The Fraud of the AGHE Part 11: Quantum Mechanics & The Sheer Stupidity of “GHE Science” on WUWT

We will get to quantum mechanics at the end of this.  You know, I can hardly even write this article.  What I have to write about is so intellectually offensive, so incredibly stupid, that it is painful that I even need to do it.  We are going to see very quickly here, just how amazingly mentally degenerate, and how incredibly stupid the people are who believe in the greenhouse effect and who support the climate alarm debate.

On this fairly recent post at “Watts Up With That”, guest contributor Willis Eschenbach developed an explanation for the Atmospheric Greenhouse Effect using steel shells, with one inside the other.  What Willis does is so intellectually offensive, so mentally incompetent, that it beggars the imagination.

Willis takes an inner sphere radiating at a certain temperature as a blackbody, and then surrounds the sphere with a metal shell in order to trap the radiation.  Let’s read along, quoting from Willis’ post:

To start with, let me give a curious example of the greenhouse effect, that of the Steel Greenhouse. Imagine a planet in the vacuum of space. A residue of nuclear material reacting in the core warms it to where it is radiating at say 235 watts per square metre (W/m2). Figure 1 shows the situation.

Figure 1

This planet is at equilibrium. The natural reactor in the core of the planet is generating energy that at the planet’s surface amounts to 235 W/m2. It is radiating the same amount, so it is neither warming nor cooling.

Now, imagine that without changing anything else, we put a steel shell around the planet. Figure 2 shows that situation, with one side of the shell temporarily removed so we can look inside.

Figure 2

Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2. At that point the system is back in equilibrium. The planet is receiving 235 W/m2 from the interior, plus 235 W/m2 from the shell, and it is radiating the total amount, 470 W/m2. The shell is receiving 470 W/m2 from the planet, and it is radiating the same amount, half inwards back to the planet and half outwards to outer space. Note also that despite the fact that the planetary surface ends up much warmer (radiating 470 W/m2), energy is conserved. The same 235 W/m2 of energy is emitted to space as in Figure 1.

And that is all that there is to the poorly named greenhouse effect. It does not require CO2 or an atmosphere, it can be built out of steel. It depends entirely on the fact that a shell has two sides and a solid body only has one side.

Now, this magical system works because there is a vacuum between the planet and the shell. As a result, the planet and the shell can take up very different temperatures. If they could not do so, if for example the shell were held up by huge thick pillars that efficiently conducted the heat from the surface to the shell, then the two would always be at the same temperature, and that temperature would be such that the system radiated at 235 W/m2. There would be no differential heating of the surface, and there would be no greenhouse effect.

Let me summarize. In order for the greenhouse effect to function, the shell has to be thermally isolated from the surface so that the temperatures of the two can differ substantially. If the atmosphere or other means efficiently transfers surface heat to the shell there will be very little difference in temperature between the two.

….

Anyone who thinks that there is any actual modern physics or mathematics in this description of the greenhouse effect and who can’t immediately identify the absurd degree of pseudoscience and illogic is a complete moron.  These people are complete, unfettered idiots, and are a disgrace to mathematics.  Why am I calling names instead of of discussing the science?  Oh don’t worry, I’ll get to the actual science, math, physics, and basic logic below.  The reason why I’m calling them out as unrestrained morons is because those of us who are rational have to comprehend just how amazing their degree of stupidity is.  For us rational people who can immediately identify the fraud, we need to comprehend that the believers are incapable of comprehending and identifying the obvious fraud; if the errors are pointed out to them, as they have always been and will continue to be, their response is not to understand them and make a correction to their beliefs based on simple rational truths, but to defend their beliefs with even more sophistry and idiocy, i.e. to get even more stupid.  We have two species of humans on the planet: a) complete morons on the one hand that can not even perform basic addition, b) rational people who’s ideas and voices are subsumed by the idiocy of the intellectually degenerate mob who own popular websites.  That’s what exists in this world and these mentally degenerate cognitive midgets are the type of people who believe in the greenhouse effect.  I really don’t care that they’ll be all so upset and whatever their response is to this: they’re completely useless idiots and they’re ruining what could be a perfectly good rational planet for the rest of us.

The problem is that too many people are “doing” science today, i.e. involved in some fashion (usually self-invited!) in the scientific field and are falsely considering themselves representatives of science; stupid people believing they are smart because they know some scientific sounding words and can form sentences with them; stupid people who can make sentences but have no comprehension of logic, history, mathematics, and even well-known scientific facts in general; stupid people who can open up a website and get other stupid people to post stupid pseudoscience and be all buddy-buddy together because they feel like they’re part of the party, when they’re actually just complete idiots who’s ignorance is so profound that when you try to imagine it it feels like rape.  And you know that they enjoy that that’s what their ignorance does to rational people!  You’re going to see in the following paragraphs just why it isn’t worth having a civilized debate with these morons and why I’m speaking the way I am here: this is not a civilized debate, because there is nothing civilized about putting intellectually degenerate material on the same forum or level as what is already rationally known.

What is difficult about this is that it is hard to explain why 1+1 = 3 is wrong.  How do you explain how that is wrong?  It is obviously wrong to a rational person and it doesn’t require an explanation as to why.  But with these greenhouse idiots, not only can they not even comprehend that 1+1 doesn’t equal 3, they sometimes do use the fact that 1+1 = 2.  They don’t even comprehend that there’s a difference, and they can’t even comprehend that it is always wrong to state 1+1 = 3.

Let’s go to Willis…

Willis adds the shell around the sphere and then states “The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2“.  He didn’t show how this happens, he didn’t explain why, and he didn’t do any math to support it – he just arbitrarily doubled the output radiation and hence the temperature (according to the Stefan-Boltzmann Law) of the planet.  He just did it, arbitrarily.

Let’s look at his following statement where conservation of energy and equilibrium applies after he’s arbitrarily doubled the output from the interior: “The shell is receiving 470 W/m2 from the planet, and it is radiating the same amount, half inwards back to the planet [235 W/m2] and half outwards to outer space [235 W/m2]”.  This is his condition of equilibrium.  But wait a minute, the planet was intrinsically radiating 235 W/m2 at the start; if this was striking the shell, then his conservation of energy and equilibrium condition should have been that the shell emits half of this outward and half inward, i.e. 117.5 W/m2 either direction.  However, what he actually did was just say that the shell emits twice as much energy as it receives, i.e. a full 235 W/m2 either way, so that the interior shell now has double the energy output.  So, Willis just arbitrarily doubled the amount of energy available, so that he could add half of it back to the original 235 W/m2 in order to double it.  Just arbitrarily doubled out of nowhere.  Just made up bullshit.

And then what is strange, is that Willis stops this energy doubling process for no reason!  If at the beginning, a 235 W/m2 output comes back to double itself to 470 W/m2, increasing its own temperature, then why doesn’t the 470 W/m2 output double again from itself coming back to increase itself yet again?  Why should this process stop if the same conditions which caused this to occur in the first place still exist?  What he literally does, is he says that 1 + 0.5 = 2, i.e., the output radiation plus half the output sent back equals twice the output.  Except he just lies and says that the 0.5 portion which should be sent back is arbitrarily actually equal to 1, even though there’s no reason or possibility why this should be so.  Then once he has that, he stops this bootstrapping process cold in it tracks for no reason, and then says that 2 + 1 = 1, i.e. doubled output radiation plus a single output coming back equals a single output, ostensibly in order to “conserve energy” LOL!  What a joke!  So, it’s a bootstrapping process (as I wrote about here previously) which is created simply by lying outright on purpose, or, because you’re such an idiot that you seriously can’t do addition!  Take your pick because those are the only possibilities, and neither one gives these greenhouse advocates an out: they either know they’re lying and inventing such idiocy on purpose, or they’re really just that stupid.

At the very least, Willis’ setup would at least have obeyed *shock* addition if he didn’t arbitrarily double the initial return energy.  But he didn’t actually obey any mathematical or physics rules whatsoever.  If the outside blackbody shell was initially cold, then it wouldn’t have returned any thermal energy at the start at all!  Not only did Willis just double the output energy arbitrarily, in order to create the GHE – that was basically the whole GHE in his setup…his arbitrary doubling of the output energy – he doubled it from a source (the shell) which would have initially had ZERO energy to radiate back in any case!

Now, in order to avoid the calculus, or, as a way to do calculus without having to do it, let us just assume that the shell has very low thermal heat capacity so that it heats up almost instantly in response to incident radiation.  If we use Willis’ own invented condition for equilibrium and energy conservation, then half of the energy of the shell is radiated outward and half inward.  At the very least, Willis could have used this to bootstrap the GHE into existence, but he didn’t even comprehend this much.  If in this condition half the energy gets radiated inward, then he could have said it would combine with the interior output and thus increase the temperature of the interior output, i.e., 1+0.5 = 1.5.  Wow, addition!  Of course, for these people who don’t know about addition and who don’t know about cause and effect and logic, they might just stop the process there.  But Willis can’t do that because he’s either too stupid, or because he knows that explaining it this way would expose how arbitrary it is to stop this process after a single go-around.  Because of course, if the output is now 1.5, then half of that will now have to be returned from the shell because this is now what the shell is receiving.  This is a geometric sequence that goes as 1.5 to the power n (1.5n) where ‘n’ is the number of cycles.  There’s no reason why ‘n’ should stop at 1, and this problem is still not nearly so bad as Willis’ arbitrary doubling of the initial energy in order to avoid explaining this, or perhaps not even understanding any of it since addition seems to be outside his area of expertise; he just arbitrarily doubled the amount of energy so that he could get the answer he wanted.

And so, even if we do try to make this bootstrap of the GHE a rational task that actually uses *shock* addition, it still breaks down and exposes itself to be a fraud because it indicates that the interior temperature will increase exponentially to infinity as the outer shell warms up.

Remember, Willis stated, his bolding, “that is all that there is to the poorly named greenhouse effect“.  He later says “this magical system works…”; yah, it’s magical alright buddy!  That’s exactly what it is.  It is blatant stupidity!  Do you not understand your own words?  You called it magic man…Get a clue!

Then look at what Willis says as a summary:

“If [the shell and interior sphere were not not at different temperatures], if for example the shell were held up by huge thick pillars that efficiently conducted the heat from the surface to the shell, then the two would always be at the same temperature, and that temperature would be such that the system radiated at 235 W/m2. There would be no differential heating of the surface, and there would be no greenhouse effect.  …  In order for the greenhouse effect to function, the shell has to be thermally isolated from the surface so that the temperatures of the two can differ substantially. If the atmosphere or other means efficiently transfers surface heat to the shell there will be very little difference in temperature between the two

So let’s get this right:  If there’s a pillar connecting the sphere and the shell, the pillar causes Willis’ laws of radiation to change such that radiation from the shell will no longer add (if Willis can add) to the radiation from the source.  The greenhouse can only function if there is a temperature difference between the shell and the sphere, such that the outer shell is colder, because if they were the same temperature then radiation miraculously stops adding, and the temperature would actually paradoxically drop to the initial source temperature.  If the outer shell was the same temperature as the sphere, it wouldn’t be able to cause internal self-seating and there would be no greenhouse effect even though it is still emitting energy, and more energy than if it had been colder.  A colder shell emitting radiation causes internal heating to higher temperature than the source, but a warmer shell, a shell at the same temperature as the source, doesn’t cause any heating.

What happens then if the outer shell were warmer than the interior?  By Willis’ logic I guess it should cause cooling!  Willis says that a cooler shell causes heating to a higher temperature than the source, and then that a shell at the same temperature causes no heating at the source; and so following that physics, a warmer shell must cause cooling.  But of course, we wouldn’t expect these people to be consistent, would we!

Or, how about we just give this to the greenhouse believers: both colder objects and warmer objects cause heating, but an object at the same temperature doesn’t do anything.  That’s basically what the greenhouse advocates have always wanted and have always argued: that a colder object can heat up a warmer one AND a warmer object can heat up a colder one. The only thing they’ve never asked for is that two objects at the same temperature raise each-other’s temperature.  In other words, as we’ve always said, they’re just throwing thermodynamics out the window and they want radiation from an ice cube to cause extra heating on the pavement in Las Vegas in July.

The 20’th Century and the Discovery of Quantum Mechanics

Now, what does modern physics actually have to say about radiation trapped inside a cavity?  Let’s forget that the rules of arithmetical addition are still waiting to be defined by greenhouse effect advocates in the 21st Century, and let just pretend that the 20’th Century and the rest of mathematical history actually occurred.

The idea that a cold shell outside a warm interior causes the warmer interior to heat up some more by trapping radiation inside has always been and will always be a gross violation of basic thermodynamics; as we will see it is also an embarrassing violation of quantum mechanics and the very origin and well-known history of the idea of a blackbody.

It is also specifically the pseudoscience that has been invented to explain the arbitrary difference between the kinetic temperature of the air near the surface and the equivalent blackbody temperature of the outgoing averaged radiation – an invention based on an arbitrary and physically meaningless comparison, which isn’t necessary, because the bottom of the atmospheric layer is naturally the warmest part of the total average ensemble, and because sunlight heating in day time at the ground surface actually has an average value of +49C on the sunlit hemisphere, and this heat distributes up into the atmosphere, cooling as it goes.  It is this latter fact that the invention of a cold layer heating up an already warmer layer has been meant to replace, in order to create the foundation for CO2 alarm and hence the political and societal goals related to that fraud, and of course this was done with assistance by diluting the true power of incident sunlight to a greater surface area where said sunshine doesn’t actually even exist.  Sunlight does not exist on the surface at night time and therefore all of the pseudoscience related to this proposition and subsequent to it is false.  You have to be an obscene mental degenerate in order to deny that sunlight does not have heating power on the dark hemisphere.

A cavity which traps radiation does not increase the frequency of the radiation.  It is only with an increase of the frequency spectrum of the radiation by which a radiation spectrum can induce higher temperature, and radiation can not and does not change its own frequency spectrum when trapped in a cavity.  The idea that it does is a plain violation of the whole origin of Planck’s Law and the blackbody spectrum, and hence of quantum mechanics.  Radiation trapped inside a cavity, such as between the atmosphere and surface, does not increase the frequency of the radiation spectrum, and hence does not cause a change in temperature – particularly when the actual source of radiation and input heating is an approximately 6500K spectrum, and the cavity radiation is only a 255K spectrum.  The internal cavity radiation of 255K is a result of the heating which initially occurred due to the 6500K spectrum input, and this cavity radiation can not increase its own temperature or its own frequency spectrum past what the 6500K spectrum input already did.  This is just basic thermodynamics.

In IPCC greenhouse physics and in the attendant fraudulent models, the idea that cavity radiation causes an increase in temperature requires that a hotter object emits less radiation and that a colder object emits more, violating Wien’s Law, Plank’s Law and the Stefan-Boltzmann Law (which are all just Planck’s Law).  The IPCC models stated that a warming planet from greenhouse gases should emit less energy; the data has shown that a warming planet emits more energy.  The mathematics of this invented physics also shows that at zero emission, i.e. perfect internal radiation trapping within the cavity, that at zero emission an object would have infinite temperature.  So, both an object of zero Kelvin and infinite kelvin emit zero energy.  It’s absurd.

What actually happens inside a cavity trapping radiation is simply the creation of a blackbody spectrum and standing wave-pattern field of the associated radiation spectrum.  It is these consideration which led to Planck’s creation of his law, due to the problem he solved regarding the ultra-violet catastrophe of the Rayleigh-Jeans Law, in figuring out how radiation energy would become distributed in frequency inside a cavity.  Planck’s insight was a solution which required quantization of the energy spectrum, and this led hence to the creation quantum mechanics in general.

Radiation trapped inside a cavity with a source of a known temperature does not cause run-away self-heating to infinity!  It creates a blackbody spectrum with a wavefield in all manner of superposition with constructive and destructive interference with itself; it does not cause the radiation to spontaneously increase its own temperature spectrum or the kinetic temperature of its own source.  Radiation is an electromagnetic vibration.  This vibration interacts with matter due to electrodynamics and induces vibrations of corresponding power in the matter, in the molecules and atoms.  The only power that radiation has to induce vibration in matter, and hence induce a temperature, is the power associated with its spectrum.  There can be no other expectation than this.  When the matter comes fully to “vibratory equilibrium” with the incident radiation, the radiation and the matter then simply oscillate with corresponding power – the matter emits exactly the power of vibratory energy that comes in.  That emitted radiation is of the same power as that coming in, and nowhere in this process will the radiation do more work (i.e. induce a higher temperature) than it can given what its spectrum is, and nowhere in this process will the matter spontaneously vibrate at a higher “temperature rate” than the incident radiation, because the frequency components required to do so don’t exist and are never created.

Temperature is a measure of vibration frequency and the frequencies required for a higher temperature induction are simply not there, and they’re not spontaneously created, because there’s no reason for them to be. Trapped radiation inside a cavity simply resonates – it doesn’t change its own spectrum and it doesn’t change its own temperature and it doesn’t change the vibratory frequency components of its material source.  All of this is what lead to Max Planck, blackbodies, and quantum mechanics.  To deny this is absurd.

This behavior of radiation trapped inside a cavity is not a premise I just invented – it is the foundation of quantum mechanics and comprises some of the most important parts of the scientific history of the 20’th Century.  Radiation trapped inside a cavity does not increase its own temperature spectrum and nor, commensurately, does it increase the vibration spectrum and hence temperature its own material source.  Trapped radiation combines in superposition in constructive and destructive interference and it was in solving this problem that led to the blackbody spectrum and quantum mechanics via Planck.  Radiation trapped inside a cavity simply does not change its own temperature/radiation spectrum nor does it induce higher temperature than the temperature spectrum that it is, and that is a premise which underlies quantum mechanics.

Fraud

The entire premise of climate alarm, the Atmospheric Greenhouse Effect, is a scientific travesty.  The explanation that Willis gave above is an entirely typical and standard explanation that greenhouse advocates will give.  They all would accept it, for the most part.  But none of it is true, and all of it violates modern physics in the form of Thermodynamics and even Quantum Mechanics.  They have to go as far as throwing the arithmetical rules of addition out the window, in order to create the GHE!  Comprehend the insanity!  Greenhouse effect advocates are not threatening modern physics; they’re threatening the possibility of using numbers at all.  In other words, greenhouse effect advocates are threatening the existence of mind itself.  It is no wonder they are so damned stupid.

The Slayers have systematically destroyed and exposed as scientific and mathematical fraud every single pseudoscientific plea and invention of sophistry which has been offered for the GHE.  

And do you know what the only thing is which keeps these people believing in it, which keeps them inventing bullshit after bullshit in defense of their faith?

What keeps them believing in the GHE is their obscene degree of stupidity.  Their stupidity is so profound, so complete, their faith is so devout, that they abandon basic arithmetic – addition, they abandon 1+1 = 2 – in order to continue to self-justify their belief.  They are either stricken by an obscene degree of narcissistic self-worship of their ego, that they’re driven to insane degrees of stupidity in the defense of their ego and their inability to correct themselves, or this is their job and they lie like this on purpose because they get paid for it because they’re sell-outs and cheats.

Make no mistake, some of these people who run websites promoting a debate between alarmism and skepticism make a lot of money doing it, and if this fraudulent debate ceased to exist, as it should, they would be out of funding and out of work.  It is in their interest to continue the skeptical debate over alarmism, and the last thing they want is for the debate to be over.  The debate would be over just as soon as the GHE was publicly exposed as the fraud that it is, because it is the fundamental basis of alarmism, and in their idiotic, arithmetic-violating sophistic and hysterical defense of the GHE (oh yes, they can get quite hysterical over criticism of the GHE), they are exposed as the 5’th Column back-door supporters of the alarmist agenda and fraud that it is.

It doesn’t matter if they don’t know this – it doesn’t matter to actually get any of them to admit that this is what they’re doing.  All indications may be perfectly well that they’re simply too stupid to understand the enveloping boundary conditions of this fraud.  It doesn’t matter if they know it or admit it – it is what is happening and therefore it is what they are doing.  The climate debate should not exist.  It is a fraud designed to waste time, to waste resources, to waste productivity, and to distract the masses from the things which are actually important.  The only thing the existence of the “climate debate” does is serve as a metric for how collectively stupid people are, and the people who promote the debate are merely promoting more stupidity.  Ending the fraud of the atmospheric greenhouse effect will make the entire world smarter the moment that that intellectual and scientific fraud is rendered extinct.

[Update:  As you can see from the comments below, my prediction was 100% correct.  Instead of comprehending the obvious flawed logic, people who believe in the GHE just get even more stupid in order to defend their belief.  Their math was proven to be non-existent.  Their model was shown to contradict itself.  And their reasoning was exposed as tautologous.  Their solution?  To repeat themselves several times, lol.]

Advertisements
Gallery | This entry was posted in Fraud of the Greenhouse Effect, Sophistry and tagged , , , , , , . Bookmark the permalink.

937 Responses to The Fraud of the AGHE Part 11: Quantum Mechanics & The Sheer Stupidity of “GHE Science” on WUWT

  1. Simon Conway-Smith says:

    Thank you for that Joe. Enlightening as always.

    A challenge (which I’m sure you will find a breeze), to summarise that (the physics part) in a succinct form that can be understood by your average CAGW-believing politician.

    The stupidity if most of our (UK) politicians was summarised on Wednesday by one of our members of the House of Lords, member of the Committee for Climate Change and totally self-interested warmist, Lord Deben, who claimed CO2 was an ‘alien gas’. I nearly choked!!

  2. Al says:

    I know I have said this before, but you are absolutely my hero. I shall henceforth and forevermore refer to you as the AntiGore! May he be stomped flat by his own carbon footprint!

  3. Thanks Simon,

    Yes, wonderful example of stupidity! That is precisely it isn’t it, we’ve hit the nail on the head – what we have is incredible stupidity as the fuel of the alarmist debate. It thrives on stupidity. Send that idiot “Lord” Deben my post on the Carbon Positive Campaign, lol. Alien gas…got to love how they expose themselves.

  4. LOL, thanks for the laughs and the accolades Al. Happy to be of service to rationality…and basic arithmetic!

  5. Greg House says:

    Joe, in the Willis’ sphere-shell masterpiece the main trick/mistake is, as I see it, that at the beginning he presents as a fact the very thing that he is supposed to prove to be a fact. He essentially proves that something is correct by stating that it is correct and then making some derivations. But he does not do it explicitly, of course, so no wonder some people swallow it. That thing is the “back radiation warming”, where in his fictional story the back radiation from the shell causes the sphere to radiate more.

    As for “morons” etc., I understand your feelings, but it is not so simple. If a mistake is hidden, some non-stupid people will fail to recognize it, too.

  6. Hi Greg!

    Yes great point to highlight as to the tautology of the reasoning involved, that is a very good summary of what he did.

    Yes I know not all people are stupid and it might not be correct to just call them that…but I’ve been through this so many times that the people making this stuff up ARE either stupid, or they’re lying on purpose, because they simply are unable to acknowledge basic arithmetic.

    Cheers.

  7. johnmarshall says:

    Well argued as usual Joe. Thanks.
    Another problem with this two sphere model is that the inner sphere has a surface area of less than the outer, obviously, so the radiated energy is adsorbed over a larger area thus automatically diluting it. The Willis explanation ignores the thermos flask which does not heat up its contents but slowly looses heat as it should.

  8. Pingback: Greenhouse Gas Errors Abound on WUWT Blog | johnosullivan

  9. Andrew says:

    To be fair to Willis he does clearly state the assumption that inner and outer surfaces are the same and will result in a marginal error of 0.3% or so. Joe, I have read some of your stuff before and I am very interested to understand more. Please do not resort to name calling as the layman out there will turn off and not bother to read your comments. The bigger cause here is to ensure that people understand that the fear of CAGW and GHE are greatly exaggerated. You must keep as many people onside as possible. I understand that this may be frustrating to you but keep trying to explore methods to widen this message to other scientific and media outlets.

  10. Martin Hodgkins says:

    Brilliant, There is stupidity and lying but there is also peer pressure. I really think these people would be so disgraced and would feel so ashamed at not being on the side of the good that their brains go into some sort of override and logic just doesn’t come into it.

  11. Pingback: Entering the SkyDragon’s lair | Tallbloke's Talkshop

  12. tallbloke says:

    Hi Joe.
    Much as I have my differences with Willis Eschenbach, I have to say his steel greenhouse model is theoretically correct. Where he’s gone wrong in the past is by misapplying it to the real Earth, which doesn’t have a vacuum between sky and ground, has lots of potential energy locked up in the hydrological cycle, and the dayside/nightside differential you have previously posted about.

    The colder radiating to warmer thing is easy to understand. The key is to consider the net outward radiation. At equilibrium, the outer shell has to radiate 235 to space. We can all agree on that. The inner and outer surfaces of the outer shell will both radiate, not quite equally, but near enough that we can disregard the difference. We can all agree on that too I hope. Added up, the inner and outer surfaces have to be radiating at 470 in total in order for 235 to be going to space and maintaining equilibrium. Therefore the planet surface will reach equilibrium with the outer shell by heating up until it is radiating 470 too.

    Your main complaint seems to be that the ‘back-radiation’ from the inner surface of the outer shell can’t possibly ‘heat’ the planet’s surface because that would violate the second law of thermodynamics. Quite right too… but that’s not what happens. What actually happens when the planet is first suddenly surrounded by the steel shell is this:

    The planet radiates 235 as it was doing before, but it is absorbed by the outer shell, which then radiates 117.5 outwards and 117.5 inwards. Obviously this is only half what the planet is radiating outwards, so it isn’t going to make the planet surface hotter than it already is by itself. But what it will do is add to the total amount of radiation the planetary surface is receiving. Whereas before it got 235 from the decaying radiation, and nothing from space (disregarding the CMBR), it’s now getting 235 from below, and 117.5 from above for a total of 352.5.

    In order to be at equilibrium, it’s going to rise in temperature until it’s radiating the same 352.5. But this isn’t the end of the story, because the outer shell is now receiving that 352.5 and will radiate half out and half in. So now the planets surface will still be getting 235 from within and 176.25 from above for a total of 411.25. In an iterative process, the surface will warm until it is radiating at a high enough rate to get the outer surface of the surrounding shell to radiate at 235 to space so the whole ensemble can reach equilibrium. That rate will be 470, because that is the level at which the outer shell can split its re-radiation such that 235 goes to space.

    Now, I know you will object to this on the grounds that radiation from colder objects isn’t absorbed by warmer objects due to some mysterious ‘pseudo-scattering’ process Claes Johnson didn’t manage to explain to Jeff ID’s satisfaction, but ask yourself this:

    If the ‘back-radiation’ incident on the surface is somehow ‘scattered’, where is it scattered to? if it isn’t absorbed by the surface, it has to be re-absorbed by the inner surface of the outer shell. But if that were the case, the outer shell would never rise above the temperature where it is radiating a total of 235. If that were the case only half of that would be radiated out to space. In which case equilibrium won’t be reached between the planet and space, as it was before the steel shell was added. Something, somewhere, is going to get very hot indeed if that carries on for any length of time. What would that something be? The planet’s core? Its surface? the steel shell? What else is left?

    Best to you.

    Rog TB.

  13. Hi Andrew,
    The error has nothing to do with the 0.3% problem Willis commented on. The error is in the rejection of quantum mechanics and the known behaviour of radiation trapped inside a cavity, and also basic arithmetic. This is a process of exploring methods to widen the message: if nicely telling people that 1+1 does not equal 3 doesn’t work, then maybe yelling at them will? 🙂

  14. Hi Martin,

    Yes good points, I think there is a lot to that. Maybe they’ll feel “peer-pressured” by being yelled at that it will snap them awake.

  15. Hi Roger,

    But please refer to Greg House’ comment above, and also the history of science related to the behaviour of radiation trapped inside a cavity, which lead to the origin of quantum mechanics. Radiation trapped inside a cavity does not do what is being claimed by the GHE…it does what Planck discovered it does when he founded quantum mechanics via his formulation for the blackbody spectrum.

    The sequence you describe is exactly what I described in my article, about half being sent back etc. The problem is that this sequence has no reason to stop after a single iteration, and it has no reason to stop when the outside is emitting 235 W/m2 – conservation of energy is not a “force”. Inside the shell, the forces imagined to be at work are such that the backradiation adds with itself and amplifies its own temperature. 235 W/m2 inward has to add again with the 470 W/m2, in order to produce 705 W/m2. That the outside emission might be some value does not affect what has to happen on the inside given what has already been happening on the inside.

    For your last paragraph and questions therein, I described this in the article, which I’ll quote:

    Radiation trapped inside a cavity with a source of a known temperature does not cause run-away self-heating to infinity! It creates a blackbody spectrum with a wavefield in all manner of superposition with constructive and destructive interference with itself; it does not cause the radiation to spontaneously increase its own temperature spectrum or the kinetic temperature of its own source. Radiation is an electromagnetic vibration. This vibration interacts with matter due to electrodynamics and induces vibrations of corresponding power in the matter, in the molecules and atoms. The only power that radiation has to induce vibration in matter, and hence induce a temperature, is the power associated with its spectrum. There can be no other expectation than this. When the matter comes fully to “vibratory equilibrium” with the incident radiation, the radiation and the matter then simply oscillate with corresponding power – the matter emits exactly the power of vibratory energy that comes in. That emitted radiation is of the same power as that coming in, and nowhere in this process will the radiation do more work (i.e. induce a higher temperature) than it can given what its spectrum is, and nowhere in this process will the matter spontaneously vibrate at a higher “temperature rate” than the incident radiation, because the frequency components required to do so don’t exist and are never created.

    Temperature is a measure of vibration frequency and the frequencies required for a higher temperature induction are simply not there, and they’re not spontaneously created, because there’s no reason for them to be. Trapped radiation inside a cavity simply resonates – it doesn’t change its own spectrum and it doesn’t change its own temperature and it doesn’t change the vibratory frequency components of its material source. All of this is what lead to Max Planck, blackbodies, and quantum mechanics. To deny this is absurd.

    Cheers

  16. tallbloke says:

    Hi Joe, and thanks for your interesting reply. Plenty for me to learn about. Before I do, I just want to pick up on this part of your response:

    The problem is that this sequence has no reason to stop after a single iteration, and it has no reason to stop when the outside is emitting 235 W/m2 – conservation of energy is not a “force”.

    If you follow the way I did the iterations, you’ll see that the extra amount being radiated back towards the surface is getting smaller at each iteration until no extra is added and equilibrium is (almost) reached at the nth iteration. At the first iteration, half the original amount was radiated back (117.5), at the second iteration three-quarters of the original amount was radiated back. So only an extra quarter (68.75) is added at the second interation. In sequence, the following iterations add an extra 1/8, 1/16, 1/32, 1/64….

    So there’s no instantaneous doubling, but rather a slowing down as equilibrium is reached. There is no conservation ‘force’ required, the slowing down and stopping of the increase in surface temperature is a natural outcome of the physics. The surface will rise in temperature until the same *power* being generated by the radioactive decay is being lost from the system to the heatsink of space.

  17. tallbloke says:

    I should just add that to further consolidate my response above, you could consider what would happen if the outer surface of the steel shell were to emit more energy to space than was being generated by the radioactive decay. The system as a whole would cool, because more is being lost to the heatsink than was being generated by the decay. So as you can see, the outer shell isn’t going to emit more than 235, because if it did, it’s temperature would fall, and less would be back-radiated to the surface, and the surface would cool. That would then mean less was radiated by the surface to the shell, and the whole ensemble would cool.

    A point of balance between the cooling effect of radiating to space and the warming effect of the radioactive decay is reached when the two processes are in equilibrium.

  18. The problem is that that iteration sequence has no justification…it is just made up. What law governs the 235*(1/2)^n, n>1, sequence as the return? This is arbitrary. Nowhere is it regulated that “235” is only the portion to be sent back by (1/2)^n, even though the interior output is increasing beyond this.

    It starts with 235 going out and half THAT coming back, because that is what is going out, i.e. half of 235 comes back because 235 is what goes out to the shell:

    235 + 235*(1/2) = 352.5

    So now, 352.5 is what is going out. 352.5 is what is going out so half of it has to come back, just like the “physics” of what originally occurred:

    352.5 + 352.5*(1/2) = 528.75.

    Now, 528.75 is what is going out. 528.75 is what is going out to the shell, so, half of it has to come back:

    528.75 + 528.75*(1/2).

    So the actual outgoing sequence is 235*(1.5)^n, n>=0, because the return “backradiation” is 1/2 the interior output. This is the logical sequence following the initial premise and is physically sensical following that premise.

    What you’re proposing as the outgoing sequence is 235 + 235*(1/2)^n, for n>=1, 235 for n = 0, because the return backradiation is 235*(1/2)^n, n>1, but this equation does not follow the initial physical argument and it is not sensical, and it is arbitrary. Neither is it what Willis said. This setup is just an invention with no logical or physical basis, and it violates its own initial physical premise.

  19. What does radioactive decay have to do with this? This is just making the whole mind-frame needlessly more complex assuming that the initial premises were correct in the first place, which wasn’t established and which they’re not. So the ideas here are superfluous, and not relevant.

  20. tallbloke says:

    Hi Joe,

    the radiactive decay is from Willis’ ‘nuclear core’ and the dissipating heat from the decay reaction is generating 235W/m^2 at the surface of his toy planet, as shown in fig2. As you know, 1W = 1J/s so we have a timescale as well as a magnitude for the energy flow.

    Joe said 352.5 + 352.5*(1/2) = 528.75.

    The planet surface is only going to radiate what it receives. It receives 235 from the nuclear furnace, and half of whatever the shell is kicking out. So it’s never going to be more than 470, because the shell can never be sending more than half it’s total back to the surface, nor more than the original energy out to space. So the shell maxes out at 470, and half that is 235.

    So your sum isn’t 352.5 + 352.5*(1/2)=528.75. It should be 235 + 352.5*1/2=411.25 as I said in my original comment:

    Whereas before it got 235 from the decaying radiation, and nothing from space (disregarding the CMBR), it’s now getting 235 from below, and 117.5 from above for a total of 352.5.
    In order to be at equilibrium, it’s going to rise in temperature until it’s radiating the same 352.5. But this isn’t the end of the story, because the outer shell is now receiving that 352.5 and will radiate half out and half in. So now the planets surface will still be getting 235 from within and 176.25 from above for a total of 411.25. In an iterative process, the surface will warm until it is radiating at a high enough rate to get the outer surface of the surrounding shell to radiate at 235 to space so the whole ensemble can reach equilibrium.

    The successive iteration yields less and less extra energy returning to the surface for the same reason your lukewarm coffee gets to be lukewarm much quicker than it actually ends up cold. The bigger the differential the quicker the heat exchange. As the temperature approaches equilibrium, the differential reduces and the progress to equilibrium slows down. But we’ll get there in the end.

    Cheers

    Rog TB.

  21. Roger, none of what you said makes any sense. It is just making things up and not following any logic. Like Greg House pointed out, it is starting with the assumption of the end result, and then saying things which appear to justify it but don’t actually.

    The surface radiates 235. Doesn’t matter what the source is. This is what it “gets” from the source.

    This is then what the shell gets, and the shell sends half of it back, 235*(1/2).

    Now, as the logic goes, the interior has 235 + 235/2, and so it has warmed up. This is now what it radiates.

    The surface now radiates 235 + 235/2, and this is then what the shell gets.

    The shell now receives 235 + 235/2, and it has to send half of this back, (235 + 235/2)/2.

    And that’s the sequence. It is the only logical sequence using the backradiation argument with a shell that sends half the energy it receives back. 235*(1.5)^n.

    The GHE argument is that after the first iteration, to 235 + 235/2, the shell no longer sends back half of what it actually receives, but only 1/4 of the original. There’s no justification for that and it violates the initial physical premise of what the shell should send back due to the initial conservation of energy considerations.

    The shell first sends back 1/2, then it sends back 1/4 of 235 even though 235 isn’t even what exists anymore in terms of what it receives, then it sends back 1/8 of 235 even though 235 isn’t what it receives anymore, then it sends back 1/16 of 235 even though 235 isn’t what it receives anymore, etc, etc.

    There’s no justification for this aside from the tautological result that this sequence produces what was intended in the first place, a doubling of the internal energy. It doesn’t follow any logic or physics at all.

    It is just a convenient sophistry because sum((1/2)^n), n>=0, equals 2, but this sequence has no physical basis and no physical justification, other than that is gets somewhat close to the answer desired.

  22. tallbloke says:

    The shell first sends back 1/2, then it sends back 1/4 of 235

    No. It sends back 1/2 of 235. Than after the next iteration it sends back 1/2 plus1/4. Then after the next iteration it sends back 1/2+1/4+1/8. It is reducing each time because the extra amount of outgoing energy from the surface is reducing each time too. And that’s because half of the extra is being lost to space.

  23. Stephen Fox says:

    Hi Joe and TB,
    much enjoying your exchange. I am not a scientist, so forgive me if my query is inane, but I would like to know about the assumption that the steel shell would radiate half inward, and half outward. Toward the end of your last comment TB you refer to the temperature differential of two bodies increasing the transfer of heat between them, diminishing as the two bodies approach equilibrium.
    Wouldn’t the greater differential between the shell and outer space at absolute zero mean that the bulk of the radiation be directed outward? Do all objects invariably radiate heat equally in all directions regardless of surrounding temperatures? Connected with this, an issue I also don’t quite get, does space have a temperature, since it’s a vacuum? Only objects with mass can hold heat, I hear… I seem to remember already badgering TB about this last question, Joe, so what do you think?
    Thanks for your patience.

  24. That has no logical justification or physical basis. It is completely mathematically meaningless. Those are completely meaningless sentences.

    You were trying to get to 470; this is arrived at by 235 * sum((1/2)^n), n > 0, which is what I described, and which is what you described initially.

    First it sends back 235/2. Then it sends back 235/4, then it sends back 235/8, then 235/16, etc. This is what leads to 470 at the limit and is 235 * sum((1/2)^n), n > 0.

    Now you’re describing it differently, and the result is different.

    You say “it sends back 1/2 plus 1/4”, then “1/2 + 1/4 + 1/8”. You never stated 1/2, 1/4, 1/8, etc of WHAT. If the “what” is of the original 235, then you get to the 470 limit. Now you seem to disagree with that. This is the only case where you can get to 470, with that return sequence.

    If the “what” is the current value at each iteration, which is now what you seem to be implying, this is an entirely different series and has a completely different convergence. And it is even more arbitrary. Now you’re describing a cumulative product rather than a sum: x*(1 + 1/2)*(1 + 1/4)*(1 + 1/8). The convergence of this series is x * 2.3842 = 235 * 2.3842 = 560.28.

    This whole argument has been picked up simply because the sum of the geometric sequence as usually described is a factor of 2, which then gets to 470.

    And this doesn’t even give a correct value for anything to do with the surface of the Earth in any case – it is an argument for the GHE that doesn’t even work for the Earth because that would require a convergence to 390 W/m2, not 470 W/m2.

    Not that any of this matters anyway because none of this even has anything to do with the differential equations of actual heat flow in the first place… Heat flow does not obey geometric sequences like this with a temperature heating itself up…this has nothing to do with heat flow.

    Now it appears the argument has changed but the implications for the math wasn’t even considered. Using the new argument, the series doesn’t even converge to the value originally wanted. It is just more evidence that none of this is actually being though-through at all by the people who want to believe in the GHE – it is just being made up for the sake of how the sentences sound and that they might sound like they mean something. Converted to math, they have no sensical basis whatsoever.

    Either you have 235 being returned as a diminishing function of (1/2)^n, which is an arbitrary return function with no possible justification, but which results in a doubling; or, the cumulative product is being returned as a diminishing function of (1/2)^n, which then doesn’t even converge to the value desired at all but a higher value. Either way, there is no justification for these sequences.

  25. lgl says:

    Joe,

    “What law governs the 235*(1/2)^n, n>1, sequence as the return?”

    G=A/(1-AB), http://en.wikipedia.org/wiki/Positive_feedback

    The shell sends half of the absorbed radiation back so B=0.5 and the total gain becomes 2, (A=1) i.e resulting radiation is 470 W/m2.

    Now, what are your numbers? How much radiation inward from the shell, and how much outward from the inner sphere?

  26. Hi Stephen,

    Space doesn’t have a kinetic temperature (temperature of motion of molecules etc), nor any temperature at all. The only thing that can exist in empty space is radiation, and it is the radiation which can have an equivalent temperature via the Stefan-Boltzmann Law. Space itself doesn’t have temperature, but the things IN space can, such as matter, or electromagnetic energy waves. If there is matter in space with a temperature, or electromagnetic energy in space with a temperature, then it is those things that have the temperature, not the space itself. If there were somehow no electromagnetic radiation in empty space either, i.e. the space was truly perfectly empty, then there would be no temperature to conceive of at all. If you stuck a thermometer in said space to get a kinetic reading, it would read zero, and if you opened a spectrometer to read the radiation from said space, it too would read zero. We would just say that the temperature of this space would be equal to zero, but what it would mean is that the space must be completely void of molecules and radiation…or that any molecules in that space all have zero relative motion.

    The initial assumption of half inward, half outward is for conservation of energy. The paradox in this GHE “shells” argument is that the shell actually emits twice the input at equilibrium; it is just that only half this is seen on the outside and so people think energy is being conserved. But it isn’t, because the shell is still emitting twice the energy as the input – the direction of this emission doesn’t matter! In absolute terms, the shell is emitting twice the input. The only convenience in this is the appearances and the sophistry which can be created with it.

    Yes, rate of heat transfer diminishes as the differential. For large differential, heat transfer is large, and vice versa. But what is being created, in addition to that, with this GHE argument, is that a temperature differential also causes heating on the hot side of the differential. This is not what the differential equations of heat flow describing this differential process say, nor ever have or ever will.

  27. @lgl,
    As discussed in my comments, I explain exactly what this sequence is and that it has no physical basis, in either way that it might be described, resulting in either a geometric sum or a cumulative product. There is no such thing as a positive feedback to warm-side temperature in heat flow between a temperature differential from hot to cold. None of this is a valid approach to heat-flow in the first place, and it discounts the known history of the scientific developments of radiation trapped inside a cavity which led to quantum mechanics. Almost none of the usual approach is correct.

  28. tjfolkerts says:

    Trashed for being a useless comment. – JP

  29. Martin Hodgkins says:

    If I could make an observation…….

    Perhaps you are both right and wrong. Perhaps the ‘planet’ increases in temperature until equilibrium is reached but not because of back radiation. In this very similar thought experiment from July, Leonard Weinstein explains that radiation resistance from the ‘back radiation’ causes the increase and I am inclined to believe him. He says…”This is not heating by back radiation, but is commonly also considered a radiation resistance effect.”

    http://noconsensus.wordpress.com/2012/07/20/why-back-radiation-is-not-a-source-of-surface-heating

  30. tallbloke says:

    Hi Joe,

    You said:
    If the “what” is the current value at each iteration, which is now what you seem to be implying, this is an entirely different series and has a completely different convergence.

    What I originally said was:
    “Whereas before it got 235 from the decaying radiation, and nothing from space (disregarding the CMBR), it’s now getting 235 from below, and 117.5 from above for a total of 352.5.
    In order to be at equilibrium, it’s going to rise in temperature until it’s radiating the same 352.5. But this isn’t the end of the story, because the outer shell is now receiving that 352.5 and will radiate half out and half in. So now the planets surface will still be getting 235 from within and 176.25 from above for a total of 411.25.”

    117.5 is 1/2 of the original 235. And the difference between 117.5 and 176.25 is 1/4 of 235.
    Therefore my series hasn’t changed, the progression is logical (since half of the excess is lost to space at each iteration), and I’m not “making stuff up”. My thanks to lgl for backing me up.

    I do however completely agree with you that none of this applies accurately to the Earth, it’s just a toy model, though theoretically correct for the limited definitions it is based on. Indeed I said just that in my first comment:

    I have to say his steel greenhouse model is theoretically correct. Where he’s gone wrong in the past is by misapplying it to the real Earth, which doesn’t have a vacuum between sky and ground, has lots of potential energy locked up in the hydrological cycle, and the dayside/nightside differential you have previously posted about.

    I’ll get out of the way and let others have a say.

    Best to you.

    Rog TB.

  31. Hi Martin,

    That’s the same idea as Willis’ and follows the same rules etc. Everything I’ve said applies to that as well.

    What happens inside a cavity from trapped radiation is nothing. Nothing can heat itself up with its own work. The impossibility for radiation to heat itself up or its own source inside a cavity is the basis of quantum mechanics. The only thing that will happen is that eventually the shell would come to the same temperature as the source, and then that’s it. The shell and everything inside would be the same temperature and there would be a radiation field inside corresponding to that temperature – that of the input. All the shell does is add another layer to the sphere, and it doesn’t matter if it is mechanically connected or radiatively – the laws of thermodynamics do the same thing for both. Either there would be matter with a vibration rate corresponding to the input, or a radiation with a vibration field corresponding to the input. And either way the same result is effected that the shell temperature will become equal to the input, and it doesn’t require the interior to spontaneously increase its own temperature – they just come to equilibrium.

  32. Cheers Rog…,
    There’s no justification that half of “the excess” is lost to space/returned to the interior at each iteration. There is no law or anything that can “know” to return only half of “the excess”, especially when the basis of the original return was ostensibly to conserve energy. There’s no physical justification for the subsequent sequence and neither even is there for the initial return of 1/2.
    The series is the same but reinterpreting it a few times doesn’t change the arbitrariness of it, and my analysis is exactly the same. There is no reason that 235 should be reduced by a factor of 2 each iteration – NOR does this entire discussion of “iterations” and sequences have anything to do with how heat flow actually functions or the behaviour of radiation trapped inside a cavity. We’re discussing unicorns. A meaningless simulacra of physics.

  33. Also “excess” is just a very convenient term to use because it makes it sound like there is actually some excess somewhere that can be returned. There’s no excess, there’s a deficit under this setup, but the deficit is just re-worded as an “excess” because an excess sounds useful. Rather, the deficit of what exists on the outside is what is being turned into an addition on the inside, and only because we’ve already started with the assumption that the energy should be doubled, and then the geometric sequence of turning a deficit into an excess starting with a factor of 1/2 conveniently leads to a factor of two.

  34. tallbloke says:

    Just to clarify, the factor of 1/2 is due to the fact that the shell radiates from both sides, so the back-radiation is always half of the total the shell receives.

    Cheers

  35. lgl says:

    Joe

    “The only thing that will happen is that eventually the shell would come to the same temperature as the source”

    And then the shell would radiate 235 W/m2 inward, not?

  36. Truthseeker says:

    Joseph,

    I think that the most telling point to Willis’s thought experiment is when you say …

    “Yes, rate of heat transfer diminishes as the differential. For large differential, heat transfer is large, and vice versa. But what is being created, in addition to that, with this GHE argument, is that a temperature differential also causes heating on the hot side of the differential. This is not what the differential equations of heat flow describing this differential process say, nor ever have or ever will.”

    I do not see why the radiation would be equal in all directions. Surely the potential difference between the environments inside the sphere (235K) and outside the sphere (0K) make a difference. I would expect that all of the 235K radiation received by the sphere would radiate outwards as this is the greater potential difference. The Universe is always trying to reach equilibrium (all other things being equal) and so trying to get equilibrium between the 0K outside vaccum and the 235K outer sphere is where all of the radiated energy would go. Why would the radiation received by the outer sphere act like salmon swimming against the current of the radiation coming from the inner sphere and not go towards the 0K vacuum that provides no such impediment?

    But hey, I am just trying to use practical logic here. After all I am not a scientist …

  37. @TB; Yes indeed. And it is this factor which leads to the runaway (1 + 1/2)^n self-heating.

  38. @lgl,
    Yes, and this does not cause a substance of temperature “T” to cause another substance of the same temperature “T” to heat up. They’re just in equilibrium. Read up on the history of Max Planck and the development of the blackbody equation.

  39. sunsettommy says:

    He he……

    It would seem that the shell surrounding the core would get hotter and hotter and hotter and hotter and hotter and hotter because of the 1/2 additional returning energy flow from the core is also getting hotter and hotter and hotter and hotter and hotter and hotter still the outer shell still radiates only the initial 235 into space and that is plain absurd.It would explode from the runaway cascading heat up.

    Willis writes:

    “The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2.”

    Huh?

    How did the initial 235 at the core suddenly become 470 at the outer shell? He has 235 going back to the core and 235 going into deep space.When all it was originally only 235 arriving to the inner side of the outer shell which then would radiate away to the outside only 117.5 and 117.5 back to the core.It seems that an additional 235 was created out of nothing because the outer shell in reality would be no higher than the cores level of 235.

    Following the GHE logic is this circus of cascading temperature hill that ends when the material burns up:

    The core initially starts at 235 then gets back 1/2 of 235 (117.5) from the inner side of the shell for a total of 352 at the cores surface as it would be with this unverified GHE thought excercise.But it gets worse because 1/2 of 352 (starting from the core at this higher temperature) is now 176 in return from the inner side of the shell for a total of 528 at the core that reaches the inner shell and around and round this absurdity develops until it reaches the melting point of the radiating material where the whole thing collapses.

    The mistake Willis makes it that he fails to understand that the generating radiating source is going to be the highest level of energy and everything else around that source is as warm or cooler.Otherwise he has discovered a perpetual new energy source that would only need a simple chemical reaction and bottle it in a special chamber for gigawatts of energy outflow,so how come we are still using windmills,Hydroelectric and other mundane energy sources that never seem to notice the GHE paradigm?

    Meanwhile convection and Conduction was completely left out of this one dimensional thought excercise that is in continual operation on Earth.

    I am not impressed at this kindergarten level of mathematics.

  40. Thanks Truthseeker, that’s basically about right. Cheers.

  41. tallbloke says:

    Joe says:
    self-heating.

    I think I see where the confusion has arisen. There is no ‘self heating’. What we have is the dynamic situation in which nuclear energy is being released at a steady rate, and we are tracing the flows through the various redistributions to the heatsink of space. Thus the problem is not one of amplification of the original energy in a closed system, but the rates at which it is generated, and then lost to and from the various parts of the system, before exiting the system into space.

    Putting the steel shell around the planet is analogous to putting an old fashioned activated carbon pocket handwarmer into your pocket after its been out in the cold air. It gets hotter inside the pocket, even though the carbon isn’t being consumed any faster. However, the outside of the pocket will still be just as cool as the outside of the handwarmer was when it was in the open air like the outside of the pocket is.

  42. wayne says:

    Joe, I so agree with half of what you are saying, at the q.m. level, and I really would rather speak of e/m waves. Matter only deals in complete hertz. Molecules and atoms either deal with all 2-pi or none (h·ν), standing wave are at pi nodes, the energy is quantitized. But still I’d rather stay with the dual of vibrating waves within the void between the shell and the sphere for that is what radiation really is, even travelling overtones.

    You say: “What happens inside a cavity from trapped radiation is nothing. Nothing can heat itself up with its own work. The impossibility for radiation to heat itself up or its own source inside a cavity is the basis of quantum mechanics. The only thing that will happen is that eventually the shell would come to the same temperature as the source, and then that’s it. The shell and everything inside would be the same temperature and there would be a radiation field inside corresponding to that temperature – that of the input.”

    Arrgg… I’m with you up to you saying that the sphere and shell would not have a higher temperature than the sphere with no shell. I can’t see them staying the same though I also know there is no radiation ever passing from the shell backward to the sphere. Why?

    Starting with a liquid nitrogen cooled shell snapped over that nuclear source at first there basically would be no vibrations (IR waves) coming from the shell but it would be absorbing, getting warmer. As vibrations start emitting from the shell toward the sphere the waves at every frequency from the sphere have much, much higher amplitude emanating from it’s surface, but, those small vibrations from the shell will attenuate the stronger ones coming from the sphere so the shell is not warming as fast as before and also the sphere is not cooling as easy as before. This continues until the shell and the sphere are both at the same temperature. I think we both agree to this point.

    But if the vibrations within that void were equal both ways the nuclear energy would cease to flow and you are right, if that were to happen, no energy would flow either direction at all, zero radiation, but that cannot happen. Basically the sphere will end up (assuming both perfect BBs) at a higher temperature that vibrates (radiates) at twice the rate as before… half of its amplitude actually transferring the nuclear energy outward and half of its amplitude simply being nullified, or cancelled, by the smaller vibration amplitudes emanating from the shell. Do you disagree with that description?

    Of course none of this would happen if it were not this hypothetical case of a limitless nuclear energy source inside the sphere that we are assuming that temperature does not even affect the source itself.

    The particulars in this area has bugged me quite a bit lately.

  43. Greg House says:

    Hi Roger,

    let me demonstrate how absurd the Willis’ construction is in a slightly different way.

    You will probably agree, that if the planet in his fictional story is the only source of energy, then it is impossible that more energy is radiated to the outer space than the planet produces (please, tell me, if you disagree with that).

    Now, if the shell is of almost the same radius, it is also of the same area as the surface of the planet, so we can neglect the difference and consider them to have the same area. And let us take another number to make the calculation more understandable for everyone.

    So, let us say, the planet’s “inner engine” produces constantly the same amount of energy so that in absence of any shell the planet’s surface radiates according to it’s temperature 800 to the outer space. Now comes the shell and works as you wish, that is (1) the planet warms the shell by radiation, the shell radiates then both to the planet and to the outer space, the a half to the planet and the other half to the outer space and (2) what the shell radiates to the planet is fully absorbed by the planet’s surface and then re-radiated back to the shell in addition to what the planet’s “inner engine” constantly produces. Let us focus now just on what the shell would sooner or later radiate to the outer space.

    It goes like that, step by step.

    1. Initially the shell radiates nothing, then it receives 800 from the planet and radiates a half of it to the outer space, so, the outer space receives 400.

    2. The shell radiates the same 400 back to the planet and the planet thankfully absorbs it (and gets there warmer), but not being egoistic at all the planet re-radiates it to the shell in addition to what the planet’s “inner engine” constantly produces, so, now the planet radiates 400+800=1200. Note, 1200 goes to the shell now.

    2. Well, the shell is doing what it is supposed to do and radiates a half of this 1200 back to the planet and another half to the outer space, that is the outer space receives 600 now.

    3. The same 600 goes back to the planet’s surface, is absorbed there, makes the planet warmer again and as a result the planet radiates 600+1200=1800 to the shell.

    4. The shell radiates a half of this 1800 to the outer space as usual, that means 900 goes to the outer space now.

    Well, that’s it, actually. Given that the area of the shell is practically the same as the area of the planet and that the radiation power of what is radiated to the outer space has increased from initial 800 to 900, the result is that the outer space receives more energy from the system than the system produces. I hope you understand that it is impossible in the real world. Which means that you construction is absurd, despite it looking good on paper.

  44. Precisely Tommy…precisely. Thanks for the synopsis.

  45. No TB the error does not revolve around a simple misuse of terminology. The error revolves around the violation of basic arithmetic, the history of quantum mechanics, the laws of thermodynamics, the behaviour of radiation trapped inside a cavity, the non-sensical geometric sequences which can’t actually be described mathematically at all without exposing themselves to be baseless and non-physical, etc. An activated carbon pocket warmer doesn’t get extra hot due to the presence of the pocket, etc.

  46. Hi Wayne,

    In the “void”, you just have a bunch of EM waves going all over the place. This EM field would have a blackbody spectrum corresponding to the temperature of the sphere, and this field would eventually cause the shell to have the same temperature as well, at which point it (the shell) would then radiate, at the temperature of the inner sphere. It isn’t any more complex than that.

    This is exactly and obviously how it would work for conduction, and it is the same for radiation because radiation doesn’t cause a violation of the Laws of Thermodynamics nor obey an entirely different set of rules; the rules are the same and the heat transfer processes are the same. An extended radiation field is really nothing more than a conductive process but happening at a distance; in conduction, so-called mechanical vibrating energy is “conducted” to cooler atoms/molecules by the action of electron clouds exchanging virtual photons. That’s how heat get conducted “in contact”, is with photon exchange between the electron clouds of the atoms/molecules – there is no actual physical contact but actually an exchange of photons. This “contact” process is essentially replicated at a distance with real photons, and these real photons can not cause any more heat or generate any higher temperatures than the source they are coming from. If they come back to their source, they don’t/can’t shift the material vibration frequencies to higher frequencies and thus higher temperatures, because it is impossible for them to do that by their own definition. What they do is combine in constructive and destructive interference, and if trapped create a blackbody spectrum of the requisite temperature.

  47. tallbloke says:

    Truthseeker says:
    Why would the radiation received by the outer sphere act like salmon swimming against the current of the radiation coming from the inner sphere and not go towards the 0K vacuum that provides no such impediment?

    Joe Replies:
    Thanks Truthseeker, that’s basically about right. Cheers.

    Joe thinks insulated pocket hand warmers are out of order as a comparative concept but salmon not swimming against the current is a good analogy. Newsflash: Salmon do swim against currents on the way to their spawning grounds.

    In fact, radiation going in any direction from an object doesn’t much care about the direction of radiation from other sources. Nor is it ‘resisted’ or ‘turned back’ by them. The fluxes simply interpenetrate, unless one flow is so dense that powerful magnetic fields are set up by it. Not the case where the temperatures involved are of the order we are discussing, and the fluxes so similar in strength.

  48. @TB;

    For a non-scientist Truthseeker’s explanation was fine as long as it didn’t lead him to inflating temperatures beyond their sources. If he wanted to think of it that way it was fine as an entirely meaningless analogy, given the other descriptions I’ve been putting up which he’s read. The important part is that he realized that the temperature wouldn’t go back and heat itself up, but simply heat up the shell until it was at the same temperature.

    A pocket handwarmer inside a steel shell wouldn’t create twice the temperature it was capable. If things behaved this way, it would be, and the idea of it is in the first place, perpetual motion and over-unity work production. Twice the temperature with a single unit of work instead of twice the unit of work means another 100% for free.

  49. tallbloke says:

    Joe says:
    A pocket handwarmer inside a steel shell wouldn’t create twice the temperature it was capable. If things behaved this way, it would be, and the idea of it is in the first place, perpetual motion and over-unity work production. Twice the temperature with a single unit of work instead of twice the unit of work means another 100% for free.

    You’re clearly still considering a static situation with a fixed amount of energy in a closed system. This is not how Willis’ system is defined. There is continuous production of radiative energy from nuclear processes, and we’re discussing the dynamic throughput of this energy through a system which resists the free flow with a vacuum barrier and a shell, and which dissipates eventually to space.

    Or at least, that’s what I’m discussing.

  50. Willis’ system is defined exactly like a pocket handwarmer in a shell. A pocket handwarmer is continuously putting out chemical energy (and radiative thermal energy), and this energy can’t auto-generate higher temperatures within itself and do more work than it does in the first place.

  51. tallbloke says:

    Joe: Thank you, so you agree my handwarmer analogy is apropos after all. [Edited for condescension.]

    Now. If you insulate the handwarmer while it continues to produce the same amount of energy, of course it will get hotter. Why wouldn’t it?

    Get a handwarmer and try it for yourself. 😉

  52. tallbloke says:

    My apologies, no condescension intended. Jut pleased we’re getting onto the same wavelength.
    Sleepy time for Brits, see you tomorrow.

    Cheers

    Rog TB

  53. Why WOULD it get warmer is a better thermodynamics question in line with the laws of physics and thermodynamics. Who would have ever thought something like that besides GHE people? We’re talking about the foundation of quantum mechanics here. I don’t suppose that matters though given that thermodynamics has never been important.

    If the shell had negligible thermal capacity, then it would heat up almost instantly to the same temperature as the sphere, and then nothing else would happen, because there’d no reason for it to. If the shell had a large thermal capacity, then it would slowly heat up to the same temperature as the sphere, and again, nothing more than this would happen. The energy emitted by the sphere goes into the shell until the shell is at the same temperature as the sphere, and then that’s it. The GHE has no basis and can’t even follow the arithmetic of addition. It can’t even be described mathematically at all.

  54. If you still believe in this arithmetic-violating, no-math available GHE, we’re not on the same wavelength! 🙂

  55. sunsettommy says:

    Tallbloke writes:

    “Now. If you insulate the handwarmer while it continues to produce the same amount of energy, of course it will get hotter. Why wouldn’t it?Get a handwarmer and try it for yourself.”

    He he..,you just admitted that it is not producing an increasing amount of energy thus it can’t get any hotter.The heating source is the hottest point and the surrounding air is cooler.

    Hand warmers are engineered to work on a fixed warming rate for consistency and duration of heating the area and to reach a general heating range and that is it or the user would get burned by them.

    They dont get hotter and hotter and hotter and hotter over time either or buyers of the stuff would long ago have quit using them.

  56. Imagine how easy it would be to create fire…smelt steel etc…if things worked this way!

  57. Greg House says:

    tallbloke says, (2013/03/10 at 6:24 PM): “Now. If you insulate the handwarmer while it continues to produce the same amount of energy, of course it will get hotter. Why wouldn’t it?”
    ==========================================================

    (shock) Well, it looks like there is a deeper problem here. Let me tell you a basic thing. If you claim that it will get hotter because of trapped/back radiation, you have to prove it. It is not that what you claim is a fact until proven otherwise. If you do not claim it, then your example is simply irrelevant.

    Do you have any real physical experimental proof that anything gets hotter thanks to back/trapped radiation? Until now no one I talked to about it on various blogs has been able to present anything real. On the other hand we have the R.W.Wood experiment demonstrating that trapped radiation does not have any significant effect on the temperature of the source. Thus, the “greenhouse effect” as presented by the IPCC does not have any basis in real science. Oh no, wait, Willis and you just have brilliantly debunked the R.W.Wood experiment simply by assuming that back/trapped radiation works! Great.

    Anyway, insulation in a pocket, under a blanket etc. works by suppressing convection. If you mean that in those cases trapped radiation has an effect on the temperature of the source too, prove it experimentally.

    And, by the way, that sort of insulation “warms” only if the air temperature outside is lower, than the temperature of the source, so, the blanket prevent the colder outside air from cooling what is under the blanket. But, if the air temperature outside is higher than the temperature of the source, you will get the opposite effect: preventing the isolated thing from getting warmer through contact with the warmer air. Ever thought about it?

  58. tjfolkerts says:

    WITHOUT SHELL
    PLANET:
    235 W/m^2 input to planet via conduction from nuclear heater
    -235 W/m^2 output from planet via IR radiation upward toward space
    NET: 0 W/m^2; T = 253 K
    ** Energy is conserved.
    ** The 2nd Law of Thermodynamics is upheld.

    WITH SHELL
    PLANET:
    235 W/m^2 input to planet via conduction from nuclear heater
    235 W/m^2 input to planet via IR radiation from the shell’s inner surface
    -470 W/m^2 output from planet via IR radiation upward toward space
    NET: 0 W/m^2; T = 302 K
    ** Energy is conserved.
    ** The 2nd Law of Thermodynamics is upheld.

    SHELL:
    470 W/m^2 input to shell via IR radiation from the planet
    -235 W/m^2 output from shell’s inner surface toward planet via IR radiation
    -235 W/m^2 output from shell’s outer surface via IR radiation upward toward space
    NET: 0 W/m^2; T = 253 K
    ** Energy is conserved.
    ** The 2nd Law of Thermodynamics is upheld.

    ****************************************************
    I can’t for the life of me understand how anyone is confused by this arithmetic.

  59. tjfolkerts: You can’t perform basic arithmetic or even understand a number. You just made up some sophistic sentences, and didn’t actually do any math or explain any heat flow whatsoever. You just started with your conclusion, and then ended it there. You don’t even understand how embarrassingly illogical, inept, and uneducated you are. You don’t understand, because you’re a religious fanatic.

    WITHOUT SHELL
    Planet: 235 W/m2 to outer space.

    WITH SHELL
    Planet: 235 W/m2 to the shell
    Shell: 235 W/m2 out of the shell to outer space, once the shell warms up to the input
    Between Planet and Shell: radiation field of 235 W/m2, just like if that region was filled with matter would have the corresponding temperature.

    QED. Amazing. Only confusing to religious nutjobs. And no one needed to double the energy or temperature. Magically, a 235 W/m2 imbalance can heat the shell up to 235 W/m2 when the shell starts at zero and the planet starts at 235. But when there’s a 235 W/m^2 imbalance between the planet and the shell when the planet is 470 and the shell is 235, then the differential can’t heat up the shell. This is just completely arbitrary fairy-science.

    I don’t understand how you made it past 3rd grade…elementary.

  60. Sparks says:

    All systems should reach equilibrium and no more.

  61. Max™ says:

    Say the sphere has a 100 m radius and emits 235 W/m^2 as a result of the internal power supply.
    Let’s put a shell around it which is 1 m above the surface, and let’s suppose that the outer surface of the shell emits 235 W/m^2.

    (4*pi*100^2)*235=29,530,970.94
    (4*pi*101^2)*235=30,124,543.46

    That’s an increase of 593,572.52 Watts.

    If the outer surface of the shell emits the same total power as the inner sphere, it could only emit 230.36 W/m^2.

    _______

    It is worse for the diagram Willis used, where the shell could easily have twice the radius of the inner sphere, 100 m for the sphere, 200 m for the shell.

    4*pi*100^2=125,663 m^2, A*235=29,530,970.94
    4*pi*200^2=502,654 m^2, A*235=118,123,883.77

    In order for the outer surface of the shell to emit the same total energy as the sphere it could only emit 58.75 W/m^2!

  62. You are making a few fundamental errors in your answer to Tallbloke. That is a shame, because my instinct says he is wrong but everything I remember from a couple of years studying physics and a further year including geophysics at Cambridge some 20 years ago agree exactly with what Tallbloke is saying. I would like to see a more reasoned analysis from another sceptic, to more easily judge which is right, my instinct or my memory.

    ” A pocket handwarmer is continuously putting out chemical energy”

    No it isn’t, that is the flaw in the analogy. It puts out one shot of energy during the phase change (which is relatively rapid). From then on the source of energy is simply stored heat and it cools down. This makes it impossible to establish an equilibrium. Rocks, on the other hand, continuously put out energy from nuclear fission at a rate that diminishes infinitesimally.

    “The surface radiates 235. Doesn’t matter what the source is. This is what it “gets” from the source.”

    The point you are missing is which surface, and I think that is key to your misunderstanding of Tallbloke. The surface of the closed system containing the radiation source giving this quantity of energy must radiate this amount at equilibrium. That surface in the second case is the surface of the shell, not the surface of the planet.

    Consider another thought experiment. An electric heater puts out 1 kW inside a black-body sphere of surface area 1 m^2 in a vacuum with background radiation close to 0 Kelvin, heating the sphere until it emits 1 kW per m^2. The sphere is magically surrounded by a metal shell a little above its surface which is insulated by glass-fibre wool, with air held in by an outer shell (I choose this material merely to bring in conduction as opposed to radiative transfer). At equilibrium the outer shell will still radiate 1kW. However in order to do so the heat must be conducted through an insulator, so the inner shell must be hotter than the outer. In order to conduct heat a temperature gradient must exist. Note that the surface areas are not responsible for the temperature difference, or this problem would not be consistent after a change in scale. Thus the sphere’s surface must be hotter than it was without the shell and thus be radiating more energy.

    The point (a little laboured; I apologise) is that there is nothing special about the radiation levels within the closed system matching the total outgoing radiation. They can vary, even though the outer shell must put out this constant amount once equilibrium is re-established.

    Note I am using a case where I know why the core sphere becomes warmer to illustrate the fact that it can do so, to show that it is possible and that we need to see why it is not happening in the case of the vacuum and thin, conducting shell. I am not suggesting that my example is analogous to Eschenbach’s

    My main objection to the Eschenbach’s original claim is that adding subsequent shells would double the energy flux at the planet surface every time. That to me seems unlikely, as it runs away very quickly indeed. Unfortunately it is nearly 3 in the morning, and I am too tired to think of why it is wrong, why it would not double each time or how this doubling could fit in the real world without some serious temperatures being used by my less stable but more intelligent course mates at university to destroy stuff.

  63. sunsettommy says:

    Joe Postma writes:

    “The problem is that this sequence has no reason to stop after a single iteration, and it has no reason to stop when the outside is emitting 235 W/m2 – conservation of energy is not a “force”. Inside the shell, the forces imagined to be at work are such that the backradiation adds with itself and amplifies its own temperature. 235 W/m2 inward has to add again with the 470 W/m2, in order to produce 705 W/m2.”

    Yup that is what they stumble so badly on is that they never see the picture of their repeating backradiation gambit that never stops until the core destroys itself from trying to go nova.They never seem to realize that they are preaching an endless backradiation circularity since their diagram with a shell added around the core CREATED a new heating source out of nothing where we now have the initial core getting hotter then the shell gets hotter then the core gets even hotter and so on…………….feeding off each other in a spiraling radiative storm.

    I wonder if they are Fred Hoyle fans?

  64. Max, that error was pointed out to Willis on that post at WUWT, but he just said that the error is very small and so that it doesn’t matter. He completely missed the point that the presence of the error at all, no matter how small, indicated that the entire idea was wrong.

  65. Rich,

    More thought experiments with even more shells doesn’t change anything. All systems tend to equilibrium and no more as Sparks said above.

    Rich: “Note I am using a case where I know why the core sphere becomes warmer to illustrate the fact that it can do so, to show that it is possible”

    No, you are assuming this behaviour because you don’t actually know this empirically at all, because this is a thought experiment, and it is being invented to arrive at the conclusion which is assumed in the first place. What you’re describing is no different than what Willis already did, and the discussion in these comments about the math which would describe it, and which shows itself to be impossible, applies just the same way.

  66. Tommy, exactly, they just arbitrarily stop the circularity at the value they wanted in the first place, with no justification. As we’ve seen it can’t even be modeled mathematically.

  67. Max™ says:

    With a sphere the size of the Earth, 4*pi*6371^2 = 510,064,471 km^2 surface area, an 11 km high shell has a surface area of 511,827,320 km^2, a difference of 1,762,848 km^2, if the shell does not create energy then it can only emit 234.19 W/m^2.

    414,269,515,000 Watts is “a very small error”, yes, if you compare it to the total emissions from the surface it isn’t huge, but an increase of 1.003 times the original value is still creating energy.

    Heck, he did us a favor, now we know that 1.1986515×10^14 Watts is basically 1.2027942×10^14 Watts!

    Oh, and of course that is ignoring the radiation from the INNER surface.

    Apparently the “A” in “P = ε σ A T^4” isn’t important?

  68. No nothing is important; they have a conclusion they want and so they just make stuff up to get there. It doesn’t matter if it can’t even be described mathematically or logically…if they can say something then it must be true. If you show what they say isn’t mathematically logical or possible, then they just “say” that mathematics doesn’t apply.

    [off for the night now…]

  69. tjfolkerts says:

    Joe shows that he is several steps behind in understanding when he postulates about me that: “You can’t perform basic arithmetic or even understand a number. “

    From your top post is is clear that much of what you say is (in the words of Pauli) “not even wrong”. But I’ll make one more effort for the sake of your readers.

    Here is a spreadsheet: https://docs.google.com/spreadsheet/ccc?key=0AgM8XE4GABYQdHVZTDZLblNOaTBpSkxEckxUQXdSMEE&usp=sharing

    It calculates all the numbers that you can’t fathom that I understand – including the effects of the heat capacity of the planet.

    The columns are
    A) time (1 second steps)
    B) temperature of the planet’s surface = previous temperature + (NetQ)/C
    C) nuclear power in (set to a constant 235 W/m^2)
    (ie using Q = mc ΔT for each square meter each second)
    D) IR out from the planet using SB equation (used to find Column I)
    E) IR in from the shell using SB (from Column K)
    F) Net power into the planet. If this is positive, the planet warms; if it is negative, the planet cools/

    H) temperature of the shell = previous temperature + (NetQ)/C
    I) IR in from the planet (the same as column D)
    J) IR up from the shell to space using SB
    K) IR down from shell using SB (used to find column E)
    L) Net power into the shell.
    At the top you can set the initial conditions.

    Built into the calculations are conservation of energy and all the heat flows based on current temperatures using SB. In other words, it is simple fundamental physics that anyone with an understanding of physics could verify (or try to refute). It is a simple yet effective numerical solution to the question at hand.

    ************************************************************************
    The heater at the surface is set to 235 W/m^2

    Using C(planet) = 200 and C(shell) = 20 works pretty well to illustrate what is happening and lets the system get pretty close to steady-state within 500 seconds. (Of course, these would be small systems, but the principle could be scaled up.)

    The initial temperatures are set to 254K for the planet and 211K for the shell. You can make them pretty much anything you want (I would suggest staying in the range 0-350 K). You can even start with the shell warmer than the planet. In every case, the planet approaches 2*(heater power) as the upward IR (ie ~ 302 K) and the shell approaches (heater power) as the IR upward and the IR downward (ie ~253 K).

    ****************************************************************

    What specifically do you (or anyone else) disagree with in the spreadsheet? Which calculation do you think is incorrect? Something must be wrong, since the numbers always converge toward to the results Willis and I and the entire rational scientific community predict. Surely with your mathematical skills you can easily spot the errors.

  70. tjfolkerts says:

    Please, Joe, take this discussion to the committee that granted your Master’s Degree. Maybe you respect THEM enough to listen to valid criticism.

  71. A C Osborn says:

    Joseph, over at Tallbloke’s blog thefordprefect has posted this link to an experiment that he has carried out.
    http://climateandstuff.blogspot.co.uk/2013/03/does-thermal-radiation-travel-from-cool.html

    Can you provide the physics that explain his results that the cooler plate slows the heat loss from the hotter plate?
    Is it to do with the plates being connected by atmosphere or by “Reflection” of radiation?

  72. Max™ says:

    Apparently this wasn’t appropriate over on the talkshop:

    Max™ says:
    Your comment is awaiting moderation.
    March 11, 2013 at 12:32 pm

    “You are the one proposing a free energy concept, not me.” ~tallbloke

    >.>

    Ok, I know you tried to brush off my comment about the surface area earlier, but really?

    For the record, if the sphere is the size of the Earth and the shell is 11 km above it, having the sphere emit 235 W/m^2 and the shell emit 235 W/m^2 means the shell is emitting 1.003 times as much as the sphere.

    This becomes worse the greater the radius of the shell is in comparison to the sphere, but at no point is it ever just a trivial “oh that’s unimportant” sort of situation.

    1.1986515×10^14 Watts is not “basically 1.2027942×10^14 Watts”, we’re talking about 400 Gigawatts of “free energy” being created here just from the initial conditions proposed.

    Right from the outside the thought experiment invalidates itself before any discussion of whether back radiation is capable of adding energy or merely reducing losses can take place.

    _______

    Obviously back radiation is only capable of reducing the rate of energy loss from the surface, and at no point will the system EVER reach equilibrium.

    I repeat: at no point will this system reach the form of radiative equilibrium which the argument hinges upon. It is impossible for the initial 235 W/m^2 from the sphere to be matched by 235 W/m^2 from the shell, that is flat out producing energy!

    This is made worse by the fact that the sphere is supposedly radiating the same amount of energy INWARDS, so it is producing TWICE the total output supplied to the sphere by the internal power supply!!!

    _______

    This is not science folks, that is wizardry, full stop.

    I cite my ignored reply to Tim earlier:

    “Radiation from s surface is a function of the temperature and emissivity of the surface:
    P = ε σ A T^4″ ~Tim Folkerts

    The A there matters, if you follow Willis’ argument as he lays it out, you are stating in no uncertain terms that the Area does not matter, in which case I would like to see citations that you are in fact a wizard, please.

    [Reply] Keep it polite, respectful and peaceable. I won’t publish this comment. Try again. ~tb

    S’ok, I put a more neutral version with the main point that the initial conditions generate energy from nowhere up.

  73. tjfolkerts (@2013/03/10 10:13PM), you complete unrequited idiot. Just because you set up “a spreadsheet” to calculate the numbers you WANT, doesn’t mean it has anything to do with reality. Are you really that stupid? Tim is completely unaware of the history of quantum mechanics, how it was solved by Planck considering radiation trapped inside a cavity, that such trapped radiation doesn’t heat itself up but creates a blackbody spectrum, etc etc.

    Look everybody, Tim J Folkerts made a spreadsheet! LOL Tim created a spreadsheet building in the assumption he means to “prove”. Unrequited stupidity. All the rest of us can see how tautological, illogical, and stupid doing something like that is, and then let alone believing in it. How stupid does a person have to be?

    I have a suggestion for the other commentators here: please contribute to a running list of how stupid Tim Folkerts has to be to believe in a result he assumes to be true in the first place.

    Tim Folkerts is so stupid that he:
    1) shouldn’t be able to tie his own shoes
    2) shouldn’t be able to drive a vehicle without his mother
    3) probably catches a lot of flies without realizing it

    You get the idea. Let’s all of us figure out just how dumb a person like Tim Folkerts has to be. I’m serious about that because this is actually an important research project in a lot of ways, because the better we can understand how stupid a person like Tim Folkerts is, and what the functionality of his stupidity is and how it auto-generates and sustains itself, without it killing him and with him being able to appear to anyone else as if he is a normal functioning person in society, the better we can figure out just what the source of this idiocy is and figure out a way to stop it. So if you want, please provide what you think might be a psychological breakdown of a “mind”(?) like Tim’s; a lot of it has to do with religious fanaticism and simply a very poor comprehension of logic, reason, and science, etc., and maybe that’s enough. But what I’d really like to understand is what drives this level of stupidity…I mean there has to be some very active force at work generating such drooling idiocy. Tim, would you be willing to provide us your Myers-Briggs psychological breakdown? Try this link if you don’t already know it. Did you already know it? We want to understand why you’re such a freaking retard Tim…please help us out.

    Tim, to answer your question, what specifically is wrong with the spreadsheet, is that you start with the assumption of what it is you thought you were proving. Just because you created a spreadsheet, doesn’t mean it has anything to do with reality in the best of circumstances; that you created it based on the assumptions you thought you were proving, renders it a meaningless and complete scientific joke.

    That you can’t understand that, renders yourself a complete, obscene, mental degenerate. All of the calculations are incorrect, because they’re based on the assumption of what you thought you were proving. Of course it converges to Willis’ result – you set it up that way.

    To repeat from an earlier reply:

    WITHOUT SHELL
    Planet: 235 W/m2 to outer space.

    WITH SHELL
    Planet: 235 W/m2 to the shell
    Shell: 235 W/m2 out of the shell to outer space, once the shell warms up to the input
    Between Planet and Shell: radiation field of 235 W/m2, just like if that region was filled with matter would have the corresponding temperature.

    QED. Amazing. Only confusing to religious nutjobs. And no one needed to double the energy or temperature. Magically, a 235 W/m2 imbalance can heat the shell up to 235 W/m2 when the shell starts at zero and the planet starts at 235. But when there’s a 235 W/m^2 imbalance between the planet and the shell when the planet is 470 and the shell is 235, then the differential can’t heat up the shell. This is just completely arbitrary fairy-science.

    How stupid does Tim Folkerts need to be to not understand basic heat flow, and the concept of temperature differentials?

  74. Alan Siddons says:

    If I asked somebody to poach an egg with the warmth of his hands he’d know instantly that I’m joking, for we understand implicitly that an object can’t be made hotter than its heat source. Yet most people fail to grasp that greenhouse physics is also a joke, for neither can an object be made to RADIATE more than the radiation source. The most perfect of radiators, the blackbody, can only radiate 235 W/m² in response to 235 W/m²; a blackbody cannot emit a watt more. Good lord, people who’ve been swindled by greenhouse physics need to be reminded that the ‘radiant forcing’ being described has NEVER BEEN WITNESSED. It would be a miracle if any such thing occurred.

  75. @tjfolkerts 2013/03/10 10:16PM

    I’ll ask them what radiation does when trapped inside a cavity, and if it causes itself to heat up indefinitely. THAT is what they would laugh at. Their answer will be the same as mine – read up on Planck.

  76. @Osborn,

    Very simple. All he did was slow down the cooling rate because the ambient environment was changed. When you change the environment you get different thermal behaviour.

    What did NOT happen was that the cooler plate caused the warmer plate to heat up some more to a higher temperature. That is what they need for the GHE. So what they then do is obfuscate and sophize with the looseness of the language involved here, and draw the invalid statement that a cooler object can cause a warmer object “to have a higher temperature”, because it didn’t cool down as fast. But that is not what the GHE is – the GHE is active heating, a cold thing causing itself to warm up and/or a cold thing causing a warmer thing to become hotter, to gain itself higher temperature above the actual active heating input.

    The science should be easy to understand but the language is very subtle and so it is very difficult to understand the ruse in the manipulation of the language here, and what actual functional principles are trying to be supported. It gets easily distorted along the way and you have to have a very sharp logical and scientific mind to identify it. Most people can’t. The idea to be supported is that cold can make itself hot with its own energy; the idea is not that something warm will cool down slower when it is in a warmer ambient environment – this is true by default and has nothing to do with the GHE. But therein you should be able to see how it is possible to distort the language and create unfounded conclusions from it.

  77. Thanks Alan, nice concise comment/summary. Much appreciated! Hoping to get a follow-up article from you for PSI! 🙂

  78. lgl says:

    Joe

    I can’t see you answered my question. How much radiation inward from the shell?

  79. @lgl
    I’ve answered it several times here. At equilibrium with the source, the shell emits 235 W/m2. The planet also emits that, and this is how the planet and shell are defined to be in equilibrium. Energy is balanced and conserved between the shell and the planet, and so 235 W/m2 is emitted from the shell to outer space.

    WITH SHELL
    Planet: 235 W/m2 to the shell
    Shell: 235 W/m2 out of the shell to outer space, once the shell warms up to the input
    Between Planet and Shell: radiation field of 235 W/m2, just like if that region was filled with matter would have the corresponding temperature.

    What the believers try to state, is that a 235 W/m2 imbalance can heat the shell up to 235 W/m2 when the planet starts at 235 and the shell starts at 0. But when there’s a 235 W/m^2 imbalance between the planet and the shell when the planet is 470 and the shell is 235, then magically the differential can’t heat up the shell. This is just completely arbitrary fairy-science, and believers are so stupid they don’t even realize that they started with a temperature differential to cause heating, and then their end state is exactly the same as the initial state, with a 235 W/m2 differential, but it magically stops being a differential able to cause heating.

    The shell comes to the same temperature as the source, and it isn’t any more complicated than that.

  80. Ron C. says:

    The intent was to prove that greenhouse gases in the atmosphere cause the warming of the earth’s surface by slowing the flow of heat into space and resulting in a higher temperature at the surface.
    The proof begins along the line of: Assume that the cow is a sphere . . . Let’s notice what features of the cow are being assumed away.

    Assume the surface has a constant source of heat. This sets aside several factors. Assume the output from the sun does not vary. Assume the solar energy arriving at the surface does not vary with clouds and other atmospheric changes. Assume that the surface does not cycle between day and night periods, whereby the heat goes from fully on to fully off.

    Assume that radiation is the only form of energy transfer at work. Thus, assume that conduction, convection and latent heat do not force heat to rise from the surface to space.

    Assume no differential in heat capacity between surface and shell. Which means, assume that the surface is not composed of oceans and land, equators and poles, and variable elevations and contours. Assume that the surface does not store and release energy over many time scales, and assume that the shell is not composed of air and thereby has little heat capacity.

    It becomes difficult to decide what scientific laws pertain to such an hypothetical situation.

  81. Great point Ron…exactly.

    Assume the cow is a sphere – perfect! 🙂

  82. Roger Clague says:

    The problem with Willis’s model and with the Energy Budget Theory in general is that its all about radiation. Radiation and heat are confused. Radiation is equated with heat. Mass is left out.
    The theory is supported by Nasa.

    http://earthobservatory.nasa.gov/Features/EnergyBalance/

    For example a diagram label says

    outgoing heat (longwave radiation )

    The people at Nasa are not stupid. They need money for satellites. Satellites measure radiation. So they support a theory that claims climate is all about radiation. A theory that ignores mass but pays the bills.

    I see radiation from the sun absorbed by matter. It becomes heat, motion of molecules of matter. The matter produces radiation which and later leaves the atmosphere.

    The study of heat is thermodynamics

  83. A C Osborn says:

    Joe, thanks for the response.
    Now that I know what to look for I went back and looked at the experiment again and it shows you are correct because the back wall temperature has risen from 29.3 degrees C to 35.5 degrees C.
    ie the local environment temperature has increased by approx. 6 degrees C with the addition of another heat source.

  84. This from Alan Siddons:

    Hilarious. Must have went to the same school as climate scientists.

  85. lgl says:

    Joe,

    Thanks, but that still isn’t a clear answer. You say “235 W/m2 out of the shell to outer space”, fine, but “Between Planet and Shell: radiation field of 235 W/m2” we can only guess means 235 outward from the planet and 0 inward from the shell. If that’s what you mean it’s wrong.

  86. That’s not what I mean lgl, and I don’t understand how you can’t understand it yet. I’ve perfectly and clearly stated what the conditions are. It could be any more clear. You almost understood it, and then didn’t. Your guess was wrong. The shell and the planet come to equilibrium and hence the same temperature. It does the exact same thing as if the space between were filled with matter – heat transfers outward.

  87. Simon Conway-Smith says:

    Joe,

    I hope I’ve understood this. It strikes me as being very simple. Heat flows from hot to cold, period.

    What may affect that flow is the substance between the planet and shell. In the simplistic variants, vacuum:radiation, fluid:convection, solid:conduction. Whatever of these or whatever combination and whatever the thermal capacity of these, the heat still flows in the hot-to-cold direction, with the time to equilibrium being deterministic on the substance’s heat capacity, and even when in equilibrium, heat still flows as the shell is radiating out, trying to reduce its temp, stimulating the continued heat flow from the planet that maintains the equilibrium.

  88. Brilliant, Simon. Thanks. That’s just about a perfect description of heat flow and what actually occurs. The only word I would remove for the sticklers who might not comprehend your meaning, is “stimulating”, because the process occurs spontaneously. It doesn’t need to be stimulated, it is just what occurs naturally due to mathematical nature of the temperature differential and heat flow etc. Cheers.

  89. lgl says:

    Joe

    Why can’t you just give the value, 235, 0, something in between? Besides, if the planet and the shell is at the same temperature there is no heat transfer and the planet will warm up.

  90. lgl, I’ve stated the value several times. Take a look at Simon Smith just above – he understood it just fine. He completely got it. And he’s not a trained scientist as far as I know.

    You think that if two things near each other are the same temperature, then one of them heats up indefinitely… How do you even invent that type of idea? If the planet and shell are the same temperature, this is thermal equilibrium. Ever head of it? They’re in thermal equilibrium. The shell then emits the heat to outer space. It couldn’t be more simple. I want YOU to figure out the answer and understand the physics, like Simon did. You figure it out yourself and that’s the best way to learn. There’s more than enough in my comments here to figure it out for yourself.

  91. lgl says:

    Joe
    No, all you have stated is “Between Planet and Shell: radiation field of 235 W/m2” which, to be correct, must mean there is a net energy flux of 235 W/m2 from the planet to the shell. To achieve this the planet must be much varmer than the shell.
    Yes, I have heard of thermal equilibrium, and at thermal equilibrium there is no heat transfer, the energy produced in the core can not escape and the planet will warm.
    Your self-heating ranting shows you do not understand the concept of insulation.

  92. lgl, insulation doesn’t cause self-heating and it isn’t the GHE. If insulation even was the GHE, which it isn’t, then all the non-greenhouse gases would be the cause of the GHE because they all have very low emissivity and hold on to their heat rather well, and they all contribute to holding more heat near the planet’s surface. This would have nothing to do with CO2 and greenhouse gases, and then the GHE argument falls apart.

    Thermal equilibrium does not mean that heat can not escape…that’s a completely stupid assertion. Thermal equilibrium is when the shell and planet are the same temperature. The energy from the planet has warmed up the shell to the same temperature, and this means that the shell is now the radiating surface. Simple. QED.

    Planet heats shell. Shell radiates energy to outer space. Shell doesn’t heat up planet because it isn’t a source of heat. Planet doesn’t heat itself up because nothing does because that would violate all of thermodynamics, and there’s no reason to even postulate that idea given the laws of heat flow. Please work on understanding this.

  93. Simon Conway-Smith says:

    lgl: You mis-characterise the nature of insulation. A

  94. lgl says:

    Joe
    Thermal equilibrium means no net energy transfer, read here if you really do not know: http://en.wikipedia.org/wiki/Thermodynamic_equilibrium

    and there must be a net transfer of 235 W/m2 from the planet to prevent heating.

    Noboby is saying the shell is heating up the planet. The energy comes form the core, the shell only makes it more difficult for the planet to cool (or radiate if you prefer).

  95. Simon Conway-Smith says:

    (Apologies for the truncated previous post – iPhone finger trouble)
    …A perfect insulator will simply stop heat escaping, but will NOT raise the temperature of the object it’s insulating. How can it, it’s NOT a heat source.

    An example, shine a torch through a perfectly transparent window, then against a perfectly solid (opaque) wall, does the torch get any brighter? No, of course it doesn’t. Then shine it against a perfect mirror; any brighter now? NO, not even now!

    This is really simple stuff!

  96. lgl, Thermal equilibrium doesn’t mean that no radiation leaves the shell to outer space, which is what you implied. It means that the shell and the sphere are the same temperature. The shell radiates to outer space and it gets the energy to do so from the planet. The energy radiated to outer space by the shell is immediately and spontaneously replaced with energy coming from the planet, because the shell and planet are in equilibrium. You don’t need to make it more complicated and un-understandable than that. The shell radiates for the planet.

    Yes, I covered that too: the shell doesn’t heat up the planet and the planet doesn’t heat up itself. What I said was : “Planet heats shell. Shell radiates energy to outer space. Shell doesn’t heat up planet because it isn’t a source of heat. Planet doesn’t heat itself up because nothing does this because that would violate all of thermodynamics, and there’s no reason to even postulate that idea given the laws of heat flow. Please work on understanding this.”

  97. Exactly, Simon 🙂

  98. lgl says:

    Simon

    “Thermal insulation is the reduction of heat transfer (the transfer of thermal energy between objects of differing temperature) between objects in thermal contact or in range of radiative influence. “

  99. Notice that nowhere in that description of insulation could it be implied that insulation causes run-away self-heating. The behaviour of insulation is not what the GHE is.

    Any more comments on insulation being the GHE will be trashed – it has nothing to do with the GHE and even if it did, it would violate the GHE’s own presumptions.

    insulation doesn’t cause self-heating and it isn’t the GHE. If insulation even was the GHE, which it isn’t, then all the non-greenhouse gases would be the cause of the GHE because they all have very low emissivity and hold on to their heat rather well, and they all contribute to holding more heat near the planet’s surface. This would have nothing to do with CO2 and greenhouse gases, and then the GHE argument falls apart.

  100. lgl says:

    Joe

    No, that’s not what I implied. Like I said there is no energy transfer from the planet to the shell if they have the same temperature, and then the planet will heat up, and again no, it is not the shell that heats the planet. The planet does heat itself because it has an energy source in it’s core.

  101. lgl says:

    If the Earth was perfectly insulated against space it would indeed face a “run-away self-heating” because of the 44 TW generated in its interior.

  102. mkelly says:

    Max™ says:

    2013/03/11 at 7:45 AM

    So you’re the Max over at Talkshop that supported my contention that the area mattered. Thanks.

    MKelly

  103. lgl :”there is no energy transfer from the planet to the shell if they have the same temperature, and then the planet will heat up”

    You do imply that the shell won’t be radiating outwards, because you want to stop all the energy flows so that you can say that the planet will heat itself up. That is an obvious sophistry and fraud. You sophize that if the planet and shell are in equilibrium, then no heat can flow to the shell and then the so the shell must not be radiating. Clever sophistry in its stupidity.

    You can’t stop the shell from radiating outwards. The shell and planet can perfectly well be in equilibrium and the shell will be radiating to outer space. We already went over this. The energy the shell loses will be spontaneously and instantly replaced by the planet.

  104. lgl: “If the Earth was perfectly insulated against space it would indeed face a “run-away self-heating” because of the 44 TW generated in its interior.”

    We’ve been over this. Radiation trapped inside a cavity does not heat itself up, but creates a blackbody spectrum. No source of energy can do more work on itself, i.e. raise its own temperature, then it is in the first place. Learn some thermo.

  105. Simon Conway-Smith says:

    lgl: The only way you could apply that argument is if the planet had not heated to its ‘normal/max’ temperature, which simply creates another shell (the heat source, which must be at its normal/max temperature, must be inside), meaning the EXACT same scenario applies as before. No difference!

  106. Alan Siddons says:

    Jeez, as has already been mentioned, the stupidest thing about Eschenbach’s stupid scenario is that the inverse square law is ignored, which is pretty basic stuff. A planet radiating 235 W/m² on its surface could not possibly transfer 235 W/m² to a suspended shell in the first place. Full transfer would require the shell to be in direct contact with the radiating surface! Some people apparently forget that the term “W/m²” refers to SQUARE METERS. Since the shell is larger it has more square meters and simply cannot radiate at the same intensity. For example, a shell of 1 AU radius placed around the sun would not glow with 63 million W/m² — like the sun’s surface — but a paltry 1368 or so.

  107. Simon, brilliant. I don’t know why that is so hard to understand for some people.

  108. Well exactly Alan…they really just ignore everything that might be logically or scientifically or mathematically sensical.

  109. lgl says:

    Joe

    It is you who has stopped the energy flow by putting both objects at the same temperature. It they are both at the same temp the shell will radiate 235 W/m2 towards the planet (you know the number you refuse to admit because it ruins your case) and the planet will radiate 235 towards the shell, i.e no net transfer and the planet does not get rid of the 235 Joules constantly generated every second.
    Jeez, seems you don’t even know what power and energy is. Watts is Joules per second, so please stop this “heat itself up” and “no source of energy” nonsense.

  110. Mindert Eiting says:

    Thanks Joe, for this contribution. Some time ago I gave you a very simple proof that effective back radiation does nor violate the 2nd Law. This is important as it meets a standard objection. I also showed that in a certain experiment the 2nd Law puts a restriction on the effect. Whereas the intensity of back radiation increases, its temperature effect must dwindle to zero in the direction of equilibrium (see also Tallbloke’s comments). So, there is a negative relationship between dose and effect. If I told you that swallowing one kilogram of salt is harmless but taking one gram of it is lethal, there must be something wrong. So far I could not see where back radiation violates physics. It may be in quantum mechanics as you explained. Note that even Claes Johnson describes the idea as non-physical but he does not (or cannot) explain why. If you could write on one page, without repelling your public by calling them stupid or taking a good-willing amateur as scape goat, we could have a refutation of GHE in a few pages. It must be simple, general, and convincing.

  111. lgl, you’ve delevated yourself to the status of flaming moron.

    You, again, completely ignored the fact that the shell also has to emit 235 to outer space. Why do you do that? You do it so that you can sophize around the language of thermal equilibrium. Because if you didn’t ignore the shells outer radiation it would ruin your idiotic self-heating argument.

    The shell emits 235 to outer space. Stop ignoring it. This energy will be replaced by the planet. There would be a minute degree of thermal imbalance “during the interval” but nonetheless the planet and shell would be very near thermal equilibrium and in actually it all happens continuously and not in intervals.

    You have it figured out but choose to be a complete sophist moron instead. If the shell and planet are in equilibrium, then how much heat does the shell transfer to the planet?

    None. The shell does not transfer any heat to the planet.

    However, the shell inevitably radiates to outer space, in which case it does lose heat energy. That loss of energy is replaced by the energy coming in from the planet below, because that’s the source and because there’s a small imbalance. The shell simply becomes another layer of the planet.

    You’re the one claiming that a source can heat itself up beyond its intrinsic capacity to do work – I agree you are spouting nonsense.

  112. Mindert, can think about it. This is part of the process of getting there. The outline is basically all in this post and comments. But as you can see, I’m already well aware that this isn’t about a good argument, but simply about peer-pressure and group think. It doesn’t matter if you show their own math to be wrong or their ideas absurd – that’s what they thrive on, they just get more absurd.

  113. Simon Conway-Smith says:

    lgl: Heat flow and thermal equilibrium is just like in a river, water flows between two points but the level stays the same. Equilibrium does NOT imply no flow, just the the input and output flow are the same, which can be anywhere between zero and infinity. This is the childishly simple maths Joe is referring to. Take ANY number and subtract it from itself and you get zero.

  114. Alan Siddons says:

    No source of energy can do more work on itself, i.e. raise its own temperature…

    True enough. One might as well believe that the voltage from a weak battery can be used to recharge it! Greenhouse theorists play the game that all hucksters do: Lure the victim in with assurances of getting something for nothing. It’s as if these suckers had never heard of the First Law.

  115. Simon, again, thanks so much. That’s exactly the other thing which is being glossed over in this obfuscation over thermal equilibrium.

  116. Alan, yes(!), that is exactly a perfectly valid physical analogy, that is actually physically meaningful. There are some schools of teaching which analogize thermal physics with electrical physics with Newtonian physics, in just the way you did, as a way to make learning all the concepts easier because you can transfer the same intuition “mechanically” between fields. And so indeed, what the GHE tries to say is that a low voltage battery can charge up a higher voltage battery, or that the voltage from a battery can charge up itself.
    It’s not just the 1st Law…it is primary cogitation which they have difficulty with.

  117. lgl says:

    Joe

    I’m not ignoring anything, but you are ignoring that the shell emits 235 W/m2 inwards in addition to the 235 to space. Those 470 W/m2 must be replaced by the planet.

    “None. The shell does not transfer any heat to the planet.”

    Correct, and in equilibrium the planet does not transfer any heat to the shell, which won’t work.

  118. lgl says:

    Simon

    “Equilibrium does NOT imply no flow, just the the input and output flow are the same” Which is what I just said. 235 from the planet and 235 to the planet, i.e no net energy transfer away from the planet and it will heat up, BECAUSE OF ITS INTERNAL SOURCE.

  119. lgl says:

    Alan

    That’s not an analogy at all.
    The basic feedback loop system here would be a valid analogy: http://en.wikipedia.org/wiki/Positive_feedback

    The shell will start feeding back half of the radiation which is added to that already there and the resulting output will be twice the initial input. I’ll bet the next will be it is impossible to make an amplifier like that.

  120. lgl: “I’m not ignoring anything, but you are ignoring that the shell emits 235 W/m2 inwards in addition to the 235 to space. Those 470 W/m2 must be replaced by the planet.”

    That’s a pretty stupid fancy way to double the energy.

    The shell does not transfer any heat to to the planet. QED. You know this you fool. They’re in equilibrium, hence no heat is transferred from the shell to the planet. The shell just emits to outer space and this energy gets replaced by the planet, and it doesn’t require the planet magically having twice the energy it did before.

    As I said in my update at the end of the article, you just repeat yourself instead of learning. You’re not even caring that you’re contradicting yourself with the equilibrium argument – caught in your own trap.

  121. LGL: ““Equilibrium does NOT imply no flow, just the the input and output flow are the same” Which is what I just said. 235 from the planet and 235 to the planet, i.e no net energy transfer away from the planet and it will heat up, BECAUSE OF ITS INTERNAL SOURCE.”

    No that’s not what you said at all. You have the directions all messed up and you continue to DENY that the shell radiates outward. The shell transfers no heat to the planet, but the shell loses heat to space and this is replaced by the planet to the shell. But go ahead and keep repeating yourself and denying the output from the sphere.

  122. lgl: “Alan, That’s not an analogy at all. The shell will start feeding back half of the radiation which is added to that already there and the resulting output will be twice the initial input. I’ll bet the next will be it is impossible to make an amplifier like that.”

    Wow, I mean this is fun to watch, how far idiocy goes in exposing itself. An “amplifier” in a circuit does not allow the circuit to charge up its own battery! I mean holy crap how stupid is that.

    Since you’re just repeating yourself now, your comments which have been proven wrong over and over again and are now beyond pathetic, follow ups will be trashed unless you actually start to say something intelligent.

  123. Simon Conway-Smith says:

    lgl: No, no, NO!!
    The energy flow is one-way, from planet via shell to space. There is NO flow from shell to planet! There cannot be as the shell is NOT a heat/energy source. That is a fundamental point you really need to grasp.

    Zero heat flow within a body at a higher temperature than its surroundings would require the body to contain heat energy and be completely encased by a perfect insulator. As explained, if that we’re the case, the temperature would just remain as it is, and go neither up nor down. In fact, I asked my 13 year old son this question earlier, i.e.” what happens to a body with temperature when encased by a perfect insulator,?” and he immediately replied “nothing”. Correct!

    Take the river analogy, water cannot be flowing on both directions at once, and neither can the input flow be duplicated (doubled) so there is that amount of outflow AND backflow. Where does that extra water come from? The answer is that it doesn’t come from anywhere as it doesn’t exist. It can’t magically appear out of thin air!

    This is also the fallacy of back-radiation. If a (1) unit of energy has been radiated from the surface, and ‘if’ a fraction (0.x) is radiated back (but as Joe quite adequately explains here, it doesn’t), the CAGW/warmist protagonists claim the net energy at the surface is 1.x! It’s not, it’s -1 + 0.x, which is less than 1, i.e. cooler.

    In fact, gases such as CO2, because of their IR absorbtion & re-radiation (& for water vapour only, retention) properties, they actually help COOL the surface by the additional removal of heat via conduction and convection, over and above radiation.

  124. Alan Siddons says:

    “The shell will start feeding back half of the radiation which is added to that already there and the resulting output will be twice the initial input.”

    Pertinent to this purported Magical Multiplication Mechanism, lgl, may I direct you to an essay of my own? It’s called Why conventional Greenhouse Theory Violates the 1st Law of Thermodynamics
    http://hockeyschtick.blogspot.com/2010/06/why-conventional-greenhouse-theory.html

    If you’re not interested, though, here’s the upshot: Redirecting radiant energy back to its source will add no energy to that source. And yes, this is directly analogous to a battery’s inability to recharge itself. It’s all very well to draw arrows and numbers and to imagine that radiant intensity can be amplified by feeding light back on itself, but reality vetoes such a fantasy. Just as it takes a higher temperature body to raise another body’s temperature, it takes a greater radiance to add light to a radiating body.

  125. BTW Alan here are the relations between mechanics, electricity, and thermal physics:

    a = 1/m * (F2 – F1)
    I = 1/R * (V2 – V1)
    q = k * (T2 – T1)

    These all behave analogously.

  126. And, note, that force doesn’t increase itself due to a differential; voltage doesn’t increase itself due to a differential; and temperature doesn’t increase itself due to a differential!

  127. Max™ says:

    I’m not ignoring anything, but you are ignoring that the shell emits 235 W/m2 inwards in addition to the 235 to space. Those 470 W/m2 must be replaced by the planet.” ~lgl

    Uh… if the planet supplies 235 W/m^2, the shell (being at a non-zero distance) receives energy reduced by the inverse square of the distance, and that energy is radiated outwards by the outer surface of the shell at a power which must be less than 235 W/m^2, otherwise the presence of the shell creates energy.

    The shell has two sides, it is receiving less than 235 W/m^2 on one side, how in the hell is it supposed to radiate more than it receives from both sides of the shell?

    ___________

    mkelly, indeed, my first post in the thread over there (the first post at all actually) was about the way Willis ignored area:

    “But what it will do is add to the total amount of radiation the planetary surface is receiving.” ~tallbloke

    Shouldn’t that be “it will reduce the total amount of radiation the surface loses”?

    The outer shell is not an energy source for the surface, it is heated by the surface.

    _______

    Quick note:

    Say the internally powered body has a surface area of uh, 1000 m^2, and is emitting 235 W/m^2, so the total output of the surface is 23,500 W/m^2, right?

    Suppose the outer shell has a surface area of 2000 m^2, and is receiving 235 W/m^2 from the surface of the heated body.

    If the outer shell emits 235 W/m^2 that is a total output of 47,000 W/m^2, right?

    If the outer shell had twice the surface area of the internally powered body inside, wouldn’t it only need to emit 117.5 W/m^2 * 2000 m^2 to balance the 235 W/m^2 * 1000 m^2 supplied by the interior body?

  128. They add radiant energy together in such a stupid way. “If you have one ice cube shining on you, you’ll be temperature X. If you add another ice cube shining radiation on you, now you’ll be much much hotter! Way hotter than a single ice cube.”

  129. sunsettommy says:

    lgl writes this howler:

    “I’m not ignoring anything, but you are ignoring that the shell emits 235 W/m2 inwards in addition to the 235 to space. Those 470 W/m2 must be replaced by the planet.”

    He he…. the planet Nuclear CORE as shown in the chart is only emitting 235.

    Meanwhile the emission rates between the core and the shell grows and grows and grows and grows and grows inside the shell but still only emits 235 to space…. it is a recipe for a quick meltdown and or explosion to relieve it.

    You fail to notice that the Planet core is the only source of radiant energy therefore is at the top of the energy hill that quickly drops downward when the radiation is spread outward in all directions.The shell is down the slope of the energy hill and will stay that way since it is entirely dependent on what the top of the hill provides.The shell being an absorber that can only reradiates what it received which at the most is 235 or less.

    When the core stops emitting,the hill itself goes away and everything including what the shell was receiving becomes FLAT.

  130. Max™ says:

    Hmmm, had this pop up over on talkshop:

    Yes, I see exactly what I did there. I divided both sides of an equation by the same amount (4 pi). That is perfectly legitimate algebra. There is absolutely nothing wrong. ” ~Tim Folkerts

    He took this: P = ε σ A T^4

    And replaced it with: P = ε σ (A/4*pi) T^4 when calculating the ratio of radiated power emitted by an Earth sized planet and an 11 km high shell around it.

  131. Are they over there confusing themselves about it some more?

    It doesn’t even matter about the shell radiating energy and how to calculate the energy density it receives – it does’t heat up the interior!

    But if you want to do it right you just map it from one area to the other, as you did earlier. P2 = P1*A1/A2. It’s the inverse square law when in relation to spherical surfaces.

  132. tallbloke says:

    Joseph E Postma says:
    2013/03/11 at 6:57 PM

    And, note, that force doesn’t increase itself due to a differential; voltage doesn’t increase itself due to a differential; and temperature doesn’t increase itself due to a differential!

    However, a resistor with a current running through it will get very hot if you put it in a vacuum flask…

  133. tallbloke says:

    ….and it will stop getting hotter once the outside of the vacuum flask is radiating as much energy as the resistor was in open air.

  134. Will Pratt says:

    You cannot add 235 W/m2 to 235 W/m2 and get 470 W/m2.

    All you get is 235 W/m2.

    You have done nothing to increase the amount of energy. Energy is a potential to generate heat. If you don’t increase the energy you don’t increase the potential to generate more heat.

    The simple analogy I repeated several times on Eschanbach’s sophistic nonsense post is the ice-cube analogy. This is a basic, simple to grasp refutation of the “GHE” hypothesis.

    Take a 1 ounce ice-cube radiating at 273K in an environment which is also radiating at 273K. Now place another 1 ounce ice-cube radiating at 273K beside it. What happens to the temperature of the first ice-cube?

    Answer: Nothing.

    Now add another ice-cube, add another thousand ice cubes. Try adding a million or a trillion ice-cubes. Instead of a trillion 1 ounce ice-cubes make that a trillion 1 ton ice-cubes all radiating at 273K and what happens to the temperature of the first ice-cube?

    Answer: Exactly nothing.

    235 W/m2 + 235 W/m2 = 235 W/m2

    Incidentally, the last comment on that thread which Watts censored, I had pointed out, among other things, that the sun has a surface temperature of 5700º C, while the temperature at the core is some 15,000,000º C.

    WUWT
    “The world’s most viewed site on global warming and climate change”

    Sure, whatever!

  135. Tallbloke claims that the Willis twin sphere is correct. No it is not! How does Tallbloke think a vacuum flask works? it is basically the same as the Willis model. Do the contents get hotter? Well they should given the poor model not recreating reality but reality steps in and vacuum flasks gradually cool their contents.

  136. “a resistor with a current running through it will get very hot if you put it in a vacuum flask…”

    Well sure, but that’s not the GHE either. Cheers Rog.

  137. “….and it will stop getting hotter once the outside of the vacuum flask is radiating as much energy as the resistor was in open air.”

    The temperature it gets to is dependant upon the power running through it. If the flask absorbs the thermal radiation, then it can eventually get to the same temperature as the source.

  138. Exactly Will! An entire universe of radiation from ice-cubes will not make you hotter than the ice cubes. Good one.

  139. Gareth says:

    Greg House said: “1. Initially the shell radiates nothing, then it receives 800 from the planet and radiates a half of it to the outer space, so, the outer space receives 400.

    2. The shell radiates the same 400 back to the planet and the planet thankfully absorbs it (and gets there warmer), but not being egoistic at all the planet re-radiates it to the shell in addition to what the planet’s “inner engine” constantly produces, so, now the planet radiates 400+800=1200. Note, 1200 goes to the shell now.”

    Could you please explain why all of the 400 that goes from the shell to the planet is reflected back to the shell, rather than 200 being reflected and 200 serving to raise the temperature of the planet to 1000? It only has one side unlike the shell but what is making it a perfect reflector of incoming energy in your example?

    My thinking is as follows:

    The planet begins at 800 and the shell at zero returned to the planet and zero to space.
    Then the shell returns 400 to the planet and emits 400 to space.
    The planet retains 200 of the 400 and reflects the other 200, along with the constant 800 to make 1000 now going to the shell.
    The shell then reflects 500 and emits 500 to space.
    The planet retains 250 and reflects 250, plus 800.(1050)
    The shell reflects 525 and emits 525 to space.
    The planet retains 262.5 and reflects 262.5, plus 800.(1062.5)
    The shell reflects 531.25 and emits 531.25 to space.
    etc, which tends towards a stable planetary temp of 1066.66 and a shell temp of 533.33.

    Using the Eschenbach figures it would tend towards a stable planetary output of 313.33 and a shell output of 156.66 in each direction.

  140. Alan Siddons says:

    Wow, I just noticed a recent comment about my previously mentioned 1st Law essay. And it refers to the very WUWT article that Joe is criticizing here!
    http://wattsupwiththat.com/2013/02/06/the-r-w-wood-experiment/#comment-1221185

    Here’s George Smith’s remark:

    Clausius, a Thermodynamicist of impeccable credentials, also used the second law to derive what we call “The Optical Sine Theorem”, which basically says that no optical system can create an image that has a higher radiance than the source object.

    No optical system can create an image that has a higher radiance than the source object.

    GHE believers, please try to understand that if something is producing 235 watts of energy, there’s no way to increase that energy without plugging in an additional energy source. And a mirror or ‘reradiator’ doesn’t qualify as an additional energy source. Nor does a reflection of yourself in a mirror qualify as a second you!

  141. Gareth, Greg’s post was to highlight the flaw in the reasoning, not an argument for it.

    In your own argument, you’re hiding half of the portion sent back to the planet from the shell in so-called “retention”, with only the other half, the reflected portion, going back to shell. This violates all the standard arguments for energy conservation because half of the return goes to absorption and half to reflection – the absorbed portion has to result in an intrinsic increase of energy output due to the temperature increase of the planet, and this is the way it is always typically argued. What you’re doing now is something new, and it results in a different answer than what Willis wanted and it doesn’t correspond to anything on the Earth either. Also, there is an arbitrary assumption that the return from the shell is exactly split 50/50 between absorption and reflection on the planet, and this has no basis, and isn’t compatible with the objects being blackbodies.

  142. Wow that is a good one Alan! Should send this to John @ PSI to write an article on. Because here we have an actual theorem from Clausius refuting exactly the “shell game” argument and the GHE in general.

  143. Well there’s not much out there on the “The Optical Sine Theorem”, at least at Wiki, but, George Smith’s interpretation of it is the mathematical idea, it looks like.

  144. Will Pratt says:

    Indeed Joe,

    Multiply 235 W/m2 by any figure you like and the maximum will never exceed 235 W/m2.

    This is the crux of AGW fraud. This is the fraud behind Eschenbach’s post.

    I’ve always said, there are three types of “GHE” adherents:

    Liars, fools and lying fools.

    Eschenbach is not a fool, neither is Watts, but they are most certainly both liars and quite possibly lying fools.

  145. lgl says:

    All you morons,
    on which law of physics are you mounting your claim that the shell will only radiate to space and nothing inward towards the planet?
    In this particular case it’s hilariously stupid because the shell is heated from the inside, yet the inside, which is absorbing the radiation from the planet, does not warm and start radiating, only the outside gets warm and start radiating. Can’t get more stupid than that.

  146. lgl says:

    Alan,

    “And a mirror or ‘reradiator’ doesn’t qualify as an additional energy source.”

    Why not? Is that energy just disappearing? not beeing absorbed?

  147. lgl: How stupid do YOU have to be to think that something colder will warm up something hotter? Completely idiotic. By your own argument, because the shell and planet are in equilibrium, NO heat is transferred from the shell to the planet. Your own trap trapped yourself because you have no clue about thermal physics, or numbers at all. So now you just change “heat” to radiation energy. Radiation doesn’t transfer heat energy if objects are in equilibrium, idiot.

    Didn’t you read Simon’s explanation? Of course you did but you just prefer to be an idiot…because of your religion.

    I guess we can start a list for how much of an idiot lgl is.

    lgl is such an idiot that he:
    1) thinks he can warm himself up with an icecube
    2) has never heard of cavity radiation and the origin of quantum mechanics and the blackbody spectrum
    3) should stay the heck away from science before he hurts someone with it

  148. lgl said :”Alan,

    “And a mirror or ‘reradiator’ doesn’t qualify as an additional energy source.”

    Why not? Is that energy just disappearing? not being absorbed?”

    Two icecubes beside eachother don’t warm each other up. Two light bulbs beside eachother don’t make eachother shine brighter.

  149. Simon Conway-Smith says:

    Will, It’s such a shame that both Eschenbach and Watts have fallen for this fallacy, as otherwise they have done sterling work towards debunking the whole CAGW debacle. However, they are not the only ones, folks like Monckton and Lindzen too, and also Peter Lilley MP (one of the few UK MPs who voted against the Climate Change Act, but on economic grounds, not scientific). I was at the Cambridge UK climate event last week (Wed 6th Mar, “Global Warming & Equitable Development”) at which Lindzen spoke, and he still holds to the “CO2 must have some heating effect” line.

  150. And with decades of evidence that a rise in CO2 hasn’t actually even caused a rise in temperature, (i.e., there’ been no temperature rise above 1930 values or for the last 15 years, despite a massive increase of CO2), no one seems able to make the connection that perhaps this is happening because CO2 can’t cause a temperature increase at all. All they have is the obvious fraud of starting the “anomaly” at an arbitrary cold period! Fraud. Sheesh.

  151. Simon Conway-Smith says:

    lgl, You ask why a mirror doesn’t act as an additional energy source? Why isn’t it obvious!! Isn’t the clue in the question?? Its a ‘reflector’ of energy, not an ‘originator’. IT DOES NOT ADD ANYTHING!!

    Another analogy: If you buy something from a shop for $5, hand over a $10 bill and receive $5 change, you don’t still have the original $10. What you are claiming is that you’d still have the original $10 AND the $5 change. If only that were true, but it patently isn’t, otherwise we’d have an infinite source of energy – the perpetual energy machine.

    Please, please, please think this through.

    The underlying concept is: You can’t create something out of nothing. EVER.

  152. Simon Conway-Smith says:

    Joe, ‘Occam’s Razor’ springs to mind, but so many people want to create something complex and fantastical whilst completely missing the simplest of all things, which as the razor says, is often the answer. Those pushing the CAGW/GHG/GHE/BR stuff have overlooked the simplest of all explanations of why the temperature has remained static whilst CO2 has continued to rise, and that’s that CO2 isn’t the cause!

  153. lgl says:

    Joe,
    The shell and the planet being in equilibrium is your argument, not mine. Obviously they are not since there is a net energy transfer of 235 W/m2 from the planet to the shell, got it this time?

  154. lgl says:

    Simon

    Ok, place an electrical heater inside of the shell giving 235 W/m2 towards the planet. Will the radiation from the heater be absorbed by the the planet? will that heat the planet? If so, how is the planet able to differenciate between the two radiators so that it can absorb only the radiation from the heater and dismiss the radiation coming from the shell?

    And again, how come the shell is only radiating from the outside and not the inside? If that really is your claim.

  155. No, lgl, it won’t heat the planet. Two ice cubes don’t heat each other up. Heaters don’t heat each other up with equal temperatures! If the world worked this way, we could do everything with very little. A source at 273K could be put beside another source of 273K, and these would create any amount of heat and high temperature we wanted. Doesn’t happen. And you’re a complete idiot for not getting it.

    Will soon be trashing your comments as you’re just here to waste our time at this point; although, it is fun and educational to see the bullshit you produce and destroy it.

  156. “The shell and the planet being in equilibrium is your argument, not mine. Obviously they are not since there is a net energy transfer of 235 W/m2 from the planet to the shell, got it this time?”

    The shell transfers no heat to the planet because it is either 1) cooler than the planet 2) the same temperature as the planet. The shell emits 235 to outer space and this energy comes from the planet.

  157. tallbloke says:

    John Marshall says:
    2013/03/12 at 6:42 AM

    Tallbloke claims that the Willis twin sphere is correct. No it is not! How does Tallbloke think a vacuum flask works? it is basically the same as the Willis model. Do the contents get hotter?

    Yes, if there’s a 235W/m^2 energy source in there with the coffee. Do vacuum flasks come fitted with nuclear furnaces these days?

  158. The coffee won’t get any hotter than the source, which in this case would be very cold coffee!

  159. Anthony Watts says:

    This thread is a perfect example of why WUWT is a “Slayer Free Zone” and shall remain so.

  160. Anthony, there’s very little justification in your post.

  161. Max™ says:

    “Max™ says:
    Your comment is awaiting moderation.
    March 12, 2013 at 4:09 pm

    “So no answer to my request for the magnitude of the error. I’m not surprised, because if you revealed it, you’d be exposed for making a mountain out of a molehill.” ~tallbloke

    The error is of the wrong sign entirely, 400 Gigawatts at the low end is not a molehill, because it is energy produced from nothing.

    1 Watt is “oops, musta forgot to carry a 1 somewhere”, 400 Gigawatts, whether it is 1%, or 0.1% or 0.000001% is still free energy from nowhere.

    My calculation suggested the error was between 0.003% at the low end and 0.2% or more at the high end.

    ________

    Note that is NOT including the error induced from having the surface emit at full power from the inside and outside, which doubles the energy it receives, nor is it taking into account the inverse-square law.

    ________

    Tim F. used the wrong method to calculate the ratio (by inserting the radius^2 where the area should be), but he didn’t wind up with the shell manufacturing energy as the original thought experiment does.

    [Reply] Like I said to mkelly, you guys need a new nitpick, because this one is worn down to the nub.
    Further bullshit about tiny area discrepancies resulting in tiny errors made to sound big errors or broken laws of thermo go in the bin along with this one. Patience very thin now, have a care.”

    Disappointing, tb.

  162. lgl says:

    Joe

    Trashing my comments will only prove you have realized you are wrong and want to end the discussion, bad move.
    Nobody is claiming two ice cubes will heat each other up, nor that two bodies at 273K will, but they will slow down the cooling of the other body. If you place a shell at 273K around a planet at 273K you will stop the cooling of the planet because the planet will emit and absorb the same amount of energy, net=0, no energy loss from the planet. The internal source will then heat the planet so that the cooling is restored. Unbelievable that this can be so hard to comprehend.

  163. Will Pratt says:

    Simon,

    All of the above mentioned are purposeful deliberate gatekeepers. Monckton is a Rothschild agent. His sister shadowed Princess Diana, herself a Rothschild according to Al Fayed.

    During the Thatcher government, Monckton worked hand in glove with Sir Crispin Tickell (Father GIA), himself a descendent of T.H. Huxley, a name which has prominence in my investigations into AGW fraud and links to the Father of AGW fraud, John Tyndall. Monckton and Tickell are at the heart of this fraud, make no mistake about that.

    Monckton’s role as a sceptic is nothing more than contrived “controlled opposition”. Hence the reason he defends the logical fallacies and circular arguments which under-pin the “GHE” hypothesis. He knows that if the “GHE” is established as the fraud that it is, he will be looking at jail time, as he has played a much larger role in setting this fraud up than he cares to reminds us.

    For example he actually admitted (stupidly boasted) on one WUWT thread that it was he who, while working in the Thatcher Government, developed the first computer models of “radiative transfer”. A concept which, as we are now realising through reasoned logical debate, is a violation of the basic laws of thermodynamics.

    Lindsen, Lilley, and anyone else who defends the “GHE” hypothesis are all cut from the same cloth.

    They can all be considered as Rothschild agents that have infiltrated practically every major institution the world over.

    When you connect the dots, this is obviously a Rothschild global power grab assisted by their own private army of agents.

    They must defend the “GHE” hypothesis to the end, for that is their Achilles heel. Which is why they refer to the “GHE” hypothesis as if it were established physics. It is not, it is an hypothesis, and a provably false one at that.

    If the “GHE” gets destroyed, they will be destroyed with it.

  164. lgl: I would trash your comments because you just keep repeating the same argument which has been proven to be wrong more than a dozen times now.

    You are a complete idiot. You JUST said that ice cubes cant heat eachother up…and then you said that one ice cube inside another can heat eachother up. I only keep your posts for research into stupidity at this point.

  165. Simon Conway-Smith says:

    Anthony, please explain then how you create energy out of nothing, which seems to be the claim, i.e. the planet emits 235, and the shell radiates 235 back AND 235 to space? Where does the 2nd 235 come from?
    Also:
    (i) If the planet is radiating 235, the shell cannot radiate 235 back and heat the planet,
    (ii) if the shell were to radiate zero, it would be a perfect insulator but still would not heat the planet,
    (iii) if the shell were to re-radiate 117.5 back to the planet and 117.5 out to space, it still would not heat the planet.

    ONLY if the shell had an EXTRA source of heat that raised its temperature and maintained it above 235 would there be a heat flow from the shell to the planet raising the planet’s temperature, but it would also still radiate out to space.

    Saying the shell radiates 235 inward AND 235 outward is sheer nonsense. It only has 235 to begin with. As I said to lgl, you cannot create something out of nothing – EVER.

  166. Will, thank for that analysis. Great stuff. And spot on. These people are gatekeepers. This is how “the elite” (de-elite really) get their programs through: a) create an argument about something, b) get through what they wanted that is only tangentially related to the argument.

    Ever wonder why defence of the GHE is so hysterical from the likes of Monckton, Anthony Watts, etc etc? These are probably PAID skeptics. Who funds these people’s work etc? As you point out for Monckton, the money trail is all too obvious.

    It is the GHE that they wanted to get through, even if they lost some of the alarmist debate. But they’re not even really losing the alarmist debate – they’re just as loud and as stupid as ever. They’re just waiting until the brainwahsed younger generation grows up.

    The fake skeptics, like these people who ban comments from us on their blogs, and these people who hyperventilate about the GHE with absolutely ZERO math or science or observation to back them up, aren’t skeptics at all. NO skeptic should hyperventilate at the criticism of anything. They’re religious defenders of a faith!

  167. Simon, Anthony wont answer. He’s just another hyperventilator, and I seriously doubt he could argue about any of this. He’s not an actual scientist (as a fact) – he’s a PR person that hyperventilates when someone criticizes the basis of his religion.

  168. Simon Conway-Smith says:

    lgl,

    If you place a shell at 273K around a planet at 273K you will stop the cooling of the planet because the planet will emit and absorb the same amount of energy

    You have made a HUGE assumption here, that the shell is an energy source in it’s own right. If it is, then fine, but the point is it’s not, but a passive shell that has no original energy. The only energy it gets is from the planet, so it CANNOT get to a higher temperature than the planet, nor can it raise the temperature of the planet any further. Joe’s two ice cubes and two heaters statement is entirely correct.

  169. “But they’re not even really losing the alarmist debate – they’re just as loud and as stupid as ever. They’re just waiting until the brainwahsed younger generation grows up.”

    As long as the GHE exists they WILL WIN the alarmist debate. Because it the source of the whole fraud. As long as there is a GHE, there is alarmism. And they KNOW it, and is why they hyperventilate when it is criticized.

  170. lgl says:

    Joe

    “you said that one ice cube inside another can heat eachother up”

    No I didn’t, I said the internal source of the planet would heat it up.

  171. lgl: That’s fine too. The source from the planet can heat the ice shell. Never will the ice shell heat the planet or the planet heat itself spontaneously. The presence of a temperature differential doesn’t cause the hotter side of the differential to heat up – it just warm the cooler side until the cooler side is emitting the same energy as the source. This doesn’t require the source to become hotter.

  172. Simon Conway-Smith says:

    Will, Thanks for that. I do fear for our nations who are so trapped by those who cannot understand the basics of science, yet continually lay claim to “the scientists”, but as we all know, it’s the grotesque ‘consensus’ that they cling to.

    As for the UK and Huhne, he went to prison for the wrong reason. Whilst I care that he lied to a court, this is but an atom of his culpability compared to the planet-sized crime of wilful deception and lying to the nation, namely regarding his support of CAGW and the policies and taxes forced on us by him, and now his successor Ed Davey and the uber-trougher John Selwyn Guimmer, aka. Lord Deben.

    The other person I have a severe reaction to (and repeatedly reply to his tweets, countering his fantasies with reality) is Bill McKibben. I’ve never known someone who has enjoyed human suffering so much as he, nor anyone who more aggressively and wilfully is trying to impose misery and harm on the human race. For a supposed intelligent person, you would think by now that he would rename his organisation from 350.org, as a target of 350ppm for CO2 is so unrealistic, it simply astounds me that he thinks it’s still attainable by human action.

  173. lgl says:

    Simon

    “(iii) if the shell were to re-radiate 117.5 back to the planet and 117.5 out to space, it still would not heat the planet.”

    Of cource it would. Would it just disappear?
    The 117.5 will be an addition to the 235 from the interior. All of it, 352.5, will be emitted and absorbed by the shell which will send half of it back to the planet, so in this ’round’ 176.5 will be added to the 235 and so on until the planet emits 470 W/m2 and balance towards space is restored. Until this point less than 235 has been emitted to space, some has been retained at the planet.
    This does not imply creating energy. The temperature increases because for a while some of the energy production in the core is prevented from being radiated to space.

  174. lgl: Thermal physics doesn’t behave that way. A colder source doesn’t warm up a warmer source.

  175. lgl says:

    Joe,

    This is exactly how thermal physics behave in the case at hand, the only way it can work, but you are of course free to believe your own fairytale. I have now explained this from all the angles I can think of, those who still do not understand must be quite immune to learning, so kids, get back to school, if you try real hard you might learn something.

    (and for the n-th time. Nobody is claiming a colder source warms up a warmer source. The internal source is warming it up.)

  176. lgl: A source does not heat itself up some more above the source that it is. You require to go to school, to repeat highschool chemistry at least.

  177. lgl says:

    Ok, just found one more way to explain.
    A similar case would be to put mirrors around the planet covering one half of the area, thereby cutting the radiation to space in half, sending the other half back to the planet. This would then heat the planet until it emits 470 W/m2 because half the area is blocked and only 235 W/m2 will be radiated to space. Balance is restored.

  178. lgl: Radiation can’t heat up its own source. This is still a cavity you’re describing, and it is only half as good as the previous one at that.

  179. Max™ says:

    P = ε σ A (T_hot^4 – T_cold^4)

    Why is the radiation coming back to the heat source added rather than subtracted from the energy leaving the source?

  180. Exactly. A temperature differential doesn’t mean that the hotter side of the differential has to heat up. I mean OMG.

  181. Simon Conway-Smith says:

    Oh man!!! This is so fundamental to the whole argument, that a heat source cannot heat itself up further, whether by reflection or insulation, a concept that’s child’s play, that it’s no wonder the world is in the mess it’s in.

    lgl: What balance are you trying to restore? The planet only has 235 W/m2, and that’s the only energy source, so nothing, I repeat. nothing can be done by the (passive, zero energy source) shell to increase that and raise the planet’s temperature. As has consistently been said, you cannot create something out of nothing. 235 is 235, period. No 117.5 or 470 is anywhere in sight.

    If you think it does, then why doesn’t a vacuum flask heat its contents any further, as by your wishful conjecture, it should.

    It’s lgl (+ Watts, Mann, etc. etc.) vs. the real world and verifiable physics theory (that’s been tested to destruction). Guess watt, the real world wins every time.

  182. Well that’s why he’ll switch to saying the planet heats itself up, and then when that is refuted, he’ll switch back to saying the shell heats the planet.

  183. lgl says:

    Max

    What? It’s added because it is absorbed by the surface. What is the physical mechanism for subtracting the radiation?

  184. Kristian says:

    Postma,

    I think the way you need to convey your message here is by pointing to and describing the somewhat counterintuitive (i.e. quantum) nature of how radiative heat transfer actually works, on a microscopic level (well, I’m sure you already have; but maybe you need to reiterate). People will never get it otherwise (if even then). As of now it seems impossible to get through. They’re completely stuck in their everyday linear flux budgeting Stefan-Boltzmann way of thinking.

  185. lgl: a colder thing can not warm up a warmer thing; thing of the same temperature do not warm each other up either. This is where lgl has just switched back to the shell causing heating.

  186. lgl says:

    Joe

    Using my mirror case. Explain how the planet can avoid being heated when the radiation to space is reduced to 117.5 and the core is still supplying 235, in addition to the 117.5 from the mirrors.

  187. Hi Kristian, thanks for the perspective. I do have the outline for your discussion requested above in my OP, and I could reiterate it and focus on it in another post without the ranting. Cheers.

  188. lgl: Because you don’t understand thermal physics and use this Wattage addition scheme that does the impossible.

    Just like with a full shell, the half shell will warm up to the same temperature as the source (assuming it is very close) and then emit that energy.

  189. Max™ says:

    lgl, a warm surface radiating into an environment which is cooler than it is will not radiate at full power.

    Conduction works this way, convective transport works this way, radiative transfer works this way as well.

    Only when there is a body three times warmer than the other or more can you treat it as P = ε σ A T^4.

    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2

    P = ε σ A (T_hot^4 – T_cold^4)

  190. lgl says:

    Simon

    The planet surface is not the heat source. The source is in its interior. The surface receives 235 from the interior and 235 from the shell and has to emit all the 470. How can you believe there is no radiation inwards, only outwards?
    And I’m 100% confident you can not point to one single “test to destruction” of what I’m saying. Prove me wrong.

  191. lgl: The shell receives 235 from from the planet and the shell emits 235 to outer space. This doesn’t require any 470 anywhere. The shell doesn’t send any heat back to the planet.

  192. Simon Conway-Smith says:

    lgl, I already have, it’s called the thermos flask.

  193. lgl says:

    Joe

    It isn’t a half shell, it’s a half mirror, a perfect mirror so it does not absorb and get warmer. You can even cool it to 0 K if you like, makes no difference. Try again.

  194. lgl says:

    Max

    “a warm surface radiating into an environment which is cooler than it is will not radiate at full power”

    Radiation is determined by temperature, area and emissivity only.
    You are talking about net radiation.

  195. lgl: A half mirror or a full mirror makes a cavity just the same. Radiation can not heat up its own source. Doesn’t matter how many tries you attempt to change the parameters; going from blackbodies to perfect reflectors, you have no consistency whatsoever and it is desperate obfuscation. Mirrors were never the question, and even as they are now, a source won’t heat itself up beyond the source that it is with its own radiation, and radiation can not heat up a warmer or same-temperature source. Radiation inside a cavity will produce a blackbody spectrum, not change its frequency.

  196. lgl says:

    Joe

    “The shell doesn’t send any heat back to the planet.”

    Utter nonsense. From where do you have the insane idea that a warm object will radiate only from one side and not the other?

  197. Max™ says:

    If the shell is radiating 25 W/m^2 back towards the surface, then the surface is radiating 210 W/m^2 (235 out – 25 in), if the shell is radiating 235 W/m^2 back towards the surface (impossible, since that is more energy than it could receive) then the surface is radiating 0 W/m^2 (235 out – 235 in).

    The inverse-square law means the shell can not receive 235 W/m^2, the surface area of the shell means it can not radiate more than it receives, the fact that the shell has two sides means you have to split the energy lost by the shell at equilibrium, if it loses 235 in and 235 out, that is more than twice what it could possibly receive from the planet.

    I would expect it to lose something less than half that amount, the idea that it has to emit 235 W/m^2 is ridiculous.

    The Earth receives 1366 W/m^2 from the Sun, does it emit that much?

    No?

    Didn’t think so.

  198. lgl says:

    Simon

    No you have not. A thermos flask does not contain a heat source, the planet does.

  199. lgl said: “Joe

    “The shell doesn’t send any heat back to the planet.”

    Utter nonsense. From where do you have the insane idea that a warm object will radiate only from one side and not the other?”

    OK so you DO think that a cold object heats a warmer one, or that the same temperature objects heat each other. Energy is not heat – it only causes heating if there is a differential, and the heating occurs on the cool side of the differential – the hot side doesn’t warm itself up.

  200. lgl: The flask could have an infinite heat source, a heat source at 235 W/m2 with an infinite amount of energy behind it, and it still wouldn’t heat anything to higher than 235.

  201. Kristian says:

    lgl says, 2013/03/12 at 10:21 AM:

    “If you place a shell at 273K around a planet at 273K you will stop the cooling of the planet because the planet will emit and absorb the same amount of energy, net=0, no energy loss from the planet. The internal source will then heat the planet so that the cooling is restored. Unbelievable that this can be so hard to comprehend.”

    This seems to be where the misunderstanding originates. Placing a shell at 273K around a planet at 273K (with a vacuum in between the two) will not stop the cooling of the planet.

    Consider this, lgl: As the shell’s heat GAIN from the planet to its inner surface grows steadily smaller the more the shell’s temperature approaches the planet’s, the very same shell’s heat LOSS to space through its outer surface grows steadily larger. In your ‘equilibrium state’ the shell gains 0 W/m^2 worth of heat from the planet while losing 315 W/m^2 worth of heat to space. Why wouldn’t the shell at this point cool as fast as you claim the planet would warm in the opposite situation (heat gain from nuclear power source: 315 W/m^2; heat loss to shell: 0 W/m^2)?

    Let me pose you these two questions:
    If you’ve got one radiator at 273K emitting a radiative flux of 315 W/m^2 and then place another one just next to it, also at 273K and emitting 315 W/m^2, do you consider the total flux to the surroundings from these two put together to be twice as large as the flux from just the single one? And if so, would the total temperature of the two in your opinion also be accordingly higher (325K)?

  202. lgl says:

    Joe

    So a full mirror, a perfect mirror not passing any radiation to space, will not heat the planet either? Given the internal source is still there. Is this whole thread just a joke or are you really that stupid?

  203. Alan Siddons says:

    Joe,
    As long as we’re dismantling the GHE back-radiation mechanism, do you recall the email thread where I was challenged to name any AGW “skeptic” who endorses the concept? I named Richard Lindzen, citing his depiction in this paper
    http://www-eaps.mit.edu/faculty/lindzen/198_greenhouse.pdf
    in which 240 surface watts are radiated back by a two-sided atmospheric ‘layer,’ thus producing 480 watts on the surface.

    Well, if you remember, professor Lindzen actually responded, blaming his co-author Emanuel for employing that argument. So Lindzen apparently disavowed it. But he never dropped the other shoe: If back-radiation is NOT how the GHE works, then how the hell DOES it work? He didn’t volunteer an answer.

    What do greenhouse defenders offer, then? Deflection, evasion, repetition, and even outright ostracism. But no willingness to address the physics issues. Nor are they ashamed that they can’t point to any device which draws X watts but produces 2X watts internally while also conveying X watts to an external target.

    They call it “settled science.” It’s a term that really means “stagnant brains.”

  204. lgl: Radiation can not heat up its own source to a higher temperature. Radiation doesn’t change its frequency components when reflected off a mirror or absorbed back to its source, and so therefore, it can’t cause any temperature increase.

  205. Exactly Alan, and thanks for that reminder of what happened with Lindzen. Think it through people!

  206. On WUWT blog, someone asked what exactly was the quantum effect I was referring to.

    Well, the QM effect is the creation of a blackbody spectrum – that is specifically a QM effect and only exists due to quantum mechanics. This is in fact the origin of quantum mechanics, in figuring out how the radiation from a source of thermal energy inside a cavity would distribute itself as a function of frequency. The result was finally Planck’s formulation of his law, the blackbody spectrum, which he solved by introducing quantization to the relevant equations. This formally birthed quantum mechanics. Well, Einstein’s photoelectric effect also had a role, but it was Planck’s discovery that was the really grand formalization of it all.

  207. Simon Conway-Smith says:

    Joe, It’s been an interesting discussion, terrifyingly insane in some respects, but I’m now down in Asia for the rest of the week, so here’s to lgl having an educational breakthrough. Thanks.

  208. Cheers Simon. Good luck with your trip. I’ll be southern India myself at the end of the week.

  209. Note in this discussion also that they always assume 235 W/m^2 as the input, because this is what they think the Sun inputs to the planet. The Sun does NOT input 235 W/m^2 to the planet, it inputs 1370 W/m^2 at the zenith or 872 as the integrated average. They START from an error in the first place, and then invent something to fix the error. It is no more complicated than that, what they’re doing here. Here we are arguing about 235 W/m2 when 235 W/m2 isn’t what exists in the first place.

  210. lgl says:

    Kristian

    I can not answere your question because there are unknowns. To clarify I suggest putting your two radiators at each end of a refectly insulated cylinder. Since the radiating area to space of the radiators is cut in half (very thin heaters 🙂 their temperature will have to increase to emit the same amount of energy. If your calculation says 325K thats probably it. The reason of course being one side of the radiators has lost the cooling since it receives the same flux as it emits.

  211. The point was that you’re wrong lgl – two icecubes do not make a radiation field of 325 K.

  212. lgl says:

    Joe

    So if you put a heat source in a sphere and prevent any radiation from escaping, the sphere still won’t be heated up. A real scientific breakthrough, congrats.

  213. lgl: What would be a scientific breakthrough is if such a thing could heat itself up indefinitely. We could smelt steel with a 100 Watt lightbulb if that were the case. Or heck, a 1 Watt lightbulb. You’re inventing fantasy. There is a distinct and unique QM effect in the behaviour of trapped radiation, and it is to produce a blackbody spectrum, not to shift itself to higher frequencies and higher temperature.

  214. lgl says:

    Joe
    Kristian didn’t mention icecubes, that’s just your imagination.

  215. lgl: Kristian mentioned 273K. Please use your imagination. Actually…please don’t do that any more. Try closing your mouth and learning instead. Putting the question in terms of ice cubes is the exact same question with no ambiguities.

  216. lgl says:

    Joe

    So you don’t even know that absorbtion and emission are to independent processes?
    Trapped radiation does not produce a blackbody spectrum. Trapped radiation only increases the temperature of the body. Then the temperature and emissivity ‘produces’ the blackbody spectrum.
    If there is an additional heat source the two spectra will of course not be identical, jeez

  217. lgl says:

    Joe

    And now an icecube is comparable to “one radiator at 273K emitting a radiative flux of 315 W/m^2”
    Very lively imagination indeed.

  218. lgl: Trapped radiation does not spontaneously shift its frequency to higher components. Therefore it doesn’t cause an increase in temperature.

  219. lgl: Water can be ice at 273K. This emits 315 W/m^2. Basic physics.

  220. lgl says:

    When Kristian says “one radiator at 273K emitting a radiative flux of 315 W/m^2” one must assume it’s a sustained flux i.e a radiator with an internal heat source.

  221. Bryan says:

    I have sent a copy to the patent office of a wonderful new device utilising the latest IPCC radiative physics.

    It consists of a metal planet with a metal shell one metre above it with a vacuum between.
    The shell is made from a metal that does not expand much with increasing temperature .

    Because a normal metal expands when heated the higher temperature planet will touch the shell and immediately send out a double intensity radiative pulse.
    The planet will then cool and contract recreating the vacuum gap and get twice as warm.
    This process will then repeat endlessly.

    To compare it to the Willis Eschenbach model above.

    The shell will never drop below 235W/m2 and quite often radiate at 470W/m2.

    Some sceptics might find this hard to accept.

    The usual complains about violation of the first and second law will no doubt be made.

    However the science is settled and the device must work.

  222. lgl: Doesn’t matter what it is. Use a resistor at 273K. Result is the same. You can’t get something for nothing.

  223. lgl says:

    Joe
    “Trapped radiation does not spontaneously shift its frequency to higher components. Therefore it doesn’t cause an increase in temperature.”

    And again, there is an internal heat source which you always try to ignore.

  224. Bryan, exactly, wonderful description!

  225. lgl: The heat source isn’t being ignored at all. Radiation doesn’t shift to higher frequencies when reflected back upon its source. Therefore it doesn’t cause an increase in temperature.

  226. lgl says:

    Joe

    The radiation does shift to a higher frequency when it leaves the planet because there is also energy supplied from the core, increasing the temperature.

  227. lgl: That doesn’t cause the frequency to change. You can’t just “say things” and expect them to be reality. A photon of X nanometers combining with another photon of X nanometers does not increase the frequency of either photon. They’re bosons – they pass through each other and don’t result in change. Likewise, photons of X nanometers coming back to their source emitting those X nanometers, doesn’t cause the source to emit higher frequencies, because photons combining or returning to their source can’t shift their own frequencies or the vibration frequency rate of their source. They’re the same – they don’t cause themselves to change.

  228. Wrap a frozen chicken in tinfoil and, according the some, it must eventually cook the chicken!

  229. lgl says:

    Joe
    Like I said absorbtion and emission are independent processes. It doesn’t matter what made the temperature rise. There does not have to be an absorbtion first for an emission to take place. Emission only depends on emissivity and temperature. The energy of the absorbed photon is converted to kinetic energy at the surface, the received photon does not cause another photon to be emited from the surface.

  230. lgl: The problem you have in understanding absorption and emission that you think that absorption of a photon can induce higher temperature and material vibration frequencies than the photon actually is. That is nonsensical and not possible, and there’s no reason to imagine that. Photons don’t induce or create higher temperatures than they are in the first place, because they don’t have the energy or frequency components to do so.

  231. lgl says:

    Joe
    So the energy from those ‘too low energy’ photons just magically disappears?
    Radiation is just a form of energy transfer. If radiation is absorbed temperature increases, the frequency of the photon doesn’t matter. Are you saying a warmer object will not absorb photons from a colder object? If so, again big scientific breakthrough.

  232. Max™ says:

    Nuclear ice cubes even!

    You know, Joe, I do gotta give you credit for being the first person I’ve seen to so clearly explain the basic geometrical errors involved in the GHE arguments, but now I’ve gotta wonder if you feel like you’re speaking a different language or something at times when it is ignored completely?

    I was informed by tb that my babbling about “minor errors due to the area” was beating a dead horse and would be snipped so I’ve withdrawn from the discussion there, not going to pretend it doesn’t matter.

    Though, I suppose I can get annoying with my insistence on picking at things which are often taken for granted: subtracting the contribution from ambient radiation rather than adding it, the unrealistic effects of averaging insolation over the night side of a planet, the idea that an energy budget has to be split into day and night portions, and now this bit about the shell having a larger surface area meaning it would be radiating more energy than it receives if it gave off 235 W/m^2 total after adding both sides together never mind if it emitted that much from the inner and outer surface.

    Having the planet at 255 K and the same surface area as the Earth gives 1.2229×10^17 Watts.

    Having the shell at 255 K and 11 km above the surface gives 1.2271×10^17 Watts.

    400 Gigawatts of free energy is a “minor nitpick” apparently.

    max@Funktastic:~$ calgebra
    >>> 6382/6371
    1.00172657354
    >>> 1/1.001726^2
    0.996556916705
    >>> 235*0.9965569
    234.1908715

    Inverse-square law and such, so the shell receives 234.19 W/m^2 at most at that distance.

  233. Made this reply to a comment to someone on WUWT – want to record it here for myself. Thanks to Bryan for the insightful and extremely clever comment on the “shells”:

    Yes but a coat and the physical blockage of convective cooling is not what the atmospheric GHE is – it is what a real greenhouse does, but a real greenhouse is not the atmospheric GHE. The analogy of this to trapping radiation is only an *analogy*, but it isn’t correct, because radiation trapped inside a cavity (with its source) produces a blackbody spectrum, not heat itself up. To get higher temperature the photons need to shift to higher frequency; but, photons interacting with themselves don’t change their frequency, and photons which come back to their source only interact with matter vibration frequencies which produced them in the first place, and this can’t cause a shift to higher frequencies either.

    Someone named Bryan wrote this to me:
    Consider the planet to have a high metal expansion coefficient, and the shell a small one. There is a small distance between them. When the planet heats up to 470, it will expand and touch the shell. Now the shell will heat up to 470. Whereas before we couldn’t, say, create steam to drive a generator turbine, now we can (using the relevant values). You get the idea I hope. The system might cool and stop touching but then it would just heat up again and touch again. With an intrinsic source that couldn’t create steam in the first place, with a system of shells the system can now do more work than it could have in the first place, than if the input energy was just used directly. Just by using a passive shell. Obviously that has problems of the thermodynamics type.

  234. lgl: “So the energy from those ‘too low energy’ photons just magically disappears?”

    No, but it doesn’t shift itself to to higher frequencies either. Don’t throw out what equilibrium actually means.

    lgl: “If radiation is absorbed temperature increases, the frequency of the photon doesn’t matter.”

    You’re a complete idiot. You have no science knowledge or training whatsoever. Of course you think something like this because then you can have cold things heating up hotter things. OMG so dumb.

    lgl: ” Are you saying a warmer object will not absorb photons from a colder object? If so, again big scientific breakthrough.”

    It is simple – a cooler object does not heat up a warmer object. Why are you such an idiot? How can you be this retarded? The scientific breakthrough would be if cold could heat hot, or if something could spontaneously heat itself with its own radiation.

  235. Yes Max, indeed, that is why I keep asking the question as to WHAT it is that drives these people? Some form of religion is a good bet, because you only see this type of behaviour anywhere else with creationist fundamentalists.

  236. lgl says:

    Joe
    “You have no science knowledge or training whatsoever”
    A strange statement since I’m only explaining the science known for centuries wereas you are peddling your physically totally impossible ‘homemade’ nonsense.

    A cooler object will make a warmer object warmer that it would have been without the radiation from the cooler object, but this does not mean the cooler object is ‘warming up’ the warmer object.
    This is the essence of your misunderstanding.

  237. Greg House says:

    lgl says, (2013/03/12 at 12:41 PM): “So a full mirror, a perfect mirror not passing any radiation to space, will not heat the planet either? Given the internal source is still there.”
    =======================================================

    Lgl, look, it is so simple. Please, conduct a real experiment: stand in front of a mirror (not too close to it to avoid suppressing convection significantly) and observe how it will warm you. Your face, for example. Your face has to feel the heat, right? If possible, tell us the scientific result.

  238. sunsettommy says:

    lgl,

    where did the extra 235 come from?

  239. No lgl, that is the source of your fraud. Quantum mechanics is real and the behaviour of radiation trapped inside a cavity forms its basis, and it doesn’t cause runaway self-heating. The shell-game heating idea is the dumbest idea that has ever been invented. Read up on the work of the French philosopher Jean Baudrillard and the concept of hyperreality to understand the trap you’ve fallen in to.

  240. sunsettommy says:

    lgl writes this:

    “A cooler object will make a warmer object warmer that it would have been without the radiation from the cooler object, but this does not mean the cooler object is ‘warming up’ the warmer object.”

    Then according to you putting a small bag of ice on my tummy will get me warmer since YOU say that a cooler object (ice) will make my tummy and the body as a whole a bit warmer since it does send some waves of radiation to my body adding the energy needed.But then you say the very opposite in the ending part of your convoluted paragraph.

    Charmed………………………

  241. They just say one thing and then the other back and forth as each gets refuted…mindless repetition, a product of the education system.

  242. Greg House says:

    Joseph E Postma says, (2013/03/12 at 6:31 PM): “No lgl, that is the source of your fraud. Quantum mechanics is real and the behaviour of radiation trapped inside a cavity forms its basis, and it doesn’t cause runaway self-heating.”
    =======================================================

    Not being an expert on quantum mechanics I have tried a different approach to the issue of behavior of trapped radiation at Tallbloke’s today:

    “Dear friends of back radiation’s :),

    I know from my experience on blogs that it is sometimes hard to understand, let us say, that it is at least possible that back radiation won’t warm the source. The counterargument is sometimes like “where does the back/trapped radiation go then?” meaning that radiation must find a sort of a host to settle down. This seems to be a serious obstacle to accept the possibility that back radiation won’t warm the source.

    Now, coming back the hypothetical planet-shell mutual warming process, how come nobody including the concept creators asks “where does the radiation go after leaving the shell in the direction outer space”? Because we have nothing there, so wished the creator Willis.

    Well, what I would like to suggest is this. Since we have no problem with the radiation leaving the shell in the direction outer space doomed to never finding a new home, let us think that it is exactly the same way possible to have this desperate homeless back radiation in the inner space between the shell and the planet forever, still unable to warm. Please, give it a thought.

    Then, once there is no problem of unclear back radiation’s destiny, we can all look at this back radiation warming thing and ask ourselves “WTF??”. Then we do not need such, you know, explanations as “yes, the temperature instantly drops when emission occurs from a blackbody with no heat capacity” (must be to the absolute zero, OMG!). Then we can recall the R.W.Wood experiment and dismiss the “greenhouse effect” as contradicting reality. Please, give it a thought.”

  243. Another comment of mine from WUWT which I want to save here:

    Micro: “Or, you can simply absorb more of them, can’t you? There are more coming in every second from the original heat source. If they are delayed from exiting, are they not going to start piling up?”

    Yes but photons are bosons – they don’t “pile up”, they pass right through each other. Absorption of a photon can’t induce higher temperature, i.e., material vibration frequencies, than the photon actually is. That’s not sensical or possible, by definition. Photons don’t induce or create higher temperatures than they are in the first place, because they don’t have the energy or frequency components to do so. When they come back to their own source, they only encounter matter which is vibrating at the same frequencies which produced them. These frequencies then resonate, not increase in frequency. When two photons of the same wavelength interact, they *do not* change each other’s frequency, which would correspond to a temperature increase. Likewise, if a reflected photon spectrum comes back to its source, it only encounters a material vibration frequency spectrum of the same character. These photons, being of the same vibration frequencies of the material, can not induce a higher temperature in the material because this requires a higher frequency component inputs. This is how and why radiation obeys the 2nd Law of Thermodynamics, and why radiation inside a cavity doesn’t heat itself up, but produces a blackbody spectrum, as Planck proved along ago in his discovery of quantum mechanics. It all works perfectly together to create a sane world.

    Consider “Bryan’s” comment to me which I had in one of my last posts, and understand that the system of shells being theorized here then obviously violates thermodynamics & etc. I will repeat it here:
    Consider the planet to have a high metal expansion coefficient, and the shell a small one. There is a small distance between them. When the planet heats up to 470, it will expand and touch the shell. Now the shell will heat up to 470. Whereas before we couldn’t, say, create steam to drive a generator turbine, now we can (using the relevant values). You get the idea I hope. The system might cool and stop touching but then it would just heat up again and touch again. With an intrinsic source that couldn’t have create steam in the first place, with this system of shells the system can now do more work than it could have in the first place, than if the input energy was just used directly. Just by using a passive shell. Obviously that has problems of the thermodynamics type.

    RGB: Heat doesn’t flow from cold to hot, so, the shell never sends heat to the planet, and so it never warms up the planet. The shell just will come to the same temperature as the source, the planet. Then the shell effectively becomes the new surface of the planet, and this doesn’t require the planet to become hotter. There’s no need to invent that. The presence of a temperature differential does not mean that the hotter side of the differential gets hotter – it means that the cooler side warms up until equilibrium is established with the hotter side. This does not require the hotter side to get hotter.

    Hands warmed inside gloves is a sense-perception result of reduced convective air cooling. Put your hand inside a mirror, say, a shell with a mirrored interior, and the hand will not warm up. Put it next to a mirror and the side facing the mirror will not heat up.
    Temperature and radiation is all about frequency. Hotter matter vibrates at higher frequency, producing higher frequency radiation. Commensurately, higher frequency radiation falling onto matter with lower frequency vibration components will induce higher frequency vibration in the material, thus heating it up. Because higher frequencies are being introduced. You need higher frequency radiation to induce higher temperature. This is how and why radiation obeys the laws of thermodynamics, so that radiation of a particular frequency spectrum (temperature) can not induce higher frequency vibration (temperature) in matter it falls on, including its own source as a pertinent example. This is exactly the same thing as how two objects of the same temperature can not induce higher temperature in each other via conduction; neither has the higher frequency components required to do so.

  244. Greg House says:

    Gareth says, (2013/03/12 at 8:29 AM): The planet begins at 800 and the shell at zero returned to the planet and zero to space. Then the shell returns 400 to the planet and emits 400 to space.
    The planet retains 200 of the 400 and reflects the other 200, along with the constant 800 to make 1000 now going to the shell. The shell then reflects 500 and emits 500 to space. The planet retains 250 and reflects 250, plus 800.(1050) …”
    ==========================================================

    Gareth, there is no reflection in the Willis’ small universe, the precondition is that both the planet and the shell absorb and emit. It goes back and forth, but there are some inborn defects in that thing.

    First, Willis wants his hypothetical cyclical process to stop at a certain point just like that, because it must reach equilibrium. However, his hypothetical process does not lead to equilibrium. This alone should be an alarm signal, but no, he has a “solution”: he declares the process done, just like that. Physics upside down.

    Second, they make their radiation arithmetic, assuming that the planet always retains the same temperature, but emits more and more, which “allows” them to always add the same initial 800 (initial power) to the increasing back radiation. The bodies, however, are known to emit according to their temperatures, hence their hypothetical cyclical process would lead to an endless mutual warming, so the whole thing falls apart at this stage too. There is, however, an abstract possibility to resolve this contradiction, namely by assuming that after emitting and before absorbing the planet’s temperature always drops to the initial one and then rises again and always to a higher and higher value. I offered it, hoping it would additionally demonstrate the absurdity of the initial assumption, but guess what? Roger took it indeed as a part of their process! Unbelievable. This is their physics. (http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46160)

  245. Greg House says:

    Joseph E Postma says, (2013/03/12 at 7:35 PM): “Another comment of mine from WUWT …”
    =====================================================

    Joe, could you give me the link, please? Thanks

  246. That two spectrums exist, one from a cooler body and one from a warmer, and that there is a temperature differential, does not mean that the hotter side of the differential warms up. That’s backwards. Skin can detect temperature changes on the order of what, 1/10th, 1/2 of a degree K? If backradiation could warm you up, you should be able to instantly feel it on your skin when you stand in front of a mirror. Backradiation is supposed to produce 50%, doublings, etc., of temperature increases. Talking 10’s of degrees K, hundreds of degrees K even. But it can’t actually even be detected at the skin sensitivity level of less than 1K. That is because the concept is physically impossible, given the explanation of frequency components and heat transfer etc.

  247. The presence of a temperature differential doesn’t mean that the hotter side of the differential gets hotter. That’s backwards. 100% backwards. It is not about directionality – radiation goes everywhere. It is about what the radiation is capable of doing, which is a function of its spectrum. It can’t do more than its spectrum says it can. It can’t create higher temperature than the temperature it is. This is how and why radiation obeys the laws of thermodynamics, particularly the 2nd, given that directionality is not known. Directionality might not be known, but it doesn’t need to be. What is known, what becomes known, are the frequency components when the photon and matter interact. If the radiation is of the same or cooler spectrum than the matter, it can’t induce any higher frequency components and therefore higher temperature in the matter.

  248. with ref to my post :
    http://climateandstuff.blogspot.co.uk/2013/03/does-thermal-radiation-travel-from-cool.html

    Joseph E Postma says: 2013/03/11 at 8:35 AM
    Very simple. All he did was slow down the cooling rate because the ambient environment was changed. When you change the environment you get different thermal behaviour.

    What did NOT happen was that the cooler plate caused the warmer plate to heat up some more to a higher temperature. That is what they need for the GHE.

    ———————————–
    What is the difference between the environment of the hot plate side? The area is the same. The hot plate is the same, the hotplate temperature is the same, the ambient temperature is the same. The window is the same. The main difference is that the window is passing radiation from the warm plate in one run and in the other run it is passing radiation from a 20C ambient sink.

    No true scientist says that the presence of a warm plate will heat up the hot plate.
    Providing the warm plate is above the background temperature it will slow the cooling of a hot plate. If the hot plate is being heated by a fixed energy source then when a warm plate replaces a cold background the hot plate will cool to a warmer equilibrium temperature.

  249. Gareth says:

    Greg House,

    Thank for the reply.

    One of my mistakes (I think) is in thinking that a shell receiving 800 whatevers from the planet would emit half to space and half back to the planet.

    The planet would be at 800, the emission from the planet would be 800, the shell would be at 800 and the shell would be emitting 800 at the planet and into space. The planet doesn’t heat up because the incoming 800 from the shell is equal to the outgoing 800 from the planet and the net energy emitted to space is still 800?

  250. Alan Siddons says:

    …his hypothetical process does not lead to equilibrium. This alone should be an alarm signal, but no, he has a “solution”: he declares the process done, just like that. Physics upside down.

    Just wanna say that Greg House has nailed it. In a greenhouse scenario, each of the two parties — surface and shell — must continuously adjust to what the other is radiating. The initial 235 W/m² surface induces the shell to emit 235 as well, which makes the surface emit 470, which makes the shell emit 470 too — ad infinitum. This is what Gerlich &Tscheuschner’s Falsification paper was trying to get across:

    The atmospheric greenhouse effect… essentially describes a fictitious mechanism, in which a planetary atmosphere acts as a heat pump driven by an environment that is radiatively interacting with but radiatively equilibrated to the atmospheric system.

    Interacting with but equilibrated to. Hey wait. That’s impossible.

  251. That’s about right Gareth. Remember, no heat energy is transferred from cold to hot or between two objects of equal temperature. No net heat energy flows from the shell to the planet. The best the planet can do is heat the shell to the same temperature of the planet, and then the shell essentially becomes the new surface of the planet, emitting the 800. Between the planet and shell is a radiation field of 800, and so there’s that energy there going “back and forth” between them, but it doesn’t cause heating because it intrinsically doesn’t contain the higher energy frequency components required to do so. As the shell then loses the 800 continuously to space, this is continuously replaced by the radiation field between the planet and shell, and the energy loss in that radiation field is replaced by the planet.

  252. Will Pratt says:

    thefordprefect says:
    2013/03/12 at 10:29 PM

    In your experiment, you have simply increased the mass of the system, of course it will take longer to cool down.

    However, adding CO2 to the atmosphere does not increase atmospheric mass. Your experiment is meaningless in the context of this discussion.

  253. My reply to Micro at WUWT (is everyone having problems posting there? or is it just me?):

    Micro said: “You can protest all you want that there’s no ir reflected(in a proper ir reflector), but you’re wrong.”

    My statements had nothing to do with saying that there is no IR reflection, so this objection is moot, sorry. There is IR reflection and emission from cold sources, certainly, but the nature of this radiative energy can not cause heating on its own source or a hotter source, for the reasons I’ve physically described several times. Radiation obeys the laws of thermodynamics too – not just the first one, but all of them.

    No heat energy is transferred from cold to hot or between two objects of equal temperature. Remember, we can *not* say that in a temperature differential, the hotter side has to get hotter in order to warm the cool side. Rather, the cool side just warms up. No net energy flows from the shell to the planet. The best the planet can do is heat the shell to the same temperature of the planet, and then the shell essentially becomes the new surface of the planet, emitting the 235. Between the planet and shell is a radiation field of 235, and so there’s that energy there going “back and forth” between them, but it doesn’t cause heating once equilibrium is established because it intrinsically doesn’t contain the higher energy frequency components required to do so., required to induce higher frequency material vibrations and hence higher temperature. It just can’t do it. As the shell then loses the 235 continuously to outer space, this is continuously replaced by the radiation field between the planet and shell, and the energy loss in that radiation field is replaced by the planet.

    We already discovered that the “shells game” idea violates thermodynamics when considering the physical effects of the interior sphere having a high metal expansion coefficient, and the shell a low one, and what would occur when they touched.

  254. Ron C. says:

    Thank you for this discussion, which has ranged from here to WUWT and TB Talkshop.

    I noted and understand your point about the radiation spectrum prohbiting any increase in frequency of impacted matter of the same or higher spectrum, and therefore no increase in heat or temperature.

    I am interested to know your response to something I read today on the TB site. (BTW they are searching there for the “official” physics of the GHE, and not finding anything to critique.)

    “The bonds between the atoms in a molecule are somewhat elastic, like springs, and with the atoms acting like weights at opposite ends of these springs, every molecule has at least one characteristic resonant vibrational frequency. Most molecules have many different resonant frequencies, depending on vibration modes, etc.

    CO2 molecules have a resonant frequency that corresponds to 15 micro meters (LWIR). When it encounters a photon of IR energy at that wavelength, it readily absorbs it and begins vibrating faster. Now, there are two ways it can lose that energy again. First, It can spontaneously re-emit IR energy at the same frequency, and lose that extra vibration. Second, (and more likely), when it bumps into another molecule, that extra vibration gives the collision a little extra “kick”, and both molecules move away a little faster. Faster molecules within a fluid equals higher heat energy. At this point, the original IR has become thermalized.”

    http://tallbloke.wordpress.com/2013/03/13/wikipedia-and-ipcc-ar4-the-greenhouse-effect/#more-11617

  255. lgl says:

    All you morons, again

    Try this then:
    We have two equally sized radiators, and for the argument they are thin as a paper i.e only radiating from two sides. We supply enough power to keep one of them at 300K and the other at 400K when they are far apart. Then we place them very close to each other, sealing off the narrow gap between them completly so that no energy escapes from the gap. We supply the same amount of power to the radiators as before. Now, will the temperature of the radiators change? One, both, up or down?

  256. Hi Ron,

    There is no question that radiative energy can heat up material that colder. That is the only thing the quote you provided describes – that radiative energy can heat up something colder. The implied, and false extension, is to then say, really it is just implied, that the CO2 or its reradiation has to heat up the original source of the radiation. This is not possible because the original source of the radiation is either warmer, or the same temperature, and so the CO2 or its radiation can’t warm up the source.

    Also consider that the CO2 is already vibrationally activated in ALL modes of its possible oscillations because it is being bumped around by the other 99.96% of the atmosphere, which has itself been heated by conduction with the hot ground surface. So, the IR photons interacting with CO2 only “see” a molecule which is already oscillating. The description you quoted above is applicable only for a cold CO2 molecule in complete isolation. So, IR photons “see” a CO2 molecule which is already oscillating and vibrating, hence, they don’t induce much more of that to the molecule, if at all. What the radiation field could do is scatter of off the existing oscillation because it would resonate with it. But again, just like standing in front of a mirror doesn’t warm you, the back-scattered radiation won’t warm its own source. That radiation also leaves the atmosphere within a few milliseconds anyway after a couple of scatterings because they travel at the speed of light and the atmosphere is negligibly thin compared to that…so, even if you want to talk about “photon build up” it too is negligible, although it wouldn’t cause heating above the source temperature even if it wasn’t.

  257. lgl: The cool one will warm up. The presence of a temperature differential does not mean that the hotter side of the differential has to heat up in order to warm the cool side. The cool side just warms up.

  258. Will Pratt says:

    Ron C. says:
    2013/03/13 at 8:49 AM

    Ron, 15 µm has a corresponding temperature of -80º C.

    http://www.calctool.org/CALC/phys/p_thermo/wien

    This is “coincidently” the same temperature that CO2 sublimates via phase change, from ice to gas. Therefore it is bound to absorb strongly at these wavelengths.

    The same applies to all substances. They all absorb IR strongly at their various melting/evaporating-phase change temperatures, obviously.

  259. Alan Siddons says:

    LGL, the shell doesn’t just have slightly more surface area (square meters) than the surface, it has more than TWICE because its two sides are exposed, both of which are freely radiating to their surroundings. This means that even if the shell were absorbing all of the surface’s energy it would be much cooler than the surface, radiating in the neighborhood of 117 W/m². You have to count both sides.

    Slice up a cake straight out of the oven and separate the pieces. Since you’ve increased the surface area, they’ll cool down a lot faster than the whole cake would.

  260. Yes but you see Alan, now he’s going to want to add the 117 BACK to the planet to make it hotter, even though the shell is colder. lol

  261. lgl says:

    Alan
    Finally you are beginning to grasp some of this.
    Yes, right after you place the shell around the planet it will radiate 117 to space, but then you have a problem. The core is still generating 235, the system is only loosing 117 to space. What will that lead to? Where is the other 117 ending up?

  262. lgl says:

    Joe

    And the cold one will warm to no more than 400K i presume, right?

  263. Didn’t I predict exactly what lgl would say? lol

    We already went over the sequence lgl is proposing and which Willis proposed etc., right at the beginning of these comments. The sequence results in a heating progression going as (1.5)^n, which diverges to infinity, and which is ridiculous. The other 117 gets scattered right back to the shell, warming it up until it gets to equilibrium with the planet, with both at 235 and equal temperature.

    The presence of a temperature differential does not mean that the hot side of the differential has to heat up in order to warm the cool side! How stupid do you have to be?

  264. “And the cold one will warm to no more than 400K i presume, right?”

    This has been answered numerous times, whatever your question is referring to. The planet and shell come to the same temperature assuming they are very close together.

  265. lgl says:

    Joe
    I was referring to the radiators. Will one stay at 400K and the other somewhat lower than 400K or what?
    Your problem is, if we assume each side of the radiators were 1 m2, in the inital setup the effective radiating area was 2 m2 each, so in total they radiated 3827 W.
    After they are placed close together the area to free space is only 1 m2 each, so 3827 W will now have to be radiated from only 2 m2. What temperature is required to achieve that? My calcs say 428K.

  266. lgl says:

    Joe
    “The other 117 gets scattered right back to the shell”
    Ah, more selfinvented physics.
    What about when they reach the same temperature? The shell will still absorb the 235 but the planet will not, it will still ‘scatter’ back the 235 to the shell? Why this totally different behaviour when they are at the same temperature? What utter nonsense.

  267. “My calcs say 428K”

    Your calcs are wrong. Cold doesn’t heat hot and and hot doesn’t have to become hotter to heat cold. QED.

  268. When they reach the same temperature, they’re in equilibrium. Temperature can not amplify itself and radiation can not induce a temperature higher than the temperature that the radiation is, because this would be a temperature amplifying itself.

    lgl: Stop being an idiot.

  269. Alan Siddons says:

    “When they reach the same temperature, they’re in equilibrium.”

    But that can never happen; the shell will always be colder. I would say that ‘radiative equilbrium’ is reached when the shell absorbs all the radiant energy that the surface has to offer, which amounts to 117 W/m² distributed over both sides of the shell. Given the surface area issue, the shell can’t absorb or emit more than that.

  270. lgl says:

    Joe
    “Your calcs are wrong”
    Show us the correct calcs then. Don’t they have to radiate 3827 watts? Or do you finally realize your physics is impossible?

  271. Alan, that is certainly more sensible than anything the “shell game” crowd has been thinking.

    However, consider a thick shell. If the inside is very close, it should indeed come to nearly the same, or essentially the same if it is very close, temperature as the sphere. What occurs in general is what I descried earlier:

    The best the planet can do is heat the shell to the same temperature of the planet, and then the shell essentially becomes the new surface of the planet, emitting the 235. Between the planet and shell is a radiation field of 235, and so there’s that energy there going “back and forth” between them, but it doesn’t cause heating once equilibrium is established because it intrinsically doesn’t contain the higher energy frequency components required to do so., required to induce higher frequency material vibrations and hence higher temperature. It just can’t do it. As the shell then loses the 235 continuously to outer space, this is continuously replaced by the radiation field between the planet and shell, and the energy loss in that radiation field is replaced by the planet.

  272. lgl: “Show us the correct calcs then.”

    I have numerous times: (1.5)^n. It diverges to infinity and is therefore wrong, because it violates the laws of thermodynamics. Radiation interacting with itself doesn’t increase its frequency and hence its temperature stays the same; the same thing occurs for radiation interacting with matter – it can’t induce higher temperature than it is.

  273. lgl says:

    Alan
    but then only 117 is emitted to space, 235 is still supplied from the core. Can’t you see if the system looses less than 235 to space it will heat up?

  274. Leif: I already explained what happens. Nothing can heat itself up with own radiation.

  275. lgl says:

    Joe
    “I have numerous times”
    No you have not. Tell us how much energy two 1m2 plates (two sides of course) are emiting if one is 400K and the other 300K.

  276. The question isn’t how much energy because the laws of thermodynamics are not limited to the first law. This is the mistake you can’t get passed. There’s also the 2nd and 3rd Laws. Hotter heats cooler and cooler does not heat hotter, and hotter doesn’t need to become hotter to heat cooler. The radiative reasons have been described physically several times.

    As you’ve exhausted every argument you might make several times over, and the material to educate you on the matter has been provided several times over, your comments on these matters will no longer be published, to save me the time of constantly trying to instil some education in you. Best regards.

  277. lgl says:

    Good, finally you realize you were wrong.

  278. See the parting shot from an idiot everyone? That’s what an idiot does.

  279. Simon Conway-Smith says:

    No lgl, Joe just got fed up of explaining why you are wrong time and time again, with no indication from you that you were learning anything or asking rational questions based on any learning.

  280. Alan Siddons says:

    …consider a thick shell.

    Right, that’s reasonable. But then you have the problem of energy distributed over the VOLUME of an actual mass. Joules per kilogram raises a substance’s temperature. With the same Joules per TWO kilograms, though, the temperature increase is less. Even if the shell’s mass is losing energy as fast as it’s gaining, then, its many molecules are clamoring for an equal share of energy, which thereby reduces the shell’s average temperature.

    The same applies to the introductory premise, observe, for the nuclear core must of course be MUCH hotter in order to make the surface radiate 235 W/m². By extension, the surface’s total energy (surface area × W/m²) transferred to yet ANOTHER mass will further dilute the energy available for each molecule. In effect, adding a thick shell amounts to throwing dirt over the nuclear furnace in order to reduce the surface temperature. That extra dirt will make the new surface cooler than the surface was before.

    Adding mass amounts to throwing dirt over a furnace to cool things down. Or so it seems to me.

  281. Yes indeed, when there is more mass this affects the differential equation for heat flow because the time constant is larger. Larger mass means it takes a longer time to heat up. If the volume was large so the outer shell surface area was much larger, then the outer shell would radiate at a cooler temperature – the same energy from the planet but over a larger surface area.

    On the other hand, if the shell was very thin and the gap between planet and shell very small, and the shell material made of something with a very high thermal capacity, then again the shell would heat up slowly, but it would heat up eventually to radiate at the same energy flux density, or temperature, as the sphere, since the difference in outer area of the shell vs. the sphere is in this case very small.

  282. I just talked to an electrical engineer of 40 years at the university where I work.

    I proposed this idea to him:

    “Suppose you have a electrical thermal resistor in outer space, in a circuit, and the resistor gets to some temperature, say, 80C, and puts out that equivalent amount of radiative energy.

    Now, put that resistor inside a shell with a mirrored interior (still in outer space so there is no air anywhere). I propose that the resistor will get hotter and hotter and hotter, and that with the power input required to get to 80C previously, now that same power can produce much higher temperature. This could, say, boil water for example, to drive a turbine generator, when it couldn’t do that before.”

    His reply: “Yes I see what you’re trying to do there, but it is a violation of thermodynamics. It wouldn’t work as you propose. The reason is because the returned radiation would have “phase cancellation” with the outgoing radiation, meaning that half of the radiant energy gets destructively cancelled out from superposition. The remaining half, halves, one half from outgoing and one half from returning (and then sent outgoing again), results in unity.”

    This is the same thing I’ve said elsewhere. Also note that the only way to not get phase cancellation is if one of the sources has higher frequency components of shorter wavelength, which can not combine in any phase with longer wavelengths. Only with higher frequency components can you induce higher temperature to something with lower ones. Works the same in conduction as with radiation. This is also basically all about what Claes Johnson has written about, in a lot of ways.

  283. Will Pratt says:

    Just to recap,

    235 W/m2 + 235 W/m2 = (not more than) 235 W/m2

    235 W/m2 * 10, 100, 1000 or 1,000,000 still = (not more than) 235 W/m2

    W/m2 is a measure of energy, not heat. Energy has only a potential to heat. What determines whether that potential to produce heat is realised, is the thermodynamics of the system in question.

    “CLASSICAL THERMODYNAMICS IS THE ONLY PHYSICAL THEORY OF UNIVERSAL CONTENT WHICH I AM CONVINCED WILL NEVER BE OVERTHROWN”….A Einstein.

  284. squid2112 says:

    I am glad someone included (back in comments somewhere) the topic of “No Virginia, a cold object cannot make a warmer object warmer still” … As I was going to bring that in to discussion. As I see it, Willis’ new “thought experiment” (of which I have been getting tired of lately, none of them make any sense), seems to me to be just another in a long line of attempts to get a cold object to warm a warmer object, yet again. I wish I had a quarter for every time I have seen such an attempt made, as I would no longer have to slave away at work. … just saying…

  285. Rosco says:

    The inescapable truth is that Joe is undeniably right.

    The notion of something heating itself up by “trapping” its own radiation results in a runaway mathematical series with exponential energy creation – a complete demonstration of perpetual motion which is an intellectual absurdity.

    The idea that something is “heating” up whilst radiating less – exactly what IPCC physics requires – should be an intellectual obscenity to any who think about it.

    The idea defies all of the basics of accepted science and has never been demonstrated.

    The idea that a cold object heats hotter objects (despite the absurd justification that radiative energy from cold to hot occurs but that “net” energy goes from hot to cold ???) is intellectually absurd.

    The idea defies all of the basics of accepted science and has never been demonstrated.

    Actually the reverse has been demonstrated numerous times.

    It is interesting that “climate scientists” make this claim about cold objects heating hot objects more yet have nothing than a similar thought experiment to justify it.

    Have they never heard of Pictet’s experiment relating to the apparent radiation and reflection of cold ?

    This conclusively demonstrates that the cold object did NOT cause the thermometer to heat up – the reverse was indisputably demonstrated – the thermometer dropped in temperature dramatically.

    If “climate scientists” are right it should have increased.

  286. Kristian says:

    lgl says, 2013/03/13 at 11:22 AM:

    Let me try to explain this to you, lgl.

    We place two objects in an airless box. The one object has a temperature of 300K, the other 400K. The two freely emit according to their specific temperatures. If we were to let the thermal exchange between the two objects move all the way to equilibrium, they would both ideally end up with a temperature of 360K (not 350K). At this point, there would no longer be any heat transfer between the two objects. Up until then, though, there WOULD have been heat transfer from the hot to the cooler object. And if we plotted the RATE of this heat transfer through time, we would see that it followed a curve of exponential decay – it would start off very high but end up infinitesimally low. The hot (400K) object would cool all the way to equilibrium, while the cooler (300K) object would warm. The cooler object at no point made the hot object hotter than it originally was, even though it radiated a higher and higher flux towards it as it grew hotter itself. All it did was steadily slowing down the heat transfer rate, its own warming rate and the hot object’s cooling rate, until they were both effectively zero.

    So far, so good. I hope.

    Then, we change the setup a bit. This time around the two perfectly emitting objects in the airless box are different. One object is directly connected to a constant energy source supplying a power flux sufficient to heat the object’s surface to a uniform temperature of 254K. The other object enjoys no such constant supply of energy. It hypothetically holds a starting temperature of 0 K.

    At the get-go, the warm object emits a flux of 235 W/m^2 towards the cold object and the cold object in return emits a flux of 0 W/m^2 towards the warm object.

    What happens next? Where will the ensuing exchange process take us? What will the equilibrium state be like in this scenario?

    Upon commencement, the radiative heat transfer from the warm to the cold object starts off at a prodigious rate, warming the cold object fast. But this warming once again slows down as time passes. This time the warmer object does however not cool in accordance with the warming of the colder object. The two objects do NOT meet in the middle at equilibrium. The heat transfer rate between the warm and the colder object follows the same general path as the one in the first scenario through time. But now equilibrium will not be achieved until the colder object has reached the temperature level of the warm object. At this time, heat transfer between the objects would effectively once again have come to an end.

    Why, then, isn’t the warm object cooling to meet the cold object ‘half-way’ (213K) at equilibrium in this scenario?

    Because it’s temperature/surface radiative flux is being maintained all along by its constant power supply. It will not drop in temperature as long as this source is active. It will not rise either as long as the source provides its constant power flux. This has to do with the internal (molecular) vibration (the level of KE) which relates to the intensity (frequency/wavelength) of the radiative flux received/emitted.

    A key to understanding this outcome lies in realising that in the second scenario above (WITH power supply), as the heat transfer between the two objects WITHIN the thermodynamic system (the box) grows ever SMALLER towards equilibrium, the total system energy transmission to its SURROUNDINGS (the space around) parallelly grows ever LARGER towards that same equilibrium. In the first scenario above (WITHOUT any power supplies), this total outward system flux remains the same throughout – no extra heat generated. (In reality it would actually drop; this is after all only intended to be an idealised case cleared of any such real-world clutter. It was simply meant to prove a point.)

    The heat generated and originally supplied to the warm object in other words goes into warming the COLD object (and therefore, by extension, the system as a whole) until it’s reached the constant temperature of the warm object. It does NOT (by way of ‘back radiation’ from the cold object) go into making the warm object itself warmer than it already is.

  287. roscomac says:

    At our local Mall there is an elevator with lots of mirrors.

    I imagine this enclosed space with “perfect” reflection internally.

    Any light in this enclosed space must, according to the arguments proposed by “climate scientists”, be continually reinforcing itself until infinite brightness.

    You could turn off the light and never be in dark.

    What happens to light in this scenario is an interesting thought experiment – obviously it isn’t trapped as proposed.

  288. I don’t know why they create so much bs over this when it is so obvious and so easily testable.

  289. Martin Hodgkins says:

    Joe,
    That radiator analogy from lgl (comments above) convinced me you are right. The radiators can’t get any hotter with the same power input. I am pleased about that because it has been driving me mad.
    Regards.

  290. Rosco says:

    I always find Willis Eschenbach a contradiction.

    No matter what anyone says – he starts with a constant energy of 235 W/sq metre and proposes no other energy source exists.

    His claim that reflecting this doubles the available energy ignores his original position – there is 235 W/sq metre available !

    Where does the extra come from ??

    Are the people who believe this so gullible that they ignore the basics of science ?

    Energy can neither be created nor destroyed, merely transformed.

    Can’t everyone see energy has been created out of nothing here ?

    My other favourite that I see support for from even sceptics is that insolation is 342 W/sq metre.

    Nobody questions the ~1368 W/sq metre solar constant but some how 3/4 disappears upon entering the atmosphere. Of course I recognize the averaging of a sphere to a disk but the only purpose this serves is to estimate how much energy needs to be emitted over a sphere to balance the solar radiation over a disk – and it is only a simplified estimate.

    To use this to suggest the sun can only supply sufficient energy to heat the Earth’s surfaces to minus 18 degrees C and base computer modelling to support an unproven hypothesis is so obviously absurd it is amazing people not only accept it but will go to extraordinary lengths normally not associated with intelligent discussion or society to demonize any who find the idea flawed.

    Unless “insolation” has some meaning that I cannot find it is absurd to claim that the solar radiation entering the atmosphere is 1/4 ignoring the obvious day and night – yet it stands and is defended as correct.

    Radiation that is 4 times as powerful will induce temperatures that are 1.414 times higher – it is an obvious consequence of the power of 4 of the temperature in the SB equation.

    Surely Joe has demolished this falsehood eloquently yet it continues.

  291. Cheers Rosco! As I pointed out, they don’t even understand elementary arithmetic.

  292. Martin, glad to hear.

  293. roscomac says:

    I’ll say it again – Willis has created energy out of nothing !

    Surely that is totally obvious as well as impossible ??

  294. Greg House says:

    Rosco says, (2013/03/13 at 5:04 PM): “I always find Willis Eschenbach a contradiction.
    No matter what anyone says – he starts with a constant energy of 235 W/sq metre and proposes no other energy source exists.
    His claim that reflecting this doubles the available energy ignores his original position – there is 235 W/sq metre available !
    Where does the extra come from ??

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

    As far as I understand, they mean that energy that is radiated away can be sort of caught and returned to the source (back/trapped radiation) and either slow down cooling of the source, if it is cooling, or increase the temperature of the source if the source is not cooling because of additional sort of energy, like the Willis planet. It is like you spend money, but the money spent is given to you back, so you have the same amount of money plus the things you bought. Sounds good, doesn’t it? This is a sort of fallacy that I would call “analogy fallacy”. Warmists deliver analogies, some people buy them, start then thinking in terms of those analogies and are a sort of trapped in them. They do not question the actual issue any more.

    The right approach would be to ignore analogies and speculations and ask what is really physically proven, but for many it is very hard.

  295. Kristian says:

    Rosco says, 2013/03/13 at 5:04 PM:

    “No matter what anyone says – he starts with a constant energy of 235 W/sq metre and proposes no other energy source exists. His claim that reflecting this doubles the available energy ignores his original position – there is 235 W/sq metre available! Where does the extra come from??”

    Rosco, I think you will find from what I’ve written on this thread and on the thread at Tallbloke’s that I agree with you that Joe is right in this matter.

    But here you mix up. The ‘extra energy’ comes from the constantly energy-generating core.

    You seem to forget that Watt is power, not energy. It is Joule (energy) per second. 235 W/m^2 means that 235 Joule is provided to every square metre of the surface of the core planet … per second. This means that after 2 seconds, 470 Joule have been provided. And so on. Which makes you realise that the only energy generated is still coming from the nuclear power source. The point is, it needs to be continuously shed also. Well, it is. They are simply unable to see it.

    The arithmetic/budgeting of the inner part of the system is not necessarily wrong. It’s the notion that it’s only the quantity, not the quality, of a radiative energy flux that matters when it comes to inducing a temperature. They also ‘forget’ that the flux generated in the nucleus at equlibrium escapes the system as a whole. It all balances. No energy is piling up anywhere.

  296. Kristian says:

    I should propbably rather say something like ‘energy delivered’, since energy can never simply be ‘generated’.

  297. Yes, “quality” is a good term to use in this case – “quality” is how the 2nd Law of Thermo comes into the picture for radiation, while Willis et al. only ever consider the 1st Law.

  298. Alan Siddons says:

    “quality” is how the 2nd Law of Thermo comes into the picture for radiation, while Willis et al. only ever consider the 1st Law.

    As I like to say, the 2nd Law enforces the 1st Law. If heat (or more generally ‘thermal energy’) did NOT just raise the temperature of less energetic bodies but also made the heat source warmer, this would spark the kind of infinite heating cascade that Greg House described. Such a cycle would constitute the creation of energy out of nothing, which violates the 1st Law. So the 2nd Law is how the 1st is enforced.

  299. Two objects in physical contact, one cooler one hotter, will not heat up the hotter object. Everyone knows this. But now, just separate the two objects by a short distance, and then the warmer object will warm up even more, as it also warms up the cooler object, because there’s radiation.

    If it worked like that, radiation wouldn’t obey thermodynamics, and neither does the Willis et al set up.

    Think it through people…

  300. lgl says:

    [JP: Trashed because you’re just repeating the same BS idiocy over and over again.

    Two objects in contact, one warmer and one cooler, will warm the cooler body and cool the warmer one. Separating them with a gap does not mean that the hotter one has to get hotter to warm the cooler one – the cooler one just warms, the warmer one isn’t heated by the cooler one or from itself.]

  301. Rosco says:

    What a lot of people ignore is that when radiating an object is losing energy and to maintain its equilibrium that energy needs to be replaced.

    I’ll say it again – Willis has created energy out of nothing.

  302. Loodt says:

    Hi Joseph, I’ve been watching the comments on this website grow since you first posted this entry to your blog. I admire your tenacity to argue with patience on this very fundamental issue; a basic understanding of thermodynamics. I do want to point out a few things to you. Willis has the gift of the gab, nobody can deny his ability to spin a yarn. In the tradition of great American authors he can keep an audience spellbound with his ability to sketch the great outdoors in a masterly fashion. However, Willis has no formal eduction, and he bragged on numerous occasions about his ability to out think formally eduction persons. He get could not get his head/mind around the adiabatic process that causes air to heat up under pressure, I lost all interest to read anything technical that he wrote after reading his juvenile ramblings about that topic. This steel-ball analogy made me smile, and then I skipped to something else. I am too old to fight with fools.

    Now, Anthony is driven with a desire for acceptance and formal recognition by the scientific community. His blog is clear evidence that he is doing ‘science’ as he thinks it is done at institutions of higher learning. But, like Alexander had Rasputin at his court, and the spell and hold that the latter had on the former is still being discussed and debated, Anthony is currently under the spell of Willis. And Willis, like a cocky and cheeky mongrel, likes to lash out at people far more learned and knowledgeable than himself. Did you see how Wills attacked our Piers Corbyn, an astrophysicist, with outright rude and insulting posts? Unless you have a very thick skin, and are prepared to fight dirty, stay away from Anthony’s lapdog! The man hasn’t even passed Maths, Applied Maths or Physics I.

  303. Rosco says:

    Look at Willis’s diagram. I know he didn’t claim any dimensions but it looks like the shell is twice the diameter of the steel solid.

    Plug in a radius of twice into the surface area and you need a factor of four to maintain the radiation at 235 W/sq metre at the outer shell – 4 x pi x radius squared used to be the formula for a sphere – perhaps “climate scientists are proposing amending that as well.

    Double radius equals four times surface area.

    So to achieve the claimed 235 W/sq metre exiting at the outer shell requires NOT 470 but 1880 at the inner solid.

    Outer shell 235 out + 235 backradiated = 470 x 4 times the surface area = 1880 !

    What if the inner solid is a cube ?

    The only way he can have the same value exiting the “shell” as supplied by the “core” is if the extra radius is zero – surely that demolishes this BS “science” ???

  304. Thanks for that Loodt – that explains the situation very well and confirms what had been my intuition as to the behaviours which have been observed. Spot on. Best regards.

  305. Rosco, MANY things demolish this BS science, and I am still trying to figure out why it doesn’t stop them! lol

    Willis said the distance between the shell and sphere is small and so the error was irrelevant – however, he obviously didn’t consider that the error which is exposed when you use a large gap between the sphere and shell destroys the entire argument altogether in any case.

  306. Rosco says:

    I think they are committing deliberate fraud with sleight of hand tricks.

    Their initial proposition that energy entering Earth over a disk balances energy leaving Earth over a sphere is somehow taken to mean the insolation is one quarter of the albedo adjusted solar constant ?

    This figure – ~1368 / 4 = ~342 W/sq metre reduced to ~239 in to balance the ~239 out – is then used to say that there is insufficient energy to account for surface temperatures.

    This is of course complete nonsense as this ignores their initial condition – the incoming radiation is over a disk not the whole sphere as the outgoing radiation is.

    Surely people with PhDs can see they have completely mixed up their own initial starting point ?

    The Earth only needs to radiate 239 W/sq metre average over the whole sphere to balance the average 956 W/sq metre over the disk.

    If the insolation is only say 342 W/sq metre as Trenberth et al claim then the Earth would only be radiating one quarter of this or about 85.5 W /sq metre – that is what their initial claim is.

    By the original geometrical construct the temperature induced over a disk has to be 1.414 times the temperature of the outgoing radiation.

    239 W/sq metre is the temperature associated with approximately 255 K according to Stefan-Boltzmann.

    1.414 x 255 K is approximately 360 K which is the temperature associated with 239 x 4 W/sq metre or ~956 W /sq metre.

    The insolation simply has to be 956 W/sq metre over the surface area of a disk – pi x r squared – if the outgoing radiation is 239 W/sq metre over the surface area of a sphere – 4 x pi x r squared.

    I simply fail to understand how this myth has become accepted and if you challenge the obvious mathematical failing you are some sort of heretic denier !!

    Joe has put this succintly many times but the GHE crowd continue with their misrepresentation.

  307. Bravo Rosco, exactly 🙂

    You don’t know how many climate scientists have lost their mind at me when I tell them that -18C input over a globe will NOT produce the same physical response as 1370 W/m^2 over a hemisphere – in the fake scenario sunlight can’t melt ice, in the real scenario it can. Apparently they think such a profound difference DOESN’T MATTER, and they will get downright indignant and nasty at you if you insist that sunlight only hits one side of the planet. I’ve been called so many forms of stupid for insisting that the planet is a sphere heated on one side it is mind boggling.

    You have to be lying on purpose and to be purposefully covering up a huge fraud to have to yell at someone and call them names for stating that only half the Earth has sunlight, because if people this stupid are actually thinking they’re doing science, the world should have blown up by now. At the very least, keep climate scientists away from anything remotely resembling engineering.

  308. Will Pratt says: 2013/03/13 at 8:39 AM
    In your experiment, you have simply increased the mass of the system, of course it will take longer to cool down.
    ———————————————————–
    Second experiment
    Isolated heated hot object
    Distant (well 9cm) cold plate/warmplate
    shows hot object temperature increase when external cold plate is replaced with idencticle warm plate.

    Note that 2 external plates are used (identical pressure die cast aluminium mouldings) The mass is now constant.

    The internal isolated plate is not affected by the changes in external plates other than by IR

    Not the final experiment. but geting there.

  309. A lnk to the post mentioned
    A Cool Object Reduces Energy loss from a Hot Object

    http://climateandstuff.blogspot.co.uk/2013/03/a-cool-object-reduces-energy-loss-from.html

  310. Rosco says:

    Willis’s argument is total BS and it easy to prove.

    Simply put another shell around his hypothetical setup and what happens ?

    By his own construct his shell is now radiating to the outer shell which heats up to radiate 235 to outer space plus 235 back to his existing shell – his own statement.

    The new shell is radiating equally 235 to space and 235 backradiation – this is what he said happens to his shell so it MUST be right for the new one – Willis says so !!!

    So now his original shell MUST be radiating 470 to the new outer shell otherwise the maths don’t add up !

    470 out = 235 to space plus 235 backradiated.

    And now the core MUST be radiating at 940 because his original shell is now radiating 470 out and 470 back !!!!!

    Do it again with another shell .. and another and another …..

    Surely anybody can see this is impossible BS ??

    If not engineers are totally dumb – they can improve the thermal efficiency of all thermal driven engines by simply building shells around the boilers and eventually do away with the need for the fuel at all – boy wouldn’t that reduce the carbon footprint ??

  311. Rosco says:

    Willis’ little model ought to be easy to demonstrate in experiment. I am certain those with the necessary vacuum chamber wouldn’t bother as they would have the intelligence to see through this little shell game.

    As a final postscript Willis goes on to try to demolish reality by requiring his supposed system needs to be thermally isolated to work.

    “If the atmosphere or other means efficiently transfers surface heat to the shell there will be very little difference in temperature between the two.”

    What utter BS. Why ?

    Because we know there is a substantial temperature difference between the core and shell – it is the magical 33 degrees C of the “greenhouse effect”.

    So his steel greenhouse effect – total BS anyway – cannot possibly represent Earth’s greenhouse effect because the atmosphere transfers surface heat and thermal equilibrium results – which we know is wrong.

    I cannot believe how naive people like this are !

    I thought temperature was a measue of kinetic energy.

    If the molecules at high elevations had the same kinetic energy as the molecules at ground level the temperature of each area would still be vastly different because of the vastly reduced numbers of molecules at higher elevations.

    And that is ignoring the conversion of kinetic energy to potential energy as a gas rises – which results in lower temperature by definition. This simply must be true else the molecules have gained more energy from nowhere.

    These people ignore mass all the time.

    Radiation is caused by the temperature of an object which has mass !

    Even if a gas.

    If there is radiation it simply must be in proportion to the mass – it doesn’t come from nowhere.

    At 5000 metres the temperature of the atmosphere is about – 18 degrees C. This approximates the minus 18 degrees C the so called energy balance for Earth calculation results in.

    The density of air at 5000m is about 740 grams per cubic metre. Of that 740 grams 0.06 % by weight is CO2( if CO2 is a well mixed GHG) and say 5% is water vapour.

    So 37 grams of water vapour and 0.44 grams of CO2 produce backradiation of 324 W/sq metre according to Trenberth et al.

    I simply do not believe it – never have and never will !

  312. A C Osborn says:

    Joe any comment on this?
    http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/#comment-47019
    lgl says:
    March 17, 2013 at 8:39 pm

    Westy
    My kitchen experiment was a bit more successful. 30W soldering iron heating a 0.12 m2 ‘planet’. Temp without shell, 55.7C Temp with shell, 67.4C, which means 12.4 Watts was mysteriously ‘created’ by the shell, who would have thought.
    12.4 W is of course far from 30 W, but not bad considering this was not done in vacuum.

  313. Max™ says:

    In response to this comment:

    http://tallbloke.files.wordpress.com/2013/03/ipcc-v-skydragon.png

    Claes Johnson/SkyDragon ‘Back Radiation’ Model
    If I understand the SkyDragon/Claes Johnson radiation model correctly, there is 235Wm-2 ‘back radiation’ from the under surface of the shell (as there must be by the definition of a black body) but that radiation is not absorbed by the core. Instead, it is deflected back from the core surface. So it then gets absorbed back into the shell, as shown in my right hand diagram. The consequence of this is that the energy flow rate through shell AND core is 235Wm-2. So both shell and core are at a temperature of 254K (simple application of S-B law).

    IPCC ‘Back Radiation’ Model
    In contrast, the IPCC ‘back radiation’ model allows the 235Wm-2 of ‘back radiation’ from the under surface of the shell to be absorbed by the core. The consequence of this is that the energy flow rate through the core is 470Wm-2 and the energy flow rate through the shell is 235Wm-2. So the core is at 334K and the shell is at 254K.
    ” ~David Socrates

    David, I posted this earlier, your diagram does not represent the “skydragon” explanation at all.

    …….. ^ …………… I’m fairly sure this is the version Postma would endorse.
    …….. |
    …… 235 W/m²
    ————————————- 254 K [235 in, 235 out]
    ..^..
    235 (470-235) W/m²
    ————————————- 334 K [235 in, 235 out]
    ……..^
    ……..|
    ……235 W/m²

    vs

    …….. ^ ……………. IPCC
    …….. |
    …… 235 W/m²
    ————————————- 254 K [470 in, 470 out]
    ..^ …………….. 235 W/m²
    470 W/m² ………v
    ————————————- 334 K [470 in, 470 out]
    ……..^
    ……..|
    ……235 W/m²

    The difference is only in the amount of energy said to be absorbed by the various surfaces, if radiation from a cooler surface is subtracted from the power of the radiation leaving the warmer surface, then the skydragons are right.

    If radiation from a coolder surface is added to the power of the radiation leaving the warmer surface, then the IPCC model is right.

    ___________

    I’m not sure why your diagram has a 255 shell/255 planet and a 255 shell/334 planet being compared, so I cleaned it up to more accurately demonstrate the differences between the two explanations.

  314. Rosco: Just keep adding shells and imagine the temperature you could get?! Why don’t they extend their own arguments to explore if they’re actually meaningful or not? Because they just want their faith.

  315. @ACO: He didn’t “create” energy out of anywhere. He didn’t perform the test in vacuum. There are so many confounding errors it doesn’t mean anything. He’s just reduced convective cooling, that’s all. One thing is certain – you can not get higher temperature than the work you put in. Tell them to scale this up to a power generation unit, or to bring the plans to a power engineer or an electrical engineer for approval and design, etc. If you can get energy for free, you can use it for free. They’re just confusing themselves. If not, then why don’t they solve the world’s energy problems? Because they’re amateurs and they have no clue what they’re doing.

  316. In this diagram: http://tallbloke.files.wordpress.com/2013/03/ipcc-v-skydragon.png

    there is no explanation for how radiation from a 255K source plus a 255K source can produce 334K. How does one object of temperature 255K cause another object of 255K to become 334K? It doesn’t happen. Two ice-cubes don’t heat each other up. They just “say” that 255K radiation plus 255K radiation produces 334K. That is not how reality works. The IPCC model is mentally retarded. You could melt ice to water by having two ice-walls facing each other.

    I already explained how temperature and radiation works on the wavelength “quantum” level. Photons don’t interact with themselves to increase their own frequency, and you need higher frequency to generate higher temperature. 255K radiation is just that, and it doesn’t increase its own frequency – they don’t “pile up”, they go right through each other. 255K radiation from one source plus 255K radiation from another source does not produce 334K. This is the stupidest idea that has ever existed.

  317. Kristian says:

    Rosco,

    I have also put forward another thought experiment at tallbloke’s, one where the shell would gain its heat not from a core planet, but from a hypothetical, evenly spread out nuclear source within the shell itself. The nuclear energy source supplies a constant wattage to the shell. If, as the Eschenbach apologetics insist, the shell emits half of this provided power outwards to space and the other half inwards upon itself, then the shell would only ever be able to rid itself of half the heat gained from the nuclear energy source.

    The outcome? Unstoppable runaway heating.

    To be fair, on of the proponents – David Cosserat, which BTW Joe, I feel is really misrepresenting your position on this diagramatically over at tallbloke’s – concedes that this specific shell WOULD in the end lose all its received heat through the outer surface to space (he simply arrives at that same conclusion in a more convoluted way), but for some reason fails to take the next step and see that it would ALWAYS be like this, no matter where the incoming energy is coming from. He STILL thinks the planet would necessarily heat up with the shell surrounding it.

  318. They’re just making things up at an amateur level of knowledge and understanding, and arguing towards the outcome they desire in the first place. Sophistry in other words.

  319. I have a question for everybody:

    What is the temperature of an infinite amount of heat?

    Will gather responses for a bit before approving them so that everyone gets a chance to answer independently.

  320. Max™ says:

    Well, my point with the diagram clarification is that an internally powered sphere+shell would never set up a black body spectrum with the same temperature between both surfaces, inverse square law makes that impossible.

    Assuming the two surfaces being 334 and 254 K in both diagrams is the fair way to compare the differences.

    Said difference is put simply: you, I, and modern understanding of physics all claim that radiation from a cold surface reduces the power leaving a warm surface. 235 in, 235 out.

    The IPCC models and their supporters claim that radiation incident upon a warm surface adds to the internal energy no matter what temperature the source of that radiation was. 470 in, 470 out.

    [JP: I think you meant 235 in, 470 out for the IPCC]

  321. Simon Conway-Smith says:

    Joe, Do I suspect the answer is similar to “How long is a ball of string?”.

  322. Max™ says:

    …….. ^ ……………What I think is correct
    …….. |
    …… 235 W/m²
    ————————————- 254 K [235 in, 235 out]
    ..^..
    235 (470-235) W/m²
    ————————————- 334 K [235 in, 235 out]
    ……..^
    ……..|
    ……235 W/m²

    235 from the internal power source to the planet surface which is equilbrated at 334 K and would emit 470 as a black body in a vacuum. The temperature difference between the shell and planet reduces the power that actually reaches the shell to 235, and it emits to space accordingly at 254 K.

    235 in, 235 out.
    ________________

    vs

    …….. ^ ……………. What the IPCC thinks is correct
    …….. |
    …… 235 W/m²
    ————————————- 254 K [470 in, 470 out]
    ..^ …………….. 235 W/m²
    470 W/m² ………v
    ————————————- 334 K [470 in, 470 out]
    ……..^
    ……..|
    ……235 W/m²

    235 from the internal power source to the planet surface which is equilibrated at 334 K and emits 470 no matter what.

    The temperature difference between the shell and the planet doesn’t matter, so the planet also receives 235 back from the shell, adding up to 470 in, and the 470 it emits to the shell is balanced by the 235 up and 235 down.

    470 in, 470 out.

  323. Hi Joe,

    As often happens in these heated debates, I have been misrepresented by Max and Kristian with respect to my diagram at http://tallbloke.files.wordpress.com/2013/03/ipcc-v-skydragon.png .

    All I was trying to do was to find a way of re-starting the somewhat meandering debate currently raging over at the Tallbloke site by asking people to concentrate on your actual proposition rather than inventing their own ‘models’ with all sorts of bizarre variations such as heating the shell, etc., etc.

    So I produced two diagrams which, to the best of my understanding, represent (i) the ‘back radiation’ model that IPCC and Willis adhere to; and (ii) the Claes Johnson/SkyDragon model which you adhere to.

    As I understand your proposition, it is that energy in a cavity does not enhance the temperature of that same cavity. That is a perfectly respectable scientific position to take which everyone, whether they agree with it or not, ought to be able to comprehend and debate calmly. I tried to represent your ‘standing wave’ cavity radiation diagramatically by the green u-turn energy flow loop which, in effect, does no work, because the 235Wm-2 from shell-to-core is NOT absorbed, for the reasons you have always consistently maintained (unless I have entirely misunderstood you!). So it simply does not take any effective part in the energy balance. Consequently the right hand diagram shows both shell and core at the same temperature because there is no energy enhancement, which, again, I think is your position.

    My objective in producing the diagram was NOT in the first instance to argue for the left hand or right hand model, just to secure some ground rules that we all can agree on about what the two alternative propositions actually are! So my question to you now is this: do you agree that the right hand diagram (now explained to you in non-perjorative terms) is a fair representation of your position? Or, if not, in what way does it misrepresent your position (which I am certainly not trying to do) and how would you prefer to modify it so as to more fully represent it?

    Then, and only then, can we move forward to a focussed and calm discussion that does not constantly divert onto irrelevancies (not, I would emphasise, ones of your making).

    Best regards
    David Cosserat

  324. Rosco says:

    MAX – you’ve still created energy out of nothing !

    Think about adding another shell and another etc. This is no different to the original proposition as there is a surface radiating 235 out and you stick another shell over it and you have to double the core energy again to maintain the 235 out and 235 backradiated at the outer shell.

    Surely anyone who thinks about this can see this creates double the energy each timeyou stick another shell over the ensemble.

    30 shells results in 252 GigaWatts at the inner core – all from 235 W radioactive decay.

    252 GigaWatts – i mean – c’mon !!!!

    To all those who think radiation from a cold object can cause heating of a warmer object I ask that you explain Pictet’s experiment on the reflection and radiation of cold.

    this experiment was conducted over 200 years ago and is reproducible.

    His experiment provided the exact expected thermodynamic response that sound theory predicts.

    His cold object, a container of ice water, placed at the focal point of a concave mirror caused the thermometer at the focal point of a facing concave mirroe 16 feet away to dramatically decrease its reading. Note that – 16 feet away.

    That is the expected thermodynamic response – the radiation went from the hotter object to the cold one – there was no backradiation – none at all.

    Why such a dramatic response I have no idea.

  325. Rosco says:

    The fourth root of infinity over sigma ? Now that would require some computing power to solve.

    Or something that is completely indeterminant ?

    Or ask Willis ?

  326. Rosco says:

    Perhaps I misread Max’s comment and am now totally confused.

    I still think if Willis’ proposition is valid for the addition of one shell it must be valid for the addition of any number. This is surely one of the tests that must be valid for a hypothesis – reproducibility.

    Logically that leads to an absurd consequence of unlimited energy from 235 W simply by “trapping” radiation.

    Just because it appears it is arithmetically correct does not mean it isn’t completely absurd – which is the only possible conclusion a rational person can make.

    Therefore Willis’ steel greenhouse is an absurd proposition.

    I understand why his model is with the steel shell very close to the “planet”.

    This is similar to the supposed GHG Earth where the additional 5 km or so of atmosphere to the radiating height where the temperature is 255 K – the 5 km is only a small fraction of the Earth’s radius – see the anology – the steel shell is really close to the original surface.

    I say if the steel model is BS as logic demands it is so is the Earth’s supposed greenhouse effect.

    You do not need any special physics training to see the model requires doubling the core energy for every additional shell and that is an absurd proposition.

  327. Greg House says:

    Alan Siddons says, (2013/03/14 at 8:06 AM): “As I like to say, the 2nd Law enforces the 1st Law. If heat (or more generally ‘thermal energy’) did NOT just raise the temperature of less energetic bodies but also made the heat source warmer, this would spark the kind of infinite heating cascade that Greg House described.”
    ====================================================

    I posted this description on the Tallbloke blog a weak ago (http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46123), but today, as I referred to that again, he blocked my comments and had no problem with lying straight to my face, saying that I admitted being wrong.

    My comments have never been blocked anywhere, either on WUWT or on JoNova. I even challenged green radicals on the Greenfyre’s blog and no comment of mine was blocked or deleted there. I guess, some people are getting desperate.

    Tallbloke has switched now from the “Willis universe” with a power engine to a model without power supply, so that the whole thing is just cooling and therefore it is not obvious that the cycle of mutual warming won’t work. Very “clever”. My demonstration is still valid, but references to it are unwelcome now there. Right, maybe people just forget it and he can fool them undisturbed.

    So, meet “skeptik” Roger. I have made screenshots of every comment of mine “awaiting moderation” and his remarks he inserted: http://s8.postimage.org/93gvfkyc5/image.jpg, http://s14.postimage.org/cq7v6y9jl/image.jpg, http://s8.postimage.org/jlbip2ok5/image.jpg, http://s4.postimage.org/8mn7icnfh/image.jpg

  328. Loodt says:

    In reply to the question …What is the temperature of an infinite amount of heat?..

    The answer is: as long as a piece of sting!

    Temperature is a state of mass, and without specifying the mass that contains the temperature the question is basically nonsensical.

    But, you knew that in any case.

  329. Kristian says:

    I tried this approach over at tallbloke’s. Maybe the simplest form of arithmetic might still work. We’ll see how it goes …

    A black body receiving and absorbing a continuous incoming energy flux and accordingly reemitting (as required by definition) an equal energy flux (from whichever side of the body, you insist it’s all the same) of 470 W/m^2, would that not have an emission temperature (as per the S-B equation) of 302K?

    Well, Willis’ black body shell receives, absorbs and reemits an energy flux of 470 W/m^2. But its temperature? 254K.

    Remember, the absorption always occurs before the reemission. There would be no emission without the absorption first. So why doesn’t the incoming flux of 470 W/m^2 raise the temperature of the shell to 302K? It doesn’t matter if the incoming flux is coming from the outside (a sun), the inside (emitting planet) or within (‘intrastructural’ nuclear source), the flux is still absorbed completely and then reemitted to balance. Why doesn’t the S-B equation work on Willis’ shell?

    The ‘IPCC model’ seems to claim the following:

    # CORE absorbs 235+235= 470 W/m^2, emits 470 W/m^2, S-B temperature 334K (?!, really 302K)

    # SHELL absorbs 470 W/m^2, emits 235+235= 470 W/m^2, S-B temperature 254K.

    What’s wrong with this picture?

    Well, if one wants to argue ‘Ah, but the NET incoming flux to the shell is 470-235= 235 W/m^2!’, even if the actually absorbed flux IS 470 W/m^2, there’s no getting around that, then one should consider this:

    The NET outgoing flux from the core planet is also 470-235= 235 W/m^2. Why, then, is it OK for the planet to have a temperature of 302K, but not for the shell?

    The temperature is set by the absorption of the INCOMING flux, then the emission of the OUTGOING flux is set accordingly.

  330. Rosco says:

    The discussion at tallbloke’s website is running away in a similar fashion to the runaway greenhouse effect.

    I made the argument that if the original condition is valid – ignoring all errors and geometrical effects – then when the shell is added you have the same situation as the original- except for the magical doubling of energy.

    In this new case you supposedly have a surface radiating 235 to space.

    So adding another shell requires the original shell to now radiate 470 – which is exactly what happened in the original scenario.

    This of course leads to the absurd proposition that you can create enormous energy by simply doubling it with more shells.

    No-one seems to even consider that argument, instead preferring to continue with all types of variations of the same argument why it is OK to arbitrarily double the energy with the insertion of a shell.

    I guess this proves that people really want to have faith in their personal truths and will defend them no matter what.

    I could be completely wrong and they completely right but I thought about it long enough to come to my conclusion that the shell doubling leads to absurd consequences – it appears most commentators do not even think about the adding of new shells as per the original condition they all accept – ie double energy per shell.

    Instead of even attempting to debunk this – dare I say it – obvious inconvenient truth – the discussion continues over why Willis’ proposal is theoretically OK.

    I discovered that Willis originally proposed this “steel greenhouse” in 2009 !

    That means it is accepted as “science” at WUWT for over 4 years without any questioning other than geometric ones – which debunk it anyway – and he has been allowed to spruik this gibberish twice yet critics of such chicanery are silenced.

    I thought SKS were bad enough – it is a wake up to see even sceptical websites are so entrenched in the greenhouse effect they will not allow any voice raised against it yet allow such obvious chicanery as the steel greenhouse and its magic.

    Willis always quotes that if there is a mere 170 W/sq metre insolation there is insufficient energy to prevent the oceans freezing.

    I guess some people simply cannot understand they have been conned into a religious anti science belief system.

    But I guess they say the same about me.

  331. Oh I see David, thanks for the explanation. Yes the diagram is quite useful in that way.

    So yes, the problem is in the adding of radiative energy fluxes in a direct linear way and thus leading to the IPCC model. The IPCC model can obviously be treated as infinite plane parallel “walls” which is equivalent to one shell inside the other or a planet inside a shell. Consider if there were a radiation source heating the bottom of the lower wall with 255K radiation at the position of the lower wall, and forget about the upper wall/outer shell. By the accounting of linear adding and subtracting fluxes of the IPCC, the lower wall could never even get to the temperature at which it is being heated, because of the supposed bi-directional nature of its output.

    If the lower wall received 235 W/m2 on the lower side, then supposedly it could only ever get to a temperature corresponding to 117.5 W/m2. This should intuitively indicate an error, but this invented way of adding fluxes is trusted instead. Radiative fluxes do not add with elementary-school arithmetic in their ability to perform work/induce temperature. The ability to perform work/induce vibration in matter (i.e. induce temperature) is a function of the frequency components of the radiation. It is the frequencies which induce corresponding material vibrations and hence temperature. The matter will be induced to vibrate at the frequency components of the incident radiation because this is an active forcing causing action, and the matter doesn’t care which direction the subsequent thermal energy might be emitted to. Either side of the wall will emit at the flux level corresponding to the temperature which it is at, the temperature which it has been induced. If the wall was very thin and the thermal mass was very small, both sides would be induced to be the same temperature very quickly from forcing radiation incident on one side only.

    So, the equation for heat energy radiative transfer is basically q = sigma*(T_hot^4 – t_cold^4). When T_hot and T_cold come to the same temperature, there is no net energy transfer. This is what confuses the IPCC model, because they want to insist that energy is still transferring and causing heating, just as long as you balance it all out to zero in the end. This is incorrect. There is zero net energy transfer. Hence, the upper side of the wall emits at 255K, and that energy is replaced at the bottom side by the incoming radiation, and no energy is lost from the wall on the bottom side at all. Why is no energy lost from the bottom side of the wall? Because T_hot^4 – T_cool^4 = 0.

    From this latter, you can write that on the “Skydragon” side, there is zero energy transfer between the shell and the core. They come to the same temperature, and no heat energy (i.e. ZERO energy) is transferred from the shell to the core. 235 goes from the core to the shell and 235 leaves the shell on the upper side. Between the shell and core is a radiation field, a blackbody spectrum, at 255K.

  332. For an object to be warmed, only another heat source of higher thermal energy (temperature) can achieve that. The same applies with any type of energy, e.g. light. If you take 2 torches of different brightness and shine the dimmer one at the brighter one, does the brighter one get any brighter? No it doesn’t. Or if you shine the a torch at a mirror which reflects the light back at itself, does it get any brighter? No! It cannot, as quantum physics rules that a lower frequency energy cannot raise the frequency of a higher frequency one, nor can the reflection of an energy source at a frequency raise its own frequency. If a cooler or same temperature body could raise another body’s temperature, we’d have an infinite energy generation machine, which is impossible.

    A blanket, of course, reduces convective air cooling, and this makes your skin feel warmer. It is a very poor analogy to a radiative GHE because there is no physical correspondence in the analogy – it is just an analogy, without any actual physical connection. And the reason? As pointed out: a torch cannot make itself brighter with its own radiation. Stand in front of a mirror – do you feel any warmer? NO! This is why the shell game is wrong.

  333. Alan Siddons says:

    Allow me to repeat my first point on this blog page. If you’re holding an inert substance that gets hotter than the heat source it is touching, then you are holding onto a miracle. And if you’re holding an inert substance that RADIATES more intensely than the radiation it’s receiving you’re holding a miracle too. Indeed, a miracle is what the hocus-pocus of greenhouse physics amounts to: the target is supplied the continuous power of 235 W/m² and from that it develops a continuous power of 390 W/m². In other words it is radiating more intensely than the radiation it’s receiving.

    I mean THINK of it. Think how you’d react to a salesman who claims that his Hyper-White Paint reflects more light than the light that’s falling on it. You’d run that charlatan out of town. Yet supposedly serious scientists claim that an object can absorb and EMIT more light than the light that’s falling on it. And most people take this claim for granted. The world is quite insane.

  334. And the funniest thing about it all, is that the Earth does NOT receive 235 W/m2 continuously anyway! Think of THAT! It’s all bizzaro land.

  335. Alan Siddons says:

    It’s been disgusting to witness so much “hopeful ignorance” here and on Tallbloke’s site. If physical phenomena like Eschenbach and others describe actually did exist, our ability to generate energy would be without limit. Good LORD, if a square meter on the earth’s surface can produce 390 watts from 235 with just 1% of its atmospheric gases “re-radiating” a few wavelengths on the spectrum, then imagine the power that a full-bodied continuous-spectrum Eschenbach Shell could generate. Just SHOW US, then, you silly fools. Point to a device that works as advertised. Put up or shut up.

  336. Exactly – engineer it and exploit it! We could all use such a device. I will more than happily concede everything I’ve ever said and say that I was wrong and even stupidly wrong, if they can engineer this shells game principle and create more energy than they put in. Someone already explained how to exploit it – by using a sphere and shell of metals of differing thermal expansion coefficients. If you can boil water without putting in the raw energy necessary to boil water in the first place, then you have an over-unity steam engine and you can create power with it. DO IT ALREADY! I for one would love free power.

  337. Max™ says:

    The reason I specify that the planet surface is at a higher temperature than the shell is because it is supposed to be an internally heated sphere with a shell at a non-zero distance.

    I posted a simple example for the case of a sphere with a 100 m radius and the addition of shells 1 m above the surface or the last shell.

    ________________________

    4*pi*100^2 =
    125,663.7*235 = 29,530,969.5/128,189.5 = 230.3

    4*pi*101^2 =
    128,189.5*230.3 = 29,522,041.8/130,740.5 = 225.8

    4*pi*102^2
    =130,740.5*225.8 = 29,521,204.9/133,316.6 = 221.4

    4*pi*103^2
    =133,316.6*221.4 = 29,516,295.2/135,917.8 = 217.1

    _______________________

    Whether you have four shells or one shell, if you have a shell located 4 m above an internally heated sphere with a 100 m radius which emits 235 W/m^2, then that shell will only receive 217~ W/m^2 and accordingly must emit that much or less.

  338. This is a very good point for Max to make because it indicates that the outer shell always has to be cooler than the inner sphere, as a necessary condition for equilibrium. Of course, as the difference in radii diminishes, the difference in equilibrium flux of the outer shell relative to the interior sphere diminishes.

  339. Joe,
    Thanks very much for your replies of 2013/03/19 at 6:37 PM and 2013/03/19 at 6:48 PM.

    The first thing to explain is that I am a hardline skeptic on climate change. My motive here, as elsewhere, is to find a definitive way of proving to warmists and to ‘luke warmists’ alike that they are wrong: that CO2 concentration in the atmosphere has NO EFFECT AT ALL on mean surface temperature. Not just that “sensitivity to an increase in CO2 concentration is low” – but that it is ZERO! I have come to this conclusion long ago, not by diving deeply into quantum mechanics or radiative transfer theory but because there is simply no sign of the earth warming dangerously, despite all the uninformed hype on the subject. We all form our initial prejudices in that kind of way, long before we dive deeply into a subject. But the key to success in science is always to remain open to being wrong. So I endorse very much the spirit of your comments at 2013/03/19 at 11:52 PM: I will more than happily concede everything I’ve ever said and say that I was wrong and even stupidly wrong, if they can engineer this shells game principle and create more energy than they put in. Me too! And I would do likewise if the warmists or luke warmists were ever to be able to prove their case for global warming due to increases in atmospheric CO2!

    I think that the reason there is so much heated debate over things like Willis Eschenbach’s ‘steel greenhouse’ core-warming model is because most people, even experienced scientists and engineers like me, are unskilled in the specific physics needed to analyse the issues at hand. This often leads to people on all sides in the debate making ridiculous and/or naive objections to things other people have said with the best of intetions and in all innocence. For example, nobody at all (except perhaps for a few way-out eccentrics!) would claim that ANY Willis-style shell model could really cause unlimited energy to be produced from a fixed input energy flow. So to criticise any such a model on such grounds is fine as a way of pulling people up short and making them think harder – just so long as it is intended as robust repartee and not meant to humiliate.

    So could I suggest that we make a fresh start, remembering that we skeptics really are all on the same side – not opposing warriors? In that spirit, I would first ask you to clear up the following…

    My diagrams on the TB site show what I intended to be a pictorial description of the difference between the IPCC ‘Back Radiation’ model on the one hand, and the Claes Johnson/SkyDragon model on the other. You in your response have kindly confirmed that these diagrams are quite useful (thanks!) [NOTE: in the original IPCC diagram there was an error: the left hand core temperature should have been 302K not 334K, now corrected.]

    Yet several commenters over on the TB site have said (unhelpfully – i.e. without any elucidation!) that the right hand diagram “does not represent correctly” the Claes Johnson/SkyDragon position.

    So is there anything about my right hand diagram that you disagree with, or feel could be improved? Also is there anything about the text in my accompanying blog comment that you feel is wrong or that you could improve on? Particularly the following paragraph:

    Claes Johnson/SkyDragon ‘Back Radiation’ Model
    If I understand the SkyDragon/Claes Johnson radiation model correctly, there is 235Wm-2 ‘back radiation’ from the under surface of the shell (as there must be by the definition of a black body) but that radiation is not absorbed by the core. Instead, it is deflected back from the core surface. So it then gets absorbed back into the shell, as shown in my right hand diagram. The consequence of this is that the energy flow rate through shell AND core is 235Wm-2. So both shell and core are at a temperature of 254K (simple application of S-B law).

    If so, I am more than happy to try to agree a better diagram and an improved form of words with you.

  340. Alan Siddons says:

    “the planet surface is at a higher temperature than the shell…”

    Which kind of ties in to what I was trying to get across before. Consider, for example, that the nuclear core in Eschenbach’s scenario might well be radiating 50 million W/m² by itself. If this core happened to be the surface in question, with no other mass on top of it, 50 million watts/m² would be what it emits. Dump enough rocks and dirt on it, though, and you could reduce the surface’s emission rate to almost zero — which in fact is pretty close to what the Earth’s internal heat does emit. A mass whose surface has been reduced to 235 W/m² was just convenient for Eschenbach’s argument. But introducing more mass in the guise of a dense shell will reduce the temperature even more, thus also reducing the emission. More molecules are now competing for the same amount of heat, after all.

    In my view, the 2nd Law only specifies that heat always seeks to minimize a temperature difference. Given the impediments of the inverse square law, or the amount of mass to travel through and the particular conductivity of that mass, however, the 2nd Law does NOT connote that a heat source and its target will eventually reach the same temperature. No, between Eschenbach’s hot nuclear core and an upper surface, a permanent temperature GRADIENT will exist instead — and when such a gradient is CONSTANT one can regard the two locations as being in thermal equilibrium because no further transfer is possible. Same deal with a suspended shell. In short, “thermal equilibrium” doesn’t necessarily imply the attainment of equal temperatures.

  341. I have a few comments to moderate right now and you’ve probably noticed there’s been a significant time delay in approving comments. I am currently on travel in India again this week and the time-zone difference for me is 12 hours, so I spend a lot of the first week fighting abject delirium and struggling to form coherent thoughts. Somehow I still manage to hit the ground running and write c++ GUI code, handle spacecraft equipment, perform functional test procedures of hours and hours of extremely boring documentation, analyze data and write summary reports of the findings, etc. etc. Suffice it to say, non-essential work gets sidelined and my free-time is spent either working more, or just trying to stay awake so that I can sleep at the appropriate time.
    So, just in case you were wondering. Will get to moderating comments my tomorrow morning…10 hours from now. Time to go see what the kanteena (sp?) made for dinner tonight and then pass out in a spice-augmented sweaty delirious haze 🙂

  342. Greg House says:

    Another warmist article on WUWT by Ira Glickstein: http://wattsupwiththat.com/2013/03/20/how-well-did-hansen-1988-do/.

    Quote: “…it would be reasonable to assume that Climate Sensitivity is closer to 1 ⁰C than 4 ⁰C.” And later Ira Glickstein, PhD says, (March 20, 2013 at 8:07 am): “I am pretty sure Climate Sensitivity is somewhere between 0.25 ⁰C and 1 ⁰C. If Atmospheric CO2 had zero effect, the Earth Surface temperatures would be over 30 ⁰C cooler, because the Atmospheric “Greenhouse Effect” is real.”

  343. Joe,

    I have now had time to re-read this post + comments from the beginning and, if you permit, have a few comments of my own to make which I will post as I complete them:

    Andrew says, 2013/03/09 at 11:59 PM: Please do not resort to name calling as the layman out there will turn off and not bother to read your comments. The bigger cause here is to ensure that people understand that the fear of CAGW and GHE are greatly exaggerated. You must keep as many people onside as possible. I understand that this may be frustrating to you but keep trying to explore methods to widen this message to other scientific and media outlets.

    Whilst sympathising with the frustrations vented here, I feel Andrew is absolutely right. The way forward is a clearer explanation of your position.

    Joseph E Postma says, 2013/03/10 at 8:47 AM:
    The error has nothing to do with the 0.3% problem Willis commented on.

    Agreed!

    The error is in the rejection of quantum mechanics and the known behaviour of radiation trapped inside a cavity, and also basic arithmetic. This is a process of exploring methods to widen the message: if nicely telling people that 1+1 does not equal 3 doesn’t work, then maybe yelling at them will?

    With respect I genuinely think that yelling at people is a counterproductive approach, however cathartic for you guys! The right way forward is to try harder to convince people like me who have a very strong scientific background but lack the quantum-level knowledge. If you succeed, I (and I am sure others such as TB) will be your strongest advocates.

    Remember, as a hardline skeptic I would be delighted to be convinced. But it has to be done by intellectual persuasion, not by shouting and ranting and calling people names. 🙂

  344. Simon Conway-Smith says:

    Joe, When in Bangkok last week, I avoided all those little street food stalls, and kept to the main shopping centre eateries. I do like Indian though, but have never been there to taste the real stuff. Don’t forget through, have your ice cubes ready in case it hasn’t been heated well enough 🙂

  345. Simon Conway-Smith says:

    Joe, Just been reading the SkepticalScience ‘repost’ of your debunking of the GHE hypothesis, particularly the argument you make that we have to treat the Earth as a sphere with only half being irradiated by the sun at any one time. They seem to think this is not necessary and undergrad level stuff (funny how it appears in the IPCC reports as ‘gold standard’ then), but what made me really narked was the statement…

    The atmosphere and/or ocean help smooth the diurnal temperature difference very well.

    I defy anyone to stand outside for 24 hours and say that the temperature was the same, or changed little, during all that time! Simply unbelievable!

    For them to say you “did not like a simple model of Earth’s radiative balance where we approximate the Earth as a sphere with uniform solar absorption” is to totally fail to understand your point, and their error. It’s not just that you didn’t like it, but that it’s fundamentally wrong.

  346. David Socrates @ 2013/03/20 at 5:41 AM

    Indeed, there’s been no warming above the 1930’s values even though CO2 has increased, what 50% or more since then. Hello.

    OK so for the Skydragon diagram. We need to look at the REAL, actual equations for heat flow:

    q = k*(T_hot – T_cool) {conduction}
    q = s*(T_hot^4 – T_cool^4) {radiation}

    These are directly physically analogous to other equations in physics:
    Force:
    a = 1/m * (F2- F1)
    Electricity
    I = 1/R * (V2 – V1)

    When two opposing forces are equal in magnitude, it does not mean that the forces do not exist, but it does mean that they do not do any work. Same thing with two opposing voltages of equal magnitude – it does not mean the voltages don’t exist, but it does mean that they can’t induce any current.

    And so with temperature, when you have two equal temperatures “facing each other”, it does not mean that they don’t exist or that the thermal radiation from them doesn’t exist, but it does mean that they don’t do any work or induce any temperature, or cause any heat energy flow.

    This is where the IPCC supporters think that Claes and the Slayers say that there is no such thing as backradiation or that photons must “have to know” to not be emitted in a specific direction. This is very uninformed and dis-knowledgable about physics, to say such a thing. The backradiation DOES exist, just like one force opposing the other force exists, but it doesn’t do any work. The heat energy flow equation, just like the force equation, equates to ZERO.

    What the IPCC supporters do is insist that the backradiation must perform work or heat flow, which is equivalent to saying that one of the forces must do work even through it doesn’t actually induce any acceleration/displacement. Really, they’ve just thrown the whole fundamental physics of energy, work, and the basic set of physical relations out the window.

    They WANT backradiation to “do work”, which is impossible by the definition of heat flow, because they can’t figure their way out of treating sunshine as freezing cold over the entire planet. I provided the way out (and other Slayers did too) by pointing out the fact that sunshine is NOT cold and is NOT distributed over the entire planet. (More on this in a coming moderation reply to another comment up the line here). But essentially, they want to keep the assumption of cold sunshine because this is the only vector by which CO2 can be vilified and climate alarm be created, via the manufactured GHE.

    So, on the Skydragon diagram, how much energy is sent from the shell back to the planet? Zero. Why? Because q = s*(T_hot^4 – T_cool^4), and T_hot and T_cool are either equal so that q = 0, or, the shell is the cooler meaning that the q only goes from the planet to the shell. Does the shell emit to towards the planet? Sure. Does it induce any work or heat flow? No! Push your hands together, hard. Do the opposing forces exist? Yes. Does either perform any work? No. Why? Because the displacement is zero.

    The supposed “knowledge” required by the photons is created right at the inner surface of emission: when photons are emitted at the inner side of the shell, they are immediately confronted with incoming photons from the planet of equal frequency. What happens then? They cancel each other out, in terms of net energy flow. (The photons pass through each other in fact, but the net energy they carry is balanced to zero.)

    Now, the shell will have some mass and will contain some internal energy, but, even if it was nearly mass-less, as soon as energy is emitted from the outer side of the shell and truly lost from the shell, there will be a slight deficit in the thermal energy balance between the shell and inner sphere, and so the energy gets replaced into the shell, and this happens continuously – in thermal equilibrium, the shell will lose heat energy on the outside just as fast as it is replaced on the inside. Of course as Max points out, because the shell is larger than the sphere, the actual Wattage and temperature will be less from the shell; but, the total energy will be the same.

    So, what you could draw is 235W/m2 emission from the inside of the shell, but then cross it out or something because that emission, when facing incoming energy of the same magnitude and frequencies, can’t actually induce any work or heat flow to occur. Why? Because of the basic equations of physics as shown above. The outer side of the shell CAN freely lose bonafide heat energy out to infinity, and as this energy is lost from the shell, it is continuously resupplied by the planet.

  347. Alan @ 2013/03/20 at 7:46 AM

    I would just make the clarification that increased mass in the form of increased density, while the surface areas stay the same, would just slow down the speed at which thermal equilibrium is acheived. If the surface area increases, then indeed the flux and temperature must decrease.

  348. Greg @ 2013/03/20 at 10:01 AM

    They just make stuff up as they go along as it suits them. Of course they say that NOW – their models have all failed!

  349. Simon @ 2013/03/20 at 1:06 PM

    Yes, they completely missed the point. They are directly saying that it doesn’t matter how we treat the sunshine and Earth at all. It is NOT as simple as making averages, as they like to claim and say that I don’t understand what an average is. THEY do not understand what an average is, the morons. First, 24hr sunshine at -18C is not what exists. It simply isn’t what is real, so, why would it be considered a valid starting point? Second, you can’t average-down the power input in a non-linear response system! -18C for 24 hours can’t melt ice; actual, real sunshine CAN melt ice. Isn’t that an important difference? Especially considering latent heat storage and the temperature plateau created from it? Of course it is important. In fact in my last paper I actually modeled REAL daily sunshine in real time as it actually exists, shining on an ice-ball Earth. -18C for 24 hours of course would never melt the ice ball Earth. The reality-based model melted the ice and maintained an average water temperature of +14C! And the effect of latent heat was to help hold the temperature higher than what it would have been without latent heat.

  350. Here is a plot Simon made for Max:

    Shell radiation vs. radius


    I ran a spreadsheet starting with the core at 100m radius emitting 235W/m2, all the way up to a 400m radius shell in 1m increments and plotted the radius vs. W/m2. It produces the classic inverse log (square? -JP) graph, which is what you would expect…

  351. Following up briefly on Me at 2013/03/20 at 6:14 PM:

    If you push against a wall, energy is spent, but no work, another quantification of energy, is performed. This is the nature of work and energy which seems paradoxical to the IPCC supporters and so it is rejected – they consider that any energy which is spent must result in something, i.e. work, being done. But this is not true, and it is not what physics says. You can spend lots of energy pushing against a wall, but that energy does not perform any work (another form of energy)…it doesn’t cause anything to happen.

    So, likewise, backradiation can spend lots of energy shining back to its source, but, it doesn’t perform any work. The equation for thermal physics says so, just like the one for mechanical physics. In radiative thermal terms, this lack of work means that no temperature is induced to raise. The physical reason at the quantum level is due to the frequency components analysis I’ve discussed previously. This is how and why radiation obeys the 2nd Law of thermo.

  352. Greg House says:

    Joseph E Postma says, (2013/03/20 at 10:48 PM): Following up briefly on Me at 2013/03/20 at 6:14 PM:
    If you push against a wall, energy is spent, but no work, another quantification of energy, is performed. …

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

    Hi Joe,

    I like this explanation very much. It is clear and understandable for everyone.

  353. Joe,

    Thanks very much for your carefully considered and well articulated replies of 2013/03/20 at 6:14 PM and 2013/03/20 at 10:48 PM. A number of points:

    Radiation need not do work
    Your point about radiation not necessarily “doing work” is spot on. I recently published an article on the TB site that insisted that ~333Wm-2 of radiation had to circulate round and round between the base of the atmosphere and the surface (simply as a consequence of their both being at around 288K) but that it did no work. This produced howls of rage from some skeptics who in my opinion have a quite unjustified phobia about back radiation and just deny that it exists. This despite the fact that my diagrams clearly showed the downward flow being more than balanced by the upward flow from the earth’s surface to the atmosphere, thus avoiding any possibility of a misunderstanding about a “cooler body being able to warm a warmer body”.

    Or so I thought!

    This controversy got so heated that it actually spawned another article on pyrgeometers (installed at ~190 meteorological centres around the world specifically to measure IR back radiation) where some commentators even went to the extent of arguing that the devices didn’t work and/or that pyrgeometer manufacturers were apparently in some kind of an international conspiracy with warmists. How about that for cognitive dissonance? 🙂

    So, for me it’s official – back radiation exists
    I have encountered exactly the same problem more recently in responding to others on the TB SkyDragon blog, some of whom still deny the existence of back radiation. That is one of the reasons for contacting you now. Whether or not skeptics accept your arguments, your position on the existence of back radiation puts the lie to the idea that “only warmists” support back radiation. I tried and tried in my article to explain that such people had to get over their phobia, otherwise they could not reasonably expect to be able to engage properly with sophisticated warmists without being ridiculed.

    Layer models rule OK
    I am also relieved to see you agreeing that it is legitimate when examining these energy flow issues at a simple level to use a layered ‘model’ that assumes that all radiating surfaces are of the same area, even though in a real set of concentric surfaces, this of course is not strictly the case. Why people think that, say, a 0.3% variation in surface area makes any difference to the fundamental principles under discussion is beyond me. It smacks of sophistry, trying deparately to find a flaw in a basic argument that they don’t like by putting up a nit-picking and diversionary objection.

    However…
    …even if we now set completely aside those skeptics who still deny back radiation (phew!), there is obviously still a long way to go to resolve the real issue between ‘slayers’ and ‘luke warmists’ of whom there are many distinguished examples (Lindzen, Spencer, Singer).

    The prize for you and your supporters is to win the luke warmists over. I believe that will then naturally result in the general public also being won over in increasing numbers. Otherwise you will have an ever steeper uphill task.

    All that of course assumes that you are right! As I said before, as a hard line skeptic I very much hope you are. For example, it would cut out all the tedious ‘climate sensitivity’ discussions about how much warming results from a doubling of atmospheric CO2.

    The Diagram
    I think there is still a long road ahead to convince enough people you are right…or wrong. As a very modest next step forward, can I suggest that we at least agree a simple diagram that represents the Slayer case. In your responses you have not explicitly rejected my previous diagram but you did say “…what you could draw is 235W/m2 emission from the inside of the shell, but then cross it out or something because that emission, when facing incoming energy of the same magnitude and frequencies, can’t actually induce any work or heat flow to occur.” That was essentially what I intended the u-turn arrow in my diagram to imply but I wonder whether the picture should look a bit more symmetrical, such as this one?

    Let me know what you think.

    The Physics
    You say: The physical reason at the quantum level is due to the frequency components analysis I’ve discussed previously. This is how and why radiation obeys the 2nd Law of thermo.. Please direct me to this. I will study it (yet again!) and report back.

  354. A C Osborn says:

    Joseph E Postma says:2013/03/20 at 10:48 PM Following up briefly on Me at 2013/03/20 at 6:14 PM:
    and
    Greg House says: 2013/03/21 at 6:28 AM

    Joe, what you have written does not appear to be quite true. Isn’t some Work done, the atmosphere around you would be warmed up by the excess heat that your body would generate expending all that energy.

  355. Alan Siddons says:

    “…increased mass in the form of increased density, while the surface areas stay the same, would just SLOW DOWN the speed at which thermal equilibrium is acheived. If the surface area increases, then indeed the flux and temperature must decrease.”

    Increased density delays thermal equilibrium? To the contrary, a dense material often HASTENS heat transfer because vibration (heat) spreads across its closely-packed atoms so fast — much like kinetic energy passing through a Newton’s Cradle:

    No material is absolutely conductive, however, so any material is somewhat insulative too. Surround a conductive metal rod with insulation and touch one end to the flame of a Bunsen burner. Heat will travel to the other end but it won’t get as hot as the flame. With a longer rod the other end will be cooler still, even though the surface area (the rod’s cross section) hasn’t changed. A point to stress is that the far end of a long rod won’t EVENTUALLY get as hot as a short rod. No, it will NEVER get as hot. Otherwise, insulation (i.e., something less than 100% conductive) would do no earthly good, for the same amount of heat would spill out of a thick insulator as out of thin. This kind of temperature reduction is more a matter of the distance heat has to cover rather than of surface area.

    Just as density can assist transfer, though, sparseness is able to thwart it. Indeed, the most dramatically “interrupted” gradient occurs within a material that’s nearly as light as air: Aerogel. You can comfortably press your finger on top of this tenuous stuff while on the bottom, a mere fraction of an inch away, a propane torch is blazing right at it.
    http://inhabitat.com/exciting-advances-in-insulation-with-aerogel/aerogel-insulation-2/

    Again, the surface area difference with or without aerogel is negligible. It’s simply that this uniquely disorderly material won’t let heat cross the distance, so there’s an enormous temperature difference between one side and the other — the same as you’d see with a more conductive substance over a greater distance.

    As I say, then, there’s the temperature-reducing impact of simply adding mass. A candle can bring a teaspoon full of water to a boil but never a whole pot. And there’s the diminishing return that conductive limits bring about. Neither of these instances involves a mere delay. It’s not a matter of taking more time for a previous condition to obtain but of resetting the temperature where equilibrium occurs.

    No big deal, though. I’m just trying to emphasize that these imaginary full-transfer scenarios are inherently unrealistic. But let me add that heating-by-reduced-emission scenarios are also in error. Even NASA admits that the Earth is constantly radiating away all of the radiant energy it receives from the sun. There’s no sign of atmospheric insulation. I hope Tallbloke is listening.

  356. squid2112 says:

    Joe, you are making things so incredibly perfectly clear to me it is amazing. This is awesome!

    However, I do have just one question for you. I am imagining presenting to my father your backradiation/work scenario, which I totally get, especially within the framework that the backradiation could not make a warmer object warmer still. However, when I envision presenting this argument, I anticipate the rebuttal to be “but it can make cooler objects warmer”, which would imply that there is a GHE. Not all things on the ground are necessarily going to be warmer than the air above all the time. Wouldn’t down welling long wave radiation make something cooler than itself, warmer? And if so, how does this fit within the GHE argument?

  357. Max™ says:

    Similarly, if back radiation was going to do anything, it would be reducing the work available to be done by radiation leaving the warmer surface.

    If you push on a wall while someone else pushes back the other way, does that somehow make you push the wall harder?

  358. ThePhysicsGuy says:

    Joseph,

    I participated in the WUWT discussion and had several exchanges with tallbloke. One of tb’s claims was, and I quote:
    Radiative energy doesn’t care where it is coming from or going go, or what the temperature is on either end. If I light a candle on the earth during the day, the sun ends up warmer than it would be if I didn’t light the candle. Of course the reverse is true as well, the candle ends up warmer than if there were no sun. Since NET heat flow is from the sun to the candle, no thermodynamic laws are broken … but that doesn’t mean that the light from the candle is not absorbed by the sun. It is definitely absorbed, and the sun ends up warmer because of that radiation. Physics. Don’t leave home without it.

    I am a professional civil engineer by trade, so my physics expertise is more towards Newtonian physics, but our core physics requirements at university included a semester of thermodynamics. What tb and lgl are violating is the basic Clausian statement of the Second Law of Thermodynamics…pure and simple. If a cooler body can warm up a warmer body, then that “warmed up” warmer body will in turn warm up the cooler body (the cycle continuing without end) violating both the First and Second laws since one would be creating energy out of nothing. And if a cooler body could indeed warm up a warmer body, then by the same logic, bodies of equal temperature would warm each other up because they are after all, receiving energy from each other. If this were indeed true, it would seem there would have been precise lab experiments performed by physicists demonstrating this amazing phenomenon.

    Joseph, could you comment on a post by Dr. Roger Pielke Sr. regarding the Greenhouse effect and the 2nd Law? His short explanation on why there is no violation is as follows:

    With regards to the violation of the second law, what actually happens when absorbing gases are added to the atmosphere is that the cooling is slowed down. Equilibrium with the incoming absorbed sunlight is maintained by the emission of infrared radiation to space. When absorbing gases are added to the atmosphere, more of emitted radiation from the ground is absorbed by the atmosphere. This results in increased downward radiation toward the surface, so that the rate of escape of IR radiation to space is decreased, i.e., the rate of infrared cooling is decreased. This results in warming of the lower atmosphere and thus the second law is not violated. Thus, the warming is a result of decreased cooling rates…….The bottom line here is that when you add IR absorbing gases to the atmosphere, you slow down the loss of energy from the ground and the ground must warm up.

    I don’t quite know what to make of it since atmospheric physics are beyond my expertise.

    Joe. I understand your frustration and passion for this subject, but I suggest taking the high road. Just address the errors in your methodical rational way and leave it at that……for what its worth.

  359. John in France says:

    If you push against a wall, energy is spent, but no work, another quantification of energy, is performed.
    “I like this explanation very much. It is clear and understandable for everyone.”

    I agree with Greg House (2013/03/21 at 6:28 AM) on this and as one with little science background of any account can confirm that it is a precise summing up of the situation such as this that we all need.
    So why the difficulty in getting the message through to people who are evidently not the morons you are calling them? Don’t forget we are speaking not only of people like Anthony and Willis, but Jo Nova, David Evans, Richard Lindzen, Chris Monckton, to name but very few. These people have been active in debunking the global warming / anti CO2 brigade for how many years now? Like it or not, we’re on the same side.
    The nagging question is why do Watts and his ilk so aggressively defend such an untenable theory, displaying the same attitudes and indulging in the same exclusion tactics as the pseudo scientists they decry?
    I suspect that the key is to be found in Monckton’s approach. Most of his arguments are based on the figures actually given by the IPCC and other alarmist groups; he then proceeds to take them apart and show that even by these low standards they do not in any way stack up and therefore CO2 induced global warming is no cause for concern.
    I think that adherence to the GHE at least gives them some common ground to engage with alarmists (and to show outsiders they are not cranks) and our rejection of that common ground in their eyes takes it away from under their feet.

  360. ThePhysicsGuy says:

    Joseph,

    Oops! The first quote I mentioned above regarding the candle was a Willis Eschenbach quote, (not tallbloke), made on February 6, 2013 at 2:33 pm, on the WUWT, “The R.W.Wood Experiment” post. Maybe you can make an edit for me?

  361. Kristian says:

    It’s sad seeing self-appointed ‘skeptics’ over at Tallbloke’s succumb to the exact same tactics as the alarmists employ when their dogma is challenged. They remain completely indifferent to countering arguments, just restating their axioms as truth over and over again, not addressing what the opposition is saying at all, only building strawman arguments to tear down, willfully ‘misunderstanding’ time and time again the points made by people who disagree with them. It’s frustrating. But I guess that’s their aim.

    I totally sympathise with how you feel, Joe.

  362. Max™ says:

    Crosspost, as I think this is a rather insurmountable disproof of the whole “the inner surface will emit 470 W/m^2 after adding a shell” nonsense.

    “Isn’t that a bit like saying you can add half a cup of water at 5 deg C to another half a cup of water at slightly less than 5 deg C and getting a full cup of 10 deg C?” ~Arfur

    Indeed.

    It is also like saying if you take a sphere with the previously mentioned 1 million square meter surface area (8920~ m radius) and it has an internal power supply which raises the temperature until the surface emits 235 W/m^2 (total 235 megawatts) then you could do the following:

    1. Add a shell 1 meter above the surface (so the radius of the shell is 8921~ m) and the shell will then emit 235 W/m^2 (total more than 235 megawatts) and the original surface will warm until it emits 470 W/m^2 (total 470 megawatts) with no other changes made to the system.

    2. Fill the gap between the shell and the surface with the same material as the shell and surface are made of, maintain the 235 W/m^2 emission from the shell surface and the original surface–now at a depth of 1 meter–will be at or near 302 K.

    _______________

    If this is the case, it seems to me that if we start with the original sphere+shell, assuming the sphere surface is emitting 470 W/m^2, we should be able to remove 1 meter of material inside the sphere and ask what the temperature would be afterwards.

    If it is true that adding a 1 meter gap + shell above the surface raised the surface temperature until it doubled the emissions, then adding another gap within, such that the surface is a new shell, would also double emissions, right?

    So the new surface would emit 470 and receive 470, totalling to 940 W/m^2.

    Do that again, now with 1 “outer” and 2 “inner” gaps we have a total of 1880 W/m^2 leaving the inner surface, right?

    But above I suggested that an internal power source, spherical, and 4000 m across could supply 1138 W/m^2 which would give 235 W/m^2 through a naive approximation of conduction following an inverse-square law.

    Now we find that simply removing layers of material beneath the original surface+shell would produce more energy?

    Why isn’t this being engineered and utilized?

  363. Max™ says:

    Note: that should be “235 gigawatts”, I mentally corrected it to a more sensible sounding value, but 8920~ m radius gives 1 billion m surface area.

  364. OK I’m crazy busy here in India testing spacecraft, writing code, banging my head against the laptop, etc etc. I’m just going to approve all pending comments. Either a comment will really somehow have to grab my attention for me to respond, or email me directly or keep trying to get my attention on anything you want my attention on 🙂 Cheers.

  365. @Squid 2013/03/21 at 1:40 PM

    There is nothing wrong with warmer things heating cooler things, but the atmosphere is generally cooler than the Earth and a warm atmosphere heating a cooler ground is not what the GHE is about in any case. Tell you Dad that q = k*(T_hot – T_cool), when T_hot = T_cool, means that q = 0, and never that T_hot increases itself in order to heat T_cool.

  366. Simon Conway-Smith says:

    TPG: Just ask someone; when they buy something for 3$ and hand over a $5 note, they get $2 in change. How much do they now have? By the cooler heating warmer ruse, they would now have $7!! Absurd isn’t it? When heat energy is emitted, it is not replicated, but LOST! Any back radiation, even if possible/real, is a fraction of that emitted. I offer the very simple maths calculation that back-radiationists get so very wrong, which is if 1 unit of heat energy is radiated from the surface and if 0.x is radiated back, then the net heat energy remaining at the surface is -1 + 0.x, i.e. cooler, and NOT +1 + 0.x (warmer). To clsaim +1 +0.x is to magically create energy out of nothing. You do wonder how so many ‘scientists’ can make such a fundamental maths error.

  367. Simon Conway-Smith says:

    JiF: The problem with the Moncktons of this world, however much we admire their resolve and ability to take apart and demolish the alarmists case (and long may they do so, and this is no way a criticism of their ability to do so), is that by accepting that there is some level of back-radiation/GHE, it provides a legitimacy to the alarmists case. It’s this false legitimacy, or pseudo-science, that needs to be demolished, as with that done, there is no case to be made – at all.

  368. @ThePhysicsGuy 2013/03/22 at 3:13 AM

    The quote:

    “Radiative energy doesn’t care where it is coming from or going go, or what the temperature is on either end. If I light a candle on the earth during the day, the sun ends up warmer than it would be if I didn’t light the candle. Of course the reverse is true as well, the candle ends up warmer than if there were no sun. Since NET heat flow is from the sun to the candle, no thermodynamic laws are broken … but that doesn’t mean that the light from the candle is not absorbed by the sun. It is definitely absorbed, and the sun ends up warmer because of that radiation. Physics. Don’t leave home without it.”

    assumes the intended result to be true in the first place, and so is logically, mathematically, and physically meaningless. One can not just violate the laws of thermo and hide it, and then just forget you did the deed…the deed was done! lol A torch shone another torch does not make either or both torches shine brighter! A candle does not burn brighter in the day time vs. the night time (due to solar energy “adding” to it).

    The first sentence is correct. Everything else is wrong. Look at this equation: a = 1/m * (F2-F1). F2 and F1 exist, and are real, and “press” in the direction they’re intended, but if they’re equal and opposite, they DO NOTHING. Isn’t it amazing? A real application of force, and a real expenditure of energy maintaining that force, can result in NOTHING happening.

    Now lets look at a similar equation: q = k*(T_hot – T_cool). If T_hot = T_cool, NOTHING happens, i.e. q = 0. Is it not amazing? Real temperature, with the real energy it represents, causes NOTHING to happen if they contact each other. NOTHING can happen even though energy is present and is interacting with another entity.

    In these equation, the only action that WILL and CAN occur, is the action from the stronger to the weaker, and, never does the stronger need to become stronger in order to affect the weaker – this idea is nowhere in physics…NOWHERE. The presence of a force differential does not mean that the stronger force will naturally become stronger as it affects the weaker force – it just affects the weaker force.

    Now another one: q = A*s*(T_hot^4 – T_cool^4). This works the exact same way. T_Hot does not need to become hotter to warm the cooler thing. If the temperatures are equal, the radiation DOES NOTHING. It is just as simple/amazing as the other examples: the presence of energy, this time in the form of radiation, can cause NOTHING to happen because there is no force/temperature/energy differential. By ignoring this, you can create arguments to violate thermo with radiation even though the physical analogues are strictly never considered possible of doing so.

    You said: “If a cooler body can warm up a warmer body, then that “warmed up” warmer body will in turn warm up the cooler body (the cycle continuing without end) violating both the First and Second laws since one would be creating energy out of nothing. And if a cooler body could indeed warm up a warmer body, then by the same logic, bodies of equal temperature would warm each other up because they are after all, receiving energy from each other. If this were indeed true, it would seem there would have been precise lab experiments performed by physicists demonstrating this amazing phenomenon.”

    Thank you. Exactly. Thank you for being rational.

    The quote from Pielke is identical to the “shells game” which has been discussed at length here. It is just as wrong and for the exact same reasons. He also simply does not comprehend the difference between low-power for 24hrs vs. high power as a cosine function into a non-linear system etc etc.

  369. @Max 2013/03/22 at 9:29 PM

    “Why isn’t this being engineered and utilized?”

    Indeed. Make it work people…demonstrate it!

  370. Kristian says:

    John in France says, 2013/03/22 at 3:20 AM:

    “So why the difficulty in getting the message through to people who are evidently not the morons [Postma is] calling them? Don’t forget we are speaking not only of people like Anthony and Willis, but Jo Nova, David Evans, Richard Lindzen, Chris Monckton, to name but very few. These people have been active in debunking the global warming / anti CO2 brigade for how many years now? Like it or not, we’re on the same side.

    The nagging question is why do Watts and his ilk so aggressively defend such an untenable theory, displaying the same attitudes and indulging in the same exclusion tactics as the pseudo scientists they decry?

    I suspect that the key is to be found in Monckton’s approach. Most of his arguments are based on the figures actually given by the IPCC and other alarmist groups; he then proceeds to take them apart and show that even by these low standards they do not in any way stack up and therefore CO2 induced global warming is no cause for concern.

    I think that adherence to the GHE at least gives them some common ground to engage with alarmists (and to show outsiders they are not cranks) and our rejection of that common ground in their eyes takes it away from under their feet.”

    Very well put! I think you’ve hit the nail on its head here. Tallbloke is NOT a stupid man. I am quite certain he knows his ‘official’ position on this particular subject is totally and utterly mistaken. The logical/physical flaws in his argument are simply too ridiculously elementary. But, simple and harsh truth of the matter, he feels he has to ‘adhere’ to at least the theory behind the radiative GHE not to get marginalised, losing all his ‘credibility’ and ‘impact’ as a profiled skeptic blogger.

  371. Alan Siddons says:

    Arfur’s cup of water example is apt and also hints at the arithmetic of radiant energy transfer. If 200 W/m² is radiating onto a blackbody that’s independently radiating 100, that body will proceed to radiate 200 W/m², of course, not 300. Transfer is accomplished by DIFFERENCE, not by addition. Similarly, then, if 100 W/m² radiates onto a blackbody that’s independently radiating 200, nothing happens. Otherwise you’d have an example of lesser energizing greater and a magical mutual heating cycle would commence: energy ex nihilo.

  372. Greg House says:

    Kristian says, (2013/03/23 at 7:30 AM): “Tallbloke is NOT a stupid man. I am quite certain he knows his ‘official’ position on this particular subject is totally and utterly mistaken. The logical/physical flaws in his argument are simply too ridiculously elementary. But, simple and harsh truth of the matter, he feels he has to ‘adhere’ to at least the theory behind the radiative GHE not to get marginalised, losing all his ‘credibility’ and ‘impact’ as a profiled skeptic blogger.”
    ===============================================================

    I do not know, what Tallbloke knows, but generally, if someone knows that “greenhouse effect” is BS and nevertheless promotes this BS, why not assume that the image of a “skeptic” is just designed to fool people more effectively?

    Their message is like “look, we are skeptics, and if we skeptics say that the GHE is real and CO2 does produce warming, then you can be sure that it is correct”.

  373. sunsettommy says:

    Settled science: New paper questions usefulness of core IPCC radiative forcing concept
    http://hockeyschtick.blogspot.com/2013/01/settled-science-new-paper-questions.html

  374. sunsettommy says:

    Man-made global warming theory is falsified by satellite observations
    http://hockeyschtick.blogspot.com/2013/03/man-made-global-warming-theory-is.html

  375. Nice…yes that one has been discussed here too I think.

  376. A C Osborn says:

    Greg House says:
    2013/03/23 at 8:15 AM Do you think Tallbloke could be playing devil’s advocate?
    As at times he seems to jump the fence and say something against the Back Radiation and then jump back again and says something supporting the current argument.
    Have you seen the latest TFP experiment, to my thinking it disproves back radiation and to their minds it proves it?

  377. Truthseeker says:

    Joe, you say …

    “OK I’m crazy busy here in India testing spacecraft …”

    So you ARE a rocket scientist! 🙂

  378. Hi Truthseeker,

    I don’t actually do rocket propulsion design. Rocket propulsion is an old problem and has largely been figured out…it’s not really that complex. I work with the things that go on top of the rocket 😉

    hehe

  379. Simon Conway-Smith says:

    What, like how the aerodynamics are affected when Michael Mann is tied to the nose-cone, or the heat transfer equations when he falls off back to earth from the troposphere (now *that’s* back-radiation)?

  380. Greg House says:

    A C Osborn says, 2013/03/23 at 1:19 PM: “Do you think Tallbloke could be playing devil’s advocate?”
    ============================================================

    Well, I suggest you read my previous comment here (2013/03/19 at 7:13 AM) and the screenshots I posted there, in case you missed it.

    Generally, I would divide warmists (meaning this term technically) into victims and perpetrators. Even intelligent people sometimes lack critical thinking ability and fall into demagogic traps, particularly when appeals to authority are involved in the argumentation, so, it does not seem right to me to classify them automatically as liars or morons. Often I am not sure who I am talking to, but Tallbloke does not seem to be a victim to me.

  381. Kristian says:

    Greg,

    I’ve now read your original post on Tallbloke’s, the one you linked to, and it’s very clear you made a simple mathematical mistake in you step 3). Tallbloke is actually quite right in pointing it out to you. His stepwise back-and-forth sequence of energy/heat exchange between planet and shell until the former has asymptoted to a steady-state temperature of ~302K and the latter to a temperature of ~254K is in fact internally consistent. Its only problem is that it doesn’t describe a real, physical process. It is merely a case of mathematical trickery.

    The main absurdity in his approach lies in the fact that he simply postulates that the shell, upon reception and absorption of the original 235 W/m^2 from the planet, does NOT acquire the corresponding emission temperature of 254K, but rather one of 213K. Why? Who knows why? But one explanation would be that he thinks that half the received energy (235 Joules per second per square metre) is somehow absorbed at the inner surface and the other half at the outer surface of the shell (probably some unspecified conductive property (whatever conductive thermal resistance might be doing in a model trying to emulate ‘the radiative GHE’ in the first place) lets through half and only half the amount from one side to the other). Either way, effectively what he does is splitting the received and absorbed power density flux (235 W/m^2) in two. 235 W/m^2 comes in. 117.5 W/m^2 goes out. His reasoning? The shell has two surfaces. Even though one is not facing a cold reservoir at all, only a hot reservoir. Can you believe that?! That’s the entire argument. That’s what we’re up against.

    To be clear: Only reflection serves as radiative insulation. Conductive insulation serves as … conductive insulation. Conduction holds no part in ‘the radiative GHE’. It takes part in the conductive/convective ‘GHE’, or rather ‘the real atmospheric effect’, where the atmospheric mass weighs down upon the surface, working to suppress convective cooling fuelled by the incoming radiation from the Sun. Did Willis Eschenbach ever state that the shell to some extent reflected incoming radiation, or that the shell exercised some sort of conductive resistance of any significance to the heat flow through the shell to space? If he did, he would not have modelled ‘the radiative GHE’ as is. If he did NOT say that, but rather the opposite, then he WOULD have modelled the concept behind ‘the radiative GHE’. But then there would be no restriction to the energy flow from core planet to space. And hence, his model would’ve proved that the radiative GHE does not work. It is zero.

  382. Greg House says:

    Kristian says, (2013/03/24 at 12:22 PM): “Greg,
    I’ve now read your original post on Tallbloke’s, the one you linked to, and it’s very clear you made a simple mathematical mistake in you step 3).”

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

    The math in my step 3 was 600+1200=1800, no mistake.

    If you mean physics, however, then you should have read my next comment where I addressed his point: http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46135. You were active on that thread, I wonder how you could have missed it.

    I also suggest you start thinking about why he allows you to disagree with him, but blocks my comments. Critical thinking on, please. And about why he lied straight to my face saying that I admitted being wrong.

    OK, I’d better tell you just now why your comments are welcome there, since I do not really expect you to get it: because your comments are… you know… they will do no harm to warmists.

  383. I see both Greg and Kristian making valid points.

    The cycling of energy back and forth to asymptote at the desired value is baseless and senseless. Early on here we discussed the math of that sequence and proved it to be physically impossible and logically meaningless. Of course this doesn’t stop anyone from believing in it because the goal of pretending a GHE is the intention – not actually doing a critical analysis of anything.

  384. Kristian says:

    Greg House says, 2013/03/24 at 8:12 PM:

    “The math in my step 3 was 600+1200=1800, no mistake.

    If you mean physics, however, then you should have read my next comment where I addressed his point: http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46135. You were active on that thread, I wonder how you could have missed it.”

    Greg, I’m not going to get into an argument about this one with you. We both agree his exercise is absurd. Your 600+1200=1800 should simply be 600+800=1400. That’s it. You can’t add 600 to the 400 you added in the first round. The 400 has now become 600. You have to remove the 400 and replace it with the 600. Then you get 600+800=1400 in the next step. Silly, I know, but mathematically (or like you say, ‘physically’, correct).

    If you look at it this way: The planet initially sends out 800 W/m^2. This represents radiative cooling. It is replaced by another 800 W/m^2 from the nucleus. Then it receives 400 back from the shell, absorbs it and reemits the 800 PLUS the 400. Equals 1200. Now the 1200 W/m^2 is the radiative cooling flux, but upon leaving the surface it is only replaced by the next ‘pulse’ of 800 W/m^2 from the nucleus. You might say the 400 coming in is still there. But in that case, there will only be an increase in the incoming flux from the second to the third step of 600-400=200 W/m^2. Again, 200+1200=1400. 1400 is the number. Not 1800.

  385. Greg House says:

    Kristian says (2013/03/24 at 11:30 PM): “Greg, I’m not going to get into an argument about this one with you. […] Again, 200+1200=1400. 1400 is the number. Not 1800.”
    =================================================================

    Well, as I already told you, I answered that (http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46135), but I do not see any sign that you read it. Your radiation arithmetic ignores physics. According to physics, bodies radiate according to their temperature. To justify your and Tallbloke’s approach, it would require a constant increase in temperature of the planet, then decrease, then increase again, then decrease again and so on, always a drop to the initial temperature. To radiate every time (the initial 800) + (an amount equal to back radiation from the shell) requires the temperature to drop to the initial level every time, there is no way around it. I would say, if it works this way, it is certainly a new word in physics.

    And look at that, Tallbloke did indeed say this “new word”! Here: “yes, the temperature instantly drops when emission occurs from a blackbody with no heat capacity.”! (http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46160)

    So, as you can see, your hypothetical cyclical process of warming by back radiation inevitably leads to an absurdity, one way or another. This proves that your hypothesis is wrong.

    P.S. In case you do not understand why a hypothesis is wrong if it is proven to lead to an absurdity, you might need to read about “reductio ad absurdum”.

  386. Greg I think that Kristian was simply explaining the math of their process, while maintaining in the end that it was baseless – not agreeing with it.

  387. Greg House says:

    Joseph E Postma says (2013/03/25 at 6:12 AM): “Greg I think that Kristian was simply explaining the math of their process, while maintaining in the end that it was baseless – not agreeing with it.”
    ==========================================================

    I have a quite different impression. Maybe he does not agree with the notion of warming by back radiation for some other reasons, I do not know, but here he clearly said, as Tallbloke did as well, that you should always add a new portion of back radiation to the initial 800, thus ignoring the rise in temperature of the planet (ironically, the rise in temperature is exactly what they are trying to “prove” with that planet/shell thing, LOL). So, I countered with “temperature jumping back and forth” argument, referring to my previous comment. He ignored it, no reaction. Now I have repeated this argument on this thread. Let us see, what he has to say about it. My guess is, he would ignore it and just stick to his arithmetic or maybe start counting protons, like apples and oranges.

  388. I think he was just trying to help demonstrate their math.
    He said: “Its only problem is that it doesn’t describe a real, physical process. It is merely a case of mathematical trickery. … Can you believe that?! That’s the entire argument. That’s what we’re up against. … Did Willis Eschenbach ever state that the shell to some extent reflected incoming radiation, or that the shell exercised some sort of conductive resistance of any significance to the heat flow through the shell to space? If he did, he would not have modelled ‘the radiative GHE’ as is. If he did NOT say that, but rather the opposite, then he WOULD have modelled the concept behind ‘the radiative GHE’. But then there would be no restriction to the energy flow from core planet to space. And hence, his model would’ve proved that the radiative GHE does not work. It is zero. …”

    Etc.

  389. So, the question of “what is the temperature of an infinite amount of heat”, is of course a trick question. The temperature of the heat is whatever you would like to define it as, and the concept of an infinite heat source/sink is a very common and useful one in thermodynamics. The point is that in thermodynamics, when modelling systems with infinite amounts of available heat energy, not even this much energy can be used to heat itself up to higher temperature! Is that not an amazing restriction of what energy and temperature can do? The GHE and the shells game throws these restrictions of thermodynamics out the window, and openly, proudly so…without ever actually demonstrating it to be true. Adherents just say things…and then assume it to be true.

    What do two objects in contact with each other do? Let’s use the shells game. The outer shell is actually in contact with the inner sphere. At first the inner sphere was “free”, then we came by with this cold shell and stuck it over the sphere. What happens to the sphere? What happens to the shell? Does the shell cause the sphere to heat up because it “blocks cooling”? No, two objects in contact, one hotter and one cooler, simply causes the cooler object to heat up. That’s all that would happen to the shell – it would heat up to the temperature of the surface of the sphere (assuming the thickness of the shell was negligible). Back-conduction occurs entirely analogously and truthfully as back-radiation, but, back-conduction doesn’t cause cold to heat hot! You can’t just say that back-radiation exists but not back-conduction…although they will. You could recreate the entire GHE argument with the shells game using back-conduction. But we all know intuitively that nature does not behave this way. We also know that two torches shone on each other don’t make each other shine brighter – a fundamental premise of the GHE and the shells game!! – but, in this case rational and every-day knowledge is discarded for the religious precepts that the GHE and CO2 vilification offers instead.

    It doesn’t need to be repeated – the shells game GHE is absurd, mentally retarded, grotesquely sophistical, and rationally intellectually offensive. It is amazing and disgusting that humans can be this stupid. I am always reminded of Hanlon: “Never attribute to malice that which can be adequately explained by stupidity, but don’t rule out malice.” But hey…it’s not like we don’t have thousands of years of historical precedents of the stupidity of man. Just look at religion… It’s amazing we survive at all…really, don’t you think?

  390. Greg House says:

    Joe, the whole model drops dead immediately if you ask, why they assume that the back radiation is absorbed and re-radiated and then absorbed on the other side and so on changing temperature. There is no physical proof behind this assumption. You can as well assume any absurd thing and draw a “model” on paper. You can assume 2×2=5 and convey this message all your life long. People still apply for a patent for perpetuum mobile.

    As Willis came up with that thing on WUWT, I restricted my comment to pointing out that it is nothing else than a fictional story not worth commenting on for lack of physical basis. But, since some people stick to it, I thought “why not”. Maybe reductio ad absurdum could help. So, here we are. Of course, they do not like it and will refer to other absurdities to support what is falling apart, like Tallbloke did.

  391. SHINE A GOD-DAMNED TORCH AT A MIRROR, RIGHT UP NEXT TO IT, AND THE TORCH DOES NOT SHINE BRIGHTER!

  392. You can’t even detect a fraction of a degree change in skin temperature when standing in front of a mirror! Let alone a doubling of temperature…!

  393. Greg House says:

    Joseph E Postma says (2013/03/25 at 7:39 AM): “SHINE A GOD-DAMNED TORCH AT A MIRROR, RIGHT UP NEXT TO IT, AND THE TORCH DOES NOT SHINE BRIGHTER!”
    ==========================================================

    What about 2 mirrors facing each other and a torch in the middle? (lol)

  394. Hi Joe,

    I am still waiting to hear whether you are happy that the improvement to my original diagram now adequately represents your position.

    IPCC vs Skydragon

    Cheers.

    David

  395. Max™ says:

    Yeah, I was trying to bring the discussion to the back-conduction comparison and was told by tb to “stop throwing curve-balls three at a time”, and that I should stick to one point… I’m ok with that, not so much about letting someone else choose which point I get to argue though.

    Been taking screenshots of my unedited posts when I get a chance, or resurrecting them with lazarus, but anyway, enough about that.

    The shell can not cool through radiation emitted towards the planet, so the shell can not warm the planet up, simple enough, right?

  396. Check your email David…I can’t access that link from the server I’m going through (it blocks almost everything). Send me the image via response to email and I will post it and comment.

  397. squid2112 says:

    What about 2 mirrors facing each other and a torch in the middle? (lol)

    Under the GHE scenario/principals, would this not create a runaway optical feedback? Please don’t try this at home or it could lead to a rift in the space-time continuum and destroy the universe!

  398. Oh (David) all I needed to do was use your link and insert it as an image link…will repeat here:

    IPCC vs Skydragon

    So yes that is probably all that needs to be shown. Just think for example if the shell and core were touching – you would still have exactly what you’ve shown on the RHS. Separating the shell and core with a gap doesn’t suddenly totally transform thermal physics and thermodynamic behavior and laws. I mean could you imagine if, when two objects of the same temperature are touching, nothing happens, but then if you separate them by a small distance, one of the sides starts spontaneously heating itself? Wouldn’t that be great, a wonderful discovery, a simple physical fact all young boys (and girls) leaned on their own at some point fiddling with stuff? Radiation is simply conduction at a distance, mediated in the exact same manner – just with real photons as opposed to virtual photons.

    If you could somehow incorporate the actual physically factual heat flow equation of q = A*s*(T_2^4 – T_1^4), so that q > 0 when T2 > T1, and q = 0 when T2 = T1, then this might help explain what the “loop” means. Because I can just imagine people getting upset over the loop, and so this is why it would be good to refer to actual physics and math.

  399. Because there is NO actual loop – there is in fact q = 0. If they were touching you wouldn’t actually need to show a loop, because q = 0 if they’re touching and the same temperature. Likewise the case for radiation: q = 0, and so there is no actual loop occurring. In the touching “conductive” case, yes energy is going every which way by contact, but there is no real “loop”, there is just q = 0.

    So to repeat, the loop isn’t real. Or, it is as real as it would be if the shell and core were touching. So you see the potential problem there? People will look at the diagram and take it seriously as gospel truth and then freak out about it, when in fact the “loop” was never meant to indicate anything actually occurring at all, because q = 0.

  400. Kristian says:

    Joe,

    I think the loop is not what people (the Escenbach followers) would primarily object to in the above diagram. I’m quite certain they would rather object to the red arrow beside the loop, going from the core to the shell. That one is in fact what this argument is all about. They claim that the red arrow would be zero (or not be there at all) in the no-temperature-gradient scenario. In their minds it could only be 235 W/m^2 if there were a temperature difference between the core and the shell of (in this particular case) 48K (302-254K). That is their central claim.

    I’ve tried to explain it to them using a fairly simplistic approach. To no avail. They seem (quite) unwilling to embrace any such concept: Apparent heat transfer when q is supposed to be 0.

    Could you perhaps make an attempt?

  401. ThePhysicsGuy says:

    Hi Joseph,

    I’ve seen claims that the operation of a “microbolometer” (special thermal imaging camera) proves that cool objects can indeed warm up warmer objects. See article:

    http://www.climateandstuff.blogspot.co.uk/2012/05/cool-body-can-transfer-measurable-heat.html

    So from what I understand, the claim is that the microbolometer sensor plate is “warmed up” by objects as low as -40 degrees C (the lowest temperature detection limit of the device), and the sensor plate is at ambient temperature (i.e., it is un-cooled). I am completely skeptical, but don’t know enough about the device (or enough about thermodynamics) to see how the wool is being pulled over my eyes. Are you familiar with this claim and/or the device, and could you provide comment?

    Thanks.

  402. Max™ says:

    >..>

    ……………………………………..________
    ………………………………,.-‘”……………….“~.,
    ………………………..,.-“……………………………..”-.,
    …………………….,/………………………………………..”:,
    …………………,?………………………………………………,
    ………………./…………………………………………………..,}
    ……………../………………………………………………,:`^`..}
    ……………/……………………………………………,:”………/
    …………..?…..__…………………………………..:`………../
    …………./__.(…..”~-,_…………………………,:`………./
    ………../(_….”~,_……..”~,_………………..,:`…….._/
    ……….{.._$;_……”=,_…….”-,_…….,.-~-,},.~”;/….}
    ………..((…..*~_…….”=-._……”;,,./`…./”…………../
    …,,,___.`~,……”~.,………………..`…..}…………../
    …………(….`=-,,…….`……………………(……;_,,-”
    …………/.`~,……`-………………………………./
    ………….`~.*-,……………………………….|,./…..,__
    ,,_……….}.>-._……………………………..|…………..`=~-,
    …..`=~-,__……`,……………………………
    ……………….`=~-,,.,………………………….
    …………………………..`:,,………………………`…………..__
    ……………………………….`=-,……………….,%`>–==“
    …………………………………._……….._,-%…….`
    ……………………………..,

  403. Max™ says:

    “Vacuums are good insulators. That’s why Mr Thermos got a patent.” ~tb

  404. @Kristian @ 2013/03/25 at 11:35 AM

    So, they reject the red-arrow because if the temperatures are equal then there is no heat transfer (which would be correct), but then they say that self-radiation or radiation from a cold source can heat up something warmer (which is mindbogglingly incorrect). There is no consistency anywhere in their thoughts. At one point they go nuts strictly following a definition of heat transfer, and in the next they completely reverse the directionality of heat flow. If they obey q = 0 when there is no gradient, why do they disobey the directionality of q when there is a gradient, such that the hotter side of the gradient heats up!!!?

    Look at the top of the shell – it is emitting 235. Therefore, it can not be strictly in thermal equilibrium with the core. QED. It is as simple as that. Their starting definition by which they can ignore the red 235 is incorrect, and obviously so, but they will want to hold to that strict definition so that they can create an alternative instead. As Max has been pointing out, the shell has a larger area than the core – even if it is quite thin. No matter what, the shell has larger area than the core, hence, there HAS to be a small (or large) difference in temperature between them, a small gradient. And even despite that, if the shell were a wall and the core a wall, the shell IS losing 235 on the outside, and hence this will create a differential between the core and shell, even if surface area wasn’t a factor. And this all happens continuously of course – exactly as 235 is being emitted from the shell, 235 is being filled back in to the shell. What if the shell and core (either as spherical or plane walls) were touching? It would still apply exactly as described here.

  405. @ThePhysicsGuy @2013/03/25 at 1:18 PM

    When these devices detect “cool”, it is a LACK of signal that they detect, a “negative voltage”. The devices are calibrated in this way. The cooler the object, the more negative the “q” is between the camera and source. The hotter the object, the more positive “q” is between the source and camera (positive q when the source is hotter than the camera, and vice-versa). Their interpretation is a lie.

    Does standing next to a wall of ice heat you up? Does standing next to a mirror heat you up? Nope.

  406. Kristian says:

    Joe, you said: “(…) if the shell were a wall and the core a wall, the shell IS losing 235 on the outside, and hence this will create a differential between the core and shell, even if surface area wasn’t a factor. And this all happens continuously of course – exactly as 235 is being emitted from the shell, 235 is being filled back in to the shell. What if the shell and core (either as spherical or plane walls) were touching? It would still apply exactly as described here.”

    You’re right of course. And this is what I have been trying to tell them. But they simply seem oblivious to the green arrow going out to space from the top plane (outer surface of the shell) in D. Socrates’ diagram. They seem to think that this should be zero in the event of planet and shell attaining equal temperature, because the apparent heat transfer between them at that point would have reached 0. And so the planet’s surface in their mind would no longer be able to shed the constantly provided heat input from the planet’s nucleus. In other words, they seem to think that all that matters is the rate of heat transfer between planet and shell, that the rising temperature of the shell does not matter at all, that as the heat transfer from planet to shell grows smaller, the temperature of the shell also grows higher, closer to that of the planet (that’s why the heat transfer is slowing down). And by doing so, they willfully (?) neglect the exceedingly basic fact that the shell having a temperature would have to radiate accordingly to space. And how would this constant cooling flux be replenished if not from the core planet?

  407. Joseph E Postma says: 2013/03/25 at 7:21 PM
    When these devices detect “cool”, it is a LACK of signal that they detect, a “negative voltage”.
    ———–
    Look at the drawing for a bolometer. There is an IR absober. It has IR only focussed on it (the camera lens is made from germanium – which is opaque to visible light – I have added the transmission property to the post indexed above). The IR absober temperature is measured and the this value is output to the video processor.
    Can you please explain what you mean by “detect cool” I keep stressing there are no cool rays there is only Thermal radiation (between 2u and 15um) that can get through the lens. This thermal energy adds to the energy from the ambient conditions and changes the temperature of the IR absorber – a -273C temperature adds no energy, a -20C adds more and the absorber warms above the temperature that would have occurred at -273C. a source of 1500C (within the range of the cameras calibrated sensitivity) adds even more energy so the absorber is warmer still.
    Whatever the temperature of the object that is focussed on the absorber (above -273C) energy is added to it and its temperature rises (it is not that -273C cools the object because that would require cooling rays to be focussed)

    ——-
    Does standing next to a wall of ice heat you up?

    No but it heats you up more than standing next to a wall of frozen oxygen,
    ———–
    Does standing next to a mirror heat you up? Nope

    You need to tell the manufactures of survival bags made of aluminized plastic, and the manufacturers of thermos flasks, this rather outstanding fact! IR normally lost to ambient will be reflected back and will reduce the loss of IR from your body

    .

  408. Don’t be a complete sophist moron Prefect – back-radiation doesn’t heat you up and thermos flasks etc. don’t generate increasing temperatures. Yes, there is some process going on with IR detectors, just as there is with any kind of radiation detector. But just because you can convert radiation into a signal doesn’t mean it heats up hotter things! Your own radiation from the mirror doesn’t heat you up, for example. You can detect “cold” with a thermometer or even your hand – this does NOT mean that the cool thing is heating your warmer hand or warmer thermometer, lol.

    What I should have said is that I don’t know specifically how IR detectors work, but what I do know is that they do NOT work by cold heating hot – hahaha what a stupid idea.

    Why doesn’t standing NEXT to an aluminium blanket warm you up? Crickets. Why does such a blanket only warm you up if you’re wrapped in it and the breeze is sealed off from getting inside? Because it is just trapping warm air, like a normal botanist greenhouse, and this has nothing to do with the atmospheric GHE. The behaviour of these survival blankets is almost perfect proof that the atmospheric GHE doesn’t exist and that backradiation doesn’t heat itself up. Get this crap figured out before you start spewing such vomitous mental garbage around here. You really don’t know how a thermos works? It is because convective and conductive loss to the outside flask is reduced to a minimum, occurring only at the inner-cylinder contact-point with the outer cylinder at the lip, because between the inner flask and the outer there is a vacuum, which can’t conduct or convect heat away from the surface of the inner flask.

    Never has a thermos heated itself up with its own contents so it makes ZERO sense why a thermos has EVER been presented as evidence supporting the GHEa thermos has NEVER, ever, anywhere, at anytime, in the history of thermoses, from here to the moon, heated itself up. A thermos thus proves that the atmospheric GHE doesn’t exist because a thermos has NEVER EVER done what is claimed it is analogized to do in the atmosphere! But hey why should we expect GHE faithers to actually have a rational comparison of something in reality with which to actually compare the GHE? That is how stupid such people are – that they will use an example of something WHICH DOES NOT DO SELF-HEATING to then pretend that the non-existence of self-heating proves self-heating. Mentally freaking retarded…

  409. Simon Conway-Smith says:

    I saw ‘thefordprefect’s comment and immediately thought “oh no, he’s placed himself right into Joe’s 12-bore missile target area” 🙂

    As explained to a professor friend of mine (who believes Hansen is an upright scientist, and Houghton is a saint), that the ‘blanket effect’ is a HUMAN understanding of “warming” and not the PHYSICS (i.e. scientific) one. The human understanding needs qualification and explanation – you do *feel* warmer, but that is RELATIVE to the temperature you would otherwise have been without it, as the blanket (thermos) reduces the RATE of heat loss by reducing conduction and convection.

    It really is very simple!

  410. “right into Joe’s 12-bore missile target area” – LOL, nice, love it.

  411. Greg House says:

    thefordprefect says (2013/03/26 at 6:53 AM): “Does standing next to a wall of ice heat you up?No but it heats you up more than standing next to a wall of frozen oxygen,”
    ==========================================================

    “A wall of ice dies not heat you up, but it heats you up more than…” is a contradiction.

  412. Greg House says:

    Joseph E Postma says (2013/03/26 at 7:25 AM): “Yes, there is some process going on with IR detectors, just as there is with any kind of radiation detector. But just because you can convert radiation into a signal doesn’t mean it heats up hotter things!
    ===============================================================

    Joe, I think this IR-Thermometer example should be dealt with properly, because it might sound convincing to unprepared people, including politicians and journalists. Talking to this audience we can not really argue by quantum mechanics or even by a mirror example, it would be not enough, we need to address this issue directly.

  413. Fred Kilger says:

    To maintain the radiation of 235 W/m^2 to space, these models assume (without any reason) a constant radiation source (fed from an unlimited energy source) inside of the planet.
    This constant flux source forces (in a stationary state ) constant radiation i_PH (Planet -> Shell) = i_HS (Shell -> Space), or (SB:) T_P^4 -T_H^4 = T_H^4 –T_S^4 -> T_P^4 = 2* T_H^4 –T_S^4.
    The temperature of the planet is always greater than the temperature of the shell. Putting more shells around the planet increases his temperature to arbitrary levels.

    Only the (unrealistically constructed) constant radiation source (inside of the planet) determines in these models the temperature of planet and shells.

    A constant radiation source can be applied to suns but never to real planets.
    A real planet can only radiate energy out of the energy storage of ground, sea and atmosphere. This system is never in a stationary state, all temperatures, radiation fluxes are functions of time. Radiation of energy out of a storage element always lowers the temperature of this element. Constant radiation sources do not exist.

  414. squid2112 says:

    @thefordprefect says:

    You need to tell the manufactures of survival bags made of aluminized plastic, and the manufacturers of thermos flasks, this rather outstanding fact! IR normally lost to ambient will be reflected back and will reduce the loss of IR from your body

    Ah, but it did not heat anything, as that would require additional energy to enter the system, and since you cannot simply “create” energy from nothing, the reflective properties of the aluminium plastic did NOT in fact heat anything!

  415. squid2112 says:

    Further on the aluminum plastic blanket. If what TheFordPrefect is saying were true, then I could simply wrap my house in aluminum plastic during the winter and turn my heater off, but being careful not to burn my house down because of the continuous heating that the blanket would provide. Seems like another rift in the space-time continuum coming…..

  416. …rift in the space-time continuum… lol love it.

  417. A thermos flask contents remains warm because of the vacuum stopping conduction, the flask stopping convection to ambient, and the reflective inner container reflecting the radiation from the glass (the reflective material is 1 glass wall away from the liquid, which will therefore be cooler than the liquid) back to the liquid.

    Assuming the “space blanket” is not in contact with the skin it too reflects IR back to the body It is a surface coating so will be reflecting at the same temperature as the body. neither of these 2 examples are similar to the iron greenhouse!

    The microbolometer is extremely simple in operation. It is warmed by ambient or a temperature controlled heater. It looses heat to ambient. It gains heat from whatever is focussed onto the IR receiver plate via the germanium lens+ power loss in the readout circuit+power from any heater+ heat from other sources in the camera etc.
    It periodically calibrates itself against a standard plate at fixed temperature. Thus the internal sources of heat are known. The only variable is the IR (2um to 13um) coming through the lens

    All this IR can do is to add to the energy hitting each cell, it cannot subtract. It may be coming from a 1500degC source or from a -200degC

    Perhams section 2.3 of this doc will help?
    http://download.ebooks6.com/Uncooled-Carbon-Microbolometer-Imager-download-w43166.pdf

  418. Apologies strike out
    “from the glass (the reflective material is 1 glass wall away from the liquid, which will therefore be cooler than the liquid) back to the liquid. ”
    The radiation reflected will be from the liquid and will be at the same “temperature”

  419. Rosco says:

    I just love the way people make all sorts of stupid claims about “trapping” radiation – just making stuff up and ignoring centuries of proven and demonstrable science for – well – stupidity.

    In an atmosphere energy transfer by radiation is trivial compared to diffusion/conduction to the air and convection – surely the well documented rate of cooling shown for the Moon’s surfaces demonstrate this indisputable fact ?

    These so called “radiation” blankets act primarily by preventing contact with a convecting atmospher – trapping already warmed air.

    I have yet to see a rock or any other inaminate object heat itself up by trapping radiation – or even by excellent thermal insulation – without an energy source all that happens is that objects tend to reach an equilibrium temperature/energy level with the surrounding environment.

    You could hardly claim any animal fur could possibly be considere any sort of radiation “trap” yet the same appears to be especially effective at reducing convection of the warmed air layer trapped within the fur.

    Or, perhaps it is the animal’s lack of understanding about radiation which keeps them feeling warm when in fact they are radiating all their warmth away and dying right before our eyes ?

  420. Rosco says:

    Really – who cares about the infinite number of different thought bubbles that are being submitted as “proof” that something that is absurd is real.

    Accept Willis arguments are real. Add another shell and do the math.

    A “new” shell around his original is exactly what he prescribes – a shell capable of radiating 235 to space and 235 backradiation.

    Do the arithmetic as he prescribes and then add more and more until you arrive at infinite temperature of the original planet.

    This obviously is absurd so the original proposition is absurd.

    QED.

  421. Prefect said: “neither of these 2 examples are similar to the iron greenhouse!”

    No shit. Which is why they shouldn’t be used as an analogizing example of any GHE! But, they’ll continue to do so anyway. Just amazing.

    An IR receiver does not function by cold heating hot. Interpreting it that way is pure sophist BS.

  422. That’s WHY these people and their ideas exist – because they’re so absurd! The more absurd they get, paradoxically the more people believe them, because people are so retarded. Hitler knew this all too well:

    “Make the lie big, make it simple, keep saying it, and eventually they will believe it.”

    “What good fortune for governments that the people do not think.”

    “The receptivity of the masses is very limited, their intelligence is small, but their power of forgetting is enormous. In consequence of these facts, all effective propaganda must be limited to a very few points and must harp on these in slogans until the last member of the public understands what you want him to understand by your slogan.”

    “If you tell a big enough lie and tell it frequently enough, it will be believed.”

    “The great masses of the people will more easily fall victims to a big lie than to a small one.”

    “Through clever and constant application of propaganda, people can be made to see paradise as hell, and also the other way round, to consider the most wretched sort of life as paradise.”

    “But the most brilliant propagandist technique will yield no success unless one fundamental principle is borne in mind constantly and with unflagging attention. It must confine itself to a few points and repeat them over and over. Here, as so often in this world, persistence is the first and most important requirement for success.”

    And one from Einstein:

    “Two things are infinite: the universe and human stupidity; and I’m not sure about the universe.”

    In other words, it is just as I have been saying. When confronted with the truth, GHE faithers just get even more stupid as a strategy to perpetuate the lie! Just keep on repeating it, and repeat the idea in stupider and stupider ways that make less and less sense. Such as a thermos being an example of the GHE when a thermos has NEVER warmed itself up! Do you see how stupid and retarded that is? To offer as proof something which does not even do what is claimed? I mean you got to give it to human stupidity at some level – it is a winning strategy. It’s what the controllers of society count on…and they’ve been quite successful. It also makes them pathetic of course, to find self-value and self-identification in controlling extremely stupid people. How sad and pathetic. Yah…a thermos is proof of the GHE – how RETARDED do you have to be!?

  423. Max™ says:

    ^ ___________ ^ ___________ ^
    1 ___________ 100 _______ 234.99
    Shell ________ Shell _______ Shell
    1 ___________ 100 _______ 234.99
    v ____________ v __________ v
    | ____________ | ___________|
    ^ ____________ ^ __________ ^
    235 _________ 235 _______ 235
    Core_a _____ Core_b ____Core_c

    Core_a loses 234 to shell, 1 to space, total out = 235
    Core_b loses 135 to shell, 100 to space, total out = 235
    Core_c loses 0.01 to shell, 234.99 to space, total out = 235

    In the last step, the shell is unable to cool through radiation emitted back towards the core, so it can only cool by emitting to space.

    The core is unable to cool beyond the 0.01 difference due to the inverse-square law, so the temperatures will be very similar.

  424. Joe,

    Re. your comments of 2013/03/26 at 8:31 PM above, I think you are fighting a straw man.

    As we witness everywhere, blogs are full of awkward people who seem to take a delight in talking rubbish. Fighting them is a waste of time and achieves nothing. The people that climate skeptics have to persuade are the uncommitted citizens who have been told by their governments that there is a very serious problem that needs addressing. That is why I am interested in finding simple ways to explain complex issues. I don’t subscribe to Hitler’s or Einstein’s view that the general public is stupid. Ignorant of science maybe. But not ignorant of everday life.The best way forward is by reasoned argument with reasonable people. Surely our job, here and elsewhere, is to debate the best way to explain to ‘the ordinary voter’ exactly how they have been conned over 3 decades by an unholy alliance of special interests.

    Looking at the physics is essential, but it is only part of a much bigger project, is it not?

  425. Great points David. The problem is how to utilize the techniques of propaganda for spreading truth…because you can’t just tell the truth, you have to make it hip, trendy, appeal to emotion, etc. Make it NOT look like the rational truth in other words. Make it an idea that’s just fun to have, and that makes you feel like you’re with everyone else.

  426. Mr Joseph E Postma
    If you would care to explain how a uncooled thermal imaging camera works when photographing cold objects. This would add to the understanding about how cold objects add/do not add to the energy of hot objects.

    Can you please provide an explanation.

    Here’s anothe bolometer explanation (note that this one is cooled to minimise NOISE)
    search: Detection of Light Lecture 10 bolometers
    or search this one:
    R. Westervelt Imaging Infrared Detectors II
    No where in documentation I have read does it mention that you subtract heat from the bolometer when the source is below the temperature of the bolometer – energy is always added to the bolometer.

    from the second doc.
    “Bolometers operate by sensing the temperature rise associated with the
    absorption of radiation. Bolometers are one of the oldest types of radiation
    detector, and they have many advantages. Bolometers respond to absorbed
    energy, and can be made sensitive to a very wide range of wavelengths. Because
    they sense heat rather than photocarriers, bolometers are insensitive
    to the photo carrier population and dynamics.”

  427. Alan Siddons says:

    Re: the Thermos Argument. If one asserts that the GHE works by trapping heat, then one is directly implying that the Earth radiates less energy than it gains — much like a thermos, which “bottles up” the heat it’s given. But that’s the very point, for satellites observe that the Earth radiates to space all the energy it gets from the sun. 235 W/m² or so continuously go in, 235 W/m² or so continuously go out. Obviously this is not how a thermos performs. This doesn’t bother Believers, however. Although they acknowledge that an Earth with a greenhouse effect sheds the same amount of thermal energy as an Earth without one, they argue nevertheless that the Earth is trapping heat. The Greenhouse Effect is thus like the biblical Bush That Burns Yet Is Not Consumed. Sort of a miracle.

  428. Well the first thing for you to learn is that cold things don’t heat up hot things. Try it yourself. Go around your house, the town, the country side, other countries, etc., and try to find something cold that heats up a hotter thing. Please report back on your findings. Actual findings – not just interpretations or assurances from other sources.

    Whatever the process is, you will always have the limitations of the 1st Law of Thermodynamics, and then the 2nd and 3rd etc. Cold will never heat up hot. The reading suggestions easily fall under the physically-real framework provided earlier: what is the direction of ‘q’? What you need to do is develop an explanation of the bolometer which does not make the statements that cold heats up hot or that it is a demonstration of the GHE or that it violates the laws of thermodynamics – meaning precisely that cold does not heat up hot. One thing will be clear – a bolometer is not an example of cold heating up hot, at best, it will be an example of measuring the directionality of ‘q’. Does a bolometer violate the laws of thermodynamics? No of course not, nothing does. Therefore, precisely, they are not an example of cold heating up hot. And again, what the heck does a boloemter have to do with something HEATING ITSELF UP WITH IT OWN ENERGY!? Again, this is just another retarded analogy to something which doesn’t even do what the GHE is said to do in the first place(!) – a bolometer doesn’t heat itself up with its own reflected energy, so therefore this has nothing to do with the GHE and is in fact another disproof of the GHE…

    You know one time I was told that the “bubbles from cavitation of a propeller on a nuclear submarine” were evidence of the GHE. You just have to say something retarded and then you have evidence for the GHE.

  429. Bryan says:

    thefordprefect

    Dont get mixed up between a thermopile and a bolometer.
    They work on completely different physics principles.

    A thermopile has a voltage induced if the two dissimilar materials are at different temperatures.

    A bolometer senses the change in resistance of the sensor due to temperature change.
    The sensor forms one part of an initially balanced Wheatstone bridge
    No current will flow if the external object is at the same temperature as the sensor – the null point.
    If the object is at a lower temperature; the sensor resistance loses heat and resistance goes down and the Wheatstone bridge is out of balance and a current flows through the sensor.
    If the object is at a higher temperature; the resistance increases as the temperature rises the Wheatstone bridge is again out of balance and current flows but in the opposite direction.
    If calibrated against a known temperature the current can then represent temperature.
    This change in resistance is linear near the null point but if moved too far from the null point the change in resistance is non linear and will give rise to errors.

  430. Greg House says:

    @thefordprefect

    I find the points about IR-Thermometers interesting, but at the same time we can take it closer to the GHE question. What I would like to know, is this: how much warmer gets the sensor when the IR-Thermometer is pointed to the sky, let’s say, at night? Because, you know, I would disregard the whole thermodynamics immediately, if the GHE were alleged to be like 0.000001C. Let us just clarify that before we get deeper into the issue about how it is possible that colder bodies do not warm warmer bodies but IR-sensor still detects IR from colder bodies. Let us address the core issue: how much. I am looking forward to scientifically proven numbers.

  431. Well I wouldn’t encourage the creation of more sophistry myself Greg, but I’ll let you deal with it 😉 I’ll quote Bryan’s response again because it has all the info necessary, and it basically means the thing measures the direction of ‘q’, as I said…

    A bolometer senses the change in resistance of the sensor due to temperature change.
    The sensor forms one part of an initially balanced Wheatstone bridge
    No current will flow if the external object is at the same temperature as the sensor – the null point.
    If the object is at a lower temperature; the sensor resistance loses heat and resistance goes down and the Wheatstone bridge is out of balance and a current flows through the sensor.
    If the object is at a higher temperature; the resistance increases as the temperature rises the Wheatstone bridge is again out of balance and current flows but in the opposite direction.
    If calibrated against a known temperature the current can then represent temperature.
    This change in resistance is linear near the null point but if moved too far from the null point the change in resistance is non linear and will give rise to errors.

  432. Greg House says:

    Joseph E Postma says (2013/03/27 at 9:46 AM): “Well I wouldn’t encourage the creation of more sophistry myself Greg, but I’ll let you deal with it ;)”
    ===========================================================

    Thanks Joe ;). Because, if you remember the R.W.Wood experiment, he did not engage in an endless theoretical exchange, but instead demonstrated that the whole thing is at best negligible. It was enough for the warmists back then.

    Last time I asked an allegedly knowledgeable warmist about those numbers, he suddenly did not want to discuss it any longer. Let us see, what thefordprefect can fetch.

  433. Alan Siddons says:

    Re: Bolometers and the like. I thought it was pretty simple. Warmer to cooler on the instruments induces a current so the needle registers a temperature. Cooler to warmer reverses the instrument’s current, which makes the needle register another temperature. The point is, you’re just seeing an electrical response, not a transfer of more thermal energy to less thermal energy.

  434. Loodt says:

    Hi Joseph,

    To get back to this IR meter being pointed at various objects in the house, outside, and the fridge door, I think we have to go where it all started.

    http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/

    I saw this post by Dr Roy Spencer in 2010. He placed a metal object next to his electric heater and read an increase in the temperature of heating element. There, fixed in his mind, is proof of back radiation. An Eureka moment, famous for life and for future generations, and he blogged about this new ‘Scientific”proof!

    Our good Dr is not an engineer and has never dealt with ventilation and heating problems, and evidently doesn’t know how an electric heater works.

    The hot electric metal element, heated to about 1200 to 1500 degrees Celsius is covered by a ceramic outer casing. When I grew up our heating elements had the electric wires on the outside, coiled along a ceramic core. Obviously it was too dangerous, must have started some fires, and the metallic heating element is nowadays hidden inside the ceramic cover than glows when the heater is on. If you placed a metal tube around the ceramic core, it would heat up the inner core, the metal heating element would melt, and the heater would stop working.

    This our Dr did not understand, reduce the ability of the heating element to dissipate heat and it would heat up until the metal melts.

    Ever since, we have proof of immaculate conception, and back radiation, and the disciples can out and spread the word.

    I give you further links about the musing of this esteemed Dr about this very topic:

    http://www.drroyspencer.com/2010/07/first-results-from-the-box-investigating-the-effects-of-infrared-sky-radiation-on-air-temperature/

    http://www.drroyspencer.com/2010/08/help-back-radiation-has-invaded-my-backyard/

    http://www.drroyspencer.com/2012/02/more-musings-from-the-greenhouse/

    http://www.drroyspencer.com/2012/02/yes-virginia-the-vacuum-of-space-does-have-a-temperature/

    So, my advice to you is to forget about the non-entity that is Wills, and start with Dr Roy Spencer. And being an American he can be neither a Sir, Duke, Earl, or Lord.

  435. A person that thinks that empty space has a temperature is a complete unscientific idiot. It is amazing such people can be associated with science and call themselves scientists. What did he say, that the temperature of the vacuum of space is the temperature of the CMB? What if there was no CMB…then what?

  436. Alan Siddons says:

    Correction to the above: “…you’re just seeing an electrical response, not a transfer of less thermal energy to more thermal energy.” Sorry.

  437. Greg House says:

    Loodt says (2013/03/27 at 10:22 AM): “I give you further links about the musing of this esteemed Dr about this very topic: …”
    =============================================================

    The links are dead (I wonder why…), but there is another way:

    1. http://87.248.112.8/search/srpcache?ei=UTF-8&p=first-results-from-the-box-investigating-the-effects-of-infrared-sky-radiation-on-air-temperature%2F&rd=r1&fr=yfp-t-708&u=http://cc.bingj.com/cache.aspx?q=first-results-from-the-box-investigating-the-effects-of-infrared-sky-radiation-on-air-temperature%2f&d=4677512377075159&mkt=de-DE&setlang=de-DE&w=acdjW_nG5xhPFjdP_OFleEWvs72sWiHG&icp=1&.intl=de&sig=tESnetizxiz8Uuc4W4o7ow–

    2. http://87.248.112.8/search/srpcache?ei=UTF-8&p=%2Fhelp-back-radiation-has-invaded-my-backyard%2F&rd=r1&fr=yfp-t-708&u=http://cc.bingj.com/cache.aspx?q=%2fhelp-back-radiation-has-invaded-my-backyard%2f&d=4686866851040923&mkt=de-DE&setlang=de-DE&w=Iru0kvDN3ZZTV3QO2E5VpLGtuRjL2mUC&icp=1&.intl=de&sig=be4.drVh5ok57sgc9MZhIA–

    3. http://87.248.112.8/search/srpcache?ei=UTF-8&p=http%3A%2F%2Fwww.drroyspencer.com%2F2012%2F02%2Fmore-musings-from-the-greenhouse%2F&rd=r1&fr=yfp-t-708&u=http://cc.bingj.com/cache.aspx?q=http%3a%2f%2fwww.drroyspencer.com%2f2012%2f02%2fmore-musings-from-the-greenhouse%2f&d=4774690300368764&mkt=de-DE&setlang=de-DE&w=JoqIiozDz8F7R2dHAmlc86ta4TUSQL4m&icp=1&.intl=de&sig=Iz65v9nYul9avSaB1RTrVg–

    4. http://87.248.112.8/search/srpcache?ei=UTF-8&p=Yes%2C+Virginia%2C+the+%E2%80%9CVacuum%E2%80%9D+of+Space+Does+have+a+%E2%80%9CTemperature%E2%80%9D&fr=yfp-t-708&u=http://cc.bingj.com/cache.aspx?q=Yes%2c+Virginia%2c+the+%e2%80%9cVacuum%e2%80%9d+of+Space+Does+have+a+%e2%80%9cTemperature%e2%80%9d&d=4706520609263305&mkt=de-DE&setlang=de-DE&w=0CvwrGzEBk9cR3xPdGJsw091uB0mlVxX&icp=1&.intl=de&sig=vBtIqlrReVTLTppFx6_TbA–

    I recommend saving them, just in case.

  438. A bolometer is a devive for converting radiation into heat and then measuring the effect of the heat for example by a resistance change. The bolometer does not care what wavelength it is (providing it is within its 2 to 13um range) – it can only convert the radiation to heat.
    We have a warm bolometer array (warmed from its use of power to digitise the resistance of each pixel) This will obviously be emitting as black body radiation from each pixel (it is designed that way) in all directions. Some of the pixels radiation will pass through the lens and be focussed on an object
    I think most will agree that black body objects above absolute zero will radiate energy. So whatever the temperature of the object that the radiation from the pixel hits will radiate back to that pixel. No radiation of this type can cause cooling. There is zero radiation at -273°C there is some radiation at -272°C. There is no such animal as -274°C – there is no negative radiation!
    Mr Postma the brings into play a numerical value q which depends on the temperature between the two objects. The question is where does this differencing occur.
    The pixel will warm the remote object but by a negligible amount (compare the pixel size to the object). There canl be no differencing in the space between the pixel and object. It can only occur at the bolometer.
    Assume the object is 0.3 metres away from the camera then energy leaving the bolometer pixel will not know what temperature the object is until 2ns later (2ns=2*10^-9s the time it takes to travel 2×0.3m) and similarly for the radiation leaving the object. Obviously each source of radiation does not know where it will land so the energy in transit can only depend only on the temperature of the pixel or the object.
    The bolometer is at a stable temperature so:
    energy lost [conduction, convection and radiation]=energy gained from the electronics and ambient)
    BUT this is not true! The effect of the stuff in front of lens has not been considered.
    Put a lens cap on – there is thermal radiation from the cap equivalent to its temperature.
    Point it at ice at 0°C – there is radiation being emitted at +273Kelvin.
    Point the lens at absolute zero – that would work, no radiation now!
    Absolute zero is not an option but the camera does self calibrate by inserting a warmed plate in front of the bolometer array but again this is radiating this time at a known temperature
    Consider the case when the bolometer is colder than the object then obviously the energy from the object gets focussed on the pixel and it warms – q is obviously positive
    Consider the case where the pixel is hotter than the object BUT the object is above -273°C and therefore emitting radiation. The radiation still gets focussed on the pixel. This is of course positive radiation there is no negative radiation. The bolometer does not care what wavelength of radiation is hitting it.
    Lets assume that because the pixel is hotter than the objet the object’s radiation gets rejected. But then all objects below the temperature of the pixel will get rejected and so there would be no image. This is not what happens.
    Lets assume that the cooler radiation cancels an equivalent amount of energy in the pixel (q is negative). But how does this happen there is no difference in the effect of long or short wave radiation to a bolometer it is all just energy. There is no physical mechanism for cancellation. This is not what happens

  439. Alan Siddons says, 2013/03/27 at 7:32 AM: Although they acknowledge that an Earth with a greenhouse effect sheds the same amount of thermal energy as an Earth without one, they argue nevertheless that the Earth is trapping heat. The Greenhouse Effect is thus like the biblical Bush That Burns Yet Is Not Consumed.

    Although I agree with the sentiment, you have to be careful that you are not misinterpreted. The constant flow of energy through the earth-atmosphere system from the Sun and out again to be lost to space is quite capable over time of heating the surface and atmosphere to any arbrarily higher temperature level than the surface would have if the earth had no atmosphere. It all depends on the ‘resistance to flow’ offered by the paths through which the atmosphere’s Kinetic Energy flows. In practice, it is the mean value of the atmosphere’s ‘conductivity’ that gives us the Atmospheric Thermal Enhancement and, thereby, our elevated mean surface temperature of ~288K.

    So two points:

    1. Although a thermos contains a heat source, the heat diminishes until the temperature of the contents reaches ambient. I do not think that the more sophisticated warmist scientists use the ‘thermos’ argument nowadays so criticising them using the ‘thermos’ misconception may be a bit counter-productive.

    2. Turning to the steady-state energy flow scenario, where the earth is in energy flow balance and exhibits an enhanced mean surface temperature, it is indeed the atmosphere’s ‘resistance’ to heat flow that keeps its fund of kinetic energy at an enhanced level and, therefore, at an elevated temperature. So the earth-atmosphere system is indeed ‘trapping heat’ by virtue of being a giant ‘resistive conductor’ of energy.

    I hope you are not suggesting that, in this latter scenario, the earth’s atmosphere is not ‘trapping heat’ because it obviously is (take the atmosphere away and the mean surface temperature drops to a similar level to that of the Moon). The real issue is what is the physical mechanism?

    Not GHGs because radiation is simply a transport mechanism and not a STORE of energy. It is a fallacy to assume it can ‘bootstrap’ itself into a storage role. Instead the correct answer is good old thermodynamics – specifically the resistance to energy flow up through the atmospheric column and out to space caused by the limited speed of conduction/convection of heated air and latent heat carried upwards in the water cycle.

  440. Max™ says:

    I was gonna say, it seems like someone has confused cooled and uncooled bolometers.

  441. thefordprefect – read Bryan’s answer.

  442. Greg House says:

    @thefordprefect

    You have forgotten an important scientific issue, let me repeat it. What I would like to know, is this: how much warmer gets the sensor when the IR-Thermometer is pointed to the sky, let’s say, at night? Because, you know, I would disregard the whole thermodynamics immediately, if the GHE were alleged to be like 0.000001C. Let us just clarify that before we get deeper into the issue about how it is possible that colder bodies do not warm warmer bodies but IR-sensor still detects IR from colder bodies. Let us address the core issue: how much. I am looking forward to scientifically proven numbers.

  443. Max™ says:

    The shell can cool radiatively, right?

    It can not cool by radiating towards a warmer surface, it is receiving more radiation than it loses in that direction.

    It doesn’t “know” that the surface in that direction is warmer, but across distances this short it takes a couple of nanoseconds per meter, so any photons leaving are more than replaced immediately.

    As such, no, the surface can not be treated as existing in isolation and radiating as though there are no other sources of radiation nearby, that is the first mistake Willis made, and it is commonly repeated.

    ___________

    In fact this is what will happen when you add the shell:

    Initially the shell will be cold, it will absorb close to 234.99 W/m^2, it will emit a small fraction of that amount inwards and outwards.

    Then it will warm until it is emitting say, 100 W/m^2, then it will absorb 134.99~ W/m^2 from the planet, and emit 100 W/m^2 outwards.

    Finally when it and the planet are in radiative equilibrium it will be emitting 234.99~ W/m^2, it will be absorbing a net 0.01 W/m^2 from the planet, and it will emit 234.99~ W/m^2 outwards.

    Thus the emissions outwards will balance perfectly with those from the planet with no temperature increases from unphysical greenhouse effects.

  444. Greg House says:

    Greg House says (2013/03/27 at 11:25 AM )”The links are dead (I wonder why…), but there is another way…”
    =======================================================

    The links Loodt provided (2013/03/27 at 10:22 AM) are alive now.

  445. Joseph E Postma says: 2013/03/27 at 6:53 PM
    thefordprefect – read Bryan’s answer.

    ——————
    I assume you mean the one where he describes a whetstone bridge for measuring the resistance change caused by temperature difference?

    This is jus a means of measuring the resistance change caused by thermalisation of IR. (This is certainly not the only means and definitely not the method used in an imaging array)

    What I need explained to me is what happens if the pixel is hotter than the object focussed on it. Basically what causes its TEMPERATURE to change. not how that change is measured.
    Cheers

  446. q ~ (T2^4 – T1^4)

    The direction and magnitude of q is what changes the resistance and the current. “Thermalisation” does not mean that cold is heating up hot. That’s not what thermalisation means. The sensor is only heated proper when the balance of q is positive coming into it, else q is negative and that is what happens when the pixel is hotter. Alan and Bryan probably know more about this than me, so hopefully they can comment. But what is plainly true is that nothing violates the laws of thermodynamics, and hence, precisely, a bolometer can not be an example of something cold heating up hot. Nor do they heat themselves up with their own radiation so their operation isn’t even a relevant concern in regards to the GHE in any case.

  447. Joseph E Postma says: 2013/03/27 at 7:39 PM
    q ~ (T2^4 – T1^4)

    sensor is only heated proper when the balance of q is positive coming into it, else q is negative and that is what happens when the pixel is hotter.
    ———————————–
    Apologies I still do not understand!
    are you suggesting that if the object is cooler then the pixel is cooled? q is negative?
    If so please tell me how a pixel is cooled below its ambient without assuming negative radiation

  448. q can be negative but the sensor, I believe, is attached to a large heat sink, so, it doesn’t cool even though q is negative. As more and more radiation comes in, from warmer and warmer sources, q becomes less and less negative until it is positive (when the source is hotter than the sensor). The heat sink, being very large, still would’t change. Hoping for Alan or Bryan to comment in more detail.

  449. Joseph E Postma says: 2013/03/27 at 8:05 PM
    q can be negative but the sensor, I believe, is attached to a large heat sink, so, it doesn’t cool even though q is negative.
    ———————
    tfp.
    The resistive sensor is conductively isolated as from the heat sink. For the bolometer resistance to change the sense element must change in temperature. In some bolometers there is even a mirror under the sense element to reflect IR back to the sensor.
    ————————
    jp. As more and more radiation comes in, from warmer and warmer sources, q becomes less and less negative until it is positive (when the source is hotter than the sensor). The heat sink, being very large, still would’t change.
    ——————.
    tfp.
    you use the term “radiation comes in” this as I stated in my post 2013/03/27 at 1:00 PM can only have a positive thermal effect. If you open a fridge door it feels cold because of a LACK of heat radiation (o.k. and because of cold air escaping) not because of cold radiation!
    At 0K there is no radiation and no heating. at 1K there is radiation and therefore heating.

    The bolometer measures radiation at all wavelengths (filtered usually into 2um to 14um bandwidth).
    It converts this radiation to heat The wavelength is not material to this type of bolometer.
    All changes in temperatures that the bolometer reaches are effectively referenced to zero incoming radiation i.e. the source is at 0K.
    All other temperatures of source add proportionally more radiation to the pixel.
    The pixel temperature will therefore always be above that with no incoming radiation (from 0K source)
    The temperature of the pixel will always depend on balance of heat loss and heat input.
    Heat input from the source beyond the lens is the only variable (internal heating is calibrated out).
    The focused source radiation will control the pixel temperature – higher temperature of source = higher temperature of pixel.

  450. prefect – all you have to do is use the laws of thermodynamics. Cold doesn’t heat up hot. The type of “pixels” I use in my field do not detect radiation by increasing in temperature – they convert photons to electrons. Bolometers might do something different than that and I’m not an expert on them. But the only thing you do need to know, is that cold doesn’t heat up hot. Not only that, a bolometer does not heat up via its own radiation, and so however they function has nothing to do with the GHE in any case. Therefore this is a useless discussion. We’re discussing how many teeth are in a horses mouth when the claim is that tigers can eat airplanes. Bolometers don’t heat themselves up with their own radiation, therefore they offer no support for the GHE. If a bolometer is pointed at a mirror, it does not spontaneously begin reporting higher and higher temperature.

    A bolometer senses the change in resistance of the sensor due to temperature change.
    The sensor forms one part of an initially balanced Wheatstone bridge
    No current will flow if the external object is at the same temperature as the sensor – the null point.
    If the object is at a lower temperature; the sensor resistance loses heat and resistance goes down and the Wheatstone bridge is out of balance and a current flows through the sensor.
    If the object is at a higher temperature; the resistance increases as the temperature rises the Wheatstone bridge is again out of balance and current flows but in the opposite direction.
    If calibrated against a known temperature the current can then represent temperature.
    This change in resistance is linear near the null point but if moved too far from the null point the change in resistance is non linear and will give rise to errors.

  451. Joseph E Postma quotes: 2013/03/27 at 8:52 PM
    If the object is at a lower temperature; the sensor resistance loses heat and resistance goes down and the …
    ————————————
    But the question is – Why does the temperature go down?
    Bolometers are in thermal balance with all energy inputs so power into the array is part of this balance as is the focused radiation from the source

    I think any scientist would agree that cold cannot heat up hot, However in the world of radiation balance the starting temperature should be the ambient temperature with added radiation at zero (-273°C) – “zero temp” – any change from this source temperature adds more energy and the temperature rises from its “zero temp”.

    So if there is no ghg then the earth sees TSI plus near zero radiation from space this will give be its “zero temp”.
    Add ghg and the radiation from above is perhaps from a temperature of -50°C (223K) This adds to the energy heating the earth (at perhaps -18C). The temperature will now be heated above its “zero temp”.
    This may seem like cold heating hot but really its cold heating “zero temp”.
    But what is important is that the new radiative balance allows the earth to warm to a new higher temperature above “zero temp”

    There is no breaking of any laws.

  452. “Add ghg and the radiation from above is perhaps from a temperature of -50°C (223K) This adds to the energy heating the earth (at perhaps -18C). The temperature will now be heated above its “zero temp”.”

    No, that is not correct at all. And it has nothing to do with bolometers either because bolometers do not heat themselves up or report higher temperatures from their own radiation coming back to them. Being surrounded by ice or mirrors with your own radiation coming back does not heat you up or add to your heating. This is of course what bolometers prove as well.

  453. Max™ says:

    Just to help Joe out a little: essentially tfp is claiming that the Hubble and Webb would work flawlessly in infrared wavelengths even if they weren’t cooled, because the cool sources must add energy to the ccd pixels and whatnot.

  454. [trashed]

    tfp, bolometers don’t violate the laws of thermo. Cold doesn’t heat hot. CCD’s absorb photons and create current just fine without being caused to heat up etc.

  455. Max, although this is for astro purposes this is helpful on bolometer detectors and why they are sometimes cooled (noise reduction)
    http://home.strw.leidenuniv.nl/~kenworthy/teaching/dol2011/10_DOL_Bolometers.pdf
    http://home.strw.leidenuniv.nl/~kenworthy/teaching/dol2011/11_DOL_Bolometers_part_2.pdf

  456. Reducing noise isn’t about cold heating hot.

  457. Max™ says:

    Joe also went over the actual mechanism: there is a sink tied to the device to stabilize it, the detector pixels are isolated, and the default with no radiation incoming is treated as 0 K, any level of radiation coming in will reduce that emitted by the device, this can be used to work out the temperature of a distant source.

  458. thefordprefect says:
    2013/03/27 at 10:50 PM
    [trashed]
    ————–
    well so much for open debate. []

    [JP: snipped – It’s called freedom of the press. This is my press and I am free to keep out ideas I believe are false and which have been answered repeatedly already…]

  459. Max™ says:
    2013/03/28 at 12:58 AM
    Joe also went over the actual mechanism: there is a sink tied to the device to stabilize it, the detector pixels are isolated, and the default with no radiation incoming is treated as 0 K, any level of radiation coming in will reduce that emitted by the device, this can be used to work out the temperature of a distant source
    —————
    What is the mechanism for reducing the radiation emitted by the device. There is no cancellation the bolometer does not respond to phase of radiation.
    Please explain to me how an object above absolute zero but below bolometer temperature reduces the outgoing radiation.

    But please keep it simple.

  460. Max™ says:

    Joe: I corrected David Socrates’ diagram, I think this more accurately represents your position, am I correct?

    tfp: cooling a bolometer isn’t simply to reduce noise, it is to increase sensitivity to colder surfaces.

    Btw, as far as “there is no cold waves”, while this is literally true, it is not as true as you might think, as Pictet apparently showed.

  461. tfp said: “Please explain to me how an object above absolute zero but below bolometer temperature reduces the outgoing radiation.”

    This is why I’m trashing your comments. This has already been answered repeatedly. q ~ (T2^4 – T1^4); q ~ (T2 – T1), etc etc. Outgoing q is reduced as T1 increases; as long as q is net outgoing, no actual heating (i.e. temperature increase) of the warmer source occurs from the colder source. Reducing cooling does not mean increasing temperature.

    This is a direct physical analogy:

    a = 1/m*(F2-F1). a = acceleration, m = mass, F2 = force 2, F1 = force 1. If F2 is some value, and F1 starts at zero, as F1 increases towards F2, the acceleration decreases, but, the direction of acceleration doesn’t change . The direction of the acceleration only changes when F1 becomes greater than F2.

  462. Bryan says:

    thefordprefect

    To get a handle on how a bolometer works start with the basics.
    Look up a note on the Wheatstone bridge and you will find it can be a very sensitive measuring arrangement if one of 4 resistors changes from the balance or null point.
    It can be used as a very sensitive strain gauge as well as in a bolometer.
    The sensor resistance used to be a metal but now more commonly a sensitive semiconductor is used.
    Joseph is correct when he says that a cooling arrangement (if used) is to reduce noise in the semiconductor.
    Carry out an experiment where the balance is changed to imbalance by changing the temperature of the sensor.
    Take a note of resistance with a sensitive ohmmeter
    If you record resistance against current you will find the origin at zero current and a particular resistance
    If the resistance changes up the current will go one way
    If the resistance goes down the current will go the other way

    If you graph your results you will need positive and negative current coordinates
    The plotted values show a straight sloped line through the origin as long as you don’t change the resistance too far the response will be linear.
    Since the resistance changes with temperature you can plot temperature instead of resistance against current.
    Now incorporate this into a display system and you have an accurate radiation detector.
    The use of pixels you refer to must be an amplified display system since light photons are much more energetic than the thermal radiation measured.

  463. @Max @2013/03/28 at 6:35 AM

    Well the one thing is that those could be walls, in which case the surface area differential problem of the shells game doesn’t exist. Say the core is an infinitely deep plane heat source and so we don’t care about what’s beneath it – it just holds a steady temperature and produces an output of 235 W/m2 forever – it is an infinite energy source. The “shell” is another plane then, a wall or whatever, just beside the infinite plane heat sink. It therefore gets the full 235 W/m2, and comes to the exact same temperature. What then? Then there is net 0 backwards, because q ~ (T2^4 – T1^4), and T2 is equal to T1.

    Now here’s a seemingly paradoxical situation that confuses the GHE people. If there is net zero backwards, then isn’t there also net zero “forwards”? Don’t you need a temperature differential to transmit bonafide net heat energy, and if the “shell/wall” is equal to the infinite source/sink, then how can the source heat the shell?

    But this is precisely it. The present occurs continuously and instantaneously. Heat transfer doesn’t occur in steps of one-second, but continuously instantaneously. The shell/wall is heated until its outward energy loss matches that of the source/sink. There is never any net q from the shell/wall back to the source/sink. As soon as energy is lost outwardly by the shell, it is instantaneously replaced by heat energy from the source underneath.

    A radiation field that doesn’t transmit net energy is JUST like a force that does no work. We know all too well what a “force that does no work” means – we accept that intuitively, physically and mathematically. We all accept that forces can exist, in which real energy is expended, but where no work actually gets done. I.e., push against a wall. So, it is totally legitimate to write 235 W/m^2 back from the shell, and, that it has ZERO net transfer of “temperature” and heating power. The only net transfer which can exist is that on the outside of the shell/wall, and the only quantity this value can have is the quantity which comes in to it from the bottom.

    This is an exact proof of why the shells game is wrong. If we take Willis’ end point, but the structures are infinite planes (as discussed here) rather than spherical surfaces (which btw is an entirely valid mathematical transformation in this situation in which sources and outputs are truly isotropic – unlike a planet heated on one side), then how come the bottom source plane can be twice as hot (well, in terms of flux) as the other plane and NOT continue heating up that plane? This is the left side of your (and David’s) diagram of course, just actually treating the figures as planar walls. The source is 302K. The wall receives 470 W/m2. How the heck can it be denied to heat up to the temperature it is being heated to? This situation is the most basic example of heat transfer and the heat-flow equation of q = s*(T2^4 – T1^4) is exactly what describes it. The only thing that can happen is for the source and wall to come to the same temperature, and this is defined exactly when q = 0. Therefore, the shell has to come to the same temperature as the source, and this is what exposes the shells game as a sophist trick to get something for nothing.

  464. Max™ says:

    True, I prefer to operate in the spherical paradigm because it makes it clearer that there is always a temperature difference between inner and outer, so there can be absolutely no doubt (or so I would think) that the shell can not cool by radiating towards the core.

    In the case where there is an infinite pair of walls then they would reach the same temperature and 235 in at the bottom -> equalizes temperature with the other wall -> 235 out makes physical sense.

    Amusingly I was told that I am “confusing radiation physics with arithmetic”…

    Isn’t that exactly what is being done when it is assumed that radiation down adds to radiation up?

  465. sunsettommy says:

    He he….,

    What is stopping Willis and his friends from creating a cascading shell generator power plant if what they propose really happens?

    If they can figure out a way to prevent the run away temperature increase (Siphoning off the excess between the shells?) They can replace the boring conventional Nuclear plants because they dont really work according to Willis and his friends since it does not use the concept of backradiation heating in their design.

    Imagine that if TWO outer shells generate at least 4 times the back radiation (For heating) of the original one shell and one core.Then they could use 16 shells around a single core power generator for novalike heat production that would solve all of our resource problems.

    Getting massive amounts of free energy from nothing would be something!

    Where is the free lunch power company?

  466. Kristian says:

    David Socrates, you said:

    “I hope you are not suggesting that, in this latter scenario, the earth’s atmosphere is not ‘trapping heat’ because it obviously is (take the atmosphere away and the mean surface temperature drops to a similar level to that of the Moon). The real issue is what is the physical mechanism?

    Not GHGs because radiation is simply a transport mechanism and not a STORE of energy. It is a fallacy to assume it can ‘bootstrap’ itself into a storage role. Instead the correct answer is good old thermodynamics – specifically the resistance to energy flow up through the atmospheric column and out to space caused by the limited speed of conduction/convection of heated air and latent heat carried upwards in the water cycle.”

    You saying this (on which we are in perfect agreement) is why I find it so downright weird that you can’t see that Willis Eschenbach’s thought experiment either 1) is not describing at all the radiative GH mechanism it presumably seeks to explain, but rather the conductive/convective resistance of the atmosphere, the very mechanism that you point to here, or 2) it proves that the radiative GH mechanism provides no intrinsic resistance to heat flow from surface to space and therefore ends up being completely incapable of inducing a surface warming effect.

    Quite recently, on the still ongoing thread at Tallbloke’s, you proclaimed the following:

    “Tim Folkerts doesn’t need a ‘get out of jail’ card. He (a warmist) is going with the flow of Willis’ MODEL. I (a skeptic) am going with the flow of Willis’ MODEL. How is it possible for him and me to disagree about the cause of atmospheric temperature enhancement but at the same time agree strongly on how to tackle the Will’s thought experiment? I guess it’s because from long professional experience we know how to play the ‘thought experiment’ game. You and others apparently don’t and so are reduced to irrelevant nitpicking.”

    Disregarding the extremely superior, condescending tone of your statement, a tone which I’ve noted you have a tendency to acquire and direct toward people that you disagree with, it makes me wonder: How do you really ‘tackle’ Willis’ thought experiment? If you do it the way Tim Folkerts is doing it, then there’s no justification for you coming here stating that GHGs do not ‘trap heat’ or make the surface warmer. For this effect is exactly what Willis’ thought experiment is meant to portray! If instead we’re talking about the conductive/convective resistance to energy flow, then we agree. But in that case, we certainly do not agree with Tim Folkerts and Willis Eschenbach (I can’t speak for Tallbloke, for I’m really not sure what version of the ATE he thinks he’s defending).

    Eschenbach set out with his planet/shell model to ‘prove’ the reality of the basic, physical mechanism behind the radiative GHE (that is the one at stake here, isn’t it? the one all the fuss is about). All that his model needed for it to do so, he concluded, was a shell with two surfaces. The only problem is, for his conclusion to work, some amount of the absorbed energy flux (the one coming in from the inside) at any one instant, in the steady state, can not be allowed through the shell to be radiated away from the outer surface. That is, there would have to be some kind of conductive resistance to heat flow through the shell. There would have to be a sustained temperature gradient across the shell wall, so that the inner surface is always warmer than the outer.

    If the conductivity of the shell were rather something close to perfect, the situation would look like this:

    The inner surface of the shell would always gain heat (from the continuous 235 W/m^2 incoming power density flux from the planet) and the outer surface would always lose the same amount (235 W/m^2).

    If the shell had a thermal mass, a heat capacity, then it would take time for the planet to warm it, for the KE to ‘build and spread’ throughout the shell, so to say. The temperature of the shell as a whole would take longer to reach the temperature of the planet. But the temperature of the planet (the heat source) would never during this process get any warmer than it was initially. The central flaw here is seeing the ’back radiation’ from the shell towards the planet as the shell is growing warmer, as something that makes a difference, as something that can and should be brought into the planet’s energy equation a second time, that can do thermodynamic work one more round, that can simply be added to the ‘next generation’ of energy output from the very body that already some time before expulsed it as heat loss. It cannot. It simply cannot. It is impossible. It would violate every thermodynamic law there is, not to say the least, quantum theory. [JP’s emphasis]

    Here is what happens, and what the Eschenbach followers still tend to forget or ignore: The shell starts emitting energy to its surroundings (space) as soon as it starts heating up, as a direct response to increasing levels of KE (higher temperature). The warmer the shell gets, the larger its corresponding cooling flux to space grows. That is why the heating rate of the shell slows down the closer its temperature gets to the planet’s. That is why the heat transfer rate between the planet and the shell is growing less and less the more the shell’s temperature approaches that of the planet. It hasn’t got anything to do with the ’back radiation’ from the shell somehow countering or subtracting from the ’forward radiation’ from the planet, thereby somehow reducing its cooling rate. This is a concept totally misconstrued. The cooling rate of the planet remains the same, as long as all of the energy that escapes it has a place to go (ultimately, out of the system). This is simply because its temperature never changes (constant nuclear energy source). Radiative cooling follows (is dictated by) temperature, not the other way around. The heating rate of the shell does however not remain the same. Because its temperature does change. It changes until we’ve reached q = 0. Up until that point, the energy flow from the planet has, to a smaller and smaller degree (~100% –> ~0%), been spent heightening the KE level (temperature) of the shell, and parallelly to a larger and larger degree (~0% –> ~100%), been escaping directly (as the temperature of the shell was increasing) through the outer surface of the shell as thermal radiation to space. [JP: Excellent!]

    At steady state (q = 0), the temperature of the planet and the shell is effectively equal and the transfer of heat between the two has asymptoted to ~0. But (!), the energy flow from the planet still has a place to go, and freely so: Space. From the outer surface of the shell. It just can’t warm the shell any more. There is no radiative ’trapping’ or ‘insulating mechanism’ in Willis’ model. The shell simply warms and emits accordingly to space. No radiative restriction to the energy flow from the planet through the shell to space.

    If you, David, want to argue that Willis’ model merely set out to portray ’the atmospheric effect’ in general, without necessarily emulating a specific resistance to outflowing thermal radiation, but rather to conduction of heat, then we can indeed come to terms. But then I think you better alert Tim Folkerts of our newfound agreement, because he thoroughly seems to believe that his beloved GHG warming mechanism is proved correct with this thought experiment.

    [JP: Tim Folkerts is a deranged fanatic… Great post Kristian.]

  467. Kristian says:

    Can’t see my attached image in the post above. Here’s a link to it:

    “If the conductivity of the shell were rather something close to perfect, the situation would look like this:

    – – –

    The inner surface of the shell would always gain heat (from the continuous 235 W/m^2 incoming power density flux from the planet) and the outer surface would always lose the same amount (235 W/m^2).”

  468. “Amusingly I was told that I am “confusing radiation physics with arithmetic””

    Good god is that what they said to you? It is actually more simple than that: they are just confusing arithmetic. Period.

  469. Exactly Tommy…WHY won’t they extend their design to multiple shells? Cause they know what it would show lol.

  470. Fred Kilger says:

    Joe: as mentioned in my previous comment, the main problem with the planet/shell model under
    discussion is the (non-physical) assumption of a impressed, constant radiation flux (out of the planets interior from an unlimited energy source). This (per definition) constant flux of 235 W/m^2 goes from the planet to the shell and from the shell to space regardless of the values of the temperature of planet
    and shell.
    The consequence of the impressed flux is:
    a) only the flux determines the temperatures of planet and shell
    b) the addition of a shell increases the temperature of the planet

    In case of a real planet, the radiation is determined by the temperature of the planet and is sourced
    by energy storage of ground, sea and atmosphere. Radiation of energy out of a storage
    element always lowers the temperature of this element. Constant radiation sources do not exist.
    This has the consequence:
    c) The initial temperature of the planet determines the radiation planet -> shell and shell -> space
    and the temperatures of planet and shell, all are functions of time and no constants.
    d) the temperature of the planet decreases from its initial value ( assuming no external heating from the sun)
    d) The addition of one or more shells only delays the cooling of the planet

  471. Here’s a razor I just learned about in “Omega Point”:

    The Superfluousness Razor: any theory that has superfluous features should be automatically rejected since there’s no sufficient reason why ontology should generate redundant features with no sufficient reason for their existence.

    This is obviously related to Occam’s Razor, or is in fact Occam’s Razor just stated in a different way or perhaps a more logically defined way.

    So, apply those razors here: A) heat flow obeys the known mathematical laws describing such things, following the laws of thermodynamics, and the outer shell/wall is heated to the temperature at which it is being heated, until equilibrium is established with the source.

    B) the source can not heat the wall without heating itself, with the progression of heating following no known thermodynamic laws or justifiable and sometimes even not definable mathematical sequences, and temperatures can be generated infinitely in excess of the input work which would be required to do so in scheme A.

    Which one has more superfluous features: A or B?

    No, it’s more simple than that. Which one is bonkers: A or B?

  472. Kristin says, 2013/03/28 at 8:20 AM

    Kristin,

    Whooa there! Let’s take this calmly point by point.

    1. The response I sent to Alan Siddons concerned what happens in the real earth-atmosphere system where the presence of its atmosphere, by irrefutable logic, is the cause of its enhanced surface temperature (when compared with an atmosphere-less earth). So when I am talking about that subject, I am putting forward my own point of view on what the mechanism of that enhancement is, namely the atmosphere’s effective resistance to energy flow. I’m glad we agree on that.

    2. The ongoing dialogue in the Entering the SkyDragon’s lair thread on the Tallbloke site about Willis Eschenbach’s ‘Steel Greenhouse’ Model Planet thought experiment, has developed into an interesting discussion, comparing it with the ‘SkyDragon’ Model. I depicted both models using a plane-parallel diagram. I believe this diagrammatic aproach is logically consistent with Willis’ concentric spherical shell model but much easier to draw and easier to discuss. I am pleased to see that Joe here agrees with me. I am also pleased that he supports the validity of discussing the models as they stand despite their limitations when compared with a real-world shell which has thickness and a real-world gap between core and shell.

    Unfortunately, days and days and days have gone by on the TB site with people suggesting that the reason why Willis’ model is wrong and the Skydragon model is correct is all to do with the ‘unrealistic’ assumptions of the Willis model, namely that: (i) a real shell cannot be indefinitely thin; (ii) the real gap between core surface and the shell cannot be indefinitely small; (iii) the thermal capacity of the shell cannot be arbitrarily low; and even (iv) energy can’t flow without a temperature difference (absolutely true but, ironically, in Willis’ model there is a HUGE difference between core and shell temperatures, whereas in the SkyDragon model there is NO difference!)

    The reason for my testiness (sorry about that if it offended you) was the sheer frustration of seeing all these marginal points being discussed in great detail as if they were pertinent to the matter in hand which they are not. Again as Joe has confirmed they are are marginnal and are therefore not relevant to the central discussion.

    What it came down to was a difference of opinion between those of us who were comfortable with discussing the two theoretical models as defined and those who wanted to change them to make them more realistic. On this, I sided with Tim Folkerts who also believed that contributors to the debate shouldn’t be allowed to change the model rules set originally by Willis and implicitly by Tallbloke in his article.

    You ask: How do you really ‘tackle’ Willis’ thought experiment? If you do it the way Tim Folkerts is doing it, then there’s no justification for you coming here stating that GHGs do not ‘trap heat’ or make the surface warmer. For this effect is exactly what Willis’ thought experiment is meant to portray! If instead we’re talking about the conductive/convective resistance to energy flow, then we agree. But in that case, we certainly do not agree with Tim Folkerts and Willis Eschenbach

    This is where you go haywire – by conflating my views that I offered to Alan Siddons on the real world issue of Atmospheric Thermal Enhancement with my strategy for analysing the theoretical Willis shell paradox.

    For the record, I think Willis is wrong and that a laboratory experiment would prove it so. But the fact is that, because of all the recent diversionary marginal tactics over on the TB site about thickness of shells, etc., etc., nobody there got close to explaining to Tim Folkerts WHY the Willis model is wrong and the SkyDragon model is correct.

    A final footnote: I am a hardline skeptic on global warming and have been consistently for the last 10 years since I first began to study the subject seriously. As time has gone by my position has hardened. None of that stops me finding ways to explain the skeptic position to thse people who support the warmist case but are amenable to persuasion. That is why I came to this site to ask Joe’s opinion on whether I was representing the SkyDragon position correctly, both diagramatically and in words.

  473. Joseph E Postma says: 2013/03/28 at 7:14 AM
    This is why I’m trashing your comments. This has already been answered repeatedly. q ~ (T2^4 – T1^4); q ~ (T2 – T1), etc etc. Outgoing q is reduced as T1 increases; as long as q is net outgoing, no actual heating (i.e. temperature increase) of the warmer source occurs from the colder source. Reducing cooling does not mean increasing temperature.
    ———————-
    Right so you are talking temperatures. But radiation is not temperature until thermalized.
    The real situation is:
    radiation+radiation hits pixel. Pixel converts sum of radiation to a single temperature. temperature causes resistive change. resistive change is read out by electronics.
    The pixel is not sensitive to IR wavelength. It warms by thermalisation of the radiation (frequency and phase of waves is irrelevant. The amount of radiation falling on the pixel per unit time causes a temperature rise
    So how does radiation cause a cooling?
    This paper posted by Max gives the answer at the end
    http://www2.ups.edu/faculty/jcevans/Pictet%27s%20experiment.pdf
    Thanks Max

  474. Max™ says, 2013/03/28 at 6:35 AM: Joe: I corrected David Socrates’ diagram, I think this more accurately represents your position, am I correct?

    Max my version of the diagram has been approved by Joe further up the thread. You are entitled to your own opinion but you should not have used my diagrammatic form with the same sub-heading because this may easily confuse other readers.

    It is up to Joe whether he accepts your revised version as better than mine. But my view is that your diagram is not appropriate nor indeed clear in its intent. The acid test is whether or not it confuses people trying to understand the SkyDragon position. I think it will confuse them.

    Sorry but that’s my opinion. Over to you Joe. 🙂

  475. Mitch says:

    This scenario looks like a radiation shield problem.

    link = http://www.cambridge.org/us/engineering/author/nellisandklein/downloads/examples/EXAMPLE_10.5-1.pdf
    Radiation shields are used to reduce the rate of radiation heat transfer to or from an object that
    must be thermally isolated.
    If cooling has been reduced on a process, and there isn’t any reduction in its load, then its temperature must increase.

  476. Kristian says:

    David,

    I am all for treating Willis’ model as is, so to say. And that’s exactly what I’m trying to do. That’s why I’m asking you how specifically you go about tackling it and, by extension, what insights you feel this approach is giving you.

    Because at some point we end up in the position where we need to ask: What is Willis Eschenbach trying to represent with his planet/shell model? It is of course what he sees as the most basic physical mechanism behind the radiative GHE. He is basically saying: If we put a layer around our planet made up of (or containing) a substance that absorbs and reemits IR welling up from the ground, then this layer would make the surface warmer simply by having two surfaces, one facing up and one facing back down. That is ultimately what he’s saying, David.

    And then I humbly ask you: Do you agree with his conclusion? Or don’t you?

    BTW, my name is Kristian, not Kristin. The former is a male name, the latter a female one.

  477. Kristian says, 2013/03/28 at 8:20 AM

    Further to my previous comment, just to emphasise that I think your exposition of the SkyDragon position in your last few paragraphs is spot on. Like Joe, I particularly like your rationalisation of why a real shell with thermal mass approaches the temperature of the core at an ever-decreasing incremental rate until it equilibriates with it. And then no more. That chimes well with Joe’s eaarlier description up-thread of radiation simply being ‘conduction at a distance’.

    However, in my opinion, your description sits better alongside my diagram than yours simply because mine accepts that back radiation (radiation that does no work!) exists. To see that it must exist, you only need to examine the simple case of two identical parallel plates separated by a gap and both at the same temperature of, say, 254K, suspended in a vacuum chamber that has perfectly reflective inner walls. In this thought experiment, the plates (assuming they have emissivity = 1) will continue to radiate at 235Wm-2 in all directions for ever both towards one another and towards the walls of the chamber and will remain at the same temperature for ever. It is difficult to see how the two way balancing radiation between the plates (that does no work!) can be ignored. Nor is it necessary to ignore it in order to sustain a hardline skeptic position.

    Maybe I have got the wrong end of the stick but it is my firm impression that the existence of such so-called ‘back radiation’ is part and parcel of the SkyDragon/Claes Johnson case.

  478. sunsettommy says:

    Getting off topic here:

    A misinterpreted claim about a NASA press release, CO2, solar flares, and the thermosphere is making the rounds

    http://wattsupwiththat.com/2013/03/28/a-misinterpreted-claim-about-a-nasa-press-release-co2-solar-flares-and-the-thermosphere-is-making-the-rounds/

    Anthony Watts stated here:

    “Because the “slayers” get as irrational in comments as some of the most strident AGW activists, and because it is late and I don’t want to deal with the angry dialog from some of their members who frequent here I know will happen, but would instead prefer a good night’s sleep, I’m not going to enable comments for this post. Maybe tomorrow.”

    I am not impressed.

  479. Ron C. says:

    Meanwhile in our real world society, children in schools are being taught that the greenhouse gas effect heats the earth’s surface twice as much as the sun heats it. I followed a link from an energy balance diagram at WUWT, and discovered the source, along with lesson plans.
    http://www.srh.noaa.gov/jetstream/atmos/energy_balance.htm

    This claim is unbelievable. Where do these numbers come from? What observational evidence can there be in support of this? Why are children being indoctrinated with this notion?

  480. mkelly says:

    David Socrates says:

    2013/03/28 at 1:51 PM

    “…nobody there got close to explaining to Tim Folkerts WHY the Willis model is wrong…”

    Not true. I ,and others, explained several times why the model was wrong using his own generated ratio and a radiation heat transfer equation with view factors etc. Mr. Folkerts does not want to know the model is wrong.

  481. @tfp @2013/03/28 at 3:35 PM

    Yes I read about Picktet many years ago.The results are wonderful confirmation of the laws of thermodynamics applying to radiation, and the falsity of the claims of the GHE. The results can also be understood with the idea that “radiation is conduction at a distance” as said elsewhere. I mean how amazing is the result: radiation from a cold source focused on a warmer target does not add to the temperature of the warmer target, but causes it to cool! Cold never warms up hot. Only when the thermal load is warmer than the target does it warm it up.

  482. @David @2013/03/28 at 4:11 PM

    Don’t worry about who’s diagram is who’s. The diagram itself is too simple to be patented 🙂 However, attribution should be noted or whatever, so that is a valid point, Max.

    I think Max’s diagram makes my point more clearly, because it did incorporate the idea of q ~ 0 as I was making the point of and it shows some numbers and those numbers can immediately be justified by q ~ T2^4 – T1^4 = 0 when T2 = T1.

    Your plane-parallel model is EXACTLY equivalent to Willis’ shells, and was a brilliant transformation. Sources and targets are isotropic and so the spheres can legitimately be transformed into planes – this is a mathematically exact transformation. This transformation is NOT valid when the input is non-isotropic, i.e. over half the sphere & etc.

  483. Max™ says:

    Max™ says:
    Your comment is awaiting moderation.
    March 30, 2013 at 12:28 am

    “Max says: ” Joe knows his physics ..”
    Thanks for that laugh. I needed a smile to start my day. 🙂 ” ~Tim F

    How silly of me, thinking astrophysics was in any way related to physics, I’m sure he just managed to guess his way through a masters degree in the field. *eyeroll*

    >.>

    [JP: LOL :)]

  484. @Mitch @2013/03/28 at 11:00 PM

    Changing the emissivity of an enclosure can certainly change the thermal characteristics. The temperature of the bottom of the atmosphere may in fact be nothing more complicated than the fact that O2 and N2 (the majority of the atmosphere) don’t & can’t radiate very well and therefore have low emissivity, and therefore have a higher temperature than otherwise. This result shouldn’t be thought of as a GHE.

  485. @David @2013/03/29 at 3:59 AM

    The plane-parallel model was brilliant and it took me a while to realize how much more it helped to destroy Willis’ model, and prove that Willis doesn’t actually know any physics. I should have realized it all myself right away…but whatever 🙂

  486. RonC: ” the greenhouse gas effect heats the earth’s surface twice as much as the sun heats it”

    Yes, these people and this whole “agenda” has the goal of divorcing the mind from reality. Sunshine is COLD and can’t cause any heating. It is too cold to melt ice & etc etc. They are totally destroying the mind of man, as I explained in my “religion of the GHE” series.
    On the other hand, the mind of man has never been all that rational – it is just that now we see a real push to really destroy it.

  487. BTW, Kristian, this was also very nice:

  488. Joseph E Postma says: 2013/03/29 at 7:16 PM
    From the document that you say “The results are wonderful confirmation of the laws of thermodynamics applying to radiation”

    IV. QUALITATIVE EXPLANATION OF PICTET’S EXPERIMENT
    As the radiation and reflection of cold may appear paradoxical, it may not be amiss to offer a qualitative explanation of Pictet’s experiment in modern terms.
    Let us note that every object—even a cold object—continually emits radiation. Each object also continually receives radiation that has been emitted by the objects surrounding it. The energy emitted per unit time by a given object depends both upon the object’s temperature and the properties of its surface. A highly polished metal mirror is a very poor absorber of infrared radiation; it is also, consequently, a poor radiator. Thus we may safely ignore any emission or absorption of radiant heat by the mirrors themselves. The sole function of the mirrors then consists in the reflection of the radiation that is incident upon them.
    Consider first the version of the experiment involving a flask of boiling water. This flask emits radiation in all directions, as shown schematically in Fig. 5(a). A negligible part of this radiation is received directly by the air thermometer, which has but a small cross-sectional area and which is placed at a considerable distance from the flask. In Fig. 5(b), we imagine the mirrors to be placed so that the focus of mirror A lies in the flask, and the focus of B in the bulb of the thermometer. Now we direct our attention toward that part of the radiation from the flask which happens to strike mirror A. This radiation will, after two reflections, impinge on the bulb of the thermometer. This is radiation that the thermometer did not receive prior to the introduction of the mirrors. Of course, the thermometer also receives radiation from other objects in the room. The introduction of mirror B effectively eliminates the right third or half of the room as a source of radiation for the thermometer. The effect of the mirrors is therefore to replace a part of the ambient radiation that formerly impinged on the thermometer by the more intense radiation from the flask. The thermometer therefore grows a little warmer.
    The version of the experiment involving a flask of snow may be explained in a similar way. Before the mirrors are introduced, the air thermometer receives radiation from all the objects in the room surrounding it. After the introduction of the mirrors, the radiation from the right third or half of the room is cut off from the thermometer by mirror B. This relatively high-temperature radiation is replaced by the radiation from the flask of snow. The thermometer now receives less energy per unit time than previously. The thermometer, initially at room temperature, now radiates away more energy than it absorbs, and so suffers a decrease in temperature.
    ——————-
    It is all in this last paragraph – “This relatively high-temperature radiation is replaced by the radiation from the flask of snow”
    i.e. the warm object still receives additional radiation from the snow But this is less than it would have received from the warm room. so it cools.
    The radiation from the snow is not being rejected – it is simply less than from the room.
    If you agree with this then we both agree with the operation of a bolometer!!!!

    [JP Reply: So yes then…the directionality of q, and cold radiation doesn’t heat up a warmer target.]

  489. Kristian says:

    David Socrates says, 2013/03/29 at 3:59 AM:

    “However, in my opinion, your description sits better alongside my diagram than yours simply because mine accepts that back radiation (radiation that does no work!) exists.”

    My diagram was not meant as an alternative to yours. I think yours describes the situation as a whole quite well. My diagram specifically intended to portray a situation where the shell had perfect conductivity. All my arrows represent ‘heat flow’. ‘Back radiation’ is not heat flow. And so it has no place in my diagram. That doesn’t mean it’s not there … It’s just irrelevant to what my specific diagram is meant to show.

  490. The smallest amount of energy (i.e. one quantum) that an object can absorb from IR light with a wavelength of 8µm
    Energy quantum = hν
    h is known as “Planck’s constant”, and has a value of 6.63 x 10^-34 Joule seconds (Js)
    so we need to know the frequency ν
    νλ= c
    ν = c/λ
    ν = (3.00 x 108 m/s)/(8*10-6 m) s-1
    plugging into Planck’s equation:
    E = (6.63 x 10^-34 Js)*( (3.00 x 10^8 m/s)/(8*10^-6 m) s-1)
    E (1 quanta) =2.49×10^-20 J at 8μm IR
    I small calorie raises the temp of water at 19.5C by 1°C
    4.182 joules=1cal
    I quanta of IR at 8μm will RAISE 1 gm of water by (1 / 4.18)x2.49×10^-20 °C
    =5.95×10^-21 °C!!
    NB. The absolute temperature of the water does not come into this analysis. The energy is added to whatever other energy is present and the temperature INCREASES,
    If the IR disappears the temperature falls

    [JP Reply: Well that’s a really simple-minded analysis using highschool physics. Of course, the radiation from an ice-cube doesn’t cause water to warm up, so, the source of the radiation is indeed a factor. The Laws of Thermo aren’t really learned at a mathematical level until about 3rd-year undergrad in a major in physics.]

  491. squid2112 says:

    @sunsettommy says: 2013/03/29 at 7:23 AM

    I have been reading and commenting on that post (A misinterpreted claim…) for the past couple of days now. It has devolved into “dancing with the stars”, but sorely lacking the music. I am astounded at some of the things I am reading within those comments. I am becoming increasingly disappointed in Anthony Watts and getting rather sick of reading postings by Willis Eschenbach (how many time can you be shown wrong?). I find it interesting to watch the multitude of contortions in the physical sciences. It’s like The Special Olympics of Origami in Physics. It seems that many of the participants know that the speed limit is 70mph, but they believe they can fool the traffic camera anyway. It is getting to the point, for me, that when I see a sidebar advertisement for another “free energy device”, I begin thinking of WUWT, as they are doing the same thing with the GHE hypothesis. Now, I am certainly no physicist. I don’t profess to be. However, I am a thinker, I am an extremely good researcher and have been practicing daily my entire life. I am pretty good at smelling garbage and discerning between crap and rose. I do fail to understand how some people can continue, incessantly, to try to mold crap into a rose, as if it will smell better. I have now bailed on the Anthony post you mention, as I have better things to do (like reading Noosphere).

    Cheers!

  492. Max™ says:

    Hmmm…

    We forgot an obvious counterexample to the nonsense Willis proposed: http://www.jwst.nasa.gov/sunshield.html

    According to Willis-ian thinking, the layers of the sunshield require each layer above the directly heated surface to radiate at full power, with the layers below reaching a higher temperature, so the effect of the sunshield should be to bake the telescope in full power sunlight while the closest layer should actually be glowing brighter than the sun!

    I don’t see why anyone should point out the importance of the layers being mirrored, as it seems to be ignored every time I bring it up.

    Similarly there are whipple shields on various items in space, do these not cause instant overheating issues?

    Could a whipple shield not solve any issue with keeping the crew compartment warmed?

    Hell, a whipple shield around a reactor should improve the efficiency beyond 100%, I’m a genius!

  493. Max™ says:

    More fun: http://en.wikipedia.org/wiki/Multi-layer_insulation

    The principle behind MLI is radiation balance. To see why it works, start with a concrete example – imagine a square meter of a surface in outer space, at 300 K, with an emissivity of 1, facing away from the sun or other heat sources. From the Stefan-Boltzmann law, this surface will radiate 460 watts. Now imagine we place a thin (but opaque) layer 1 cm away from the plate, thermally insulated from it, and also with an emissivity of 1. This new layer will cool until it is radiating 230 watts from each side, at which point everything is in balance. The new layer receives 460 watts from the original plate. 230 watts is radiated back to the original plate, and 230 watts to space. The original surface still radiates 460 watts, but gets 230 back from the new layers, for a net loss of 230 watts. So overall, the radiation losses have been reduced by half by adding the additional layer.
    MLI covering the heat shield of the Huygens probe

    More layers can be added to reduce the loss further. The blanket can be further improved by making the outside surfaces highly reflective to thermal radiation, which reduces both absorption and emission.” ~tha Wiki

    Hmmm…

  494. It is all in this last paragraph – “This relatively high-temperature radiation is replaced by the radiation from the flask of snow”
    i.e. the warm object still receives additional radiation from the snow But this is less than it would have received from the warm room. so it cools.
    The radiation from the snow is not being rejected – it is simply less than from the room.
    If you agree with this then we both agree with the operation of a bolometer!!!!

    [JP Reply: So yes then…the directionality of q, and cold radiation doesn’t heat up a warmer target.]
    ————————————
    But q you have stated is the difference in temperatures. Here they are talking about radiation. Radiation does not have a plus and minus. So there is NO directionality of radiation. zero radiation does nothing. Added radiation gets thermalized whatever its value or source.
    A hot object in a -273°C universe will be colder than the same object after receiving on quanta of energy from a hot object or one at -272°C
    A hot object at 100°C in a 100°C universe will not cool add an object at 99°C that blocks 10% of the hot objects view of the 100°C universe will cause the hot object to cool. I would postulate that a cold -273°C point source blocking 0% of the 100°C universe would NOT cause the hot object to cool.

    [JP Reply: q ~ (T2^4 – T1^4), so yes, q does have a direction even with radiation.]

  495. Greg House says:

    thefordprefect says (2013/03/29 at 10:41 PM): “i.e. the warm object still receives additional radiation from the snow But this is less than it would have received from the warm room. so it cools.
    The radiation from the snow is not being rejected – it is simply less than from the room.”

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

    Very clever, congratulations. Fool us all the time.

    In your example the cold bottle indeed acts as a shield preventing radiation from some warmer bodies from reaching the thermometer, therefore the thermometer can not stay as warm as it was before that shield was introduced. But your example does not demonstrate that if the cold shield had been even colder, the thermometer would have shown an even lower temperature. Only such a demonstration could prove that radiation from a colder body affects the temperature of a warmer body.

    I asked you twice, by the way, about how exactly IR of colder targets warms the sensor of an IR thermometer. No answer, I wonder why.

  496. Ron C. says:

    I was initially confused by a statement in Evans and Popp on Pictet’s experiment.

    “It follows that cold and heat are relative terms. The rays from a particular object will be either frigorific or calorific, according as they impinge on other objects either warmer or colder than itself. Imagine three identical bodies, A, B and C. Let A be at the temperature of freezing water, B at the temperature of 72F, and C at 112F. The rays emitted by B will be calorific with respect to the colder body A, but firgorific with respect to C. Moreover, they will be just as efficacious in heating the former as in cooling the latter.”

    This sounds like rays from a colder body have the ability to lower the temperature of a warmer body, assuming both bodies absorb and emit the relevant wavelengths. However, I think that the modern view is that all radiation conveys some energy, and that the lower temperature is attributed to more energy emitted than absorbed. If radiation from a colder body can in fact reduce the kinetic energy, and thus the temperature of a warmer body, I would like to know about it.


    [JP reply: Yes cause then we could make a “freeze ray”! Certainly, such experimentation could be conducted to further limits and other interpretations might be found.]

  497. Kristian says:

    thefordprefect, you say:

    “But q you have stated is the difference in temperatures. Here they are talking about radiation. Radiation does not have a plus and minus. So there is NO directionality of radiation.”

    You really do seem to have a serious blind spot here. Joe and others have been (rather patiently, I must say) trying to explain this very, VERY basic point to you now for quite some time. But you simply do not get it.

    Let me try.

    q is the heat transfer rate between the hot and the cold object. q will be positive as seen from the cold object and negative as seen from the hot. Seen from the cold object q can be regarded as the warming rate, seen from the hot object q can be regarded as the cooling rate. Yes, thermal radiation flows both ways between the two objects, but these flows can never be individually isolated and hence, neither of the two can ever be measured directly. They will both have to be inferred from q, the net of the two opposing flows and the only thing here being actually physically measurable. From this the individual flows will then have to be calculated using formulas with baked-in assumptions about ideal relations between temperature/emissivity and emitted radiative flux (like Stefan-Boltzmann).

    If you point your measuring device toward a warmer object, then q will be positive (net radiation coming in, i.e. more is coming in than going out). (How positive depends mainly on the temperature difference between your device and the warmer object.) This positive q will manifest itself inside the device as a positive voltage. From this one can start estimating the incoming flux and the temperature of the object pointed at. But the only thing really ever measured was the positive heat transfer (q) through the induced positive voltage. If you then point your measuring device toward a colder object, then q will be negative (net radiation going out, i.e. more going out than coming in) and so forth …

    The important thing to remember here, tfp, is that as long as the device is artificially kept at a constant temperature, it will not itself cool. Pointing it toward a cooler object will not make it cooler. (And it most certainly will not make it warmer!) We are only talking about q here. Is there more radiation going out from the device than coming in, or is more coming in than going out?

    Don’t let the K&T people fool you into thinking thermal radiation from the atmosphere in any way matters in setting the surface temperature of the Earth. Believing their ploy that DLR will somehow add heat to the ground is the first step of falling into their trap.

  498. Alan Siddons says:

    I second Kristian’s line of thought. A bolometer operates on the principle of a current that’s established between dissimilar metals at different temperatures. Whereas a simple Seebeck Effect-based thermocouple shifts the current according to the temperature difference, a bolometer imposes a bias voltage, which CAUSES a temperature difference between two dissimilar metals. This is known as the Peltier Effect, and in this case a change in electrical RESISTANCE makes the meter go up or down. The bolometer isn’t responding to photons however; its circuitry just registers temperature differences inside it.

  499. Joseph E Postma says, 2013/03/29 at 7:35 PM: @Mitch @2013/03/28 at 11:00 PM. Changing the emissivity of an enclosure can certainly change the thermal characteristics. The temperature of the bottom of the atmosphere may in fact be nothing more complicated than the fact that O2 and N2 (the majority of the atmosphere) don’t & can’t radiate very well and therefore have low emissivity, and therefore have a higher temperature than otherwise. This result shouldn’t be thought of as a GHE.

    Spot on. The following is my take on this whole issue:

    O2 and N2 together constitute 99% by volume of the atmosphere so this is where almost all of the vast FUND of kinetic energy is held.

    It is maintained at its current enhanced level of thermalisation by the energy flow from the Sun which exactly balances the outgoing energy flow to space. The incoming energy is transformed to KE by the following processes :

    (1) direct thermal transfer from the heated surface by conduction/convection
    (2) latent heat transfer from the heated surface, subsequently thermalised at cloud level during precipitation
    (3) radiation transfer from the surface, absorbed immediately by GHGs (mainly water vapor) at the base of the atmosphere, and thence transferred via diffusion to O2 and N2
    (4) from the Sun’s incident IR radiation, absorbed by GHGs (mainly water vapour and cloud droplets at cloud level) and thence transferred by diffusion to O2 and N2.

    Once the Sun’s incoming radiation is thermalised, it does indeed become trapped, because it is subject to the much slower rate at which energy is able to ascend the atmospheric column by convection. So the atmosphere as a whole presents an overall average effective impedance to kinetic energy flow.This is much as an electrical resistor presents a constant impedance to a constant current source (AKA ‘Sun’s energy flow’) resulting in the development of a fixed potential difference (AKA ‘enhanced temperature’) across it.

    Then at the top of the atmosphere the KE stored mainly in O2 and N2 is transferred by diffusion to the GHGs that are cooling while emitting radiation to space, thus completing the energy flow ‘circuit’.

    So although GHGs (mainly water vapour) are vital for converting the incoming Sun’s rays to atmospheric KE, and although they are vital for converting the atmospheric KE at the top of the atmosphere to radiation lost to space, in no sense is the atmospheric warming effect due to radiation being in any way ‘trapped’.

    It can’t be said too many times: radiation is an energy transport mechanism, not an energy storage device. End of story.

    [JP reply: oooh I like that last sentence a lot!]

  500. Alan Siddons says:

    Whoa, David, I gotta disagree with your 2013/03/27 post. You say, for instance, that “it is the mean value of the atmosphere’s ‘conductivity’ that gives us the Atmospheric Thermal Enhancement.” To the contrary, conduction and convection by gases only serve to COOL a heated surface. In other words, a sunlit landscape would be hotter in a vacuum because only ONE means of transferring heat is available, radiation. Granted, heated air moves to cooler zones over the Earth, thereby mitigating temperature extremes the same as the oceans do, but to no extent does air ‘enhance’ a temperature or provide better insulation than a vacuum can.

    You also say, “The constant flow of energy through the earth-atmosphere system from the Sun and out again to be lost to space is quite capable over time of heating the surface and atmosphere to any arbitrarily higher temperature.” And you offer ‘resistance to flow’ as the mechanism. Yet that’s the Greenhouse Premise in a nutshell, or at least one of them. Let me repeat a simple point, then: An inert body cannot get hotter than the heat source it’s in contact with, nor can it emit more radiant energy than what it’s receiving. This alone puts a heavy constraint on “any arbitrarily higher temperature.”

    To be clear, though, such a temperature/emission rule allows a lot of freedom. For example, you probably know from experience how hot a piece of shiny chrome can get on a sunny day. The damn thing’ll burn your hand. But why, if chrome is so reflective, i.e., non-absorptive? Well, because chrome is also a poor emitter. The small amount of radiant energy that it does absorb it can’t easily get rid of, and as a result it reaches a higher temperature than most any object around it.

    Now, there’s currently some disagreement among us as to whether irradiated chrome or other substances can actually become hotter than a perfectly absorptive/emissive blackbody, but let’s assume an equal temperature to illustrate a point. Say a blackbody in a vacuum is absorbing 1000 W/m² from the sun. This body’s surface will therefore emit the same 1000 and be at a temperature of 91° Celsius. We’ll have it that our piece of polished chrome is just as hot. Fine. But at the same temperature it will only be emitting about 60 W/m²! Even if the chrome did get hotter than a blackbody, then, it would need to be at 463° in order to emit the 1000 W/m² that a blackbody does at 91°. The point is that an ideal blackbody ALONE is able to radiate all the energy it’s receiving. Although other bodies may (or may not) get hotter , they cannot radiate as much energy. The same applies to a steel shell. So there’s the rule again: An inert body cannot get hotter than the heat source it’s in contact with, nor can it emit more radiant energy than what it’s receiving.

    But contrast this with the foul physics that Anthony Watts, Roy Spencer, Chris Monckton and other alleged “skeptics” would have us swallow. That 235 W/m² absorbed by a surface is CONVERTED to 390 W/m² of emitted energy due to certain trace gases radiating within a convectively cooling atmosphere! One might as well believe that dangling a thousand charcoal briquettes from the ceiling will take the winter chill off your apartment — because, after all, those black briquettes are absorbing and re-emitting the infrared rays that keep you warm. What utter nonsense.

    In sum, a piece of chrome teaches a valuable lesson: to the extent that a body FAILS to radiate it’s able to hold onto heat. So how can Climate Clowns insist that radiating molecules in the atmosphere make the Earth WARM? Surrounded by the vacuum of space, the Earth’s only means of cooling OFF is by radiation.

  501. Max™ says:

    I was right, I knew I recognized the arrangement of Willis’ thought experiment from somewhere.

    I thought this problem seemed familiar for some reason.

    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 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 radiatied to the surroundings?

    (b) How is the total power radiatied affected by additional heat shields?
    (Note that this is a crude model of a star surrounded by a dust cloud.)
    UC Berkley

    Solution:
    (a) At radiative equilibrium, J – J₁ = J₁ or J₁ = J/2. Therefore T₁⁴ = T⁴/2, or T₁

    (b) The shield reduces the the total power radiated to half the initial value. This is because the shield radiates a part of the energy it absorbs back to the black sphere.

  502. @Max just above;
    Look at problem 1026 on the UC Berkley link. The temperature of the blackbody does not increase when surrounded by the shell – rather, and simply, the problem just determines a new Q. The final answer for the temperature of the shell is precisely what Max has been going on about with the surface areas. The temperature of the interior blackbody sphere does not increase. It is precisely Willis’ shells game, but the interior sphere does not heat itself up! Because the problem was worked out by an actual real physicist. Fini, Willis & WUWT, you dumb asses.

  503. Kristian says, 2013/03/30 at 2:13 AM: My diagram was not meant as an alternative to yours. …All my arrows represent ‘heat flow’. ‘Back radiation’ is not heat flow.

    Ok, so we are in agreement. 🙂

    However, normally these kind of diagrams do mean radiation flow. So I suggest you differentiate your diagram as such if you use it again.

    Cheers,
    David

  504. Alan Siddons says, 2013/03/31 at 9:50 AM: Whoa, David, I gotta disagree with your 2013/03/27 post.

    Hi Alan, thanks for responding. This subject is crucial, so I am grateful for the dialogue.

    You say: …conduction and convection by gases only serve to COOL a heated surface. In other words, a sunlit landscape would be hotter in a vacuum because only ONE means of transferring heat is available, radiation and … to no extent does air ‘enhance’ a temperature or provide better insulation than a vacuum can.

    I find it difficult to reconcile those statements with the following:

    1. The Moon’s mean surface temperature, as determined empirically by the Diviner Orbiter project, is around 197K.

    2. The Earth’s mean temperature is generally agreed to be around 288K.

    If it is not the atmosphere’s resistance to energy flow to space, what is the cause of this temperature enhancement?

    Further, you say:An inert body cannot get hotter than the heat source it’s in contact with, nor can it emit more radiant energy than what it’s receiving. This alone puts a heavy constraint on “any arbitrarily higher temperature.”

    I agree that what you say is true for purely radiative exchange of energy. But I was pointing out that, where there is resistance to energy flow through a conductive medium, that medium can heat up arbitrarily.

    For example suppose we have a really feeble 1W electric element that is heating a cube of heat-conductive material encased in an almost perfect insulating jacket. The cube will go on heating up until the temperature gradient between the cube and the ambient surroundings allows exactly 1W of heat to flow through the almost-non-conductive jacket. So by adjusting the conductivity k of the insulating jacket we can achieve any arbitrarily large steady-state temperature ‘enhancement’ we wish (unless the insulation melts first!)

    In a less dramatic way, that MUST be what is happening when we add back in the atmosphere to an atmosphere-less earth. As we know empirically, the mean surface temperature will climb up up from ~197K to a steady-state ~288K and then stick there. At that precise point, the leakage of energy flow to space will exactly balance the incoming absorbed insolation and the earth system is then in steady state.

    So the atmosphere must surely be offering a mean heat-conductivity k of just the right amount to achieve that 288K result.

  505. Max™ says:

    Whoops, was supposed to be linked to here, for the record: didn’t notice I hadn’t scrolled to the right spot when I copy-pasted it, doh.

  506. squid2112 says:

    @Max™ : Genius! … That is great stuff Max!

    @Joe: And the truth shall set you free!

    Case closed.

    [JP: Indeed!]

  507. @David 2013/03/31 at 1:27 PM

    Well I would point out the actual day-time lunar temperature of ~100C degrees rather than the mean temperature, to indicate the actual power of solar heating on a bare surface. In this regard, the atmosphere seems to keep the surface of the Earth a lot cooler when under actual direct heating. Then, factor in the low emissivity of air, and the latent heat in water and water vapour which both helped reduce the terrestrial day-time high, and that heat releasing over night and preventing as much temperature drop as there could have been. Also factor in the natural lapse rate gradient and the fact that the average temperature of the thermal ensemble can not be found at the bottom of the atmospheric column in any case, but has to be found somewhere in the middle, and the bottom of the atmosphere therefore has to be warmer than the thermal average (the thermal average presumably being 240 W/m^2 or -18C). So I myself don’t know that “thermal resistance” is an accurate term, when all these other more direct physical factors all have the result of smoothing the temperature oscillation to a much smaller range than the moon, with the atmosphere actually causing a lot of cooling in the day time, and with the bottom of the atmosphere necessarily having to be the warmest part of the grand thermal ensemble, and that the atmosphere has low emissivity, and with stored latent heat presenting about a centuries worth of trapped solar energy which keeps much of the surface from wanting to get below 0C. Perhaps the low emissivity could be called “resistance”, but, it also doesn’t need to be re-labelled that because it is already just called low emissivity.

  508. So apparently over at the WUWT thread on the NASA article, Roy Spencer was doing the shell game with his two plates or some such.

    These people are depressingly stupid…I abuse them here because Mr. Roy has sent me personal emails attacking my intelligence and character in the past, so he too can get a life, and get a clue. This is what these morons do – attack us personally in private, and in public, and then they say that WE (PSI, Slayers etc) are the ones who are nasty. Again I ask the question, why do certain so-called skeptics with a clear lack of actual physics training shield the GHE from the slightest criticism with such piss and vitriol?

    For Roy’s two plates…whatever it is he is pretending with them. A source held at a constant temperature, because it is either an infinite heat sink or it has a continuous input has the same result – it will only heat something else to its same temperature, and it will not heat itself. A source as an infinite sink, or a source as a constant temperature with continuous input holding it at that temperature, are identical. It is why physicists invented the idea of an infinite heat sink in the first place. Again, look at problem 1026 from the Max’s UC Berkley link – this problem can either be plane parallel walls, or “plates”, or it can be Willis’ shells game. Look at the result – nothing heats itself up. The energy from the source doesn’t heat itself up some more in order to heat a cooler thing.

    That is the essence of all these people’s arguments – that in order for something warm to heat up something cooler, the hotter thing has to heat up some more! You either have been trained in physics and know that that is ridiculous, or you’re a fraud. I can easily “forgive” the people who don’t have the training; I don’t have much tolerance for the frauds who pretend they should know better.

    A whole paradigm has been built with the GHE and it extends much further than simple alarmism; there’s emotional attachment to it, of various kinds and degrees, etc. It satisfies certain emotional archetypes that most people will not realize they’re being satisfied with. At the basic scientific level, yes, the GHE is the basis of alarmism, and so as long as the GHE is protected, shielded, denied from criticism, babied, blocked, treated with kid gloves, etc. etc., alarmism will forever exist and the useless debate over climate change will continue forever. So tell me, in who’s interest is it to deny the possibility of GHE criticism, and, why do certain people get so nasty when you try to do it, given the fact that the GHE is the basis of alarm? If we’re critical of alarm we must automatically be critical of the GHE, because they are one and the same. But this isn’t what some people want to allow. Haven’t certain people actually made careers for themselves via the alarmist debate? Would it actually be in their interest to have the alarmist debate disappear by removing its basis, the GHE?

    There is simply no consistent, rational, mathematically valid, or scientific explanation of the GHE. There isn’t because the GHE isn’t actually a thing, but a desire. We’ve seen some explanations for the GHE presented recently, and they’re plainly wrong. People either get it, or they don’t. When one model of the GHE is shown to be wrong, another wrong one is presented instead. That behaviour specifically indicates the total absence of science, and the presence of religion, or at least, propaganda.

  509. Will Pratt says:

    Joseph E Postma says:
    2013/03/31 at 10:37 AM

    “Willis & WUWT, you dumb asses.”

    Like I said further up the thread, they are not fools, but they certainly are liars and quite probably lying fools.

    Fake opposition are easy enough to recognise. They claim scepticism while at the same time, they underpin the very logical fallacies of “GHE” hypothesis on which AGW fraud is based.

    Such people are the gatekeepers, the enablers, of AGW fraud.

  510. Precisely.

    They claim scepticism, while at the same time they perform the very same logical fallacies which they pretend to criticize in alarmism…

  511. You create the appearance of a debate, while you push through the policies the debate is supposed to be debating about, by having one aspect of the debate that none of the fake opposition will ever debate about. “We all agree there is a GHE – the question is ‘how much warming’?” Well, that’s not a debate at all, is it – it is outright and blatant support for alarmism. Subtle trick. Just watch them all get their knickers in a knot and hyperventilate when you say that “the sun heats the Earth on one side”. Have you ever witnessed grown men hyperventilate at that statement, in any rational or sane context?

    It’s just so obvious at this point.

  512. Kristian says:

    Max™ says, 2013/03/30 at 8:33 PM:

    “More fun: http://en.wikipedia.org/wiki/Multi-layer_insulation

    Yes, quite pertinent. The one decisive quoted sentence is: “The original surface still radiates 460 watts, but gets 230 back from the new layers, for a net loss of 230 watts.”

    So they do what we do. They separate between temperature –> emission/cooling flux as per the S-B equation of the hot surface, and q, which is the net radiation going from the hot to the cold surface, the heat transfer rate. The folks at Tallbloke’s seemingly have a real hard time grasping this fundamental distinction. The back radiation from the cold layer can never add to the emission flux (and by implication the temperature) of the hot layer. That is set by whatever maintains its temperature in the first place. What it does do is reduce the heat transfer rate between the two layers. It is of paramount importance to keep these two ‘effects’, as it were, apart.

    Notice also what they are saying here: “Now imagine we place a thin (but opaque) layer 1 cm away from the plate, thermally insulated from it, and also with an emissivity of 1.” This is specifically describing a ‘radiation shield’, that is, it insulates the inner layer radiatively by reflecting part of the received flux and so will have a hard time reaching the same temperature as the inner layer, because it doesn’t absorb its entire flux. If the opacity however was just a matter of thermal mass, the obstruction to radiative flow would only be temporary, until the outer layer had reached the same temperature as the inner.

    The radiative insulation enables us to reduce the heat input to the inner layer (the electric bill, so to say) to maintain a steady temperature, because less heat is ultimately lost from the system to space per unit time with the insulating layer than without.

    This also means that if we didn’t ‘turn down’ the heat input to the inner layer in this setup, it would start warming and consequently emit more radiation toward the outer layer. It is however NOT the back radiation (or the reflected radiation) that forces it to warm by subtracting from the forward flux. It is the fact that (again, at balance) the system’s output (flux to space) is smaller than the input (flux to the inner layer). Much of the energy delivered to and at balance emitted from the surface of the inner layer simply can not find its way out of the system and since it’s still being continuously supplied with new (hot) energy through conduction from its internal (?) heat source, this new energy will start piling up, forcing the surface to raise its temperature in order to emit more and faster.

    How then does all this pertain to Willis’ planet/shell model? Well, first of all, it shows us quite clearly that we can not and do not add (or subtract, however you see it) the back radiated flux from the shell to (from) the forward flux from the planet. It does not in itself make a difference to the heat budget (or the temperature) of the planet. What matters to the planet is the outgoing flux from the shell. The amount/rate of energy emitted from the planet will remain the same (as will the temperature of the planet) as long as all of it has a place to go, ultimately out of the system. In Willis’ system there is no restriction whatsoever to why it shouldn’t all ultimately be able to leave the system. There is no radiatively insulating (reflecting) layer anywhere. The continuous flow of energy/heat from the planet simply warms the shell until it has a temperature dynamically equal to the planet’s so that at this point it can escape the system through the outer surface of the shell rather than from its own surface. The outer surface of the shell has effectively superseded the planet’s surface as the system’s new boundary with its surroundings.

    If they want Willis’ model to exhibit a steady-state temperature gradient from core to outer surface, they will have to introduce some kind of insulating factor. It is not enough for the shell to have two surfaces. Pure geometry won’t do the trick. The shell can only lose heat through its outer surface. And its heat loss will have to balance its heat gain. Otherwise it would simply warm until it did. There is no way around it. Therefore it’s absurd to think, absorbing the entire flux from the planet, it would not eventually reach the same temperature as the planet. We need specific restriction to (preferably radiative) heat flow out of the system. As far as I can see, there exists no such restriction in Willis’ model. The shell is not stipulated as a ‘radiation shield’ or a ‘heat shield’, a radiative (or a conductive) insulator.

  513. Will Pratt says:

    Joseph E Postma says:
    2013/04/01 at 1:37 AM

    That is exactly the purpose of all these “official” sceptics. It appears to be based on the “Hegelian Dialectic”. Later adopted and adapted by Marx and Engels in London around the same time that John Tyndall first began formulating his version of the “GHE” fallacy.

    In both cases the underlying “philosophy” can be dismissed as nothing more than a cheap confidence trick dressed up as science.

  514. Kristian says:

    Max™, about the UC Berkeley problem:

    The shell receives a certain flux from the inner sphere, but can only emit half of it to space as heat loss. So heat gain is 2X heat loss. How can that be a stable condition, a state of equilibrium for the shell? If the inner sphere is sustained by a constant heat source, and barring distances and area differences, the shell will eventually have to raise its temperature to match that of the inner sphere in order to balance heat gain with heat loss. There is no way around it.

    If, that is, it does absorb the entire flux.

    They call their shell a ‘heat shield’ and liken it to a ‘dust cloud around a star’. I guess we know a thing or two about interstellar dust clouds, Joe probably much more than me. But mostly we seem to know very little for certain about their absolute properties. One thing <em<is for certain, though: A dust cloud around a star would never absorb all the radiation coming in from the star. It would mostly scatter or reflect it, or simply let it right through. So it could never ever even approach the temperature of the star just from absorbing its light. As I said, Joe will definitely know more about this than I do.

    (Willis’ shell is neither a ‘heat shield’ nor a scattering dust cloud.)

    But I wonder how they got to their solution. It appears to be simply asserted, much like with Eschenbach. There is no experimental or observational proof presented. Not even any mathematical or physically derived proof. There is just the same claim as Willis provides, that the shell will be cooler than the sphere because it radiates some of its received flux back, i.e. simply because it has two sides. Pure geometry once again. This is only ’proving’ Willis’ point by referring to someone else having the exact same idea about this as he does. Maybe he even got it from here …

    The basic assumption here is simply asserted, not backed up in any way. The math itself is fine. But it is based on circular reasoning: ’This shell will radiate only half of the flux from the sphere to space and hence be cooler than the sphere. Why? Because it has two sides and will radiate equally to each side.’ Can you show this to be the case at equilibrium? Therewith, the problem is posed and solved. But the solution is arrived at simply by appealing to the original premise. Nothing else.

    I second your ‘Hmmm’ …

  515. Kristian says:

    Shite, I see now that I wrote:

    “The back radiation from the cold layer can never add to the emission flux (and by implication the temperature) of the hot layer. That is set by whatever maintains its temperature in the first place. What it does do is reduce the heat transfer rate between the two layers.

    No, it doesn’t. It is the outgoing flux from the outer surface, not the back radiation from the inner that reduces the heat transfer rate …! Sorry.

  516. Max™ says:

    The interior is a black body cavity with a power source, if there were no losses through the exterior it would equalize to the same temperature everywhere, as it is half the energy reaching the shell is able to escape.

    My earlier mistake was to assume that the black body condition would hold even with losses to the environment, and thus to think that the shell and sphere would wind up at the same temperature.

    I now think the shell will wind up at a temperature where it emits around half what the sphere does.

    It’s not exact because of effects like whether the interior of the shell is able to radiate towards itself in any significant amount (rather than towards the shell) while all the radiation emitted by the sphere is absorbed by the shell.

  517. I thought we’d concluded this? The q from the shell is 0 back to the sphere, and so it reaches the same temperature (or close enough if the radius isn’t much larger) as the sphere and emits outward. This problem was solved in the UC Berkley link.

  518. Max™ says:

    I thought the UC Berkley link said it would reduce the losses to the environment?

  519. I was referring to problem 1026 which is an exact replica of Willis’ shells game. The interior sphere doesn’t heat up and the shell simply attains the temperature required by its outer surface area to radiate the same amount of energy as the sphere. As you had pointed out long ago, this means the outer shell will be a little cooler because it has larger outer surface area.

  520. Kristian says:

    Max™, you said: “The interior is a black body cavity with a power source, if there were no losses through the exterior it would equalize to the same temperature everywhere, as it is half the energy reaching the shell is able to escape.”

    Huh? If there are no losses to the system’s surroundings and yet the central power source is continually pumping out energy, what do you think would happen?

    The losses are necessary to maintain a dynamic equilibrium. As soon as the shell loses energy to the system’s surroundings, it is replaced by the same amount of energy from the power source. Steady state. Effectively equal temperatures planet/shell.

    Have they gotten to you over at Tallbloke’s, Max?

  521. Ron C. says:

    In recent discussion at WUWT on the greenhouse effect, one commenter (Mosher) provided a link to an interesting presentation that he said was a credible description of the greenhouse effect. It comes from course materials at U. Wisconsin.
    http://www.aos.wisc.edu/~aos121br/radn/radn/sld001.htm

    The basic notion is that additional CO2 raises the ERL (Effective Radiating Level), thereby raising the temperature profile of the troposphere including at the surface. I am unsure of some of the points, e.g. that the atmosphere is mostly opaque to IR radiation. I wonder whether additional CO2 increases the cooling power of the atmosphere, rather than warming.

    I find the material includes much good information, and would like to know what parts are questionable.

  522. Alan Siddons says:

    Two comments that might help the discussion.

    1. Perhaps we can agree that Willis’s shell can’t be considered both a blackbody kind of absorber and also a radiant insulator, for a proper insulator would of course be reflective. As a fully efficient absorber, the shell could only act as a radiant conveyor. With an emissivity of 1 its surface would radiate to space the same amount of energy that the surface emits, although with less energy on a watt per square meter basis. Indeed, were the shell in direct contact with the surface (and not appreciably thick), it would emit the same 235 W/m² that the surface did before. A good reflector, on the other hand, is a poor emitter. Even if it reached a temperature similar to the surface’s, then, it could never radiate the same amount of energy to space. This presents quite a problem for the GHE religion, for it must answer how such a shell manages to emit all of the surface’s energy and still be considered an insulator.

    2. You can’t trap light in a bottle. That is, if a mirrored chamber prevents a luminous body from emitting to its surroundings, the light doesn’t build up and get brighter and brighter inside. This may seem strange but it’s true. A ‘standing wave’ is created by the mirrors instead. This is akin to the 2nd Law, wherein nothing happens when a heated body encounters another body at the same temperature, because no heat transfer is possible. So too, when a luminous source encounters its own light, there’s no transfer of light to the source and thus no increase. Sadly for the GHE religion, then, you can’t construct a Radiant Bomb by sealing a flashlight inside a thermos.

  523. Will Pratt says:

    Ron C. says:
    2013/04/01 at 7:01 AM

    Ron, Steve Mosher often repeats this same tired circular argument.

    Increased heat from the sun can raise the ERL of the atmosphere because it causes the atmosphere to expand. That is an actual fact which is confirmed by the radiosonde data.

    Increases in CO2 cannot cause the atmosphere to warm and therefore expand. That is an incorrect, unsubstantiated hypothesis known as the “greenhouse effect”.

    Therefore his statement is by definition a circular argument. A logical fallacy. Just one of the logical fallacies on which the “greenhouse effect” hypothesis is entirely based.

    Logical fallacies are all these idiots have.

    It is important to remember that John Tyndall never once recorded, or even attempted to measure, a temperature increase in any of the gases he experimented with. I think that even he understood the concept of Local Thermal Equilibrium and knew well enough that certain IR frequencies would simply be scattered rather than thermalised.

  524. Joseph E Postma says: 2013/04/01 at 3:42 AM
    I thought we’d concluded this? The q from the shell is 0 back to the sphere, and so it reaches the same temperature (or close enough if the radius isn’t much larger) as the sphere and emits outward. This problem was solved in the UC Berkley link.

    Joseph E Postma says 2013/04/01 at 3:52 AM
    I was referring to problem 1026 which is an exact replica of Willis’ shells game. The interior sphere doesn’t heat up and the shell simply attains the temperature required by its outer surface area to radiate the same amount of energy as the sphere. As you had pointed out long ago, this means the outer shell will be a little cooler because it has larger outer surface area.

    There is J1 flux back to the core from the shell.
    There is J1 flux out of the shell to space

    The UC Berkley link shows back radiation!! this does not agree with your 3:42 statement
    Yet you seem to suggest this as an explanation of your theory in your 3:52 post?

    Please explain. Thanks

  525. Greg House says:

    Ron C. says (2013/04/01 at 7:01 AM): “In recent discussion at WUWT on the greenhouse effect, one commenter (Mosher) provided a link to an interesting presentation that he said was a credible description of the greenhouse effect.”
    =======================================================

    Ron, virtually nobody would care about “greenhouse effect”, if the IPCC had not issued their reports about the “greenhouse effect” and if politicians did not base their policies on those reports.

    What Mosher presented is not the “greenhouse effect” as presented by the IPCC. Actually, they’d better call it differently in order to avoid confusion. Any person is, of course, entitled to invent his/her own private “greenhouse effect”, but only the IPCC’s one is politically relevant, and it is about “back radiation warming”. Demonstrating that the IPCC’s “greenhouse effect” is crap does make sense, because it destroys the foundation of cutting CO2 emissions policy. Shifting to the Mosher’s one dies not.

  526. @tfp 2013/04/01 at 8:38 AM

    It shows backradiation which does not cause any increase in temperature. Please look at what it says and sketches carefully. The backradiation does not increase the temperature of the source. The only thing the shield does is decrease the rate of emission – and this does not increase the temperature of the source. The result couldn’t be any more clear: T1 = T/4root(2), and so there was no change to T, the source temperature. Answers a) and b) say nothing about the source becoming warmer, with the presence of backradiation, as would be expected from thermodynamics.

    Then, on problem 1026, just below the one you refer to, it describes the Willis’ shells game idea, and again, the source doesn’t become heated, but this time the shell just radiates the equal amount of energy that the source provides.

    Both the problem 1023 and 1026 are consistent in that the presence of backradiation does not increase the source temperature – that is very explicitly stated in the results. Back radiation can exist, but q can be equal to zero. q being equal to zero does not mean that backradiation doesn’t exist. Backradiation can exist and not cause any heating.

  527. Will Pratt says:

    Two comments that may help Mr Siddons.

    1. Under absorption of ambient IR, dark substances get hot, bright reflective substances do not.

    2. However if you actively heat a light coloured or reflective substance to the same temperature as a dark substance, just like the dark substance, the light or reflective substance will emit according to its temperature.

  528. Well emissivity does factor in there Will. If the emissivities are different then they won’t emit the same.

  529. Greg House says:

    Alan Siddons says: “2013/04/01 at 7:58 AM): “…Sadly for the GHE religion, then, you can’t construct a Radiant Bomb by sealing a flashlight inside a thermos.”
    ==============================================================

    Alan, this is a very good illustration, I like it.

  530. Alan Siddons says:

    I disagree, Greg. This notion of a higher altitude of emission (the “effective radiating level”) is part and parcel of standard greenhouse theory. The idea is that GHGs make it warmer UNDER the ‘blanket’ they provide by making it cooler ABOVE that blanket. That’s Roy Spencer’s view, for instance, and this is why you’ll hear occasionally that an increased greenhouse effect on the surface forces the stratosphere to become colder.

    Sorry that Mister Pratt failed to understand my comments.

  531. Will Pratt says:

    Joseph E Postma says:
    2013/04/01 at 9:08 AM

    “Well emissivity does factor in there Will. If the emissivities are different then they won’t emit the same.”

    If their temperatures are the same, their emissivities will be the same.

  532. If that were true then why is emissivity even a factor? How come emissivity isn’t always thought to just be a function of temperature? If for all objects of equal temperature, emissivities are equal, but emissivity is not equal for different temperatures, then emissivity would simply be a universal function of temperature, and be independent of surface properties. Something is wrong with that. Are you referring to Kirchhoff’s Law in some fashion?

    Anyway I’m on India time so I’ll be back in 8 or 9 hours…time for sleep.

  533. Will Pratt says:

    “then emissivity would simply be a universal function of temperature, and be independent of surface properties”.

    Are you confusing emissivity with reflectivity Joe?

    emissive |iˈmisiv|
    adjective technical
    having the power to radiate something, esp. light, heat, or radiation.

    That is power as in W/m2.

    [Reply: Yes, and it (emissivity) is a function of surface properties, and is not equal for equal temperatures for different surfaces, unless they’re blackbodies or by happenstance.]

  534. Greg House says:

    Alan Siddons says: (2013/04/01 at 9:34 AM):”I disagree, Greg. This notion of a higher altitude of emission (the “effective radiating level”) is part and parcel of standard greenhouse theory.”
    =======================================================

    My point is that this “standard” or not theory is not the “greenhouse effect” as presented by the IPCC. If you wish, the IPCC standard explanation is “back radiation warming”. Here, e.g.: http://www.ipcc.ch/publications_and_data/ar4/wg1/en/faq-1-3.html.

    Shifting discussion to other alleged effects does not make sense to me. What does make sense is making clear what the IPCC maintains and demonstrating that it is an unscientific piece of crap.

  535. Joseph E Postma says: 2013/04/01 at 9:01 AM
    It shows backradiation which does not cause any increase in temperature. Please look at what it says and sketches carefully. The backradiation does not increase the temperature of the source. The only thing the shield does is decrease the rate of emission – and this does not increase the temperature of the source. The result couldn’t be any more clear: T1 = T/4root(2), and so there was no change to T, the source temperature. Answers a) and b) say nothing about the source becoming warmer, with the presence of backradiation, as would be expected from thermodynamics.
    ————–
    So if what you say is correct then all your diagrams showing 235watts passing unchanged from the core to the shell must be incorrect.

    A core generating 400 watts receives 200 watts back radiation from the shell the shell transmits 200 watts radiation to space is the what you are claiming the Berkley question shows, The only way this could be true is if the generator and back radiation cancel and we are then back to positive and negative radiation.

    This still has the problem that you do not admit to the back radiation?

    I think you will find the Berkley model is showing a non generating core and the effect of a single “multi” layer insulator

    If the source were still generating 400 watts the 400 watts must be emitted to space I think all would agree.
    If you substitute that in the equations then J1=400 and J-J1=400 and therefore J=800

    [JP Reply: In problem 1026 the same amount of energy is leaving the shell to outerspace that it is receiving from below from the sphere. You do see that, correct? You do see that problem 1026 exists and you’ve read it? The result is that the same amount of energy is emitted from the shell as it receives from below. Also, backradiation doesn’t cause heating of the sphere in either problem. You see that too? There does seem to be a contradiction between problems 1023 and 1026 in terms of the total energy output – there is no contradiction in that backradiation or radiant trapping does not cause the source to become warmer. The correct answer for thermal equilibrium is when the outer shell/wall/etc reaches the same temperature as the source, with it emitting outwardly all the radiation it receives from below, as we’ve discussed here. “Positive and negative” radiation would just be the usual ‘q’ term we’ve discussed at length. Admitting to backradiation? q ~ T2^4 – T1^4.

    I am not sure why problems 1023 and 1026 seem to describe the same idea but give different results – in terms of halving the energy. They do give the same result in that backradiation or radiant trapping does not cause the source to warm up. Let me solve that problem for you, because it certainly can and is causing confusion.]

  536. Max™ says:

    Brb guys, gonna go get a vacuum chamber, black body paint, a spherical power source with constant and unwavering output which will fit within the chamber… another shell to go around the power source… and I need to find some way to mount it I guess.

    Or I could just make use of the known examples of stars in dust clouds which don’t start into feedback loops because of reradiation.

    Ah well, that’s that I guess, after seemingly agreeing with me, then misinterpreting my statement and example, only to later claim the example was ambiguous/inconclusive, I’ve now been excluded from the discussion for evading questions which I answered several times.

    Poor form, tb.

  537. Max™ says:

    Just to set the record straight here.

    I was asked by tb about what will happen after the shell is added and equilibrium is reached here in this comment.

    This comment was actually intended to be a direct response to his question: ” (a) At radiative equilibrium, J – J₁ = J₁ or J₁ = J/2. Therefore T₁⁴ = T⁴/2, or T₁”, apparently I should have been more specific.

    I responded specifically this question here, and less clearly in the posts above it with the UC Berkeley examples: “Yeah, at radiative equilibrium 235 is radiated by the sphere, 117.5 back towards the sphere, 117.5 outwards.”

    This was a request for feedback from tb: “If the situation described would cause the core to heat up until it was emitting 470 and the shield 235 in/235 out, why would the inward radiation not cause it to heat up again, and again, and again?” which was missed.

    This was a response to clarify a couple of points, and attempting to explain my above missed points, while again looking for clarification.

    This comment about tb’s question from lgl went “unanswered” due to my “evasion” while I was sleeping.

    This comment was my respone after I awoke, which was not passed through moderation.

    This comment was one of several times when I’ve answered this question which I have now been “retired from posting” for not answering.

    [JP Reply: I don’t understand why they a) insist on ignoring problem #1026, b) don’t realize that either problem shows that NO heating of the source occurs. Hello? Either of those problems totally destroys Willis’ shells game – I don’t know why you think you should use them to support you – they DO NOT heat the source – hello???….nope no one there]

  538. Greg House says:

    Will Pratt says (2013/04/01 at 1:24 AM) “…Such people are the gatekeepers, the enablers, of AGW fraud.”
    ==============================================================

    Gatekeeper at work (http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-4/#comment-49184): “Reluctantly, I’ve retired Max from the thread. Hopefully he use the spare time to try a simple experiment to test his hypothesis that adding a steel shell around a heat source will make no difference to the surface temperature of the heat source.”. They do not need to prove their “back radiation warming” point experimentally, they have already done that by Willis’ fictional story!

  539. Rosco says:

    I noted a reference to Roy Spencer on his space is “cold” bullshit. He repaets this stupidity ad infinitum.

    Space has little or no substance therefore physical concepts are meaningless.

    These idiots repeat this to convince suckers the atmosphere acts like a blanket keeping us toasty warm.

    Space has absolutely no constant properties at all – some regions are subject to enormous amounts of gravity, magnetic and other radiation whilst some areas are so remote from any objects there is a minimum amount of anything.

    The space surrounding the Earth is constantly subjected to radiation at levels that would quickly kill an exposed human being or any other animal life for that matter.

    Too try to hoodwink the illinformed that near Earth space is “cold” is deception that we should all condemn- it has absolutely zero in common with deep space – absolutely zero and the dishonesty displayed by these idiots is simply astounding – who would believe anything they say ??

    Or perhaps they are truly stupid to miss this obvious criticism of their space is cold message to further their greenhouse beliefs when near Earth space is awash with powerful solar radiation continually and the only place to avoid it is in the “shadow” of a planet.

    I treat everything these people claim as junk – I won’t call it science ’cause it isn’t.

    How do these first class fools get their PhDs anyway ?

  540. “Hopefully he use the spare time to try a simple experiment to test his hypothesis that adding a steel shell around a heat source will make no difference to the surface temperature of the heat source.”

    Well that’s what the UC Berkely physics problem shows! I got a suggestion: how about THEY perform an experiment to prove that there sphere does the opposite of what the physics text shows?! As if they deem themselves immune from scientific experimentation? Good god the hubris…and illogic.

  541. Rosco, because they’re paid sophist frauds. Or, they’re fools, or lying fools, or whatever.

  542. Max™ says:

    Hey Joe, assuming you were someone with some knowledge of astrophysics, what do you think would be the result if you put a perfectly black dyson sphere around a (record cold) Y-brown dwarf that had a surface temperature of say, 254K?

    ________

    Hrm…

    [Reply] Max, the final straw was the “The temperature of the inner surface is not determined by the radiation it emits” response. Everything radiates according to its temperature and emissivity. I’m trying to draw this discussion to a useful conclusion. and comments such as this just put a spanner in the works. ~tb

    >.>

    Wait, so I got “retired” because I pointed out that emissions do not determine temperature? Indeed, I can not think of a way to produce a temperature change just from emission changes, though the converse is of course true.

  543. Max, it wouldn’t heat up the star some more. Dyson would have thought of that, yes? Being the genius he is? Did he say it would cause the inner star to heat itself and blow itself up? No he did not. It would have destroyed his sphere and we would have realized that.

    TB is being petty, and gatekeeping. The UCB links show that backradiation does not cause heating of the source. QED.

    Have they produced a mathematical equation that describes what they desire? At the beginning of this thread here, they tried, and I destroyed them. So then they want back to word arguments. The UCB link shows the actual mathematics of how heat flow works; where is their mathematics? They need to show equations, not just word descriptions. Word descriptions are meaningless in science – you have to show the equations. So far, they have no equation, so QED, they have nothing. UCB already shows what the equations are, so doubly they have nothing.

    Max, you won this argument right from the beginning in pointing out the surface area difference, and then in finding the UCB links. Great job man…don't let idiocy get you down. You're always welcome to bask in the light of rationality here.

  544. Alan Siddons says:

    “Shifting discussion to other alleged effects does not make sense to me.”

    Well, it makes sense to me that if a Ron C asks about the ‘Effective Radiating Level’ version of the Greenhouse Effect, he shouldn’t be told that this is irrelevant to the discussion. The British and Australian governments swear by it, after all. It’s part of the Met Office’s sales pitch. Not everything revolves around the IPCC, which didn’t invent back-radiation theory, anyway: Balfour Stewart did in his 1871 An elementary treatise on heat. Download it
    http://archive.org/download/elementarytreati00stewrich/elementarytreati00stewrich.pdf
    and you’ll see him playing the same shell game, only his version has a shell made of glass rather than steel. Same stupid assumption, too, that the glass will radiate twice as much energy as the surface. Quoting him, “Now let R’ denote the radiation of this envelope outwards into space, then R’ will also approximately denote the radiation of the envelope inwards towards the sphere, since as the envelope is very thin, both its surfaces may be imagined to be of the same temperature. Hence the radiant heat which leaves the envelope will be 2 R’…”

    The greenhouse paradigm is wide and very deeply rooted, Greg. Every branch of this horrible tree deserves attention.

  545. Max™ says:

    Huh, there’s an aspie question, why would someone choosing to shut you out be able to get you down?

    I assume there is some sort of social transaction I’m missing going on here… hell, are there social ramifications in being shut out of the thread? Lemme think, we’ve got what… group beliefs, ideological defense, hmmm, would the need to save face come into play too?

    Asked the woman about, she’s my social wizard/expert/translator, she said that the “wrong” here is being silenced, I’m annoyed that I made a strong point, it was almost agreed with, and then the near-agreement was reversed, the point was misinterpreted, downplayed, and dismissed… and I’m not able to defend it!

  546. lol…yah, indeed.

    Again, just ask them to show their equations. They never show any. Spencer never shows any, Willis doesn’t show any, TB doesn’t show any, no one shows any equations. They just describe what they want…which is meaningless. The UCB links show the equations, and they show what they show – no heating of the source. Fini. If they don’t have the math to explain what they’re pretending – and they don’t because UCB actually has the math – then whatever they’re doing is childish amateurish make-pretend. I mean seriously, just think about how much of a joke this all is, that they can’t even put what they’re talking about into math, into an actual equation of heat flow, or any equation at all. Truly it is a great victory for us.

  547. Greg House says:

    Alan Siddons says (2013/04/01 at 8:50 PM): “Well, it makes sense to me that if a Ron C asks about the ‘Effective Radiating Level’ version of the Greenhouse Effect, he shouldn’t be told that this is irrelevant to the discussion.”
    ===========================================================

    Well, as I said, other versions including this ERL word salad are certainly interesting, but as far as I know, the IPCC summaries for policy makers based on their reports based on their< (“greenhouse effect”=”back radiation warming”) concept are the basis for the policies of cutting CO2 emissions, and not Mosher’s or whoever other hypothesis.

    So, for practical political reasons such a shift in a debate would be a distraction and let warmists off the hook.

    To put it in a simple way, they are on the hook now with their 150 years old and 100 years ago already experimentally debunked concept of back radiation warming. They can only obfuscate the matter by providing a lot of propaganda on blogs and in the press. Let me ask you something: you do not really suggest governments should be told that they do not need to cut the CO2 emissions because Mosher is wrong on something, do you? So, yes, it has been about and around the IPCC and this will not change. Demonstrate the governments and the people that the IPCC has been wrong from the very beginning, and warmists are done.

    [JP reply: Hi Greg, of course Mr. Siddons doesn’t think things so simple. But the points you made are good: having multiple versions of the GHE is exactly what these people use to defend it! When you bust the IPCC version, which has been repeatedly, well then all the alarmist community needs is Mosher, Roy, Willis, etc., to come up with other versions equally senseless and equally void of mathematics and physics. Their confusion and their refusal to abide by a strict definition of the GHE is exactly what gives them their strength of obfuscation and sophistry. The shell game was roundly debunked right here, and what do they do? Just pretend it didn’t happen. Mosher et al THINK that their versions are equivalent to the IPCC. I agree with you that debunking the IPCC version is the goal, however, it has already been done, so, we keep trucking along.]

  548. M Postma
    Question 1023 refers to a fixed temperature body.
    It is not a power source.
    So your response to my question is not valid:
    A core generating 400 watts receives 200 watts back radiation from the shell the shell transmits 200 watts radiation to space is the what you are claiming the Berkley question shows, The only way this could be true is if the generator and back radiation cancel and we are then back to positive and negative radiation.

    This still has the problem that you do not admit to the back radiation?

    If the source were still generating 400 watts the 400 watts must be emitted to space I think all would agree.
    If you substitute that in the equations then J1=400 and J-J1=400 and therefore J=800

    [JP Reply: In problem 1026 the same amount of energy is leaving the shell to outerspace that it is receiving from below from the sphere. You do see that, correct? You do see that problem 1026 exists and you’ve read it? The result is that the same amount of energy is emitted from the shell as it receives from below.
    =======
    tfp: so this agrees with my 1023 comment j1=400 J=800
    =======
    Also, backradiation doesn’t cause heating of the sphere in either problem.
    =======
    tfp the core is constant temp according to 1023.
    =======
    …there is no contradiction in that backradiation or radiant trapping does not cause the source to become warmer.
    ===========
    If the core is generating 400 watts and 200 watts (according to 1023) is coming back into it It must still be generating 400 watts so the shell has to emit 400 watts to prevent thermal runaway.
    Remember the core is a generator not a fixed temperature

    [JP Reply: Right, the shell emits the equivalent of 400 Watts outwards (equivalent depending on its radius/surface area), and that’s what the core generates, so, no backradiation heating and all energy balances out.]

  549. Will Pratt says:

    [Reply: Yes, and it (emissivity) is a function of surface properties, and is not equal for equal temperatures for different surfaces, unless they’re blackbodies or by happenstance.]

    Sorry Joe but that is so vague as to be meaningless to me. If you heat two pieces of steel to glowing red hot, it makes no difference if of one was a piece of polished steel and the other an oil blackened piece of steel, you have changed their emissivity by actively heating them to red hot.

    You need to specify the conditions when you talk about emissivity.

    Anyway I am not here to argue in circles just to keep the debate going. That is how the Hegelian Dialectic works. To keep debate in flux while moving closer to the goal.

    Greg House says:
    2013/04/01 at 4:58 PM

    Yes Greg, Roger Tattersall is a liar and a gatekeeper of AGW fraud. He has come direct from WUWT and his apparent “spat” with Watts and Eschenbach was pure play acting. It was obviously used to split the sceptics on the WUWT forum who were waking up to Watts and the fallacies of the “GHE”. The tallbloke talk it to death shop is just another AGW fraud gatekeepers rabbit hole. That has always been obvious to me.

    The biggest mistake that genuine sceptics are making and keep making is to think that they can come to blog sites and get closure on an issue that we all know is a scam. Such sites are not there to provide closure, they are there for quite the opposite reason.

  550. Well lets make some closure shall we? I know how to do it and I’ll begin writing the report.

  551. Strike that. All you got to do is ask them to show their math and their heat flow equations. It’s that simple. They don’t exist. They don’t exist.

    They think that science is “just saying something”. Unfortunately that is what a lot pf people think science is, but it isn’t.

  552. 1}} Joseph E Postma says: 2013/04/01 at 9:01 AM @tfp 2013/04/01 at 8:38 AM
    It shows backradiation which does not cause any increase in temperature. Please look at what it says and sketches carefully. The backradiation does not increase the temperature of the source. The only thing the shield does is decrease the rate of emission – and this does not increase the temperature of the source. The result couldn’t be any more clear: T1 = T/4root(2), and so there was no change to T, the source temperature. Answers a) and b) say nothing about the source becoming warmer, with the presence of backradiation, as would be expected from thermodynamics.
    …Both the problem 1023 and 1026 are consistent in that the presence of backradiation does not increase the source temperature – that is very explicitly stated in the results. Back radiation can exist, but q can be equal to zero. q being equal to zero does not mean that backradiation doesn’t exist. Backradiation can exist and not cause any heating.
    —————-
    2}} thefordprefect says: 2013/04/01 at 9:55 PM
    If the core is generating 400 watts and 200 watts (according to 1023) is coming back into it It must still be generating 400 watts so the shell has to emit 400 watts to prevent thermal runaway.
    Remember the core is a generator not a fixed temperature

    [JP Reply: Right, the shell emits the equivalent of 400 Watts outwards (equivalent depending on its radius/surface area), and that’s what the core generates, so, no backradiation heating and all energy balances out.]
    =========
    You cannot have it both ways
    in 1 you say back radiation exists from the inside of the shell and halves the energy loss from the core (400w to 200w) and the remaining 200 w goes to space from the outside of the shell
    in 2 you say there is no backradiation to the core reducing the energy loss so 400 watts from the core hits the shell and the shell then radiates 400w to space. If there is back radiation then the 400watts from core to shell gets split between back and fore radiation

    All the problems in the document show that a shell will radiate from both sides equally (assuming emissivity is the same)

    I assume therefore that you are saying that this is wrong?
    If so can you please explain what the difference is between the inside of the hot shell and the outside of the hot shell that prevents the inside radiating.

    If you are saying that the radiation from the inside does nothing to the warmer core then you are saying the physics in the paper is wrong. But then we already agree that multilayer insulation is a proven concept and this relies on back radiation reducing the net radiation from the warmer core.

    so

    does multilayer insulation work
    Does MLI rely on backradiation. If not then how does it work radiatively.
    If back radiation does not exist then presumably the questions in the document are wrong.
    If back radiation does exist then what happens to it when it approaches an object hotter than the one it left.

    I am very confused and would really appreciate a summary of the physics you believe is happening.
    cheers.

  553. also why does the inside of a shell not radiate like the outside of a shell. The heat does not know which side it is leaving?
    If it does leave the inside where does it go.
    The steel green house is based on earth dimensions so the shell to a reasonable accuracy is the same size as the core.

    Ps. Please try not use your T^4 equations here as we are talking radiation which is not the same!

  554. I’ll jut answer this one. tfp: “Try not to use T^4 equations, because we are talking radiation, which is not the same!”

    And with that, I am sorry, but I can not help you any longer. You do not have the education level required to discuss this on an informed basis. Real science and mathematics is a meritocracy, and if you’re not smart enough, it isn’t an insult, it is just a statement of fact. If you go to university for a physics degree, you will start to understand your own questions by the end of the 2nd year, in the North American system. If you have been through such a system, it has failed you. Your questions have been answered repeatedly now, and if you didn’t understand the answers before, you won’t now.

    You’re just not smart enough, at this point in time. It is a telling example of the type of mind that believes in the GHE. There is nothing I can do about it, and nothing I can do to help you. You are simply where you are at, and you likely won’t be any better until your next life or something, or if you actually go get a degree in physics. Go actually get a degree in physics…even just an undergrad. If you have one, something went very wrong. It is clear you do not have one, since you don’t use any math and equations in your discussions.

    Radiated energy does not equate to net heat transfer or even net energy transfer. The equation of heat flow for radiation, from physics, from actual physics textbooks and from actual universities and actual physics degrees, is q ~ (T2^4 – T1^4). If T2 = T1, then q = 0, and nothing heats up, even though there’s all that radiation. It works the same way in conduction.

    A temperature does not spontaneously heat itself up. You have the UCB links Max provided which precisely show that the sphere does not heat itself up. Please work those questions out for yourself on paper as well. If you can’t do that, then you will never be able to understand thermodynamics. You could understand thermodynamics if you simply adopted the knowledge and intuition that a temperature does not heat itself up.

  555. Alan Siddons says:

    Re: the ERL and other alternative ‘explanations.’ Absolutely right, that’s the tactic — these various arguments serve to keep moving the goal posts. After you’ve discredited back radiation as any sort of heating mechanism, they’ll just tell you that you don’t understand how the GHE really works. “This doesn’t involve active surface heating by radiation but rather the emission delay engendered by a higher Effective Radiating Level.” And on and on it will go. It’s futile arguing with them. A better strategy is to inform the man in the street about this scam.

  556. Kristian says:

    tfp, you say: “does multilayer insulation work
    Does MLI rely on backradiation. If not then how does it work radiatively.
    If back radiation does not exist then presumably the questions in the document are wrong.
    If back radiation does exist then what happens to it when it approaches an object hotter than the one it left.”

    Multilayer insulation works because each layer absorbs less than the entire flux from the layer beneath it. The rest of the flux is reflected back. Less than perfect absorption from the source, lower potential temperature than the source, less heat loss to the next layer and ultimately to space.

    What the (simply asserted) solutions to the UCB problems seem to forget is that the shell can only ever lose heat through its outer surface, to space. Space is its cold reservoir. The planet is its hot reservoir. Heat gain comes from the hot, heat loss goes to the cold. Always. This basic thermodynamic fact is ignored again and again.

    This means that as the shell gains e.g. 800 W/m^2 from the planet within, in the UCB (and Willis’) setup it can only rid itself of 400 W/m^2 through heat loss. Everyone (?) should be able to see intuitively that this can never be a steady-state situation for the shell. It can not gain twice as much as it expels and maintain a fixed temperature. The temperature must rise. As long as the central planet is provided with a constant heat input, it will simply heat the shell more an more an more, albeit at a steadily slowing pace (q is getting smaller and smaller, the warmer the shell gets), until the temperature of the shell is effectively and dynamically equal to the planet’s. Never during this gradual heating process of the shell will the core planet warm (or cool) itself. Why on earth should it? It is simply an absurd idea. It is the shell (the cold object) that’s being heated, not the planet (the hot object).

    At the point where the shell has attained a temperature ~equal to the planet’s, a flux of 800 W/m^2 goes OUT to space from the shell and 800 W/m^2 goes INWARDS from the shell. But only the former flux is heat loss for the shell. So there is perfect balance. And q is 0. The planet can no longer make the shell any warmer. Still, it continuosly sustains the steady-state temperature of the shell. As soon as 800 W/m^2 leaves its outer surface, a new 800 W/m^2 comes in to replace it from the planet inside.

    This course of events will inevitably be the case as long as the shell at all times absorbs the entire incoming flux from the planet. In Willis’ model it does! If it DOESN’T however (like with reflective radiative insulation), then it will end up cooler than the planet.

    Hope this helps …

  557. [JP Reply: Trashed because we’ve already answered you. q from the shell to the planet is 0. ZERO. There is no heat loss from the shell to the planet. Even if the shell is emitting on the inside, there is no heat loss to the planet. The only direction the shell can lose heat is outwards, and hence it loses the equivalent of 800 W/m2 outwards.]

  558. Temperatures are not all
    wiki:
    Thermospheric temperatures increase with altitude due to absorption of highly energetic solar radiation. Temperatures are highly dependent on solar activity, and can rise to 2,000 °C (3,630 °F).

    How does this gell with your T^4 equations. q to earth must be staggeringly big.!!!
    wiki:
    Even though the temperature is so high, one would not feel warm in the thermosphere, because it is so near vacuum that there is not enough contact with the few atoms of gas to transfer much heat. A normal thermometer would read significantly below 0 °C (32 °F), because the energy lost by thermal radiation would exceed the energy acquired from the atmospheric gas by direct contact. In the anacoustic zone above 160 kilometres (99 mi), the density is so low that molecular interactions are too infrequent to permit the transmission of sound.

    [JP Reply: The thermosphere doesn’t violate the Stefan-Boltzmann Law, and nothing about this violates T^4. It fits exactly and precisely into it. If you don’t like T^4, you must really hate the Stefan-Boltzmann Law.]

  559. Greg House says:

    thefordprefect says (2013/04/02 at 6:20 AM): “Does MLI rely on backradiation. If not then how does it work radiatively.”
    ===========================================================

    Very simple. Apparently, multi-layer insulation in vacuum prevents satellites from overheating by the Sun. Back radiation neither slows down cooling of the Sun (source) not does it warm the Sun up.

    If you mean that back radiation affects the temperature of the source (the IPCC position and the one of some so called “skeptics”) I suggest you present a real experimental proof for that.

  560. Alan Siddons says:

    Here’s another angle on a radiating shell. In real life, thermographers often attach black tape to the target of interest in order to make it radiate more like a blackbody. The tape increases the target’s emissivity, thereby making it easier for thermographers to assess its true temperature. If this black tape were attached to a self-heated steel ball the same thing would occur: it would become brighter in infrared and thus LOSE energy faster than it did before. So would suspending a shell of absorptive/emissive tape around the ball change the result? Yes. Now deprived of direct contact, the radiatively efficient shell could no longer absorb the steel’s HEAT, only its radiation. The shell would therefore emit the same amount of energy that the steel ball is emitting, albeit over a larger surface area. In short, the self-heated ball is in the same thermal condition with or without a suspended shell. There’s no such thing as a free lunch.

  561. Greg House says: 2013/04/02 at 9:05 AM
    If you mean that back radiation affects the temperature of the source (the IPCC position and the one of some so called “skeptics”) I suggest you present a real experimental proof for that.
    ==========
    I linked the experiment I did – the post got trashed
    But just in case here is the link again. I would suggest posting there as I think my replies would get trashed here!
    http://climateandstuff.blogspot.co.uk/2013/04/the-copper-greenhouse-new-test.html

    [JP Reply: You get trashed comments for asking the same question which has been answered dozens of times.]

  562. Greg House says:

    thefordprefect says (2013/04/02 at 10:13 AM): “I linked the experiment I did – the post got trashed
    But just in case here is the link again. I would suggest posting there as I think my replies would get trashed here!
    http://climateandstuff.blogspot.co.uk/2013/04/the-copper-greenhouse-new-test.html

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

    Very clever. Now, let me formulate the question more precisely. The IPCC referred to the back radiation warming effect (“greenhouse effect”) in all their reports and explanations. A good example is this: http://www.ipcc.ch/publications_and_data/ar4/wg1/en/faq-1-3.html, published in 2007. So, my question is: what is the real scientific experimental proof for back radiation warming effect published prior to the IPCC reports? Please, present clear references and description of the experiment that is supposed to prove specifically affect on temperature of the source.

  563. Greg House says:

    thefordprefect says (2013/04/02 at 10:13 AM) “…”
    [JP Reply: You get trashed comments for asking the same question which has been answered dozens of times.]

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

    By the way, since he refereed to IR thermometers as a proof for back radiation warming, I asked him 3 times on this thread the same question about how much exactly IR from a colder target warms the sensor. Still no answer. I wonder why.

  564. Ron C. says:

    Thanks for this discussion. I learned a lot and can now respond to those who advocate backradiation. My university physics basics needed some refreshing, and that was provided here.
    Thanks to those who commented regarding the ERL version of the warmist theory. I do think that other than backradiation, many warmists (and luke-warmists, especially) prefer to argue the “warming due to delay” kind of theory, of which ERL is a variation. This is the notion that more CO2 is slowing the cooling of the planet surface, and as the solar heat continues, the system temperature rises as a result.
    I take the point made above that a higher ERL is a result, not a cause of higher temperature in the troposphere. But I would like a better grasp of the fallacy in the delayed cooling notion.

  565. Kristian says:

    Joe,

    Looking at Max’ two problems, 1023 and 1026, it is quite evident that they are all the time implicitly referring to the inner BB sphere as a cooling one, that is, one that is somehow heated to a temperature T but then as the process is set off immediately starts losing heat to its surroundings, its temperature dropping as a consequence. A continuously heated object will not cool, after all. So any cooler nearby object could not affect its cooling rate, only the heat transfer rate from the warm to the cool object.

    This being the case is shown quite clearly in problem 1023 when they state that Q” = Q’. What they are basically saying with this is that the cooling rate of the planet will always match the heating rate of the shell. And that is exactly what will happen if you put two (unpowered) objects with different temperatures next to each other: The heat transfer rate between them (q) would be exactly the same seen from the cold object as seen from the hot, only with opposite sign. q seen from the cold object would classify as ‘heating rate’, q seen from the hot object would classify as ‘cooling rate’. In such a setting, in an infinite vacuum at 0 K, the temperatures of the planet and the shell would finally converge at 0 K (Tshell = Tplanet / 2^0.25). It would simply take longer for the planet to reach 0 K surrounded by the shell than not.

    Now, what if the hot object didn’t cool because it was being constantly powered? Then the expression Q” = Q’ couldn’t hold true.

    It seems the appropriate expression in such case would rather be Q’ = Q – Q” … Here Q’ is the heat transfer rate between planet and shell. It might still be seen as the heating rate of the shell, but significantly NOT as the cooling rate of the planet. The planet after all maintains a steady cooling rate to match the heating rate from its power source. It does not change. It is Q = Q’ + Q”, Q” being the cooling rate of the shell and by extension of the system as a whole. At dynamic equilibrium, Q” equals Q and Q’ is 0.

    [JP Reply: Well the easiest way to look at the inner planet as a constant source is as an “infinite heat sink” – this is a concept used every single day in thermal physics because it provides the idea of something continuously heated at a constant temperature internally etc. We just avoid having to discuss what is causing the heating and just say the heat source is “infinite”. Now, your analysis is good about the energy loss & cooling from the planet being the gain of the shell, and that’s fine, but now, we can truly think of the planet as constantly producing the same intrinsic output. It can lose energy but this doesn’t result in cooling because the energy is replaced. How do you heat up the infinite heat sink? Only, as ever with something that is hotter – although it will take a long time because of its size but the very pertinent point is that the direction of heat flow, q, is from the heat sink to a cooler object and the cooler object can never send heat into the sink (and thus heat it) unless the cooler object/shell is warmer than the sink, because only then will q be directed into the sink. As we’ve been pointing out, the shell can basically never send heat, q, back to the planet, and so the 800 W/m2 “both ways” idea simply isn’t a problem – this is a mathematical physics fact, because this is how we define heat/energy transfer, with q. At a continuous infinitesimal level on the knife edge of the present, as parcels of energy leave the shell, they get replaced from underneath by the planet.]

  566. Ron C. says:

    While waiting in moderation, I had these thoughts about “warming due to delayed cooling.”

    Several points could be made. Some say that present concentrations of CO2 have already saturated the frequency of IR that is absorbed. Others say (Miskolczi, I think) that at the TOA increased CO2 is offset by decreased H2O, and vice-versa, resulting in a fairly constant optical depth, so no change in cooling. Another point is that radiative transfers are instantaneous, and even if more CO2 increased the number of such transfers, it couldn’t delay cooling from surface to TOA even for a minute.

    [Reply: “Delayed cooling” is one of their sophist tricks. Delayed cooling doesn’t mean heating. This is explicitly stated in those UCB problems Max posted. Delayed cooling means cooling more slowly. They sophize with it and expect you to accept, and most people do, that delayed cooling means getting warmer. But that’s a cheat of logic.]

  567. Will Pratt says:

    Joseph E Postma says:
    2013/04/02 at 4:44 AM

    Closure . . . . .

    Joseph E Postma says:
    2013/04/02 at 5:19 AM

    . . . . . not !

    [Reply: LOL :)]

  568. Max™ says:

    ” Question 1023 refers to a fixed temperature body.” ~tfp

    No, it does not.

    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 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 radiatied to the surroundings?

    (b) How is the total power radiatied affected by additional heat shields?
    (Note that this is a crude model of a star surrounded by a dust cloud.) UC Berkley

    Solution:
    (a) At radiative equilibrium, J – J₁ = J₁ or J₁ = J/2. Therefore T₁⁴ = T⁴/2, or T₁

    (b) The shield reduces the the total power radiated to half the initial value. This is because the shield radiates a part of the energy it absorbs back to the black sphere.

  569. Rosco says:

    I am amazed that this is still going with people trying to prove it is OK to arbitrarily double energy by adding a spherical steel shell around a spherical radiating object.

    Cannot they understand that this amounts to a function of 2^n times the radiation ?

    If this proposition is valid for one shell then adding another or any number of extra shells – ignoring geometrical errors – ultimately results in infinite energy as the number of shells increases from a finite source and that result is absurd.

    I read Willis’ original proposition on the steel greenhouse and he proposed adding one shell doubled the energy but adding another only increased the original energy emitted from the planet by 3 ???

    So one shell gets heated to equilibrium and radiates 235 in and out but an extra one surrounding this ensemble only gets heated to the extent it radiates 162.5 in with 235 out ???

    How the hell did he achieve the 3 result ?

    Just make stuff up I reckon !

    Someone on WUWT said I simply failed to understand the science and mathematics and recommended I study the hypothesis more closely – it is a valuable model for how things work in greenhouse land ???

    [Reply: Morons. There IS NO MATHEMATICS which describes the idea! The math is 2^n, that’s it! Just because they use numbers they think they’re doing math. They don’t actually have an equation that can be written for it, because I did that for them at the beginning here and none of them could be physically justified or looked anything like actual heat flow equations.]

  570. Max™ says:

    Uh, I don’t think the shell will reach the same temperature as the ball originally reached.

    It has to emit with twice the surface area before factoring in the inverse square law changes on radius.

    If x energy is supplied from a sphere with area A to a surrounding shell with area > 2A then the total radiation from each side of the shell shouldn’t be x, I’d think it would need to be x/2.

    If the sphere was at 254K (235 W/m^2) and the shell reaches any temperature above 213.36K (117.5 W/m^2) then it would be emitting more energy than it receives, wouldn’t it?

    [Reply: Well what if the shell and planet were touching? Then obviously the shell comes to the same temperature. Solution 1026 says the same thing will happen radiatively but with the radius factor reducing the energy density a bit. This radiation problem is not trivial at all, and I think this non-triviality is the door-way to the sophistry and accidents of thought which can be created with it. We’ve been discussing ‘q’ a lot, but what if there’s an outside ambient radiation field? Then q to the outside won’t be the full 235 W/m2, just like q to the inside is 0 if there’s radiative equilibrium on the inside. So if the outside radiation field corresponded to, say, 213K or 117.5 W/m2, then, q from the shell to the outside will only be 235-117.5 = 117.5.
    Fine. However, what was the sphere outputting to the outside originally without the shell in the first place? If the sphere was an infinite source, at 255K, then the sphere’s q to space was actually only 117.5 W/m2 as well. Remember, that is the actual math of heat flow, this q parameter due to a difference in temperature; if there is no difference in temperature, then q = 0, and this does not mean that anything heats itself up! So, the shell simply becomes the new surface emitting the equivalent q (heat transfer) of 117.5 W/m2 which the planet had been; if the shell is very close to the planet’s size, then it is very close to the same temperature as the planet.]

    ]

  571. Rosco says:

    I remain completely amazed that people think radiation has very much to do with heat loss in an atmosphere.

    Radiation is demonstrably a very slow heat loss mechanism – simply note the rate of “cooling” data from the Moon surveys.

    At the Moon’s maximum surface temperature the rate of cooling (which is a maximum at higher temperatures) is about 1 K per hour.

    In an atmosphere we rely on forced atmospheric convection or water circulation and forced convection in all our machinery to maintain a safe operational temperature.

    If we relied on radiation we would have no viable thermal powered machinery at all.

    Even heat sinks in electrical equipment relies on convection – its why they have the slots in the enclosure and why they insert fans in equipment which uses larger amounts of current.

    You can’t even talk of measuring radiation unless you thermally isolate the experimental apparatus to exclude conduction or convection.

    I’ve seen no end of stupid comments where some one throws some sort of object over a heated object and notes the temperature increases and claims backradiation as the cause.

    Haven’t they heard of thermal insulation whereby conduction to and convection in the free atmosphere is suppressed and hence an increased heating capability is falsely witnessed when all that has happened is the object being insulated is suppressed from losing energy at the rate it does minus the insulation.

    Pots of water do not get hotter by backradiation when a lid is placed on them – the heated water evaporating or boiling is trapped in the covered pot and, with it, the heat that would have been lost without the lid.

    Climate science has obviously caused a major problem in academia with real science that has been known and exploited for hundreds of years being corrupted by an unproven hypothesis being taught as fact.

    Heaven help us !

  572. Greg House says: 2013/04/02 at 10:54 AM
    http://www.patarnott.com/atms749/pdf/LongWaveIrradianceMeas.pdf
    Fig 2 and 3 both show downward LW IP being measured 280w/sqm+ how would you explain this? where is it coming from?

    This test I did shows that cool object radiating to a hot object can reduce the heat loss and thus will increase temperature of the hot object.

    There is little chance that the greenhouse effect can be measured directly. The test setup is too large.

  573. Greg House says:

    thefordprefect says:
    2013/04/02 at 7:15 PM

    Greg House says: 2013/04/02 at 10:54 AM
    http://www.patarnott.com/atms749/pdf/LongWaveIrradianceMeas.pdf
    Fig 2 and 3 both show downward LW IP being measured 280w/sqm+ how would you explain this? where is it coming from?
    =============================================================

    I did not say that, it is your words. Could you please format your quotes properly?

  574. Alan Siddons says:

    Ron,

    Here’s the standard line of thinking about the ERL as I see it.

    1. You first accept that the Earth would be far colder (minus 18°) without radiating trace gases (GHGs).
    2. Then you swallow the notion that the atmosphere of this colder Earth would be isothermal, i.e., always the same cold temperature (minus 18°) as you ascend in altitude.
    3. Then you convince yourself that GHGs “tamp down” the photons that are trying to escape from the Earth’s surface. The higher up you go, though, the less effective is the tamping down process until all the photons escape. Where enough photons escape as to conform to the radiation from a minus 18° blackbody, THIS is the “Effective Radiating Level,” a term which is meant to convey that the special GHG zone has formed a SECOND SURFACE that corresponds to what a minus 18° Earth would emit. Above that point the temperature gets even colder — in other words, not isothermal at all.

    So what’s the upshot? Near the Earth’s surface “lingering photons” have made the air warmer than it would have been, and above the ERL colder than it would have been. The flaw, of course, is that what’s being described is merely an adiabatic slope, a feature common to ALL planets with an atmosphere — with or without GHGs!

  575. Greg House says:

    thefordprefect says (2013/04/02 at 7:15 PM): “http://www.patarnott.com/atms749/pdf/LongWaveIrradianceMeas.pdf
    Fig 2 and 3 both show downward LW IP being measured 280w/sqm+ how would you explain this? where is it coming from? … There is little chance that the greenhouse effect can be measured directly. The test setup is too large.”

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

    Your answer has nothing to do with the question. Let me remind you: “The IPCC referred to the back radiation warming effect (“greenhouse effect”) in all their reports and explanations. A good example is this: http://www.ipcc.ch/publications_and_data/ar4/wg1/en/faq-1-3.html, published in 2007. So, my question is: what is the real scientific experimental proof for back radiation warming effect published prior to the IPCC reports? Please, present clear references and description of the experiment that is supposed to prove specifically affect on temperature of the source.”

    And the other thing. How much exactly IR from a colder target warms the sensor of an IR thermometer, for the 4th time? It is your point that there must be some warming there, and you seem to be an expert on IR thermometers. Or did you make it up?

  576. Alan Siddons says:

    “You can’t even talk of measuring radiation unless you thermally isolate the experimental apparatus to exclude conduction or convection.” — Rosco

    Which is why Prefect’s experimental conclusions are bogus.

  577. Max™ says:

    Downward longwave exists, that isn’t the point.

    We’re not saying it doesn’t exist, we’re saying it isn’t going to raise the temperature of a heat source.

    [Reply: Yes. Q!]

  578. Kristian says:

    Max™ says, 2013/04/02 at 6:31 PM:

    “It has to emit with twice the surface area before factoring in the inverse square law changes on radius. If x energy is supplied from a sphere with area A to a surrounding shell with area > 2A then the total radiation from each side of the shell shouldn’t be x, I’d think it would need to be x/2.”

    Don’t confuse W with W/m^2, Max.

    In these problems, the shell does not receive a power flux. It receives a power density flux. Area is already factored in. The only way to reduce a power density flux is to increase the distance from the source (inverse square law). It does not matter how large the area irradiated by that flux is, as long as it’s at the same distance from the source. The power density flux reaching the Moon from the Sun is 1361 W/m^2, exactly the same as for Earth, even though on Earth it needs to be spread across a much larger area. Why? Because both bodies are 1AU from the emitting heat source (the Sun). If you took away the Moon, it doesn’t mean the power density flux hitting Earth would suddenly rise.

  579. Max™ says:

    I’ve brought up exactly the same point on tb’s thread, but intensity (power density of radiation in W/m^2) is not a conserved quantity either, only total energy is for this case.

  580. Kristian says:

    It doesn’t matter. If the shell absorbs the entire incoming flux from the planet and this flux is continuous and constant, then eventually the shell will have built enough KE, in other words, contain all the energy it needs to emit according to its final steady state temperature, which will be effectively equal to the planet’s. If the shell has a thermal mass and receives only one parcel of energy worth of 235 W/m^2, then of course the shell will not be able to attain the temperature of the planet. This however is not such a situation.

  581. The emission of the shell will be affected by its increased surface area – to conserve 235 W/m2 coming from the planet, a larger surface area shell will emit at a lower flux density, and hence lower temperature.

  582. Max™ says:

    If the shell emits 235 W/m^2 in and out then it is emitting twice what it receives, isn’t it?