# ‘To heat a planetary surface’ for dummies; Part 3

We’re still discussing Willis Eschenbach’s ‘Steel Greenhouse’.

How come the warming EFFECT of putting the shell around the sphere is real but Eschenbach’s “back radiation” EXPLANATION of how it comes about is wrong?

Simply put, it’s because the effect doesn’t violate the 2nd Law of Thermodynamics, but the explanation does.

In Part 1 and Part 2 we established some fairly basic principles of thermodynamics that we can now put to use in analysing Eschenbach’s explanation of why and how the radiating central sphere needs to warm with the steel shell surrounding it:

“In order to maintain its thermal equilibrium, the whole system must still [after the steel shell is placed around the sphere] radiate 235 W/m2 out to space. To do this, the steel shell must warm until it is radiating at 235 watts per square metre. Of course, since a shell has an inside and an outside, it will also radiate 235 watts inward to the planet. The planet is now being heated by 235 W/m2 of energy from the interior, and 235 W/m2 from the shell. This will warm the planetary surface until it reaches a temperature of 470 watts per square metre. In vacuum conditions as described, this would be a perfect greenhouse, with no losses of any kind.”

The first part of this paragraph simply describes the necessary conditions for reaching a new dynamic equilibrium upon putting the steel shell up around the radiating sphere. Nothing mysterious about it at all.

But then (in the bolded part) Eschenbach starts ‘explaining’ how he sees this new state of dynamic equilibrium to be accomplished.

And this is where any connection to basic, ordinary physics – and hence, to the real world – appears to be lost.

Let’s parse what he’s saying:

• Of course, since a shell has an inside and an outside, it will also radiate 235 watts inward to the planet.

Now will it indeed? Says who? The people adhering to the ‘bidirectional flow’ concept of radiative heat transfer do. This concept was however only ever meant to be a theoretical, descriptive model. It is not observed reality. It is a SUGGESTION, an ASSUMPTION, of how things might work, and nothing else. I feel the need here to reiterate the words of quantum theory master Niels Bohr:

“There is no quantum world. There is only an abstract quantum physical description. It is wrong to think that the task of physics is to find out how nature IS. Physics concerns what we can SAY about nature …”

Nature reveals to us the unidirectional transfer of energy as ‘heat’ from a hot object to a cold one. This spontaneous transfer (or, specifically, its effect) can be directly observed, measured, detected, sensed. It is, then, by definition, a real, directly verifiable physical phenomenon.

However, in order to deconstruct this phenomenon, to describe exactly what goes on in detail when energy is transferred from a hot to a cold object as heat (in our case, radiative heat), to describe (and – hopefully – be able to explain) the inner nature of a heat transfer, so to say, we have to resort to mental models. Because then we enter the realm of the invisible, the unobservable. We cannot possibly say in this case how nature really IS. We can only guess and theorise about it, say something ABOUT nature based on what it does in fact reveal to us.

The ‘bidirectional flow’ concept of radiative heat transfer (originally Pierre Prévost’s ‘radiative theory of exchanges’ (1791)) is such a mental model (a stubborn one at that!), an attempt to describe in an orderly and (‘macroscopically’) understandable, instructive fashion what might happen on a microscopic (quantum) level when we observe energy being transferred as radiative heat across a vacuum between two surfaces at different temperatures.

That doesn’t mean we KNOW that what this model describes is reality, the real situation, that the suggested outline of what’s going on is what actually physically happens. It is merely a (convenient, useful) model of how it perhaps could happen.

Note, we can only ever physically observe (detect) the actual ‘heat’ being transferred. And temperatures changing as a result. Nothing else. Everything else is pure theory. People tend to forget (or ignore, or deny) this.

Let’s find out, then, if this model in the end turns out to be a good description of reality. What’s Eschenbach saying next?

• The planet is now being heated by 235 W/m2 of energy from the interior, and 235 W/m2 from the shell.

Well, yeah, if you are one of those believing that there is in fact “back radiation” constituting a real flux (transfer) of energy from the warm shell to the (even warmer) sphere, then the sphere will necessarily receive (and absorb) ‘235 W/m2 of energy from the interior, and 235 W/m2 from the shell’.

Finally:

• This will warm the planetary surface until it reaches a temperature of 470 watts per square metre.

What Eschenbach is doing here is this: He takes the heat flux equivalent from the nuclear power source inside the sphere (235 W/m2) and ADDS to this the “back radiation” flux from the shell (235 W/m2). From this he simply ends up with [235+235=] 470 W/m2. And acquires a corresponding blackbody temperature for the sphere surface amounting to 301.7K, ~1.19 times the absolute temperature of the shell temperature and of the initial temperature of the sphere itself.

Can he do this?

Of course he can’t.

I will, however, in the following try to show that the ultimate problem here is not Eschenbach’s use of the ‘bidirectional flow’ model. It is the ‘bidirectional flow’ model itself.

The first catch with a ‘two-stream’ concept such as this becomes apparent as soon as one attempts to apply it to a decidedly thermodynamical problem, one which strictly deals with the transfer of energy between systems to change ‘internal energies’ and ‘temperatures’ of those systems.

The ‘bidirectional flow’ model aims to portray a single heat transfer as a ‘net’ of TWO separate and opposite ‘energy transfers’ in one.

Such a conceptual idea doesn’t square well (it is in fact completely and directly at odds) with basic thermodynamic principles, principles that are firmly founded on what we can actually observe in nature. We simply do not observe a two-stream energy transfer between two objects. Anywhere. At any time.

The ‘bidirectional flow’ model hence conveys a certain mental embroidering of reality that goes beyond what reality itself reveals, a mere attempt at making sense of a phenomenon that by its inscrutable nature alone would appear to boggle a human mind.

It is so very easy at this point to lose one’s focus, to let words and mental visualisations cloud one’s actual vision.

So I want you to hold on to this fundamental knowledge, based on observed reality:

• A thermodynamic ‘energy transfer’ between two systems (or a system and its surroundings) is always unidirectional. There is only ONE transfer of energy going on between two systems at any one time. Furthermore, such an ‘energy transfer’ can only come in two varieties: as ‘heat’ [Q], a spontaneous transfer of energy by virtue of a temperature difference between the two systems, always and only from the warmer to the cooler system; or as ‘work’ [W], by one system doing work on the other.

Take note:

• By the 1st Law of Thermodynamics [ΔU = Q – W], a transfer of energy to or from a system (as ‘heat’ [Q] or ‘work’ [W]) changes the ‘internal energy’ [U] – and thus normally the temperature – of that system. This effect (changing system internal energies and thus temperatures) is exclusive to the two varieties of thermodynamic energy transfer. If a thermodynamic process brings about a change in a system’s ‘internal energy’ [U] (and thus temperature), you have transferred energy to or from it as ‘heat’ [Q] and/or ‘work’ [W]. There is no other way. Transfer heat TO it or let its surroundings do work ON it, the internal energy increases. And temperature rises. Transfer heat FROM it or let IT do work on its surroundings, the internal energy DEcreases. And temperature drops.

Eschenbach and the rGHE “back radiation” proponents seem utterly oblivious to these simple facts of thermodynamics. Or they completely ignore them (sweep them under the rug) and hope that no one will notice.

So where, ultimately, does the ‘bidirectional flow’ concept go wrong? What specifically in the end makes splitting up a unidirectional ‘heat’ transfer into two separate and opposing ‘energy’ transfers violate the 2nd Law of Thermodynamics?

I will try my best to explain:

Let’s return to Eschenbach’s ‘Steel Greenhouse.’ Hopefully we can all agree that q below would be the ‘radiative heat flux’ from the warmer sphere to the cooler shell:

q = σ(Tsphere4 – Tshell4)

According to the ‘bidirectional flow’ concept of radiative heat transfer, the radiative heat is simply the ‘net’ of the two opposing flows (fluxes) of radiative energy between the two objects involved in the heat transfer. In other words, neither of the two temperature terms on the righthand side are in themselves supposed to be ‘heats’; rather, they are considered ‘radiative emission fluxes’. Only the ‘net’ – the vector sum – of the two apparently constitutes what is called a ‘radiative heat transfer’. So, the ‘bidirectional flow’ concept tries to separate between ‘radiative emission fluxes’ and ‘radiative heat fluxes’. ‘Energy’, apparently, moves both ways, but ‘heat’ moves only one.

Awkward, perhaps, but a fair enough principle. As long as you manage to stick to it. But does the ‘bidirectional flow’ concept really pull off the task of keeping the opposing ‘radiation fluxes’ and the resulting (net) ‘heat flux’ apart as separate entities all the way? That’s the question.

Well, already from the onset it starts losing some of its sense and coherence. Because two separate, opposing fluxes could never make one unidirectional flux in between them. The first flux would simply transfer energy from the warmer system to the cooler and the second flux from the cooler system to the warmer. The ‘heat’, then, could only ever be the ‘net result‘ of this two-way transfer: Since the cool object transfers less energy by radiation to the warmer object than the other way around, then after each cycle of radiative exchange there will be an energy surplus (an increase in ‘internal energy’) in the cooler object (large flux IN minus small flux OUT) and an energy deficit (a decrease in ‘internal energy’) in the warmer object (small flux IN minus large flux OUT), meaning, the cooler object would warm and the warmer object would cool. There has been a net transfer of energy from the warmer to the cooler object. This would be the ‘heat’. Not itself an actual flux, that is. Just a net result of two (opposing) fluxes.

Peculiar? Maybe. Maybe not. At least such a view doesn’t appear to violate any laws of thermodynamics. Yet.

But this is precisely where the problem starts emerging. For the ‘bidirectioners’.

The size, the magnitude, of this net transfer of energy (q, the ‘heat’) in a radiative heat transfer between two systems at different temperatures depends very much on the size, the magnitude, of that temperature difference between them. It’s right there in the Stefan-Boltzmann radiative heat transfer equation above. So, large difference, large net transfer of energy, large q; small difference, small net transfer of energy, small q. If the difference is zero, then q is also zero. Because then each object transfers as much energy to the other as the other transfers to it. According to the ‘bidirectional flow’ concept …

This too I presume we can all agree on.

Further, the q is, at the same time, the net energy loss of the warmer object per unit time per unit area (J/s/m2) AND the net energy gain of the cooler object per unit time per unit area. Which means it is simultaneously the Qout (the energy lost as ‘heat’) of the warmer object AND the Qin (the energy gained as ‘heat’) of the cooler object. The warmer object being the central sphere or the surface of the Earth, the cooler object being the surrounding shell or the atmosphere.

I hope this is also clear to everyone.

However, the warmer object (the sphere, Earth’s surface), as we all know by now, does not only have a Qout. It doesn’t just lose energy. It also itself has a Qin. Coming from ‘behind’, so to say, from its power/heat source, basically its very own ‘hot reservoir’. Through a different, separate heat transfer:

Figure 1. QH is the Qin of the sphere, its energy input or gain. QC is the Qout of the sphere, its energy output or loss. To keep the sphere system in a dynamic equilibrium, the sphere’s internal energy, and thus temperature, in a steady state, these two ‘heats’ must be of equal size (as much energy coming IN as what goes OUT at any time).

The point here is that as the temperature difference between the sphere and the shell grows less (from the shell warming), then q (equal to QC in Figure 1; the sphere’s Qout, energy lost as ‘heat’ to the shell, the cold reservoir) also naturally becomes smaller.

At the same time, the sphere’s Qin (equal to QH in Figure 1; energy gained as ‘heat’ from the hot reservoir) remains the same as before.

It should go without saying, then, that in this situation, the energy gained by the system (the sphere) can no longer be matched by the energy lost: Qin is now larger than Qout. Which means the ‘net heat’, the Q in the First Law equation (defined as ‘the energy transferred to the system as heat’, basically the balance between the system’s heat IN and heat OUT), is no longer zero, like it was in the initial state of dynamic equilibrium, before the shell started warming. It would now be positive again, as it was during the original heating of the sphere.

And so, some of the energy constantly provided to the sphere by the internal power source would once again pile up inside the sphere’s mass (not as much energy can escape as what comes in), increasing its U and hence its T. Towards a new dynamic equilibrium, reached only when the sphere’s Qout once more matches its Qin.

Basic thermodynamics. Basic energy accounting.

q needs to be preserved. The sphere thus needs to warm.

The problem for the ‘bidirectioners’ appears just here. According to the ‘bidirectional flow’ concept, the radiative emission fluxes are simply dependent on each object’s surface temperature, so as long as the sphere’s blackbody surface stays at 255K, it will always send out a ‘radiative emission flux’ of 240 W/m2, matching the ‘heat flux’ equivalent from its internal power source. So as much energy escapes the surface of the sphere per unit time as radiation as what enters at the core of the sphere per unit time.

So how come, as the shell warms, the sphere warms also, if its radiative output stays the same as always?

Because as the shell warms, the q (the ‘radiative HEAT flux’ from the sphere to the shell) decreases, which means the sphere’s Qout decreases. And according to the ‘bidirectional flow’ concept, how does the q decrease if the radiative emission flux from the sphere stays the same? By the opposing radiative emission flux from the shell increasing, of course:

q = σ(Tsphere4 – Tshell4)

(The larger the Tshell4 term with the sphere temp kept constant, the smaller the q. Simple as that.)

So if q needs to be preserved (and it does), then the Tsphere4 term needs to increase correspondingly.

But how does the sphere’s surface temperature rise? It can only happen by a direct increase in its internal energy (+U). Energy from somewhere will have to pile up.

But from where?

We know already that the energy in from the power source at its core is constant, it never changes. No help there. We also know that the emissive power of the sphere (its radiative output) is completely and only due to its surface temperature. In the ‘bidirectional flow’ model. So unless it actually cools, its output rate will not diminish. No help there either.

That leaves only one alternative.

Only the now ‘extra’ energy in from the shell is available to pile up at/below the surface of the sphere. The “back radiation” flux from the warming shell. That’s the only difference. The only ‘new’ energy.

In the ‘bidirectional flow’ model.

So in effect, an extra ‘heat input’ has been added to the sphere that wasn’t there before the shell came into place. A ‘radiation flux’ merely, you say. No, ‘radiative heat‘. A physical transfer of energy to the sphere, directly and all by itself increasing its ‘internal energy’, thus raising its temperature and, consequently, its corresponding ‘radiative emission flux’ out. Back towards the shell. [235+235=] 470 W/m2.

That’s a transfer of energy as ‘heat’. By thermodynamic definition.

From cool shell to warm sphere …

This is what Eschenbach is saying:

q (the sphere’s Qout) needs to be conserved. And he’s correct. Simply because its Qin is conserved. q needs to match the Qin in order for the sphere’s Q (‘net heat IN/OUT’) to remain at zero; Q=0 means ΔU=0 and so no further temperature rise.

He then says that as the shell warms, q naturally decreases, forcing the sphere to warm also to make it grow back to where it was. And he’s still correct.

What he then does is simply using the ‘bidirectional flow’ model to explain how this further warming of the sphere comes about. And within the framework of this particular view of the world, he would actually still be correct. For there it is specifically the “back radiation” from the shell to the sphere that makes the ‘internal energy’ of the sphere increase. It is the “back radiation” from the shell being absorbed by the sphere as ‘extra’ energy, in addition to the energy from the internal power source, that raises its surface temperature and thus enables it to radiate out more energy than before, than with only the internal power source providing the energy.

So Eschenbach, when he claims that, at the new dynamic equilibrium, the shell radiates 235 W/m2 to space from its outer surface and 235 W/m2 to the sphere from its inner surface, forcing the sphere to radiate 470 W/m2 to the shell to maintain its Qout (sphere>shell q) at 235 W/m2 ([470–235=] 235 W/m2), to match the steady 235 W/m2 input equivalent from the internal power source, then he is well within his rights to do so.

The ‘bidirectional flow’ model allows him to; in fact, demands him to …

So the basic problem with Eschenbach’s explanation of the ‘Steel Greenhouse’ (and with the climate establishment’s common explanation of the rGHE (see below)) lies not so much with Eschenbach himself, but rather with that fundamental idea of a ‘bidirectional flow’ in a radiative heat transfer, the model concept of a two-way radiative flux exchange making up a ‘net sum’ called the ‘heat’.

It is a problematic concept to say the least. Any physical explanation ending up violating the 2nd Law of Thermodynamics is …

The ‘bidirectional flow’ model in effect allows you to pretend that there are two separate, simultaneous heat transfers going on between the sphere and the shell, one where the sphere gets to add to the internal energy of the shell, thus warming it, and one where the shell gets to add to the internal energy of the sphere, thus warming it. This model basically treats both terms on the righthand side of the radiative heat transfer equation as real, separate, thermodynamically working ‘radiative heat fluxes’ (as if the two objects were thermally isolated from one another):

q = σ(Tsphere4 – Tshell4)

The thing is, even though we know from basic thermodynamic principles that this cannot be, ‘heat’ by definition flows only one way between systems and in nature invariably and exclusively from hot to cold, the fundamental flaw lying at the heart of this (admittedly, deeply ingrained) conceptual idea only becomes a real and obvious problem as you connect a system to both a (steady) hot and a cold reservoir, essentially when you both (constantly) heat and insulate it at the same time. At other times* we need not worry too much about it. We can sort of avoid having to be directly confronted with it. The concept ‘works’. It ‘explains’ stuff. It is ‘practical’. (So why discard it?)

*Normally, in radiative heat transfer textbooks, the bodies in question are either all unpowered and simply start at different temperatures, moving towards thermal equilibrium between each other, or they are somehow kept at constant temperature (meaning, the power input varies). I can’t recall having seen any example where a central object surrounded by a passive insulating shell is rather heated by a constant power input (meaning, the temperature is free to change). This would be the instance where the inherent flaw in the ‘bidirectional flow’ concept emerges as a direct explanatory problem.

But in the case of such a three-body arrangement (where the central system is constantly heated by the first system and insulated by the third), adhering to the ‘bidirectional flow’ explanation of things, the only reason the central system warms beyond the equilibrated state with its heat source (its hot reservoir) alone, is the additional energy input from the “back radiation” flux from the insulating layer, its heat sink (its cold reservoir). The cool system heating the warm system. There is no way around it. This is the corollary of the ‘bidirectional flow’ concept. You are simply forced to break the 2nd Law of Thermodynamics.

The situation with the heated central sphere and the surrounding steel shell insulating it (Eschenbach’s ‘Steel Greenhouse’) is the exact equivalent to the rGHE “back radiation” idea of how the surface of the Earth warms beyond its pure solar radiative equilibrium temperature; also a three-body setup, sun > sfc > atm:

Figure 2. (Derived from Stephens et al. 2012.)

Evidently, the Sun here could itself only possibly warm the surface as far as 165 W/m2 (232K), so you need the addition of the 345 W/m2 down from the cooler atmosphere to warm it to 289K, radiating a corresponding blackbody emission flux of 398 W/m2. In other words: The entire rise in surface temperature from 232 to 289K (57 degrees) is specifically due to the absorption of the atmospheric extra radiative energy input to the surface, the DWLWIR (“back radiation”) ‘flux’, and nothing else. Note that there is NO restriction whatsoever to the outgoing radiation from the surface at any point in the cycle. As per the ‘bidirectional flow’ concept. The warming, then, is ONLY caused by more energy coming IN, not in any way by less energy going OUT. Increased energy input. More ‘heat’ to the surface. And that extra heat is NOT from the hot Sun, but from the cool atmosphere …

Conclusion:

• If you somehow insulate a constantly heated object, then the object WILL equilibrate at a higher temperature than if you DIDN’T insulate it.
• However, it is NOT energy from the insulating layer doing the extra warming. Cold cannot heat hot. It is STILL only the energy from the power source constantly heating the object that can make it warmer.

– – –

• What the insulating layer does, though, is warming – by absorbing the heat flux from the object it insulates – to a temperature beyond the heated object’s original surroundings, the ultimate heat sink now outside the insulating layer.
• This increases the temperature ‘potential’ facing the central heated object, which in turn reduces the DIFFERENCE between the temperature ‘potential’ of the heated object itself and that of its surroundings, compared to what it used to be before the insulating layer started warming.
• Since it is the difference in system temperature ‘potentials’ that generates a ‘heat flux’ between the systems, the size of this difference matters: The larger the difference, the larger the heat flux. And the smaller the difference, the smaller the heat flux.
• If the INCOMING heat flux (the Qin) to an object is kept constant, but its OUTGOING counterpart (the Qout) is reduced, the object will naturally warm, from the Qin energy not being able to escape as fast as before, thus partially piling up inside of it instead. Until we reach a new (and higher) surface temperature to restore the balance.

– – –

• An insulating layer simply impedes the energy ESCAPING the heated/powered central object as heat, thus indirectly forcing it to warm further. The insulating layer, however, does this merely through its temperature being higher than the heated object’s original surroundings. It doesn’t do it by radiating some of the heated object’s own emitted energy back to it as recycled ‘heat’ to be absorbed a second time to warm it some more. This would violate the laws of thermodynamics.

– – –

• The source of today’s ubiquitous and deep-seated belief that a warm surface can warm even further from simply absorbing “back radiation” from a cool atmosphere is the (archaic) ‘bidirectional flow’ concept of radiative heat transfer …

Next up: How do you heat a planetary surface, then? If not by the Earth’s own thermal radiation, a result of its temperature rather than a cause of it … How does the atmosphere insulate the surface?

## 88 comments on “‘To heat a planetary surface’ for dummies; Part 3”

1. Bryan says:

Oculaer says

“The source of today’s ubiquitous and deep-seated belief that a warm surface can warm even further from simply absorbing “back radiation” from a cool atmosphere is the (archaic) ‘bidirectional flow’ concept of radiative heat transfer”

Its not clear here whether you agree with Willis that if you accept two way transfer of radiation you must also accept its identical to a two way transfer heat transfer.

Willis admits he has had no formal training in physics.
A person calling a whale a fish is similarly just not properly informed.
Willis is simply unaware of the technical meaning the term ‘heat’ in thermodynamics.

Nobody with a physics background would say the spontaneous flow of photons from a colder to a hotter object is a heat transfer.

You have presented a long exposition of the classical one way heat transfer,
Fine.

However you seem to think that the two way transfer of energy and the one way transfer of heat somehow violates the second law.

It appears that you are uncomfortable with photons and photon flow.
Claes Johnson is also of that opinion.
Good luck to Claes but physics history tells us that to establish a new paradigm the old paradigm must fail to address some critical problem.
So far Claes has not done that.

Every physics textbook on the planet accepts the photon explanation.

• okulaer says:

I’m a bit stunned by your comment, Bryan. Because you argue not as if you’ve misread my post, but rather as if you haven’t read it at all.

You say: “Nobody with a physics background would say the spontaneous flow of photons from a colder to a hotter object is a heat transfer.”

OF COURSE they wouldn’t! That’s my whole point. No one is saying so. This is the grand self-deception. People just don’t see it. No one CALLS it heat. Because everyone knows that would violate the 2nd Law. But everyone still keeps treating the postulated “back radiation flux” AS IF IT WERE heat! A complete blind spot to what they are in fact doing. They expect it to give a result upon (re)absorption by the warmer surface exactly as if it were an extra, separate heat flux, namely by making it directly warmer. Not less cool, Bryan. Warmer. In absolute terms. From 232K to 289K. All because of the 345 W/m2 “back radiation flux” from the cooler atmosphere. And nothing else. ADDED to the solar heat flux. As if the two were of the exact same kind.

How you can possibly fail to see this is quite frankly beyond me.

This is not about semantics, Bryan. And I thought I explained exactly why in my post. Apparantly I still wasn’t clear enough.

Yes, if you ‘accept two way transfer of radiation you must also accept its identical to a two way transfer heat transfer.’ This is what my post is about.

Plenty of radiative heat transfer textbooks already employ terminology to this effect, that there is heat going both ways, only more from hot to cold than the other way around, so resulting in a ‘net heat transfer’. Not ‘net energy’. ‘Net heat.’

Connect the warmer object to a constant power source, and you end up having the flux from the cooler insulating object heating it an extra bit.

This is not about the EFFECT, Bryan. It’s real. It violates no physical laws. It’s about the EXPLANATION.

Do you agree with Eschenbach? Is the central sphere absorbing 235 W/m2 from its internal power source PLUS another 235 W/m2 from the shell, thus emitting a flux back to the shell of 470 W/m2 at 302K? If you don’t agree, how would you explain it differently, still adhering to the ‘bidirectional flow’ principle? I am actually very curious to know.

• Please indicate where a Thermal photon from a lower temperature surface is transiting to a higher temperature surface and has ever been detected or observed? Please indicate where “any” photon in transit has ever been detected or observed? The term photon was invented as a scam just like back radiation. Some folk that should know better accepted the scams. There are demonstrations of the photoelectric effect every day. In every case the demonstration of the action of an EM wavelet was sufficient to equal or exceed the work function of the matter absorbing that action and the resultant WF event. Only that event has ever been detected. If the frequency of that wavelet is to low that event ceases. It is never true that the lower frequency wavelet/s cannot produce events with a much lower work function like specific heat.

2. mkelly says:

Willis further makes a claim (not stated as such) that 4*pi*r^2 equals 4*pi*(r+y)^2.

The inner surface of the shell is at a lower temperature than the sphere by virtue of its surface area. He skips over that part. So no heat can pass from she’ll to sphere.

• okulaer says:

The sphere has a higher temperature than the shell at dynamic equilibrium because its Q_out (equal to the q from sphere to shell) needs to be conserved in order to match the sphere’s Q_in from its internal power source.

But you are of course right that the larger surface area of the shell would make the difference even larger. However, this point is not actually relevant to the insulation concepts discussed.

3. Bryan says:

Oculaer says

“Plenty of radiative heat transfer textbooks already employ terminology to this effect, that there is heat going both ways”

Name one PHYSICS textbook that says heat is going two ways.
I don’t think you can.

Because in some non physics textbooks the writers are sloppy with language is a matter to be regretted.
For instance the term centrifugal force is used in some descriptions of earth satellite motion is to be found in non physics explanation’s.
A physics textbook would indicate that the centrifugal force in this case is a pseudo force and would avoid using the term.
I prefer to stay within the framework of physics and have no trouble accepting the description using photons.
The hotter surface radiation is of higher intensity and includes higher frequency photons than the colder surface.
The net of the two flows is called heat and the transfer is always spontaneously from higher to lower temperature.
Your method of classical single heat flow and ignoring the two streams is also valid.
However to say categorically that the colder surface does not emit photons or that these photons are not absorbed by the hotter surface is wrong.

I will write a longer note with my take on this particular topic but it will take a little longer to write up.

• okulaer says:

Bryan, you say: “Name one PHYSICS textbook that says heat is going two ways. I don’t think you can.”

You need to pay attention to what I say. Here is what I said: “Plenty of radiative heat transfer textbooks already employ terminology to this effect, that there is heat going both ways, only more from hot to cold than the other way around, so resulting in a ‘net heat transfer’. Not ‘net energy’. ‘Net heat.’”

They of course never come out and state directly that ‘heat transfers both ways’. That would be absurd. But they do IMPLY that it does by using terminology like “heat exchange”, “net heat transfer” and “net heat flux”. It is inconsistent, un-physical and confusing. Confusing people like Science of Doom (and he’s certainly not alone!) to believe that it does if fact happen like this: radiative heat transfers both ways, only more from hot to cold than from cold to hot, so the NET HEAT moves from hot to cold.

Here are a couple of authoritative textbook examples:

Further, you say: “The hotter surface radiation is of higher intensity and includes higher frequency photons than the colder surface.”

The mean photon frequency of each opposing radiation flux is of NO importance to the ‘bidirectional’ interpretation of the Stefan-Boltzmann radiative heat transfer equation, Bryan: q = σT_h^4 – σT_c^4. And I think you know that. The photons in the flux IN from the cooler object to the warmer one replace exactly the photons in the flux OUT from the warmer object to the cooler one, save the ‘surplus’ in intensity (joules per second per square metre) OUT. It is ONLY the difference in flux intensity of the two opposing radiation fluxes that makes the ‘heat flux’.

If anything, your suggestion confirms that the ‘unidirectional’ interpretation of the S-B radiative heat transfer equation (q = σ[T_h^4 – T_c^4]) is the only valid one, because here it is only the TEMPERATURE difference between the objects in question that matters and the radiative transfer of energy from hot to cold is directly determined by this difference alone, so we need not concern ourselves with whether or not a counter-flux replacing (much or most of) the ‘warmer’ flux out from the hot object is ‘energetic’ enough to really manage to replace it in that way.

You say in your next comment, Bryan (below): “There is a wise guideline in science that tells us that theories that cannot be supported by experiment cannot be justified.”

Indeed. And the ‘bidirectional flow’ concept – the ‘theory of exchanges’ – is just one such theory. It cannot be supported by experiment and so cannot really be justified other than hypothetically. Take it from the horse’s mouth, from the discoverer of the T^4 relationship between temperature and radiative heat (yes, ‘heat’) emission from a blackbody himself, Josef Stefan in 1879:

“After studying Tyndall’s data, Stefan wrote, “From weak red heat (about 525°C) to complete white heat (about 1200°C) the intensity of the radiation increases from 10.4 to 122, thus nearly 12-fold (more precisely 11.7). The ratio of the absolute temperature 273+1200 and 273+525 raised to the fourth power gives 11.6.”

(…)

He [Stefan] wrote that his analysis had a, “hypothetical nature and reasoned support for [it] was impossible, so long as measurements are not made of radiation to surroundings at absolute zero, or at least a very low temperature.”

We can only ever observe the ‘heat’, Bryan. Never the two assumed fluxes INSIDE the heat transfer. Which is something any experimentalist, engineer or physicist would know as a fundamental empirical fact of life, the universe and everything.

The point is that the idea of a “back radiation flux” ultimately violates the 2nd Law of Thermodynamics. Not in and by itself. But as soon as the surface starts warming beyond the pure solar radiative equilibrium temperature (or the sphere starts warming beyond its equilibrated temperature with its internal power source alone), the ‘bidirectional’ EXPLANATION of what’s going on can only lead you to the conclusion that it is the energy back from the cooler atmosphere (or shell), being reabsorbed by the warmer surface/sphere, and ONLY that, that increases the internal energy – and thus temperature – of the surface/sphere. Effectively, extra HEAT from cooler to warmer. It is never PROCLAIMED to be heat, of course. But it is expected to give a result exactly AS IF IT WERE heat. That’s the persistent mental block that people live with right there. They cannot see what the ‘bidirectional’ principle in the end makes them do. Because they don’t want to see it. It all sounds so plausible and right, so neat. So it cannot be wrong. Not in a million years.

I appreciate your input, Bryan (like in the following comment here), but I would like to ask you again: Do you agree with Eschenbach? Is the central sphere absorbing 235 W/m2 from its internal power source PLUS another 235 W/m2 from the shell, thus emitting a flux back to the shell of 470 W/m2 at 302K? If you don’t agree, how would you explain it differently, still adhering to the ‘bidirectional flow’ principle?

I would very much like to see your answer to this. It would illuminate a few things …

• Heat Transfer Textbook 4th edition, by John H. Lienhard V and John H. Lienhard IV.
Dover (2011) IBSN-1`3: 978-0-486-47931-6 \$34.95
Lots of the view factors are screwy and wrong also!

4. Bryan says:

The steel greenhouse model is supposed to have some relevance to the greenhouse effect on Earth.

Most people on seeing it for the first time think it defies common sense and hence there must be something wrong with it.

To get it to ‘fly’ you must abandon common sense and make a number of unphysical assumptions.

We will do this but also point out the gross departure from reality on occasion.

The core or planet is heated by a constant heat source of 240W/m2.

A constant heat source in reality is very hard to supply .

That will for example work away through all changes in temperature.

The planet (core) and the shell are superconducting so that each has its own isothermal temperature.

This is to satisfy the requirement for central core heating to reach the surface.

In reality a central heating core through rock would take millions of years to reach the surface.

One model suggests millions of tiny nuclear reactors scattered evenly through the core but we will stick with whichever methods achieves an isothermal core.

The shell also needs to be isothermal for simplicity so that both inner and outer radiate equally.

A metal core and shell both have emissivity values of 1 despite metals having values nearer zero in the infrared but it simplifies the maths.

The shell is almost touching the surface to maximise the radiative power/area from the shell.

The input energy = P x t where P = 240w/m2 and t is time in seconds

All other units are from the MKS system and have largely been omitted for clarity

At all the interfaces net energy flow(heat) is from higher to lower temperature.

Output energy = dU1 + dU2 + Radiative losses to space from outer shell surface

U1 = gain in internal energy of core = Cc.Mc.dTc

U2 = gain in internal energy of shell = Cs.Ms.dTs

Cc = specific heat capacity of planet (core) = 2000 (value for rock)

Cs = specific heat capacity of planet (core) = 2000 (value for rock)

Mc = mass of Earth = 6 x 10^24

Ms = mass of shell ( lets make it one thousandth the Earth mass )

dT = delta T is temperature change.

Radiative losses will occur as temperature rises from zero to final temperature .

The instantaneous value is proportional to the temperature raised to the fourth power.

Readers can supply their own values of specific heat ( for example water = 4200 ) but I don’t think it will make a lot of difference.

For temperature change in the first instance we will choose one degree Kelvin.

This modest value allows us to leave the radiative value of outer core as minimum to be calculated later if required.

This gets the model into a rough scale of magnitude.

Input energy for whole planet surface = 240.A.t

A = area of Earth Surface = 5.1 x 10^14

t = time in seconds = bit we are trying to find out

U1 = gain in internal energy of core = Cc.Mc.dTc = 2000 x 6 x10^24

U2 = gain in internal energy of shell = Cs.Ms.dTs = 2000 x 6 x 10^21

Time this all takes in seconds = energy output/power input = 12.002 x 10 ^27/ 1.224 x 10 ^17

Time in seconds = 9.8 x 10 ^10

Time years = 3107 years for a one Kelvin temperature rise.

Keen readers might want to go on and add radiative losses from the outer shell surface but that will only increase the time beyond 3107 years.

This is well beyond anything that can be determined by experiment.

A bit like saying if you empty a bucket of hot water in the Ocean then the Ocean temperature will rise

There is a wise guideline in science that tells us that theories that cannot be supported by experiment cannot be justified.

Its enough to say that if a constant power source is acting then further insulation around the source is added then a temperature rise will occur.

Why would this model be pushed to support a conjecture that a tiny rise in the atmospheric CO2 fraction will push the Earth into dangerous climate territory?

• “The instantaneous value is proportional to the temperature raised to the fourth power.”

Not at all, that value is the “lowest” temperature that the planet “must” maintain to radiate that much power in all direction with “no” opposing radiance at any wavelength . if anything interferes with such exitance like low emissivity or an interfering “shell” that surface temperature must increase to compensate.

5. markstoval says:

“Most people on seeing it for the first time think it defies common sense and hence there must be something wrong with it.”

Upon first seeing it, I though it was pure bull dropings to claim that the shell would warm up the planet. I still do.

The thing that I saw that was not addressed (or I missed it — getting old here) was that as the shell gets warmer it radiates ever more towards deep space and not 50/50 — half always being radiated back towards the planet. In other words, the heat exchange is going from the planet to the shell to deep space. Nothing is warming the planet as far as I can see.

• “In other words, the heat exchange is going from the planet to the shell to deep space.”

Correct Just like Dr. Kirchhoff said in his law of radiative equilibrium.

“Nothing is warming the planet as far as I can see.”
True but anything like lowering emissivity could! The Earth’s atmosphere does not as the atmosphere is a better emitter to space than the surface can be. Gravity keeps the surface temperature above the radiative atmosphere

• okulaer says:

Mark, you say: “Upon first seeing it, I though it was pure bull dropings to claim that the shell would warm up the planet. I still do.”

The shell isn’t warming the sphere. This should be abundantly clear from my posts. If you still think that’s what I’m claiming, I urge you to read my posts on this subject carefully one more time, or … I can’t help you.

There is NO energy transferred from the shell to the sphere. The energy transfer is unidirectional and goes like this: internal power source >> sphere >> shell >> space.

The thing is that the radiative heat does not escape as easily (joules per second per square metre) from the sphere and out with a warm shell around it as it did with only the vacuum of space around it, courtesy of the radiative heat transfer equation:

q = σ(T_sphere^4 – 0) vs. q = σ (T_sphere^4 – T_shell^4)

As you can hopefully gather from this, the temperature (thus, the ‘radiative potential’) facing the sphere’s surface would be higher with a warm shell than with a ‘cold’ space. Hence generating a smaller heat flux (q).

Since the q from sphere to shell equals the sphere’s Q_out (and the shell’s Q_in), then if q (the transfer of energy as radiative heat away from the surface of the sphere through the intervening vacuum to the inner surface of the shell) gets smaller (upon the warming of the shell), then the sphere’s Q_out is no longer able to match its Q_in from its internal power source, which always remains the same.

When Q_in is larger than Q_out for the sphere system, then its Q (‘net heat IN/OUT) becomes positive again, like it was during the sphere’s original warming process. Which means that energy naturally piles up inside the system (this is how any real object warms to begin with, after all), increasing the internal energy [U] and raising the temperature as a result, until the system’s Q_out once more matches its Q_in (as much energy OUT as heat per unit of time as what comes IN as heat).

I wonder why this perfectly logical and basic process is so hard to grasp … It is pure energy budgeting. If more comes in than what goes out, then there’s an accumulation of the incoming energy. Note: of the INCOMING ENERGY. It is the energy from the sphere’s internal power source that piles up, that makes it warmer, NOT returning energy from the shell.

Mark, you also repeat the argument: “(…) as the shell gets warmer it radiates ever more towards deep space and not 50/50 — half always being radiated back towards the planet.”

Exactly! You present this as if this is a flaw in my reasoning. But this is exactly what I’m pointing out! In the ‘bidirectional flow’ model, the shell always radiates half back to its heat source, the sphere, making it even warmer. In the ‘unidirectional’ way of seeing things, this does not happen. The energy ONLY transfers from hot to cold, from sphere to shell to space.

But the shell radiating ever more energy as heat to space from its outer surface does not concern what’s happening on its inner surface. Only its rising temperature concerns what’s going on on its inner surface, where the heat comes IN. The heat out to space is a pure result of its temperature rising. Only at the point where the shell has reached a temperature where it can radiate as much energy per unit time to space from its outer surface as what comes in from the sphere to its inner surface, will the shell stop warming. We have reached a dynamic equilibrium.

The warmer the shell gets, the more heat it radiates to space and the harder it is for the incoming heat flux from the sphere to warm it further, because less and less of the energy transferred stays within the mass of the shell.

However, the sphere NEEDS to warm (ideally) from 255 to 303K (1.19x) in order to be able to feed a constant heat flux to the shell, at dynamic equilibrium finally matched by the shell’s own heat flux out to space.

• markstoval says:

“Exactly! You present this as if this is a flaw in my reasoning. But this is exactly what I’m pointing out! In the ‘bidirectional flow’ model, the shell always radiates half back to its heat source, the sphere, making it even warmer. In the ‘unidirectional’ way of seeing things, this does not happen. The energy ONLY transfers from hot to cold, from sphere to shell to space.”

I don’t know why you think I was trying to find fault with your reasoning. I am just presenting my own understanding of the situation. I am not totally untrained in this area but I am little better than the “man in the street” as far as I am concerned. So, as I follow the back and forth of real skeptics I like to give my understanding and see who shoots it down and why. I enjoyed Will J. commenting on my take on it for example.

I have been reading this blog a long time. I would not have been reading it if I were an enemy. Don’t be paranoid.

Peace. 😉

• okulaer says:

Sorry, Mark 🙂

I think I’ve grown more cynical than paranoid over the last couple of years of dicussing subjects like this with all kinds of people, mostly with rabid, blindly dogmatic warmists. I simply find it a bit hard to decipher what in my argument as laid out it is that you object to, because it sure does look to me as if you don’t agree with the sphere’s need to warm with the shell surrounding it. I just can’t get my head around your reasoning why.

I’m most interested in hearing people’s thoughts about this, by all means. Differences may indeed be subtle. But I need to see coherent (counter-) arguments to be able to meet them properly.

I just don’t get the “but the shell is losing heat to space, so therefore the sphere cannot warm” argument. Or “the heat moves only from the sphere to the shell, so therefore the sphere cannot warm” argument.

If there had been only the heat transfer between the sphere and the shell involved in this process, OF COURSE the sphere couldn’t warm. But there is also a (constant) heat transfer between the internal power source and the sphere going on. Postma in his argumentation, for instance, seems oblivious to this fact … He sees the need for the q between the sphere and the shell to move all the way to zero for equilibrium. But that’s absurd. That would be the case only for a SINGLE, ISOLATED heat transfer.

• “you don’t agree with the sphere’s need to warm with the shell surrounding it. I just can’t get my head around your reasoning why”.

It is the phrase “to warm” that is the Climate Clown intentional ambiguity. In the vernacular the phrase can mean to feel warmer, or to increase temperature as well as the scientific add sensible heat energy to. I think it would be more clear with, “The sphere must spontaneously increase in temperature in order to maintain the same outward flux in the presence of the shell at some intermediate temperature.”. Here there is no need for any mass or energy only a one way power (flux) that need be maintained by the creation (somehow) of that power as sensible heat which must result in temperature for the generation of such thermal EM flux. The whole thing about the so called “conservation of energy” is nonsense, especially with power transfers between Sol Earth and Space through and with a dispersive atmosphere.
The whole thing is not easy, and the easiest way to confound yourself, is to use their language.
The Clowns speak of energy budget, again deliberate miss direction. It is but a balance of flux in and flux out, rather than the maintenance of some superficial thermalization. By the same token any average of “temperature” is truly and intentionally meaningless, in the presence of the highly nonlinear relationship between EMR and temperature. Averaging flux to derive something meaningful, would be only for the competent 3% never for the incompetent 97%.

• okulaer says:

“I think it would be more clear with, “The sphere must spontaneously increase in temperature in order to maintain the same outward flux in the presence of the shell at some intermediate temperature.””

Sounds good to me 🙂

6. Kristian,
While your efforts are fine and correct they are not convincing.
Start with this (q = σ[T_h^4 – T_c^4]) Wrong!
The S-B equation is P/A =epsilon σ(T_h^4 – T_c^4)
P/A is not energy it is “flux from a surface” A vector quantity
The only vector on the right side is the parenthesis. This is the needed vector addition of the two opposing “radiances” proportional to some power of temperature. This is the difference in radiative potential as scaled by sigma and the dimensionless “emissivity/epsilon”. This difference in radiative potential is the only force that produces any flux again the difference.
Read any book prior to 1960 on electromagnetic field theory.

7. Bryan says:

Kristian

My initial idea was to split the problem into two parts.
With the core(planet)_ and shell in place from the start;

The first was with the heater to start at zero Kelvin and find out how long it takes to increase by one degree Kelvin the radiation from the outer shell face could be ignored for simplification
Answer 3107years see post above

Once the temperature rises the radiative losses to space rise very considerably
My next part would be to repeat the calculation at 250K for shell ie near its new Equilibrium

Input energy for one second = P . A. t = 1.224 x 10^17 Joules (specified at start)

Near equilibrium, core temperature = 1.19 shell temperature.

Output energy = dU1 + dU2 + Radiative losses to space from outer shell surface

U1 = gain in internal energy of core = Cc.Mc.(298 – x)

U2 = gain in internal energy of shell = Cs.Ms.(250 – y)

Cc = specific heat capacity of planet (core) = 2000 (value for rock)

Cs = specific heat capacity of planet (core) = 2000 (value for rock)

Mc = mass of Earth = 6 x 10^24

Ms = mass of shell ( lets make it one thousandth the Earth mass )

dT = delta T is temperature change.

Radiative losses to space from outer shell surface = A . # . T ^4

A = surface area of Earth
# = SB constant

Radiative losses = 1.13 x10^17 Joules/s

x = 1.19y
Solve for x and y

Result. My 12 figure calculator could not find any MEASURABLE increase in temperature of Shell and Core

Conclusion
If specific heat capacities are included the Willis Steel Greenhouse model never reaches anything like the temperatures claimed even if you could wait around for the age of the universe

Why did SoD, Willis and other presenters of similar models fail to include heat capacity?

8. okulaer says:

Bryan, I do see your point. However, the ‘Steel Greenhouse’ is only meant to be a conceptual model. The concept being illustrated is ‘insulation’, in this particular case ‘radiative insulation’.

Radiative insulation works. In principle. But it is nothing like the real Earth system. And there is no “back radiation flux” from the cooler atmosphere (or shell) making the warmer surface (or sphere) even warmer upon reabsorption.

The atmosphere does NOT insulate the surface of the Earth by radiative means. I will get to this in the next post in this series …

But what about the ‘bidirectional flow’ concept, Bryan? How do you see it work for the ‘Steel Greenhouse’ if not warming the sphere? How will the “back radiation flux” from the shell NOT warm the heated sphere in such a model? Could you please give me a summary of your position?

9. Bryan says:

okulaer

” Could you please give me a summary of your position?”

Two approaches (classical equilibrium thermodynamics and photon approach) give the same results.
So use either
The photon approach is the orthadox physics apprach.
There is overwhelming evidence that electromagnetic radiation is quantised.

http://hyperphysics.phy-astr.gsu.edu/hbase/qapp.html#c3

There are even experiments involving a single photon

There are differences between the two streams which could be detected with an IR spectrometer.
Particularly if there is a large difference in temperature.
Pointing at the hotter surface and isolated from the colder a more intense beam and higher frequency would be observed.
Then reverse spectrometer and look at colder surface a less intense and also without the higher frequencies.

My main objection to the general climate science approach is not the term back-radiation but the quantity which the IPCC ascribe to it .
Atmospheric radiation is thin and band widths with many frequencies missing.
The so called atmospheric window and so on .

Yet the diagram below shows it almost as strong as the surface radiation.

They apparently think that the SB equation can be used for gases.
There is also questions about Kirchhoff’s Law and SB equation .

The video is also very good

http://hockeyschtick.blogspot.co.uk/2014/05/new-paper-questions-basic-physics.html

• okulaer says:

Thanks, Bryan.

You say: “Two approaches (classical equilibrium thermodynamics and photon approach) give the same results.”

If what you refer to here is the ‘unidirectional’ and the ‘bidirectional’ approach respectively, then they both indeed give the same EFFECT (result), but they EXPLAIN how this effect comes about in very different ways. This is specifically what my post discusses.

The unidirectional approach explains it like this: Less energy manages to escape the surface of the high-temperature sphere as heat per unit time with a lower-temperature (though still warm) shell facing it than with the coldness of space facing it. Its energy OUTPUT is reduced. With a constant energy INPUT from its internal power source, this creates an imbalance. Energy from the SOURCE piles up inside the sphere, thus forcing it to warm.

The bidirectional approach, however, explains it like this: Just as much energy manages to escape the surface of the sphere (after equilibration with its internal power source) with a warm shell facing it as with space facing it. Only its temperature determines its radiative OUTPUT. The energy INPUT from the internal power source of course also stays the same. The two are of equal size. However, there is more energy now coming IN to the sphere’s surface than before, the “back radiation” from the warm shell. THIS energy piles up at/below the surface, forcing it to warm.

See the difference, Bryan? I sure hope you do. The EFFECT is the same. But the former EXPLANATION do not violate any laws of thermodynamics. It claims a reduced energy OUTPUT from the surface. Hence, it is the heat SOURCE energy that piles up, doing the extra warming. The latter rather claims an increased energy INPUT to the surface. Hence, it is the heat SINK energy that piles up, doing the extra warming.

The ‘bidirectional’ approach claims an increase in U – hence a rise in T – of the warmer sphere SOLELY as a result of the extra transfer of energy from the cooler shell. This is a (hidden) claim of an extra HEAT TRANSFER to the sphere, Bryan. Heat from cold to hot.

No need to give me the “No one’s claiming it’s heat” argument. Of course no one’s claiming it’s heat. I’m not saying they do. I’m saying they TREAT it like it’s heat. A transfer of energy from one system to another, all by itself and directly, in absolute terms, increasing the internal energy [U] and temperature of the absorbing system, is a transfer of HEAT, Bryan. If not ‘work’. It is Q_in. I’m sorry you can’t see this. That’s why they SAY there’s only one heat transfer, but in reality they TREAT the exchange as if there are TWO separate ones, moving past each other in the opposite direction, BOTH heating where they end up.

This doesn’t work, Bryan. Such an explanation of a thermodynamic effect cannot stand.

And I can’t see where you address this point … You seem to avoid it. Rather talking about how ‘photons’ do exist. OF COURSE they exist. Well, they are still conceptual entities of the quantum theory. I prefer to call them ‘EM wavefronts’ or ‘radiative energy packets’, ‘quanta’. But ‘photons’ will do.

What I can’t get people to grasp, it seems, is that even though there is radiation going on everywhere, even though ‘photons’ fly around everywhere, in all directions, it doesn’t mean a cool object TRANSFERS energy by radiation, sends a radiative FLUX, to a warmer object in its vicinity. You need to distinguish between the realms of the microscopic and the macroscopic. The world of quantum mechanics and the world of thermodynamics.

There are good analogies to this. Individual electrons INSIDE an electric current fly around in all directions at tremendous velocities, they don’t give a damn what the ‘system’ as a whole is up to. But the current, the general flow of electrons, moves ONE WAY only, from high voltage (potential) to low voltage (potential).

Likewise, individual air molecules certainly fly around at will, wherever they want, until they collide with other air molecules and bounce somewhere else. They could care less what the bulk air is doing. But if you have a high pressure and a low pressure zone (and you ignore friction/turbulence), then a wind will blow in ONE DIRECTION only, from high to low potential.

If you look at the water in a cup standing on a table, it looks completely still. And it is. In macroscopic terms. In microscopic terms, however, the water molecules move about constantly and incessantly. Not as freely as in air, of course, but just as randomly. If you pour the water out in the kitchen sink, the individual molecules will still move in whatever direction, up, down and to the sides. But the bulk of the water falls ONE WAY only – down, under gravity’s pull. From high to low potential.

Yes, ‘photons’ don’t carry mass, so will not bounce off each other, but that’s not the point here. The point is that microscopic entities (molecules, electrons, photons) flying around in all directions doesn’t translate into general macroscopic movement. In a heat transfer, energy is only being TRANSFERRED one way, from hot to cold, because that’s where the general movement of energy through the radiation field is going, from high potential to low potential.

• That’s pretty good.,

My generalization of the Clausius 2LTD is “Stuff don spontaneously go uphill’! Scientific Laws. are never unfalsified theory, they come strictly by carefully observing what don happen never.

• You that keep insisting that “energy” the accumulation of power (Q) , must be identical to force times distance: “work”(W) and be equatable. This is mathematical nonsense. Your algebraic symbols have no meaning whatsoever. The two may be related somehow,, but are never linear nor commensurate. At best the two are orthogonal. At worst for understanding, they are the complex of each other in four dimensions

• Interesting your site deleted the word conjugate after the word complex. is complex conjugate a dirty phrase.as some claim?

• markstoval says:

“There are good analogies to this. Individual electrons INSIDE an electric current fly around in all directions at tremendous velocities, they don’t give a damn what the ‘system’ as a whole is up to. But the current, the general flow of electrons, moves ONE WAY only, from high voltage (potential) to low voltage (potential).”

I like this analogy. I wonder what Dr. Brown or the other regulars at WUWT would say to this one. (yes, I really have never believed that “back radiation” can warm up the planet)

• markstoval says:January 28, 2015 at 9:36 am

“There are good analogies to this. Individual electrons INSIDE an electric current fly around in all directions at tremendous velocities, they don’t give a damn what the ‘system’ as a whole is up to. But the current, the general flow of electrons, moves ONE WAY only, from high voltage (potential) to low voltage (potential).”

I like this analogy. I wonder what Dr. Brown or the other regulars at WUWT would say to this one. (yes, I really have never believed that “back radiation” can warm up the planet)

Mark,
The equations describe an earthling algebraic concept. This physical always pisses on such concept. This planet does whatever. If earthlings ever discover this “whatever” this planet is already doing a way way different whatever!

10. “Yes, ‘photons’ don’t carry mass, so will not bounce off each other”

Wavelets have no mass but definite inertia, a wee part of the action, absorption by mass transfers this inertia. Normal reflection from mass transfers twice that inertia as that wavelet is proceeding at the speed of light along an opposing vector with the same inertia or -inertia, depending on your consideration of inertia (to keep on doing), a scalar or vector, your choice, do not change horses. I wish the quantum folk would carefully write down “what don happen never”.
If such has any finite probability, such is not a law.

• okulaer says:

Yes, I tend to agree. They do carry ‘momentum’. So there is some kind of ‘interaction’ possible. Basically, no one knows what’s really going on inside a radiation field, so the only thing we can really be sure of is what we can derive from direct observations. And in a heat transfer the ONLY thing we observe is the UNIDIRECTIONAL transfer of energy from hot to cold. We do not and cannot observe (detect) individual ‘radiative emission fluxes’ passing each other on the way to their respective targets … This can never be any more than pure conjecture, a guess.

• “Yes, I tend to agree. They do carry ‘momentum’. So there is some kind of ‘interaction’ possible.”

Please be careful with the term “momentum” Which is the function (velocity x mass). Constant velocity but zero mass. Zero momentum. The alternate concept is constant velocity and energy with an attitude. “inertia”, “This energy wish to do what I are doing”!! Stop me via absorption, I will be pissed. Reflect me to go in the opposite direction I will be doubly pissed. .

“Basically, no one knows what’s really going on inside a radiation field, so the only thing we can really be sure of is what we can derive from direct observations. And in a heat transfer the ONLY thing we observe is the UNIDIRECTIONAL transfer of energy from hot to cold. We do not and cannot observe (detect) individual ‘radiative emission fluxes’ passing each other on the way to their respective targets … This can never be any more than pure conjecture, a guess.”

Indeed such is a contradiction of all of Maxwell’s equations. I wish the Climate Clowns would admit they all have no clue, and are totally incompetent in this issue. Something to do with my dreams.

11. Bryan says:

Oculaer

One last attempt to make the orthodox physics interpretation of photon two way flow clearer.

Two metal blocks A and B sit separated inside a vacuum filled adiabatic enclosure.

Adiabatic enclosure consists of a perfect reflector face surrounded by a perfect insulator

Initially both at the same temperature. The zeroth law of thermodynamics applies.
Both emit and absorb equal amounts of radiation.

Neither one is said to heat the other.

One block (A) has a power supply which is now switched on causing the temperature of the block to rise.
This in turn means that it will emit more radiation.

A will now heat B causing its temperature to rise.
B will in turn emit more radiation but this back radiation is caused by A.

Now comes the clincher

If B were not there at all the temperature A would be even higher.

So B cannot be said in any meaning of the word as a cause of heating A

• okulaer says:

But this is a pointless example, Bryan. You are only avoiding the issue. To me it seems like you know that what I say is true, but try your utmost to find a way not having to admit it.

I will ask you one last time: Do you agree with Eschenbach? Is the central sphere absorbing 235 W/m2 from its internal power source PLUS another 235 W/m2 from the shell, thus emitting a flux back to the shell of 470 W/m2 at 302K? If you don’t agree, how would you explain it differently, still adhering to the ‘bidirectional flow’ principle?

Yes or no? Explain yourself. Don’t evade and divert.

• January 28, 2015 at 11:45 am

“Oculaer One last attempt to make the orthodox physics interpretation of photon two way flow clearer.”

We all hope so! No photons ever observable only the effect of a wave packet on a mass with lower radiance

“Two metal blocks A and B sit separated inside a vacuum filled adiabatic enclosure.
Adiabatic enclosure consists of a perfect reflector face surrounded by a perfect insulator.”

Aah! a thought problem, guaranteed to produce nonsense for those that refuse to think.

“Initially both at the same temperature. The zeroth law of thermodynamics applies.
Both emit and absorb equal amounts of radiation.”

The Zeroth law has nothing to do with radiation whatsoever. Kirchhoff wrote, that for any surface, the emissivity and absorptivity of that surface are the same, at each frequency and in each direction. Nothing at all of radiation. Your post normal textbooks screw that over royally also. In your thought problem all radiances are the same, even the walls if your enclosure.
No radiative flux exists at all within this enclosure. Maxwell’s equations demand such.

“Neither one is said to heat the other.”
“One block (A) has a power supply which is now switched on causing the temperature of the block to rise. This in turn means that it will emit more radiation.”

Block A generates flux exactly Power/surface area of A, All must be absorbed by block B. There is nowhere else for this power to go. Both temperatures must continue to increase until that temperature destroys what ever that power source may be. All power generation ceases. At that point all temperatures and radiances will equalize and all radiative flux will return to zero.

“A will now heat B causing its temperature to rise.
B will in turn emit more radiation but this back radiation is caused by A”

Block B never emits anything, with a lower radiance it can only absorb. Radiance in every direction is higher than that of itself, until that power ceases..

“Now comes the clincher”
“If B were not there at all the temperature A would be even higher.”

Slightly! The exact temperature that destroys the power supply. Then all radiative flux cease, and there is no need to equalize anything

“So B cannot be said in any meaning of the word as a cause of heating A”

That is true, but your post demonstrates that you refuse to think.

• Bryan says:

Will Janoschka

Just so you know
.
I consider your posts to be little more than psycho babble.

This is a perfect example .

The adiabatic enclosure sufficient for the purpose of demonstration could consist of expanded polystyrene box with polished aluminium as an inner lining.

Any experimenter would realise that.

The power supplied to block A would be calculated enough for the experiment to produce a result not an infinite amount sufficient to melt the block.

Any experimenter would realise that.

Your posts show that you have never attempted an experiment since leaving school which seems to be around 1960

The result of the experiment would show that colder object cannot heat warmer objects.

The result of your ill tempered posts means that I for one will leave you to stew in your profound ignorance.

• okulaer says:

“The result of the experiment would show that colder object cannot heat warmer objects.”

We all know that, Bryan. That’s not the issue. The question is, if you insulate a constantly heated object, it will become warmer than if you didn’t insulate it. Why? Is it because the insulating layer impedes the energy OUTPUT from the heated object? As per the UNIdirectional principle of heat transfer. Or is it because the insulating layer increases the energy INPUT to the heated object? From cold to hot. As per the BIdirectional principle of heat transfer.

What will it be, Bryan?

• Bryan says:

” The question is, if you insulate a constantly heated object, it will become warmer than if you didn’t insulate it. Why? Is it because the insulating layer impedes the energy OUTPUT from the heated object? As per the UNIdirectional principle of heat transfer.”

No I prefer to say impedes the heat output but not the energy output as far as radiative transfer is concerned.
This is radiative insulation

Again we are talking about radiative transfer for your next question

“Or is it because the insulating layer increases the energy INPUT to the heated object? From cold to hot. As per the BIdirectional principle of heat transfer.”

Yes

Now that I have answered your questions perhaps you will answer mine about radiation from the colder object in the direction of the hotter object

Does the cold object radiate towards the warm object
Yes or no?

When it arrives at the hotter object is the radiation absorbed
Yes or no?

If the answer is no where does the radiation from cold object go?

• January 29, 2015 at 10:05 am
Will Janoschka Just so you know
.
“I consider your posts to be little more than psycho babble. This is a perfect example .”

“The adiabatic enclosure sufficient for the purpose of demonstration could consist of expanded polystyrene box with polished aluminum as an inner lining.”

Ok I have done that. What have you ever done?

“Any experimenter would realize that. The power supplied to block A would be calculated enough for the experiment to produce a result not an infinite amount sufficient to melt the block.”

Indeed I would provide some exitance for such power that would not destroy everything. As I have done experimentally. I truly hate AW Shits, when I am trying to measure.

“Your posts show that you have never attempted an experiment since leaving school which seems to be around 1960”

Let me guess, Bryan You be some snot nosed kid with a PHD lecturing to innocent children, with no knowledge of what you are lecturing except that as is written in the Bible Physics textbook.

I still experiment with many AW shits, the only way to learn.

“The result of the experiment would show that colder object cannot heat warmer objects.”

There is no need for experiment of what is known. Except for idiots such as Bryan.

“The result of your ill tempered posts means that I for one will leave you to stew in your profound ignorance.”

I do not stew, I ponder the complete incompetence of all current institutions of higher learning.

• okulaer says:

Bryan, you say:

“No I prefer to say impedes the heat output but not the energy output as far as radiative transfer is concerned. This is radiative insulation.”

This is warmist claptrap, Bryan, I’m sorry. We’re constantly going in circles here. You switch from ‘energy’ to ‘heat’ and back again to ‘energy’ as suits your argument. Frankly, it doesn’t become you.

By doing so, you neatly at all times avoid the issue at hand, just like the warmists are doing. You seriously don’t see what you’re doing, Bryan?

Yes, we all know that the heated object warms as its HEAT output is reduced. But how do we EXPLAIN this reduction in heat output, Bryan? That’s where I’m trying to direct your focus. You’re making it a tough task. Unnecessarily tough. It’s like you’re playing a game of dodging the argument. (I’m afraid I will soon simply have to interpret this as a concession on your part that I’m right and you’re wrong.)

Listen, and please pay attention …

What specific ‘energy transfer’ is really doing the extra warming of the heated object as we insulate it?

(1) Is it an extra ‘energy transfer’ from its ‘heat source’ (the internal one)?

(2) Or is it an extra ‘energy transfer’ from its ‘heat sink’ (the insulating layer)?

(3) Or, is there no extra ‘energy transfer’ at all? Just less energy transferred away from the heated object, making the energy from the ‘heat source’ pile up inside it?

It think we can both agree that (1) is not it. That leaves (2) and (3). (2) is the BIdirectional explanation. (3) is the UNIdirectional explanation.

Since (2) clearly and directly violates the 2nd Law of Thermodynamics, we can be certain that that one’s not it either.

That leaves (3). The UNIdirectional explanation. The only coherent one. The only one consistent with the laws of thermodynamics.

– – –

“”Or is it because the insulating layer increases the energy INPUT to the heated object? From cold to hot. As per the BIdirectional principle of heat transfer.”

Yes”

Thanks, Bryan. Was that so hard?

And with that confirmation, you just admitted that the ‘bidirectional flow’ concept of radiative heat transfer directly ends up violating the 2nd Law of Thermodynamics. And so there isn’t much point in continuing this discussion …

You need to unchain from your dogma, Bryan.

“Does the cold object radiate towards the warm object
Yes or no?”

No. There is no radiative flux transferring energy from the cool object to the warm object. If you’d read my posts on this subject, you would know my position on this.

“When it arrives at the hotter object is the radiation absorbed
Yes or no?”

It doesn’t “arrive” and so isn’t absorbed.

“If the answer is no where does the radiation from cold object go?”

The radiation from the cold object moves as a heat flux towards its ‘heat sink’. DOWN the potential slope. Not UP towards its ‘heat source‘.

The energy transferred as radiative heat from the warmer object is absorbed by the cooler object. It accumulates within the cooler object, making it warmer in the process, and gradually and progressively escapes as radiative heat out towards its ‘cold reservoir’ on the outside. That’s where it ‘goes’.

‘Energy transfer’ is UNIdirectional, Bryan. That’s the only way, and it is what we observe in nature.

• “Does the cold object radiate towards the warm object
“Yes or no?”

No never, the cold object can only absorb EMR, according to its radiance.

“When it arrives at the hotter object is the radiation absorbed
Yes or no?”
There is no such radiation only your fantasy.

“If the answer is no where does the radiation from cold object go?”

Just how do you expect “no radiation” to go anywhere?

• Bryan says:

Oculaer

If you think that the colder object stops radiating, then I’m afraid I agree with you further discussion is pointless.

But you will have to ask yourself why would all the current physics textbooks have got it wrong.

All I can suggest is you read

Heat and Thermodynamics by Zemansky first published 1937.

This book was the only book recommended by Feynman in his lectures.

Its ultra orthodox classical thermodynamics .
Because it ran into several editions it can be bought cheap on Amazon.

In it you will find on radiation;

‘the difference between the radiation absorbed and the radiation emitted by a body is called heat and it is always spontaneously transferred from a higher to a lower temperature object’

So the hotter body emits more radiation that it absorbs the colder body absorbs more than it emits.

The heat transfer is from the hotter to the colder.

• okulaer says:

“If you think that the colder object stops radiating, then I’m afraid I agree with you further discussion is pointless.”

Bryan, THAT’S the dogma. That any object MUST and WILL radiate in all directions regardless of its surrounding conditions. There is NO experimental evidence this is a physical necessity. It is ALL theory. An assumption. THAT’S the ‘bidirectional flow’ principle. It is a theory without real-world empirical support, Bryan.

My contention is that the surrounding conditions DOES matter!

I know what the textbooks are saying. This is my whole point. I’m specifically criticising them for uncritcally accepting and adhering to this archaic concept introduced by Pierre Prévost in 1791, a remnant of the caloric theory. Everyone KNOWS it hasn’t been and cannot be verified. And still they treat it like an established FACT.

People need to THINK.

Again, if you read my post, you would see what I write about this whole thing. The ‘bidirectional flow’ concept SEEMS to be working fine as long as you stay away from the particular case of constantly heating an object and then insulating it. And NOWHERE in ANY of these physics textbooks will you find an example such as this. The central sphere is ALWAYS kept at a constant TEMPERATURE.

Bryan, I KNOW you’re a sceptic. I know you have a sceptical mind. Try to be sceptical at EVERYTHING you’re being told, not just parts of it … Think it through for yourself. Don’t just buy it.

If you feel there’s no more point discussing this topic, then fine. Thanks for visiting.

12. Mindert Eiting says:

Kristian said ‘If what you refer to here is the ‘unidirectional’ and the ‘bidirectional’ approach respectively, then they both indeed give the same EFFECT (result), but they EXPLAIN how this effect comes about in very different ways.’

At this point (like Bryan)I have to ask a question. Isn’t it a miracle that both explanations have the same result?

Elsewhere I used a financial metaphor. On your bank account you have steady streams of income and expenses. Both may equal each other in the frame of one month. Without additional income you can increase the amount of money on your account by delaying your expenses. Actually, you create debt, to be paid out later. This is an example of the conservation law applied at budgeting.

So you may decide to delay your monthly payment to your landlord. He will not like that but that’s not the issue. The bidirectional explanation for the extra money on your account, is that your landlord paid money back.

In the world of finance this may happen in case of errors but also here applies a kind of second law that money flows from debtors to creditors only. Weird as the explanation may be, there is a parameter involved. Did the landlord pay all your money back? Of course not, just fifty percent. Why that amount? Well, he did it in steps. After receiving those fifty percent back, you decided to pay him immediately fifty percent of the returned amount, and he returned fifty percent of what he got next, etc. Simple math may show the end result, but why?

In the unidirectional explanation you may decide to pay nothing to your landlord, or just one half of your debt. Let’s say your debt is x dollars. The bidirectional explanation must show the reason why the landlord paid back those x dollars in the game above. Why did his generosity precisely equal the amount about which you made your decision? Turning the question around, the bidirectional game may have the amount y as result. Why should you decide in the unidirectional game that x=y?

• okulaer says:

Hi, Mindert. You ask:

“At this point (like Bryan)I have to ask a question. Isn’t it a miracle that both explanations have the same result?”

No. Because we use the same equation, only interpreted differently. Bryan interprets the Stefan-Boltzmann radiative heat transfer equation like this:

q = σT_sphere^4 – σT_shell^4

That is, it is interpreted to describe a bidirectional exchange of emission fluxes.

I interpret it like this (and this is how the equation is actually and originally formulated):

q = σ[T_sphere^4 – T_shell^4]

That is, describing only a unidirectional transfer of energy from hot to cold.

The difference is subtle and mathematically/algebraically, they’re equivalent.

But regarding the physical understanding behind them, they are very different.

Being able to do something mathematically, doesn’t mean that you’re describing something that’s physically real.

– – –

Note, Mindert, that it is only the UNIdirectional transfer that is actually observed in nature, that is actually, physically detectable. The bidirectional interpretation is just that … a theoretical interpretation of reality.

• “The difference is subtle and mathematically/algebraically, they’re equivalent..”

Only the value, not the concept. Ask any mathematician of such claimed equivalence. Mathematicians respect parenthesis above all. Parenthesis are used only when required, to preserve rather than confuse. Both Dr. Stefan and Dr. Boltzmann understood this! Unlike the Climate Clowns.

13. Bryan says:

Oculaer says

“But this is a pointless example, Bryan. You are only avoiding the issue. To me it seems like you know that what I say is true”

Far from it
.
It is the simplest example you can find.

A hotter object and a colder object isolated from all other sources of energy.

Does the cold object heat or warm the hotter object …….No, quite the reverse!

Are the two objects exchanging energy……..Yes

Does the hotter object heat the colder …….Yes

So while the energy exchange is bidirectional the heat transfer is always one way;
spontaneously from higher to lower temperature.

My interpretation is in every physics textbook dealing with thermodynamics.
I cannot make it much simpler than that.
I afraid that if you don’t ‘get it’ we will have to disagree on this point.

• okulaer says:

“It is the simplest example you can find.”

No, Bryan. It is a contrived example to avoid the actual issue at hand. The simplest example you can find to illustrate what the ‘bidirectional flow’ principle entails is … the sphere/shell example.

Please answer the question I’ve asked you now three times around, Bryan. And stop evading.

“My interpretation is in every physics textbook dealing with thermodynamics.”

I’ve explained exactly at what point the ‘bidirectional flow’ concept gives absurd results and why. I’ve read the textbooks, Bryan. I know what they say. And I know about the examples they provide. They do not address the problem I’m pointing out. And neither are you, Bryan.

If you believe in this ‘bidirectional flow’ concept so much, why do you refuse to defend it directly? Why are you only circling around it …? Seems to me you’re caught by a dogma.

It is you who appear unwilling to ‘get it’, Bryan.

• okulaer says:

Sorry, I like you Bryan, but when you’re acting like this, you can be a bit annoying …

• “My interpretation is in every physics textbook dealing with thermodynamics.
I cannot make it much simpler than that.”

Read any physics textbook published prior to 1960. There you will find that the algebraic equation only describes the physical with understanding. The equation never is the physical.

14. Mindert Eiting says:

Thanks, Oculaer. I remember Claes Johnson said something similar. If you only have a difference in mathematical notation, this suggests to me that the theory is incomplete up to the observational level. After all, violation of the Second Law must have a price (like creating a perpetual mobile of the second kind). In my example, accounting at volume level cannot tell the difference but a check of the sequence of all bookings will. The SB equation seems symmetric in time (as most laws of nature), whereas the Second Law is not. However, I do not want to cause here philosophical distraction.

I have begun reading your former posts. Many are quite good. My I give you the tip to compile the best of them in a separate section, some with a bit of editing?

15. Thank you Mindert
“The SB equation seems symmetric in time (as most laws of nature),
whereas the Second Law is not. However, I do not want to cause here philosophical distraction.”

The S-B equation must be symmetric in time as such is a property of time.” (power)!
The Clausius 2LTD is independent of time (orthogonal). Spontaneously, stuff does not go uphill.

16. Kristian,
Hard to keep up. A sidebar column of latest posts would be helpful . WordPress will maintain that at no cost to you

• okulaer says:

Hmm, yeah, you’re probably right 🙂

17. markstoval says:

From an above comment exchange:

okulaer says:
January 27, 2015 at 4:57 pm

Sorry, Mark 🙂

I think I’ve grown more cynical than paranoid over the last couple of years of dicussing subjects like this with all kinds of people, mostly with rabid, blindly dogmatic warmists. I simply find it a bit hard to decipher what in my argument as laid out it is that you object to, because it sure does look to me as if you don’t agree with the sphere’s need to warm with the shell surrounding it. I just can’t get my head around your reasoning why. …

Hi, sorry to take so long to respond to that question, but my life at present affords me precious little “free” time to do this sort of thing. (sick mother and my students come before all else — and then there is the wife … )

To answer your question: I admit to being guilty of being skeptical of the central sphere heating up beyond what it was before the shell was added. I do not deny that it may be so, but I am unconvinced that it must be so.

The way I see it, when the shell is added it receives all the heat that the central ball can send to it and starts radiation to deep space as it heats up from whatever initial temperature it was at when added to the system. As the steel shell heats up it radiates to space just as the steel planet did. There is no loss in ability of the system to shed heat. In fact, since the shell has a larger diameter than the ball I think it may have an even greater ability to emit heat.

The shell could have been a top layer of the planet itself that was raised up a short distance so as to have a vacuum between it and the ball. If looked at that way, why would the shell lose any ability to emit the heat generated from the heat source?

Insulation works to prevent convective cooling and air circulation. I am led to believe that in this radiative example there is no convection/conduction before or after; so, I have trouble seeing any “insulation” happening.

I am not tied to the above in any dogmatic fashion, and would love to see some group do a tightly controlled experiment of the steel ball and shell to settle the question. Any correction is my understanding that you may offer will be greatly appreciated.

~Mark

• okulaer says:

Mark, good to see you’re hanging in there. I do realise (to my slight surprise) that my solution to the sphere/shell problem – that I feel myself is so obvious – comes off as quite a bit less intuitive to people than what I had expected.

However, reading your comment, I’m afraid I will once again have to simply repeat what I’ve already pointed out several times (first time in the top post itself). Only, I will try a slightly different approach this time, by posing a few questions, just to see if we can locate the ‘bone of contention’ here. I hope you can find your time to answer them.

1) No shell in place. Do you agree that upon initial equilibration with its internal power source, after having warmed to reach a final steady state temperature, the sphere emits as much energy per unit time from its surface OUT into space through radiant heat (q = σTh^4) as what it absorbs coming IN from its internal power source at its core, meaning effectively that its Q_out equals its Q_in to leave its Q (its ‘net heat IN/OUT’, as in: ΔU = Q – W) finally at zero (no further change (increase) in U and thus no further temperature rise)?

2) Shell in place. Do you agree that, according to the radiative heat transfer equation (q = σ[Th^4 – Tc^4]), as the shell starts absorbing the radiant heat coming in from the sphere, making its temperature (Tc) rise, then q (the radiant heat flux from sphere to shell) will grow smaller … provided that Th is not ‘allowed’ to rise also?

3) Do you agree that q in the equation above is exactly the same quantity/entity as the energy lost through radiant heat by the sphere (its Q_out) AND at the same time the energy gained through radiant heat by the shell (its Q_in)?

4) Do you agree that the energy input to the sphere from its internal power source – its effectve Q_in – remains constant throughout?

Please tell me which of these points/questions you do not agree with and specifically why. If you actually agree to all of them, I can’t see how you could possibly claim that the sphere wouldn’t warm further to a new steady state (dynamic equilibrium) with the shell up around it …

Cheers 🙂

• okulaer says:

BTW, Mark. You say: “Insulation works to prevent convective cooling and air circulation. I am led to believe that in this radiative example there is no convection/conduction before or after; so, I have trouble seeing any “insulation” happening.”

I’m not sure where you got this idea from, this restricted interpretation of the concept of insulation. What you are talking about is simply conventional insulation as we use it in our daily lives: building insulation, clothes, blankets and so forth. We live in a world full of air. The heat loss we mostly need to worry about (to impede) is thus primarily convective.

In places where there is no air, radiative heat loss is the main mechanism we have to ‘fight’. We then use ‘radiative insulation’.

The concept of ‘thermal insulation’ is simply about impeding the heat loss from a body, that is, to reduce its rate of heat loss. And that’s it. You can do this by obstructing the convective heat loss, the evaporative heat loss, the conductive heat loss or the radiative heat loss. They’re all heat loss mechanisms. Somehow suppress their smooth/free operation and you slow the cooling rate of the body being insulated. If the body is then also simultaneously heated, then it will warm.

What IS true, of course, is that on Earth, you won’t accomplish much by only reducing a body’s radiative heat loss, leaving all the other heat loss mechanisms alone. But, yeah, theoretically (I know) there should be an effect …

18. Mark,
Just do the whole equation for any n shells, with an isopower in the smallest, and shell(n) being any isotherm. Hint: work backward from the only temperature you have.

Psupply = a[1]σ(T[1]^4 – T[2]^4) = a[2]σ(T[2]^4 – T[3]^4) = a[3]σ(T[3]^4 – T[4]^4)….. a[n-3]σ(T[n-3]^4 – T[n-2]^4) = a[n-2]σ(T[n-2]^4 – T[n-1]^4) = a[n-1]σ(T[n-1]^4 – T[n]^4)

T[n] is the isotherm, and each smaller shell must have a higher temperature as it has less area thus more flux for the same Psupply I addition each rethermalization reduces the “effective” emissivity of the next smaller shell by one half. a[1] must be at the highest temperature for equilibrium. This also demonstrates a one way flux, but does not remove desired stupidity.
Remove any shell a[2]..a[n-1} and all inner shells must “decrease” in temperature by 1/1.19 because one rethermalization has been removed. There is only one power flow from the center to outer where all is absorbed to maintain that isotherm. Homework problem in 1964!
I measured such with three shells in 1967 to determine what wide band emissivity we could get with a fast black oxide on brushed aluminum.

19. markstoval says:

@ okulaer

To address your questions to me up above:

#1 — I can agree that the shell by itself with its internal heat source will indeed reach a state where it is radiating to deep space the same amount of energy that it is receiving from the internal heat source.

#2 — This is where we have difficulty (and so 3 & 4 are moot at present). Let me offer a quote that represents my understanding.

The most general statement of the radiative heat flow equation is

Q’ = A*(F1 – F2)

where F1 and F2 are a local flux from the two different sources. That is, the fluxes correspond to a specific single location, not their own locations independently. By convention, the location which the fluxes correspond to are that at the location represented by F2. That way, if Q’ is positive, then it denotes positive heat flow into the location specified by F2, and so that location would rise in temperature. If Q’ is negative, it denotes heat flow out of the location F2, and so if F2 is not a powered source of heat, then it will fall in temperature. However, whether or not the heat flow out of F2 towards F1 can actually cause an increase in temperature at location F1, requires its own analysis with all of the terms switched to the new location F1. …

In my understanding, as the shell receives heat flux from the sphere, Q will be positive as the sphere is warmer than the shell. As the shell gains heat, the difference will decrease but still be positive, and possibly at some point the shell will become as warm as the sphere (but maybe not as the surface area of the shell is greater) and Q will tend towards 0. I see no problem with Q becoming less as the shell and sphere become more equalized.

I can’t for the life of me see the passive shell heating up the sphere. Sorry. I can barely see the shell reaching the sphere’s level of heat given that is is radiation directly into deep space with a larger surface area than the original no-shell situation of the sphere. Again, sorry.

Please note: I am very aware that this issue has caused passions to flare up on all sides. I am no expert and you may discount my thinking on this — but I have never been able to see the shell heating the sphere.

• okulaer says:

Mark,

I see you’re quoting Postma. The problem with Postma’s position is that it focuses solely on the heat transfer between the sphere and the shell. And like I’ve pointed out before, if you do that, then you can trick yourself into believing that the sphere will only heat the shell without the need to warm itself. Of course it won’t. Not within a single, isolated heat transfer. In a single heat transfer, the hot end ALWAYS cools and the cold end ALWAYS warms. And then Postma simply declares that since the hot end in this case is heated constantly, then it will not cool.

This is completely nonsensical. No, it will not cool. It will WARM!

The hot end (the sphere) is constantly heated, not kept at a constant temperature. If energy could not escape as heat from the surface of the sphere, the heat input from the internal power source would just continue piling up inside the sphere forever, at the rate of 240 joules per second. Making the sphere hotter and hotter.

You say: “In my understanding, as the shell receives heat flux from the sphere, Q will be positive as the sphere is warmer than the shell. As the shell gains heat, the difference will decrease but still be positive, and possibly at some point the shell will become as warm as the sphere (but maybe not as the surface area of the shell is greater) and Q will tend towards 0.”

OK. But then you do agree, Mark, to my point 2): Shell in place. According to the radiative heat transfer equation (q = σ[Th^4 – Tc^4]), as the shell starts absorbing the radiant heat coming in from the sphere, making its temperature (Tc) rise, then q (the radiant heat flux from sphere to shell) will grow smaller … provided that Th is not ‘allowed’ to rise also?

My q is the same as Postma’s Q’, Mark.

You continue: “I see no problem with Q becoming less as the shell and sphere become more equalized.”

Really!? You see no problem with the heat flux from the sphere to the shell moving towards zero while the heat flux from the internal power source to the sphere keeps coming in at 240 W? No, because you don’t see this part. You, like Postma, focus only on the heat transfer between the sphere and the shell, in complete isolation. I’m afraid you can’t do that …

The heat moving from sphere to shell is the Q_out of the sphere, Mark. It is the energy it loses as heat from its surface per unit time. If it loses less energy as heat per unit time, but gains as much as before from its heat source, then what!?

How you (and Postma) are not able to see this glaring inconsistency in your argument, I cannot fathom.

It is all a matter of simple thermodynamic principles. Of energy budgeting. Accounting for the energy being transferred TO the system (the sphere) as heat at any one time versus the energy being transferred FROM the system as heat at any one time. It is exceedingly basic. The principle of insulation.

Postma appears utterly oblivious to this simple fact: The temperature of an object is NOT determined by the incoming energy (through ‘heat’ [Q] and ‘work’ [W] transfer) alone, but by the balance between its absorbed and its emitted energy (through ‘heat’ and ‘work’ transfer).

Only a blackbody radiating its entire heat loss directly into a perfect vacuum at 0 K will ever end up at a steady-state temperature exactly set by its mean incoming radiative heat flux (like this: 240 W/m2 IN >> 255K >> 240 W/m2 OUT). As soon as that object faces something warmer than 0 K, then its heat output is determined by the DIFFERENCE in temps between them, not by the blackbody’s own temp alone. And this means it will have to end up warmer in the final steady state than with only space surrounding it, because its outgoing heat at dynamic equilibrium must STILL balance its incoming (which doesn’t change).

– – –

I don’t get this absolute fear of acknowledging that the sphere will need to get warmer at steady state with the shell up around it than with only the vacuum of deep space surrounding it. There is nothing mysterious or abhorrent about this proposition at all. In fact, it follows from basic thermodynamic principles. Insulation. Impede the heat loss (ANY heat loss) from a constantly heated object and it will necessarily become warmer. The point is: IT WILL NOT BECOME WARMER FROM THE (COOLER) INSULATING LAYER TRANSFERRING EXTRA ENERGY TO IT AS IF IT WERE HEAT. IT WILL BECOME WARMER FROM THE ENERGY CONSTANTLY COMING IN FROM ITS (WARMER) HEAT SOURCE AS HEAT NOT BEING ABLE TO ESCAPE AS FAST AS BEFORE WITH THE INSULATING LAYER FACING IT.

20. Will Janoschkas says:

“#1 — I can agree that the shell by itself with its internal heat source will indeed reach a state where it is radiating to deep space the same amount of energy that it is receiving from the internal heat source.”

Can you also agree that not the shell, but the sphere, itself with its internal heat source must indeed reach a state (temperature) where it is radiating, to its own environment that same amount of power that it is receiving from the internal power source, independent of the temperature of that environment? When there is “no” opposing radiance to the generation of flux, the sphere will be at the lowest, not the highest, temperature needed to emit such flux.

“#2 — This is where we have difficulty (and so 3 & 4 are moot at present). Let me offer a quote that represents my understanding.”

The most general statement of the radiative heat flow equation is

” Q’ = A*(F1 – F2) where F1 and F2 are a local flux from the two different sources.”

Ahh, here is the problem the “assumption” of opposing flux! This is what is currently taught is Physics, but never in EE as such contradicts Maxwell’s equations of radiative field and radiative flux. There is no such thing as radiative heat. There is only EMR , with a wide band field strength
accumulating at each frequency and proportional to T^4 of such surface, for a black Lambertian surface. That field strength expressed as W/m^2 is not a flux but a field strength vector. It has both magnitude and direction, and must be treated algebraically as a vector, especially within parenthesis.

The correct S-B equation for the “maximum” thermal EM flux “from” any black flat surface(a) into a enclosing hemisphere(b) and is: P/Aa = σ(Ta^4 – Tb^4) where:

P/Aa is the power per unit area (flux) “from” black flat surface(a) a vector (from).
Ta Is the absolute temperature of surface(a) A field potential vector from(a) in the direction of (b).
Tb Is the absolute temperature of hemisphere(b) A field potential vector from(b) in the direction of (a).
σ is Stefan’s constant, a scalar representing the conversion of the vector sum within the parenthesis. The two T^4 are not the actual EM opposing field strengths but represent such all scaled, including the PI steradians effective radiative solid angle of any flat surface.
The S-B equation is a vector equation that only indicates the “maximum” possible aggregate thermal EMR vector flux, always in the direction toward the lower temperature surface. at each frequency and in each direction. The use of such an equation to actually determine physical EM flux. is always a mathematical error as this physical, as this physical has no black Lambertian surfaces.

“That is, the fluxes correspond to a specific single location, not their own locations independently. By convention, the location which the fluxes correspond to are that at the location represented by F2. That way, if Q’ is positive, then it denotes positive heat flow into the location specified by F2, and so that location would rise in temperature. If Q’ is negative, it denotes heat flow out of the location F2, and so if F2 is not a powered source of heat, then it will fall in temperature. However, whether or not the heat flow out of F2 towards F1 can actually cause an increase in temperature at location F1, requires its own analysis with all of the terms switched to the new location F1. … ”

Thank you Mark, for your interpretation from thermodynamics, of why the two stream approximation is always nonsense, and why your “F2” does not increase the temperature of A(1). there is no “F2”. Because of the rethermalization of EMR by A(2), the effective emissivity of A(1) is reduced, and A(1) “must increase the temperature of itself” to maintain the same flux P/A(1).

21. markstoval says:

@ okulaer

I don’t get this absolute fear of acknowledging that the sphere will need to get warmer at steady state with the shell up around it than with only the vacuum of deep space surrounding it. There is nothing mysterious or abhorrent about this proposition at all. In fact, it follows from basic thermodynamic principles. Insulation. Impede the heat loss (ANY heat loss) from a constantly heated object and it will necessarily become warmer. The point is: IT WILL NOT BECOME WARMER FROM THE (COOLER) INSULATING LAYER TRANSFERRING EXTRA ENERGY TO IT AS IF IT WERE HEAT. IT WILL BECOME WARMER FROM THE ENERGY CONSTANTLY COMING IN FROM ITS (WARMER) HEAT SOURCE AS HEAT NOT BEING ABLE TO ESCAPE AS FAST AS BEFORE WITH THE INSULATING LAYER FACING IT.

Well, there is no “fear” as I don’t have any reputation on the line in all this; only an interest in getting it right. I blog about radical libertarian issues, not climate issues. (when I get the time that is) I have an experiment in mind that I’ll describe a few days from now. I would like your take on it.

But on to the above quote from you. I understand that you think that the sphere will warm because the shell is stopping heat from making it out to the cold of deep space. I understand that you are saying that the system will have more heat since the heat can’t get out as easily as before. (I hope I have stated your position fairly)

I don’t see the shell impeding heat from leaving the system. I see heat being transferred to the shell every bit as fast as the sphere wants to shed it. I see the shell receiving heat on one side and releasing it to deep space on the other side without any problem. You see, if it did make heat “pile up” in the sphere then the heat would just keep building without bound. I guess the sphere would become infinitely hot if less heat leaves the system as enters it.

The way I see it, if the shell touched the sphere then it would be part of the sphere radiating into deep space all the energy as before. Does having a vacuum between sphere and shell really cause the sphere to heat up? How much? Until it melts?

I don’t mean to irritate you. I can see by the all caps that we are close to being done. I will leave a note about an experiment that I think could be done that would show who is right here. (I don’t think you have science without experiments)

I am looking forward to you next post even if I never see this one aspect your way. After all, I can admit I might be wrong even as I am not convinced I am.

• Mindert Eiting says:

What a horrible model the shell game is. Perhaps a simpler one may suggest some conceptual errors, Mark. Without any knowledge of thermodynamics, I would explain the issue to primary school children as follows.

Water is flowing from a mountain into the sea, as a river from high to low. You may build a dam in that river. Before the dam a reservoir will be formed with a certain width and height. The width of the lake should be treated as fixed as well as the height of the dam.

Why the reservoir? While making the dam, you prevent mountain water from streaming downhill, The river below will dry up for a short time and you deprive the sea from its mountain water. However, up-hill the mountain water keeps coming as before. If the content of the lake would not increase, a conservation law would be violated.

After finishing the dam, the height of the water level of the lake must increase but this cannot go on for ever. After some time the water level of the lake will equal the height of the dam and water will flow over. From now on the amount of water from the mountain equals the amount arriving at the sea.

This must be the case. We gave the dam a fixed height as promised and therefore the reservoir a fixed content. If input and output would not equal each other, we would violate the conservation law again. You may finish the lesson by destroying the dam and all piled-up water will flow in an instant into the sea.

Making the dam and reservoir is the same as insulation. The higher the dam the more insulation. In the next lesson you may make the existing dam a bit higher and demonstrate that more water piles up before the dam. At the new ‘dynamic equilibrium’ the lake contains more water than before whereas the sea receives again as much water as is coming from the mountain.

Do you want an infinite amount of water in the reservoir? Make the dam as high as the top of the mountain and see what happens. Whether the shell is made of iron or rubber, it must have a heat capacity above zero, like the water capacity of the reservoir.

• okulaer says:

“I understand that you think that the sphere will warm because the shell is stopping heat from making it out to the cold of deep space.”

No. The shell is making less energy escape the sphere as heat to its surroundings (which used to be space, but which is now the shell). The sphere no longer ‘feels’ deep space. Its thermal connection with it is broken. The shell is now its cold reservoir (heat sink), not space. The shell is able to warm. Space wasn’t (and isn’t). Therefore, with the shell facing the sphere, there’s a smaller temperature difference between ‘heat source‘ (the sphere) and its ‘heat sink‘ (used to be space, is now rather the shell). This produces a smaller ‘heat flux’ from source to sink than before. Just like in conductive and convective heat transfer: steeper temp slope leads to faster transport of energy as heat down the slope. This is exactly what the radiative heat transfer equation is saying also: q = σ[Th^4 – Tc^4].

“I understand that you are saying that the system will have more heat since the heat can’t get out as easily as before.”

Yes, the sphere+shell will necessarily accumulate more energy (‘internal energy’ [U], not ‘heat’ [Q]; ‘heat’ is the energy transported between systems at different temperatures, not the energy held inside them) than the sphere alone did.

“I don’t see the shell impeding heat from leaving the system. I see heat being transferred to the shell every bit as fast as the sphere wants to shed it.”

But only in your last comment you admitted that as the shell warms, then Q will grow smaller, maybe even reach close to zero. And you even had no problem with this. Now all of a sudden you’re saying you can’t see this happening!?

I don’t think you understand how this situation works at all, Mark. You just ‘feel’ that this and that should be happening …

“I see the shell receiving heat on one side and releasing it to deep space on the other side without any problem.”

Precisely! But only at the new dynamic equilibrium. When the sphere’s absolute temperature has become ~1.19 times higher than it initially was: 255K >> 303K.

At this point, the shell receives 240 W of radiant heat on its inner surface (it did throughout its period of warming) and releases exactly the same – 240 W of radiant heat – to deep space from its outer surface – balance.

It never did up until this point, though. It started at 0 W (assuming its initial temp was that of space). And ended at 240 W. By warming from 0 to 255K.

When the shell finally matches it heat input (from the sphere) with its heat output (to deep space), it cannot warm any further. We have reached dynamic equilibrium: Q_in = Q_out. Its temperature (and its heat output) is in a steady state.

So, the shell was never in radiative balance until its temperature finally got high enough to enable it to emit as much to space as it got in from the sphere (240 W). Ideally, disregarding differences in radii, this temp would be 255K, the same as the sphere’s temp before the shell was put in place and started warming; in reality, of course, it would be ever so slightly lower* (larger surface + the inverse square law).

*If you compare the radius of the Earth (6371 km) and the radius of a shell surrounding Earth, separated by perhaps 5 kilometres of air (or vacuum in our case) (6376 km), this difference would be highly negligible. The surface area of the shell (facing space) would be 0.157% larger than the surface area of the Earth.

The point to this is, the heat in to the shell from the sphere is 240 W the entire time. It was 240 W before the shell was put around it. And it was 240 W when the shell reached its steady state temperature and balanced its heat input with its heat output. It was always 240 W.

Why? Not because the shell needed it to be, but because the sphere needed it to be.

Why? Because the heat input to the shell equals the heat output from the sphere (the q on the lefthand side of the radiative heat transfer equation above). And the sphere’s Q_out needs to balance … its Q_in. Its Q_in coming from the sphere’s own heat source, its mysterious internal power source. The Q_in is constant. It is always equal to 240 W. So this is the power the sphere needs to put out. And this is finally what the shell needs to put out as well. Upon new equilibration.

So we’re back to the radiative heat transfer equation and its very simple concept of temperature slope between hot and cold reservoir: The steeper the slope, the larger/faster the heat transfer. And the other way around.

So with the shell warming beyond space, it’s easing the temperature slope back toward the sphere, reducing the rate of heat transfer away from the sphere in the process. And this becomes a problem for the sphere: Q_in > Q_out. Q_in accumulates inside the sphere. Increasing its U. Raising its T. The slope steepens again. Before the shell warms a little bit more. And the whole process repeats itself. Tiny little steps. We however never get to see these separate steps. It’s all one smooth operation. The sphere output seemingly just stays the same throughout. 240 W. But its temperature goes up. Gradually.

“You see, if it did make heat “pile up” in the sphere then the heat would just keep building without bound. I guess the sphere would become infinitely hot if less heat leaves the system as enters it.”

You apparently haven’t paid much attention, Mark.

Why should this be an infinite process!? It simply moves to a new dynamic equilibrium. When the sphere reaches 303K and the shell reaches 255K, the warming stops. Balance. Steady state.

I spent the main part of the top post explaining this, Mark. You know, with all the diagrams and figures. I must say I’m a bit disappointed that you – at such a late stage – would simply bring in this regurgitated drivel from Postma.

“I don’t mean to irritate you.”

People simply disagreeing with me doesn’t irritate me. Disagreement is what makes the world go round.

What does annoy me a bit is when these people simply appear not to read what I write. When it seems it doesn’t matter what I say, they’ve just made up their mind and keep repeating their arguments as if I had never even made an attempt to counter them.

Forcing me to repeat myself endlessly … Which is what I feel I’m doing here now.

“I will leave a note about an experiment that I think could be done that would show who is right here. (I don’t think you have science without experiments)”

Thanks.

And please don’t believe that I think Eschenbach’s ‘Steel Greenhouse’ is some kind of model of a real ‘atmospheric radiative greenhouse effect’. Such an effect does not exist. It’s a chimaera. What I was trying to do is actually show that the whole “back radiation” idea violates the 2nd Law of Thermodynamics. Using Eschenbach’s ‘Steel Greenhouse’ was simply a way to conceptualise it … I’m not trying to ‘do science’ through this exercise. You’re absolutely right: Science in the end is about observation and experiment, not about speculating on a blackboard.

“I am looking forward to you next post even if I never see this one aspect your way. After all, I can admit I might be wrong even as I am not convinced I am.”

Glad you’re interested. I’m afraid not too many people are …

Stay sceptical. But be rational about it. Nullius in verba!

22. Mindert,
One dam is to simple for the shell game.
Build Two dams as high as needed one right after the other near bottom of the first dam put in an orfice of a fixed diameter, than a larger pipe “level and througn” the wall of the second dam. Perforate the pipe so water can leak into the second reservoir.
In order for all the water to exit through the orfice the first reservoir. must fill to a certain level above the orfice to create a pressure head. The second reservoir will fill to the level of the ofice and perforated pipe The water from there drains through the large pipe to the sea.
Now put in second orfice at the wall of the second dam.
In order to create the same “head” for orfice #2, the water in the second reservoir must rise to the level that was in the first reservor Here comes the shell game. As the second reservoir fills the “head” for the first orfice decreases becaust of back pressure, not back flow. This in turn results the level of the first reservoir. to increase to maintain that “head” also. But the river, not any back flow did this Automagically!

Call the level in the first reservoir Tsphere^4.
Call the level in the second reservoir Tshell^4.
Shell game over! What caused what?
.

23. markstoval says:

on that steel greenhouse

Note to Okulaer — An Experiment of the metal “greenhouse” that can be done

As a recap, let me tell why I disagree with Okulaer in his analysis of the steel “greenhouse” warming up due to a shell. I posted here that I had an experiment in mind and would post it when time allowed me to. Today is that day. So, here goes.

Part One:

1) Please imagine a very large sphere with a constant heat source at its center which is located in deep space with no other objects near it. Sit and wait until the system stabilizes and measure the surface temperature. Let us call that temperature X. (not very creative)

2) Now imagine that there are some pockets or “bubbles” in the sphere that are vacuums.

In this case I can see no reason to suspect that the vacuum “bubbles” will effect the equilibrium surface temperature at all. As radiation will travel at the speed of light, I suspect that the surface temperature remains X in this case.

Part Two:

1) Now imagine the same large sphere with a constant heat source at its center which is located in deep space with no other objects near it.

2) Now imagine that the pockets or “bubbles” in the sphere that are vacuums are morphed to separate the mass of the sphere into 100 shells.

I see no reason why the system would not have exactly temperature X on the surface just as in part number one. The reason I think this is that the transfer of heat in the system is not impeded at all by vacuum. One hundred shells or just one sphere seems to me to be the same.

Part Three:

1) If one hundred shells are the same as one sphere, then one sphere and one shell are also of no real difference and the temperature will remain the same.

Now there has been a lot of argument, emotion, ad homs, anger and so on over this issue for some time. I think it time to do an experiment to see what reality tells us about the situation.

We can not use the same temperature as deep space (unless we imagine one hell of a large budget experiment) so let us start with a large commercial freezer. If one of those could get down to 0 F degrees then that would be enough for this demonstration I think.

We do not have an anti-gravity device to suspend the steel ball in the freezer, but we could use an aluminum ball and set it on top of a column of Styrofoam like they use to package up computers or other devices when they are shipped by freight. Sure, the sphere will now be touching some mass and there will be a tiny bit of conduction, but it would be so small that we can ignore that.

Next, drill a hole in the sphere and insert a heat source. Then attach a very sensitive thermometer to the surface of the sphere — perhaps inside a drilled hole for better measurement.

Let the system come to a steady state and measure the temperature. Wait some time and measure is over and over to make darn sure it is in a steady state.

Now take a metal shell that is open and flat on its lower side — I am thinking that the part cut away from the shell needs only be enough so you can set it over the sphere. You set it down on a large ring of Styrofoam.

Now measure the temperature and look for a rise in temp. I say it will not be there.

Now redo the experiment with the freezer and a vacuum pump to get as much air out of the freezer as your budget will allow. I think you will see no temperature rise on the surface of the sphere.

===========

Okulaer, what do you think of this amateurish little experiment to see if you are right on the sphere heating up?

~ Mark

• Mark, here are some real numbers for your consideration: The two power columns are a 5.67 ohm resistor with 1 amp and 2 amps of current. the first set is temperatures and flux for a single shell of different surface area radiating to space. All others are nested shells and how the shells affect the equilibrium temperature of the inner shells, with a constant power source.

``` Each shell to space P(W) is constant power outward. None is inward N Dia(cm) A(m^2)_ P(W)_ T(k) __ W/ ^2 __ P(W)_ T(K)___ W/m^2__ S csa(m^2) 1 05.66____0.01__ 5.67_ 316.23_ 567.0__ 22.68_ 447.21_ 2268___ 1 0.0025 => 1K 2 07.88____0.02__ 5.67_ 265.90_ 283.5__ 22.68_ 376.06_ 1134___ 2 0.0050 => 1K 3 11.28____0.04__ 5.67_ 223.60_ 141.75_ 22.68_ 316.23_ 0567___ 3 0.0100 => 1K 4 15.76____0.08__ 5.67_ 188.03_ 070.865 22.68_ 265.91_ 0283.5_ 4 0.0200 => 1K 5 22.56____0.16__ 5.67_ 158.11_ 035.43_ 22.68_ 223.61_ 0141.75 5 0.0400 => 1K```

``` 5 shell to space N Dia(cm) A(m^2)_ P(W)_ T(k) __ W/ ^2 __ P(W)_ T(K)___ W/m^2__ S csa(m^2) 1 05.66____0.01__ 5.67_ 373.09_ 567.0__ 22.68_ 527.63_ 2268___ 1 0.0025 2 07.88____0.02__ 5.67_ 311.67_ 283.5__ 22.68_ 440.06_ 1134___ 2 0.0050 3 11.28____0.04__ 5.67_ 257.18_ 141.75_ 22.68_ 363.71_ 0567___ 3 0.0100 4 15.76____0.08__ 5.67_ 208.09_ 070.865 22.68_ 294.28_ 0283.5_ 4 0.0200 5 22.56____0.16__ 5.67_ 158.11_ 035.43_ 22.68_ 223.61_ 0141.75 5 0.0400 => 1K 4 shell to space N Dia(cm) A(m^2)_ P(W)_ T(k) __ W/ ^2 __ P(W)_ T(K)___ W/m^2__ S csa(m^2) 1 05.66____0.01__ 5.67_ 370.04_ 567.0__ 22.68_ 523.35_ 2268___ 1 0.0025 2 07.88____0.02__ 5.67_ 305.85_ 283.5__ 22.68_ 433.53_ 1134___ 2 0.0050 3 11.28____0.04__ 5.67_ 247.46_ 141.75_ 22.68_ 349.96_ 0567___ 3 0.0100 4 15.76____0.08__ 5.67_ 188.03_ 070.865 22.68_ 265.91_ 0283.5_ 4 0.0200 => 1K 3 shell to space N Dia(cm) A(m^2)_ P(W)_ T(k) __ W/ ^2 __ P(W)_ T(K)___ W/m^2__ S csa(m^2) 1 05.66____0.01__ 5.67_ 363.71_ 567.0__ 22.68_ 514.37_ 2268___ 1 0.0025 2 07.88____0.02__ 5.67_ 294.28_ 283.5__ 22.68_ 416.18_ 1134___ 2 0.0050 3 11.28____0.04__ 5.67_ 233.61_ 141.75_ 22.68_ 316.23_ 0567___ 3 0.0100 => 1K 2 shell to space N Dia(cm) A(m^2)_ P(W)_ T(k) __ W/ ^2 __ P(W)_ T(K)___ W/m^2__ S csa(m^2) 1 05.66____0.01__ 5.67_ 349.96_ 567.0__ 22.68_ 494.92_ 2268___ 1 0.0025 2 07.88____0.02__ 5.67_ 265.91_ 283.5__ 22.68_ 376.06_ 1134___ 2 0.0050 => 1K ```

```#5 shell to space no others N Dia(cm) A(m^2)_ P(W)_ T(k) __ W/ ^2 __ P(W)_ T(K)___ W/m^2__ S csa(m^2) 1 05.66____0.01__ 5.67_ 321.06_ 567.0__ 22.68_ 454.04_ 2268___ 1 0.0025 5 22.56____0.16__ 5.67_ 158.11_ 035.43_ 22.68_ 223.61_ 0141.75 5 0.0400 => 1K ```

Please tell us why the numbers turn out that way with emissivity = 100%
Please redo the above with all shell emissivities 90%. Why are they soo different?

24. ``` Each shell to space P(W) is constant power outward. None is inward N Dia(cm) A(m^2)_ P(W)_ T(k) __ W/m^2 __ P(W)_ T(K)___ W/m^2__ S csa(m^2) 1 05.66____0.01__ 5.67_ 316.23_ 567.0__ 22.68_ 447.21_ 2268___ 1 0.0025 => 1K 2 07.88____0.02__ 5.67_ 265.90_ 283.5__ 22.68_ 376.06_ 1134___ 2 0.0050 => 1K 3 11.28____0.04__ 5.67_ 223.60_ 141.75_ 22.68_ 316.23_ 0567___ 3 0.0100 => 1K 4 15.76____0.08__ 5.67_ 188.03_ 070.865 22.68_ 265.91_ 0283.5_ 4 0.0200 => 1K 5 22.56____0.16__ 5.67_ 158.11_ 035.43_ 22.68_ 223.61_ 0141.75 5 0.0400 => 1K```

``` 5 shell to space N Dia(cm) A(m^2)_ P(W)_ T(k) __ W/m^2 __ P(W)_ T(K)___ W/m^2__ S csa(m^2) 1 05.66____0.01__ 5.67_ 373.09_ 567.0__ 22.68_ 527.63_ 2268___ 1 0.0025 2 07.88____0.02__ 5.67_ 311.67_ 283.5__ 22.68_ 440.06_ 1134___ 2 0.0050 3 11.28____0.04__ 5.67_ 257.18_ 141.75_ 22.68_ 363.71_ 0567___ 3 0.0100 4 15.76____0.08__ 5.67_ 208.09_ 070.865 22.68_ 294.28_ 0283.5_ 4 0.0200 5 22.56____0.16__ 5.67_ 158.11_ 035.43_ 22.68_ 223.61_ 0141.75 5 0.0400 => 1K 4 shell to space N Dia(cm) A(m^2)_ P(W)_ T(k) __ W/m^2 __ P(W)_ T(K)___ W/m^2__ S csa(m^2) 1 05.66____0.01__ 5.67_ 370.04_ 567.0__ 22.68_ 523.35_ 2268___ 1 0.0025 2 07.88____0.02__ 5.67_ 305.85_ 283.5__ 22.68_ 433.53_ 1134___ 2 0.0050 3 11.28____0.04__ 5.67_ 247.46_ 141.75_ 22.68_ 349.96_ 0567___ 3 0.0100 4 15.76____0.08__ 5.67_ 188.03_ 070.865 22.68_ 265.91_ 0283.5_ 4 0.0200 => 1K 3 shell to space N Dia(cm) A(m^2)_ P(W)_ T(k) __ W/m^2 __ P(W)_ T(K)___ W/m^2__ S csa(m^2) 1 05.66____0.01__ 5.67_ 363.71_ 567.0__ 22.68_ 514.37_ 2268___ 1 0.0025 2 07.88____0.02__ 5.67_ 294.28_ 283.5__ 22.68_ 416.18_ 1134___ 2 0.0050 3 11.28____0.04__ 5.67_ 233.61_ 141.75_ 22.68_ 316.23_ 0567___ 3 0.0100 => 1K 2 shell to space N Dia(cm) A(m^2)_ P(W)_ T(k) __ W/m^2 __ P(W)_ T(K)___ W/m^2__ S csa(m^2) 1 05.66____0.01__ 5.67_ 349.96_ 567.0__ 22.68_ 494.92_ 2268___ 1 0.0025 2 07.88____0.02__ 5.67_ 265.91_ 283.5__ 22.68_ 376.06_ 11m4___ 2 0.0050 => 1K ```

```5 shell to space no others N Dia(cm) A(m^2)_ P(W)_ T(k) __ W/m^2 __ P(W)_ T(K)___ W/m^2__ S csa(m^2) 1 05.66____0.01__ 5.67_ 321.06_ 567.0__ 22.68_ 454.04_ 2268___ 1 0.0025 5 22.56____0.16__ 5.67_ 158.11_ 035.43_ 22.68_ 223.61_ 0141.75 5 0.0400 => 1K ```

25. Mindert Eiting says:

As I wrote, the unidirectional and bidirectional approach may have different results but all approaches in which the temperature develops proportional to a temperature difference, amount to a power function with one parameter for heat capacity. If you allow me, I will demonstrate that for one model, showing that you are right, Kristian.

The most simple case is a closed system without input. Here we may apply Newton’s law for the temperature of a body in a certain surrounding. The Wikipedia version gives the surrounding temperature but we can always transform the temperature scale, in Fahrenheit fashion, such that it is zero. With this scale, the law reads

T(t) = T(0) exp -kt, [1]

with T(t) temperature of the body at time t, T(0) body’s initial temperature, and k>0 a constant for heat capacity and time unit. Note that we could violate here the Second Law by taking a negative capacity.

The bidirectional game can be translated into a money game between me and my landlord. At any time t, I pay my landlord an amount and he pays me back a proportion, 0<= p <= 1, of that amount. I forward him the returned amount, and he repeats his behaviour, etc.

Let T(0) be the initial amount I owe him and let one loop between us define one unit of time, then

T(t) = T(0) p^t. [2]

If the unidirectional and bidirectional approach give the same result, we should equate [1] and [2], which gives

k = – LN(p). [3]

It is only a matter of symbols. The pay-back proportion nicely expresses heat capacity, as it should do in a temperature game. We may try several other bidirectional games but the essence is the proportionality feature.

With regard to this, it is only a matter of elegance to avoid bidirectional games as in Newton's case. All talk about back radiation amounts to heat capacity. Insulation increases heat capacity. Quite peculiar that the specific heat capacity of CO2 is less than that of normal air. More CO2 in the atmosphere should de-insulate the earth? See also Claes Johnson's most recent post.

26. Mindert,

Consider a thermal resistance, with no thermal mass between a constant power source and any isotherm. That power source will have a constant flux, producing an upward temperature gradient from the isotherm and the power source. No energy, or power accumulation, is involved only a constant power flux. Add a similar thermal resistance in series. The flux, and the gradients will remain as constants, but the temperature differential (potential difference) will double because of the doubling of thermal resistance, with no back anything.
It is the introduction of the radiantly opaque shell, that doubles the thermal radiative resistance (impedance). This requires the doubling of radiative potential difference, that is the “difference” in two absolute temperatures each raised to the fourth power.
Electromagnetic radiative potential, or pressure, is a “force” in four-space, not in linear-space, hence the raising to the fourth power. This is similar to gravitational “pressure”, which is also a “force” in four-space. Neither fits well with linear thermodynamics.
The best quantum folk, and the best stringy folk, can mathematically run rings around any engineer. However they lack the skill required to nail that stuff together so you can actually poke at it, to see what ugly crawls out.

• Mindert Eiting says:

Thanks, Will, but may I proceed as a poor soul, with only Newton at my side? Bidirectional people cannot deny the proportionality p in the escapement of heat. Therefore, they have to use a return-rate q. We still have one confounded parameter pq in Newton’s equation (with k = -LN(pq)). Only because of this confounding both approaches give the same result. Kristian can forget the citation of Niels Bohr. In an experiment in which the return-rate could be manipulated, the approaches should give different results.

I would do the experiment on the dark side of the moon and manipulate the return-rate of IR with a mirror at a geostationary spot above the surface. The terrain getting back radiation should cool at a slower rate than a neighbouring area. The temperature difference between the areas can be used for powering a heat engine.

With the sun below the moon horizon, we can also use the mirror for shedding sunlight on an area. Again the heat engine will run. You may explain why the engine only runs on hot radiation (2nd Law).

I was raised in an experimental tradition and noticed that the shell game is not an experiment. It is a disgusting construction and compares with a homoeopathic demonstration with manipulated water. If you give that to a group of ill people, the fact that some of them become less ill after some time, does not prove anything because ill people usually become more healthy if they do not die. So you have to attack the confounding of effects by giving normal water to a control group.

27. D Appell says:

“Cold cannot heat hot.”

1) All objects radiate electromagnetic radiation, per the Stefan-Boltzmann equation
2) This radiation carries energy (whether you think of it as a wave or a photon)
3) When that radiation strikes another objects and is absorbed, the object gains that energy (and hence, temperature per the Stefan-Boltzmann equation).

The hot object is thus warmer than it would be in the absence of the cold object, because its net loss of radiative energy is lower.

• Bryan says:

D Appell says

“The hot object is thus warmer than it would be in the absence of the cold object, because its net loss of radiative energy is lower.”

This is incorrect

Since you have chosen the bidirectional model lets see if your statement makes any sense.
The way Physics would deal with your conjecture is to have a hotter and colder object interact radiatively but be isolated from the rest of the universe.

This can be done by considering two metal blocks A and B sit separated inside a vacuum filled adiabatic enclosure.

Adiabatic enclosure consists of a perfect reflector face surrounded by a perfect insulator.
For the purposes of a demonstration a polystyrene box with the inner face lined with polished aluminium foil would work.

Initially both at the same temperature. The zeroth law of thermodynamics applies.
Both emit and absorb equal amounts of radiation.

Neither one is said to heat the other.

One block (A) has a power supply which is now switched on causing the temperature of the block to rise.
This in turn means that it will emit more radiation.

A will now heat B causing its temperature to rise.
B will in turn emit more radiation but this back radiation is caused by A.

Now comes the clincher

If B were not there at all, the temperature A would be even higher.

So B cannot be said in any meaning of the word as a cause of heating A.

This is the approach that you will find in all physics textbooks dealing with heat transfer.
Only in climate ‘science’ do they distort language to reach your conclusion.

28. D Appell says: February 11, 2015 at 1:05

It was only a matter of time until the self appointed Dr. David Appell reared ugly!

(“Cold cannot heat hot.”)

“1) All objects radiate electromagnetic radiation, per the Stefan-Boltzmann equation”

As per the Stefan-Boltzmann equation, all thermal electromagnetic flux is strictly limited by any opposing thermal “radiance”. This is the difference enclosed by the very necessary mathematical parentheses, of the demonstrated S-B equation.

“2) This radiation carries energy (whether you think of it as a wave or a photon)”

The allowed thermal electromagnetic flux transports only “action”, (energy within a time interval). .

“3) When that radiation strikes another objects and is absorbed, the object gains that energy (and hence, temperature per the Stefan-Boltzmann equation).”

Thermal electromagnetic flux, transporting “action”, that encounters “mass” may have that action differentiated into absorption, transmission, or reflection. Mathematically the proportions are
reflectivity, transmissivity, absorptivity always summing to unity. Please note that for any mass at any frequency, area, and direction, your emissivity value is precisely that value of the fractional absorptivity. The “action” absorbed by “mass” may be converted to any other action not involving sensible heat or temperature.
Many forms of conversion of Solar transmitted “action” have nothing to do with temperature.
Other forms may be electrical, charging of battery, electrical conversion to linear or angular momentum. On this Earth most forms of conversion are chemical. Reduction of oxidic process.
to useful hydrogen, and carbon, along with the automagical construction of trees. This is obvious to squirrels. Why never to incompetent Climate Clowns, Physicists, Meteorologists, Politicians, or “The Media”.

“The hot object is thus warmer than it would be in the absence of the cold object, because its net loss of radiative energy is lower.”

So what? Can your non described term “warmer” ever have understanding for those digging in the artifacts of stupid earthlings?

29. Mindert Eiting says:

1) OK. 2) OK. 3) My car drives at 100 mph and yours at 10 mph. After a collision my car continues at less than 100 mph.

• Indeed with linear thinking. Your car with 100 kg x 100 mph has 10000 of something called linear momentum. In a collision with a 2000 kg cement mixer at 10 mph that has twice that the same but opposing linear momentum,. After collision all momentum “must” be conserved! What is left of you and your car is travailing backward, and the cement mixer is slowed some. The whole mess is proceeding at less than 5 mph in some direction. What is your point?

• Mindert Eiting says:

When I hit him from behind, his speed increases and mine decreases, whatever our masses are. Let’s use tennis balls in order to stay on track, Will. He says in point 3 that my ‘fast’ tennis ball will gain the energy from the ‘slow’ tennis ball, and must proceed faster than before. Is kinetic energy not a relative concept? I could set in this example the speed of the ‘slow’ tennis ball arbitrarily at zero, and so its kinetic energy. What absolute amounts of energy go in and out the SB equation, so that only brackets can avoid the absurd?

30. Mindert Eiting says: February 13, 2015 at 12:40 pm

“What absolute amounts of energy go in and out the SB equation, so that only brackets can avoid the absurd?”

The S-B equation has nothing to do with energy! The result Is a power per unit area, an EMR flux to or from a unit area and its hemispherical environment. The terms inside the parenthesis, (Kristian uses brackets) are the temperature part of opposing “radiances” If the environment has higher temperature than the area That equation is still correct but the flux includes a minus sign indicating that the calculated flux is to the given area rather than from that area.
One way of considering the physical effect is to consider “radiance” as a pressure (force per unit area) The gas mass flow through a limiting orifice is proportional to the difference in pressure on either side of that orifice The flow of gas do not consist of two opposing gas mass flows from the two absolute pressures to a pressure of zero Pa. resulting in some “net” flow. The mass flow between two pressures like that of thermal EMR flux are always self cancelling with no potential difference quite independent of the absolute of the ;pressure.
See the chart Will Janoschka says: February 8, 2015 at 1:42 am
Each temperature number is carefully calculated stepwise using only the whole S-BB eq1uation from the coldest 1 Kelvin (space) to the highest (that containing the fixed power source. Note there is no mass or specific heat, as none is required for this equation only the equilibrium temperatures and flux is calculated. Likewise no time or time integral of power is ever needed

• Mindert Eiting says:

Thanks, Will. I try to understand where all this misery comes from. In my last comment I took resort to an example of tennis or billiard balls. The less energetic object can be represented by the edge of a pool table. After shooting a ball against the edge, it will (with a change of direction), continue its course at less speed. Part of its kinetic energy is transferred to the edge. When we make a film of this affair and spool it backwards, we may see the ball at low speed first, touching the edge, and continuing its course at a higher speed. This is absurd as the ball gained energy from the edge.

I have the strong lay-man intuition that by spooling the film of radiative thermodynamics backwards, we should see similar appell-events. We may shoot a hot ball in the direction of a zero-Kelvin edge. The ball continues slightly cooler, leaving behind an 1K edge. On the backwards film we see a hot ball approaching an 1K edge, and next continuing its route as a still hotter ball leaving behind a zero-Kelvin edge. This seems to me a clear appell-event.

31. Mindert Eiting says: February 14, 2015 at 1:35 pm

In regard to your car and my cement mixer, please remember that for an elastic collision, both energy and momentum is supposed to be conserved. The two momenti are vectors but the energies are not, direction maters. I will try to get back after pondering, with much beer!

32. Mindert Eiting says:

Will, I have already some beer but I do not want to frustrate your efforts. I my world cool beer may enter a hot body and make it hotter still.

Let’s distinguish back radiation (which is just radiation) and its presumed effect. Back radiation should slow down a cooling rate. Opponents like Kristian say that the mechanism behind the phenomenon violates the Second Law. I will show that the mechanism does not matter. The theory fails on the observational level. Therefore, we need the best representation of the theory, actually better than found in many texts by its proponents.

There is nothing in the claim about an open or closed thermodynamic system. Therefore, we make take a closed system and apply Newton’s law for the temperature of a body in a certain surrounding. The usual version gives the surrounding temperature but we may transform linearly the temperature scale, such that the surrounding temperature is zero by definition (in order to avoid difficulties with extreme temperature differences, we may begin with Kelvin to the fourth power). With this scale the law reads

T(t) = T(0) exp -kt, [1]

with T(t) temperature of the body at time t, T(0) body’s initial temperature, and k a parameter for heat capacity and time unit. The greater k, the faster the body cools or warms. For t sufficiently large, T(t) becomes zero, i.e. the surrounding temperature.

The name ´cooling law´ is misleading. For T(0) >0 Equation [1] gives the temperature of a cooling body, for T(0) 0. [2]

The parameter may represent heat capacity or anything else influencing the rate in [1]. For a cooling body, anything making it cooling at a slower rate, amounts to making k smaller. So we should treat the parameter as a product of two parameters,

k = rs, [3]

in which r is a back radiation effect, and s a heat capacity effect. Then the bidirectional claim amounts to

0< r 0 the body cools at a slower rate than with r=1, and for T(0)<0 it warms at a slower rate than with r=1. Obviously, [4] does not violate [2] and therefore does the bidirectional theory all the justice it deserves. The back radiation effect does not violate the Second Law in this expression.

Equation [3] is a triviality. We may write k as a product of all parameters we want. We need on the observational level an experimental setting in which we may vary r independently of s. It may be noted that [4] only sets a limit, which may be the cause of arbitrary amounts of energy exchange in so-called energy budgets. The closer r to zero, the stronger the impeding effect of back radiation. There must be a formal relationship between r and the proportion (p) of radiation coming back.

Proportionality, as applied by Newton him self, amounts to a power function. Take as example a money game between you and your landlord. You owe him an amount of T(0). At any time you pay him an amount and he pays you back a proportion p. You forward him that amount and he repeats his behaviour. So your account shows T(0), T(0)*p, T(0)*p*p, etc. Let one loop define the time unit, then

T(t) = T(0) p^t. [5]

Equate this with Equation [1], while substituting r for k, to obtain

r = -LN(p). [6]

The proportionality principle of [6] is the best we have to offer to the bidirectional theory. From [1], [3], and [6] we get

T(t) = T(0) p^(st), with 0<p0 the body cools and for T(0)<0 it warms till zero, which is the surrounding temperature. The higher p, the slower the cooling or warming, which is precisely what we want. Time and again the proponents try to explain this and their point is as valid as can be, except for one aspect.

With the function of [7] we may perform a thought experiment in a closed thermodynamic system. On the dark side of the moon we may do that experiment with satellite screens on lunar-stationary spots. We may put them in a vertical position or flap them into a horizontal position. In the latter case the screen reflects a proportion, p, of the moon-photons back to some surface area and lets (1-p) through into space. The value of p is a variable property of the screen. Herewith we can create in the course of time, many surface areas with same T(0) and s, but different temperatures according to [7]. Finally, we may let the screens slowly revolve around their axes in full concert. This does not add energy to our closed system.

In this way we may force the spatial temperature variance of the surface. However, variance is a measure of entropy. If we have in a container hot fluid at one side and cold fluid at the other side, we have low entropy. Consequently, [7] permits violation of the Second Law, prescribing that in a closed system the entropy only increases.

For a short diagnosis we may notice that in [7] p can be taken as a variable over space and time. Take Equation [7] and add to time a spatial variable. Whatever the spatial variance of the T(0) may be, the spatial variance of the T(t) is strictly decreasing in time for p a constant (i.e. times p^2st). Let the initial variance be zero, i.e. a homogeneous temperature T(0) all over the surface, then by manipulating p, the variance of T(t) can be forced above zero. This falsifies the bidirectional theory. The theory must take p as a constant, but does not, as we are told that it may increase till alarming levels.

• Mindert Eiting says:

Sick and tired of HTML. Insert with the formula in words

… claim amounts to

r between 0 and 1 [4]

In language, the back radiation effect [4] makes that for T(0)>0 the body cools …

• Mindert Eiting says:

And the editor simply deleted part of [7] and some text after the unequal sign

T(t) = T(0) p^(st), with p between 0 and 1 [7]

Apparently, Equation 7 does not violate the Second Law. For T(0) >0 the body cools and …

• Mindert Eiting says:

Around Eq 2, the text is also mutilated because of an unequal sign. Read …
for T(0) less than 0, the temperature of a warming body (i.e. when we put an item from the fridge on the table).

The Second Law of Thermodynamics says here,

k greater than 0 [2]

33. Mindert Eiting says:

With excuses for some clumsy expressions because of HTML, here is the formal proof.

My text up to Equation 7 is a treatment of the back radiation effect. The equation gives Newton´s law with two parameters, s for heat capacity, and p the proportion of back radiation. In order to see that the equation is valid, substitute r = -LN(p), which returns Newton’s Law.

T(t) = T(0) p^(st), with p a constant between 0 and 1. [7]

For fixed t, p^(st) is fixed. Multiplying a variable with a constant makes that the variance is multiplied by the constant-squared. Without index for the space variable we get

VAR T(t) = (VAR T(0)) p^(2st). [8]

Because p is between 0 and 1, VAR T(t) becomes zero for t sufficiently large. This is a direct consequence of the Second Law, as expressed in [7]. Equation [8] also shows that VAR T(t) does not exceed VAR T(0). Variance is the complement of entropy.

It suffices to take VAR T(0) = 0. If this were not true for all places, take a subset of places with equal temperatures at the start. Here T(0) is constant over space and so

VAR T(t) = VAR (p^(st))* T(0)^2. [9]

By definition of variance

VAR (p^(st)) is positive, if p varies over space. [10]

Unless T(0) = 0, we have positive variance at time t, violating [8], saying that for the subset of places the variance at time t must be zero.

This completes the proof. The parameter p must be a constant for all places. We get Newton’s Law in return with one parameter for heat capacity. The back radiation effect does not exist as the proportion of radiation coming back, is not allowed to vary.

• That is a proof of what? You never define what “is” radiation! So a back anything is fantasy.
You seem to demonstrate that Newton’s Laws are self consistent over the domain of Newton’s Laws, DUH!!!
The S-B equation has nothing to do with heat, heat capacity or time. The S-B equation allows for the calculation of the “maximum” thermal electromagnetic flux between a flat surface and a surrounding hemisphere at any two temperatures, excluding zero Kelvin that cannot be raised to a power greater than one.
The S-B equation is considered correct for both both magnitude and direction of such maximum electromagnetic flux, including that with no temperature difference. This maximum has never been achieved, but the asymptotic behaviour with increasing emissivity, and temperature difference, is well documented. No bi-directional electromagnetic flux at any frequency has ever been demonstrated. The resulting increase in constant power source temperature with decreasing emissivity is well documented with no reference to any opposing electromagnetic flux. 🙂

• okulaer says:

Haven’t paid much attention to your discussion, been a bit busy lately. But there’s a new post out now …

Will, you say: “The S-B equation has nothing to do with heat (…)”

The S-B equation has everything to do with heat. Radiative heat. That’s what the Q is. The ‘net’ and unidirectional radiative energy transfer from the warmer object to the cooler object. This energy transfer is equal to the thermodynamical phenomenon called ‘heat’. You know, from the 1st Law: ΔU = Q – W. The change in a system’s ‘internal energy’ [U] is equal to the net transfer of energy as ‘heat’ [Q] to/from it, minus the net transfer of energy as ‘work’ [W] to/from it.

• Mindert Eiting says:

Will, the back radiation effect must have a relationship with magnitude, if the bidirectional theory is correct. It does not matter what radiation is, as we only consider a temperature effect. Take the most simple case of a closed system at which Newton’s law applies. We may use it to get heat capacity and the required proportion in the root. For large p the body slowly cools and for small p fast. Without energy input we may distribute back radiation over the surface in differing amounts. Consequently, we induce temperature differences, and so increasing variance. The Second Law says that the variance in a closed system always decreases, whereas the bidirectional theory allows you to increase it. Hence the Second Law is violated. You say back anything is fantasy, and I wanted to prove that. I think that the idea of transfer from cold to warm is not even needed. Entropy suffices.

34. okulaer says: February 19, 2015 at 1:11 pm

“Haven’t paid much attention to your discussion, been a bit busy lately. But there’s a new post out now …Will, you say: “The S-B equation has nothing to do with heat (…)”

“The S-B equation has everything to do with heat. Radiative heat.”

Kristian,
There is no such energy. All of relativistic EMR has no Energy Q, which implies some integral of power over time. Power may be accumulated, somehow, in mass or angular momentum of mass.
There is no such accumulation within the concept of electromagnetic power transfer between locations in 3-D space.

EMR requires no mass and no time. EMR has only power and reciprocal time, known as cycles per second, or radians per seconds. Each cycle has one Planck unit of “action”, equivalent to your integral of energy over time, or better for understanding, energy within a unit of time.

“That’s what the Q is. The ‘net’ and unidirectional radiative energy transfer from the warmer object to the cooler object. This energy transfer is equal to the thermodynamical phenomenon called ‘heat’. You know, from the 1st Law: ΔU = Q – W. The change in a system’s ‘internal energy’ [U] is equal to the net transfer of energy as ‘heat’ [Q] to/from it, minus the net transfer of energy as ‘work’ [W] to/from it.”

That is but your illusion of some accumulation of power in mass that you call energy. Specific heat times mass times accumulation of power, now what you call the temperature of that mass.
“That’s what the Q is.”. BS, no possible “Q” in then relativistic with no mass and no time. 🙂

35. David A says:

Okulaer, thank you for your excellent posts, in particular your posts on the satellite vs. surface data sets

I am your basic layman on science subjects like thermodynamics, with perhaps even less math then most, but perhaps greater philosophical or deductive reason applied then the mean. As such I dared develop a law of physics that may be useful to understanding some thermodynamics.

Two PHD’s, Dr. Brown and Ira Glickstein of WUWT, both accepted in general my law;
===================================================================
David’s Law

“Only two things can change the energy content of a system in a radiative balance; either a change in input, or a change in residence time of some aspect of the energy within the system.”
===================================================================

Now in our case we are defining our “system” as the earth, including land, oceans and atmosphere. “The residence time depends on both the materials encountered, and the WL of the watt under consideration.” (This is an important corollary to David’s Law and critical for further thought)

I think you refer to this residence time law more then once in your post so I hope you will find “David’s Law” useful. (-;

“But how does the sphere’s surface temperature rise? It can only happen by a direct increase in its internal energy (+U). Energy from somewhere will have to PILE up.”

“If the INCOMING heat flux (the Qin) to an object is kept constant, but its OUTGOING counterpart (the Qout) is reduced, the object will naturally warm, from the Qin energy not being able to escape as FAST as before, thus partially piling up inside of it instead. UNTIL we reach a new (and higher) surface temperature to restore the balance.”

I have a number of thoughts on how this may be useful.