spectrum emitted by GHGs?

August 6, 2010 2:26 pm

Jeff Id: Great post. Nice to have the greenhouse effect explained in a clear way.
Now consider that the three primary suppliers of long-term surface temperature measurements (GISS, Hadley Centre, and NCDC) only include surface air temperature measurements and extrapolations for approximately 30% of the Earth’s surface. The remaining 70% is ocean, the temperature of which is measured at the top “x” feet.
The oceans have their own “greenhouse effect”: they absorb downward shortwave radiation to depths of approximately 100 meters and absorb longwave radiation at only the top few millimeters, but the oceans can only release heat at the surface. The amount of heat released from the oceans varies due many factors. One of these includes the direct impacts of ENSO, which increases the surface temperature of the tropical Pacific through changes in trade wind strength and changes in the strengths of surface and subsurface ocean currents. Those changes cause warm waters to be drawn from the Pacific Warm Pool and spread across the surface of the equatorial Pacific. (This rise in temperature of the tropical Pacific SST directly impacts the measured global temperature, because “x” % of the ocean rose.) Also included are the indirect impacts of ENSO outside of the tropical Pacific, which are caused by changes in atmospheric circulation. Sea surface temperature rises when wind speed decreases, and vice versa.
Just a few things that always seem to be overlooked when there are discussions of greenhouse effect.
Regards

latitude
August 6, 2010 2:27 pm

Michael J. Dunn says:
August 6, 2010 at 12:48 pm
Experimental proof, anyone? The predicted tropospheric “hot spot” does not exist. As Tom has pointed out, it cannot exist. This is a key falsification of the “global warming” theory.
=======================================================
Most excellent Michael, thank you for taking the time.
I’m tired of this “CO2 warming” business.
CO2 doesn’t warm anything, it insulates.

latitude
August 6, 2010 2:28 pm

Michael, I meant “all” of your post, not just the line I quoted.

Gail Combs
August 6, 2010 2:33 pm

ShrNfr says:
August 6, 2010 at 11:52 am
The last time I saw a first derivative of temperature with respect to CO2 concentration from a source I could respect, it was of the order of 0.00125 to 0.00250 degrees C per ppm of CO2…. Do you have the source handy?
…. The earth is not a greenhouse. There are no glass plates on the roof. Sometimes it takes that other greenhouse gas up with it and makes these funny things called clouds which make your “window” pretty darn reflecting instead of transparent. (Hint, that is why deserts are hot by and large).
________________________________________________
Actually thanks to the lack of H2O deserts are very hot during the day and very cold at night. It would be a great place to check the average temperature and then compare it to the average temperature of a humid area at the exact same latitude.
It would also be a great place to do some CO2 experiments similar to the open field “CO2 fertilization” experiments to see if a large increase in CO2 causes an actual change in temperature.

A Crooks of Adelaide
August 6, 2010 2:34 pm

Man, I’m way out of my comfort zone here but I thought the absorbtion was by photon capture by the electrons in the Carbon atom, and the energy went into boosting them into higher energy shells, not into kinetic energy of the molecule. Isn’t that what your absorbtion diagram is showing? They then re-emit the photon at the same energy level as they de-excite and this photon continues on its way either of into space, off to be captured by the next C atom or back down to the ground.
I have to admit I’m a geologist and so science isn’t my strong point.

Michael J. Dunn
August 6, 2010 2:37 pm

Can comment only briefly, so apologies.
Crazydung: Thanx and a Hatlo Hat Tip to Tom Vonk.
Jeff: Think of the “absorptive” medium as a “re-radiant” medium. Yes, the laser cavity will get warm, because it is solid material. The gaseous medium will care less, because it is shedding photons as fast as it is picking them up. (Good old spontaneous emission and the Einstein A and B coefficients. Takes me back to grad school!) Or, think of it as a Roaring 20s mirrored ball: redistributing all the energy over 4 pi steradians, but not retaining any. (This is the basis for an even longer discussion, but gas lasers earn their keep by creating flows that are in severe disequilibrium and things are happening so fast that the processes are called “rate chemistry.” In fact, a laser’s inversion state is characterized by a negative temperature. Strange but true.)
Pamela Gray: The molecular weight of CO2 does not matter. Molecular diffusion in the atmosphere is so pervasive and effective that the atmosphere is essentially homogeneous to the upper stratosphere (where ozone production starts to mess with this picture). Parcels of air rise when they are heated by the ground (or ocean). Sometimes the buoyancy is influenced by the addition of water vapor, which is lighter than air. (It is the buoyancy of water vapor that holds clouds up; all the water droplets are actually heavier than air. The droplets maintain an equilibrium with the water vapor and are held up as an aerosol.) Air is best considered as 4/5 nitrogen and 1/5 oxygen, with seasoning in the form of argon, CO2, H2O, and trace gases. Excepting water vapor, the trace gases have no effect on air’s thermodynamic properties or behavior.
Khunkat: I think I have to defend Jeff’s graph. Yes, if the Earth were situated at the surface of the Sun, the Sun’s radiance would outstrip Earth’s all across the spectrum…which is the result that you get when you apply the Stefan-Boltzmann equation right out of the parking lot. But the Earth is 93 million miles away, which attenuates the radiance, so that the portion of the solar spectrum in the far infrared is much lower than the right-in-our-faces Earth emission spectrum.
There’s a lot of physics out there. All of it on the side of truth.

jorgekafkazar
August 6, 2010 2:39 pm

Steven mosher says: “…Like you I believe in moving the discussion FORWARD….”
Amen!!
“…And just for good measure we will throw in some appeal to authority: Spenser….”
I particularly like his Faerie Queen.

August 6, 2010 2:49 pm

One thing which I mentioned on tAV was that the perfect insulating cylinder isn’t quite as non-physical as it might seem.
If you think of the atmophere which is basically in equilibrium and define a 1 meter cylinder boundary from ground to orbit. Energy which escaped from the side of the cylinder will come back in from adjacent air on the other side of the cylinder. (Infinite insulation).
If you think of energy re-striking the ground, an equal energy is coming back up, also infinite insulation.
So you have a perfectly insulated cylinder with a perfect emitting window at the top. — not so bad an example I think.
DeWitt Payne gets credit for it though.

Gail Combs
August 6, 2010 2:51 pm

Pamela Gray says:
August 6, 2010 at 12:42 pm
Gail, the Dewey denigrating article you linked to is a bit out of date, or if currently being spread, is somewhat out of touch. While I understand that historically, whole language become the preferred/only (and very unfortunate) method of reading instruction, it is no longer the case….
____________________________________________-
As a victim of the reading method described I am glad it got trashed. However the historic facts of Dewey’s effect on the education of most of us who are now adults is still visible. It was for the history I linked to the article.
“For 10 years, William Schmidt, a statistics professor at Michigan State University, has looked at how U.S. students stack up against students in other countries in math and science. “In fourth-grade, we start out pretty well, near the top of the distribution among countries; by eighth-grade, we’re around average, and by 12th-grade, we’re at the bottom of the heap, outperforming only two countries, Cyprus and South Africa.”
http://www.enterstageright.com/archive/articles/0804/0804textbooks.htm
…the U.S. ranks 21st out of 29 Organization for Economic Cooperation and Development (OECD) countries in mathematics scores, with nearly one-quarter of students unable to solve the easiest level of questions….In 2000, 28 percent of all freshmen entering a degree-granting institution required remedial coursework
http://www.edreform.com/_upload/CER_JunkFoodDiet.pdf
Our education system is still crap and it traces back to John Dewey. If you have a better reference I would be happy to see it.

George E. Smith
August 6, 2010 3:00 pm

Jeff,
The only place I would have any disagreement with you is in the numbers. I have the same set of atmospheric absorption spectra you have; but I don’t know definitively whether they are calculated or measured; but not too important.
Some people have criticised the incoming and outgoing BB spectra arguing the vertical scales are wrong (including me).
Of course the direct solar BB spectrum is orders of magnitude higher than depicted; BUT that spectrum is attenuated by the inverse square law down to the extra=terrestrial TSI of 1366 W/m^2 or to a somewhat distorted (AM = 1) ground level spectrum at about 1000 W/m^2 .
Your 255 K earth emission source would give only 240 W/m^2, and one would then claim that the 1kW/m^2 only falls on 1/4 of the surface while the 240 is emitted from the entire surface.
I would disagree the 1000 W/m^2 certainly only falls on a portion of the surface; but the effect of that on that portion of the surface is quite different from having 250 W/m^2 fall on the entire surface.
And the 255 K is way too low to correctly represent the outgoing. Even at the mean surface temperature which is allegedly 288 K the outgoing would be 390 W/m^2; but in reality significant amounts of the surface are a lot hotter than 15 deg C so they radiate a more intense and shorter wavelength spectrum than your 255 K source which would peak at 11.4 versus 10.1 for the 288 K source or as low as 8.7 for a hot 60 deg C desert surface; which will emit 1.8 times what a 288 K source emits.
These are small differences I know; but I don’t think they should be ignored because the match between the actual GHG absorption spectra; and the real surface radiation emission spectra is critical for calculating the correct amounts of energy intercepted by GHG.
But I am quite in agreement with your end conclusion that it is all moot because I believe that cloud feedback is ALWAYS negative and highly so as is hinted at in Wentz et al; “How Much more rain Will Globalk Warming Bring ?” SCIENCE for July 7 2007.
Rmember when we talk about cloud feedback what we mean is something like:- If average global cloud cover increases from its present value by say 1% for the next 30 years (climate time scale) will that heat or cool the earth; and the answer is unquestionably it will cool the earth. Wnetz et al results hint that cloud cover; at least the precipitable component of cloud cover in terms of area, optical density, and persistence time should increase by about 7% for a one deg C increase in average global surface temperature; since their measurments show that Global evaporation, total atmospheric moisture and global precipitation ALL do increase by 7% per deg C Temperature rise; and I don’t know about you; but I like to have clouds with my precipitation; I’m kinda funny that way.
So I think the negative cloud feedback is huge, and totally swamps any effect that CO2 or any other GHG could have.
ONLY water (H2O) exists in the atmosphere in all three ordinary phases of matter; and it is the physical and chemical and other properties of water that totally regulate the range of temperatures on earth to the extent that even solar fluctuations get washed out by cloud modulation over the long haul.
I’d like to see the incoming solar spectrum curve at a 4x amplitude relative to the outgoing since the day time surface temperatures reached would never be reached with only 1/4 of the insolation.
On average, absolutely nothing ever happens. Mother Gaia does not do statistical mathematics; she works in real time.

Stephen Wilde
August 6, 2010 3:01 pm

Bob Tisdalr said:
“The oceans have their own “greenhouse effect”: they absorb downward shortwave radiation to depths of approximately 100 meters and absorb longwave radiation at only the top few millimeters, but the oceans can only release heat at the surface.”
Thanks Bob. That is my ‘Hot Water Bottle Effect’ in a nutshell and it’s hugely greater than the so called Greenhouse Effect.

Stephen Wilde
August 6, 2010 3:05 pm

George E. Smith said:
“ONLY water (H2O) exists in the atmosphere in all three ordinary phases of matter; and it is the physical and chemical and other properties of water that totally regulate the range of temperatures on earth to the extent that even solar fluctuations get washed out by cloud modulation over the long haul.”
Thanks, George. That would support my contention that the regulatory mechanism is the speed of the hydrological cycle.

A Crooks of Adelaide
August 6, 2010 3:06 pm

Further to my post above …
If it was as simple a transfer to kinetic energy as you imply, then why wouldn’t the atoms absorb at all wave lengths and just result in atoms with different speeds? Your explanation doesn’t explain the limits in the absorbtion bands. And if its not an energy transfer into kinetic energy but into molecular energy, does that mean the atoms actually dont increase in temperature?
My devil’s advocate position would say that the Carbon atoms would absorb all the photons they could until all the electrons occupy all the available higher orbits and then the rest would simply have to pass through.
I might add that intuitively, I think your system ought to warm, but I’m not too happy with your explanation.

p. solar
August 6, 2010 3:12 pm

@pamela:
Does it rise all by itself? YES. It’s called gas diffusion. Denser gases will only stay at the bottom of a volume for a limited time, until they diffuse.
, nice simple explanation but …
>>
If feedback is high and positive as the models predict, then …
>>
AFAIK the models ASSUME high positive feedback, they do not predict it.
“climate sensitivity” is purely a fiddle factor to make make naive models which do not attempt to take “internal variability” into account, do what they want them to do.
The trouble is “internal variability” means things like changes in ocean currents (PDO etc.) and changes in cloud formation.
Having removed nearly all that causes climate variation , they have to scale up the real CO2 effects, predicted by real physical models, and get a very bad fit to climate data.
“climate sensitivity” is an arbitrary , fictitious fiddle factor to make the simplistic model fit the outcome they assumed before starting the model.
They then rerun the model without the fictitious man made warming and it does not fit at all. This is presented as “proof” of the initial models validity.
http://www.metoffice.gov.uk/climatechange/guide/quick/doubts.html
>>
When natural factors alone are considered, computer models do not reproduce the climate warming we have observed. Only when man-made greenhouse gases are included do they accurately recreate what has happened in the real world.
>>
this is doubly misleading since the models do not “accurately recreate” anything.

Stephen Wilde
August 6, 2010 3:12 pm

Gail Combs kindly referred us to this
Research Article: Automated Observations of the Earthshine
http://www.hindawi.com/journals/aa/2010/963650.html
“For a decade, we have been measuring the Earth’s reflectance by observing the earthshine, which is sunlight reflected from the Earth to the Moon and retroflected to the nighttime Earth. …..The earthshine observations reveal a large decadal variability in the Earth’s reflectance [7], which is yet not fully understood, but which is in line with other satellite and ground-based global radiation data…”
I would be inclined to bet that the variations in Earthshine and the consequent albedo changes will eventually be found to be directly related to the average latitudinal positions of the cloud bands of the various jet streams and the ITCZ. The variations being due to changes in the angle of incidence of solar shortwave energy as those clouds move poleward and equatorward beyond normal seasonal variability over centuries.

Gail Combs
August 6, 2010 3:18 pm

Roy Clark says:
August 6, 2010 at 12:47 pm
There should be no doubt that the downward LWIR flux from both CO2 and H2O in the atmosphere exchanges LWIR flux with the surface and helps keep the surface warmer than it otherwise would be….
___________________________________________________________
Roy how about an article about this with explanations that can be understood by lay people. As others have mentioned Al Gore had great influence because he kept the message simple. We need to refute him with science, but science that the majority of people can understand.

jorgekafkazar
August 6, 2010 3:24 pm

Pamela Gray says: “…While I understand that historically, whole language bec[a]me the preferred/only (and very unfortunate) method of reading instruction, it is no longer the case….“Read Naturally” [is] often touted as a computer based reading fluency intervention but is at the top of my bad list. It is a whole language holdover that needs to be entirely removed from our schools, burned till nothing is left but ashes, and the authors tarred and feathered.”
Yes, one of my pet peeves. Another tragedy is that the “look-say” method of reading instruction has been paralleled by the “think-say” practise of speech, in which every single thought of the individual is immediately expressed, whether it has any merit or not.

Stephen Wilde
August 6, 2010 3:25 pm

Jeff Id said:
“For those who are discussing convective heat transfer, of course you are right that this happens but it is related to the magnitude of the warming effect rather than whether the effect exists.”
I don’t think that’s quite right because it doesn’t clearly define ‘warming’.
If extra downward IR from more CO2 causes more (or more accurately, earlier /accelerated ) evaporation then the surrounding environment cannot warm because evaporation is a net cooling effect.
Instead the additional latent heat in the evaporated water vapour makes the body of air containing that water vapour lighter so that it rises with an increase in convection but because all the extra energy is in latent form there need be no (possibly cannot be any) discernible temperature increase.
I think that resolves a lot of confusion. There is ‘warming’ of a sort but only by way of more energy in the air in the form of latent heat in water vapour. So you do not necessarily need higher temperatures to create faster convection.
I’m glad I just thought of that because it disposes a problem I’ve had for some time with warmists who say there must be some warming from extra downward IR over water.
Clearly not so, simply because water vapour is lighter than air and more water vapour in a given volume of air will provoke faster convection (and thus a faster hydrological cycle) with no discernible warming at all.

Stephen Wilde
August 6, 2010 3:35 pm

Scott said:
“according to Roy Spencer, all of the supposed warming in recent decades can be accounted for with only a 2% change in cloud cover.”
Well I reckon you could get that from the latitudinal shift in the cloud bands that we have actually observed over the period 1970 to 1995.
Since then the cloud bands have been going back equatorward again and hey presto albedo is increasing.
http://wattsupwiththat.com/2007/10/17/earths-albedo-tells-a-interesting-story/
This climate stuff looks to me like turning out to be quite simple after all. The only thing we need to ascertain is the precise cause of those latitudinal shifts and I’ve already set out my ideas on that here and elsewhere.

Gail Combs
August 6, 2010 3:37 pm

donald penman says:
August 6, 2010 at 1:01 pm
So the idea that co2 does not warm the atmosphere is to be dismissed with a thought experiment,I am open to the idea that co2 does warm the atmosphere but have not seen any proof yet….
______________________________________________________________
Donald go back to Tom Vonk’s recent post and read it very carefully. A photon packet of energy is absorbed by a CO2 molecule and an electron goes from the rest to the excited state. This does not change the velocity of the molecule unless it collides with another molecule and the energy is translated into velocity. velocity = heat.
The second part is for every excited CO2 molecule that collides with another molecule and increases its velocity, there is an equal number of “high speed” molecules colliding with CO2 and bumping the electron into an excited state.
The net effect is essentially zero AT LOCAL EQUILIBRIUM.
Tom’s post: http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/

August 6, 2010 3:40 pm

Gail Combs says:
August 6, 2010 at 11:53 am
A “perfectly transparent” window material is a plate of salt (NaCl) it is used in chemistry to hold the test sample in Infrared Spectrophotometers.
It was used by Woods in 1909 to show the true” greenhouse effect” found in greenhouses was from the normal glass allowing the sun’s energy (high wavelengths) to pass into the greenhouse but trapping the infrared energy (lower wavelengths) so it could not be re-radiated out. A greenhouse built with salt glass did not become warm.
************************************
Gail, this is completely untrue. Please read the old 1909 Woods paper. The greenhouse made with salt was almost identically warm as the greenhouse made with glass. Woods showed that the way a greenhouse works is by preventing convection – nothing at all to do with blocking IR radiation. The idea that, somehow the glass ‘blocks’ the infrared emissions, and that this is the mechanism that warms greenhouses, was utterly debunked by Woods in 1909. Unfortunately, the idea that, as you put it, greenhouses work by “normal glass allowing the sun’s energy (high wavelengths) to pass into the greenhouse but trapping the infrared energy (lower wavelengths) so it could not be re-radiated out” is a complete myth. The atmospheric ‘greenhouse effect’ is not in any way like the way greenhouses get warm: if find it incredible to think that people still believe this after its debunking over 100 years ago.

Jordan
August 6, 2010 3:41 pm

I have no issue with moderate GH warming, but would like to put up another thought experiment. (This could be tested in practice.)
I have a long tube full of CO2 (looking like a telescope). It has a transparent window at one end and the other end (the closed end) is a solid black surface. The sides are well insulated so it can only receive and radiate through the window.
I launch the tube to a position above our atmosphere and point it into to the sun. This raises the temperature of the black surface at the closed end.
I have an identical tube, except that it is empty (or filled with a non-GHG). I launch this and place it alongside the first tube, pointing into the sun. The temperature of its black surface also rises.
The physics discussed on this thread suggest the closed end of the first tube will rise to a higher temperature compared to the second tube. So I use this “potential difference” to drive an engine: the closed end of the tube full of CO2 is the hot reservoir and (purely for for demonstration) the closed end of the second as the engine heat sink. Both pointing directly into the sun.
The engine would appear to violate a thermodynamic concept that we cannot expect a practical engine to operate if the sun is the ultimate heat source and heat sink.
However, the CO2 has introduced a frequency shift between the incoming radiation and the outgoing radiation. We could call this machine a “thermal diode”?
Question for the physicists: does this engine do any useful work?

Kevin Kilty
August 6, 2010 3:43 pm

Jeff has produced a good explanation of the basic physics, and my only complaint is that the diagram shows all radiation leaving via the distant end of the tube, which is not precisely what happens (it is what happens net after the tube reaches equilibrium).
However, there is the devil in the details still. People have already begun to add clouds, ocean surface, and other complications to the issue. Here are a couple of thoughts I have and if someone can add details to these I’d appreciate it.
1) Jeff’s diagram is probably based on calculations using MODTRAN. I assume MODTRAN has been validated with measurements, but maybe it has not. Undoubtedly it could be validated only in certain bands, or perhaps on average over large portions of spectrum. Does anyone know?
2) Jeff did not mention the detailed calculations which probably depend on the assumption of local thermodynamic equilibrium (LTE), a topic on Tom Vonk’s thread yesterday. LTE allows one to apply the Planck function to the distribution of radiation at any assumed temperature. However, the Planck function applies to cavity radiation, and in the case of the atmosphere there is no cavity. LTE does not strictly apply unless radiation is fully coupled to molecular states of translation, rotation, etc, and obviously this is not the case within the atmospheric window near 10um, and maybe is not the case in many other regions of the IR spectrum. Anyone know?
3) The window I mention in 2) is to some degree a thermostat. Discussion is generally limited to the impact of CO2, but water vapor dimers and trimers may have bands within the window. Thus a moistening of the atmosphere, and increasing polymerization of water vapor may have a big impact. I’ve alluded to this before in regard to El Nino. I’ve been looking for detailed info on the internet, but there is little. Anyone know of work along these lines?

David, UK
August 6, 2010 3:49 pm

Many thanks to Jeff for a nice, easy-to-understand-for-laypeople piece! The inappropriately-named GH-effect does of course exist – but so of course do a zillion other influencing factors, which collectively surely must add up to a self-regulating climate in the long term. Or else the earth would not have survived this long without melting. This layperson is more worried about corrupt governments (is there any other kind?) than anything nature can throw at us – as devastating as nature can be.

George E. Smith
August 6, 2010 3:59 pm

“”” Michael J. Dunn says:
August 6, 2010 at 2:37 pm
Can comment only briefly, so apologies.
Crazydung: Thanx and a Hatlo Hat Tip to Tom Vonk.
…………………………..
Khunkat: I think I have to defend Jeff’s graph. Yes, if the Earth were situated at the surface of the Sun, the Sun’s radiance would outstrip Earth’s all across the spectrum…which is the result that you get when you apply the Stefan-Boltzmann equation right out of the parking lot. But the Earth is 93 million miles away, which attenuates the radiance, so that the portion of the solar spectrum in the far infrared is much lower than the right-in-our-faces Earth emission spectrum. “””
Michael; while the spectrum of the solar radiation is approximately the 5780 K (or pick your own number) that Jeff postulates; the Stefan-Boltzmann emittance of course is attenuated by the square of the ratio of sun radius to earth orbit radius to finally yield thge 1366 W/m^2 at our orbit location.
I disagree with those who want to divide this number by 4 after applying some albedo reduction to get some puny 240 W/m^2 all over the earth surface. That insolation level will never raise a desert surface temeprature to +60 deg C or even as high as +90 deg C for some black asphalt surfaces; and it is those much hotter than 288 K sureal surface temepratures that are responsible for the greatest earth cooling effect during the heat on the noonday sun.
Mother gaia does NOT wait till after sunste to start cooling her place down; she cools it 24/7 as the saying goes; in fact Mother Gaia probably invented the term 24/7 because she operates always in real time with real instantaneous vlaues and as I have said ; does NOT do statistical mathematics.
Working with average numbers instead of instantaneous values gets you on average that nothing ever happens.
It is the hottest dryest desert surfaces on earth that are cooling the planet via radiation; not the Arctic, and Antarctic wastelands that are puny radiators by comparison.

Gail Combs
August 6, 2010 4:01 pm

Alexander K says:
August 6, 2010 at 1:16 pm
Gail Combs says ‘don’t you believe that stuff (about a good education), Alexander.
Sorry Gail, but I meant a non-dumbed down education…
_______________________________________________________
On that we certainly agree. My Hubby’s Grandfather came over from Europe, did not speaking any English, had a 6th grade education and still read his way through the entire public library… There is Education and then there is “education”

Editor
August 6, 2010 4:04 pm

joshv says:
August 6, 2010 at 10:44 am

“From the perspective of 15 micrometer wavelength infrared laser, the CO2 filled air is black stuff.”
I believe this is incorrect. Chemical bounds within CO2 absorb the energy of specific photons, and at equilibrium emit them at the same rate, though not necessarily in the same direction, as they are whizzing around, smacking into each other. It’s not black stuff, it’s “white” stuff, like a cloud. Does a 15 micrometer detector pointed at the earth from space see a black ball?

The detector doesn’t see the IR photons radiated by the ground – they’re absorbed but eventually IR photons from CO2 higher up are released and make it to the detector, so yeah white will do. If you could afford a broad spectrum color camera, then you’d see the Earth mostly okay, but covered with a yellow (or whatever color it displays for 15 um) haze – and a brighter haze in humid regions where H2O emissions join in.
Back to the laser – the laser shove photons in – they don’t make it out. The laser might call it black. However, the whole gas does glow in that IR yellow and gets brighter when the laser is on. Sort of translucent/milky, except that’s due to scattering photons, not absorbing and emitting photons.

George E. Smith
August 6, 2010 4:13 pm

“”” kuhnkat says:
August 6, 2010 at 1:27 pm
Jeff Id,
You ignore the FACT that the incoming IR from 1-4 microns is much larger than the outgoing through the earth’s complete power range among other things. It is really hard to compute RTE’s when you ignore large segments of the flux and concentrate on one area.
The graph you show has the sun’s output scaled to under 10 -6. Even Science of doom only used a graph at 10 -6 for the sun to get it on the same graph. Naughty naughty. “””
Well I don’t think Jeff really ignored anything. 98% of the solar energy is contained between about 250 nm and 4.0 microns, with only 1% beyond each end of that range; actually less than that at the UV end But water vapor intercepts a considerable amount of that 1-4 micron incoming, in fact H2O starst at around 750-60 nm or so. CO2 on the other hand hardly kicks in to the 4 micron band and as I have said there is less than 1% of the solar energy beyond that.
But you have to take into account the inverse square law attenuation of the Sun’s st3efan-Boltzmann like emission to arrive at the 1366 W/m^2 TSI at earth’s orbit.

Gnomish
August 6, 2010 4:33 pm

Gail Combs says:
August 6, 2010 at 1:14 pm
NOPE.
Absorption of IR causes an electron to go from the ground state to the excited state. VELOCITY of the molecule is what we call heat. You missed a step.
____________________________________________________________
Gail- you have it precisely backwards. It is important to be able to distinguish between heat and temperature. Please fix your idea. VELOCITY of the molecule is what we call TEMPERATURE.
Heat is something completely different and is NOT measured with a thermometer at all.

George E. Smith
August 6, 2010 4:35 pm

“”” Stephen Wilde says:
August 6, 2010 at 3:05 pm
George E. Smith said:
“ONLY water (H2O) exists in the atmosphere in all three ordinary phases of matter; and it is the physical and chemical and other properties of water that totally regulate the range of temperatures on earth to the extent that even solar fluctuations get washed out by cloud modulation over the long haul.”
Thanks, George. That would support my contention that the regulatory mechanism is the speed of the hydrological cycle. “””
I’m not up on your contention; and don’t know quite what you mean by “the speed of the hydrological cycle”
From my point of view it is a quasi static situation with some average amount/density and persistence of cloud cover globally. If the CO2 content of the atmosphere is raised; which would tentd to raise atmospheric temepratures and surface temepratures; that simply leads to more evaporation on a persistent basis; which leads to more clouds in order to get more precipitation and that means more clouds which swamp the effect of the CO2.
Likewise cosmic rays or volcanic dust or arosols which enhance cloud formation simply allow more cloud cover at lower temepratures, so the temperature falls.
All one has to do is do the two mental experiments I have proposed which I call the “Birdseye” experiment after the inventor of quickfrozen foods; and the “Venus Experiment which is its opposite.
In the former one simply cools the entire atmosphere and surface down to zero deg C; unless it is already colder, and then removes the remainder of the atmospheric H2O down to the last molecule, and drop them all on the surface in whatever phase was originally there.
Now Peter Humbug says he already did this on his Playstation except he apparently didn’t drop the temperature to help get rid of the water; he just excommunicated it all; and in his experiment he says he got it all back in three months. I prefer to drop the Temperature to zero but do not freeze the oceans (they don’t freeze at zero anyway).
Absent water vapor in the atmosphere, and hence no clouds, the earth albedo is vastly reduced, and the ground level insolation soars in the mother of all forcings.
Evaporation begins on a massive scale, and the earth returns to some stable state where cloud cover balances against further warming.
IN the Venus experiment we heat the atmosphere and surface to something very high maybe +60 deg C, and we install clouds from ground to say +20 km all over the earth from pole to pole; again without melting all the ice that exists.
Now the ground level insolation is virtually zero so precipitationa nd cooling will begin on a massive scale; and it will rain I am sure for 40 days and 40 nights, until eventually some sunlight will make it to the ground and start to warm the place up again till you reach some new stable temperature where cloud cover balances the situation.
I’m not aware of any Physics that would cause the condition reached fromt he Birdseye experiment to be different from that reached after the Venus experiment. In short; there can be no tipping point that leads to thermal runaway.
Now all that presupposes that earth’s orbit does not shift enough to kick in a new ice age.
Carl Sagan went to his grave having never detected so much as one single digit (bit) of scientific data evidence of some extra-terrestrial intelligence; what a waste.
I wouldn’t want to go to mine having wasted my life stepping into and out of cloud shadows, to try and find a cloud shadow that warmed me up instead of cooling me down.
And I wouldn’t grant so much as a brass razoo to any researcher to try the same study.

Gnomish
August 6, 2010 4:40 pm

JeffID
Sometime chat with somebody in industry who actually uses lasers for delivering heat to materials (as for cutting shapes- maybe ask at Tap Plastics).
If you don’t remove the vaporized work material, the laser can not deliver the energy to the surface to do the desired work.
Therefore, blowers are used and the laser may be pulsed to allow vapor removal.
If this is not done, one is putting the watts into a cloud and wasting it.

Gail Combs
August 6, 2010 4:45 pm

Spector says:
August 6, 2010 at 1:50 pm
RE: Gail Combs says: (August 6, 2010 at 1:05 pm) “The photon absorbed and the photon emitted should be exactly the same energy value.”
That only applies if there is absolutely no change in the energy of the entity that absorbed the photon in the first place. If you get $20 and it is stolen, then you cannot spend it. If the robber only takes $10, then you might be able to spend the other $10.
In the molecular world, the robbery may take place as a collision between two molecules. Such a collision may also result in the donation of energy.
_____________________________________________________________________
I guess I was not clear. I thought that is what I said.
case# 1. Photon excites CO2, CO2 emits photon of same wavelength and returns to rest state.
case# 2. Photon excites CO2, CO2 collides with another molecule of air and returns to the rest state. The 2nd molecule absorbs the energy in any of a variety of ways including and increase in velocity (increase in kinetic energy) or heat.
However I was under the impression the excitation and emission energy for a specific molecule, and for a specific type of energy (rotational or what not) had to be in discrete “packets” that conform to the wavelengths shown in figure #2.

Scott
August 6, 2010 4:56 pm

Stephen Wilde says:
August 6, 2010 at 3:35 pm
Hi Stephen. Thanks for the link…I was unaware of that information. What I really liked was the first comment on the article you linked though…the perfect question to greenhouse gas warming. 🙂
-Scott

Derek B
August 6, 2010 5:07 pm

Great post, Jeff. Just one weak point at the end: that the uncertainty over feedbacks “makes true and high quality results from Anthony’s surfacestations project so critically important”. The problem is that it is extremely hard to know where all the heat goes, so the time taken to reach equilibrium is also unknown. If e.g. there is more mixing with deep ocean than expected then the observed warming at the surface will be at a lower rate but persist much longer. To observe the actual warming with any confidence will take several times the density of metering that we have today, and fifteen or twenty years of data from them. If the alarmists are even half right, that’s time we don’t have. So, much as we distrust them, we are reduced to depending on models and basing policy on risk minimisation. Meanwhile, we do know that there was significant surface warming over the 20th century as a whole. Yes, there are several candidate explanations, but none of them can claim to be more convincing than the known rise in CO2.

Gnomish
August 6, 2010 5:07 pm

Heat is measured in :
joules
calories
BTUs
ergs
watt hours
dyne meters
electron volts
newton meters
poundal feet
NOT DEGREES

Gail Combs
August 6, 2010 5:15 pm

Pamela Gray says: “…“Read Naturally” [is] often touted as a computer based reading fluency intervention but is at the top of my bad list. It is a whole language holdover that needs to be entirely removed from our schools, burned till nothing is left but ashes, and the authors tarred and feathered.”
___________________________________________
jorgekafkazar says:
August 6, 2010 at 3:24 pm
Yes, one of my pet peeves. Another tragedy is that the “look-say” method of reading instruction has been paralleled by the “think-say” practise of speech, in which every single thought of the individual is immediately expressed, whether it has any merit or not.
_______________________________
I got stuck with that “look-say” method of reading. UGH I do have a very high reading speed but I can not pronounce or spell words correctly, I can not read out loud and I also have a problem connecting the idea to the spoken word in conversation. I got some phonetics from my parents at home thank goodness, otherwise I would have had a much harder time of it in school.
It also screwed me up completely with trying to learn a language. Especially when I got stuck in a “total immersion” experimental French class with no text books and no English spoken. I learn through sight not hearing so it was a complete disaster (my only F)

Gnomish
August 6, 2010 5:17 pm

We don’t live in a thermometer. We live in a dynamic heat pump.

August 6, 2010 5:23 pm

Jeff Id,
With all due respect, your post did not address the same topic as Tom Vonk but was titled “A reply to Vonk . . . “.
WUWT?
John

August 6, 2010 5:27 pm

Derek,
Don’t misinterpret my post for an alarmist, believer or even we should do something about it one. It’s just the initial physics, the rest is up for disagreement.
This is one of the key’s to ‘settled science’. They use the line and revert to the basics, if skeptics don’t respond intelligently to the basics, their understanding is weak and not worth the time.
Once the basics are agreed to, the rest is up for discussion, because the rest of the physics are not certain. It’s called uncertainty in climate, but I call it unknowainty because there is a difference.

August 6, 2010 5:31 pm

John Whitman :
A quote from Tom Vonk’s post:
“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.
They are not ‘wrong’, but their magnitude and effects are definitely in question.

George Steiner
August 6, 2010 5:31 pm

If a CO2 molecule encounters a photon at 15 nm then immediately emits a photon at 15 nm, did the CO2 molecule beacame warmer? Became colder? Became neither warmer nor colder?

NoMoreGore
August 6, 2010 5:48 pm

Stephen Wilde says:
August 6, 2010 at 3:35 pm
Stephen,
I Really liked your hot water bottle post. The atmosphere just transports heat. The oceans are the repository.

Merrick
August 6, 2010 5:48 pm

Gail Combs:
Nope. Absorption of an IR photon causes a vibrational excitation. Electronic excitations require visible or ultraviolet photons.
And, no, velocity and heat are not interchangeable. Heat is rigorously defined only for systems in thermodynamic equilibrium. That means that not only is there energy in kinetic (external) energy of the atoms and molecules but that there is also energyin thermally available rotations and excitations.
Doesn’t anyone other than physical chemists understand this?

Kevin
August 6, 2010 6:04 pm

Ok, that’s a nice example showing that a CONFINED volume of a gas will selectively absorb and then remit specific wavelengths of light. While a bit fancier than the plastic soda bottle filled with CO2 in front of a light bulb it does not show anything new. Yes indeed gases can selectively absorb certain wavelengths of light, then the gas warms, then the gas emits energy at the same wavelength in all directions, then in a final and very important step, the gas cools by an amount equal to the energy reemitted.
Some more salient observations are;
No point on the surface of the Sun, on the surface of the Earth, or in the gaseous atmosphere of the Earth ever reaches an “equilibrium” temperature. As any point reaches a new higher temperature it immediately starts to cool faster. If this faster cooling is not overwhelmed by another larger warming force it will cause the temperature of that location to start to drop.
As the Sun illuminates the Earth’s surface (with mostly visible light) the surface “races” to catch up to the new “equilibrium” temperature. However, due to the thermal capacities and thermal conductivities (combined they determine the speed of heat) in the materials involved the surface never reaches the “equilibrium” temperature predicted by Kirchhoff’s Law.
As the Earth’s surface “races” to catch up to the new higher “equilibrium” temperature the gases in the atmosphere are also “racing” to catch up to the new higher “equilibrium” temperature at their location.
After the Sun sets the Earth’s surface temperature now switches direction and begins “racing” towards the new lower “equilibrium” temperature, likewise for the gases in the atmosphere.
The whole idea of an “equilibrium” temperature is very useful for a Pizza oven with a thermostat, but it is useless in understanding a system that consists of many different materials each of which respond differently to energy inputs. The only common denominator is that as each material warms it will respond by cooling faster. Some materials (i.e. seawater) radiate across a very broad spectrum, but mostly in one direction. Other materials (i.e. “greenhouse” gases) absorb and radiate selective wavelengths in all directions. Still other materials (i.e.”non-greenhouse” gases) exchange energy with any nearby material mostly via conduction and convection.
The only important question is;
Does the substitution of very small quantities (tens of parts per million) of “non-greenhouse” gases with “greenhouse” gases change the speed at which the temperature at any location “races” to meet up with its new ”equilibrium” temperature ?
My postulated answer is;
The replacement of “non-greenhouse” gases with “greenhouse” gases actually works to increase the SPEED OF HEAT through the atmosphere since more heat/energy travels at the speed of light (i.e. Infrared Radiation) versus at the Speed of Heat (much slower). The result of this is that each location more closely approaches its “equilibrium” temperature during each “Sunlight” .vs. “Sans Sunlight” cycle (i.e. Each Day). This does not cause any permanent increase in the “average” surface temperature of the Earth.
Now you might argue that this causes higher daytime and lower nighttime temperatures, but I think if you do some simple calculations you will find that this effect is so small that we probably could not spend enough money to measure it. For example, calculate the thermal capacity in Joules per cubic meter of sea water and compare it to the thermal capacity in Joules per cubic meter of the atmosphere when it contains 4% water vapor. Or for another analogy, try docking the Queen Mary using a couple of Radio Controlled Hobby Boats, yes, in theory it can be possibly be accomplished, but I don’t want to be around to see the attempts……….
In summary, I do not deny that some gases in the atmosphere selectively absorb and reemit energy (in the form of infrared light) backwards towards the Earth’s surface. I do deny that this can in any conceivable way cause the Earth’s surface to assume a new higher “equilibrium” temperature.
Cheers, Kevin.

Steve Fitzpatrick
August 6, 2010 6:21 pm

Kevin,
Steve Mosher’s comment (August 6, 2010 at 10:06 am ) was addressed to Jeff, but I think it was really written for you.

It's always Marcia, Marcia
August 6, 2010 6:21 pm

Again, I will repeat, co2 does not cause warming in earth’s atmosphere because of it’s effects on H2O. Roy Spencer’s work shows a negative feedback from H2O. Yes co2 does have warming effects on its own and it would stay a warming effect if not for far more powerful forces at work in the atmosphere that keep co2’s effects easily in check. H2O comprises 95% of GHG’s. It is the largest factor in GHG’s and it is the factor that should be getting the most attention in studies.
My oh my how we’ve fretted over CO2 for far too long.

August 6, 2010 6:21 pm

Stephen Wilde: You wrote, “Well I reckon you could get that from the latitudinal shift in the cloud bands that we have actually observed over the period 1970 to 1995.” And continued, “Since then the cloud bands have been going back equatorward again and hey presto albedo is increasing.”
And as always when you write this, I will ask, have you found a dataset that illustrates this latitudinal variation in clouds from 1970 to 1995 and back toward the equator since then?

August 6, 2010 6:28 pm

The complexity of Jeff’s post, obscures a simple reality. Realize that the radiation which is reflected in the 15 nm range is very close to constant. Certainly increasing or decreasing the amount of CO2 in the atmosphere does nothing to increase the amount of 15 nm radiation being reflected from the earth. The real issue is, how much does incremental CO2 increase the greenhouse gas effect. Dr. Heinz Hug has done experiments which indicate that doubling CO2 to 700 PPM, would likely result in increasing the temp. or the earth by only a measly .015 deg C.
http://www.john-daly.com/artifact.htm
I thought exercise which can be used to understand why this true might be helpful. Picture a sunny room on a bright day. Now attempt to completely darken the room by adding layers of shades. The first shade leaves only 30% of the incoming light, the next shade leaves only 30% of the light passing through shade 1, which leaves only 9% of the original light level. The next shade leaves 2.7%, shade four leaves .81% etc.. by shade seven, it is nearly pitch black since only .022% of the original light now enters the room. Adding more shades will simply not make much difference to the brightness of the room since it is already 99.98% dark after shade seven. This is the similar to what happens when CO2 is added to the atmosphere since most of the 15 nm range of IR energy available to CO2 is absorbed by the first 300 ppm of CO2. Raising CO2 from .03% to say .07% of the atmosphere will simply have no real impact on global warming. Vegetation however, would flourish.

August 6, 2010 6:34 pm

Jeff Id says:
August 6, 2010 at 5:31 pm
John Whitman :
A quote from Tom Vonk’s post:
“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.
They are not ‘wrong’, but their magnitude and effects are definitely in question.

——————-
Jeff Id,
I think there is a misunderstanding of the thrust of Tom Vonk’s post and the thrust of your reply to it. Maybe my solution is to read them both for the third time? OK. I will.
John

Gail Combs
August 6, 2010 6:38 pm

Derek B says:
August 6, 2010 at 5:07 pm
…So, much as we distrust them, we are reduced to depending on models and basing policy on risk minimisation. Meanwhile, we do know that there was significant surface warming over the 20th century as a whole. Yes, there are several candidate explanations, but none of them can claim to be more convincing than the known rise in CO2.
______________________________________________________
You left out the other side of the risk equation: The climate gets a lot colder from natural causes. Funny how everyone always over looks that part of the risk equation.
Woods Hole Oceanographic Institution states:
…It ignores recent and rapidly advancing evidence that Earth’s climate repeatedly has shifted abruptly and dramatically in the past, and is capable of doing so in the future… Fossil evidence clearly demonstrates that Earth vs climate can shift gears within a decade… Thus, world leaders may be planning for climate scenarios of global warming that are opposite to what might actually occur…
Peer-reviewed papers:
Lesson from the past: present insolation minimum holds potential for glacial inception (Science speak for we have the correct conditions for an Ice Age)
Solar energy reached a summer maximum (9% higher than at present) ca 11 ka ago and has been decreasing since then, primarily in response to the precession of the equinoxes. The extra energy elevated early Holocene summer temperatures throughout the Arctic 1-3° C above 20th century averages….
So what if, as these papers show, the only thing PREVENTING a return to full blown Ice Ages conditions is mankind’s increasing CO2????
During the 20th century the sun has been very active according to this paper and NASA However in the 21st century this is no longer true according to the Solar Dynamics Observatory Mission News
This shows how temperatures in Greenland have gradually fallen during the Holocene (present on right) http://jonova.s3.amazonaws.com/graphs/lappi/gisp-last-10000-new.png
This shows how temperatures have fallen during the Holocene in the Antarctic -10,000 yrs of Vostock Ice Core data (present on right)
http://2.bp.blogspot.com/_cHhMa7ARDDg/SsVwd55PJ8I/AAAAAAAABKY/52SrhXN4C3c/s1600-h/Vostok-10Kd.jpg
And finally the real kicker, the Vostock Ice Core shows we are at the end of the Holocene time wise. The sharp pointy upspikes are the interglacials everything else is ice cube city. You will note we have had a long run compared to most of the rest of the interglacials. (present on the left)
http://cdiac.ornl.gov/trends/temp/vostok/graphics/tempplot5.gif
Don’t you think this evidence should also be considered before any decisions are made to eviscerate our present technological society? I think rapid conversion to nuclear power is the best solution for both risks, not wimpy solar and wind power.

August 6, 2010 6:45 pm

Stephen Wilde wrote, “Thanks Bob. That is my ‘Hot Water Bottle Effect’ in a nutshell and it’s hugely greater than the so called Greenhouse Effect.”
No reason to thank me for my earlier comment about the oceans having their own greenhouse effect. That’s something I always keep in mind from an old John Daly post. Took me a few seconds to find it again. Here ya go:
http://www.john-daly.com/deepsea.htm
Scroll down to “DO THE OCEANS WARM THE PLANET?”

Merrick
August 6, 2010 6:48 pm

Wow. Now I know that solid state gain media get hot in lasers but gaseous gain media in laser doesn’t get hot – because the Einstein A and B coefficients ensure that the gaseous media radiate the photons oput as fast as they come in.
And laser media in a state of population inversion have a negative temperature.
Sorry. That’s just wrong. Please point me to any sources that state that gas lasers don’t require cooling and that lasing media in population inversiom are characterized by negative temperature. Negative temperature is an unrelated concept related to very cold materials with highly aligned spin states.

Gail Combs
August 6, 2010 6:58 pm

Merrick says:
August 6, 2010 at 5:48 pm
Gail Combs:
Nope. Absorption of an IR photon causes a vibrational excitation. Electronic excitations require visible or ultraviolet photons.
And, no, velocity and heat are not interchangeable. Heat is rigorously defined only for systems in thermodynamic equilibrium. That means that not only is there energy in kinetic (external) energy of the atoms and molecules but that there is also energyin thermally available rotations and excitations.
Doesn’t anyone other than physical chemists understand this?
_________________________________________
Doesn’t anyone other than physical chemists understand this?
Not many. Only physicists, physical chemists, a smattering of other chemists and the avid readers who devour anything connected to climate. That is why I suggest a few of these “simplified physics” articles were a very good idea.
SO thank you for the clarification. My chemistry teacher for physical chemistry (thermo) had a heart attack and we got stuck with an organic chemist who hated P chem and did not understand it. That class, my one and only, was forty years ago. That is why I asked in my first post for correction if I got stuff incorrect.
Hopefully others will learn from my putting both feet in my mouth.

August 6, 2010 7:02 pm

George Steiner,
At the Air Vent Pat Frank left a calculation which showed 30 milliseconds is typical before release of the energy. More than enough time for a collision to occur. This means most of the energy is released through kinetic interaction.

JimboW
August 6, 2010 7:06 pm

Jeff,
Thank you for this post.
As many others have pointed out above, it is very important for contributors here to call out sub-standard science /arguement, rather than remain silent and so leave the impression that the readership is in general agreement with something which is easily discreditted. Otherwise you end up looking like realclimate.

Oliver Ramsay
August 6, 2010 7:21 pm

George Steiner says:
August 6, 2010 at 5:31 pm
If a CO2 molecule encounters a photon at 15 nm then immediately emits a photon at 15 nm, did the CO2 molecule beacame warmer? Became colder? Became neither warmer nor colder?
—————–
I know the answer to that one!
It became warmer, warmed adjacent N2 molecules, radiated back to Earth and sent what was left to the sun.

Smoking Frog
August 6, 2010 7:27 pm

Gail Combs: I don’t see any chance of refuting Al Gore with “science that the majority of people can understand,” if that means science that isn’t oversimplified so far as to be incorrect. On the other hand, my alternative may be equally unrealistic. I think the majority of people should be skeptical (of both sides, or all sides) as any reasonable person would be of something he doesn’t understand.
I know that doesn’t answer the question of what happens when people vote; despite their skepticism, they may vote for the wrong side; but I just don’t see that a reasonable but ignorant person should have any attitude other than the skepticism I’m talking about. How could he? I know more science and math than the majority of people, and I’ve been following AGW for nearly 20 years, but I can’t swear that the warmists are wrong. I think they’re wrong, but if they’re right, that’s the way life is.
Don’t get me wrong. There are things that are within the reach of the majority, if the media would present them. They’re not all scientific. The majority don’t know beans.

August 6, 2010 7:35 pm

Interesting. But empirically, addition of carbon dioxide to the atmosphere simply does not change the transmittance of the atmosphere in the infrared as Ferenc Miskolczi has shown. He used NOAA weather balloon database to show that the global average annual infrared optical density of the atmosphere has not changed for 61 years and has the value of 1.87. This means that constant addition of CO2 to the atmosphere for the last 61 years has had no effect on how much infrared radiation the atmosphere absorbs. Where does this fit in with your theory?

Editor
August 6, 2010 7:35 pm

OK.
It appears there may be a problem with the recent-increase-in-CO2-causes-catastrophic-global-warming theory …
From the above graph, CO2 at 380 ppm has an absorption peak between 10.5 and 11 microns that will absorb some of the outgoing radiation from the earth. This absorbed energy is then assumed to increase the vibration (motion) of the (very few) CO2 molecules in the atmosphere, which then hit nearby air (O2 and N2 molecules) and cause them to increase the motion (temperature). Fine, an increase in CO2 would be assumed to cause an increase in net energy in the air (although that increased heat energy cannot be measured yet, and the increase will decrease logarithmically as CO2 increases linearly.)
However, this graph also shows that O2 and O3 at 209460 ppm (537 times the amount of CO2!) have a combined absorbtion peak in the same range of outgoing radiated energy (between 9.8 and 10 microns).
Why then, is CO2 is considered a massive threat towards heating the earth when O2 (being subject to the same heat exchange and radiation equations) is 537 times more prevalent? Is it because only CO2 can be used to control the world’s capitalist monies by taxation, restrictions, and international policies of government control?
(Yes, the ratio of peak absorption for CO2 is higher than that for O2/O3 … But it isn’t 537 times higher. Water – conventionally considered a greater green house gas than CO2 – is also conveniently ignored in this graph. Again, because the UN cannot tax water vapor.)
I add a caveat: Since 2003, the amount of O2 in ppm has decreased slightly. Since worldwide temperatures have reamined the same, the amount of volcanism has remained near constant, but CO2 has increased and O2 decreased, what is the net reflective and absorptive difference in the earth’s heat imbalance?
The CAGW alarmist cannot explain the cyclical changes in the earth’s tempeorature. It doesn’t fit their cherished, simplistic theory that REQUIRES the earth temperature respond linearly to a linear change in CO2 levels. Therefore, they had to invent – and still have to cling to – Mann’s-made global warming via the hockey stick and by ignoring changes since the depths of the LIA in the mid-1600’s.

August 6, 2010 7:42 pm

Arno,
Here is another reminder that the science is not settled:

The following article by Dr. Zbigniew Jaworowski, M.D., Ph.D., D.Sc., was published, after peer review, in March 2007. Dr. Jaworski was one of the first to point out the loss of scientific integrity in the field of global warming research.

Article and link.

Amino Acids in Meteorites
August 6, 2010 8:07 pm

stevengoddard says:
August 6, 2010 at 11:36 am
CRS, Dr.P.H.
Thanks for the Hansen link.

I hope you are going to make a post, or posts, on it!

Jim D
August 6, 2010 8:39 pm

RACookPE1978 says:
August 6, 2010 at 7:35 pm
“However, this graph also shows that O2 and O3 at 209460 ppm (537 times the amount of CO2!) have a combined absorbtion peak in the same range of outgoing radiated energy (between 9.8 and 10 microns).”
This would be entirely O3, which is of order 1-10 ppm in the stratosphere. It is quite an effective GHG too. As mentioned elsewhere O2 contributes almost nothing at these wavelengths.

Alan McIntire
August 6, 2010 8:42 pm

George E Smith made a good point regarding the “average” 240 watts from the sun.
In fact, earth is receiving, and radiating away, heat at higher temperatures during the daytime than during the nighttime. As the earth heats up, earth’s radiation band shifts towards the left, and a smaller FRACTION of earth’s radiation is absorbed by CO2- negative feedback.
Regarding those greenhouse models, I checked out this article regarding multi
layer atmospheres,
http://www.geo.utexas.edu/courses/387H/Lectures/chap2.pdf
came across Newton’s law of cooling, and posted this at
“Science Of Doom”
http://scienceofdoom.com/2010/07/24/the-amazing-case-of-back-radiation-part-two/#comment-5086
My simple model assumed we get the same amount of radiation from the
atmosphere both day and night, and we get all our radiation from the sun during
the day, the same amount all day.
Obviously a lot of latent heat goes into the atmosphere during the daytime , otherwise the earth’s surface would get much hotter during the day than it actually does

JimF
August 6, 2010 8:47 pm

So, the science isn’t really settled? Gosh, this earth is a complicated place. 😉

Amino Acids in Meteorites
August 6, 2010 8:54 pm

For those who still haven’t seen Roy Spencer’s talk on negative feedback here it is in 2 parts:
Part 1

Amino Acids in Meteorites
August 6, 2010 8:54 pm

Part 2

Jim D
August 6, 2010 8:55 pm

A point missed in the thought experiment is that the 1000 W coming out is all at 15 microns because that is the only wavelength CO2 could emit at (unless the temperature has pushed the envelope in Figure 2 far enough left to make other bands possible). This is an important point that gases emit only at their allowed wavelengths within their black-body temperature envelope, and with the intensity consistent with that envelope. It is like it is trying to be a black body but only has a few windows to output its photons. This is known as low emissivity (where perfect emissivity of 1 is a black body).
The intensity of the 15 micron emission of a region of gas is defined by the temperature and concentration of the CO2 in that region.

Barry Moore
August 6, 2010 9:02 pm

Congratulations Jeff probably one of the best posts on the physics involving the interaction between CO2 and outgoing LW radiation and it clearly shows the competition between water vapour and CO2 in the 15um wavelength range. But since the outgoing LW radiation energy is all absorbed the heating effect on the atmosphere is the same irrespective of which molecule absorbs the energy.
The IPCC do however repeatedly talk about the reradiated energy which finds its way back to the surface and thus increases the radiation absorbed by the earth.
Your post does not cover this claim by the IPCC and I realize one can not cover all the bases in one post so this is not a criticism.
In actual fact the absorption of IR by a molecule if looked at in detail is considerably greater at the actual spectral line frequencies and decreases at the frequencies on either side of the spectral line, thus you have a mini Planck curve around each spectral line. There are many spectral lines in the absorption zones shown in your diagram and as the concentration of the gas increases the Planck curve gets wider which is known as line broadening. It is this line broadening effect which forms the theoretical basis for the IPCC formula relating CO2 increase with increased temperatures I believe the forcing factor due to increased water vapour is strictly empirical. However at the current levels the Planck curves all overlap so the line broadening effect has no impact.
With regard to the reradiation back to earth the CO2 molecule will only reradiate energy which is exactly at the spectral lines so although the intermediate wavelengths travel further vertically since their absorption rate is less the resonant frequencies travel a much shorter distance downwards before they are absorbed. Thus a minuscule amount of reradiated energy gets back to earth and the concentration of CO2, once the mini Planck curves have overlapped, has no effect on the energy absorbed and therefore temperatures.

Kevin Kilty
August 6, 2010 9:28 pm

Jordan says:
August 6, 2010 at 3:41 pm
I have no issue with moderate GH warming, but would like to put up another thought experiment. (This could be tested in practice.)
I have a long tube full of CO2 (looking like a telescope). It has a transparent window at one end and the other end (the closed end) is a solid black surface. The sides are well insulated so it can only receive and radiate through the window.
I launch the tube to a position above our atmosphere and point it into to the sun. This raises the temperature of the black surface at the closed end.
I have an identical tube, except that it is empty (or filled with a non-GHG). I launch this and place it alongside the first tube, pointing into the sun. The temperature of its black surface also rises.
The physics discussed on this thread suggest the closed end of the first tube will rise to a higher temperature compared to the second tube. So I use this “potential difference” to drive an engine: the closed end of the tube full of CO2 is the hot reservoir and (purely for for demonstration) the closed end of the second as the engine heat sink. Both pointing directly into the sun.
The engine would appear to violate a thermodynamic concept that we cannot expect a practical engine to operate if the sun is the ultimate heat source and heat sink.
However, the CO2 has introduced a frequency shift between the incoming radiation and the outgoing radiation. We could call this machine a “thermal diode”?
Question for the physicists: does this engine do any useful work?

This is an interesting suggestion — i’ll give it a shot. The two objects as described are essentially blackbody cavities (The classical construction of a blackbody is a deep slender cavity drilled into any material–the specific material doesn’t matter). Radiation goes in and bounces around, and some gets absorbed, and what comes out can only come out the window in each case. Most of the radiation is absorbed in each cavity and only a little comes out the window at first. The cavity filled with CO2 will heat up more quickly than the empty cavity. However, at equilibrium what each is made of, and filled with, will not matter at all. As long as each can only view the sun through the window, each will achieve the same temperature, and as long as each is only exchanging radiation with the sun, each will reach the surface temperature of the sun. In short, each will be perfectly absorptive and emittive. There will be no temperature difference to exploit in an engine.
In the realistic case, of course, each tube will radiate into a large solid angle, as it isn’t possible to prevent conduction through a real layer of insulation. Then there may be a temperature difference at equilibrium, but this would be no different than the scheme of connecting a stainless steel panel to a white painted aluminum panel through a heat engine. The steel panel will get quite hot in sunlight, and the aluminum will stay cool. The heat engine produces work. You don’t even need to put this contraption in orbit.

Mike Blackadder
August 6, 2010 9:40 pm

At first I thought this was a straightforward post. Michael Dunn has me thinking about this a bit more. I think I’ll want to read Tom’s post before commenting further.
George Smith, I thought that I should argue a point with you regarding intensity of incoming solar radiation:
“I disagree with those who want to divide this number by 4 after applying some albedo reduction to get some puny 240 W/m^2 all over the earth surface.”
It seems to me that the divide by 4 is straightforward. Solar radiation is averaged over the surface of the earth (area 4 pi radius^2). Projection of solar radiation onto earth is (pi radius^2) (ie. this is the area of space that Earth blocks the solar radiation). So I think that you need to divide by 4.

kuhnkat
August 6, 2010 10:40 pm

George E. Smith, and there is still more incoming in the near infrared than outgoing in the earth’s blackbody.
About 50% of the suns’ blackbody is in its infrared. About 45% is in the visible.
Maybe you can tell me if the Climate Models use the 1-4 micron range as visible energy? If not, their accounting is rather shoddy.

Al Tekhasski
August 6, 2010 10:43 pm

As much as I dislike aggressive and ignorant AGW climatards, I hate when the opposite side makes authoritative but wrong or half-accurate statements. Let’s see:
“What you need to understand about Tom Vonk’s post is that he was explaining that the atmosphere does not warm by any radiative absorption process. (And it doesn’t.”
Oh yes it does. What we need to understand that the term “warm” is usually consonant with “up” (and “cool” with “down”), which emphasizes some directional process. Therefore, if we start with a cool gas mixture and begin to shine the 15um laser, the mixture will warm up. Period. Obviously, when/if the system reaches a STATIONARY STATE, it stops warming up, and stays flat. Only then we can say that radiation does not warm it, which is kind of trivial. Even then we need to understand that Tom Vonk is wrong, he does not realize that LTE is not an absolute equilibrium, but is only a first-order approximation of mechanics of continuous medium at a mesosciopic level.
In second approximation, the media allows for temperature gradients, macroscopic motions, eddies, and diffusion transport. That’s exactly what happens with radiation in highly absorbing bands, where the media is “optically thick”. For this reason the propagation of radiation in this opaque media mathematically resembles diffusion, and is called “diffusion approximation”, or Rosseland approximation.
Therefore, it takes time to interact with temperature fluctuations of the size of mean free path of photons. Therefore, the other statements as
“So, temperature equilibrium is almost instantly established” and “Radiative equilibrium is established throughout the atmosphere faster than the air can gain temperature”
are highly inaccurate bunk, it is valid only for nearly transparent (“optically thin”) bands.
“And the air is worthless as a heat sink”.
Not if it flows at a speed. Every computer uses moving air to “sink” the heat from CPU and other parts. As a matter of paradox, the _effective_ thermal conductivity of atmosphere is about 100 times better than copper.
Also, transition from Fig.6 to Fig.7 will take quite a measurable time due to finite rate of diffusion of radiation and finite thermal capacity of air inside, so it does take time to warm up before starting to emit 1000W through that small window. Of course, the term “instantaneous” is quite stretchable…
More, the radiation from Jeff’s chamber might be pretty close to black body, because the can itself looks like a deep cavity, and inner surface of the cavity will experience multiple reflections, will warm up, and will emit close to black body. Hollow cavity is a standard model for blackbodies in physics.
It is also interesting that even with plain air at 375ppm of CO2 and the chamber length of 10cm, the chamber will have to absorb about 350W. Assuming the window area of 10cm2, it is equivalent to 350 kiloWatts/m2, such that the air (and chamber walls) will heat up to about 1500K before reaching a stationary state. Pretty stiff, is not it?
Cheers,
– Al

asmilwho
August 6, 2010 11:04 pm

Hallo,
I’m interested in other work that Tom Vonk has done as a physicist.
So if anyone can point me to his university, or papers he has had published , that would be great, thanks

pwl
August 7, 2010 12:47 am

Hmmm… interesting article… how does Ferenc Miskolczi’s Saturated Greenhouse Effect Theory fit into this?
Ferenc Miskolczi’s Saturated Greenhouse Effect Theory: C02 Cannot Cause Any More “Global Warming”
http://pathstoknowledge.net/2010/01/13/ferenc-miskolczi%E2%80%99s-saturated-greenhouse-effect-theory-c02-cannot-cause-any-more-global-warming
It would really be great to have Ferenc Miskolczi or one of his collegues address the issues and points raised by Tom Vonk and Jeff Id.

Stephen Wilde
August 7, 2010 1:47 am
Stephen Wilde
August 7, 2010 1:51 am

My post doesn’t seem to have registered so here it is again. Mods please delete one if it comes up twice.
Bob Tisdale:
Some of these links may help you.
http://search.orange.co.uk/all?q=jet+stream+shifts+1970+to+2000&brand=ouk&tab=web&p=searchbox&pt=todayweb_hp4&home=false&x=21&y=15

joletaxo
August 7, 2010 2:08 am

Great experiment but
What is the pressure in the “chambers” from whom will depend the rate off conductive occurence between the CO2 molecule, and the molecule off the other gazes?
We know for sure(maby) that the other molecule will not trap the laser beam, so the only way too exchange energy is let Us say”mecanical” and we foud again the “laps rate off transmission”.
An amount off energy is trapped by the CO2 molecule, and reemitted immediately,but a part is exchange with the molecule off the other gazes by conduction depending off the pressure in the chamber,and then reemitted as IR, in all directions,weekening the total beam.
The exemple off the cutting laser is perfect.Iron has a structural organisation off molecule that enchance the rate off transmission and the energy ,captured in surface odd the sheet is transmitted further by conduction,so the cutting effect
In the atmosphere, things are quite different:CO2 is a trace gaz, and the “mecanical” process is very week.And in the upper atmosphere, tings are going to be more critical, because pressure is very low.There in praticaly no chance off enchancing the t° off the other gazes there.
Back to the black board, because things appear to be unclear….

John Marshall
August 7, 2010 2:08 am

I have read it but has he actually done the experiment? No! It was a thought experiment. There is one glaring error in that you cannot get a perfect insulator. An insulator will delay heat flow but will not stop it. ie. the 2nd law of thermodynamice holds true. What happens when something heat up quickly, ie it has a low specipic heat so requires less heat to raise its temperature through 1 degreeK than an object that heats slowly. This is easy to check- get two saucepans with the same weight of 1-water and 2 cooking oil and heat on the stove, check the time to raise the temperature from ambient to, say, 50C. The oil will take the shorter time. Now check cooling time and the oil will cool faster. This second part is important because it is a good example of the 2nd law- entropy must increase. The temperature must equal out so heat can only flow from hot to cold.
The atmosphere, containing some CO2 is heated by incoming solar radiation, the CO2 directly and the other air molecules by conduction/radiation from the CO2 and conduction with the ground and long wave radiation from the ground. Warm air rises and will cool adiabatically as it rises. So where is the back radiation commimg from? Not from the rising air because this is now cooler than the surface. We have established that hot to cold is OK but not the reverse.
This has been the problem with the Greenhouse Theory and the lost heat which Mann described as a travesty that it could not be found. The models, which believe the Greenhouse Theory and are embedded with it, show a warm lower troposphere which does the reradiation back to the surface but measurement by radiosonde or dropsonde cannot find it. If you cannot measure a temperature difference then that temperature difference is not there and the reradiation cannot take place.
In the atmosphere the rising air, though cooler than the surface, is still warmer than that above so it will radiate to the cooler zones. Cloud, which plays a vital part in temperature control in the atmosphere, is another story and one that we have yet to master and one that is not covered by any models which is one reason why they are always wrong.
Radiative adsorbtion and black body radiation is all very well but the theory of GHG’s must obey the two basic laws of thermodynamics. It does not.

P Wilson
August 7, 2010 2:13 am

It is difficult to justify this assertion. The wavelength at which c02 captures radiation is in the subzero region around its peak of 15microns and shoulders slightly less, and at the shoulders c02 has to compete with oxygen and nitogens’s wavelnths of heat absorption. These figures give around 7% heat absorption for c02 of outgoing radiation, and that is fixed according to saturation windows than quantity. C02 suspends heat loss to space a tiny fraction of IR, but nothing significant. It cannot send it back to terrestrial levels, as air presure, climate and other factors maintain a higher temperature at ground level than the regions of the troposphere where c02 is the most active in its heat absorbing height (where the troposphere is subzero). In other words, c02’s subzero active region cannot add heat below, where it is already warmer.
The experiment sends flags. Applied to AGW it would have to replicate air pressure, temperature and various other climatic factors, again. to be verifiable.
Its true that a c02 molecule’s stretching mode would allow it to transfer energy to other atmospheric molecules, such as the ghg water vapour, but this requires so much energy that it doesn’t occur even at 300K, with the c02 absorbtion bands, and there’s some 3,000 other molecules apart from c02 in a given volume of air, making collisions between thermally excited c02 molecules very unlikely.

Denis Purdy
August 7, 2010 2:36 am

I greatly appreciated the graphic showing the absorption spectra of various atmospheric gases as compared to the incoming and outgoing radiation. However there was no spectrum for methane. Here in New Zealand we are constantly told by warmists that we NZers have one of worst GHG emission rates per capita due to the levels of methane produced by our cows and sheep and because “methane absorbs 15x as much IR”.
This may be true, and if methane was a GHG in isolation might be important. But it seems to me that if methane’s absorption spectrum significantly overlaps those of the other gases, then then the argument about adding extra coats of paint on a window that is already largely opaque applies.
I have tried to search the web for a graph showing the absorption of methane compared to other gases; thus far I have come up with nothing. Has anybody seen a graph they can point me at.

kwik
August 7, 2010 2:37 am

beegdawg007 says:
August 6, 2010 at 6:28 pm
Ah, yes! A very good post from Jeff Id. Now, next step is a real experiment with a laser, and 0.04% CO2. Then step up to 0.08%CO2. Measure.

Liam
August 7, 2010 2:38 am

Figure 2 seems to show that just about all the out-going radiation that CO2 can absorb is already being absorbed.

August 7, 2010 3:05 am

Isn’t the main flaw in Jeff ID’s argument that we should look at what happens if we go from a 280ppm CO2 chamber to a 560ppm CO2 chamber? Or more general from a non-zero CO2 to a higher CO2 chamber? Anyone able to explain how that would alter Jeff’s “gedankenexperiment”?

August 7, 2010 3:26 am

To me the two disputants appear to argue past each other slightly, so I’m grateful for Merick’s comment above. To me, it goes to the heart of the matter.
But, knowing that as a layman I am easily led astray, I found Merick’s statement that “air temperature is NOT an expression of the average kinetic energy in air” disquieting. That statement doesn’t appear critical to his overall point, but, since I had always thought temperature is indeed the measure of average TRANSLATIONAL molecular kinetic energy, it makes me wonder whether something else in his comment is wrong, too. Has anyone detected anything?

August 7, 2010 3:35 am

Gail Combs says:
the CO2 laden air will absorb IR close to the ground, it will then heat up,
____________________________________________________________
NOPE.
Absorption of IR causes an electron to go from the ground state to the excited state. VELOCITY of the molecule is what we call heat. You missed a step.
Gail, an interesting assertion, and whether I was right to miss this step depends on the relative conversion rate from energised electrons to vibrational modes.
My assertion was based on an article I read that IR given off by CO2 in the atmosphere travels only a few meters before being reabsorbed. The point of the article was that a change in CO2 levels meant a difference of a few meters in the distance CO2-IR travelled.
Now if we assume 10m average distance travelled, then 10km of atmosphere would represent 1000 average path distances.
The big question we have is: “would we obtain near enough to heat equilibrium in substance with 1000 average path distances to say” the IR causes the atmosphere to heat up. (for a 10km slice)
My hunch is that for all practical purposes you can assume the atmosphere heats up by IR, if however someone can puts some figures on that hunch I would be extremely grateful and if necessary start baking a humble pie.

August 7, 2010 3:43 am

Sorry, I didn’t put that well.
The question is:
When IR is absorbed by CO2, what percentage of that IR is re-emitted as IR, what percentage is converted to vibrational & motion modes, and what percentage to other “pathways”.
Using my engineers rule of thumb, if the relative proportions of IR-IR to IR-‘heat’ is not greater than 100-1, (i.e. if more than 1% of events are converted to heat) then after 1000 absorption-re-emission events, the relative proportion of IR-quantised energy will be an order of magnitude less than that of motion energy (heat)

Spector
August 7, 2010 3:44 am

We are given that the concentration of gaseous water or free H2O molecules (I am avoiding the standard term ‘water vapor’ because in common usage a vapor is a fog) in the atmosphere ranges, according to the EPA, from 30,000 ppm to 40,000 ppm and the concentration of CO2 is about 390 ppm. Thus it seems reasonable to assume that for every single CO2 radiation photon that escapes to outer space or is received from the surrounding atmosphere in a given volume of atmosphere per unit time, there may be roughly 77 to 103 equivalent H2O radiation photon events.
Even considering the fact that the lower concentration of CO2 may allow CO2 photons to escape the atmosphere at lower altitudes, I would still think that H2O must be by far the most dominant greenhouse gas in the atmosphere and must play a primary role in determining the altitude and temperature of the tropopause.

Jordan
August 7, 2010 3:57 am

Thanks for your answer Kevin Kilty. You gave a detailed explanation, but I’d like to follow with some questions and clarifications.
The two tubes may be akin to blackbody cavities, where one is filled with CO2 (to some pressure) and the other with a transparent gas (or we can say it is empty). We can say that equilibrium can is achieved in each tube when the OLR at each window matches the incoming solar radiation.
Staying at the windows for a second, I assume that the tube full of CO2 has an emitting surface at the window, whereas the empty tube does not. I’m not sure whether that has any bearing on the matter.
So what do we expect to happen at the other end of the tubes according to GHG theories?
OLR emitted from the deep end of the empty tube gets free passage straight through the cavity and out the window. It can equilibriate in much the same way as the moon. Perhaps reaching a similar temperature to the surface of the moon in daylight.
This thread discusses CO2 as a resistive medium to OLR. The GHE effect appears to demand that equilibrium can only be achieved at the window of the tube full of CO2, if the temperature at its closed end is much higher. (We could almost decide what temperature we want by design of length of tube and maybe CO2 pressure?).
The argument of “OLR resistance” seems to suggest that the closed end of the tube full of CO2 could be sustained at higher temperature compared to temperature at its window. Certainly by radiative arguments – I’m not sure whether conduction will/could alter this.
Could we stick with a scenario where the sides are well insulated and conduction loss is negligible (incoming solar is quite powerful so we need not get too concerned if there is some leakage through the sides).
That leaves the question: when pointed into the sun, what are the temperatures at the closed ends of the tubes to radiate identical amounts of OLR at the windows?
If there is a significant temperature difference at the closed end of the tubes, we have potential to drive an engine and do work. Flow of heat through the engine helps the OLR to bypass the resistive medium in the tube full of CO2.
Why am I sceptical about this?

Dave Springer
August 7, 2010 4:33 am

joshv says:
August 6, 2010 at 10:44 am
“From the perspective of 15 micrometer wavelength infrared laser, the CO2 filled air is black stuff.”
I believe this is incorrect. Chemical bounds within CO2 absorb the energy of specific photons, and at equilibrium emit them at the same rate, though not necessarily in the same direction, as they are whizzing around, smacking into each other. It’s not black stuff, it’s “white” stuff, like a cloud. Does a 15 micrometer detector pointed at the earth from space see a black ball?

Yes, at 15um the detector sees “black”. This is because the optical depth of the atmosphere at 15um is far less than the actual depth of the atmosphere.
Astronomy 101: Composition of cold clouds of gas backlit by continuous spectrum have an absorption line (a gap where energy goes to zero) at the characteristic absorption frequency of the gas.

David, UK
August 7, 2010 4:34 am

Derek B says: August 6, 2010 at 5:07 pm: “If the alarmists are even half right, that’s time we don’t have. So, much as we distrust them, we are reduced to depending on models and basing policy on risk minimisation. Meanwhile, we do know that there was significant surface warming over the 20th century as a whole. Yes, there are several candidate explanations, but none of them can claim to be more convincing than the known rise in CO2.”
I smell a troll! Still pushing the precautionary principle, Derek? We are only “reduced to depending on models” because the models have been so elevated to a level of respect that far exceeds their validity. Too many assumptions, too many unknowns, too many “what ifs,” too many ultra-wide error bars, too much cherry-picking. Just because the alarmists have been “reduced to depending on models” doesn’t mean the rest of us have to be.

David, UK
August 7, 2010 4:36 am

Dang, why can’t this site have a “preview” button so we can check when the HTML hasn’t worked? Sorry.
[Fixed. Sorry, WordPress doesn’t support a preview function. ~dbs, mod.]

Merrick
August 7, 2010 4:55 am

Barry Moore on line broadening:
Barry, I haven’t read the IPCC report and doubt I’ll have time to do that soon. You seem to quote the IPCC report as claiming that the increased concentration of CO2 will result in line broadening. If so, the IPCC is incorrect and I hope you’ll look more closely at the actual science.
Many factors contribute to line broadening. Two of the most important effewcts are temperature (via Doppler) and *pressure* broadening. Not cencentration, but pressure. If a sealed vessel has it’s concentration of a gas increased so that the pressure also increases then you will see the effect. For non-polar molecules this is a pretty rigorous rule. In the same sealed vessel, if you added more N2 instead of CO2 to your original CO2 you’d still observe pressure broadening. It’s definitely not a concentration based effect
Since, even with increased CO2 in the atmosphere we are predicting no change in atmospheric pressure, the IPCC is simoly wrong if it in fact states this. Can you please point me to the section and page where this is discussed?

August 7, 2010 5:09 am

Mike Haseler says:
August 7, 2010 at 3:43 am
The question is:
When IR is absorbed by CO2, what percentage of that IR is re-emitted as IR, what percentage is converted to vibrational & motion modes, and what percentage to other “pathways”.

At tAV, (comment 47) Pat Frank calculated the decay half life for radiative emission as 30 milliseconds, and the collision time between molecules at 10^-8 seconds – so almost none of the gas decays and re-emits (sorry to those who don’t like the word).
I didn’t know it was that much difference myself.

Merrick
August 7, 2010 5:18 am

Liam – that is correct – all of the outgoing ~15um radiation that can be absorbed is. It’s the re-radiation by the warmed gas that matters. If there were 100 ppm less CO2 it would pretty much all still be absorbed by the atmosphere, but the altitude at which, say 90%, of the CO2 had been absorbed would be higher than at current concentrations. And if the concentration were 100 ppm higher than today that altitude would be lower. Does that make sense?
Now, go back to the point of this post: that warmed CO2 re-emits and *some* of it re-emits toward earth. Remember the inverse square law and that CO2 will be re-absorbing some of the re-emitted light on the way down and it’s easy to see that if on average the ~15 um radiation is absorbed higher in the atmosphere (low concentration case) then less of that reradiated energy makes it back to the ground and if on average the ~15 um radiation is absorbed lower in the atmosphere (high concentration case) then more of that reradiated energy makes it back to the ground.the difference might be small, but there clearly is a difference.
That part of the physics can be summarized just that simply. If people are going to argue over the finer details of how that reradiation happens I wish they’d be more careful about those details, but the conclusion of this very small part of the much more complicated picture can really be summed up that simply.
Many are going to immediately respond back regarding LTEs and forcings and negative feedbacks. Yes – of course they are all very important points to take into consideration. But this small portion of the topic seems still to be poorly understood by most of the folks here and it sure would be nice if we could get a significant plurality up to speed on at least this part.
A boy can dream, can’t he!

Roger Clague
August 7, 2010 5:28 am

John Marshall asks ‘has he actually done the experiment?’ The answer is NO!
IR exert R.W. Wood did something similar 100 years ago and showed that CO2 does not trap heat.
Philosophical magazine , 1909, vol 17, p319-320
A bottle of monatomic argon reacts to IR radiation the same as a bottle of CO2.

old construction worker
August 7, 2010 5:47 am

“As it heats, emissions from the window increase in energy according to Planck’s blackbody equation. Eventually the system reaches a new equilibrium temperature where the output from our window is exactly equal to the input from our laser – 1000 watts. Equilibrium! – (Figure 7)”
questions, questions, always more questions:
What happens when the diameter of the chamber is increased ( keeping the 50% CO air mixture)? At what diameter will the laser beam “seen” on the out put side of the chamber? Or does (figure 7) the diameter of the chamber mater?
If you had “heat” sensor on the walls of the chamber would they always detect the same amount of “heat” on the walls as the diameter increased?
If the “heat” on the wall goes from X degrees to same degrees of “room” temperature on the outside of the walls wouldn’t indicate the 50% CO2 air mixture close to the walls not doing a lot of re radiation of “heat”?

August 7, 2010 5:50 am

This post is more acceptable to me then Tom’s and the experiment at least proves that there is a warming effect.
On the other hand, 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….
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 info 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.
So all gases other than nitrogen also cause a cooling effect!!My question was en is: what is the net effect of the cooling and warming of each the gases in the atmosphere other than nitrogen?
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?

Roger Clague
August 7, 2010 5:58 am

Mr Id’s thought experiment is an assertion in drag. A bit like computer climate models.

Dave Springer
August 7, 2010 6:08 am

PJP says:
August 6, 2010 at 10:22 am
Other questions, to which the answer almost certainly exists, but I have no idea where to look:
The absorption spectrum of CO2 shown indicates some maximum absorption (attenuation of the 15 micrometer radiation). How does this change as the concentration of CO2 increases?
I think this is vitally important to answer. At some concentration the gas will become completely opaque to 15 micrometer radiation. Adding more will make no difference.
Are we already at that point? What is that point?

Very perceptive question. Alarmingly few people know enough to ask it.
The troposphere is opaque to 15um radiation upwelling from the earth’s surface in tens of meters. That is indeed vitally important. Adding more CO2 would have the effect of bringing total absorption closer to the surface – more kinetic energy in a smaller air mass makes the sensible temperature higher. The nub is that convection keeps the air near the surface well mixed at distances greater than tens of meters so at this point decreasing the optical depth of the atmosphere at 15um won’t have any effect – total radiated energy from the surface at 15um will be evenly distributed in the first few hundred meters of atmosphere regardless of how much more CO2 gets added.

Ninderthana
August 7, 2010 6:31 am

The most fundamental underpinning of modern science is observation.
When there is uncertainty about the existance or non-existance of a phenomenon, scientists usually try to perform an experiment (i.e. fair test) or make an observation that will either refute or verify the existance of a phenomenon.
In the journal Energy and Environment Volume 21, Number 4 / August 2010
Dr. Ferenc M. Miskolczi has published an article:
“The stable stationary value of the earth’s global average atmospheric Planck-weighted greenhouse-gas optical thickness ”
http://multi-science.metapress.com/content/nm45w65nvnj3/?p=a52f8c4dc8a1411392c30af7d57b34f2&pi=1
which conclusively shows from observation that the mean infra-red opacity of the atmosphere has NOT SIGNIFICANTLY CHANGED over the last 61 years.
Hence, while it may be true that increasing levels of CO2 have increased the mean infra-red opacity of the atmosphere – it also true that some other infra-red absorber (most likley water vapor) has effectively counter-balanced the increase in opacity caused by CO2.
Any reasonable scientist who has read Dr. Ferenc M. Miskolczi’s paper would have no trouble dismissing CAGW on scientific grounds.

John of Kent
August 7, 2010 6:44 am

Jeff, you can put what you want in your post but it still does not address the basic problem in physics that makes the alleged back radiation greenhouse effect impossible. A colder object (lower atmosphere) CANNOT heat a hotter object. Heat always moves from hotter to colder, this is like claiming you can make a river flow uphill. Not without a pump (i.e. WORK input you cannot).
Eventually sceptics/skeptics/realists, including Lindzen and Spencer are going to HAVE to accept that the greenhouse effect from back radiation is impossible.

John of Kent
August 7, 2010 6:56 am

This is the important question:-

The question is:
When IR is absorbed by CO2, what percentage of that IR is re-emitted as IR, what percentage is converted to vibrational & motion modes, and what percentage to other “pathways”.

I read on another site that it has been calculated that it takes only about 0.4 second for an IR photon to leave the atmosphere from the surface. It is all the absorptions and re-emmissions by IRIG’s (Infra Red Interacting Gasses) that slow the IR photon down. At the speed of light, if you assume the atmosphere is 100miles thick, it should take only approx 0.0005376 of a second to leave the atmosphere.
All the IRIG’s do is scatter IR and warm the atmosphere a little. This scattered IR cannot warm the ground. See 2nd law.
The real greenhouse effect is due to the slowness of convection which is the main cooling mechanism.

Jim D
August 7, 2010 7:27 am

A lot of people are fixated on the idea of photons coming in and going out of CO2 molecules, re-emission, etc. I am going to suggest a paradigm shift that leads to a more accurate way of thinking. CO2 intrinsically emits around 15 microns with a strength that depends on its temperature and concentration. The emission intensity only depends on these things, and is equal in all directions obviously.
Of course it absorbs too at those wavelengths, so you may ask what is the net effect. This depends on the background radiation. When you look up at CO2 against a cold sky, it is emitting more than absorbing, because it is warmer than the background. When you look down with the earth as a warm background, it is a net absorber. Either way it is absorbing and emitting at those wavelengths.
The important part is that from the ground perspective it is a net emitter, making for a warming effect.
From the space perspective it is a net absorber making for a reduction of outgoing radiation that also has consequences for the net earth energy budget.

August 7, 2010 7:29 am

Very few skeptics doubt that atmospheric CO2 is a source of warming. What they suggest is that, compared to other sources, its effect on Earth’s temperature is so minuscule as to be background noise. The insulating atmospheric effect (the term “greenhouse effect” is incorrect and misleading) created by water vapor is much more pronounced.
By the way, as Dr. Ference Miskolczi’s findings (which are routinely ignored) have shown, the total infrared optical thickness of the atmosphere, and its theoretical value is 1.87, has not changed, despite a 30 per cent increase in CO2 over the last 61 years.
(Says Miskolczi: “I performed these computations using observations from two large, publicly available datasets known as the TIGR2 and NOAA. The computations involved the processing of 300 radiosonde observations, using a state-of-the-art, line-by-line radiative transfer code. In both datasets, the global average infrared optical thickness turned out to be 1.87, agreeing with theoretical expectations.”)
His results “show an apparent warming associated with no apparent change in the absorption properties. Change in absorption properties cannot have been the cause of the warming.”
Please see:
http://www.examiner.com/x-32936-Seminole-County-Environmental-News-Examiner~y2010m2d12-Former-NASA-scientist-defends-theory-refuting-global-warming-doctrine
http://www.friendsofscience.org/assets/documents/E&E_21_4_2010_08-miskolczi.pdf

August 7, 2010 7:37 am

Jeff Id, in your lead off post you say;
“None of the statements in this post violate any of Tom’s equations; the difference between this post and his, is only in the assumption of energy flow from the Sun to Earth and from Earth back to space. His post confused equilibrium with zero flow and his conclusions were based on the assumed zero energy flow. The math and physics were fine, but his conclusion that insulating an energy flow doesn’t cause warming is non-physical and absolutely incorrect.”

————————–
Jeff Id,
I did my homework and read both your post and Tom Vonk’s post one more time (3rd time).
Thank you for your excellent and educational post.
I find that your above paragraph from your post is not an accurate re-statement of what Tom Vonk said.
I liked your post immensely, except for it stating to be a reply to Vonk. Yours is a good educational tool for understanding our atmosphere by itself when all references to Vonk are removed.
I found that Tom Vonk explicitly addressed a specific narrow aspect concerning the interaction of gases in the atmosphere among themselves and also with IR radiation. His study did not pretend to address anything more widely of the whole earth system. His caveats explicitly stated processes and conditions in the whole earth system he did not address.
John

Wayne
August 7, 2010 7:47 am

Radiation is not only governed by Plank theory, and it is very much dependent on View Factor, any body with radiative heat transfer course training, knows why the sun feels hot to earth at noon time than sunset time, the angle of sun radiation, or technically – View factor, is a key factor of radiative heat transfer
CO2, a 400 ppm quantity in the air – which means out of 100 gas molecule in the air, there is less than 0.04 molecule are CO2, so the effective radiation heat transfer due to such small View factor is questionable, need to be carefully calculated, no mention the physical parameter of absorptivity, emissivity of CO2 gas is highly temperature dependent, common knowledge is that it is significant of absoption at combustion temperature level, which is 3000 F range. atmosphere is far low temperature than combustion
by the way, the experiment of 50% CO2 in the can does not reflect the real air composition which only has 0.04% CO2 in volume. You have to compare apple to apple, right?

cal
August 7, 2010 8:21 am

Merrick’s last post is correct but I think his description is looking at the effect from the wrong perspective if one wants to be quantitative. I am sorry if I am boring anyone by repeating the same point I have made several times before but the key issue is not where the 15 micron photons are first absorbed (which is only a few feet off the ground) but where they are last absorbed before being radiated into space. If say the concentration of CO2 doubles the point where this final radiation takes place has to move upwards until the partial pressure of CO2 is half that at the original level so that the photon mean free path is the same.
This view is more informative because it is not possible to discuss the energy balance of the earth from bottom up. It is too complex. For this reason I did not find either Tom’s or Jeff’s papers very convincing. The only reasonable approach is to view the world (with its atmosphere) as a whole. It is then quite simple.
All the energy aborbed by the earth has to be radiated into space.
The earth’s albedo will change due to clouds and the disribution of land and water and as the earth rotates but if we neglect these (or averaged them over time) the energy absorbed and that radiated is constant, equal and opposite.
The absorbtion is mainly by land and sea because the atmosphere is a poor absorber of sunlight.
The radiation out is from three main agents. The earth’s surface, CO2 molecules in the atmosphere and H2O molecules in the atmosphere. These three account for 99% of the energy radiated into space.
The energy radiated will depend on the temperature of these radiating elements, and their concentrations in the case of CO2 and H2O.
The earth’s surface radiates directly into space at wavelengths around 10 micron where the atmosphere is nearly transparent. The temperature is therefore around 288K for these wavelengths. Photons between 13 and 18 micron are radiated by CO2 molecules from at or just below the tropopause (altitude 16km over the tropics and 8km over the poles). The tropopause is at 220K. Most of the remaining energy is radiated by H20 from various altitudes and temperatures depending on wavelength.
Thus is should be clear that photons emitted by CO2 and H20 are being radiated from molecules at a lower temperature than the surface temperature of the earth. If the atmosphere did not contain these molecules that energy would indeed be radiated from the surface. In this scenario the earths temperature would not have to be so high in order to radiate the same total energy. This is the unambiguous argument to show that the (badly named) greenhouse effect really does exsist.
However it does not follow that increasing CO2 will automatically increase surface temperature. The argument made by the Hadley centre is that an increase in CO2 increases the effective radiating altitude (as discussed above) and this implies a drop in temperature. A lower temperature means less radiation and therefore the surface has to warm in order to restore the energy balance. However all the satellite data I have seen suggests that the radiating temperature is 220K which is the lowest temperature found in the atmosphere. Unless the tropopause is cooling ( it may be but I have seen no evidence) it could easily be argued that an increase in CO2 will lead to increased radiation due the higher CO2 concentrations and therefore the surface has to cool to compensate.
When it is not clear whether an effect is positive or negative it is hard to say that the science is settled!

August 7, 2010 8:23 am

Jeff Id says:
At tAV, (comment 47) Pat Frank calculated the decay half life for radiative emission as 30 milliseconds, and the collision time between molecules at 10^-8 seconds – so almost none of the gas decays and re-emits (sorry to those who don’t like the word).
So basically, the IR given off by the warm earth is absorbed in around 10m (at peak absorption) and then the molecule doesn’t re-emit for around 30msecs, during which time it’s get banged 3million times which is far more than is necessary to convert the energy held by the excited electron into motion/heat energy.
Another interesting facet is that if the mean path of the IR is around 10m, then that IR energy takes around 30seconds to go through 10km of atmosphere. Or if we were to decide the really “thick” bit of the atmosphere was 30km, then within 100secs the IR will at a layer which has a relative “open window” into space.
Now, what is the effect of CO2 high in the atmosphere? It collides roughly every 10^-8 seconds, and in any one of those collisions, it can act as a vector to enable that heat energy to be lost via IR.
CO2 is a bit like sticking holes in a bag of dry, fine sand. The hole itself can be very small, but eventually (like a sand hour-glass), the conduit will allow the heat to escape.
So, one CO2 molecule in a vast array of gas can be like one small hole in a sand-bag, two – like two holes. Put enough CO2 in the atmosphere and you get a dramatic cooling effect.
PAUSE FOR REFLECTION
This is precisely the argument of the global warmers, but turned upside down. CO2 is a cooling gas, too much CO2 in the atmosphere will lead to catastrophic cooling of the atmosphere and “unless we all stop burning so much fossil fuel – it will inevitably lead to global cooling and we could see up to 6C warming, the sea level may fall making many ports unnavigable. The cooling temperatures would certainly lead to a rise of winter illnesses like flu (23,000 die in the UK each winter).
Global cooling by excessive CO2 is a real and imminent threat to national security and those who do not accept the simple truth of basic science that CO2 cools the atmosphere are clearly are in the pay of fossil fuel companies.
Moreover, the recent 21st century cooling clearly shows the CO2 induced global cooling effect is happening “faster than we had imagined” and …. etc.

Stephen Wilde
August 7, 2010 8:39 am

Cal said:
“Unless the tropopause is cooling ( it may be but I have seen no evidence) it could easily be argued that an increase in CO2 will lead to increased radiation due the higher CO2 concentrations and therefore the surface has to cool to compensate.”
The tropopause doesn’t appear to cool . Instead it’s height changes. It moves up when the troposphere is warming and down when the troposphere is cooling.
Contemporaneously the air circulation systems move poleward when the tropopause is rising and the troposphere warming.
Similarly the air circulation systems move equatorward when the tropopause is falling and the troposphere is cooling.
Both the changes in the height of the tropopause and the changes in the latitudinal positions of the air circulation systems are direct evidence of a change in the speed of the hydrological cycle as I have said boringly often before.
Thus a bit more energy in the atmosphere from more CO2 or any other GHG just provokes more ocean surface evaporation and a miniscule compensating increase in the speed of the hydro cycle for probably no change in surface temperature.
If a significant change in radiative balance could be caused then a higher equilibrium temperature would be achieved but if that happened there would also have to be a change in the optical depth of the atmosphere and Miskolczi seems to have shown as best we can discern from the available date that the optical depth has not changed for over 60 years despite more CO2 in the air.
AGW theory now has a serious case to answer.

Steve Keohane
August 7, 2010 9:03 am

It got too late to comment last night, but thank you everyone! There is a lot of great science in the comments above, too many names to mention. This is what makes WUWT the best site on the net.

Bill Illis
August 7, 2010 9:06 am

People get caught up in the absorption and emission bands of CO2 and H20 and do not understand there is radiation ocurring all over the spectrum. CO2 has specific frequencies where it strongly interacts with EM radiation but there is also base black-body radiation (that is not as strong or intense sometimes) but there is two kinds of radiation to be concerned about – the strong Absorption and Emission bands and the more general black-body radiation.
Let’s take the Modtran results for the sensor at 50 metres high, the height we are concerned about since we live on the surface (no one lives at 20 kms high).
Here is the chart of the radiation “looking down” at 50M (for the tropics).
http://geodoc.uchicago.edu/tmp/rad.07105553.gif
It is nearly a perfect black-body curve at 20C, the average surface temperature in the topics. There are no bands, everything is radiating as if it were 20C.
The chart “looking up”. Now we see the same black-body curve again – no CO2 Absorption bands are evident here – except in the Atmospheric Windows, the intensity of the emissions has fallen (still “back-radiating” as some like to call it) but the intensity in the windows has now dropped to something like -20C.
http://geodoc.uchicago.edu/tmp/rad.07105841.gif
Let’s move the sensor altitude to 20 kms high where the temperature is -60C.
Looking up, we see that CO2 is still quite active here. There is some activity in the H20 and methane bands but nothing is going on in the window regions, The radiation at these spectra is already on its way to space.
http://geodoc.uchicago.edu/tmp/rad.07105142.gif
Looking down, we see the windows are emitting like the blackbody curve of 20C, CO2 is emitting as if it were -60C, and H20 is emitting as if it were -18C and so on.
http://geodoc.uchicago.edu/tmp/rad.07110031.gif
So one needs to consider there are a number of things going on here.

Spector
August 7, 2010 9:53 am

RE: cal says: (August 7, 2010 at 8:21 am ) “The argument made by the Hadley centre is that an increase in CO2 increases the effective radiating altitude (as discussed above) and this implies a drop in temperature.”
I believe the argument they are making is that increased CO2 in the atmosphere will increase the depth of the CO2 ‘cloud’ around the planet in which CO2 photons cannot escape to outer-space. This, in *their* view, will necessarily force the tropopause to move a higher level to re-establish the previous radiative equilibrium and the natural adiabatic lapse rate will translate this increased altitude to higher temperatures on the ground.
However, if the level and temperature of the tropopause are now largely being determined by H2O, then I think there should be little change at all until we really add enough CO2 to the atmosphere to overwhelm the radiative significance of H2O.
I suspect that the temperature of the tropopause is determined primarily by the equilibrium established by the amount of incoming solar energy that can be absorbed primarily by H2O and CO2 at that altitude.

Mike Blackadder
August 7, 2010 9:53 am

“The difference between this result and Tom Vonk’s recent post, is that he confuses equilibrium with zero energy flow.”
After reading Tom’s post I agree with Jeff that this illustrates where Jeff made an error.
I don’t know for certain that Michael Dunn is wrong arguing (basically) that radiation absorption/emission is so fast that it would be equivalent to scattering of radiation which would not alter the temperature of the atmosphere, however this is not the argument that Tom made in his original post and I suspect it is incorrect.
So essentially the surface is a source of LW radiation and some of that radiation will be absorbed by greenhouse gases and some will pass through atmosphere and escape to space. If we increase the concentration of greenhouse gases then the atmosphere will absorb more of the radiation passing through (unless concentrations are already high enough to impede radiation to an extend that convection dominates) and this extra energy absorbed by atmosphere will translate into higher temperature of the whole air mass through collisions. The response to more absorption is to reach higher equilibrium temperature and therefore higher rate of emission.

August 7, 2010 10:06 am

Stephen Wilde says:
Both the changes in the height of the tropopause and the changes in the latitudinal positions of the air circulation systems are direct evidence of a change in the speed of the hydrological cycle as I have said boringly often before.
Stephen, I was very interested in your assertion that the latitude of the air circulation systems may change. I agree with you that an increase in CO2 will lead to little change in surface temperature because the effect is to change the effectiveness of the convective cooling system. However, one possible outcome from that increased effectiveness is that we could potentially see a change in the structure of the Hadley cells (which I assume is the same as you said).
To put it so it is understandable: the sun heats the equators, causing air to rise, it then is forced out from the equator at high altitude until it looses enough heat (via infra-red emission into space) for that air mass to descend and repeat the cycle. In the process you get rising moist air over the equator and descending dry air over the Sahara, the Sahara being situated at a position where the air has been exposed to the IR window into space just long enough for the air to loose enough heat to start descending.
Similarly, over the poles, you get excess cooling. Air masses cool excessively forming a bulk of dense cold air which then spreads out from the poles. As it moves away from the poles it picks up heat until it rises (above the UK) – it then is sucked into the polar region by the cooling-descending air in that region.
The result is an area of high pressure at the sahara and low pressure over the UK, and we therefore get an intermidiate Hadley cell bringing air from the high pressure over the Sahara latitudes up to the UK latitudes. The result is the classic three Hadley cell (6 when you count the south hemisphere)
In theory, however, it is possible to have any number of Hadley cells. They are afterall only cooling convective currents and so long as there is an odd number in each hemisphere you will get the dominant rising air over the equator and descending air over the poles. (In theory you could get only one cell!). If however you change the rate of COOLING, so that the equatorial air mass looses heat quicker by some added vector (CO2), the result could be to shift the high pressure-dry zone of the Sahara further toward the equator. As each cell is really a convection cycle of heat-uplight-IR emission, the addition of CO2 would reduce the average time it takes for the air mass to cool, thereby tending to reduce the size of each convective cooling-cycle cell.
So, whilst CO2 is unlikely to have much effect on temperature, it could conceivably affect the number of Hadley cells. The result would be to move the Sahara Southward (and shrink it), to move the typical UK weather Northward (an area of continual weather fronts) around iceland. We would then see two further zones: one of typical dry climate with high pressure at the North of the Med, and another south of the med with continual weather fronts as hot southerly winds and colder northerly winds meet.
In reality, Hadley cells are not fixed, and more than likely at certain times, we already get a five Hadley cell structure to the atmosphere.

Mike Blackadder
August 7, 2010 10:18 am

However, I suppose that this explanation is missing something from Tom’s post. Increasing CO2 concentration would have cooling effect until the point where the CO2 is sufficiently ‘excited’. During this transient period collisions would have the net effect of transferring energy from non-greenhouse gases to CO2 (in form of vibrational excitation) until these modes are in equilibrium. Still, I don’t think that this transient condition has any impact on the final equilibrium condition where the higher CO2 concentration must result in greater absorption of surface LW and higher equilibrium temperature to restore balance of radiation in and out of the air mass.

August 7, 2010 10:43 am

On the issue of the “greenhouse” effect, many are tying themselves in knots by unnecessarily delving into confusing details. Try thinking of the problem this way:
If it is easier for solar energy to reach the Earth’s surface than it is for heat energy to be irradiated away from the Earth, then it will be warmer at the Earth’s surface.
If the resistance to inflow and outflow of energy were equal, it would be cooler at the Earth’s surface (Eg. the Moon, on average, is cooler than the Earth.)
1. Solar energy is mostly non-infrared.
2. Heat radiated away from the Earth is mostly infrared.
3. The atmosphere allows non-infrared energy to pass through more easily than infrared. Energy gets in easier than out.
4. Therefore there is a “delay” in heat escaping the Earth as a whole system (planet including atmosphere), hence heat builds up near the surface until equilibrium between inflows and outflows are reached.
5. The more resistant the atmosphere is to the infrared flow away from the Earth, the higher the equilibrium temperature.
6. If greenhouse gases increase, the infrared outflow resistance increases, hence the heat content of the system rises to a higher equilibrium
(7. There are undoubtedly negative feedbacks in the Earth system. Otherwise, the Earth’s system would not have oscillated by only a few percentage point in temperature (in degrees Kelvin) over the last billion or more years. A system dominated by positive feedbacks is inherently unstable and gyrates wildly. A system dominated by negative feedbacks acts like the Earth has.)
The details of the mechanisms of radiation absorption, reemission, conversions to and from the kinetic energy of moving molecules and the potential energy of excited electrons, saturation effects, bandwidth broadening, etc. are critical to quantitative evaluation and predictions of the the equilibrium temperature for various changes in albedo, GHG concentrations, etc.
The details are not important for understanding the simple existence of a greenhouse effect due to Earth having an atmosphere.

Kevin Kilty
August 7, 2010 10:59 am

Let me clarify point by point.

#
#
Jordan says:
August 7, 2010 at 3:57 am
… We can say that equilibrium can is achieved in each tube when the OLR at each window matches the incoming solar radiation.
Staying at the windows for a second, I assume that the tube full of CO2 has an emitting surface at the window, whereas the empty tube does not. I’m not sure whether that has any bearing on the matter.

Your original suggestion involved “ideal” things of all sorts and it is difficult to decide how these ideal things will work. In the case of the CO2 filled cavity the window will become hot through conduction with the CO2 gas. In the vacuum cavity it will become hot through contact with the structure of the tube, except the tube as stated has no conductivity (i.e. well insulated, you said) so what happens then? Look, it doesn’t matter because the window is ideal. It has perfect transmission, no reflection, and no absorption, so it has emissivity of zero. Doesn’t matter what its temperature is. It does not radiate.

So what do we expect to happen at the other end of the tubes according to GHG theories?

Distant end of the CO2-filled tube heats more rapidly, perhaps, but at equilibrium both have the same temperature, as I will explain in a moment.

OLR emitted from the deep end of the empty tube gets free passage straight through the cavity and out the window. It can equilibriate in much the same way as the moon. Perhaps reaching a similar temperature to the surface of the moon in daylight.

No, won’t be anything like the Moon. The sun is just a small object at the distance of the Moon, the materials of the moon do not absorb completely, absorptivity and emissivity are not the same value because the emission and absorption take place in different parts of the spectrum, and the moon radiates into a hemisphere.
The tubes, in contrast, as I have imagined them are perfectly insulated except for a small aperture. They gain and lose energy only through the aperture, and I assume the view through the aperture is only that of the Sun. So equilibrium will occur when the incoming and outgoing spectra are the same, and the tubes have reached a temperature equal to the sun. The OLR is not LR; it is thermal radiation at 5800K.

This thread discusses CO2 as a resistive medium to OLR. The GHE effect appears to demand that equilibrium can only be achieved at the window of the tube full of CO2, if the temperature at its closed end is much higher. (We could almost decide what temperature we want by design of length of tube and maybe CO2 pressure?).
The argument of “OLR resistance” seems to suggest that the closed end of the tube full of CO2 could be sustained at higher temperature compared to temperature at its window. Certainly by radiative arguments – I’m not sure whether conduction will/could alter this.
Could we stick with a scenario where the sides are well insulated and conduction loss is negligible (incoming solar is quite powerful so we need not get too concerned if there is some leakage through the sides).

I have assumed this scenario, except, as I tried to explain, it has some awful ambiguities.

That leaves the question: when pointed into the sun, what are the temperatures at the closed ends of the tubes to radiate identical amounts of OLR at the windows?
If there is a significant temperature difference at the closed end of the tubes, we have potential to drive an engine and do work. Flow of heat through the engine helps the OLR to bypass the resistive medium in the tube full of CO2.
Why am I sceptical about this?

You are sceptical because you ought to be; the engine you describe would violate the second law. I have tried to explain why these two tubes, if allowed only interaction with the Sun, will reach the same temperature, and never be able to drive a heat engine. Each intercepts exactly the same energy from the Sun as does the other, at equilibrium each radiates back toward the sun the same energy, there is zero net to flow through the heat engine and no temperature difference to drive it.
If you want to make a heat engine, then you must allow the two tubes to exchange energy with more than just the Sun. This prevents them from reaching the same temperature, allows one of the two to supply net energy to the other through an engine, which the temperature difference will drive. As you have described this device, its CO2 filling doesn’t provide enough difference between the tubes to make much of an engine here even if you keep the two tubes from reaching equilibrium.
Now, if you really want a simple heat engine then make two radiating panels of different materials; one that gets very hot in the Sun and one that stays cool.

1DandyTroll
August 7, 2010 11:07 am

@winterkorn
‘The details are not important for understanding the simple existence of a greenhouse effect due to Earth having an atmosphere.’
But the details are of utmost import to understand why other planets, and some of their moons as well, in this same solar system also has warmed slightly just like earth the blue marble of the fried gaia.
And, personally, I’m getting pretty sure the other bodies around sol that has warmed slightly haven’t done so due to some thousands of a percentage increase of green house gases on earth. But of course, what if….

cal
August 7, 2010 11:08 am

Bill Illis says
People get caught up in the absorption and emission bands of CO2 and H20 and do not understand there is radiation ocurring all over the spectrum. CO2 has specific frequencies where it strongly interacts with EM radiation but there is also base black-body radiation (that is not as strong or intense sometimes) but there is two kinds of radiation to be concerned about – the strong Absorption and Emission bands and the more general black-body radiation.
I thought I had covered this clearly in my last post. The radiation from the surface approximates to a black body ( it is not exact – if it were the sea would not be blue!).
So of the three radiating elements I mentioned the surface is a near black body radiator and the gases are absorbing and radiating at their characteristic wavelengths but with intensities that are dependent on temperature.
I like your charts. I have not seen the ones “looking up” before. However I am not sure what your commentary is trying to say. You keep using the words “as if” it were such and such a temperature whereas that is exactly what it is.
For example when looking down from 50M all the radiation at all wavelengths will reach the sensor from the surface or from gas molecules in between which will be very close to the same temperature as the surface. This is exactly the point that Tom was alluding to. Any radiation absorbed by CO2 for example will be re radiated. Since the temperature decline within 50M is small you will not see that the re radiation is slightly lower. So no absorbtion bands are to be expected. Water vapour bands are even less likely to be seen.
Looking up from 50M you will see energy radiated downwards from excited molecules in the atmosphere above. There will be very little in the atmospheric window around 10 micron since these wavelengths are not absorbed and therefore cannot be re radiated downwards.
At the other wavelengths you will see energy radiated downwards mainly by H2O and CO2. Because they are such powerful absorbers the level at which they will radiating downward will ( on average ) be not far above the 50M where you have placed your sensor. So the fact that they are radiating at -20C seems entirely plausible to me.
If you look down from 20Km you will see exactly the picture I was descibing where all the radiation is either from the surface at about 288K ( higher during the day in the tropics) or from CO2 and H2O at the temperatures you quote.
Looking up you will mainly see radiation characteristic of CO2 since this is by far the dominant radiating gas above the tropopause.
All the data you showed seem exactly as one might expect.
Maybe I missed something that you were trying to say.

Dave Springer
August 7, 2010 11:20 am

The abstract physics discussions about how insulators insulate (or not for those who somehow believe insulators don’t insulate) and quantum mechanical accounting for every particle and photon is interesting it must ultimately agree with the observations.
If we accept the temp record from 1880 to 2000 as reasonably accurate and break it into two 60 year intervals we find temperature rose 0.4c in each interval.
If we accept an anthropogenic CO2 driven increase of 75ppm during that time we find that it rose 25ppm in the first period and 50ppm in the second.
If we accept that CO2 acts as insulator and does raise surface temps but as with any other insulator it becomes exponentially less effective per unit of increase then everything makes sense and we can make some predictions for the next 120 years.
Let’s assume we don’t repent our wicked ways and continue doubling how much fossil fuel we burn every 60 years.
Years +CO2 +temp
1880-1940 +25ppm +0.4c
1940-2000 +50ppm +0.4c
2000-2060 +100ppm +0.4c
2060-2120 +200ppm +0.4c
As near as I can tell from the observations and by accepting the CO2 vs. temp correlations as causation (for the sake of argument) there will be a CO2 doubling to about 650ppm in 120 years and a 0.8c temperature rise because of it.
This is not inconsistent with IPCC projections. In fact it is their best case scenario and it happens to be right on the money for the ten years from 2000-2010 that we didn’t analyze above.
Now we must compare the upsides to the downsides.
If we examine all nations that underwent industrialization during this 1880-2000 period in history, compare their fossil fuel comsumption to gross domestic product, we find another correlation – growth in fossil fuel consumption goes hand in hand with economic growth.
Slowing economic growth is a huge downside but that’s what limiting fossil fuel consumption will do absent a comparably cost/effective alternative energy source.
As well, scientific and engineering advances are driven by growing wealth. If there’s less economic growth there will be less wealth available to fund research and development, including research and development of alternative energy sources.
There’s another big downside but it will happen if we strangle our growing energy needs.
A warmer earth with a higher CO2 content is a greener earth. This is just indisputable testimony from the geologic column. Personally I prefer green plants and animals to bare rock and ice so I consider this an upside to continued exponential increase in fossil fuel consumption.
We can expect sea level to rise about 12″ from thermal expansion. This is the only downside. 120 years to deal with it makes it seem almost insignificant.
So the CAGW crowd basically wants to throw civilization under the bus to stop a “problem” that’s actually a great benefit to both the biosphere and standard of living for each human member of it.
I suspect the CAGW crowd is suffering a mass “messiah complex”. They want to save the world and be recognized for it. But there’s a big problem for them if the world doesn’t need saving. Thus we get the climategate chucklemonkey writing in 2008 “We can’t explain the lack of warming since 1998 and it’s a travesty that we cannot”.
Actually the IPCC did explain it. It just happens to be their best case scenario where nothing scary happens – the world doesn’t need saving – in fact the world gets better by burning more fossil fuels.
So there. I feel much better now.

Stephen Wilde
August 7, 2010 11:26 am

Mike Haseler says:
August 7, 2010 at 10:06 am
Hello, Mike. Your post shows that you’ve got my point and that you see some of the implications.
As regards CO2 specifically though the fact is that we see quite large circulation changes from natural climate variability so any effect from CO2 is likely to be unmeasurable particularly since the response of the hydro cycle is highly scaleable by which I mean that the energy shifting efficiently of the latitudinal shifts would increase geometrically as the shifts become more extreme. After all over billions of years they have prevented permanent disruption of the system through ice ages and much warmer periods than we have now.
I haven’t yet thought through the precise consequences beyond the latitudinal movement of the jets but some of the possibilities are as you suggest.

Mike Blackadder
August 7, 2010 11:52 am

Continuing my conversation with myself….
Thinking about it more I’m thinking Tom may be right and Jeff (and I) might be wrong. Maybe additional concentrations of CO2 does not directly warm the surrounding air:
1) Presumably in actual atmosphere the air does not radiate like a blackbody because greenhouse gas concentrations are small (ie. If there were no greenhouse gases the atmosphere certainly would not radiate like a black body.)
2) The Energy Equipartition law suggests that statistically energy will be distributed evenly across the different kinetic modes available for a molecule. Air is a mixture of different molecules, and not all molecules have vibrational modes available to interact with radiation. Larger concentration of CO2 impacts the characteristics of the air mass as a whole since there is now larger proportion of available vibrational energy state relative to translational.
3) Therefore, an air mass with larger CO2 concentration and SAME equilibrium temperature would be expected to absorb and radiate a larger amount of radiation.
I’m sure someone can tell me if I’ve made errors in my logic above (ie. my interpretation of black-body/ energy equipartition law or otherwise). If not then it seems the argument I made earlier that the temperature of the air mass would have to increase in order to balance the change in absorbed radiation is wrong. Moreover, I don’t see why an injection of CO2 would result in a net transfer of energy from vibrational excitation to translational. Greater opportunity for vibrational to translational conversions would be equal to greater opportunity for translational to vibrational conversions.
Note: Even if Tom is correct, this does not mean that greenhouse gases have no warming effect on earth or atmosphere. The GHGs still warm the surface and warming of the surface results in atmosphere reaching higher equilibrium temperature. It’s just that higher absorption of radiation in atmosphere (due to higher CO2 concentration) does not DIRECTLY warm the surrounding air.

Rob Z
August 7, 2010 12:14 pm

This is a valuable “Assume the horse is a perfect sphere” experiment. The blogger says: My statement is – CO2 does create a warming effect in the lower atmosphere.” is incorrect. The blogger merely proves that CO2 absorbs 15um wavelength radiation and relaxes. Whoop de doo. Congratulations!! The “physicist” fails in his “perfect sphere” experiment by failing to ask the question, “is there any way that CO2 would cause cooling to be faster or cause net cooling in general?” The answer is an emphatic YES! The second question is, “Will the cooling forces (excluding convection) overwhelm the radiative heating mechanism. The answer again is YES. How do we know this?
Convective cooling using air is substantially more effective than conductive cooling using air. It’s why we blow air on hot soup to cool it faster…more air…more cooling (CO2 at higher concentrations than 400ppm).
Conductive cooling using air is substantially better than radiative cooling with no air. Radiative transfers of energy are slow, it’s why things stay hot or cold in a vacuum. FOR A LONG TIME!!!
If you were to add a small amount of CO2 and only CO2 to the interstitial space (area between the surface contacting the coffee and the outside surface) of a thermos for your hot coffee… guess what …it get’s colder faster. Go do the experiment in a glass thermos or any thermos. Since your coffee is radiating at LWIR, the CO2 gas doesn’t keep the coffee warmer. That is basic physics and thermodynamics. Also, since there is more Oxygen than CO2 in the atmosphere…and replacing Oxygen with a CO2, a gas with a lower heat capacity at constant pressure, the atmosphere will be more conductive and cool faster.
Jeff hedges his bets…and says….”I’ve not said anything about the magnitude or danger or even measurability of the effect. I only assert that the effect is real, is provable, it’s basic physics and it does exist.”… This is disingenuous. The conductive effects of added CO2 to the atmosphere will outweigh the absorptive effects. The net effect is cooling. When the physical effect is not measurable in the real world, is it necessary to account for it? If a measured effect contains both mechanisms…and one outweighs the other, do you use the net effect in the calculation? If yes, do you make conclusions based on the net effect or the only one of the components? If CO2 can absorb heat and do conductive/convective cooling and the net result is cooling, do you claim that CO2 causes heating if the net effect is cooling?? The “green house effect” is bogus. Don’t believe me… convince a nursery to pump in CO2 (800ppm final concentration ought to be enough) to keep it warm at night in the green house instead of turning on the heat when the temperature approaches freezing outside. Should be able to bump the temp up by 5C. For good measure…have Jeff bring his laser.

Al Tekhasski
August 7, 2010 12:33 pm

Mike Blackadder says: “Therefore, an air mass with larger CO2 concentration and SAME equilibrium temperature would be expected to absorb and radiate a larger amount of radiation.”
Equilibrium temperatures of chamber “C” are not the same for different CO2 concentrations. Yes, at larger CO2 concentration an air mass will absorb larger amount of radiation. The parcel of air will warm up to a level of temperature when absorbed and emitted radiation are equal, regardless if it is a black body or not, at which point the radiation will stop warming the air. What might be so confusing in this simple schema?

Dave Springer
August 7, 2010 12:39 pm

@Cal
For what it’s worth the following is what you see in the infrared spectrum looking upward and downward:
http://www.sundogpublishing.com/fig8-2.pdf
It’s from the 2006 textbook “A First Course in Atmospheric Radiation” by Grant Petty.
All the figures from the textbook may be viewed at:
http://www.sundogpublishing.com/AtmosRadFigs.html

August 7, 2010 1:21 pm

Mike Blackadder
1 “Presumably in actual atmosphere the air does not radiate like a blackbody because greenhouse gas concentrations are small (ie. If there were no greenhouse gases the atmosphere certainly would not radiate like a black body.)”
Not true, the collisions are more numerous than radiation, you will get blackbody radiation.
3 – “Therefore, an air mass with larger CO2 concentration and SAME equilibrium temperature would be expected to absorb and radiate a larger amount of radiation.”
This is just confusing. If the equilibrium temperature is dependent on the absorbed radiation, then what you are saying is that — if your air has the same energy input with or without CO2……
“I don’t see why an injection of CO2 would result in a net transfer of energy from vibrational excitation to translational. ”
C02 is far more likely to collide than radiate. So the CO2 will absorb the energy, the molecules vibrate and collide, and you get heat. More CO2 means more capture. Maybe only slightly more, but still more.
winterkorn says:
August 7, 2010 at 10:43 am
“If it is easier for solar energy to reach the Earth’s surface than it is for heat energy to be irradiated away from the Earth, then it will be warmer at the Earth’s surface. ”
Nice.

August 7, 2010 1:40 pm

Stephen Wilde: You replied, “Some of these links may help you.
http://search.orange.co.uk/all?q=jet+stream+shifts+1970+to+2000&brand=ouk&tab=web&p=searchbox&pt=todayweb_hp4&home=false&x=21&y=15”
Thanks for the link. But I don’t need links to search engine outputs. I’ve researched that. Specific data upon which you rely was what I was asking for. On the page you offered, there was little to document the global latitudinal variations you commonly refer to.
I will again suggest you learn to use the KNMI Climate Explorer:
http://climexp.knmi.nl/selectfield_obs.cgi?someone@somewhere
For example, it would allow you create Hovmoller diagrams of CAM-SOPI precipitation data from 10S-20N, at different ocean longitudes, from 1979 to present. The following are samples, using 12-month averages to eliminate seasonal variations. I’ve thrown some lines on them to make it easier to identify the variations. Here’s one that captures the ITCZ over the Pacific at 180 long (also catches the SPCZ on occassion):
http://i36.tinypic.com/sbnfcg.jpg
And the ITCZ at 120W:
http://i35.tinypic.com/qprp1v.jpg
And the ITCZ for the Atlantic at 30W:
http://i33.tinypic.com/k5324k.jpg
ENSO appears to be the cause of the greatest year-to-year variability, as one would expect.

sky
August 7, 2010 1:57 pm

Sanctus simplissimus! The thought experiment of passing a laser through gases in a container (which itself may absorb IR) confuses signal transmitivity with the flow of thermal energy in the atmosphere, which in the geophysical setting is demonstrably more through non-radiative mechanisms than through terrestrial IR radiation. In that setting there is no radiative algebra that can determine the temperatures throughout the atmosphere, as the AGW proponents invariably imply. Like temperature itself, radiative intensity is a NON-CONSERVATIVE state variable, and not a flow variable! For a solid physical understanding of the problem, the conservation of enthalpy under the inviolable direction of entropy needs to be mastered. This is almost never seen in discussions on either side of the climate debate.
Like all strong absorbers, CO2 is also a strong emitter. Its effect thus cuts both ways. It’s not a generator of any energy. Those who are inclined to believe that CO2 has a rare ability to trap or store thermal energy, squirelling it away like gold coins, forget that its specific heat is less than that of aluminum. For them, I suggest finding a large aluminum post on a cold winter day and see for themselves how much their temperature is increased by standing next to it. In fact, if they would put their tongues on said post they would never again attempt to argue that a colder body, unsustained by any independent power source, raises the temperature of the warmer body.

Mike Blackadder
August 7, 2010 2:52 pm

Jeff, thanks for the reply. Bear with me for a minute, I’m still trying to sort this out.
“Not true, the collisions are more numerous than radiation, you will get blackbody radiation.”
So you are saying that we will get a fixed amount of radiation from a mass of air at given temperature regardless of the CO2 concentration? Just to clarify. Say you have two packets of air in a vaccuum and both are at the same temperature, but one packet has concentration CO2 like earth and the other twice the amount; would the packet with higher CO2 not emit more radiation to space? Also I don’t see why many collision events have any impact except that it gives opportunity for the air temperature to reach equilibrium in response to radiation events. In other words a CO2 molecule would presumably transfer vibrational energy to translational and vice versa many times between absorbing or releasing a photon.
I said: ‘I don’t see why an injection of CO2 would result in a net transfer of energy from vibrational excitation to translational. ‘
Jeff: “C02 is far more likely to collide than radiate. So the CO2 will absorb the energy, the molecules vibrate and collide, and you get heat. More CO2 means more capture. Maybe only slightly more, but still more.”
My point (and I believe Tom’s point) is that the converse is also true. More CO2 means that there are more molecules with the additional vibrational degrees of freedom. CO2 that isn’t excited that collides with another particle will result in translational energy converted to vibration of the CO2. Therefore, if we increase CO2 concentration and we don’t increase the amount of energy in the air packet then the temperature is lower (ie. if CO2 has no vibrational excitation there would be a transient effect of net translational to vibrational conversion until reached an equilibrium).
I suppose that where this thinking goes wrong is in the event that there is a very large amount of radiation. If everytime a CO2 molecule transfers vibrational energy to another molecule that it immediately absorbs another photon that with the many collisions we are effectively pumping radiation into the system. I could see in that case why there would be a net transfer of energy from vibrational to translational. That seems to be the case in the example that you’ve given and perhaps is true in the atmosphere.

Stephen Wilde
August 7, 2010 3:08 pm

Thank you Bob, very helpful.
When I retire from the day job I’ll make a start. In the meantime I suspect that new observing techniques and closer attention to the relevant observations may make such work unnecessary.
Have you given more thought to extending your work to mesh in with global rather than regional events ?

cal
August 7, 2010 3:37 pm

Dave Springer
Thanks for the interesting links
Rob Z
I am sorry but I cannot agree with your view that conduction is more powerful than radiation. A few things are known for certain and one of these is how much energy would be radiated by the earth at its current surface temperature if there was no atmosphere. This is far more than it currently radiates. Therefore if there were no atmosphere it would cool until radiation balance is achieved. So despite what you may feel intuitively the nett effect of the atmosphere is not a cooling one.
Your example of the thermos flask is not valid. The thermos is designed to remove all forms of heat loss. So the there are two walls of poorly conducting material where the inner surfaces are silvered to reduce radiation and the gas is removed to reduce conduction. I am pretty sure that the thermos would work better with silvered surfaces and 0.0003 bar of CO2 in the gap than it would with blackened surfaces and a perfect vacuum.
More importantly we can do the sums for the real thing. The maximum temperature gradient across the atmosphere is 8K per kilometre.The conductivity of air is 0.024W/mK so the energy lost by conduction is only a tiny fraction of a watt per square metre compared with the hundreds of watts lost by radiation.
In any event the issue is not really what the atmosphere is doing now but what it would do if you added more CO2. As I have said previously this will never be answered using a bottom up analysis. The micro models used by climate scientists today may work when we study gross changes that occur for short periods during weather events they don’t work for subtle changes over long periods relating to climate. Indeed I suspect subtle effects like an increase in any trace gas or aerosol will not be understood using any type of analysis until we have hard data from satellites correlating actual emissions with actual temperatures as the earth responds over time to changes in clouds, wind, convection currents, ocean temperatures etc. Then we can do real science.

August 7, 2010 3:46 pm

Stephen Wilde says: “Have you given more thought to extending your work to mesh in with global rather than regional events ?”
ENSO and AMO are regional phenomenon with global effects. Have you forgotten my posts about OHC? I present and discuss things regionally because one can identify causes of variability–SLP, ENSO, etc. With global perspectives, one loses that ability, and assumptions made at the global scale may very well be incorrect, because of the noises created by the regional variations at different time scales.

Gnomish
August 7, 2010 4:22 pm

Rob Z says:
August 7, 2010 at 12:14 pm
You got it.
Any increase of heat capacity of the working fluid improves the efficiency of a heat pump.

August 7, 2010 5:33 pm

Mike,
from this link:
http://en.wikipedia.org/wiki/Planck%27s_law
you can see that intensity of emission I is based only on temperature. I really have a hard time following the rest of your post. Every time I think I’m getting it, I loose it.
Maybe an even simpler explanation.
The laser passes through clear stuff — it gets absorbed by black stuff. The CO2 is black stuff. If we consider all the other minor aspects of what happens, the black stuff is still black. The laser cannot pass — so it makes heat.
The rest of the consideration about re-emission time, how much light passes through and the rest only determines the level of heating, not whether it happens or not.
I hope that is helpful.

Spector
August 7, 2010 6:33 pm

RE: Gail Combs: (August 6, 2010 at 4:45 pm) “I guess I was not clear. I thought that is what I said.”
I am sorry, I thought that was an isolated general question…

Bill Illis
August 7, 2010 6:41 pm

There is another dimension to consider here – “Time”. How fast does the radiation exit the Earth system? How much time does it spend moving from molecule to molecule.
When the Sun sets each night, the average surface radiation level falls from 418 W/m2 at the peak to 364 W/m2 just at sunrise or a decline of 10C.
So over the 12 hours, an average 4.5 W/m2 is lost to space each hour (obviously slower in the CO2 and H20 bands and faster in the windows.)
But if the Sun did not come up tomorrow, at 4.5 W/m2 loss per hour, it would only take 80 hours – a little more than 3 days – before the surface temperature in the centre of the continents approached that of the Cosmic Background Radiation – give or take some lag from water bodies.
Or one could also say, the average time that the energy represented by a visible spectrum photon from the Sun spends in the Earth system before it is lost to space is 40 hours – either a surprisingly long number or a surprisingly short number depending on your perspective. But at the speed of light and the average relaxation time for excited molecules, this means the photons are spending an extremely long period of time bouncing around from molecule to molecule. Maybe someone wants to take those numbers and calculate how many molecules each photon ends up in before it escapes to space – lots of Zeros in that number.

Steve Fitzpatrick
August 7, 2010 7:35 pm

Jeff,
You and Willis are ‘the men’! Thanks for the post and comments.

Mike Blackadder
August 7, 2010 8:04 pm

Jeff,
I think I see the problem.
So we have a perfectly insulated chamber with windows at the ends to let radiation in and out, and I will add that it is surrounded by a vaccuum.
A 15um laser is firing through the gas in the chamber and we’ll consider how this effects the temperature of the gas depending on mix of N2 and CO2. Note: We’ll get to the case of no CO2 at the end.
Scenario 1) 1% CO2: Turn the laser on. With 1% CO2 we’ll assume that the gas does NOT appear completely black to the laser. So some laser light passes right through the chamber. The CO2 absorbs X photons/sec. Before being able to re-emit it gets in millions of collisions and will tend to transfer their vibrational energy to the cooler N2 gas (and also the collisions will result in translational energy of the CO2 molecules as well). So the air heats up. So the air is absorbing X photon/sec but is emitting less than X photon/sec. Eventually the gas will warm up to an equilibrium temperature where it is emitting X photon/sec and absorbing X photons/sec.
Here’s the important part. This only happens at the point when the following collision events are equally probable: CO2* + N2 -> CO2 + N2′, and CO2 + N2′ -> CO2* + N2. If this is not the case then a net amount of absorbed radiation is being translated into heating of the case or vice-versa.
Scenario 2) 2% CO2. Turn the laser on. Still some laser light passes through the gas without being absorbed but now a greater amount is absorbed: Y photons/sec. The same thing happens when the gas is cool, where there are more CO2* + N2 -> CO2 + N2′ than the CO2 + N2′ -> CO2* + N2, and so the CO2 doesn’t get a chance to release photons before transfering it to heating of the gas. We know that it will warm up and reach equilibrium when emitting Y photons/sec.
The problem is you assume this is a higher gas temperature than scenario 1, but it isn’t. In fact it is only when collision event CO2*+N2 -> CO2 + N2′ is equally probable to CO2 + N2′ -> CO2* + N2, which is the same temperature. The CO2 and N2 molecules involved in this collision have no knowledge of gas concentrations surrounding them. There is no reason why N2 suddenly has to be a higher temperature in order to energize CO2 as often as it deenergizes CO2.
Finally Scenario 3) 0% CO2. Turn on laser. Hopefully this helps clear things up. No laser light is absorbed by the gas, and all laser light passes right through. The gas doesn’t heat up.
Question: What’s the gas temperature? Before you say 0 K, remember the gas is in a perfect insulator surrounded by a vaccuum. The truth is the gas temperature could be anything. This illustrates how you’re only considering one side of the problem. In Scenario 1 & 2 we know the gas is cool when we start the laser because the gas contained CO2 which was emitting heat out of the chamber (net CO2 + N2′ -> CO2* + N2) transactions. Without the CO2 this can’t happen.
Where did I go wrong this time?

Mike Blackadder
August 7, 2010 8:09 pm

Re: 0% CO2 case. I forgot to mention. It has been suggested that adding CO2 to the chamber results in warming of the gas. This is not necessarily true since we don’t know what the temperature of the gas is without CO2. Adding CO2 could just as easily result in cooling of the gas, even with the 15um laser running.

Gnomish
August 7, 2010 9:06 pm

I find JeffID’s model quite good for the purpose of illustrating that radiation can be absorbed and converted to kinetic. Vonk examined in detail the mechanism of energy transfer and conversion.
The effect of an energy absorbing molecule is to absorb energy and quickly share it with everything else.
The study of the radiative physics is interesting and educational. It is the determinative property for heating something from above and it is the way heat is given up by a radiator.
In a heat driven system, however, it is really only significant at the end points because phase change and convection completely dominate the system between those terminals.
Neither Vonk nor JeffID are offering any explanation of conduction or convection.
The reason radiation has seized the field as the straw man of the argument is because the only thing that the taxable substance can do that distinguishes it in any way is be darkish in infrared. The fact that is has a profoundly underwhelming effect (and the opposite of what is claimed) on a convective system is obscured by the prestidigitation.
CO2 can’t hold a place on stage with water, but they’ve already got water meters.
You know that privately they are laughing like maniacs that they will tax your air and have succesfully guided the dialectic to elevate a trivial entity to the throne of a demonic pantheon at your expense.
By Avogadro, the ideal gas constant has the same value for all gases, so PVT = PVT regardless of the molecule.
1 mole of gas molecules, any kind = 24.45 litres at 298K ( 24.85C, 76.7F) and 101.325 kPa (14.696 psi, 1 atm)
In 1000 liters of air, at STP there are 1000/24.45 = 40.9 moles. As in a previous illustration, let the water component be 1% = 10 liters = 4.09 moles. It doesn’t matter what the other gases are for this.
At the critical temperature of water vapor, this 10 liters condenses and occupies 4.09 * 18g * 1g/cc = 736.20 cc.
So the 1000 liters of air would now be 990.74 liters.
Insofar as PVT = PVT for gases, if the pressure alone changes, it means (using 24.85C and 1 atm)
PVT start is 14.696 psi*1000*(24.85C+273.15K)
final P is
14.696 psi*0.990.74, or 1% pressure drop
To get a 1% pressure drop by changing the temperature alone you need to do from 298 to 292.04 = 5.96 degrees.
http://img534.imageshack.us/img534/4959/hadleypump.jpg
http://www.engineeringtoolbox.com/spesific-heat-capacity-gases-d_159.html
Gas or Vapor kJ/kg
Air 0.287
Carbon dioxide 0.189
Water Vapor 0.462
Steam 1 psia.
120 – 600 oF
That’s what it takes to change the temperature 1 degree K.
When CO2 changes from 1 to -1 C, a change of 2 degrees C, it radiates 2(0.189 kJ/kg) = 0.378 .
http://en.wikipedia.org/wiki/Enthalpy_of_vaporization
When water vapor changes from 1 to -1 (and condenses) it radiates 2257 kj/kg + 2(0.462 kJ/kg) = 2257.853776 kJ/kg.
It does this every single time you see a cloud.
But CO2 has no phase change so it carries no heat – the numbers:
All gases at the same temperature have the same number of molecules per unit volume. (Avogadro)
Water, being light, masses 18g/mole and CO2 masses 44 g/mole
Using 1 mole of air, just to make math easy:
We lowball the water in the atmosphere at 1% of the molecules
So, in a mole of atmosphere, we have 0.01 moles of water = 0.18g
now we highball the CO2 at 500ppm which is 0.0005, or 1/2000 of a mole of CO2.
1/2000 * 44g/mole = 0.000484 moles of CO2 = 0.021296g
So in our mole of air with but 1% H2O and a generous 500ppm CO2-
the water condensing radiates 0.18g * 2257.853776 kJ/kg = 406.41367968 J
while the CO2 radiates 0.021296g * 0.378 kJ/kg = 0.008049888 J
the ratio of 0.008049888/406.41367968 = .00001980712855516645290496438242332
or as much to say that water vapor in the example carries 50486.873814890343815963650674393 times more heat than the CO2 does.
And that’s just rain. If it turns to snow- multiply by 5-6.
Meanwhile, Venus is a ball of active volcanoes with a dry heat pump to radiate it poorly.
That is why Earth’s climate doesn’t resemble that of Venus.
Forget about CO2.
All things radiate as blackbodies (or maybe a bit grayish) and noting that while water does not change temperature as it changes phase, it radiates many hundreds of times more energy in the process than any other gas.
Therefore, the blackbody spectrum may not change a whit, but –
There are a number of things that water gas does which are scarcely mentioned. It seems to be considered nothing but a personal assistant to CO2.
However there are many things that water does which define the atmosphere, the lapse rate and the thermal equilibrium.
In the first place, it evaporates. When it does, 3.7 teaspoons of liquid becomes one liter of gas. This happens without temperature change. No change occurs in the black body spectrum.
The expansion increases the local pressure above what a dry gas can under the same conditions.
At the same time, water is much lighter than any other gas in our atmosphere (except the traces of He and H), , massing a measly 18g/mole – so it rises straight up, shifted by coriolis effect as it billows wider and wider.
When it finally condenses, at the same temperature as the surrounding gas, it radiates the one spectrum throughout its phase change, indicating no higher temperature while it radiates 406.41367968 J and changes back to 3.7 teaspoons from (a bit less than, now) a liter of gas, producing a local low pressure drop of 1% that draws the atmosphere below up to fill it. (To get a 1% pressure drop by changing the temperature in a dry gas you need to do from 298 to 292.04 = 5.96 degrees.)
If water is but one percent of the volume, (using a sample volume of 100 liters that started at STP) the constituents would radiate their share as well, depending on the specific heat-
um… well, the other gases don’t radiate any more than they gain from below or sideways, or the temperature would actually drop- but for a one degree drop:
N2 (89.3g = 78%) 1.039 kJ/kg = 92.J
O2 (13.1g= 20%) 0.915 kJ/kg = 12.0J
CO2 (0.02g = 500ppm) 0.189 = 0.008J
H2O (18g = 1%) 0.462 kJ/kg = 406.41367968J
Water does more work than everything else combined – without having to change its blackbody spectrum.
(Compared to the CO2, water moves 50,000 times more energy from surface to space.)

Gnomish
August 7, 2010 9:13 pm

Increasing the heat carrying capacity of the atmosphere improves the efficiency of the system. Adding CO2 helps convection cool mo bettah. I didn’t make up the physics.

Gnomish
August 7, 2010 10:21 pm

You see, our atmosphere is not in a tube. Any heat on the surface is just at one end of a constantly moving conveyor to the infinite heat sink of space.
It’s not sitting there gaining heat and the more heat you add, the faster the conveyor moves. A bigger differential from source to sink increases the flow rate.
Swapping one non-phase-change gas in the working fluid will only infinitesimally affect the heat capacity and flow rate.
It all ends up in space at the end of the conveyor unless it’s stored in the mass of the planet.
You have to actually increase input to warm the convection fluid. Everything is wrung out of it in a matter of days.

Jordan
August 8, 2010 3:34 am

Kevin Kilyy – thanks for sticking with my questions.
I tend to agree when you say the suggested machine would be a violation of the second law.
But I also want to test the consequences of the argument that GHG has a resistive medium to LWR, and (on that argument) what we might do if we allow SWR to enter a chamber with free passage, then resist the passage of outgoing LWR.
The empty/transparent tube is only included to drive this thought experiment to an extreme. To show that there is either something very different happening in the two chambers, or there is not.
I appreciate that the concept is highly idealised, but I wanted to remove confounding factors in order to test the principle when I described it here.
Please note the windows are only for containment. Your point about perfect tranmission is well made, but I did not wish to create the impression that the windows have any bearing on the tranmission of radiation. When I mentioned an emitting surface at the window of the chamber containing CO2 I meant the outer boundary of the CO2, not the window itself.
The main reason why I mentioned the window-end of the CO2 chamber is because the last molecules are the emitting surface for that chamber due to its optical depth to LWR. I mentioned it to emphasise the notion of a possible temperature gradient in that tube which would not exist in the non-GHG or empty space in the other tube.
I accept your comment about equilibrium temperature on the surface of the moon. But I was thinking more in terms of what SB would demand for incoming solar radiation at circa 1500 W/sqm. Even in the closed chambers mentioned and where the image of the sun fills the apeture (that’s a good point which I failed to mention), I did not expect this to cause a rise to the same temperatire of the surface of the sun becuase of the 1/r^2 “dilution” of solar flux. Doesn’t SB suggest something more like 400K?
I understand that there are some awful ambiguities in reaching for the idealised physics. And the machine does seem to be a viloation of the second law. But these points leave little room for the argument that CO2 is a resistive medium, which could result in a potential difference at the closed ends of the tubes.
It is also important to mention that the engine does not permit the CO2 filled tube to reach its own equilibrium. If there is a temperature difference at the closed ends, it uses that potentital to remove energy from the CO2 filled tube.
I’m not trying to deny the physics – just testing them. And thanks for the discusson, please stick with it.

August 8, 2010 3:39 am

Stephen Wilde says: Hello, Mike. Your post shows that you’ve got my point and that you see some of the implications.
As regards CO2 specifically though the fact is that we see quite large circulation changes from natural climate variability so any effect from CO2 is likely to be unmeasurable particularly since the response of the hydro cycle is highly scaleable

Stephen, whenever I see these radiation transmission graphs, I really should come down like a tonne of bricks on the idea that there is ONE such graph. Cloud cover is by far the biggest factor effecting heat loss. This is so patently obvious to anyone who has gone out on a cloudless night – for heaven’s sake even the Romans knew this as they used to use the clearness of the night sky in the Sahara to make ice in massive insulated pits they would open up on clear, still nights to let it freeze.
Anyone can check out the massive change in radiative properties by buying a cheap IR thermometer, going out on a cloudy and sunny day/starry night and noting the difference. The sky’s radiative temperature (as measured by the emissivity curve of the IR thermometer) about (winter) 4C and -40C. In fact it is so cold on clear nights it goes off the end of the scale (-40C).
Quite obviously, and never admitted by the global warmers, the single biggest factor effecting the climate is the extent of cloud cover.
What goes up comes down
By the laws of physics clouds should not exist. Afterall, water is heavier than air and so it will fall down to earth. Therefore, in a static model of the atmosphere there can be no cloud, because if the air were static, the water would fall out of it.
So, clouds are by their very nature part of the dynamic convective cooling system of the world. They are also have a dramatic effect on temperature. It’s not difficult to prove that if that effect were to cause temperatures to drift from “normal” then, we’d already have had run-away global warming or global cooling and none of us would be here.
So, clouds are part of the heat regulatory system of the atmosphere, and so the key measurement I personally would use to monitor so called “climate change” (yuck) would be the extent of cloud cover.
That is to say, that in a self-regulatory system like the atmosphere, the best way to measure a change is not in the variables being regulated (temperature), but in the variable doing the regulation (cloud cover). Depending on the quality of that regulation, you could have imperceptible changes in temperature but the system could be under huge stress, because the regulation was highly effective. So, the place to look is at the thing doing the regulation of temperature not the temperature itself.
And, then when you start to think about human’s influence of the cloud cover, you begin to see where the real problems can lie:
1. Soot and dust from smoke
2. Aircraft con trails
3. And the level of high energy ionising particles which can form the nucleus on which water droplets can form.
And where is the biggest change? It has been the dramatic rise in smoke levels during the industrial revolution, followed by dramatic reductions in smoke levels from clean air legislation in the 1970s.
I would strongly suggest that the main reason for “global warming” is the environmentalists who (rightly) insisted that we stop pouring so much soot into the air in the 1970s – after which we appear to see a clear rise in temperature which so scared the warmers
Ironic isn’t it. We are all suffering from this global warming hysteria because the world had the good sense to remove smoke from the atmosphere which appears to have resulted in apparent warming.
And, If you want proof that dust can cool the planet, JUST MENTION TO A GLOBAL WARMER THEIR LAST SCARE REGARDING THE NUCLEAR WINTER!

Jordan
August 8, 2010 3:51 am

Ouch! Sorry for the typo, that last post from me should have been to Kevin Kilty

Gnomish
August 8, 2010 4:45 am

http://pielkeclimatesci.wordpress.com/2010/08/06/comments-on-a-new-paper-on-climate-sensitivity-by-lin-et-al-2010/
“Until the climate community moves away from the surface temperature trends, with its inaccuracy and unnecessary complexity, and replaces it with the diagnosis of ocean heat content changes in Joules over time, the policy community will continue to be misled as to the actual warming and cooling of the climate system.”

stephen richards
August 8, 2010 7:22 am

At tAV, (comment 47) Pat Frank calculated the decay half life for radiative emission as 30 milliseconds, and the collision time between molecules at 10^-8 seconds – so almost none of the gas decays and re-emits (sorry to those who don’t like the word).
Presumably at 1ATMOs

stephen richards
August 8, 2010 7:26 am

This has been the best thread on any site for some long time. Very few trolls have bothered to contribute so the flow of the thread has been maintained and the contributions have been intelligent, informed and very useful.
Thanks from an old physicist.

August 8, 2010 7:29 am

Jeff Id:
My statement is – CO2 does create a warming effect in the lower atmosphere.

WHEN a known greenhouse gas whose percentage can and does vary daily and weekly and with that variance _does_ cause noticeable local, overnight and daily temperature variances which vary proportional to that GHG, a veritable living experiment observable with minimal cost, time or equipment avails itself and supports the above JI statement.
What? Has no one else has observed basic outdoor temperatures in a meteorological context while also observing humidity/moisture/water content of the surrounding air mass and noted some correlation? … and now I have to ask the question, would these conditions in the longer term affect an observed average?
There is also another concept simply involving energy ‘flux density’ as it relates to the electromagnetic exchange of incoming (solar) energy to outgoing (earth LWIR) energy; the introduction of any mechanism which delays, temporarily ‘stores’ via low percentage reflection or re-radiation or ‘back’ radiation or ‘back’ scatter to use a radio/RADAR term is going to necessarily increase the ‘energy flux’ density in said system, and, in this case, it will cause some small increase in a metric we call temperature
No?
.

August 8, 2010 8:36 am

This is a great post and I thank Jeff and Tom for presenting things so clearly. To me the whole human-caused warming theme boils down to Figure 5 which is filled with oxygen. Clearly, the tank case will get warmer due to convection and radiation (photons scattering as shown at the exit). There should be a Figure 5b where 390PPM of CO2 is added and allowed to displace some oxygen. Now, will the case get warmer still? Yes, it will. Will the warming be measurable? I very much doubt it…even with extremely accurate equipment and very careful procedures. If you can’t measure it in this lab experiment, then you’ll never measure it in our atmosphere. Sorry, Dr. Mann, this case is closed.
As a correction to John Marshall above, it’s Kevin ‘Travesty’ Trenberth (not Mann).

August 8, 2010 8:53 am

Henry Haseler
I must say, I liked your last post. It is good. It is what I had been thinking also. Can I quote you?
Henry@wintercorn
Due to the elimination of CFC’s and many human activities, there has been an increase in ozone. Ozone eliminates a large portion of UV radiaton, where the sun’s intensity is very high.
So, don’t you think global cooling is imminent?

Dave Springer
August 8, 2010 8:59 am

Jeff Id says:
August 7, 2010 at 1:21 pm
C02 is far more likely to collide than radiate. So the CO2 will absorb the energy, the molecules vibrate and collide, and you get heat. More CO2 means more capture. Maybe only slightly more, but still more.

Only if all the energy isn’t already being absorbed. The effective optical depth of the troposphere at 15um is far less than the actual depth so it is indeed all being absorbed. It can’t absorb more than 100% of the emission.
Atmospheric transmission at sea level (1000mb) of 15um is nearly 0% across a distance of just one meter. At 20 kilometers (100mb) across 1 meter of air transmisson rises to 50%. textbook reference: http://www.sundogpublishing.com/fig9-13.pdf
Now compare to zenith transmission (entire CO2 column from sea level to 20 kilometers) where absorption is 100% across a much broader swath centered on 15um.
http://www.sundogpublishing.com/fig9-12.pdf
This is the “shoulder broadening” that is often talked about that happens with increased CO2.
Now, near as I can tell from actual observations, if we accept those and accept the correlations as causations, each CO2 doubling beginning at 280ppm effectively raises the average temperature at the surface (absent positive or negative feedbacks) pretty darn near to the IPCC published number of 1.1c.
While I can’t prove it, because correlation is not causation, it appears to be credible enough. CO2 does act as an insulator. That’s just basic classical mechanics. Explaining the exact mechanism at the quantum scale is interesting, evidently quite controversial, but ultimately irrelevant because we’re dealing with scales far outside the quantum realm and at that scale classical mechanics explains things quite well.
So if, for the sake of argument, if we accept the 1880-2000 data:
1) temperature rise of 0.8c
2) CO2 rise of 75ppm
then accept
3) the CO2 increase was due to anthopogenic emission
4) correlation between more CO2 and rising temp is actually causation
and then note (H/T to Vonk)
5) the temperature rise was 0.4c for the first 25ppm added from 1880 to 1940
6) the temperature rise was 0.4c for the next 50ppm added from 1940 to 2000
then it is reasonable to reach a tentative conclusion that the shoulder broadening effect of adding more CO2 follows (like many other energy distribution phenomenon) and inverse square law – i.e. for each incremental temp increase of 0.4c it requires twice as much additional CO2 as the previous increment.
This makes sense in many different ways from the basic physics of insulators straight on through to the observations of increasing CO2 correlating with rising temperature through in inverse square rule.
Whatever positive or negative feedbacks are associated with rising or falling CO2 must be presumed to have been operating over the entire period of 1880 to 2000 so while arguing about them might be interesting the net effect is included in the actual observations.
We can also look much further back in history when atmospheric CO2 was 10 to 20 times greater than today yet temperatures during those periods was only several degrees C higher. This is yet more observational evidence that we have an inverse square law governing the relationship between CO2 and temperature.
We can’t possibly burn enough CO2 to reach those prehistoric levels and even if we could it would only make the earth green from pole to pole as it was during the Eocene thermal optimum 50 million years ago. I prefer plants and animals to rocks and ice so when it comes to fossil fuels I say “Burn baby, burn! And in the meantime have a plan to find and develop an economical alternative energy source for day when the fossil reserves become economically unrecoverable”.

Dave Springer
August 8, 2010 9:13 am

I wish I could edit these things. Where I referred to CO2/temp correlation following an inverse square rule I should have written that was an exponential rule. Double the CO2 ppm to get the same incremental rise in temperature not square the ppm.

Dave Springer
August 8, 2010 9:22 am

stephen richards says:
August 8, 2010 at 7:22 am
At tAV, (comment 47) Pat Frank calculated the decay half life for radiative emission as 30 milliseconds, and the collision time between molecules at 10^-8 seconds – so almost none of the gas decays and re-emits (sorry to those who don’t like the word).
Presumably at 1ATMOs

Even if it does re-emit it’s irrelevant because near total absorption of 15um at 1ATMOS occurs in a distance of 1 meter. Even at 0.1ATMOS absorption is 50% across a distance of 1 meter:
http://www.sundogpublishing.com/fig9-13.pdf

cba
August 8, 2010 9:25 am

Jeff,
great explanation. Also hats off to Dewitt P and his contributions to your effort. He was of great help to me a couple of years back while getting my feet wet.
I think though that the problem with Tom’s explanation was a little different, having to do with difference of temperatures of the BB curve versus the temperature of the LTE example. In any case, it leads to a conclusion that is problematic in that if co2 did not transfer energy to other molecules in the sample after and change in conditions, then the LTE condition would become lost and one could end up with different temperatures for the various types of molecules present. I think part of Tom’s purpose for writing the post was that there seems to be a delusion among some CAGW fanatics that there is some big energy reservoir in the sky just like the oceans where energy is sucked up over time and getting ready to pop out in some deluge of heat.

cba
August 8, 2010 9:36 am

PJP says:
August 6, 2010 at 10:22 am
Other questions, to which the answer almost certainly exists, but I have no idea where to look:
The absorption spectrum of CO2 shown indicates some maximum absorption (attenuation of the 15 micrometer radiation). How does this change as the concentration of CO2 increases?
——————–
Actual attenuation is going to be directly related to the total number of the co2 molecules in the air column. Other effects are going to be secondary and involve the width of each spectral line and that is controlled by such things as the concentration of co2 in the gas (co2 partial pressure), temperature of the gas and pressure of the gas. As the total pressure drops, one has narrower line widths. The line width and height can be thought of as the effect at each wavelength and in order to capture any significant amount it must be over some range of wavelength. The 15 um band is composed of thousands of individual lines that overlap. At much lower pressures there may be little to no overlap and one winds up with portions of the band that are transparent.

cba
August 8, 2010 10:12 am

Scott says:
August 6, 2010 at 10:24 am
Can some back-of-the-envelope numbers using Beer’s Law and molar absorptivities/extinction coefficients be provided in this analysis? If not, where can I find something like that? That is where part of my skepticism lies, as anyone who’s run IR in organic chemistry can tell you that it doesn’t take a long pathlength before the CO2 absorption at 2350 wavenumbers (cm-1, around 4.25 microns) becomes saturated. Another fun thing I just realized is that CO2′s other main absorption band maxes out right around 666 cm-1…interesting.
Anyway, the main point of my comment is that I want someone to direct me to where I can get molar absorptivities for CO2 and hopefully water too. If no one has run the actual numbers with respect to saturation, I can start working on that (lots of numerical integrations, I know).
—————————————
Something like the HITRAN database can provide raw information but it’s a tremendous effort to get something fairly complete out of that. One can also get fair mileage out of Archer’s online modtran calculator. Some lines will have extinction path lengths measured in the cm distances. A short distance away in the wavelength, one can have path lengths of 1km or more.
This area tends to not be where the problems and arguments (and errors of the warmers) with the warmers resides. It is also only valid for clear sky conditions which is less than half of the Earth’s surface at any one time. The problem lie in the nature of cloud effects, albedo and absorption, in the total sensitivity effect present today in the amount of how much change in power results in how great a change in temperature, and ultimately, just how much change in temperature get’s conveyed into feedback that affects the amount of power getting through the atmosphere. There’s plenty there to indicate in a fairly robust fashion just how little effect occurs and there’s plenty there to suggest where the errors and fallacies of the co2 warmers exist.

Doug McGee
August 8, 2010 10:36 am

Wow, it seems only yesterday this site was promoting G&T’s alleged refutation of CO2 as a GHG. I guess this is how blog science works – baby steps towards what most everyone already knows.

Roger Clague
August 8, 2010 11:09 am

Doug McGee says
‘Wow, it seems only yesterday this site was promoting G&T’s alleged refutation of CO2 as a GHG. ‘
This site does not promote points of view. It allows discussion. There are plenty of contributers, including me, who agree with G and T. There are those who disagree with them.
No consensus here, thankfully, only civilised debate. Join the fun.

Foley Hund
August 8, 2010 11:16 am

I see it like this: Diluting the atmosphere H2O with a minimal green house gas as is CO2, then the concentration of retained heat should therefore decline with the declinge H2O concentration.

wayne
August 8, 2010 11:50 am

Jeff Id says:
August 7, 2010 at 5:33 pm
Mike, …
Maybe an even simpler explanation.
The laser passes through clear stuff — it gets absorbed by black stuff. The CO2 is black stuff. If we consider all the other minor aspects of what happens, the black stuff is still black. The laser cannot pass — so it makes heat.
_______
Jeff, oh I love your example to Mike. That’s a great classic view. So answer this if you can:
How much more black do you have to add to an already totally black gas system, the atmosphere, to have it absorb more of the 15 µm radiation from your laser than it already did? It seems the black gases at the exiting glass cannot get any warmer than it takes for the black (co2) to emit 1000 W at that temperature per fourth power law, SB with emissivity. Look deep right there.
You see, additional black does not decrease the amount of energy, 1000 W, flowing through the system after it has reached total blackness and a equalization temperature. Now, if some original laser radiation is still leaving through the exit glass, meaning true blackness has not yet been achieved, then more black it seems would warm the N2, but once total blackness (path length short enough to cause multiple interactions) has been achieved, more black does nothing, absolutely nothing, or so it seems. The temperature at the exit glass of the black content will always be the exact temperature it takes to eject 1000 W of energy, no more, no less. I think you said that above.
Now, if your entrance glass is identical to the exit glass, would 500 W be radiated out of both glasses? I can see where more would exit from the entrance glass than the exit glass being closer to the energy source. However, that doesn’t mimic the world very well. Better would be to change the laser for a special and fictional selective hot plate that emits at only ~15 µm IR that exactly covers the entrance to the container, i.e. the earths surface. Now, will the hot plate get hotter as you add more black to an already totally black air at the ~15 µm radiation inside the container?
I have to stop and think a bit on that one myself, my first thought is no, the hot plate would get warmer and warmer only until total blackness was initially achieved, from that point on, same temperature throughout the container for you cannot trap heat at the source end in the gases.
I might see that if there is such a property as “radiative conductivity”, i.e. the speed of radiation energy transfer, though I’ve never come across it, and that property was less than one, then would more black to an already black gas increase the temperature near the source end making the source end hotter than exit end? Seems non-physics to me. Once again you would be trapping heat at one end.
That’s like asking, “Can radiation energy transfer through a gas be slowed down in a macro sense when path length is short compared to the total distance traveled (don’t speak of individual photon jumps (q.m.), be symmetric)?”. The minute you turn the plate on it would be for a short period but would quickly equalize throughout, this is radiation heating, not thermal conduction, or that is my initial thought.
This might be what Tom Vonk and Mike Blackadder were trying to convey but looked at it from your “blackness” example.
Interesting. Your thoughts? My words are somewhat rough so tweak if necessary.
_____
Mr. Vonk – I learned much from your article. Keep it up.

LightRain
August 8, 2010 12:00 pm

Why not take a real laser, lock it on a satellite that can measure the incoming 15 uM during the day, night, through clouds and clear conditions and see how much is absorbed. Double the output power and test again. What are the results?

Kevin Kilty
August 8, 2010 12:01 pm

Jordan says:
August 8, 2010 at 3:34 am
…But I was thinking more in terms of what SB would demand for incoming solar radiation at circa 1500 W/sqm. Even in the closed chambers mentioned and where the image of the sun fills the apeture (that’s a good point which I failed to mention), I did not expect this to cause a rise to the same temperatire of the surface of the sun becuase of the 1/r^2 “dilution” of solar flux. Doesn’t SB suggest something more like 400K? …

If the outgoing radiation from one of the tubes is allowed to radiate into a hemisphere, and has unit emissivity, then the equilibrium temperature is 394K (@1366W/m2). On the other hand, if the aperture can only see the sun, which is what I assumed, then the temperature is far higher, but could never exceed the surface temperature of the Sun in any case, as then we would have another machine of second kind. At any rate, this is a example that small differences in assumed conditions lead to very different results. Radiation problems in engineering are difficult enough without adding gasses into the problem. You will note that there is a lively discussion on this thread about the effects of gasses.
Gasses have a finite number of internal states that allow them to absorb radiation only at specific frequencies. These frequencies are widened into “bands” through the Doppler effect (velocity of molecules), collisions, and bound associations of two or more molecules (dimers), but until temperature/pressure is extremely high, gasses do not really approach a thermal emission spectrum. Rather they have some band gaps and the occasional lonely line or group. This is a large contributor to the complexity of radiation calculations in gasses — there is no simple rule like the Planck function for cavities to specify how molecules radiate. People use simplifying assumptions like LTE to justify the Planck function, but radiation in real gasses is not as simple as cavity radiation.
By the way, SB only applies if the entire spectrum is present in the emitted power. In the case of a window through which radiation must pass, such as that near 10 um in the Earth’s atmosphere, then radiation passing away from the surface follows more nearly a temperature to the 4.5 power. Darn those gasses!

Doug McGee
August 8, 2010 12:31 pm

Roger Clague :

This site does not promote points of view. It allows discussion. There are plenty of contributors, including me, who agree with G and T.

Sorry to hear that. It makes attempting debate pointless. Even a debate must begin with a concession of what is known and not known, but you want to argue over if the podium is a podium or not. Sad.

Jordan
August 8, 2010 12:51 pm

Thanks again for your replies Kevin Kilty.
Sad to admit how little I know about optics, but your comments were very helpful.
I kinda got your point about the temperature rise in the cavity when the sun fills the aperture. However I found such an unusual point hard to believe, and I pressed the point about SB. Your explanation of SB holding for emission to a hemisphere sorted it out.
If I could conclude: the machine doesn’t work, because we cannot apply SB to the emitting surface at the basie of either tube. My mistake was assume we could. We expect both tubes will rise to a much higher temperature than implied by SB and the existence of CO2 in one of the tubes has no material effect.
Therefore there is no potential difference in the tubes and the machine doesn’t work. Also confim the rule that a practical machine cannot sink energy into its own heat source.

Spector
August 8, 2010 1:57 pm

cba says: (August 8, 2010 at 10:12 am) “Something like the HITRAN database can provide raw information but it’s a tremendous effort to get something fairly complete out of that. One can also get fair mileage out of Archer’s online modtran calculator. Some lines will have extinction path lengths measured in the cm distances. A short distance away in the wavelength, one can have path lengths of 1km or more.
It might be interesting to also calculate the incoming extinction altitude for each line. This would be the altitude at which the remaining atmosphere to the edge of outer-space contains the same total mass per unit area of the absorbing gas as is contained over the surface level extinction distance.
I would be curious to see how many of these altitudes were above or below the equatorial tropopause level. As a first cut, one could ignore possibility that the gas concentrations might be altitude dependent.

wayne
August 8, 2010 3:18 pm

LightRain says:
August 8, 2010 at 12:00 pm
Why not take a real laser, lock it on a satellite that can measure the incoming 15 uM during the day, night, through clouds and clear conditions and see how much is absorbed. Double the output power and test again. What are the results?
___
Since out atmosphere is totally black to ~15 µm IR radiation, due to CO2 and H2O, it’s not clear and transparent. The laser beam would be absorbed very high. You would just see the bright smear where it hit near the top of the atmosphere from the satellite and on the ground you would never see it at all. Same if doubled but brighter bigger smear as seen from the satellite. Much as a visible laser in dense fog, it wouldn’t travel far before scattered and absorbed no matter how bright and powerful.
Or that is the way I see it.
It’s black because the path length, you can view that as the distance any photon of that frequency would travel before encountering a molecule that can interact with it, is very short, I have read 10 meters for CO2 at these concentrations today. (anyone, if I off on that description of path length please let me know, still seeking correct information on that for CO2 and H2O)

Mike Blackadder
August 8, 2010 3:22 pm

Jeff,
I know this thread is petering out, but do you have any reply to my last comment?
Is there a flaw in your original post or did I make some error?

Jim D
August 8, 2010 3:26 pm

Regarding the first linked graphic posted by Dave Springer:
August 7, 2010 at 12:39 pm
Looking down from 20 km you see the 15 micron band at about 225K. This would almost entirely be due to CO2 at that temperature as the H2O content at those temperatures is much less than CO2 (maybe about 100 versus 390 ppm). This is why CO2 is highly relevant to the budget of outgoing longwave radiation, which is part of earth’s top-of-atmosphere energy budget. This point seems to have been missed when people talk about H2O dominating at the ground.

Brian Macker
August 8, 2010 4:08 pm

” The math and physics were fine, but his conclusion that insulating an energy flow doesn’t cause warming is non-physical and absolutely incorrect.”
You are correct and he is wrong.

Jim D
August 8, 2010 4:16 pm

I will attempt my answer to Mike Blackadder.
The laser intensity at 15 microns is equivalent to the emission of a black body at some (probably high) temperature. This is the temperature towards which the CO2 will go. Yes, for a low enough intensity (cold) laser, the CO2 could cool instead, because it would initially emit more than it receives.
The atmospheric-layer analogy is a hot laser at one end (bottom) and a cold laser at the other (top). In the real atmosphere, cooling wins out in almost all tropospheric layers, except perhaps a very thin layer near the ground.
This may seem surprising after all this discussion about GHG warming, but that is the way it is, and no, it does not contradict anything in AGW to say that the net radiative effect in the IR is atmospheric cooling (the shortwave warming is weaker). This net clear-sky cooling is compensated by convection in the troposphere that has a net warming effect. Without net radiative cooling, convection could not exist.

Gnomish
August 8, 2010 4:22 pm

“Without net radiative cooling, convection could not exist.”
Technical quibble:
Change of density is sufficient for convection.

Gnomish
August 8, 2010 4:25 pm

A more accurate statement might be:
Heating from below causes convection.
Radiation is not required at all, even though that’s the only way heat gets to the bottom of the system from outside of it.

Gnomish
August 8, 2010 4:28 pm

The 3 forms of heat flow in a gas are conduction, convection and radiation.
They are distinct processes.
The abstract isolation of radiative processes is worth a study, but it does not answer as a substitute for the other ones.

Gnomish
August 8, 2010 4:30 pm

So, the elephant is really not so much like a laser. You’ve only felt the tip of the trunk.

Jim D
August 8, 2010 4:33 pm

Re: Gnomish.
I should have said convection could not exist in the mean state that would result without radiation. The atmospheric profile would become stable and stay that way. The troposphere is often described as being in a convective-radiative equilibrium.

Al Tekhasski
August 8, 2010 4:55 pm

Jim D says: “Looking down from 20 km you see the 15 micron band at about 225K. This would almost entirely be due to CO2 … This is why CO2 is highly relevant to the budget of outgoing longwave radiation ”
Here is an interesting point that most are missing. Where this irradiance comes from? From 5-6km “average area” that has temperature this 225K as AGW theorists assume, or from stratosphere? As Dave Springer noted, even at 20km the opacity of air at many lines in 15um band is huge (optical thickness is few meters), which means that even at 20km all (14-16um) radiation still gets re-absorbed, re-emitted, and still diffused across the “black” media. In other words, 20km is not yet the TOA for this band. When CO2 is increased, this boundary would go even higher, where “higher is warmer”, and therefore the 15um band works in opposite direction to commonly assumed warming.

Jim D
August 8, 2010 5:26 pm

Re: Al Tekhasski: August 8, 2010 at 4:55 pm
Let’s say that that the downward-looking emission height for 15 microns corresponds to 10 km. Yes, some parts of the earth have almost no lapse rate at 10 km (e.g. higher latitudes where this will be in the stratosphere), others like the tropics and subtropics have this elevation in the troposphere with a negative lapse rate. Adding CO2 raises the effective emission height, and so would have different effects in these two situations, but the global average will be dominated by the lower latitudes that cover a larger area.

Al Tekhasski
August 8, 2010 5:45 pm

No. You are downplaying the effect. The TOA for 14-16um band seems to be way above 20km, since optical thickness is just 2-3m even as 100mb as Dave Springer said.
Consider the following example: Let the OLR to have only two bands.
Let a narrow band (say, 14-16um) have strong absorption, and another, wider area (say, 3x of that, the “transparency window”) with very weak absorption. Their average “emission height” is, say, at 6km. According to the standard theory of averages, the ”average emission height” will go up with increase of CO2, where “higher= colder”. Colder layer emits less, and therefore the global imbalance of OLR would occur forcing climate to warm up.
However under a more careful view, this average of 6km is made of 0km emission height for 3/4 of IR range, and 24 km emission height for the remaining 1/4. Increase of 0 km band gives you zero change in OLR, while increase in 24km emission height will give you MORE OLR, because the temperature gradient in stratosphere is opposite to one in troposphere, so “higher = warmer”. As result, the warmer layer would emit more, and the energy imbalance would be POSITIVE implying COOLING, or just exactly opposite to what the standard “averaging” theory says.
In reality the spectrum is more complex, many different trends coexist, but the above example suggests that it seems very likely that warming and cooling effects cancel each other in the first approximation. Therefore, the effect of sensitivity to CO2 increase is a second-order effect, and must be much harder to calculate accurately.

Gnomish
August 8, 2010 5:57 pm

Jim D says:
August 8, 2010 at 4:33 pm
Re: Gnomish.
I should have said convection could not exist in the mean state that would result without radiation. The atmospheric profile would become stable and stay that way. The troposphere is often described as being in a convective-radiative equilibrium.-
———————————-
I understand. It just one aspect, though. In fact, convection can occur with no change of temperature at all.
Understanding is considerably enhanced by highlighting that seemingly obscured point.
Delivering heat to the mass of the planet via CO2 radiation is a lot like trying to boil lake superior by shining a heat lamp on the mississippi. You might warm the gulf a bit, though.

Al Tekhasski
August 8, 2010 6:04 pm

sorry for screwed terminology, I meant to say that optical thickness 1 correspond to 2-3m, (or beam is attenuated by e=2.7 x). BTW, if anyone knows how to formally define “effective emission height” at a given frequency, please speak up.

old construction worker
August 8, 2010 6:13 pm

One for the books:
While discussing CO2 drives the climate with a friend of mine, her mother said ” We didn’t have this problem until they started to poke holes in the atmosphere with rockets.” All we could do was chuckle.

Jim D
August 8, 2010 6:25 pm

Re: Al Tekhasski: August 8, 2010 at 5:45 pm
This seems to suggest it is an oversimplification to assign individual heights to the center of the 15 micron band and the other bands and windows. The wavelengths that change emission height most are likely to be the sides of the bands, which sample a full atmospheric profile, and the stratospheric increase of temperature with height is quite weak in terms of K/km, so maybe that is how the emission decreases as CO2 increases, which has been established taking all these line-by-line factors into account.

Spector
August 8, 2010 6:33 pm

RE: Jim D: (August 8, 2010 at 3:26 pm) “Looking down from 20 km you see the 15 micron band at about 225K. This would almost entirely be due to CO2 at that temperature as the H2O content at those temperatures is much less than CO2 (maybe about 100 versus 390 ppm). when people talk about H2O dominating at the ground.”
Do you have a source for this value? I believe I found a statement in one of the documents at the EPA NEPIS website that said that water vapor content of the atmosphere was 3 to 4 percent. That would be about 70 to 100 times that current concentration of CO2. If this statement is only at the ground, then I agree; it is beside the point. My primary curiosity at this point is obtaining a better understanding of the tropopause, which seems to the primary heat exhaust point of our thermal convection system.
Bi-Molecular Water
I note that there has been little discussion of the concentration of or effect of bimolecular water in the atmosphere at any level. BMW should have a completely different array of vibration and atmospheric absorption/radiation bands than mono-molecular water does and I believe the continual clumping and unclumping of these two species should provide a well-known and powerful radiation signature.

Kevin Kilty
August 8, 2010 6:34 pm

I guess this thread is petering out, so you and I may be the only two left here soon. I hadn’t time to read all that much, but there have been some very insightful postings. It’s interesting how many different ways there are to look at this issue.

Jordan says:
August 8, 2010 at 12:51 pm
Thanks again for your replies Kevin Kilty.
Sad to admit how little I know about optics, but your comments were very helpful.
I kinda got your point about the temperature rise in the cavity when the sun fills the aperture. However I found such an unusual point hard to believe, and I pressed the point about SB. Your explanation of SB holding for emission to a hemisphere sorted it out.
If I could conclude: the machine doesn’t work, because we cannot apply SB to the emitting surface at the basie of either tube. My mistake was assume we could. We expect both tubes will rise to a much higher temperature than implied by SB and the existence of CO2 in one of the tubes has no material effect.

Yes, essentially the point.

Therefore there is no potential difference in the tubes and the machine doesn’t work. Also confim the rule that a practical machine cannot sink energy into its own heat source.

Indeed you are correct. In a heat engine the source of heat cannot also be the sink. Failure to recognize this has lead to the pursuit of many second law machines. A recent example liberated money from thousands of “investors” and sent a fellow to prison in Washington State for securities fraud. Literally people bought small shares in this venture to the tune of millions of dollars in total!

Spector
August 8, 2010 7:04 pm

RE: Gnomish says: (August 8, 2010 at 4:25 pm) A more accurate statement might be:Heating from below causes convection. Radiation is not required at all, even though that’s the only way heat gets to the bottom of the system from outside of it.
That is true as it stands, however to have convective cooling, the heat transported upward must leave the planet, radiation being the only ticket out, or you have the equivalent of a stove with a plugged chimney.

Mike Blackadder
August 8, 2010 7:25 pm

Jim D,
Thanks for your reply.
“The laser intensity at 15 microns is equivalent to the emission of a black body at some (probably high) temperature. This is the temperature towards which the CO2 will go.”
My conclusion is actually the same as yours. Except I’m pointing out that CO2 concentration doesn’t matter. CO2 concentration doesn’t change the equilibrium temperature (at least not in the case of perfectly insulated chamber surrounded by a vaccum). I argued this point in my last post. I haven’t seen anyone try to refute that argument.
If I can add CO2 or take it away and it doesn’t change the temperature then isn’t Tom right to say that CO2 has no net effect warming the gas?

Jim D
August 8, 2010 7:33 pm

Re: Spector: August 8, 2010 at 6:33 pm
At 225 K and 200 mb pressure, I figured (and may be wrong) the ice saturation value is less than 200 ppm. Taking an average as 50% RH relative to ice gives less than 100 ppm, so that is where it comes from. The 3 to 4 per cent is an atmospheric average, but it varies strongly with temperature being restricted by the saturation upper limit that decreases with temperature.

Jim D
August 8, 2010 7:46 pm

Mike Blackadder says:
August 8, 2010 at 7:25 pm
“If I can add CO2 or take it away and it doesn’t change the temperature then isn’t Tom right to say that CO2 has no net effect warming the gas?”
Interesting point. It doesn’t change the equilibrium temperature of the CO2 when you go from 1 to 2 per cent, as that is still determined by the laser intensity. What it does change is the emission intensity at 15 microns of that CO2, which depends on the number of molecules at that temperature.

Jim D
August 8, 2010 7:49 pm

correction to my 7:33 pm
3 to 4 per cent is the extreme value, found near the ground in the tropics. The atmospheric average is about 0.4 per cent.

Gnomish
August 8, 2010 7:58 pm

Aye, true. Radiation is the dominant heat transfer process at the radiator end. It’s also the delivery system for the boiler.
See calculations for the vast energy lofted to the radiator by convection with no temperature change and no BB radiation spectrum shift that is known as phase change.
If the water is in the air as gas, no temperature change was required. The density of H2O gas is the lowest of any major constituent of our atmosphere. It goes uP.
It’s also really important to note that degrees can not be converted to watts.
Air on the surface is not static or else the CO2 would just puddle there by density.
It is a continuous flow – away.

Gnomish
August 8, 2010 8:01 pm

CO2 is a diffuser and disperser = conductor of heat just as much as every other gas.
I sure don’t know how it earns any sort of prize for insulation.

Gnomish
August 8, 2010 8:05 pm

If you put black carbon particles in your coolant fluid, your engine will run cooler, guaranteed.

Al Tekhasski
August 8, 2010 8:06 pm

Jim D says: “maybe that is how the emission decreases as CO2 increases, which has been established taking all these line-by-line factors into account.”
Do you mean it was established by using entire two atmospheric profiles, tropical and extratropicals, out of dozens of climatic zones, and not counting any polar areas? Like in Myhre at al.?

Mike Blackadder
August 8, 2010 8:37 pm

Jim D,
“It doesn’t change the equilibrium temperature of the CO2 when you go from 1 to 2 per cent, as that is still determined by the laser intensity. What it does change is the emission intensity at 15 microns of that CO2, which depends on the number of molecules at that temperature.”
Exactly. Thanks again for the reply. That was the point that I thought I was having an argument about and you’ve helped to clear that up.
So essentially the effect of higher CO2 concentration is that it allows the gas temperature to adjust faster in response to changes in radiation forcing, not to change the equilibrium temperature for a constant radiation forcing.

Jim D
August 8, 2010 8:42 pm

GCMs routinely integrate over global areas, and this is one of the things that is easy to test by just doubling the CO2, keeping everything else the same, and seeing its instantaneous effect on global outgoing longwave. In fact, this can be tested without even running the GCM forward in time, just specifying a global analysis in it, then compare with doubling the CO2. The radiation schemes and global weather analyses are certainly good enough to test this, which is why expect it has been done many times already.

Al Tekhasski
August 8, 2010 9:22 pm

Jim_D, my example establishes that theoretically speaking the effect of increasing CO2 across multilayer- structured atmosphere has a _negative_ effect on warming. In particular, it is painfully clear that the effect is definitely negative if the atmosphere has an inverted sounding profile, which is known to frequently happen in polar areas. Are you saying that the estimation of “radiative forcing” from CO2 doubling is most accurate if you exclude polar areas? You do realize that entire AGW scare runs on the basis of this estimation, around that notorious 3.7W/m2, which RC called once as “best accurate estimate” of CO2-based radiative forcing? Are you being serious?

Jim D
August 8, 2010 10:07 pm

Re: Al Tekhasski: August 8, 2010 at 9:22 pm
I am saying that I am sure this is part of the global average effect that is normally given. Polar areas are actually quite small, and latitudes should be weighted as cosine latitude to give the correct area weight. Less than 10% of the global area is in the Arctic and Antarctic circles, 40% is tropical and 50% is the rest.

Spector
August 9, 2010 1:33 am

RE Jim D: (August 8, 2010 at 7:33 pm) “At 225 K and 200 mb pressure, I figured (and may be wrong) the ice saturation value is less than 200 ppm. Taking an average as 50% RH relative to ice gives less than 100 ppm, so that is where it comes from. The 3 to 4 per cent is an atmospheric average, but it varies strongly with temperature being restricted by the saturation upper limit that decreases with temperature.”
“correction … 3 to 4 per cent is the extreme value, found near the ground in the tropics. The atmospheric average is about 0.4 per cent.”

A correction on my part, I find the original source of my high H2O concentration levels was from a seemingly general statement from: [denatured link]
http : // www . netl . doe . gov /KeyIssues/climate_change3 . html
“Water vapor is present in the atmosphere in concentrations of 3-4% whereas carbon dioxide is at 387 ppm or 0.0386%.”
If this were true at the tropopause level it would suggest that for every molecule of CO2 in a given volume of upper air that might radiate to outer space, there should be a lot more H2O molecules ready to do the same thing. I now do suspect that writer of this DOE document may have made an unwitting error without realizing that this statement, if true in the upper atmosphere, had the potential to pull rug out from under the CO2 Global Warming Theory.

August 9, 2010 5:04 am

Mike,
I want to answer because you’re putting a lot of work into this but it’s so long worded that I’m having trouble following, I’ll try to answer in bold.
Mike Blackadder says:
August 7, 2010 at 8:04 pm
Jeff,
I think I see the problem.
So we have a perfectly insulated chamber with windows at the ends to let radiation in and out, and I will add that it is surrounded by a vaccuum.
A 15um laser is firing through the gas in the chamber and we’ll consider how this effects the temperature of the gas depending on mix of N2 and CO2. Note: We’ll get to the case of no CO2 at the end.
Scenario 1) 1% CO2: Turn the laser on. With 1% CO2 we’ll assume that the gas does NOT appear completely black to the laser. So some laser light passes right through the chamber. The CO2 absorbs X photons/sec. Before being able to re-emit it gets in millions of collisions and will tend to transfer their vibrational energy to the cooler N2 gas (and also the collisions will result in translational energy of the CO2 molecules as well). So the air heats up. So the air is absorbing X photon/sec but is emitting less than X photon/sec. Eventually the gas will warm up to an equilibrium temperature where it is emitting X photon/sec and absorbing X photons/sec.
Here’s the important part. This only happens at the point when the following collision events are equally probable: CO2* + N2 -> CO2 + N2′, and CO2 + N2′ -> CO2* + N2. If this is not the case then a net amount of absorbed radiation is being translated into heating of the case or vice-versa.
Scenario 1, not all laser energy absorbed, eventually we reach equilibrium
Scenario 2) 2% CO2. Turn the laser on. Still some laser light passes through the gas without being absorbed but now a greater amount is absorbed: Y photons/sec. The same thing happens when the gas is cool, where there are more CO2* + N2 -> CO2 + N2′ than the CO2 + N2′ -> CO2* + N2, and so the CO2 doesn’t get a chance to release photons before transfering it to heating of the gas. We know that it will warm up and reach equilibrium when emitting Y photons/sec.
Scenario 2, more but not all laser energy absorbed, eventually we reach equilibrium
The problem is you assume this is a higher gas temperature than scenario 1, but it isn’t. In fact it is only when collision event CO2*+N2 -> CO2 + N2′ is equally probable to CO2 + N2′ -> CO2* + N2, which is the same temperature. The CO2 and N2 molecules involved in this collision have no knowledge of gas concentrations surrounding them. There is no reason why N2 suddenly has to be a higher temperature in order to energize CO2 as often as it deenergizes CO2.
I think I’m following your logic now. Maybe, but it’s still confusing especially in the lst sentence. If more of the laser light passes straight through the gas, I assert it will be cooler. If less passes through, it will be warmer. Otherwise, where would the energy go?
Finally Scenario 3) 0% CO2. Turn on laser. Hopefully this helps clear things up. No laser light is absorbed by the gas, and all laser light passes right through. The gas doesn’t heat up.
Question: What’s the gas temperature? Before you say 0 K, remember the gas is in a perfect insulator surrounded by a vaccuum.
initial temperature is room temp – ambient as written on the cans
The truth is the gas temperature could be anything. This illustrates how you’re only considering one side of the problem. In Scenario 1 & 2 we know the gas is cool when we start the laser because the gas contained CO2 which was emitting heat out of the chamber (net CO2 + N2′ -> CO2* + N2) transactions. Without the CO2 this can’t happen.
the gas temp was the same as the room so no net energy loss (cooling)over time, At ambient temp the second law of thermo states that the net energy transfer through the perfect window is zero. No cooling.
==========================================================
I’m sorry if I didn’t keep up with all the comments here, feel free to leave a comment at the Air Vent if I’ve missed something important.
Thanks to Anthony for carrying this post, I hope people found it helpful.

TomVonk
August 9, 2010 5:39 am

Jeff Id
I have no issues with this post and agree with almost everything .
The only (minor) issue is when you state : Where he goes wrong is equating that assumption to AGW.
I do not think that I used the term AGW once and I neither analysed radiative transfer nor considered transient regimes . The purpose of my post was very exactly and exclusively to give arguments for the statement written at the beginning .
Like you rightly say I considered a volume in LTE what , per definition , excludes transients at microscopical level .
If you wish , in the frame of your laser experiment , my post would cover what happens inside the cylinder for all t > t0 where t0 is the time at which the cylinder emits by the transparent side 1000 W .
It is for those t that the “CO2 does not heat the N2” even if it continues to absorb the whole 1000 W .
My purpose was to say that in those conditions there can’t be a net energy transfer from the vibrational modes to the translational modes because else the cylinder would continue to heat and emit less than it absorbs .
A minor technical point is that like Merrick says , temperature is indeed not merely an average of the kinetic energy but an average of energy what includes non kinetic energy too .

cba
August 9, 2010 5:50 am

Spector,
A single line has something like a gaussian distribution. Variations used include Lorentz distribution and Voigt distribution. That boils down to the situation that where on this curve one is looking at to determine the path length. What’s more, as one goes higher, the line width is going to get narrower, reducing further the path length at that wavelength and that also increases the attenuation closer in to the line center due to reduced pressure. All this is compounded by the temperature which affects the likelihood that there will be an emitted photon going outward to replace the one captured. when the temperature of the gas reaches that of the surface emitting the continuum, there will be enough molecules in the gas sample that will radiate outward as there is a likelihood that the incoming photon is absorbed. At that point, one no longer sees absorption lines occurring. In fact, if the gas temperature increases beyond that of the original BB radiating surface, one would see emission lines of the gas.

jae
August 9, 2010 7:10 am

So, the OCO aids in thermalizing the atmosphere, which stores heat for a while, but does not “trap” heat.

R. de Haan
August 9, 2010 7:39 am

Mean Free Path of Photons through the Troposphere and Time of Crossing Path of Photons by Dr. Nasif S. Nahle, Scientific Research Director at Biology Cabinet
“The carbon dioxide (CO2) is a very important molecule for life on Earth. Carbon dioxide is taken in from the atmosphere by photosynthetic organisms, which use the molecules of CO2 on building more complex substances where the energy transferred from the photons to the molecules of chlorophyll is stored.
The CO2 is diluted in the atmosphere in a concentration of 0.038%, and it has been proposed by the IPCC as a main driver of the climate on Earth. Contrary to what the IPCC proposes, the physics of the thermal energy transfer indicates smashingly that the CO2 is not capable of changing the temperature of the atmosphere in a significant way”.
http://climaterealists.com/?id=6111

Gnomish
August 9, 2010 8:56 am

http://climaterealists.com/?id=6111
good find!!
And so, in this one thread there is every bit of info needed to put CO2 in its place.
It’s not much sizzle and certainly no steak.

Al Tekhasski
August 9, 2010 9:49 am

“Mean Free Path of Photons through the Troposphere and Time of Crossing Path of Photons by Dr. Nasif S. Nahle, Scientific Research Director at Biology Cabinet”
How could anyone take seriously and refer to such an incoherent drivel as this? The guy obviously is not familiar with distinction between scattering and selective absorption, and with anything else. I am so sorry for his students.

Gnomish
August 9, 2010 12:25 pm

Drench them in your pity, Al.
Emo is so progressive.

Spector
August 9, 2010 1:45 pm

RE cba: (August 9, 2010 at 5:50 am) “Spector, A single line has something like a gaussian distribution. Variations used include Lorentz distribution and Voigt distribution. That boils down to the situation that where on this curve one is looking at to determine the path length.”
Perhaps going line-by-line might be getting a little too detailed – I understand that each line observed may represent a convolution of structural theory and real-world processes.
What I am hoping to see is the development of a theory that goes beyond Planck’s law of black-body radiation to cover radiation from molecular trace gases embedded in an otherwise largely transparent gas matrix and where the radiating zone may have an effective average depth and radiating population density. I assume one could develop a function indicating the expected radiation rate and absorption cross-section for each vibration mode of any given polar molecule or polar molecule-pair (dimer) as a function of temperature and how much of that radiation escapes to outer-space.

Eli Rabett
August 9, 2010 3:26 pm

Sorry Spector, the spectra are sharp enough and their positions irregular enough that you have to do a line by line calculation (to say nothing about the collisional continua. The weeds are very high on this one.

George E. Smith
August 9, 2010 3:49 pm

I have noted from time to time that posters here cite “Beer’s Law” or the “Beer-Lambert Law” which are two names for the same thing.
This is the E = E0. e^-alpha.x form; that assumes a linear relationship between input energy/power/intensity/whatever; and the corresponding output after some thickness (x) of absorbing medium; and embodying the concept that any subsequent equal absorber thickness will simply treat its reduced input in the same way.
In the geometrical Optics field; it is also fashionable to point out that this is “Internal Absorption” only, and that at the (presumed) two ends of the absorptive medium; where a change of medium occurs; there will inevitably be a Fresnel reflection/rfraction split involving Polarisation and incidence angles ets. Well in an atmospheric absorption case; we tend to dispense with the Fresnel losses at the end; and concentrate on the internal absorption.
There is one very fundamental aspect of Beer’s Law that seems to be constantly overlooked and ignored. In its derivation, it is tacitly assumed that the absorbed optical energy simply disappears from the known universe; never to be seen again.
It will sometimes be noted that the absorption coefficient (alpha) is likely to be a function of wavelength; so depending on the input spectrum and the absorbing medium, the output energy spectrum is likely to differ from that of the input; and sometimes quite markedly; and deliberately;when used in spectral filtering applications. But one thing that the user of Beer’s Law can be absolutely sure of, is that the Output Spectrum WILL NOT conatain ANY energy having spectral components that were not present in the input source. It is a lossy system; and nothing is added that was not previously present.
Well you see there is the rub; the atmosphere simply DOES NOT obey Beer’s Law. And a great many kinds of Optical Media DO NOT obey Beer’s Law. Well they don’t when the output and input are simply energy/power/intensity/whatever and no limitations are placed on the spectral correspondence of input and output.
I have at my desk; some very common Optical materials, and if I place them in the path of a HeCd blue Laser beam at 441.6 nm wavelength; and pass the ouput through a grating monochromator tuned to 441.6; I can easily get an extinction of four to five orders of magnitude from a thin 3 mm sample.
If I remove the monochromator, and simply measure the output power/energy/intensity/whatever with a broadband detector I get all kinds of power output; so far I have never got out more than I put in but I get an amazing amount out; easily enough to see; and it no longer is blue. Could be yellow/orange.red/whatever.
Those sharp cut optical filter glasses, and lots of other media; including the atmosphere are “fluorescent” or some other “escent” that converts some fraction of the input to an output at a totally different wavelength or output spectrum.
So it is advisable to forget about Beer’s Law when talking about atmospehric absorption.; well it may still be true that the actual absorption of the input photons; if you can assign a Maxwell’s Demon to account for all of them; but it will get a bit tricky sorting them out if the output spectrum actually overlays the input spectrum; even though they are different.
Beer’s law did no contemplate that the absorbed energy would ever see the light of day again (pun intended).
And in the case of the atmosphere where specific absorptions are a product of various GHG molecular species; while output emissions are the product of a thermal continuum spectrum that is due simply to the Temperature of the atmosphere; we will actually have a cascade of successive absorption and emission events that are far more complex than Beer’s law to analyse.
So far as I know (and you know what that means) the atmosphere does radiate a blackbody like (thermal) continuum spectrum of emissions based on its Temperature; and likely following a roughly 4th power with Temperature Law; but not necessarily with a full BB Planck spectrum; but the absorption of radiation in the same general spectral range (LWIR) is more of a molecular line/band spectrum; characteristic of the absorbing molecular species.
So I don’t think it is correct to say that the atmosphere; or gases in general “absorb” blackbody radiation; they do pick out pieces of it with molecular selectivity; while the total atmosphere should radiate based on just the Temperature. Classical Physics places the source of this emission as being the acceleration of electric charge due to the atomic or molecular motions.
My Quantum Mechanics stopped short of learning what the non-classical explanation for BB radiation is. My recollection is that Jeans and others were messing around with the equipartition principle; and the number of degrees of freedom; which is some whacking great number involving Avogardo’s number or maybe Factorial Avogadro’s number; and ultimately Planck assigned 0.5kT to each mode.
Evidently Jeans must have learned from Planck; because I seem to recall that Jeans did a similar computation to establish the low Temperature Specific Heats of solids using a similar argument (no I wasn’t there at the time).
Hey there has to be something extra you have to learn to get a PhD; so I didn’t learn that; OK !

August 9, 2010 4:24 pm

George E. Smith
Doesn’t absorbed light in an optical medium simply turn to heat and then become radiated as a Planck blackbody ish curve?

kwik
August 9, 2010 4:33 pm

After reading through it all…. it all boils down to the most difficult decision;
-Which physical law can be used where?
I think I go with Claes Johnson;
http://www.nada.kth.se/~cgjoh/simpleclimate.pdf

cba
August 9, 2010 5:11 pm

Spector,
The approach required is more than just the temperature or a single molecule. Ultimately, you’ve got to create a spectrum from all of these lines and conditions because that’s what you’ve got. For fidelity, you’ll need to do the analysis by layer as well.
Note that there is a response to me from Tom Vonk on the other thread that includes a reference to a 200 page paper concerning the characteristics of IR in the atmosphere. It has all sorts of details that would prove helpful if this is the sort of thing you want to do.

cba
August 9, 2010 5:18 pm

Jeff Id says:
August 9, 2010 at 4:24 pm
George E. Smith
Doesn’t absorbed light in an optical medium simply turn to heat and then become radiated as a Planck blackbody ish curve?
——-
Of course. the reason why the stuff works well is that it’s normally associated with visible light and room temperature experiments. If you consider only the absorption or separate the absorption concept from the emission concept, it is of value in describing the absorption when emission also exists. It becomes radiative transfer when both are combined.

Spector
August 9, 2010 5:23 pm

One point I would like to make about figure 2 in this article is that it is a standard representation of earthshine modulating (greenhouse) gases in the atmosphere, [apparently derived originally from articles by J. N. Howard, 1959: Proc. I.R. E 47, 1459; and R. M. Goody and G. D. Robinson, 1951: Quart. J. Roy. Meteorol, Soc. 77 153] and its primary purpose is to show how black-body radiation from the Earth’s surface (earthshine) is influenced by these gases, both individually and in total. Even though it may be hiding the fact that these bands are actually composed of many fine hair-lines, it is a good illustration of the process of surface radiation blocking by greenhouse gases.
What it does not do, (and was not intended to do) in my opinion, is give a picture of the upper-air radiation emission process that should be crucial for the proper understanding of the operation of the convective thermal control system in our atmosphere.
Even though one can see a wide gap in H2O absorption spectrum over the earthshine thermal radiation band, I speculate that H2O plays a random game of dollars and cents over this range, continually emitting more 20 micron photons as the temperature increases and occasionally emitting 6 micron photons to make up the difference. I believe the same might be said of CO2 from 4 to 15 microns. While this is probably quite a simplified description of what really happens and perhaps subject to error, I hope it is useful.

Al Tekhasski
August 9, 2010 5:35 pm

Jeff,
If you have only CO2 and other gases are optically inert, your 15um beam will be absorbed by CO2, and re-distributed as heat to all other molecules (called “buffer gas”). However, since these other molecules have nothing to emit on their own, the emission will occur only on CO2 lines, including 2.5 and 4um (and all other) lines, but from the entire gas chamber volume. Whole thing must heat up to a temperature quite higher than the equivalent blackbody to keep the overall balance. In the case of small chamber, the walls will be involved, and only then you will have a wider and continuous spectrum.
It would be interesting however to see how the 1kW 15um beam will look in open air. I think it will look as a plasma blowtorch.

August 9, 2010 5:39 pm

In his essay, Jeff ID states a proposition. When skeptics fail to admit that this proposition is true, Jeff says, “…it undermines the credibility of their otherwise good arguments.” In a response, stevengoddard states his concurrence with Jeff on skeptics’ loss of credibility if they fail to accept Jeff’s proposition as true.
However, skeptics do not undermine the credibility of their arguments by failing to admit that a proposition is true if this proposition is false. Is Jeff’s proposition true or is it false? In the following remarks, I examine details of Jeff’s argument in an attempt at answering this question.
As stated in the second paragraph of Jeff’s essay, Jeff’s proposition is that “CO2 does create a warming effect in the lower atmosphere.” Before the truth or falsity of a proposition may be adjudicated, this proposition must be stated in such a way for the the variable which is called the “truth-value” of this proposition to take on the value of “true” or “false” but not both values simultaneously. This requirement sets the associated requirement for the proposition to be stated unambiguously. Jeff’s statement of his proposition is somewhat ambiguous. In order to move forward, I’ll disambiguate it while trying to be faithful to Jeff’s intent.
In technical writing, the “effect” of a statistical event is the state of nature that is produced by the “cause” of this event. Here, the cause seems to be a rise in the CO2 concentration in the lower atmosphere and the effect seems to be a rise in the temperature at an (x, y, z) space point in the lower atmosphere. Also, in climatology, a “warming” of the atmosphere designates increasing temperature at a space point. If this is what Jeff has in mind then Jeff’s proposition can be expressed in disambiguated form as:
“A rise in the CO2 concentration in the lower atmosphere causes a rise in the temperature at a space point in the lower atmosphere.”
Going forward, I’ll assume the above statement is an accurate representation of Jeff’s proposition.
At paragraph 11, Jeff announces his intention of proving this proposition true. His method of proof is generalization from the result of a thought experiment.
The result from Jeff’s thought experiment supports the proposition that “A rise in the CO2 concentration of the experimental apparatus causes a rise of the temperature at a space point in this apparatus.” Jeff’s proposition that “A rise in the CO2 concentration in the lower atmosphere causes a rise in the temperature at a space point in the lower atmosphere” results from generalization of the previously stated proposition.
Is this generalization true? A single empirical counter example would prove it false.
As the CO2 concentration rises monotonically with time, the temperature at every space point in the lower atmosphere must rise monotonically in order for Jeff’s generalization to be true. However, it has sometimes been observed that the temperature at a space point falls. Thus, Jeff’s generalization is false.
In circumstances such as these, economists sometimes preserve a statement of a cause and effect relationship from falsification by addition of the clause “ceteris paribus” (other things being equal) to the description of this relationship. Using this approach, Jeff could modify his proposition to read “A rise in the CO2 concentration in the lower atmosphere causes a rise in the temperature at a space point in the lower atmosphere, ceteris paribus,” thus preserving it from falsification.
A downside from avoidance of falsification by addition of a “ceteris paribus” clause is that the achieved generalization is not falsifiable, for if the realized effect differs from the stated one, the “other things” may not have been equal. That it is not falsifiable places the achieved generalization outside science, under Karl Popper’s criterion.

George E. Smith
August 9, 2010 5:47 pm

“”” Jeff Id says:
August 9, 2010 at 4:24 pm
George E. Smith
Doesn’t absorbed light in an optical medium simply turn to heat and then become radiated as a Planck blackbody ish curve? “””
Well yes Jeff, of course it does; and of course I knew that you knew that; but that is almost never pointed out in normal Optics courses; since the absorbed and radiated spectra are so completely different.
Where we are facing problems in the atmospheric radiation question is that the spectra being emitted (thermal continuum) and that being absorbed cherry is picked “bands” from some other thermal continuum from the surface or other atmospheric layers; with almost the same Temperature signatures.
Which is why people go astray in talking about the “Saturation” of the CO2 bands. No matter how big a chunk of the available radiation is grabbed by CO2 layers near the ground; or by a thinner layer if the CO2 amount goes up as in say doubling; the re-emission of that energy from the warmed atmosphere; simply has to try and make it through the next layer of CO2 containing atmosphere.
The “saturation” argument is as deceptive as the second law argument; and I’m quite sure you knew that too.
When you get a nice 20 mWatt HeCd laser beam of 2.8 eV photons absorbed in a slab of red glass; which doesn’t get very warm; you don’t think about hunting for it by chasing 120 meV photons; and with room temperature all around you they are damn hard to find even if you think of it.

Jim D
August 9, 2010 5:47 pm

Al Tekhasski :
I want to post a graphic here that may be of interest in discussing temperature inversion effects.
http://climateaudit.files.wordpress.com/2008/01/daly_spectra.gif
Note in this downward spectrum, the minimum around 660 /cm is interrupted by an upward spike, and generally looks like it is folded over in that region. I believe this is the inversion effect. The spike is a CO2 spike almost exactly at 15 microns, that far exceeds other CO2 absorption around it. The spike may be responding to the inversion, causing the folding effect. What happens as you double CO2 is that this spike will amplify upwards, but meanwhile the broader band will widen, and the latter would be the dominant effect in this region (which is Guam).
This (top) figure in the following link shows the basic CO2 spectrum with the spike at 15 microns surrounded by shoulders.
http://how-it-looks.blogspot.com/2010/03/infrared-spectra-of-molecules-of.html

Gnomish
August 9, 2010 6:15 pm

It would be interesting however to see how the 1kW 15um beam will look in open air. I think it will look as a plasma blowtorch.
———————————-
There are lots of videos on youtube of high powered co2 lasers in operation.
You can’t see the beam at all, of course, because it’s IR. If you make something incandesce, you can see the light from that.
http://cgi.ebay.com/New-150W-co2-laser-tube-water-cool-power-supply-/130417546927?pt=LH_DefaultDomain_0
Everyone should have one or more!

Spector
August 9, 2010 7:21 pm

Regarding the proposition that CO2 does create a warming effect in the lower atmosphere, I believe that all depends on what is meant by a warming effect and whether it is linear progressive or not. Here are a few selected results I obtained from the MODTRAN online radiation calculator utility for clear tropical air for the surface temperatures required to obtain a fixed output flux of 292.993 W/m2.

   ATM CO2       Surface Tmp
     ppm             °K
    0.000         293.56
    1.094         294.67
   12.375         296.79
   70             298.96
  140             299.85
  280             300.73
  396             301.18
  560             301.63
  792             302.10
 1,120            302.58
 2,240            303.61
 8,960            306.10
71,680            312.40

I believe these represent the raw CO2 effect on the radiation budget of the atmosphere. This is without regard to the action of any natural temperature regulation or feedback effects. The MODTRAN results should not be confused with real data – they are only computer calculated projections. As I understand it, there has only been about a 0.6 deg C increase in average global temperatures since 1880.
In the end, it all depends on what Jeff means by a ‘warming effect.’

cba
August 9, 2010 8:10 pm

spector
I believe you have it right there. One of the options is the 1976 std atm. which is typical even if it’s not an average. It provides a nice standard. The calculator is of course for clear sky only and it is radiative only.

Al Tekhasski
August 9, 2010 8:24 pm

Gnomish says:
“There are lots of videos on youtube of high powered co2 lasers in operation. You can’t see the beam at all, of course, because it’s IR. If you make something incandesce, you can see the light from that.”
This is a 10.6um laser. Absorption at 15um is 35,000 times stronger. Air at 10.6 um is practically transparent at 400ppm CO2. That’s why these lasers are in use, and that’s why you don’t see the beam, only when it hit something. At 15um, the entire (99%) of 1kW laser beam will be absorbed in 1m. If the beam is 1mm in diameter, there must be quite substantial power density, and very small heat exchange. But you are right, it probably will not glow yet.

Al Tekhasski
August 9, 2010 8:30 pm

Jim D,
Re http://climateaudit.files.wordpress.com/2008/01/daly_spectra.gif
This chart can have a different interpretation, that the entire band is already “inverted”, and the spike is higher because it emits from warmer area of stratosphere.

Jim D
August 9, 2010 8:59 pm

I also recommend this site for how the 70 km downward-looking spectrum changes with increasing CO2.
http://chriscolose.wordpress.com/2010/02/18/greenhouse-effect-revisited/
Note how the spike is first downwards then “bounces” up with higher concentrations. This is expected as it saturates in the inverted stratospheric temperature profile.

Stephen Wilde
August 9, 2010 9:07 pm

Spector said:
“In the end, it all depends on what Jeff means by a ‘warming effect.’ “.
Yes, that raises a point that I made in an earlier post in this thread.
The ‘warming’ effect of increased downward IR from more CO2 in the air quickly gets gets translated into increased evaporation from water surfaces and of course evaporation has a net cooling effect so the temperature of the surrounding environment should actually drop all other things being equal.
Anyway the energy from the extra CO2 is still present but it gets converted to latent heat which does not register on thermometers.
So it could be somewhat misleading to assert that the extra CO2 causes warming in the usual sense. What it actually does is increase the energy content of the atmosphere but not in a way that necessarily results in any measurable warming at or near the surface. At least not netted out globally if water surfaces predominate as they do.
Then we have to bear in mind that water vapour is a very light gas so despite the lack of detectable warming there will nonetheless be an increase in convection just because the extra water vapour is lighter that the surrounding air.
So the hydrological cycle speeds up despite a lack of measurable warming and again that increases the speed of upward energy transport for another net cooling effect at the surface on top of the earlier net cooling effect of evaporation alone.
Then cloudiness increases to reduce solar shortwave into the body of any water and any rainfall is cooler than the surface temperature so the warming of the oceans from solar input slows down too.
So we see a whole raft of cooling processes that can be initiated by more CO2 (or any other GHGs) in the air with the warming effect not registering on thermometers because all the extra energy goes into faster evaporation and gets hidden as latent heat.
I’m satisfied that in theory, in the absence of a hydrological cycle GHGs of whatever nature are capable of making an atmosphere warmer than it otherwise would be.
But what is the net result of the presence of so much water with all it’s phase changes? It appears logical to me that the hydrological cycle effectively negates ANY such attempted warming effect from the atmosphere by converting ALL the extra energy from GHGs to latent form thereby speeding up the hydrological cycle to maintain an equilibrium temperature set not by GHGs but instead by the density and pressure differentials between oceans, air and space and crucially also by the properties of water.
Forget the greenhouse effect because the Hot Water Bottle Effect reigns supreme.
Unless of course it can be shown empirically that there is extra non latent (sensible) heat left over on a netted out basis globally from the extra GHG induced downward IR radiation. But that couldn’t be so unless the planet was less than 50% water covered because evaporation has that net cooling effect and one would need the greater effect of warming over a larger global land area to more that offset the cooling evaporative effects over the oceans. We do not have that additional necessary factor of a surplus of land surfaces.

Gnomish
August 9, 2010 9:37 pm

http://www.john-daly.com/artifact.htm
It looks like this was tested-
maybe 10 meters is closer for the extinction at 15um?
http://lasersparkpluginc.com/uploads/CO2_Absorption_Data.pdf

Gnomish
August 9, 2010 9:45 pm

Indeed. In 43F ambient you can make ice by evaporative cooling.

Gnomish
August 9, 2010 9:53 pm

It seems a reply disappeared, so rerun:
It looks like the CO2 in a tube experiment has been quantified:
http://www.john-daly.com/artifact.htm
and then there’s this-
http://lasersparkpluginc.com/uploads/CO2_Absorption_Data.pdf
so maybe 10 meters for extinction at 15um ?

Al Tekhasski
August 9, 2010 10:23 pm

Jim D,
Ok, the sequence of spectra looks pretty convincing. In fact, the 15um peak begins to invert at 25ppm, and the side bands start to fold back above 100ppm. So, MODTRAN calculates that, despite of the fold back with negative contribution to flux, the band broadening has bigger effect. However, please note that actual peak spacing in side bands around 15um is about 1.8cm-1, while the MODTRAN resolution is 2 cm-1. This means that the picket-fence spectrum structure gets averaged, and most of the inversion effect is therefore calculated incorrectly. So my concern still applies.
I am aware that there have been several publications that claim that LBL codes based on HITRAN confirm the overall result of 4W/m2 per CO2 doubling. But given the fundamental differences between band-averaged spectra and their fine line structure, I am sorry, I cannot believe in these results. It is simply impossible.

Dave Springer
August 10, 2010 6:35 am

Spector says:
August 9, 2010 at 7:21 pm
Here are a few selected results I obtained from the MODTRAN online radiation calculator utility for clear tropical air for the surface temperatures required to obtain a fixed output flux of 292.993 W/m2.
ATM CO2 Surface Tmp
ppm °K
0.000 293.56
1.094 294.67
12.375 296.79
70 298.96
140 299.85
280 300.73
396 301.18
560 301.63
792 302.10
1,120 302.58
2,240 303.61
8,960 306.10
71,680 312.40
In the end, it all depends on what Jeff means by a ‘warming effect.’

It’s very important that this be understood.
The actual CO2 and temp record (assuming they are trustworthy) in the small range from 280-396ppm agree with predictions from statistical thermodynamics.
It appears any feedbacks, either positive or negative, are either absent or cancel each other out, given that the calculated rise in temperature due to additional insulation by CO2 matches the observed rise in temperature.
In short then there is nothing to fear from exponentially rising anthropogenic emission of CO2 and in fact we can expect great benefit from it as global economic output increases and the biosphere becomes more productive.
The only thing we have to fear is fear itself. ~Franklin Delano Roosevelt

Dave Springer
August 10, 2010 7:44 am

@Al Tekhasski
“If you have only CO2 and other gases are optically inert, your 15um beam will be absorbed by CO2, and re-distributed as heat to all other molecules (called “buffer gas”). However, since these other molecules have nothing to emit on their own
That’s quite wrong. Unless they are at zero kelvin they will emit blackbody radiation with peak energy corresponding to temperature. In effect the CO2 converts upwelling 15um radiation from the surface into thermal (kinetic) energy which radiates equally in all directions. This is how the insulation happens – the net effect is the energy in the 15um band is slowed down on its way out the door in the upper atmosphere. All insulators work this way. They slow down either or all of convective, conductive, or radiative transport of energy. The roof and insulation of a building stop virtually all convective and radiative transfer from inside air to outside air and force the energy to flow through conduction alone while a material that has poor conductive properties (like fiberglass batting)is used to minimize that path as well.

Al Tekhasski
August 10, 2010 11:15 am

Dave Springer says: “That’s quite wrong”, with regard to my statement that other [optically-inert] molecules have nothing to emit on their own.
Dave, you are confusing rarefied gas spectroscopy with solid-state spectroscopy. Please.

Jim D
August 10, 2010 5:24 pm

Re: Al Tekhasski: August 9, 2010 at 10:23 pm
It needs to be shown to me that better spectral resolution changes the results. Unless all the unresolved peaks are significantly broad and higher than the value shown, they are not going to change the mean behavior shown by MODTRAN. If there was an error of this type due to resolution alone, it would have been noticed by now, and hyped by the non-AGW people, but they seem silent on this issue. MODTRAN is successful in reproducing satellite-observed spectra. That is the bottom line that matters, unless you also doubt the satellites, which maybe you do.
Gnomish, the arguments about 10m extinction distances fall apart when you see how this applies to a very small fraction of the CO2 effect (the spike). Most of the CO2 doubling effect is coming from the troposphere, not the stratosphere, as is seen by the cold effective temperatures in observed emission.

Jim D
August 10, 2010 6:18 pm

Regarding CO2 1 K per doubling.
People show that we expect 0.4 K from CO2 alone since the beginning of the industrial age, and we actually have 0.6 K. They then infer a small positive feedback by neglecting any other possible effect on temperature than CO2, while at the same time arguing that there are all kinds of other effects, but that is beside the point. The point is that the rapid rise in aerosols, which I think was due to the oil age that started in the 40’s, did cause cooling, especially downstream of industrial areas (see global dimming). By increasing the albedo due to haze and cloud-particle effects, it can account for one degree of cooling and the reduced solar radiation observed, stunted tree growth in Siberia(?), and the fact that more cooling occurred in the northern hemisphere from the 40’s to 70’s as the haze cloud spread to its maximum size (due to limited lifetime of haze in the atmosphere and limited industrial areas). If you add that degree, you get 1.6 K, and a feedback factor of 4. My opinion is that you can’t just choose to neglect the effect of global dimming in this period.

Al Tekhasski
August 10, 2010 8:05 pm

Jim D, August 10, 2010 at 5:24 pm
(1) At 100mb, peaks in P- and R-branches are 300 times higher than their average. Check with spectralcalc.com. It is pretty much clear that “average emission height” is not where it is believed to be, and change in it does not have the warming effect.
(2) This is technically not an error with resolution. The error is in more subtle mathematics of averaging order.
(3) There is no surprise that static under-resolved calculations match the shape of under-resolved observations. However, the _change_ in absolute value of emission integral is completely different thing. Convincing observations of this change are lacking.

Jim D
August 10, 2010 8:54 pm

Al Tekhasski :August 10, 2010 at 8:05 pm
Thanks for the spectralcalc.com link. Neat tool, even just the free access part.
Obviously the combination of their strength and sparseness has to be represented somehow at lower resolution. I can’t claim to know much about how MODTRAN does that, but I imagine someone has done the same integrated calculations with HITRAN and compared them, and I would like to know how that turned out, and am unconvinced there could be much difference relative to the CO2 doubling signal.

cba
August 11, 2010 6:45 am

Al Tekhasski says:
August 10, 2010 at 8:05 pm
Jim D, August 10, 2010 at 5:24 pm
(1) At 100mb, peaks in P- and R-branches are 300 times higher than their average. Check with spectralcalc.com. It is pretty much clear that “average emission height” is not where it is believed to be, and change in it does not have the warming effect.
———-
I’ve got a radiative transfer model with variable resolution using HITRAN. Currently, I’m running about 10nm wavelength bin widths. Initially, I used 1 nm widths. Some lines fit inside 1nm but most do not. I usually run it from 0.1 to 75 um. Actually, over very limited bandwidths, a few Angstroms, I’ve run it down to 0.005 Angstrom resolution also – amazing how you can see the equipment smearing present on some of the highest resolution stellar spectrums ever done in comparison.
The net result though is that 1nm merely made it time consuming to process the data. While there was still a few W/m^2 of power at wavelengths longer than 75um (difference between the sum of the bins and stefan’s law), there was somewhat less difference between the 10nm and 1nm resolution. My model also broke down the atmosphere into over 50 layers with the bottom being 1km slabs, each with their own temperature, pressure, and composition.
effects of each layer are substantially different in the troposphere. Lower, one has much broader lines and it’s easy to see just looking at the patterns in the array of raw numbers in the excel spread sheets.
Results for a co2 doubling comes out to about 3.6 W/m^2 at the tropopause and about 2.6 at the 120km altitude. This is clear sky only and radiative only and excludes feedbacks. The 120km altitude is likely inaccurate as LTE is a problematic assumption above the stratosphere, although it appears valid for co2. By assuming a uniform rise in T for the entire column along with constant relative humidity, one can also get an idea of what the ‘primary’ feedback (more h2o vapor) is going to do. An assumption of 5 deg C would result in the absolute humidity increase of 30% – which is a long way from a doubling of h2o vapor. An increase of of 2 deg C results in only a 13% increase. Despite h2o having a stronger effect per doubling than co2, one finds that the 5 deg C rise with a co2 doubling results in a total increase in radiative absorption of under 6 W/m^2, which is less than twice what the co2 only effect would be. A 2 deg C rise would result in an increase to less than 5W/m^2.
If one goes back to simple averages to get an idea of sensitivity, one finds that a 288k temperature body of emissivity = 1 would emit 391 w/m^2 and that planet Earth receives 341 w/m^2 solar incoming power – of which 0.3 is reflected away, leaving 239 w/m^2 to be absorbed. For balance that means only 239 w/m^2 can be emitted away. For clear sky, that means 391-239 = ~150 w/m^2 average absorption of outgoing surface power (including cloudy conditions). With a 33 deg C rise from the theoretical BB object radiating 239 w/m^2, one can get the average sensitivity as 33/150 = 0.22 K rise per W/m^2 increase.
Applying this sensitivity, one has a contribution for a co2 doubling, including the ‘primary’ feedback of h2o vapor that amounts to just over 1 deg C (0.22 x ~5w/m^2). And that leaves us missing almost another degree’s worth of power absorption still needed to achieve that amount of T rise.
Not that the clouds are quite complex in contributions. We’ve got about 62% cloud cover of all types, each with their own effects. During the day, they contribute substantially to the albedo, sky = 0.22 while surface = 0.08 of the 0.30 total. At all times, they absorb most of the outgoing IR and reradiate a full continuum from their tops (at their characteristic temperature – which is much cooler than the surface). The net result is to provide a slight cooling effect – but not as much as one would assume. That is, the more clouds, the cooler – but not by a bunch.
In clear skies, there’s about 280 w/m^2 escaping (much more than permissible for balance with current albedo, but there would also be a loss of 0.22 worth of albedo which means the balance point would shift to 313 w/m^2 – requiring an increase of T. Note that 280 w/m^2 in clear skies means that there’s 391-280 = ~110 w/m^2 of actual GHG absorption and for the surface emission, only 280/391 = 0.7 fraction of surface emissions escape. For this ‘world’ with no clouds, balance would require (313-280 )/ 0.7 = 47 w/m^2 additional surface emissions. Using stefan’s law, with 391 + 47 w/m^2, we get a final T of 296.5k which is a rise of 8.3 deg C due to a complete loss of cloud cover.
Note that the spectrum of radiated power is quite complex. Absorption from h2o vapor and co2 isn’t total because it will radiate outbound at some altitude (and temperature). No two wavelengths will have exactly the same altitude as the path length varies even depending upon where on an individual spectral line one is looking.
If you want a laugh, go read Hansen’s paper published in some national geographic publication around 1998. In it, he establishes an altitude using stefan’s law and the whole power emission of an assumed black body and then tries to show what the emission temperature change with lapse rate will be from the change in that altitude and how it must heat up to permit the level of original emission again.

Al Tekhasski
August 11, 2010 12:00 pm

cba, that was a very comprehensive report, thanks.
However, you are considering too many things at once. We have here an entirely theoretical discussion about effect of “instant” doubling of CO2 without any feedbacks or else. So please let’s start from basic conditions:
(a) we are considering, say, Standard USA temperature profile, with fixed lapse rate across troposphere, and fixed temperature gradient in stratosphere;
(b) we are looking at total OLR in the energy-containing range from 50 to 2500 cm-1, or just between 600 and 800 cm-1, where the IR windows end and CO2 absorption peaks.
(c) the question is: what is the _change_ in outgoing radiance in this range if CO2 concentration is instantly doubled, say from 400 to 800ppm.
Now few technical questions for you:
1. Is this true that you can do this in Excel?
2. What do you mean “50 layers”? What are the assumed conditions between slabs?
3. How do you account for variable temperature-pressure broadening? Which waveform shape do you assume?
4. For strongly opaque lines, how do you account for scattering?
More questions to follow…

cba
August 11, 2010 5:27 pm

Al Tekhasski says:
August 11, 2010 at 12:00 pm
(a) we are considering, say, Standard USA temperature profile, with fixed lapse rate across troposphere, and fixed temperature gradient in stratosphere;
(b) we are looking at total OLR in the energy-containing range from 50 to 2500 cm-1, or just between 600 and 800 cm-1, where the IR windows end and CO2 absorption peaks.
(c) the question is: what is the _change_ in outgoing radiance in this range if CO2 concentration is instantly doubled, say from 400 to 800ppm.
Now few technical questions for you:
1. Is this true that you can do this in Excel?
2. What do you mean “50 layers”? What are the assumed conditions between slabs?
3. How do you account for variable temperature-pressure broadening? Which waveform shape do you assume?
4. For strongly opaque lines, how do you account for scattering?
I use and think only in terms of wavelength rather than the per cm frequency of the spectrometry people. 25/cm is way down in the realm of radio mm microwave. Most of the spectral broadband emission occurs at 75 um and below. For dealing with situations of visible light, I go down to the relatively hard uV which is well beyond the IR cut off.of global T emissions.
Although not limited to it, I almost always use the 1976 std US atmosphere as it’s a well known, common, and rather typical or close to what one would consider as an average atmosphere. From there, I modify co2 and or h2o concentrations.
As such, an instant doubling (usually 330 to 660ppm as 330ppm is the 1976 value) results in a change of in radiation of a 3.6x w/m^2 decrease over the range of essentially 0 to 75um with a surface emission of 386 w/m^2 (288.2k emissions over 0-75um). And this value is at the tropopause of 11km. Drop another w/m^2 for 120km and take your risk of non LTE effects possibly occurring higher up.
I use excel for a portion but not all. I wrote a program to read the HITRAN processed output database from their javahawks application which creates a 296K 1 atm pressure database of lines. My program takes the line data, corrects it to the desired temperature and pressure and builds the contribution to the spectrum by wavelength bin for each line enabled. Most lines take up multiple bins but in all cases, the contribution to the entire bin is calculated. Calculations and adjustments are made according to the HITRAN documentation, Rothman 1996, appendix A found on the official HITRAN site. Input to my program includes a list of molecules to select and a number of layers to create along with the pressure and temperature for each layer.
Once my program has run, it creates an output file containing the transmission factor by wavelength per cm length for each layer. That is imported into a spreadsheet where additional processing is performed. This includes the BB spectrum and thicknesses for each layer. It is the excel program that determines the actual radiative transfer calculations are done by wavelength bin. Options include summing (integration) over wavelengths and graphing by wavelength.
The radiative transfer consists of the attenuation by wavelength within a layer and emissions generated by the gases in that layer. Lower levels are 1km thick while intermediate levels are 2.5km thick and high altitudes are 5km thick.
A temperature and pressure are assigned to each layer. I don’t recall if these are the bottom of the layer conditions or somewhere a portion of the way through the layer. It is an approximation but each layer has just one temperature assigned and one pressure.
there are no assumptions concerning between slabs. Each slab has assigned values and thicknesses and the radiation must go through each to reach a higher altitude or escape the Earth. Most 1-d models use about half that many or less from what I understand.
I use the line broadening approach provided in the Rothman 1994 paper appendix. Seems like it is a Gaussian curve that includes a peak shift due to pressure as well as width calculations based upon T and P. This allows for a reasonably precise calculation on the fly for each line to determine the amount of line contribution to the absorption of each bin that can execute in fast calculations.
Scattering is not handled in the program at present. We’re mostly dealing with fairly far IR here and it is 1/wavelength^4 as I seem to recall making it quite small. There’s way too many other parameters associated with the real world conditions that are just variable or unknown that have far greater effects upon the system than does a co2 doubling. There is also nothing on dimers involved or for that matter, aerosols in general. If you really try to deal with all these sorts of things, the requirements of what is required to in the way of starting data and conditions exceeds what could be known about the system. As you said above, it’s easy to try to consider too many things at once.

Jim D
August 11, 2010 5:53 pm

cba,
I agree with Al that the question, at this stage, is simply what happens to the integrated outgoing longwave in CO2 doubling for a standard atmosphere, looking down from about 70 km. I don’t have the software to do this, but testing the sensitivity to resolution would be interesting, and the band of interest is 600-800 /cm.
Your result of only 2.6 W/m2 at 120 km is a little surprising, which is why I would like to see if that varies with spectral resolution. This amounts to a CO2 forcing at the top of the atmosphere of about 0.7 K, instead of nearer 0.9 as others say, but maybe this is something that would vary according to the mean sounding chosen.

Spector
August 11, 2010 7:04 pm

Just for reference; it appears that the savi.weber.edu HITRAN plotting tool has recently gone offline.

Al Tekhasski
August 11, 2010 7:48 pm

cba,
Could you please calculate just two numbers? After doubling in CO2, what is the change in OLR in two narrow bands, (a) 679.7-679.8, and (b) 680.7-680.8 cm-1, in your model? All looking down from 120km? Thanks.

cba
August 11, 2010 8:14 pm

jim d
I don’t recall there being a substantial difference with a change in resolution. Now with 1 nm resolution, I can only do 65um bandwidth and it is extremely difficult as a recalc in excel can take almost an hour on a larger and much faster computer as I am using now. I went to 10nm simply to cut down on the difficulties. Going to a much narrower bandwidth will have a substantial effect as there are contributions in addition to just the 15um band.
The lower power blocking at higher altitudes is due to the fact that even though the pressure is dropping, the temperature can be significantly hotter than the surface.
Note that the 10nm model gives 2.66 w/m^2 at 70km difference between 330 ppm and 660ppm of co2. Note that the modtran calculator is good out to about 70km and you can adjust the co2 values looking down and selecting the 1976 std atm US.
note that I ran the modtran calculator at http://geoflop.uchicago.edu/forecast/docs/Projects/modtran.orig.html for the 1976 atmosphere option, 70km and a doubling from 330 to 660ppm and wound up with 2.86 w/m^2 absorption – about 0.2 w/m^2 difference from my calculation. Note though that they go out to 100um compared to my 75um maximum wavelength and that they are using a totally different system from HITRAN.
Note that 70km is about the upper limit for the modtran calculator. above that I don’t think they actually do calculations beyond that despite an absence of warnings.
for whatever reasons, the 11km tropopause measurement differs a bit more in the wrong direction. The modtran calculator provides 3.4 w/m^2 while my calculations gives 3.7w/m^2 with the same differences in bandwidth mentioned above. I do not claim that my model is more accurate than the modtran calculator but it does provide a value in the ballpark of what the modtran calculator produces and also what the general climatology claim is and it is a bit closer to this claimed value than the modtran calculator produces for this situation.

Bryan
August 12, 2010 7:34 am

Merrick
….”Many are going to immediately respond back regarding LTEs and forcings and negative feedbacks. Yes – of course they are all very important points to take into consideration. But this small portion of the topic seems still to be poorly understood by most of the folks here and it sure would be nice if we could get a significant plurality up to speed on at least this part.
A boy can dream, can’t he!”……….
I have found your posts on this topic rational and interesting.
However in the nature of things they must be partial to “stay on topic”
Any chance of Anthony inviting you to do a post giving the big picture on this topic?

cba
August 12, 2010 8:34 am

Al Tekhasski says:
August 11, 2010 at 7:48 pm
cba,
Could you please calculate just two numbers? After doubling in CO2, what is the change in OLR in two narrow bands, (a) 679.7-679.8, and (b) 680.7-680.8 cm-1, in your model? All looking down from 120km? Thanks.
———
These are 14.710 – 14712 um and 14.689 to 14.691 um amount to 1/5 of the resolution I’m using. I made a special run for you for these at high resolution, 0.01 nm bin size. It was a bit more time consuming than expected.
at 2x co2, the power transmitted from 120km outward is:
14.689.00 to 14.691.00 um 0.018268837 W/m^2
14.710.00 to 14.712 um 0.018976876 W/m^2
at 1x co2:
14.689.00 to 14.691.00 um 0.017229193 W/m^2
14.710.00 to 14.712 um 0.019007363 W/m^2
tropopause
1x co2
14.689 to 14..691 um 0.0138603 W/m^2
14.710 to 14712 um 0.019035344 W/m^2
2x co2
14.689 to 14.691 um 0.0138603 W/m^2
14.710 to 14.712 um 0.019007948 W/m^2
I’m not sure what you hope to ascertain by these but they do seem to behave a bit differently from each other.

Jim D
August 12, 2010 5:37 pm

Al,
I don’t think you are going to get those two lines to contribute anything significant compared to the dense 667/cm spike. They are too widely spaced.

Al Tekhasski
August 12, 2010 10:00 pm

cba, it looks like your program calculates things in right direction.
In my example, the narrow 0.1cm-1 range (b) contains a medium-absorbing peak, one from the “R-branch” area of CO2 spectrum. The range (a) is between peaks, with weak absorption. This entire area is a part of CO2 15-um absorbing band, it has this typical fine structure everywhere.
From your results, the “weak” band (14.7113um) produces reduction in OLR by 0.00003W/m2, which is “consistent” with NEGATIVE imbalance, as per AGW theory.
Now, peak spacing in this area is 1.4cm-1. Therefore there are approximately 13 “weak” areas for each “strong” area. When averaged, the entire area will show NEGATIVE imbalance approximately 14*-0.00003= -0.0004W/m2. You can try to reproduce this number directly using 2cm-1 spectral resolution.
However, from your other result, the “strong band” (14.69um) produces an INCREASE in OLR, +0.001W/m2. The actual averaged OLR (1.4cm-1 band) is therefore +0.001-0.0004 = +0.0006 W/m2. The result is POSITIVE imbalance.
So, your result is a solid illustration that high-resolution spectrum produces OPPOSITE result to medium-resolution spectrum, at least in most important IR absorbing areas. Just as I said earlier, the overall official imbalance of -3.7W/m2 for CO2 doubling seems to be wrong, or highly inflated. Thanks.

cba
August 13, 2010 6:49 am

Al Tekhasski,
I don’t agree with you on that conclusion on the basis of the fundamental physics. While at 120km, we should still be in LTE for co2, we’re probably not in LTE for NO and some other molecules. The atmosphere is above is quite thin (rather quite a good vacuum) and the temperatures are quite above mean surface T. That places the optical path at fairly large lengths and the likelihood of photon escape to be fairly good, especially in the wings where lower pressure reduces the likelihood of absorption and emission compared to lower levels.
These details are smeared into whatever resolution I’m dealing with, whether it’s 10 nm or 0.01 nm. Ultimately, the effect of the whole spectral line is considered, regardless of how many bins it fills and as the pressures decrease and the line narrows towards a zero bandwidth width line, it fits into one bin as a fraction of the bin. But then, we’re integrating the result over all the bins anyway.
While one might argue that an integration over dx provides a more accurate result than does a summation over delta x bins, that’s true only if the delta x bin values are averages that are approximate rather than integrated or if we’re dealing with a range that is not an integral multiple of the bin size. I take the integrated result of the line’s area under the curve in a non numerical methods approach.
If there were approximations involved (as in my first attempt), the results would be highly resolution dependent and without sufficient resolution, the results would be in error. However, I integrated the line function and attributed the appropriate amount of integrated result to each bin for each line of each molecule and these are using 39 molecules plus isotopes. The approach is no different for any resolution.
Since it’s using radiative transfer, each layer has an emission as well as absorption and the net is going to depend upon the Temperature spectrum incoming and on the temperature of the layer. The likelihood of emission is the same as the likelihood of absorption times the energy distribution (which is easiest to envision as the the planck BB spectrum curve for the temperature of the layer). Multiply one by the other at each wavelength and the emission curve is determined. That has the effect of showing that a gas cloud in front of a radiating surface will have emission lines if the gas cloud is hotter and absorption lines if it is cooler than the radiating surface and will have no lines if it is the same temperature.

Al Tekhasski
August 13, 2010 7:43 pm

cba, which “fundamental physics” you are talking about? I don’t think LTE has nothing to do with anything, you can try 70km. I think the result will be even better (for me). Also, we seems to have established that other molecules are out of the picture, so please forget them.
What was demonstrated here is that (a) the order of integration is critically important, and (b) spectral resolution has to be better than 1nm, otherwise the result is wrong.
When you integrate a line over your bin, you smear it, it’s effective emission height is much, much lower. It makes the whole difference when atmosphere has different temperature gradients with height. It was clearly demonstrated with your own numbers, the entire effect from CO2 doubling gets REVERSED, at least in this 14-16um zone.
So, what is left? The side edges of the band. However, these areas are literally “gray areas” of radiation physics, they fall right between “thin” and “thick” approximations. Would you agree that areas without good physical approximation could be prone to substantial computational uncertainty, softly speaking?

cba
August 14, 2010 8:20 am

Al,
power is power per bandwidth. just because a particular bandwidth of 2nm shows increased emission while a nearby 2nm bandwidth shows increased absorption doesn’t mean that my program isn’t taking both into account in fewer bins when at 10nm bin resolution. The difference between 11km (tropopause) and 120km for total co2 absorption from a doubling is proof of that. At 11km, it’s 3.7w/m^2 over the 75um range. At 120km, it’s 2.6w/m^2. Clearly there was an increase in emissions higher up that made up for part of the difference from lower down.
As mentioned before, I actually integrated the width function before applying it in the software. Rather than smearing it, I have the area under the curve. Each line has a function of absorption vs wavelength and essentially, at low pressures, it’s very narrow and tall and at higher pressures it’s broader. To find out just how much that line blocks out of the continuum spectrum, one must integrate this function over the spectrum – that is over all of the spectrum that is affected by that line. One then has the total absorption involved. An integration over an entire range can be broken down into two integrations over two halves of that range or 4 integrations over the four quarters of the the range. Hence, when I sum the bins, I am combining the integrations already done into a final integration over the whole range of interest. BTW, these are done at each of the 50+ layers using an average P and T for that layer.
well, my model isn’t capable of being subtracted out from the high resolution measured spectral lines and show a perfect subtraction that allows simple computer processing of the result to eliminate atmospheric lines from the result. Actually, it almost can and can readily permit one to identify what lines originate in our atmosphere versus what lines are a genuine part of the incoming spectrum. And some of the problem can be attributed to the limiting resolution of the measurement apparatus and some can be attributed to the difference in a model typical atmosphere and in what the atmosphere was actually like during the measurements.
To make matters more interesting in this application, we are not dealing with two lines but thousands or tens of thousands of lines total. Inaccuracies of a few % which are random are going to be diminished to very small values by an averaging effect leaving only some sort of systematic error remaining.
there’s plenty of room for tremendous error overall. I doubt there’s much error in this for several reasons. First, the database has been developed over 30 + years for practical applications, more than a few are probably classified still. Second, when I can take a super high resolution slice of spectrum and compare with the calculations of the model and have to do averaging to smear it down to the resolution of the instrumentation and have the two appear visually identical, I’m pretty sure that you’re talking mole hills and not mountains. BTW, it wasn’t a solar spectrum and it is one of only three ever done at that high a resolution.
Where the inaccuracies lie are primarily in the fact that these are clear sky calculations. Over half the world is covered in clouds of one description or another at any one time. That totally scrambles the results right there. Cloud cover, cloud albedo, and total albedo toss in massive uncertainty. Clouds themselves do that by themselves on both the albedo front and on the IR blocking front.

Al Tekhasski
August 14, 2010 11:24 am

cba, you wrote:
“power is power per bandwidth. just because a particular bandwidth of 2nm shows increased emission while a nearby 2nm bandwidth shows increased absorption doesn’t mean that my program isn’t taking both into account in fewer bins when at 10nm bin resolution.”
My example of integration (based on your excellent data) shows that your program calculates integrals wrongly. I suggested one narrow bit (0.1cm-1) at a peak absorption, and another in between peaks, with weak absorption. The overall spectrum structure is periodic, 1 strong bin, 13 weak bins, all based on HITRAN database and spectralcalc plotting routine. Then I integrated your emission results per a band of 14 bins, 13 at low absorption, and 1 at high. The result was POSITIVE change in OLR, while if I spread all stuff over 14 bins evenly, the result will be NEGATIVE, the one you are assuming as being correct. Since the spectrum structure is not random but nearly periodical, it is obvious that the result applies to thousands of nearly identical bins across the entire 14-16um band, and likely across all other spectral regions. This result of CORRECT integration is based on the same spectral data, 30 years of 150 years, I don’t care.
What I do care is that the order of averaging in all (yours included) OLR calculations does make profound effect on the global OLR integral. Averaging spectrum into bins that are wider than typical absorption peaks of CO2 before calculating path emission is mathematically incorrect procedure. Under atmospheric pressures of interest, 100-200ppm, these peaks have typical width of 0.1cm-1 (less than 1nm). Therefore, all results that use wider that 1nm spectral resolution in LBL calculations must be incorrect. Your data and my analysis prove this.
Re: “… have to do averaging to smear it down to the resolution of the instrumentation and have the two appear visually identical, I’m pretty sure that you’re talking mole hills and not mountains.”
Our joint result is based on detailed accurate physics: strong bin emits from stratosphere, and weak bins are transparent to IR and do not change much. The idea of smearing the spectrum down to instrumental resolution is a highly unwise one: the fact that instruments see the world through “murky lens” does not mean that the underlying physics becomes different. And the physics (as confirmed by your calculations) says that OLR increases with increase in CO2.
Regarding clouds: no, clouds do not scramble any of the above results, because spectrum lines behave independently. If a particular spectrum band of CO2 is masked under background emission “noise”, change in CO2 will not affect OLR anyway. If certain absorption lines outstand into stratosphere, the effect of CO2 increase results in higher OLR, or global cooling. Again, we are considering a hypothetical situation when all other things are equal. Therefore, masking and scrambling of areas between strong CO2 absorption peaks only enhances my result.
The inescapable conclusion is: the 3.7W/m2 radiative forcing from CO2 doubling is likely a very overstated number. For the same reasons the same conclusion is likely for all IPCC forcings from all other GH gases.

Al Tekhasski
August 14, 2010 11:57 am

cba: “The difference between 11km (tropopause) and 120km for total co2 absorption from a doubling is proof of that. At 11km, it’s 3.7w/m^2 over the 75um range. At 120km, it’s 2.6w/m^2. Clearly there was an increase in emissions higher up that made up for part of the difference from lower down.”
This means that the optical path area above tropopause is quite important, and the direction of change is consistent with my concept of importance of stratospheric cooling for global radiative forcing. It also means that you program works correctly, even for the set of input parameters with very coarse resolution. Now if you re-calculate everything with 1nm resolution, I expect the difference will be much bigger, up to a reversal of entire result.

cba
August 14, 2010 1:31 pm

Al,
just to make sure you understood what I provided you, here it is again. The data I provided came from a narrow band run of 0.01 nm bin widths and the results were summed to provide 2nm width bands. The gas mix was the full 39 HITRAN data base molecules.
I do have some 1nm runs on another computer. I’ll have to try to dig them up. Also, they only go to 65um. I didn’t think there was any difference in the total power absrption.
as for my real project, it’s not about IR or outgoing emissions. It’s about atmospheric effects on incoming and it’s about identifying lines as being atmospheric or not.

cba
August 14, 2010 3:46 pm

Al,
I checked the other computer where I have 1nm resolution.
Both are adapted to 0.2 to 65.5um
the rest of the conditions are the same as mentioned previously
Here’s the following results:
1nm
11km 70km 120km
3.47 3.39 3.39 W/m^2
10nm
3.69 2.63 2.71 W/m^2
I don’t have the details as to whether the 1nm has the integration upgrades as it’s old and obsolete. However, you’ll note that the differences indicate it is still in the ball park.

John Millett
August 14, 2010 9:01 pm

“Yes, Tom is right, these conversions go in both directions. The energy moves in and out of CO2 and other molecules, but as shown in cavity C above, the gas takes finite measurable time to warm up and reach equilibrium with space (the window), creating a warming effect in the atmosphere.”
In the thought experiment, isn’t the CO2-heavy gas introduced into the cavity at ambient temperature and already in temperature equilibrium with the window? If so, in the time lapse referred to the gas doesn’t heat up, rather it is induced by the two-way excitation process, on which you agree with Tom, to emit at a rate to balance the laser input, though in a broader spectrum and in a diffused pattern?