## No Math

I already debunked the steel greenhouse idea, but here I will simplify it all to one crystal clear concept.  Won’t bother with quotes from these people etc. since that has all been covered previously.  Let us just look at their pictures:

Image Source: Willis Eschenbach

Image Source: Willis Eschenbach

Do you see what they did there?  Look at what is being radiated outward originally by the sphere core, and then what is being radiated outward afterward by the shell.  Look at the numbers and the units.

They conserved energy flux density, not total energy!  Energy flux density, W/m2, is not a conserved quantity.  Only the total energy measured as power, W, is. And of course W, a Watt, is a Joule per second (J/s), which is the flux multiplied by the surface area of the emitter.  They didn’t factor in the surface areas of the objects at all, but doing that is essential if you intend to conserve energy.

Radiative flux decreases as the inverse square of the distance from the source.  Total radiative power doesn’t!  The inverse-square law of radiative (and gravitational for ex.) flux is one of the most basic and fundamental laws in science.  The people who promote this steel greenhouse thing either don’t know it, or they’ve heard of it but don’t know how to apply it.  The scientific incompetency of the people who believe in the greenhouse effect and climate alarm should be enough to indicate that the entire ideology must be wrong.  And it does.

They actually do no math at all, and no physics at all, and then assume the result they want in the first place.  It is really awful to do that.

## The Math

Due to the inverse square law, the shell only receives a flux given by

1]                                                  Fsh = Fsp(Rsp/Rsh)2

where Fsh is the flux at the shell, Fsp is the flux from the sphere, and R is the radius of either the sphere or the shell.  This equation for the flux at the shell conserves the total energy.  This necessarily requires that the flux is not conserved.  The flux immediately begins decreasing in intensity as the radiation moves away from the sphere. Rsp is always smaller than Rsh and hence Rsp/Rsh is always less than 1, and therefore Fsh is always less than Fsp.  And that ratio less than one decreases at its square.  That’s the inverse square law.

Even if we just pretend to go along with their doubling of the original core flux Fsp, then we still simply arrive at Fsh = 2*Fsp(Rsp/Rsh)2 and so the flux at the shell is still, obviously, a function of distance from the sphere.  Either way it is impossible that the shell could outwardly emit the same flux as the sphere.

The flux that the shell absorbs on its inside corresponds to the temperature that it will develop.  Using the standard Stefan-Boltzmann equation gives

2]                                                  Tsh = (Fsh/σ)1/4

The flux received on the inside of the shell determines the temperature the shell can achieve, and assuming a thin shell then this temperature is what determines the outward flux on the outside of the shell, which will be the same flux as received on the inside, and thus, equation 1].

You can go back to the steel greenhouse debunking thread and read the original silly argument made for it, but here I will walk you through what actually happens:

First we start with the sphere core radiating 235 W/m2 over its surface area, which means it has a surface temperature of -19.20C (using the Stefan-Boltzmann equation and the usual ideal blackbody assumption).

The flux from the sphere decreases as the inverse square of the distance ‘r’ from the surface of the sphere, and so as a function:

3]                                           Fsp(r) = 235*(Rsp/r)2

where r > Rsp.

Now we add a shell around the sphere.  The shell has a radius at r = Rsh and so the flux from the sphere at the distance of the shell is:

4]                                     Fsp(Rsh) = 235*(Rsp/Rsh)2

The shell absorbs this radiation on its interior, and with the simplifying assumption of it being of negligible mass and thickness, then the temperature induced by this flux is the same on the inside and the outside, and so the temperature of the shell is:

5]                                      Tsh = (235*(Rsp/Rsh)2/σ)1/4

This temperature will be less than the temperature of the sphere because the radii ratio in the equation is less than 1, and the temperature of the sphere is given by (235/σ)1/4.  The shell then radiates on the outside at this temperature, and thus, this exterior radiation will fully emit all of the energy that is absorbed from on the inside from the shell.

And that’s what happens.  That’s it.

The shell’s temperature is maintained by the sphere.  The shell’s temperature is created by the sphere, and it requires all of the sphere’s energy to maintain it, because the same amount of energy that the shell receives on its interior, it loses on its exterior.

The shell is not an independent ambient system around the sphere.   It doesn’t provide heat or energy to the sphere that it is being heated by – the only heating that is being done is from the sphere, to the shell, and in thermal equilibrium, the sphere supplies exact amount of energy that the shell emits on its exterior.  Since the shell emits all of the energy from the sphere, then it is not possible, for at least this reason, for the shell to heat the sphere.  Of course there are many other reasons from the viewpoint of the various explanations thermodynamic physics could provide for that.

In the silly steel greenhouse idea, the shell causes the sphere to heat up some more because the shell is said to emit to the sphere.  However, if the sphere now has an increased temperature, then it emits more energy as required, and thus, the interior of the shell has to see more energy coming from the sphere, and thus, the shell has to warm up some more again, and thus, the sphere would see yet a higher “ambient” environment it itself created, and thus would have to warm up some more again, etc. The runaway heating problem can not be avoided by arbitrarily stopping the physics after a single iteration.

By the direct math and physics of this steel greenhouse scenario, it debunks the climate greenhouse effect itself.  When you do the math and physics correctly, you would never even invent the steel greenhouse idea in the first place.

The steel greenhouse is the debunk of the climate pseudoscience greenhouse effect.

## End with some Physics

Just a quick comment that was posted by Rosco, which helps make the point from the previous post on the nature of light.

Remember the Planck curves of the two temperatures, and that only the hotter one has microstates activated at high frequency?

Recall that Q is the difference between the two Planck curves, i.e., it is the heat flow between the two objects. If these two objects were facing each other, then the hotter source object loses high energy, high frequency photons that the cooler object neither can nor does replace.  That’s what the black, middle curve shows – what frequency components the cooler curve is missing, particularly at the highest frequencies.  Since it is missing those frequencies, that is why radiation from a cool object can not heat up a warmer object.  The hotter source radiation is certainly going to activate some higher frequency microstates and energies in the cooler object, and thus raise its temperature, but it is impossible for the cool object to do that in reverse because those are the very frequencies it lacks, let alone having even higher ones.

Photons and the thermal microstates that they correspond to are not equal when the frequencies and energies are different.  The terrestrial spectrum emission doesn’t, and can’t, have the same effect that the solar spectrum originally had on heating the planet.

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### 57 Responses to The Steel Greenhouse Debunks the Climate Greenhouse Effect

1. Derek Alker says:

Yup, the difference in length (circumference) between the earth’s surface and earth’s effective surface of emission is not taken into account. Presumably a 2 parallel plane model was used to hide this fact.
After that, a multitude of sins follows…

2. squid2112 says:

I wish you could get this through the thick skull of folks like Tony Heller (StevenGoddard) as I believe he is doing tremendous harm to the attempts at stopping this fraud. Heller is hell bent on trying to convince people that “there is a greenhouse effect” and that “the greenhouse is very strong” and “if it weren’t for the greenhouse effect our entire planet with be like Antarctica” … Unfortunately, I can no longer post on his site as he has banned me, or I would be reposting much of what you write here. Comically, he banned me whilst complaining about the very things he was engaging in, the very things he purported to ban me FOR … what a dickehad. He complained about everyone questioning him, while he continued to fuel the flames, posting more absurd posts and continuing rude and arrogant comments further complaining about responses to his shit.

I am very saddened this week as I now question the very credibility of Tony Heller, bringing into question many of his prior posts. I can no longer assume anything that he says is fact, and I will have to fact check those items that I find interest in, just to be sure they are indeed factual. Heller has lost my trust.

Keep up the great work here Joe! … Thank you for this post. I think this is your best presentation refuting the “Steel Greenhouse” thought experiment (originating with Willis Eschenbach on WUWT) and a great debunking of the so-called atmospheric “Greenhouse Effect” itself. Your presentation here simply cannot be refuted. Thank you!

3. How would the entire planet be like Antarctica when sunlight can heat the surface all by itself to almost 100C on a clear day? Oohp, but, can’t acknowledge the Sun now can we…

4. squid2112 says:

How would the entire planet be like Antarctica when sunlight can heat the surface all by itself to almost 100C on a clear day?

I know, I am dumbfounded by the stupid crap I have been reading on his blog the past few days. Heller has been full on attacking anyone who even slightly questions the validity of the “greenhouse effect” hypothesis. I have rarely, perhaps never, seen such twisting, turning and mental gymnastics in an attempt to convince everyone that a blanket doesn’t “keep” you warm, but actually “makes” you warm. Even my wife, who is not a scientist, and has little to no understanding of “physics” or schooled sciences, completely and purely understands that a blanket doesn’t “make” you warm, but instead helps to “keep” you warm. This is so fundamental and obvious that even a 5 year old child can understand this. And yet, Heller somehow believes he has the power to make other people “believe” as he does, all the while condemning people for acting as he does. Truly incredible stuff. The stupidity has left me breathless.

No matter. My time is MUCH better spent reading the works by folks such as yourself, and actually LEARNING something. I can ALWAYS learn things from you and I value the factual information and maths that you present in each and every post and comment. I am grateful that you have begun posting some new things again. Thank you!

5. squid2112 says:

Say, I had a thought last night, something that has kind of been bugging me about these “warming from back radiation” discussions and sort of falls into the “Steel Greenhouse” topic (kind of).

I noticed that in virtually every single discussion of this sort, the conversation begins with “I have a sphere that is powered by “X” watts of energy…”, etc…

My thought is, why must I have any watts of energy powering such sphere? … Let me explain…

If I have an ambient temperature (in my kitchen for example) of 26C (~80F), and I place a sphere on the counter, which is at thermal equilibrium to its surroundings (ambient temperature), and I then surround it with material, any material of any shape or form, any number of layers (10 layers of blankets perhaps?), can I increase the temperature of the sphere above the ambient?

To me this is a simple as you can make your “Steel Greenhouse”, and the obvious answer, that my lovely wife blurted out immediately, is, and always will be, a resounding NO!

This tiny and extremely simplistic little version of the “Steel Greenhouse” is irrefutable, testable and provable beyond any shadow of doubt. This did not require any fancy dandy calculations or wattages or any other magical stuff. One could very easily measure this, and each and every time you carried out this experiment, you would find, empirically, that the sphere could never, under any circumstances, achieve a temperature greater than the ambient. It simply cannot be done in this universe! … Case closed … There is no “Greenhouse Effect” in this universe.

6. Thanks Joe for posting this issue again with even simpler explanations. As squid2112 says though, many people will not be shaken off from their greenhouse belief; to them that is the holy grail that is part of the dreaded settled science and their egos can not allow them to be wrong. Pity them for they will be trampled on as and when the truth does finally dawn on them. With regards the blanket analogy, I’d like to remind everyone that the magical “CO2 enhanced greenhouse effect” is actually not just a blanket, it’s an electric blanket, sending energy back to itself and becoming even warmer … – HT to Alan Siddons. We battle on against wilful ignorance.

7. Tom says:

Another killer piece Joe. Are you going to run it past your local academic hierarchy?

8. I’ve got much, much bigger things I’m teaching them about. This stuff is peanuts compared to what I’m doing to my local academics 🙂

Maybe I’ll tie things off with this one day though, because connections can be made philosophically.

9. Rosco says:

The simple algebraic manipulation of the Stefan Boltzmann as is performed by the advocates of the Greenhouse effect CAN be completely wrong if done in a careless manner as the greenhouse effect advocates routinely do and therefore produce absurd results such as this steel greenhouse nonsense.

Of course this was designed to “prove” the atmospheric greenhouse effect although what 2 steel structures heated internally in a vacuum has to do with a surface heated externally and an atmosphere is beyond me.

When confronted with the reality that radiation exchanges between 2 objects always means the hotter object is emitting more than the cold object and hence the NET exchange is always hot -> cold as the SB equation explicitly says they revert to even more ludicrous claims about photons.

““In other words, a photon being emitted by the cooler star doesn’t stick its finger out to see how warm the surroundings are before it decides to leave.”

So I guess they are claiming ALL PHOTONS ARE CREATED EQUAL ???

This is arrant nonsense yet again by the greenhouse advocates.

A photon has specific energy and is defined to be the product of Planck’s constant times the frequency or h.f or h.c/lambda in wavelength terms.

You can plot the photon flux for 235 W/sqm emission and 470. It resembles the Planck curve above and the exact same argument Joe used applies.

The 470 W/sqm emission has significant emission of large numbers of high energy – short wavelength – photons. The 235 emission has virtually zero photons emitted at these wavelengths.

So the net effect is the hotter object not only emits more photons over the entire spectrum but it also emits high energy photons that the cold object CANNOT replace because it emits close to zero or zero at these short wavelengths.

Thus it is impossible for a cold object to “heat” a warmer object just as the Stefan-Boltzmann equation explicitly says.

One of the classical mistakes made in apply “trivial” algebra – as Dr Robert Brown (PhD) describes it – is careless thinking.

Many students of physics apply such simple algebraic manipulation when asked about how to change the variable in Planck’s equation from frequency to wavelength.

If you apply Dr Robert Brown (PhD)’s “trivial” algebraic methods you get the WRONG answer even though the algebra is completely correct !

This is the result of careless application of the rules of maths as is Dr Robert Brown (PhD)’s supposed proof of the steel greenhouse effect.

For transforming the Planck curve variables you cannot just substitute – you have to acknowledge the calculus relationship as well as the algebraic one.

Dr Robert Brown (PhD) made the childish error of assuming he could legitimately equate the difference between two real fluxes as a discrete flux itself – the black curve shown in Joe’s diagram.

And of course it isn’t a discrete flux in itself – it is just a number.

Dr Robert Brown (PhD) couldn’t understand his “proof” is nothing more than saying that

P(net) = P2 – P1 or that P(net) + P1 = P2.

Unless you know P2 explicitly you cannot know P(net) and there are infinite solutions.

The sequence Dr Robert Brown (PhD) claims “proves” the 2:1 relationship is absurd – it is a simple statement of the obvious.

He “forced” his solution onto the problem by breaking every rule of thermodynamics yet he is so – what do I say – evangelical in his Greenhouse belief ? – that he cannot recognise this.

10. Got another post coming on this issue Rosco. Be out tonight hopefully. It’s a good one, really showing their math and physics errors explicitly, on their terms.

11. Rosco says:

Joe said

“really showing their math and physics errors explicitly, on their terms.”

I have never said I have any special knowledge in how to sum radiation fluxes – which is the crux of the whole “greenhouse effect” mantra !

If summing radiation fluxes using simple “trivial algebra” as Dr Robert Brown (PhD) claims turns out to be WRONG – as I believe it is – these analyses demonstrate that the whole of climate science collapses into a morass of pseudo science !

There is compelling evidence “on their terms” that they do not correctly apply what science we have as accepted science whilst they offer childish mantra.

I now appreciate what Claes Johnson wrote with his “how to fool yourself with a prygeometer” and it wasn’t until I programmed some Planck curves myself that it really hit home.

The SB equation simply churns out a number and to do all the algebra they do on it to arrive at ludicrous conclusions such as the steel greenhouse makes little sense.

12. Exactly right Rosco. Claes Johnson is correct.

13. paul says:

Hi joe,
I think you’l find the philosophical connection and the much bigger picture is exactly what the more intuitive readers are thinking.
Great work !

Paul

14. I hope so Paul. Cheers.

15. JWR says:

This discussion has been solved by C. Christiansen, Annalen der Physik und Chemie, Leipzig,1883
When we consider a convex body 1 with surface area A1 and temperature T1 inside a concave surface 2 with area A2 and temperature T2, then the heat flow for T1>T2 is given by
q(1–>2)= sigma12*(T1^4-T2^4) where
1/sigma12= 1/sigma1+A1/A2*(1/sigma2-1/sigma)
sigma= Stefan Boltzmann constant for a black body
sigma1=eps1*sigma , eps1 emission coefficient of surface 1.
sigma2=eps2*sigma , eps2 emission coefficient of surface 2
q(1–>2) heat flux at the surface of 1 into the diirection of 2, in Watt/m^2.
The total heat from 1 to 2 is therefore A1*q(1–>2) Watt which is absorbed by surface 2 and per unit area , surface 2 receives : A1/A2*q(1–>2), Watt/m^2.
The relation of Christiansen of 1883 reflects the second law: inside the parenthesis we have always (T1^4-T2^4) , no heat flow when T1=T2, a conclusion of Fourier against the Prevost claim.
It suggests to look always for pairs of surfaces, the surface conditions and the magnitude of the area of the surfaces and to define an effective SB coefficient sigmq12.
Do not use the two-way formulation as advocated in astrology,
The heat exchange between the convex and the concave surfaces depend on both temperatures (within the parenthesis) and on both surface conditions and magnitudes, taken in the term sigma12 outside the parenthesis.
Two extremes follow: when the concave surface is very close A1/A2 goes to 1 and the relation of two parallel plates results.
When A2>>A!, the the heat flow only depends on T2 and not on sigma2 that means not on eps2.See also numerical examples for A1/A2=1: two parallel plates.

16. Willis Eschenbach says:

As the author of the two drawings, I have to laugh. I’m accused of ignoring the difference in areas, which of course I highlighted in the text … which is likely why my text has been omitted. Here’s the missing text:

… let me note that because the difference in exterior surface area of the shell and the surface is only 0.03%, I am making the simplifying assumption that they are equal. This clarifies the situation greatly. Yes, it introduces a whopping error of 0.3% in the calculations, which people have jumped all over in the comments as if it meant something … really, folks, 0.3%?

Yes, the whole claim that I’ve made some huge error is based on a simplification that makes a 0.3% error in the results … seriously? That’s the only thing wrong with my exposition?

w.

17. Willis, it doesn’t matter if you want to say the difference in some particular example is 0.03%. It could be 0.01%, and it could be 50%. What if you were 100% off, would it matter then? And if the surface areas are equal, then there is in fact no shell at all.

The point is that you haven’t factored for the difference in area at all, and yes, it does make a difference because the scenario needs to work properly for any difference in area.

Thus, you’ve tried to conserve the numerical value of the flux, but the flux from the sphere has to be reduced with distance from the sphere, and so the shell can not have the same initial flux as the sphere, ever. That is, the shell has to be a lower temperature than the initial output flux from the sphere. You didn’t indicate this anywhere, but assumed you can make the shell the same temperature as the sphere. This is simply impossible and it makes your exposition 100% incorrect, not slightly off.

Secondly, the shell doesn’t pass any heat to the sphere since it is passive, and, it is cooler than the sphere. The shell doesn’t lose any energy with internal emission since internally it is an enclosed space. The only loss of power from the shell occurs on its exterior.

These difference couldn’t be any more important, because they mean that what you created is incorrect. The sphere heats the shell, then the shell emits the power it receives from the sphere on its own exterior. The sphere doesn’t get hotter with the presence of a passive shell.

First law of thermodynamics: When energy passes, as work, as heat, or with matter, into or out from a system, its internal energy changes in accord with the law of conservation of energy.

Does the shell pass energy as work, heat, or matter, to the sphere? Alternatively, does the sphere pass additional energy as work, heat, or matter to itself?

No. Your scenario is not mathematically or physically logical in the first place, and subsequently, it violates the First Law of Thermodynamics.

Would you care to submit a retraction or correction, for the good and sake of the community?

18. Conservation of energy requires 0% difference. If you have a scenario where you can be 0.03% off, then it is 100% wrong, not “slightly off”. Furthermore, if said scenario can equally produce results that are 25%, 50%, 100%, 200%, etc., off, then it becomes even more ridiculous to justify being only 0.03% off in one particular case – this doesn’t save the rest. Saying that the difference goes to 0% at the limit of getting rid of the shell entirely, i.e., making the shell the surface of the sphere itself as their areas become equal, is what is called sophistry.

19. Derek Alker says:

Joseph Postma writes –
“Thus, you’ve tried to conserve the numerical value of the flux, but the flux from the sphere has to be reduced with distance from the sphere, and so the shell can not have the same initial flux as the sphere, ever.”

Hmmm, yet sunlight is input in to Willis’s model whilst allowing for the distance from the sun. A rather inconvenient contradiction in Willis’s (never peer reviewed) models logic……

20. Willis Eschenbach says:

Joe, small differences of less than 1% are often ignored in general analyses such as this one. It makes absolutely no difference to the underlying principles. All that happens is that because the two areas are different by 0.3%, the final equilibrium temperatures will be slightly different (<0.3%) than they would be otherwise. But the physical principles remain.

So if you want to, you could adjust the numbers so that they agree exactly. For a difference in area of 0.3%, that would reduce the out-and-inbound radiation of the shell by 0.3%, giving it a final value of 234.295 W/m2 instead of 235 W/2, and giving the surface a final value of 469.295 W/m2 instead of 470 W/m2. Now the numbers balance exactly to the last decimal … you happy now? Because that has done nothing to show my example is incorrect.

On the other hand, if you wish, you can make the difference as small as you like. Simply increase the radius of the sphere, and reduce the distance between the shell and the sphere to 1 mm and the situation is still the same. In that case the difference in area will be minuscule.

But none of that changes anything regarding how the greenhouse effect works. It doesn't make the drawing incorrect. Physically, the situation is the same, with a radiating surface, a vacuum, and a surrounding shell. From a physical point of view, it will still give you ~ twice the radiation at the surface as is radiated outwards from the shell.

I gotta say … if that's the only problem you find with the example, you've failed to show anything wrong.

Finally, Derek, your comment makes no sense at all. Point out to me where "sunlight is input to Willis's model". There's no sunlight in it at all. Please read the example before uncapping your electronic pen, it just makes you look foolish.

w.

21. Willis, your error of 0.3% comes in because your analysis is wrong, not because it is an approximation. You made no approximations anywhere in any case, but you did arrive at a result that was wrong. Therefore it is your method that is wrong, not any non-existent approximations.

See here for the math actually getting worked out:

There’s a noticeable lack of such mathematical analysis in your conjecture, but you do make various assumptions to support the result you intend to arrive at.

Did you overlook the part about the First Law of Thermodynamics?

First law of thermodynamics: When energy passes, as work, as heat, or with matter, into or out from a system, its internal energy changes in accord with the law of conservation of energy.

Does the shell pass energy as work, heat, or matter, to the sphere? Alternatively, does the sphere pass additional energy as work, heat, or matter to itself? No.

The sphere emits 235 W/m2. The shell receives less flux than that and so is warmed to a lower temperature. It then emits externally (internal emission is not lost) at that lower temperature over a larger external surface area, thus conserving total power.

Thus, the sphere doesn’t get warmed by the shell, because that would violate conservation of energy and the laws of heat flow – the sphere only warms the shell! And the sphere doesn’t have to warm up more to warm the shell.

The shell loses no energy by internal emission since it is an enclosure. That is your error. And it sends no heat to the sphere either since it is cooler than the sphere. That is also your error.

The shell emits only on its exterior the energy it receives from the sphere. It doesn’t lose energy internally. That is your error. Thus, the exterior of the shell emits the power it receives from the sphere, and it is impossible for the shell to cause the sphere to become warmer than its own internal power source.

Please read the linked post and see the correct math and physics Willis. The steel greenhouse debunks the greenhouse effect by your own attempt to prove it.

22. CW says:

Willis E….a jack of all trades, and a master of none. nuff’ said….

23. Willis Eschenbach says:

CW says:
2014/12/05 at 10:06 AM

Willis E….a jack of all trades, and a master of none. nuff’ said….

When someone like Joe Postma and CW starts attacking my history, I know I’ve won the argument. If they actually had scientific arguments, they’d bring them up …

Game over, guys.

w.

PS-Joe, your claim that

The shell emits only on its exterior the energy it receives from the sphere.

is hilarious. Do you actually consider these statements before writing them down? If you could build a heated shell that only emits on its exterior, there’s no limit to what you could do … but tell us, dear Joe, what physical principle keeps it from emitting on the interior? Good intentions? Pixie dust? What?

24. Willis, when your argument has been debunked by a simple reference to the First Law of Thermodynamics, and with some further mathematics and physical analysis to precisely demonstrate the flaw, and with a subsequent discovery that you have no scientific or mathematical training to speak of, it doesn’t mean you won the argument. When you can’t recognize the error of conserving the numerical flux value as opposed to the total energy, it doesn’t mean you’re winning anything other than demonstrating your actual incompetence.

Indeed Willis, the shell does not lose any energy on its interior. You don’t understand that? I didn’t say that it didn’t emit on the interior, I said that it didn’t lose energy on its interior.

To quote, I said “The shell loses no energy by internal emission… It doesn’t lose energy internally.“.

Let me ask you this: If we get rid of the sphere, and just say that it is the shell itself which has an interior (to its shell material) power source of 235 W, and if the shell has a surface area of 1 square meter, then what does the flux output and temperature need to be of the shell to conserve energy?

25. CW says:

Willis E. is indeed NOT a scientist….he has no clear understanding of thermodynamics, and to make matters worse, he expects a scientific response to the nonsense he proposes in his greenhouse model….JP has illustrated with much patience his errors, but, as he has limited mathematical skills, he continues with his charade…he should just stay on WUWT, and not try to pose ridiculous physical models on scientific blogs. I am not a physicist, but, I do follow the mathematics. Any physical model has mathematical underpinnings, and if you can not understand basic equations of convective heat transfer or heat transfer by radiation, then, do not try to build credibility with a model that contradicts first principles. Willis is really just a blog bully…a poser!

26. CW says:

In correct format: Poser = Sophist

27. Willis Eschenbach says:

Joseph E Postma says:
2014/12/05 at 8:05 PM

Willis, when your argument has been debunked by a simple reference to the First Law of Thermodynamics, and with some further mathematics and physical analysis to precisely demonstrate the flaw, and with a subsequent discovery that you have no scientific or mathematical training to speak of, it doesn’t mean you won the argument.

Joe, I fear I didn’t see where you “debunked” my examply by a “simple reference to the first law”. Perhaps you could restate it so I can understand what you are referring to.

Next, whether or not I have “scientific or mathematical training” is an ad-hominem argument. The only relevant question is, are my claims true. The fact that you keep reverting to ad-hominem arguments makes me question your understanding.

When you can’t recognize the error of conserving the numerical flux value as opposed to the total energy, it doesn’t mean you’re winning anything other than demonstrating your actual incompetence.

Since the areas are equal to within an arbitrarily small number, I’ve used the simplifying assumption that they are equal. However, I also showed above that we can play it your way, and that it changes nothing about the strength of my argument.

Indeed Willis, the shell does not lose any energy on its interior. You don’t understand that? I didn’t say that it didn’t emit on the interior, I said that it didn’t lose energy on its interior.

To quote, I said “The shell loses no energy by internal emission… It doesn’t lose energy internally.“.

Your actual statement that I was referring to was:

The shell emits only on its exterior the energy it receives from the sphere.

This means that it does NOT emit energy on the interior, only on the exterior. It says nothing about “losing energy”. However, it seems you now would like it to mean something else.

Let me ask you this: If we get rid of the sphere, and just say that it is the shell itself which has an interior (to its shell material) power source of 235 W, and if the shell has a surface area of 1 square meter, then what does the flux output and temperature need to be of the shell to conserve energy?

I don’t understand how this is different from Figure 2 above—there’s a shell, and an interior power source. What is the difference? I must be missing something in your description.

w.

28. Willis Eschenbach says:

CW says:
2014/12/06 at 8:16 AM

Willis E. is indeed NOT a scientist….he has no clear understanding of thermodynamics, and to make matters worse, he expects a scientific response to the nonsense he proposes in his greenhouse model….JP has illustrated with much patience his errors, but, as he has limited mathematical skills, he continues with his charade…he should just stay on WUWT, and not try to pose ridiculous physical models on scientific blogs. I am not a physicist, but, I do follow the mathematics. Any physical model has mathematical underpinnings, and if you can not understand basic equations of convective heat transfer or heat transfer by radiation, then, do not try to build credibility with a model that contradicts first principles. Willis is really just a blog bully…a poser!

In other words, CW, you can’t point out a single error in my exposition, so you are joining Joe in an ad-hominem attack on my credentials.

You seem to think that if the janitor writes “E=MC^2” on the blackboard it’s wrong, but when Einstein writes it, it’s correct … CW, my credentials are totally immaterial. All that matters is, are my claims true or not?

And so far, you haven’t even attempted to show that my claims are not true.

So, given that you are such a genius, perhaps you can explain why my model is wrong.

w.

29. I fear I didn’t see where you “debunked” my examply by a “simple reference to the first law”. Perhaps you could restate it so I can understand what you are referring to.

First law of thermodynamics: When energy passes, as work, as heat, or with matter, into or out from a system, its internal energy changes in accord with the law of conservation of energy.

Does the shell pass energy as work, heat, or matter, to the sphere? Alternatively, does the sphere pass additional energy as work, heat, or matter to itself? No.

Therefore the presence of the shell can’t cause the sphere to increase in temperature beyond its actual active internal heat source.

Next, whether or not I have “scientific or mathematical training” is an ad-hominem argument. The only relevant question is, are my claims true. The fact that you keep reverting to ad-hominem arguments makes me question your understanding.

However, your claims aren’t true, and a very likely reason is that you have no scientific training. You lack of qualification isn’t an ad-hominem in the reality that you factually lack scientific training and your claims are erroneous. A fact is a fact; whether it makes you feel good is something entirely different.

Since the areas are equal to within an arbitrarily small number, I’ve used the simplifying assumption that they are equal. However, I also showed above that we can play it your way, and that it changes nothing about the strength of my argument.

Actually it changes everything about the argument, and there’s no valid reason to make such an assumption in the first place since it is totally unnecessary. It changes everything about the argument and it exposes your seeming complete lack of understanding of the error, since you are so unwilling to acknowledge it.

Your actual statement that I was referring to was:

The shell emits only on its exterior the energy it receives from the sphere.

This means that it does NOT emit energy on the interior, only on the exterior. It says nothing about “losing energy”. However, it seems you now would like it to mean something else.

Your misinterpretation of my comment which had been put into obvious context more than once is either an indication of your willing choice to interpret it in a sophistic way, or, an inability to interpret it in the obvious way it should have been and in the context which was provided. The shell loses no energy internally – it only loses energy externally. This changes everything about your argument and exposes its error. You do not understand that the shell does not lose energy by internal emission?

I don’t understand how this is different from Figure 2 above—there’s a shell, and an interior power source. What is the difference? I must be missing something in your description.

Get rid of the sphere, it is the shell itself which has its own power source. The shell itself produces the power. There’s no sphere inside the hollow cavity of the shell. The power source is 240W, the surface area is 1m^2, and the shell is negligible thickness. What’s the temperature of the shell? What is its surface flux?

30. But sure Willis, it doesn’t need to be about you and your competence. It is whether your idea is correct or not, and even competent people can make mistakes sometimes.

So let’s forget all that and get back to this, if you’re not there already:

Get rid of the sphere, it is the shell itself which has its own power source. The shell itself produces the power inside its material. There’s no sphere inside the hollow cavity of the shell. The power source is 240W, the surface area of the shell is 1m^2, and the shell is negligible thickness. What’s the temperature of the shell? What is its surface flux?

31. CW says:

I do not need to give you reasons why your model is incorrect. I know absolutely that I can learn nothing from you regarding physics or mathematics. The reason I visit JP’s blog is the ability of JP and others on this blog to provide “real” insights into the physical world, via mathematics and the laws of physics.
Also, you cannot accept the fact that mathematical rigor is very important to physics—it is the language of nature, and you need to go back to the basics. I have studied mathematics for many years, and still spend hours everyday learning. My son is an astrophysicist, and I will leave his comments regarding your “model” out of this response–just understand it was not “positive.”

32. Willis Eschenbach says:

Joseph E Postma says:
2014/12/06 at 3:24 PM
<blockquote

>I fear I didn’t see where you “debunked” my examply by a “simple reference to the first law”. Perhaps you could restate it so I can understand what you are referring to.

First law of thermodynamics: When energy passes, as work, as heat, or with matter, into or out from a system, its internal energy changes in accord with the law of conservation of energy.

Does the shell pass energy as work, heat, or matter, to the sphere? Alternatively, does the sphere pass additional energy as work, heat, or matter to itself? No.

Therefore the presence of the shell can’t cause the sphere to increase in temperature beyond its actual active internal heat source.

Huh? Replace the shell with an insulative blanket. Would you say that if you put an insulative blanket around the sphere, it wouldn’t cause the sphere to heat up?

You put on a jacket when you go outdoors, and as a result, you end up warmer than you would be without a jacket … according to your logic, this would violate conservation of energy.

Please explain where that logic is wrong, thanks.

w.

33. Willis Eschenbach says:

Joseph E Postma says:
2014/12/06 at 3:37 PM

But sure Willis, it doesn’t need to be about you and your competence. It is whether your idea is correct or not, and even competent people can make mistakes sometimes.

So let’s forget all that and get back to this, if you’re not there already:

Joe, it’s hard to “forget all that” when so far you seemingly can’t make a post that doesn’t contain an ad hominem. So how about YOU forget all the personal attacks, it’s not me that’s making them.

Thanks,

w.

34. Willis Eschenbach says:

A further comment. The difference between the shell and the sphere is that the shell has (approximately to an arbitrary accuracy) twice the surface area of the sphere, half inside and half out. When it receives radiation, it radiates half of it back in, and half of it outward. it will warm (in the example above) until it is radiating 235 W/m2 outward. At that point, it reaches equilibrium.

Now, the question is, what happens to the energy that is radiated inwards at that point? It is absorbed by the surface. At that point the surface is absorbing 235 W/m2 from the shell, in addition to the 235 W/m2 coming from the nuclear core. As a result, it radiates 2 * 235 W/m2.

w.

35. “Huh? Replace the shell with an insulative blanket. Would you say that if you put an insulative blanket around the sphere, it wouldn’t cause the sphere to heat up?

You put on a jacket when you go outdoors, and as a result, you end up warmer than you would be without a jacket … according to your logic, this would violate conservation of energy.

Please explain where that logic is wrong, thanks.”

Well first of all you are shifting reference frames to sense-perception-based analogies which additionally have to do with a different scenario of physics, i.e., that of convective air-circulation restriction as opposed to radiation effects with blackbodies and vacuums.

Secondly, let us indeed put a blanket in the form of another layer of steel on the steel greenhouse. Or shrink the radius of the shell until the shell is touching the sphere. What would happen if you put another layer of passive cold steel on the exterior of the steel sphere?

It won’t warm the sphere up. It will reduce the flux on the outside of the sphere since the sphere will now have a larger surface area, and thus, the new sphere will be cooler. At the position inside the sphere where the old exterior used to be, it will be the same temperature, still assuming uniform outward flow of power from the center of the sphere.

If we use a blanket as a new layer instead of a layer of steel, assuming the blanket has the same emissivity and absorptivity of the steel, the effect will be similar and it still won’t make the sphere get hotter.

You feel warmer when you put on a jacket because you no longer have cold air against your skin wicking heat out of your body. But this has nothing to do with your steel greenhouse example because there is no air in it.

36. So how about YOU forget all the personal attacks

Just jumping the gun, trying to figure out what makes this greenhouse thing so confusing for people. All explanations need to be examined. This isn’t our first rodeo here. Sorry it’s a bad habit…I’ve been lied to so many times by bad people I’ve come to assume this. First time talking directly with you though.

Can you answer the question yet, or no?

Get rid of the sphere, it is the shell itself which is its own power source. The shell itself produces the power inside its material. There’s no sphere inside the hollow cavity of the shell. The power production is 240W, the surface area of the shell is 1m^2, and the shell is negligible thickness. What’s the temperature of the shell? What is its surface flux?

37. The difference between the shell and the sphere is that the shell has (approximately to an arbitrary accuracy) twice the surface area of the sphere, half inside and half out. When it receives radiation, it radiates half of it back in, and half of it outward.

Willis, the shell doesn’t lose energy in internal emission. There is no loss of energy via internal emission since it is an enclosed cavity. Any internal emission is regained, not lost, because it is a closed cavity. Internal emission is trapped inside, and thus doesn’t leave the shell. The shell only loses energy on its outside. For one, internal emission by the shell at any point is replaced instantly by emission from all other internal opposing points in view; for two, the shell has a power source inside it in the form of a sphere which emits at higher flux, thus necessitating that heat only flows one way, i.e., from the sphere to the shell, even with a negligible gap. The shell radiates internally but it doesn’t lose energy internally, thus it only loses energy on one side, its exterior.

it will warm [] until it is radiating 235 W/m2 outward. At that point, it reaches equilibrium.

Yes that is correct, with the proviso that it loses 235 W (not W/m2) outward only, and no energy internally at all. As the shell only loses energy externally, then it simply needs to warm up to the flux level impinging on its interior from the sphere, which has a power of 235 W, and then energy is conserved.

Now, the question is, what happens to the energy that is radiated inwards at that point? It is absorbed by the surface.

Thermal absorption of energy occurs when there is heat flow, but since the flux from the shell is lower, or at best almost equal, to the flux from the sphere, then the sphere does not absorb heat from the shell, and thus doesn’t rise in temperature. Rather, the shell has been heated by the flux from the sphere, and then it emits that same amount of energy on its own exterior, conserving energy.

At that point the surface is absorbing 235 W/m2 from the shell, in addition to the 235 W/m2 coming from the nuclear core. As a result, it radiates 2 * 235 W/m2.

The sphere can’t absorb heat energy from the shell because the shell is cooler, or at best, almost equal in temperature. The equation for heat flow shows Q’ = A*(Tsh4 – Tsp4), approximating equal area as you like to do, and thus, Q’ from the shell to the sphere is negative since Tsh < Tsp, or at best, Q' = 0 if we use your approximation. Q' = 0 between the sphere and shell when they are in equilibrium, and the total power flows through the system in this configuration to be emitted on the outside of the shell.

You can't make the assertion that the sphere absorbs all of the energy from a cooler source as heat when the the heat flow equation specifically says that this is not what can happen. If the sphere was 300K and you surrounded it with a passive shell at 0K, it would heat the shell, and then the shell would emit that same energy on its own exterior; the shell wouldn't make the sphere get hotter because it can’t, because it’s not a source of heat to the sphere.

You should have a look at the math from here.

38. Again, think of this thought experiment to help clarify things:

Get rid of the sphere, it is the shell itself which is its own power source. The shell itself produces the power inside its own material. There’s no sphere inside the hollow cavity of the shell. The power production is 240W, the surface area of the shell is 1m^2, and the shell is negligible thickness. What’s the temperature of the shell? What is its surface flux?

39. Willis Eschenbach says:

Joseph E Postma says:
2014/12/06 at 7:33 PM

So how about YOU forget all the personal attacks

Just jumping the gun, trying to figure out what makes this greenhouse thing so confusing for people. All explanations need to be examined. This isn’t our first rodeo here. Sorry it’s a bad habit…I’ve been lied to so many times by bad people I’ve come to assume this. First time talking directly with you though.

I don’t care if you usually hang out with dishonest people, that’s your business. I’m not one of them.

Can you answer the question yet, or no?

Ignoring the unpleasant phrasing, yes, I’m happy to answer the question. Your question is:

Get rid of the sphere, it is the shell itself which is its own power source. The shell itself produces the power inside its material. There’s no sphere inside the hollow cavity of the shell. The power production is 240W, the surface area of the shell is 1m^2, and the shell is negligible thickness. What’s the temperature of the shell? What is its surface flux?

The surface flux of the shell is 240 W/m2 … so what? That wasn’t my example, it’s a totally different physical configuration which has nothing to do with the greenhouse effect.

Here’s my question for you in regards to the two drawings of mine that you posted in the head post (without attribution as to the artist, I note) …

When the shell in figure 2 radiates inwards, the radiation is absorbed by the surface of the sphere. This is additional radiation striking the surface which is not present in figure 1. So here’s the question:

Does this additional radiation a) cool the surface, b) warm the surface, or c) make no difference to the temperature of the surface of the sphere? Please explain your answer.

w.

40. squid2112 says:

Willis, with all due respect, please stop with the tired and worn out bullshit of the “blanket” analogy. I am so sick and tired of hearing that worn out crap. If you do not realize that a blanket (or any other material) cannot “make” you warmer, but instead simply “keeps” you warmer, than I cannot accept anything else you have to have to offer as being credible.

If I place a blanket around the banana sitting on my counter, is it going to warm up? … give me a freaking break. I am so sick and tired of the stupid “blanket” analogy. Anyone bringing it up is instantly marked as a dumbass in my book.

41. “The surface flux of the shell is 240 W/m2 … so what? That wasn’t my example, it’s a totally different physical configuration which has nothing to do with the greenhouse effect.”

Actually it is exactly just like your example. In your example, the shell has to lose power on both its interior and exterior, and so the power is divided by two surface areas, i.e., the flux is divided by two. To quote you:

“The difference between the shell and the sphere is that the shell has (approximately to an arbitrary accuracy) twice the surface area of the sphere, half inside and half out. When it receives radiation, it radiates half of it back in, and half of it outward.”

And so you divide the power output of the shell by two. Now you say that the surface flux of the shell is 240 W/m2, i.e. not divided by two, which is a direct contradiction to your steel greenhouse argument. So, is this you lying, or is it you not following your own logic? See how this makes it difficult for us to tell what your motive is? Never mind that. How do you justify concluding two mutually exclusive things for what are the same scenarios? The people I hang around with don’t do things like this.

“Here’s my question for you in regards to the two drawings of mine that you posted in the head post (without attribution as to the artist, I note) …

When the shell in figure 2 radiates inwards, the radiation is absorbed by the surface of the sphere. This is additional radiation striking the surface which is not present in figure 1. So here’s the question:

Does this additional radiation a) cool the surface, b) warm the surface, or c) make no difference to the temperature of the surface of the sphere? Please explain your answer.”

Yes I already did explain that. Here it is again:

The shell doesn’t lose energy in internal emission. There is no loss of energy via internal emission since it is an enclosed cavity. Any internal emission is regained, not lost, because it is a closed cavity. Internal emission is trapped inside, and thus doesn’t leave the shell. The shell only loses energy on its outside. For one, internal emission by the shell at any point is replaced instantly by emission from all other internal opposing points in view; for two, the shell has a power source inside it in the form of a sphere which emits at higher flux, thus necessitating that heat only flows one way, i.e., from the sphere to the shell, even with a negligible gap. The shell radiates internally but it doesn’t lose energy internally, thus it only loses energy on one side, its exterior.

It loses 235 W (not W/m2) outward only, and no energy internally at all. As the shell only loses energy externally, then it simply needs to warm up to the flux level impinging on its interior from the sphere, which has a power of 235 W, and then energy is conserved.

Thermal absorption of energy occurs when there is heat flow, but since the flux from the shell is lower, or at best almost equal, to the flux from the sphere, then the sphere does not absorb heat from the shell, and thus doesn’t rise in temperature. Rather, the shell has been heated by the flux from the sphere, and then it emits that same amount of energy on its own exterior, conserving energy.

The sphere can’t absorb heat energy from the shell because the shell is cooler, or at best, almost equal in temperature. The equation for heat flow shows Q’ = A*(Tsh^4 – Tsp^4), approximating equal area as you like to do, and thus, Q’ from the shell to the sphere is negative since Tsh < Tsp, or at best, Q' = 0 if we use your approximation. Q' = 0 between the sphere and shell when they are in equilibrium, and the total power flows through the system in this configuration to be emitted on the outside of the shell.

You can't make the assertion that the sphere absorbs all of the energy from a cooler source as heat when the the heat flow equation specifically says that this is not what can happen. If the sphere was 300K and you surrounded it with a passive shell at 0K, it would heat the shell, and then the shell would emit that same energy on its own exterior; the shell wouldn't make the sphere get hotter because it can’t, because it’s not a source of heat to the sphere.

You should have a look at the math from here.

I’ll gladly add captions to the images to cite attribution to you.

42. Willis Eschenbach says:

Thanks for your reply, Joe, but it doesn’t answer the question. I’ll ask it again.

When the shell in figure 2 radiates inwards, the radiation is absorbed by the surface of the sphere. This is additional radiation striking the surface which is not present in figure 1. So here’s the question:

Does this additional radiation a) cool the surface, b) warm the surface, or c) make no difference to the temperature of the surface of the sphere? Please explain your answer.”

I’m not interested at this point in your exegesis about the whole system, we can get to that once we settle this very simple question.. You need to pick a), b), or c), because those are the only possibilities. I can’t figure out from your “explanation” above which one you claim is the case.

Now, remember that in the situation shown in figure 1, the sphere is only heated by the internal nuclear reaction, there is no external radiation hitting in. But in figure 2, there is new, additional radiation hitting the sphere which is not present in figure 1.

So I ask again—will the new, additional radiation a) warm, b) cool, or c) make no difference to the temperature of the sphere? Start by picking a letter, please, and then explain your reasonong.

Thanks for your offer to credit the figures, much appreciated.

w.

43. Willis Eschenbach says:

squid2112 says:
2014/12/06 at 11:02 PM

Willis, with all due respect, please stop with the tired and worn out bullshit of the “blanket” analogy. I am so sick and tired of hearing that worn out crap. If you do not realize that a blanket (or any other material) cannot “make” you warmer, but instead simply “keeps” you warmer, than I cannot accept anything else you have to have to offer as being credible.

If I place a blanket around the banana sitting on my counter, is it going to warm up? … give me a freaking break. I am so sick and tired of the stupid “blanket” analogy. Anyone bringing it up is instantly marked as a dumbass in my book.

Thanks, Squid. I was not using the “blanket analogy” to claim that it is anything like the greenhouse effect. It is not. My apologies for the lack of clarity.

I was using the blanket analogy to point out the fault in the reasoning that a cold object can’t leave a warm object warmer than it would be without the cold object. For example, when you go out you wear a coat. Why? Because it slows heat loss, despite being colder than you are.

In the same way, a cold atmosphere can leave a planet warmer than it would be without the atmosphere, despite being colder than the planet. The atmosphere does so in the same way the coat does, by slowing heat loss, despite being colder than the planet.

And yes, you are totally correct that the physical mechanisms for slowing heat loss are totally different. One works by back radiation, the other works by a combination of insulation and reducing air flow.

However, my point was that in both cases, a cold object (a coat or an atmosphere) can leave a warm object (a human, a planet) warmer than it would be without the existence of the cold object, by slowing heat loss.

I appreciate your objection, I hope this clarifies my point, and yes, you’re right, it only works on objects warmer than their surroundings … as you say, it won’t warm a banana.

w.

44. Rosco says:

I just love it when these clowns start talking about jacket making us warmer.

The simple truth is that no matter which way you look at it the “Steel Greenhouse” hypothesis breaks every law of thermodynamics and it does so egregiously !

In the initial analysis of the sphere alone there is 235 X A joules per second energy generated by the sphere and emitted to space in total.

The proposal then states the shell has very little internal energy initially and is “heated” by the emissions from the sphere and thus its internal energy and hence temperature increases.

Standard physics that I see in ALL of the texts – not my physics – standard physics that I quoted when I wrote about this a year or more ago says that the radiation shield – the shell – reduces external emissions by 50 % and I have references supporting this.

NONE of them claim this back radiation increases the temperature of the sphere except proponents of the “Steel Greenhouse” hypothesis !

The Stefan-Boltzmann equation EXPLICITLY says the NET energy exchange between 2 objects at the same temperature is ZERO !

It also EXPLICITLY says the ONLY way one object HEATS another is if there is a NET exchange of energy from the one that is HOTTER to the cooler.

The “Steel Greenhouse” hypothesis REVERSES this STANDARD SCIENCE and has NET energy flowing from the cooler object to the hotter object!!

ANY analysis of Planck curves including the FACT that sigmaT^4 is the area under a Planck curve at temperature T proves radiation from a cold object cannot heat a warmer object – ALL PHOTONS ARE NOT EQUAL AND COOOLER OBJECTS DO NOT EMIT HIGH ENERGY PHOTONS THAT A WARMER OBJECT DOES AND THEREFORE CANNOT REPLACE THIS LOST ENERGY.

Pictet demonstrated this centuries ago !!!

Besides that look at the energy balance in their FINAL scenario –

470 x A NOW emitted by the sphere

PLUS

470 x A NOW emitted by the shell (as 235 X A out to space Plus 235 x A inwards to the sphere.)

This represents a TOTAL of 940 X A joules per second !!

FOUR TIMES the original available energy !!!

The only original source of energy is the 235 x A joules per second from the sphere !!!

AND THEY CLAIM THIS IS NOT A BREACH OF EVERY LAW OF THERMODYNAMICS ????

Fuck me – that is gross stupidity !!!

45. Thanks for your reply, Joe, but it doesn’t answer the question. I’ll ask it again. I’m not interested at this point in your exegesis about the whole system, we can get to that once we settle this very simple question.. You need to pick a), b), or c), because those are the only possibilities. I can’t figure out from your “explanation” above which one you claim is the case.

Willis my answer was very clear and repeated several times over. Did you not understand it? You aren’t able to comprehend the answer? It goes to a specific letter choice, and of course it is always good to explain what happens to the whole system since that is exactly what the context is: the system of the sphere and shell and what they do. The “very simple question” is precisely the “exegesis” provided for you, clearly explained.

I note that you haven’t explained your own contradiction with the shell-only:

In your steel greenhouse example, the shell has to lose power on both its interior and exterior, and so the power is divided by two surface areas, i.e., the flux is divided by two. However now you say that the surface flux of the shell is 240 W/m2, i.e. not divided by two, which is a direct contradiction to your steel greenhouse argument. So, is this you lying, or is it you not following your own logic?

46. Willis the blanket argument is only one of obfuscation as you even agree that the underlying physics have nothing to do with each other. It’s not a valid analogy at all. A real greenhouse warms by stopping convection, which is what a jacket does too, and these have nothing to do with trapping radiation, even though radiation is in fact present. If heat was trapped by radiation then the greenhouse could get warmer than its radiative heating input, but they don’t, both empirically confirmed and because that would violate the First Law of Thermodynamics: a blanket does not pass energy as work, heat, or matter to the thing it covers, therefore the thing it covers doesn’t get warmer. When you put on a jacket it is a sense-perception-based phenomena of not feeling the cold actively wick heat out of your body as fast via conduction and convection. In the steel greenhouse, there is no such similar cooling on the sphere at all, and so the analogy is 100% false. The idea of trapping thermal radiation in radiatively absorptive systems to get a higher temperature is false. The only way you can reduce thermal radiative emission is by lowing the emissivity.

In the same way, a cold atmosphere can leave a planet warmer than it would be without the atmosphere, despite being colder than the planet. The atmosphere does so in the same way the coat does, by slowing heat loss, despite being colder than the planet.

The cold atmosphere is touching the ground surface, therefore it can not warm the surface and can only cool it, like it does your skin. This is why we wear jackets and build greenhouses, to prevent the open atmosphere from causing cooling on the surface. And since the atmosphere is cooler than the surface, then the surface heats the atmosphere, and the surface isn’t heated because it heats the atmosphere. You also leave out the existence of the lapse rate gradient which necessarily makes the surface-air warmer than the cooler average which is necessarily higher in altitude, and this is independent of any greenhouse analogy. You’re misattributing things.

Anyway, the only thing you need to do is to explain the contradiction you’ve put yourself in with the other analysis. If your contradiction goes uncorrected I will be required to write about it more widely, because aside from the analysis I’ve provided here and elsewhere debunking the steel greenhouse and climate greenhouse effect, it directly compromises you with your own words.

47. Willis Eschenbach says:

Joseph E Postma says:
2014/12/08 at 1:25 PM

Thanks for your reply, Joe, but it doesn’t answer the question. I’ll ask it again. I’m not interested at this point in your exegesis about the whole system, we can get to that once we settle this very simple question.. You need to pick a), b), or c), because those are the only possibilities. I can’t figure out from your “explanation” above which one you claim is the case.

Willis my answer was very clear and repeated several times over. Did you not understand it?

I’m sorry, Joe, but I didn’t understand it, which was why I simplified the question.

When the shell in figure 2 radiates inwards, the radiation is absorbed by the surface of the sphere. This is additional radiation striking the surface which is not present in figure 1. So here’s the question:

Does this additional radiation a) cool the surface, b) warm the surface, or c) make no difference to the temperature of the surface of the sphere? Please explain your answer.”

But even after I simplified it, I re-read your reply several times, and I still can’t tell whether you think the surface absorbing additional radiation will heat it, cool it, or leave it unchanged. Please pick a letter and explain just that part of the question. You keep dragging in a bunch of other stuff. I’m trying to find out what you think piece by simple piece, so please select a), b), or c), and let us know what your reasoning is.

[JP: And I explained to you, that you’re making assumptions which you shouldn’t. The radiative heat flow equation shows that the sphere absorbs no heat from the shell, and therefore, it doesn’t warm up due to the presence of the shell. Rather, simply, the sphere heats the shell, and the shell then sheds the requisite energy on its own exterior. So to answer you directly, Willis, your assumptions are wrong, hence your question is wrong. What happens is that the shell can not cause the sphere to warm up (are you able to figure out which letter this is?), and the reason is because it sends no heat to the sphere, and this is supposed to educate you about why the framing of your question is in error.]

I note that you haven’t explained your own contradiction with the shell-only

There is no contradiction. There are two very different physical setups, which are heated in two very different locations, and which result in two very different outcomes. This is what we would expect, isn’t it?

[JP: Not at all Willis. In both cases, the shell is being supplied with energy. In your steel greenhouse, the shell emits and loses this energy on both its interior and exterior and so the flux is divided by two. With the shell only, now you say that the shell doesn’t emit and lose energy on its interior. It’s a direct contradiction.]

So, is this you lying, or is it you not following your own logic?

It appears you must hang out with a bunch of liars, the way you bandy the term around so lightly. It may surprise you to know that some of us are honest men, and we deeply resent people calling us liars without provocation or evidence.

You said above that you were going to stop this kind of personal attack, yet you haven’t … but that doesn’t make you a liar.

So are you going to stop the personal attacks? Because if you’re going to attempt to attack my honesty every time we disagree, I’m not interested.

w.

[JP: Well, is it you lying or is it you not following your own logic? You’re still in a contradiction of your own answers.

In the steel greenhouse, the shell emits and loses energy due to its surface temperature on both its interior and exterior surfaces. Since emission is intrinsic to a surface’s temperature then this has to happen whether there’s anything inside the shell cavity or not. In your steel greenhouse, the internal emission and loss of energy from the shell goes back into the sphere, dividing the flux from the shell by two. With the shell only, you now claim that the shell doesn’t emit and lose energy on its interior. The reason you wish to avoid this is because of the answer it exposes if we follow your original steel greenhouse logic: any interior point of the shell surface will absorb the emission from the rest of the interior of the shell, and so the internal emissions from the shell will heat itself up by doubling its own flux on its interior. Of course, firstly, this violates any thermodynamic analysis, since if the difference in temperature of any points on the interior is zero, then no radiative heat absorption can occur between those points. Of course, you steadfastly avoid any reference whatsoever to the radiative heat flow equation in your steel greenhouse as well. And secondly, any point on the interior of the shell would now see double the flux, and would thus need to absorb that again unto itself, etc., which exposes the run-away self-heating fallacy of the original steel greenhouse.]

48. Willis Eschenbach says:

Joseph E Postma says:
2014/12/08 at 1:56 PM

Willis the blanket argument is only one of obfuscation as you even agree that the underlying physics have nothing to do with each other. It’s not a valid analogy at all.

Joe, read what I said. I never said the underlying physics were the same. It was in reference to your claim that:

Therefore the presence of the shell can’t cause the sphere to increase in temperature beyond its actual active internal heat source.

Your argument seemed to be that a cold object can’t leave a warm object warmer than it would be without the cold object, so I gave you the example of a cold blanket leaving your body warmer than without the cold blanket.

[JP: And as you just agreed, and as I just explained, the underlying physics and physical situations are not the same in any way. Thus, it is obfuscatory.]

I’m sorry for the confusion. I was not making the claim that a blanket and the greenhouse effect have the same “underlying physics”. Squid also was confused by my words, so I accept that the error was mine. In response to him I said:

Thanks, Squid. I was not using the “blanket analogy” to claim that it is anything like the greenhouse effect. It is not. My apologies for the lack of clarity.

I was using the blanket analogy to point out the fault in the reasoning that a cold object can’t leave a warm object warmer than it would be without the cold object. For example, when you go out you wear a coat. Why? Because it slows heat loss, despite being colder than you are.

Hope this clears it up,

w.

[JP: A blanket/jacket is in fact identical to the real greenhouse effect found in botanist’s greenhouses, using the same underlying physics of preventing convective cooling. This isn’t about some sophism that a cool jacket or greenhouse ceiling prevents convective cooling, and thus a “cold object causes a warmer object to warm up than it would otherwise”. That usage of words and reasoning is semantic sophistry, and has nothing to do with the underlying physics actually occurring which leads to the “higher temperature than otherwise”, nor the physics of radiative thermal heat transfer, which does not follow such a scheme at all. When the scenario is radiative thermal energy transfer, which is the only thing this climate greenhouse business is about, and has no relation to the function of convective circulation reduction as in real greenhouses and jackets/blankets, then a cold object of course does not transfer its radiative thermal energy to a warmer source, and so, the presence of a cold object does not cause a warmer source to warm up some more. Of course, a cold jacket/blanket or greenhouse ceiling doesn’t transfer its thermal energy to a warmer source either by any means of thermal transfer. When you put on a cold jacket it feels cold. Something else is doing the heating and you need to know what it is doing, rather than just saying that cold caused hot to get hotter. With a real greenhouse, or a jacket, cold does not cause hot to get hotter. The whole underlying problem here is this dual and contradictory definition of a greenhouse effect: the real one for actual greenhouses and blankets, and the semantic sophistry one of climate science that is opposite how a real greenhouse functions and which further doesn’t actually exist and which a real greenhouse demonstrates is false. It’s from this contradiction that the sophistry and false analogy is able to be created and thus defended with that false, fake, lying, dishonest language, while doing its best to avoid any actual reference to the physics and principles of the Laws of Thermodynamics (such as the first, which debunks the steel greenhouse and climate greenhouse effect, but leaves the real greenhouse effect intact), and its true mathematics. Which is what we see here with Willis. Whether he does it on purpose, knowing, or because he doesn’t know any better, is unknown. Either way, these people will need to be held responsible for their crimes.]

49. Squid2112 says:

Joe, I would like to thank you for pointing out, with extreme clarity, that the “blanket analogy” exactly describes an actual greenhouse and not the atmosphere. I have heard these warmists proclaim over and over again, how the way a “greenhouse” works isn’t really how the atmosphere works, and that the term “greenhouse” is just a term used, however, they will then spin right around and say “but it works like a blanket” … ROFL … and yet, a “blanket” is precisely how a real “greenhouse” works.

You just can’t make up stupid like that.

To Willis, I thank you for your demeanor. Your response to me was very pleasant indeed, despite the fact that I was really calling you a dumbass. I do appreciate the respect you have shown me, very much different than when I was originally arguing these points with you and others in the comments on the original Steel Greenhouse posting on WUWT, where you and Anthony would throw wild punches at practically any question I asked, or any criticism I gave, which is the reason I no longer consider WUWT a credible source of information, nor do I frequent there anymore.

In closing to Willis, it would be nice of people such as yourself, to have the capacity to realize their errors and respectfully admit to them. Your “Steel Greenhouse” thought experiment is a bust. It doesn’t work. It has been shown 1000 ways from Sunday how it does not work, and yet you insist that it does, spinning the most incredible mental gymnastics in the process. Would it not be easier to admit to the mistakes in this experiment, learn by them, and thank those people, such as Joe and others, for their hard work at determining the real truths? Afterall, isn’t truth the whole objective here? I might remind everyone, this “Steel Greenhouse” thought experiment was born out of “Yes Virginia, a colder object can make a warmer object warmer still” by Dr. Roy Spencer. He is wrong, and has been proven to be wrong time and again, by a plethora of people and facts. The “Steel Greenhouse” is simply a spin of the same failed topic from Spencer. Accept the errors, learn by them and seek the truth.

Joe, thank you so much for continuing to exercise patience and clearly illustrating the truth here! I appreciate it more than you can know.

50. Willis,

From memory of the original article and what you have said here, you made the assumption of a small distance between the sphere and the shell to make the maths easier. Let me try and say what is being explained to you here in another way.

Let’s assume that there is a large distance between the sphere and the shell. After all if the principle you are explaining is valid, it should not matter what the actual values are. Let’s give some values to some things to make everything clearer.

The surface area of a sphere is 4 x pi x r^2. So if the sphere has a radius of 10m then the surface area of the sphere is 4 x 3.1416 x 10 x 10 = 1256.64 m2. At 235 W/m2 the sphere has a total energy output of 295310.4 Watts (joules per second). Now let’s say that the shell is 10m from the sphere. The surface area of the shell is 4 x 3.1416 x 20 x 20 = 5026.56 m2. Dividing the 295310.4 Watts that the sphere is producing by the surface area of the shell we get an incoming flux of 58.75 W/m2. What temperature will that flux make the shell? How is that going to make the sphere hotter and produce more energy? Let me know when you can get an ice block to make a candle flame burn hotter by being next to it.

It is energy that has to be conserved, not energy flux. Just because there is radiation does not mean there is heat.

You have tried to use an analogy of two solid objects and a vacuum with an internal heat source to explain an effect that apparently occurs with a solid and fluid planetary body with a free flowing atmosphere and an external heat source. What could go wrong? Apparently almost everything.

51. Great stuff Lorenzo, exactly. Of course, Willis thinks that if he violates the Law of Conservation of Energy “just a little bit”, then it’s OK, because “approximations are used all the time in physics”. What an idiot! He knows not what he does.

52. Willis Eschenbach says:

Lorenzo from Oz says:
2014/12/23 at 5:38 AM

Willis,

From memory of the original article and what you have said here, you made the assumption of a small distance between the sphere and the shell to make the maths easier. Let me try and say what is being explained to you here in another way.

Let’s assume that there is a large distance between the sphere and the shell. After all if the principle you are explaining is valid, it should not matter what the actual values are. Let’s give some values to some things to make everything clearer.

The surface area of a sphere is 4 x pi x r^2. So if the sphere has a radius of 10m then the surface area of the sphere is 4 x 3.1416 x 10 x 10 = 1256.64 m2. At 235 W/m2 the sphere has a total energy output of 295310.4 Watts (joules per second). Now let’s say that the shell is 10m from the sphere. The surface area of the shell is 4 x 3.1416 x 20 x 20 = 5026.56 m2. Dividing the 295310.4 Watts that the sphere is producing by the surface area of the shell we get an incoming flux of 58.75 W/m2. What temperature will that flux make the shell?

Lorenzo, you are almost there. The only part you are missing is that the shell is radiating from an area which is twice as large as it is absorbing on. This is because the shell has two sides. So it is absorbing over an area of A (5026.56 m2) but radiating over an area of 2A.

[JP: Willis, the shell doesn’t lose radiation on the inside; the shell is an enclosure, so any internal emission doesn’t lead to loss of energy. Energy is only lost on the outside. The shell does NOT radiatively shed energy over 2 surface areas, only 1.]

At this point the shell is absorbing 295,310.4 watts. But it is only radiating half of that to space, with the other half being radiated inwards, back to the sphere.

So … at that point the sphere is absorbing half the amount radiated by the shell, and the other half is going out to space. Note that the amount radiated inwards doesn’t magically disappear. It warms the sphere, which by the immutable laws of physics must then emit more radiation.

[JP: Willis, again, please familiarize yourself with the equation for radiative transfer of heat energy. No heat is transferred from the shell to the sphere, nor from inside the shell to other parts of itself (the shell) inside.]

This system is obviously not in equilibrium, because the shell is not radiating to space the amount of energy being generated in the center of the sphere. Surrounding the planet with a shell has reduced the amount of energy sent to space, since half of it is being radiated inwards.

At what point does the system reach equilibrium? Well, it reaches it when the shell is radiating to space the amount it receives from the sphere. This, of course, means it is radiating the same amount inwards. And that inward radiation warms the surface of the planet.

[JP: Except a real greenhouse should then behave like this, but it doesn’t. A real greenhouse doesn’t warm by internal self-emission/absorption. Nothing does that in fact. And heat can only transfer from warm to cool, thus, the cooler shell can not warm the warmer sphere. The shell gets the energy from the sphere, the shell warms up to the equivalent temperature, the shell then emits that energy on its exterior, thus conserving energy. This reality conserves energy whereas you violate conservation of energy. The system is in equilibrium when the shell reaches the temperature of the flux impingent upon it from the sphere; the shell doesn’t increase the flux from the sphere because it is not a power source for the sphere, nor is it a heat source for the sphere, nor does it transfer energy to the sphere. See the heat flow equation. I’ve provided the links which prove the math previously.]

There is no violation of the laws of physics. It’s simple. Put anything around the sphere which slows the radiation to space, and the sphere will heat up.

[JP: Actually your method violated the law of conservation of energy, and then you just claimed that it was OK to do that because you were violating it only a little bit. That is quackery. And your statement is merely a claim anyway, with no proof, and which debunks itself when analysed.]

It is energy that has to be conserved, not energy flux. Just because there is radiation does not mean there is heat.

You seem to be a bit confused. “Heat” is a NET flow of energy from a warmer object to a cooler object. This, of course, is what is happening in both figures in the head post. In the first figure, the net flow of energy is from the planet to space. In the second, there is a net flow of energy from the warmer planet to the cooler shell, and another net flow from the shell to space. Thus, there is no second law violation.

[JP: There is first law violation, let alone second. Since there is no heat flow from the shell to the sphere, then the sphere can’t be heated by the shell. Heat is net energy flow, and it requires heat to raise something’s temperature. Since there is no heat flow from the shell to the sphere, then the shell doesn’t heat the sphere.]

Let me know when you can get an ice block to make a candle flame burn hotter by being next to it.

I fear you are asking the wrong question there. You need to be asking “compared to what”?

For example, does the shell make the planet hotter? Well … compared to what? Compared to no shell it does, because there is a flow of radiation from the shell to the planet, and no flow of radiation from space to the planet. As a result, the planet ends up warmer than it would be without the shell.

[JP: Heat flow, Willis, not just radiation emission irrespective of the heat flow equation. There is no heat flow from the shell to the sphere since the shell is cooler, and since that is what the math physics equation also says for radiative heat transfer. The sphere heats the shell. The shell can by definition not raise the temperature of the sphere. You seem mixed up with definitions and concepts here, which is common for amateurs without actual education in these matters.]

So yes, a candle flame will burn hotter when next to an ice block [JP: Wrong. The flame just heats the ice.], compared to when it is next to a block of say dry ice (solid CO2) at a much colder temperature. It’s simple physics. And if I shine a flashlight at the sun, the sun ends up warmer. [JP: Making stuff up. Where’s your math? You’ve never used any…right I forgot.] Warmer than what? Warmer than it would be if it were not absorbing the energy in the photons from the flashlight. Strange but true. The photons don’t know where they came from, they just add energy to whatever they hit. Of course the flow of HEAT (the net energy flow) is from the sun to the flashlight … but the sun ends up warmer in the process. [JP Sophistry. You just spin wild speculative BS.]

I hope that this clears up the confusion. If not, let me know, and I’ll give it another shot.

[JP: It clears up how much of a sophist you are, and that you’re working for the climate alarmists protecting their greenhouse pseudoscience. We see what you are Willis.]

My best to you,

w.

53. Opinionated says:

Just to clarify it was me posting as “Lorenzo from Oz”. I was staying with my brother at the time and was using his computer to comment, hence the name change.

Willis, you say …

Lorenzo, you are almost there. The only part you are missing is that the shell is radiating from an area which is twice as large as it is absorbing on. This is because the shell has two sides. So it is absorbing over an area of A (5026.56 m2) but radiating over an area of 2A.

At this point the shell is absorbing 295,310.4 watts. But it is only radiating half of that to space, with the other half being radiated inwards, back to the sphere.

So … at that point the sphere is absorbing half the amount radiated by the shell, and the other half is going out to space. Note that the amount radiated inwards doesn’t magically disappear. It warms the sphere, which by the immutable laws of physics must then emit more radiation.

Again Willis you show your ignorance of basic physics with these statements. I was showing how the flow of energy was actually occuring and you have magically changed things to suit your narative. As the shell receives the energy it will heat up according to its absorbtion and emissitivity characteristics and will ultimately achieve a much lower temperature than the sphere because it is receiving about a quarter of the energy per unit of surface area. As the shell temperature rises, the differential between the sphere and shell reduces which will affect the net flow of heat between the two, but when a new equilibrium is reached the total system will emit at a flux equal to the total energy divided by the total surface area of the exterior.

You continue with …

So yes, a candle flame will burn hotter when next to an ice block compared to when it is next to a block of say dry ice (solid CO2) at a much colder temperature. It’s simple physics. And if I shine a flashlight at the sun, the sun ends up warmer.

Wow. I am speachless at this level of ignorance. We are talking about the candle flame, not the ambient temperature of the surrounding area. The candle will burn at the temperature dictated by the chemical reaction of the materials that are being used. Not all photons are created equal. Any photons being radiated from the ice cube will not have the energy to excite the molecules of the candle flame to reach a higher state. This is the fundamental mechanism why heat flows from hot to cold. Radiation is real, but it does not automatically mean more heat. It can also mean scattering and reflection which is what occurs in your sphere and shell world which is ultimately why the total system will emit energy flux at the total energy level divided by the total surface area.

I have no confusion. You seem to be full of it.

54. Manxman says:

I am really disappointed when these comments end,
Take Willis or the SS [Cook et al] positions, they always boil down, to sophistry, as a layman sometimes it is hard to see.

Here with Joes inserted bolded replies, the sophistry is easy to discern, frankly i admire the professionalism of the old school educated scientific honesty.

I recognise without fully understanding the equations, or complexities of the conversations taking place, integrity, and for me integrity slices through all their sophistical bullshit like a scalpel.

You know Joe its a pity Willy the wood-worker didnt stay longer, same as the SS bloggers,
The slice and dice of their sophistry is a joy to behold, and i hope you all never tire of it.

.