Spencer slays with sarcasm

Heh. In response to a ridiculous claim making the rounds (I get comment bombed at WUWT daily with that nonsense) which I debunked here: A misinterpreted claim about a NASA press release, CO2, solar flares, and the thermosphere is making theĀ rounds

Dr. Roy Spencer employs some power visual satire, that has truth in it. He writes:

How Can Home Insulation Keep Your House Warmer, When It Cools Your House?!

<sarc> There is an obvious conspiracy from the HVAC and home repair industry, who for years have been telling us to add more insulation to our homes to keep them warmer in winter.

But we all know, from basic thermodynamics, that since insulation conducts heat from the warm interior to the cold outside, it actually COOLS the house.

Go read his entire essay here. <Sarc> on, Roy!

UPDATE: Even Monckton thinks these ideas promoted by slayers/principia/O’Sullivan are ridiculous:

Reply to John O’Sullivan:

One John Oā€™Sullivan has written me a confused and scientifically illiterate ā€œopen letterā€ in which he describes me as a ā€œgreenhouse gas promoterā€. I do not promote greenhouse gases.

He says I have ā€œcarefully styled [my]self ā€˜science adviserā€™ to Margaret Thatcher. Others, not I, have used that term. For four years I advised the Prime Minister on various policy matters, including science.

He says I was wrong to say in 1986 that added CO2 in the air would cause some warming. Since 1986 there has been some warming. Some of it may have been caused by CO2.

He says a paper by me admits the ā€œtell-tale greenhouse-effect ā€˜hot spotā€™ in the atmosphere isnā€™t thereā€. The ā€œhot spotā€, which I named, ought to be there whatever the cause of the warming. The IPCC was wrong to assert that it would only arise from greenhouse warming. Its absence indicates either that there has been no warming (confirming the past two decadesā€™ temperature records) or that tropical surface temperatures are inadequately measured.

He misrepresents Professor Richard Lindzen and Dr. Roy Spencer by a series of crude over-simplifications. If he has concerns about their results, he should address his concerns to them, not to me.

He invites me to ā€œthrow outā€ my ā€œshredded blanket effectā€ of greenhouse gases that ā€œtrapsā€ heat. It is Al Gore, not I, who talks of a ā€œblanketā€ that ā€œtrapsā€ heat. Interaction of greenhouse gases with photons at certain absorption wavelengths induces a quantum resonance in the gas molecules, emitting heat directly. It is more like turning on a tiny radiator than trapping heat with a blanket. Therefore, he is wrong to describe CO2 as a ā€œcoolantā€ with respect to global temperature.

He invites me to explain why Al Gore faked a televised experiment. That is a question for Mr. Gore.

He says I am wrong to assert that blackbodies have albedo. Here, he confuses two distinct methods of radiative transfer at a surface: absorption/emission (in which the Earth is a near-blackbody, displacing incoming radiance to the near-infrared in accordance with Wienā€™s law), and reflection (by which clouds and ice reflect the Sunā€™s radiance without displacing its incoming wavelengths).

He implicitly attributes Margaret Thatcherā€™s 1988 speech to the Royal Society about global warming to me. I had ceased to work with her in 1986.

He says that if I checked my history I should discover that it was not until 1981 that scientists were seriously considering CO2ā€™s impact on climate. However, Joseph Fourier had posited the greenhouse effect some 200 years previously; Tyndale had measured the greenhouse effect of various gases at the Royal Institution in London in 1859; Arrhenius had predicted in 1896 that a doubling of CO2 concentration would cause 4-8 K warming, and had revised this estimate to 1.6 K in 1906; Callender had sounded a strong note of alarm in 1938; and numerous scientists, including Manabe&Wetherald (1976) had attempted to determine climate sensitivity before Hansenā€™s 1981 paper.

He says, with characteristic snide offensiveness, that I ā€œcrasslyā€ attribute the ā€œheat-trapping properties of latent heat to a trace gas that is a perfect energy emitterā€. On the contrary: in its absorption bands, CO2 absorbs the energy of a photon and emits heat by quantum resonance.

He says the American Meteorological Society found in 1951 that all the long-wave radiation that might otherwise have been absorbed by CO2 was ā€œalready absorbed by water vaporā€. It is now known that, though that is largely true for the lower troposphere, it is often false for the upper.

The series of elementary errors he here perpetrates, delivered with an unbecoming, cranky arrogance, indicates the need for considerable elementary education on his part. I refer him to Dr. Spencerā€™s excellent plain-English account of how we know there is a greenhouse effect.

The Viscount Monckton of Brenchley (April 18, 2013)

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April 24, 2013 10:06 am

Here in the UK most houses are built with a double skinned wall with breeze blocks (I think in the USA you call them ash blocks) forming the inner wall, then an air gap, then bricks on the outer wall.
The bricks are porous and absorb water when it rains, which then evaporates when it is windy and dry. This arrangement keeps the inside of the house warm in winter and cooler in summer because of the still air in between the two. This system has worked well for years.
There are companies that will fill the cavity with foam or fibreglass insulation and claim that as a result your house will be warmer. It can also be damper, as the rainwater passes from the outer wall, by capillary action to the breeze blocks and then to the plaster on the inside wall. Mildew then develops and the house is damp, also water vapour from breathing, cooking, showering etc condenses on the coolest parts of the interior of the house, which is usually the parts of the walls that the cavity wall insulation could not get to, causing more mildew and dampness. The house feels cold and damp and is a health hazard to those with asthma.
This is all very expensive to put right and these companies usually go out of business on a regular basis to avoid legal action.
So this article is less sarcastic than Roy thinks!
By the way “insulation conducts heat” is that not an oxymoron or at least a contradiction?

Jim Hodgen
April 24, 2013 10:14 am

One weakness of sarcasm is that it requires honest comparison between norms and the offered information in order to get it. Thus it frequently leaves the target of the pointed information completely clueless…. but a great fallback is the entertainment value for those that have the ability to make the comparison.
Thank you Dr. Spencer for the analogy… and thank you Anthony for the entertainment.

Mark Bofill
April 24, 2013 10:15 am

Cue the Slayers to come in and …explain… this again.
cringe Oh noes, not again!
I’ve quit arguing this point with people. When I get this urge, I remind myself that I’ve found it’s much more productive and fulfilling to beat my head against concrete pylons instead.

Tom J
April 24, 2013 10:16 am

I mean, c’mon, duh. A home is a home, not a greenhouse. Got it? The Earth is a greenhouse. That’s the difference. A greenhouse is for growing plants. A home is for growing people. So the Earth is not a home. Uh, wait a minute. Uh, we grow here too. So the Earth is a home, I guess, but, uh, mmm…
Quick, John Holdren, help me out here.

alex
April 24, 2013 10:22 am

Sarcasm apart. What did the guy wanted to say?
Surely more insulation helps you keeping the house warmer. What was the question?
More clothes you have on, the warmer you feel.

Stuart Elliot
April 24, 2013 10:22 am

All materials conduct heat. Insulation conducts heat, but the heat transfer is slowed compared to any material which insulates less.
The first commenter touches on the problem that arises when moisture is not managed inside a sealed house. The examples he cites are caused by contractors not understanding the right way of insulating a dwelling. It’s not the insulating material wot dunnit, it was the act of changing only one thing in a system that was more complicated than it looked.
Kind of like “climate” “policy”…

April 24, 2013 10:29 am

Insulation in your house prevents draft leakage and convective loss to the outside, by trapping material molecules. This simply makes it easier for the furnace to hold temperature. A gas, CO2, doesnā€™t trap itself. It doesnā€™t prevent its own convection and drafting etc. And, such insulation doesnā€™t raise the temperature above the temperature of the input, of the furnace.
So then the analogy is created that trapping photons, via CO2, has s a similar heat trapping effect, even though trapping photons does occur in a real greenhouse but actually has nothing to do with the temperature inside a real greenhouse, because the temperature inside a real greenhouse is caused by preventing convective cooling when the input is sunshine at, say, +80C at noon. The analogy is then further extended to say that trapped photons can cause a -18C input (the incorrect and absurd flat earth assumption for the solar input), to become +15C, even though this doesnā€™t actually occur inside a real greenhouse nor with the insulation in your home. In fact, trapped radiation inside a cavity simply produces a blackbody spectrum of the temperature of the source, and this is of course the origin of quantum mechanics; trapped radiation doesnā€™t raise the temperature of its own source in any case. Photons are bosons and can pile on top of each other without noticing, other than equal amounts of constructive and destructive interference which results in no net change. This is markedly different behaviour from molecules which, when packed on top of each other, can not simply pass through but must find smaller and smaller volumes of space in which to situate themselves, thus increasing pressure and temperature. Shine two flashlight directly into each other and seal them perfectly, and they will not explode or even shine any brighter, unlike what would happen if you were compressing more and more gas into the sealed chamber.
Youā€™re being lied to by these sophists. Either you understand it or you donā€™t. That being said, it is quite difficult to understand, but within that difficulty is where the sophists create their garbage. 50% (and more) increase of CO2 over the last few decades and no increase in temperature above 1930ā€²s values, or above the warm periods before that, or above the first entire half (or more) of this whole interglacial; ice core records which show that the Earth enters an ice-age when CO2 level is high(!). Youā€™re being lied to by sophists.

Josh C
April 24, 2013 10:34 am

Close!
But the real cause of warming in the house is the insulation traps the CO2’s and that leads to ‘Housing Warming.’

April 24, 2013 10:40 am

What this infra-red thermography shows is convective heat stratification within the home and the thermal exit paths. The higher section of the windows and roof are heated more than the lower section. Home insulation provides a physical barrier to convective flow. The home repair industry does not add ‘radiative barriers’ of Carbon Dioxide to the hot spots to ‘redirect’ the radiative flow. Since the premise is meaningless, there can be no ‘satire’. Roy would benefit from reviewing a sample of satire. I recommend, “Amazing! New! Wrongco Proxy Crock” posted at Canada Free Press.

April 24, 2013 10:41 am

If things behaved this way, if trapping heat could raise the temperature above the temperature creating the heat, then ALL of thermodynamic research in the 1800’s would have revolved around exploiting this and today we would have devices which could create an internal extremely high temperature (to use for doing useful work with) with only a minuscule, tiny input; say, a AAA battery and tiny resistor could be used to generate 5000K inside a shell, and then you could smelt some steel or something with a AAA battery. Instead they created steam engines powered by coal and developed things like the Carnot Cycle. They never ever developed anything about trapping heat to create higher temperature than the input. What they found when they tried to do so, was what was eventually codified into the Laws of Thermodynamics.

April 24, 2013 10:45 am

One can hope that folks will follow anthony’s and roy’s lead here by calling out skydragon nonsense for what it is. The real debate is over how much warming GHGs will cause.
That is the debate. That science is not settled. Skeptics with smart arguments will get published, see Nic lewis and Troy masters. Its the central debate..

Luis Dias
April 24, 2013 10:45 am

Yeah, this sarcasm just went over my head. The argument was very funny, but I really don’t know what the hell was that about.

tadchem
April 24, 2013 10:52 am

The difference between a greenhouse and the Earth is convection. A greenhouse sitting in the sunlight gets hotter because it has no convection – the hot air gets trapped beneath the cover provided by whatever media is used to put a lid on everything.

Stephen Wilde
April 24, 2013 10:54 am

But is CO2 an insulator ?
It is supposed to block outward longwave from the surface but in doing so increases convection, evaporation and radiation to space (by providing a radiative window not supplied by non GHGs).
Evaporation has a net cooling effect of 5 to 1 (enthalpy of vaporisation) and so is a hugely powerful negative system response.
I am not yet convinced that CO2 has any net insulating effect at all once the negative system responses have been accounted for.
Note that I do not deny the thermal characterisatics of GHGs, merely do I question the sign and power of the system response.
The insulated house analogy is entirely inappropriate because one is simply interposing a less effective conductor between the air inside the house and the air outside the house.

April 24, 2013 10:58 am

Kind of like the “thermos Joke”:
http://www.funniestcleanjokes.com/joke/funny-thermos-joke

Paul Westhaver
April 24, 2013 11:02 am

I can actually work that one out from first principles. Having studied a great deal of thermodynamics, I am equipped to tell you that Roy’s statement is 100% true.
The angels are in the details.
If you have an electric water heater, how thick do you make the insulation jacket?
As Roy points out, making the insulation thicker does not mean that you will reduce the total heat loss.
Heat loss happens by conduction, convection and radiation. The models for heat loss for an electric water heater are very well known and accurate and reliable. As you increase the diameter of the heater you increase the surface area, thereby increasing the conduction, the convection and the radiation even though you’ve increased the resistance of heat flow from the tank. Insulation is a conductor… in fact.
I recall it was a question on one of my Thermo II course exams.
Common sense does not prevail. Understanding the underlying models and experimental results wins. Real science is great!

Paul Westhaver
April 24, 2013 11:06 am

šŸ™‚

Paul Westhaver
April 24, 2013 11:10 am

It is a problem of diminishing return. At what point does the cost of insulation yield a saving that is less than the cost of insulation. It depends on the cost of heat and the cost of insulation.

April 24, 2013 11:36 am

Let’s say you have a pot of water that is being boiled (~373 K) and a block of ice (~273 K) radiating towards each other.
Correct: P = ĪµĻƒA(Thā“ ā€“ Tcā“)
Incorrect: P = ĪµĻƒAThā“ ā€“ P = ĪµĻƒATcā“
Doing it the right way gives 782.63~ W/m^2 from the pot to the ice block.
Doing it the wrong way gives 1097.59 W/m^2 from the pot to the ice, and 314.96 W/m^2 from the ice to the pot, so now the pot is receiving input+314.96 W/m^2, and we then equate the output to the input to say the pot is receiving 1097.59+314.96=1412.55 W/m^2!
So now we claim it has to emit 1412.55 W/m^2 which equates to a temperature of 397.28 K!
So just by putting a block of ice next to a pot of boiling water we raised the temperature of the boiling water by 24 K!
It is the difference between this:
P = ĪµĻƒA(Thā“ ā€“ Tcā“)
and this:
P = ĪµĻƒA(Thā“ + Tcā“)

crosspatch
April 24, 2013 11:47 am

If your home is heated from the outside by a big lamp, adding insulation will make it cooler.

John West
April 24, 2013 11:55 am

Joseph E Postma says:
ā€if trapping heat could raise the temperature above the temperature creating the heatā€
I donā€™t recall that ever being said:
http://www.asterism.org/tutorials/tut37%20Radiative%20Cooling.pdf
Letā€™s say I turn the heat on to some body for 12 hours out of every 24, obviously, if I add insulation the body will cool slower during the off heat 12 hours and could be warmer at the initiation of heat for the next heat on 12 hours; in this manner of reducing heat loss the body gets warmer not that the insulation directly heats the body.

A C Osborn
April 24, 2013 11:58 am

crosspatch says:
April 24, 2013 at 11:47 am
If your home is heated from the outside by a big lamp, adding insulation will make it cooler.
That is precisely what I asked Roy Spencer, his response was
The house would be slower to heat if that heat was applied to its outside, it would warm quicker if it the source of heat was inside. In the case of the climate system, most sunlight is absorbed at the surface, so it is ā€œheated from withinā€.”
So CO2 does not act as an Insulator at all because if it did it would prevent the Radiation from heating surface in the first place.
So we are back to the Back Radiation issue.
The House is a crap analogy and he knows it, I am glad Anthony finds it funny, but for all the wrong reasons.

Roy Spencer
April 24, 2013 12:02 pm

Joe Postma, yes you probably CAN design a system for generating 5,000K temperatures using just a AAA battery. All you need to do is resistance heat a very small object with little heat capacity, and insulate it very well. The problem would be the insulation, because it, too,would heat up. Remember, 10,000,000 K temperatures are created at the core of the Sun with heating rates less than what the human body produces (per unit mass).

DaveG
April 24, 2013 12:04 pm

Roy,
Many thanks for coming out all guns blazing on this. It should be greatly entertaining as well as informative to all who follow this issue. What PSI finds amusing is how you have tripped yourself up by your blatant contradictions. e.g. You say:
ā€œOne of the first things you discover when putting numbers to the problem is the overriding importance of infrared radiative absorption and emission to explaining the atmospheric temperature profile. These IR flows would not occur without the presence of ā€œgreenhouse gasesā€, which simply means gases which absorb and emit IR radiation. Without those gases, there would be no way for the atmosphere to cool to outer space in the presence of continuous convective heat transport from the surface.
Indeed, it is the ā€œgreenhouse effectā€ which destabilizes the atmosphere, leading to convective overturning. Without it, there would not be weather as we know it. The net effect of greenhouse gases is to warm the lowest layers, and to cool the upper layers.
The greenhouse effect thus continuously ā€œtriesā€ to produce a lapse rate much steeper than the adiabatic lapse rate, but convective overturning occurs before that can happen, cooling the lower troposphere and warming the upper troposphere through a net convective transport of heat from lower layers to upper layers.ā€
http://www.drroyspencer.com/2011/12/why-atmospheric-pressure-cannot-explain-the-elevated-surface-temperature-of-the-earth/
So, Roy, what youā€™re claiming is that the enormous heat from GHGs leads to convective heat-mixing, which makes the lapse rate shallow. But the RADIATIVE GHE would make the lapse very steep.
Good lord. Thatā€™s just awful. You are saying that without GHGā€™s, the atmosphere wouldnā€™t be able to cool???!!!! Donā€™t you mean it would be HOTTER without GHGā€™s??
Convection occurs automatically because warm air is less denseā€¦duh! It gets heated and then rises. GHGā€™s donā€™t cause convectionā€¦how could 0.04% force the rest? So backwards.
And then why is the lapse rate for dry air SIMPLY what it is if you calculate it using U = gh + CpT. Whereā€™s the GHG effect? How come when you factor in condensation heat release to the equation, then you just get exactly the wet rate? You are referring to an effect on the lapse rate that is non-existent.

April 24, 2013 12:11 pm

Roy Spencer says: “yes you probably CAN design a system for generating 5,000K temperatures using just a AAA battery”
Wow. That is of course ridiculous. Actually let me rephrase that.
Please make it so and prove it. šŸ™‚

Roy Spencer
April 24, 2013 12:12 pm

DaveG has just posted a comment that is actually from John O’Sullivan, who Anthony has banned.
But I will provide the same think I posted on my blog to John’s Comment:
“Without greenhouse gases, the UPPER atmosphere would indeed be much warmer, but the LOWER atmosphere would be much cooler. Quit confusing the two, like you did with your recent misinterpretation of the NASA upper-atmosphere study. Itā€™s been almost 50 years since Manabe and Strickler (1964) showed how the upper atmosphere is cooled by CO2, which is PART OF the greenhouse effect. Nothing new there.
Adding more insulation to your house makes the outside of your house cooler, and the inside warmer (in winter). Greenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, thus making the warm side warmer, and the cool side cooler.”

April 24, 2013 12:15 pm

Roy Spencer says: “Remember, 10,000,000 K temperatures are created at the core of the Sun with heating rates less than what the human body produces (per unit mass).”
Good god man. You forgot about gravity. Fusion is not the GHE.
Let me repeat this for everybody: Nuclear fusion is not the GHE. The solar core and nuclear fusion have no relevance or support to the supposed GHE. It is arguments like this which should destroy any credibility to the faith…oh wait. šŸ™‚

April 24, 2013 12:16 pm

Remember, 10,000,000 K temperatures are created at the core of the Sun with heating rates less than what the human body produces (per unit mass).” ~Roy
Uh, the power leaving the core is low because everything is around the same temperature deep within the Sun.
You wouldn’t want a chunk of fusing core material dropped next to you, it wouldn’t be fun for anyone but Superman, who could shrug off a thermonuclear bomb.
____________________________
For the record, the temperatures aren’t created by the heating rates you’re talking about, the temperatures are created by the mass of the rest of the star compressing the core material enough for it to undergo fusion.
If a hunk of stellar material were replaced with non-fusing material at around the same temperature as your body it would be crushed rather dramatically until it reached a temperature where radiation pressure could support the rest of the star above it.

Gary Hladik
April 24, 2013 12:19 pm

Anthony, you weren’t getting enough traffic at WUWT, so you bring this up again? šŸ™‚
Maxā„¢ says (April 24, 2013 at 11:36 am): “Letā€™s say you have a pot of water that is being boiled (~373 K) and a block of ice (~273 K) radiating towards each other.”
Max, your thought experiment is quite complicated, involving phase changes, atmospheric cooling & heating via conduction/convection, etc.
What’s your take on the simpler thought experiment described here:
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/

A C Osborn
April 24, 2013 12:26 pm

Gary Hladik says: “Whatā€™s your take on the simpler thought experiment described here:”
Only someone who believes in the Back Radiation theme would say that because it gets cooler in the box than the surrounding atmosphere that Proves Back Radiation because without it the box would go down to 3 degrees K.
To most people it would suggest that the back Radiation is actually cooling the box.
So what was your own take on it, or did you just accept Roy’s conclusion without actually thinking about it?

April 24, 2013 12:29 pm

“Adding more insulation to your house makes the outside of your house cooler, and the inside warmer (in winter).”
Preventing drafts is not what a gas can do…a gas is a gas and is drafty. CO2 doesn’t prevent drafts. Putting something beside a heater simply heats up the thing…not the heater. Heat simply conducts into the thing.

A C Osborn
April 24, 2013 12:35 pm

Steven Mosher says:
April 24, 2013 at 10:45 am
The real debate is over how much warming GHGs will cause.
That is the debate. That science is not settled. Skeptics with smart arguments will get published, see Nic lewis and Troy masters. Its the central debate..
Absolute B***Sh** as usual, if you can’t get an absolute measure of how much is natural and how much is supposed CO2 or other GHGs you can never ever work out the sensitivy, it is all complete guesswork, hence the “pause” in warming, which by any other name is quickly going in to cooling.
Even Anthony has to show all the Cold Records being set, but where are the explanations, where is it being shouted from the MSM roof tops.
Even the Met Office has had to back track yet again.

A C Osborn
April 24, 2013 12:43 pm

Isn’t it amazing what difference a few years can make, it was all about CO2 is the control knob, it overpowers everything.
Now we have it is Coal burning causing cooling, the Oceans eating the heat, it,s black dust, anything but CO2 doesn’t do what they said it does.

Gary Hladik
April 24, 2013 12:46 pm

[snip we don’t link to that book promoting website -mod]

Stephen Richards
April 24, 2013 12:56 pm

Roy’s example is appalling as many of the comments here have shown.
REPLY: no, what is appalling is the principia fools that believe this rubbish about the greenhouse effect being “bogus” as they say right on their web page.

This sort of stuff is the rational skeptics worst enemy. Like arguing with vaxxers – Anthony

Alberta Slim
April 24, 2013 1:00 pm

Two questions;
1. What happens if the house has the R15 [or whatever] spun glass insulation taken out and R15 sealed bags of CO2 put in?
Will the house [assuming constant heat supply] cool off faster or slower?
2. I thought gasses[atmosphere] expanded to absorb extra heat.
The greenhouse gets hotter because the air cannot expand.
What stopping the atmosphere from expanding?

Jim S
April 24, 2013 1:00 pm

My understanding is that when you ā€œaddā€ insulation to a wall cavity, itā€™s the billion pockets of isolated, trapped air (gasses) that slow down the convection of heat from one side of the wall to the other. The insulating material actually serves as a ā€œthermal bridgeā€ that allows ā€œhigherā€ conductive heat-loss across the cavity ā€“ but the conductive heat-loss is more than off-set by the reduction in convective heat-loss.
The presence of the solid, insulation material actually speeds up heat-loss.

Gary Hladik
April 24, 2013 1:07 pm

Joseph E Postma says (April 24, 2013 at 10:41 am): “…say, a AAA battery and tiny resistor could be used to generate 5000K inside a shell, and then you could smelt some steel or something with a AAA battery.”
This reminds me of Willis Eschenbach’s “steel greenhouse”:
http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/
http://wattsupwiththat.com/2013/02/06/the-r-w-wood-experiment/

more soylent green!
April 24, 2013 1:12 pm

More insulation keeps my house cooler in the summer, when I’m running the air conditioner, or warmer in the winter, when I’m running the heat.
So adding more insulationing to my home can and does keep it cooler.

Gary Hladik
April 24, 2013 1:17 pm

A C Osborn (April 24, 2013 at 12:26 pm), I think you’re confusing the “Yes, Virginia” thought experiment with something else. The “Yes, Virginia” thought experiment is the one with the chilled vacuum chamber containing two facing plates, one heated, one not.
I have thought quite a bit about the “Yes, Virginia” thought experiment, which I think distills the issue of so-called “back radiation” down to its essence, in contrast to Max’s messy thought experiment. I’ve also thought a lot about the criticism of this thought experiment by the so-called “slayers”, and I find their arguments…unconvincing, to put it politely.
Mods: My apologies for linking to a prohibited site. I’ll try to remember that in future.

April 24, 2013 1:18 pm

Sorry if this is a double post.
The solar input occurs only over a single hemisphere. The intensity distribution of this input goes as the cosine function about the solar zenith. The integrated average value of this intensity has a temperature value of +49C. Solar power has an average heating power of +49C continuously on the sunlit hemisphere. This is very different from the flat-earth approximation of -18C. This is a rational consideration, because +49C continuous input is physically real, and will have a very different effect on the system than a global -18C. +49C can do a lot of things that -18C can’t.

davidmhoffer
April 24, 2013 1:19 pm

Joseph E Postma says:
April 24, 2013 at 10:29 am
Insulation in your house prevents draft leakage and convective loss to the outside, by trapping material molecules
>>>>>>>>>>>>>>>>>>
Wow. Three absolutely false statements in a single sentence. Way to go Joe.
REPLY: Yeah, I wanna see those molecule traps. Joe needs to take a time out before he claims the sun revolves around the Earth or something – Anthony

April 24, 2013 1:21 pm

Max is correct I suppose. Nobody never measured 324W coming from the night sky AFAIK. And whoever believes, that this “back ratiation” from “greenhouse gases” by itself increases our night time temperature from -150C (like on the Moon) to +15 here, well, keep believing. Simple thermal retention of the N2+O2 explains bulk of the (wrongly calculated) “+33K” myth. It is as foolish as forever trying to calculate doubling sensitivity from CO2 and latest positive phase of AMO. You get some numbers in all cases; but it is just virtual number.

Roy Spencer
April 24, 2013 1:21 pm

Jim S, it doesn’t matter to the point being made whether we are talking about conduction, convection, or radiation. ANYTHING that reduces heat loss from a heated object can increase its temperature. And that “anything” is usually at a COOLER temperature than the heated object itself. No laws of thermodynamics are broken, as alleged by the slayers.

April 24, 2013 1:28 pm

By trapping material molecules I was of course referring to preventing drafts, which prevents warm air escape to the outside. Warm air is composed of molecules. Preventing drafts is of course one of the main functions of insulation, in your house.

davidmhoffer
April 24, 2013 1:33 pm

Juraj V. says:
April 24, 2013 at 1:21 pm
Max is correct I suppose. Nobody never measured 324W coming from the night sky AFAIK.
>>>>>>>>>>>>>>>
I suggest you read what has been measured:
http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/

April 24, 2013 1:35 pm

” it doesnā€™t matter to the point being made whether we are talking about conduction, convection, or radiation. ANYTHING that reduces heat loss from a heated object can increase its temperature.”
It only increases “the” temperature, it doesn’t increase the temperature of the source. Real greenhouses aren’t heated to higher than the source, nor does the insulation in a house cause an increase of the temperature of the gas flame in the furnace.
Hereā€™s a physics textbook example of trapping radiation from an active source inside a cavity.
http://books.google.no/books?id=PfadZy35Wh0C&pg=PA442&lpg=PA442&dq=blackbody+sphere+surrounded+shell+radiation&source=bl&ots=TDbus0Dwu4&sig=3Aj5S6SlUB55MY9ry_MBXzTBm84&hl=no&sa=X&ei=d6VxUfOHC8Os4ASGsYCACQ&ved=0CEUQ6AEwBDgU#v=onepage&q=blackbody%20sphere%20surrounded%20shell%20radiation&f=false
Problem 27.: ā€œA sphere of radius R is maintained at a surface temperature T by an internal heat source (Figure 3). The sphere is surrounded by a thin concentric shell of radius 2R. Both objects absorb and emit as blackbodies. Show that the temperature of the shell is T/(8^1/4) = 0.595 T. (Hint: Both the inner and outer surfaces of the shell emit as blackbodies.)ā€
The surface temperature of the sphere is maintained by its internal heat source. And thatā€™s that. There is no “then the sphere has to heat up some more because radiation is trapped” etc etc. Of course, trapping photons in a cavity from a thermal source only produces a blackbody spectrum; it doesn’t cause the photons to change frequency to higher temperature. Photons are bosons and can pile on top of each other without caring, aside from equal constructive and destructive interference which results in no net change. This is very different than the behaviour of trapping more and more molecules into a confined space.
REPLY: and none of this matters – Anthony

Gary Hladik
April 24, 2013 1:35 pm

Alberta Slim says (April 24, 2013 at 1:00 pm): “Two questions;
1. What happens if the house has the R15 [or whatever] spun glass insulation taken out and R15 sealed bags of CO2 put in?”
If the R-value of the new material is the same as the old, shouldn’t it insulate the same?
http://en.wikipedia.org/wiki/R-value_(insulation)
“2. I thought gasses[atmosphere] expanded to absorb extra heat.
The greenhouse gets hotter because the air cannot expand.
What stopping the atmosphere from expanding?”
Nothing. Like regular houses, greenhouses generally aren’t airtight. Air convection is inhibited, not prevented entirely. Note that heating your house with, say, a gas heater doesn’t raise the house internal pressure and make your ears pop.

Roy Spencer
April 24, 2013 1:37 pm

ALL:
It is important to remember that there is NO WAY to determine the temperature of anything based upon the rate of energy input alone, for example the Earth absorbing an average of ~240 W/m2 from the Sun. Temperature is a function of BOTH energy input (typically not temperature dependent) AND energy loss (typically VERY temperature dependent), neglecting issues related to heat capacity which mainly affect the time required for the system to equilibrate. The temperature of anything heated will increase until the rate of energy *loss* equals the rate of energy *gain*. So, temperature can be increased by increasing INPUT, or decreasing OUTPUT.

April 24, 2013 1:42 pm

“The temperature of anything heated will increase until the rate of energy *loss* equals the rate of energy *gain*”
When energy input equals energy output, then the system is in equilibrium. In this state the system will not be hotter than the input. Of course, the input isn’t -18C, but is +49C, on the Earth, and, the Earth isn’t hotter than this. In the textbook example of trying to control the output radiatively, this only creates a blackbody spectrum inside the cavity which is created to do that. When you shine a flashlight at a mirror, into a mirror, or two flashlights at each other with a good seal between the, it doesn’t cause the flashlights to shine brighter. Equal temperatures can not heat each other.
REPLY: Joe, take a time out. This is becoming absurd with your thread bombing – Anthony

Sidney Somes
April 24, 2013 1:44 pm

The position that Co2 causes heating at the Earth’s surface seems to be contradicted by millions of years of temperature records showing that temperature precedes Co2 increases, by about 800 years. The so-called “runaway greenhouse effect” would have happened millions of years ago when Co2 was ten times the level that it is today. So perhaps our greenhouse physics is missing something, like DATA.

April 24, 2013 1:49 pm

Ok, have it be two black bodies of arbitrary material, large enough that the area we’re interested in is radiating towards the other body, one at 373 K and one at 273 K, radiating towards each other.
I’m confident that the radiation field between them can be described by the following equation:
P = ĪµĻƒA(Thā“ ā€“ Tcā“)
rather than:
P = ĪµĻƒA(Thā“ + Tcā“)
max@Funktastic:~$ calgebra
>>> (5.6703*10^-8)*(373^4 – 273^4)
782.63220248
>>> (5.6703*10^-8)*(373^4 + 273^4)
1412.55397668
Both of those can not be right, and if the colder black body raises the temperature of the warmer black body then the second one is correct, which would be roughly the same power as that emitted by a 297.28 K black body.
If the radiation from the colder body is added to that from the warmer body, that is akin to a block of ice raising the temperature of a pot of nearly boiling water by over 24 K!

April 24, 2013 1:49 pm

Anthony I am not thread bombing, I am responding to comments, and on topic. I am being perfectly pleasant. Cheers.

John West
April 24, 2013 1:54 pm

Obviously weā€™re still having trouble with the concept of NET heat exchange despite Max’s efforts:
The surface radiates as described by the Stephan-Boltzmann law:
Surface radiation = ĪµĻƒ T^ 4
The atmosphere radiates as described by an empirically derived equation:
Down-welling atmospheric radiation = (1+ KC^2) x8.78 E ā€“ 13T^5.852 x RH^0.07195
Therefore the NET heat loss from radiation of the surface is:
NET Radiation = [ĪµĻƒ T^ 4] ā€“ [(1+ KC^2) x8.78 E ā€“ 13T^5.852 x RH^0.07195]
Therefore, as long as [(1+ KC^2) x8.78 E ā€“ 13T^5.852 x RH^0.07195] is greater than 0, the NET radiation MUST be LESS THAN the ā€œgrossā€ (or Stephan-Boltzmann calculated) radiation that would be the NET if there were no radiation from the atmosphere (i.e. no GHE). This is measurable, known, understood stuff; like the world is round(ish).
Please read these 5 pages for details:
http://www.asterism.org/tutorials/tut37%20Radiative%20Cooling.pdf

Robert Austin
April 24, 2013 1:58 pm

The best way to understand the role of greenhouse gases is to do a thought experiment as to the temperature structure without greenhouse gases including water. I such a case, ignoring airborne aerosols, cooling of the earth would be occur only by black or grey body radiation from the earth’s surface. So the energy balance equation would show a much colder earth’s surface and there would be no significant lapse rate structure to the atmosphere. On the other hand, a significant portion of outgoing energy is radiated to space from the top of the troposphere in an atmosphere with greenhouse gases and the temperature at the top of the troposphere is much lower than the surface temperature. So the greenhouse gas containing atmosphere is like an insulating wall in that a greenhouse wall radiates from the exterior surface of the wall whereas a non greenhouse wall would be akin to a thin membrane wall essentially radiating from the interior wall surface.

CodeTech
April 24, 2013 2:06 pm

I have triple glazed windows in my house. The gas they put inside the windows is Nitrogen. Seems to me if CO2 was a great insulator, they’d be using that instead…

jono1066
April 24, 2013 2:06 pm

Sorry but I couldnt help it.
take house wall with a cavity down the middle, standard thermal calcs for houses says `and then ignore the thermal R value (U value in good old Blighty) of the outer wall` this is specifically due to the radiatiive value of the surface and the convective value of the air in the cavity once heated above ambient and the normal movement of air through the cavity due to the house `breathing`. This is true in both directions, bricks heated from the outside by the sun or from the house heated from inside. Just think of it as a plate heat exchanger. Reducing the radiative value (emmissivity) of the cavity surface (polish it) and preventing to a great degree simple convection/pressure driven air flow allows the walls (lets say normal heating from the inside) to change the temperature gradient across them, hence the cavity side of the ash block increase in temperature towards but never past the temperature inside the house, Thermal stabilisation will occur at some point with the cavity side of the inner wall always below the temperature inside the house. Convection is the biggest problem for double glased windows hence the normal limit of 19 to 22m between inner and outer pane after which thermal resistance goes down. Styrofoam beads approx 3 to 5mm diameter are very highly packed closed cell honeycombe nature and convection is very difficult to initiate within such a small volume CO2 filled eps would not noticeably change the insulating/resistance value. Adding insulation to your house allows you to maintain a specified and stable thermal gradient from any one side to the other with a lower thermal input. Incorrectly applied insulation works directly and absolutely measurably through the `placebo` effect.
Did I get the job ?

Kelvin Vaughan
April 24, 2013 2:07 pm

Just build 2 greenhouses of identical size. Enhance the CO2 in one to 1000ppm. Keep them at the same temperature over a year. Check if the enhanced CO2 greenhouse used less KWh over the year.

davidmhoffer
April 24, 2013 2:11 pm

Kelvin Vaughan says:
April 24, 2013 at 2:07 pm
Just build 2 greenhouses of identical size. Enhance the CO2 in one to 1000ppm. Keep them at the same temperature over a year. Check if the enhanced CO2 greenhouse used less KWh over the year.
>>>>>>>>>>>>>>>>>
Sure, that would provide measurable data. Of course to get measurable data large enough to actually measure with even very accurate instrumentation, you’d need to build the greenhouses several kilometers tall.

crosspatch
April 24, 2013 2:14 pm

In the case of the climate system, most sunlight is absorbed at the surface, so it is ā€œheated from withinā€.ā€
So CO2 does not act as an Insulator at all because if it did it would prevent the Radiation from heating surface in the first place.

But the sun does emit infrared that does get blocked from reaching the surface by “greenhouse” gases. It works in both directions.

April 24, 2013 2:15 pm

Just in case anyone is curious, with my cleaned up example above (two black bodies, 373 K and 273 K, respectively), the radiation field from the warmer to colder body is roughly the same as that emitted by a black body at 342.758 K, and while I can’t quite justify it as being certain, I would not be surprised if the two bodies reached equilibrium around that temperature, i.e. with the warmer body being cooled and the colder body being warmed.
Seems reasonable enough for the pot of nearly boiling water + block of ice as well, getting warm water wouldn’t seem odd, but getting hotter water than you started with by adding ice would be a very interesting result, wouldn’t it?

crosspatch
April 24, 2013 2:16 pm

And it is true that per unit of mass, the sun produces very little heat. It’s about that produced by the average compost heap. It’s just a very big compost heap and so it produces a lot of total heat that takes a long time to get to the surface.

davidmhoffer
April 24, 2013 2:21 pm

CodeTech says:
April 24, 2013 at 2:06 pm
I have triple glazed windows in my house. The gas they put inside the windows is Nitrogen. Seems to me if CO2 was a great insulator, theyā€™d be using that insteadā€¦
>>>>>>>>>>>>>>>>>
Seems to me they’d put in whatever gas experimentation proved provided the best result. Probably argon btw, not nitrogen. What they used and why they used it says absolutely zero about co2 as an insulator other then they found something they think works better.

Selgovae
April 24, 2013 2:29 pm

Gary Hladik mentions Roy Spencer’s “Yes, Virginia” thought experiment. I have a question about this that’s been been bugging me for some time. Perhaps someone can put me out of my misery. There appear to be two views about “back radiation”: that energy can radiate from a cooler material (atom I guess) to a warmer one or that it can’t. I’d always held the latter view, for no particular reason other than that’s how I thought things worked. But in either case, isn’t the net effect more or less the same? This is how I interpret things:
Case 1: Radiation flows between both warmer and cooler plates. The effect is the result of the net radiation flow, and so the warmer plate becomes warmer than it would have been if the cooler plate wasn’t there.
Case 2: Radiation only flows from the warmer plate to the cooler plate. But at some point, the atoms at the surface of the cooler plate reach the temperature of the warmer plate. The warmer plate can no longer radiate to the cooler plate until the cooler plate’s surface atoms cool again, presumably by a combination of conduction within the plate and radiation from the other side. Those times when the warmer plate can’t radiate lower the rate of heat loss from the warmer plate, and so the warmer plate becomes warmer than it would have been if the cooler plate wasn’t there.
Gently, please!

Gary Hladik
April 24, 2013 2:32 pm

Maxā„¢ says (April 24, 2013 at 1:49 pm): “Ok, have it be two black bodies of arbitrary material, large enough that the area weā€™re interested in is radiating towards the other body, one at 373 K and one at 273 K, radiating towards each other.”
Max, what’s your take on the “Yes, Virginia” thought experiment?
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/

Greg House
April 24, 2013 2:39 pm

“Spencer slays with sarcasm”
=======================================================
Sarcasm is indeed there, but there is also a problem with the scientific point.
First, the analogies with “cold keeps warm”, blankets, houses etc. a very misleading. A blanket would keep us cold, not warm, if the air temperature outside the blanket is higher, than our body temperature.
Second, even on the theoretical level the concept of “back radiation warming” is physically absurd and therefore impossible, because if the source has been kept initially at a constant temperature, then “back radiation warming” will inevitably lead to an endless mutual warming of the source of radiation and the thing that provides back radiation, without an additional input of energy, which is physically absurd. And everybody can make a simple experiment: just stand in front of the mirror and enjoy the “back radiation warming”. Must be 33C or more, if the “greenhouse effect” as presented by the IPCC exists. Please, not too close to the mirror, to avoid the effect of suppressed convection.
Coming back to the theoretical level, it goes like that. You have initially a body kept at a certain temperature by itā€™s internal source of energy. Now you put another colder body at the absolute zero temperature, let us say, in vacuum close to the warm body.
The warmer body will start warming the colder body immediately. Then, according to the ā€œback radiation warmingā€ concept, the back radiation from the colder body will increase the temperature of the warmer body. Actually, already on this stage we should start screaming and crying ā€œhow come?ā€, but let us proceed. So, the now even warmer warm body will warm the colder body even stronger, and the colder body will repay by sending even more back radiation to the warmer body, thus increasing itā€™s temperature even further. The warmer body will get warmer again. So will the colder body in turn. And so on.
This is the mutual endless warming without any additional input of energy I meant previously, and I hope it is easy to understand how physically absurd it is. This proves that the concept of ā€œback radiation warmingā€ is physically absurd. The ā€œgreenhouse effectā€ as presented by the IPCC can not exist.
REPLY: No, it doesn’t. This argumentum ad infinitum that the slayers push about “cold can’t heat warm” is the absurd part, and nothing but a strawman argument. The greenhouse effect (a misnomer) isn’t so much about reheating the atmosphere with reradiated LWIR from top to below, it is about slowing the progress of LWIR to the top of the atmosphere. Without CO2 or other GHG’s the LWIR would proceed quickly to space, and the Earth would cool faster, and have a lower average temperature at night. The “backradiation” has a limited scope of effect, but it is there as part of the trasnfer process. GHG’s are not unidirectional, they are omnidirectional in the LWIR process.
GHG’s act as a LWIR transfer regulator from the surface to the top of the atmosphere where it is radiated into space. They slow the transfer. Without GHG’s nightitme temperature would drop quickly as LWIR is lost directly to space.
A simple proof of this has to do with the differences between nighttime temperatures between moist and dry climates at the same latitude/altitude where water vapor is the dominant GHG. See Knappenberger et al 1996 for example.
Dry desert climes cool faster at night and have a greater diurnal temperature variation than moist climes. Water vapor is the difference. CO2 has a smaller effect, but an effect nonetheless. – Anthony

Mark Bofill
April 24, 2013 2:49 pm

Selgovae says:
April 24, 2013 at 2:29 pm
————–
Well, look at it this way. Radiation has no way of knowing the temperature of bodies out in existence someplace it may eventually encounter. The sun shines outwards in all directions equally (possibly some random variations, I’m not a solar expert, but darn close I think), regardless of whether or not there is a hotter surface out there someplace in space that some of its radiation may eventually encounter.
So yes, I’d expect radiation to propagate from a source through space with complete disregard for the temperature of the object it may eventually hit.

Noelene
April 24, 2013 3:08 pm

So there is a consensus on the greenhouse effect?

Kristian
April 24, 2013 3:15 pm

davidmhoffer says, April 24, 2013 at 2:21 pm:
“Seems to me theyā€™d put in whatever gas experimentation proved provided the best result. Probably argon btw, not nitrogen. What they used and why they used it says absolutely zero about co2 as an insulator other then they found something they think works better.”
http://gaia.lbl.gov/btech/papers/29389.pdf
Interesting quotes:
ā€œLow-emittance coatings are much more effective at reducing infrared radiation heat transfer than IR absorbing gasses. Gasses for gas-filling should be chosen for their low conductivity and high kinematic viscosity in order to effectively reduce conductive/convective heat transfer. The effective use of infrared absorbing gasses is thus limited to horizontal windows heated from above, or to thin gaps where low-emittance coatings cannot be used.ā€
ā€œ(ā€¦) even though the gas absorption/emission dampens the natural convection, the absorbing gasses being used as gas-fills have lower kinematic viscosities than air and some of the other low-conductivity gasses (argon, krypton) being used in windows (fig. 6). And, from Glaserā€™s results for vertical windows it can be seen that the convective transfer becomes significant at around 9 mm for SF6, while there is practically no convective transfer through an air-filled window at gapwidths up to 20 mm under these conditions. In fact, air outperforms SF6 at gapwidths greater than 9 mm in a vertical window and the benefits from infrared absorption by SF6 have been negated by the magnitude of the convection.ā€
ā€œFig. 3 shows that the effect of the infrared radiation properties of CO2 is unnoticable (ā€¦)ā€

April 24, 2013 3:18 pm

“Max, whatā€™s your take on the ā€œYes, Virginiaā€ thought experiment?” ~Greg
The heated bar will raise the temperature of the unheated bar, adding another object to be warmed reduces the energy density, there is no way it can result in a higher temperature without creating energy.
150 Fahrenheit is 338.7 K and would emit 746.2~ W/m^2 as a black body
100 Fahrenheit is 310.9 K and would emit 529.7~ W/m^2 as a black body
max@Funktastic:~$ calgebra
>>> (5.6703*10^-8)*(338.7^4)
746.219894665
>>> (5.6703*10^-8)*(310.9^4)
529.771907497
>>> (5.6703*10^-8)*(338.7^4 – 310.9^4)
216.447987168
>>> (5.6703*10^-8)*(338.7^4 + 310.9^4)
1275.99180216
From the heated bar:
746.2 to shell, 216.4 to unheated bar
From the unheated bar:
529.7 to the shell, -216.4 to heated bar
If the power supply is constant then consider the case where the unpowered bar begins at the same temperature as the powered bar.
I see no reason why a body at given temperature should raise the temperature of another body at the same temperature, so accordingly I see no reason why a cooler body should raise the temperature of a warmer body.

OldWeirdHarold
April 24, 2013 3:24 pm

“Roy Spencer says:
April 24, 2013 at 12:02 pm
Joe Postma, yes you probably CAN design a system for generating 5,000K temperatures using just a AAA battery. All you need to do is resistance heat a very small object with little heat capacity, and insulate it very well. The problem would be the insulation, because it, too,would heat up.”
=====
It’s called an incandescent flashlight.

Reply to  OldWeirdHarold
April 24, 2013 3:29 pm

“Itā€™s called an incandescent flashlight.”
It wasnā€™t about a AAA battery lighting a flashlight filament to 5000K. Thereā€™s nothing wrong with that. It is about the resistor (lets use the filament) becoming even hotter still if you shine the flashlight at a mirror. This does not increase the brightness (temperature) of the filament. Trapping the radiation does not increase the temperature of the filament.

Gary Hladik
April 24, 2013 3:24 pm

Selgovae says (April 24, 2013 at 2:29 pm): “This is how I interpret things: [snip]”
Mark covered the basic problem above (i.e. postulating clairvoyance by inanimate–or animate–objects), but here are a couple more thoughts on your case 2:
a) The temperature of the unheated plate (“plate 2”) never reaches that of the electrically heated plate (“plate 1”). Plate 2 is receiving energy from plate 1 on one side but radiating energy from two sides.
b) Plate 1’s radiation is based on its temperature
http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
so if its radiation decreases, its temp must have decreased. But as you know the temp doesn’t decrease, so your postulated decrease in radiation can’t happen.

pat
April 24, 2013 3:26 pm

Spencer “with sarcasm”, Ban Ki-Moon without satire!
25 April: Bloomberg: UN Says Clean Energy Funding Too Low to Halt Climate Harm
UN Secretary General Ban Ki-moon said the funds flowing to renewable power and efficiency arenā€™t sufficient to avert environmental calamities and that investors must move more quickly to back new energy technologies.
ā€œClimate change is a threat to economies large and small and to the stability of the ***global financial system***,ā€ Ban said today in a speech at a Bloomberg New Energy Finance conference in New York. ā€œThe climate clock is ticking. The longer we delay the greater the cost. We only have one planet Earth. We have no plan B.ā€ā€¦
The UN has led an effort to bring more than 190 nations together to develop carbon-reduction policies, and Ban said he received a personal commitment from U.S. President Barack Obama two weeks ago to work toward a binding climate agreement in 2015.
ā€œHeā€™s assured me that the U.S. will lead by example,ā€ Ban saidā€¦
He told bankers and fund managers gathered at the conference they must ā€œlead by exampleā€ and that heā€™s doing the same.
ā€œI have been dutifully, faithfully turning off lights in my hotel,ā€ Ban said. ā€œSometimes itā€™s very difficult these days, all different hotels have a very different system of lighting. Normally I stay in a suite so there are many, many lights.ā€ā€¦
http://www.bloomberg.com/news/2013-04-24/un-says-clean-energy-funding-too-low-to-halt-climate-harm.html
——————————————————————————–

davidmhoffer
April 24, 2013 3:29 pm

Kristian;
ā€œFig. 3 shows that the effect of the infrared radiation properties of CO2 is unnoticable (ā€¦)ā€
>>>>>>>>>>>>>>>>
I wouldn’t expect it to be noticeable. Not at that gap width. The effect we’re talking about in terms of the atmosphere is from an air column 14 km high. Further, the article you quote refers to suppression of convection and conduction as being the primary factors for the application, and so they recommend materials that perform the best against those parameters.
Again, not using CO2 in a window has nothing to do with how it behaves in a 14 km high atmospheric air column and everything to do with what delivers the best result in a window. I could just as easily argue that they don’t build cargo ship hulls out of wood as proof that wood doesn’t float. It does float, there’s just other ways to float your boat that are more effective.

Greg House
April 24, 2013 3:32 pm

REPLY: No, it doesnā€™t. This argumentum ad infinitum that the slayers push about ā€œcold canā€™t heat warmā€ is the absurd part, and nothing but a strawman argument. The greenhouse effect (a misnomer) isnā€™t so much about reheating the atmosphere with reradiated LWIR from top to below, it is about slowing the progress of LWIR to the top of the atmosphere. Without CO2 or other GHGā€™s the LWIR would proceed quickly to space, and the Earth would cool faster, and have a lower average temperature at night.
GHGā€™s act as a LWIR transfer regulator from the surface to the top of the atmosphere where it is radiated into space. They slow the transfer. […] ā€“ Anthony

==============================================================
Anthony, about “the slayers”, I do not refer to any slayers, I did not even read their book, however I read some of their articles. I do not care who exactly said what and what else right or wrong they said etc., I only like good argumentation. Only the quality of the argumentation counts.
Your second point is about what “greenhouse effect” means. This issue is a little bit confusing, and not because it is a misnomer, but because there are a few different versions around. When I say “the greenhouse effect as presented by the IPCC”, I do it specifically to avoid confusion, like in my previous comment. The point is, that the “greenhouse effect” as presented by the IPCC is exactly about warming by back radiation: http://www.ipcc.ch/publications_and_data/ar4/wg1/en/faq-1-3.html. Then I say that the warming by back radiation is physically impossible, therefore the “greenhouse effect” as presented by the IPCC does not exist.
If some people have other version of “greenhouse effect”, then we could discuss it, but my point is that the “greenhouse effect” as presented by the IPCC does not exist. I hope you can agree with me on that.

April 24, 2013 3:36 pm

” Itā€™s called an incandescent flashlight.” ~OldWeirdHarold
Tungsten filaments reach around 3,600 K, not quite 5,000 K.
I don’t know of a material that will survive at 5,000 K and remain solid, the surface of the sun is around 5,700~6,000 K for comparison.

April 24, 2013 3:42 pm

Kelvin Vaughan says:
April 24, 2013 at 2:07 pm
Just build 2 greenhouses of identical size. Enhance the CO2 in one to 1000ppm. Keep them at the same temperature over a year. Check if the enhanced CO2 greenhouse used less KWh over the year.
Wrong.
The “greenhouse’ effect has nothing whatsoever to do with these types of experiments.
Energy leaves earth Via one route and one route only: radiation to space.
Earth radiates to space from the ERL.
When you add GHGs you raise the ERL.
When you raise the ERL the earth radiates to space more slowly.
That results in a surface that is warmer than it would be otherwise.
This effect cannot be tests with woods like experiments EVER because the effect depends upon changing the ERL and no experimental set up or lab can duplicate the conditions required: the full atmospheric column.
in 20 or so simple slides
http://www.aos.wisc.edu/~aos121br/radn/radn/sld001.htm

April 24, 2013 3:42 pm

Noelene:
Your post at April 24, 2013 at 3:08 pm goes to the crux of this thread’s subject. It asks

So there is a consensus on the greenhouse effect?

The answer is YES but the consensus is not what the media tends to present.
Anyone who understands radiative physics knows
the radiative properties of greenhouse gases (GHGs; mostly water vapour but also carbon dioxide, methane, etc.) in the air provide a greenhouse effect (GE) without which the Earth would be much colder.
That is the overwhelming consensus and is irrefutable physics.
However, that fact does not – of itself – mean additional GHGs in the air will discernibly raise global temperature. Any such rise in global temperature will depend on the feedbacks in the climate system. Those who believe in anthropogenic (i.e. man-made) global warming (AGW) think feedbacks are strongly positive so will greatly increase any such temperature rise. But AGW-skeptics (or climate realists) either think the feedbacks are weakly positive so will have little effect, or they are negative so will reduce any such temperature rise.
If the feedbacks are weakly positive or negative then AGW would be so small as to be indiscernible because natural climate variability is much larger.
In other words, the argument about AGW is really an argument about the feedbacks.
However, there are a few people (often known as ‘Dragon Slayers’ or ‘Slayers’) who claim that radiative physics is wrong so the GE does not exist. Of course, all science is open to amendment – that is the nature of science – but the assertions made by the Slayers are preposterous.
Unfortunately, the Slayers are vociferous, and the ludicrous nature of their assertions brings discredit (i.e. guilt by assumed association) on all AGW-skeptics.
Spencer is an AGW-skeptic, and his sarcasm is aimed at the silly notions of the Slayers. Hence, the title of this thread is “Spencer slays with sarcasm”.
I hope this clarifies the matter.
Richard

April 24, 2013 3:50 pm

“When you add GHGs you raise the ERL.
When you raise the ERL the earth radiates to space more slowly.”
Missing hot spot. Earth is radiating more these days, not less. For something warming to be emitting less radiation is a violation of the Stefan-Boltzmann Law. Only if you reduce the emissivity can the temperature be raised. CO2 does not reduce emissivity. O2 and N2 have zero emissivity and therefore trap and hold heat much better than any other gas and hold temperature much better than otherwise. CO2 collisionally absorbs energy from O2 and N2 and then causes it to be lost.

Alberta Slim
April 24, 2013 3:52 pm

Greg House says:
April 24, 2013 at 2:39 pm
ā€œSpencer slays with sarcasmā€…………………………
I totally agreewith Greg. “Back-welling” or “back-radiation” from CO2 is not possible.
It may help slow thing down. Also, no one [Dr Spencer?] has told me why the atmosphere will
not expand with extra heat from the earth’s suface. This would satisfy the “Conservation of Energy” with no increase in Temperature
Question 3. If in the thermograph, above, the triple glazed windows were filled with CO2,
what color would we observe? If CO2 is “backwelling” into the house the color
should be other than red. No???

Greg House
April 24, 2013 4:07 pm

Alberta Slim says (April 24, 2013 at 3:52 pm): “I totally agreewith Greg. ā€œBack-wellingā€ or ā€œback-radiationā€ from CO2 is not possible.”
===========================================================
I did not say that ā€œback-radiationā€ from CO2 was not possible. My point was that back radiation would not affect the temperature of the source.

Gary Hladik
April 24, 2013 4:12 pm

Maxā„¢ says (April 24, 2013 at 3:18 pm): “I see no reason why a body at given temperature should raise the temperature of another body at the same temperature, so accordingly I see no reason why a cooler body should raise the temperature of a warmer body.”
Thanks for taking a look at the “Yes, Virginia” thought experiment, Max. Your initial numbers are close to what “Nullius In Verba” calculated for the one-plate case at a certain Site-Which-Must-Not-Be-Linked, although he went on to calculate the equilibrium temperature of the two-plate system, in which the temp of the heated bar is higher.
Radiative physics as illustrated by the “Yes, Virginia” thought experiment is fundamental to modern science, see numerous textbook examples in the comment thread under Willis’s R W Wood Experiment article:
http://wattsupwiththat.com/2013/02/06/the-r-w-wood-experiment/
In fact, if the “Yes, Virginia” version of radiative physics is overturned, it would be worth a Nobel Prize at least, and would earn wealth, fame, and adulation for whoever brought about this scientific revolution. All from performing a relatively simple thought experiment “for real” and getting a different result from what Dr. Spencer predicts.
My question is, since this is a relatively simple experiment to do (no particle accelerators required), why have none of the so-called “slayers” performed it (or one like it) and taken their place alongside Newton, Planck, Einstein, and Hawking?
Any thoughts, Max?

KevinK
April 24, 2013 4:15 pm

Max wrote (re ā€œYes Virginiaā€¦..ā€);
ā€œIf the power supply is constant then consider the case where the unpowered bar begins at the same temperature as the powered bar.ā€
NOTE the word ā€œpower supplyā€, this is distinct from an ā€œenergy supplyā€, hence the distinct
nomenclature.
Yes a power supply will cause (under the very best of thought experiment conditions) the first bar to reach a higher temperature since itā€™s view factor to the colder vacuum chambers wall is reduced. HOWEVER it will consume more energy from the power supply while doing so. The power supply will have to provide enough energy to ā€œchargeā€ both metals bars with heat. It will draw more current from the energy supply (your friendly local electric company). Yes, I know electric utilities are often called XYZ electric POWER Company, but they are in fact selling you energy. If you do Dr. Spencerā€™s thought experiment you should expect a higher energy bill.
If you replace the power supply with a battery (a store of energy, or energy supply) the first bar will reach a higher temperature (again under the very best of thought experiment conditions), but more energy will be drawn from the energy supply (since it has to ā€œchargeā€ both bars with heat). Thus the battery discharges more quickly. So the bars get warmer for a while, then start cooling down when the energy supply is exhausted. Which results in no increase in the average temperature.
1 bar results in a lower temperature for a longer time, 2 bars results in a higher temperature for a shorter time, same amount of energy used in both cases. No trapped heat, no extra energy, no net energy gains.
The sunlight arriving during one day is represented by the initial charge in the battery, once itā€™s gone, itā€™s goneā€¦.
UNITS do matter when doing energy budgets (w/m^2 are units of power density, NOT energy).
Engineers that routinely calculate ā€œnet energy gainsā€ either get fired or bumped up into management.
Cheers, Kevin.

Jim Brock
April 24, 2013 4:15 pm

Well, insulation does in a sense cool the house. It does transmit heat from the hot (inside) side of the insulation to the cool (outside) of the house. Some transmits more, some less. That is what the R factor measures. BUT it transmits less heat than the uninsulated version. This keeps the house warmer than the alternative, but still cooler than if it did not conduct heat at all (the posited but non-existent perfect insulator). Even vacuum bottles transfer heat from the hot side to the cool interior, just slower than the alternative.

Greg House
April 24, 2013 4:16 pm

Steven Mosher says (April 24, 2013 at 3:42 pm): “Earth radiates to space from the ERL. When you add GHGs you raise the ERL. When you raise the ERL the earth radiates to space more slowly. …”
=============================================================
This is not what the IPCC considers to be the “greenhouse effect” and what they base their policy recommendations upon. They mean that “greenhouse gases” warm the surface by back radiation. Apparently they do not support your “effect”.

JimF
April 24, 2013 4:18 pm

There certainly seems to be a lot of animosity in the comment train. I enjoyed Postma’s paper “The Model Atmospheric Greenhouse Effect” because I learned a lot about black body radiation, etc., and I fully subscribe to his idea that the Sun’s energy hitting the Earth’s surface isn’t something divided by four, but the whole thing (at some point on the surface).
My guess – based on my experience in geological arguments – is that both sides of this argument are blowing hot air at least in part. It further seems to me that a great deal of what passes for knowledge is simply stating one’s case, louder with each iteration. CAGW is utter BS in my mind, and the condemnation of CO2 by many is dangerous. If one believes the assessments of the amount of CO2 in the Earth’s atmosphere over time, and understands its role in the life as we now know it of this planet, then we are in a precarious position. Much of the CO2 endowment of Earth seemingly is now in a state that cannot benefit plants, and that ain’t good.

davidmhoffer
April 24, 2013 4:33 pm

Greg House;
They mean that ā€œgreenhouse gasesā€ warm the surface by back radiation. Apparently they do not support your ā€œeffectā€.
>>>>>>>>>>>>>>>>>>>>>
Back radiation increases the ERL. Same thing explained from a different perspective.

Gary Hladik
April 24, 2013 4:38 pm

KevinK says (April 24, 2013 at 4:15 pm): “If you replace the power supply with a battery (a store of energy, or energy supply) the first bar will reach a higher temperature (again under the very best of thought experiment conditions), but more energy will be drawn from the energy supply (since it has to ā€œchargeā€ both bars with heat). Thus the battery discharges more quickly.”
So if instead of using an electric outlet or a battery to heat the first plate, we use an embedded radioactive heat source
http://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator
the plutonium (for example) will decay more rapidly when the second plate is introduced?

davidmhoffer
April 24, 2013 4:38 pm

JimF says:
April 24, 2013 at 4:18 pm
There certainly seems to be a lot of animosity in the comment train.
>>>>>>>>>>>>>>>>
It is indeed unfortunate that the “information age” has given way to the “misinformation age”. In today’s world, it is mind boggling to me that the CAGW nonsense can live side by side with the “back radiation doesn’t warm things” nonsense. The actual physics seems to take a back seat to both arguments, and yes that generates some animosity.

richard verney
April 24, 2013 4:45 pm

crosspatch says:
April 24, 2013 at 2:16 pm
And it is true that per unit of mass, the sun produces very little heat. Itā€™s about that produced by the average compost heap. Itā€™s just a very big compost heap and so it produces a lot of total heat that takes a long time to get to the surface
///////////////////////////
Is that so? Per unit of mass USED? The issue is not how much mass the sun has, but rather how much of that mass it uses to produce the energy in which we all bask.
Is it not the position that sun only uses a little of its mass to produce its energy and thats why its still burning after 4.5 billion years and will burn for another 4.5 biliion years or so?

Greg House
April 24, 2013 4:45 pm

davidmhoffer says (April 24, 2013 at 4:33 pm): “Back radiation increases the ERL.”
========================================================
Again, this is not what the IPCC supports. You can invent whatever effect you want, but only the IPCC’s one is politically relevant, because governments and agencies refer to the IPCC reports and recommendations, not to davidmhoffer or Steven Mosher.
And again, the IPCC “greenhouse effect” is physically impossible, as I demonstrated earlier on this thread, so their proposed policy of cutting CO2 emissions is not justified by their reports. I do not think we should ignore that.

April 24, 2013 4:50 pm

UNITS do matter when doing energy budgets (w/m^2 are units of power density, NOT energy).” ~KevinK
In discussions of radiation W/m^2 is intensity, if we’re going to nitpick. Watts are power, or the rate at which energy is supplied per second, Joules are energy or work done.
Again though I will go back to the case of a heated or powered bar at 338 K (~150 Fahrenheit) and an unpowered bar at 338 K.
Will the presence of another body at the same temperature cause the temperature of the first body to increase?

April 24, 2013 4:53 pm

“and the ludicrous nature of their assertions”
It is not ludicrous to state that the distribution of solar energy is a cosine function about the solar zenith, and that this distribution has an integrated average of +49C, as the input. What is ludicrous is that people deny that the Earth is round and think that +49C vs. -18C doesn’t matter. I do detect some ludicrousness when it is said that the mere statement of these facts is supposed to be ludicrous.
The cosine function is real. An integrated average is real. Sunshine is really hot. You can’t average energy input into a geometry where it doesn’t exist. These are all rational facts. It is ludicrous to ridicule those of us who state them and want to explore what effect and changes it has on the usual assumptions. As we have shown, the changes are not insignificant. Both the wet and dry lapse rate can be calculated precisely without any reference to GHG radiation, for example. Latent heat itself holds the temperature higher than it would otherwise be, for example. This is science. It is ludicrous to ignore it.

Downdraft
April 24, 2013 4:53 pm

Which brings to mind the age old question, “How does a Thermos bottle know whether you put cold or hot things in it, and that it is supposed to keep cold things cold and hot things hot?” Maybe the Slayers can work on that one next.

Gary Hladik
April 24, 2013 4:55 pm

Greg House says (April 24, 2013 at 4:45 pm): “Again, this is not what the IPCC supports.”
Hi, Greg! I figured you’d show up sooner or later. Welcome.
Now it’s a party! šŸ™‚
“And again, the IPCC ā€œgreenhouse effectā€ is physically impossible, as I demonstrated earlier on this thread…”
Anthony refuted your argument verbally, scienceofdoom refutes it mathematically here
http://scienceofdoom.com/2010/05/08/radiation-basics-and-the-imaginary-second-law-of-thermodynamics/

tjfolkerts
April 24, 2013 4:58 pm

@ Greg April 24, 2013 at 2:39 pm
ā€œSo, you have initially a body kept at a certain temperature by itā€™s internal source of energy.ā€
Just to be more definite, make it a blackbody sphere far from any star, so the surrounding are already ~ 0 K. Furthermore, letā€™s make that heater 240 W for each square meter, so the temperature is 255 K (like the numbers for earth).
ā€œNow you put another colder body at the absolute zero temperature, let us say, in vacuum close to the warm body.ā€.
Letā€™s make that cold object a thin shell completely surrounding the sphere.
ā€œThe warmer body will start warming the colder body immediately.ā€
Yes, but all this time, the cool outer shell will be radiating less than 240 W/m^2 out to space (because it is colder than 255 K). This means energy will be building up inside the shell, warming the shell. Eventually the shell will warm to 255 K. At this point the shell will be radiating 240 W/m^2 and a steady-state will be achieved. (Actually, the shell must be slightly less than 255 K since it will have slightly larger surface area, but this is a minor correction).
ā€œActually, already on this stage we should start screaming and crying ā€œhow come [the inner object warms, and not just the shell around it]?ā€
I would be screaming ā€œhow could you doubt the inner object will warm?ā€.
Suppose the inner sphere did NOT warm above 255 K. The shell and the sphere around it would both be 255 K. But heat (240 W/m^2) would be flowing with no temperature gradient. Why in the world would 240 W/m^2 of heat be flowing from one object at 255 K to another object at 255 K?
Max was right with
P = ĪµĻƒA(Thā“ ā€“ Tcā“)
There must be a difference between Th (the inner sphere) and Tc (the shell) for this 240 W/m^2 of power to flow. Th will be 302 K in this particular case.
There is no ā€œrun-awayā€ effect.
* The inner sphere warms from 255 K to 302 K.
* The outer shell warms from 0 K to 255 K.
* 240 W/m^2 of heat moves from the inner sphere to the outer shell
* 240 W/m^2 of heat moves from the outer shell to space.
* No laws of thermodynamics are broken.

Reply to  tjfolkerts
April 24, 2013 5:04 pm

“No laws of thermodynamics are broken.”
Except that the source is now twice as hot as its own heat input. Of course, we have seen the textbook examples which show that this does not occur. A shell surrounding a heated sphere simply gets heated by the sphere. This does not cause the sphere to heat up some more. This was shown in worked problems in physics textbooks which I am sure you saw. The supposition of trapped radiation causing self-heating was refuted.
REPLY: Best advice: stop digging – Anthony

davidmhoffer
April 24, 2013 4:59 pm

Greg House;
And again, the IPCC ā€œgreenhouse effectā€ is physically impossible,
>>>>>>>>>>>>>>>
The only thing physically impossible is educating you, as a rather long list of PhD physicists, engineers, and chemists recognized around the world for their knowledge have repeatedly demonstrated on this blog.

Selgovae
April 24, 2013 4:59 pm

Mark, Gary, thanks for the responses.
I don’t get the clairvoyance argument. That model suggest photons are being fired off like tennis balls and that they ‘travel’ in space until they find a destination. That’s at odds with my mental image of relativity and the speed of light. Surely they know where they’re going (in a sense ‘going nowhere’ as they have no distance to travel or time to take), and therefore can’t leave until they have a destination.
“The temperature of the unheated plate (ā€œplate 2ā€³) never reaches that of the electrically heated plate (ā€œplate 1ā€³).”
Is that true? I’m thinking only of atoms on the surface nearest to plate 1. Won’t some of them temporarily reach the same temperature as plate1, thus slowing down the emission from plate 1 until they cool down again?

steveta_uk
April 24, 2013 5:01 pm

Max, your assumptions were wrong.
You started with a block of ice and a pot of boiling water. So you know the temps at the outset, 0C and 100C.
You then try and add energy radiated from the ice to the boiling water, and assume this means the pot is even hotter.
Why? That energy is already included in the energy being radiated by the boiling pot. The energy output from the pot is exactly as you calculated. If you know that this must include the energy from the block of ice (again, as you calculated, with the somewhat unphysical assumption that all that energy goes the the pot, and all other directions are at absolute 0) then you could work out the reduced heating requirement to keep the pot at 100C when the ice is present, compared to open space.
I.e. as per Dr Spencer, the block of ice has reduced the heating requirements of the pot.

davidmhoffer
April 24, 2013 5:02 pm

tjfolkerts;
There is no ā€œrun-awayā€ effect.
* The inner sphere warms from 255 K to 302 K.
* The outer shell warms from 0 K to 255 K.
* 240 W/m^2 of heat moves from the inner sphere to the outer shell
* 240 W/m^2 of heat moves from the outer shell to space.
* No laws of thermodynamics are broken.
>>>>>>>>>>>>>>>
Thank you sir for injecting some reality into the discussion.

richard verney
April 24, 2013 5:10 pm

I can remember an old design of 3kw electric fire.
It consisted of three 1 Kw elements running horizontally, ie., one of the elements was positioned midway between the upper and lower elements. The back of the fire was a concave mirror. It had a simple grill over the front of a not sufficiently tightly spaced grill to stop prying fingers! The fire could be switched low ( the bottome 1kw element), meduium (2 of the 1kw elements) or high (all 3 of the 1kw elements).
When switched on high all 3 elements were on. The middle element would recieve ‘heat’ from being sandwiched between the upper and lower elements, and of course, in addition, that reflected from the concave mirror. The middle element never looked noticably hotter than the other two elements (ie., it did not appear white hot and the other element red hot, I recall that they all had a similar glow).
So the question is why did the middle element not get noticably hotter and take on a much brighter appearance?

Greg House
April 24, 2013 5:11 pm

davidmhoffer says (April 24, 2013 at 4:38 pm): “In todayā€™s world, it is mind boggling to me that the CAGW nonsense can live side by side with the ā€œback radiation doesnā€™t warm thingsā€ nonsense.”
============================================================
Who said ā€œback radiation doesnā€™t warm thingsā€??:evil:
It is ā€œback radiation doesnā€™t affect the temperature of the source“, davidmhoffer, the source.

CodeTech
April 24, 2013 5:17 pm

Actually, the only reason I mentioned the gas being used in my windows was to be sarcastic. Here, they use nitrogen for the same reason they put nitrogen in car tires. It’s not as prone to pressure changes at different temperatures, and it’s dry.
In fact, the best gas to use in windows would be NO gas – a vacuum. Unfortunately that would tend to cause the glass to bow inward.
When it comes to windows in a winter climate the “radiative” factor is essentially meaningless. We’re rarely trying to keep solar energy out, and for that my windows have a tinted film. The idea is to have dead layers that slow thermal transfer.

steveta_uk
April 24, 2013 5:21 pm

CodeTech, they could make double glazing work even better by adding some glass wool as well as dry nitrogen between the panes to stop convection.
Might make it difficult to see out, I suppose.

April 24, 2013 5:24 pm

richard verney says:
April 24, 2013 at 5:10 pm
*post snipped due to exceeding the awesome quota*
So the question is why did the middle element not get noticably hotter and take on a much brighter appearance?

Good point, richard, why doesn’t the middle element get hotter/brighter?
” Max, your assumptions were wrong.
You started with a block of ice and a pot of boiling water. So you know the temps at the outset, 0C and 100C.
You then try and add energy radiated from the ice to the boiling water, and assume this means the pot is even hotter.”
~steveta_uk
Are you referring to my post at 1:39 PM?
Ok, have it be two black bodies of arbitrary material, large enough that the area weā€™re interested in is radiating towards the other body, one at 373 K and one at 273 K, radiating towards each other.
Iā€™m confident that the radiation field between them can be described by the following equation:
P = ĪµĻƒA(Thā“ ā€“ Tcā“)

rather than:
P = ĪµĻƒA(Thā“ + Tcā“)
max@Funktastic:~$ calgebra
>>> (5.6703*10^-8)*(373^4 ā€“ 273^4)
782.63220248
>>> (5.6703*10^-8)*(373^4 + 273^4)
1412.55397668
Both of those can not be right, and if the colder black body raises the temperature of the warmer black body then the second one is correct, which would be roughly the same power as that emitted by a 297.28 K black body.
If the radiation from the colder body is added to that from the warmer body, that is akin to a block of ice raising the temperature of a pot of nearly boiling water by over 24 K!

For clarification, I am stating that this is the correct calculation:
P = ĪµĻƒA(Thā“ ā€“ Tcā“)
As follows:
>>> (5.6703*10^-8)*(373^4 ā€“ 273^4)
782.63220248

tjfolkerts
April 24, 2013 5:29 pm

TIM: ā€œNo laws of thermodynamics are broken.ā€
JOE: “Except that the source is now twice as hot as its own heat input.”
No!
The “heat input” is not “255 K”. The heat input is “240 W/m^2” to the shell and this was fixed in the example. This input will be whatever temperature is needed (within some broad engineering limits). (Similarly, sunlight will provide energy to earth’s surface, not matter what the surface temperature might already be.)
If the “heat input” were held at 255 K by some thermostat, then the power required to hold the inner shell @ 255 K would drop. Specifically, the power required would drop to 240/2 W/m^2 = 120 W/m^2 and the shell would only warm to 214 K.
There is no law of thermodynamics that says “a heater with a fixed power will be the same temperature no matter what ‘heat load” you impose.”

tjfolkerts
April 24, 2013 5:32 pm

DANG! That should have read “The heat input is ā€œ240 W/m^2ā€³ to the INNER SPHERE… “ in my last post!

joeldshore
April 24, 2013 5:32 pm

Joseph Postma says:

You canā€™t average energy input into a geometry where it doesnā€™t exist.

This is just silly sophistry. If you don’t understand the concept of an average, you don’t have to use it. Just compute energy balance using Total energy in = Total energy out.

Both the wet and dry lapse rate can be calculated precisely without any reference to GHG radiation, for example.

Nice strawman. That is because the wet and dry adiabatic lapse rates represent stability conditions: A lapse rate larger than the appropriate adiabatic lapse rate is unstable to convection, which then is very efficient at lowering the lapse rate back down to the adiabatic lapse rate. Go read any elementary book on atmospheric science and stop pretending that the few things you say that are correct are actually new.

Mark Bofill
April 24, 2013 5:36 pm

Selgovae,
If you want to believe that about electromagnetic radiation, that it doesn’t radiate without some sort of ‘quantum entanglement’ handshake with a receiver at some future space time coordinate, be my guest I guess. I can’t disprove that, am not aware of any theoretical foundation or empirical data to support the idea, and have no particular interest in the notion.
Another way to tackle it might be this. Suppose you shine a flashlight at the sun. Without having performed the experiment myself, I’m utterly confident it would radiate, even though the surface of the sun is considerably hotter than the filament of the bulb. Take the experiment to space if you’ve got an issue with the atmosphere, I’m sure the results will be the same.

April 24, 2013 5:38 pm

“The heat input is ā€œ240 W/m^2”
Well this isn’t really relevant to the planet Earth in any case because the input is not 240 W/m^2 but 610 W/m^2, as an integrated average over the projection factor on the hemisphere. So, any of these argument starting at 240 W/m^2 are irrelevant, because they don’t correspond to reality. Of course, the atmosphere isnā€™t a source of heat and so all the arguments created to try to make it a source of heat or a cause of heating are wrong. The error originates when we assume that sunshine is -18C instead of +49C.
Again, we have seen the text book examples which show that the interior sphere does not become heated. We do not find this in worked textbook examples. If you have a constant power input creating thermal radiation inside a cavity, this just produces a blackbody spectrum, it doesn’t cause a shift in the frequency spectrum of the radiation and thus higher temperature.

davidmhoffer
April 24, 2013 5:39 pm

Greg House;
Who said ā€œback radiation doesnā€™t warm thingsā€??:evil:
It is ā€œback radiation doesnā€™t affect the temperature of the sourceā€œ, davidmhoffer, the source.
>>>>>>>>>>>>>>>>
Ah yes, the source has a little sign at the front door that says “If you have ever been here before, leave your joules of energy at the door”. The sign is probably easy to find, it is right next to Joe P*stma’s “molecule traps”.

KevinK
April 24, 2013 5:44 pm

Max,
Thanks for that kind (cough cough, condescending) explanation of the most basic units of physics (Watts (not Anthony) and Joules). I do note that those namesakes did not deem it necessary to trademark their names, maybe a little less hubris back in the day, oh how I yearn for simpler times.
You wrote;
ā€œWill the presence of another body at the same temperature cause the temperature of the first body to increase?ā€
Certainly a totally fair question;
If the first body is NOT supplied energy from an external source, and it is radiating towards a colder object (Royā€™s imaginary vacuum chamber walls) then NO the presence of a second body (at the same temperature or lower) WILL NOT cause the temperature of the first body to increase
HOWEVER, If the first body is supplied energy (not power) from an external source, and it is radiating towards a colder object, then YES anything that changes the ā€œview factorā€ MAY cause its temperature to increase. I use the qualifier MAY because there are other secondary factors to consider, for example if a conductive path exists that can cool the second body faster than the first body can warm it.
So like many ā€œsimple questionsā€ it is necessary to ask follow up questions to understand ALL the conditions present before pronouncing the final outcome of what is still just a thought experiment. Thatā€™s why I prefer real hard observations before accepting any Hypothesis based on the say so of MAXā„¢, or Roy, or Anthony (not the historical one), or Lord Monckton.
BTW; as an engineer, ā€œnitpickingā€ on my part keeps the people that use my products alive and uninjured. And I am proud to be one of the finest pickers of nits in these parts.
Cheers, Kevin

April 24, 2013 5:45 pm

Molecule traps are ridiculed but radiation traps are ok. Strange. When you stop a draft in your house, you’re stopping molecules – the gas – from moving around. This prevents heat loss, it prevents warm gas (molecules) from escaping with the draft to outside the house. Trapping radiation inside a cavity, however, just makes the radiation constructively and destructively interfere with itself, causing no net change; it actually creates a blackbody spectrum at the temperature of the source.

Gary Hladik
April 24, 2013 5:45 pm

Selgovae says (April 24, 2013 at 4:59 pm): “I donā€™t get the clairvoyance argument. That model suggest photons are being fired off like tennis balls and that they ā€˜travelā€™ in space until they find a destination. Thatā€™s at odds with my mental image of relativity and the speed of light. Surely they know where theyā€™re going (in a sense ā€˜going nowhereā€™ as they have no distance to travel or time to take), and therefore canā€™t leave until they have a destination.”
Take a look at the Hubble Extreme Deep Field composite
http://en.wikipedia.org/wiki/Hubble_Extreme_Deep_Field
The light from some of those galaxies “left” about 13.2 billion years ago. If those photons stayed put until they had a destination (the Hubble telescope), some of them could see over 13 billion years into the future, which is what I call “clairvoyance”! šŸ™‚
ā€œIā€™m thinking only of atoms on the surface nearest to plate 1. Wonā€™t some of them temporarily reach the same temperature as plate1, thus slowing down the emission from plate 1 until they cool down again?”
As a layman I’m uncomfortable discussing the temperature of individual atoms, thinking instead of temperature as a macroscopic property. But note that even the plate 2 layer of atoms facing plate 1 is getting radiated energy from one direction and losing it in two: by radiation toward plate 1 and by conduction to the rest of plate 2.
Besides, talk about clairvoyance! Now a photon can’t leave until it knows where its destination is and what the destination temperature is! Photon emission must be a nightmare process:
“This is Photon Traffic Control. Photon 10244759800321, destination carbon atom 8876635549, flight duration 983,376,455,987.44776 years, has been delayed indefinitely. Target temperature will be a balmy 7.02 degrees Kelvin at arrival time, which unfortunately is 0.01 degrees warmer than we are here. Photon 7466453553…”
The really sad thing about being a photon is that you never get to vacation in a warmer spot…

April 24, 2013 5:52 pm

Photons do not actually experience time or space in the way we expect of material entities. At the speed of light there is no lapse of time and all space is shrunk to zero length. It is an interesting relativistic thing to think about. Photons are in a sense in contact with everything at all “times”. Just apply the Lorentz equations with v = c. You get zero time and zero distance. This is life as a photon.
In any case, you can shine a flashlight at the Sun and the the beam will be there. But the flashlight doesn’t heat the Sun. Only hotter things heat cooler things.
Anthony, I am truly sorry if you think I thread bomb. That’s not what I am trying to do. I am just trying to state some things in a thread which regards me and reply to comments which are useful to make some points. I write fast, true, and can respond quickly, but it is not meant to “bomb”. Best regards.

Selgovae
April 24, 2013 5:52 pm

Mark,
“Suppose you shine a flashlight at the sun. ”
Under my notion, that causes some experimental difficulties. How do I measure whether the flashlight is radiating at the sun? I’m not confident at all about what actually happens. But I like the ā€˜handshake with a receiver at some future space time coordinate’ idea. Although I’m not sure ‘future’ is the right word, perhaps ‘different’.
Anyway, that’s probably enough for my feeble brain today. Thanks for the comments.

April 24, 2013 5:53 pm

“How do I measure whether the flashlight is radiating at the sun?”
That is a very good point!!! Insightful. Anything you put between you and the Sun to measure becomes the target. And with the strange nature of photons….hmmm. Smart fellow šŸ™‚

KevinK
April 24, 2013 5:58 pm

Gary Hladik wrote;
“So if instead of using an electric outlet or a battery to heat the first plate, we use an embedded radioactive heat source
(redacted wiki link here)
the plutonium (for example) will decay more rapidly when the second plate is introduced?ā€
No, a radioactive heat source is not the same thing as an electric (actually electro-chemical) storage battery. Names matter, you canā€™t just willy nilly throw in different parts in your thought experiment and claim the same outcome. A radioactive heat source is infinite (for the time frames of interest) and is more like a battery that never discharges (something the current US DOE has inappropriate dreams about).
Different thought experiment, different expected outcome.
Cheers, Kevin.

Ivan
April 24, 2013 6:00 pm

So, this is actually an attack on Mishcolczy if I am not wrong? I think the only sensible reaction that any normal skeptical person will have here is: Wow, let’s see what got them all worked up so much?

Gary Hladik
April 24, 2013 6:00 pm

richard verney says (April 24, 2013 at 5:10 pm): “So the question is why did the middle element not get noticably hotter and take on a much brighter appearance?”
Have you actually done the calculations so you would know if the difference should be “noticeable”? Please show your work.

Gary Hladik
April 24, 2013 6:03 pm

KevinK says (April 24, 2013 at 5:58 pm): “Different thought experiment, different expected outcome.”
So if it’s radioactively heated, you wouldn’t expect plate 1 to get warmer when plate 2 is introduced?

April 24, 2013 6:13 pm

“If you donā€™t understand the concept of an average, you donā€™t have to use it. Just compute energy balance using Total energy in = Total energy out.”
An average applies to where something exists, to what is real. Especially with power inputs. The power from the sun is not ever averaged over the entire globe, but only a hemisphere. This is what you average, and you can do it using calculus – an integrated average. There is no integrated average for solar energy over the globe because that doesn’t exist. You have to do the average correctly. 240 W/m^2 is the output, not the input. Yes the input and output balance in total energy, but not in terms of power flux, not in terms of the work that they can do. The input flux can do way more work than the output flux. The input flux is much higher than the output flux. This is very important, because -18C or 240 W/m^2 input can’t melt ice into water, while the actual real-time solar input and physically real integrated average (of +49C) can do so. It makes a big difference. It is a very simple fact that real sunlight can melt ice into water all by itself, while -18C or 240 W/m2 never could. Then, with the high heat capacity of water, it can stay liquid for a very long time once the sun has melted it, much longer than a single night. Then the latent heat also acts as a barrier to dropping back below zero. This all holds the temperature higher. These are interesting scientific facts.
The wet and dry lapse rates are calculated perfectly without regard to GHG radiation. Convection is of course natural and occurs whenever the near-surface air is heated. Certainly the system will tend to whatever lapse rate is appropriate given the water vapor concentration and strength of gravity (the latter of which doesn’t change of course).

Mark Bofill
April 24, 2013 6:17 pm

How do you know it’s radiating? Measure the power consumption I’d say.
Look, if you want to turn this into a philosophical question, you can’t prove much of anything. Decartes and his malicious demon trying to trick him, or whatever the argument was. That’s fine, but I don’t think it’s got much to do with science. Stick with Occams Razor instead, it gets you places.

April 24, 2013 6:21 pm

I do note that those namesakes did not deem it necessary to trademark their names, maybe a little less hubris back in the day, oh how I yearn for simpler times.” ~KevinK
My last name has a T, M, and vowel sounds, and I noticed many sites will allow unicode for usernames, as Max is often taken, I use this one. Strictly speaking you read it as “Max Thyme”, so it is little different than your Kevin K moniker.
I condescend when condescended towards, seems fair enough right?
Anyways, enough snippiness, we are both adults, my point was raised because of various examples I find in textbooks, such as this one which do not support the idea that adding a passively heated object would raise the temperature of an actively heated object.
A classical example is a star in a dust cloud, the dust cloud will be receiving radiation from the star and will thus be warmed, correct?
There are some who would claim that the dust cloud would be warmed until it emits the same total amount to space as the star would have, and there are some who would claim that the star will emit approximately half of what it receives inwards and outwards, rather than the same intensity it receives in both directions.
I suppose I should show my hand and call though: if adding an object the same temperature at some distance could or would raise the temperature of an actively heated object, what would happen if the second object were brought into contact with the first?

April 24, 2013 6:23 pm

“How do you know itā€™s radiating? Measure the power consumption Iā€™d say.
Look, if you want to turn this into a philosophical question”
It is quite a real problem, not just philosophical. You provided a good answer in terms of power consumption. That’s as philosophical as required.

Rosco
April 24, 2013 6:43 pm

Roy Spencer wrote:
“ALL
It is important to remember that there is NO WAY to determine the temperature of anything based upon the rate of energy input alone, for example the Earth absorbing an average of ~240 W/m2 from the Sun. Temperature is a function of BOTH energy input (typically not temperature dependent) AND energy loss (typically VERY temperature dependent), neglecting issues related to heat capacity which mainly affect the time required for the system to equilibrate. The temperature of anything heated will increase until the rate of energy *loss* equals the rate of energy *gain*. So, temperature can be increased by increasing INPUT, or decreasing OUTPUT.”
I find many things wrong with this statement.
1. One assumes that the 240 W/sq m “average” over a sphere is derived from 960 W/sq m over a disk – this is the fundamental basis of all the initial mathematical derivations of the greenhouse theory I have seen.
Using Stefan-Boltzmann – 240 W/sq m equates to a maximum temperature of about 255 K.
Using Stefan-Boltzmann – 960 W/sq m equates to a maximum temperature of about 360 K
Which temperature more accurately predicts the potential heating power of the Sun ?
Remember that one “disk shaped” hemisphere of the Earth is always illuminated – it is simply the portion of the Earth that changes form day to night.
2. Your statement appears to imply that you can heat something up beyond the power of the input radiation by reducing output. This is in contradiction of the results of Planck et all and is reflected in the Stefan-boltzmann equation and its application by climate science.
3. I remember that you once described a post I made where I stated my belief that the period of rotation of a planet was significant as :-
“This post is a good example of how some people have polluted the minds of others with pseudo-science. ”
I’m not sure I’ve polluted anyone’s mind – they are free to take it or leave it – as I am free to find little credibility in many of your arguments.
Are you now saying that the period of a planet is now crucial ?
The Moon and Mercury both represent heated surfaces radiating to space. Both have significant temperature swings from high to extremely high down to about 100 K – perhaps lower.
The Moon has been mapped and loses “heat” at about 1 K per hour average from noon to sunset – a time of about about 177 Earth hours to decrease from about 390 K to about 220 K and a further 120 K, maybe more, in the 354 Earth hour long night.
The Earth obviously loses less energy in 12 hours of night than the Moon in about 354 Earth hours or Mercury in its 2111 odd Earth hour long night.
Is not this the same as saying the Earth’s output is decreased as compared to the others ?
Venus is a totally different kettle of fish but I find it amazing that climate science can keep a straight face claiming Venus’ 730 K or 16,101 W/sq m – yes 16.1 KW/sq m – can be caused by the albedo adjusted solar input of 132 W /sq m – lecture 3 on the Greenhouse Effect UCLA.
4. When you say “So, temperature can be increased by increasing INPUT, or decreasing OUTPUT.” I assume you refer to the “constant” input of solar radiation.
Even so the Stefan-Boltzmann determines that the temperature is determined by the radiation input so the question remains – can one increase the radiative input by adding the radiation from a cold atmosphere to the solar radiation to create a higher temperature ?
If so which radiative input – the more realistic 960 W/sq m during the day exclusively or the mythical averaged 240 W/sq m ignoring day and night ?
Remember a hemisphere of the Earth is always in day – under the 960 W/sq m (not my figures by the way but simply the albedo reduced solar constant over a disk as used in calculating the “effective radiating temperature of Earth) – it is simply rotating which part is under the Sun at its zenith.
I find “averaging” the solar input as meaningless for any real purpose – the Sun is either heating the surfaces of a part of the Earth during the day or not during the night.
Adding support to this is the real observation that my solar panels only produce during the day and they require an input of approximately 1000 W/sq metre to produce anything like their rated capacity.
At 240 W/sq m illumination at 15.6% efficiency (quoted for 1000 W/sq m – significantly less for lower power but why argue) and 1.2 sq m panels my 14 panel array would produce about 630 W per hour.
I know they produce a “feed in” of up to 7 KW hrs a day and supply up to 5 Kw hours of reduced energy from the grid.
I don’t have a 19 hour long day at 240 W/sq m solar input – we rarely have more than 14 hours sunlight a day being almost within the tropics.
Where I live the clear sky solar power at noon is calculated approximately to be Cos 27 degrees S x 1367 x 0.7 or about 852 W/sq m.
842 x 1.2sq m per panel x 14 panels results in a maximum of 2.2 Kw per hour and requires 6 to 8 hours of varying levels of solar radiation due to the angle of the Sun.
This result seems much more realistic to me.
Anyway – that’s my 2 cents worth.

CodeTech
April 24, 2013 6:49 pm

By the way, if you shine your flashlight at the sun you won’t reach it. You need to lead. Slightly to the right. But your beam WILL hit the sun. And be just as effective as the rest of the light bouncing off the planet at increasing the sun’s overall temperature… unless you have a mighty bright flashlight.

tjfolkerts
April 24, 2013 6:53 pm

Joe Postma says: Well this isnā€™t really relevant to the planet Earth …
Well, the discussion is specifically relevant to the situation described above (introduced April 24, 2013 at 2:39 pm and refine April 24, 2013 at 4:58 pm). Quite clearly this is not the earth. But to the extent that correct physics is relevant everywhere, it is relevant to earth.
” … the input is not 240 W/m^2 but 610 W/m^2, as an integrated average over the projection factor on the hemisphere. “
No, the input from the sun (accounting for albedo) is ~ 0.7 * 1370 W/m^2 = ~ 960 W/m^2
Averaged over a hemisphere, that is 480 W/m^2
Averaged over a whole sphere that is 240 W/m^2
Your 610 W/m^2 is averaged over a CYLINDER, not a hemisphere (or equivalently, it only applies at the equator, but no where else). You are too big by a factor of 4/Ļ€ for a hemisphere.
“The error originates when we assume that sunshine is -18C instead of +49C.
Sunshine has (approximately) the spectrum of a ~ 5800 K blackbody. As such, the only truly meaningful “temperature” of the photons is 5800 K. Assuming either -18C or 49C is wrong.
* Focus sunlight onto an object from all sides and that object will be 5800 K (but sunlight can never be focused to warm the object above 5800 K because that would violate the 2nd Law of Thermodynamics).
* If that sunlight comes from only a small part of the sky (ie a disk ~ 0.5 degrees across as seen at 1 AU, with the rest of the sky @ 2.7 K) ) and is spread over a hemisphere with albedo 0.7, the object will be ~ 88 C
* if is is spread out over a cylinder, the object will be ~ 49 C
* if is is spread out over a sphere, the the object will be ~ -18 C
Sunlight is 5800 K. Geometry and albedo combine to limit its warming ability. (But if you make the rest of the sky warmer than 2.7 K, the combined effect of this sunlight and “atmospheric light” will combine to make the new temperature -18 C < T < 5800 K, ie somewhere between the temperature of the cool atmosphere and the warm sun).
“Again, we have seen the text book examples which show that the interior sphere does not become heated.”
Here is one “textbook example” (Sorry I don;t have the reference for the original).
http://i341.photobucket.com/albums/o396/maxarutaru/bunnytaru/lol%20censorship/heatsourceinshellatequilibrium_zps1c9d662a.png
This will get a little subtle, so hopefully people will not get confused — the situation in this textbook and the situation Greg introduced are SIMILAR, but not the same.
The “black sphere” (eg a star) in this example DOES stay the same temperature BUT that was part of the assumptions for this example. What DOES change is the power input. This star maintains the same temperature but with HALF the power when the shell is added.
That means that if the power was “turned back up” to the original value (corresponding to Greg;’s example), the interior would necessarily warm up.
Sorry, but this particular textbook supports me and not you. Do you have a “textbook example” you can cite that supports your claim?

joeldshore
April 24, 2013 6:55 pm

Joseph Postma says:

Well this isnā€™t really relevant to the planet Earth in any case because the input is not 240 W/m^2 but 610 W/m^2, as an integrated average over the projection factor on the hemisphere. ..
The error originates when we assume that sunshine is -18C instead of +49C.

Cut the sophistry, Joe. The Earth absorbs an amount of energy each second equal to (pi*R^2)*L_solar*(1-albedo)*(1 second) where the albedo is about 0.3, the solar constant L_solar = 1361 W/m^2, and the radius of the Earth is R = 6371 km. That makes for 1.21 x 10^17 Joules.
If the Earth were at an average temperature of 49 C, its emission of energy each second would be [at least because of Holder’s Inequality] (4*pi*R^2)*sigma*T^4*(1 second) where T = 322 K and the Stefan-Boltzmann constant sigma = 5.67 x 10^-8 W/[m^2*K^4]. That makes for 3.11 x 10^17 Joules. [If instead we use the correct average temperature of 255 K (-18 deg C), we get an emission of 1.22 x 10^17 Joules, which within the accuracy of the calculation, gives energy balance.]
In other words, you are spewing nonsense in an apparent attempt to confuse people who don’t know any better.

Reply to  joeldshore
April 24, 2013 8:07 pm

“That makes for 1.21 x 10^17 Joules.”
And these Joules are absorbed over a hemisphere with the intensity distribution going as the cosine of the solar zenith angle. This is reality.
“If the Earth were at an average temperature of 49 C”
I did not ever say that the Earth was at +49C. I said this is the equivalent temperature value of the average input. This too is reality. This is very different from the usual assumption of a -18C or 240 W/m2 input. It makes a big difference in what the input can and can not do. The total energy is conserved, the input is equal to the output in terms of total energy, but not in terms of power, not in terms of what the input and output power flux can do. The input can do a lot more work than the output. The input can generate very high temperatures on the surface. The atmosphere is then heated by the surface conductively/convectively and radiatively. That’s all that needs to happen because the input is so hot. Hot heats cold. You can go outside and feel the hot input on a sunny day. Latent heat carries a lot of this input around the planet via water vapor and liquid water.

April 24, 2013 7:07 pm

Here’s the original, Tim: that up there is just a snapshot I took to embed.
There was nothing about the power input being changed in that example, go to problem 1026 to see another examination of this problem.
As for the integration over a sphere, you’re treating his 610 value as a projection, it is integrated from the peak value at the subsolar point (1366~ before albedo) to the value at the edges (0~) of the terminator.
After albedo it works out to around 476 W/m^2, or as I prefer to express it, 1.22×10^17 Joules absorbed by one hemisphere, and ~5.11×10^16 Joules emitted by each hemisphere.

tjfolkerts
April 24, 2013 7:10 pm

@ Max April 24, 2013 at 6:21 pm
Your “textbook example” is similar to the one I just discussed. It says that to maintain a constant temperature, an object needs less power by a factor of
R^2 / (R^2 + r^2)
If the radius of the shell (R) is about the same as the radius of the sphere (r), this becomes 1/2. So only 1/2 as much power is needed to keep the inner sphere warm with the shell in place as was required without the shell. This example has the advantage of also showing what would happen is the shell was significantly larger that the sphere inside.
And as I just said, if the power was ā€œturned back upā€ to the original value (corresponding to Gregā€™s example), the interior would necessarily warm up above its original temperature.

KevinK
April 24, 2013 7:12 pm

Max, fair enough, I took the ā€œTMā€ as a slight bit of ā€œargument from authorityā€, my apologies, I did not realize it was a simple contraction from a common name.
And my generic observation about using the right units was not ā€œaimedā€ at anyone specific, just a ā€œlook out that you donā€™t step in that same mud puddle I just soiled my pants withā€ kind of observation. I have been tripped up many times when I mismatched the correct units for a calculation. One of the most valuable lessons back in college was to ā€œreconcile your unitsā€ always, always, alwaysā€¦..
So, in summary, I believe that the proponents of the ā€œGHEā€ have some valid points; for one, a photon leaves its source without regard to where itā€™s going. And yes a colder object (or gas cloud) can affect the temperature of a warmer object (ONLY by slowing the velocity of energy flowing away from the warmer object, thatā€™s what the fiberglass in the walls of your house accomplishes)).
But the ā€œdeniersā€ of the GHE also have some valid points; the proponents of the GHE hypothesis have not demonstrated that the ā€œeffectā€ actually reduces the velocity of heat/energy (they alternate between IR EMR and molecular rate of vibration (i.e. temperature)) through the system. In fact I maintain that the ā€œGHEā€ only causes heat/energy to make multiple passes through the system at the speed of light (there are several flavors of the ā€œspeed of lightā€, the fastest is in a vacuum, and the slowest is in the ā€œflintā€ glasses) which is known to be MUCH faster than the speed of heat through any gases (GHG or otherwise) or the Oceans or Icesheets or soil in the complex Earth/Atmosphere system.
The much ballyhooed ā€œGHEā€ only delays the transit of heat/IR by a few tens of milliseconds. That is why you can measure it with an IR sensor with no time reference (Dr. Spencerā€™s ā€œlook I can measure itā€ posting) and not realize that it is simply ā€œpassing throughā€ at the speed of light and causes no ā€Higher Equilibrium Temperatureā€.
Cheers, Kevin.

April 24, 2013 7:19 pm

“Do you have a textbook example?”
Here are some textbooks examples of how a continuous source of heat as a sphere trapped inside a shell doesn’t cause an increase of temperature of there sphere (with continuous input):
http://books.google.co.in/books?id=dQGC0ifkE34C&pg=PA24&lpg=PA24&dq=concentric+sphere+black+body&source=bl&ots=Zh6N1e35jc&sig=m-7nVWch4_zv-l3ISR5k7bluSUQ&hl=en&sa=X&ei=_ldYUd7EFZLU9ATVzYHoDw&redir_esc=y#v=onepage&q&f=false
Also problem *1023* above it. Oh, I see that’s the one you referred to. The problem says nothing about halving the input to the sphere, nor does the other one. You’re making that up. The problem specifically refers to a continuous input which means the sphere is heated to a constant temperature. You could think of it as an infinite heat sink. There is nothing in the two examples about the input being reduced – the input is constant.
“Averaged over a whole sphere that is 240 W/m^2”
You can not average the input over a sphere because the input does not occur over a sphere. The input does not occur over a sphere, it only ever occurs over a hemisphere. The 610 W/m^2 average input is the integrated average intensity over the hemisphere. The energy comes in over a hemisphere, and so this is the correct and only meaningful way to average it, as an integrated average over that geometry.
“the only truly meaningful ā€œtemperatureā€ of the photons is 5800 K. Assuming either -18C or 49C is wrong.”
Tell that to the GHE crowd. Of course, we are speaking about the heating potential of sunlight at the distance of the Earth and given its input distribution on the Earth. Which results in the integrated average as discussed. Of course the raw heating potential from sunlight for a flat blackbody is +121C.
The only way to get higher temperatures than this from sunlight at the Earth is to re-condense it using a magnifying glass. GHG’s do not re-condense incoming sunlight, hence they can not create 5000K out of sunlight. GHG’s only interact with the outgoing IR, and as you have pointed out they can not create a higher temperature than their source – the surface of the Earth.
Of course, the actual input is +49C as a temperature and this is higher than the surface temperature. Total energy balances out but the input itself is much stronger than the output – the input can heat the Earth to high temperature, melt ice etc etc.

April 24, 2013 7:27 pm

Steven Mosher [April 24, 2013 at 3:42 pm] says:
Wrong.
The ā€œgreenhouseā€™ effect has nothing whatsoever to do with these types of experiments.
Energy leaves earth Via one route and one route only: radiation to space.
Earth radiates to space from the ERL.
When you add GHGs you raise the ERL.
When you raise the ERL the earth radiates to space more slowly.
That results in a surface that is warmer than it would be otherwise.
This effect cannot be [seen? measured?] tests with woods like experiments EVER because the effect depends upon changing the ERL and no experimental set up or lab can duplicate the conditions required: the full atmospheric column.
in 20 or so simple slides

http://www.aos.wisc.edu/~aos121br/radn/radn/sld001.htm

Well that’s awfully convenient. Greenhouse Theory hypothesis cannot be experimentally reproduced so we’ll just have to take the word of climate kooks instead. Experiments replaced by models and slideshows. Yep. The natural conclusion of the pop-science movement ever since Sagan, Ehrlich, Hansen and Mann lowered the bar to unimaginable levels.
P.S. some C02 related experiments can be reproduced Steven …

LdB
April 24, 2013 7:27 pm

Roy I am so glad that scientists on the AGW side of climate science is taking on the pseudoscience junkies and there is no greenhouse effect garbage it had been worrying me for sometime.
The fact all these pseudoscience types ignore is the ability to challenge it is way out the league of climate science stupidity because it meshed in Planck’s law which forms part of the foundations of Quantum Mechanics.
Planck already worked out classic physics of Boltzmann, Wein and Kirchoff were simplifications based on classic physics. Using quantum mechanics he showed you could derived the classic physics law it is an important piece of science history like E=MC2.
A reasonable short version link on the subject:
http://csep10.phys.utk.edu/astr162/lect/light/radiation.html
Wikipedia carries the full version of derivations of the classic laws from Planck’s laws
http://en.wikipedia.org/wiki/Planck%27s_law_of_black-body_radiation
The dragon slayers like all pseudoscience lunatics miss the fact we know QM is right we have tested it extensively and continue to test it everyday of every year and to date it has never been shown wrong.
I have stated this before we also know that the greenhouse gases are quantum active because all of the gases involved can also be used as the medium in a standard laser tube (CO2, N2O, Water Vapour). If you want absolute proof that you can optically pump the gases involved it doesn’t get more basic than a laser because there are only 3 requirements for a laser tube
A laser is constructed from three principal parts: (http://en.wikipedia.org/wiki/Laser_construction)
An energy source (usually referred to as the pump or pump source),
A gain medium or laser medium, and
Two or more mirrors that form an optical resonator.
So dragonslayers there is your absolute proof that the substances must be a gain medium because they can be used as a laser medium.
All of this is way beyond doubt and argument especially when your argument omits the fact that your alternative explaination is falsified by Quantum mechanics.

KevinK
April 24, 2013 7:29 pm

Whoops, I inadvertently typed the ā€œDā€ word, apparently my latest comment is ā€œtrappedā€ā€¦ā€¦ā€¦ā€¦ā€¦.. ha ha ha
I still donā€™t see much ā€œslayingā€ going on here, Dr. Spencerā€™s ā€œThe Sun/Earth system is just like your House/Furnaceā€ analogy is (for lack of a better match) still a severely ā€œurine deprivedā€ analogy.
Cheers, Kevin

Gary Hladik
April 24, 2013 7:30 pm

tjfolkerts says (April 24, 2013 at 7:10 pm): ‘And as I just said, if the power was ā€œturned back upā€ to the original value (corresponding to Gregā€™s example), the interior would necessarily warm up above its original temperature.’
Aaaaaand we’re back to the steel greenhouse:
http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/

Genghis
April 24, 2013 7:37 pm

Roy Spencer says:
April 24, 2013 at 1:37 pm
ALL:
It is important to remember that there is NO WAY to determine the temperature of anything based upon the rate of energy input alone, for example the Earth absorbing an average of ~240 W/m2 from the Sun. Temperature is a function of BOTH energy input (typically not temperature dependent) AND energy loss (typically VERY temperature dependent), neglecting issues related to heat capacity which mainly affect the time required for the system to equilibrate. The temperature of anything heated will increase until the rate of energy *loss* equals the rate of energy *gain*. So, temperature can be increased by increasing INPUT, or decreasing OUTPUT.
———
Kirchhoffs law says that emission = absorption. Absorption at the TOA equals apx 341 W/M ^2 which if we plug into the S-B equation gives us 5Ėš C for any albedo value given Kirchhoff’s law.
It just so happens that the average ocean temperature by mass is apx equal to 5ĖšC and the atmospheric temperature by mass is apx equal to 5ĖšC. Which clearly demonstrates the accuracy of Kirchhoff’s law and the S-B equation. 341 W/M^2 in = 341 W/M’2 out, which essentially never changes! Which means that the average temperature never changes either!
But we aren’t really talking about average temperatures here are we? We are talking about transient boundary conditions where the surface of the oceans temp varies from 0 to 30Ėš and the temperature of the air immediately over the ocean has the same temperature.
But we aren’t really talking about that either are we? In fact the whole AGW meme is over the response time, the rate at which temperature lags radiation inputs.. And that is the question isn’t it? Because the lag time can range from a few seconds for a photon at 15 microns to travel out of the atmosphere impeded by CO2 ( and yes warming the atmosphere temporarily) to millenia if the photon is headed into the dark recesses of the ocean.
While we don’t know the exact response times, we do have clues don’t we? We know that CO2 levels have been much higher in the past and there were no hockey sticks, so we can falsify Hansen’s AGW theory out of hand. That doesn’t mean the GHG effect isn’t real, it is, we are just a little uncertain about its limits.

KevinK
April 24, 2013 7:58 pm

Gary Haldik wrote;
KevinK says (April 24, 2013 at 5:58 pm): ā€œDifferent thought experiment, different expected outcome.ā€
“So if itā€™s radioactively heated, you wouldnā€™t expect plate 1 to get warmer when plate 2 is introduced?”
Frankly I have not thought about that one yet, few of us folks out in the real world have access to radioactive heat sources, although I do scan ebay once in a while to see if I can get rid of this dā€”n utility company that wants my money in exchange for their energy.
But if it is radioactively heated (an effective ā€œunlimitedā€ heat source, anybody else remember back in the 50s and 60s when electricity was going to be ā€œtoo cheap to meterā€ ?) I would expect that the ā€œcycle timeā€ concerns would go away. And then plate 1 and plate 2 (and all the unicorns in the neighborhood) would all be warmer and cozier. So as soon as the Sun stays UP all day the GHE should show up. I could of course model that for say about $2 million US (oh wait I forgot about all the travel to warm exotic places to WARN everybody about the terrible dangers presented by warm cozy unicorns) Better make that $1 billion.
Cheers, Kevin.

Reed Coray
April 24, 2013 8:02 pm

Dr. Spencer, whom I greatly admire, wrote:
Roy Spencer says: April 24, 2013 at 12:12 pm
Greenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, thus making the warm side warmer, and the cool side cooler.ā€

I disagree with the above statement. I am going to apply what you wrote to a vacuum thermos bottle. Place coffee in a thermos bottle and place the thermos bottle in a heat sink at a fixed temperature lower than the temperature of the coffee. The thermos bottle has a “warm side” (the wall of the chamber that holds the coffee–i.e., the inside wall) and a “cool side” (the wall in contact with the heat sink–i.e., the outside wall). In a vacuum thermos bottle, the volume between these two “walls” is empty–i.e., is a vacuum. According to what you wrote, relative to a vacuum thermos bottle inserting a greenhouse gas into the vacuum space will “reduce the rate at which heat flows from the higher temperature to the lower temperature.” If true, the performance of the thermos bottle (i.e., the time required for the temperature of the coffee to reach the temperature of the heat sink) is enhanced (lengthened) by the presence of the GHG. I don’t believe it. If this were true, then why does anyone manufacture vacuum thermos bottles? Wouldn’t CO2 (a greenhouse gas) thermos bottles be preferred?
Although you didn’t explicitly say so, if what you say is true, shouldn’t increasing the amount of GHG placed in this space result in improved thermos bottle efficiency? Do you really believe that relative to a vacuum thermos bottle, a thermos bottle with high pressure CO2 gas (or for that matter, CO2 gas at any pressure) in the region between the warmer (inside) wall and the cooler (outside) wall will reduce the rate thermal energy leaves the coffee, and therefore reduce the rate at which the temperature of the coffee declines?

joeldshore
April 24, 2013 8:03 pm

Genghis says:

Kirchhoffs law says that emission = absorption. Absorption at the TOA equals apx 341 W/M ^2 which if we plug into the S-B equation gives us 5Ėš C for any albedo value given Kirchhoffā€™s law.

(1) No Kirchhoff’s Law says emissivity = absorptivity, not emission = absorption.
(2) Kirchhoff”s Law is true only at a fixed wavelength. An object can have a completely different absorptivity for radiation from the sun, which is primarily at visible and near-IR and UV wavelengths, as it has emissivity for radiation at terrestrial temperatures, which is primarily at far-IR wavelengths.
(3) The 341 W/m^2 value is the amount that is incident at the TOA, not the amount that is absorbed.
And, no, we are not talking about “lag times”. We are talking about the ***rate*** at which energy is emitted and absorbed.
It is really painful to read all of these misconceptions. It is fine to not understand things, but please don’t trumpet your misunderstandings as wisdom.

April 24, 2013 8:09 pm

“explaination is falsified by Quantum mechanics”
A laser is not a demonstration of the GHE. QM in fact shows that radiation trapped inside a cavity with a continuous source simply creates a blackbody spectrum. This is the very origin of QM, of Planck’s Law. Trapped radiation does not change its own frequency distribution and hence its temperature.

April 24, 2013 8:09 pm

So dragonslayers there is your absolute proof that the substances must be a gain medium because they can be used as a laser medium.” ~LdB
So… your argument is that because you can excite CO2 molecules and then stimulate them to emit photons… therefore greenhouse effect?
>.>

davidmhoffer
April 24, 2013 8:10 pm

LdB;
The dragon slayers like all pseudoscience lunatics miss the fact we know QM is right we have tested it extensively and continue to test it everyday of every year and to date it has never been shown wrong.
>>>>>>>>>>>>>>>>
Well said. Every day engineers design and build everything from mundane kitchen appliances to nuclear reactors using the exact laws of physics applied in exactly the same way that Spencer and tjfolkerts and Joel Shore and others are explaining. The things they design work. When they don’t, it is always because there was a design error. There has never been a case otherwise. P*stma and company would have us believe otherwise.

joeldshore
April 24, 2013 8:13 pm

Reed Coray says:

Although you didnā€™t explicitly say so, if what you say is true, shouldnā€™t increasing the amount of GHG placed in this space result in improved thermos bottle efficiency?

No. Different effects dominate in different circumstances. Heat transfer is a process that can involve conduction, convection, and radiation. For the Earth, the only significant communication of energy out into space is via radiation. For a thermos bottle, conduction and convection of heat are more important (and CO2 probably doesn’t have much effect on radiation because the effect on reducing radiation from the Earth is a subtle interplay between the absorption and re-emission of radiation by greenhouse gases and the temperature structure of the atmosphere which controls how much they emit). Hence, in a thermos, conduction and convection would be increased by putting CO2 in and radiation would hardly be affected at all.

KevinK
April 24, 2013 8:18 pm

LdB;
OH GOD LORD, not the atmosphere acts like a laser argument AGAINā€¦..
So dragonslayers there is your absolute proof that the substances must be a gain medium because they can be used as a laser medium.
Absolute proof that the substances must be a gain mediumā€¦ā€¦ā€¦
Where to start;
1) The ā€œgain mediumā€ in a laser DOES NOT, EVER, NEVER, NO WAY, NO HOW provide energy gain, PERIOD, END OF SENTENCEā€¦ (moderator please add lots of periods if you have nothing else to do)
2) EVERY laser known to mankind CONSUMES FAR MORE ENERGY INPUT than it produces in coherent EMR (light) output, typically with a ratio of about 5%.
3) For example, a typical semiconductor laser diode produces say 5 milliwatts of coherent optical power and consumes something like 50 milliamps times 2.5 volts or 125 milliwatts of electrical power, for an efficiency of 5/125 = 4%. Go to DigiKeyā„¢ or RadioShackā„¢ and check the specs for an ā€œoff the shelfā€ laser diode, these are typical values.
SHOW US THE OPTICAL RESONANT CAVITY IN THE ATMOSPHEREā€¦ā€¦ā€¦ (again lots more periods needed here)
It turns out that making optical resonant cavities is in fact VERY DIFFICULT (hence there are very few ā€œgarage shopsā€ that make lasers).
A gas CAN be a gain medium (power gain, NOT energy gain) when properly confined and excited, but this is HARD to do.
Cheers, Kevin.

LdB
April 24, 2013 8:19 pm

@Reed Coray
You can’t use classic physics to try and understand any of this CLASSIC PHYSICS is wrong or at best a gross simplification you are dealing with a quantum effect.
The anti-greenhouse position is basically saying a non gain medium does not exist it was an argument that the advent of the laser disproved. There are mediums that have a gain factor because of quantum mechanics.
You can’t talk your way around this with classic physics garbage its a simple test can a medium have a gain factor QM said yes it can and the laser was invented.
The fact CO2 laser exist tells you that CO2 has a medium gain there is no other option or the CO2 laser wouldn’t work.
It is that simple and basic and no amount of stupid classic physics with thermos flasks is going to save you … simple question posed is CO2 a quantum gain medium answer yes …. end of story.
No possibly way to talk around it unless you want to alternatively explain how a laser works.

peterg
April 24, 2013 8:22 pm

If there was a gas layer where the major energy inputs were ozone absorbing shortwave (solar radiation, especially UV) and convective inputs from below, and the energy outputs were primarily CO2 molecules radiating long wave to outer space, then extra CO2 could well have a cooling effect, because more CO2 molecules would cause more radiation. This effect would probably be in the nature of a negative feedback rather than an overall cooling.

April 24, 2013 8:37 pm

The fact CO2 laser exist tells you that CO2 has a medium gain there is no other option or the CO2 laser wouldnā€™t work.” ~LdB
Ok, are you really equating the process of exciting CO2 molecules and stimulating emission of photons… to the description of the greenhouse effect?

Tim Folkerts
April 24, 2013 8:38 pm

Joe says: “Here are some textbooks examples of how a continuous source of heat as a sphere trapped inside a shell doesnā€™t cause an increase of temperature of there sphere … There is nothing in the two examples about the input being reduced ā€“ the input is constant. “
SIGH.
No, those examples have a constant TEMPERATURE with a DECREASING heat source –exactly the opposite of what you are claiming. The whole POINT is to calculate the decrease in power required. The heat flow required to hold a constant temperature drops (from 240 W/m^2 to 120 W/m^2 for a thin shell close around the inner sphere held at 255 K using my specific numbers). If the power was turned back up to 240 W/m^2, the inner sphere would warm up (to 302 K from 255 K using my numbers).
” The 610 W/m^2 average input is the integrated average intensity over the hemisphere.”
No. That is still the integrated average at the equator. You are over-estimating the heating power of the sun. (Interestingly, this will STILL only get the equator up to a 24 hr average temperature ~ -2 C. That is warmer than -18C, but not nearly warm enough to account for the surface temperature.)
“The only way to get higher temperatures than this from sunlight at the Earth is to re-condense it using a magnifying glass.”
An, but we don’t have to rely on just sunlight to warm things. The very hot sun @ 5800K only covers about 0.001% of the sky. If the rest of the celestial sphere were @ 2.7 K, then the we get those temperatures we both know well (~ 361 K @ noon @ the equator (with vanishingly small heat capacity) or ~ 255 K averaged over the whole earth over 24 hr).
But we could put something ELSE over the other 99.999% of the celestial sphere surrounding us. Even something quite cool that covers ~ 100,000 times as large of a solid angle as the sun will integrate out to be a significant amount of added power to the ground.
And of course, that “something else” is clouds and GHGs that can radiate some energy IN ADDITION TO the sunlight.

Scott
April 24, 2013 8:38 pm

Thanks so much for posting the link to this Anthony. For those of you who wonder why Roy would write such a post, go look at the comments on most of his articles for the last several months–they’re basically dominated by people saying the GHE doesn’t exist…even when the topic has little to do with the GHE. It’s very unfortunate because Roy has some VERY good material at his site. His blog entry immediately before the one linked here is, in my opinion, the number one argument skeptics have on their side…yet the comments were overrun by GHE naysayers, really ruining the discussion. šŸ™
-Scott

April 24, 2013 8:43 pm

Tim, I’m not sure if you’re getting the same thing from those examples, are you talking about problem 1023 and 1026 from the links I gave?
Oh, and LdB: http://upload.wikimedia.org/wikipedia/commons/4/48/Commercial_laser_lines.svg
Xenon, Neon, Helium, Argon, Flourine, CO2 and various other elements can be used in a laser, would we have a greenhouse effect if we replaced CO2 with Ruby or Sapphire? (joking of course, but if you wanted to compare it more fairly, work out what a thin shell of ruby or sapphire around the planet would do)

Mooloo
April 24, 2013 8:47 pm

It is ā€œback radiation doesnā€™t affect the temperature of the sourceā€œ, davidmhoffer, the source.
Sources don’t have a temperature. They have an energy flow. (That’s what being a source means. Otherwise they would just be a hot thing.)
Their temperature is determined by the ability of the heat to get away. If I put a electric element into a block of copper and a similar mass block of concrete, the concrete will reach a higher temperature at equilibrium. Because it’s an insulator and copper is a conductor. The energy flow in will be the same, but the concrete will need to reach a higher temperature before its energy flow out matches the energy put in.
If you cannot tell the difference between temperature and energy you should leave any thoughts about discussing physics behind.
If I understand the sky-dragon position, a cold mirror should not reflect radiation back at the source. After all, by their “logic” it is only the temperature difference that matters. So how is it that mirroring is an effective means of insulation?

April 24, 2013 8:52 pm

“those examples have a constant TEMPERATURE with a DECREASING heat source ”
They say nothing about a decreasing heat source. You are making things up. They can be thought of as infinite heat sources (usually called a sink). You are changing the meaning of the questions. Radiation trapped inside a cavity doesn’t cause the cavity to heat up above the temperature of the source in any case, it just creates a blackbody spectrum at the temperature of the source.
“No. That is still the integrated average at the equator.”
It is the integrated average over the entire hemisphere. Which is the same as over the equator because an equatorial line is symmetric rotated about the solar zenith. You can literally feel hot sunlight and you can feel very hot rocks, surface, etc., which is created by the sunlight. This then heats the air, and the air circulates the heat about. Then you have the natural lapse rate making sure that the bottom of the atmosphere is the hottest part of the atmosphere, and the bottom of the atmosphere is not representative of the whole ensemble temperature in any case. The majority of the gas (N2 and O2) has very low emissivity and so can hold a much higher temperature than otherwise. Plus the heating only occurs mainly at the bottom of the atmosphere.
“that ā€œsomething elseā€ is clouds and GHGs that can radiate some energy”
GHG’s do not radiate at 5800K nor even at 286K. They never radiate at a warmer temperature than the surface and so can not heat the surface. GHG’s do not increase the sky-area of solar radiation and so they do not contribute to the solar spectrum at 5800K or the heating from such.

LdB
April 24, 2013 8:56 pm


Yes if we had ruby and sapphire in the atmosphere yes they too would greenhouse gases.
Here is a crash coarse in QM and the stupidity that is classic physics.
The classic physics view of temperature is it is the sort of the speed of the molecules in the substance. QM scientists 100 years ago realized that is a really bad approximation it is a number of complex quantum properties that make up an effect namely it appears to make a column of liquid expand in a thin tube. See the funny thing in QM there is not even a property called temperature because it is a mixture of different quantum properties.
Now under classic physics absolute zero is the temperature at which all motion of molecules stops and you can’t go any colder that’s not quite so under QM and here is the write up in nature magazine
http://www.nature.com/news/quantum-gas-goes-below-absolute-zero-1.12146
So before you start lecturing me about you massive understanding of temperature and how it works perhaps you can explain how we get a temperature colder than absolute zero to show us all you really really understand temperature and your classic physics works in all world cases.
See the problem classic physics has it is wrong has been for 100 years.
There isn’t an easy way to understand greenhouse effect under classic physics because its a quantum effect and nor can you explain a laser or temperatures below absolute zero using classic physics.

davidmhoffer
April 24, 2013 9:00 pm

Scott;
His blog entry immediately before the one linked here is, in my opinion, the number one argument skeptics have on their sideā€¦yet the comments were overrun by GHE naysayers, really ruining the discussion. šŸ™
>>>>>>>>>>>>>>>>
You just triggered a thought. Dr Spencer helped design, implement and operate a network of satellites that accurately (verified by weather baloons, weather stations, etc) measures temperatures all over the globe and at many altitudes. Yet he is besieged by nay sayers insisting he doesn’t understand atmospheric physics. How do they suppose he got the satellites working correctly? A gigantic cosmic coincidence?

LdB
April 24, 2013 9:06 pm

To all you dragonslayers who want to keep arguing this stupidity here is a straight forward effect to explain away using your great physics.
We call it laser cooling that is cooling using something you would consider hot
http://en.wikipedia.org/wiki/Laser_cooling
It is routinely done now as the normal way to bring things down to absolute zero and on very large scales
http://web.mit.edu/newsoffice/2007/super-cool.html
So use your classic physics and explain what is happening .. small open challenge.

April 24, 2013 9:07 pm

Negative temperatures aren’t news, LdB, and I’m not operating from a classical position.
I’m not sure why you think I’m somehow ignorant of quantum mechanics, but I could also ask you which quantum paradigm you’re working within.
Spoiler alert: just because someone disagrees with you, it doesn’t mean they are ignorant of modern physics.

tjfolkerts
April 24, 2013 9:07 pm

Max asks: “Tim, Iā€™m not sure if youā€™re getting the same thing from those examples, are you talking about problem 1023 and 1026 from the links I gave?”
Yes, I’m talking about the same examples. Specifically, 1026 says …
* The object is “at absolute temperature T” and is held at that temperature throughout the problem.
* The goal is to “find the factor by which this radiation shield reduces the rate of cooling”. Now “reduces the rate of cooling” is a bit ambiguous, but then the solution starts out with “the rate of energy loss”, so what they are finding is reduction in heat loss by radiation– which would be equal to the input of heat to maintain the constant temperature T.
* The only thing that is changing is that the temperature of the surroundings goes from T(0) to T(1). We have been assuming that T(0) was ~ 0 K, but that is not necessary.

Rosco
April 24, 2013 9:13 pm

Per problem 1023 from the link above –
for the radiation the limit approaches 0.5 for R approximately equal to r and it approaches 1 if R >>> r and the.
At least this “slays” Willis’ steel greenhouse completely !
The shell is always at a lower temperature than the enclosed object with the maximum being 84 %
In fact I am still not sure there isn’t something wrong with the whole concept of using a “nearby” heat shield.
This model has extra heat shields eventually resulting in almost no radiation escaping –
1/2 + 1/4 + 1/8 + 1/16 and we are already up to 93.75 %.
But at least it is progressing away from infinite energy by adding more shells unlike Willis’ proposal.

Greg House
April 24, 2013 9:13 pm

REPLY: …The greenhouse effect … is about slowing the progress of LWIR to the top of the atmosphere. Without CO2 or other GHGā€™s the LWIR would proceed quickly to space, and the Earth would cool faster, and have a lower average temperature at night. The ā€œbackradiationā€ has a limited scope of effect, but it is there as part of the trasnfer process. … GHGā€™s act as a LWIR transfer regulator from the surface to the top of the atmosphere where it is radiated into space. They slow the transfer. Without GHGā€™s nightitme temperature would drop quickly as LWIR is lost directly to space. […] ā€“ Anthony
===============================================================
OK, Anthony, as I said, this is not exactly the IPCC concept, but I would like to address it anyway. For the sake of clarity, when I say “Earth” I mean the solid Earth. So, there is the Earth and the atmosphere around it.
Now, the first problem in your approach is the idea that the radiative cooling of the Earth surface (let us say at night to simplify it) is somehow connected to what happens to the radiation after it leaves the Earth surface, like meeting some “greenhouse gases”. Well, if we ignore conduction/convection, then it is simple: what is gone is gone. The radiation is gone and the surface cools accordingly. If the radiation meets something else and warms it, fine, but this fact alone does not affect the temperature of the surface. To affect the temperature of the surface there must be additional energy coming to it.
Now, at the first glance, the “greenhouse gases” are doing exactly that: providing additional energy by back radiating towards the Earth. The problem is, however, that back radiation can not warm the source, this is physically impossible. So we have this situation: back radiation is there, but no effect on the temperature of the source is possible.
That’s it, actually. How long and hard the way of the radiation from the surface to space is, with all this bouncing or delays or whatever, does not matter. All this happens outside the surface and gives the surface no additional energy.
Sometimes at this point people start asking “but where does the back radiation go?” or “how it is possible that it does not warm?”. My answer usually is like “why should I care and explain everything?” I only care about effect on the temperature of the source, and this is exactly something that leads to an absurd, as I demonstrated earlier, so this is sufficient. We can, of course, start speculating about it and make up theories, but why should we? Anyway, it is clear, that the well known IPCC radiation arithmetic has nothing to do with reality. The radiation simply does not obey the laws of arithmetic. It is a different, special thing. All this talk about photons “not knowing” anything is misleading. If you like, the fact is that something we call “photons”, despite being a countable noun, does not obey the laws of arithmetic and can not affect the temperature of the source. This is the reality. No “greenhouse effect”.

LdB
April 24, 2013 9:20 pm


The explain laser cooling which is the essence of your stupid argument that hot can’t cool by definition laser cooling does exactly that.
And if you understand that you can have temperatures colder than absolute zero then your realize that classic physics can be wrong because it is a bad simplification and therefore you should not be surprised that classic physics can’t be used to understand the greenhouse effect because its a quantum problem … that is if you really aren’t ignorant.

April 24, 2013 9:23 pm

* The object is ā€œat absolute temperature Tā€ and is held at that temperature throughout the problem.
Uh? Problem 1026 says no such thing.
Verbatim:
A spherical black body of radius r at absolute temperature T is surrounded by a thin spherical and concentric shell of radius R, black on both sides. Show that the factor by which this radiation shield reduces the rate of cooling of the body (consider space between spheres evacuated, with no thermal conduction losses) is given by the following expression: aR^2/(R^2 + br^2), and find the numerical coefficients a and b.Problem 1026

April 24, 2013 9:28 pm

LdB, I’m not engaging in childishness with you, sorry bucko.

April 24, 2013 9:29 pm

“So how is it that mirroring is an effective means of insulation?”
So what you are saying is that if you took a whole bunch of mirrors and had them focus on a point you could cook something right?
At night too?

TomR,Worc,MA,USA
April 24, 2013 9:32 pm

Noelene says:
April 24, 2013 at 3:08 pm
So there is a consensus on the greenhouse effect?
===========================================
I see what you did there.
Not that I agree with you, but well done.
TR

LdB
April 24, 2013 9:42 pm


It isn’t a childish argument its the fact the dragonslayers make at the centre of the argument that heat can’t cool.
Problem is laser cooling does exactly that it is hot put your hand infront of the beam if you want to see how hot and it cools to almost absolute zero.
In there world it isn’t possible but it happens a fact they like you have to ignore because they can’t explain it and there whole argument goes up in smoke.
The hard sciences are that explicit about this topic on in AGW does this sort of garbage get air time.

April 24, 2013 9:56 pm

Roy Spencer says:
April 24, 2013 at 12:12 pm re: carbon dioxide and insulation
===================================================
The effect is the same but the mechanism is completely different. Insulation does not absorb and radiate photons. The IR photons inside a house don’t even make it past the paint on the sheetrock, much less reach the insulation. Funny how the ocean comes to mind.
House insulation is just a down sleeping bag. A tiny bit of thermal mass and a huge barrier to convection.

Gary Hladik
April 24, 2013 10:21 pm

KevinK says (April 24, 2013 at 7:58 pm): “But if it is radioactively heated (an effective ā€œunlimitedā€ heat source, anybody else remember back in the 50s and 60s when electricity was going to be ā€œtoo cheap to meterā€ ?) I would expect that the ā€œcycle timeā€ concerns would go away. And then plate 1 and plate 2 (and all the unicorns in the neighborhood) would all be warmer and cozier.”
Thanks, Kevin, for that, er, picturesque response.
Back to the battery-operated plate 1: Why do you say the battery output would increase when plate 2 is introduced?

Konrad
April 24, 2013 10:26 pm

Dr Spencer and other lukewarmers are wrong. Radiative gases cool our atmosphere at all concentrations above 0.0ppm. The problem is not with physics of radiative gases, but with fluid dynamics and gas conduction
For the record I am not a ā€œSlayerā€, my following comments are based on my own empirical experiments. These experiments disprove the radiative green house effect. I have described simple versions that other readers can build and run for themselves here –
http://wattsupwiththat.com/2013/04/05/a-comparison-of-the-earths-climate-sensitivity-to-changes-in-the-nature-of-the-initial-forcing/#comment-1267231
I would urge other readers to try these experiments for themselves. There is far to much cut&paste and linking on climate blogs and not enough empirical work. These experiments relate to the ā€œDo notsā€ of climate modelling –
A. Do not model the ā€œearthā€ as a combined land/ocean/gas ā€œthingyā€
B. Do not model the atmosphere as a single body or layer
C. Do not model the sun as a Ā¼ power constant source without diurnal cycle
D. Do not model conductive flux to and from the surface and atmosphere based on surface Tav
E. Do not model a static atmosphere without moving gases
F. Do not model a moving atmosphere without Gravity
G. Do not model the surface as a combined land/ocean ā€œthingyā€
H. Do not apply SB equations to a moving gaseous atmosphere (the davidmhoffer rule šŸ˜‰
Does the surface emit IR? Yes.
Does the atmosphere absorb IR and heat? Yes.
Does the atmosphere radiate some IR back to the surface slowing it’s cooling rate? Yes.
Is there a radiative GHE? Yes.
Is the NET effect of radiative gases in our atmosphere warming? no.
To understand the role of radiative gases in the atmosphere, the conditions in a non radiative atmosphere should first be considered. Radiative gases are critical to strong vertical convective circulation in the troposphere. It should be noted that above this level there are few radiative gases and little vertical convective circulation. The role of energy loss at altitude in convective circulation is demonstrated in Experiment 3. It should also be noted that pneumatic heating and cooling of rising and descending air masses does not represent energy gain or loss. IR emission to space is the primary way gases at altitude lose energy.
If there were no radiative gases in our atmosphere, gases would still be heated by surface conduction. These gases would convect to altitude, but they could not lose energy and buoyancy and descend. Full convective circulation would stall. Experiment 4 demonstrates that a gas column heated at the base and cooled at the top has a lower average temperature than a gas column heated and cooled at separate locations at the base. In Experiment 4 box 1 has a higher ā€œsurfaceā€ temperature and a lower gas temperature. In box 2 the average ā€œsurfaceā€ temperature is lower yet the average gas temperature is higher. Box 2 models a non radiative atmosphere. Experiment 4 also demonstrates why an atmosphere heated by surface conduction will have its temperature set by surface Tmax not surface Tav.
Experiment 5 demonstrates that for a moving gaseous atmosphere in a gravity field, the surface is better at conductively heating the atmosphere than it is at conductively cooling it.
Dr Spencer has not correctly modelled the critical role radiative gases play in convective circulation or the importance of this to atmospheric temperatures. The land surface under a non radiative atmosphere may have a lower Tav, however the moving gases in the atmosphere above will have a higher average temperature than present. Build experiment 4 big enough (limiting equalisation through gas conduction) and you can find out just how dramatic the difference would be. If atmospheric temperatures are higher for a non radiative atmosphere, then radiative gases must act to cool the atmosphere at all concentrations above 0.0ppm.

Gary Hladik
April 24, 2013 10:34 pm

Reed Coray says (April 24, 2013 at 8:02 pm): “Although you didnā€™t explicitly say so, if what you say is true, shouldnā€™t increasing the amount of GHG placed in this space result in improved thermos bottle efficiency?”
No. By introducing a gas to the space between layers, you’ve added conductive cooling to radiative cooling of the inner shell, which would overwhelm any reduction in radiative cooling by such a thin layer of gas.
Think of the top of our atmosphere as the inner layer of the vacuum thermos bottle. With nothing but a vacuum “outside”, the only significant cooling mechanism is radiation. Convection and conduction are major energy transport mechanisms within the atmosphere, but don’t function in a vacuum.

peterg
April 24, 2013 10:52 pm

I do find all this denial of basic radiative physics fairly boring. The water vapour cloud feedback loop cannot be discussed without presuming the basics of radiation physics, and when assertions based on these laws get challenged, the discussion becomes a bit pointless.

Gary Hladik
April 24, 2013 10:56 pm

Maxā„¢ says (April 24, 2013 at 9:23 pm): “Uh? Problem 1026 says no such thing.”
Out of curiosity, if the problem says there is a constant non-zero power input to the inner sphere, and the shell reduces the rate of cooling by (for example) roughly 20%, what do you think would happen to the sphere’s equilibrium temperature after the shell is added? Would it be higher, lower, or the same?

April 24, 2013 11:05 pm

I would be interested in anyone’s comment on this:
I think the classical 33Ā° analysis falls apart because it assumes albedo is the input and solves for the greenhouse effect. This is not how heat is managed on this planet. Albedo is the output. Try it this way instead:
Start with the albedo of areas including everything but clouds first. Include areas with mist / haze, but not clouds (no small task). This will be considerably darker and the SB temperature much higher. Solve for the SB equilibrium temperature. What did you get? A:__________
Note: This is what the situation would be if the earth were at a radius from the sun farther than it is, not including the effect of increasing icecaps. We are too close to the sun to not employ evaporative cooling which is the main mechanism of heat regulation. This really needs to be done as an integral over one rotation.
The clouds are naturally produced to balance the rest of the energy input such that the surface temperature lands where it is. The first answer (A) is the natural temperature of the earth without the main cloud effects.
Now take the measured surface temperature (B). A-B is the cooling effect of water vapor + the warming effect of all other GHGā€™s. We donā€™t know the exact magnitude of either, just the net from this exercise.
But we can measure the effect of water vaporā€™s GHG effect in many areas (like buoys) where we measure LW, T and humidity at the same location all the time. So we know what it is with varying temperatures and humidity which we can uses to estimate humidityā€™s role on the received LW radiation from the atmosphere, and deduce what ā€œall otherā€ effects total to. I donā€™t think anyone has so far integrated the moisture data well enough to describe what local effects (storms) are occurring based on the shapes of the curves over time.
Once people give up on the absurd idea that albedo is an input to the equation, the rest of the results should be fairly easy to estimate. Water is going to be the major net driver in the tropics (due to reflection and carrying water droplets aloft where they can radiate out with far less optical thickness), but water vapor will also be a warmer in other conditions. CO2 will be shown to be almost meaningless in the face of H20ā€²s effect.
Albedo is the earthā€™s output, it is a reaction to the input (solar energy), and internal variation.
So let’s try it. Using the 20% that NASA estimates is reflected from clouds, the results should look like this:comment image
As you can see, treating albedo as an output (since clouds are generated to reflect and transport heat to space), now you only need to account for 16.6Ā°K after the effect of cloud reflection alone is accounted for. It’s a start and I suspect this is one of the major links that supports why sensitivity data is without question coming in at half or less of the flawed classical analysis. Note also, this is not a feedback effect. It is entirely the result of the initial albedo assumption being wrong.comment image

Alec M
April 24, 2013 11:11 pm

I wrote a comment on the original Spencer article describing how most people misinterpret the S-B equation in Climate Alchemy. It’s because the atmosphere is semi-transparent so cannot be considered a grey body. This is one of the three mistakes made by Houghton. The others are to fail to understand you can’t apply the two-stream approximation at an optical discontinuity and the Earth does not emit IR as if it were an isolated black body in a vacuum.
Also, there is no net surface IR from the Earth’s surface in the main GHG IR bands. This means there can be no CO2-AGW from this cause. There is no ā€˜back radiationā€™ [because for a normal temperature gradient it is annihilated at the surface]. Pierrehumbert does a good job of saving the blushes of Climate Alchemy by his argument about the ā€˜CO2 biteā€™ in OLR. However, this heating, ~3 W for doubled CO2 is offset by a process involving clouds. Iā€™m writing a paper on this.
The Earthā€™s climate is controlled by a negative feedback control system involving the optical properties of clouds so temperature is independent of CO2 concentration. Forget surface IR, that is mostly irrelevant for CO2.

Konrad
April 24, 2013 11:25 pm

peterg says:
April 24, 2013 at 10:52 pm
ā€œI do find all this denial of basic radiative physics fairly boring.ā€
———————————————————————————–
The reason the AGW hypothesis fails is not because of any problem with radiative physics. The radiative physics is fine* The problem is that the fluid dynamics of the atmosphere and the surface to atmosphere conductive flux were modelled incorrectly.
Radiative gases are critical to tropospheric convective circulation, without which our atmosphere would heat dramatically due to surface conduction.
The reality of an atmosphere in a gravity field in which the gases are free to move is that the surface is far better at conductively heating the atmosphere than it is at cooling it.
*Except for that little mistake of assuming that the oceans could have their cooling rate slowed by incident LWIR. Not a huge mistake, only 71% of the earthā€™s surface, probably nothing to worry about šŸ˜‰

Gary Hladik
April 24, 2013 11:26 pm

Rosco says (April 24, 2013 at 9:13 pm): “At least this ā€œslaysā€ Willisā€™ steel greenhouse completely !”
Eh? The problems described are in complete agreement with the “Steel Greenhouse” article. Far from “slaying” the steel greenhouse, this reference work vindicates Willis completely.
Check out Figure 1 here:
http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/
The shell (approximately the same size as the planet) radiates–gasp!–exactly half as much as the planet, just as problem 1023 says it should! Coincidence? šŸ™‚

Gary Hladik
April 24, 2013 11:39 pm

Konrad says (April 24, 2013 at 10:26 pm): “Dr Spencer and other lukewarmers are wrong. Radiative gases cool our atmosphere at all concentrations above 0.0ppm.”
So that would mean that, just as absence of positive correlation between temperature and CO2 argues against a significant warming effect, the absence of a negative correlation between temperature and CO2 argues against a significant cooling effect?

davidmhoffer
April 24, 2013 11:39 pm

Konrad;
The problem is that the fluid dynamics of the atmosphere and the surface to atmosphere conductive flux were modelled incorrectly.
>>>>>>>>>>>>>>>>
So you’re saying that the satellite temperature data is wrong?

Pierre-Normand
April 24, 2013 11:54 pm

Greg House said: “Again, this is not what the IPCC supports. You can invent whatever effect you want, but only the IPCCā€™s one is politically relevant, because governments and agencies refer to the IPCC reports and recommendations, not to davidmhoffer or Steven Mosher.”
The IPCC AR4 WG1 glossary defines the Greenhouse Effect thus: “Greenhouse gases effectively absorb thermal infrared radiation, emitted by the Earthā€™s surface, by the atmosphere itself due to the same gases, and by clouds. Atmospheric radiation is emitted to all sides, including downward to the Earthā€™s surface. Thus greenhouse gases trap heat within the surface-troposphere system. This is called the greenhouse effect. Thermal infrared radiation in the troposphere is strongly coupled to the temperature of the atmosphere at the altitude at which it is emitted. In the troposphere, the temperature generally decreases with height. Effectively, infrared radiation emitted to space originates from an altitude with a temperature of, on average, ā€“19Ā°C, in balance with the net incoming solar radiation, whereas the Earthā€™s surface is kept at a much higher temperature of, on average, +14Ā°C. An increase in the concentration of greenhouse gases leads to an increased infrared opacity of the atmosphere, and therefore to an effective radiation into space from a higher altitude at a lower temperature. This causes a radiative forcing that leads to an enhancement of the greenhouse effect, the so-called enhanced greenhouse effect.”
This seems to be exactly what Mosher and Hoffer have been saying. The backradiation and the ERL arguments are equivalent and both are explicitly endorsed by the IPCC.

Kristian
April 24, 2013 11:59 pm

tjfolkerts says, April 24, 2013 at 9:07 pm:
“Yes, Iā€™m talking about the same examples. Specifically, 1026 says ā€¦”
http://books.google.no/books?id=dQGC0ifkE34C&pg=PA24&lpg=PA24&dq=concentric+sphere+black+body&source=bl&ots=Zh6N1e35jc&sig=m-7nVWch4_zv-l3ISR5k7bluSUQ&hl=en&sa=X&ei=_ldYUd7EFZLU9ATVzYHoDw&redir_esc=y#v=onepage&q=concentric%20sphere%20black%20body&f=false
What Problem 1026 is telling us is specifically that if R ~ r, the Q = Q’ + Q”. This is the exact equivalent to the Carnot cycle: W = Qh – Qc, where Q=Qh, Q”=Qc and Q’=W:
http://i1172.photobucket.com/albums/r565/Keyell/Carnot_zps4049e783.jpg
At any point in time, the heat flux leaving the surface of the sphere (Q) corresponds directly to (depends on) its specific emission temperature, which in turn depends fully and solely on (is set by) the heat flux provided to it by its heat source (so Q equals the heat input).
Q then goes half into warming the shell and keeping up its equilibrated temperature (Qā€™), half from the new layer to space (Qā€): Q = Qā€™ + Qā€. The entire flux comes from the sphere. It is simply split up upon absorption by the shell.
No need to invent secondary heating of the sphere to achieve radiative balance, beyond what the input from its heat source can manage on its own. Before the shell was emplaced, the power from the sphereā€™s heat source went into warming the sphere only (one object). After, the same amount of power goes into warming the sphere AND the shell (two separate objects). Hence the smaller flux to space. Itā€™s that simple ā€¦ Q is not reduced. The radiative flux to the cold reservoir (the surroundings) (before, Q, after, Q”) is reduced. This does not increase the surface temperature of the sphere. It maintains the temperature of the shell (Q’).
If the sphere were held at a constant temperature it would be because it received a constant energy supply from its heat source. With a variable heat source, the temperature would not be constant.
Let’s compare Problem 1026 with Problem 1023:
http://i341.photobucket.com/albums/o396/maxarutaru/bunnytaru/lol%20censorship/heatsourceinshellatequilibrium_zps1c9d662a.png
1023 specifically asks: “(…) what is the effect of the shell on the total power radiated to the surroundings?
As you can see, J (Q) doesn’t change. J is simply replaced by J1 as ‘the total power radiated to the surroundings’ as the shell is put around the sphere. The only thing that changes is that ‘the total power radiated to the surroundings’ is halved. The rest goes into heating and maintaining the temperature of the shell, something that J didn’t have to do before the shell was emplaced. Perfect radiative balance: Q = Q’ + Q” — J = (J – J1) + J1.
The sphere/shell is also specifically likened to a star and a surrounding dust cloud (b). A star is pretty constantly heated, isn’t it?
There are more examples of course showing the exact same thing, that the source of the shell’s or the insulating layer’s incoming (warming) radiative heat flux (across a vacuum) does not heat up some more when surrounded or covered by the shell/insulating layer. The ‘insulating effect’ simply amounts to reducing ‘the total power radiated to the surroundings’ (Q –> Q”). There is no energy ‘piling up’ at the surface of the sphere/inner layer. Q (J) remains unchanged.
But we can get back to those later …

LdB
April 25, 2013 12:29 am

@Konrad
The greenhouse effect can heat or cool depends on the energy input and outputs and I am not sure any scientist ever said anything different.
I can also make a laser beam which has an extremely high temperature heat or cool things depending on how we up the relationship and frequency of the beam..
The problem with classic physics it is familiar only with heat only heating and so you struggle with the problem because your understanding of temperature is sort of simplistic and wrong. Roy is trying to bring a QM effect back into sort of classic physics so you can understand it but at some point you can’t.
Try understanding laser cooling and reading about it because sort of shows your problem and understanding of heat and temperature.

Konrad
April 25, 2013 12:35 am


The cooling effects of changing CO2 levels from 0.03 to 0.04 percent would be unmeasurable with current technology.

Gary Hladik
April 25, 2013 12:54 am

Konrad says (April 25, 2013 at 12:35 am): “The cooling effects of changing CO2 levels from 0.03 to 0.04 percent would be unmeasurable with current technology.”
Heh. I suppose the same could be said of the alleged warming effect. šŸ™‚

April 25, 2013 1:15 am

LrB, the childishness was you slinging terms like ignorant around when you don’t seem to recognize the problem with your own argument.
A laser is not a spontaneous transfer from hot to cold, thermal radiation is a spontaneous transfer from hot to cold.
A laser involves doing work, and the production of waste heat.
Look into the Clausius statements of the laws of thermodynamics, as well as Carnot efficiency before calling others ignorant, perhaps?
___________
The shell (approximately the same size as the planet) radiatesā€“gasp!ā€“exactly half as much as the planet, just as problem 1023 says it should! Coincidence? šŸ™‚” ~Gary
The shell in 1023 does this without the implication that the inner body would increase in temperature.
Essentially it demonstrates that the steel greenhouse involves these steps:
1. Get an internally heated body
2. Add a shell around it which radiates half as much inward and outward as the inner body
3. A wizard does something
4. The temperature of the internally heated body increases

Rosco
April 25, 2013 1:36 am

I cannot agree that radiation would quickly escape to space if GHGs did not slow the process.
It is inconceivable that greenhouse gases at minus 18 degrees C and trace concentrations could radiate as powerfully as the solid hot Moon surfaces.
NASA have clearly demonstrated that the heated Moon’s surface radiating effectively into a vacuum loses energy at a slow rate. From 390 K at lunar noon to 220 K at lunar sunset is a 170 K change in temperature in approximately 177 Earth hours – a rate of 1K per hour.
The Moon cools far more slowly once the Sun sets – from 220 K to about 100 K in 354 Earth hours.
Although there is less data for Mercury, the night on Mercury is some 2111 Earth hours long so cooling from ~740 K to ~90 K is easily understood.
There is no quick reduction in temperature of heated surfaces radiating into a vacuum.
Surely we observe a similar phenomenon here on Earth ?
The poles receive the least direct Solar radiation of anywhere on Earth and it is the fact that zero direct solar radiation is received for some months that is surely responsible for the extreme cold.
These are observable facts – there is no need of convoluted arguments to explain simple truths.
Anywhere in the Universe that is cold is so because of low levels of radiation. There is evidence that cooling from a warm state to a cold state by radiating to space takes significant time – NASA says so.
I believe them.
Besides the vacuum space at the Earth’s orbit cannot by any stretch of the imagination be thought of as cold with ~1370 W/sq m of solar radiation continually emitted by the Sun.
NASA prove the cold Earth orbit space myth in their “When we left the Earth” series when the solar panels and “parasol” – NASAs words – fitted to Skylab failed to deploy.
The astronauts that arrived to fix the problems found the temperature inside Skylab exceeded 70 degrees C.

Selgovae
April 25, 2013 1:49 am

Joseph E Postma wrote, “In any case, you can shine a flashlight at the Sun and the the beam will be there. But the flashlight doesnā€™t heat the Sun.”
Joseph, I have trouble with the idea that “the beam will still be there”. Does that imply that the source will lose energy (now in the beam?) and thus cool. In which case, where does the energy go? I find it easier to envisage that the beam will not be there and thus the source gets warmer as it sheds energy at a lower rate.
BTW, I’m not sure that shining a flashlight at the sun is a good thought experiment. I’d assume that some energy would probably be absorbed by the sun, just based on the color of the light from the flashlight. Pointing a finger at the sun might be a better example.

Konrad
April 25, 2013 1:53 am


There are few problems with Dr Spencers method of measuring the microwave emissions from O2 molecules. It is far more accurate than surface station data.
This system may however start to over read as global cooling occurs over solar cycle 25. Reflection from increasing snow and ice cover may be an issue. However correlation with radiosonde balloons during overpass is an eazy fix.

Rosco
April 25, 2013 1:56 am

The other thing I want to comment on is the explanation given by people about what happens when you place some sort of insulation over a heat source – commonly putting a lid on a pot of boiling water.
They claim the back-radiation increases the heat.
Well I say bunkum – you can agree or disagree but at least consider the following :-
Exposed to the conduction/convection of the atmosphere no heat source will ever achieve its heating potential. Reduce that energy loss by effective insulation and the heat source will achieve near its potential. Radiation loss reduction may or may not contribute to this effect.
We know, or should know, that “trapping” warmed air close to our bodies, by reducing convective loss of that warmed air, is the principal manner in which we stay warm.
I see people saying stupid things like the back-radiation from the ice of igloos keeps the occupants warm.
I say rubbish.
The igloo traps the warmer air inside, maybe even warming it enough to melt the ice a little. The occupants do not shed all their clothing and romp inside naked.
There is absolute proof about the fact that reducing convective heat loss is the principal means by which we stay warm – radiation trapping has little to nothing to contribute.
Animals have only one mechanism to keep warm – their fur or feathers. There is no radiation barrier to prevent radiative losses – none.
Animal fur reduces the loss of the warmed air by reducing convection – there is no other mechanism possible.
We know this is true as we understand the phenomenon of wind chill which acts to destroy the trapping of the warmed air.
As a qualified Environmental Health Officer I know I am on firm ground in discussing simple biology.

LdB
April 25, 2013 2:23 am


And now laser cooling is not spontaneous .. please describe how you arrived at this great piece of physics another of your great theories. You and Konrad look more and more ignorant the more you comment.
If you want to talk about work so lets put the two on even footing you have this big massive energy source in the sky we call it a sun it produces energy. I have a laser unit plug into a power socket so we both have a power source to do work they are equivalent get it functionally identical.
Now going down to the exchange level the transfer in laser cooling has to be instantaneous or you couldn’t get to zero degree kelvin that is in your classic garbage world how you define ABSOLUTE ZERO. Any non instant transfer would mean you would have motion try making an exchange of energies with a collision and have absolutely no movement.
The problem I have with you whackjobs is the same as Roy you are running up against some impossible physics for anything in classic physics to challenge because QM showed that classic physics is wrong 100 years ago. Instead of listening like all pseudoscience junkies instead of listening that you keep digging looking more and more stupid as you go.

April 25, 2013 2:34 am

If animals were relying on reducing radiative losses, to further support Rosco’s point, then evolution would probably begin producing nearly perfect mirrors, as that is by far the best radiative insulation.

April 25, 2013 2:40 am

Again, LrB, I live in a universe governed by relativistic and quantum physics, I know they are experimentally confirmed, and I know I do not live in a classical universe, learn how to read, kiddo.
Let me bold it for you:
I am not operating within a classical paradigm, the universe only resembles a classical one in the weak field regime, at small scales it is quantum, at large and fast scales it is relativistic, learn how to read before you go off on a rant next time.
A laser is not a spontaneous process, I know masers form naturally, but I’ve never seen a CO2 laser randomly appear and function.

April 25, 2013 2:54 am

TomR,Worc,MA,USAs
At April 24, 2013 at 9:32 pm you say

Noelene says:
April 24, 2013 at 3:08 pm

So there is a consensus on the greenhouse effect?

===========================================
I see what you did there.
Not that I agree with you, but well done.
TR

I answered Noelene and you may be interested in that answer. However, you may not have seen it because it was held in moderation for an hour and the thread had moved way past it when it appeared out of moderation.
However, P*stma saw my answer to Noelene because he quoted from it at April 24, 2013 at 4:53 pm and was unjustifiably offended by it.
In case you are interested, my answer to Noelene was at April 24, 2013 at 3:42 pm and this link jumps to it
http://wattsupwiththat.com/2013/04/24/spencer-slays-with-sarcasm/#comment-1286437
Richard

Gary Hladik
April 25, 2013 3:03 am

Maxā„¢ says (April 25, 2013 at 1:15 am): “The shell in 1023 does this without the implication that the inner body would increase in temperature.”
Really? OK, let’s assume the inner sphere has a constant heat source, as in the Steel Greenhouse, and is in thermal equilibrium with the shell as in problem 1023. Sphere temp is at T1. We agree that the power radiated by the system to its surroundings is half the power radiated by the inner sphere, as in both problem 1023 and Willis’s Figure 1, yes? (If not, why not?)
Now snatch away the “insulating” shell, real sudden-like. The system is suddenly cooling twice as fast, as per problem 1023, but its power input is unchanged! Oh no! What happens to the temperature of the inner (now alone) sphere? Increase? Decrease? Remain the same?
Let the system come to equilibrium, sphere at temp T2. Now add the shell back. Let the system come to equilibrium. Inner sphere is now at temp T3. What’s the relationship of T1, T2, and T3?
a) T1 = T2 = T3
b) T1 = T3 > T2
c) something else (specify)
and why?

Sergey
April 25, 2013 3:17 am

There are two different mechanisms of radiative cooling of gases: by emitting quants of wavelength corresponding to quantum resonance of moleculae and by emitting at all wavelengthes (so called blackbody radiation). The first mechanism prevails for greenhouse gases, but the second is fairly universal and applies also to nitrogen and oxigen. And even greenhouse gases such as water vapour and carbon dioxide do not necessary emit at their quantum resonance wavelength when excitated by short-range radiation but can transfer excessive energy by collisiom with moleculae of nitrogen and oxigen without emitting anything. How much energy is re-emitted by carbon dioxide and how much simply heat atmosphere without emitting anything is everybody’s guess, because quantum mechanical calculations are too cumbersome to perform even using supercomputers. And heating of atmosphere enhances convection and make convective layer depth higher, which can compensate or even overcompensate backscattering by CO2. So the real contribution of greenhouse effect to surface warming in real atmosphere with convection, evaporation, condensation and radiation cooling can not be calculated, measured or estimated by any other means. It can be arbitrary small or even negative.

steveta_uk
April 25, 2013 3:19 am

Greg House says: April 24, 2013 at 9:13 pm

Now, at the first glance, the ā€œgreenhouse gasesā€ are doing exactly that: providing additional energy by back radiating towards the Earth. The problem is, however, that back radiation can not warm the source, this is physically impossible. So we have this situation: back radiation is there, but no effect on the temperature of the source is possible.

(my bold).
Here you are just making an assertion without any backing.
For a simple example of how you are wrong, why are you assuming that the back radiation is from a colder source? Ground frosts at night are very common – they are caused by the surface losing energy by radiation, mostly directly to space, in which case the air close to the surface can clearly be warmer than the surface, esp. when the surface is a good radiator (e.g. a black car).
So even if you don’t believe in the “cooler can warm a warmer thing” idea, you don’t need to. Clearly GHG is providing energy to the surface.

A C Osborn
April 25, 2013 3:28 am

LdB says:
April 25, 2013 at 2:23 am
You are abusing Konrad who has carried out actual experiments which he has published on various blogs, so he does know what he is talking about.
Empirical results which show that theories are wrong are the way that science is supposed to be done.
I suggest that you also take a look at TheFordPrefect’s experiments which show that if Back Radiation exists at all it is a minute amount of energy, nothing like the 50% bandied about by the climatologists.

richard verney
April 25, 2013 3:34 am

Rosco says:
April 25, 2013 at 1:56 am
“…I see people saying stupid things like the back-radiation from the ice of igloos keeps the occupants warm.
I say rubbish.
The igloo traps the warmer air inside, maybe even warming it enough to melt the ice a little. The occupants do not shed all their clothing and romp inside naked.
There is absolute proof about the fact that reducing convective heat loss is the principal means by which we stay warm ā€“ radiation trapping has little to nothing to contribute….”
//////////////////////////////////////////////////////
I have lost count the number of times that I have pointed out that Igloos work by reducing convection and thereby by trapping heat.
The same is so of space blankets/insulatuion blankets. To have any substantial effect these need to be wrapped around the user thereby reducing convection and trapping heat. If they worked primarily by re-radiating heat, one could construct a space blanket much like a toilet roll. it could be say 6ft high and 2ft in diameter. The user could stand inside the mirrored roll and enjoy all the back radiated heat. But it would not be effective because of convection; significant heat loss would occur vertically. In fact from the radiative point of view, the diameter could be 3ft or 10ft, and the user would theoretically enjoy the same levels of re-radiated body heat, yet as the diameter size of the roll increases, it becomes less and less effective since convection plays an increasingly important roll.
Where heat can be lost through conduction or through convection or through radiation, then radiation is weak, and is overwhelmed by the other processes.
Today in sunny Spain it is cold (like most of Europe the last 3 months have been cold). I have no heating on in the house. My granite work top (in the kitchen) is cold to the touch. If I place a plastic/poluthene bag on top of the granite and place my hand on top of that bag, it does not feel cold. If I place a sheet of aluminium foil on the granite work top and place my hand on top of the foil, it feels cold.
In short, whatever miniscule radiative effect the aluminium foil has, it does not overcome conduction. The cold surface of the granite is conducted. Whereas the plastic bag is a good insulator and keeps my hand warm (even though plastic is a poor radiator)
If I suspend the aluminium foil horizontally in the air, and place my hand above it, possibly I can feel some reradiated heat when my hand is about 2mm above the foil, but the effect is so slight that I am unsure whether it is real or placebo.
But the point is that in the real world, redatiated heat (energy) is overcome by conduction and convection and this is so in the workings of the Earth’s atmosphere (not also forgetting the phase changes of water and associated latent heat).

April 25, 2013 3:35 am

Joseph E Postma says:
April 24, 2013 at 3:29 pm
ā€œItā€™s called an incandescent flashlight.ā€
It is about the resistor (lets use the filament) becoming even hotter still if you shine the flashlight at a mirror. This does not increase the brightness (temperature) of the filament. Trapping the radiation does not increase the temperature of the filament.
Not my prefered item of interest (radiation… a very long time ago) but as far as I remember, if you can concentrate some of the outgoing light (including IR) via a concave mirror back onto the filament, the latter wouldn’t survive for long.
Or with an experiment: make that the outher wall of the bulb is coated with silver and test how long it takes before the filament is burned up, compared to a non-coated bulb… In the case of non-halogen bulbs, only radiation is at work, no conduction or convection.

gbaikie
April 25, 2013 3:37 am

There is no evidence that CO2 in the atmosphere acts like insulation
in a house.
Fiberglass insulation in a house reduces the heat lost from convection.
What fiberglass does prevent the movement of gas molecules- which
inhibits convection of heat. If you inhibit to loss of heat from convection all
gases [including CO2] make good insulator of heat. Fiberglass
also inhibts warmth from entering the house. So insulated house
is easier to be cooler during warmer weather.
Insulation doesn’t make something warmer, it prevent the loss
or transference of heat. A thermos does not make coffee warmer than
the temperature of the coffee which was poured into the thermos.
So the thermos preventing a transfer of heat, so coffee stays warm
for a longer period of time.
To make an analogy of CO2 to insulation is correct in sense
that there is also no way for CO2 to cause the surface or air
at the surface to become hotter. Or an asphalt road in sunlight will
not become hotter because of CO2, What affect the temperature
of this road is the intensity of the sunlight and the loss of heat
from convection of air. If the air is already warm, there will less
heat loss from convection of air. Warmed air and intense sunlight
[sun near zenith and clear skies] will cause it reach it’s max
temperature.
And we have not seen any increases in max surface temperature
from an increase in global CO2 content, nor should this be
expected.
What is expected by some is that CO2 will cause there to less
loss of heat. So generally since it doesn’t cause the day to be
warmer, it could cause the nights to become less cold.
But there is no evidence which clearly indicate that CO2 does
cause less cooling- e.g. there no clear evidence that nights are warmer
due to CO2.
Whether CO2 cause such warming and thereby causes an increase
in average global temperature has not be proven.
But what should rather obvious is that increasing levels of CO2 or
other greenhouse gas are not going make earth unbearable hot
[unless a person presently already thinks it’s unbearable hot].
Or there no reason to assume that greenhouse gases could cause
CAGW like conditions as is commonly described by it’s faithful- or no
reason to assume that Earth could become in anyway similar to
Venus due to greenhouse gases.
No reason to believe greenhouse gases will cause massive widespread
droughts, or numerous other “predicted” drastic changes in climate.
To say the danger from CO2 has been over hyped is an understatement.
Instead it has been series of fabricated stories delivered by mass media
which as had the intention to scare the public. Which has caused many people
a lot of unnecessary stress and concern.
The public has already been conned out hundreds of billions of tax dollars
and has already had numerous coercive and oppression laws designed to
solve a scary problem which is totally fictional.

LdB
April 25, 2013 3:50 am


Then you haven’t looked very hard because the atmosphere and sun form a natural form of a optical pump which is the key component of a laser.
You only need 3 requirements an energy source we call it the sun, for a man made laser we use RF, microwaves, electric discharge anything will generally do its an efficiency issue.
You need an quantum active medium something like CO2, N20 or water vapor will generally will generally do.
Then all you need is a partial mirror which means a different input and output characteristics something like what the atmosphere does and we measure.
All pretty basic and not hard to understand .. see your problem it is that basic.

Sergey
April 25, 2013 3:51 am

A question to Roy: Why we can see the Sun? Why it radiates? Its atmosphere certainly does not contain any greenhouse gases, being almost pure hydrogen with a small traces of helium. What we see is blackbody radiation at 6000 C, and every gas emits according Plank formulae even when its temperature is near 0 K. So it sounds weird that without greenhouse gases Earth atmosphere would not emit anything into space.

April 25, 2013 3:54 am

Now snatch away the ā€œinsulatingā€ shell, real sudden-like. The system is suddenly cooling twice as fast, as per problem 1023, but its power input is unchanged! Oh no! What happens to the temperature of the inner (now alone) sphere? Increase? Decrease? Remain the same?” ~Gary
FIrst off, 1023 isn’t about the cooling rate.
To see what would happen though, consider the case where the shell is in contact with the sphere. If the gap was filled with the same material as sphere and shell are made of, then the volume of material goes up while the energy supply remains the same as it was originally, right?
Two spheres with the same power input/composition which only differ in size won’t reach the same temperature, will they?
So now you remove a section from the larger sphere, such that it is the same size as the other sphere, what happens?
Now remove the same amount of material from the larger sphere, except for a thin shell at the original surface, such that the inner sphere is the same size as the smaller sphere, what happens?
Let the system come to equilibrium, sphere at temp T2. Now add the shell back. Let the system come to equilibrium. Inner sphere is now at temp T3. Whatā€™s the relationship of T1, T2, and T3?
a) T1 = T2 = T3
b) T1 = T3 > T2
c) something else (specify)
” ~Gary
Interestingly, that is almost the set up for problem 1024, fancy that.
If T_sphere = T_background =T_shell at some point, shouldn’t it remain that way?
If T_sphere > T_background, adding the shell would give:
T_sphere > T_shell > T_background

Myrrh
April 25, 2013 4:05 am

Joseph E Postma says:
April 24, 2013 at 8:07 pm
I did not ever say that the Earth was at +49C. I said this is the equivalent temperature value of the average input. This too is reality. This is very different from the usual assumption of a -18C or 240 W/m2 input.
AGWScienceFiction’s Greenhouse Effect attributes this -18°C to the temperature of the Earth without its variation of “greenhouse gases”, that is, without “longwave infrared imbibing gases”, i.e. “radiated heat imbibing gases mainly water vapour and carbon dioxide”, but, it is crucial to understand this point in order to see the AGW sleight of hand in play, that -18°C figure comes from traditional real world physics and is the temperature of the Earth without any atmosphere at all, that is, also without the bulk of the atmosphere which is nitrogen and oxygen. The comparison in traditional real world physics is with the Moon, which has no atmosphere.
The equivalent temperature of the Moon corresponding to the Earth without any atmosphere at all, is around -23&degC.
In other words, it is the bulk of our atmosphere of around 98% nitrogen and oxygen which is primarily the real world greenhouse gases without which the Earth would be -18°C – the comparison is with the Moon which has no atmosphere.
AGWSF has by sleight of hand science fraud misappropriated this figure and claims it is the temperature of the Earth only without its version of “greenhouse gases, but with the rest of the atmosphere of mainly nitrogen and oxygen in place”.
From this AGWSF claims that “its greenhouse gases bring up the Earth’s temperature 33°C to 15°C. This is an illusion.
Firstly because they have based this on their science fraud of misattribution of the -18°C figure, but also because there is no mechanism they can show for “their greenhouse gases” having the power to raise global temperature 33°C from -18°C to 15°C.
And thirdly, because they have excised the Water Cycle to create their AGWSF Greenhouse Effect Illusion.
Here’s how. In real world traditional science the temperature of the Earth with the atmosphere in place, but without water, think deserts, the temperature of the Earth is given as 67°C.
In other words, in real world traditional physics, it is the bulk atmosphere of mainly voluminous and heavy under gravity nitrogen and oxygen which makes the real thermal blanket around the Earth.
The bulk atmosphere of mainly nitrogen and oxygen which with the inherent properties and process of real gases keeps the Earth from the extreme temperature differences of the Moon, because these real gases not only trap heat through gravity keeping them in place, but also cool the Earth through convection, because in the real world in real physics hot air rises and cold air sinks.
Any meteorologist who doesn’t understand this doesn’t know how we get our winds and weather. Winds are convection currents in our heavy fluid gas atmosphere created by differential heating of volumes (sometimes called packets) of the real gas air. As volumes of heated air rise taking away heat from the Earth, volumes of colder air sink, flowing beneath. Winds are volumes of air on the move. Hot air rises because real gases expand when heated and so become less dense and so lighter than air; these are areas of low pressure, because less dense they weigh less under gravity. Colder heavier more dense air will spontaneously sink flowing down into the low pressure area of less dense hotter air.
Winds blow from high to low. Winds blow from high pressure to low pressure. From denser heavier to less dense lighter. Condensed heavier colder masses of real gas sink, expanded lighter hotter masses of real gas rise. Just as currents are created in the ocean, as volumes of heated ligher water rise and colder heavy volumes sink.
So, with our heavy ocean of real gas atmosphere of mainly nitrogen and oxygen in place, but without water, the extremes of the atmosphere less Moon are avoided and we get a temperature of 67°C.
Putting back water which comprises some 7/10ths of the surface, this 67°C is brought down to 15°C. This is called the Water Cycle.
AGWSF has excised the Water Cycle to create the illusion of “33°C warming by its greenhouse gases”.
The AGWSF Greenhouse Effect is an illusion.
It makes a big difference in what the input can and can not do. The total energy is conserved, the input is equal to the output in terms of total energy, but not in terms of power, not in terms of what the input and output power flux can do. The input can do a lot more work than the output. The input can generate very high temperatures on the surface. The atmosphere is then heated by the surface conductively/convectively and radiatively. Thatā€™s all that needs to happen because the input is so hot. Hot heats cold. You can go outside and feel the hot input on a sunny day. Latent heat carries a lot of this input around the planet via water vapor and liquid water.
Water has a very high heat capacity, as heated water becomes even less dense and so even more lighter than air it evaporates more quickly, this fluid gas water vapour taking the heat into the colder heights as it rises where it releases it and condenses back to liquid water or ice, and colder returns to the surface as rain.
The Water Cycle cools the Earth from 52°C from the 67°C temperature the Earth would be without it, but with the bulk of the heavy real gas atmosphere of nitrogen and oxygen still in place.
It is the properties and process of real gases with volume, weight, attraction, which create the lapse rates in conjuction with gravity acting on the real mass; gravity stronger nearer the surface pulling the molecules of gas together, condensing them so they are heavier, allowing them to expand as gravity is weaker away from the surface so they become less dense and so lighter.
AGWSF’s the Greenhouse Effect Illusion is created out of the imaginary “ideal gas”, without mass and therefore nothing for gravity to act on. That’s why they don’t have gravity in their world and don’t understand real world physics. They have no weather at all because their ideal gases with no volume weight attraction cannot expand and condense to create weather. They are climate scientists of a non-existant climate.
All they have is an imaginary world where their ‘atmosphere’ “is empty space with ideal gases zipping through it at great speeds under their own molecular momentum bouncing off each other in elastic collisions and so spontaneously diffusing to mix thoroughly so they can’t be unmixed”. Their massless, weightless “ideal gas molecules” not subject to gravity can only create pressure by bouncing off some imaginary invisible wall of a “container”.
The AGWSF Greenhouse Effect world does not have an atmosphere, they go straight from the surface to empty space. So their “radiation only” world.
Without heat transfer by conduction and convection.
They have nothing to convect.
There is no internal consistency, how could there be, in their fake fisics – how can their “ideal gas carbon dioxide molecules accumulate in the atmosphere for hundreds of years”, when they have no atmosphere and there is no, unknown to real world physics, “invisible container” around the Earth keeping them in? Whatever atmosphere they might have had has long gone into outer space, and so all their gases and weather.
Their “ideal gases travel at great speeds miles apart from each other through empty space under their own molecular momentum” because they don’t have real gases with volume and weight containing them. They have no sound in their world – without gravity acting on real gases with volume they have no medium through which sound can travel.
The only hot air they have is their irrational arguments which they can’t hear because they have empty space around them in their impossible world.
There is no “33°C warming by the AGW greenhouse gases”.
It is an illusion created by science fraud and fake fisics in all its parts.

Kristian
April 25, 2013 4:06 am

Gary Hladik says, April 25, 2013 at 3:03 am:
“Now snatch away the ā€œinsulatingā€ shell, real sudden-like. The system is suddenly cooling twice as fast, as per problem 1023, but its power input is unchanged! Oh no! What happens to the temperature of the inner (now alone) sphere? Increase? Decrease? Remain the same?”
The system is radiating out power to its surroundings twice as fast as soon as the surrounding shell is removed, Gary. The sphere isn’t. Q (or J if you like) remains unchanged. The central sphere has been putting out the same power density flux all along, because that is what its emission temperature dictates, the emission temperature in turn being dictated by the heat supplied to the surface of the sphere from its heat source, and only that.
Read my comment
http://wattsupwiththat.com/2013/04/24/spencer-slays-with-sarcasm/#comment-1286834
The flux leaving the surface of the sphere into the vacuum between it and the surrounding shell (Q) will be the same as long as its internal heat source is active and constant. This flux is simply intersected by the shell, making use of some of it (heating the shell and after equilibration, maintaining its temperature) and letting the rest through to the surroundings. In other words, half goes to the shell, half goes to space. The flux from the sphere is the same all along, but the heat transfer between the sphere and the shell (Q’) is only 1/2Q, as is the heat transfer between the shell and space (Q”). Together they make up Q: Q = Q’ + Q” (Q = (Q – Q”) + Q”). The only thing you need to remember is that the entire energy/heat flux ultimately comes from the sphere, the shell simply ‘splits’ it upon absorption, a sort of intermediate reservoir.

DennisA
April 25, 2013 4:23 am

How many angels were there on the head of that pin? My brain hurts. This is my simplistic layman approach to it and I know people will jump all over it, but so what.
Is CO2 increasing?
Year in year out.
Is temperature increasing in tandem?
It seems to be getting colder.
Does it get hot when the sun comes out?
Always seems to.
Does it get colder when the sun is blocked by cloud cover?
Always seems to
Did it get colder in the 60’s and 70’s when CO2 was also increasing.
It really did. I was around then.
Was it warmer than now in the MWP, the Climatic Optimum and the Roman Warm Period?
I wasn’t around then but a lot of credible people say so.
Was CO2 higher then than today?
Apparently it was a lot lower.
Did the LIA happen?
There’s a lot of historical evidence it did.
Were those temperatures normal?
Hope not.
Is it warmer now?
Thankfully
Has there ever been less ice in the Arctic than there is now?
History says yes
Did polar bears die out
History says no.
Did the Industrial Revolution start in 1850-1880?
Ask Sir Clement Clerke and others from 1678, using coal furnaces known as cupolas, or Abraham Darby, who was using coke to fuel his blast furnaces at Coalbrookdale in 1709.
So is CO2 from human emissions warming the planet out of control?
Well, at the end of the day and not to put too fine a point on it, I find it very difficult with the evidence of my own subjective observations, to actually believe that it is.

joeldshore
April 25, 2013 4:40 am

gbaikie says:

What is expected by some is that CO2 will cause there to less loss of heat. So generally since it doesnā€™t cause the day to be warmer, it could cause the nights to become less cold.

The notion that CO2 can’t cause the days to become hotter is nonsense.
There is “cooling” going on even during the daytime in the sense that the Earth is always radiating heat. It is just that the warming due to the sun is larger than the cooling due to the Earth radiating. If you increase GHGs, you can change that balance and cause it to become warmer during the day too. There are reasons why nighttime temperatures are expected to warm more than daytime temperatures under rising GHGs, but both are expected to rise (and have been rising).

tjfolkerts
April 25, 2013 4:49 am

Kristian says: “What Problem 1026 is telling us is specifically that if R ~ r, the Q = Qā€™ + Qā€.”
More specifically, Problem 1026 is telling us Q’ = Q” = 1/2 Q for R ~ r. In other words, the inner sphere originally emits thermal IR at a rate Q. After adding the heat shield, the inner sphere emits Q/2 IR to the shell. The shell in turn also emits Q/2 to space. So with the aptly named ‘heat shield’, we have reduced the heat loss to 1/2 of what it was, so only 1/2 as much power needs to be supplied to the the sphere to maintain the postulated steady temperature T.
“This is the exact equivalent to the Carnot cycle:”
No, not really. There is no work being done in Problem 1026, and no cycle. Other than that fact that Problem 1026 and the Carnot cycle both have something warm and something cool, there is really not much else that the two have in common.
Even “Q” itself means quite different things. The “Q” for a Carnot Cycle is the total heat transfer. The “Q” in Problem 1026 is the RATE of heat transfer.
“At any point in time, the heat flux leaving the surface of the sphere (Q) corresponds directly to (depends on) its specific emission temperature … “
The heat, ie the net transfer of thermal energy depends both on the temperature of the surface and the temperature of the surroundings. . You are ignoring the “J1” = J/2 that is clearly shown coming back to the inner sphere from the shell due to the temperature of the shell..
The NET loss from the planet is J/2
The NET loss from the shell is J/2.
The power required from a heater to maintain this situation is J/2.
“A star is pretty constantly heated, isnā€™t it?”
One would think so. But clearly this textbook problem assumes something else — that the TEMPERATURE is constant and the power adjusts to maintain that temperature. They could have instead assumed that the power input was constant, in which case the temperature of the star would have gone up after adding the heat shield.

A C Osborn
April 25, 2013 4:49 am

richard verney says:
April 24, 2013 at 5:10 pm
“I can remember an old design of 3kw electric fire.
So the question is why did the middle element not get noticably hotter and take on a much brighter appearance?”
Richard, come on that is the real world, not the Climatological world.
It should have been about 50% hotter than the 2 outside ones, as they only received half the Radiation (from the other elements) as the middle one.
So all you Physicists please explain, as I notice you didn’t try to when Richard posted it.

April 25, 2013 4:56 am

Tim, you’re mixing two problems, I’ve pointed this out several times now.
1026, regarding the effect adding a shell has on the rate of cooling
1023, regarding the effect a shell has on the power radiated to the surroundings
1023 is the one described as analogous to a star in a dust cloud, with the implication that it has a constant input.
1026 is the one describing a body cooling with a shell added, with the implication that it does not have a constant input.

Bryan
April 25, 2013 5:25 am

A number of posts advocating the greenhouse theory include the idea of a source of constant power.
That is that backradiation plus constant power force the source to a higher temperature.
Now electrical engineers are aware of electrical constant power sources.
But these are specially designed circuits and are not perfect.
I am not aware of any naturally occurring spontaneous constant power sources.
An increase in temperature of the source would imply it was capable of spontaneously increasing the energy QUALITY of its output
If a theory depends on such a violation of the Second Law then it must be false.

April 25, 2013 5:39 am

There’s no link to the thing you were ridiculing either here or on Spencer’s article. That makes it harder to share in the joke. I guess there would be a lot less to talk about on WattsUpWithThat if the greenhouse effect wasn’t true. And there would be a lot less books for Roy Spencer to sell.
In any case, the Tyndall experiment does not prove the existence of an atmospheric greenhouse effect as Monckton claims, merely that a cold gas target will absorb heat from a hot source. In the atmosphere CO2 molecules (and H2O etc) emit as much as they absorb, so there is no greenhouse effect. (My own post on it here.)

Dry desert climes cool faster at night and have a greater diurnal temperature variation than moist climes. Water vapor is the difference. CO2 has a smaller effect, but an effect nonetheless. ā€“ Anthony

Regarding the desert at night, it is kept warmer because the water vapour changes the adiabatic lapse rate, not because of IR trapping. Also, humidity makes it feel warmer to a human.
IR trapping cannot make the source warmer than it otherwise was, unless you create energy in violation of the first law of thermodynamics.
Likewise, backradiation cannot make an object warmer than it otherwise was, unless heat moves spontaneously from a cooler to a warmer pool in violation of the second law. Another way of phrasing the second law is: EMR from a cooler object cannot be made to warm a warmer object further still.

April 25, 2013 6:03 am

Mosher writes “Earth radiates to space from the ERL.”
Agreed.
And then writes “When you add GHGs you raise the ERL.”
Maybe.
And then writes “When you raise the ERL the earth radiates to space more slowly.”
Why? Those slides state the following
“Surface and lower atmosphere cool by infrared radiation to space from upper troposphere (ERL= effective radiating level, such that total CO2 above is fixed)”
and
“Infrared radiation leaves earth for space from upper troposphere (ERL). Amount increases with temperature at ERL (immediate). Height of ERL is such that total CO2 above it is constant.”
And the fundamental assumption is that “total CO2 above it is constant” meaning the amount of energy leaving is dependent on opacity. The problem I have with that is that if say the CO2 levels double, then the total number of molecules at the ERL that have sufficient energy to radiate remains the same but the number of CO2 molecules that have sufficient energy to radiate doubles.
Therefore the amount of radiation heading upwards has doubled. So whilst the opacity has increased, so has the number of packets of energy trying to get past so the assumption that “total CO2 above is constant” is intuitively wrong for me.

Yorkshireman
April 25, 2013 6:07 am

A C Osborn says:
April 25, 2013 at 4:49 am…..
My approx. back-of-the-envelope calculations:
(1 cm diameter elements spaced at 5cm)
Assume elements are long enough to disregard effects at ends.
Each element receives ~ 3% of its neighbours radiation. (diameter / 2 * pi * spacing)
Centre element receives ~ 3% more energy input (power) than outer elements or 1.03 times.
Assuming no convection/conduction, using 4th root, centre element is 1.007 times as hot (in Kelvin).
If element temperature is several hundred degrees C, then difference is a few degrees,
This is probably too small to be noticeable by eye.

tjfolkerts
April 25, 2013 6:18 am

Max says: “Tim, youā€™re mixing two problems … “
BOTH problems have a “star” at a fixed temperature, and then add a “heat shield”. BOTH calculate how much less power is required after adding the shield. The only real difference is that 1023 assumes the shell is “nearby” so the R ā‰ˆ r, while 1026 allows the outer shell to be any radius. 1023 is a limiting case of 1026, and both say the total power required is 1/2 as much when the shell is added nearby.

LdB
April 25, 2013 6:40 am

@Sergey
You are sort of getting it and instead of answering your question I will pose two questions which answer you question but with you doing the thinking.
If I looked at the earth in only the infrared section what do I see. Here is a hint shot from Galileo spacecraft at a distance of 1.32 million miles (http://space.about.com/od/pictures/ig/Earth-Pictures-Gallery/Global-View-of-Earth-in-the-Ne.htm) Earth in the far infrared it would be even brighter … why????? You might want to look in contrast at the ultraviolet shot of earth from space shot from rosetta spacecraft (http://www.redorbit.com/news/space/1809886/revealing_earths_ultraviolet_fingerprint/)
So your question why is the infrared image so much brighter than the ultraviolet image when viewed from a distance of space even though the sun emits predominately ultraviolet or visible spectrum.
Second question for you to work on since you are concerned by the suns blackbody temperature. The temperature of the corona of the sun is several million degrees not 6000K so the black body emission temperature discrepancy is actually far far worse than what your poor attempt to portrait the problem does. So how do we resolve a temperature of several million degrees on the corona to a black body temperature of 6000 degree and why isn’t a problem for science.

Greg House
April 25, 2013 7:16 am

steveta_uk says (April 25, 2013 at 3:19 am): “For a simple example of how you are wrong, why are you assuming that the back radiation is from a colder source?”
============================================================
Very simple: per definition. Radiation from a warmer source is not back radiation per definition. You can call it “direct radiation”, if you wish, but a new term is not necessary, because there has been no confusion about it. Well, with the exception of your comment.

beng
April 25, 2013 7:41 am

It’s bad enough to deal w/warmers that inflate known GHG effects to absurd levels. Now add to that slayers that don’t even understand that much (but think they do).

Joel Shore
April 25, 2013 8:07 am

Noelene says:

So there is a consensus on the greenhouse effect?

Actually, the fact that one gets so much argument over something that is such basic physics should totally refute the argument that there is no consensus about AGW because you can find lots of arguments on the web about it.
The mistaken arguments of Monckton and others revered by those at WUWT are a little less silly than the Slayer’s arguments…but not that much.

joeldshore
April 25, 2013 8:17 am

TimTheToolMan says:

And the fundamental assumption is that ā€œtotal CO2 above it is constantā€ meaning the amount of energy leaving is dependent on opacity. The problem I have with that is that if say the CO2 levels double, then the total number of molecules at the ERL that have sufficient energy to radiate remains the same but the number of CO2 molecules that have sufficient energy to radiate doubles.

No…You don’t get it. If the total number of CO2 molecules doubles, the atmosphere becomes more opaque and the ERL moves up to a new level where the number of CO2 molecules is less (so the total opaqueness above it is still the same). What you are not understanding is how things work in a gas where you have both absorption and radiation.
Look at plots of the Earth’s emission as observed from space and you will see that the CO2 absorption lines result in suppression of emission at these wavelengths, exactly as theory and models predict. The whole field of remote sensing is based on us having a correct understanding of how this works. So, if you are confused about it, accept that fact that you are confused and need to learn more rather than continuing to believe you are right in the face of overwhelming evidence that you are not.

Stanb999
April 25, 2013 8:29 am

Dry desert climes cool faster at night and have a greater diurnal temperature variation than moist climes. Water vapor is the difference. CO2 has a smaller effect, but an effect nonetheless. ā€“ Anthony
——————————————————————————————————————-
Are your sure it isn’t just the H2O being present that changes the dynamic of temperature flow from desert to “moist” areas? Check out the properties of H2O. It is an amazing element. Raise 1 pound of water 1 degree. 1 BTU. Raise that same pound the last degree to steam. 900 BTUs. Free water from it’s solid state… about 140 btu’s..
Of differently. It takes a tremendous amount of energy for the earth to have a temperature different than that of liquid water for a given pressure. The land cools till condensation, Dew/frost. It heats till evaporation caused clouds… This affect has nothing to do with the IR capability of h2o vapor.

tjfolkerts
April 25, 2013 8:32 am

Greg, the “definition” of “back radiation” deals only with direction, not temperature. “Back radiation” is simple radiation directed back toward earth’s surface. By extension we could use your term “direct radiation” or “emitted radiation” or “forward radiation” for the thermal IR leaving any surface. Then “back radiation” or “absorbed radiation” would be the IR absorbed by the surface.
It would be very odd to my thinking to reserve “back radiation” for radiation from a cooler object toward a warmer object. If there is both a warmer and a cooler object around, then a surface would simultaneously be emitting “back radiation” and “direct radiation” from the same atoms. Why use two different names for exactly the same process producing exactly the same spectrum of photons?

Matt in Houston
April 25, 2013 8:38 am

@ Dr. Spencer
” ANYTHING that reduces heat loss from a heated object can increase its temperature.”
This is categorically wrong. It cannot heat anything unless it is at a higher temp than the “heated object”. If there is some insulating object (ie any matter) or gas it REDUCES the RATE of cooling from the “heated object”, under no circumstance can a cooler object HEAT a a HEATED OBJECT.
I don’t think that Dr. Spencer meant his statement exactly the way he stated it though.
Perhaps it can be re-stated in a more thorough fashion Dr. Spencer?
If anyone already addressed this comment, I apologize, I am running out the door and had to post without reading everything in here. Great discussion though and I hope Dr. Spencer and Mr. Watts will try to invite Dr. Pierre Latour to come in and write a response article on this matter. His writings on this subject are articulate, principled and clear.

April 25, 2013 8:50 am

Joel Shore says:
“Noā€¦You donā€™t get it.”
What Joel Shore doesn’t get is the plain fact that the planet is not responding to the rise in CO2 in the way that every climate alarmist was predicting [until the planet showed that it was not cooperating with the endless, incessant, and wrong predictions of runaway global warming].
Who should we believe? Joel Shore? Or Planet Earth?
Because they cannot both be right.

Roy Spencer
April 25, 2013 8:57 am

Matt in Houston, no it is not wrong. You are being misled about basic thermodynamic principles. I meant my statement exactly as stated: “ANYTHING that reduces heat loss from a heated object can increase its temperature”. Furthermore, that “thing” is almost always at a lower temperature than the heated object. The examples are literally all around you, even the clothes you wear.

Greg House
April 25, 2013 8:57 am

tjfolkerts says (April 25, 2013 at 8:32 am): “Why use two different names for exactly the same process producing exactly the same spectrum of photons?”
=====================================================
a) because it causes confusion and b) because it makes possible to mislead via the trick “the same spectrum of photons”/”warming effect”, although the warming effect is physically impossible.
The notion of “photons” alone is misleading enough, because people fall into the trap and start counting them without any basis in real science. The idea about “all radiation being equal” has no basis in real science either and misleads people as well.

April 25, 2013 9:47 am

Most of science is merely the appearance of confidence. If you make something appear confident, people will follow it, totally independent of any rationality. In fact they’ll create sophistry, and will fully believe themselves, just to defend the apparition of confidence so that they can feel like they’re part of it or that the confidence is part of them. Then, they’ll get really upset about questions which threaten the foundation of the confidence. As we see. Just look at how people call it ridiculous to say that the Earth is heated on one side only by the Sun, and that this is the average solar input. People with PhD’s literally call this ludicrous. They call it ludicrous that the Sun heats the Earth on one side only, and that you can mathematically quantify this and that it physically makes a big difference to the usual assumptions. No one wants to even discuss it calmly, kindly, or openly, they just obfuscate and change goal posts, and all manner of other things. Mostly they get really upset and call people names. They expose themselves for the most part quite uncomfortable with what the scientific method is actually supposed to be about. But they have the appearance confidence, and the masses of number who follow confidence. It is not possible to have a civilized discussion with people who think it is ludicrous to talk about an Earth heated on one side only with a very high forcing temperature. There is no civility to be had between the idea that the Earth is heated on one side, and the idea that it is not. From there, the only thing that happens is that abuse gets thrown about from those who think it is ludicrous to speak of the Earth being heated on only one side. And then the people who say that the Earth is heated on one side, somehow, magically, become blamed for the abuse that other people created. This is all very fascinating.

Reed Coray
April 25, 2013 9:56 am

I want to thank joeldshore (April 24, 2013 at 8:13 pm), LdB (April 24, 2013 at 8:19 pm) and Gary Hladik (April 24, 2013 at 10:34 pm) for responding to my comment. I’ll reply to each response. But before doing that, I want to make it clear– I am not trying to “slay” the “greenhouse effect dragon”!” For one thing, I’m not sure exactly what the “greenhouse effect dragon” is. If the “greenhouse effect” is the statement that some gases in the Earth’s atmosphere absorb/radiate electromagnetic energy in sub-bands of the IR, then I believe in the greenhouse effect. If the “greenhouse effect” is the statement that greenhouse gases in the Earth’s atmosphere will affect the temperature as a function of location on the Earth’s surface, then, again, I agree. If the “greenhouse effect” is the statement that greenhouse gases in the Earth’s atmosphere will increase the “average temperature” (whatever that is) of the Earth’s surface, then I don’t have sufficient knowledge to arrive at an informed opinion. My original comment was a response to Dr. Spencer’s statement: “Greenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, …
Response to joeldshore.
I agree that increasing the amount of CO2 in the vacuum region of a thermos bottle won’t result in improved thermos bottle performance. However, based on the rest of your response I believe you agree with the statement: “adding any amount of CO2 gas to the region between the inner and outer walls of a vacuum thermos bottle will NOT improve thermos bottle performance.” If my belief is valid, then isn’t Dr. Spencer’s immediately above statement incorrect?
Response to LdB.
Who mentioned CLASSIC PHYSICS verses quantum mechanics? The experiment is independent of any particular theory. Injecting a fixed amount of greenhouse gas into the volume between the inner and outer walls of a vacuum thermos bottle will have one and only effect on thermos bottle performance–where thermos bottle performance is defined as the time interval required for the temperature of the coffee to reach the temperature of the heat sink into which the thermos bottle is placed. It will either (a) have no effect (equal time intervals), (b) improve performance (increase the time interval), or (c) degrade performance (decrease the time interval). If either (a) or (c) is true, then Dr. Spencer’s statement is wrong. If (b) then, like all physics’ statements/theories, the experiment doesn’t prove Dr. Spencer’s statement/theory is true under all conditions, but the experiment cannot be used to demonstrate that Dr. Spencer’s statement/theory is wrong.
Next, it seems to me that your introduction of “gain factors” is obfuscating the issue at hand–to wit, will adding CO2 to the vacuum space of a thermos bottle decrease the rate of cooling? And to your comment: “It is that simple and basic and no amount of stupid classic physics with thermos flasks is going to save you ā€¦ simple question posed is CO2 a quantum gain medium answer yes ā€¦. end of story. No possibly way to talk around it unless you want to alternatively explain how a laser works” is both irrelevant and nonsensical. The issue of whether CO2 is a “quantum gain medium” may be relevant to a theoretical understanding of the phenomenon, but has no bearing on the experiment itself. And if the experiment shows that CO2 does NOT reduce the rate of cooling, then I have no need to explain “how a laser works”. The onus, if there is one, would be on you to show why the thermos bottle experiment doesn’t agree with your “quantum gain medium” position.
Response to Gary Hladik.
If I understand what you wrote, you and I agree that adding more CO2 won’t improve thermos bottle performance. Then you mentioned thinking of the top of the atmosphere as “the inner layer of the vacuum thermos bottle”. By “inner layer” did you mean the wall of the chamber that holds the coffee, or the wall of the thermos that is in contact with the heat sink? I agree that with the exception of matter leaving the Earth (e.g., the velocities of many helium gas atoms at room temperature exceed “escape velocity”) radiation is the only mechanism for energy to leave the earth/earth atmosphere system. However, there is a difference between the “inner wall” of the thermos (i.e., the wall of the chamber that holds the coffee) and the top of the atmosphere. In particular, in the case of a vacuum between the walls, radiation escapes to the outer wall only from the inner wall. However, in the case of greenhouse gas in the region between the two walls, in addition to thermal conduction and convection between the inner and outer walls, there is no single “surface” from which radiation to the outer wall takes place. To various degrees, the greenhouse gases in the space between the walls will act like an “inner wall” in that a portion of the radiation from greenhouse gases at all locations between the walls will find its way to the outer wall. As I see it, the issue of an Earth atmosphere greenhouse effect is one of determining the Earth surface temperature for two scenarios: (a) a surface surrounded by a non-greenhouse gas atmosphere such that most if not all radiation that escapes to space originates from the surface, or (b) a surface surrounded by an atmosphere containing greenhouse gases such that radiation escapes to space from both the surface and the greenhouse gases in the atmosphere. As I mentioned at the outset, I have insufficient knowledge to determine these temperatures. I’m open to temperature (a) being both higher and lower than temperature (b).
Finally, if by “inner wall” you mean the thermos wall in contact with the heat sink, then my response is: “let the heat sink be the vacuum of cold space.” The nature of the heat sink will have no effect on whether the CO2 thermos outperforms the vacuum thermos.

A C Osborn
April 25, 2013 10:44 am

Can someone point me to where the actual measured CO2 PPM for the upper troposphere or troposphere are stored?

April 25, 2013 10:45 am

Reed Coray:
I appreciate your challenging statements made. I write to answer a question you put because it comes between two people and – being neither of them – I may be able to help.
In your post at April 25, 2013 at 9:56 am you quote Roy Spencer having said

ā€œGreenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, ā€¦ā€

Spencer’s statement is correct.
(Any statement from him is likely to be correct, and if it were shown to be wrong then he would admit it.)
But it seems you have been confused about that correct statement by – in this case, also correct – statements by joeldshore.
(Your confusion should not dismay you because confusion is a normal outcome of interaction with joeldshore.)
You say

I agree that increasing the amount of CO2 in the vacuum region of a thermos bottle wonā€™t result in improved thermos bottle performance. However, based on the rest of your response I believe you agree with the statement:

ā€œadding any amount of CO2 gas to the region between the inner and outer walls of a vacuum thermos bottle will NOT improve thermos bottle performance.ā€

If my belief is valid, then isnā€™t Dr. Spencerā€™s immediately above statement incorrect?

Your “belief” and Spencer’s comment are both correct. I explain as follows.
The “region between the inner and outer walls of a vacuum thermos bottle” contains a good vacuum. And a good vacuum is a very good insulator. Any gas will not be as effective an insulator as a good vacuum. Therefore, introducing any gas “between the inner and outer walls of a vacuum thermos bottle” reduces its insulation. The gas replaces a good insulator (vacuum) with a less good insulator (gas) and, thus, reduces the performance of the thermos bottle.
The atmosphere is a mixture of gases.
Most of the gases in the air (e.g. nitrogen and oxygen) do not “insulate” the Earth’s surface from flow of radiation from the surface to space. (They are similar to a gas being between the inner and outer walls the thermos bottle).
But greenhouse gases do “insulate” the Earth’s surface from flow of radiation from the surface to space. (They are like a good vacuum being between the inner and outer walls of the thermos bottle).
So, the Earth’s surface is better “insulated” when the gases of the atmosphere include greenhouse gases.
I hope this helps.
Richard

Gary Hladik
April 25, 2013 10:49 am

Reed Coray says (April 25, 2013 at 9:56 am): “By ā€œinner layerā€ did you mean the wall of the chamber that holds the coffee…”
Yes. This wall touches the coffee on its inside surface and on the outside is surrounded by the vacuum.
“However, in the case of greenhouse gas in the region between the two walls, in addition to thermal conduction and convection between the inner and outer walls, there is no single ā€œsurfaceā€ from which radiation to the outer wall takes place.”
Correct. The earth’s upper atmosphere, however, does have something like a “surface” from which radiation finally escapes to space, i.e. the “effective radiating layer” which Steve Mosher mentioned above (April 24, 2013 at 3:42 pm).
“As I see it, the issue of an Earth atmosphere greenhouse effect is one of determining the Earth surface temperature for two scenarios: (a) a surface surrounded by a non-greenhouse gas atmosphere [snip] or (b) a surface surrounded by an atmosphere containing greenhouse gases”
Dr. Spencer covers that case. I thought it was a pretty good read:
http://www.drroyspencer.com/2009/12/what-if-there-was-no-greenhouse-effect/

Gary Hladik
April 25, 2013 10:59 am

Matt in Houston says (April 25, 2013 at 8:38 am): “…I hope Dr. Spencer and Mr. Watts will try to invite Dr. Pierre Latour to come in and write a response article on this matter. His writings on this subject are articulate, principled and clear.”
His writings are handwaving. Might I suggest you contact him yourself and ask him why he has not performed the “Yes, Virginia” experiment to prove Dr. Spencer (and innumerable textbooks) wrong and win a Nobel Prize in the process?
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/

April 25, 2013 11:38 am

BOTH problems have a ā€œstarā€ at a fixed temperature, and then add a ā€œheat shieldā€. BOTH calculate how much less power is required after adding the shield. The only real difference is that 1023 assumes the shell is ā€œnearbyā€ so the R ā‰ˆ r, while 1026 allows the outer shell to be any radius. 1023 is a limiting case of 1026, and both say the total power required is 1/2 as much when the shell is added nearby.” ~Tim
Here, let’s go ahead and see if this is the case, shall we?
“Consider a black sphere of radius R at temperature T which radiates to distant black surroundings at T = 0K.
(a) Surround the sphere with a nearby heat shield in the form of a black shell whose temperature is determined by radiative equilibrium. What is the temperature of the shell and what is the effect of the shell on the total power radiated to the surroundings?
(b) How is the total power radiated affected by additional heat shields?
(Note that this is a crude model of a star surrounded by a dust cloud.)
Solution:
(a) At radiative equilibrium, J – Jā‚ = Jā‚ or Jā‚ = J/2. Therefore Tā‚ā“ – Tā“/2, or Tā‚ = āˆœTā“/2 = T/āˆœ2
(b) The heat shield reduces the total power radiated to half of the initial value. This is because the shield radiates a part of the energy it absorbs back to the black sphere.” ~Problem 1023
ā€œA spherical black body of radius r at absolute temperature T is surrounded by a thin spherical and concentric shell of radius R, black on both sides. Show that the factor by which this radiation shield reduces the rate of cooling of the body (consider space between spheres evacuated, with no thermal conduction losses) is given by the following expression: aRĀ²/(RĀ² + brĀ²), and find the numerical coefficients a and b.
Solution:
Let the surrounding temperature be Tā‚€. The rate of energy loss of the black body before being surrounded by the spherical shell is
Q = 4Ļ€rĀ²Ļƒ(Tā“ – Tā‚€ā“).
The energy loss per unit time by the black body after being surrounded by the shell is
Q’ = 4Ļ€rĀ²Ļƒ(Tā“ – Tā‚ā“), where Tā‚ is the temperature of the shell.
The energy loss per unit time by the shell is
Q” = 4Ļ€RĀ²Ļƒ(Tā‚ā“ – Tā‚€ā“).
Since Q” = Q’, we obtain
Tā‚ā“ = (rĀ²Tā“ + RĀ²Tā‚€ā“)/(RĀ² + rĀ²).
Hence Q’/Q = RĀ²/(RĀ² + rĀ²), i.e., a = 1 and b = 1.ā€ ~Problem 1026

davidmhoffer
April 25, 2013 11:47 am

Anthony
Might I suggest yet one more resource page on WUWT? This topic simply brings out a level of misinformation that I for one find aggravating (which I promptly make worse by getting involved in the discussion).
Might it make sense to have a resource page with a small set of seminal articles on the topic? Willis’ Steel Greenhouse, Ira Glickstein’s series, perhaps one or two more. Putting the credible articles all in one place would be an excellent resource for those new to the discussion and would limit (I’d hope) the confusion and argument by circular reasoning that dominates threads like this one (and I think does more harm than good.)
REPLY: Yes, I’ll consider it. – Anthony

Gary Hladik
April 25, 2013 12:10 pm

Maxā„¢ says (April 25, 2013 at 3:54 am): “First off, 1023 isnā€™t about the cooling rate.”
Ah, I think I see the source of confusion here. Actually problem 1023 is all about the cooling rate. Based on its temp, the sphere radiates (cools) at rate J, but thanks to the shield the system radiates (cools) at rate J1, where J1 = J/2.
In fact problem 1023 is much like problem 1026, which mentions “cooling” explicitly. The only difference is that in 1026 the radius of the shell is variable, but it reduces to problem 1023 when R is very close to r.
I notice you didn’t answer my question, but I’ll be happy to answer yours.
“To see what would happen though, consider the case where the shell is in contact with the sphere. If the gap was filled with the same material as sphere and shell are made of, then the volume of material goes up while the energy supply remains the same as it was originally, right?
Two spheres with the same power input/composition which only differ in size wonā€™t reach the same temperature, will they?”
At equilibrium, the surface of the larger sphere will be cooler than the smaller, because it’s radiating the same power from a larger surface. To calculate temp, use the Stefan-Boltzmann calculator
http://calculator.tutorvista.com/stefan-boltzmann-law-calculator.html
Set emissivity to 1, choose arbitrary constant power input, vary the surface area, and solve for temp.
“So now you remove a section from the larger sphere, such that it is the same size as the other sphere, what happens?”
Surface temp goes up, matches the second identical sphere.
“Now remove the same amount of material from the larger sphere, except for a thin shell at the original surface, such that the inner sphere is the same size as the smaller sphere, what happens?”
Surface temp of the sphere goes up, because according to problem 1026 (your reference), “this radiation shield reduces the rate of cooling of the body”. Same power input, reduced cooling, the sphere’s surface temperature must go up until at equilibrium the shell is radiating as much power as the inner sphere receives.
Congratulation, Max! You’ve used a standard text to derive the Steel Greenhouse! The wizard is slain! Willis would be proud of you! I know I am. Well done.

joeldshore
April 25, 2013 12:29 pm

Gary Hladik says:

Matt in Houston says (April 25, 2013 at 8:38 am): ā€œā€¦I hope Dr. Spencer and Mr. Watts will try to invite Dr. Pierre Latour to come in and write a response article on this matter. His writings on this subject are articulate, principled and clear.ā€
His writings are handwaving.

“Handwaving” is a rather nice euphemism. “Complete and utter nonsense” is a more accurate summary of what Pierre Latour has to say on the subject.
Joseph Postma says:

Just look at how people call it ridiculous to say that the Earth is heated on one side only by the Sun, and that this is the average solar input. People with PhDā€™s literally call this ludicrous. They call it ludicrous that the Sun heats the Earth on one side only, and that you can mathematically quantify this and that it physically makes a big difference to the usual assumptions. No one wants to even discuss it calmly, kindly, or openly, they just obfuscate and change goal posts, and all manner of other things. Mostly they get really upset and call people names.

No…We don’t deny that it heats the Earth from one side only. What we deny is that you can ignore half the Earth in your average. And, if this concept is too confusing for you, then don’t work with averages at all…Look at the total energy in and out as I explained in this post: http://wattsupwiththat.com/2013/04/24/spencer-slays-with-sarcasm/#comment-1286615
The reason that we are not so calm and kind is we have been round and round on this before (e.g., a couple years ago on Judith Curry’s blog) and I really think at this point, you should know better. You either have one heck of a monstrous mental block or you really do understand but have, for some bizarre reason, decided that you would rather engage in sophistry. Either way, it is not a pretty sight.

Gary Hladik
April 25, 2013 12:35 pm

Kristian says (April 25, 2013 at 4:06 am): “The system is radiating out power to its surroundings twice as fast as soon as the surrounding shell is removed, Gary. The sphere isnā€™t.”
When the shell is snatched away (along with its insulating effect), the sphere is the system.
“Q (or J if you like) remains unchanged.”
No, but let’s go on.
“The central sphere has been putting out the same power density flux all along, because that is what its emission temperature dictates, the emission temperature in turn being dictated by the heat supplied to the surface of the sphere from its heat source, and only that.”
No. As Dr. Spencer points out, you need to know both the heating rate and the cooling rate. According to problem 1023, the system was in equilibrium with a cooling rate of J1, which is half of the sphere’s cooling rate J. Snatch away the shell and instantly the cooling rate doubles. Same power input, double cooling rate, the temp must go down. This is basic thermodynamics.
In fact the temp will go down until at equilibrium the sphere’s cooling rate is J1, which was (surprise!) the original (constant) power input.

gbaikie
April 25, 2013 1:05 pm

“joeldshore says:
April 25, 2013 at 4:40 am
gbaikie says:
What is expected by some is that CO2 will cause there to less loss of heat. So generally since it doesnā€™t cause the day to be warmer, it could cause the nights to become less cold.
The notion that CO2 canā€™t cause the days to become hotter is nonsense.
There is ā€œcoolingā€ going on even during the daytime in the sense that the Earth is always radiating heat. It is just that the warming due to the sun is larger than the cooling due to the Earth radiating. If you increase GHGs, you can change that balance and cause it to become warmer during the day too. There are reasons why nighttime temperatures are expected to warm more than daytime temperatures under rising GHGs, but both are expected to rise (and have been rising).”
As I said there people who believe this, but there is no evidence which supports it.
The lunar surface during the day is much hotter than Earth is during the day, and the
Moon has no greenhouse effect.
According to greenhouse theory, the greenhouse effect adds 33 C to *average temperature*.
And Earth’s greenhouse effect can’t be making it warmer during the day, as Earth is colder
than the Moon [which has no greenhouse effect] during the day.
During the night Earth is far warmer than the Moon during it’s night. So the only large
effect possible of Earth’s greenhouse effect, or this +33 C which is suppose to added
from the greenhouse effect can only be in keeping the night times warmer.
So there is no evidence of it.
And the theory which believer believe in, does not support the idea that
greenhouse gases increase daytime temperature, yet still many believers
hold this view. [due to being brainwashed].

April 25, 2013 1:08 pm

Again, I just posted the actual text from the problem at 12:10 PM, unaltered, ctrl+F > cooling:
ā€œConsider a black sphere of radius R at temperature T which radiates to distant black surroundings at T = 0K.
(a) Surround the sphere with a nearby heat shield in the form of a black shell whose temperature is determined by radiative equilibrium. What is the temperature of the shell and what is the effect of the shell on the total power radiated to the surroundings?
(b) How is the total power radiated affected by additional heat shields?
(Note that this is a crude model of a star surrounded by a dust cloud.)
Solution:
(a) At radiative equilibrium, J ā€“ Jā‚ = Jā‚ or Jā‚ = J/2. Therefore Tā‚ā“ ā€“ Tā“/2, or Tā‚ = āˆœTā“/2 = T/āˆœ2
(b) The heat shield reduces the total power radiated to half of the initial value. This is because the shield radiates a part of the energy it absorbs back to the black sphere.ā€
~Problem 1023
Note: I missed your question I’ll check and see what it was real quick… hmmm, not sure which one you meant.
Surface temp of the sphere goes up, because according to problem 1026 (your reference), ā€œthis radiation shield reduces the rate of cooling of the bodyā€. Same power input, reduced cooling, the sphereā€™s surface temperature must go up until at equilibrium the shell is radiating as much power as the inner sphere receives. ~Gary
Ah, so you like 1023 when it agrees with you, but now it isn’t important?
According to 1023 the shell radiates half as much power as the inner sphere receives.
The problem specifically states that it is a simple model of a star in a dust cloud, hence it is roughly analogous to an internally heated sphere and shell.
Perhaps you can find me a Hubble image of a dust cloud glowing as bright as the surface of the stars embedded within, to support this “the sphere’s surface temperature must go up until the shell is radiating as much power as the inner sphere receives” conjecture?

Myrrh
April 25, 2013 1:24 pm

A C Osborn says:
April 25, 2013 at 10:44 am
Can someone point me to where the actual measured CO2 PPM for the upper troposphere or troposphere are stored?
AIRS has it. But it will not release the measurements for the lower or upper troposphere.
However, the scientists working on the project concluded, from all the measurements, that Carbon Dioxide in the atmosphere was lumpy and not at all well-mixed, to their astonishment as they had been brainwashed by this AGW fictional fisics meme, and so insignificant compared with water vapour in the atmosphere.
They released this conclusion at the same time as releasing cherry picked mid troposphere graphics which didn’t show the lumpiness and insignificance of carbon dioxide which they had found to arrive at their conclusion.
Their elementary sleight of hand worked and a flurry of discussions followed as if the conclusion related to the graphics released, which it didn’t. Although to some extent these showed it not well mixed, it didn’t show how lumpy it was.. Instead showing a spread within a few ppm, lumpy suggest areas with zilch.
Carbon dioxide is a real gas, not an “ideal gas” as claimed for the AGW’s Greenhouse Effect.
It has weight because of mass subject to gravity, it is one and a half times heavier than air, this means it will always sink displacing air unless work is being done to change that. Heavier than air gases sink, lighter than air gases rise. It will not readily rise from the ground into the atmosphere, just like the dust on your desk. If it is heated it will as all gases do, expand becoming less dense and so lighter than air will rise, but on releasing this heat it will again condense, real gases condense, and so becoming again heavier than air will sink back to the surface.
Winds, which are volumes of air on the move, that is, convection currents created by differential heating, can move carbon dioxide from one place to another, but, when the wind stops real carbon dioxide will not defy gravity but will sink to the surface. If large amounts of carbon dioxide get into the the great trade winds it can get carried great distances, over the sea to Hawaii for example, but most winds are local and short lived and will move any carbon dioxide around only at a local level. Our great wind systems do not cross hemispheres, so even if carried by these it will not become “well mixed in the atmosphere”. This is what makes carbon dioxide lumpy together with the fact that some areas just don’t produce much and some use all or most of what they produce locally.
And that’s the carbon dioxide which doesn’t come down as rain, all rain is carbonic acid.
AGWSF doesn’t have rain in its “carbon cycle”, firstly because it has eliminated the Water Cycle to create its “Greenhouse Effect”, and secondly because the AGWSF “carbon dioxide” is not real, it is an “ideal gas with no mass, volume, weight, attraction not subject to gravity”, and so not the real carbon dioxide and real water vapour which are greatly attracted to each other in the atmosphere. Real carbon dioxide is in this fully part of the Water Cycle with the residence time of water in the atmosphere of 8-10 days.
There was some Japanese study a short while back which confirmed the lumpiness of carbon dioxide, I think over China, and I recall something about the graphic being pulled and some ambiguous one put in its place.

Peter C
April 25, 2013 1:30 pm

To Gary Hladik,
Why not actually try this for yourself, as an experiment? That way you would know what happens. I have attempted a similar experiment. My result was that the inner black body did not heat up when it’s outgoing. Radiation was reflected back apron it’s self.
My experimental result may be challenged as simplistic and flawed in some respects but in my view it is better than arguing from theory. Maybe some can do a better experiment. Until they do it is my view that cooler objects DO NOT WARM hotter objects, even acting as insulators.

Kristian
April 25, 2013 1:39 pm

tjfolkerts says, April 25, 2013 at 4:49 am:
“You are ignoring the ā€œJ1ā€³ = J/2 that is clearly shown coming back to the inner sphere from the shell due to the temperature of the shell..”
In what way am I ignoring it? Did you see the J = (J – J1) + J1?
“The NET loss from the planet is J/2
The NET loss from the shell is J/2.
The power required from a heater to maintain this situation is J/2.”

Nope. The heater will still have to provide J. J = J/2 + J/2. It has to maintain both losses. They both originally trace back to the surface of the sphere, after all … as J.

Kristian
April 25, 2013 1:52 pm

ā€œYou are ignoring the ā€œJ1ā€³ = J/2 that is clearly shown coming back to the inner sphere from the shell due to the temperature of the shell..ā€
Ok Tim, so let’s see what this entails …
At first you have the sphere alone in space, powered by an internal heat source providing the surface of the sphere with a constant and evenly distributed power density flux (the heat input) of say 400 W/m^2. This flux brings the surface temperature to a steady 290K and this temperature in turn sets the outgoing flux from the surface of the sphere (the heat loss) to an equal 400 W/m^2. We have energy balance and a steady surface temperature corresponding to the input/output – Stefan-Boltzmann.
Then we surround the sphere concentrically with the shell, a narrow gap of vacuum between the two bodies (R ~ r). What happens?
Well, the internal heat source of the sphere still supplies the surface with its constant flux of 400 W/m^2 which in itself could still only bring the surface temperature to the same 290K as before … and no further. These 290K would in turn by necessity (dictated by the laws of physics) also still produce the same emission flux of 400 W/m^2, output equalling input, as before.
In other words, nothing relating specifically to the sphere and its internal heat source has changed – same input (400 W/m^2), same corresponding emission temperature (290K), same corresponding output (400 W/m^2).
So what has changed with the shell in place? The surface of the sphere now gets 200 W/m^2 in return as ‘back radiation’ from the shell as it’s heated to a temperature of 244K by the outgoing flux from the sphere. That is, it now receives an additional 200 W/m^2 which weren’t there before.
Now, you want to add this flux of 200 W/m^2 to the already supplied flux to the surface of the sphere of 400 W/m^2 from the internal heat source, according to you in the end bringing the new system to a steady state where the shell is at a temperature of 290K and the surface of the sphere is at 345K. At this point the sphere radiates 800 W/m^2 to the shell but gets 400 W/m^2 in return (to a net of 400 W/m^2) and the shell radiates 400 W/m^2 to space, equal to the still constant internal input to the surface of the sphere.
So what’s wrong with this picture?
Raising the temperature of an object or a surface takes absorption of heat. Objects are heated by positive heat transfer. Heat only spontaneously transfers from a warmer to a cooler object.
So how is the 200 W/m^2 flux from the cool shell to the warm surface of the sphere ever going to be an inward positive transfer of heat?
If you want to claim that the 290K sphere warms up to a higher steady-state temperature (by 19%) in the presence of the 244K shell, then you are in effect saying that the shell operates as a second, independent heat source for the sphere.
Why? Because nothing else has changed. The internal heat source still provides but the 400 W/m^2 to the surface of the sphere. This flux will in itself not warm it past the 290K it did before. The sphere also still emits its corresponding 400 W/m^2 flux from its surface based on this temperature.
The 400 W/m^2 from the internal heat source still goes straight out from the surface. It doesn’t in any way linger.
So if the temperature then still rises, then this must somehow be caused by something else. It must be caused by extra absorbed heat from somewhere. A positive transfer of (a gain in) thermal energy.
Well, the only other body in this system is the 244K shell. The only thing thatā€™s different is the additional 200 W/m^2 flux from the cool shell to the warm sphere.
You want to claim that this radiative flux rather somehow slows the cooling rate of the sphere, thus heating it indirectly? Fine, but then you should also explain and show how specifically this is accomplished without the 200 W/m^2 flux of 244K spectrum radiation itself transferring HEAT to it and thereby raising the warmer sphereā€™s kinetic energy level beyond what it would be otherwise ā€“ and hence its temperature? Does it somehow disallow half of the 400 W/m^2 of 290K BB spectrum radiation from leaving the surface of the warm body? No. You’re just adding the 200 to the 400, aren’t you? Effectively as extra transfer of heat to the sphere.

Gary Hladik
April 25, 2013 1:59 pm

Maxā„¢ says (April 25, 2013 at 1:08 pm): “Note: I missed your question Iā€™ll check and see what it was real quickā€¦ hmmm, not sure which one you meant.”
Not important. You asked essentially the same question in your own way when you derived the Steel Greenhouse.
“Ah, so you like 1023 when it agrees with you, but now it isnā€™t important?”
1023 and 1026 are essentially the same problem. In fact, as tjfolkerts agrees (April 25, 2013 at 6:18 am) 1026 reduces to 1023 when R is close to r. Now 1023 is cuter, but I like both 1023 and 1026 equally. šŸ™‚
“The problem specifically states that it is a simple model of a star in a dust cloud, hence it is roughly analogous to an internally heated sphere and shell.”
Right, problem 1026. That’s why they made R larger than r, since you don’t see many dust clouds a kilometer or so from a star’s “surface”. In fact in the case of Zeta Ophiuchi
http://www.spitzer.caltech.edu/images/5517-sig12-014-Massive-Star-Makes-Waves
R would be about half a light year, and even then the dust isn’t a perfectly opaque shell.
“Perhaps you can find me a Hubble image of a dust cloud glowing as bright as the surface of the stars embedded within, to support this ā€œthe sphereā€™s surface temperature must go up until the shell is radiating as much power as the inner sphere receivesā€ conjecture?”
Conjecture? Since when are the First Law of Thermodynamics and the Stefan-Boltzmann Law “conjecture”? You don’t believe your own reference? 1023 says the shell cools at a rate J1, half the rate J of the inner sphere. If the system is in equilibrium (no temp change), J1 must be the same as the power input to the sphere. The sphere cools at rate J = 2*J1, but it receives J1 from its power source plus J1 from the shell (look at the diagram), so it’s in equilibrium, too.
Now take away the shell. The sphere (temporarily) radiates/cools at rate J, but now receives only J1 = J/2 from its power source. It must cool until it radiates/cools at rate J1 (plug your own figures for J and J1 into the Stefan-Boltzmann calculator).
Add the shell back, and the sphere’s temp goes back up because, as problem 1023 says, at equilibrium the shell cools/radiates to the surroundings at half the radiation/cooling rate of the sphere, and from Stefan-Boltzmann a surface radiating at J = 2*J1 is hotter than a surface radiating at J1. Easy peasy.

April 25, 2013 2:02 pm

Myrrh says:
April 25, 2013 at 1:24 pm
Myrrh,
The language used by the AIRS people is quite confusing: calling a variation of a few ppm’s (+/- 2% of the full range) over the seasons as “not well mixed”, while about 20% of all CO2 in the atmosphere is exchanged with CO2 from other reservoirs over the same seasons. That only shows that CO2 in the atmosphere is well mixed and quite rapidely. Any huge change in CO2 at ground level is mixed within days to weeks for the same altitude and latitude band, it takes weeks to months to mix it over a full hemisphere and 1-2 years between the hemispheres.
Anyway, over the full globe, except for the first few hundred meters over land, the measured CO2 levels are within +/- 2% of the full range. That is for over 95% of all air…

Reed Coray
April 25, 2013 2:07 pm

richardscourtney says: April 25, 2013 at 10:45 am
Thanks for trying to remove any confusion that might exist. Maybe I can’t see the forest for the trees; but Dr. Spencer did not qualify his statement by caveating that it applies to the Earth and its atmosphere. In fact, he parenthetically included a reference to a house. Taken at face value, Dr. Spencer’s statement:
ā€œGreenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, ā€¦ā€
imply that everything else being equal, if two objects exist at different temperatures the presence of a greenhouse gas between the objects will, relative to the absence of that greenhouse gas, reduce the rate heat flows from the high-temperature object to the low-temperature wall. If this statement is true, then a CO2 thermos should outperform a vacuum thermos. I haven’t conducted the experiment; and it may turn out that, yes, a CO2 thermos outperforms a vacuum thermos. I don’t believe that is the case–i.e., I believe the vacuum thermos will outperform the CO2 thermos. If the vacuum thermos outperforms the CO2 thermos, then Dr. Spencer’s statement without caveats is not correct. The presence of a greenhouse gas does NOT reduce the rate heat flows from the thermos hotter inner wall to the thermos cooler outer wall.
I’m probably nitpicking here, but from my perspective many people on both sides of the AGW fence arrive at conclusions based on statements made by “experts”. I agree that Dr. Spencer is an expert. But if Dr. Spencer makes a statement that applies only under certain conditions but when making that statement does not identify those conditions, then I’m concerned that people will inappropriately apply that statement to conditions not covered by the caveats. Occasionally the process snowballs out of control leading to incorrect conclusions with societal impact.
Put into yes/no questions, in your opinion
(a) For hot coffee inside the chamber of a vacuum thermos bottle located in a room at a lower temperature than the coffee, is there one or more hot objects and one or more cold objects?
(b) Relative to the outer wall, is the thermos bottle chamber wall a hot object?
(c) Relative to the chamber wall, is the thermos bottle outer wall a cool object?
(d) Will heat flow from the chamber wall to the outside wall?
(e) Will the rate of heat flow from the chamber wall to the outside wall be reduced if CO2 gas is injected into the space (a vacuum prior to CO2 gas injection) between the inside wall and the outside wall?
If the answers to these questions are (a) yes, (b) yes, (c) yes, (d) yes, and (e) no, then do those answers constitute a counterexample to the statement: “Greenhouse gases reduce the rate at which heat flows from higher temperatures to lower temperatures?”
If your answer to the last questions is “No”, then I’d appreciate hearing your rationale.
Thank you for your time.

joeldshore
April 25, 2013 2:24 pm

gbaikie says:

As I said there people who believe this, but there is no evidence which supports it.
The lunar surface during the day is much hotter than Earth is during the day, and the
Moon has no greenhouse effect.
According to greenhouse theory, the greenhouse effect adds 33 C to *average temperature*.
And Earthā€™s greenhouse effect canā€™t be making it warmer during the day, as Earth is colder
than the Moon [which has no greenhouse effect] during the day.

And the theory which believer believe in, does not support the idea that
greenhouse gases increase daytime temperature, yet still many believers
hold this view. [due to being brainwashed].

No…They believe it because they understand physics.
Look, you can only get so far by comparing the Earth and the moon. The fact that one has a greenhouse effect and the other doesn’t is only one difference. There are other important differences: The moon has very little atmosphere at all and does not have water. This means that the effective heat capacity or “thermal inertia” is much smaller so temperatures vary more dramatically between day and night. Furthermore, the moon has a much longer day than the Earth…about 30 times longer, so this also favors larger temperature variations. Neither of these differences should affect the average temperature (or, more precisely, the average of T^4), which is determined by radiative balance, but they do affect the range of temperatures.
For heaven’s sake, think things through before making statements about people being brainwashed. You might find that you can actually learn from people who have thought about these things much more than you apparently have.

April 25, 2013 2:25 pm

Myrrh says:
April 25, 2013 at 1:24 pm
It has weight because of mass subject to gravity, it is one and a half times heavier than air, this means it will always sink displacing air unless work is being done to change that. Heavier than air gases sink, lighter than air gases rise. It will not readily rise from the ground into the atmosphere, just like the dust on your desk.
Think one moment further: why should the dust on your desk or the sand of the Sahara travel thousands of km through the air, with a density some 100 times that of air and CO2 at only 1.5 times heavier would sink out? Lookup “Brownian motion” works as good for dust and feathers or sand as for CO2 molecules (or even for far heavier CFC molecules reaching the stratosphere…).

Gary Hladik
April 25, 2013 2:26 pm

Kristian says (April 25, 2013 at 1:52 pm): “So whatā€™s wrong with this picture?
Raising the temperature of an object or a surface takes absorption of heat. Objects are heated by positive heat transfer. Heat only spontaneously transfers from a warmer to a cooler object.
So how is the 200 W/m^2 flux from the cool shell to the warm surface of the sphere ever going to be an inward positive transfer of heat?”
Aaaaaand the imaginary second law of thermodynamics strikes again!
http://scienceofdoom.com/2010/10/07/amazing-things-we-find-in-textbooks-the-real-second-law-of-thermodynamics/

Matt in Houston
April 25, 2013 2:26 pm

Thank you Dr, Spencer for your response to my comment. I think it is likely that we agree in real principle but we are not communicating clearly in the context of your comment. A cooler object cannot HEAT a warmer object. This violates the Second Law of thermodynamics. However if we take an object of high temperature X and surround it with some insulative gas (ie ~GHG) it’s rate of cooling will necessarily decrease due to the lower rate of radiative heat transfer. However it cannot have it’s temperature go up as a result of adding some cool gas surrounding it…To be more clear, I will go back and reread all of the source material that led into this discussion so that we are not talking past each other or that I am misunderstanding the context of your statement. I appreciate your efforts in all of these matters Dr., Spencer I believe this is one of the most critical paths to leading to a proper understanding of Atmospheric dynamics.
I can see that many folks here as usual have flown off the handle on this topic. This is not the way to a clear picture of what really happens from a systematic description of the real physical world. Not once in any of my mathematics, chemistry or physics courses, have I seen quantitative mathematics described as “hand waving” or “complete and utter non-sense”. Perhaps I am biased having come from an aerospace background to Dr. Latour’s statements on this matter, but the folks who responded to my comment have not made any statement of principle to refute anything he has said, so I take it the assault on character is the only response they are willing to expend energy on. This is not a solution. IF you have a clear statement to be made, please enlighten us. I am more than willing to listen to a good rationale response.
My goal is to understand the physics properly- this is to say, what is reality- exactly, in cold calculated mathematical terms that are capable of predicting system behavior in order to design and operate hardware inside that system, be it outer space, under the ocean or inside the earth’s crust on the end of a drilling rig. Too many folks here are just bent on smashing people down and playing king of the mountain. This is unfortunate. I have an engineering degree, I spent a decade building hardware that has to fly in space. I have a grasp of the physics, but too many people here are playing the angry genius game and it is pathetic. IF you want to right the ship, SHOW US that you are right in the Scientifically principled sense please. Stop insulting people and SHOW US.

Kristian
April 25, 2013 2:35 pm

Gary Hladik says, April 25, 2013 at 12:35 pm:
“”Kristian says (April 25, 2013 at 4:06 am): ā€œThe system is radiating out power to its surroundings twice as fast as soon as the surrounding shell is removed, Gary. The sphere isnā€™t.ā€
When the shell is snatched away (along with its insulating effect), the sphere is the system.”

Yes. It is NOT the system when the shell is in place. Hence my quoted words.
“As Dr. Spencer points out, you need to know both the heating rate and the cooling rate.”
Absolutely not. When it comes to radiative heat transfer (BBs in vacuums) the surface temperature of an object is set by the absorbed heat flux. This flux sets a specific corresponding emission temperature which in turn directly determines the emitted heat flux, equal in power density to the absorbed one. ABSORBED HEAT –> EMISSION TEMPERATURE –> EMITTED HEAT. The emitted heat is a function of the temperature which is a function of the absorbed heat. The emitted heat flux does not in any way affect the temperature of the emitting object, no more so than the temperature dictates the absorbed heat flux. In other words, at steady state (dynamic equilibrium) ‘the cooling rate’ is always equal to ‘the heating rate’: Q = Q’ + Q” = Q/2 + Q/2 (J = (J – J1) + J1 = J/2 + J/2 if you will …).
“According to problem 1023, the system was in equilibrium with a cooling rate of J1, which is half of the sphereā€™s cooling rate J. Snatch away the shell and instantly the cooling rate doubles. Same power input, double cooling rate, the temp must go down. This is basic thermodynamics.”
Apparently not basic enough. You keep mixing up the system and the sphere. The flux leaving the surface of the sphere is always J, with or without shell. The ‘system’ flux to the surroundings is J with sphere alone and J1 (J/2) with shell surrounding it. How can this be? J from the sphere is intersected by the shell, splitting it, like a Carnot engine coming in between a hot and a cold reservoir. The shell (or the engine) ‘takes’ some of the heat flux for itself. How? The shell gets heated, the engine does mechanical work. The rest? Goes out to the cold reservoir as heat loss.
You see, before the shell came in place, the power from the sphere’s internal heat source went into heating the sphere only (one object). With the shell, that same power flux now goes into heating and maintaining the temperature of TWO objects, not just one. Hence the reduced system flux to the surroundings.

Gary Hladik
April 25, 2013 3:03 pm

Kristian says (April 25, 2013 at 2:35 pm): “The flux leaving the surface of the sphere is always J, with or without shell. The ā€˜systemā€™ flux to the surroundings is J with sphere alone and J1 (J/2) with shell surrounding it. How can this be? J from the sphere is intersected by the shell, splitting it, like a Carnot engine coming in between a hot and a cold reservoir. The shell (or the engine) ā€˜takesā€™ some of the heat flux for itself. How? The shell gets heated, the engine does mechanical work. The rest? Goes out to the cold reservoir as heat loss.”
So according to you, at equilibrium the power input to the system (including shell) is J, and the system output (from shell) is J1 = J/2. Unless you’re somehow destroying energy (and violating the First Law), the “system” temperature must rise until the system output is J. You’ve violated your own condition that the system is in equilibrium.

April 25, 2013 3:04 pm

Conjecture? Since when are the First Law of Thermodynamics and the Stefan-Boltzmann Law ā€œconjectureā€? You donā€™t believe your own reference? 1023 says the shell cools at a rate J1, half the rate J of the inner sphere. If the system is in equilibrium (no temp change), J1 must be the same as the power input to the sphere. The sphere cools at rate J = 2*J1, but it receives J1 from its power source plus J1 from the shell (look at the diagram), so itā€™s in equilibrium, too.” ~Gary
1023 says:
(a) At radiative equilibrium, J ā€“ Jā‚ = Jā‚ or Jā‚ = J/2. Therefore Tā‚ā“ ā€“ Tā“/2, or Tā‚ = āˆœTā“/2 = T/āˆœ2
(b) The heat shield reduces the total power radiated to half of the initial value. This is because the shield radiates a part of the energy it absorbs back to the black sphere.
Take the Sun and a Dyson sphere 1 AU wide.
At 1 AU the power radiated by the Sun, J = 1366 W/mĀ², so the sphere would radiate Jā‚ = 683 W/mĀ² inward and outwards.
So your position is then that the Sun, a ball of hydrogen and helium so massive it has a self-sustaining thermonuclear fusion reaction in the core… is going to be warmed by the 683 W/mĀ² radiated back inwards from the shell towards the surface until the temperature rises sufficiently for the shell to receive 2732 W/mĀ² and then emit 1366 W/mĀ² outwards?
But… if 683 W/mĀ² coming back in from the shell was able to raise the temperature of the Sun… wouldn’t 1366 W/mĀ² coming back in from the shell do the same thing?

While we’re at it, what is the “rate of cooling” of a main sequence star?

Gary Hladik
April 25, 2013 3:05 pm

Peter C says (April 25, 2013 at 2:44 pm): [snip]
Peter, check out Willis’s article on the R W Wood experiment:
http://wattsupwiththat.com/2013/02/06/the-r-w-wood-experiment/

April 25, 2013 3:08 pm

Whoops, meant “Dyson sphere with a radius of 1 AU”.

Kristian
April 25, 2013 3:08 pm

Gary Hladik says, April 25, 2013 at 2:26 pm:
“Aaaaaand the imaginary second law of thermodynamics strikes again!”
Please, read the whole post before commenting.

Kristian
April 25, 2013 3:11 pm

Gary Hladik says, April 25, 2013 at 3:03 pm:
“So according to you, at equilibrium the power input to the system (including shell) is J, and the system output (from shell) is J1 = J/2. Unless youā€™re somehow destroying energy (and violating the First Law), the ā€œsystemā€ temperature must rise until the system output is J. Youā€™ve violated your own condition that the system is in equilibrium.”
You’re still not reading what you’re commenting on, Gary.

joeldshore
April 25, 2013 3:14 pm

Matt in Houston says:

A cooler object cannot HEAT a warmer object. This violates the Second Law of thermodynamics.

Correct.

However if we take an object of high temperature X and surround it with some insulative gas (ie ~GHG) itā€™s rate of cooling will necessarily decrease due to the lower rate of radiative heat transfer.

Correct.

However it cannot have itā€™s temperature go up as a result of adding some cool gas surrounding it.

Incorrect. The temperature of the Earth is determined by the need to balance the energy coming in and the energy going out. If you surround the Earth by some cool gas, then you reduce the energy going out and the Earth responds by heating up until it is again emitting as much energy out as before.
One way to look at this: That gas may be cool but absolute zero (or 3 K) is much colder. That gas is helping to keep the Earth warm by insulating it from the much colder outer space.

Not once in any of my mathematics, chemistry or physics courses, have I seen quantitative mathematics described as ā€œhand wavingā€ or ā€œcomplete and utter non-senseā€.

I have invested a lot of time and effort trying to have a rational discussion with Pierre Latour (and Joseph Postma and company). It is not possible. They are engaged in active deception and sophistry (even if it is self-deception so that they actually believe what they are saying, which is hard for me to believe, but could be true given the ability of otherwise intelligent people to be incredibly obtuse when they actively don’t want to understand something). After a while, one has to call a spade “a spade”.

April 25, 2013 3:15 pm

Reed Coray:
I am replying to your post at April 25, 2013 at 2:07 pm.
I tried to help by removing a confusion. Clearly, your post I am answering shows I failed. Sorry.
The matter has nothing to do with the words of any authority: it is about how the world works.
And I will not get bogged-down in detailed discussions of coffee pots, thermos flasks, and etc.. Instead, I will try to explain the over-arching principle which applies to all the illustrations. I will then present the simplest analogy I have imagined.
1.
Heat flows from a hotter object to a colder one. So, the hotter object gets colder (i.e.its temperature falls) unless it is supplied with a source of heat.
2.
When supplied with a source of heat the temperature of the hotter object will rise unless it loses heat to a colder object.
3.
If the temperature of the hotter object does not change (i.e. is constant) then the rate at which the hotter object loses heat is a function of how much heat is supplied to it and the temperature difference between it and the colder object.
4.
A restriction to the transfer of heat from the hotter to the colder object reduces the rate at which heat is lost from the colder object. But the supply of heat remains constant. Therefore, the hotter object gets even hotter (i.e. its temperature rises) until its temperature difference with the colder object again establishes a loss of heat equal to the heat input.
As an analogy, consider a bath tub supplied with water by a tap (US: faucet) and with no plug in its exit to a drain.
In this analogy,
the bath is like the hotter object,
the water from the tap is like the input of heat,
the drain is like the colder object,
the flow of water through the plug-hole to the drain is like the transfer of heat, and
the level of water in the bath is like the temperature of the hotter object.
If the water supply from the tap is constant then the bath establishes a level of water which remains constant.
Now inhibit the loss of water through the plug-hole by covering part of the plug-hole.
(This is akin to inhibiting the loss of heat from the hotter object by putting insulation between it and the colder object).
The water level in the bath rises until the difference in pressure difference between water in the bath and in the drain pushes the water through the reduced plug-hole at the same rate as the supply of water from the tap.
(This is akin to insulation raising the temperature of the hotter object).
It does not matter how or with what you inhibit water flow from the bath to the drain: if the flow through the plug hole is restricted then the water level of the bath rises unless its water supply is reduced.
Similarly,
It does not matter how or with what you inhibit heat flow from the hotter object to the colder object: if the flow through the the route to the colder object is restricted then the temperature of the hotter object rises unless its heat supply is reduced.
And this is true for all heat flows from hotter to colder objects.
So, insulation in the walls of a house raises the temperature in the house in the same basic way as greenhouse gases in the air raise the temperature of the Earth’s surface.
I hope the issue is now clear. If not, then I apologise because I do not think I can simplify it further.
Richard

joeldshore
April 25, 2013 3:19 pm

Oh yeah…I should perhaps add one more point to my last post: The strict sense in which a colder object cannot heat a warmer object is that the heat (net energy flow) must be from the hot object to the colder object. So, no, adding some cold gas around the Earth and causing the temperature to rise does not mean that the colder gas has heated the Earth: the net flow of energy is still from the Earth to the colder gas. You have just made it harder for energy to leave the Earth, i.e., the Earth has to maintain a higher temperature at its surface in order to radiate away the energy at the rate it is receiving it from the sun.

Konrad
April 25, 2013 3:31 pm

AGW is a physical impossibility. Those on this thread debating the physics of radiative gases are looking in the wrong place. Radiative physics is fine. The critical mistakes in the ā€œbasic physicsā€ of the ā€œsettled scienceā€ are in fluid dynamics and gas conduction.
Under a non radiative atmosphere the average land surface temperature will be lower. However a non radiative atmosphere would be dramatically hotter.
In a non radiative atmosphere there would be no strong vertical convective circulation.
An atmosphere in a gravity field which the gases are free to move is more easily conductively heated than it is conductively cooled by the surface.
Some of the critical mistakes of the climate scientists are
1. Failing to consider the critical role of radiative gases in tropospheric convective circulation.
2. Failing to correctly model conductive and convective energy transfer in moving gases in a gravity field.
3. Failing to understand that conductive heating of the atmosphere by the surface is set by surface Tmax not surface Tav.

joeldshore
April 25, 2013 3:32 pm

Reed Coray says:

Maybe I canā€™t see the forest for the trees; but Dr. Spencer did not qualify his statement by caveating that it applies to the Earth and its atmosphere.

But if Dr. Spencer makes a statement that applies only under certain conditions but when making that statement does not identify those conditions, then Iā€™m concerned that people will inappropriately apply that statement to conditions not covered by the caveats. Occasionally the process snowballs out of control leading to incorrect conclusions with societal impact.

I think Roy Spencer’s statement is fine. If one tries to phrase any statement in such a way that it cannot possibly be misinterpreted (if that is even possible), then one ends up writing something that is more like a legal document than a scientific one and one needs a lawyer to interpret it.
And, I fail to see how Roy’s statement is going to lead to incorrect conclusions with societal impact unless there is some push that I am not aware of in our society to put greenhouse gases into Thermos bottles. As regards the greenhouse effect on the Earth, Roy’s statement is correct.

April 25, 2013 3:34 pm

jkoeldshore writes “Noā€¦You donā€™t get it. If the total number of CO2 molecules doubles, the atmosphere becomes more opaque and the ERL moves up to a new level where the number of CO2 molecules is less (so the total opaqueness above it is still the same). ”
I think I do get it. And I described it to you but you justy parrotted back the standard reasoning why the ERL increases based on a constant opacity. If the there is more radiation because there are more CO2 molecules with sufficient energy at the same temperature then surely that means more radiation from that altitude escapes. Afterall the amount of captured radiation above is a probability based on the number of molecules (ie that opacity)…and as far as I can tell operates for each molecule trying to escape. Add more and more must escape surely?
If this reason is wrong then how about explaining the physical reason I’m wrong rather than parroting back something you’ve heard. How about showing me YOU understand why because if the number of molecules above must be constant then there must be a physical explanation for that.

Gary Hladik
April 25, 2013 3:51 pm

Matt in Houston says (April 25, 2013 at 2:26 pm): “I have a grasp of the physics, but too many people here are playing the angry genius game and it is pathetic. IF you want to right the ship, SHOW US that you are right in the Scientifically principled sense please. Stop insulting people and SHOW US.”
Why do you think that Latour & co. refuse to SHOW us they’re right by performing an actual experiment? Note that the burden of proof is on them, because all the textbooks and all the physicists say they’re wrong. Why do they handwave instead of doing the work that (in their minds) would get them a Nobel Prize, for starters? Why do you let them get away with handwaving when a relatively simple experiment would prove them right?

Gary Hladik
April 25, 2013 3:59 pm

Kristian says (April 25, 2013 at 3:08 pm): “Please, read the whole post before commenting.”
I did. Did you?
Kristian says (April 25, 2013 at 3:11 pm): “Youā€™re still not reading what youā€™re commenting on, Gary.”
Oh, but I did. I read very carefully, multiple times, because I couldn’t actually believe I was reading what I thought I was reading. Here, your words:
“The ā€˜systemā€™ flux to the surroundings is J with sphere alone and J1 (J/2) with shell surrounding it. How can this be?”
Indeed! How can this be unless 1) you’re destroying energy, or 2) the system isn’t in equilibrium? What do you think I’m missing here? Specifically.

davidmhoffer
April 25, 2013 4:06 pm

TimTheToolMan;
If the there is more radiation because there are more CO2 molecules with sufficient energy at the same temperature then surely that means more radiation from that altitude escapes.
>>>>>>>>>>>>>>>>>>>>
Joel Shore steered you correctly. Your assumption that more CO2 molecules = more radiation from a given altitude is incorrect. The energy into the system always equals the energy radiated out. Adding more CO2 molecules changes the average altitude at which any given photon escapes, and hence the amount radiated at any given altitude DOES change. But the average across the atmospheric air column as a whole stays exactly the same (at equilibrium).

tjfolkerts
April 25, 2013 4:17 pm

TimTheToolMan,
Maybe this will help.
Suppose that back when the CO2 concentration was 300 ppm, that 1/2 of the IR photons from 12 km up were able to escape to space — the rest being absorbed by CO2 molecules above 12 km. (these are not meant to be accurate numbers, but rather are hypothetical numbers for the sake of discussion).
Now bump the CO2 up to 400 ppm. There will indeed be more photons being created at 12 km altitude by more CO2 molecules. BUT less than 50% of them will escape to space because there are a lot more CO2 molecules above 12 km than there were before. We might have to go up to 12.5 km altitude before 1/2 of the emitted photons can now escape. The escaping photons now come from a higher altitude on average, And that higher altitude will be colder, so the net effect is LESS photons emitted to space from the atmosphere.

Matt in Houston
April 25, 2013 4:37 pm

joeldshore says:
April 25, 2013 at 3:19 pm
Thanks for the clarification, I was about to start hand waving and jumping around. =)
I appreciate the response. And I agree with it, however that was not the context of the comment I was making, hence the confusion. I suspect the conditional assumptions that people have in their minds is a large part of the problem in sorting all of this out into a standard model that accurately describes the mechanics of the earth system. Assumptions in these problems must be crystal and precise or they cannot be reworked by others to the same ends= more confusion and jumping around.
I don’t know what your relationship with Dr. Latour is but I believed him to have a cordial relationship discussing this matter with Dr. Spencer, which was why I asked for them to invite him. My readings of his material have not shown me any animosity in the discussion, however I know some folks are just not meant to be “friends”, lol. C’est la vie.
As I stated previously my end goal, as I would hope everyone here has, is an accurate and precise understanding of the physics of the system. Basic physics (including Quantum mechanics) allows us to form boundary condition problems (ie, lots of simplistic assumptions to make the problem tractable) to test the hypotheses that people are bandying about so we can attempt to verify and validate them. This is not really a simple thing, the earth atmosphere system is not a simple static system and many years have been spent by men far more brilliant than I (not that I am), attempting to resolve them down to principled mathematical descriptions, however in the case of the earth it is intractable in reality at this point as far as I can tell (GCMs anyone?) or someone would have demonstrated the right from the wrong in an unquestionable fashion. This discussion is all good stuff in my opinion, it is the only way to a clear understanding of reality. IT is what we need, hopefully in a more cordial manner than not. Hahahaha!

April 25, 2013 4:39 pm

Indeed! How can this be unless 1) youā€™re destroying energy, or 2) the system isnā€™t in equilibrium? What do you think Iā€™m missing here? Specifically.” ~Gary
Well, in my above Sun/1 AU radius Dyson sphere example, the sphere is gaining a net of 683 W/mĀ² from the Sun and losing 683 W/mĀ² to space, isn’t it?
Oh, I forgot, is it supposed to make the Sun hotter when it does that?

joeldshore
April 25, 2013 4:43 pm

Matt in Houston says:

A cooler object cannot HEAT a warmer object. This violates the Second Law of thermodynamics.

I originally pronounced this correct, but the more that I think about it and the problems it seems to cause for some people, the more I think it is best said to be “correct but ambiguous and prone to misinterpretations”. A better statement of the 2nd Law is the one that Flanders and Swann use in their song ( http://www.youtube.com/watch?v=Cb2kBFqrZx8 ) : “Heat cannot of itself pass from one body to a hotter body”.
The reason this statement is better is it makes it clearer that the issue is with the direction of the net flow of energy, whereas the statement about a cooler object not heating a warmer object can be misinterpreted to mean that a cooler body cannot in any way cause a hotter body’s temperature to increase over what it would be in the absence of the cooler body. That interpretation is flat-out wrong.

Myrrh
April 25, 2013 4:52 pm

Ferdinand Engelbeen says:
April 25, 2013 at 2:25 pm
Think one moment further: why should the dust on your desk or the sand of the Sahara travel thousands of km through the air, with a density some 100 times that of air and CO2 at only 1.5 times heavier would sink out?
As I said, one of the ways for carbon dioxide to move is from one place to another within the wind system, this is work being done to change the norm, just as, carbon dioxide can become lighter than air when heated because heated real gases expand becoming less dense therefore lighter than air under gravity, the heat is doing the work of changing the norm of carbon dioxide being heavier than air.
It is not always windy, it is not always hot…
It is simply a fact that AGW has substituted the imaginary “ideal” gas, pre Van der Waals, for the real gases of our real atmosphere. Out of the descriptions of the imaginary “ideal gas”, AGW has created an imaginary atmosphere.
The “ideal gas” has no mass, real gases have mass, therefore there is nothing for gravity to work on in the imaginary AGW atmosphere.
That’s why the imaginary AGW atmosphere is “empty space with ideal gases miles apart from each other travelling at great speeds under their own molecular momentum bouncing off each other in elastic collisions and so thoroughly mixing”.
Empty space is not an atmosphere.. The AGWScienceFiction’s Greenhouse Effect is built on imaginary fisics, it does not have an atmosphere, it goes straight from the surface to empty space.
It does not have any weather, weather cannot be created out of “ideal gas” with no mass, no volume, no weight, no attraction, not condensable and not subject to gravity.
Winds and weather are created by gases which have real properties and processes under gravity, which expand when heated and become less dense and so lighter because gravity gives real gases weight. Real gases condense when cold, becoming heavier. This is how we get our areas of high and low pressure – high pressure colder heavier, low pressure hotter lighter. The real Earth has a real gas atmosphere, a heavy volume of fluid gas (gases and liquids are fluids), weighing down on us, exerting pressure on us, because being pulled by gravity, of a stone per square inch, a one ton weight on your shoulders, constraining the speed of molecules. Empty space populated by massless ideal gas has no weight because it has no volume and so no voume of other gases to constrain its own molecular momentum.
Using concepts of properties and process from real world traditional physics to explain this fictitious ‘atmosphere’ to pretend that it is real, simply can’t be done. There is no joined up logic in the meme explanations from AGWSF.
I’ve given the example that they cannot make the claim that “carbon dioxide accumulates for hundreds and thousands of years”, because their gases are not subject to gravity, they have long gone into outer space…
So Likewise in your “Brownian motion” meme:
Lookup ā€œBrownian motionā€ works as good for dust and feathers or sand as for CO2 molecules (or even for far heavier CFC molecules reaching the stratosphereā€¦).
You look up Brownian motion.. And tell me how ideal gases without volume in empty space have anything to do with it. And tell me how Brownian motion can move carbon dioxide hundreds of miles when its scale is nanometres..
What is capable of moving feathers and dust and carbon dioxide hundreds of miles and the scent from the bottle opened in the classroom or ink poured into a glass of water, is not diffusion by ideal gas mollecular momentum nor Brownian motion of particles constrained by the voluminous fluid they are in jiggling them around, they are moved by convection currents. Volumes of fluid on the move.
Convection currents are volumes of gas and liquids on the move, called winds in our atmosphere and currents in our ocean.
The ideal gas of AGW has no volume, there is nothing to move en mass..
The fictional world of the AGW Greenhouse Effect has no sound because its ideal gases have no volume. They cannot form a medium through which sound can travel.
There is no sound in empty space.
So they have no wind or weather in their world, because they have no atmosphere..
Our atmosphere is the volume of real gases around our Earth pulled in by gravity.
We do not have empty space around us, but a heavy voluminous ocean of the real gas Air, that expands and condenses.
So which world are you living in?

joeldshore
April 25, 2013 4:55 pm

Matt in Houston: I don’t really disagree with anything in your latest comment except this statement:

however in the case of the earth it is intractable in reality at this point as far as I can tell (GCMs anyone?) or someone would have demonstrated the right from the wrong in an unquestionable fashion

I think this notion that science that goes against strongly-held beliefs will be accepted if only the evidence is strong enough is shown to be incorrect if one looks at the resistance by a large part of our society (including some who are scientists) to evolution / Big Bang Theory. I think there is reasonable hope that this will happen within the scientific community, e.g., no major scientific organizations dispute evolution, the overwhelming majority of peer-reviewed literature supports it, etc. However, in the larger community, strong resistance continues, including claims of persecution and bias against scientists who disagree with evolution, etc. Does any of this sound familiar?

Reed Coray
April 25, 2013 5:18 pm

richardscourtney says: April 25, 2013 at 3:15 pm
What you said is simple enough, and I agree with everything you said. If relative to heat leaving the Earth/Earth atmosphere system greenhouse gases increase the thermal insulation properties of the atmosphere without affecting the rate heat enters the Earth, then I agree, greenhouse gases will cause the temperature of the Earth in energy-rate equilibrium to be higher than the temperature of the Earth in energy-rate equilibrium without the greenhouse gases. The questions (issues) are then “(a) Does the presence of atmospheric greenhouse gases have an effect on the rate energy enters the Earth, and if so which way and by how much; (b) Does the presence of atmospheric greenhouse gases increase/decrease the thermal insulation of the atmosphere, and if so which way and by how much; and (c) considering both of these effects, is the energy-rate equilibrium temperature of the Earth higher or lower in the presence/absence of greenhouse gases? I don’t know the answers to any of these questions. But the statement that “Greenhouse gases reduce the rate at which heat flows from higher temperatures to lower temperatures” cannot by itself be used to argue that atmospheric greenhouse gases increase atmospheric thermal insulation because the statement is not universally true.

tjfolkerts
April 25, 2013 5:22 pm

Max, your Dyson Sphere would correspond to “Problem 1026”, not to “Problem 1023”. The results from Problem 1023 — where the shell is assumed to be “nearby” cannot be applied to the case of a Dyson Sphere where the shell is “far far away” from the sun.
So basically your entire set of conclusions are doomed before you even start.

Gary Hladik
April 25, 2013 5:28 pm

Maxā„¢ says (April 25, 2013 at 4:39 pm): “Well, in my above Sun/1 AU radius Dyson sphere example, the sphere is gaining a net of 683 W/mĀ² from the Sun and losing 683 W/mĀ² to space, isnā€™t it?
Oh, I forgot, is it supposed to make the Sun hotter when it does that?”
Why do you think the sun wouldn’t absorb the incoming energy?
Look at the diagram for problem 1023. Do you see the J1 arrow pointing from the shell toward the inner sphere? What do you think happens to that J1 when it hits the inner sphere? Does it somehow disappear?
Max, you’re not having second thoughts about your own reference, are you?

April 25, 2013 5:47 pm

tjfolkerts writes “Now bump the CO2 up to 400 ppm. There will indeed be more photons being created at 12 km altitude by more CO2 molecules. BUT less than 50% of them will escape to space because there are a lot more CO2 molecules above 12 km than there were before.”
Make the numbers easier. Bump the CO2 levels from 300ppm to 600ppm and now there are twice the molecules creating radiation at 12km altitude and twice the molecules above. My assumption is that is twice the opacity.
So twice the photons making their way through twice the opacity is an equivalent amount of energy leaving from the same altitude.
Your statement above “less than 50% of them will escape to space because there are a lot more CO2 molecules above 12 km than there were before” is correct. With a 33% increase in CO2 (300ppm to 400ppm), I’d expect 33% increase in radiation out at that same altitude. That is indeed less than half.
Where is this reasoning broken?

April 25, 2013 6:04 pm

tjfolkertds says:
“…less than 50% of them will escape to space…”
That’s wrong. Due to the curvature of the earth, the higher the altitude, the more likely that 50%+ of all emitted photons will escape to space.
Molecular density does not matter, because it all averages out. Eventually, the last photon will either hit the surface, or go out into space. More than half of all photons will escape to space once the emitting molecules are any altitude above the surface.
Gedanken: Imagine a molecule at 25,000 miles altitude. The earth would appear as a distant ball. How many molecules, on average, will be emitted toward the surface from that altitude? Answer: far less than 50%
Climate alarmists are constantly trying to justify their belief that CO2 causes any measurable global warming. But the earth itself is proving them wrong. Who to believe? Earth? Or alarmists?

davidmhoffer
April 25, 2013 6:33 pm

TimTheToolMan
With a 33% increase in CO2 (300ppm to 400ppm), Iā€™d expect 33% increase in radiation out at that same altitude. That is indeed less than half.
Where is this reasoning broken?
>>>>>>>>>>>>>>>>>
It is broken on the assumption that increasing the concentration increases the radiation out from a given altitude. It doesn’t. Several people have responded explaining why.

joeldshore
April 25, 2013 6:43 pm

TimTheToolMan says:

So twice the photons making their way through twice the opacity is an equivalent amount of energy leaving from the same altitude.

You are missing the whole point. The amount of energy leaving is not determined by this tradeoff between number and opacity. The amount of energy that a substance emits is determined by its temperature. For emission from the atmosphere, the opacity just determines where in the atmospheric column the emitted photons (i.e., those that successfully escape to space) come from. And, as the opacity increases, this emission has to occur at higher and higher levels of the atmosphere where, because of the lapse rate in the troposphere, it is colder and hence (by the Stefan-Boltzmann Law) less emission occurs.
And, this isn’t some sort of theoretical notion. It is FACT, verified by satellite observations ( http://www.barrettbellamyclimate.com/userimages/Iraq2.jpg ), and without it working as we understand it to work, the whole field of remote sensing would not be possible. It is, among other things, completely inconsistent to simultaneously hold the opinion that Spencer and Christy’s satellite measurements of the tropospheric temperatures have any validity and to believe that the greenhouse effect does not work in the way we are explaining it does. In fact, you can’t even really believe in the IR satellite photos that you see on the weather reports on TV without accepting these basics of radiative transfer in the atmosphere.

April 25, 2013 6:44 pm

davidmhoffer writes “It is broken on the assumption that increasing the concentration increases the radiation out from a given altitude. It doesnā€™t. Several people have responded explaining why.”
No they haven’t. I expect we’ll both agree that increasing the concentration increases the radiation rate (both up and down) at that altitude. tjfolkerts seems to agree with that and its pretty basic physics.
But when it comes to how much gets past the increased opacity above, nobody has explained anything, They’ve simply stated that more CO2 above stops the radiation without addressing the fact there is more radiation trying to get through. Your post is no exception.

tjfolkerts
April 25, 2013 6:54 pm

dbstealy, you are addressing a different issue. Geometry does indeed play a role in the fraction of the IR photons that are emitted “away from the earth”, and here molecular density does not matter. But even 20 km up, the horizon is only about 4.5 degrees downward.
But molecular density tells us if those photons emitted upward will indeed escape before hitting another CO2. Near the surface, 0% of the upward photons will escape.
(And now that I thin about it, the “nearly horizontal photons” are really not important. A photon that is heading “sideways” will go miles through “fairly dense CO2” and hence has a high chance of being absorbed. The photons heading “nearly vertical” will be heading straight for the thinner higher atmosphere and hence has a good chance of escaping. )

tjfolkerts
April 25, 2013 7:10 pm

TimTheToolMan,
It is not obvious whether the increased emission or the increased absorption will be “stronger”. Let me give two “plausibility” arguments.
1) Increasing the concentration of CO2 near the surface will have no effect on the outgoing radiation, since 100% of the photons created from the thicker CO2 will be absorbed before escaping. Conversely, if you go way up high where 99% of the photons currently escape, the doubled concentration will double the created photons, but most (maybe 95%) will escape. So we now have ~ 1.9 times as many photons escaping form this altitude. Overall, this pretty clearly is raising the effective radiating level.
2) More detailed calculations like MODTRAN support the conclusion that more CO2 reduces the outgoing radiation. http://forecast.uchicago.edu/Projects/modtran.html
At this level of discussion, i am not sure there is a “better” answer than these.

maximo
April 25, 2013 7:13 pm

[snip. ~mod.]
Insulation neither cools nor heats, it simply retards the transition from one state to the other. ie. Greatly slowing warming during hot weather and greatly slowing cooling during freezing weather.

April 25, 2013 7:18 pm

Why do you think the sun wouldnā€™t absorb the incoming energy?
Look at the diagram for problem 1023. Do you see the J1 arrow pointing from the shell toward the inner sphere? What do you think happens to that J1 when it hits the inner sphere? Does it somehow disappear?
” ~Gary
I think it replaces some of the photons leaving the Sun, but I don’t think it is going to make any sort of noticeable difference, the Sun isn’t heated at the surface, after all.
The star is heated by the action deep inside, and the temperature of the outer layers is determined by the photons that spend ages trickling out from the interior where the fusion reactions take place, not photons that return from whatever is illuminated by the star.

A Dyson sphere as wide as our orbit around the Sun, at this distance it receives 1366 W/mĀ², and it radiates half of that in and out, so you’ve got a shell of returning photons that started out at 683 W/mĀ² making it back to the surface of the Sun.
That replaces half of what escapes the surface, and the other half escapes to space from the exterior of the shell, seems like the system would be in balance there.
J = 1366 W/mĀ² | ā†’ |
Jā‚ = 683 W/mĀ² | ā† | ā†’
Even if one were to argue that the Sun should heat up, what mechanism do you propose would do this? What imbalance would this be correcting? Wouldn’t that create a greater imbalance if the surface had to heat up until the exterior of the shell radiated at the same power as the surface originally did?
Heck, if the net Surface > Shell is 683 W/mĀ², then wouldn’t the exterior already radiate at the same power when it emits 683 W/mĀ² to space?
Your argument sounds reasonable, but I don’t see how you justify the “and then the surface heats up until the exterior of the shell emits as much as the surface originally did” step.

April 25, 2013 7:30 pm

tjfolkerts writes “It is not obvious whether the increased emission or the increased absorption will be ā€œstrongerā€.”
Thats the way I see it and I’m always skeptical of these kinds of blanket statements that cant be backed up with a reasonably straightforward physical explanation. This one (increased ERL with increased CO2) appears to rely on non-linearity in CO2 absorbtion with concentration, a non-intuitive result.

Konrad
April 25, 2013 7:39 pm

The ERL hand waving fails because the gases in our atmosphere move. Hot gases are radiating more strongly than the gases at the altitude they are rising through.
The AGW nonsense also fails because the gases in our atmosphere move. Climate pseudo scientists are rubbish at fluid dynamics.

davidmhoffer
April 25, 2013 7:41 pm

TimTheToolMan;
Theyā€™ve simply stated that more CO2 above stops the radiation without addressing the fact there is more radiation trying to get through. Your post is no exception.
>>>>>>>>>>>>>>>>>>
But there isn’t.
There’s 240 w/m2 going in, and 240 w/m2 going out. Double the CO2. There’s STILL 240 w/m2 going in and STILL 240 w/m2 going out once equilibrium is re-established. There is NOT more radiation trying to get through.
Let me try and explain it another way. Suppose you have one row 4 of co2 molecules 1 meter above earth surface. Release 8 photons. 4 hit CO2 molecules, get absorbed, and re-emitted to space. The other 4 go straight on through. What is the effective radiating height?
4 escaped from ground which is zero, and 4 escaped from 1 meter. Average is 1/2 meter.
Now double the CO2. one row of 8. Release 8 photons, but now each and every one gets absorbed and re-emitted to space. Average emission height is now 1 meter. But there are still only 8 photons released.
Now think of that but with millions of layers. Any given photon might go just one layer before being absorbed and re-emitted, or it might go through several. It might get to space in 8 leaps straight up, or perhaps 4 up 3 down 1 up 2 down 6 up 1 down 3 up. Or millions of those. As the density of the CO2 molecules increases, the distance any given photon can travel before hitting a CO2 molecule goes down. The number of times any given photon will be absorbed and re-emitted before escaping will increase. The chances of any given photon escaping from a given altitude will be reduced at all altitudes, but will be reduced MORE the lower the altitude is.
So when CO2 doubles, you have a period where equilibrium is broken because any given photon has to make that many extra bounces up, down, and sideways until it escapes. But that’s temporary, the system re-established equilibrium with the exact same w/m2 coming in as before, and the exact same w/m2 leaving as before, but the place they leave from on average is a higher altitude.

April 25, 2013 7:47 pm

davidmhoffer writes “Now double the CO2. one row of 8. Release 8 photons, but now each and every one gets absorbed and re-emitted to space. Average emission height is now 1 meter. But there are still only 8 photons released.”
The thing is that there aren’t 8 photon released when you double the CO2. There might be 8 released at at ground level, but up where ERL happens the radiation is supplied by the GHGs not the ground. So now there are 16 photons emitting from twice the GHGs that are ready to emit and the argument follows from there.

April 25, 2013 7:55 pm

TimTheToolMan,
I think you’re missing Hoffer’s point [and mine]. The point is that when all is said and done, the average photon either heads for space, or hits the surface. It doesn’t matter if there are a dozen photons, or a zillion. Their final direction averages out.
The higher the altitude of emission, the more likely it is that any particular photon will head out to space. Simple as that. The curvature of the earth, along with the altitude, requires it.

Greg House
April 25, 2013 7:57 pm

richardscourtney says (April 25, 2013 at 3:15 pm): “So, insulation in the walls of a house raises the temperature in the house in the same basic way as greenhouse gases in the air raise the temperature of the Earthā€™s surface.”
==============================================================
This is so very wrong.
First, “greenhouse gases” in the air do not raise the temperature of the Earthā€™s surface, anyway not by radiating back radiation towards it, because such an effect of back radiation is physically absurd and therefore impossible, see my previous comments on this thread.
Second, insulation in the walls of a house prevents the air temperature in the house from increasing, decreasing or has no effect at all, depending on certain factors. Middle school stuff and practical experience of everyone.
In terms of insulation, the “greenhouse gases” only insulate the Earth surface from some portion of solar IR by blocking it. That is all. No warming of the surface by back radiation from “greenhouse gases” is possible.

April 25, 2013 8:05 pm

dbstealy writes “The higher the altitude of emission, the more likely it is that any particular photon will head out to space.”
I agree with you. This discussion is about precisely locating the average height at which that happens and looking at what happens to that average height when the concentration of greenhouse gasses changes.
AGW theory relies on that height increasing so its an important point to understand.

davidmhoffer
April 25, 2013 8:08 pm

TimTheToolMan;
The thing is that there arenā€™t 8 photon released when you double the CO2. There might be 8 released at at ground level, but up where ERL happens the radiation is supplied by the GHGs not the ground.
>>>>>>>>>>>>>>>
For the love of *** no!
ALL the photons are from the ground in the first place. You can double, quadruple or millionuple the CO2, ALL the photons in the system originate from the ground. All that changes is the number of times they bounce around before escaping. The GHG’s do not, I repeat DO NOT add additional photons to the system.
I’ll try again. Imagine many layers of CO2. At any given instant there are (for illustrative purposes) 8 leaving the ground, 8 bouncing around in the middle, and 8 escaping to space.
Double the CO2. Now there’s a lot more bouncing around to do before escaping to space, so equilibrium is broken. During this time, at any given instant, there are 8 photons leaving the ground, some number greater than 8 bouncing around in the middle, and some number less than 8 escaping. The number bouncing around in the middle will continue to increase until equilibrium is established again.
Once the new equilibrium state has been established, at any given moment in time, there are 8 photons leaving the ground, 16 photons bouncing around in the middle, and 8 photons escaping to space. So the density of photons in the middle does go up, but the number escaping to space once equilibrium is established is exactly the same. But the average height that they escape from is higher.

davidmhoffer
April 25, 2013 8:16 pm

TimTheToolMan;
Think of it like a lake with a river at one end and a dam at the other. The river flows at 240 m3/2. How much water goes over the dam? Assuming equilibrium and no other factors, 240 m3/2.
Raise the dam by 1 meter. At first it chokes off the flow of water altogether, maybe all the way to zero. How much water is coming in though? 240 m3/s, same as before. What happens to the lake? The level starts to rise. As the level rises, water starts to flow over the dam again. How much does the lake level rise? 1 meter. At which point the amount of water going over the dam is now 240 m3/s, same as before, and same as the amount coming in from the river.
Did the dam create new water? No. Is there more water in the lake than there was before? Yes. But the dam didn’t create it, the dam being raised just held the water back until the depth increased by the same amount as the dam was raised. What is the average height at which the water is escaping from the lake over the dam? 1 meter higher than it used to be. But at no pint did we create “new” water, the only water in the system comes from the river.

davidmhoffer
April 25, 2013 8:17 pm

AAAAARGH/ m3/s not m3/2
more bourbon…. need more bourbon…

April 25, 2013 8:26 pm

davidmhoffer writes “The GHGā€™s do not, I repeat DO NOT add additional photons to the system. ”
Of course they do. Photons from the surface are absorbed by the GHGs low down in the atmosphere just above the surface (first 100m or so) and the energy from that photon temporarily held by the GHG is almost always transferred to the rest of the atmosphere (O2 or N2)by collision and heats it.
Thats the “capture radiation from the earth’s surface and warm the atmosphere” part of the process.
Meanwhile the atmosphere as a whole consists of molecules with various amounts of energy each. The distribution of those energies is described by the Maxwell-Boltzmann distribution and at any time a number of molecules have sufficient energy to radiate a photon. Or course many of those molecules are O2 or N2 molecules which cant radiate but a portion of them are GHGs and they do radiate. So if you double the number of GHG molecules then you’re doubling the number of molecules in the atmosphere that can and do radiate both up and down.

tjfolkerts
April 25, 2013 8:29 pm

Max says: “A Dyson sphere as wide as our orbit around the Sun, at this distance it receives 1366 W/mĀ², and it radiates half of that in and out, so youā€™ve got a shell of returning photons that started out at 683 W/mĀ² making it back to the surface of the Sun.”
No, the Dyson sphere with a 1 AU radius will absorb 1366 W/m^2 from the sun and emit ~ 1366 W/m^2 outward and ~ 1366 W/m^@ inward. Of course, most of the 1366 W/m^@ inward will hit some other part of the shell, so the net heat transfer from the shell to the shell is ~ 0.
See “Problem 1026” for more details.

davidmhoffer
April 25, 2013 8:33 pm

TimTheToolMan;
So if you double the number of GHG molecules then youā€™re doubling the number of molecules in the atmosphere that can and do radiate both up and down.
>>>>>>>>>>>>>>>>>>
Yup. But what they radiate up and down is the exact same 240 w/m2 as before doubling. So the w/m2 get spread across a greater number of co2 molecules. But there is still only 240 w/m2 coming in and 240 w/m2 going out.

April 25, 2013 8:51 pm

davidmhoffer writes “So the w/m2 get spread across a greater number of co2 molecules.”
In the sense there is no net change, this is true. So if you put more CO2 above a particular altitude then its harder for an individual photon to escape from that altitude. This is the basis for the increased ERL in AGW theory. But if more photons are heading upwards then this must compensate for that increase in opacity.
Remember the original statement for the pro-AGW reference provided by Mosher
“ā€œInfrared radiation leaves earth for space from upper troposphere (ERL). Amount increases with temperature at ERL (immediate). Height of ERL is such that total CO2 above it is constant.ā€”
CO2 above is constant? That is a non-intuitive argument as far as I’m concerned even though I can see why many would find it acceptible. with no further thought.

Alberta Slim
April 25, 2013 8:52 pm

joeldshore says:
April 25, 2013 at 3:19 pm
“Oh yeahā€¦I should perhaps add one more point to my last post: The strict sense in which a colder object cannot heat a warmer object is that the heat (net energy flow) must be from the hot object to the colder object. So, no, adding some cold gas around the Earth and causing the temperature to rise does not mean that the colder gas has heated the Earth: the net flow of energy is still from the Earth to the colder gas. You have just made it harder for energy to leave the Earth, i.e., the Earth has to maintain a higher temperature at its surface in order to radiate away the energy at the rate it is receiving it from the sun……………”
My question;
First your statement: “You have just made it harder for energy to leave the Earth………”
The GHGs are acting like insulators or a blanket then. Correct?
If so, insulation and blankets are solids. they dont expand.
GHGs are gasses that expand, and rise, to absorb the extra heat, and the volume if the atmosphere increases with no increase in temperature. The expanded atmosphere has a greater radiating area to space that handles the extra energy flow.
What is incorrect here? The gasses do not expand? Or what?

davidmhoffer
April 25, 2013 9:21 pm

TimTheToolMan;
But if more photons are heading upwards then this must compensate for that increase in opacity.
>>>>>>>>>>>>>>
It seems to me that you are thinking of photons purely as particles. For the purposes of illustrating why the ERL increases in altitude, that’s good enough. Going to the deeper levels and details though, they are as much waves as particles, and at that point easier to think of in terms of w/m2 (an energy flux). Neither is ideal, but then photons are not particles or waves either, so no explanation that treats them as one or the other can possibly be completely accurate.
As for Mosher’s original statement, sure, his explanation leaves much to be desired. But him providing a poor explanation doesn’t mean the effect he describes is wrong or that the other explanations you got are incorrect.

Bill from Nevada
April 25, 2013 9:44 pm

Everytime you ask a Greenhouse Gasser for some instrumental readings where the giant infrared light in the sky can be seen, you’re told it’s too secret to measure.
The infrared astronomy field? It’s too hard to measure the giant infrared light in the sky.
The ground based DOWN WELLING RADIATION SENSING FIELD?
They checked your story. For fourteen years.
After fourteen years from the mid ’90s to abou 2010 with hundreds of thousands of readings,
there’s LESS atmospheric infrared down welling at night.
Ask the Greenhouse Gas crowd. Those instruments don’t work right.
Neither does the entire infrared astronomy field. They can’t find the infrared light, so something’s wrong. “They aren’t looking right.”
Al Gore did an experiment online where he put CO2 into a jar beside one with regular atmospheric air. The CO2 jar didn’t get as warm. High Definition photographic artifacts proved he swapped thermometers.
Anthony Watts repeats the experiment online in an unbroken feed. The CO2 filled jar didn’t get as warm. By an amount quite similar to Gore’s.
It doesn’t matter. Just because no instrument can see it, doesn’t mean it’s not there.
The giant infrared light in the sky.
No instruments can find it.
NOAA tried. I told you. They didn’t find any CO2/infrared correlation at night.
http://journals.ametsoc.org/doi/abs/10.1175/2011JCLI4210.1?journalCode=clim
Then there’s the people trying to claim that in spite of the fact that the mathematics for equilibrium show energy migrating from more concentrated to less, that
the math
lies.
On and on and on the stories go. Refrigerators prove thermal energy flows both ways. Go look in a thermodynamics book and it says clearly: refrigerators make use of the fact heat moves from more concentrated to less.
It boils down to the fact that every time you see a Greenhouse Gas person you see
(1) them saying just because the instruments don’t show it, doesn’t mean they don’t show it.
(2)them demanding you put aside the experimental and instrumental evidence showing it’s not as they say, and change the subject to a bright colored picture they put up and demanded you admit, proves it.
(3)them trying to take you with them, down the road of “thermodyamic equations lie when they say energy migrates from more concentrated to less.
Thermodynamic equations, Carnot, Clausius: all that means nothing.
But the fact is that if there were any correlation between infrared light and CO2 we would know.
It’s a giant infrared light.
And we’ve got enough to check for the giant infrared light in the sky a thousand times over.

gbaikie
April 25, 2013 9:44 pm

“Noā€¦They believe it because they understand physics.
Look, you can only get so far by comparing the Earth and the moon. The fact that one has a greenhouse effect and the other doesnā€™t is only one difference. There are other important differences: The moon has very little atmosphere at all and does not have water. This means that the effective heat capacity or ā€œthermal inertiaā€ is much smaller so temperatures vary more dramatically between day and night. Furthermore, the moon has a much longer day than the Earthā€¦about 30 times longer, so this also favors larger temperature variations. Neither of these differences should affect the average temperature (or, more precisely, the average of T^4), which is determined by radiative balance, but they do affect the range of temperatures.”
The greenhouse theory does not consider length of day or heat capacity. Wiki Greenhouse effect:
“If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 Ā°C. However, since the Earth reflects about 30% of the incoming sunlight, this idealized planet’s effective temperature (the temperature of a blackbody that would emit the same amount of radiation) would be about āˆ’18 Ā°C. The surface temperature of this hypothetical planet is 33 Ā°C below Earth’s actual surface temperature of approximately 14 Ā°C. The mechanism that produces this difference between the actual surface temperature and the effective temperature is due to the atmosphere and is known as the greenhouse effect.”
https://en.wikipedia.org/wiki/Greenhouse_effect
So idea that ideal blackbody could be in a vacuum or not is not relevant, and the general assumption would be that things in space are in a vacuum and this ideal blackbody is in a vacuum. So in model it’s in a vacuum and whether it’s rotating or spinning on two axis is not considered a factor.
Or there is no modifier for having to do with whether a planet has thick atmosphere or a vacuum,
and whether planet has a lot of water or no water.
The only modifier is distance to Sun and how much the body reflects. If body doesn’t reflect sunlight or is a black body like the Moon it is suppose to have a average temperature at Earth distance from the sun of 5.3 Ā°C. Since the Moon is slightly reflective, it’s average temperature
should be around 5 C. And if the Moon was reflective as Earth, then it’s average temperature should according to this theory be āˆ’18 Ā°C..
If the Moon was an reflective as Earth and Moon had same average temperature a earth [about 14 C] then the only reason it could 33 C warmer then “it should be” is because it must have a greenhouse effect.
As this is according to Greenhouse theory the only way it can be warmer. So theory does not allow for a body to be warmer because it has faster rotation or because it has water [unless water is vapor form which makes a “greenhouse gas”].
As I said you don’t even follow [or know] the theory you purport to support.
“For heavenā€™s sake, think things through before making statements about people being brainwashed. You might find that you can actually learn from people who have thought about these things much more than you apparently have.”
Hmm. I am curious, do you think the people in North Korea have been and are being brainwashed. Also do you think the German people were brainwashed by the Nazi government?
Would know what brainwashing was if you saw it?
brainĀ·washĀ·ing
“[breyn-wosh-ing, -waw-shing] Show IPA
noun
1.
a method for systematically changing attitudes or altering beliefs, originated in totalitarian countries, especially through the use of torture, drugs, or psychological-stress techniques.
2.
any method of controlled systematic indoctrination, especially one based on repetition or confusion: brainwashing by TV commercials.
3.
an instance of subjecting or being subjected to such techniques: efforts to halt the brainwashing of captive audiences.”
http://dictionary.reference.com/browse/brainwashing
So are willing to admit that TV commercials a form of brainwashing- even if they are funny and amusing?
And do you know what social conditioning is?
Wiki:
“Social conditioning refers to the sociological process of training individuals in a society to respond in a manner generally approved by the society in general and peer groups within society. The concept is stronger than that of socialization, which refers to the process of inheriting norms, customs and ideologies. Manifestations of social conditioning are vast, but they are generally categorized as social patterns and social structures including education, employment, entertainment, popular culture, religion, spirituality and family life. The social structure in which an individual finds him or herself influences and can determine their social actions and responses.
Social conditioning represents the environment and personal experience in the nature vs. nurture debate. Society in general and peer groups within society set the norms which shape the behavior of actors within the social system.”
Social conditioning or brainwashing is what Al Gore and others who “lie for the sake of the cause” are trying to do- this is the only point of saying such things as the “science is settled”.
This the only point of suggesting people should imprisoned or executed because they don’t believing in global warming [or are unwilling to do enough to stop it].
Get it?

wayne
April 25, 2013 9:51 pm

“No, the Dyson sphere with a 1 AU radius will absorb 1366 W/m^2 from the sun and emit ~ 1366 W/m^2 outward and ~ 1366 W/m^@ inward. ”
Are you sure you mean it that way TimF? Seems you have 1366 input and 1366 x 2 output.

Kristian
April 25, 2013 9:54 pm

Gary Hladik says, April 25, 2013 at 3:59 pm:
“Kristian says (April 25, 2013 at 3:11 pm): ā€œYouā€™re still not reading what youā€™re commenting on, Gary.ā€
Oh, but I did. I read very carefully, multiple times, because I couldnā€™t actually believe I was reading what I thought I was reading. Here, your words:
ā€œThe ā€˜systemā€™ flux to the surroundings is J with sphere alone and J1 (J/2) with shell surrounding it. How can this be?ā€
Indeed! How can this be unless 1) youā€™re destroying energy, or 2) the system isnā€™t in equilibrium? What do you think Iā€™m missing here? Specifically.

Sigh.
No, evidently your reading is not very ‘careful’, Gary. Careful in your ‘misunderstanding’ of what you’re reading, yes. But not in actually taking in the meaning of the words in front of you. Because the answer is right there for all to see in the very posts you’re commenting on. So your misrepresentation of them seems to me to be simply an exercise in building strawmen to take down.
As a matter of fact, for your comment on April 25, 2013 at 2:26 pm, you need only read on, just past your quote of choice, to see it. In your comment on April 25, 2013 at 3:03 pm, it’s even in the very paragraph you’re quoting (and, rephrased, in the one directly following it, which you are not).

davidmhoffer
April 25, 2013 10:14 pm

Bill from Nevada;
Everytime you ask a Greenhouse Gasser for some instrumental readings where the giant infrared light in the sky can be seen, youā€™re told itā€™s too secret to measure.
>>>>>>>>>>>>>>>>>>>>>>>
Oh bull.
http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/
scroll down and you will find the exact instrumental readings you claim don’t exist.
Or you can go here for ERBE data showing satellite data of the same:
http://eos.atmos.washington.edu/erbe/
The hardest part about gaining any new idea is sweeping out the false idea occupying that niche. As long as that niche is occupied, evidence and proof and logical demonstration get nowhere.
~ Robert A Heinlein

Konrad
April 25, 2013 10:59 pm

TimTheToolMan says:
April 25, 2013 at 8:51 pm
——————————————–
Tim,
It should be obvious from responses to your comments that you have struck a nerve. You are essentially correct. The energy that radiative gases are radiating to space includes not just intercepted surface IR, but energy from surface conduction and release of latent heat. The atmospheric cooling effect is an almost linear function of CO2 concentration in the atmosphere, whereas the warming effect is an inverse logarithmic function of CO2 concentration.
For a hypothetical static atmosphere this would mean that initially CO2 would warm but after a certain CO2 concentration (point of no concern) the two curves would cross and the net radiative effect of added CO2 would be cooling. However for a moving atmosphere this does not hold true. For an atmosphere in which the gases can move, CO2 cools at all concentrations above 0.0ppm.
The reason the defenders of the AGW faith are fighting so hard to keep the ERL game alive is not to save the AGW hypothesis. They know the jig is up. The mistake at the foundation of the AGW hypothesis is in the basic physics of gas conduction and convective circulation. So basic it is embarrassing. At this time they are looking for a less embarrassing exit that sounds suitably ā€œsciencyā€. Sadly they are being assisted in their exit plans by equally embarrassed Lukewarmers who endorsed the mistakes.
If you want evidence that you are on the right track, look no further than Steve Moshers early comment on this thread –
ā€œThe real debate is over how much warming GHGs will causeā€
The defenders of the AGW faith want the exit answer to be ā€œCO2 causes warming but we overestimated how much.ā€ The truth is that radiative gases act to cool our atmosphere at all concentrations above 0.0ppm.
There are no planets or moons in our solar system that have managed to retain an atmosphere without radiative gases. They are that important to atmospheric cooling.

April 25, 2013 11:12 pm

No, the Dyson sphere with a 1 AU radius will absorb 1366 W/m^2 from the sun and emit ~ 1366 W/m^2 outward and ~ 1366 W/m^@ inward. Of course, most of the 1366 W/m^@ inward will hit some other part of the shell, so the net heat transfer from the shell to the shell is ~ 0.” ~Tim
Uh… as wayne pointed out, that’s 1366*2 from the sphere, isn’t it?

Bill from Nevada
April 25, 2013 11:47 pm

Hoffer you do well to quote science fiction authors.
“These plots are annual and seasonal means of data obtained from the NASA Earth Radiation Budget Experiment (ERBE) scanner instruments onboard the ERBS and NOAA-9 satellites during January 1985 through December 1986.”
I and anyone else who was really serious about this would be talking about you can replicate – showing a correlation between CO2 and atmospheric infrared, in the earth-radiant frequency.
Or CO2 and global temperatures.
Nothing reinforcing your claim that there’s CO2 associated thermal behavior, OR atmospheric infrared in attributable is there in your spam.
Hoffer you’re the one I saw making the claim that refrigeration is based on the laws of thermodynamics allowing heat to flow two ways, I believe.
You’re also the one I know
I saw say that when the thermodynamic equations are done proving heat travels one way, there is actually energy flowing both ways, but that mathematics can’t show that.
F for grabbing graphs that mean nothing,
F for bloviating about it THEN grabbing graphs that mean nothing.
And F for thinking the math of thermodynamics is lying when it tells you heat travels one way to equilibrium.

April 25, 2013 11:48 pm

So I missed joeldshore’s post which amongst other things said this “The amount of energy that a substance emits is determined by its temperature.”
That is the crux of his argument and is correct. The amount of radiative energy radiated by an object is given by S-B and does indeed relate only to temperature. However the number of molecules that will be radiating is determined not by the temperature alone but by their concentration according to Maxwell-Boltzmann. And it is this that joeldshore is missing in his argument.

Bill from Nevada
April 26, 2013 12:10 am

Hoffer :
when you told someone the math of thermodynamics lie
when they say *heat flow is one way*
you entered into that special realm called “No, I don’t have any credentials but I do a lot of blogging, pretending to be an expert on the internet.”
I’ve seen you try MULTIPLE TIMES, telling people refrigerators are built on the principle that heat flows simultaneously two directions between two entities.
I saw you say it on the RW Wood thread.
But the fact is, thermodynamics texts point out the refrigerator as a proof of energy flowing ONE way.

April 26, 2013 12:42 am

In response to joeldshore I wrote “The amount of radiative energy radiated by an object is given by S-B and does indeed relate only to temperature. ”
But was too quick to agree with joeldshore. Normally the amount of energy radiated varies “only” with temperature but according to Stefan-Boltzmann it actually varies with the product of temperature and emissivity. In the case of adding GHGs to the atmosphere, the emissivity is actually increasing so we can expect more radiation to propagate throughout the atmosphere (up and down for no net effect).
I should have realised earlier but it has to be this way for the result to be in line with the requirements of Maxwell-Boltzmann’s distribution of energies…

Myrrh
April 26, 2013 1:28 am

davidmhoffer says:
April 25, 2013 at 8:08 pm
TimTheToolMan;
The thing is that there arenā€™t 8 photon released when you double the CO2. There might be 8 released at at ground level, but up where ERL happens the radiation is supplied by the GHGs not the ground.
>>>>>>>>>>>>>>>
For the love of *** no!
ALL the photons are from the ground in the first place. You can double, quadruple or millionuple the CO2, ALL the photons in the system originate from the ground. All that changes is the number of times they bounce around before escaping.

Your: “ALL the photons are from the ground in the first place.”
Only in the make believe world of AGWScienceFiction fisics.
In the real world we get real heat photons from the Sun – this is what we feel as heat,
this is the electromagnetic radiation wave of thermal infrared, also known as HEAT, also known as thermal infrared also known as heat radiation also known as radiant heat also known as longwave infrared.
AGWSF has completely excised that direct longwave infrared from the Sun in order to claim that the only “downwelling longwave infrared from the atmosphere is from backradiation by greenhouse gases”. This is science deliberately faked.
AGWSF has given the properties and processes of longwave infrared from the Sun to the shortwaves classified in traditional physics as Light not heat – mainly visible, a small amount of uv and 1% near infrared. The latter two hardly mentioned so the AGWSF claim is that Visible light from the Sun heats the Earth’s land and water – this is patently absurd to anyone with real world basic physics..
Visible light interacts with matter on the electronic transition level, this is not capable of moving whole molecules into vibration which is what it takes to heat matter.
Visible light from the Sun can’t even get to the electrons of water, water transmits visible light unchanged.
Visible light is bounced all over the sky by the electrons of the molecules of nitrogen and oxygen which absorb visible light and then emit it, this is called reflection/scattering and is how we get our blue sky.
If visible light is such a powerful energy as the AGWSF claims that it heats matter on being absorbed, how much is blue visible light from the Sun heating the atmosphere as the electrons of the gas air absorb it?
As if that wasn’t absurd enough, AGWScienceFiction gives two reasons why there is no direct thermal energy in transfer from the Sun, which is longwave infrared also known as thermal infrared to differentiate between the heat it is and the non thermal shortwave infrared which are classed in with Light, not with Heat; classed in with Reflective, not Thermal.
We cannot feel Light as Heat.
The heat we feel from direct from the Sun is thermal infrared, longwave infrared. AGWSF claims we get no direct heat from the Sun and explains this firstly by claiming “there is an invisible barrier at TOA like the glass of a greenhouse which stops longwave infrared from the Sun but allows visible from the Sun to get through and heat the Earth’s surface”.
What is this invisible barrier around the Earth stopping the Sun’s direct heat, longwave infrared, from reaching us? It is unknown in traditional real word physics.
The second even more absurd explanation given by AGWSF for why they claim “no longwave infrared from the Sun reaches us”, is by saying “the Sun produces insignificant amounts of longwave infrared and so there is nothing of this to reach us”.
The heat we feel from the Sun is longwave infrared. We cannot feel visible light, we cannot feel the shortwaves of visible, uv and near infrared which AGW claim is heat energy. We cannot physically feel these shortwaves as heat.
AGWSF is actually in real world physics saying there is no heat direct from the Sun.
The explanation they give for why “the Sun produces no significant longwave infrared” is that “the peak energy from the Sun is visible and that makes the Sun 6000°C, so no significant longwave infrared which is the electromagnetic wave of heat energy is produced, it’s all changed to mainly visible”.
Those mindlessly regurgitating this AGWSF meme of “shortwave in longwave out” can’t see how ludicrous that explanation. How ludicrous that a three hundred mile band of visible light, the first layer of atmosphere around the Sun, defines the temperature of Sun, which is millions°C and continually radiating out this heat.
The hotter something is the more heat it radiates, like the incandescant lightbulb which radiates 95% heat, longwave infrared, and only 5% visible light. The 300 mile wide band of visible light around the Sun is as insignificant as it is in the lightbulb. The Sun is a massive blazing STAR! It is radiating massive amounts of HEAT!
Heat is Longwave infrared.
This sleight of hand trickery has been put into place by AGWSF in order to push AGW, it has had the effect of dumbing down basic real world physics for the majority population.
Because the AGWSF meme that “visible light from the Sun heats the Earth’s surface” has been brainwashed through the general education system so the majority population without traditional physics thinks Light is Heat, is why you think that the only source of heat downwelling in the atmosphere comes from the heated Earth radiating out, from the upwelling of waste heat being “backradiated”.
NASA used to teach traditional physics that the heat we feel from the Sun is longwave infrared and that we cannot feel shortwave infrared, which is not hot, is not thermal.
You cannot ignore the following, you must address this because it completely contradicts your AGWSF Greenhouse Effect Energy Budget.
NASA used to teach:
ā€œFar infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared.
“Shorter, near infrared waves are not hot at all ā€“ in fact you cannot even feel them. These shorter wavelengths are the ones used by your TVā€™s remote control.ā€
You live in a cold world – with no direct longwave infrared from the Sun which is radiant heat which is the Sun’s thermal energy in tranfer, and your visible light from the Sun cannot raise the temperature of matter.
Deal with it.

April 26, 2013 1:41 am

“Climate pseudo scientists are rubbish at fluid dynamics.”
Heat Transfer too. They would fail a basic engineering exam (multimodal heat transfer). Logic is not their strong side as well.

Nullius in Verba
April 26, 2013 1:43 am

Bill,
If you think the maths of thermodynamics says heat flows only one way, how do you suppose the body sending the heat know the temperature of the receiving body to decide when to send it? That would appear to require faster-than-light propagation.
So far as I know, refrigerators are based on the same thermodynamics as any heat transfer, they’re principally used as an example to show that heat engines are reversible. A heat engine transfers heat from warm to cold, doing work on its surroundings. A refrigerator transfers heat from cold to warm as a result of the surroundings doing work on it. It’s a demonstration that heat can go both ways, but not explicitly a demonstration that it does so simultaneously (although it does).
That thermodynamic heat flow is the net effect of bidirectional heat flows is shown by equilibrium behaviour combined with the requirement for the laws of physics to be local – with action-at-a-distance banned. Locality requires that events at a point can only be caused by conditions at that point. The transfer of heat between one point and another takes time. So the decision to send it can only depend on conditions at the sender location. For it to decide how much to send depending on the temperature difference would require that it somehow know what the temperature is at the receiver location. But if the receiver location is not emitting any heat in its direction, there must be some other means of signalling what the remote temperature is. What is it?
And what evidence is there for it, when simply having all objects emit heat according to their own local temperature, and for the heat emitted to rise as the temperature does, gives exactly the observed result?

Gary Hladik
April 26, 2013 2:10 am

Maxā„¢ says (April 25, 2013 at 11:12 pm): “<iā€œNo, the Dyson sphere with a 1 AU radius will absorb 1366 W/m^2 from the sun and emit ~ 1366 W/m^2 outward and ~ 1366 W/m^@ inward. Of course, most of the 1366 W/m^@ inward will hit some other part of the shell, so the net heat transfer from the shell to the shell is ~ 0.ā€ ~Tim
Uhā€¦ as wayne pointed out, thatā€™s 1366*2 from the sphere, isnā€™t it?”
tjfolkerts is correct. I plugged these numbers into problem 1026:
R = 1 AU ~ 1.5 x 10^11 m)
r ~ 7 x 10^8 m
T = 5778 K
T0 = 0 K
and solved for T1, the temperature of the shell. The answer is ~394 K. I entered that into an online Stefan-Boltzmann calculator and got ~1366 watts per square meter. Since the shell has two sides, it radiates 1366 watts per square meter from each side of the shell.
Intuitively this makes sense: at 1 AU the sun subtends an angle of only about 0.5 degrees, so a vastly greater proportion of the shell’s “IN” radiation misses the sun and hits…the shell again. The shell is, in effect, getting a “double dose” of the sun’s radiation, part from the sun, and part from itself.
In problem 1023, R is very close to r, so the shell only “sees” the sphere and not itself. The shell’s inner radiation heats the inner sphere and not itself, at least not directly.

Gary Hladik
April 26, 2013 2:27 am

Maxā„¢ says (April 25, 2013 at 7:18 pm): “Why do you think the sun wouldnā€™t absorb the incoming energy?
I think it replaces some of the photons leaving the Sun, but I donā€™t think it is going to make any sort of noticeable difference…”
Not “noticeable” on such a hot object, but not zero, either.
“The star is heated by the action deep inside, and the temperature of the outer layers is determined by the photons that spend ages trickling out from the interior where the fusion reactions take place, not photons that return from whatever is illuminated by the star.”
The surface matter of the sun is heated by conduction, convection, and radiation from the interior of the sun. So why would the surface matter of the sun not interact with and be heated by radiation from outside the sun?
As tjfolkerts has pointed out, a shell at 1 AU won’t send much radiation to the sun, so if it helps, put the shell closer, as in problem 1023. Now we’re getting roughly half the sun’s output back. Significant, right? Now move the shell outward incrementally. At what distance does the sun start “rejecting” incoming radiation?
BTW, of course we’re ignoring any effects such an imaginary shell would have on solar magnetic fields, flares, etc.

Gary Hladik
April 26, 2013 2:49 am

Kristian says (April 25, 2013 at 9:54 pm): “As a matter of fact, for your comment on April 25, 2013 at 2:26 pm, you need only read on, just past your quote of choice, to see it. In your comment on April 25, 2013 at 3:03 pm, itā€™s even in the very paragraph youā€™re quoting (and, rephrased, in the one directly following it, which you are not).”
I did read all of that, and it still doesn’t change the fact that, supposedly at equilbrium, power J is input to the system and power J1 = J/2 is radiating out. Either the temperature of the system rises (in which case it wasn’t at equilibrium), or energy is being destroyed. If the system is at equilibrium, you have to explain where the “extra” power J1 = J/2 is going. Note that if the input power is actually J1, inputs and outputs balance nicely and the system is indeed at equilibrium.
BTW, rather than telling me to read on to find “it”, why not just repeat “it”? That makes it a lot easier to figure out what you’re trying to say.

April 26, 2013 3:15 am

Reed Coray:
I am replying to your post addressed to me at April 25, 2013 at 5:18 pm
http://wattsupwiththat.com/2013/04/24/spencer-slays-with-sarcasm/#comment-1287852
You post concludes saying

I donā€™t know the answers to any of these questions. But the statement that ā€œGreenhouse gases reduce the rate at which heat flows from higher temperatures to lower temperaturesā€ cannot by itself be used to argue that atmospheric greenhouse gases increase atmospheric thermal insulation because the statement is not universally true.

That changes the subject.
I attempted to resolve a confusion and you say that is now achieved. This pleases me.
In fact, I think you can answer all the questions you pose for yourself.
1.
Consider that atmospheric water vapour was not a greenhouse gas (GHG) and none of the atmospheric trace gases were (GHGs). Then the albedo of the Earth would be unchanged and the total heat input (from the Sun) to the Earth+atmosphere system would be unchanged. And all of that heat would be provided to the Earth’s surface. All that energy would radiate from the surface to space unhindered by GHGs.
[Radiation from Earth surface –> space]
2.
In reality, water vapour and some of the trace gases in the air are (GHGs). A little of the mostly short-wave radiation from the Sun is absorbed in the atmosphere by GHGs. But GHGs only absorb a little of this short-wave radiation. Importantly, the total heat input (from the Sun) to the Earth+atmosphere system is the same as in Case 1. Most of the radiation from the Earth’s surface is long-wave radiation which is efficiently absorbed in the atmosphere by GHGs. This inhibits the passage of radiation from the Earth’s surface to space so the temperature of the Earth’s surface rises (for the same reason that an insulated house is warmer than an uninsulated house).
[Radiation from Earth surface –> atmosphere (i.e. effectively insulation) –> space]
I really do think you now have sufficient basic information to think this through for yourself. And answering the questions you have posed yourself will be more cogent for you than being ‘told’ their answers by anybody.
Please be assured that I have tried to help.
Richard

Kristian
April 26, 2013 3:30 am

Nullius in Verba says, April 26, 2013 at 1:43 am:
“If you think the maths of thermodynamics says heat flows only one way, how do you suppose the body sending the heat know the temperature of the receiving body to decide when to send it? That would appear to require faster-than-light propagation.”
“That thermodynamic heat flow is the net effect of bidirectional heat flows is shown by equilibrium behaviour combined with the requirement for the laws of physics to be local ā€“ with action-at-a-distance banned.”
“And what evidence is there for it, when simply having all objects emit heat according to their own local temperature, and for the heat emitted to rise as the temperature does, gives exactly the observed result?”
Yes, this seems very much to be where the confusion arises. SoD’s confusion, for instance, about what the second law of thermodynamics is actually saying and not saying.
Heat is always energy. Energy is NOT necessarily always heat. There is a very specific physical definition of ‘heat’ and it has been proposed and conceptually refined over the years in order to distinguish between energy flows that can and cannot warm an object. Only energy in the form of heat can make an object warmer. Here is HEAT defined by Borgnakke & Sonntag (‘Fundamentals of Thermodynamics,’ 2009):
“If a block of hot copper is placed in a beaker of cold water, we know from experience that the block of copper cools down and the water warms up until the copper and water reach the same temperature. What causes this decrease in the temperature of the copper and the increase in the temperature of the water? We say that it is the result of the transfer of energy from the copper block to the water. It is from such a transfer of energy that we arrive at a definition of heat.
Heat is defined as the form of energy that is transferred across the boundary of a system at a given temperature to another system (or the surroundings) at a lower temperature by virtue of the temperature difference between the two systems. That is, heat is transferred from the system at the higher temperature to the system at the lower temperature, and the heat transfer occurs solely because of the temperature difference between the two systems.”
Heat, like work, is a form of energy transfer to or from a system.

The last line relates heat and work and thus relates the sphere/shell problem to the Carnot cycle. Q = Q’ + Q” is the equivalent to W = Qh – Qc:
http://i1172.photobucket.com/albums/r565/Keyell/Carnot_zps4049e783.jpg
Combine this with what Wikipedia says:
“Heat in physics is defined as energy transferred by thermal interactions. Heat flows spontaneously from hotter to colder systems. When two systems come into thermal contact, they exchange energy through the microscopic interactions of their particles. When the systems are at different temperatures, the result is a spontaneous net flow of energy that continues until the temperatures are equal. At that point the net flow of energy is zero, and the systems are said to be in thermal equilibrium. Spontaneous heat transfer is an irreversible process.”
When energy is transferred to a body purely as heat, its internal energy increases. This additional energy is stored as kinetic and potential energy of the atoms and molecules in the body. Heat itself is not stored within a body. Like work, it exists only as energy in transit from one body to another or between a body and its surroundings.

There is no net heat flow. There is only net energy flow. The net energy flow is the heat flow. Heat is specifically the surplus energy transferred from a warmer body to a cooler one. There is no ‘heat’ going the other way. There is ‘energy’ going the other way, but this can do nothing to make the warmer object warmer still. Only heat can heat.
There is also a fairly interesting distinction between ‘thermal energy’ and ‘heat’:
http://en.wikipedia.org/wiki/Thermal_energy

David
April 26, 2013 3:52 am

Roy Spencer says:
April 24, 2013 at 1:37 pm
ALL:
It is important to remember that there is NO WAY to determine the temperature of anything based upon the rate of energy input alone, for example the Earth absorbing an average of ~240 W/m2 from the Sun. Temperature is a function of BOTH energy input (typically not temperature dependent) AND energy loss (typically VERY temperature dependent), neglecting issues related to heat capacity which mainly affect the time required for the system to equilibrate. The temperature of anything heated will increase until the rate of energy *loss* equals the rate of energy *gain*. So, temperature can be increased by increasing INPUT, or decreasing OUTPUT.
=======================================================\
Well Dr Spencer, you could have just qu0ted David’s Law…
“At its most basic only two things can effect the energy content of any system in a radiative balance. Either a change in the input, or a change in the ā€œresidence timeā€ of some aspect of those energies within the system.”
It therefore possibly follows that any effect which increases the residence time of LW energy in the atmosphere, (for instance; green house gases) but reduces the input of SW energy entering the oceans, (atmospheric heating causing more W/V and or cloud cover) causes a net reduction in the earthā€™s energy balance, proportioned to the energy change involved, relative to the residence time of the radiations involved.

tjfolkerts
April 26, 2013 4:15 am

Kristian says: “Only energy in the form of heat can make an object warmer.”
Rubbing your hands together is work, not heat.
Running electric current through a resistor is work, not heat.
Compressing a piston is work, not heat.
All of these clearly make an object warmer.

Nullius in Verba
April 26, 2013 4:20 am

Kristian,
Yes, that’s a technical distinction some people choose to make, and it is confusing. I’m not sure whether it is more confusing to use the technical sense or the everyday sense. Which conveys understanding better to people who don’t know the physics?
It depends as well where you draw the boundaries. Objects can have non-uniform temperature.
If you want to use the technical definition, then you can make a warm object warmer without transferring any heat. Adiabatic changes are those with no heat transfer, but warm objects can get warmer as a result of them. You can argue that only heat can heat, but then heating does not mean the same thing as raising the temperature. Heating is not warming.
That seems to me even more likely to confuse.

joeldshore
April 26, 2013 5:43 am

TimTheToolMan says:

But was too quick to agree with joeldshore. Normally the amount of energy radiated varies ā€œonlyā€ with temperature but according to Stefan-Boltzmann it actually varies with the product of temperature and emissivity. In the case of adding GHGs to the atmosphere, the emissivity is actually increasing so we can expect more radiation to propagate throughout the atmosphere (up and down for no net effect).

What you are missing is that emissivity is proportional to absorptivity (and both depend on the thickness of the gas layer). When it emits more, it also absorbs more. At the end of the day, what changes as you change the concentration of GHGs is where the emission occurs…and that, through the temperature, determines how much is emitted.
Let’s consider the case of a solid rather than a gas. Now, if you make this solid twice as thick, do you think you get twice the radiative emission because you have twice the number of emitting molecules? No, you get the same because the amount depends on the surface area. Now imagine the entire atmosphere is solid and that it somehow still has the same temperature gradient (say 8 K per kilometer to keep the math simple) that we observe with the lapse rate, with a temperature of 288 K at the bottom. If this solid atmosphere were initially 5 km high, then the emission would occur from an upper surface that is at 248 K. If it then became 1 km thicker, so it extended up to 6 km high, its upper surface would be at 240 K and it would emit less energy to space.
I know this is all somewhat counterintuitive but the fact that you don’t understand what is going on is not going to change the reality of what happens: http://www.barrettbellamyclimate.com/userimages/Iraq2.jpg and the fact that entire fields of technology are based on atmospheric radiation transfer working the way that we say it does.

joeldshore
April 26, 2013 6:01 am

gbaikie says:

The greenhouse theory does not consider length of day or heat capacity.
..
So idea that ideal blackbody could be in a vacuum or not is not relevant, and the general assumption would be that things in space are in a vacuum and this ideal blackbody is in a vacuum. So in model itā€™s in a vacuum and whether itā€™s rotating or spinning on two axis is not considered a factor.
Or there is no modifier for having to do with whether a planet has thick atmosphere or a vacuum,
and whether planet has a lot of water or no water.

Which part of what the following sentence that I wrote did you not understand: “Neither of these differences should affect the average temperature (or, more precisely, the average of T^4), which is determined by radiative balance, but they do affect the range of temperatures”?
Let me try again: In your post that I was responding to, you talked about diurnal temperature ranges (e.g., the fact that it can during the day get much hotter on the Moon than on the Earth). That goes beyond considerations of radiative balance relevant to the greenhouse effect, which only constrains what the average temperature (average of T^4) must be over the planet. I explained to you that the moon can get hotter during the day (and colder at night) and still have an average surface temperature compatible with its emission temperature being what it has to be given its albedo and absence of a greenhouse effect.

As this is according to Greenhouse theory the only way it can be warmer. So theory does not allow for a body to be warmer because it has faster rotation or because it has water [unless water is vapor form which makes a “greenhouse gas”].
As I said you donā€™t even follow [or know] the theory you purport to support.

Theory does not allow the average of T^4 to be larger…but it does allow the maximum diurnal temperature to be warmer. And greenhouse theory does not predict that diurnal temperature range, which will indeed depend on details like rotation rate and thermal inertia. So, it is you who apparently do not understand the basics of the theory.

Hmm. I am curious, do you think the people in North Korea have been and are being brainwashed. Also do you think the German people were brainwashed by the Nazi government?
Would know what brainwashing was if you saw it?

The point is that you decided other people were brainwashed because, as it turns out, they understand science that you do not. Perhaps you need to entertain the idea that you are the one who is brainwashed.

joeldshore
April 26, 2013 6:02 am

I said “What you are missing is that emissivity is proportional to absorptivity”.
Actually, I should have said “is equal to” not simply “is proportional to”.

gbaikie
April 26, 2013 6:09 am

Question:
Make a large cube in vacuum of space and have the cube made from smaller cubes.
And have these small cube tightly stacked together.
So 1 km width, length, high cube comprised of a billion 1 meter cubes.
The amount energy this 1 km cube will radiate depends upon it’s temperature [say 300 K]
and it’s total surface area [6 square km].
Now, change it so all the small cubes are at certain distance from each other [say 10 meters
apart]. So have 1 billion 1 meters cubes with a 10 meter space between them and forming much larger cube. So get a cube shape which if looking at one side or any side of it, one mostly see space between them and mostly just see 1 meter cubes which are on outside of cube.
But depending angle looking at it, it can look like it’s solid- say looking at from angle the cube looks like pyramid. So depending angle looking at it, one see part cubes sides which in the interior. But for most part it’s just going to be the 1 meter cubes on the exterior radiating most of the energy.
So the cube which widely space cubes will radiate more energy than the tightly packed cube,
but I would guess it should not be a lot more energy.
Also with tightly stacked cubes one have heat conducted from interior to exterior, whereas if there is distance between them that only radiate energy can be transferred.
So question what would the differences be between tightly stacked blocks and spaced blocks be in terms how much and how they radiate energy?
For example it seems each of the blocks on the exterior of the widely spaced cube should radiate
more energy as compare to if tightly spaced and it also seems like the interior blocks would retain heat longer [as compared if tightly stacked].

Kristian
April 26, 2013 6:19 am

Nullius in Verba and tjfolkerts,
Yes, sorry, a bit sloppy there. I did not include the concept of ‘work’ as a heating process simply because that wasn’t really part of the distinction I was trying to make. The ‘separate’ energy flows radiated between two BBs at different temperatures across a vacuum versus the net of the two (Q) was what I was getting at.
Bottom line, though: ONLY the net is a transfer of heat. The individual energy flows are NOT.
About ‘thermal energy’ vs. ‘heat’:
Thermal energy is the part of the total potential energy and kinetic energy of an object or sample of matter that results in the system temperature. This quantity may be difficult to determine or even meaningless unless the system has attained its temperature only through warming, and not been subjected to work input or output, or any other energy-changing processes.”
“Microscopically, the thermal energy may include both the kinetic energy and potential energy of a system’s constituent particles, which may be atoms, molecules, electrons, or particles in plasmas. It originates from the individually random, or disordered, motion of particles in a large ensemble, as [a] consequence of absorbing heat.”
“Heat, in the strict use in physics, is characteristic only of a process, i.e. it is absorbed or produced as an energy exchange, always as a result of a temperature difference. Heat is thermal energy in the process of transfer or conversion across a boundary of one region of matter to another, as a result of a temperature difference.”
“When two thermodynamic systems with different temperatures are brought into diathermic contact, they spontaneously exchange energy as heat, which is a transfer of thermal energy from the system of higher temperature to the colder system.”

http://en.wikipedia.org/wiki/Thermal_energy
I find this definition/distinction very sound and easy to relate to. I don’t see how it confuses you, Nullius in Verba.

beng
April 26, 2013 6:46 am

***
davidmhoffer says:
April 25, 2013 at 10:14 pm
***
David, your example is fine. The best IMO is the graph of IR leaving the earth. The integral under the curve is the total IR emission, and that must stay the same in equilibrium (since the sun’s input is constant). There’s a big “chunk” taken out by CO2. If the CO2 “chunk” is increased by increased CO2, then other parts of the curve MUST increase to maintain equilibrium (maintain the same total area under the curve). The surface emission is where it will increase — higher surface temp, while the TOA temp decreases (where CO2 emits). Same total IR leaving, just cooler at TOA & warmer at surface. Slayers need to understand this simple energy conservation law.
This does NOT address feedbacks/cloud & convection changes, etc — another story.

davidmhoffer
April 26, 2013 7:24 am

But the fact is, thermodynamics texts point out the refrigerator as a proof of energy flowing ONE way.
>>>>>>>>>>>>>>>>>>
This is a classic example of not understanding the difference between heat flow and energy flux.
As for the half dozen attack comments toward me overnight, I think it clear that a sincere attempt to help people understand the physics that is used to design and build everything from nuclear reactors to, yes, refrigerators, is unappreciated by those determined to wallow in their own ignorance, ignoring the math and data presented while accusing me of ignoring the math and data!
The ranks of the skeptics have become sickeningly infiltrated by those whose belief system trumps the facts, and who quote the laws of physics as if they say the opposite of that they do. These people are no better than the Phil Jones and Michael Mann’s of the world. They are blind because they choose to be blind.
Dropping thread. What a testament to the willful self imposed ignorance of the human race.

Kristian
April 26, 2013 7:46 am

Gary Hladik says, April 26, 2013 at 2:49 am:
“I did read all of that, and it still doesnā€™t change the fact that, supposedly at equilbrium, power J is input to the system and power J1 = J/2 is radiating out. Either the temperature of the system rises (in which case it wasnā€™t at equilibrium), or energy is being destroyed. If the system is at equilibrium, you have to explain where the ā€œextraā€ power J1 = J/2 is going. Note that if the input power is actually J1, inputs and outputs balance nicely and the system is indeed at equilibrium.
BTW, rather than telling me to read on to find ā€œitā€, why not just repeat ā€œitā€? That makes it a lot easier to figure out what youā€™re trying to say.”

Gary, according to your own words, you’ve read all the relevant passages multiple times and still you’re on repeat as if you hadn’t read it even a single time.
You’re completely stuck on the notion that the ‘system’ heat loss to space MUST equal the input to the sphere with or without the shell. As if the conditions in the two situations were equal. Even when it’s blatantly obvious (and stated several times) that with the shell in place, the steady input flux from the sphere’s internal heat source has more ‘work’ to do on its way to space than without the shell intercepting it. The entire flux from the surface of the sphere (J) cannot reach space, because it ā€™has toā€™ heat the shell on the way. Ergo J = J/2 + J/2, Q = Q’ + Q”, 400 W/m^2 = (400 – 200) + 200.
Q’ stays within the system. Q” escapes it. There is still perfect radiative balance.
Look at it this way: J is the entire and the only ‘real’ flux in the system. The two J1s are simply components of J, manifestations of the splitting of J upon interception by the shell (which has no other heat source than the sphere: J).
There are in effect TWO thermodynamic systems to consider here: the sphere and the shell. The sphere’s heat budget looks like this:
Hot reservoir (internal heat source): J –> || Sphere’s surface: J –> || Cold reservoir (vacuum between sphere and shell)
J = 400 W/m^2 (290K)
The shell’s heat budget looks like this:
Hot reservoir (sphere’s surface): (J – J1 =) J1 –> || Shell’s outer surface: J1 –> || Cold reservoir (space)
J1 = 200 W/m^2 (244K)
Letā€™s say the sphere’s surface is provided with a constant heat input from its source of 400 W/m^2. This warms the surface to an emission temperature of 290K. This temperature in turn spontaneously and naturally produces a flux emitted as outgoing heat loss of 400 W/m^2. This radiative flux is not something you can suppress or disallow the way you can suppress convective heat loss. It is dictated by the laws of physics.
Since the heat input is constant, the surface of the sphere will not cool in the meaning ‘temperature dropping’. As it would to cooler surroundings if the input were cut. This also means that there is no actual ‘cooling rate’ to reduce. The surface of the sphere would remain at 290K just as much with surroundings at 0 K as at 289K, it doesn’t matter. Neither of these two surrounding temperatures is capable of transferring heat to the sphere.
You are mixing Q (the heat transfer between two systems) into the concept of what sets the surface temperature of the warrmer object. If the warmer object is heated by a constant heat source, then the absorbed heat from that heat source is what sets the surface temperature of that object. And the heat loss flux will automatically correspond directly. Q doesnā€™t help. Q (or in the case of the shell, Q’ (J1)) simply reflects the speed by which the warm object can heat the cooler one, or at dynamic equilibrium how much net energy is being transferred per unit time from the warmer object in order to maintain the temperature of the cooler object.

April 26, 2013 8:50 am

Myrrh says:
April 25, 2013 at 4:52 pm
Myrrh, diffusion works for all molecules, no matter their specific weight. Diffusion is in fact Brownian motion at molecular scale. Wind and convection is what brings CO2 to long distances and height, up to the stratosphere. But the movements of real molecules of real gases in the real world is what keeps CO2 in the mixture.
In stagnant air, as is the case for e.g. air in open pores in snow, the heavier molecules and isotopes have a tendency to collect at the bottom of the column. For the Law Dome ice core, that resulted in about 1% increase of CO2 at closing depth, 72 meters below surface, after 40 years of settling time. For which is corrected in the measurements of CO2 in the ice cores.
In open air, the air movements even over a few years are more than sufficient to mix all CO2 everywhere and keep it mixed…

tjfolkerts
April 26, 2013 9:14 am

Kristian, I applaud your enthusiasm, but Gary’s explanations are much better than yours here.
You say:

Look at it this way: J is the entire and the only ā€˜realā€™ flux in the system. The two J1s are simply components of J, manifestations of the splitting of J upon interception by the shell (which has no other heat source than the sphere: J).

Analogies are fraught with peril, but money can often be a very good analogy for energy.
Suppose “Mister Planet” has an income of Q = $400 one day. This is the “entire and the only ā€˜realā€™ payment into the system” (note: the income from the outside is NOT “J”). He in turn pays “Miss Shell” $400 that day = “J”. Miss Shell ‘splits J upon interception’ into J1 = $200 to Mrs. Space, and another J1 = $200 back to Mr. Planet.
Well, the next day, Mr Planet has an income of $400 (Q) AND another $200 (J1),. So now he pays Q + J1 = J = $600 to Miss Shell, who sends half to Space ($300) and half back to Planet ($300).
The next day: Q + J1 = $400 + $300 = $700 income and J = $700 passed on to Shell, who passes $350 back and $350 to Space. This is an infinite series approaching $400 + $400 = Q + J1 in to the planet and J = Q + J1 = $800 out from the planet (ie steady state with no temperature change). The shell has $800 in from the planet, $400 out to the planet, and $400 out to space (ie steady state with no temperature change).

April 26, 2013 10:25 am

Hmmm, I ran the calculation like this:
>>> ((3.84*10Ā²ā¶)/((5.6*10Ā²Ā³)*(5.6*10ā»āø))ā°ā‹…Ā²āµ
332.65
(Total Solar Output/(Total Surface Area*Ļƒ))ā°ā‹…Ā²āµ
Doing that I got J/2 = Jā‚ for 1366 and 683, or thereabouts, as expected from problem 1023.
Hmmm…
Initially it would be:
1366ā‡’ā‡0 | 0ā‡’, Net 1366 to shell, 0 to space, 0 emitted by shell
Then as it began heating the shell:
1366ā‡’ā‡341 | 341ā‡’, Net 1025 to shell, 341 to space, 683 emitted by shell
1366ā‡’ā‡683 | 683ā‡’, Net 683 to shell, 683 to space, 1366 emitted by shell
That fits problem 1023, and it seems like it should be stable.
If Net 683 to shell, 683 to space isn’t stable then perhaps:
1366ā‡’ā‡1366 | 1366ā‡’, Net 0 to shell, 1366 to space, 2732 emitted by shell
If Net 0 to shell, 1366 to space is stable, I don’t see how you would get the star to heat up without:
*star heats up until the shell receives 1400*
1400ā‡’ā‡1366 | 1366ā‡’, Net 34 to shell, 1366 to space
1400ā‡’ā‡1400 | 1400ā‡’, Net 0 to shell, 1400 to space
*star heats up until the shell receives 1500*
1500ā‡’ā‡1400 | 1400ā‡’, Net 100 to shell, 1400 to space
1500ā‡’ā‡1500 | 1500ā‡’, Net 0 to shell, 1500 to space
*star heats up until the shell receives… etc, etc, etc*
Though it’s worth noting the scale of the shell radiation in comparison to the solar radiation:
>>> (5.6*10ā»āø)*(5778ā“)
62,416,322.5146
>>> (5.6*10ā»āø)*(5778ā“ – 395ā“)
62,414,959.2618
>>> (5.6*10ā»āø)*(5778ā“ – 332ā“)
62,415,642.1521
*sings*
Around 62 million watts per square meter, around 62 million watts… you take one in and emit it right back, about 62 million watts left after all…

Oh, using 1026 I get that a shell at that distance reduces the rate of cooling by 0.999978222696

paulinuk
April 26, 2013 10:51 am

Why does the sun emit near blackbody radiation. I ask this question because the Sun is primarily made of hydrogen and a bit of helium which aren’t greenhouse gasses. Why does plasma radiate?
Now imagine a significant mass of nitrogen in outer space, held together by gravity which undergoes nuclear fusion. Does it emit any radiation? According to “ARGUMENT BY AUTHORITY” it’s not a GHG and doesn’t have any way to radiate heat to space does it? Does it then reach infinite temperature? What if we lower it’s temperature to 15c. Does it radiate then? What if we replace the iron core in Willis’s thought experiment with this Nitrogen ball surrounded by a metal shell. Any takers?
I thought that all matter with a temperature greater than 0K emitted radiation, the wavelength determined solely by it’s temperature, and NOT spectral emissions.

Nullius in Verba
April 26, 2013 10:57 am

“I find this definition/distinction very sound and easy to relate to. I donā€™t see how it confuses you, Nullius in Verba.”
It doesn’t confuse me. I already have an intuitive understanding of what’s going on, whatever you choose to call the components of the system, and so it’s just a matter of relabelling. You can call it ‘Wakalixes’ if you like. But it’s a lot harder to simultaneously *construct* an intuitive understanding while struggling to figure out the usage of words being used to describe it, especially when they’re used in similar but not identical ways to their everyday meanings. It’s a poor teaching method.
In a textbook, where you can set out the definitions systematically, building up the complexity, it’s a valid way to describe/label a mathematical exposition to aid memory. What you’re actually teaching is the mathematics, you have to read the definitions from the maths, and the conventional names for them used in thermodynamics are a convenience because they’re easier to pronounce than mathematical symbology. But to explain a bit of physics casually without that carefully constructed framework, it’s confusing.
I should perhaps have asked what definitions were being assumed, or tried to write it in a way that avoided the confusing terms. To always say ‘energy’ rather than ‘heat’, for example. I don’t know if it helps, though.
It seems to me that the ambiguous terminology is sometimes used as a tactic and excuse to avoid answering the point, and people switch definitions freely to make whatever claim they want to at that moment. They’re not using it in a technical sense for a specialist concept – they’re just equivocating. I don’t have any patience for it. Whatever you call the things, the physics works the way it works, and even if people were using them in the technical sense they still couldn’t reach those conclusions by any valid logic. When somebody says heat flow is bidirectional, or that in a refrigerator heat flows from cold to hot, it’s perfectly obvious which convention they’re using and what they mean. To argue with that on the basis of language is avoiding the point; it’s not what the argument is about. To try to claim that this means a cold object can’t warm a hot object is invalid, whichever and whatever convention is used.
Anyway, when I said I thought it was even more likely to confuse, I meant it in the same sense as your “this seems very much to be where the confusion arises”. It wasn’t intended as a confession. šŸ™‚

joeldshore
April 26, 2013 11:18 am

@paulinuk:
Here’s the basic deal: Isolated stationary atoms and molecules have emission spectra consisting of lines. So, that means that a tenuous gas of said objects still has the same except the lines start to broaden some due the molecular speeds (“Doppler broadening” http://en.wikipedia.org/wiki/Doppler_broadening ) and due to the interactions between molecules (“pressure broadening” or “collisional broadening”). As the molecules get closer together, i.e., the gas gets denser and denser, the lines become broader and broader. By the time that you get to densities of the liquid and solid state, you are really no longer talking about lines at all but rather a continuous spectrum.

I thought that all matter with a temperature greater than 0K emitted radiation, the wavelength determined solely by itā€™s temperature, and NOT spectral emissions.

For dense matter, like solids and liquids, this is essentially true. I say “essentially” because there is still a unitless factor between 0 and 1 called emissivity which is a function of wavelength. (In a sense, even a tenuous gas can be thought of in this way, except in such a case the emissivity will be an extremely strong function of wavelength and will be essentially zero except where there are the spectral emission lines.)

tjfolkerts
April 26, 2013 11:52 am

@ Max April 26, 2013 at 10:25 am
What exact problem are you doing? It looks like you are putting a 1 AU shell around the sun, but then trying to compare that to “Problem 1023”, which only works for a shell near the inner sphere. Let’s assume you have replaced the sun with a 0.999 AU sphere @ 394 K radiating 1366 W/m^2.
For Problem 1023 and the 0.999 AU sun, then your results do work:
“1366ā‡’ā‡683 | 683ā‡’, Net 683 to shell, 683 to space, 1366 emitted by shell”
The one thing you seem to be overlooking is that the “net FROM the sphere” is now only 1366 – 683 = 683. So we can how power the 0.999 AU sun with only 683 W/m^2 rather than 1366 W/m^2 that would have been needed with no shell around it.
Exactly as Problem 1023 suggests. šŸ™‚
[If you want a “tiny sun”, then your “ā‡683” is wrong. Of the 683 emitted by the shell inward, ~682.99 will hit other parts of the shell, and only ~0.01 will hit the sun. The “net to shell” will be 1366 – 0.01 ~ 1366, while the “net to space” will still be 683, so the shell will keep warming up.]

Gary Hladik
April 26, 2013 12:06 pm

Myrrh says (April 26, 2013 at 1:28 am): “We cannot feel Light as Heat.”
Myrrh is here! Now it’s really a party! šŸ™‚
As for “feeling” light, stand in sunlight coming in through a glass window. Feel the heat? Most glass blocks infrared, yet you still feel the sun.
http://www.physicsforums.com/showthread.php?t=388609
As an experiment, you can buy a pane of glass guaranteed to transmit no infrared, put it between you and the sun, and see if you still feel the heat.
“…and your visible light from the Sun cannot raise the temperature of matter.”
Then a green laser shouldn’t be able to pop a balloon, right?

And a blue laser shouldn’t be able to burn paper, right?

Myrrh, doesn’t it bother you that your bizarro physics is so easily refuted?

joeldshore
April 26, 2013 12:09 pm

I hadn’t noticed this til now…
Christopher Monckton says (regarding John O’Sullivan):

He says a paper by me admits the ā€œtell-tale greenhouse-effect ā€˜hot spotā€™ in the atmosphere isnā€™t thereā€. The ā€œhot spotā€, which I named, ought to be there whatever the cause of the warming. The IPCC was wrong to assert that it would only arise from greenhouse warming. Its absence indicates either that there has been no warming (confirming the past two decadesā€™ temperature records) or that tropical surface temperatures are inadequately measured.

Perhaps if I wait long enough, Monckton will eventually admit he is wrong and I am right about everything that we disagreed on as he has now implicitly done about the “hot spot”. He had long argued that it was a fingerprint of warming due to greenhouse gases and had even badly misinterpreted contour maps in the IPCC report to bolster his claim, despite my patient attempts to try to set him straight. Now, he has reversed his position with no admission of his having been wrong and I right.
Of course, he still gets some things wrong:
(1) The IPCC has never asserted that “it would only arise from greenhouse warming”. He misinterpreted the IPCC report as saying this despite attempts by myself (and probably others) to explain to him why he was wrong.
(2) He misses the most obvious explanation for the hot spot being “missing”, which is remaining issues with artifacts in the multidecadal trends in some of the satellite and radiosonde analyses. In fact, we know this is the case because the analyses don’t even agree with each other, besides some of them disagreeing with the expected behavior.
(3) He ignores the fact that amplification of temperature fluctuations & trends as one goes up in the tropical atmosphere (the more technical description of the “hot spot”) has in fact been confirmed for temperature fluctuations on the monthly to yearly-timescales where artifacts in the data are not a problem. It is only for the long-term trends, where artifacts are a big issue, that there has been some difficulty in confirming such amplification.

Bryan
April 26, 2013 12:11 pm

Nullius in Verba says
” When somebody says heat flow is bidirectional, …… itā€™s perfectly obvious which convention theyā€™re using and what they mean.”
Is it ?
A similar example would be a person who says ‘a whale is a fish’
Both show ignorance of the technical definition of the word.
It depends on whether they would understand the correct definition if it was pointed out to them.
This WUWT thread is attracting a reasonable level of scientific interest so it is worth sticking to the correct technical definition of the words.

Alberta Slim
April 26, 2013 1:06 pm

richardscourtney says:
April 26, 2013 at 3:15 am
“This inhibits the passage of radiation from the Earthā€™s surface to space so the temperature of the Earthā€™s surface rises (for the same reason that an insulated house is warmer than an uninsulated house).,……………….”
Consider this:
Solar energy at the earth’s surface evaporates water [Heat of Vaporization]
The air, the water vapour and the CO2 with it heat up, expands and rises. [part of the water cycle].
The air, water vapor, and CO2 rises in the atmosphere and give off heat at higher altitudes.
The clouds form as the water vapour condenses. [Heat of Condensation], and possibly precipitation. The heat at the surface is of the earth is transferred from the surface to the upper atmosphere by conduction, convection and some radiation.
That does not look like “trapping” heat to me.

April 26, 2013 1:10 pm

Alberta Slim:
re your post at April 26, 2013 at 1:06 pm .
You may be right, and to some degree you certainly are. But that changes the subject.
It is not relevant to what I was trying to explain.
Richard

joeldshore
April 26, 2013 1:54 pm

Alberta Slim says:

The air, water vapor, and CO2 rises in the atmosphere and give off heat at higher altitudes.
The clouds form as the water vapour condenses. [Heat of Condensation], and possibly precipitation. The heat at the surface is of the earth is transferred from the surface to the upper atmosphere by conduction, convection and some radiation.
That does not look like ā€œtrappingā€ heat to me.

So, what you are saying is that convection (and accompanying evaporation/condensation) is another way to transport heat away from the surface of the Earth. However, there are a couple of problems:
(1) This moves heat up in the atmosphere but not out into space. The only way to transfer energy out into space is via radiation.
(2) What convection does do is raise the temperature of the upper atmosphere relative to the surface. Since an important element of the greenhouse effect is that the temperature at the height at which the radiation escapes to space is lower than the temperature at the surface, why can’t convection “fix” this, i.e., why can’t it just transport heat up in the atmosphere until the temperature no longer decreases as you go up in altitude? The answer is that it turns out that the atmosphere is unstable to convection only as long as the lapse rate (decrease in temperature with height) is greater than what is called the “adiabatic lapse rate”. This means that convection can reduce the lapse rate to a certain degree but no further. If it could eliminate the lapse rate entirely, then the greenhouse effect would disappear…but it can’t, so it doesn’t.

Reed Coray
April 26, 2013 2:12 pm

richardscourtney says: April 26, 2013 at 3:15 am
Thank you for the dialog. I have some concerns regarding your latest comment. Specifically, your second point includes the statement
In reality, water vapour and some of the trace gases in the air are (GHGs). A little of the mostly short-wave radiation from the Sun is absorbed in the atmosphere by GHGs. But GHGs only absorb a little of this short-wave radiation. Importantly, the total heat input (from the Sun) to the Earth+atmosphere system is the same as in Case 1. Most of the radiation from the Earthā€™s surface is long-wave radiation which is efficiently absorbed in the atmosphere by GHGs. This inhibits the passage of radiation from the Earthā€™s surface to space so the temperature of the Earthā€™s surface rises (for the same reason that an insulated house is warmer than an uninsulated house).
[Radiation from Earth surface –> atmosphere (i.e. effectively insulation) –> space]
(emphasis mine).
First, it is true that greenhouse gases including water vapor will absorb a small percentage of the incoming solar energy. [As an aside, I’m not sure exactly how much energy is absorbed, so I’m not sure if “small” is the correct adjective. In particular, for a blackbody at a temperature of 5780 Kelvin (the sun’s equivalent blackbody temperature), approximately 51 percent of the radiated energy is in the IR band (0.3 Terahertz to 430 Terahertz). The absorption bands of the various atmospheric greenhouse gases may correspond to a relatively small fraction of the total IR band, but at first glance it appears that greenhouse gases may act as an “insulator” (albeit a weak one) for incoming solar radiation.] However, water vapor forms clouds, and clouds reflect incoming solar energy. As I understand it, the average albedo of the Earth is 0.3; and most of this albedo is due to sunlight reflection from clouds. As such, although it is technically correct to say greenhouse gases absorb a small fraction of the incoming solar radiation, it is not correct to say greenhouse gases have minimal impact on the solar radiation absorbed by the Earth/Earth-atmosphere system. One way to see this is to replace the greenhouse gas water vapor with the greenhouse gas CO2. An atmosphere containing only the greenhouse gas CO2 will affect the total radiation absorbed by the Earth/Earth-atmosphere system differently than an “equivalent” amount of the greenhouse gas water vapor. This definitely makes the type of greenhouse gas an issue when discussing the total absorbed solar energy.
Second, if radiation from the Earth’s surface is the only way heat leaves the Earth/Earth atmosphere system, then ignoring cloud formation, I agree. Greenhouse gases would act “like an insulator” to outgoing radiation. However, in addition to absorbing IR radiation, greenhouse gases emit IR radiation. Furthermore, convection and conduction transfer the heat present in all gases near the surface of the earth higher altitudes where greenhouse gases can more efficiently radiate to space. It may very well be that the net effect of all these processes is an increase in the Earth surface temperature; but that is anything but obvious. As other commenters have pointed out, for a given Earth surface temperature fluid dynamics as well as radiation affect the rate energy leaves the Earth/Earth atmosphere system. These issues (and potentially other issues for which I am not even aware) are why I say “I don’t know the answer to any of the questions posed in my April 25, 2013 at 5:18 pm comment.
Reed

Gary Hladik
April 26, 2013 3:08 pm

Kristian says (April 26, 2013 at 7:46 am): “Gary, according to your own words, youā€™ve read all the relevant passages multiple times and still youā€™re on repeat as if you hadnā€™t read it even a single time.”
Indeed, I have read it multiple times, and I understand exactly what you’re saying, probably better than you do. I just don’t buy it, because it contradicts known physics.
“Youā€™re completely stuck on the notion that the ā€˜systemā€™ heat loss to space MUST equal the input to the sphere with or without the shell.”
That’s true, I am completely stuck on the First Law of Thermodynamics. Funny thing, so is the rest of the universe. šŸ™‚
“Qā€™ stays within the system. Qā€ escapes it. There is still perfect radiative balance.”
No. For one thing, you’ve left out the so-called “back radiation” J1 = J/2 from the shell toward the sphere (see the diagram for problem 1023). Since the shell is very close to the sphere (problem 1026 with R ~ r), J1 has nowhere to go but the sphere. What happens to it?
“There are in effect TWO thermodynamic systems to consider here: the sphere and the shell.”
With ONE input (e.g. a radioisotope inside the sphere), and ONE output (the shell). Either power in = power out, or the temperature changes. There are no other choices, unless you’re creating/destroying energy.
“The surface of the sphere would remain at 290K just as much with surroundings at 0 K as at 289K, it doesnā€™t matter. Neither of these two surrounding temperatures is capable of transferring heat to the sphere.”
Which is the entire problem. Bottom line is, you’re a disciple of the “imaginary second law of thermodynamics”, as scienceofdoom puts it (or “i2L” as I like to call it). From your comment April 26, 2013 at 3:30 am (“SoDā€™s confusion, for instance, about what the second law of thermodynamics is actually saying and not saying.”), I take it you’ve read his textbook examples of the REAL Second Law,
http://scienceofdoom.com/2010/10/07/amazing-things-we-find-in-textbooks-the-real-second-law-of-thermodynamics/
but still prefer to make up your own science. This is discouraging, but I’ll attempt to lead the horse to water one more time:
It turns out that a solar cooker, for example here:
http://solarcooking.org/plans/funnel.htm
can also act as a refrigerator at night, reaching temperatures below ambient atmospheric and even producing ice when air temp never appoaches zero degrees C. If you read the refrigeration section of the article, you’ll note the instruction to point the “cooker” at the open night sky rather than warmer objects like clouds, trees, walls, etc. This refutes your claim that radiation from the “surroundings” has no effect on an object’s temperature.

AndyG55
April 26, 2013 3:44 pm

What Roy’s image of the thermal release from a house shows is that heat ALWAYS rises.
This means that the atmosphere must ALWAYS act as a surface cooling mechanism.
ie.. IT IS NOT A BLANKET AND IT DOESN’T HEAT DOWNWARDS

April 26, 2013 3:59 pm

joeldshore writes “What you are missing is that emissivity is proportional to absorptivity (and both depend on the thickness of the gas layer). When it emits more, it also absorbs more.”
But that’s precisely NOT what I’m missing. So at the ERL in the case of more GHG it is both emitting more and absorbing more…for no net effect. So why does the ERL have to change with more CO2?
I’m not saying it doesn’t, I’m wondering why it “does”. And as per one of my earlier posts the only reason I can come up with is that opacity is not linearly proportional to distance/concentration which is entirely non-intuitive and furthermore I haven’t ever seen that result documented.
It worries me that the particular result has been derived from measurements of the atmosphere (and entered into modtran) when there is actually another reason for it.

April 26, 2013 4:19 pm

One more thing joel, you said “What you are missing is that emissivity is proportional to absorptivity” but in fact emissivity is precisely equal to absorptivity for the same substance. So there is possibility of different emission/absorption from say H2O to CO2 but not from CO2 to CO2.

April 26, 2013 5:04 pm

joeldshore try this explanation. Maybe it’ll help you see where I’m coming from with this and perhaps you can spot the flaw (if there is one) with this reasoning.
Taken from the point of view of a CO2 molecule at the ERL ready to emit. The photon emitted is heading up and above it is a haze of CO2. But at the ERL that photon has a 50/50 chance of escaping (by definition) and so maybe it makes it, maybe it doesn’t.
Now double the CO2 levels. Again taken from the point of view of the CO2 molecule at the (original) ERL , the photon released upwards sees a doubly thick haze of CO2 and so now it only has a 25% chance of escaping. But with double the CO2, right alongside the first CO2 molecule is a second CO2 molecule also ready to emit. And the photon produced also has a 25% chance of escaping. So as independent events, together there is still a 50% chance of one of them escaping.
This represents no net effect and the ERL doesn’t need to change for the same amount of radiation to escape. Where is the flaw? Please dont parrot back the standard speil, Use my example to illustrate where it goes wrong.
One might argue for example that the second CO2 molecule might actually radiate downwards instead and the answer to that is to simply choose another one that is radiating upwards. When the CO2 levels are doubled, the second half equally radiates both up and down and you can (on average) pair up ups and downs from each half.

Nullius in Verba
April 26, 2013 7:20 pm

“But with double the CO2, right alongside the first CO2 molecule is a second CO2 molecule also ready to emit. And the photon produced also has a 25% chance of escaping. So as independent events, together there is still a 50% chance of one of them escaping.”
And if you quadruple the CO2 level, there would be a 100% chance of one escaping? And what happens after that?
It helps to think of a more intuitively familiar situation like a layer of fog, or a glass full of diluted milk. To infra-red eyes, that’s what the Earth’s atmosphere would look like. It is somewhat fuzzy, but it is as if the ‘visible’ surface radiating to space were the top of the fuzzy layer. If you add some milk to the water, you can see into it about an inch. If you add more milk, the original surface is obscured.
For optically thin materials (i.e. that you can see right through) it does work somewhat as you say. But as the material becomes optically denser, the surface becomes more “solid” and the light emitted approaches a maximum, but the light escaping diminishes in geometric progression. Eventually the substance may be considered ‘opaque’, and the absorbing surface layer is only a few tens of thousands of molecules thick. All the light that escapes comes only from the surface layer. All the lines of sight from deeper layers to the outside are blocked.
If the material was isothermal, then this would make no difference. Every line of sight ends in some absorbing molecule, so all that changes is the thickness of the translucent surface layer. The light emitted (once thick enough to be entirely opaque) is the same. But when the temperature varies, the thickness of the translucent layer comes to matter. A thin layer only emits at the upper surface temperature. A thicker layer emits at a range of temperatures.

gbaikie
April 26, 2013 7:51 pm

“First, it is true that greenhouse gases including water vapor will absorb a small percentage of the incoming solar energy. [As an aside, I’m not sure exactly how much energy is absorbed, so I’m not sure if “small” is the correct adjective.”
This graph:
http://en.wikipedia.org/wiki/File:Solar_Spectrum.png
Shows how much of the solar spectrum is blocked from incoming solar energy.
Graph is linked form
http://en.wikipedia.org/wiki/Sunlight
In the graph the yellow is how solar energy reaches the top of Atmosphere and the red
is how much reaches the surface at sea level. Assuming sun is at zenith and it’s a clear day.
The graph indicate 5 chunks taken out by the H20. And a little bit by oxygen, CO2, and ozone [O3].
The vertical part of graph relates to amount watts square meter per nanometer of spectrum.
Also note a large amount spectrum is missing in visible light spectrum- which due no gas absorbing this spectrum of light but instead the sunlight scatter/reflected/diffused going thru Earth’s thick atmosphere.
So sunlight is 1360 watts per square meter at top of atmosphere and by time reaches earth surface it’s around 1000 watts per square meter- losing about 360 watts per square meter.
Or more correctly prevent 360 watts per square meter from directly reaching the sensor which measuring how sunlight is reaching the surface- some that sunlight could be scattered so reaches other parts of earth [or arriving at different angle]. So all it’s indicating is the energy is
not directly reaching the sensor at the ground [pointed at the sun].
Anyhow one can make a guess of how much watts of sunlight is blocked by H20.
So looks like about 50 nm and at about 1/4 watt per square meter- so about 12.5 watts for first two chuncks taken out. So total of about 25 watts. Third chunk remove seem like most amount
and it’s less than 1/4 watt per square meter removed but about 100 nm. So it about 25 watts.
And the two chunks look less than 25 watts in total. So somewhere around 100 watts per square
meter is blocked by H2O. This energy is not reflected but is absorbed and would be re-radiated
in some direction.

April 26, 2013 8:42 pm

Nullius writes “And if you quadruple the CO2 level, there would be a 100% chance of one escaping? And what happens after that?”
No. Still 50%. There will be 4 tries to escape each with a 12.5% chance of succeeding because the amount of CO2 above has again doubled and so the chance for each escape has halved. Each are independent events.

Nullius in Verba
April 26, 2013 9:13 pm

Hmm. Yes. I see what you mean.
Independent probabilities don’t quite add like that. n molecules each with escape probability p result in a probability (1-p)^n that none will escape, and hence 1 – (1-p)^n that at least one will. With two molecules it’s 44%, not 50%. With 4 molecules 41%. However, since there is a chance of more than one photon escaping it works out the same on average.
However, that’s only the case for optically thin materials. The total cross section of all the molecules cannot exceed 1.

joeldshore
April 26, 2013 9:33 pm

TimTheToolMan: Notice that your argument only works if we were right at the level where there was a 50-50 chance for the radiation to escape to space.
Let’s consider emission from other levels: For example, emission from a level lower in the atmosphere where there was originally only a 25% chance of escape. Now, doubling CO2 reduces that to only a 6.25% chance of escape per photon, so even with double the photons emitted, you are still down in total emission from this level by a factor of two.
On the other hand, consider a level higher in the atmosphere where the chance of escape was originally 90%. Doubling the CO2 reduces that to 81%, and since you have double the photons emitted, you are now actually up in emission from this level.
So, what is the net effect? The net effect is that you have more emission from levels higher in the atmosphere and less from those lower in the atmosphere. This would be a “wash” if the atmosphere were isothermal, but it’s not: The levels lower in the atmosphere are warmer and emit more while those higher in the atmosphere are colder and emit less.
Hence, the result is more “weight” from levels of the atmosphere that emit less and less “weight” from levels of the atmosphere that emit more.

paulinuk
April 26, 2013 9:36 pm

Joel Shore said:
“Hereā€™s the basic deal: Isolated stationary atoms and molecules have emission spectra consisting of lines. So, that means that a tenuous gas of said objects still has the same except the lines start to broaden some due the molecular speeds (ā€œDoppler broadeningā€ http://en.wikipedia.org/wiki/Doppler_broadening ) and due to the interactions between molecules (ā€œpressure broadeningā€ or ā€œcollisional broadeningā€)”
I say:
Thanks for the reply but this only complicates things for me. It doesn’t mention gas density in Thermal Doppler Broadening: the Doppler shift is related to the spectral frequency, velocity (or temperature) and mass of the particle.
If you accelerate electrons in a radio antenna they emit radio waves and the same is true if you accelerate protons. If a neutrally charged atom is accelerated, due to it having a temperature, then won’t it too emit radio waves because it contains internal charge? This radiation has got nothing to do with spectral line broadening because electrons and protons don’t have a spectrum as far as I know!

joeldshore
April 26, 2013 9:54 pm

Nullius in Verba says:

For optically thin materials (i.e. that you can see right through) it does work somewhat as you say.

Of course, optical thickness depends on wavelength and there are wavelengths in the atmosphere for which it is optically-thin. However, you still don’t end up with more emission at these wavelengths as the density of optical absorbers/emitters increases. The reason is because you have the Earth below (sort of the ultimate optically-thick backstop) and so you always end up in net subtracting from the Earth’s emission. [I believe if you did have an optically-thin gas with no as-warm or warmer solid emitter behind it then you could increase the emission by adding more gas.]

paulinuk
April 26, 2013 10:07 pm

Here’s an experiment: Inject Nitrogen gas at 1atm and at 400K into a large chamber full of N2 at 1atm and at 288K. Observe the hot plume with an infra-red camera as it enters the chamber. It should be invisible as “N2 can’t radiate heat”???. If the plume is visible then there’s something terribly wrong with atmospheric radiation theory.

April 27, 2013 12:16 am

Nullius writes “However, thatā€™s only the case for optically thin materials. The total cross section of all the molecules cannot exceed 1.”
The definition of ERL from Mosher’s link (which seems good enough to be a working definition) is
“ERL: The lowest level in the atmosphere from which infra red radiation is able, on average, to escape upwards to outer space without being reabsorbed”
“On average escaping” means 50% escape and so that defines the opacity quite clearly. If instead of using the ERL, we used ground level then the chance of a photon escaping would be very low. And doubling the amount of CO2 above it would still only halve the chance but when the chance is minuscule, it hardly matters conceptually does it.

gbaikie
April 27, 2013 2:13 am

“Hereā€™s an experiment: Inject Nitrogen gas at 1atm and at 400K into a large chamber full of N2 at 1atm and at 288K. Observe the hot plume with an infra-red camera as it enters the chamber. It should be invisible as ā€œN2 canā€™t radiate heatā€???. If the plume is visible then thereā€™s something terribly wrong with atmospheric radiation theory.”
The hot air could heat the walls- that should be visible.
Not sure if replace hot N2 air with hot CO2, you would see the gas.

RACookPE1978
Editor
April 27, 2013 2:44 am

No.
You don’t need any physical experiment or “Einsteinium thought games” to find a fundamental problem with these infinitely-continuing CHG discussions:
From above:

So sunlight is 1360 watts per square meter at top of atmosphere and by time reaches earth surface itā€™s around 1000 watts per square meter- losing about 360 watts per square meter.

For simplicity, let’s assume those values are correct.
Now. If every second of every day, 360 of Trenberth’s 1360 watts DON’T penetrate a supposedly “completely transparent” atmosphere, but since their energy does arrive at top-of-atmosphere but doesn’t arrive at bottom-of-atmosphere, then how do those 360 watts get “lost” from the earth’s overall system?
Clearly, they ARE absorbed by the oxygen, nitrogen and argon and dust and aerosols, and ARE re-radiated by those same gasses and dusts and solids. However, just as clearly, they ARE also ignored by the CAGW community because their theory does not allow such inconsistent, inconvenient maths.

Nullius in Verba
April 27, 2013 3:18 am

“Now. If every second of every day, 360 of Trenberthā€™s 1360 watts DONā€™T penetrate a supposedly ā€œcompletely transparentā€ atmosphere…”
The term isn’t “completely transparent”, it’s just “cloudless”.
And in what sense are they being “lost” or “ignored”? They appear in the Kiehl-Trenberth diagram, in which the numbers add up. It’s simply re-radiated, ultimately to outer space. It doesn’t actually matter that much whether it was stopped on the way in or the way out, all that matters is what level it is radiated out from. And the CAGW community go on about aerosols quite a lot.

April 27, 2013 4:16 am

joeldshore writes ” emission from a level lower in the atmosphere where there was originally only a 25% chance of escape. Now, doubling CO2 reduces that to only a 6.25% chance of escape per photon”
If you’re at a level where there is 25% chance for a photon to escape and you double the CO2 then you halve the chance, not quarter it. Hence it becomes 12.5% chance per photon and there are two of them still totalling 25% chance to escape for independent events.
Are you sure you have your probabilities right ? Or am I the one making a mistake in this?

Nullius in Verba
April 27, 2013 4:20 am

“If youā€™re at a level where there is 25% chance for a photon to escape and you double the CO2 then you halve the chance, not quarter it.”
You square it. Probabilities multiply. Think of it as two 25% chances, one after the other.

April 27, 2013 4:55 am

Nullius writes “You square it.”
Of course. Silly me, clearly a blonde moment. So its molecules radiating NOT at the ERL that make the differences. sheesh.

Nullius in Verba
April 27, 2013 5:53 am

If it’s any comfort, I had the same moment. I knew the probabilities didn’t add, but when I saw that it worked out fairly close for p = 0.5, I got myself mixed up for a bit.
You can see a certain depth into any translucent material. The upper part of that layer isn’t so strongly obscured, but isn’t dense enough to be very visible, either. As it gets thicker, it both gets more visible and more obscured by the material above it. The obscuration increases exponentially faster so it soon disappears from sight. But in the shallow surface layer, you can initially get some partial cancellation from the two effects. I’ve learnt something from having to think about it. Thanks.

joeldshore
April 27, 2013 6:04 am

RACookPE1978 says:

Clearly, they ARE absorbed by the oxygen, nitrogen and argon and dust and aerosols, and ARE re-radiated by those same gasses and dusts and solids. However, just as clearly, they ARE also ignored by the CAGW community because their theory does not allow such inconsistent, inconvenient maths.

Yes, there was actually a meeting held in a super-secret location organized by Maurice Strong in which all the groups that do atmospheric modeling pledged to not include in their models things that are inconvenient like this. It’s amazing that you have found this out!
Boy, and that guy that you all hate because he claimed a correlation between belief in AGW being a hoax and belief in conspiracy theories was clearly completely out-to-lunch!

April 27, 2013 8:22 am

How long do you think it takes the energy represented by one surface photon to escape the Earth’s atmosphere? Unimpeded by collision, it would be nearly instantaneous, but slowed by what could be many millions of collisions and absorption/emission cycles? Sure, there is a delay which is enormous from a photon POV, but it will be small from perception POV. There would be a distribution, but what’s the median value of the distribution? Five milliseconds? Ten? You cannot build a practical heater from this delay. And, as should be obvious, anything capable of increasing the Earth’s surface temperature by 33C would be a very practical and useful heater.
Put a gallon jug of water in your front yard and assign yourself the trivial challenge (trivial compared to Mother Nature’s hard work of a 33C increase) of increasing its average temperature by 10C or even 5C using “back radiation”.. You academic climatologists can’t do it. Not even close. Oh, I forgot, the GHE heating machine only works when the the air is rarefied and cold. Right.
Don’t take it from me, take it from Maria Bykova, a PhD chemist from Novosbirtk, RU. “There is no mechanism,” she said to me. You AGW academics and true believers are running a flea circus.

Kristian
April 27, 2013 8:40 am

tjfolkerts says, April 26, 2013 at 9:14 am:
“Kristian, I applaud your enthusiasm, but Garyā€™s explanations are much better than yours here.”
Your post is written in such a ridiculously condescending tone that I really shouldn’t be bothered
answering it. But never mind.
We (hopefully) all agree that the flux from the surface of the sphere cannot be suppressed or disallowed in any way by a countering flux. It could not physically obstruct or block it, reducing it that way. No, we are simply subtracting it budgetwise (OUT-IN) at the surface of the sphere.
So in this case we have:
400 –> 400
Planet alone in space.
400 –> || 400 –> <– 200 || 200 –>
Just after the shell is emplaced.
The heat loss from the sphere (and from the shell) is now too small (400-200=) 200 W/m^2. So it has to increase.
The surface temperature of the sphere rises to 321K, resulting in an outward flux of 600 W/m^2 and a net flux (heat loss) toward the shell of (600-200=) 400 W/m^2:
400 –> || 600 –> <– 200 || 200 –>
But remember, the incoming to the shell is still physically the 600 W/m^2 leaving the sphere. It’s not been reduced. So it will split into:
400 –> || 600 –> <–300 || 300 –>
For a net heat gain for the shell of (600-300=) 300 W/m^2 and a corresponding heat loss (from the outer surface) of 300 W/m^2.
Still no radiative balance, though. So the sphere needs to heat up some more:
400 –> || 700 –> <– 300 || 300 >
Well, we have the same thing going, the shell will warm up further and then the sphere will have to warm up even more and so on …
Let’s go straight to the equilibrated situation, then:
400 –> || 800 –> <– 400 || 400 –>
Now compare this to mine:
Sphere – 400 –> 400
Shell – (400-200=) 200 –> 200

What is the difference? Not much really. There is just one thing to notice. I am not adding any heat to the sphere from the shell. I am only adding heat to the shell from the sphere.
How so?
Let’s deconstruct the process.
For the shell it works out fine for Tim. The heat received is (800-400=) 400 W/m^2 to the inner surface, from the sphere. The corresponding heat ejected from the outer surface to space is 400 W/m^2. The same. We have energy AND heat balance. This is exactly equivalent to my shell’s steady state.
What about the sphere? I have heat balance. Why? Because the incoming flux from the internal source (the heat gain) is exactly matched by the resulting outgoing flux (the heat loss). I also have energy balance. Why? Because the outgoing flux of 400 W/m^2 is transferred to the other thermodynamic system, the shell, in whose budget it is fully accounted for (in a reduced heat gain and heat loss). We cannot BOTH subtract the outgoing flux from the shell’s inner surface from the incoming flux from the sphere (reducing the shell’s heat gain) AND from the outgoing flux from the sphere’s surface (without taking into account the still existing opposite flux from the outer surface of the shell). This would constitute double counting. Or rather, you would turn Q = Q’ + Q” into Q = 2Q’, skipping Q” and rather doubling Q’. This is what Tim does.
What you actually have for the flux leaving the surface of the sphere with the shell in place, i.e. between sphere and space, is:
400 out from sphere, minus 200 in from the shell, plus 200 out from the shell. You cannot include the minus without the plus: Q = (Q ā€“ Qā€™ā€™) + Qā€™ā€™, J = (J ā€“ J1) + J1.
You can see this quite clearly if you compare with the Carnot cycle:
http://i1172.photobucket.com/albums/r565/Keyell/Carnot_zps4049e783.jpg
If youā€™re standing in place of the shell (or the Carnot engine), facing inwards toward the surface of the sphere (the hot reservoir) and with space (the cold reservoir) at your back, what do you see? You see the incoming flux from the sphere/hot reservoir (Q = J = Qh). This is absorbed in its entirety and split upon re-release. When the shell is heated to its steady state temperature, it sends out J1 (J/2) from its inner surface and a similar flux J1 (J/2) from its outer. The outwardflowing J1 flux (Qā€™ā€™ = Qc) is the heat loss of the shell. But what about the inwardflowing J1 flux? Where does that fit into it all? It reduces the heat gain of the shell: J ā€“ J1 = J1, Q ā€“ Qā€™ā€™ = Qā€™ = Q/2. In other words, it reduces the net energy coming in to the shell (J ā€“ J1), not the originally incoming flux from the sphere itself (J). So hereā€™s the take-home message: The inward J1 reduces the heat gain to the shell. The outward J1 increases its heat loss. J, though, is not touched by any of the J1s. If it were, then there would be no J to come in to the shell in the first place. It wouldā€™ve been reduced (halved) even before arrival.
Tim agrees. Heā€™s quite intent on letting the full outward flux from the sphere (800 W/m^2) hit the inner surface of the shell even after heā€™s cut the heat loss flux from the surface of the sphere in half ((800-400=) 400 W/m^2).
It is actually easier to follow this line of thought with the Carnot cycle (above):
There you would have:
Qh
V
Engine –> W (Qh/2)
V
Qc (Qh/2)

Half of Qh from the hot reservoir is used for work in/by the engine (W), half is shed to the cold reservoir (Qc). Nothing at all at any time is ā€™flowing backā€™ to the hot reservoir to reduce its heat loss (Qh = J) or, worse, to somehow add to its heat gain. W is the equivalent to Qā€™ (J ā€“ J1, Qh ā€“ Qc). Energy is being continuously utilized by the intercepting ā€™layerā€™ and this cannot ever reach the cold reservoir. Hence the reduced Qc/Qā€™ā€™.
I have a hard time seeing how this simple circumstance could be such an impossible thing to comprehend.
After the incoming net flux of energy to the shell is reduced by half by the inward J1, there is still no net flux flowing from the shell to the surface of the sphere, meaning going IN. The resulting net flux is still positive for the shell. It is still moving outwards ā€¦ ultimately from the outer surface of the shell to space: Q = Qā€™ + Qā€™ā€™ = (Q ā€“ Qā€™ā€™) + Qā€™ā€™. Note from this, the heat transfer rate to the shell from the sphere is reduced as the heat transfer rate to space from the shell is increased.
Tim does have an apparent heat balance for the sphere’s surface:
400 –> || 800 –> <– 400
(400 IN; (800-400=) 400 OUT).
But upon closer inspection his problem readily reveals itself. How? Well, how on earth did he get to the 800 outgoing W/m^2 when the heat input is only 400 W/m^2? The 800 W/m^2 after all has to come from a rise in surface temperature from 290K to 345K (a 19% increase). How is this accomplished?
We know already that the internal heat source itself only supplies 400 W/m^2 to the surface. This corresponds to an emission temperature of 290K. We also know that the incoming 400 W/m^2 from the shell can not in any way disallow or suppress/block the resulting BB emission flux from the surface of the sphere toward the shell of 400 W/m^2.
So what are we left with?
The incoming 400 W/m^2 from the shell. Extra HEAT. Only these could have provided the heat to raise the surface temperature of the sphere past the Stefan-Boltzmann dictated emission temperature from the internal heat source input.
So in the end, there is no subtraction, no reduced heat loss from the sphere. There is only addition of extra heat from the shell to the sphere’s surface. This is the only way to raise its temperature beyond 290K.
And this is of course a complete and blatant violation of the second law of thermodynamics. And it goes directly against what Spencer (and all the rest) is claiming, that the reason why the sphere is heated is because its heat loss is reduced by the shell, not because its heat gain is increased by it. A cooler object simply cannot increase the heat gain of a warmer one.

Kristian
April 27, 2013 8:43 am

Gary Hladik says, April 26, 2013 at 3:08 pm:
“Indeed, I have read it multiple times, and I understand exactly what youā€™re saying, probably better than you do.”
Haha, you’re quite a piece of work, aren’t you Gary? No, you’ve understood diddly-squat. You think (and claim) you understand what I’m saying, but quite clearly you’re not. You’re simply wallowing in your own self-satisfied ‘I’ve already got the answer, the Truth, so you don’t even have to try to prove me wrong’ kind of mind-set. Completely blinkered. Well, good for you. I guess it keeps you happy.
“Thatā€™s true, I am completely stuck on the First Law of Thermodynamics. Funny thing, so is the rest of the universe. :-)”
Only, this is not about the first law of thermodynamics. šŸ™‚
The first law of thermodynamics does not say in any way or fashion that the heat loss of one thermodynamic system needs to balance the heat gain of another. How come you’re so willing to misrepresent it?
Do you know the concept of thermodynamic systems, Gary? And how it relates to the first law of thermodynamics?
Do you know anything about the Carnot cycle? About ‘work’ and ‘heat’? And how they relate?
One starts to wonder.
“ā€œQā€™ stays within the system. Qā€ escapes it. There is still perfect radiative balance.ā€
No. For one thing, youā€™ve left out the so-called ā€œback radiationā€ J1 = J/2 from the shell toward the sphere (see the diagram for problem 1023). Since the shell is very close to the sphere (problem 1026 with R ~ r), J1 has nowhere to go but the sphere. What happens to it?”

For crying out loud! Q’ is J – J1! There’s your J1. If I ‘left out’ the back radiation J1 = J/2, then Q’ would be the same size as Q and Q” would be 0. We would have || J –> || 0 –>. J = J + 0. How’s that for a radiative balance?
“ā€œThere are in effect TWO thermodynamic systems to consider here: the sphere and the shell.ā€
With ONE input (e.g. a radioisotope inside the sphere), and ONE output (the shell). Either power in = power out, or the temperature changes. There are no other choices, unless youā€™re creating/destroying energy.”

No, no, no! The sphere has one heat input and one heat output (400 –> 400), the shell has one heat input and one heat output ((400-200=) 200 –> 200). How hard is this to get?
“ā€œThe surface of the sphere would remain at 290K just as much with surroundings at 0 K as at 289K, it doesnā€™t matter. Neither of these two surrounding temperatures is capable of transferring heat to the sphere.ā€
Which is the entire problem. Bottom line is, youā€™re a disciple of the ā€œimaginary second law of thermodynamicsā€ (…)”

Nope. You’re a disciple of the ones who believe such an ‘imaginary law’ exists, having ultimately created it for themselves, as a strawman to focus on and cling to instead of dealing with what the real objections to what they’re doing are actually saying.
“(…) I take it youā€™ve read his textbook examples of the REAL Second Law, http://scienceofdoom.com/2010/10/07/amazing-things-we-find-in-textbooks-the-real-second-law- f-thermodynamics/ but still prefer to make up your own science.”
Yes, I’ve read them. SoD is the one imagining things. But I can see the source of his confusion.
Regarding that, I also take it you’ve read my ‘modern’ physics references of how ‘heat’ is defined as opposed to ‘thermal energy’? How is the shell tranferring heat to the sphere, Gary, increasing its content of thermal energy, elevating its kinetic level, raising its temperature beyond what the input from the internal heat source can muster?
Do tell me.

Nullius in Verba
April 27, 2013 9:01 am

“Put a gallon jug of water in your front yard and assign yourself the trivial challenge (trivial compared to Mother Natureā€™s hard work of a 33C increase) of increasing its average temperature by 10C or even 5C using ā€œback radiationā€”
Sigh. The greenhouse effect as implemented in the climate models doesn’t work by back radiation. It’s not the mechanism the science proposes. So you’re completely wasting your time attacking it. Even if you disproved the existence of back radiation (which you can’t, because it does exist) it would have absolutely no impact on AGW, because it’s totally irrelevant. All the argument does is discredit scepticism to no effect.
Increasing the temperature of a jug of water with backradiation is easy – it’s even easier to do it with a whole swimming pool of water. Just put it at the bottom of a solar pond.

Kristian
April 27, 2013 10:14 am

Gary and Tim,
The idea that you people seemingly subscribe to, is that the half of the outgoing flux from the surface of the sphere that does not end up as heat loss to space from the outer surface of the shell, has the power to do work twice. First it is spent warming the shell. Then it sort of reawakens to do a second warming job. It goes back to the sphere whence it originally came, and warms that some more too. (Of course leaving the shell bereft of thermal energy in the process ā€“ i.e. it cannot maintain its temperature AND warm the shell at the same time.)
This is loony accounting.
Inserting an opaque layer between a heated surface and space will make the layer warm, not the surface warmer than before. The surface is the layerā€™s source of heat. The layer simply ā€™takesā€™ some of the heat from the surface that used to go straight out to space for itself, spending it on heating up. There is still perfect radiative balance between the surface and its surroundings (heated layer + space). There is also radiative balance between the heated layer and its surroundings (surface + space). The whole situation is simply different from before the layer was inserted.
How come this fundamental knowledge of the inner workings of thermodynamic systems completely bypass seemingly grown and, one must assume, well-educated people when it comes to the magic of the radiative GHE?

Myrrh
April 27, 2013 11:14 am

Ferdinand Engelbeen says:
April 26, 2013 at 8:50 am
Myrrh says:
April 25, 2013 at 4:52 pm
Myrrh, diffusion works for all molecules, no matter their specific weight. Diffusion is in fact Brownian motion at molecular scale. Wind and convection is what brings CO2 to long distances and height, up to the stratosphere. But the movements of real molecules of real gases in the real world is what keeps CO2 in the mixture.
Ferdinand – AGWScienceFiction doesn’t have real gas molecules. So Brownian Motion is not applicable.
They use Brownian motion “explanations” because they don’t have convection, if they explained convection they wouldn’t have the AGW Greenhouse Effect, so they have to avoid it. They avoid it by saying their gases are ideal, and then the rest is just adding confusion by giving explanations which ‘appear’ to explain well mixed, but when you break it down you can see it’s not applicable.*
What I am trying to explain here is that AGWSF has substituted “ideal” gas for real gas, ideal gas has no properties – it does not have volume, it does not have attraction, it does not have weight, it is not subject to gravity – because it is not real. Ideal gases are imaginary hard dots of nothing with no mass so nothing for gravity to act on. It remains a useful beginning point in calculations but only makes sense when all the missing properties and processes are added back in. This is a great source of confusion in these discussions, when those from traditional physics and applied science don’t know this is what they have done in AGW fisics…
AGWSF have empty space instead of our real world atmosphere which is composed of real gases, not ideal.
Here, real world gases with volume is how we can hear sound: http://www.mediacollege.com/audio/01/sound-waves.html
“Note that air molecules do not actually travel from the loudspeaker to the ear (that would be wind). Each individual molecule only moves a small distance as it vibrates, but it causes the adjacent molecules to vibrate in a rippling effect all the way to the ear.”
This is not the “empty space with molecules miles apart” of the AGW SF ideal gas narrative, but a medium created by the juxtapositions of molecules with volume squashed together by gravity.
They have no sound in their impossible world, so they can’t hear anyone of this..
This is our real atmosphere, real gases with volume under gravity. These real gases expand when heated and become lighter than air and rise, they condense when they get cold, become heavier than air and sink. This is basic bog standard meteorology, this is how we get our winds, by differential heating of volumes (packets) of air.
This is how we get our rain, by heated water evaporating and rising because less dense lighter than air carrying that heat away from the surface, and when it reaches the colder heights the real gas water vapour releases its heat and condenses into liquid water or ice, and precipitates out, it rains.
That is the Water Cycle, the “greenhouse gas” water vapour takes heat away from the surface into the colder heights of the troposphere, this is how we get our weather. Rain is carbonic acid, because real gases have attraction and water and carbon dioxide are greatly attracted to each other – water has a residence time of 8-10 days in the atmosphere, and carbon dioxide in that is fully part of it.
I am also trying to point out that there is no internal joined up logic in their explanations, they use “ideal gas diffusion by their own molecular momentum travelling at great speeds through the empty space atmosphere of ideal gas bouncing off each other in elastic collisions and so thoroughly mixing”, and, they use Brownian Motion which is applicable to particles in volumes of fluid, gas or liquid, and ideal gases have no volume, and, they use “turbulent mixing by winds” – none of these are applicable in the real world of real gases.
I’ve just come indoors, there is not a breath of wind outside – as I look out of the window the tops of the trees are still, as I look into the sky the clouds are not moving. Brownian motion hasn’t moved the dust on my desk.. The major wind systems do not cross hemispheres, there is no “global mixing by turbulent winds”. They don’t have winds anyway! Because winds are volumes of atmosphere moving, and they have only empty space. Because to get wind you have to have real gases which can get lighter and heavier than air which separate out.., which means heat transfer by convection, etc. etc.
These are AGWSF “memes” to explain “well mixed” which their fictional fisics based on ideal gas is incapable of producing.
They have no rain in their carbon cycle because it spoils their AGW narrative, not that any of them have noticed the whole of the real Water Cycle is missing.., but they couldn’t have it anyway with their gases which go straight from the surface to empty space..
How do they get clouds with all the ideal gas molecules miles apart from each other bouncing off each other at great speeds?
Their Earth and its atmosphere is a complete and utter fiction, it is created by sleights of hand mixing and removing and renaming and lots of other techniques, manipulations using terms from physics.
*AGWSF introduced this into the general education system some time ago, they began by teaching new infant/junior level teachers who weren’t specialised, and it’s these which began the classroom “explanations of well mixed” by opening bottles of scent and saying it was Brownian motion which spread the smell, and by pouring coloured ink into water to show that this was “how carbon dioxide thoroughly mixes by diffusion so it can’t be unmixed”. These, in the real world of real gases, are by spread convection currents.
“Separation diffusion from convection in gases
“While Brownian motion of multi-molecular mesoscopic particles (like pollen grains studied by Brown) is observable under an optical microscope, molecular diffusion can only be probed in carefully controlled experimental conditions. Since Graham experiments, it is well known that avoiding of convection is necessary and this may be a non-trivial task.
“Under normal conditions, molecular diffusion dominates only on length scales between nanometer and millimeter. On larger length scales, transport in liquids and gases is normally due to another transport phenomenon, convection, and to study diffusion on the larger scale, special efforts are needed.
“Therefore, some often cited examples of diffusion are wrong: If cologne is sprayed in one place, it will soon be smelled in the entire room, but a simple calculation shows that this can’t be due to diffusion. Convective motion persists in the room because the temperature inhomogeneity. If ink is dropped in water, one usually observes an inhomogeneous evolution of the spatial distribution, which clearly indicates convection (caused, in particular, by this dropping).”
http://en.wikipedia.org/wiki/Diffusion
The confusion is deliberate, only some/one who knew real world physics very well indeed could have come up with this complex, subtle, scam.

April 27, 2013 11:53 am

Kristian says:
April 27, 2013 at 8:43 am
And it goes directly against what Spencer (and all the rest) is claiming, that the reason why the sphere is heated is because its heat loss is reduced by the shell, not because its heat gain is increased by it. A cooler object simply cannot increase the heat gain of a warmer one.
As I said before, not my preferred subject of discussion (too long ago…), but what happens if you simply wrap the sphere in some kind of insulation? With a constant heat source, the sphere temperature simply will go up much higher than without insulation, until a new equilibrium is reached where energy source = energy loss.
Something similar happens with the shell: that hinders the radiation to the outside world, no matter if that is by direct transfer of heat (via a real insulator) or indirect via radiation. In all cases, all the energy of the source must be radiated to the outside world, or one of the parts of the system will heat up, either the sphere or the shell or both. As the system is in equilibrium when energy out = energy in, the shell need to emit the same amount of energy as the source within the sphere, nothing less.
No, no, no! The sphere has one heat input and one heat output (400 ā€“> 400), the shell has one heat input and one heat output ((400-200=) 200 ā€“> 200). How hard is this to get?
The 200 radiated from the shell to the sphere is additional to the 400 from the inner source. The sphere thus has two heat inputs. The sphere receives 400+200=600, but only emits 400. Thus the sphere heats up. That is what happens.
The sphere doesn’t know that the source of the outside 200 is cooler than itself, energy is energy, no matter if that is from vibrating molecules or from radiation. Still no violation of any physical law, as the radiated energy flows from warmer objects to colder objects are larger than reverse. All what happens is that the total energy loss of the warmer object is reduced, like as if it were insulated.
Some molecules in a material have higher vibration (thus a higher “temperature”) than others, even in a cooler object compared to a warmer. That doesn’t violate any law, as the average energy flow is from a warmer object to a cooler object.

Myrrh
April 27, 2013 12:04 pm

Gary Hladik says:
April 26, 2013 at 12:06 pm
Myrrh says (April 26, 2013 at 1:28 am): ā€œWe cannot feel Light as Heat.ā€
As for ā€œfeelingā€ light, stand in sunlight coming in through a glass window. Feel the heat? Most glass blocks infrared, yet you still feel the sun.
..
As an experiment, you can buy a pane of glass guaranteed to transmit no infrared, put it between you and the sun, and see if you still feel the heat.

But of course, all these companies making windows to block out the heat rays from the Sun but maximising visible light to bring down the costs of air conditioning, shucks, they must be ignoramuses. They should be selling blinds to block out visible light to keep the room cool..
You do the experiment.
ā€œā€¦and your visible light from the Sun cannot raise the temperature of matter.ā€
Then a green laser shouldnā€™t be able to pop a balloon, right?
So the Sun is a laser??!
Unenhanced Visible light is what we get from the Sun. Visible light from the Sun works on the tiny electronic transition level, not on the bigger molecular vibrational level.
Rub your hands together, that is mechanical energy vibrating the molecules in your skin heating them up which is kinetic energy. Kinetic energy is heat. Average kinetic energy is temperature.
Visible light from the Sun cannot do this. Radiant heat from the Sun can and does do this.
The radiant heat from our millions of degrees hot Sun heats up our skin and penetrates into our bodies several inches, heating up the water in us, heating up our blood and flesh and bones by vibrating the molecules.
We cannot feel visible light from the Sun, we do feel radiant heat from the Sun and this radiant heat is the Sun’s thermal energy in transfer by radiation, which is thermal infrared aka longwave infrared.
Near infrared is not thermal. Thermal infrared is thermal.
Thermal means “of heat”, longwave infrared is the electromagnetic wave of heat. We can feel it is hot.
We cannot feel the smaller photons of near infrared and it is classed in with Light and not Heat, with Reflective and not Thermal.
Please read again the quote I gave from NASA, which comes from traditional real world physics. It contradicts you.
The heat we feel from the Sun is thermal infrared, longwave infrared.
Visible light from the Sun is not capable of heating matter. It gets bounced around all over the sky in reflection/scattering by the electrons of the molecules of nitrogen and oxygen when these electrons absorb it and are briefly energised and then coming back to their ground state they emit the same energy they took in, hence our blue sky. The more energetic blue visible light from the Sun is bounced around more. Visible light is not big enough to move the whole molecules of nitrogen and oxygen into vibration, which, is, what, it, takes, to, heat, something, up.
This real heat longwave infrared from the Sun has been excised from the AGWScienceFiction’s Greenhouse Effect Illusion Energy Budget and the shortwaves of mainly visible they have put in its place cannot heat up matter- you have no heat from the Sun in your world.
Which means you have no global warming problem..

Tim Folkerts
April 27, 2013 12:10 pm

Kristian,
No matter how many times you repeat your arguments, it does not make them right. Here are a couple specific things to consider:
1) You seem to be using a heater that maintains 290 K, not 400 W/m^2. For the bare planet, both are the same. But when you add a shell, you are in effect saying that the heater cannot get any warmer than 290 K even with other power inputs.
Try this … Planet A has 400 W/m^2 heaters that warm Planet A to 290 K. Planet B has 200 W/m^2 heaters that warm Planet B to 244 K. Now suppose I take these “cooler heaters” from Planet B and move them to Planet A. What will be the new temperature of Planet A? The energy from these “cooler heaters” will help warm Planet A to 321 K. The 400 W heaters, with this power boost, can now reach a higher temperature. Just like the 400 W/m^2 heater in the “shell scenario” can warm the planet above 290 K with a boost from the photons that is absorbs.
Adding a 400 W/m^2 heater to another 400 W/m^2 heater IS an 800 W/m^2 heater, and can provide the same heat as a single 800 W/m^2 heater, whether those are two electric heaters or two nuclear heaters, or one electric heater and one IR heater.
2) The Carnot cycle has NOTHING to do with this example. There is no cycle. There is no work being done. Drop the Carnot cycle discussion.
3) Draw a picture of your system. Every “surface” (above the heaters) should have a net outward flux of 400 W/m^2 reaching it from below and leaving it from above. For example, the inner surface of the shell has 800 W/m^2 of power up from the surface and 400 W/m^2 back down = 400 W/m^2 net upward flux at the inside of the surface and 400 W/m^2 conducted upward through the shell. This will also guarantee that every “object” has a net flux of 0 (exactly as much enters as leaves).

April 27, 2013 12:12 pm

Kristian says:
April 27, 2013 at 10:14 am
it cannot maintain its temperature AND warm the shell at the same time
If the shell receives 400 and emits 200 to space and 200 back to the sphere, it is in perfect thermodynamic equilibrium and the shell doesn’t warm or cool.

April 27, 2013 12:27 pm

Myrrh says:
April 27, 2013 at 12:04 pm
Myrrh, it may be a long time ago that I have been involved in radiation, but no matter the wavelength of any radiation, if it is absorbed by the skin (or an object), it is felt as heat (or will heat up that object), simply because its radiation energy is transferred into vibrational energy, thus heat. Only if all (visible or invisible) wavelengths are reflected in a mirror or by absolute white objects, these don’t heat up.

Myrrh
April 27, 2013 12:36 pm

Reposting minus possible problematic words..
Ferdinand Engelbeen says:
April 26, 2013 at 8:50 am
Myrrh says:
April 25, 2013 at 4:52 pm
Myrrh, diffusion works for all molecules, no matter their specific weight. Diffusion is in fact Brownian motion at molecular scale. Wind and convection is what brings CO2 to long distances and height, up to the stratosphere. But the movements of real molecules of real gases in the real world is what keeps CO2 in the mixture.
Ferdinand – AGWScienceFiction doesn’t have real gas molecules. So Brownian Motion is not applicable.
They use Brownian motion “explanations” because they don’t have convection, if they explained convection they wouldn’t have the AGW Greenhouse Effect, so they have to avoid it. They avoid it by saying their gases are ideal, and then the rest is just adding confusion by giving explanations which ‘appear’ to explain well mixed, but when you break it down you can see it’s not applicable.*
What I am trying to explain here is that AGWSF has substituted “ideal” gas for real gas, ideal gas has no properties – it does not have volume, it does not have attraction, it does not have weight, it is not subject to gravity – because it is not real. Ideal gases are imaginary hard dots of nothing with no mass so nothing for gravity to act on. It remains a useful beginning point in calculations but only makes sense when all the missing properties and processes are added back in.
This is a great source of confusion in these discussions, when those from traditional physics and applied science don’t know this is what they have done in AGW fisics…
AGWSF have empty space instead of our real world atmosphere which is composed of real gases, not ideal.
Here, real world gases with volume is how we can hear sound:
http://www.mediacollege.com/audio/01/sound-waves.html
“Note that air molecules do not actually travel from the loudspeaker to the ear (that would be wind). Each individual molecule only moves a small distance as it vibrates, but it causes the adjacent molecules to vibrate in a rippling effect all the way to the ear.”
This is not the “empty space with molecules miles apart” of the AGW SF ideal gas narrative, but a medium created by the juxtapositions of molecules with volume squashed together by gravity.
They have no sound in their impossible world, so they can’t hear anyone of this..
This is our real atmosphere, real gases with volume under gravity. These real gases expand when heated and become lighter than air and rise, they condense when they get cold, become heavier than air and sink. This is basic bog standard meteorology, this is how we get our winds, by differential heating of volumes (packets) of air.
This is how we get our rain, by heated water evaporating and rising because less dense lighter than air carrying that heat away from the surface, and when it reaches the colder heights the real gas water vapour releases its heat and condenses into liquid water or ice, and precipitates out, it rains.
That is the Water Cycle, the “greenhouse gas” water vapour takes heat away from the surface into the colder heights of the troposphere, this is how we get our weather. Rain is carbonic acid, because real gases have attraction and water and carbon dioxide are greatly attracted to each other – water has a residence time of 8-10 days in the atmosphere, and carbon dioxide in that is fully part of it.
I am also trying to point out that there is no internal joined up logic in their explanations, they use “ideal gas diffusion by their own molecular momentum travelling at great speeds through the empty space atmosphere of ideal gas bouncing off each other in elastic collisions and so thoroughly mixing”, and, they use Brownian Motion which is applicable to particles in volumes of fluid, gas or liquid, and ideal gases have no volume, and, they use “turbulent mixing by winds” – none of these are applicable in the real world of real gases.
I’ve just come indoors, there is not a breath of wind outside – as I look out of the window the tops of the trees are still, as I look into the sky the clouds are not moving. Brownian motion hasn’t moved the dust on my desk.. The major wind systems do not cross hemispheres, there is no “global mixing by turbulent winds”. They don’t have winds anyway! Because winds are volumes of atmosphere moving, and they have only empty space. Because to get wind you have to have real gases which can get lighter and heavier than air which separate out.., which means heat transfer by convection, etc.
etc.
These are AGWSF “memes” to explain “well mixed” which their fictional fisics based on ideal gas is incapable of producing.
They have no rain in their carbon cycle because it spoils their AGW narrative, not that any of them have noticed the whole of the real Water Cycle is missing.., but they couldn’t have it anyway with their gases which go straight from the surface to empty space..
How do they get clouds with all the ideal gas molecules miles apart from each other bouncing off each other at great speeds?
Their Earth and its atmosphere is a complete and utter fiction, it is created by sleights of hand mixing and removing and renaming and lots of other techniques, manipulations using terms from physics.
*AGWSF introduced this into the general education system some time ago, they began by teaching new infant/junior level teachers who weren’t specialised, and it’s these which began the classroom “explanations of well mixed” by opening bottles of scent and saying it was Brownian motion which spread the perfume, and by pouring coloured ink into water to show that this was “how carbon dioxide thoroughly mixes by diffusion so it can’t be unmixed”. These, in the real world of real gases, are by spread convection currents.
A good explanation:
“Separation diffusion from convection in gases
“While Brownian motion of multi-molecular mesoscopic particles (like pollen grains studied by Brown) is observable under an optical microscope, molecular diffusion can only be probed in carefully controlled experimental conditions. Since Graham experiments, it is well known that avoiding of convection is necessary and this may be a non-trivial task.
“Under normal conditions, molecular diffusion dominates only on length scales between nanometer and millimeter. On larger length scales, transport in liquids and gases is normally due to another transport phenomenon, convection, and to study diffusion on the larger scale, special efforts are needed.
“Therefore, some often cited examples of diffusion are wrong: If cologne is sprayed in one place, it will soon be smelled in the entire room, but a simple calculation shows that this can’t be due to diffusion. Convective motion persists in the room because the temperature inhomogeneity. If ink is dropped in water, one usually observes an inhomogeneous evolution of the spatial distribution, which clearly indicates convection (caused, in particular, by this dropping).”
http://en.wikipedia.org/wiki/Diffusion
The confusion is deliberate, only some/one who knew real world physics very well indeed could have come up with this complex, subtle, manipulation.

Gary Hladik
April 27, 2013 1:06 pm

Kristian says (April 27, 2013 at 8:40 am): “[major snippage] It is actually easier to follow this line of thought with the Carnot cycle (above):”
Perhaps it is. I see Ferdinand Engelbeen has “taken one for the team” and addressed some of your points, so let’s go with the Carnot Cycle as an analogy to problem 1023 (I actually like tjfolkert’s money analogy, and may use it in future, if only I can come up with a plausible T^4 spending algorithm, where T = “treasury”.) šŸ™‚ I see tjfolkerts has tried another approach now, but despite his disapproval, I think the Carnot analogy can be useful.
Equilibrium state 1: The “Hot Reservoir” is analogous to the sphere of problem 1023, and we’ll assume it has a constant power input of 400 watts, i.e. Qin = 400 = Qh before we fiddle with the analogy. The “Engine” corresponds to the shell, and at the start the engine/shell (50% efficient) is losing Qc = 200 as “waste” and doing W of useful work, e.g. powering a 200 watt light bulb in Detroit. Do we all agree that the “system” input is 400 W and the “system” output is 400 W (Qc + W = 400)? (If anyone is confused, just put a “black box” around the reservoir and engine, then look at the black box input and output.)
OK, now we disconnect Detroit and instead connect the engine output to the reservoir/sphere. Now the reservoir/sphere is getting 600 W (Qin = 400 + W). Since there are no losses shown between reservoir/sphere and engine/shell, the engine/shell (50% efficient) can now do 300 W of useful work and loses 300 W to “waste”. But now the reservoir/sphere is getting 700 W (now W = 300, Qin still 400). And so on until we reach…
Equilibrium state 2: Qin is still 400, Qh = 800, W = 400 (now routed to the reservoir/sphere) and Qc is 400. Qc (output of engine/shell) is the only output from the “system”, but unlike Kristian’s flawed analysis of problem 1023, system input = system output, in accordance with the First Law of Thermodynamics. (Again, if anyone is confused, put a black box around the reservoir/sphere and engine/shell). There is no perpetual motion machine and no magical destruction of energy as in Kristian’s view of problem 1023.

April 27, 2013 1:16 pm

Myrrh says:
April 27, 2013 at 12:36 pm
Myrrh, a long discours to prove what?
In the real world, real measurements show that CO2 rapidely mixes with air and is distributed all over the world within +/- 2% of full scale, from ground level to the lower stratosphere (except for the first few hundred meters over land). Even when some 20% of all CO2 in the atmosphere is exchanged with other reservoirs over the seasons. No matter if that happens via diffusion, convection, winds, storms,…
I call that well mixed.
For the AGW effect, the +/- 2% differences don’t matter at all, as even a CO2 doubling has little effect…

Gary Hladik
April 27, 2013 1:25 pm

Myrrh says (April 27, 2013 at 12:04 pm): “So the Sun is a laser??!”
Green light is green light. The laser is more intense than green light from the sun, but according to you, green light has no heating ability, so intensity shouldn’t matter. BTW, the sun can also pop a balloon through a magnifying glass, even if you put a pane of glass over the magnifier (or if the magnifier is made of IR-blocking glass).
BTW, how do you see “visible” light if it has no energy to initiate chemical reactions in your retina?

Myrrh
April 27, 2013 1:38 pm

Ferdinand Engelbeen says:
April 27, 2013 at 12:27 pm
Myrrh says:
April 27, 2013 at 12:04 pm
Myrrh, it may be a long time ago that I have been involved in radiation, but no matter the wavelength of any radiation, if it is absorbed by the skin (or an object), it is felt as heat (or will heat up that object), simply because its radiation energy is transferred into vibrational energy, thus heat. Only if all (visible or invisible) wavelengths are reflected in a mirror or by absolute white objects, these donā€™t heat up.
It must be a very long time ago and you’ve misremembered, in fact, I’d say you have in the meantime picked up the AGWmemes ..
The AGWScienceFiction meme is put in place to confuse light with heat from the Sun in order to be able to claim that all downwelling longwave infrared from the atmosphere comes from their “backradiation from the upwelling longwave infrared of the heated surface”, the meme “shortwave in longwave out”.
To this end, to avoid mentioning longwave infrared heat from the Sun and ‘convince’ that shortwaves have the same power to heat matter, they have also put in place the meme “all electromagnetic energy is the same and all creates heat on being absorbed”.
This is simply not true, firstly as I’ve given, it takes the vibration of whole molecules to heat something up and visible light isn’t even, in the real world, heating up the molecules of nitrogen and oxygen in the atmosphere where their electrons actually do absorb visible. Water is a transparent medium and visible doesn’t even get in to play with the electrons, but is transmitted through unchanged, so it can’t be heating the ocean.
But crucially, what they have done with this meme is taken out all the differences between wavelengths in order to push their meme that “all create heat on being absorbed”, the properties and the processes for which these were given different names.., so they can’t explain photosynthesis or sight, for example. In real world physics energy converts to other energies and not heat; in photosynthesis this is conversion to chemical energy, in sight to nerve impulses. Their physics really is faked, it can’t explain the natural physical world around us.
There’s a bit on this under http://en.wikipedia.org/wiki/Transparency_and_translucency
If you scroll down to:
“Electronic: Transitions in electron energy levels within the atom (e.g., pigments). These transitions are typically in the ultraviolet (UV) and/or visible portions of the spectrum.
Vibrational: Resonance in atomic/molecular vibrational modes. These transitions are typically in the infrared portion of the spectrum.
UV-Vis: Electronic transitions”
The second and third possible outcomes are how visible gets absorbed by the electrons of nitrogen and oxygen and reflected/scattered, and how visible is transmitted through transparent mediums such as water, not even getting to the electrons because not absorbed at all.
Further:
“Infrared: Bond stretching
“Normal modes of vibration in a crystalline solid
“The primary physical mechanism for storing mechanical energy of motion in condensed matter is through heat, or thermal energy. Thermal energy manifests itself as energy of motion. Thus, heat is motion at the atomic and molecular levels.

“When a light wave of a given frequency strikes a material with particles having the same or (resonant) vibrational frequencies, then those particles will absorb the energy of the light wave and transform it into thermal energy of vibrational motion. Since different atoms and molecules have different natural frequencies of vibration, they will selectively absorb different frequencies (or portions of the spectrum) of infrared light. Reflection and transmission of light waves occur because the frequencies of the light waves do not match the natural resonant frequencies of vibration of the objects. When infrared light of these frequencies strikes an object, the energy is reflected or transmitted.”
Near infrared is reflected, it is in the category Reflected not Thermal. As used in near infrared cameras, they work on the same principle as visible light cameras, they capture the reflected near infrared. But again, as the NASA quote tells it, these are not hot anyway. It takes a great deal of heat to heat up matter, we know this every time we cook or stand in front of a fire to get warm.., how can something that is not hot warm us up?
AGWSF has effectively destroyed our natural understanding of what heat is. This is, truly, tragic.

April 27, 2013 2:48 pm

Insulation inhibits or resists heat transfer. It helps keep your house temperature stable. Warmer than the cold outside in winter, and cooler than the hot outside in summer.

Myrrh
April 27, 2013 3:37 pm

Ferdinand Engelbeen says:
April 27, 2013 at 1:16 pm
Myrrh says:
April 27, 2013 at 12:36 pm
Myrrh, a long discours to prove what?
That the AGW Greenhouse Effect is an illusion created out of fake fisics.
In the real world, real measurements show that CO2 rapidely mixes with air and is distributed all over the world within +/- 2% of full scale, from ground level to the lower stratosphere (except for the first few hundred meters over land).”
Gosh, let’s see it..
Since Carbon Dioxide is heavier than air it will be found mainly in the first few hundred metres..
.. and of course it is being continually washed out of the atmmosphere with the rain cycle, every 8-10 days..
Ever wondered why it is a trace gas when so much of its produced? Double it it’s still a trace gas.
Here, read about real carbon dioxide: http://hvo.wr.usgs.gov/volcanowatch/archive/2005/05_06_02.html
“Carbon dioxide (CO2) is a colorless, nearly odorless gas that is denser than air. While toxic at very high concentrations, it can kill at lower concentrations by displacing oxygen, causing asphyxiation.
..
“Usually the large amounts of carbon dioxide released by Kilauea get dispersed by winds so we can breathe nice, healthy, oxygen-rich air on the caldera floor. Because CO2 is heavier than air, it doesn’t readily rise into the atmosphere and, instead, tends to pool in low areas.”
Denser than air, means that it is heavier and so displaces air. Does not readily rise in air, because it is heavier than than air. The dust on your desk will not spontaneously diffuse into the atmosphere because it is heavier than air and does not readily rise in air. This is real gases and particles under gravity.
AGWSF’s pretend ideal gas carbon dioxide is not subject to gravity.
Real gases and real processes and real heat in conjunction with gravity is how we get our weather.
Even when some 20% of all CO2 in the atmosphere is exchanged with other reservoirs over the seasons.
The atmosphere is not a “reservoir”, carbon dioxide is heavier than air it cannot defy gravity to accumulate for hundreds and thousands of years, even if it wasn’t being washed out continually and brought back to the surface as rain.
Rain is carbon dioxide – next time it rains go stand in it, be thankful that this natural Water Cycle exists as your feel the carbon dioxide drenching you as carbonic acid.
Mildly disinfecting you..
All the AGWSF fisics is faked. They are charlatans who claim to teach the Carbon Cycle and miss out rain.
“Reservoirs” is an AGWSF meme, to confuse, to distract from this natural Carbon Cycle which in traditional real world physics tells you carbon dioxide gets washed out of the atmosphere in the Water Cycle – which residence time in the atmosphere is 8-10 days.
That’s how rocks get weathered, how your iron garden furniture rusts..
This is an example from Columbia University – the AGW corruption of education. No mention of rain. Either the writer is regurgitating this nonsense mindlessly, or he knows that carbon dioxide forms carbonic acid in the atmosphere and not just at surface water, and is deliberately lying.
http://www.columbia.edu/~vjd1/carbon.htm
No matter if that happens via diffusion, convection, winds, storms,ā€¦
They don’t have any of that.., no winds, no weather, no storms, because no convection. They don’t even have diffusion as their ideal gas molecules bounce each other into outer space, because, since you’re talking real world, there isn’t an invisible container around our real Earth keeping their ideal gas molecules from flying off forever as they are not subject to gravity, so your carbon dioxide measurements are irrelevant to them.
I call that well mixed.
AIRS concluded that “Carbon Dioxide was not at all well-mixed, but lumpy” – so where did you get your data contradicting them?
And, their conclusion was based on all their data, including top and bottom of troposphere which they haven’t released.

Myrrh
April 27, 2013 3:57 pm

Sincere apologies, I missed out a close italics when I split one of your paragraphs – it’s late here, time for bed.
Ferdinand Engelbeen says:
April 27, 2013 at 1:16 pm
Myrrh says:
April 27, 2013 at 12:36 pm
Myrrh, a long discours to prove what?
That the AGW Greenhouse Effect is an illusion created out of fake fisics.
In the real world, real measurements show that CO2 rapidely mixes with air and is distributed all over the world within +/- 2% of full scale, from ground level to the lower stratosphere (except for the first few hundred meters over land).”
Gosh, let’s see it..
Since Carbon Dioxide is heavier than air it will be found mainly in the first few hundred metres..
.. and of course it is being continually washed out of the atmmosphere with the rain cycle, every 8-10 days..
Ever wondered why it is a trace gas when so much of its produced? Double it it’s still a trace gas.
Here, read about real carbon dioxide:
http://hvo.wr.usgs.gov/volcanowatch/archive/2005/05_06_02.html
“Carbon dioxide (CO2) is a colorless, nearly odorless gas that is denser than air. While toxic at very high concentrations, it can kill at lower concentrations by displacing oxygen, causing asphyxiation.
..
“Usually the large amounts of carbon dioxide released by Kilauea get dispersed by winds so we can breathe nice, healthy, oxygen-rich air on the caldera floor. Because CO2 is heavier than air, it doesn’t readily rise into the atmosphere and, instead, tends to pool in low areas.”
Denser than air, means that it is heavier and so displaces air. Does not readily rise in air, because it is heavier than than air. The dust on your desk will not spontaneously diffuse into the atmosphere because it is heavier than air and does not readily rise in air. This is real gases and particles under gravity.
AGWSF’s pretend ideal gas carbon dioxide is not subject to gravity.
Real gases and real processes and real heat in conjunction with gravity is how we get our weather.
Even when some 20% of all CO2 in the atmosphere is exchanged with other reservoirs over the seasons.
The atmosphere is not a “reservoir”, carbon dioxide is heavier than air it cannot defy gravity to accumulate for hundreds and thousands of years, even if it wasn’t being washed out continually and brought back to the surface as rain.
Rain is carbon dioxide – next time it rains go stand in it, be thankful that this natural Water Cycle exists as your feel the carbon dioxide drenching you as carbonic acid.
Mildly disinfecting you..
All the AGWSF fisics is faked. They are charlatans who claim to teach the Carbon Cycle and miss out rain.
“Reservoirs” is an AGWSF meme, to confuse, to distract from this natural Carbon Cycle which in traditional real world physics tells you carbon dioxide gets washed out of the atmosphere in the Water Cycle – which residence time in the atmosphere is 8-10 days.
That’s how rocks get weathered, how your iron garden furniture rusts..
This is an example from Columbia University – the AGW corruption of education. No mention of rain. Either the writer is regurgitating this nonsense mindlessly, or he knows that carbon dioxide forms carbonic acid in the atmosphere and not just at surface water, and is deliberately lying.
http://www.columbia.edu/~vjd1/carbon.htm
No matter if that happens via diffusion, convection, winds, storms,ā€¦
They don’t have any of that.., no winds, no weather, no storms, because no convection. They don’t even have diffusion as their ideal gas molecules bounce each other into outer space, because, since you’re talking real world, there isn’t an invisible container around our real Earth keeping their ideal gas molecules from flying off forever as they are not subject to gravity, so your carbon dioxide measurements are irrelevant to them.
I call that well mixed.
AIRS concluded that “Carbon Dioxide was not at all well-mixed, but lumpy” – so where did you get your data contradicting them?
And, their conclusion was based on all their data, including top and bottom of troposphere which they haven’t released.

April 27, 2013 4:13 pm

Myrrh says:
April 27, 2013 at 1:38 pm
Myrrh, you must read everything that is written in your references, not only what you like.
About UV-vis light photons aborbed at electron level:
Several things can happen then to the absorbed energy as it may be re-emitted by the electron as radiant energy (in this case the overall effect is in fact a scattering of light), dissipated to the rest of the material (i.e. transformed into heat),…
In the case of an opaque colored material, only the color frequencies are re-emitted, the rest is dissipated as heat.
Even water is only partially transparant to radiation: from a fraction of a mm for heat waves to several hundred meters for blue light waves. But there it ends: the ocean water surface is heated mostly by visible light, less by heat waves… See for different depths reached by different frequencies sheet 6 of:
http://www.geo.arizona.edu/geo4xx/geos412/OcSci07.PhysProps.pdf

April 27, 2013 4:42 pm

Myrrh says:
April 27, 2013 at 3:57 pm
Since Carbon Dioxide is heavier than air it will be found mainly in the first few hundred metres..
In the first few hundred meters over land, you can find much higher and much lower levels, depending of nearby sources and sinks. In a forest: high levels at night (up to 600 ppmv), low levels during the day (down to 250 ppmv, a lot below the 400 ppmv in the bulk of the atmosphere). Simply because the mixing time is longer than the source or sink speed. The same for volcanic vents.
That is for 5% of all air mass where CO2 is not well mixed.
If you measure in Hawaii at sealevel or at 3,400 m height at Mauna Loa, you will find the same levels of CO2 within 1 ppmv.
Rain is carbon dioxide
Rain is water, with very little CO2 in it, even less if SO2 and NOx are present as stronger acids. Further, CO2 in rain comes from where the clouds were formed, not from near ground. Thus that CO2 needs to go up first… And where most water evaporates (equatorial oceans), CO2 goes in the atmosphere too from the deep ocean water upwelling.
AIRS concluded that ā€œCarbon Dioxide was not at all well-mixed, but lumpy
Have a look at their scale of CO2: +/- 8 ppmv worst case, +/- 2% of full scale, I call that well mixed, taken into account that 20% of all CO2 in the atmosphere moves in and out within a year. They may call that “lumpy”, just to show how good that their satellites are…

Kristian
April 27, 2013 5:33 pm

Gary, Tim (and Ferdinand),
Yawn,
Interesting to see what we’re up against. Complete blockage.
I’m done here šŸ™‚

Gary Hladik
April 27, 2013 5:38 pm

Kristian says (April 27, 2013 at 5:33 pm): “Iā€™m done here :-)”
I’m not surprised. Being proved wrong by your own analogy is tough to take. You can take this as a learning experience, or hide under a rock.

joeldshore
April 27, 2013 5:57 pm

Kristian says:

Interesting to see what weā€™re up against.

Yeah…You are up against people who actually understand some physics. Must be a bummer.

April 27, 2013 6:04 pm

Pay no attention to joelshore. He is just miserable because the planet is falsifying his religious belief system.

joeldshore
April 27, 2013 6:22 pm

Kristian,
Just for the sake of curiosity, is it your belief that what you are arguing for in this thread is in agreement with what is said in widely-used and respected introductory physics textbooks or, rather, is it your belief that the textbooks have it wrong too?

joeldshore
April 27, 2013 6:46 pm

Kristian says:

Youā€™re completely stuck on the notion that the ā€˜systemā€™ heat loss to space MUST equal the input to the sphere with or without the shell. As if the conditions in the two situations were equal. Even when itā€™s blatantly obvious (and stated several times) that with the shell in place, the steady input flux from the sphereā€™s internal heat source has more ā€˜workā€™ to do on its way to space than without the shell intercepting it. The entire flux from the surface of the sphere (J) cannot reach space, because it ā€™has toā€™ heat the shell on the way. Ergo J = J/2 + J/2, Q = Qā€™ + Qā€, 400 W/m^2 = (400 ā€“ 200) + 200.
Qā€™ stays within the system. Qā€ escapes it. There is still perfect radiative balance.

This is kind of fascinating from a PER (Physics Education Research) point of view. PER has found that students, even once they have learned Newton’s Laws, still tend to fall back on an Aristotlean viewpoint, which includes notions that a force must be supplied to keep an object moving.
Our friend Kristian has come up with an analogous view with energy: Energy must in net be supplied just to keep an object at a certain temperature. And, energy can just sort of get used up and disappear…It doesn’t have to be conserved.

gbaikie
April 27, 2013 9:02 pm

“RACookPE1978 says:
April 27, 2013 at 2:44 am
No.
You donā€™t need any physical experiment or ā€œEinsteinium thought gamesā€ to find a fundamental problem with these infinitely-continuing CHG discussions:
From above:
So sunlight is 1360 watts per square meter at top of atmosphere and by time reaches earth surface itā€™s around 1000 watts per square meter- losing about 360 watts per square meter.
For simplicity, letā€™s assume those values are correct.”
Well it is correct. There some variations- so it’s not precise.
One can get a bit more than 1000 watts per square at the surface at sea level when sun is at zenith. One will also get a lot less, if sun is not at zenith. Since sun never at zenith in Germany
it will always be less than 1000 watts per square meter in Germany. Or at 9 am it will a lot less than 1000 watt per square meter [anywhere].
“Now. If every second of every day, 360 of Trenberthā€™s 1360 watts DONā€™T penetrate a supposedly ā€œcompletely transparentā€ atmosphere, but since their energy does arrive at top-of-atmosphere but doesnā€™t arrive at bottom-of-atmosphere, then how do those 360 watts get ā€œlostā€ from the earthā€™s overall system?”
As said roughly 100 watts is absorbed by greenhouse gases, mostly H2O. Though some of it could due to water droplets rather than H2O gas
“Clearly, they ARE absorbed by the oxygen, nitrogen and argon and dust and aerosols, and ARE re-radiated by those same gasses and dusts and solids. However, just as clearly, they ARE also ignored by the CAGW community because their theory does not allow such inconsistent, inconvenient maths.”
They are absorbed by dust and aerosols- anything solid generally absorbs all wavelengths- so few feet of thick smoke or miles of very thin smoke does absorb visible wavelengths.
But oxygen, nitrogen, and argon are fairlly transparent gases to visible light [as is CO2- as is most types of gases]. But things which are transparent are also reflective- window pane can quite reflective, as can a calm lake be seen to perfectly reflect mountains when sunlight at the right angle.
So a clear atmosphere can reflect some sunlight, but it probably diffusing and scattering most of it.
If in a house with lights on and it’s dark outside, the windows will reflect the lights in the house- the windows will appear be somewhat mirror like. Of course if there a light outside then you see whatever is lit outside [and not see the reflection of inside lights]. Point is a window always has some reflection, but with the more light outside it’s overwhelms such reflections.

gbaikie
April 27, 2013 9:25 pm

” Ferdinand Engelbeen says:
April 27, 2013 at 12:27 pm
Myrrh says:
April 27, 2013 at 12:04 pm
Myrrh, it may be a long time ago that I have been involved in radiation, but no matter the wavelength of any radiation, if it is absorbed by the skin (or an object), it is felt as heat (or will heat up that object), simply because its radiation energy is transferred into vibrational energy, thus heat. Only if all (visible or invisible) wavelengths are reflected in a mirror or by absolute white objects, these donā€™t heat up.”
Reflective material or white objects to do absorb a small amount light [or sunlight] and do warm up. Reflective material also don’t emit heat [emit IR light] very well. And since they don’t absorb and don’t emit much, they can get quite warm- if in a vacuum [in atmosphere the loss of heat from convection of air not effected by how shiny something is].

Gary Hladik
April 27, 2013 11:53 pm

joeldshore says (April 27, 2013 at 6:46 pm): “Our friend Kristian has come up with an analogous view with energy: Energy must in net be supplied just to keep an object at a certain temperature. And, energy can just sort of get used up and disappearā€¦It doesnā€™t have to be conserved.”
I hoped he might see his error if the problem could be presented in more familiar terms, i.e. the reservoir/engine diagram he suggested as an analogy, but he gave up before working through it.
*sigh* “How do I reach these keedz?”

April 28, 2013 2:03 am

dbstealey says:
April 27, 2013 at 6:04 pm
Pay no attention to joelshore. He is just miserable because the planet is falsifying his religious belief system.
I do disagree with Joel Shore on the real effect of a CO2 doubling, but that doesn’t mean that I should disagree with him on every other item. In fact, the skeptic community is its own enemy and lacks credibility due to attacking those items where the science is rock solid: that CO2 levels increased due to human emissions and that an increase of CO2 in the atmosphere induces some increase in temperature, all other things being equal.
Where the real discussion is, is in the feedbacks of this initial increase in temperature. That makes a hell of a difference in the result of a CO2 doubling near the end of the current century: largely beneficial (what I expect from a temperature increase around 1Ā°C) to problematic (up to 4.5Ā°C according to the high IPCC range). Nature seems to agree with me (or more likely, nature doesn’t listen to me, simply does what it likes to do).

Myrrh
April 28, 2013 3:15 am

Ferdinand Engelbeen says:
April 27, 2013 at 4:13 pm
Myrrh says:
April 27, 2013 at 1:38 pm
Myrrh, you must read everything that is written in your references, not only what you like.
About UV-vis light photons aborbed at electron level:
Several things can happen then to the absorbed energy as it may be re-emitted by the electron as radiant energy (in this case the overall effect is in fact a scattering of light), dissipated to the rest of the material (i.e. transformed into heat),ā€¦
In the case of an opaque colored material, only the color frequencies are re-emitted, the rest is dissipated as heat.

Of course I’ve read it, but I can now spot the AGWScienceFiction’s indoctrination of fake fisics memes.
You can’t rescue your fake fisics by scrabbling around trying to find any mention of “heat”. The Sun is not a laser, real world industries make windows to maximise visible light and keep out longwave infrared from the Sun in order to keep rooms cool… So, “dissipated as heat” means what in context that visible light is not hot is not a thermal energy and can’t move whole molecules into vibration WHICH IS WHAT IT TAKES TO HEAT SOMETHING UP?
You’re not doing yourselves any favours if you keep avoiding confronting the fakery. You have lost all sense of heat, of the difference between hot and cold and what something hot does.
Because you have no grasp of properties and processes, because of the AGW meme that “all electromagnet energy is the same and all create heat when absorbed” you can’t follow my argument.
I’m trying to point out how the sleights of hand have been done by giving you information from real world traditional physics. So first of all, read again that NASA quote I gave, it contradicts the AGWSF Greenhouse Effect Energy Budget, of the KT97 and all its ilk, do you understand the enormity of what I’m saying here?
Traditional physics still teaches that the heat we feel from the Sun is the longwave infrared, the thermal infrared wavelength/photons. Traditional physics still teaches that near infrared is not thermal, it is not hot. What kind of heat do you think something that is not hot is giving to matter on being “absorbed”?
Visible light from the Sun is actually, physically real world being absorbed by the electrons of the molecules of nitrogen and oxygen in our atmosphere, which is practically all of our atmosphere, how much are these molecules being heated by absorbing visible light? Why isn’t this figure in the GreenhouseEffect energy budget?
The different wavelengths/photons/particles have different properties and processes. You ignore that this is first of all a blatant lie because the processes of sight and photosynthesis show that visible light energy doesn’t convert to heat but to electric impulses and to chemical energy of sugars in these different processes.
Do you really think that “dissipates into heat” is capable of the intensely heating the physical matter of land and water at the equator, physically raising their temperature, physically making them so hot that they heat the air above them to such a great extent that massive volumes of the hot real gas air expand and rise and move to the cold poles – the beginning of our all our great wind and weather systems?
Visible light does not work on the thermal vibrational level which is what it takes to heat matter. Visible light is too tiny to move whole molecules into vibration. This is simply a real physical fact.
Your page 6 of the link you gave – what do you think it is saying by “absorption is wavelength specific” when traditional physics teaches that water is a transparent medium for visible light because it does not absorb visible but transmits it through unchanged, because visible doesn’t even get in to play with the electrons of the molecules of water? Do you really think the scrabble for nanoscale examples of “dissipated heat” from other causes which is so totally insignificant is going to change the basic physical reality that visible light is unable to enter into molecules of water? That’s what transparent means, it doesn’t even get in to the electrons to be reflected/scattered, but is transmitted through unchanged. As in diamonds and glass.
Because this fake fisics manipulation has been successfully introduced into the general education system unless you know what traditional physics, that actually works, says about it you are going to get confused by the presentation of information skewed by these memes designed to confuse, which is now ubiquitous. These sleights of hand are now the “official” version and words like “light”, your page 6, and “absorbed” are part and parcel of this, playing with meaning to confuse.
Visible light does not get absorbed by the molecules of water, but “absorbed” is used by AGWSF to confuse to fit in with the basic fake fisics meme that “all electromagnetic radiation creates heat when absorbed”. Here, in the ocean, the general word absorbed is used to give the impression that visible light from the Sun is actually physically heating the water, but it physically can’t, visible light attenuates for a variety of reasons. It isn’t physically absorbed in the physics sense by the molecules of water.
And even when it is being physically absorbed by particles to be reflected/scattered, this isn’t physically moving the whole molecules of the particles into vibration which is what it takes to heat them up.
Look again at your page 6, the heading is “light” in which they have included heat waves, the box below is specifically about visible light and it says “near shore absorption is greater for green and yellow due to particulate matter”
There would be no life in the ocean if water absorbed visible light – no photosynthesis. The beginning of our Carbon Life Food Chain.
When you can see the tricks that are being played by this clever tweaking of real physics, then the “explanation” that “blue visible light heats the deeper water because it travels further” will make you laugh, or cry.
As I’ve said, this AGWSF fisics is now the official version for the general public, to university level and taught by all the science bodies and in encyclopaedias and so on. It is not easy to penetrate that to see how the fakery was created.
Even water is only partially transparant to radiation: from a fraction of a mm for heat waves to several hundred meters for blue light waves. But there it ends: the ocean water surface is heated mostly by visible light, less by heat wavesā€¦ See for different depths reached by different frequencies sheet 6 of:
http://www.geo.arizona.edu/geo4xx/geos412/OcSci07.PhysProps.pdf

Water absorbs heat, is an excellent absorber of heat energy. Water in the ocean absorbs the direct heat from the Sun which is the thermal infrared, the longwave infrared, the Sun’s direct radiant heat which we can feel is hot, which heats up matter, which we can feel heating us up – which has been excised from the AGWSF’s GE Energy Budget.
Visible light from the Sun is not hot, it is not a thermal energy, we cannot feel it as heat, it does not warm us up, it is not heat, but light.
Leaving aside the arguments about what visible can and cannot do for the moment, my main point is that the AGWSF’s Greenhouse Effect Energy Budget has substituted shortwave light for longwave infrared heat from the Sun and claims visible heats the land and water and longwave infrared heat doesn’t get through TOA or that the Sun produces insignificant amounts of it, as I said to DavidHoffer here: http://wattsupwiththat.com/2013/04/24/spencer-slays-with-sarcasm/#comment-1288275
You cannot ignore that this is not just me saying it, I have given a quote from a traditional physics page from NASA which says the heat we feel from the Sun is longwave infared, thermal infrared, and that we cannot feel the shortwave near infrared as thermal.
This is utterly at odds with the AGWSF energy budget. This is not me saying it, this is NASA saying it. You cannot ignore this. Either NASA is right here and you are wrong, or you are right and NASA is wrong. NASA here challenges your claim.
You, and you generic here, must deal with this.
NASA used to teach:
ā€œFar infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared.
ā€œShorter, near infrared waves are not hot at all ā€“ in fact you cannot even feel them. These shorter wavelengths are the ones used by your TVā€™s remote control.ā€
Deal with it.

April 28, 2013 4:23 am

Myrrh, I’m often mystified by your thoughts. Can I ask a simple, direct question?
If you created a huge, hundred meter diameter focusing lens with a topside coating that reflected all light bandwidths except green (say 500-550nm which pass through unimpeded) and focused all the green light of direct sunlight onto a spot 1-foot-wide and put your hand at that focus, what would happen to your hand? This would be an interesting experiment–I’d be very happy to let you go first.

Kristian
April 28, 2013 5:28 am

Gary Hladik and tjfolkerts,
You are both so fundamentally wrong on this issue and at the same time so utterly (and proudly) incapable of seeing it that it’s quite pointless to continue this ‘discussion’. So I won’t. I will only leave you with this:
Repeating the definition of ‘heat’ from Borgnakke & Sonntag (‘Fundamentals of Thermodynamics’ 2009) provided earlier on this thread:
Heat is defined as the form of energy that is transferred across the boundary of a system at a given temperature to another system (or the surroundings) at a lower temperature by virtue of the temperature difference between the two systems. That is, heat is transferred from the system at the higher temperature to the system at the lower temperature, and the heat transfer occurs solely because of the temperature difference between the two systems. (…) Heat, like work, is a form of energy transfer to or from a system.
Interesting, don’t you think? HEAT can only be transferred across the boundary of a system. It is never contained WITHIN a system.
So why are you so bent on treating the sphere and the shell as ONE thermodynamic system? When they very obviously are not. If there is a heat transfer occurring between the sphere and the shell, then by thermodynamic definition, the two constitute separate systems.
The first law of thermodynamics only applies to single (isolated) systems. So why on earth are you still insisting on the clearly flawed assumption (and frankly, flawed on an embarrassingly basic level) that one system (the shell) MUST lose as much heat to space as a different system (the sphere) gains from its heat source in order for there to be radiative balance?
No such precondition exists!
http://en.wikipedia.org/wiki/Thermodynamic_system
Of a system and its surroundings: The system is the part of the universe being studied, while the surroundings is the remainder of the universe that lies outside the boundaries of the system. It is also known as the environment, and the reservoir. Depending on the type of system, it may interact with the system by exchanging mass, energy (including heat and work), momentum, electric charge, or other conserved properties. The environment is ignored in analysis of the system, except in regards to these interactions.”
So, for the surface of the sphere, our first system, the surroundings (its reservoirs) are made up of everything that’s outside or beneath it, in other words: Everything that’s NOT the surface of the sphere.
It’s hot reservoir lies beneath (within). It is the internal heat source, its provider of heat, supplying the surface with a constant flux of 400 W/m^2.
It’s cold reservoir lies outside of it. It is made up of 1) the vacuum between it and the shell, 2) the shell, and 3) space outside the shell.
The surface of the sphere sheds a constant (outgoing) flux of 400 W/m^2 into the vacuum between it and the shell, to balance the incoming flux from the hot reservoir (Qh). This flux is J or Q (Qc).
The surface of the sphere however also gets a flux in return from its surroundings, the inward J1 from the shell. So the final expression for the heat balance between the sphere surface system and its reservoirs (surroundings) is:
J –> || J –> <– J1 || J1 –> or Qh(sphere) = Qc(sphere) = Q’ + Q” = J = (J – J1) + J1 = (400 – 200) + 200 = 400 W/m^2
There is no requirement anywhere for Q” to equal Qc(sphere) at steady state. Q” is merely part of Qc(sphere). Q” is Qc for the shell, not for the sphere.
So, what about the shell, our second system? Its surroundings are made up of … everything outside and inside of it, everything that’s not the shell.
The shell’s hot reservoir is the surface of the central sphere (ultimately, the sphere’s internal heat source).
The flux received from the sphere by the shell is 400 W/m^2. However, at steady state, the shell radiates back 200 W/m^2, so the net transfer of energy (the ‘heat’ transfer) to the shell from the sphere is (400-200=) 200 W/m^2. This is the equilibrated heat gain flux of the shell.
The cold reservoir of the shell is space outside of it. It receives a flux from the outer surface of the shell of 200 W/m^2, balancing exactly the incoming heat flux to the inner surface. The heat balance expression for the shell becomes:
(J – J1 =) J1 –> || J1 –> or Qh(shell) = Qc(shell) = Q’ = Q” = J1 = (400 – 200) = 200 W/m^2

Kristian
April 28, 2013 5:34 am

dbstealey says, April 27, 2013 at 6:04 pm:
“Pay no attention to joelshore. He is just miserable because the planet is falsifying his religious belief system.”
Heh. Yes, I’m not sure if he’s only comical or if we should somehow feel sorry for him … šŸ˜‰

Kristian
April 28, 2013 5:39 am

joeldshore says, April 27, 2013 at 6:22 pm:
“Kristian,

Just for the sake of curiosity, is it your belief that what you are arguing for in this thread is in agreement with what is said in widely-used and respected introductory physics textbooks (…)”
Yup. Of course. Not ‘belief’ though …

Kristian
April 28, 2013 5:42 am

joeldshore says, April 27, 2013 at 6:46 pm:
“This is kind of fascinating from a PER (Physics Education Research) point of view. PER has found that students, even once they have learned Newtonā€™s Laws, still tend to fall back on an Aristotlean viewpoint, which includes notions that a force must be supplied to keep an object moving.
Our friend Kristian has come up with an analogous view with energy: Energy must in net be supplied just to keep an object at a certain temperature. And, energy can just sort of get used up and disappearā€¦It doesnā€™t have to be conserved.”

Riiight …

Kristian
April 28, 2013 6:15 am

Can anyone here provide me with a single physics textbook example (or better yet, an actual account of a controlled experiment conducted in a vacuum chamber) where it is shown, or even remotely discussed, that supplied with a constant energy input, the central sphere will heat up beyond the original input energy temperature with the shell in place around it? A single one where this is even considered as a topic. Shouldn’t this phenomenon after all be a highly interesting subject to delve into, especially seen from an engineer’s point of view, giving us a tool to create incredibly high temperatures from a small original input simply by applying layers/shells of radiative insulation like that around the source. If this phenomenon is so natural and real and important, why won’t any physics or engineering textbooks let their students learn anything about it? Why are all such examples (and by that, I mean ALL) only concerned with the equilibrated temperature of the shell and the reduction of heat loss to space? Why is this adding back of energy to create even higher source temperatures only a topic in the realm of the radiative GHE?
“How many perfectly absorptive shells, separated by vacuums, must we put around a sphere (or a satellite) in space with a constant input of say 400 W/m^2 for it to reach a steady state surface temperature of say 10,000 Kelvin? And how many fewer such shells do we need if their reflectivity is say 0,9?”
Why can’t we find any such questions or problems being posed in any physics or engineering textbooks …?

Joel Shore
April 28, 2013 6:37 am

Kristian says:

Yup. Of course. Not ā€˜beliefā€™ though ā€¦

Okay, then let me ask you a few questions:
(1) Why do you think that both Robert Brown (rgbatduke) and I, two people who teach physics at the university level and who have very different views on whether AGW is a big concern, both say that you are wrong (along with a lot of other people here, of course)?
(2) Why do you think that the algebra-based physics textbook that we use here at RIT (and is used at many schools around the country) http://www.amazon.com/College-Physics-Strategic-Technology-MasteringPhysics/dp/0321815114/ have a whole page discussing the greenhouse effect and global warming in which they say, among other things: “Because it’s easier for visible radiant energy to get in than for infrared to get out, the earth is warmer than it would be without the atmosphere. The additional warming of the earth’s surface because of the atmosphere is called the greenhouse effect…Carbon dioxide is a powerful absorber of infrared radiation. Adding more carbon dioxide makes it even harder for emitted thermal radiation to escape, increasing the average temperature of the earth. The net result is global warming…Global warming is one of the most serious challenges facing scientists, engineers, and all citizens in the 21st century”?
(3) Just in case you think it is a fluke, why do you think that the calculus-based physics textbook that we use here at RIT (and is used at many schools around the country) http://www.amazon.com/University-Physics-Modern-MasteringPhysics-Package/dp/0321675460 have a discussion of the greenhouse effect and global warming in which they say among other things: “Molecules of CO2 in our atmosphere have the property that they absorb some of the infrared radiation coming upward from the surface. They then re-radiate some of the absorbed energy but some of the re-radiated energy is directed back down toward the surface instead of escaping into space. In order to maintain thermal equilibrium, the earth’s surface must compensate for this by increasing its temperature T and hence its total rate of radiating energy (which is proportional to T^4).”? It then goes on to discuss the increase in CO2 so far and the projected increase and resulting warming expected and the consequences of this, concluding with the statement, “Coping with these threats is one of the greatest challenges facing 21st-century civilization.”

tjfolkerts
April 28, 2013 7:42 am

Kristian …

HEAT can only be transferred across the boundary of a system. It is never contained WITHIN a system.

Yes, that is obvious.

So why are you so bent on treating the sphere and the shell as ONE thermodynamic system?

We aren’t. And we never have.

The first law of thermodynamics only applies to single (isolated) systems. ā€œ
Conservation of energy applies everywhere at all times. The First Law is conservation of energy applied to thermodynamics. But for this problem, it is useful to apply the First Law to the sphere and the shell separately.

So why on earth are you still insisting on the clearly flawed assumption (and frankly, flawed on an embarrassingly basic level) that one system (the shell) MUST lose as much heat to space as a different system (the sphere) gains from its heat source in order for there to be radiative balance?

We are simply assuming conservation of energy for the shell. The only loss of energy for the shell is radiation to space. The only gain of energy for the shell is thermal IR from the sphere. Since the energy of the shell is staying the same (ie the temperature is staying the same), then the loss must equal the gain. (This, of course, is after the system has come to its steady-state conditions.)

The surface of the sphere sheds a constant (outgoing) flux of 400 W/m^2 into the vacuum between it and the shell, to balance the incoming flux from the hot reservoir (Qh). This flux is J or Q (Qc).

Let’s be more careful with definitions, because frankly I am having a hard time following your nomenclature. Let’s use ā€œQā€ for ā€œheat fluxā€ = ā€œnet rate that thermal energy is being transferred across each square meter of a boundaryā€. Fortunately in our case, there are no non-thermal energy exchanges (eg no work, no chemicals moved ā€¦). Let’s use ā€œJā€ for the flux of photons, which can go both ways. (This seems to be pretty close to the way you are using the letters, and pretty close to the way they are used in many other places.)
FOR THE SPHERE
Qh(sphere) enters the sphere from below. Qc(sphere) leaves from above. I believe this would be
Qh(sphere) ā€“> || Qc(sphere) ā€“>
in your nomenclature, where clearly Qh(sphere) = Qc(sphere) , by conservation of energy for a steady-state situation.
At the outer surface of the planet, Qc(sphere) consists only photons. In your notation, we could summarize the situation at the outer surface of the planet as
| J ā€“> <ā€“ J1
We have an outward flux of ā€œJā€ and an inward flux of ā€œJ1ā€, which gives a net flux of J-J1. From this we can immediately deduce Qc(sphere) = J-J1.
So we have found your error:
Qh(sphere) = Qc(sphere) = J [Kristian]
Qh(sphere) = Qc(sphere) = J-J1 [Correct]
FOR THE SHELL
(J ā€“ J1 =) J1 ā€“> || J1 ā€“> or Qh(shell) = Qc(shell)
This I agree with. (J-J1) enters the shell from below and J1 leaves from above. (J-J1) = J1, or J1 = Ā½ J. (Assuming the shell is ā€œcloseā€ to the sphere).
Put it all together and we have
Qh(sphere) = Qc(sphere) = J-J1 = Ā½ J = J1 = Qh(shell) = Qc(shell)

The net flow is 400 W/m^2 enters the sphere and 400 W/m^2 leaves the sphere and 400 W/m^2 enters the shell and 400 W/m^2 leaves the shell. (When said that way it almost seems too obvious.) Between the sphere and shell, the flow is J = 800 W/m^2 of photons up and J1 = 400 W/m^2 of photons down = 400 W/m^2 net.
Conservation of energy and simple algebra once again lead us to the correct answer. Physics works. šŸ™‚

tjfolkerts
April 28, 2013 8:06 am

DANG! Missed a formatting tag above. About 8 lines down that should be …

The first law of thermodynamics only applies to single (isolated) systems. ā€œ

Conservation of energy applies everywhere at all times. The First Law is conservation of energy applied to thermodynamics. But for this problem, it is useful to apply the First Law to the sphere and the shell separately.

So why on earth are you still insisting on the clearly flawed assumption (and frankly, flawed on an embarrassingly basic level) that one system (the shell) MUST lose as much heat to space as a different system (the sphere) gains from its heat source in order for there to be radiative balance?

We are simply assuming conservation of energy for the shell. …

Bryan
April 28, 2013 9:40 am

Kristian
Have you read the recent post on pyrgeometers on tallblokes?
It verifies earlier points we both made there.
Its rather uncomfortable reading for backradiation warmers.
http://tallbloke.wordpress.com/2013/04/26/pyrgeometers-untangled/

joeldshore
April 28, 2013 9:41 am

Kristian says:

Can anyone here provide me with a single physics textbook example (or better yet, an actual account of a controlled experiment conducted in a vacuum chamber) where it is shown, or even remotely discussed, that supplied with a constant energy input, the central sphere will heat up beyond the original input energy temperature with the shell in place around it?

I’ll do you one better and give you two very popular introductory physics textbooks (the ones we use at RIT) that talk about the greenhouse effect and global warming in particular:
Knight, Jones, and Field, “College Physics: A Strategic Approach”, 2nd edition has a whole page discussing the greenhouse effect and global warming in which they say, among other things: “Because it’s easier for visible radiant energy to get in than for infrared to get out, the earth is warmer than it would be without the atmosphere. The additional warming of the earth’s surface because of the atmosphere is called the greenhouse effect…Carbon dioxide is a powerful absorber of infrared radiation. Adding more carbon dioxide makes it even harder for emitted thermal radiation to escape, increasing the average temperature of the earth. The net result is global warming…Global warming is one of the most serious challenges facing scientists, engineers, and all citizens in the 21st century.”
Young and Freedman, University Physics, 13th edition has a discussion of the greenhouse effect and global warming in which they say among other things: “Molecules of CO2 in our atmosphere have the property that they absorb some of the infrared radiation coming upward from the surface. They then re-radiate some of the absorbed energy but some of the re-radiated energy is directed back down toward the surface instead of escaping into space. In order to maintain thermal equilibrium, the earth’s surface must compensate for this by increasing its temperature T and hence its total rate of radiating energy (which is proportional to T^4).”? It then goes on to discuss the increase in CO2 so far and the projected increase and resulting warming expected and the consequences of this, concluding with the statement, “Coping with these threats is one of the greatest challenges facing 21st-century civilization.”
Why do you think that both Robert Brown (rgbatduke), myself, and TJ Folkerts, three people who teach physics at the university level and who have a range of views on whether AGW is a big concern, all say that you are wrong (along with a lot of other people here, of course)?

Phil.
April 28, 2013 9:52 am

RACookPE1978 says:
April 27, 2013 at 2:44 am
From above:
“So sunlight is 1360 watts per square meter at top of atmosphere and by time reaches earth surface itā€™s around 1000 watts per square meter- losing about 360 watts per square meter.
For simplicity, letā€™s assume those values are correct.”
Now. If every second of every day, 360 of Trenberthā€™s 1360 watts DONā€™T penetrate a supposedly ā€œcompletely transparentā€ atmosphere, but since their energy does arrive at top-of-atmosphere but doesnā€™t arrive at bottom-of-atmosphere, then how do those 360 watts get ā€œlostā€ from the earthā€™s overall system?

The problem is with this incorrect assertion, the 1360 watts/m^2 which arrives at the top of the atmosphere is only capable of arriving at the bottom of the ā€œcompletely transparentā€ atmosphere at a very limited location for a very limited time. This can only occur at the equator and only at the local noon, every where else and at all other times it will be less than that because of the difference in the angle of incidence at the location. So that 1360 will actually be:
1360*cos(dlat)* cos(dlon) which when integrated over the whole illuminated hemisphere accounts for the division by 4.
Clearly, they ARE absorbed by the oxygen, nitrogen and argon and dust and aerosols, and ARE re-radiated by those same gasses and dusts and solids. However, just as clearly, they ARE also ignored by the CAGW community because their theory does not allow such inconsistent, inconvenient maths.
They are not absorbed by oxygen, nitrogen and argon, except by oxygen above the troposphere in the UV, absorption by water in the near IR occurs in the troposphere, as well as elastic scattering by particulates and aerosols. Contrary to your assertion, that loss is taken into account, it’s in the Trenberth diagram for instance. 77/342 is scattered some of which is from clouds and 67/342 is absorption, which amounts to 225/1360 and 196/1360 which more than covers your 360/1360. Absorption by the water in the troposphere isn’t lost to the system since in the most part is transferred to the atmosphere as heat. Absorption by ozone and oxygen in the stratosphere doesn’t make it to the troposphere and will be lost via radiation as will elastic scattering.

Kristian
April 28, 2013 10:34 am

tjfolkerts says, April 28, 2013 at 7:42 am:
“So we have found your error:
Qh(sphere) = Qc(sphere) = J [Kristian]
Qh(sphere) = Qc(sphere) = J-J1 [Correct]”

Tim, this is not treating the sphere and the shell as two separate systems. This is treating them as one. The error is yours and yours alone. You’re ignoring the outward J1 from the shell, the other part of J. As I showed you in the very post you’re flailingly trying to refute.
You’re wrong and you know it.
Regarding that, did you read my post right above yours?
Care to answer?
And why not calculate the effect of multiple shells/layers on the sphere surface temperature while you’re at it?
Satellites often have MLIs with 20 highly reflective, vacuum separated insulating layers covering the craft. According to your logic with the back-adding of energy and the need for the heat loss from the outermost layer to space to exactly match the original input to the instruments inside, ideally, if you let the energy input run constantly (let’s say 400 W/m^2 from within to the surface of the instruments), then 4-5 such layers only would be enough to melt the device long before equilibration. With 20 such layers, the steady input to maintain the interior instrument temperature of 290K (400 W/m^2) would only need to be 3,8 x 10^-24 W/m^2! Fantastic, isn’t it?! Why do we never hear of such amazing real-world facts? And of how to utilize them?

Kristian
April 28, 2013 10:49 am

joeldshore says, April 28, 2013 at 9:41 am
That’s very much NOT what I was asking for. Read my post (all of it) again, then you might find out.
What you’re giving me here is just the same tired old arguments from assertion and assumption from the specific realm of the radiative GHE. No real-world back-up whatsoever. And theoretically, exceedingly sloppy. I sure hope you’re not teaching thermodynamics.
“Why do you think that both Robert Brown (rgbatduke), myself, and TJ Folkerts, three people who teach physics at the university level and who have a range of views on whether AGW is a big concern, all say that you are wrong (along with a lot of other people here, of course)?”
I would prefer it if you just provided me with what I’m asking for in the post you’re quoting rather than this huffing and puffing with arguments from authority.

joeldshore
April 28, 2013 11:02 am

Kristian: Let me remind you how this transpired.
I said:

Just for the sake of curiosity, is it your belief that what you are arguing for in this thread is in agreement with what is said in widely-used and respected introductory physics textbooks (ā€¦)ā€

You replied:

Yup. Of course. Not ā€˜beliefā€™ though ā€¦

Now that I have provided conclusive evidence that your “not ‘belief'” was completely wrong, you’ve decided these are “the same tired old arguments from assertion” or “arguments from authority”.
So, basically, you won’t accept our attempts to explain to you the basic concept of conservation of energy and you won’t accept authorities on the subject even though you claimed to be quite sure that the authorities would agree with you. Clearly, you just want to wallow in your own ignorance. Have fun!

Kristian
April 28, 2013 11:04 am

Bryan says, April 28, 2013 at 9:40 am:
“Kristian
Have you read the recent post on pyrgeometers on tallblokes?
It verifies earlier points we both made there.
Its rather uncomfortable reading for backradiation warmers.
http://tallbloke.wordpress.com/2013/04/26/pyrgeometers-untangled/

Yes, I have. Funny how these old ‘truths’ that have been allowed to proliferate without questioning just proves plain wrong and silly as soon as someone simply care to try and find out what actual science they rest upon.
Problem is, most of the time, the proponents need to find out for themselves. They can’t simply be told. As we both know …

April 28, 2013 11:40 am

Myrrh says:
April 28, 2013 at 3:15 am
Myrrh,
Only one question: as no visible light reaches the ocean depths below a few hundred meters, what happens with the energy contained in that light between zero and 200 meter depth were it all is absorbed by water and/or suspended matter?

tjfolkerts
April 28, 2013 11:51 am

Kristian says: “I would prefer it if you just provided me with what Iā€™m asking …”
The problem, Kristian, is that even when I *do* provide exactly what you are asking for and walk you through the steps, you invent objections like “you are treating it as one system” even though I was following exactly your arguments (often evening using your notation).
Point to ONE SPECIFIC equation that you object to in my post. Explain why you think it is wrong.

Bryan
April 28, 2013 12:47 pm

Kristian
Joel Shore says
“Young and Freedman, University Physics, 13th edition has a discussion of the greenhouse effect and global warming in which they say among other things: ā€œMolecules of CO2 in our atmosphere have the property that they absorb some of the infrared radiation coming upward from the surface.”
Interestingly Joel and i had an exchange about this about two years ago at WUWT.
I have the 9th edition of the same book (1996).
No mention of the Greenhouse Effect.
Apparently this theory is very new or has been included to sell books to Climate ‘Science’ students.
During the posts then several people (from all sides) pointed out that the textbook had made several mistakes in the Greenhouse entries
It was a real ‘cut and paste job’ which ruined an otherwise good textbook..
It is surely shameful that a good textbook is ruined by including passages of second rate pseudoscience to suit the AGW faith inquisition

April 28, 2013 1:18 pm

Ferdinand Engelbeen says:
“I do disagree with Joel Shore on the real effect of a CO2 doubling, but that doesnā€™t mean that I should disagree with him on every other item. In fact, the skeptic community is its own enemy and lacks credibility due to attacking those items where the science is rock solid: that CO2 levels increased due to human emissions and that an increase of CO2 in the atmosphere induces some increase in temperature, all other things being equal.”
Ferdinand, I agree completely with what you wrote here.
ā€¢ CO2 causes some warming. We do not know how much, but it is too small to be measured independent of natural warming
ā€¢ The excess CO2 in the atmosphere is the result of fossil fuel burning
ā€¢ About 4 ppm per year is added to the atmosphere, and 2 ppm is retained; the rest is absorbed by the biosphere
ā€¢ Excess CO2 is also emitted due to natural global warming. There is plenty of empirical evidence proving that fact
Regarding Joel Shore, his physics is fine ā€” as far as it goes. But obviously there is something missing, something he is not aware of, because after a ā‰ˆ40% rise in CO2, global temperatures have stopped rising for the past decade and a half. The universal prediction was that rising CO2 would cause global warming. In the ’90’s there were endless predictions stating that rising CO2 would trigger runaway global warming. Skeptics were scoffed at.
But those predictions were completely wrong. All of them. Therefore, something major is missing from the analysis. I think there is a different effect between radiative physics in a closed laboratory container, and radiative physics in a convecting atmosphere. There are also negative feedbacks, both known and unknown. Clouds are not properly accounted for, etc.
When a conjecture such as rising CO2 = runaway global warming is found to be totally wrong, per empirical observations over many years, honest scientists will admit they got it wrong and try to figure out why. But many mainstream climate scientists still try to defend their failed predictions instead of working with skeptical scientists to find out where they went wrong.
Finally, there is no scientific evidence demonstrating that rising CO2 is harmful, but there is a lot of real world evidence showing that the added CO2 is beneficial. But when the government hands out more than $100 BILLION in federal grants since 2001, plenty of scientists get the message: demonize “carbon”! Because no one will get much of that money if they tell the truth: that at current and projected concentrations, CO2 is globally harmless, and it is beneficial to the biosphere. More is very likely better.
I am always ready to change my mind. But if I do, it will be based on empirical measurements, not on the current narrative. It is a real shame that alarmist scientists like Michael Mann and lots of others refuse to acknowledge that rising CO2 has not had the predicted effect, and therefore their conjecture has been falsified.

tjfolkerts
April 28, 2013 1:42 pm

Kristian says: “Satellites often have MLIs with 20 highly reflective, vacuum separated insulating layers covering the craft.”
And why do they do that? Exactly for the reason we are talking about! The heat loss is decreased dramatically!
“With 20 such layers, the steady input to maintain the interior instrument temperature of 290K (400 W/m^2) would only need to be 3,8 x 10^-24 W/m^2!”
Wherever you got “3,8 x 10^-24 W/m^2”, it is not even close to right. For blackbody shells
1 layer = 1/2 as much heat loss
2 layers = 1/3 as much heat loss (not 1/4)
3 layers = 1/4 as much heat loss (not 1/8)

20 layers = 1/21 as much heat (not 1/ 2^21) So 20 blackbody shells would result in about 5% as much heat loss.
Of course, no one would use black body surface for such a thing, since reflective coatings are MUCH more efficient. 20 layers of highly reflecting shells (emissivity = 0.01) would lower the heat output by a factor of ~ 0.00025, or ~ 0.1 W/m^2. That is certainly quite an improvement (about 100x better than the black body shells) , but now where near the 0.00000000000000000000000001 times better that you claim.
(Real efficiency would be a little less, but still quite impressive for such thin, light-weight insulation.)
Details at http://en.wikipedia.org/wiki/Multi-layer_insulation

April 28, 2013 1:43 pm

Kristian says:
April 28, 2013 at 5:28 am
The surface of the sphere however also gets a flux in return from its surroundings, the inward J1 from the shell. So the final expression for the heat balance between the sphere surface system and its reservoirs (surroundings) is:
J ā€“> || J ā€“> or Qh(sphere) = Qc(sphere) = Qā€™ + Qā€ = J = (J ā€“ J1) + J1 = (400 ā€“ 200) + 200 = 400 W/m^2

Kristian,
I am just trying to follow your reasoning, but I think something goes wrong in the first part of the equation where you insist that the outgoing radiation of the sphere = J.
The incoming energy for the sphere is J + J1. The outgoing energy is J (=Qc, or call it Jsphere). Thus the sphere is not in energy balance. That means that the sphere must heat up. With as result an increase of the outgoing radiation from the sphere to the shell, until Jsphere (Qc) = J + J1. But that also increases J1, as the shell receives more energy from Jsphere, thus the shell heats up too. At the moment that J1 = 400 W/m^2 (and thus Jsphere = 800 W/m^2), every part of the system (sphere, shell) and the whole system (from heat source within the sphere to space) is in steady state equilibrium.

joeldshore
April 28, 2013 1:46 pm

Kristian says:

With 20 such layers, the steady input to maintain the interior instrument temperature of 290K (400 W/m^2) would only need to be 3,8 x 10^-24 W/m^2!

Not sure how you got your result…But, you just motivated me to work out the formula one finds using the correct laws of radiative physics and conservation of energy for N opaque shells having a reflectance R and an absorbance [and emittance] (1-R) surrounding a perfect blackbody emitter (and neglecting issues of differences in radii, i.e., assuming the radii of the shells are essentially the same as the radius of the object at the center) . [Yes…I guess I really don’t have anything better to do this afternoon!]
The result is that the temperature T_0 of the blackbody object at the center and the temperature T_N of the outer shell are related by
T_0^4 = [2*N/(1-R) – (N-1)] * (T_N)^4.
Hence, the input sigma*T_N^4 necessary to maintain a temperature of T_0 = 290 K for N = 20 would be ~19 W/m^2 for 20 blackbody (R = 0) shells and ~0.512 W/m^2 for 20 shells with a very good reflectance R = 0.95. Even if you had 20 shells with a really fantastic reflectance of R = 0.99, the necessary input would still be ~0.100 W/m^2.
So, your calculation is not even close to correct.

joeldshore
April 28, 2013 1:58 pm

tjfolkert says:

20 layers of highly reflecting shells (emissivity = 0.01) would lower the heat output by a factor of ~ 0.00025, or ~ 0.1 W/m^2. That is certainly quite an improvement (about 100x better than the black body shells) , but now where near the 0.00000000000000000000000001 times better that you claim.
(Real efficiency would be a little less, but still quite impressive for such thin, light-weight insulation.)
Details at http://en.wikipedia.org/wiki/Multi-layer_insulation

Damn…I just worked my butt off to calculate this and I could have looked up the answer on Wikipedia!?! Well, at least I am glad to see that they got the same answer as I did.
Bryan says:

I have the 9th edition of the same book (1996).
No mention of the Greenhouse Effect.

It was just added in the 13th edition.

Apparently this theory is very new or has been included to sell books to Climate ā€˜Scienceā€™ students.

It is not that theory is very new but that it sometimes takes physics textbooks a while to get around to including practical physics applications from other fields.

During the posts then several people (from all sides) pointed out that the textbook had made several mistakes in the Greenhouse entries
It was a real ā€˜cut and paste jobā€™ which ruined an otherwise good textbook..
It is surely shameful that a good textbook is ruined by including passages of second rate pseudoscience to suit the AGW faith inquisition

You have the perfect unfalsifiable religion: If a respected physics authority disagrees with your notions of physics, clearly that authority has been corrupted and can no longer be trusted! Have fun believing your garbage-science.

April 28, 2013 2:06 pm

Ferdinand Engelbeen says:
April 28, 2013 at 1:43 pm
Kristian,
The html language deleted part of the equation that I copied from your message, but I suppose that you already knew that…

CEH
April 28, 2013 2:07 pm

joeldshore says:
April 25, 2013 at 12:29 pm
“The reason that we are not so calm and kind is we have been round and round on this beforeave been round and round on this before
(e.g., a couple years ago on Judith Curryā€™s blog) and I
really think at this point, you should know better.
You either have one heck of a monstrous mental block or you really do understand but have,
for some bizarre reason, decided that you would rather engage in sophistry.
Either way, it is not a pretty sight.”
Sir, itĀ“s your comment that is not a pretty sight.
The fact that you “have been round and round on this before” does not give you the right to answer with denigrating comments.
Besides, instead of putting up strawmen and attacking them, why donĀ“t you go to his site and read the post
“The Fraud of the Atmospheric Greenhouse Effect Part 2: Moving to Reality”.
Anthony: Do you employ different mod standards depending upon whom is being denigrated?
Period.

tjfolkerts
April 28, 2013 2:13 pm

HIGH FIVE to Joel! šŸ™‚
We each independently came up with the same results for the heat flow for multilayer shells with adjustable emissivity (although I admit I simply went to Wikipedia and used their result for emissivities other than 1, rather than re-deriving it myself).

joeldshore
April 28, 2013 2:17 pm

dbstealey says:

I am always ready to change my mind. But if I do, it will be based on empirical measurements, not on the current narrative.

That was good for a laugh! Do you really have that little self-awareness as to be deluded into thinking that your conclusions are based on empirical measurements? No they are not…They are dictated by your ideology, which is why they are in such stark contrast to the conclusions of respected scientific authorities like the National Academy of Sciences, which are able to more objectively look at the empirical evidence.

joeldshore
April 28, 2013 2:27 pm

CEH says:

Besides, instead of putting up strawmen and attacking them, why donĀ“t you go to his site and read the post
ā€œThe Fraud of the Atmospheric Greenhouse Effect Part 2: Moving to Realityā€.

What strawman have I put up? Postma clearly says that the correct temperature no-greenhouse temperature to compare to is +49degC. Where is the strawman? I use that temperature and show that assuming an average temperature of that on the Earth does not even come close to giving radiative balance. If you use the accepted no-greenhouse temperature of -15degC, you do get radiative balance.
All his day-side, night-side stuff is just a distraction. Radiative balance only constrains the average surface temperature…It does not have much to say about diurnal or spatial temperature distributions, which are determined by other issues (thermal inertia, length of the day, …)

tjfolkerts
April 28, 2013 2:57 pm

CEH says: “why donĀ“t you go to his site and read the post
ā€œThe Fraud of the Atmospheric Greenhouse Effect Part 2: Moving to Realityā€.

Joe Postma makes numerous basic physics and mathematics errors on his blog. In that particular post, there are two primary errors — one rather obvious and one rather subtle.
OBVIOUS: He correctly deduces that sunlight spread out over the entire sphere would be 240 W/m^2, or an effective blackbody temperature of 255 K = -18 C. This result is widely quoted and widely known. He then tries to spread out the sunlight over only 1/2 of the sphere. But instead of simply doubling the intensity of the sunlight to 480 W/m^2 (30 C), he increases in by a factor of 4/pi to 611 W/m^2 (49 C). Basically, he is calculating the results only for the warmest location — the equator — rather than the whole sphere.
SUBTLE: His results require that the sunny side of the sphere has a low heat capacity so that the planet will warm up quickly toward the daytime equilibrium temperatures he calculates. But then he requires that the night side has a large heat capacity so that the planet cools slowly. This is not easy to show without much more detailed calculations, so it is easy to gloss over. Unless you are willing to do a detailed calculation (ie either an actual integration or a numerical integration using some reasonable estimates for heat capacity), this error will not be easy to spot.
(It is interesting to note that this error is even hinted at in his drawing. The night side gets cooler from dusk to dawn, but the day side does not get warmer from dawn to dusk. Right at “dawn” the temperature is implied to jump up to a warm value immediately — the same temperature it is at dusk.)

Bryan
April 28, 2013 3:04 pm

Joel Shore says
“You have the perfect unfalsifiable religion:”
Which is a strange claim for an advocate of the theory of catastrophic consequences for mankind if more CO2 enters the atmosphere.
Indeed the ‘science’ behind AGW theory is quite happy to accept the theory predicts colder weather,warmer weather,wetter weather,dryer weather and so on .
Nothing can contradict the IPCC religion.
This was the very reason that Ivar Giaever ( Nobel Prize in Physics) cited when he resigned from the APS.
He said that the IPCC promoted pseudoscience because it could not be tested within the framework of science.
Perhaps Archbishop Joel could provide us with a test that would falsify the theory of the CO2 driven Greenhouse Effect.
I bet he cant!

April 28, 2013 3:18 pm

Bryan:
re your post at April 28, 2013 at 3:04 pm .
I make this suggestion as a sincere kindness.
Please, for your own benefit, apply some introspection.
Yes, Joel Shore is a deluded fanatical AGW proponent.
But that does not justify you being a deluded AGW opponent.
The radiative physics is clear, and Shore is right about that while you are wrong.
The empirical data is also clear and Shore is wrong about that while you are right.
The empirical data shows the feedbacks in the climate system are negative so AGW cannot be a sufficiently large effect for it to be discernible from natural climate variability. Simply, AGW is not a problem and it cannot be a problem, but that is because the climate system is as it is. It is not because radiative physics is fundamentally wrong.
I commend you to read my above post which you can jump to by use of this link
http://wattsupwiththat.com/2013/04/24/spencer-slays-with-sarcasm/#comment-1286437
Richard

April 28, 2013 4:16 pm

joelshore says:
“…They are dictated by your ideology, which is why they are in such stark contrast to the conclusions of respected scientific authorities like the National Academy of Sciences…”
Ah. Yet another example of psychological projection by Joel Shore, with a dollop of Appeal to Authority.
The central fact remains: Planet Earth is mocking Joel Shore’s religious belief system: global temperatures continue to decline, even while CO2 continues to rise. Shore’s belief system is being falsified every day, and his response is to accuse truth-tellers of being ‘ideological’. As if.
The fact is that Lefitst ideology totally rules Joel Shore. Everything is political to him, so when the science proves he is wrong, his response is to turn the debate political. Sadly, he has a lot of company in his asinine beliefs. But the truth is coming out, proclaimed by the ultimate Authority ā€” Planet Earth.
Who ya gonna believe? Planet Earth? Or Joel Shore?

Onar ƅm
April 28, 2013 4:35 pm

I have actually had the pleasure of following the debate about the cooling effects of CO2 on the climate sceptics mailing list for many years, long before it ever became an issue in the popular press. Due to the long arguments pro and con there (including my own contributions) I can perhaps add a useful perspective to the debate.
It all started by someone asking the question: “how warm would a planet be WITH an atmosphere, but WITHOUT greenhouse gases?” (I think even Spencer has touched upon this question). I found a very surprising result. Radiative physics teaches us that without greenhouse gases the *surface* temperature of the earth would approach something close to a black body. However, the *atmosphere* would be much much warmer than today! Here is how the mechanism works:
1. During the day the hot solar heated surface would heat the air above through conduction. This would cause convection, transporting massive amounts of heat up into the atmosphere. In a greenhouse world this would have been radiated away into space, but without greenhouse gases there is very little radiation, and so the heat becomes trapped in the atmosphere.
2. During the night the radiating surface would cool the air above through conduction. This would NOT cause conduction because cool air does not rise. Therefore the cooling at night can only be transported upwards into the atmosphere through layer to layer conduction, which obviously is muc slower than convective heating.
3. The asymmetry between night time cooling and daytime heating would cause the atmosphere to heat dramatically, even though the average temperature on the surface would be that of a black body.
CONCLUSION: greenhouse gases warm the earth by cooling the atmosphere. The greenhouse gases cause the total amount of heat energy stored in the atmosphere to FALL, and radically changes the temperature profile of the atmosphere. Thus, we can think of greenhouse gases acting like a kind of heat engine, transporting heat from the top of atmosphere down to the surface. In this process of heating the surface, there is net energy loss to space (through radiation).

tjfolkerts
April 28, 2013 5:19 pm

Onar ƅm, I question your conclusions about an atmosphere without any GHGs (although at some lever this is a moot point since ALL known atmospheres in the solar system do indeed have radiative gases. There would also have to be no aerosols and no clouds)
1) Convection must be symmetric in the sense that rising gas in one place must be offset by falling gas in another. So for every place that warm ground is warming the cool atmosphere and causing it to rise, there is another place where the cool ground is cooling the warm descending air.
2) Convection tends to stop when the actual lapse rate gets smaller than the adiabatic lapse rate. If the atmosphere was “warm all the way up”, then convection would stop near the surface. This suggests that convection will not warm the upper reaches of the atmosphere. Only the very inefficient conductive warming could do this.
Throw in rotation and we have a very messy fluid dynamics problem. I don’t know the answer, but I suspect it is not as simple as the whole atmosphere being as warm as the warmest place on the surface (as I have heard people conclude). It is fun to think about though.
*************************************
I do agree with your main conclusion though …
“greenhouse gases warm the earth by cooling the atmosphere [in the upper troposphere in particular]. ” šŸ™‚

joeldshore
April 28, 2013 5:37 pm

Onar ƅm: I agree that there are lots of strange things about the situation of absolutely no greenhouse gases that make it a sort of strange and singular limit, in many ways more complicated to think about than when you do have them! (There are also practical questions of how you remove all greenhouse gases since this removes clouds and that, plus the temperature changes, causes changes in albedo…and so on and so forth.)
However, these complications have no real bearing on the question of what the effect is of increases in greenhouse gases from our current state (despite the views of people like Konrad to the contrary).

joeldshore
April 28, 2013 6:10 pm

Well, Richard S Courtney and I may disagree about just about everything (including what the empirical data show), but I agree with him that the question of what the feedbacks are and what the climate sensitivity is are at least the questions where there is still some room for legitimate scientific disagreements.
The arguments about the basic radiative physics (or…and here I know that he will disagree…the clear fact that the rise in CO2 is anthropogenic) just serve to make AGW skeptics look even worse to the scientific community as a whole than they already do, which is why some like Spencer try to hard to combat this blatant incorrect physics, to little avail with those committed to believing nonsense.

Nullius in Verba
April 28, 2013 11:09 pm

Onar ƅm,
It’s an interesting thought experiment, that I’ve given some thought to myself.
It’s essentially the situation in the oceans, only inverted. Water near the poles cools, and sinks, which drives the convection cycle. Because there’s no way for the deep water to radiate to space, and it’s convectively stable with warm on top of cold, the deep water remains uniformly at the coldest surface temperature, until it rises to the surface where contact with the surface warms it. (It’s a bit more efficient because water is transparent,) The warm water then rises and flows across the surface back to the poles.
Likewise in a GHG-free atmosphere, the air would rise at the equator into a relatively warm atmosphere of nearly constant potential temperature, driving the convection cycle. It remains warm until it descends to the polar surface, where contact cools it, and it flows across the surface back to the equator.
The difference is that air is far more compressible. The adiabatic lapse rate in the oceans is about 0.1 C/km, and indeed the oceans warm about 0.4 C as you descend 4 km. (You could call it a sort of deep-ocean greenhouse effect. After all, liquid water is a ‘greenhouse’ material!) The adiabatic lapse rate in dry air would be around 10 C/km, and so the air would cool with altitude much as it does now. And warm as it descends, but only by as much as it cooled. The surface would be much colder on average, and less uniform equator-to-pole, and the atmosphere would be cold at altitude, more uniform equator-to-pole, but a lot warmer than you would expect given the coldness of the surface below it.
It’s an interesting thought experiment.

Gary Hladik
April 28, 2013 11:19 pm

Kristian says (April 28, 2013 at 6:15 am): “Can anyone here provide me with a single physics textbook example (or better yet, an actual account of a controlled experiment conducted in a vacuum chamber) where it is shown, or even remotely discussed, that supplied with a constant energy input, the central sphere will heat up beyond the original input energy temperature with the shell in place around it?”
Such examples seem to be few and far between, but after a short search I found one that seems to come close (problem 7.12 on page 245):
http://books.google.com/books?id=J2KZq0e4lCIC&q=sphere+shield#v=snippet&q=sphere%20shield&f=false
The low emissivity of the shield would seem to make this more of a reflection than an emission problem, and I’m not clear how Kristian views the ability of a reflection to heat its source.
Unfortunately the problem analysis is not part of the preview and I don’t have the book (it’s a bit pricey). So if anyone still reading this thread (yes, both of you) has the book–or an earlier edition?–perhaps you can tell us how it all comes out. šŸ™‚

Bryan
April 29, 2013 12:50 am

richardscourtney says at April 28, 2013 at 3:18 pm
Bryan re your post at April 28, 2013 at 3:04 pm .
“Yes, Joel Shore is a deluded fanatical AGW proponent.”
…………I agree
“But that does not justify you being a deluded AGW opponent”.
“The empirical data is also clear and Shore is wrong about that while you are right.”
Richard my post that you refer to was largely about the views of Ivar Giaever ( Nobel Prize in Physics).
He said that Climate ‘Science’ as accepted by the IPCC was pseudoscience because it could not be falsified.
Do you disagree with him and if so why?
I also challenged Joel Shore to give some test that could falsify the theory he advocates.
It come as no surprise to me that he cannot.
I did not give any opinion about radiative physics in this thread so its a puzzle to me as to where you find some disagreement.
My own opinion is that the radiative properties of CO2 while important at furnace temperatures are negligible at atmospheric temperatures.
This is in line with the findings of R.W. Wood and are not very different from your own views.

paulinuk
April 29, 2013 12:53 am

If the air consisted solely of non- GHG and couldn’t radiate heat, the surface would get to the average blackbody temperature of an airless planet and at the boundary of outer space would at the average BB temperature of space.
If however non-GHG could radiate heat, then the downwelling IR from the air would add to the downwelling IR from the star and raise the surface temperature until such time that the outgoing radiation balances the incoming radiation, the temperature being determined by the stephan-boltzman equation.
Can someone contact the National Radio Astronomy Observatory and tell them where they’ve gone wrong. They do actually measure real things out there in the universe.
From
http://www.nrao.edu/index.php/learn/radioastronomy/radiowaves#blackbody
“Thermal Emission
Blackbody Radiation
Thermal emission is perhaps the most basic form of emission for EM radiation. Any object or particle that has a temperature above absolute zero emits thermal radiation. The temperature of the object causes the atoms and molecules within the object to move around. For example, the molecules of a gas, as in a planet’s atmosphere, spin around and bump into one another. When the molecules bump into each other, they change direction. A change in direction is equivalent to acceleration. As stated above, when charged particles accelerate, they emit electromagnetic radiation. So each time a molecule changes direction, it emits radiation across the spectrum, just not equally. As a result, the amount of motion within an object is directly related to its temperature.”

April 29, 2013 1:30 am

Bryan:
At April 29, 2013 at 12:50 am you say to and ask me.

Richard my post that you refer to was largely about the views of Ivar Giaever ( Nobel Prize in Physics).
He said that Climate ā€˜Scienceā€™ as accepted by the IPCC was pseudoscience because it could not be falsified.
Do you disagree with him and if so why?

Hmmm. “Largely about”? I answered its part which I understood concerned the discussion in this thread.
I would need to know what Ivar Giaever specifically said before I would be willing to agree or dispute any individual statement in it.
Much IPCC so-called science is falsifiable; e.g. total loss of Himalayan glaciers within 35 years. But the reliance of much IPCC so-called science on not-validated computer models is pure pseudoscience. None of the computer models has any demonstrated forecasting skill but they are being used to predict and project the future: this is so wrong that it makes astrology look good.
However, this modelling activity is falsifiable and it has been falsified in several ways because its predictions have been observed to NOT emulate reality; e.g. the tropospheric ‘hot spot’ is missing, and the “committed warming” has vanished, and etc..
The problem is that each time the models are falsified the AGW-advocates ‘move the goal posts’, the problem is not that the models are inherently not falsifiable. In other words, we are confronted with pseudoscientists conducting pseudoscience.
Dealing with the pseudoscientists can only be accomplished by defeating them with science. It cannot be accomplished by combating them with alternative pseudoscience.
AGW-is bunkum. It takes a small and trivial piece of scientific information (i.e. the existence of the radiative GHE) and exaggerates it into a problem which does not – and cannot – exist in reality then uses politics to impose harmful policies in response to the non-problem.
Simply, the AGW-hypothesis is pure Lysenkoism. Like the original activities of Trofim Lysenko, AGW can only be overcome by championing adherence to the realities of true science and the scientific method.
Richard

Myrrh
April 29, 2013 2:26 am

Ferdinand Engelbeen says:
April 27, 2013 at 4:42 pm
Myrrh says:
April 27, 2013 at 3:57 pm
“Since Carbon Dioxide is heavier than air it will be found mainly in the first few hundred metres..”
In the first few hundred meters over land, you can find much higher and much lower levels, depending of nearby sources and sinks. In a forest: high levels at night (up to 600 ppmv), low levels during the day (down to 250 ppmv, a lot below the 400 ppmv in the bulk of the atmosphere). Simply because the mixing time is longer than the source or sink speed. The same for volcanic vents.
That is for 5% of all air mass where CO2 is not well mixed.

Ferdinand – that’s the whole point about carbon dioxide, it’s heavier than air because it has mass and therefore subject to gravity and therefore has weight because gravity gives weight, because it is a real gas and not the pretend ideal gas of AGW, it will be predominantly near the surface in higher concentrations – where it is being produced, and, recycled by flora, on land and in the ocean. Differing figures, but around 70% of all the oxygen produced is said to be by photosynthesis in the ocean. Plants when they are not involving themselves in photosynthesis breathe in oxygen and breathe carbon dioxide, just like us..
It remains a trace gas. Double it, it is still a trace gas.
What world do they live in who think a trace gas can act like a thermal blanket?
If you measure in Hawaii at sealevel or at 3,400 m height at Mauna Loa, you will find the same levels of CO2 within 1 ppmv.
Sorry, I just don’t believe those figures.
Hawaii is one of the world’s greatest producers of carbon dioxide! A major hot spot in a warm ocean actively creating volcanic islands. Active volcanos constantly venting, thousands of earthquakes a year, and, Mauna Loa is the world’s biggest active volcano, and, that’s besides the lush vegetation in and out the ocean breathing out the stuff. The station of Mauna Loa is slap bang in the middle this great carbon dioxide production.
Do you know how they measure there? Completely arbitrarily! They decide what is and is not from the great volcanic production – it’s not possible to tell that. It’s ludicrous, it is completely unscientific.* And Hawaii has had a huge increase in air traffic over the last decades, day and night.
If anyone really thinks Keeling went to Hawaii from Antarctica because he thought “background carbon dioxide levels could be measured anywhere in the world and they would be the same”..
“Rain is carbon dioxide”
“Rain is water, with very little CO2 in it, even less if SO2 and NOx are present as stronger acids. Further, CO2 in rain comes from where the clouds were formed, not from near ground. Thus that CO2 needs to go up firstā€¦ And where most water evaporates (equatorial oceans), CO2 goes in the atmosphere too from the deep ocean water upwelling.”
What? All rain is carbonic acid. Of course it comes from where the clouds are formed, that’s why it takes out all and any of the carbon dioxide in the atmosphere with it…
..the residence time of water in the atmosphere is 8-10 days, carbon dioxide is fully part of that Water Cycle.
Globally, every time it rains it rains carbonic acid, that is carbon dioxide being recycled every 8-10 days in the Water Cycle.
AGWSF’s Greenhouse Effect doesn’t have the Water Cycle. They have excised it from their models.
Well, actually, with their ideal gases they can’t have it anyway.
Real gases have attraction, water and carbon dioxide are irresistably attracted to each other.
These from a wiki page – http://creationwiki.org/Carbonic_acid
“Carbonic acid’s ancient names were ‘acid of the air’ and ‘aerial acid'”
“Acid rain or normal rain has dissolved carbonic acid in it has a pH of 4.5 to 5.6.”
Note, normal natural clean rain which is carbonic acid is never called acid rain in science.
That small piece goes on to say that: “The carbonic acid in the ocean is increasing, causing many animals, like plankton, to die and is corroding to shells and coral. The acid rain falls and drains into the ocean making it have more acid than it has had previously. The ocean is at a pH of 8.1, which is about 0.1 lower than at the beginning of the Industrial Revolution.”
This is typical AGWSF meme scaremongering by twisting science.
A pH of 8.1 would not be made more acidic but less alkaline. That’s besides the blatant lies about causing plankton to die and corroding shells and corals. Plankton is also plant life, algae, which feed on carbon dioxide, and the animal life which feed on plankton would also benefit.
A pH of 8.1 would not be made more acidic but less alkaline – it would have to go through neutral first, which is pH 7, before it became acidic. And, each whole number up and down the scale is logarithmic, a pH of 9 is ten times more alkaline than a pH of 8, so their “0.1 lower than at the beginning of Industrial Revolution” is brainwashing scary for the unwary.
So much misinformation in such a small piece, composed of AGWSF fake fisics memes.
“AIRS concluded that ā€œCarbon Dioxide was not at all well-mixed, but lumpy”
Have a look at their scale of CO2: +/- 8 ppmv worst case, +/- 2% of full scale, I call that well mixed, taken into account that 20% of all CO2 in the atmosphere moves in and out within a year. They may call that ā€œlumpyā€, just to show how good that their satellites areā€¦
Nope, what you are looking at is a cherry picked bite from mid troposphere, they haven’t released any of the upper and lower troposphere data, not any of it.
Their conclusion is based on all their data, the complete picture from top to bottom of troposphere.
Of course carbon dioxide will be shown to be lumpy and particularly at the bottom of the troposphere because that is where it is produced, and where it sinks down because it is a real gas and heavier than air.
This is just the same ol’ same ol’ manipulation of real data, our temperature records have been totally screwed by them – even claimed to be “lost”. You take these on trust mistakenly, is my opinion.
*Well worth reading from Timothy Casey re volcanic production: http://carbon-budget.geologist-1011.net/
“1.2 The Location of CO2 Monitoring Station in regions enriched by volcanic CO2”
One of the conclusions from the AIRS team was that they would have to go away and look at wind systems..
Which reminds me too, that when I checked out the Mauna Loa claim that it is a “pristine measuring site uncontaminated by local production”, supposedly capturing clean unpolluted “background” from trade winds, I think I found one of the winds coming in their direction which made for great wind surfing was coming from China and not from the opposite direction, I can’t be sure because I didn’t have the time then to really check if I got it right, and then time passed..
Anyway, the AIRS team were truly shocked to find carbon dioxide was not well mixed, because they had been educated in the AGW memes. They were astonished to find it was lumpy and not well mixed at all.

Myrrh
April 29, 2013 3:46 am

kencoffman (@kencoffman) says:
April 28, 2013 at 4:23 am
Myrrh, Iā€™m often mystified by your thoughts. Can I ask a simple, direct question?
If you created a huge, hundred meter diameter focusing lens with a topside coating that reflected all light bandwidths except green (say 500-550nm which pass through unimpeded) and focused all the green light of direct sunlight onto a spot 1-foot-wide and put your hand at that focus, what would happen to your hand? This would be an interesting experimentā€“Iā€™d be very happy to let you go first.

Well, I’m not mystified by your response..
Visible light from the Sun is not a thermal energy, it cannot heat matter. It works on the tiny electronic transition level not on the big molecular vibrational level which is what it takes to move matter into heat.
We cannot feel visible light from the Sun, it is not hot. What we feel as heat from the Sun is longwave infrared which is thermal infrared; of heat infrared and not of reflective light infrared, which is the not hot shortwaves of infrared.
See the NASA quote I posted from traditional physics.
The AGWScienceFiction’s GreenhouseEffect Illusion’s Energy Budget has taken out the direct real heat from the Sun, which is the Sun’s thermal energy in transfer by radiation, and claims that non-thermal shortwaves do the heating.
Idiocy by the standard of traditional physics.

April 29, 2013 4:00 am

Myrrh says:
April 29, 2013 at 2:26 am
Myrrh, it is hard to discuss with somebody who reject real world facts.
– The CO2 levels and trends at Cape Kumukahi near sealevel and at Mauna Loa at 3,400 m height are near the same. See:
http://www.esrl.noaa.gov/gmd/dv/iadv/graph.php?code=KUM&program=ccgg&type=ts
and
http://www.esrl.noaa.gov/gmd/dv/iadv/graph.php?code=MLO&program=ccgg&type=ts
Indeed the Mauna Loa data are “cleaned” from outliers with a simple algorithm: when there is downwind air from the volcanic vents, the variability within an hour is larger than 0.25 ppmv. These data are marked and not used for averaging. But that doesn’t influence the yearly averages and trends with more than 0.1 ppmv. The unadjusted raw data still are available. Raw and cleaned data plotted together for MLO and SPO:
http://www.ferdinand-engelbeen.be/klimaat/klim_img/co2_mlo_spo_raw_select_2008.jpg
The measurements at the South Pole started before Mauna Loa but couldn’t go on as the base was about closed. Only flask samples were taken during a few years. Later on, the South Pole base started again with continuous measurements. These show similar levels as Mauna Loa and many other places.
As far as I know, all emissions of China are at ground level. According to you, these can’t reach Mauna Loa at all.
CO2 measurements are made in lots of places, by different organisations from different countries. Hundreds of people are involved. If you think that the data are manipulated, why is nobody of these people complaining about that, even not after retirement?
Rain is 99.65% water. The maximum solubility of CO2 in water is 3.5 g/l or 0.35%. That is CO2 + carbonic acid + bicarbonate + carbonate, where CO2 + carbonic acid is less than 1% of the carbon mixture. See:
http://www.engineeringtoolbox.com/gases-solubility-water-d_1148.html
The creation website you refer to has many errors, including that carbonic acid is insoluble in water???
Together with vaporizing water, CO2 is released from the oceans and follows the same path from equator to poles and from near surface to the stratosphere. Simply think about wind speed and convection, which are much faster than the settlement of CO2 in stagnant air: 1% increase at the bottom of a 70 m column in 40 years…
Timothy Casey is completely wrong about the influence of volcanoes. Especially about the 13C/12C ratio: magmatic CO2 has a lower 13C/12C ratio than subduction CO2, but still is higher than the ratio in the atmosphere. The drop in ratio as seen in the atmosphere can only be from fossil fuel burning or from vegetation decay. But vegetation is growing, as the oxygen balance proves.
Further, you have no idea what the AIRS people know or don’t know about CO2 levels in the lower troposhere and in the stratosphere. They call the mid-stratosphere CO2 levels “lumpy” because they can see a variability of +/- 2% of full scale. That is fine for their instrumentation but of no interest for the effect of CO2 on temperature or for not calling CO2 “well mixed” over the world.
Further, over 70% of the near surface, that is over the oceans, similar levels as at 3400 meter height are found, as good as at 12,000 meters. Regular commercial flights sampled CO2 already over 50 years ago between Scandinavia and the VS. That shows the following seasonal curves:
http://www.ferdinand-engelbeen.be/klimaat/klim_img/seasonal_height.jpg
The maximum difference of CO2 between sealevel and 12 km height was 6 ppmv…
Only in the first few hundred meters over land, CO2 is not well mixed, but very variable with lower values when photosynthesis is at work and higher values when there is inversion at night.

April 29, 2013 4:04 am

Myrrh says:
April 29, 2013 at 3:46 am
I still am awaiting your response to what happens with the visible light that enters the oceans and is absorbed in the first few hundred meters. Something says me that there must be conservation of energy?

Myrrh
April 29, 2013 4:40 am

Ferdinand Engelbeen says:
April 28, 2013 at 11:40 am
Myrrh says:
April 28, 2013 at 3:15 am
Myrrh,
Only one question: as no visible light reaches the ocean depths below a few hundred meters, what happens with the energy contained in that light between zero and 200 meter depth were it all is absorbed by water and/or suspended matter?
“Absorbed” does not necessarily mean the energy is turned to heat. Unless you think an ice cube is hot..
Lot’s of things. Attenuation, dissipation; scattering, absorption by photosynthesis and pigments along the way, used up in all the eyes in the ocean for seeing by conversion to nerve impulses.., and simply, just stops. Oh and I’ve mentioned in my last post to you, some 70% of our oxygen is produced by photosynthesis in the oceans, which is conversion to chemical energy, sugars..
Visible light is created, what is created is not eternal, some things just stop because they run out of energy like the petrol in your car runs out, the energy of visible light is also using up its own energy in motion, it’s its own petrol. The atmosphere slows down visible light, blue being more energetic gets absorbed by the electrons and scattered more, hence our blue sky, the ocean slows down visible light from the Sun around 14 times more than air, visible light uses up its energy by this.
Visible light is slowed down in the atmosphere from all its encounters with the electrons, of course, still at huge speeds, but each absorption by an electron which is energised and then returns to its ground state, takes time. Visible light separates out into the colours in the first place because they are each affected differently by the same medium, as in prisms and in rainbows. In transparent mediums such as glass prisms and water droplets, and so in the ocean, visible light is not absorbed at all but transmitted through unchanged, but, each encounter it has trying to get into to play with the electrons delays it more than simple absorption by the electrons in the fluid gas air, it tries ‘several’ times to get in to reach the electron in the volume which is a molecule of water, in which the electrons dance, as it can’t it then gets ‘passed along’, transmitted, this delays visible considerably more.
Where in the atmosphere the longer wavelengths of visible have greater ability to pass through because they are bigger than blue, their size slows them down more in the denser medium of the ocean, so they attentuate more quickly. Blue and violet slip through more easily.
Check out how they deal with attentuation in optic cables.
Hmm, first page I pulled out: http://www.eetimes.com/design/communications-design/4139197/Performing-Fiber-Optic-Cable-Attenuation-Measurements-A-Tutorial
Be careful here, I don’t know what he means by ” OH water molecules called “high water” are another deformity incorporated into the material during the manufacture that results in absorption”. Is this water the same as H20 in transmitting property? If so is he using “absorbed” in the general sense knowing that water is a transparent medium and so attenuation is because not absorbed, but slowing down greater, or is this another example of AGW memes so ingrained that he doesn’t know what he’s talking about..? What motivates me to keep going in these arguments is the deliberate confusion created in basic science, not only in the general population in order to create the AGW scare, but because it is producing people working in, and especially teaching about, such industries who would not be able to design the products from scratch..
Short piece on attenuation:
“Attenuation
“This article is about attenuation in physics. For other uses, see Attenuation (disambiguation).
In physics, attenuation (in some contexts also called extinction) is the gradual loss in intensity of any kind of flux through a medium. For instance, sunlight is attenuated by dark glasses, X-rays are attenuated by lead, and light and sound are attenuated by water.
“In electrical engineering and telecommunications, attenuation affects the propagation of waves and signals in electrical circuits, in optical fibers, as well as in air (radio waves).”
Continued on: http://en.wikipedia.org/wiki/Attenuation
Another short piece:
“Attenuation and Dispersion in Fiber-Optic Cable
Correct functioning of an optical data link depends on modulated light reaching the receiver with enough power to be demodulated correctly. Attenuation is the reduction in power of the light signal as it is transmitted. Attenuation is caused by passive media components, such as cables, cable splices, and connectors. While attenuation is significantly lower for optical fiber than for other media, it still occurs in both multimode and single-mode transmission. An efficient optical data link must have enough light available to overcome attenuation.
Dispersion is the spreading of the signal in time. The following two types of dispersion can affect an optical data link:
“Chromatic dispersionā€”Spreading of the signal in time resulting from the different speeds of light rays.
Modal dispersionā€”Spreading of the signal in time resulting from the different propagation modes in the fiber.”
http://www.juniper.net/techpubs/en_US/release-independent/junos/topics/concept/fiber-optic-cable-signal-loss-attenuation-dispersion-understanding.html
Look like I should have used dispersion rather than dissipation..
Lots about the properties of light in Optics, haven’t found anything in Thermodynamics proclaiming the great heating power of visible light..

Phil.
April 29, 2013 5:06 am

Myrrh says:
April 29, 2013 at 3:46 am
Visible light from the Sun is not a thermal energy, it cannot heat matter. It works on the tiny electronic transition level not on the big molecular vibrational level which is what it takes to move matter into heat.

More of your unscientific nonsense, the electronic transition is the large one and the vibrational transition which is the small one (although not as small as the rotational transition). See here and learn:
http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/molec.html#c2
We cannot feel visible light from the Sun, it is not hot. What we feel as heat from the Sun is longwave infrared which is thermal infrared; of heat infrared and not of reflective light infrared, which is the not hot shortwaves of infrared.
Which has all to do with the properties of our skin, not the properties of the light!

April 29, 2013 5:21 am

Friends:
I strongly commend that posts from Myrrh be ignored.
It is not possible to address such complete nonsense as this irrational twaddle posted by Myrrh at April 29, 2013 at 4:40 am.

Visible light is created, what is created is not eternal, some things just stop because they run out of energy like the petrol in your car runs out, the energy of visible light is also using up its own energy in motion, itā€™s its own petrol. The atmosphere slows down visible light, blue being more energetic gets absorbed by the electrons and scattered more, hence our blue sky, the ocean slows down visible light from the Sun around 14 times more than air, visible light uses up its energy by this.

Richard

Phil.
April 29, 2013 5:36 am

Be careful here, I donā€™t know what he means by ā€ OH water molecules called ā€œhigh waterā€ are another deformity incorporated into the material during the manufacture that results in absorptionā€. Is this water the same as H20 in transmitting property? If so is he using ā€œabsorbedā€ in the general sense knowing that water is a transparent medium and so attenuation is because not absorbed, but slowing down greater, or is this another example of AGW memes so ingrained that he doesnā€™t know what heā€™s talking about..?
It’s clear that you don’t know what’s being talked about because you have no clue about the interaction of light with matter!
In the case of fiber optics the presence of OH- ions incorporated in the silica matrix leads to absorption, particularly at the 1380 cm-1 transition, thus removing energy from the signal. The presence of such impurities (at the ppb level) defines the performance of the fiber and specifies the useful wavelength windows for it (i.e. not 1380 cm-1).

April 29, 2013 5:46 am

richardscourtney says:
April 29, 2013 at 5:21 am
Richard, I agree. It is hopeless.

Bryan
April 29, 2013 5:52 am

richardscourtney says
“Much IPCC so-called science is falsifiable; e.g. total loss of Himalayan glaciers within 35 years. But the reliance of much IPCC so-called science on not-validated computer models is pure pseudoscience.”
I agree but will they admit it?
Rhetorical question, because the answer is always no.
An excuse will be found to explain any departure from reality.
Historically CO2 fraction of atmosphere always lagged temperature.
Recent history shows that despite a large increase in CO2 fraction over the last 16 years temperatures have remained steady or decli ning.
Thats why I thought it as bit rich coming from Joel Shore that a test of falsification should be the mark that separates science from pseudoscience (or religion).
Yet Joel cannot give a test for AGW science.
We are therefore entitled to class it as pseudoscience (or religion).

Myrrh
April 29, 2013 6:25 am

Phil. says:
April 29, 2013 at 5:06 am
Myrrh says:
April 29, 2013 at 3:46 am
“Visible light from the Sun is not a thermal energy, it cannot heat matter. It works on the tiny electronic transition level not on the big molecular vibrational level which is what it takes to move matter into heat.”
More of your unscientific nonsense, the electronic transition is the large one and the vibrational transition which is the small one (although not as small as the rotational transition). See here and learn:
http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/molec.html#c2

Says nothing about electronic transitions being the bigger and vibrational being the smaller..
You’re full of AGWScienceFiction memes – the electron is considerably smaller than the whole molecule which includes the electrons; the whole molecule has volume, it is the whole volume of the molecule which has to be moved in order to heat it.
Visible light is tiny, much tiner than thermal infrared –
it’s much tinier than the shortwave non thermal infrared, the shortwave infrared is microscopic, the longwave infrared around the size of a pin head.
You have lost all sense of scale in these fake fisics memes, as the wavelengths get shorter they also get smaller.
All electromagnetic energy is not the same.. That’s why they have been given different names, and why real phyisics has been able to understand what they can and cannot do because they have different properties and processes from each other.
A big as a house radio wave is not the same as a very very very tiny gamma ray. They do not have the same effect on matter.
Sometimes joining up in sets, like gamma is ionising and some uv is ionising and so in the ionising set, but other uv not.
We can use the energy one non-ionising uv wavelength to convert to Vitamin D..
AGWScienceFiction has taken out all the properties and processes of energy and matter and dumbed this down to a series of lies, all electromagnetic energy does not create heat on being absorbed – photosynthesis proves that. Photosynthesis is the conversion of visible light to sugars, chemical energy not heat energy.
So come on, tell us how much visible light is heating the atmosphere by being absorbed by the electrons of the molecules of nitrogen and oxygen? Which is how we get our blue sky.
“We cannot feel visible light from the Sun, it is not hot. What we feel as heat from the Sun is longwave infrared which is thermal infrared; of heat infrared and not of reflective light infrared, which is the not hot shortwaves of infrared.”
Which has all to do with the properties of our skin, not the properties of the light!
What? Our skin doesn’t create the radiating heat wave coming from the Sun – it measures it. It measures it by the effect the longwave infrared has on it. This thermal infrared is absorbed by the skin, matter, and because heat is a powerful energy it moves the molecules of the skin into vibration, so heating them up. We can feel this.
Rub your hands together, that is powerful mechanical energy causing the molecules in your skin to vibrate, which is kinetic energy which is heat, this is how powerful the direct heat energy from the Sun – and that is the bigger thermal infrared.
We cannot feel visible light, it is classed Reflected not classed Thermal in the class distinction Reflected/Thermal. It is classed as Light not Heat, in the class distinction Light/Heat.
Also, the direct heat energy from the Sun, which is longwave infrared which we feel as heat, penetrates several inches into our bodies and heats us up inside – heating up the solid and liquids in us, our flesh and bone and blood – especially because we are made up of a lot of water and water is a great imbiber of heat energy.
Look to real world industries to understand this. Humid air is cooler than dry air because of water’s known great capacity for absorbing heat energy before showing any change in temperature; this real world physics knowledge is used to keep rooms cool, put out a pan of water… And, conversely, if you have instead the problem of keeping a room warm in a cold wet climate then de-humidify your room – because the water sucks out the heat you’re putting in making the room feel colder..
So why do you think, there’s a post above, that real world industries make windows that minimise infrared and maximise visible to keep rooms cool and save on air conditioning costs?
Are they daft? Shouldn’t they know visible is the great heat energy which heats the land and water of the Earth..? Oh, maybe they’re still doing traditional science.
But if the people who created this technology were to disappear – those thinking AGW fake fisics is real wouldn’t have a hope in hell of understanding how to repeat it.

Phil.
April 29, 2013 7:38 am

Myrrh says:
April 29, 2013 at 6:25 am
Phil. says:
April 29, 2013 at 5:06 am
Myrrh says:
April 29, 2013 at 3:46 am
ā€œVisible light from the Sun is not a thermal energy, it cannot heat matter. It works on the tiny electronic transition level not on the big molecular vibrational level which is what it takes to move matter into heat.ā€
More of your unscientific nonsense, the electronic transition is the large one and the vibrational transition which is the small one (although not as small as the rotational transition). See here and learn:
http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/molec.html#c2
Says nothing about electronic transitions being the bigger and vibrational being the smaller..

So not only can’t you comprehend what you’re reading but you don’t understand graphs, no big surprise! In a graph showing energy on the y-axis the longer vertical lines representing electronic transitions correspond to greater energy change than the smaller vibrational transition.
Also from the text: “Vibrational transitions occur between different vibrational levels of the same electronic state. ”
So come on, tell us how much visible light is heating the atmosphere by being absorbed by the electrons of the molecules of nitrogen and oxygen? Which is how we get our blue sky.
How we get our blue sky is elastic scattering, no absorption!

joeldshore
April 29, 2013 7:40 am

Bryan says:

Historically CO2 fraction of atmosphere always lagged temperature.
Recent history shows that despite a large increase in CO2 fraction over the last 16 years temperatures have remained steady or decli ning.

(1) Trends over such a period are not statistically-significant different from zero but they also are not statistically-significant different from the long-term (post-1975) trend.
(2) The long-term trend (post-1975 to now) is not significantly different from the trend from 1975 through 16 years ago (March 1997) as this plot shows: http://www.woodfortrees.org/plot/hadcrut4gl/from:1975/plot/hadcrut4gl/from:1975/trend/plot/hadcrut4gl/from:1975/to:1997.25/trend [In fact, the trend over the full period is slightly higher, although that detail is in the noise and depends on the exact endpoint.] Hence, there is no compelling reason to believe that we have fallen off the long-term trend that we have been following since ~1975.
These sorts of things are why respected scientific authorities like National Academy of Sciences think that the pseudoscience lies with “AGW skeptics” arguments and not with AGW. Unless you learn not to make such pseudoscientific arguments, you will not convince the scientific community of your arguments…But maybe that’s not really the goal?

joeldshore
April 29, 2013 7:42 am

Bryan says:

Historically CO2 fraction of atmosphere always lagged temperature.
Recent history shows that despite a large increase in CO2 fraction over the last 16 years temperatures have remained steady or decli ning.

(1) Trends over such a period are not statistically-significant different from zero but they also are not statistically-significant different from the long-term (post-1975) trend.
(2) The long-term trend (post-1975 to now) is not significantly different from the trend from 1975 through 16 years ago (March 1997) as this plot shows: http://www.woodfortrees.org/plot/hadcrut4gl/from:1975/plot/hadcrut4gl/from:1975/trend/plot/hadcrut4gl/from:1975/to:1997.25/trend [In fact, the trend over the full period is slightly higher, although that detail is in the noise and depends on the exact endpoint.] Hence, there is no compelling reason to believe that we have fallen off the long-term trend that we have been following since ~1975.
These sorts of things are why respected scientific authorities like National Academy of Sciences think that good scientific arguments lie in favor of AGW and. Unless you learn not to make such unscientific arguments, you will not convince the scientific community of your arguments…But maybe that’s not really the goal?

Phil.
April 29, 2013 7:51 am

richardscourtney says:
April 29, 2013 at 5:21 am
Friends:
I strongly commend that posts from Myrrh be ignored.
It is not possible to address such complete nonsense as this irrational twaddle posted by Myrrh at April 29, 2013 at 4:40 am.

Agreed, I learned that about Myrhh several years ago and usually do ignore him, but every so often it’s necessary to refute some of his nonsense lest someone might actually think he knows what he’s talking about!

April 29, 2013 9:10 am

joeldshore says:
April 29, 2013 at 7:42 am
(2) The long-term trend (post-1975 to now) is not significantly different from the trend from 1975 through 16 years ago (March 1997)
The trend 1910-1945 is as steep as the trend 1976-2000, but CO2 levels in these periods increased resp. 10 ppmv and 50 ppmv. There is no fivefold increase in warming speed after 1946. Further, there was a slight cooling trend 1945-1975, which is caused – according to climate models – by the increased emissions of sulfate, cooling aerosols. But as I still remember from these times, London was fequently kept in green/black killing fog, thus warming black carbon aerosols were more important than cooling sulfates…
The current standstill seems quite identical to the 1945-1975 standstill, thus may extend over 30 years. There is a quite good correlation between the PDO cycle and global temperatures in the same periods. The PDO is a cycle of 60-80 years and now is in a cold phase…
The problem with current models is that they all are based on the increase over 1976-2000 and the near full attribution of all warming on CO2. There is no room in the models for natural variations like ENSO, PDO, NAO,… But if these are causing the variation around the trend, one has to halve the influence of CO2.

Tim Folkerts
April 29, 2013 9:17 am

Bryan says:

Historically CO2 fraction of atmosphere always lagged temperature.

But history is only a good guide if conditions are similar. Historically, animals have not been digging up and burning ~ 80,000,000 barrels of oil and 20,000,000 tons of coal per day. Conditions relating to CO2 *are* different than they were the last few times the earth transitioned in and out of glacial periods.
There is an old saying that the definition of insanity is doing the same thing over and over and expecting different results.
There should be a corollary saying that it is also insane to do something completely unprecedented and to expect that things will continue to behave the same way as before.

April 29, 2013 9:33 am

Tim Folkerts:
Your post at April 29, 2013 at 9:17 am is misguided.
Atmospheric CO2 concentration is so low that plants struggle because they evolved at higher concentrations. And plants need atmospheric CO2 so everything up the food chain from plants needs it. Atmospheric CO2 is literally the ‘stuff of life’.
There are known benefits of higher atmospheric CO2 concentration: all the biosphere benefits. And there are no known problems.
Then, of course, there is the little fact that nature emits 34 CO2 molecules for each CO2 molecule emitted from human activity. So, you are saying you are afraid that a trivial increase to CO2 emissions will change CO2 from being the ‘stuff of life’ into being the harbinger of Armagedon.
Riiiiight.
Richard

davidmhoffer
April 29, 2013 9:44 am

richardscourtney;
could you please contact me via email?
davidDOThofferATmts.net

April 29, 2013 9:46 am

Ferdinand Engelbeen says:
“The trend 1910-1945 is as steep as the trend 1976-2000…”
They are almost identical, despite the big difference in atmospheric CO2.

Phil.
April 29, 2013 9:55 am

Ferdinand Engelbeen says:
April 29, 2013 at 9:10 am
joeldshore says:
April 29, 2013 at 7:42 am
(2) The long-term trend (post-1975 to now) is not significantly different from the trend from 1975 through 16 years ago (March 1997)
The trend 1910-1945 is as steep as the trend 1976-2000, but CO2 levels in these periods increased resp. 10 ppmv and 50 ppmv. There is no fivefold increase in warming speed after 1946.

Nor should one have been expected, the response to CO2 increase at present atmospheric levels is logarithmic not linear.

April 29, 2013 10:14 am

davidmhoffer:
re your message to me at April 29, 2013 at 9:44 am.
You may want to check your inbox.
If you don’t find my email then my email address is
richardscourtneyATaolDOTcom
Richard

Phil.
April 29, 2013 10:30 am

richardscourtney says:
April 29, 2013 at 9:33 am
Tim Folkerts:
Your post at April 29, 2013 at 9:17 am is misguided.
Atmospheric CO2 concentration is so low that plants struggle because they evolved at higher concentrations. And plants need atmospheric CO2 so everything up the food chain from plants needs it. Atmospheric CO2 is literally the ā€˜stuff of lifeā€™.
There are known benefits of higher atmospheric CO2 concentration: all the biosphere benefits. And there are no known problems.
Then, of course, there is the little fact that nature emits 34 CO2 molecules for each CO2 molecule emitted from human activity.

And the other little fact that you omit mentioning that nature absorbs about 34.5 CO2 molecules for each CO2 molecule emitted from human activity, for a net natural sink effect.

Phil.
April 29, 2013 10:51 am

dbstealey says:
April 29, 2013 at 9:46 am
Ferdinand Engelbeen says:
ā€œThe trend 1910-1945 is as steep as the trend 1976-2000ā€¦ā€
They are almost identical, despite the big difference in atmospheric CO2.

But not such a big difference in log(CO2), about a 5% increase.

joeldshore
April 29, 2013 10:54 am

Phil. says:

And the other little fact that you omit mentioning that nature absorbs about 34.5 CO2 molecules for each CO2 molecule emitted from human activity, for a net natural sink effect.

And hence, we see a prime example of why the AGW skeptics don’t come off well with actual scientists. The sorts of arguments that Richard used here aren’t designed to convince knowledgeable scientists. They are designed to confuse people who don’t understand…particularly those who want to believe a certain thing.

April 29, 2013 11:02 am

joelshore says:
“And hence, we see a prime example of why the AGW skeptics donā€™t come off well with actual scientists.”
Therefore, Lindzen, Christy, Michaels, Spencer, etc., etc. are not “actual scientists” according to joelshore. Shows just how screwed up he really is.
joelshore is miserable because the planet is falsifying his belief system. It refuses to do what the alarmist crowd incessantly predicted it would do.
Most of us are ‘AGW skeptics’ for the simple reason that AGW cannot be measured. AGW amounts to an assertion. A conjecture. If it cannot be quantified with empirical measurements, it is only an unverifiable belief. Like a religion.

tjfolkerts
April 29, 2013 11:18 am

Richard,
I didn’t say anything about whether more CO2 is good or bad. Putting words in my mouth and then attacking those words is a very ineffective debate tactic.
I was simply pointing out a logical fallacy ….

* In the past, global warming caused increased CO2 levels.
* Therefore, global warming also caused the recent increase in CO2

Am I “misguided” because I object to this logic?

April 29, 2013 11:28 am

joeldshore:
At April 29, 2013 at 10:54 am you quote Phil. (i.e. another AGW-fanatic like you) trying to be clever by altering my words and you say of the alteration by Phil.

And hence, we see a prime example of why the AGW skeptics donā€™t come off well with actual scientists. The sorts of arguments that Richard used here arenā€™t designed to convince knowledgeable scientists. They are designed to confuse people who donā€™t understandā€¦particularly those who want to believe a certain thing.

NO! We see a prime example of how you misrepresent AGW skeptics.
Actual scientists like me don’t do that sort of thing, Shore.
Only people like you who pretend to be scientists do it.
This is what Phil. said as a supposed correction of my statement.

And the other little fact that you omit mentioning that nature absorbs about 34.5 CO2 molecules for each CO2 molecule emitted from human activity, for a net natural sink effect.

You quoted those words as mine. But even Phil had sufficient honesty to quote what I actually said, so you cannot have made a simple misunderstanding.
You pretended Phil’s words were mine then slagged them off.
And I actually said simple truth in this post at April 29, 2013 at 9:33 am. Phil also quoted this post before making his suggested alteration which you have pretended were my words.

Tim Folkerts:
Your post at April 29, 2013 at 9:17 am is misguided.
Atmospheric CO2 concentration is so low that plants struggle because they evolved at higher concentrations. And plants need atmospheric CO2 so everything up the food chain from plants needs it. Atmospheric CO2 is literally the ā€˜stuff of lifeā€™.
There are known benefits of higher atmospheric CO2 concentration: all the biosphere benefits. And there are no known problems.
Then, of course, there is the little fact that nature emits 34 CO2 molecules for each CO2 molecule emitted from human activity. So, you are saying you are afraid that a trivial increase to CO2 emissions will change CO2 from being the ā€˜stuff of lifeā€™ into being the harbinger of Armagedon.
Riiiiight.

Richard

April 29, 2013 11:35 am

tjfolkerts:
I do not have a clue what you are talking about in your post addressed to me at April 29, 2013 at 11:18 am.
You say

Putting words in my mouth and then attacking those words is a very ineffective debate tactic.

What words did I put in your mouth, when and where?
Or are you trying the Joel Shore tactic of pretending I said other than I did?
Richard

tjfolkerts
April 29, 2013 11:44 am

ā€œAnd hence, we see a prime example of why the AGW skeptics donā€™t come off well with actual scientists.ā€

Therefore, Lindzen, Christy, Michaels, Spencer, etc., etc. are not ā€œactual scientists according to joelshore.ā€

It is somewhat amusing to note that this whole thread started precisely because some “AGW skeptics donā€™t come off well with actual scientists”. The “actual scientist” Roy Spencer was mocking “AGW skeptics” at PSI who have no clue about physics (or more precisely, they have just enough clue to be dangerous). PSI is so far over the line that even other skeptics are calling them out!
So, yes, there are enough really uninformed, really vocal “AGW skeptics” that they paint “AGW skeptics” as a whole in a bad light. (And, of course, there are plenty of “alarmists” who paint the other side in a bad light.)

Phil.
April 29, 2013 11:47 am

richardscourtney says:
April 29, 2013 at 11:28 am
joeldshore:
At April 29, 2013 at 10:54 am you quote Phil. (i.e. another AGW-fanatic like you) trying to be clever by altering my words and you say of the alteration by Phil.

I didn’t alter what you said, I quoted it in italics as I normally do, and pointed out your omission where you pretend that the contribution of ‘nature’ is an increase of 35 molecules/molecule of CO2 emitted by human activity whereas it is in fact a net sink of 0.5 molecules/molecule.
This is what Phil. said as a supposed correction of my statement.
“And the other little fact that you omit mentioning that nature absorbs about 34.5 CO2 molecules for each CO2 molecule emitted from human activity, for a net natural sink effect.”

which is an actual correction of your statement.

tjfolkerts
April 29, 2013 11:51 am

richardscourtney asks rather emphatically “What words did I put in your mouth, when and where?”
Ummm …. how about the post I was referring to @ April 29, 2013 at 9:33 am?

” So, you are saying you are afraid that a trivial increase to CO2 emissions will change CO2 from being the ā€˜stuff of lifeā€™ into being the harbinger of Armagedon.”

In fact, I said nothing of the sort.

April 29, 2013 11:53 am

Phil. says:
April 29, 2013 at 10:51 am
But not such a big difference in log(CO2), about a 5% increase.
That is exactly the problem: the temperature rise 1910-1945 is mostly attributed to natural factors in the models, where the 10 ppmv rise in CO2 has not much influence. The temperature rise 1976-2000 is near solely attributed to the 50 ppmv rise in CO2, which is a 5% log[CO2] increase. Still about 5 times even in log increase, compared to the previous period. But the temperature increases (speed and height) are quite identical. The inbetween period and the period since 2000 are even more interesting: slight cooling and flat temperatures with rising CO2 levels.
The first cooler period is argumented away by a nice tuning knob: human induced aerosols, of which nobody can tell for sure if they cool the world or warm the world. The brown haze over India probably warms the world. But the reduction of over 60% in SO2 emissions in the Western world didn’t show any temperature effect at places where the largest effect should be found: downwind the largest emitters.
The current period of standstill can’t be argumented by aerosols, as these didn’t change much over the recent decades: what the Western world cleaned up now is extra emitted in China, but even there one begins to clean up the emissions, because they have more and more problems.
Thus the main problem is what causes the current (and previous) standstill in temperature increase and what is its attribution to the warming in the other periods and what are the consequences for the sensitivity for 2xCO2.

April 29, 2013 12:12 pm

Phil.:
re your assertions in your post at April 29, 2013 at 11:47 am.
Firstly, I made clear that your alteration was NOT a misrepresentation of my words by you. Indeed, I wrote to Shore saying

You quoted those words as mine. But even Phil had sufficient honesty to quote what I actually said, so you cannot have made a simple misunderstanding.
You pretended Philā€™s words were mine then slagged them off.

So, your complaint is without foundation when you write

I didnā€™t alter what you said

And NO! Your alteration was not a “correction”. It was a distortion.
I wrote

Then, of course, there is the little fact that nature emits 34 CO2 molecules for each CO2 molecule emitted from human activity. So, you are saying you are afraid that a trivial increase to CO2 emissions will change CO2 from being the ā€˜stuff of lifeā€™ into being the harbinger of Armagedon.

You are saying you are “correcting” that by changing it to

ā€œAnd the other little fact that you omit mentioning that nature absorbs about 34.5 CO2 molecules for each CO2 molecule emitted from human activity, for a net natural sink effect.ā€

One could argue about the difference between “34” and “about 34.5” but your alteration emphasises my point about the relative magnitudes of anthropogenic and ‘natural’ emissions.
More importantly, your alteration to my second sentence changes its subject from threat of dangerous effects to the subject of carbon cycle sinks. An alteration which changes the subject is not a “correction”.
Very importantly, your change to my second sentence alters my statement about assertion of dangerous effects of anthropogenic CO2 to be an assertion of no dangerous effects. I agree there would be no dangerous effects but I was talking about the view of Folkerts (which I thought you shared).
Richard

April 29, 2013 12:25 pm

tjfolkerts:
Thankyou for correcting my misunderstanding with your post at April 29, 2013 at 11:51 am which says

richardscourtney asks rather emphatically ā€œWhat words did I put in your mouth, when and where?ā€
Ummm ā€¦. how about the post I was referring to @ April 29, 2013 at 9:33 am?

ā€ So, you are saying you are afraid that a trivial increase to CO2 emissions will change CO2 from being the ā€˜stuff of lifeā€™ into being the harbinger of Armagedon.ā€

In fact, I said nothing of the sort.

OK. I understand that.
It was not that “I put words in your mouth”
but
I did the equally reprehensible practice of misrepresenting your view.
Please accept my apology.
I genuinely did understand – on the basis of posts you have made on other WUWT threads – that you are a supporter of the catastrophic AGW hypothesis. I now know that my understanding was wrong.
I apologise for misrepresenting your view and in future I will ensure that people recognise I know you think AGW will not be a serious problem.
Richard

April 29, 2013 12:27 pm

Sorry for the formatting error in my post at April 29, 2013 at 12:25 pm
All quotation should have ended after
“In fact, I said nothing of the sort.”
Sorry.
Richard

April 29, 2013 12:30 pm

richardscourtney says:
April 29, 2013 at 12:12 pm
The fundamental difference in our dispute about the cause of the increase of CO2 in the atmosphere is exactly about the definition of natural “emissions”. While these are huge, that doesn’t matter for the increase in the atmosphere, as long as the natural sinks equal the natural sources. In this case it is proven that the natural sinks exceed the natural sources, thus nature is a net sink for CO2, not a net source.
Even if the natural emissions were 1000 times larger than the human emissions, that doesn’t increase the levels in the atmosphere, as long as the natural sinks also are 1000 times larger than the human emissions, or more.
The natural emissions are part of a natural cycle. What counts is the net result of the full cycle and that is a net sink after a year for every single year in the past 50 years. The height of one part of the cycle is not of the slightest interest for the atribution of the rise in CO2 and distracts from the real cause.
If the rise is dangerous or beneficial is an unrelated question to the attribution…

Phil.
April 29, 2013 12:31 pm

Ferdinand Engelbeen says:
April 29, 2013 at 11:53 am
Phil. says:
April 29, 2013 at 10:51 am
But not such a big difference in log(CO2), about a 5% increase.
That is exactly the problem: the temperature rise 1910-1945 is mostly attributed to natural factors in the models, where the 10 ppmv rise in CO2 has not much influence. The temperature rise 1976-2000 is near solely attributed to the 50 ppmv rise in CO2, which is a 5% log[CO2] increase. Still about 5 times even in log increase, compared to the previous period.

No the forcing is proportional to log(CO2) so 5% change not 5x change!
But the temperature increases (speed and height) are quite identical. The inbetween period and the period since 2000 are even more interesting: slight cooling and flat temperatures with rising CO2 levels.
The first cooler period is argumented away by a nice tuning knob: human induced aerosols, of which nobody can tell for sure if they cool the world or warm the world. The brown haze over India probably warms the world. But the reduction of over 60% in SO2 emissions in the Western world didnā€™t show any temperature effect at places where the largest effect should be found: downwind the largest emitters.

Don’t forget atmospheric nuclear bomb testing which was conducted from 1945-63, they injected massive amounts of particulates into the stratosphere, rather like multiple volcanoes.
The current period of standstill canā€™t be argumented by aerosols, as these didnā€™t change much over the recent decades: what the Western world cleaned up now is extra emitted in China, but even there one begins to clean up the emissions, because they have more and more problems.
Thus the main problem is what causes the current (and previous) standstill in temperature increase and what is its attribution to the warming in the other periods and what are the consequences for the sensitivity for 2xCO2.

ENSO effects mainly see: Foster & Rahmstorf
http://iopscience.iop.org/1748-9326/6/4/044022

Phil.
April 29, 2013 12:53 pm

richardscourtney says:
April 29, 2013 at 12:12 pm
Phil.:
re your assertions in your post at April 29, 2013 at 11:47 am.
Firstly, I made clear that your alteration was NOT a misrepresentation of my words by you. Indeed, I wrote to Shore saying
You quoted those words as mine. But even Phil had sufficient honesty to quote what I actually said, so you cannot have made a simple misunderstanding.
You pretended Philā€™s words were mine then slagged them off.
So, your complaint is without foundation when you write
I didnā€™t alter what you said

Actually it does have a foundation, you assert that I altered your statement, I did not.
And NO! Your alteration was not a ā€œcorrectionā€. It was a distortion.
I wrote
Then, of course, there is the little fact that nature emits 34 CO2 molecules for each CO2 molecule emitted from human activity. So, you are saying you are afraid that a trivial increase to CO2 emissions will change CO2 from being the ā€˜stuff of lifeā€™ into being the harbinger of Armagedon.
You are saying you are ā€œcorrectingā€ that by changing it to
ā€œAnd the other little fact that you omit mentioning that nature absorbs about 34.5 CO2 molecules for each CO2 molecule emitted from human activity, for a net natural sink effect.ā€
One could argue about the difference between ā€œ34ā€³ and ā€œabout 34.5ā€³ but your alteration emphasises my point about the relative magnitudes of anthropogenic and ā€˜naturalā€™ emissions.

No it doesn’t, it shows that you are being disingenuous by only addressing part of the natural processes, in fact, using your figures, about 34 mole/mole of CO2 is generated naturally while about 34.5 mole/mole is removed naturally leaving the net effect of nature to be -0.5 mole/mole, which is contrary to your point about relative magnitudes.
More importantly, your alteration to my second sentence changes its subject from threat of dangerous effects to the subject of carbon cycle sinks. An alteration which changes the subject is not a ā€œcorrectionā€.
I didn’t alter your statement I added one of my own pointing out your omission, so my statement most assuredly is a ‘correction’.
Very importantly, your change to my second sentence alters my statement about assertion of dangerous effects of anthropogenic CO2 to be an assertion of no dangerous effects. I agree there would be no dangerous effects but I was talking about the view of Folkerts (which I thought you shared).
Again no change was made to any of your statements, they were quoted accurately.

April 29, 2013 12:54 pm

Friends:
I write to say that with one exception I agree everything in the post by Ferdinand Engelbeen at April 29, 2013 at 12:30 pm.
The exception is his saying

The height of one part of the cycle is not of the slightest interest for the atribution of the rise in CO2 and distracts from the real cause.

As Ferdinand knows, I absolutely disagree with that sentence.
The important point in this discussion is – as Ferdinand says –

If the rise is dangerous or beneficial is an unrelated question to the attributionā€¦

Exactly so.
Richard

April 29, 2013 1:39 pm

Phil.:
I strongly object to your claiming in your post at April 29, 2013 at 12:53 pm that I was being “disingenuous”.
My standards are much higher than yours and I am offended that you assume I would behave as you do.
Richard

joeldshore
April 29, 2013 1:55 pm

Richard S Courtney says:

You quoted those words as mine. But even Phil had sufficient honesty to quote what I actually said, so you cannot have made a simple misunderstanding.
You pretended Philā€™s words were mine then slagged them off.

No, I quoted those words as Phil.’s http://wattsupwiththat.com/2013/04/24/spencer-slays-with-sarcasm/#comment-1291419 (Hint: The “Phil. said:” at the beginning is a subtle clue.)
And, then remarked on the fact that you had, as Phil. noted, left out this important piece of information…a piece of information that is more than a little relevant to the issue.

Phil.
April 29, 2013 2:01 pm

richardscourtney says:
April 29, 2013 at 1:39 pm
Phil.:
I strongly object to your claiming in your post at April 29, 2013 at 12:53 pm that I was being ā€œdisingenuousā€.
My standards are much higher than yours and I am offended that you assume I would behave as you do.

I speak as I find Richard, why else would you just compare the sources but not the sinks, you give totally the wrong impression about the influence of anthropogenic CO2 on the atmospheric concentration. Since I’m sure that you are well aware of the relative magnitudes of the sources and sinks I believe the omission to be deliberate, hence disingenuous. That’s not something I do so your standards aren’t there yet! You also led off with your usual ad hom and made false accusations which you still haven’t acknowledged let alone apologized for, low standards indeed!

joeldshore
April 29, 2013 2:36 pm

Richard S Courtney says:

Phil.:
I strongly object to your claiming in your post at April 29, 2013 at 12:53 pm that I was being ā€œdisingenuousā€.
My standards are much higher than yours and I am offended that you assume I would behave as you do.

So, under these high standards that you speak of, what exactly does mentioning only one side (the slightly smaller side) of a nearly-equal two-sided process of emission and absorption of CO2 by nature constitute? It seems to me that “disingenuous” is perhaps a tad euphemistic.

April 29, 2013 2:38 pm

Phil. says:
April 29, 2013 at 12:31 pm
No the forcing is proportional to log(CO2) so 5% change not 5x change!
The increase in forcing during the period 1976-2000 indeed is 5%, but compared to the period 1910-1945 (where the increase in forcing was about 1%) still 5 times in increase rate of the forcing. But the net result over both periods is near identical, which is quite strange if CO2 has a huge effect.
Donā€™t forget atmospheric nuclear bomb testing which was conducted from 1945-63, they injected massive amounts of particulates into the stratosphere, rather like multiple volcanoes.
Interesting thoughts, but I think that most surface tests were done on islands with relative small amounts of vegetation or other debris that could cause some stratospheric cooling. Compared to the Pinatubo eruption, human bombs seems to be of little impact, both in force and stratopheric debris injection. Most debris from the Pinatubo was falling out within a few months for the heavier particles, the SO2 aerosols lived for only a few years.
ENSO effects mainly see: Foster & Rahmstorf
http://iopscience.iop.org/1748-9326/6/4/044022

I see one fundamental problem with the Foster & Rahmstorf approach: ENSO is not an “exogenic” factor, not a forcing in the same sence as volcanoes or insolation. ENSO may be part of the cause of the trend itself (together with the PDO, NAO,…) and of the warming and standstill episodes. In times of positive PDO, more warm El NiƱo’s are prevalent, in times of a negative PDO, El NiƱo’s are more suppressed. The point is that El NiƱo’s show a residual warming while La NiƱa’s show little residual cooling.
Further, what causes ENSO and/or PDO and other oceanic oscillations?
After writing the above, I searched WUWT about this work and found that Bob Tisdale had similar objections:
http://wattsupwiththat.com/2012/01/14/tisdale-on-foster-and-rahmstorf-take-2/

tjfolkerts
April 29, 2013 3:45 pm

Richard,
I appreciate the apology.
I know it can be tough to keep track of who said what and “believes” what in a rapid-fire discussion like this. I have done it myself a time or two.
However, now you go a little TOO far the other way by one again trying to state what you think that I think: “and in future I will ensure that people recognise I know you think AGW will not be a serious problem.”
I would prefer that you just let me speak for myself (or quote my words). I am somewhere between the two positions you have attributed to me.
*****************************************************************
I am quite sure that …
1) CO2 does help plants grow and is beneficial everything else being equal
2) CO2 does warm the planet everything else being equal.
There is a middle ground of science for which I am less sure. For example,
* I am unsure what the climate sensitivity is for a doubling of CO2. The “raw” sensitivity seems to be right at 1 C/doubling. Depending on whom you ask, the feedbacks can change this number up or down between ~ 0 and ~ 4.5 C/doubling. I would seriously doubt it is 0.0, and I would seriously doubt it is above 4.5. I have seen “credible” scientific papers giving a wide range of estimates based on a wide variety of analyses. I keep my mind open.
* I am unsure how much the current weak sunspot cycle is influencing climate. There seems to be a broad but imprecise correlation between sunspots and temperature, so the current weak cycle could be foreshadowing a natural cooling cycle. Or it might not.
There is an even larger economic/ecologic/political question about how “catastrophic” this will be. My opinion is that global warming is, to a significant but not exclusive degree, due to human influences on land use and the atmosphere. But there is considerable uncertainly in the trajectory of future temperatures and even greater uncertainly in the impacts for specific regions and species.
I could throw out my own “guesstimates” on many of these topics, but I tend not to. I prefer to restrict my comments to the basic physics of IR radiation, lapse rate, 2nd Law of Thermodynamics, etc. These I know pretty well and these are pretty certain. That is why I tend to only comment in the “science” threads and not the political or op/ed threads.

Gary Hladik
April 29, 2013 5:13 pm

tjfolkerts says (April 29, 2013 at 3:45 pm): [snip]
Well said.
In fact, I hope man/nature is actually turning up the thermostat just a bit, so I won’t have to. Have you seen what it costs to heat a home these days? šŸ™‚

Konrad
April 29, 2013 7:45 pm

tjfolkerts says:
April 29, 2013 at 11:44 am
ā€œPSI is so far over the line that even other skeptics are calling them out!ā€
—————————————————————————————–
Yes, the PSI thing did not really work too well at all. It seams that sceptics are not really the forces of ā€œanti scienceā€ after all.
Other Assault Clowns have failed for similar reasons. All sceptics know that incident SW heats solids and liquids. Sceptics also know that incident LWIR can slow the cooling of most materials.
I guess Dr. Spencer likes D.C. around as he can then group any who challenge AGW into the same basket. Traffic volume and high scientific literacy means that the PSI thing was never going to work at WUWT. It would have needed a significant percentage of commentators linking to or repeating the junk. Next to no WUWT readers bought into it.

Myrrh
April 30, 2013 3:20 am

What rise in carbon dioxide?
From a post I’ve just made elsewhere:
The Callendar/Keeling fraud to pretend to track increasing levels of the not possible to tell apart man-made from volcanic carbon dioxide* by cherry picking a low start point is contradicted by measurements by real scientists.
This AGWScienceFiction’s Greenhouse Effect illusion has fraud written all the way through it.
* http://carbon-budget.geologist-1011.net/
“Both tectonic and volcanic CO2 are magmatic and depleted in both 13C & 14C. In the absence of statistically significant isotope determinations for each volcanic province contributing to the atmosphere, this makes CO2 contributions of volcanic origin isotopically indistinguishable from those of fossil fuel consumption. It is therefore unsurprising to find that Segalstad (1998) points out that 96% of atmospheric CO2 is isotopically indistinguishable from volcanic degassing. So much for the Royal Society’s unexplained “chemical analysis”. If you believe that we know enough about volcanic gas compositions to distinguish them chemically from fossil fuel combustion, you have indeed been mislead. As we shall see, the number of active volcanoes is unknown, never mind a tally of gas signatures belonging to every active volcano. We have barely scratched the surface and as such, there is no magic fingerprint that can distinguish between anthropogenic and volcanogenic sources of CO2.”
And more, real rational empirical proxy data via plant stomata:
http://debunkhouse.wordpress.com/2010/03/28/co2-ice-cores-vs-plant-stomata/
ā€œPlant stomata suggest that the pre-industrial CO2 levels were commonly in the 360 to 390ppmv range.ā€
See also: http://www.geocraft.com/WVFossils/stomata.html

April 30, 2013 3:44 am

Myrrh says:
April 30, 2013 at 3:20 am
Myrrh, please read what I have said before: magmatic CO2 from volcanoes is lower in 13C than subduction CO2. But still higher in 13C/12C ratio than what is already in the atmosphere. Thus it is impossible that volcanic CO2 is the cause of the 13C decline in the atmosphere.
Stomata data have far more problems than ice cores for CO2 levels: by definition stomata data are obtained from land plants, where the stomata index (SI – stomata density divided by total cell density) of one year is based on the average CO2 levels in the growing season of the previous year. But CO2 over land is highly variable and in average somewhat higher (40-50 ppmv) than background. That is taken into account by calibrating the SI data over the past century with direct measurements and CO2 levels in firn and ice cores. So far no problem.
The main problem is that nobody can tell you what the bias was 500 or 1000 years ago, with lots of changes in landscape over the centuries in the main wind direction. Even the main wind direction may have been changed between the MWP and the LIA and back…
Anyway, if the average levels in stomata data don’t agree with the average levels in ice cores, then the stomata data are certainly wrong.

April 30, 2013 4:23 am

Phil. and joeldshore:
I strongly object to your clearly deliberate misrepresentation of me as method to make the false and untrue assertion that I have been “disingenuous”.
I was NOT disingenuous – I was factual – when I wrote

Then, of course, there is the little fact that nature emits 34 CO2 molecules for each CO2 molecule emitted from human activity.

And I asserted that the increase of the anthropogenic emission is a trivial increase to the total emission: IT IS.
There was no need for me to have mentioned the sinks. It was not relevant to my point.
You may agree with Ferdinand that the anthropogenic emission has caused the recent (measured at Mauna Loa since 1958) rise in atmospheric CO2 concentration. But whether or not the rise is anthropogenic in part or in whole cannot be known on the basis of existing information (Ferdinand disagrees with this statement).
However, if such a trivial increase to total CO2 emissions as the anthropogenic emission can have a catastrophic effect on climate then it is a wonder that no such catastrophe has happened in the 4.5 billion years since the Earth gained an oxygen-rich atmosphere. And THAT WAS MY POINT.
Importantly, as Ferdinand says concerning atmospheric CO2 concentration

If the rise is dangerous or beneficial is an unrelated question to the attributionā€¦

The assertion from you two weasels that I would be disingenuous is laughable.
Richard

joeldshore
April 30, 2013 5:41 am

Richard: Let me explain it to you this way. If one of my physics colleagues who is not up on the details of the subject of AGW came to me and said he had heard from AGW skeptics that for every one CO2 molecule we emit, nature emits 34 and asked me if this is true and I explained to him what was actually the case regarding both emissions and absorptions, do you think his reaction would be:
(1) Boy, those AGW skeptics are making a good point
OR
(2) Boy, those AGW skeptics are certainly using deceptive arguments

However, if such a trivial increase to total CO2 emissions as the anthropogenic emission can have a catastrophic effect on climate then it is a wonder that no such catastrophe has happened in the 4.5 billion years since the Earth gained an oxygen-rich atmosphere. And THAT WAS MY POINT.

Who says there haven’t been events that were, or would today be considered catastrophic in the past? We know of at least 6 major extinction events in the past, some of which may have been due to climate change. We also know that sea levels have varied by 100’s of meters. Maybe that’s not a catastrophe from your point of view, but with modern civilization have built many significant population centers very close to sea level, I think it is of some relevance.
Furthermore, it is unclear when in the past there has been this rapid a rise in CO2 levels. The input by man might not sound like much compared to the rates that the biosphere, ocean mixed layer, and atmosphere trade back and forth, but it is fundamentally different to be taking carbon from a source that has been locked away for millions of years and liberating it into these reservoirs and doing so year in and year out. The fact is that CO2 levels have never been as high as they are now for at least 3/4 of a million years and likely several million years and that there is no evidence of nearly as fast a rise in CO2 even during the many glacial – interglacial cycles that the ice core record covers.

Phil.
April 30, 2013 7:55 am

richardscourtney says:
April 30, 2013 at 4:23 am
Phil. and joeldshore:
I strongly object to your clearly deliberate misrepresentation of me as method to make the false and untrue assertion that I have been ā€œdisingenuousā€.

Object all you like, you deliberately told half a story in an attempt to mislead, that is being disingenuous.
I was NOT disingenuous ā€“ I was factual ā€“ when I wrote
Then, of course, there is the little fact that nature emits 34 CO2 molecules for each CO2 molecule emitted from human activity.
And I asserted that the increase of the anthropogenic emission is a trivial increase to the total emission: IT IS.

No it is not because it’s an irrelevancy, what counts is the net change, to represent a balance sheet with only the earnings and not the expenditures is fraud, that’s what you attempted to do. You got caught out, face up to it.
Tell it like it is: annual anthropogenic emission is ~double the annual growth in CO2, how is that trivial?
Rate of growth of savings=annual salary-expenditures+bonus
Salary=$35,000, expenditures=$35,500, bonus=$1,000
you can tell someone that the bonus is trivial wrt the salary but the growth of the savings is crucially dependent on that bonus!
You may agree with Ferdinand that the anthropogenic emission has caused the recent (measured at Mauna Loa since 1958) rise in atmospheric CO2 concentration. But whether or not the rise is anthropogenic in part or in whole cannot be known on the basis of existing information (Ferdinand disagrees with this statement).
I do agree with Ferdinand and have posted to that effect for years, your assertion that we don’t have enough information is incorrect.
Rate of change must equal sources-sinks, that cannot be wished away no matter how much you wish it could.
The assertion from you two weasels that I would be disingenuous is laughable.
Again more of your ad hominem attacks, you’re a rather unpleasant individual.
I would urge you to pay attention to the Sermon on the Mount:
“3 And why beholdest thou the mote that is in thy brother’s eye, but considerest not the beam that is in thine own eye?
4 Or how wilt thou say to thy brother, Let me pull out the mote out of thine eye; and, behold, a beam is in thine own eye?
5 Thou hypocrite, first cast out the beam out of thine own eye; and then shalt thou see clearly to cast out the mote out of thy brother’s eye.”

April 30, 2013 8:20 am

Phil.:
I will be generous and assume your post at April 30, 2013 at 7:55 am demonstrates you have severe reading difficulties and need to undertake remedial classes in reading comprehension.
Your twaddle is – not unusually from you – a pack of lies.
My post at April 30, 2013 at 4:23 am fully explained why the “sinks” ARE NOT RELEVANT to my statement.
You are not merely “a rather unpleasant individual”: you are an anonymous coward who hurls abuse and lies from behind the shield of anonymity. I merely tell the truth as I see it.
As for quoting the Sermon On The Mount at me, I strongly suspect that my knowledge and understanding of it are at least superior to yours. But that is also not relevant to your having stated lies.
Richard

Phil.
April 30, 2013 8:39 am

Ferdinand Engelbeen says:
April 29, 2013 at 2:38 pm
Phil. says:
April 29, 2013 at 12:31 pm
“No the forcing is proportional to log(CO2) so 5% change not 5x change!”
The increase in forcing during the period 1976-2000 indeed is 5%, but compared to the period 1910-1945 (where the increase in forcing was about 1%) still 5 times in increase rate of the forcing. But the net result over both periods is near identical, which is quite strange if CO2 has a huge effect.

The forcing in the early period is according to you 1.01 which defines the slope, in the second period it is 1.05, an increase in slope from 1.01 to 1.05 is not a factor of 5!
“Donā€™t forget atmospheric nuclear bomb testing which was conducted from 1945-63, they injected massive amounts of particulates into the stratosphere, rather like multiple volcanoes.”
Interesting thoughts, but I think that most surface tests were done on islands with relative small amounts of vegetation or other debris that could cause some stratospheric cooling. Compared to the Pinatubo eruption, human bombs seems to be of little impact, both in force and stratopheric debris injection. Most debris from the Pinatubo was falling out within a few months for the heavier particles, the SO2 aerosols lived for only a few years.

The nuclear tests went on for more than a decade, even if the cloud is mostly water that results in a massive injection of water into the stratosphere which has significant impact and takes much longer to fall out than solid debris. To penetrate the stratosphere requires an explosion greater than a megaton:
http://en.wikipedia.org/wiki/File:Nukecloud.png
Between them the US and Russia exploded 23 devices greater than 4 Mton, the Russian ones were all in the Arctic where the impact on the stratosphere would be greater because of the lower tropopause (one of those was over 50 Mton).
I don’t think the testing hypothesis can be as simply dismissed as you think.
“ENSO effects mainly see: Foster & Rahmstorf
http://iopscience.iop.org/1748-9326/6/4/044022
I see one fundamental problem with the Foster & Rahmstorf approach: ENSO is not an ā€œexogenicā€ factor, not a forcing in the same sence as volcanoes or insolation. ENSO may be part of the cause of the trend itself (together with the PDO, NAO,ā€¦) and of the warming and standstill episodes. In times of positive PDO, more warm El NiƱoā€™s are prevalent, in times of a negative PDO, El NiƱoā€™s are more suppressed. The point is that El NiƱoā€™s show a residual warming while La NiƱaā€™s show little residual cooling.
Further, what causes ENSO and/or PDO and other oceanic oscillations?
After writing the above, I searched WUWT about this work and found that Bob Tisdale had similar objections:
http://wattsupwiththat.com/2012/01/14/tisdale-on-foster-and-rahmstorf-take-2/

Objections which have been addressed I believe, however whether the method is absolutely correct is not the point, what is clear is that ENSO correlates with global temperature so a period starting with an El NiƱo and ending with a La NiƱa will show a reduction in trend. Clearly this must be taken into account in order to determine whether there is still underlying GHG warming.

Phil.
April 30, 2013 8:59 am

richardscourtney says:
April 30, 2013 at 8:20 am
Phil.:
I will be generous and assume your post at April 30, 2013 at 7:55 am demonstrates you have severe reading difficulties and need to undertake remedial classes in reading comprehension.
Your twaddle is ā€“ not unusually from you ā€“ a pack of lies.
My post at April 30, 2013 at 4:23 am fully explained why the ā€œsinksā€ ARE NOT RELEVANT to my statement.

Actually so far you have made no attempt to explain your reasons you just assert that it’s true.
This is typical of you when challenged on an assertion, you make ad hominem attacks, bluster but never address the point. Just as you have done here.
You are not merely ā€œa rather unpleasant individualā€: you are an anonymous coward who hurls abuse and lies from behind the shield of anonymity. I merely tell the truth as I see it.
No abuse from me, that’s your stock in trade, and no lies from me either, you however still persist in the lie that I altered your statement for example.
As for quoting the Sermon On The Mount at me, I strongly suspect that my knowledge and understanding of it are at least superior to yours.
Not supported by your behavior here, I’m urging you to follow the advice, that you know it and understand it aren’t relevant to your not doing so!
But that is also not relevant to your having stated lies.

No lies from me, I note that you are unable to produce one. Posting a statement you disagree with is not lying!

April 30, 2013 10:13 am

Phil.:
I have reached the conclusion that you lie so often that you have forgotten what a lie is.
I will not take up several pages by itemising each of your lies in this thread. I merely point to some in your most recent post at April 30, 2013 at 8:59 am because it demands that I provide examples.
You lie

Actually so far you have made no attempt to explain your reasons you just assert that itā€™s true.

I explained it repeatedly. For example, in my post at April 30, 2013 at 4:23 am where I wrote

However, if such a trivial increase to total CO2 emissions as the anthropogenic emission can have a catastrophic effect on climate then it is a wonder that no such catastrophe has happened in the 4.5 billion years since the Earth gained an oxygen-rich atmosphere. And THAT WAS MY POINT.

You lie

This is typical of you when challenged on an assertion, you make ad hominem attacks, bluster but never address the point. Just as you have done here.

That is not only a lie, it is also a clear example of psychological projection.
(a)
Your statement that I am addressing is bluster but I do not bluster because I have no need to do it.
(b)
You throw ad homs. all the time. For example, you said to me, “youā€™re a rather unpleasant individual” but I did not reply with an ad hom.. Instead, I made a factual response saying,
“you are an anonymous coward who hurls abuse and lies from behind the shield of anonymity”.
(c)
I always address a point as I have here. Indeed, this argument is your refusal to address a point and, instead, changing the subject (from harmful warming to carbon sinks).
I could go on but there is no point. You are not worth bothering with and others can see the truth of the matter.
Richard

tjfolkerts
April 30, 2013 11:18 am

I have usually found it wise to (try to) adhere to a variation of “Hanlon’s Razor” during online discussions:

“Never attribute to malice (or dis-ingenuousness or lies) or that which is adequately explained by stupidity (or misunderstanding or brevity).”

Calling a statement “half the story” is neutral; calling it “disingenuous” ascribes motives that cannot be independently verified.
Calling a statement “incorrect” is neutral; calling it “a lie” ascribes motives that cannot be independently verified.
Attacking someone’s motives is almost certain to lower the quality of a discussion, while addressing their words almost certainly raises the quality of the discussion. Whether or not someone is intentionally lying really has no bearing — the ideas stand or fall on their own merits.
So for example, we can address this statement by Richard:
And I asserted that the increase of the anthropogenic emission is a trivial increase to the total emission: IT IS.
Richard has already stipulated that 1/35 ~ 3% of the CO2 emissions into the atmosphere are anthropogenic (which sounds about right). So the questions becomes “is this 3% ‘trivial’?”
“Trivial” could mean “statistically insignificant increase to the total emissions each year”. While a 3% is a fairly small value, the other numbers for the carbon cycle seem well enough known that 3% is measurable within a given year.
“Trival ” could mean “statistically insignificant increase to the total emissions over time” but that is countered by graphs like this:
http://www-das.uwyo.edu/~geerts/cwx/notes/chap01/Image18.gif
For centuries, nature was “emitting 34 units” and “absorbing 34 units” of CO2 in a pretty steady balance, not varying much from 280 ppm total. Then the total started to up slowly. And then faster. And then faster yet. Coinciding almost perfectly with human emissions (and from what I understand, supported by isotope ratios). The net result is ~ 40% increase in the last 200 years, over half of which was in the past 50 years due to that extra “1 out of 34” anthropogenic CO2 molecules.
So the effect is statistically significant in each year AND has a long-term effect.
“Trivial” could also mean “no actual impact”. But we have from Richard himself that this change has impacted plant growth in a positive way (other factors being equal). Furthermore, most people ascribe at least SOME warming to CO2, which impacts heating and cooling costs, as well as agriculture and sea level. Warming even 1 degree C would significantly change all of these.
So the effect is statistically significant in each year AND has a long-term effect AND has real impacts (both positive and negative).
In what way is this anthropogenic CO2 “trivial”?

April 30, 2013 11:26 am

Mr. Folkerts says: “For centuries, nature was ā€œemitting 34 unitsā€ and ā€œabsorbing 34 unitsā€ of CO2 in a pretty steady balance, not varying much from 280 ppm total.”
Assuming this to be true, which I do not, what then explains the Holocene Optimum temperature being so much higher than today?

April 30, 2013 12:04 pm

tjfolkerts:
Thankyou for your post at April 30, 2013 at 11:18 am which asks me

In what way is this anthropogenic CO2 ā€œtrivialā€?

I thought – and still think – I explained that in the context in which I stated it.
And I expanded on that explanation in my post at April 30, 2013 at 4:23 am where I wrote

However, if such a trivial increase to total CO2 emissions as the anthropogenic emission can have a catastrophic effect on climate then it is a wonder that no such catastrophe has happened in the 4.5 billion years since the Earth gained an oxygen-rich atmosphere. And THAT WAS MY POINT.

Simply, it is a very small variation in emission which could have been expected to have happened previously but did NOT have catastrophic effects. The change to the emission is a trivial alteration to the climate system.
The argument broke out because Phil. (he says deliberately) changed my statement so it ignored the putative climate effects and switched the subject to the carbon cycle (i.e. “sinks”) then said I was “disingenuous” because I stated what I wanted to say instead of what he wanted said.
Your post I am answering also discusses the possible effect on the climate system.
OK. I will briefly answer that. But it is not relevant to this thread and it is not what I was talking about: it is what Phil. said he preferred me to talk about..
You assume the ice core data indicates a stable ~280 ppmv atmospheric CO2 concentration prior to industrialisation. That is a dubious indication for several reasons, most importantly that the temporal resolution of the ice cores cannot show it. Also, the stomata data do not agree it. There are reasons to doubt both the ice core and the stomata data and it is pure prejudice to rely on either one then to ignore the other.
Ferdinand asserts that the seasonal variation in atmospheric CO2 should be ignored. But our investigation of the carbon cycle indicates it is very informative.
(ref. Rorsch A, Courtney RS & Thoenes D, ‘The Interaction of Climate Change and the Carbon Dioxide Cycle’ E&E v16no2 (2005) )
The dynamics of the seasonal variation indicate that the natural sequestration processes can easily sequester ALL the CO2 emission (n.b. both natural and anthropogenic) of each year. But the annual increase indicates that the sequestration processes don’t sequester all of the emissions.
Furthermore, the considerations of seasonal variation and all other available data strongly suggest that the anthropogenic emissions of CO2 will have no significant effect on atmospheric CO2 concentration. The main reason is that the rate of increase of the anthropogenic production of CO2 is very much smaller that the observed maximum rate of increase of the natural consumption of CO2. Indeed, the dynamics of the seasonal sequestration show it is more than two orders of magnitude smaller.
So, the question is why don’t the natural sequestration processes sequester all the emissions (both natural and anthropogenic) when it is clear that they can?
The answer has to be that the equilbrium of the carbon cycle has changed and the system is adjusting to the new equilibrium. Some of the rate constants in the system are years and decades so the system takes decades to adjust to a new equilibrium. And the rate of adjustment is not constant because it is affected by variations e.g. of temperature.
There are many possible causes of such an alteration to the equilibrium. One possibility is the anthropogenic emission, but there are several natural possibilities. A few hours ago on another thread I stated one natural possibility to Phil.. It is here
http://wattsupwiththat.com/2013/04/30/usa-todays-breathless-co2-announcement-not-quite-there-yet/#comment-1292546
Richard

tjfolkerts
April 30, 2013 5:01 pm

mkelly,
1) I was basing my conclusion on the graph I referenced (and various comments I have heard in the past). Of course, all measurements are subject to some uncertainties (and potentially to biases of the person presenting the data). All in all, I have pretty good confidence in ice cores, but it would have been really nice if plant stomata had been in better agreement. I guess we need one MORE technique to help judge between the two.
2) I don’t know the origins of the Holocene warming! It is clear that there are drivers of the climate besides CO2 that are important. For one thing, precession changed the seasonal input of light significantly over the last ~ 10,000 years, which could have had an impact. (It is an eternal challenge in climate studies that the only REALLY good data we have covers only the last ~50 years. That makes it tough to come to firm conclusions, esp[especially among those of us (me included) who don’t have the luxury of lots of free time to become experts on all the dat aand all the techniques). )

May 1, 2013 3:27 am

The dynamics of the seasonal variation indicate that the natural sequestration processes can easily sequester ALL the CO2 emission (n.b. both natural and anthropogenic) of each year.
The net result of each year over the past 50 years is that the natural sequestration processes don’t sequester all CO2 emissions (whatever the source). That is the fundamental error in your reasoning. Each year there is a residual increase in the atmosphere which is about halve the human emissions in quantity.
Thus while the quantities involved are huge, the seasonal exchange is limited in quantity, depending on temperature changes over the seasons. But that temperature dependency is not more than 5 ppmv/Ā°C. This results in a natural variability not more than halve the human emissions as can be seen in the trend over the past 50+ years.
Thus the human emissions are highly relevant to the increase in the atmosphere.

May 1, 2013 4:07 am

Ferdinand:
I stated why the seasonal CO2 variation is – in my view and that of some others – very important.
You have stated why the seasonal CO2 variation is – in your view and that of some others – not important.
I suggest that on this thread we leave discussion of the carbon cycle at that. We are discussing carbon cycle issues on another thread where it is relevant to the thread.
Richard

Kristian
May 1, 2013 11:41 am

joeldshore says, April 28, 2013 at 11:02 am:
“Now that I have provided conclusive evidence that your ā€œnot ā€˜beliefā€™ā€ was completely wrong, youā€™ve decided these are ā€œthe same tired old arguments from assertionā€ or ā€œarguments from authorityā€.
It’s not something I have ‘decided’. They ARE tired old arguments from assertion and assumption. Nothing is being shown, quantified or substantiated at all about the claimed effect. It is just asserted as real. Like an incantation.
No, when I stated that what I’m arguing for on this thread is in agreement with what basic physics and physics textbooks are saying, I was specifically NOT referring to the kind of slapdash twaddle you decided to present, but to the actual PHYSICS that these textbooks actually handle. Like basic thermodynamics.
“So, basically, you wonā€™t accept our attempts to explain to you the basic concept of conservation of energy (…)”
Sigh. You people are utterly entrenched in your belief that the sphere and the shell is and should be treated as ONE thermodynamic system. Hence your wildly confused idea that the heat loss from the SHELL has to balance the heat gain to the SPHERE at dynamic equilibrium. This is your entire argument: The sphere needs to heat up so that the shell can put out a flux to equal the sphere’s input. Completely ridiculous! And you try to lecture me on the concept of conservation of energy. You evidently don’t even know what the 1st law of thermodynamics is actually saying and what it refers to!

Kristian
May 1, 2013 11:45 am

tjfolkerts says, April 28, 2013 at 11:51 am:
“The problem, Kristian, is that even when I *do* provide exactly what you are asking for and walk you through the steps, you invent objections like ā€œyou are treating it as one systemā€ even though I was following exactly your arguments (often evening using your notation).
Point to ONE SPECIFIC equation that you object to in my post. Explain why you think it is wrong.”

Good grief!
No. You specifically did NOT follow my argument. You dropped the outward J1 and ended up with the heat loss for the sphere as being J – J1. That is, right there and then you jumped back to treating the sphere and the shell as ONE thermodynamic system. You didn’t follow my argument at all. I explained it to you quite carefully in that very post what would happen if you treat the sphere and shell as two separate systems. But you simply chose to ignore it. As you tend to do …
I already pointed to ONE SPECIFIC equation in your post, Tim. The one where you claimed you ‘found’ my error, when the error in fact was all yours.

Kristian
May 1, 2013 11:52 am

tjfolkerts says, April 28, 2013 at 1:42 pm:
“Wherever you got ā€œ3,8 x 10^-24 W/m^2ā€³, it is not even close to right.”
This is only getting better and better!
I distinctly stated: According to your logic with the back-adding of energy and the need for the heat loss from the outermost layer to space to exactly match the original input to the instruments inside, ideally, if you let the energy input run constantly (letā€™s say 400 W/m^2 from within to the surface of the instruments), then 4-5 such layers only would be enough to melt the device long before equilibration. With 20 such layers, the steady input to maintain the interior instrument temperature of 290K (400 W/m^2) would only need to be 3,8 x 10^-24 W/m^2!”
I did not set out to portray a real scenario. I set out to represent YOUR scenario, drawn from your specific arithmetical logic used to explain why the planet in the infamous planet/shell model would have to warm up with the shell surrounding it.
The first BB shell cuts the outward flux in half (1/2). The second shell in turn cuts this half flux in half (1/2 * 1/2). The outward flux is now down to one quarter of the original. The third shell then cuts this flux in half again (1/4 * 1/2). And so forth and so on …
With a highly reflective layer/shell, about 90% of the flux from the sphere is reflected back, so only 10% is absorbed. Of the 10%, 5% is ideally reradiated inwards and 5% outwards. That means, only 5% of the original flux from the sphere passes through even the first layer. That’s 1/20. So with two such layers, you have a reduction of the original flux of 1/400, three layers give 1/8000 the outward flux. And so on and so forth …
That’s how I came to the fantastic 3.8 * 10^-24 W/m^2 for the input to maintain a steady interior temperature of 290K and a flux to space from the outermost sheet of the MLI of … yes, 3.8 * 10^-24 W/m^2.
Absurd? Yes, indeed! But the original logic is yours, not mine.
Or, could you please tell me in what way this is a misrepresentation of your planet/shell logic? Can you explain to me the reasoning behind your 1/2, 1/3, 1/4 … succession? As per the planet/shell model.
– – –
“And why do they do that? Exactly for the reason we are talking about! The heat loss is decreased dramatically!”
Yes, indeed! The heat loss to space is reduced by the MLI. The temperature of the interior instruments providing the original flux does NOT need to rise as a result.

Kristian
May 1, 2013 11:56 am

Ferdinand Engelbeen says, April 28, 2013 at 1:43 pm:
“The incoming energy for the sphere is J + J1.”
This is a fundamental mistake you make here.
Yes, I know this is what is claimed in the upside-down world of the radiative GHE.
But you see, J1 is not incoming energy. It is part of the outgoing energy flux J from the sphere. All it can do is REDUCE the outgoing. It cannot INCREASE the incoming.
Incoming is heat gain. Outgoing is heat loss. ‘Back radiation’ (J1) can only reduce radiative heat loss. It can not increase the heat gain.
And I’ve specifically accounted for the reduction in Q’ by virtue of the ‘back-radiated’ J1. You cannot then also add J1 to J in the next step. J1 is not HEAT.
The expression is right there for the sphere:
J = (J – J1) + J1 = (400 – 200) + 200 = 400 W/m^2
I don’t see what’s not to understand here …

joeldshore
May 1, 2013 12:03 pm

Kristian says:

No, when I stated that what Iā€™m arguing for on this thread is in agreement with what basic physics and physics textbooks are saying, I was specifically NOT referring to the kind of slapdash twaddle you decided to present, but to the actual PHYSICS that these textbooks actually handle. Like basic thermodynamics.

That makes no sense. The textbooks I am referring to are using this as an example applying the basic concepts of thermodynamics (and electromagnetic radiation). Why would they have a clearly incorrect application of physics (as you seem to think this is) as one of their basic examples? Look, what you have is a bizarre religious conviction that violates the laws of physics but you are going to refuse to drop it no matter what evidence we present because you are wedded to your religious convictions.

Hence your wildly confused idea that the heat loss from the SHELL has to balance the heat gain to the SPHERE at dynamic equilibrium. This is your entire argument: The sphere needs to heat up so that the shell can put out a flux to equal the sphereā€™s input.

Yes, because by the setup of the problem, the shell is the only one that is emitting to the surroundings and the sphere is the only one receiving thermal energy from its surroundings (or from conversion from other forms of energy if you want to imagine a sphere being heated by internal processes). Usually, one attacks an argument by explaining what it wrong with it and by explaining how your notions obey conservation of energy. You have done neither.
Considered as two separate systems, each one also has to obey energy conservation and doing that calculation yields the exact same result.

Kristian
May 1, 2013 12:04 pm

Gary Hladik says, April 28, 2013 at 11:19 pm:
“Kristian says (April 28, 2013 at 6:15 am): ā€œCan anyone here provide me with a single physics textbook example (or better yet, an actual account of a controlled experiment conducted in a vacuum chamber) where it is shown, or even remotely discussed, that supplied with a constant energy input, the central sphere will heat up beyond the original input energy temperature with the shell in place around it?ā€
Such examples seem to be few and far between (…)”

Such examples seem to be nonexistent, Gary. And why do you think that is? Doesn’t that fact make you even just a little bit discomforted in your self-assured position?
“(…) but after a short search I found one that seems to come close (problem 7.12 on page 245):”
I’m sorry to tell you this, Gary, but it does not come close. Here is what it says:
“Initially, the sphere is at 200K and is suddenly irradiated with a laser providing an irradiation of 100W (normal to beam) to raise its temperature rapidly to its melting point (2741K). Determine the time required to reach the melting point.”
This is about heating rates (the other side of ‘cooling rates’). The sphere is heated by the laser input and nothing else. It is specifically not mentioned anywhere that the surrounding shell adds to the input from the laser to make the final temperature of the sphere hotter than it would have been with only the laser and the sphere present. All it says is the following:
“A spherical aluminum shield (with a circular cutout) is placed around the sphere as shown, to reduce heat loss from the sphere.”
Yes, all the shell does and can do is expedite the heating rate of the sphere so that we get to the final temperature in less time.
As soon as the sphere were to be no longer heating (or, as in other cases, cooling), the shell could not do anything further. It could not (and can not) increase the temperature of the sphere beyond the source input.
– – –
“(…) Iā€™m not clear how Kristian views the ability of a reflection to heat its source.”
Gary, it wouldn’t surprise me in the least if you actually walk around thinking that reflected light (like from a mirror) can indeed heat its source.
Then switch on a flashlight and seal it in a thermos bottle and watch the whole thing melt down (or explode, whatever your preference) in no time!
The humanity …

Kristian
May 1, 2013 12:07 pm

So, no one was able to come up with a single real-world experiment or even a physics or engineering textbook example that theoretically deals with the heating of a sealed source of constant power beyond the input from its heat source … The only one even making an attempt was Gary Hladik.
I wonder why.
But cocksure and patronising they remain! Nothing new in that department. Even though their pet hypothesis is fundamentally based on no real, observed, theoretically quantified physical mechanism at all. Only assertions and assumptions of an alleged effect somewhere …

joeldshore
May 1, 2013 12:40 pm

Kristian says:

There are more examples of course showing the exact same thing, that the source of the shellā€™s or the insulating layerā€™s incoming (warming) radiative heat flux (across a vacuum) does not heat up some more when surrounded or covered by the shell/insulating layer. The ā€˜insulating effectā€™ simply amounts to reducing ā€˜the total power radiated to the surroundingsā€™ (Q ā€“> Qā€). There is no energy ā€˜piling upā€™ at the surface of the sphere/inner layer. Q (J) remains unchanged.

When designing a spacecraft, why would one care about reducing the total power radiated to the surroundings if one isn’t concerned with temperature control of the object inside? Are we worried about satellites causing warming to outer space?
JPL says ( http://www2.jpl.nasa.gov/basics/bsf11-4.php ):

The image above shows the Cassini spacecraft fitted with its multilayer insulation thermal blanketing. Their gold color results from a reflective silvery aluminum coating behind sheets of amber colored Kapton material. MLI reflects sunlight to shade the spacecraft against overheating, and retains internal spacecraft heat to prevent too much cooling.

Are they misguided in thinking that their insulation prevents too much cooling?

joeldshore
May 1, 2013 6:30 pm

Kristian says:

So, no one was able to come up with a single real-world experiment or even a physics or engineering textbook example that theoretically deals with the heating of a sealed source of constant power beyond the input from its heat source

I’ve given you examples of discussions of exactly that…and for the specific case of the greenhouse effect in case there was any ambiguity…in both of the major textbooks that we teach with here at RIT. Why should we waste our time trying to find still more examples when it is clear that your point-of-view has nothing to do with any evidence that we can provide?
At some point, you can’t force people to abandon ignorance when they are too invested in remaining ignorant.

Bryan
May 2, 2013 1:11 am

Kristian
Its quite possible that some post year 2000 physics textbooks are now including some AGW pseudoscience.
Its a shameful episode.
This is due to commercial pressures because of the vast army of students of Climate Nonsense now at University and also the Jesuit like activities of fanatics like Joel with the new Green Inquisition.
I mentioned the howling mistakes (even from an IPCC perspective) in the previous textbook cited by Joel.
Its also worth noting that Joel appears to know about the Popper test of falsification for eligibility of a scientific theory.
Yet Joel cannot apply this test to CO2 driven catastrophic AGW.
Come on Joel give some test that could falsify this theory or agree with me that it is pseudoscience.

Kristian
May 2, 2013 4:47 am

joeldshore says, May 1, 2013 at 12:03 pm:
“Yes, because by the setup of the problem, the shell is the only one that is emitting to the surroundings and the sphere is the only one receiving thermal energy from its surroundings (or from conversion from other forms of energy if you want to imagine a sphere being heated by internal processes).”
And this precisely goes to show that you simply cannot grasp the very basic concept of thermodynamic systems, their surroundings and the total of fluxes entering and leaving their boundaries. This is scary coming from a working physicist.
“Considered as two separate systems, each one also has to obey energy conservation and doing that calculation yields the exact same result.”
I’ve already shown upthread how this works out perfectly. But you apparently ignore or refuse to take in what opposing arguments to yours are actually about. What they are and are not saying. So then it’s hard to keep up a factual, rational discussion.
Frankly, I’m getting quite fed up with your bluster … That seems to be all you’ve got.

Kristian
May 2, 2013 4:54 am

joeldshore says, May 1, 2013 at 6:30 pm:
“Iā€™ve given you examples of discussions of exactly thatā€¦and for the specific case of the greenhouse effect in case there was any ambiguityā€¦in both of the major textbooks that we teach with here at RIT. Why should we waste our time trying to find still more examples when it is clear that your point-of-view has nothing to do with any evidence that we can provide?”
Uh, you’ve done no such thing. You’ve given me exactly ZERO examples of what I was ACTUALLY asking for. You’ve only tried your best to divert. Read my post and my request one more time, Joel! It’s not that hard.

joeldshore
May 2, 2013 12:13 pm

Oh…Kristian and Bryan: You can also add what is probably the most popular textbook for a first upper-level undergraduate course in thermal physics / statistical mechanics for physics majors (“Thermal Physics” by Kittel and Kroemer) to your list of banned textbooks. They have committed the heresy of also including a simple discussion of the greenhouse effect: http://books.google.com/books?id=c0R79nyOoNMC&pg=115#v=onepage&q&f=false
I guess teaching physics in a way approved by Bryan and Kristian is going to become as difficult as teaching biology in a way approved by Jerry Falwell & Pat Robertson!

Gary Hladik
May 2, 2013 12:40 pm

Kristian says (May 1, 2013 at 12:07 pm): “So, no one was able to come up with a single real-world experiment or even a physics or engineering textbook example that theoretically deals with the heating of a sealed source of constant power beyond the input from its heat source ā€¦ The only one even making an attempt was Gary Hladik.”
Well, let’s be very clear on what I didn’t find. I didn’t find fully-computed textbook examples of problem 1023 aka “steel greenhouse” that supported either “Spencer-type” physics or “Kristian-type” physics. Kristian is of course free to link experiments or computed textbook examples that support his view (good luck).
What I did find is pretty close to problem 1023, and I only hope the differences won’t confuse Kristian. To start, look at this discussion of thermocouples:
http://eyrie.shef.ac.uk/eee/cpe630/comfun2.html
You’ll see (near Figure 1.2), “The greatest problem with measuring gas temperatures is combatting radiation loss. This cannot be ignored and should always be estimated to make sure the difference is not excessive. There are a number of commonly used methods. The simplest is to reduce the size of the thermocouple…
The next simplest method is to surround the probe with a radiation shield [see Fig 1.2] The gas is free to pass through the shield and the shield and thermocouple are heated. The thermocouple bead radiates to the shield which is much hotter than the surrounding walls. Thus the radiative loss and hence temperature error is significantly reduced.” (Thanks to “Phil.” who provided this link in an earlier thread at WUWT.)
Here’s a computed textbook example, showing thermocouple temps before and after a radiation shield is added (the topic starts on page 297, example 9.25 is on page 298):
http://books.google.com/books?id=eQ6yfl3cUKsC&pg=PA298&dq=radiative+heat+transfer+thermocouple+shield&hl=en&sa=X&ei=1f6BUd_pLOPHiwLj2YGoCg&ved=0CDsQ6AEwAA#v=onepage&q=radiative%20heat%20transfer%20thermocouple%20shield&f=false
The same example is also in the 2011 edition, but the preview excludes part of the solution.
Note that the bead temperature is higher after the radiation shield is added, even though the steady-state conductive heat flow to the bead is lower because of the smaller temperature difference between bead and gas stream. The difference is the “back radiation” from the cooler shield, which Kristian claims should have no effect on the warmer bead.
This is also textbook confirmation of Dr. Spencer’s comment (April 24, 2013 at 1:37 pm): “Temperature is a function of BOTH energy input (typically not temperature dependent) AND energy loss…”, which Kristian has disputed (April 25, 2013 at 2:35 pm).
A second computed example (Problem 12-089) is from this book:
http://www.amazon.com/Heat-Transfer-Practical-Yunus-Cengel/dp/0072458933/ref=sr_1_26?s=books&ie=UTF8&qid=1367520360&sr=1-26&keywords=Yunus+Cengel
The solution is here:
http://www.belgeler.com/blg/1k20/heat-chap-12-088
This one is tricky, because when the shield is removed the gas temp is raised to keep the thermocouple reading constant. However, it’s clear that the shield keeps the thermocouple closer to the gas temp.
Now these examples aren’t experiments, but thermocouples manufactured according textbook principles are used in any number of practical applications, e.g. jet engine design. If Kristian thinks the textbooks are wrong, I suggest he refrain from flying. šŸ™‚
I don’t expect Kristian to accept these examples (by now he’s dug in so deep he’s tap dancing in Lucifer’s attic), but he certainly can’t claim any longer that he’s in accordance with accepted physics. If he still thinks he is, Kristian should be able to find computed textbook examples that
support his view (again, good luck).
Kristian, the ball’s in your court.

tjfolkerts
May 2, 2013 12:59 pm

Konrad asks: “Can you explain to me the reasoning behind your 1/2, 1/3, 1/4 ā€¦ succession?”
Clearly I cannot. I have tried once already. Joel has tried. I linked to the wikipedia article for MLI that discusses the problem. There are lots of places on the internet that explain this problem (this one looks good — http://cybele.bu.edu/courses/gg612fall99/gg612lab/lab1.html). The fact that you can’t even understand the physics of a single layer is further proof that I cannot explain this to you in a way that you understand what I am saying.
But at least readers will see what people on “both sides” of this issue think and will hopefully be able to decide for themselves what is correct.

Gary Hladik
May 2, 2013 1:22 pm

tjfolkerts says (May 2, 2013 at 12:59 pm): “There are lots of places on the internet that explain this problem …”
For example Leonard Weinstein at The Air Vent:
https://noconsensus.wordpress.com/2012/07/20/why-back-radiation-is-not-a-source-of-surface-heating/
where he derives equation 8: Tā€™/To=(N+1)^0.25
T’ is temperature of sphere with shell(s)
To is sphere temp without shell(s)
N is number of shells
For people who dislike the term “back radiation”, Weinstein explicitly does not attribute “heating” of the sphere to “back radiation”, although in the comment thread we see the distinction, while perhaps scientifically rigourous, is basically semantic for us laymen.
This article also explains why Kristian’s comment (May 1, 2013 at 11:52 am) about multiple shells is merely demolishing a straw man. Kristian, when disputing someone’s calculations, you should at least understand how the disputed calculation is being done.

Gary Hladik
May 2, 2013 1:52 pm

Kristian says (May 1, 2013 at 11:56 am): [snip]
Summarizing, Ferdinand Engelbeen writes (April 28, 2013 at 1:43 pm): ā€œThe incoming energy for the sphere is J + J1.ā€
Kristian writes: “This is a fundamental mistake you make here… But you see, J1 is not incoming energy. It is part of the outgoing energy flux J from the sphere. All it can do is REDUCE the outgoing. It cannot INCREASE the incoming.”
It just occurred to me that Kristian’s claim is analogous to matter/anti-matter annihilation. You see, half the sphere’s outgoing energy per second (half of 400 watts = 200 watts) is “annihilated” by the “anti-energy” radiated by the inner surface of the shell. The outer shell surface of course radiates “normal” energy at 200 watts, but the inner surface of the same shell radiates “anti-energy”.
Of course the difference between “anti-energy” and anti-matter is that matter/anti-matter annihilation yields energy equivalent to the original reactants, whereas energy/”anti-energy” annihilation yields…nothing. This is potentially a great way to reduce the cooling requirements of, say, a nuclear power plant. 400 watts of waste heat in a pipe, add a shell–presto!–only 200 watts to the cooling tower! Add enough shells and you could cool the plant with a garden hose! We’ll call it the “Kristian Kooler!” šŸ™‚
Kristian says (May 1, 2013 at 12:04 pm): “Gary, it wouldnā€™t surprise me in the least if you actually walk around thinking that reflected light (like from a mirror) can indeed heat its source.”
Thanks for the clarification. So if I have an incadescent light bulb with constant power input and surround it with a perfectly reflecting shell, i.e. none of the energy going into the bulb escapes the shell, I could pour power into the bulb until Doomsday and the temperature inside the shell would not increase? Am I understanding you correctly?

joeldshore
May 2, 2013 2:26 pm

Kristian says:

Yes, indeed! The heat loss to space is reduced by the MLI. The temperature of the interior instruments providing the original flux does NOT need to rise as a result.

So, let’s say the instruments were generating 100 W initially. Now, because of our heat shield, only 10 W is escaping to space. What happens to the other 90 W? Does it just disappear into the ether?

tjfolkerts
May 2, 2013 2:44 pm

Kristian says: I donā€™t see whatā€™s not to understand here ā€¦
The arithmetic, of course, has no flaws. The problem comes from your lack of clear definitions of the terms you use and what the equations mean.
Most specifically you say:
“J1 is not incoming energy. ….
ā€˜Back radiationā€™ (J1) … ”
So what is “J1”?. Because “back radiation” IS EXACTLY “incoming IR energy ” from the sphere (or the atmosphere for earth). You can’t have “back radiation” that is not “incoming energy”.
And what is “J” for that matter? It seems that it represents both “IR flux from the surface” and “heat input from the heaters” in your equations. While these might be the same, there is no a priori reason that these two very different quantities must be the same. That is something that must be proven. (And in this case, the somewhat surprising answer is that the surface radiates TWICE as much IR as the heaters directly provide.)

Bryan
May 2, 2013 4:21 pm

Joel Shore says
“You can also add what is probably the most popular textbook for a first upper-level undergraduate course in thermal physics / statistical mechanics for physics majors (ā€œThermal Physicsā€ by Kittel and Kroemer ) ”
1980 edition?
Joel your link does not give any indication of the conclusion of the discussion.
Please outline relevant details.
My old school textbook A Level Physics by Nelkon & Parker 1973
(pitched about first year American University Standard);
also discuss the Greenhouse Effect.
But dismiss it as being falsified by R W Wood.
By the way what has happened to the falsification test for CO2 driven catastrophic greenhouse effect?
You might as well admit that you hold a belief similar to a religious conviction that is however not subject to scientific scrutiny.

tjfolkerts
May 2, 2013 4:39 pm

Oops … I was typing too fast. That should, of course been …
“Because ā€œback radiationā€ IS EXACTLY ā€œincoming IR energy ā€ from the sĢ¶pĢ¶hĢ¶eĢ¶rĢ¶eĢ¶ shell (or the atmosphere for earth)…. “

Gary Hladik
May 2, 2013 5:03 pm

Bryan says (May 2, 2013 at 4:21 pm): ” My old school textbook A Level Physics by Nelkon & Parker 1973 (pitched about first year American University Standard); also discuss the Greenhouse Effect. But dismiss it as being falsified by R W Wood.”
Could you quote it directly? I found a text-only version of the 3rd Edition (1970) online here
http://archive.org/stream/AdvancedLevelPhysics/NelkonParker-AdvancedLevelPhysics_djvu.txt
It explains that a (real) greenhouse warms by inhibiting convection. It only mentions R W Wood in a discussion about “optical resonance” of mercury vapor.

Gary Hladik
May 2, 2013 6:13 pm

Gary Hladik says (May 2, 2013 at 5:03 pm): “I found a text-only version of the 3rd Edition (1970) online here…”
And a scanned version here
http://archive.org/details/AdvancedLevelPhysics
It’s not searchable, but it has figures and everything. R W Wood appears on p 359, the real greenhouse on p 360. Two more tidbits:
On p 344 we find the famous Herschel experiment, in which he passed sunlight through a prism and bathed a thermometer in the various resulting colors. Both visible and infrared light raised the thermometer temp, falsifying (again) Myrrh’s bizarre claim that visible light can’t heat a surface. šŸ™‚
A thought experiment on p 356 illustrates “Prevost’s Theory of Exchanges” by moving a warm body C to an enclosure D at lower temperature, or to enclosure F at a higher temperature: “It seems unreasonable to suppose that C stops radiating when it is transferred to F; it is more reasonable to suppose that it goes on radiating but, while it is cooler than F, it absorbs more than it radiates”. There are no computed examples, but it is a pre-2000 reference to so-called “back radiation”. šŸ™‚

joeldshore
May 2, 2013 7:25 pm

Bryan says:

Joel your link does not give any indication of the conclusion of the discussion.

Well, I know google books can be a little finicky about previews, but it works fine for me. Basically, it goes through the simple 1-blackbody shell model illustrating the greenhouse effect and then refers the reader to a source which I think might be the first IPCC report for more details (although the edition year is 1980, it seems that the discussion of the greenhouse effect and Bose Einstein condensation were added in a 1994 reprinting…Let the conspiracy theories begin!!!).

My old school textbook A Level Physics by Nelkon & Parker 1973
(pitched about first year American University Standard);
also discuss the Greenhouse Effect.
But dismiss it as being falsified by R W Wood.

Well, I certainly can imagine it possible that a book of that era could mistakenly say something like that. On the other hand, I found the 1970 edition here http://archive.org/stream/AdvancedLevelPhysics/NelkonParker-AdvancedLevelPhysics#page/n371/mode/2up/search/greenhouse , searched it and all I could find was a single paragraph on greenhouses; it does talk about the fact that greenhouses work by inhibiting convection (p. 360) but it doesn’t say anything about the atmospheric greenhouse effect. Since you often seem to get very confused on basic distinctions like this, perhaps you are misremembering a bit?

By the way what has happened to the falsification test for CO2 driven catastrophic greenhouse effect?

That is not really the topic of this current thread. But, just to digress a bit, let’s look at all the previous falsification tests that it has passed since Arrhenius first proposed it a century ago:
(1) Scientists thought the oceans could absorb all the CO2 we emitted and hence that it would not be building up in the atmosphere. It was not until the modern measurements in the lat 1950s, along with the theoretical work of Revelle establishing why the oceans would not absorb it all, that it was understood what actually happens.
(2) Scientists thought that the effect would saturate. It was not until the modern era of radiation transfer theory (and its subsequent verification by the entire field of remote sensing and, in particular, measurements of the radiation seen by satellites from space) that it was understood that such saturation does not occur.
(3) When first proposed, the notion of the water vapor feedback, while plausible, did not have empirical evidence to back it up. Now, there is very good evidence that the feedback operates as predicted ( http://www.sciencemag.org/content/323/5917/1020.summary ) , especially for temperature fluctuations on timescales on the order of months. [It is harder to verify long-term trends in water vapor because of instrumental and satellite changes in radiosondes over time, although the most reliable data do appear to show it, and it is also hard to explain how a feedback operating on the timescales it does could work on the monthly to yearly timescales and fail over the longer timescales.]

joeldshore
May 2, 2013 7:32 pm

Whoops…After submitting, I see that Gary has beaten me to it in finding the 1970 edition of Nelson & Parker.
I also meant to mention at the end of my last post that what I wrote is just a few things I thought of off the top of my head and not meant to be an exhaustive list.
The point is that scientists are testing aspects of AGW every day and the notion that it is not falsifiable is hokem. That said, some people may be disappointed that their grade school notions of falsifiability don’t really apply quite so neatly in the real world, where empirical data and analysis of that data are subject to error to and for nearly all major theories one could dig up data that seems to contradict the theory. It is a constant process of trying to reconcile theory and empirical data / experiment and, once theories are well-established, they are…for good reason…not going to be tossed out every time some piece of empirical data seems to disagree with them.

Kristian
May 2, 2013 10:38 pm

Gary Hladik says, May 2, 2013 at 12:40 pm:
“Well, letā€™s be very clear on what I didnā€™t find. I didnā€™t find fully-computed textbook examples of problem 1023 aka ā€œsteel greenhouseā€ that supported either ā€œSpencer-typeā€ physics or ā€œKristian-typeā€ physics. Kristian is of course free to link experiments or computed textbook examples that support his view (good luck).”
Gary, EVERY textbook example supports my position on this. NONE of them supports yours. Why? Because my claim is that is isn’t an issue, because it isn’t an actual physical phenomenon. It isn’t a thing to be discussed. Because it doesn’t happen. It isn’t a thing. Your claim is that it truly and clearly and obviously and naturally is. SO WHY ISN’T IT EVER MENTIONED OR DISCUSSED AT ALL?!!

Kristian
May 2, 2013 10:51 pm

Gary Hladik says, May 2, 2013 at 12:40 pm:
Gary, kudos for at least trying. Joel and Tim here are doing nothing but frantically waving their hands. No substance or argumentative value whatsoever.
But your thermocouple example just does not cut it. The thermocouple is trying to measure the REAL temperature of the gas. The REAL temperature of the gas is always WARMER than what the thermocouple can measure. The only thing we can do to help the thermocouple is try and make it come closer to the REAL temperature. The heat shield does nothing to heat the thermocouple, Gary. I can see why you would think that. It is the GAS heating it. But it cannot heat it all the way to its REAL temperature because of the simultaneously ongoing radiative heat loss. Reduce the heat loss and the thermocouple comes CLOSER to measure the REAL temperature of the gas. The issue here, though, is: Can the heat shield do anything to help the thermocouple become WARMER than the gas ‘input’?
THAT is what you need to show, Gary. Because THAT is the underlying claim of this whole discussion.
You fail to grasp what my argument is actually about. What it IS saying and what it’s NOT.

Kristian
May 2, 2013 11:14 pm

Regarding the point I was trying to make in my last post, I want to quote a very relevant and pertinent observation made by commentator CementAFriend on the thread at Tallbloke’s a while back named ‘Entering the SkyDragon’s Lair’.
He said:
“I have not read all the comments but it appears many who are commenting are mentioning supposed models or thought bubbles without having any real experience. The only thing that counts is actual measurement and then relate that to equations, determined from real experience, which could explain the process. This comparison process tells you if your measurements are reasonably accurate or the equations your are attempting to verify are wrong.
I have had experience with measurement in furnaces. My experience is as follows:
a) with a poorly insulated furnace there will be heat loss from the walls which results in a flame temperature less than the theoretical and lower energy in the exhaust gases.
b) as the insulation is improved there is less heat loss from the walls, slightly higher flame temperatures, and higher energy in the exhaust gases
c) even with perfect insulation it is not possible that the flame temperature exceeds that of the theoretical temperature ie energy can not be created to exceed the energy input

When the theoretical (potential) temperature of an object (like 290K with input of 400 W/m^2 for a blackbody) is attained, no amount of heat shields and insulating layers can raise that temperature further. ONLY THE SOURCE OF HEAT for that particular blackbody could do that.

Kristian
May 2, 2013 11:53 pm

tjfolkerts says, May 2, 2013 at 12:59 pm:
“Konrad [my name is Kristian] asks: ā€œCan you explain to me the reasoning behind your 1/2, 1/3, 1/4 ā€¦ succession?ā€
Clearly I cannot. I have tried once already. Joel has tried. I linked to the wikipedia article for MLI that discusses the problem.”

Wikipedia article for MLI? You specifically stated:
“For blackbody shells
1 layer = 1/2 as much heat loss
2 layers = 1/3 as much heat loss (not 1/4)
3 layers = 1/4 as much heat loss (not 1/8)
ā€¦
20 layers = 1/21 as much heat (not 1/ 2^21) So 20 blackbody shells would result in about 5% as much heat loss.”

Let’s take it from the top, Tim. I am not asking you to show me what actually happens in the real world, with real-world logic and equations. I want you to justify the 1/2, 1/3, 1/4 … succession within the framework of your distinctive planet/shell (the Eschenbach setup) logic. You specifically use the halving of the output from the shell’s outer surface to explain the need for the planet to heat up so that the shell output to space grows to equal the input to the planet. Using this logic, if you suddenly were to put TWO shells rather than ONE around the planet, both separated by vacuums, wouldn’t the output to space from the outer shell then (initially, in your case) be one quarter (1/2 * 1/2 – gaining half from the inner shell, splitting it in two and thereby ejecting half of the half from its outer surface) rather than one third of the input to the central planet? And so on …

Bryan
May 3, 2013 1:31 am

joeldshore
“although the edition year is 1980, it seems that the discussion of the greenhouse effect and Bose Einstein condensation were added in a 1994 reprinting”
Its a little odd to have have significant additions without a new edition.
My equivalent textbook Heat and Thermodynamics by Zemansky does not mention the Greenhouse Effect
Its difficult to know which Greenhouse Effect you are advocating.
Vaughn Prat (Stanford) claims he has experimental evidence( large temperature differential) of a Greenhouse Effect in a shoebox sized enclosure.
De Witt Payne has spent the best part of two years trying to do the same but with no results.
Both these efforts would attempt to counter R W Wood and the conclusion of Nelkon and Parker that this Greenhouse Effect is really a result of trapped convection.
Vaughn Prats experiment was badly set up and he no longer defends it.
I don’t know of anyone else who believes in a measurable Greenhouse Effect in a shoebox, perhaps you do?
The other more scientifically respectable version requires a troposphere sized column to display the effect .
Testable predictions of this model include
1. Hot Spot
2. Increasing near surface air temperature
3. Increasing height of ERL
4. Decreasing temperature at this height.
None of these things are currently happening, does that not make you think that the CO2 driven AGW theory is false.
If after 30years (say) of negative indicators for your theory would you abandon it?

Bryan
May 3, 2013 3:03 am

Joel
G&T cite the Kittel book (2000 print) as only referring to the IPCC report.
Rather than presenting a full exposition of the theory.
Indeed they doubt if such a properly presented rigorous version of this conjecture exists
Read Page 35
Falsification Of the atmospheric CO2 greenhouse effects within the frame Of Physicsā€ by Gerhard Gerlich and Ralf D. Tscheuschner; International Journal of Modern Physics B, Vol. 23, No. 3 (2009)
http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf

Kristian
May 3, 2013 4:40 am

Gary Hladik says, May 2, 2013 at 1:52 pm:
“Kristian writes: ā€œThis is a fundamental mistake you make hereā€¦ But you see, J1 is not incoming energy. It is part of the outgoing energy flux J from the sphere. All it can do is REDUCE the outgoing. It cannot INCREASE the incoming.ā€
It just occurred to me that Kristianā€™s claim is analogous to matter/anti-matter annihilation. You see, half the sphereā€™s outgoing energy per second (half of 400 watts = 200 watts) is ā€œannihilatedā€ by the ā€œanti-energyā€ radiated by the inner surface of the shell. The outer shell surface of course radiates ā€œnormalā€ energy at 200 watts, but the inner surface of the same shell radiates ā€œanti-energyā€.”

Gee whiz. Trying out the tactic of putting words in my mouth, building a strawman for you to ridicule. Nice.
Gary, the only HEAT input (that is, positive transfer of thermal energy to produce a higher body temperature) to the sphere’s surface is from its internal heat source. This input is equal to J. This the sphere then emits as HEAT loss, to balance this heat GAIN, from its surface into the vacuum surrounding it. Nothing is annihilated.
The point is rather that YOU need half of this outgoing flux to be disallowed from leaving the surface of the sphere in the first place (or ‘annihilated’) to be able to say that the ‘back radiation’ flux (J1) is only reducing the heat LOSS of the sphere’s surface, not increasing its heat GAIN. You see, according to the Stefan-Boltzmann law, the sphere HAS TO emit 400 W/m^2 from its surface at 290K. It gains 400 W/m^2 from its internal heat source, attains a corresponding BB emission temperature of 290K, and emits accordingly. Nothing of this changes with the ‘incoming’ J1. It cannot stop or restrict J from continuously leaving the sphere’s surface.
So if the sphere’s surface temperature rises beyond this equilibrated emission temperature, then it means it has gained HEAT (a positive (net) transfer of thermal energy) from somewhere else, this being the cooler shell. You are transferring HEAT from a cooler to a warmer object. Only problem then is that the HEAT transfer between the sphere and the shell in fact will still always go from sphere to shell, not the other way around. The back radiation from the shell (J1) reduces this heat transfer rate, diminishing the sphere’s ability to heat the shell all the way to 290K (the shell does receive the full 400 W/m^2 after all and absorbs it, but can only reach a steady temperature of 244K). But it does NOT reverse it.
That is what J1 does, it reduces the shell’s temperature by reducing the incoming J -> J – J1. But in no way or fashion is the sphere’s HEAT input (to set its temperature) ever going to be J + J1!
Once again. J is the flux leaving the surface of the sphere. This is intercepted by the shell, heating the shell, but not to 290K, only to 244K. Why? Because the shell splits the 400 W/m^2 of J and turns it into two times J1, one going inwards, one continuing outwards.
So J leaving the sphere in the end becomes (J – J1) + J1 = J. Both the shell AND space are part of the sphere surface’s surroundings. As is its internal heat source. You cannot leave out the flux continuing to space, keeping the flux going back.
I realise this is all Greek to you so that we’ll just keep going around in circles forever. But at this stage I’ve really said all I can say about the matter. There apparently is no way to make you see your total abuse of the laws of thermodynamics here …

Kristian
May 3, 2013 5:18 am

This is how energy is conserved in the dual sphere/shell system:
All the energy from the original (internal) power source (400 W/m^2) is turned into heat at the sphere’s surface. This creates a temperature of 290K.
All the energy (400 W/m^2) is then emitted from the sphere’s surface.
All the energy (400 W/m^2) is received and absorbed by the surrounding shell.
HALF of the energy (200 W/m^2) then goes into heating the shell to a temperature of 244K. The other HALF of the energy (200 W/m^2) is emitted from the shell’s outer surface to space.
All the energy is accounted for.
But then you want to suggest that the inward J1 is somehow an additional bunch of energy to all this that cannot leave to space but instead ends up being extra HEAT input to the sphere, warming it some more.
You cannot have both heating of the sphere to only 244K (200 W/m^2 incoming HEAT) and emission of (200+200=) 400 W/m^2 outgoing HEAT as if it had a temperature of 290K. You double count the inward J1.

tjfolkerts
May 3, 2013 5:56 am

“HALF of the energy (200 W/m^2) then goes into heating the shell to a temperature of 244K. The other HALF of the energy (200 W/m^2) is emitted from the shellā€™s outer surface to space.”
Wow! Is the mistake really that simple?
When the system has not yet reached steady-state, then indeed some of the energy goes into heating the shell (and heating the sphere as well). This is not necessary half — if the shell is cold then nearly all the energy goes into heating the shell; if the shell is warm, very little actually goes into raising the temperature of the shell.
But once the object has already reached a steady-state condition where the temperature is not changing, then no more energy goes into “heating it to” that temperature. This a the very familiar equation from high school chemistry (and physics)
Q = mc Ī”T
Note specifically the “Ī””, ie the change in temperature. If the shell starts at 244 K and ends at 244 K, then Q = 0 for “holding the shell at 244 K”. There is still the 200 W/m^2 lost to space, so there is still a need for 200 W/m^2 of heating from the planet. So there is still a need for a 200 W/m^2 heater in the planet (but NOT a 400 W/m^2 heater).

joeldshore
May 3, 2013 6:31 am

Bryan, first of all, I have noticed your failure to even address the issue of whether you can find any support for your claim about what your edition of Nelson & Parker in light of the evidence from the 1970 edition that Gary and I have found. Or, do you now admit that you have completely misrepresented what they wrote?
Bryan says:

Its difficult to know which Greenhouse Effect you are advocating.
Vaughn Prat (Stanford) claims he has experimental evidence( large temperature differential) of a Greenhouse Effect in a shoebox sized enclosure.

I guess you could certainly illustrate some aspects of CO2’s absorption of infrared radiation by such an experiment but to truly replicate the greenhouse effect, you need to have system big enough that there is a significant lapse rate, i.e., convection can never equalize the temperature because of the fact that an adiabatic parcel of air cools as it moves up in the atmosphere and expands.

Testable predictions of this model include
1. Hot Spot
2. Increasing near surface air temperature
3. Increasing height of ERL
4. Decreasing temperature at this height.
None of these things are currently happening, does that not make you think that the CO2 driven AGW theory is false.

First of all, the things that you are trying to list all go beyond evidence for the greenhouse effect and address the issue of global warming. Second of all, you have badly garbled them:
1. Even Monckton, the originator of the completely false claim that the “hot spot” is a unique signature of warming due to greenhouse gases now admits (in this very thread) that he was wrong, saying:

The ā€œhot spotā€, which I named, ought to be there whatever the cause of the warming. .

(You are also wrong about the extent of empirical evidence for it, but since that is irrelevant, I won’t pursue it further.)
2. Air temperatures have been increasing. Only people who don’t understand statistics are fooled by the claim that is hasn’t warmed for 16 years. (The trend over this carefully cherrypicked period still has large enough error bars that it is statistically-indistinguishable from both no trend and the ongoing trend since ~1975.)
3. I believe there is in fact evidence for the increasing height of the ERL. I certainly know of no evidence that shows that it hasn’t increased.
4. I have no idea what you are talking about on this one. Can you explain to us why a decreasing temperature is predicted at the ERL?
The empirical evidence for the greenhouse effect is actually overwhelming: It is simply impossible to explain how the Earth’s surface can emit 390 W/m^2 but the Earth as seen from space only emits 240 W/m^2 (as required for radiative balance given what it absorbs from the sun) without this difference being due to the absorption by IR-absorbing elements in the atmosphere. The fact that the spectra of the emissions as seen from space corresponds essentially exactly with what is predicted using radiative transfer theory using the known absorption bands of the various IR-absorbing elements is then “icing on the cake”.

joeldshore
May 3, 2013 6:42 am

Kristian: The error that tjfolkerts has pointed out is exactly what led me to explain how you have what I would call (in analogy with the mechanics case) an Aristotlean view of energy. Aristotle thought that it was necessary to have a net force to keep an object moving at a constant velocity; Newton recognized that you don’t need a net force although in practice (because there is most commonly frictional force slowing things down), you do need a force to balance the friction force so that the net force on the object is zero.
In your view of energy, you think that net energy must be input just to keep an object at a certain temperature. The correct view is that the net energy input to keep the object at a constant temperature is zero. Since the object will be emitting thermal energy, you do need an input to balance that emission. But, that’s not what you are proposing. You are proposing that the energy input to the shell has to exceed the energy that it outputs in order to keep it at a constant temperature.
Or maybe you just haven’t been understanding this whole time that we have been dealing with the steady-state temperature of the objects and not the transients that occur when steady-state has not yet been reached?

joeldshore
May 3, 2013 6:54 am

Kristian says:

I realise this is all Greek to you so that weā€™ll just keep going around in circles forever. But at this stage Iā€™ve really said all I can say about the matter. There apparently is no way to make you see your total abuse of the laws of thermodynamics here ā€¦

You are dealing with people who know statistical mechanics, the theory that underlies and gives meaning to thermodynamics. This is why we are able to distinguish between what the Second Law of Thermodynamics actually says and your magical view of the 2nd Law (which essentially leads to violations of the First Law because you have energy flows that magically can’t influence the temperature of an object).
I suggest you read this article and try to really digest it, with a little bit of self-reflection: http://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect You won’t, of course, be able to for exactly the reasons that Dunning and Kruger explain.

Bryan
May 3, 2013 8:26 am

joeldshore
Can you read?
Look at Nelkon &Parker page 360
The Greenhouse is explained by keeping plants warm through stopping convection.
No mention whatsoever about radiation from the glass having any influence.
These people were quite aware that the the Greenhouse Theory had been falsified by R W Wood.
That then was the consensus view.
Remember in the 1970s climate alarmists like yourself were predicting a new ice age.
Thats why you will find no reference to a greenhouse effect in physics textbooks of the period.
………………………………..
Joel why don’t you present properly the theory you are trying to support?
I said above, highlights and possible testable features include
“2. Increasing near surface air temperature
3. Increasing height of ERL
4. Decreasing temperature at this height.
None of these things are currently happening, does that not make you think that the CO2 driven AGW theory is false.”
You replied
“3. I believe there is in fact evidence for the increasing height of the ERL. I certainly know of no evidence that shows that it hasnā€™t increased.
4. I have no idea what you are talking about on this one. Can you explain to us why a decreasing temperature is predicted at the ERL?”
Read this link
http://www.realclimate.org/index.php/archives/2004/12/why-does-the-stratosphere-cool-when-the-troposphere-warms/
I notice that you will still be a believer in the Greenhouse Theory even if the near surface temperature declines for the next 30 years.
Well that’s fanaticism for you !
What about 300 years of near surface temperature declines .
Still a faithful believer ?

joeldshore
May 3, 2013 8:50 am

Bryan says:

Can you read?
Look at Nelkon &Parker page 360
The Greenhouse is explained by keeping plants warm through stopping convection.
No mention whatsoever about radiation from the glass having any influence.

Let me quote what you originally claimed:

My old school textbook A Level Physics by Nelkon & Parker 1973
(pitched about first year American University Standard);
also discuss the Greenhouse Effect.
But dismiss it as being falsified by R W Wood.

You have had it explained to you countless times that the “greenhouse effect” is a name given to an effect that is known not to be the same exact mechanism by which a greenhouse works (analogous in some ways but not in others, in particular that one involves radiation and the other convection). Yet you still made a false claim about what your physics textbook said. It is hard to avoid the conclusion that you intentionally misled us with your statement. It is hard to imagine your credibility slipping any lower than it already was, but I am afraid it has.

Remember in the 1970s climate alarmists like yourself were predicting a new ice age.

…Except that more papers in the peer-reviewed literature were predicting warming than cooling and the National Academy of Sciences report from the mid-1970s said that the science was at a state where one could identify the various competing factors influencing the climate but could not yet predict which would win out, which is very different from what they are saying today. So, in fact, what we learn from this is that organizations like the NAS are very cautious and trustworthy.

Read this link
http://www.realclimate.org/index.php/archives/2004/12/why-does-the-stratosphere-cool-when-the-troposphere-warms/

That doesn’t address the temperature at the ERL, which is well within the troposphere.

I notice that you will still be a believer in the Greenhouse Theory even if the near surface temperature declines for the next 30 years.

Since the greenhouse theory is based on physics which has incredibly strong empirical support, I don’t see what temperature declines would have to do with it. They would certainly have some impact on understanding of what the climate sensitivity is but not on the greenhouse effect itself.
As for whether I would modify my opinions regarding the climate sensitivity if temperatures were to decline for the next 30 years, sure I would if that were actually to come to pass. Why don’t we check back in 30 years and see who had a better understanding of climate and what the future holds in store?

joeldshore
May 3, 2013 8:55 am

By the way, in regards to what you claimed about Parker and Nelson, it is worse than not distinguishing between the greenhouse effect and the mechanism by which a greenhouse operates. You also invented the fact that they discussed the effect on radiation at all but dismiss it in light of Wood’s experiment. Even in regards to greenhouses only, they did not mention the notion of radiation playing a role and then dismiss it and did not mention Wood at all in this context. Rather, they just talked about how a greenhouse works by limiting convection.
Your whole claim about Parker and Nelson bears essentially no resemblance to reality.

Bryan
Reply to  joeldshore
May 3, 2013 9:58 am

Joel who is this Nelson, you refer to?
I am talking about a book by Nelkon and Parker and gave the page number.
The Greenhouse is not heated by radiation but by suppressed convection.
This observation was made by R W Wood and was the accepted view at that time.
What problem do you have with that?

joeldshore
May 3, 2013 10:55 am

Sorry for the name typo. The problem that I have with that is that you made the completely false claim “My old school textbook A Level Physics by Nelkon & Parker 1973 (pitched about first year American University Standard); also discuss the Greenhouse Effect. But dismiss it as being falsified by R W Wood.”
Now you have changed that claim to a claim that the authors said something that nobody disputes, despite your attempt to imply otherwise, and which is mentioned in many discussions of the greenhouse effect, including that on Wikipedia.
You basically completely invented what Nelkon and Parker said…and what they actually said had absolutely zero bearing in the atmospheric greenhouse effect.

Bryan
May 3, 2013 2:31 pm

Joel
I like to look on a positive outcome, yet you say….
“Now you have changed that claim to a claim that the authors said something that nobody disputes”
What about Vaughn Pratt and De Witt Payne?
You now agree with R W Wood , Nelkon and Parker and all others who have studied the topic that radiation has nothing (or very little) to do with the heating effect in a glass house or greenhouse .
Suppressed convection is the cause.
“You basically completely invented what Nelkon and Parker said”
I gave the textbook and page number and since the book is linked online we can let readers decide what part radiation plays in glasshouses or atmospheric greenhouse analogs

joeldshore
May 3, 2013 2:51 pm

Bryan says:

You now agree ,,,.

I don’t see where me or anybody else here has ever disagreed with the statement that in real greenhouses, the primary mechanism for warming is the blocking of convection.

Gar
I gave the textbook and page number and since the book is linked online we can let readers decide what part radiation plays in glasshouses or atmospheric greenhouse analogs

No…You gave us the name of the book and the year. It took research by Gary and I to find that we could actually access it online and when we did, we found that it didn’t say what you claimed it said. So, basically, at the end of the day, your claim that a physics textbook you used “dismissed” the atmospheric greenhouse effect was not true. It simply did not discuss the atmospheric greenhouse effect whatsoever, which is not particularly surprising.

Gary Hladik
May 3, 2013 3:17 pm

Kristian says (May 2, 2013 at 10:51 pm): “Reduce the heat loss and the thermocouple comes CLOSER to measure the REAL temperature of the gas.”
Interesting. So adding a radiation shield does nothing to the temperature of a radiating sphere, but adding a radiation shield DOES affect the temperature of a radiating thermocouple. And Kristian writes this with a straight face???
Back to our old friend problem 23, only instead of looking at it as a steel greenhouse, let’s consider it as a temperature measuring device. Our blackbody sphere alone (in a vacuum chamber whose walls are maintained at 0 degrees K) radiates 400 W/m^2. To simplify, let’s make the sphere area 1 square meter, so it’s radiating 400 W (at roughly 290 degress K).
We never specified exactly how the sphere is heated, but note that the solution to problem 1023 doesn’t depend on the method of heating. We only require that the sphere radiate 400 W/m^2, i.e. 400 W. So let’s say it’s hollow, and powered by a miniature nuclear reactor cooled by some circulating gas; the gas is the only thing in contact with the sphere. If the inside sphere wall has some insulation, the gas must necessarily be hotter than the surface of the sphere. Then the outside temperature of the sphere will be an imperfect measure of the inside gas temperature, just as temperature of the unshielded thermocouple in example 9.25 is an imperfect measure of the gas surrounding it. What to do?
Hey, let’s add radiation shields to both sphere and thermocouple. Note that our concentric shield around the sphere is so close to it (though separated by a vacuum) that its inner and outer areas are very close to 1 square meter.
We’ve added a radiation shield to both our measuring devices. Kristian says the thermocouple’s temperature increases with shield present and approaches the surrounding gas temp, making it a more accurate measure of the gas temp. With regard to the sphere (our device for measuring the internal gas temperature), Kristian says that in the presence of its snug radiation shield the sphere’s surface temperature…
A) Increases
B) Decreases
C) Remains the same
making it a…
1) More accurate
2) Les accurate
3) Unchanged
measure of the internal gas temp.
Kristian?

joeldshore
May 3, 2013 3:26 pm

Kristian says:

This is how energy is conserved in the dual sphere/shell system:
All the energy from the original (internal) power source (400 W/m^2) is turned into heat at the sphereā€™s surface. This creates a temperature of 290K.

And, here in your first line is already a major error in your thinking. A certain amount of heat absorbed by the body (in this case from its internal power source) does not create a certain temperature when there are other sources of energy around. That is nonsense that Slayers want you to believe. The temperature of an object is determined by the balance between the total amount of energy it absorbs and energy it emits. If the object starts at a certain temperature at which it is absorbing more energy than it is emitting, this net absorption of energy causes its internal (“thermal”) energy and hence its temperature to increase until it gets to the point where it is emitting as much energy as it is absorbing. If, on the other hand, it starts at a certain temperature at which it is emitting more energy than it is absorbing, this net emission of energy causes its internal (“thermal”) energy and hence its temperature to decrease until it gets to the point where it is absorbing as much energy as it is emitting.
If you don’t understand this first basic concept, there is no reason to proceed any further.
For an object that is surrounded by empty space at absolute zero, its net emissions is equal to its gross emissions and can be calculated from the Stefan-Boltzmann equation P = sigma*T^4. However, for an object surrounded by a shell at a temperature T_0, then the net emission from the surface is
sigma*(T^4-T_0^4).
Those of us not allergic to the concept of back-radiation consider the second term to be the energy emitted from the surroundings absorbed by the object (and, in fact, this is what it is and how it is described in all the textbooks). But, even if you don’t want to think of it this way, the fact is that you will find this equation is nearly every physics textbook that talks about thermal radiation (unless it talks about radiative emission in such a cursory manner that it never discusses net emission given an object’s surroundings).
Note that the temperature T^4 of an object whose emissions are described by the equation above depends not only on whatever power it is absorbing from its internal energy source but also on the temperature of the surroundings T_0.

Bryan
May 4, 2013 1:13 am

joeldshore its pretty obvious that you have reading difficulties and further cannot separate real issues from trivia.
I said
“I gave the textbook and page number and since the book is linked online we can let readers decide what part radiation plays in glasshouses or atmospheric greenhouse analogs”
You said
“Noā€¦You gave us the name of the book and the year.”
I said above on May 3, 2013 at 8:50 am
“Can you read?
Look at Nelkon & Parker page 360”
Since there is an online link readers can check for themselves how much part trapped radiation plays in the heating of a greenhouse.
Answer is nothing or negligible according to Nelkon and Parker in line with the experimental results of R W Wood whose views were accepted by all physicists at the time as far as I know.
All subsequent reliable tests confirm Woods findings which must be a bummer for alarmists but you cannot argue with experimental results.
I’m off now for a sailing trip for the next 3 days so my educational efforts here will have to be suspended.

joeldshore
May 4, 2013 5:51 am

Bryan: Everyone can plainly see the sort of deception and sophistry that characterizes your posts here. You badly misrepresented what a textbook said and now you make statements that continue to badly misrepresent the arguments of those who understand the greenhouse effect. You don’t educate; you simply confuse.

joeldshore
May 4, 2013 6:36 am

Kristian says:

There is no net heat flow. There is only net energy flow. The net energy flow is the heat flow. Heat is specifically the surplus energy transferred from a warmer body to a cooler one. There is no ā€˜heatā€™ going the other way. There is ā€˜energyā€™ going the other way, but this can do nothing to make the warmer object warmer still. Only heat can heat.

Kristian: If you really think about this statement, maybe we can take you from your view of physics as some sort of arbitrary and capricious magic to the true beauty of physics. What exactly is it that prevents the energy going the other way from having any effect? Energy is energy and if there is a flow of energy that energy is going to have an effect on the temperature, which is a measure of the internal (or “thermal”) energy of an object. So, is there a little sign on the warm object that says: “No energy flows from colder objects allows?” Do the photons get turned back at the boundary or does there energy they have magically disappear?
The whole beauty and mystery of the 2nd Law of Thermodynamics is exactly the opposite: It is the story about how laws like the fact that heat flows only from warmer to colder objects, facts that give a direction to the arrow of time, arise at the macroscopic scale from completely reversible behaviors on the microscopic scale. The whole idea is that at the level of individual photons, there are no arbitrary and capricious laws. Photons go where they want; deposit their energy when they interact with matter, irregardless of the temperature (since in fact, there isn’t really such a thing as “temperature” at the level of individual atoms or molecules they interact with).
From a real physics point-of-view, the fact that heat flows from hot to cold doesn’t arise from arbitrary and capricious laws about how energy behaves differently in different situations. It arises from the fact that because the rate at which objects emit radiation increases with their temperature, the energy flow from the hot object to the cold object is always greater than the energy flow from the cold object to the cold object and hence, in an interaction involving these two objects only, the hot object will get colder and the colder one hotter until they come to equilibrium.
Now, let’s imagine there are 3 “objects”: The sun, the Earth, and space. The sun is at its temperature T_S, space is at absolute zero (well, really about 3K, but let’s not quibble) and the Earth’s temperature is determined by the need to balance the energy it receives from the sun and the energy it emits back out into space. Note that the heat, or net energy flow, is from sun to Earth to space.
Now, add a 4th object: an IR-absorbing atmosphere. Now, the temperature of the Earth (and the temperature of the atmosphere) will be determined by the more complicated energy balance involving all 4 objects. The claim that the temperature of the Earth in this case can’t be higher than in the case of 3 objects above has absolutely no foundation in the 2nd Law of Thermodynamics (or any other law of physics). It is simply made-up nonsense that relies on some sort of capricious view of the laws of physics that is much less beautiful and imaginative than the real laws of physics.

Konrad
May 5, 2013 7:38 pm

Shore
You have been very reluctant to discuss the role of radiative gases in tropospheric convective circulation. Radiative gases are of course critical for continued convective circulation and atmospheric cooling in the troposphere. This has been established science for some time. A simple explanation of the role of radiative gases in convective circulation can be found here –
http://www.st-andrews.ac.uk/~dib2/climate/tropics.html
Without radiative cooling at altitude and convective circulation below the tropopause, our atmosphere would heat dramatically.
However at AGW supporter site Scienceofdoom we find this attempt to write the role of radiative gases in convective circulation out of atmospheric science –
http://scienceofdoom.com/2012/12/23/clouds-water-vapor-part-five-back-of-the-envelope-calcs-from-pierrehumbert/
– which includes the following summary of Pierrehumberts wholly un-empirical 1995 claims –
ā€œSo increasing the emissivity from zero (increasing ā€œgreenhouseā€ gases) cools the climate to begin with. Then as the emissivity increases past a certain point the warm pool surface temperatures start to increase again.ā€
You have often offered up Pierrehumberts work as justification for the AGW hypothesis, yet every time I have raised the issue of the role of radiative gases in cooling a moving atmosphere you have run back to static atmosphere two shell models. Do you support Pierrehumberts 1995 claims? Is there any other scientist you are aware of making such claims before 1990?

tjfolkerts
May 6, 2013 3:21 am

Konrad,
Before we think about moving on to a very different topic, do you now agree that your interpretation of energy flows and temperatures for the shells we have been discussing was completely wrong? Do you agree with Joel and Gary and me that a “bare” blackbody planet with heaters set to 400 W/m^2 will be 288 K, while the same planet with the same 400 W/m^2 heaters but now surrounded by a blackbody shell will be radiating 800 W/m^2 of thermal IR radiating @ T=345 K (and receiving 400 W/m^2 of “back-radiation”) at the planet’s surface? (Or conversely, that we could lower the heaters to 200 W/m^2 and still keep the planet’s surface @ 288 K when the shell is in place?)

Konrad
May 6, 2013 5:37 am

Tim,
As I have argued at all times, the two shell radiative model is irrelevant to an atmosphere in which the gases are free to move. Again lets review the ā€œdo notsā€ of atmospheric modelling –
1. Do not model the ā€œearthā€ as a combined land/ocean/gas ā€œthingyā€
2. Do not model the atmosphere as a single body or layer
3. Do not model the sun as a Ā¼ power constant source without diurnal cycle
4. Do not model conductive flux to and from the surface and atmosphere based on surface Tav
5. Do not model a static atmosphere without moving gases
6. Do not model a moving atmosphere without Gravity
7. Do not model the surface as a combined land/ocean ā€œthingyā€*
If you do not understand why these rules are important, then you do not understand the basic physics of convection and gas conduction. It appears that Pierrehumbert did realise radiative cooling at altitude is critical to tropospheric convective circulation. It appears that Pierrehumbert did realise that radiative gases cool our atmosphere at all concentrations above 0.0ppm. That is why he tried to get away with this –
ā€œSo increasing the emissivity from zero (increasing ā€œgreenhouseā€ gases) cools the climate to begin with. Then as the emissivity increases past a certain point the warm pool surface temperatures start to increase again.ā€
A ā€œcertain pointā€ was it? How many ppm is that? Empirical evidence? Not likely. This attempt to write the role of radiative gases in convective circulation out of atmospheric science will not stand up to scrutiny. It is very clear why you and Joel hate discussing the role of radiative gases in convective circulation. You and Joel can try running back to your two shell models but it will not work. In a moving atmosphere, radiative gases act to cool the atmosphere at all concentrations above 0.0ppm. AGW is a physical impossibility for a moving atmosphere.
You know Pierrehumberts ā€œit cools and then it warmsā€ thing is tripe. I have given you plenty of opportunities to offer it up in response to questions of the role of radiative gases in a moving atmosphere. It seems even AGW believers are too embarrassed to repeat it.

joeldshore
May 6, 2013 6:06 am

Konrad: I am happy to discuss radiative gases in convective circulation and have explained it several times before – When you add more radiative gases under the current conditions, the condition on convection is determined by the fact that the lapse rate stays very close to the appropriate adiabatic lapse rate. For places where that is the dry adiabatic lapse rate, that means the lapse rate doesn’t change. For places where that is the saturated adiabatic lapse rate, that means the lapse rate decreases slightly. This is a negative feedback in all of the climate models that has the effect of taking back some of the amplification caused by the water vapor feedback.
But, the net effect of adding greenhouse gases is a warming of the troposphere.

tjfolkerts
May 6, 2013 6:22 am

Konrad, I owe you an apology.
The discussion was about shells and heat transfer, and I was focused on that discussion and mistook one “K” for another (Konrad for Kristian). Kristian was the one so adamantly (and incorrectly) proclaiming that the shell would not matter. Kristian is the one who hopefully has realized his mistakes.
Now on to a few other points:
1) This is a quote I like from George Box: “all models are wrong, but some are useful” The simplified models (like uniform heating and blackbody shells) are USEFUL for gaining understanding of various concepts.
2) Even when gases are free to move, the radiative shell models are not irrelevant. Convection certainly makes major modifications to the results of a simple shell model, but energy flow is ALWAYS relevant to temperature.
3) The Science of Doom article looks interesting — thanks for pointing it out. I haven’t had time to do more than skim through it. It DOES say that is is a simple model — that ignores several of your points about what a model MUST have, so it is certainly too simple for precise modeling of temperatures. I hope to have a chance to look further at it sometime soon.

Kristian
May 6, 2013 9:51 am

It is simply stunning to see you people still so fundamentally confused and at the same time so self-assuredly certain about the shell having to make the sphere warmer than what its constant internal input can manage.
And this after you have failed to come up with even a single physics or engineering textbook example anywhere that describes this (one would think) important phenomenon or even mentions it as an eventuality or interesting corollary. Without having even one single controlled real-world vacuum chamber experiment to show for who proves this effect to be a real one as per your theory.
Without caring to answer me when I ask you why the alleged necessary heating of the sphere and the apparently well-established mechanism behind is never discussed anywhere in physics. Why it is always only how much cooler than the sphere the shell will end up being, and hence accordingly how much the flux to space will be reduced, that seems to be the subjects of interest and relevance. Why is it stated over and over again that the temperature of the sphere will remain constant, and yet never is it explained why and how, even with a word, that this constant temperature can only be sustained by a reduction in its internal power input to exactly match the ‘back radiation’ from the sphere? How the surface temperature of the sphere (like a star) is maintained constant by an equally constant internal input flux of, by necessity, a set magnitude, as long as it’s alone in space, but how this abruptly changes when the shell is put in place around it, by a sudden and perfectly balanced reduction in the internal input power. Wouldn’t it be prudent to tell easily bewildered young student minds that their possibly natural intuitive first assumption (by Occam’s razor) that a still constant surface temperature for the sphere would cohere with the internal input also still being the same, is wrong? And explain them why? Somewhere. Anywhere?
Why is it never discussed how much the internal power to the sphere will or could be reduced or how much warmer the sphere will need to be if it isn’t, but rather kept constant, with so and so many shells surrounding it, absorptive shells or reflecting shells? Why don’t we see such a problem posed? Anywhere. Ever. Why does this claimed effect of yours remain such an elusive, such a total non-subject, apparently completely uninteresting to even mention in passing, by engineers and physicists alike?
You know why? Because it doesn’t exist. Because it’s not a physical thing. The shell cannot across a vacuum heat a blackbody sphere at its core whose surface is heated by a constant internal power source, beyond this input. Because the ‘back radiation’ is NOT extra HEAT. It originally comes from the sphere itself, emitted as heat loss. Letting heat loss turn into heat GAIN for the originally emitting object is letting the object be its own heat source.
The shell can only reduce final heat loss to space (Q –> Q”) by splitting Q from the planet in half (two sides), Q = Q’ + Q”. It cannot touch Q itself. Q’ and Q” are both Q/2. Q’ sets a limit to the maximum (equilibrated) temperature to which the sphere can possibly heat the shell. Q” is a function of this limit.
All the shell can do is reduce radiative cooling of the sphere when and if the sphere is ACTUALLY cooling (which as a blackbody it won’t be – once you shut off its internal power input, the last bit of energy escapes it as emitted radiation never to be replaced and the temperature drops directly from 290 to 0 K, because there is no thermal mass and perfect conductivity).
Insulation reduces COOLING RATES. Or increases HEATING RATES. Or decreases heating rates. It does not however, under any circumstances, raise the theoretical input temperature of the insulated object. That’s an outrageous idea! That’s like saying that if you could perfectly focus the solar rays with a magnifying glass so that the focus area becomes 100% as hot as the surface of the Sun itself (the absolute limit for focusing efficiency), then you could just insulate it in some SW- and NIR-transparent but FIR-opaque material and make it even hotter. Or better yet, just point an oven at the area and it will surpass the temperature of the Sun. Well, you don’t even need that. The atmosphere surrounding your focussing area is itself warmer than the vacuum of space around the Sun. So the area should upon perfect focus already be able to surpass the surface temperature of our star.
Let us get this sorted out, once and for all. Why doesn’t Spencer (or any of you) explain exactly how the ‘back radiation’ from the cooler shell is NOT raising the warmer sphere’s temperature by increasing its HEAT GAIN, but rather by reducing its HEAT LOSS? How does it accomplish this without actually transferring HEAT to the sphere, a positive inward transfer of ‘thermal energy’? After all, the heat loss flux balancing the internal heat gain isn’t restricted or disallowed. The constant internal power source in itself provides the same input to the surface as before, in itself producing the same potential surface temperature, in itself inducing the same, S-B-dictated emitted flux. All that changes in this setup is after all the ‘incoming’ back radiation flux. Are you for real then calling this an additional HEAT flux? I’m sure you are. Because you’re not subtracting fluxes, are you? You’re adding. To gain a larger outgoing flux from the sphere’s surface. Like F. Engelbeen stated it: J + J1.
So you’re hiding cleverly behind what appears to be a non-2nd-law-violating statement that the shell helps making the sphere warmer not by providing extra HEAT to it (’cause that would be absurd, right?), but rather by reducing its heat LOSS. But what you’re actually saying is indeed the HEAT thing. You in effect turn the cooler shell into a second independent HEAT source for the warmer sphere’s surface.

Kristian
May 6, 2013 9:58 am

Gary Hladik says, May 3, 2013 at 3:17 pm:
“Kristian says (May 2, 2013 at 10:51 pm): ā€œReduce the heat loss and the thermocouple comes CLOSER to measure the REAL temperature of the gas.ā€
Interesting. So adding a radiation shield does nothing to the temperature of a radiating sphere, but adding a radiation shield DOES affect the temperature of a radiating thermocouple. And Kristian writes this with a straight face???”

You simply do not get it, do you? Once again you simply do not read what I write. (BTW, you saw the thing about the thermocouple setup being in a medium, not in a vacuum, right? Clue: convective heat loss.)

tjfolkerts
May 6, 2013 12:04 pm

Kristian says …
1) “Why is it never discussed … ”
Undergraduate physics books don’t often address this specific case because it is a rather specialized case.
Engineering texts don’t often discuss this specific case because practical applications usually involve convection and conduction as well.
But astronomy texts (including one linked to above) and climate texts do discuss it. Google (one-layer greenhouse effect model) and you will get over a million hits of people discussing this.
2) You are arguing semantics for the most part when you discuss “heat”. While the word is often confusing, the actual physic is not unclear (and unfortuantely, “heat” is used colloquially in several different meanings in different contexts, which can add to the semantic confusion).
You ask for an explanation. First, read a good thermodynamics text until you understand that “heat” (in the technical thermodynamics sense) is the NET transfer of energy from warm to cool. That will clarify much of the ideas that you are trying to grasp and prevent you from incorrectly assuming that we would think things like “Are you for real then calling this an additional HEAT flux?”. Until you talk “the same language” of physics that we do, you won’t understand what we are saying, or be able to explain to us what you are thinking.
3) “Thatā€™s like saying that if you could perfectly focus the solar rays …”
This is actually some of the best, most interesting physics you have discussed. The maximum temperature in this case is the temperature of the hottest object providing the thermal energy. So with sunlight from a 5800 K, the hottest theoretical temperature WOULD be 5800 K.
But the heaters here are not limited to 288 K, but rather limited to 400 W/m^2. From an engineering standpoint, this is not possible at all temperatures, but we could, for example, use a 400 W light bulb, where the limiting temperature would be ~ 3000 K. The “theoretical input temperature” (as you call it) is ~ 3000 K, not 288 K. This would effectively limit the planet’s surface to 3000K no matter how may layers you added if you used light bulbs as the heaters.
But of course, this is not an issue for our planet, with temperatures an order of magnitude cooler.
4) To make a couple analogies, you seem to be thinking (among other errors) of the heater as a ….
* a constant voltage source rather than a constant current source. A constant voltage source can only charge a capacitor to a fixed voltage. A constant current source can charge it to any voltage (until you hit engineering limits of the devices).
* a constant pressure air pump rather than a constant volume air pump. A constant pressure pump can only fill a tank to a fixed pressure. A constant volume pump can fill it to any pressure (until you hit engineering limits of the devices).

joeldshore
May 6, 2013 12:17 pm

Kristian says:

And this after you have failed to come up with even a single physics or engineering textbook example anywhere that describes this (one would think) important phenomenon or even mentions it as an eventuality or interesting corollary.

Absolutely untrue. You just keep finding excuses to ignore them. I have given you three examples, two being qualitative discussions of global warming in two of the most commonly-used intro physics textbook and one being a quantitative discussion of global warming involving exactly that shell model in what is one of the most (if not the most) commonly-used textbook for a first upper-level physics majors course in thermal physics: http://books.google.com/books?id=c0R79nyOoNMC&pg=115#v=onepage&q&f=false
To be fair, I’ll admit that this sort of example is not discussed as often as it ought to be. I think the reason for that is largely that they tend to take the problem through to a certain point and then assume that once they have explained (for example) how multi-layer insulation reduces heat transfer, you could use that result to calculate steady-state temperatures given some situation. But, it really should get more attention, if for no other reason than to disabuse people of incorrect notions that they might have about what the Second Law says.

Why is it never discussed how much the internal power to the sphere will or could be reduced or how much warmer the sphere will need to be if it isnā€™t, but rather kept constant, with so and so many shells surrounding it, absorptive shells or reflecting shells? Why donā€™t we see such a problem posed? Anywhere. Ever.

http://books.google.com/books?id=c0R79nyOoNMC&pg=115#v=onepage&q&f=false [Hint: Repeating a false claim over and over again does not make it truer.]

Insulation reduces COOLING RATES. Or increases HEATING RATES. Or decreases heating rates. It does not however, under any circumstances, raise the theoretical input temperature of the insulated object.

There is no such thing as a “theoretical input temperature” unless the object is being controlled by a heater that automatically adjusts the power it supplies to maintain a certain temperature. The temperature of an object is determined by the balance of the energy coming in and the energy going out and no amount of bluster from you will change that fact.
You believe in what Robert G Brown (another PhD physicist) sarcastically calls “the Postma statement of the second law of thermodynamics, which is ‘Cold objects can never ever make warm objects change their temperature in any way, regardless of what else is going on.’ ” ( http://www.drroyspencer.com/2013/04/how-can-home-insulation-keep-your-house-warmer-when-it-cools-your-house/#comment-75875 ) However, as us physicists know, the Postma statement of the Second Law is a complete fiction. It has absolutely no basis in physics.

Thatā€™s like saying that if you could perfectly focus the solar rays with a magnifying glass so that the focus area becomes 100% as hot as the surface of the Sun itself (the absolute limit for focusing efficiency), then you could just insulate it in some SW- and NIR-transparent but FIR-opaque material and make it even hotter.

Yes, it is actually rather instructive to look at exactly this question: Why could we never make the “radiative insulation” so good that the Earth gets hotter than the sun? The answer is much more interesting and creative than the sort of arbitrary and capricious laws of nature that you imagine: The reason is simply that what the greenhouse effect is relying upon is the atmosphere selectively blocking the terrestrial radiation and not the solar radiation. However, as the Earth gets warmer and warmer, its spectrum shifts closer and closer to the spectrum of the sun and so such selectivity becomes impossible! [You might think that you could still make it work by, say, having the Earth have a higher absorptivity than the sun does. But, then Kirchhoff’s Law of Thermal Radiation ( http://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation ) tells you that the material that absorbs more at a certain wavelength also emits more at that wavelength.]

Why doesnā€™t Spencer (or any of you) explain exactly how the ā€˜back radiationā€™ from the cooler shell is NOT raising the warmer sphereā€™s temperature by increasing its HEAT GAIN, but rather by reducing its HEAT LOSS? How does it accomplish this without actually transferring HEAT to the sphere, a positive inward transfer of ā€˜thermal energyā€™?

It accomplishes it just as you said…by reducing its heat loss (at a fixed temperature), so that it must increase its temperature in order to again be in radiative balance where it is emitting as much heat as it is absorbing. Again, you are invoking the Postma statement of the 2nd Law rather than the actual 2nd Law. The actual statement of the 2nd Law says that heat (the net energy flow) must be from the hotter object to the colder object. And, indeed it is, since you will see when you do the calculation that the radiative emission from the object to the shell is greater than the radiative emission from the shell to the object. Hence, the net flow of energy, or the heat, is from the hotter object to the colder object.

So youā€™re hiding cleverly behind what appears to be a non-2nd-law-violating statement that the shell helps making the sphere warmer not by providing extra HEAT to it (ā€™cause that would be absurd, right?), but rather by reducing its heat LOSS.

What we are doing is applying the real 2nd Law of Thermodynamics to the problem rather than the Imaginery 2nd Law of Thermodynamics because the real 2nd Law is the one that has been confirmed by countless experiments and is also understood to derive from more basic microscopic considerations through statistical mechanics. The Imaginary 2nd Law is an invention of people who want to fool you…and seem to be doing a pretty good job of it!

joeldshore
May 6, 2013 12:39 pm

Just as an after-thought, I think this is a good general rule when dealing with heat flow problems: The Second Law is never really a rule that you have to apply in an arbitrary and capricious manner. It is simply something that will naturally be obeyed if you use correct physics along the way: The increase in radiative emission with temperature, together with Kirchhoffā€™s Law of Radiation ( http://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation ) guarantees that the 2nd Law will be obeyed. If you find yourself enforcing it independently, then you are probably enforcing some version of the Imaginary Second Law of Thermodynamics rather than the real Law of Thermodynamics.

Reply to  joeldshore
May 6, 2013 1:52 pm

Nobody has a problem with increasing radiative emission with temperature as per Stefan-Boltzmann, nor is there a problem with Kirchhoff’s Law.
The problem, and violation of rationality and transition to religion, is with the claim that a colder object’s radiation will cause a temperature increase in a warmer object, by whatever semantical argumentative method is invented to pretend such an idea such as back-radiation, or self-heating, or simply slowed-cooling etc. An object held at a constant temperature with a constant input with 100% efficiency in converting input power into thermal kinetic energy will not be further heated (i.e. its temperature caused to increase) by the radiation from a colder object placed in the vicinity. Rather obviously, the cooler object simply becomes heated (i.e. temperature caused to increase) by the constant-input, warmer, source. Just as you can not create higher than the solar spectral temperature when focusing sunlight, even if you “insulate it and prevent heat loss”, likewise you can not take a 255K spectral source and cause it to heat itself with its own spectrum beyond the spectrum that it is. The 255K spectral source has nothing to do with the solar spectral input (n.b. they are different spectra…) and so in no way can it be thought to be amplifying or re-condensing the solar spectrum to a potential of its temperature, such as with a magnifying glass.

tjfolkerts
May 6, 2013 3:29 pm

Joseph says: “An object held at a constant temperature with a constant input … “
So which is it — constant temperature or constant power input? This is like saying “I have a power supply that provides a constant voltage and a constant current” or “I have a pump that pumps water at a constant pressure and flow rate”. As long as the load doesn’t change, then both can be true. But when the load changes then one (or both) of the parameters MUST change.
If the temperature is constant, then the power input will vary depending on the surroundings.
If the power input is constant, then the temperature will vary depending on the surroundings.

You are quite correct that if the heater is set to a constant TEMPERATURE, then no amount insulation or back-radiation will warm the object above the temperature of the heater. But this scenario (and the earth as well) does not have a constant temperature for the heater. The “shell models” have heaters that (within engineering limits that are not being tested) hold a constant POWER, so that the temperature adjusts according to the surroundings — which includes IR from nearby objects. The earth has that 5800 K input (NOT 255 K or 303 K or 322 k or 361 K or 394 K or any similar temperature that someone might try) we are talking about, so (in principle), the sun could warm an object to 5800 K. Of course the earth never gets to temperatures like that because it is insulated poorly and the sunlight is rather spread out. But having the sun provide some energy and the nearby objects providing other energy — both of which get absorbed by the earth — does not violate the 2nd Law of Thermodynamics in the least.
A bare, blackbody planet with 400 W/m^2 heaters will be 288 K (to use the numbers discussed up-thread). If you change the surroundings by adding a blackbody shell, then you can either …
* keep the TEMPERATURE constant @ 288 K, but lower the power to 200 W/m^2
* keep the the POWER constant @ 400 W/m^2, but raise the temperature to 345 K.

(The real earth is, of course, much more complicated due to atmosphere, rotation, variable emissivity, variable albedo …. But none of that changes the fact that IR radiation from other objects (even colder ones) can help raise the temperature of an object with an independent, constant power source.)

Reply to  tjfolkerts
May 6, 2013 9:10 pm

Folkerts asked: “So which is it ā€” constant temperature or constant power input?”
A constant power input which creates a constant temperature with 100% efficiency in converting input power to thermal kinetic energy. When a cooler object is brought near this constant power input, constant temperature source, the cooler object will be warmed. The warming of the cooler object and the presence of the cooler object does not require the warmer source to have to heat up – the warmer source simply heats up the cooler object. The source has no knowledge to adjust itself to be cooler – it just holds the temperature it has with the constant input it has, and thus warms up cooler objects. IR radiation from a colder object does not warm up a warmer object – the warmer object simply warms the cooler.
The Solar energy at the distance of the Earth can only heat an object to the solar spectral temperature (5800 K) if the solar energy is re-condensed, i.e. focused with a magnifying glass or mirror. GHG’s do not perform this function, even in the case of the terrestrial radiation, and so the analogies here are moot. All we get here at the Earth is direct solar heating at the flux density it arrives at and then is attenuated to (equivalent to +121C [raw] at maximum under the Sun, but distributed as the cosine function over a hemisphere and with albedo attenuation), followed by redistribution of that heat and high-temperature forcing by water and atmosphere and cooling at night time.
Joel Shore, it seems to me that capriciousness is in saying that a source will know to power itself down in order to hold a constant temperature because it is wished that radiation from a cold object can heat up a warmer object. There is nothing arbitrary about the fact that a colder object never warms up a warmer object – that is a very consistent rule. In conduction it never happens, and likewise in radiation it doesn’t happen. This is elegant and beautiful because the classical laws of thermodynamics and statistical mechanics are universally applicable even among different modes of physics (matter/conduction & photons/radiation). I say that cold things never heat up hotter things either by conduction or by radiation because such a rule is consistent and truthful; there is nothing arbitrary or capricious about stating and applying such a rule. Kristian has done just a brilliant job explaining how the physics actually works, for example. I’ll have more on these concerns and the related questions and physics in a future blog post.

Joel Shore
May 6, 2013 7:23 pm

Joseph: Did you ever take a course in statistical physics? It seems strange that someone with a background in astrophysics wouldn’t and yet it seems even stranger that someone who did would write what you just did in your post.
You are teaching people to think that physics (and thermodynamics in particular) is some set of arbitrary and capricious laws when it is actually elegant and beautiful. And, you are doing this for what purpose?

Phil.
May 6, 2013 9:08 pm

Kristian says:
May 6, 2013 at 9:58 am
Gary Hladik says, May 3, 2013 at 3:17 pm:
ā€œKristian says (May 2, 2013 at 10:51 pm): ā€œReduce the heat loss and the thermocouple comes CLOSER to measure the REAL temperature of the gas.ā€
Interesting. So adding a radiation shield does nothing to the temperature of a radiating sphere, but adding a radiation shield DOES affect the temperature of a radiating thermocouple. And Kristian writes this with a straight face???ā€
You simply do not get it, do you? Once again you simply do not read what I write. (BTW, you saw the thing about the thermocouple setup being in a medium, not in a vacuum, right? Clue: convective heat loss.)

That would be convective heating of the thermocouple and radiative heat exchange with the shield.

May 6, 2013 9:19 pm

Folkerts asked: “So which is it ā€” constant temperature or constant power input?”
A constant power input which creates a constant temperature with 100% efficiency in converting input power to thermal kinetic energy. When a cooler object is brought near this constant power input, constant temperature source, the cooler object will be warmed. The warming of the cooler object and the presence of the cooler object does not require the warmer source to have to heat up – the warmer source simply heats up the cooler object. The source has no knowledge to adjust itself to be cooler – it just holds the temperature it has with the constant input it has, and thus warms up cooler objects. IR radiation from a colder object does not warm up a warmer object – the warmer object simply warms the cooler.
The Solar energy at the distance of the Earth can only heat an object to the solar spectral temperature (5800 K) if the solar energy is re-condensed, i.e. focused with a magnifying glass or mirror. GHG’s do not perform this function, even in the case of the terrestrial radiation, and so the analogies here are moot. All we get here at the Earth is direct solar heating at the flux density it arrives at and then is attenuated to (equivalent to +121C [raw] at maximum under the Sun, but distributed as the cosine function over a hemisphere and with albedo attenuation), followed by redistribution of that heat and high-temperature forcing by water and atmosphere and cooling at night time.
Joel Shore, it seems to me that capriciousness is in saying that a source will know to power itself down in order to hold a constant temperature because it is wished that radiation from a cold object can heat up a warmer object. There is nothing arbitrary about the fact that a colder object never warms up a warmer object – that is a very consistent rule. In conduction it never happens, and likewise in radiation it doesn’t happen. This is elegant and beautiful because the classical laws of thermodynamics and statistical mechanics are universally applicable even among different modes of physics (matter/conduction & photons/radiation). I say that cold things never heat up hotter things either by conduction or by radiation because such a rule is consistent and truthful; there is nothing arbitrary or capricious about stating and applying such a rule. Kristian has done just a brilliant job explaining how the physics actually works, for example. I’ll have more on these concerns and the related questions and physics in a future blog post.

Konrad
May 6, 2013 9:52 pm

joeldshore says:
May 6, 2013 at 6:06 am
ā€œBut, the net effect of adding greenhouse gases is a warming of the troposphere.ā€
——————————————————————————————————-
Joel,
No, you are still not getting it. Pierrehumbert realised what the problem was. Without radiative gases, convective circulation would cease and the stagnated warm poorly radiative gases at altitude would super heat due to limited absorption of IR and UV. Average tropospheric temperatures would be so much hotter than present that we would probably lose most of our atmosphere to space. This is why he cooked up the ā€œradiative gases cause cooling but after a ā€œcertainā€ concentration they cause warmingā€ bafflegab.
The net effect of radiative gases is cooling of the atmosphere. I have shown this by empirical experiment. To make the AGW hypothesis pig fly, Pierrehumbert tried to change this to ā€œThe net effect of adding radiative gases to the atmosphere is initially cooling, but after a magical point that just happens to coincided with current politics, it causes warming.ā€
Joel, I will ask you directly –
A. Do you support Pierrehumberts 1995 claims?
B. Is there any other scientist you are aware of making such claims before 1990?

joeldshore
May 7, 2013 5:35 am

And, we see in Joseph’s latest post the usual sophistries of the Sl*yers:
(1) According to the Sl*yer’s, temperature is determined by the rate of energy in; the rate of energy out doesn’t matter. Insulation doesn’t exist.
(2) They always talk about “the fact that a colder object never warms up a warmer object”, but never define their terms. So, one can’t accuse them of being wrong: It is indeed true in a sense that the colder atmosphere does not warm the Earth; the net flow is from the Earth to the atmosphere. The sun warms the Earth…but the atmosphere influences the rate at which energy leaves the Earth (for a given Earth temperature) and hence what temperature the Earth must be at in order for there to be balance between the energy it receives from the sun and the energy it radiates back out into space.

In conduction it never happens, and likewise in radiation it doesnā€™t happen.

Really, so insulation doesn’t exist? So, at the microscopic level, if you have a temperature gradient in, say, a gas (higher on the left and lower on the right) then in any collision between molecules, the energy transfer is always to the right?

This is elegant and beautiful because the classical laws of thermodynamics and statistical mechanics are universally applicable even among different modes of physics (matter/conduction & photons/radiation).

It is not elegant at all. It posits bizarre rules that there is no way to enforce on the microscopic scale. It relegates thermodynamics from a beautiful explanation of how the statistics of many particles can lead from symmetry in time on the microscopic scale to asymmetry on the macroscopic scale to a bunch of arbitrary and capricious laws on the microscopic scale. And, worst of all, it’s wrong.

Reply to  joeldshore
May 7, 2013 9:41 am

Shore said: “According to the Sl*yerā€™s, temperature is determined by the rate of energy in; the rate of energy out doesnā€™t matter. Insulation doesnā€™t exist.”
Insulation keeps a space warm by preventing, typically, draft losses, air leaks, and reducing conductive and convective loss. Insulation does not make the source itself warmer, but allows a space the much hotter source is warming, to be warmer, because youā€™re trapping the warm gas instead of letting it escape. The atmosphere does this by a different process because N2 and O2 have zero emissivity, and also latent heat release from water/vapour keeps much of the planet warmer than otherwise. In the GHE, however, the source-induced temperature actually has to be amplified by some process, because the source is incorrectly treated as 240 W/m^2, and so a self-warming mechanism is created based on this incorrect initial assumption, that amplifies 240 W/m^2 input to a higher temperature than 240 W/m^2 can actually create. But as we all know by now, 240 W/m^2 is not the actual input, and the actual input is much hotter and creates much warmer temperatures than 255K and drives far more energetic processes than 240 W/m^2 could ever create. On Earth, the loss rate is equal to the input rate, and this can only equate to a higher temperature than the averaged input if the emissivity is low, which is a result of N2 and O2.
Temperature is actually determined by the total content of energy within a mass. Equilibrium is reached when the input equals the output, and in all conductive cases the mass receiving energy cannot be made a higher temperature than the input. We cannot conduct a higher temperature into a target than the temperature the source is. It can happen radiatively only when the emissivity is less than the albedo, but low-emissivity induced higher temperatures (such as you will get with O2 and N2) are not a GHE. If all things are equal, say, an assumed 240 W/m^2 fully absorbed input and an output emissivity of unity, then 240 W/m^2 can never create a temperature higher than 255K, even with as much insulation as you want. Only low emissivity can turn 240 W/m^2 into an induced temperature warmer than 255K. Of course on Earth, we also have water and latent heat holding things warmer and also the natural lapse rate distribution of temperature in the atmosphere which dictates that the bottom of the atmosphere has to be warmer than the top, and thus with the average in the middle (255K), and independent of the presence of any GHG’s. In fact, when the supposed GHG water vapour is present, average surface temperatures are lower, not higher.
Shore said: “They always talk about ā€œthe fact that a colder object never warms up a warmer objectā€, but never define their terms.”
Those terms are very well defined. A colder object has a lower temperature than a warmer object. A warmer object has a higher temperature than a colder object. A colder object never heats up a warmer object. A warmer object generally heats up a colder object. This is all very clear and concise.
Shore said: ā€œthe atmosphere influences the rate at which energy leaves the Earth (for a given Earth temperature) and hence what temperature the Earth must be at in order for there to be balance between the energy it receives from the sun and the energy it radiates back out into space.ā€
The atmosphere can only influence the rate at which energy leaves the Earth if it has lower emissivity than the albedo (or less than 1 emissivity with the albedo-corrected value of 240 W/m^2 for the input). It is non-GHGā€™s which supply the low, less than unity, emissivity. This means the atmosphere will hold a higher temperature than it would if emissivity were actually unity. Also we have constant latent heat release preventing the temperature from dropping too low overnight which also results in a higher average diurnal temperature than there would have been otherwise.
Shore said: ā€œSo, at the microscopic level, if you have a temperature gradient in, say, a gas (higher on the left and lower on the right) then in any collision between molecules, the energy transfer is always to the right?ā€
Heat transfer is from hot to cold, in this example from left to right. Individual gas molecules can bounce from right to left (cold side to hot side), but this is never considered as back-conduction heating or as causing a general increase in temperature on the hotter side. Or if two solid surfaces were brought into contact, one cold and one hot, the colder surface would be ā€œbouncingā€ (at the molecular level) against the warmer surface, but this would not heat the warmer surface because the heat flow is from hot to cold only ā€“ the warmer surface is more energetic, just as the warmer gas is more energetic. The heat flow is from hot to cold, meaning that the cold side rises in temperature; this does not mean the hot side also rises in temperature, the hot side simply warms up the cold side. Net heat or net energy flow from hot to cold means that the cold side warms up; any energy flow from cold to hot does not manifest as heating (general temperature increase) on the warm side because there is always more energy flowing from hot to cold. Net heating means that heating, temperature increase, is one way, meaning that temperature only rises in one direction, even if energy flow may be two-way. It does not mean that the cold side heats up the hot side and that the hot side heats up the cold side simultaneously, just as long as more heating is occurring from hot to cold – Thatā€™s a capricious and arbitrary and even sophistic reinterpretation of such thermodynamics, and is bizarre and inelegant. Statistical mechanics proves that any symmetry on the microscopic scale (such as a ā€œcold-sideā€ molecule bouncing in the ā€œhot-sideā€ direction) are always outweighed by the majority of the asymmetric transfers of energy or energy flow from higher energy states (high temperature) to lower energy states (low temperature). During thermal interaction there is no possible general increase in the warmer side’s higher energy states (higher temperature) being caused by the cooler side’s lower energy states, simply because such a thing is essentially statistically impossible.

Konrad
May 7, 2013 5:56 am

@ Joel
No sceptic every bought into the PSI crud. Appearing to fight it is just a waste of everyoneā€™s time.

tjfolkerts
May 7, 2013 7:11 am

Konrad, I empathize with your conclusion, but think it is worth at least occasionally fighting the “PSI crud” (and not everything they say is “crud”, but unfortunately just enough is to “be dangerous”).
1) The PSI people are skeptics, so at least SOME skeptics agree with PSI.
2) There is the idea that “a stitch in time saves nine”. If left completely unchecked, bad physics can easily spread among those who are not really trained in physics. Just like skeptics are valuable for pointing out the follies of the more outrageous alarmists, physicists are valuable for for pointing out the follies of the more outrageous PSI proponents.

joeldshore
May 7, 2013 7:40 am

Konrad: I would have to read Pierrehumbert’s paper in more detail before having a strong opinion, but as I have noted below, an atmosphere with no ability to radiate is a bizarre singular limit and, in such a limit, it might be possible that there would be some strange things like a discontinuity between the temperature of the surface and the temperature of the atmosphere.
However, two things that we do know are the following:
(1) The surface temperature of the Earth in the limit of removing greenhouse gases has to go to the colder temperature that is required by radiative balance (255 K assuming unchanged albedo from the current state). As greenhouse gases are added, the surface warms.
(2) In the current situation, addition of greenhouse gases warm the lower atmosphere. Claiming this is not true is really not that much different than making the claim that the Sl*yers do: To believe otherwise goes against all understanding and calculations of radiative and convective processes in the atmosphere that are well-confirmed by the entire field of remote sensing.

joeldshore
May 7, 2013 12:16 pm

Joseph Postma says:

Those terms are very well defined. A colder object has a lower temperature than a warmer object. A warmer object has a higher temperature than a colder object. A colder object never heats up a warmer object. A warmer object generally heats up a colder object. This is all very clear and concise.

No..In the case of more than two objects, your terminology becomes unclear. There is no debate in the case of two objects interacting with no other heat inputs. The question that comes up is when we now have 3 objects (sun, Earth, atmosphere). Everyone agrees on the following:
(1) Heat (net macroscopic energy flow) is always from hot to cold.
(2) With two objects radiative interacting and no other sources of energy (from a third object or a conversion of some other form of energy into heat), the warm object will cool and the cool object will warm.
Where things get imprecise is when we have 3 objects, sun, Earth, and atmosphere. You then try to characterize any change in the atmosphere that causes an increase in the temperature of the Earth as being a colder object heating a warmer object. However, that is a misleading use of terminology: The heat is flowing from the sun to the Earth to the atmosphere, so the atmosphere is not heating up the warmer object. The sun is heating up the Earth. However, in the presence of an IR-absorbing atmosphere, the sun is able to heat up the Earth to a higher temperature because the temperature of the Earth is determined by that temperature necessary for there to be a balance between the heat it absorbs and emits.

Heat transfer is from hot to cold, in this example from left to right. Individual gas molecules can bounce from right to left (cold side to hot side), but this is never considered as back-conduction heating or as causing a general increase in temperature on the hotter side.

Simple example that I could give to intro physics students: Letā€™s say we have an insulated aluminum rod, 10 cm long, 1 cm^2 cross-sectional area, so itā€™s a 1-D conduction problem. You attach the left end of the rod to a heater that puts energy into the rod at the constant rate of 10 W/m^2. The other end of the rod, you attach to a piece of dry ice fixed at -78 C. The question is: What is the steady-state temperature of the left end of the rod? Now, take the right end of the rod and attach it to just a regular piece of ice (T = 0 C) instead of dry ice. Is the steady-state temperature of the left end of the rod change from what you found before?

Heat transfer is from hot to cold, in this example from left to right… During thermal interaction there is no possible general increase in the warmer sideā€™s higher energy states (higher temperature) being caused by the cooler sideā€™s lower energy states, simply because such a thing is essentially statistically impossible.

This entire argument of yours is a complete strawman and I think you know it. We don’t disagree on what happens for the energy transfer between any two objects at different temperatures (and the resulting change in temperature of the two objects if this is the only thing going on).
The point that you are trying to argue, rather, is that the rate of heat flow between two objects is completely unrelated to the temperature difference between them. (Of course, you never make this claim directly because people would see right through it, but both of us know that mathematically that is the only way that you could possibly be correct.)

Reply to  joeldshore
May 7, 2013 8:40 pm

Shore said: ā€œThere is no debate in the case of two objects interacting with no other heat inputs.ā€
Glad you agree that cold things do not heat up warmer things. The presence of any number of objects of course does not change this thermodynamic law. If there is a third object which we can think of as an insulating layer, then the insulating layer can only come to the same temperature as the 2nd object which is being heated by the 1st object input, because the source of heat for the 3rd object insulating layer is the 2nd object, not the 1st object source (the 3rd object insulating layer doesn’t interact with the 1st object source; its heat source is the 2nd object). If the insulating layer however reduces the emissivity of the surface of the 2nd object, then only in this case will higher temperatures than the input be achieved for the 2nd object.
Shore said: ā€œYou then try to characterize any change in the atmosphere that causes an increase in the temperature of the Earth as being a colder object heating a warmer object.ā€
Thatā€™s not what I say, that is what arguments for backradiation heating say. I agree that you are pointing out that it is an incorrect argument and statement of the problem. The colder atmosphere does not heat the surface with backradiation. This has been our position all along. So now that backradiation heating has actually been acknowledged as a wrong statement, hopefully other GHE advocates will stop using that argument, such that cold can heat hot while hot heats cold because the cold radiation has to be emitted, has to be absorbed, and thus has to cause heating on a warmer object, as long as more warming occurs on the colder object then both the hotter and colder object can become heated, which is of course wrong and sophistic. The cold-source backradiation does not cause the heating.
It would be helpful if you told the back-radiation heater people to stop spreading false physics ā€“ you might imagine it becomes difficult to have to track and argue against all of these conflicting theories on what the GHE actually is and how it actually functions. We also have people saying that it is in the raising of the mean radiating level which creates more depth for adiabatic heating which is the GHE, and this is still different from your current argument and the backradiation heating argument. The argument you are using now is that reducing emission from the surface results in a higher surface temperature. This would be true if the reduction in the ability to emit was caused by lowering the emissivity.
Shore said: ā€œThe heat is flowing from the sun to the Earth to the atmosphere, so the atmosphere is not heating up the warmer object. The sun is heating up the Earth. However, in the presence of an IR-absorbing atmosphere, the sun is able to heat up the Earth to a higher temperature because the temperature of the Earth is determined by that temperature necessary for there to be a balance between the heat it absorbs and emits.ā€
And this then returns to the idea that the colder atmosphere causes a warmer surface to become warmer still than the input. But of course, you cannot insulate something to be warmer than the actual input, and in the GHE the input is only 240W/m^2 or 255K. Insulating a 240W/m^2 or 255K input cannot create a higher temperature than 255K; only reducing the emissivity can. Arguments then turn to discussing the spectral temperature of incoming sunlight and focusing it with a magnifying glass which can produce extremely high temperatures, but this of course has nothing to do with GHGā€™s because they do not magnify the incoming solar spectrum and so that discussion is a red herring. GHGā€™s interact with the terrestrial output in their GHE function, and they donā€™t even magnify that radiation, and, if they did, it is acknowledged that they wouldn’t create a warmer temperature than the terrestrial radiation spectral temperature in any case because the spectral temperature sets the limit. Hence, GHGā€™s canā€™t raise the temperature above the spectral source and thus the spectral source, the surface temperature, has to have its own cause.
In order for the improved insulation/reduced radiative loss from the surface to actually cause an increase in global temperature at the surface, requires that sunshine no longer be treated at 240 W/m^2 or 255K but as its actual incoming real-time value; the problem here for GHE orthodoxy is that with real-time high-temperature forcing from the Sun acknowledged, we suddenly have all sorts of high-energy, high-temperature phenomena occurring and being produced that the 255K input framework could never produce ā€“ not without that frameworkā€™s invention of a GHE. So, this is why real-time high-temperature input power from the Sun is never acknowledged, because such a physical fact is always detrimental to GHE orthodoxy and makes it fall apart. Real-time high-power high-temperature input from the Sun can do all sorts of things by itself and on its own that 240 W/m^2 could never do on its own; hence, other stuff needs to be invented to augment the 240 W/m^2 255K framework in order to make it mimic what the Sun is actually doing all by itself. The stuff that is invented is backradiation heating & etc., in general, the GHE.
That aluminum rod problem is not really relevant to an atmosphere whose majority components have near-zero emissivity. You can also use that problem with a constant input and consider what happens if the other end of the bar has unit emissivity or near-zero emissivity. You attach the left end of the rod to a heater that puts energy into the rod at the constant rate of 10 W/m^2. The outer cylinder of the rod is perfectly insulated against any heat or energy loss. The other end of the rod has the same surface area as the input side (0.01 m^2) and has unit emissivity and can radiate freely to outer-space. Then solve the problem with the other end of the rod having an emissivity of 0.1. This is much more like the atmospheric problem because of the low emissivity of the atmosphere which results in a higher temperature than with unit emissivity.
Shore said: ā€œThe point that you are trying to argue, rather, is that the rate of heat flow between two objects is completely unrelated to the temperature difference between them.ā€
Nowhere have I stated such a thing or argued that. Iā€™m not going to defend statements that I never made. I have always stated that heat flow rate is proportional to the temperature difference. Please do not create strawman arguments by putting words into other peopleā€™s mouths. My own statement which you called a strawman is a statement of fact and it applies with any number of objects and any sources and sinks of heat ā€“ it is statistically impossible for cold to heat hot in any combination of sources and sinks. If you agreed with my statement then there is no reason to call it a strawman. Your aluminum rod discussion is a proper red herring however because it doesn’t actually bear on the atmosphereā€¦the atmosphere and heat flow within it doesn’t behave like or have the same properties as an aluminum rod, nor is heating constant, nor is the top of the atmosphere held fixed at different temperatures. In fact, GHE orthodoxy usually says that the top of the atmosphere gets colder with more GHGā€™s, meaning that the bottom of the atmosphere will be cooler too for the same rate of energy flow as per the way you set up the aluminum rod example. My modification of your aluminum rod example is an improvement and fix to it, but it becomes apparent that low emissivity of bulk-quantity gases are more relevant to holding higher temperatures than trace gases with high emissivity. Of course, when the ā€œGHGā€ water vapour is present, cooler average temperatures are observed, not warmer. And of course, with nearly a 100% increase in atmospheric CO2 levels, the atmosphere and surface is still actually the same temperature and even significantly cooler than past warming cycles which had much less atmospheric CO2, and so the postulated effect of GHE warming by whatever method is argued is simply non-existent in reality. A 100% increase in CO2 levels has not produced a statistically significant signal of warming beyond past climate variations.

Konrad
May 8, 2013 2:21 am

joeldshore says:
May 7, 2013 at 7:40 am
———————————————————————————-
Joel,
As I have clearly stated many times before radiative gases do heat moving gases (temporarily) near the surface by intercepting surface IR. IR emission from these gases can slow the cooling of land surface. However for a moving atmosphere, surface to atmosphere radiative exchange is largely irrelevant for average tropospheric temperatures. IR radiation to space at altitude has a far more significant effect on temperatures.
Have a look at a standard NASA atmospheric energy budget cartoon. While there are many inaccuracies in the figures, they are workable. The net energy gained by the atmosphere by intercepting both surface IR and solar UV/SW/IR is less than half of the energy the atmosphere is emitting as IR to space. Even without taking convective circulation into account, the net effect of radiative gases in our atmosphere is cooling.
If a gas column in a gravity field is maintaining a stable temperature but exhibiting strong vertical convection then it can mean only one thing. Energy is flowing through the gas and energy must be entering the gas at a lower point than it is exiting. The speed of the convective circulation depends on the energy imbalance between low and high altitude. Radiative gases add to the net energy entering the lower atmosphere and do over twice that work emitting energy to space at higher altitude. Adding radiative gases to a moving atmosphere simply increases the energy imbalance between low and high altitude and therefore speeds up convective circulation.
Radiative gases cool the atmosphere by IR emission to space at altitude. The net effect of radiative gases on lower atmospheric temperatures is also cooling as these gases are critical for tropospheric convective circulation. To understand this, this image of developing convective circulation in a fluid should help –
http://wattsupwiththat.files.wordpress.com/2013/02/rayleigh-benard-circulation.jpg
– Look at the average near surface temperatures at 4s before breakaway occurs, then at 5s when circulation is established.
You may claim that ā€œan atmosphere with no ability to radiate is a bizarre singular limitā€, but it should be clear that I have shown through empirical experiment that without radiative cooling at altitude the atmosphere would be far hotter. As radiative gases increase from 0%, the atmosphere cools and convective circulation develops. To claim that radiative gases heat a moving atmosphere, you would have to claim as Pierrehumbert does, that at first radiative gases cause cooling, but after a ā€œcertain pointā€ they cause warming. Are you going to support Pierrehumbert’s pseudo science? How many ppm of water vapour would that be before cooling stops and warming begins?
There is no magical point at which radiative gases stop cooling and start warming. Radiative gases act to cool our atmosphere at all concentrations above 0.0ppm.

Konrad
May 8, 2013 2:44 am

tjfolkerts says:
May 7, 2013 at 7:11 am
ā€œThe PSI people are skepticsā€
—————————————————————-
Pull the other one, it has bells on šŸ˜‰

joeldshore
May 8, 2013 5:47 am

Konrad says:

Adding radiative gases to a moving atmosphere simply increases the energy imbalance between low and high altitude and therefore speeds up convective circulation.

Sounds great until one realizes the consequences of what you are saying, which is that the high altitude would have to heat up, reducing the lapse rate…However, we already know that convection can only drive the temperature down to the lapse rate and no further. This is the sort of science one gets if one says lots of nice words but never works out the mathematical consequences of those words.

Radiative gases cool the atmosphere by IR emission to space at altitude.

Sorry…The IR gases also absorb. And, we know the net result from both satellite measurements from space and radiative transfer calculations that agree excellently with these satellite measurement. And, that result is that the radiation to space at the wavelengths absorbed by the IR gases is lower than at the wavelengths where they don’t absorb. At the end of the day, the IR gases are reducing the emission of radiation to space, forcing the surface temperature (and the temperature throughout the troposphere) to rise.
Another conjecture shot to h*ll by empirical data.

joeldshore
May 8, 2013 6:17 am

Joe Postma says:

If there is a third object which we can think of as an insulating layer, then the insulating layer can only come to the same temperature as the 2nd object which is being heated by the 1st object input, because the source of heat for the 3rd object insulating layer is the 2nd object, not the 1st object source (the 3rd object insulating layer doesnā€™t interact with the 1st object source; its heat source is the 2nd object).

Neat trick, Joe. If the 3rd object is at the same temperature as the 2nd object, how are you going to get heat to flow from the 2nd object to the 3rd object (which it must do to escape back out into space)? If the 3rd object is at a lower temperature, how are you going to get it to radiate enough to get back out into space the same amount of energy as was received by the system?
It’s no wonder we have never seen a simple mathematical model (like the shell model) from you on this point: You can’t create one that satisfies the basic laws of physics that constrain us.

So now that backradiation heating has actually been acknowledged as a wrong statement, hopefully other GHE advocates will stop using that argument, such that cold can heat hot while hot heats cold because the cold radiation has to be emitted, has to be absorbed, and thus has to cause heating on a warmer object, as long as more warming occurs on the colder object then both the hotter and colder object can become heated, which is of course wrong and sophistic.

Joe, Sorry…You can’t just ignore the sun. Did you take an introductory physics course? Have you ever seen the equation (Net Intensity emitted) = sigma*epsilon*(T^4-T_0^4) ? Do you understand that the second term is generally understood as being radiation from the cooler surroundings that are absorbed by the object? If you don’t like that interpretation, you don’t have to believe it: It won’t affect the fact that the rate of heat flow from an object depends on the temperature of the surroundings, which is all you need (along with conservation of energy) to prove that your claims here are nonsense.

It would be helpful if you told the back-radiation heater people to stop spreading false physics ā€“ you might imagine it becomes difficult to have to track and argue against all of these conflicting theories on what the GHE actually is and how it actually functions. We also have people saying that it is in the raising of the mean radiating level which creates more depth for adiabatic heating which is the GHE, and this is still different from your current argument and the backradiation heating argument.

They are not conflicting theories. They are different levels of detail. The purely radiative picture ignores the fact that there is also convection. In the presence of convection, the lapse rate is driven down to the adiabatic lapse rate, which will tend to reduce the temperature difference between the surface and the mean radiating level. To say the cause of the rise in surface temperature above that possible in the absence of an IR-absorbing atmosphere is due to adiabatic heating is not correct and I haven’t seen anyone claim that. The point is that convection is reducing the greenhouse effect from what it would be with a steeper lapse rate in the absence of convection, but because the atmosphere is only unstable to convection for lapse rates steeper than the adiabatic lapse rate, convection can only reduce the greenhouse effect so far.

But of course, you cannot insulate something to be warmer than the actual input, and in the GHE the input is only 240W/m^2 or 255K.

Unfortunately, if you repeat pseudoscientific nonsense often enough, it doesn’t become science. There is not a maximum temperature associated with a certain input. You are making the assumption that you are radiating back out to space at 0 K, i.e., you are ignoring the T_0 part of the equation (Net Intensity emitted) = sigma*epsilon*(T^4-T_0^4) that you can find in any intro physics textbook in order to calculate T.

That aluminum rod problem is not really relevant to an atmosphere whose majority components have near-zero emissivity.

You are changing the subject. It is a conduction problem, not a radiation problem. I understand why you want to change the subject since it shows that the Postma statement of the 2nd Law is utter nonsense. Of course the temperature on the left end of the rod depends on the colder temperature on the right end: The constant heat input on the left end will enforce a certain temperature gradient through the rod. For the numbers that I gave, the difference in temperature between the ends works out to somewhere around 40 C…I don’t have the paper on which I worked it out in front of me now to give a more precise value.

Shore said: ā€œThe point that you are trying to argue, rather, is that the rate of heat flow between two objects is completely unrelated to the temperature difference between them.ā€
Nowhere have I stated such a thing or argued that. Iā€™m not going to defend statements that I never made. I have always stated that heat flow rate is proportional to the temperature difference. Please do not create strawman arguments by putting words into other peopleā€™s mouths.

As I explained, everybody who can do basic math to enforce energy conservation, can see that this is exactly what you are in effect saying. Of course, as I noted, you never actually say it because people would immediately see that your claims are nonsense. But, there is no way for you mathematically to get the results you are claiming without making that assumption. That’s why the Sl*yer arguments are most appealing to people who have no ability to check the math.

joeldshore
May 8, 2013 7:46 am

I said: “However, we already know that convection can only drive the temperature down to the lapse rate and no further.”
That should read: “However, we already know that convection can only drive the temperature down to the (appropriate) adiabatic lapse rate and no further.”

tjfolkerts
May 8, 2013 9:34 am

Konrad,
I really like your approach — doing experiments yourself and then also looking at other papers that have dealt with some of the some issues. My main concern is that you may not be incorporating enough details some times.
1) Your experiments seem to be using fairly small boxes (but I don’t have a good description handy, so it is tough to comment too exactly). This could do many things —
* eliminate the importance of the lapse rate;
* introduce boundaries that create friction and turbulence;
* introduce boundaries that are about the same thermal conductivity as the air;
* change the various “time constants” for the system to reach steady-state by many orders of magnitudes;
* eliminate “day” and “night”.
Any one of these creates what could be significant differences between your “model” and the real atmosphere.
2) The diagram of Rayleigh-BĆ©nard convection you linked to comes from this paper:http://www.crss.ucsb.edu/music/LEVEL0/ConferencesOpen/Conferences.2004/ICAPP4241.pdf
* they use a thin layer of liquid, not a thick layer of gas.
* they use huge heat inputs (44,000 w/m^2) and small time-scales (~ 10 seconds)
* they start with an isothermal sample, and then suddenly start to heat it.
Again, any of these could make it very difficult to draw analogies between this system and the earth’s atmosphere. What I see is that the heat input at the bottom does indeed warm the fluid, but convection LIMITS how much the fluid at the bottom warms, which is pretty much in line with most interpretations I have seen for the effect of earth’s atmosphere’s convection.

Konrad
May 9, 2013 12:39 am

joeldshore says:
May 8, 2013 at 5:47 am
——————————————————————–
ā€œSounds great until one realises the consequences of what you are saying, which is that the high altitude would have to heat up, reducing the lapse rateā€
– No. Radiative gases allow energy loss at altitude, cooling the upper atmosphere, increasing buoyancy imbalance and increasing the speed of tropospheric convective circulation. Remember that the speed of tropospheric convective circulation is 0.0 m/s in the absence of radiative gases. Pierrehumbert’s magical point at which these gases stop cooling and start warming does not exist.
Further to the issue of lapse rate, the lapse rate below the troposphere is a product of convective circulation. If this circulation were not occurring the troposphere would be isothermal (prior to super heating and being lost to space). Convective circulation in the troposphere depends on radiative gases. Here is a simple experiment which will demonstrate the role of vertical circulation in generating the tropospheric lapse rate. PSI Assault Clown Doug tried not to understand, but you should be able to see why it works –
http://www.drroyspencer.com/2013/04/how-can-home-insulation-keep-your-house-warmer-when-it-cools-your-house/#comment-76518
ā€œAt the end of the day, the IR gases are reducing the emission of radiation to space, forcing the surface temperature (and the temperature throughout the troposphere) to rise.ā€
– Only for the land surface, not the atmosphere. Without radiative gases the atmosphere loses it’s primary cooling mechanism. Further to this my empirical experiments show why the surface is effective at conductively heating the atmosphere, but ineffective at cooling it. Radiative gases increase the emission of energy to space from the atmosphere, drive convective circulation and cool the surface during the day.
As to surface temperatures, radiative gases raise land surface Tmin. They have no significant direct radiative effect on surface Tmax. However these gases have an indirect effect reducing surface Tmax by driving convective circulation. While land surface Tmin may be higher under a radiative atmosphere, land surface Tav is lower as is tropospheric Tav. Radiative gases cool our atmosphere at all concentrations above 0.0ppm.
Joel, there is nothing wrong with radiative physics. The AGW hypothesis fails because the role of radiative gases in the fluid dynamics of the atmosphere was not properly considered. This has been compounded by the mistake of assuming that Tav for a moving non-radiative atmosphere would be set by surface Tav.
At some point you need to acknowledge that tropospheric convective circulation depends on radiative gases.

Konrad
May 9, 2013 2:39 am

tjfolkerts says:
May 8, 2013 at 9:34 am
ā€œWhat I see is that the heat input at the bottom does indeed warm the fluid, but convection LIMITS how much the fluid at the bottom warms, which is pretty much in line with most interpretations I have seen for the effect of earthā€™s atmosphereā€™s convection.ā€
——————————————————————————————————————
Tim,
Your interpretation is essentially correct. The problem for the AGW hypothesis is that convective circulation below the tropopause is dependant on radiative gases. The lower atmosphere is being heated by surface conduction, intercepted surface IR and release of latent heat. All this energy, not just intercepted surface IR, is being radiated from the upper atmosphere by radiative gases. This drives convective circulation.
Surface Tav and Tmax* is reduced by convective circulation. There would be no tropospheric convective circulation without radiative gases. The benefits of radiative gases in the atmosphere are three fold –
A- Plants can breathe and grow.
B- Surface temperatures are liveable. (increased Tmin, reduced Tmax)
C- Most of our atmosphere does not boil off into space, so no ones breathing privileges need be revoked. (look at molecular temps in the mesosphere, thatā€™s what happens without radiative gases)
I acknowledge the limitations of small scale experiments. Experiments 4 & 5 do suffer from scale effects due to the speed of gas conduction, hence my advice to build them a large as possible. Surface friction is however required for experiment 4 if built small, you will notice surface friction controls added in the photo of the version I built only 250mm tall (not needed in 1m version). However the basic physics being demonstrated in experiments 1 to 5 are sound. The instructions given are for others to replicate with easily available resources. I have access to materials others may not have such as vacuum metallised Mylar filled with q-cell insulation as opposed to EPS foam. I do not expect other bloggers to have access to such materials or high torr vac pumps etc**. The instructions I post are modified accordingly, and yes I have tested each of the modified set-ups.
*Land surface Tmin is increased by radiative gases.
** You need a very good vac pump to do the two shell experiment (hint ā€“ donā€™t use spheres, use plates and heat the primary plates with sw radiation from the back, not with a resistor.). -950 mb will not do it. You can get close with a delrin piston roughing pump and a carbon rotary vane secondary. What is really needed is a secondary turbomolecular. Out of my price range but I wish I had this –
http://www.creare.com/services/fluid/minivac.html. If only I had ā€œClimate Scienceā€ money, imaginary ā€œBig Oilā€ dollars don’t cut it šŸ˜‰