# ‘To heat a planetary surface’ for dummies; Part 2

For something – anything – to acquire a temperature above absolute zero (0 K), it somehow needs to be able to warm. The only real requirement for something to be able to warm is for it to possess a ‘thermal mass’, or simply ‘mass’. A thermal mass provides the thing in question with what is (a bit awkwardly) called a ‘heat capacity’, meaning a capacity to absorb and store energy from some energy source (external or internal).

We already know, from basic thermodynamic principles, how energy can be transferred to (or from) an object. It can be transferred in the form of ‘heat’ [Q] or in the form of ‘work’ [W]. Whenever energy is transferred to an object, the ‘internal energy’ [U] of that object increases as a result, which simply means that the object in question has absorbed (energy isn’t ‘transferred’ to a system until it’s actually become ‘absorbed’ by it) the energy to store it inside its mass, as microscopic kinetic and potential energy of its atoms and molecules.

We already know, from the first post in this series, how system ‘internal energy’ [U] relates to system ‘temperature’ [T]. We know that a system with a high ‘heat capacity’ will warm more slowly than a system with a low ‘heat capacity’, both systems absorbing equal energy inputs, the high-heat-capacity system simply storing a larger portion of the absorbed energy as internal/molecular PE rather than as internal/molecular KE (determining the temperature). Both systems, however, will warm, only at different rates. U and T invariably move in the same direction. Unless there is an ongoing phase transition. Then U will increase and T will not. There is no process, though, where U increases and T decreases. The two correspond.

OK. We know that to make an object warm, we must make it accumulate ‘internal energy’. If it doesn’t, it cannot warm.

Before we reach the steady state (dynamic equilibrium), which is the point above which the temperature of a warming object will no longer rise, some of the incoming energy (the energy transferred to the object) will always accumulate within the mass of the object, raising its internal energy. This energy is then held inside the object, it stays there, in the end making up the object’s static, equilibrated fund of energy, its baseline U, determining its final, steady-state temperature.

The corollary of this storing up process of incoming energy for the object to warm, is that not as much energy will be able to leave the system as what comes in. Until such time as you actually do reach the steady state. The reason you will have no more warming upon reaching the steady state is, after all, that energy OUT finally equals energy IN. If there is no ‘work’ involved, then we have a situation where Qout finally equals Qin. Remember this. It is the heats that need to balance.

At this point, it is time to discuss Willis Eschenbach’s so-called ‘Steel Greenhouse’, as a suitable model to demonstrate and utilise the basic thermodynamic concepts we have learned.

Remember first that the ‘Steel Greenhouse’ model is a strictly radiative one. Except from the ultimate, constant energy source, the nuclear one inside the central sphere, heating it through conduction, there is no heat loss or heat gain by other means than by ‘thermal radiation’ between systems/surroundings in this setup.* All arrows represent radiative heat only. Meaning, no air (mass) and hence no real temperature surrounding or between our systems. This is very much a hypothetical situation simply designed to be instructive on certain basic concepts. Eschenbach and I very much disagree on exactly what these concepts are …

*However, there is also the point of conduction through the shell. Which we will ignore for the moment, but will get back to in the next post or the post after that. It is, after all, an essential point to how the real Earth system actually works …

What I very much want to be perfectly clear on, is that it’s not the described EFFECT that I’m objecting to. It is a very real, very natural effect and violates no thermodynamic laws whatsoever. No, it’s strictly Eschenbach’s cringeworthy attempt at an EXPLANATION of how the effect comes about, that I object to. And that quite frankly anyone in their right, critical mind should object to. At once, instinctively, loudly. I’m honestly amazed how people can seriously buy into his utterly inane fabulations, trying to pass them off as well-established Truths of ‘basic physics’.

Well, let’s get to it.

We start off with the basic setting – the central sphere without a surrounding shell:

Figure 1.

We assume first that the sphere is a perfect blackbody and that its surroundings are a perfect vacuum at (an equivalent) 0 K.

We then assume that there is an energy source of some sort (most suitably a nuclear one, but it doesn’t really matter) at the core of the sphere supplying it with a constant input of energy equivalent to a 240 J/s/m2 flux. This input would be defined as the Qin of the central sphere. There is no point in assigning a particular temperature to the energy source. Only its energy output is of interest. (The same can be said about the Sun as the energy source of the Earth’s surface.) The energy thus transferred at a constant rate from the central source to the sphere around it is absorbed and slowly accumulates within the mass of the sphere. It spreads out through conduction, from molecule to molecule, until the energy accumulating is evenly distributed throughout the sphere, from its core (where the energy enters) to its outer surface (from where energy is finally able to escape the sphere). In this state, the sphere has become isothermal, and its surface temperature has reached a level where it is finally able to radiate away a flux equal to the incoming at its core: 240 W/m2. At this point, the sphere cannot warm any further. It has reached its steady-state temperature, a state of dynamic equilibrium – as much energy exits the system per unit of time as enters.

Since the sphere in this initial situation directly abuts a perfect vacuum at 0 K, it is to be considered a ‘pure emitter’, an object with its entire heat loss effectuated through the emission of EMR. Since it is also defined as a blackbody (and ‘ideal emitter’), an object with an emissivity of unity, its steady-state, omnidirectional radiative heat loss flux from its surface can be directly determined by the most simplified version of the Stefan-Boltzmann equation:

q = σT4

Where q is the radiative ‘heat flux’ (equal to [Q dot]/A, radiative ‘heat transfer rate’ or ’emissive power’ [P] over area); J/s/m(W/m2). Whenever the pure emitter/blackbody conditions above are fulfilled, a q of 240 W/m2 demands a surface T of 255K or -18°C:

Figure 2. Graph originally from ‘The Engineering ToolBox’. The red circle, however, is my addition. It marks the spot where a surface temperature of -18°C would converge with the blue radiative heat emission flux curve, at an intensity of 240 W/m2.

The thing to bear in mind here is of course that some fraction of the energy constantly being transferred to the mass of the sphere from its core energy source will always stay stored within it, gradually raising its ‘internal energy’ [U] towards its final value in the process, rather than being transferred straight back out again by radiation from its outer surface, on the way to this final (steady-state) temperature. If this didn’t happen, it would simply never be able to reach it. It wouldn’t and couldn’t warm at all.

And only upon actually reaching this temperature will the heat loss flux from the surface of the sphere have grown all the way to become 240 W/m2, finally equalling the incoming energy input.

It is the heats that need to balance. The intensity of the incoming heat flux (qin) minus the intensity of the outgoing heat flux (qout) at any one particular point in time determines the incremental net heat (δQ) transferred to the system, in our case the sphere. This value, together with the total ‘heat capacity’ of the system (how much goes into internal potential energy vs. kinetic), in turn determines its instantaneous heating rate, the rate with which its temperature rises towards its final steady-state value.

Only in the most basic state, surrounded by nothing but a perfect vacuum at absolute zero (Figure 1), will the steady-state radiative heat flux radiated isotropically from the blackbody surface of our sphere into its perfect heat sink (cold reservoir) be precisely determined – through the Stefan-Boltzmann equation – by its own steady-state temperature alone.

As soon as its cold reservoir warms above absolute zero, however, this direct, simple relationship is broken. Now the radiative heat flux emitted by the sphere at dynamic equilibrium will be determined by the difference between the steady-state temperatures of two separate, opposing systems (sphere vs. cold reservoir surrounding it). Which means that even though the final radiative heat flux emitted from the sphere’s surface might very well end up the same, its steady-state surface temperature emitting this flux in this case would NOT. Because now the steady-state temperature of its facing heat sink would have to be taken into account as well.

Hold this thought. (It’s much less complicated than it sounds 😛 )

Moving on …

Figure 3.

We now place the steel shell around the central sphere.

At the first instant (Figure 3), the shell is still hypothetically at 0 K (its radiative heat flux to space therefore 0 W/m2), but as soon as it starts absorbing the incoming radiative heat flux from the sphere now trapped within it, energy starts accumulating inside its mass, and its internal energy [U] and its temperature [T] starts rising. And with an increasing temperature comes an increasing radiative heat output (to the void outside of it, NOT in any way back towards the sphere, its energy (heat) source!).

The shell will also, like the sphere, necessarily have a ‘thermal mass’. It needs to for it to be able to warm. A hypothetical shell with zero ‘heat capacity’ is a meaningless entity. It is also completely useless in a model like Eschenbach’s ‘Steel Greenhouse’, because such an object could not warm, it could accordingly never acquire a temperature, and would thus be totally helpless at demonstrating the concept of insulation, which is what the steel shell around the central sphere ultimately, as we shall see, turns out to be – an insulating layer.

So, this is where the fun begins …

Figure 4.

After some time, the shell’s temperature has risen to 50K. It is thus emitting a radiative heat flux to space (its Qout) worth of 0.35 W/m2. Not by any means a whole lot. But still larger than zero.

What, then, happens to the radiative heat flux going from the central sphere to the surrounding shell as the latter starts warming?

If the central sphere weren’t itself heated by the constant input from its nuclear energy source, the sphere/shell heat flux would grow smaller and smaller as the shell grew warmer and warmer. In the end, it would reduce to zero, at which point the temperatures of the sphere and the shell would be exactly equal. What the shell would do in this case, is extend the sphere’s total cooling time, that is, reduce its cooling rate from start to finish. Like proper insulation.

With the central sphere being constantly heated by the input from its internal energy source, the end result turns out quite differently. Now, none of the bodies will cool. Temperatures will rise at both ends.

Remember what we said about individual heat transfers: ‘They can only ever result in heating at one end, cooling at the other.’

This still holds. It is the first situation outlined above. Switch off the internal energy source of the central sphere and you’re down to one heat transfer only, and the sphere inevitably starts cooling.

But now we have two heat transfers working in tandem.

Still, don’t forget what we said about one system being involved in two separate heat transfers: ‘You need to keep the individual heat transfers apart in order to see which one is doing what.’

That system here is the central sphere:

Figure 5.

Always keep in mind that there is no transfer of energy going on from the sphere to its energy source at its centre; likewise that there is no energy transfer going on from the shell to the sphere. In thermodynamics, energy transfers are always unidirectional. In the case of heats [Q], they always move spontaneously from warmer to cooler only. If you want to claim that there is something moving the other way also, then this something would by definition NOT be heat.

Also, according to basic thermodynamic principles like the 1st Law, how do you change the internal energy [U] – and thus normally the temperature [T] – of a system like the sphere? There are only two ways. You transfer energy to or from it by way of heat [Q] or work [W]. No other way. In our situation here, there is no work being done anywhere, so the ONLY way to change the U and thus the T of the sphere system is through the transfer of energy as heat.

So what do we have? Disregarding work, if you transfer energy to the sphere and this energy transfer to the sphere ends up increasing its internal energy and consequently its temperature … then you have transferred HEAT to the sphere. There is absolutely no way around it. Now, take a look at Figure 5 above. Wherefrom would such an energy transfer possibly arrive? From one place only. From the left. From the energy source inside of it. (In the case of the surface of the Earth, it would come from the Sun and from the Sun only. From a hotter place!) It simply could not come from the right, from the shell. Because the shell is cooler than the sphere and thus cannot transfer any energy as heat to the sphere. The sphere is rather the one transferring energy as heat to the shell, warming it in the process. The shell is NOT an energy source to the sphere. Not in any way. It could never be! It is its energy recipient. Its cold reservoir. Its heat sink. The sphere is the energy source of the shell. And this relationship does not change even when the shell starts warming. It remains cooler than the sphere.

So in the particular heat transfer between the sphere and shell (the right-hand one in Figure 5), the sphere is always losing internal energy (−U), thus cooling, the shell always gaining (+U), thus warming, by way of the unidirectional transfer of energy from the former to the latter as heat.

People, you absolutely NEED to take this to heart!

Still, when the shell is in place and starts warming, the sphere starts warming too. Why? How? What’s going on?

This is where people get confused. And this confusion is exactly what the ‘climate establishment’ capitalises on.

Is the shell now somehow the thing that’s doing the extra warming? The sphere, after all, was already equilibrated with its energy source at its core, at 255K, before the shell was brought in, its total steady-state content of internal energy all originating directly from this source. So one would think that this is the maximum temperature (and hence, the maximum content of internal energy) that could ever be achieved by the sphere with only the 240 W/m2 input from its inner energy source.

And one would seemingly be right.

However, does this in any way mean that, when the sphere evidently starts warming beyond this maximum temperature with the shell up around it, it is somehow energy from the shell adding on to that equilibrated internal energy content of the sphere, further raising its temperature?

NO!!!! Most certainly NOT!

The ‘extra’ energy now accumulating inside the sphere, raising its temperature in the process, is of course still directly drawn from the core energy source, and only therefrom. It could only ever be. Nothing returns to accumulate a second time!

We’ll get back to this in the next post …

The interesting circumstance here is that, when the sphere has achieved heat balance with its energy source (its Qout equals its Qin), it wants to stay there. At all costs. Any process disrupting this balance will have to be dealt with immediately. And it is, by physical necessity.

Since, as we pointed out earlier, the conceptual model of the ‘Steel Greenhouse’ is basically a purely radiative one, the Stefan-Boltzmann equation for radiative heat transfer applies. We already found the equilibrated radiative heat loss for the blackbody sphere alone in space. Now that the blackbody steel shell is emplaced around it, we need to expand the formula a bit:

q = σ(Th4 – Tc4)

Since q (the sphere’s Qout) needs to remain at 240 W/m2 (to balance the sphere’s Qin), we immediately realise that, in order to pull this off, the temperature of the sphere (Th) will have to rise as the temperature of the shell surrounding it (Tc) rises. This is what we discussed a bit earlier.

But by how much? We rearrange to solve for Th:

Th = 4√((q/σ) + Tc4)

Th = 4√((240/σ) + 504) = 255.1K

A rise in sphere temperature of a mere 0.1K (!) will suffice to retain balance with a shell temperature of 50K.

So why and how does the sphere’s temperature increase? It all happens automatically and instantly as the opposing shell temperature rises.

We know from before that as long as a system’s Qin is larger than its Qout, then some of the energy supplied by the energy source (providing the Qin) will always naturally accumulate inside the system, increasing its internal energy [U] and, accordingly, its temperature [T]. This will in turn progressively increase the system’s output (Qout), up to the point where the outgoing energy finally balances the incoming and the accumulation stops.

So what happens if you then, after initial equilibration, manage to reduce the outgoing heat flux back again from its attained steady, maximum level, somehow retard its energy loss per unit of time, while keeping the incoming constant?

Well, the same thing would happen as during the original warming process. Same general mechanism. Only now with the final, equilibrated U and T turned from end points to starting points. In other words, there would be further warming.

This is the basics of how insulation works.

The process is technically a stepwise one (reduced Qout → accumulation of internal energy from the constant (and now once again larger) Qin → raised T → Qout increases, back to balance), but outwardly it appears as a continuous, smooth rise without discernible lag. As long as the surrounding (insulating) layer keeps warming, this process will go on.

When the shell temperature (Tc) rises, its ‘radiative heat potential’ (σTc4) increases, and so the difference in heat potentials between the two opposing surfaces – the sphere surface and the shell surface – is reduced [refer to the radiative heat transfer equation just above]. This in turn leads to a smaller actual transfer of energy as radiative heat per unit time per area [q] from the warmer sphere to the (still) cooler shell than before (when the shell was even cooler).

Notice how the shell’s σTc4 is a mere radiative heat potential when facing the warmer surface of the sphere inside (heats do not oppose each other!), but an actual radiative heat flux (a real transfer (output) of energy) from the outside of the shell, facing its cold reservoir, the 0 K vacuum of space.

As the shell continues to warm from (most of) the energy continuously transferred to it as heat from the warmer sphere piling up inside its mass, the sphere is also forced to warm to maitain its heat balance with its own energy source:

Figure 6.

With a shell temperature of 100K, how much warmer than its initial equilibrated temperature is the sphere forced to become?

Not very much:

Th = 4√((q/σ) + Tc4)

Th = 4√((240/σ) + 1004) = 256.5K

1.5 degrees warmer.

At this point, the shell emits an isotropic radiative heat flux to the 0 K vacuum surrounding it worth of 5.67 W/m2. Still not hugely impressive; a flux 16.2 times as intense as the one at 50K, but still 42.3 times less intense than the incoming heat flux from the sphere on the inside.

But, the shell grows ever warmer:

Figure 7.

It is now at 200K. Its Qout has grown to be 90.73 W/m2. We’re getting there. The slope of its output is ever steepening. And as the slope of its output steepens, its warming rate is progressively declining. Because a smaller and smaller portion of the energy absorbed (from the 240 W/m2 of Qin) manages to remain inside the mass of the shell from one second to the next. Because more and more is shed (Qout) at the same time. The amount staying after each ’round’ moves towards zero. The steady state of dynamic equilibrium is drawing near.

So how warm is the sphere at this point? The difference in opposing radiative heat potentials needs to stay the same. So when the shell temp goes up, the sphere temp naturally must go up, from the resulting internal piling up of energy supplied by its central source (its Qin). But the temperature rise is by no means equal in absolute terms:

Th = 4√((q/σ) + Tc4)

Th = 4√((240/σ) + 2004) = 276.3K

The sphere is now 21.3 degrees warmer than originally. The shell, though, is 200 (!) degrees warmer.

Finally we reach the steady state, when neither the shell nor the sphere can warm any longer:

Figure 8.

At this final stage, the shell has reached the initial temperature of the sphere and thus emits a similar flux to space (240 W/m2), the same size as the incoming flux equivalent from the ultimate energy source at the core of the central sphere. We have established radiative heat balance.

So, when the shell is at 255K in the steady state, how warm must the sphere be?

Th = 4√((q/σ) + Tc4)

Th = 4√((240/σ) + 2554) = 303.3K

The sphere at radiative equilibrium has warmed a grand total of 48.3 degrees, from 255 to 303.3K. This is its new steady-state temperature.

The final temperature of the sphere is close to 1.19 times its initial temperature. It might sound like a random observation, but in fact, this would always be the case with a blackbody arrangement such as this. It doesn’t matter how hot or cold it was to begin with. It doesn’t matter how large or small the energy input. The central, heated object, insulated by a surrounding one, will always reach its steady-state at a temperature 1.19 times higher than its original one. Meaning, it will also always end up being 1.19 times warmer (in absolute temperature) than its surrounding shell.

1.19 (or 1.189207115…..) is simply equal to 4√2.

At radiative equilibrium in this particular situation (disregarding differences in radii*), the ‘radiative heat potential’ of the sphere (σTh4) is, per the Stefan-Boltzmann equation, exactly twice the size of the ‘radiative heat potential’ of the shell (σTc4), meaning, the actual radiative heat flux moving from the sphere to the shell is at this point of exactly the same intensity as the actual radiative heat flux moving from the shell to space (as can be seen from Figure 8 above).

*The inverse square law says that the intensity of a radiative flux diminishes the further away from its source it gets. Also, different radii means different overall areas from which the energy is emitted, so the fluxes would actually not be the same, only the total amount of energy. We have ignored any such mathematical complications to our conceptual model, simply because they would only distract from the concepts we want to explore.

This relationship leads us to the conclusion that the final (radiative equilibrium) temperature of the sphere raised to the fourth (Tf4), is exactly twice the initial temperature of the sphere raised to the fourth (Ti4):

Tf4 = 2 * Ti4

Which in the end takes us here:

Tf = 4√2 * Ti

In the upcoming post, we will explain exactly why and how Willis Eschenbach’s “back radiation” explanation of the extra warming of the sphere must be wrong. How to properly resolve the workings of ‘radiative heat transfer’?

And we will start exploring the real mechanisms responsible for heating a planetary surface under a massive atmosphere. Recall that point about conduction through the shell …

## 56 comments on “‘To heat a planetary surface’ for dummies; Part 2”

1. geran says:

Just a couple minor quibbles:

1) Willis used 235 W/m^2, where as you used 240. (This is info just so someone will not get confused trying to study the issue. I actually prefer your 240, as it relates to the 255 K.)

2) My interpretation of the scenario presented by Willis was that the energy supplied to the sphere (240/235 W/m^2) would be maintained constant. That would only change your final state to: sphere temp = 255 K, and shell temp = 214.4 K. It’s all about how you interpret the original scenario. Your work is correct.

Final two cents–I don’t believe the “steel greenhouse” is originally from Willis. I did some research years ago and I believe it came from some college/university in the Pacific Northwest. I have searched several times since, trying to find what may be the original source, but it appears “removed’. Maybe they took it down out of embarrassment!

Anyway, thanks for your efforts, from just another “Dummy”. 🙂

• okulaer says:

Hi, geran 🙂

1) Yes, you’re right. I believe the 235 W/m2 comes from the original K&T97 budget.

2) Mmm, but the energy supplied to the sphere is kept constant. It remains 240 W/m2 throughout. That doesn’t mean the sphere is unable to attain a temperature higher than 255K. That only has to do with how much energy is accumulated within its mass. If you limit its Q_out, but keeps its Q_in constant, then more energy (from the source) will necessarily pile up inside the sphere than if you didn’t limit its Q_out.

Like I pointed out, the warming EFFECT of putting up a shell (an insulating layer) around a constantly heated object is very real. However, Eschenbach (and the ‘climate establishment’) tries to EXPLAIN this effect through “back radiation heating”, which is clearly ridiculous. I will elaborate on this in the next post.

2. Rosco says:

Aren’t you conserving flux rather than energy ? You have simply stated the sphere needs to heat up but haven’t really justified this assertion.

The sphere can continue to emit 240 x A watts because you have failed to account for the portion of this which must be absorbed by the shell to supply the energy it requires to warm.

The radiation from the sphere is actually performing work on the shell – increasing the shell’s internal energy.

There is no logical reason why the sphere must increase in temperature to achieve this.

Say I could somehow surround a sphere like this with a thin shell of ice – are you saying the ice would make the sphere hotter rather than the ice simply increase in temperature itself until it melted ?

The sphere can emit 240 x A (A = surface area) watts or joules per second as per it’s internal energy source.

This energy is absorbed by the shell and warms it. When the shell warms to a point where it is emitting 120 x 2A watts over its double surface area there is a balance of energy.

240 x A watts from the sphere is completely required to cause an emission of 120 x 2A watts from the shell as geran stated above.

Why is any of what I have just written limiting the energy out from the sphere ?

All of the 240 out from the sphere is “used” in maintaining the shell’s temperature.

At this point – 240 x A from sphere causing 120 x 2A in the shell represents conservation of energy.

Without the 240 from the sphere the shell will quickly cool.

So the issue is the 120 back radiation as Willis claimed.

People perform all sorts of ridiculous algebraic manipulations on the Stefan-Boltzmann equation but it does not describe radiation emissions – it simply relates the total power emitted by an object at a given temperature.

There is no real evidence that the reverse is true – i.e. there is no real evidence that simple algebraic sums can be used to calculate temperatures.

Plot some Planck curves for various temperatures and see if the sums such as 239.7 + 239.7 = sigmaT^4 where T = ~303 K actually work.

To my mind they don’t because at 303 K the object is emitting high energy photons that the cooler object cannot and hence never replace so to my mind back radiation is a false construct.

You can extract temperature as a variable from Planck’s equation and see if any curves produced by such combinations actually equal sigmaT^4.

And they don’t because sigmaT^4 is a unique relationship only completely described by Planck’s equation.

Energy shields are used in all sorts of applications but no matter how much I search I never see the claim they cause a steel greenhouse effect.

I have even seen the claim that more shields leads to infinite increases in radiation flux at the sphere – 2 = double, 3 = 3times, 4 = 4 times etc. – each extra shield increasing the flux by the original amount.

One classic reason I do not believe this is the case of Apollo 13.

The astronauts had to retreat to the smaller module to conserve oxygen. Despite travelling through an energy flux of ~1370 W/sqm. of solar radiation and there being 3 humans in a shielded environment each at ~37 C temperature they had to endure days of 4 degree C internal temperatures because they had to turn down the heat to conserve energy for returning home.

There was certainly no steel greenhouse effect in this case and their heat shields nearly killed them by very efficiently repelling the sun’s radiation.

Anyway that is my 2 cents worth – I see no compelling reason why conserving flux is realistic and certainly not preferable to conserving energy.

After all TOA of the Earth is continually illuminated by 1370 W/sqm. but radiates only 239 w/sqm.

And geran is right – the steel greenhouse is not an original concept by Willis – he copied it without acknowledging this – never trust a plagiarist !

• okulaer says:

“Aren’t you conserving flux rather than energy? You have simply stated the sphere needs to heat up but haven’t really justified this assertion.”

Then I’m afraid you haven’t read my post properly, Rosco.

“The sphere can continue to emit 240 x A watts because you have failed to account for the portion of this which must be absorbed by the shell to supply the energy it requires to warm.”

I haven’t ‘failed to account for’ anything. Read the post.

“The radiation from the sphere is actually performing work on the shell – increasing the shell’s internal energy.”

No it’s not. It’s heating the shell, increasing the shell’s internal energy: ΔU = Q.

“There is no logical reason why the sphere must increase in temperature to achieve this.”

Yes there is, and I have provided it in the very post you’re commenting on.

“Say I could somehow surround a sphere like this with a thin shell of ice – are you saying the ice would make the sphere hotter rather than the ice simply increase in temperature itself until it melted?”

The ice wouldn’t make it hotter. The ice would simply reduce its power output, like all insulation does. The energy source of the sphere is what would make it hotter. Because it supplies it with a constant input of energy that can no longer escape as fast as before. Thus more of the energy supplied by the source will accumulate inside the mass of the sphere – making it warmer. All of this was explained in the post, Rosco. There is nothing mysterious about it.

The sphere can emit 240 x A (A = surface area) watts or joules per second as per its internal energy source.

This energy is absorbed by the shell and warms it. When the shell warms to a point where it is emitting 120 x 2A watts over its double surface area there is a balance of energy.

240 x A watts from the sphere is completely required to cause an emission of 120 x 2A watts from the shell as geran stated above.

Why is any of what I have just written limiting the energy out from the sphere?

All of the 240 out from the sphere is “used” in maintaining the shell’s temperature.

At this point – 240 x A from sphere causing 120 x 2A in the shell represents conservation of energy.

Without the 240 from the sphere the shell will quickly cool.

So the issue is the 120 back radiation as Willis claimed.”

Er, no Rosco. Again I would admonish you to actually read the post properly before you comment on it. In fact, you should read Part 1 also. You very obviously didn’t. It seems to me you came here with very particular preconceptions about what you think I’m claiming. And then you’re attacking those ideas rather than what I actually write. That’s called ‘straw man argumentation’.

“People perform all sorts of ridiculous algebraic manipulations on the Stefan-Boltzmann equation but it does not describe radiation emissions – it simply relates the total power emitted by an object at a given temperature.”

What are you trying to say? It relates the radiative heat emitted by a ‘pure emitter’.

“There is no real evidence that the reverse is true – i.e. there is no real evidence that simple algebraic sums can be used to calculate temperatures.”

True, what is always actually measured (like in a pyrgeometer) is the ‘heat flux’ in to or out of the sensor plus the sensor temperature. Everything else is calculated from these two inputs.

“Plot some Planck curves for various temperatures and see if the sums such as 239.7 + 239.7 = sigmaT^4 where T = ~303 K actually work.

To my mind they don’t because at 303 K the object is emitting high energy photons that the cooler object cannot and hence never replace so to my mind back radiation is a false construct.”

I don’t think you’ve understood at all what my post is about, Rosco. This is NOT how I go about. I specifically do NOT add together two 240 fluxes to get from 255 to 303K. That’s the whole “back radiation heating” nonsense. Newsflash, I do not even remotely endorse that ridiculous construct. Read the post.

“I see no compelling reason why conserving flux is realistic and certainly not preferable to conserving energy.”

I know perfectly well, Rosco, that energy is what needs to be conserved, not flux. The model, however, is a purely conceptual one, at the most simplistic level, where the conservation of flux becomes equivalent to the conservation of energy. To avoid unnecessary distractions from the relevant concepts actually being discussed. I know about the inverse square law. I know what happens when you increase emitting area. I pointed out specifically in the post that I have ignored these points as they would rather confuse and complicate things than help people understand those particular basic concepts I want them to understand. You may disagree with this approach. Feel free. I don’t mind. But don’t pretend it somehow invalidates my discussion. That’s just daft.

Thanks for the interest.

3. Rosco says:

I think you have also made the mistake all of the advocates of this steel greenhouse effect make.

You write q = σ(Th^4 – Tc^4) but then you manipulate this q as if it is an actual flux equal to sigmaT^4 –

AND OF COURSE IT ISN’T.

It is actually q(net) and has no temperature profile at all and is certainly not equal to sigmaT^4.

Plot some Planck curves using these values T = 255.1 (roughly 240 W/sqm) and T = 303.3 K and also add another column for double the column for the T = 255.1 curve.

Use wavelength and multiply each result in each column by pi.

You will quickly see that doubling every value for T = 255.1 K does not produce the 303.3 K curve.

You will also see that the 303.3 K curve has high energy values that the 255.1 curve does not.

Use a sum of histograms to confirm the area under the curve equals the correct flux values – which it will if you have say 5000 data points from wavelength 1 in increments of say 0.05 micrometre.

You can also plot the difference curve for these 2 temperatures using the solution for extracting T as the variable – you already have the columns for the Planck curve values versus wavelength.

For the T = 225.1 Planck curve the plot of t versus Planck equation is a straight line parallel to the horizontal axis at, obviously, 255.1 K. So is the plot for the T = 303.3 K curve versus Planck values.

This result is trivial but confirms the relationship and the Stefan-Boltzmann relationship for sigmaT^4.

But plot the difference curve for as you write q = σ(Th^4 – Tc^4) and the result is not a constant temperature curve.

To my mind this confirms absolutely that q = σ(Th^4 – Tc^4) cannot be manipulated as you do in all your algebra because it is not a flux with equality to sigmaT^4

Remember the Stefan-Boltzmann equation is the integral of Planck’s curve over all wavelengths by pi.

I cannot see how it is legitimate to use algebraic manipulations of flux values if the result does not produce a valid Planck curve because this is the only case where the result is equal to sigmaT^4.

Perhaps I am wrong and I do not claim to know how this works but I think I can show how it doesn’t work.

The only way I can see whereby the higher frequency (shorter wavelengths) emissions can be induced is if there is a hotter source – I cannot see you have explained how this occurs.

This is all a supposed justification for the “natural greenhouse effect” of Earth and how it doesn’t breach thermodynamic rules.

I see the equation of 239.7 solar plus 239.7 atmospheric = sigmaT^4 at 303 K taught in University lectures.

Plot the Planck curves for T = 255.1 K and double it and you do not get the 303.3 K curve.

Then plot the curve for the solar temperature of say 5778 K – NASA reference – scaled by inverse square law to Earth.

The plot has an area of 1370 W/sqm under the curve.

But it completely overshadows the 255.1 K curve in the high energy areas so all the surface heating is due to the solar radiation with little to zero from the atmospheric temperatures.

Claiming 239.7 (solar) + 239.7 (atmosphere) = 479.4 at 303.3 K is simply ludicrous – look at the Planck curves.

You have to plot the solar scaled curve logarithmically to see the curve it is so much more powerful.

The surface may be heated to 303.3 K but it is the solar radiation doing it.

• okulaer says:

“I think you have also made the mistake all of the advocates of this steel greenhouse effect make.

You write q = σ(Th^4 – Tc^4) but then you manipulate this q as if it is an actual flux equal to sigmaT^4.”

q is an actual flux, Rosco. Always. It is the RADIATIVE HEAT FLUX from sphere to shell. The one we always measure (detect), remember? It is an actual physical entity. The terms that are NOT actual fluxes are the ones on the righthand side of the equation, the two opposing σT^4 terms. These merely represent ‘radiative heat potentials’, the fluxes the two opposing surfaces, thermally isolated from one another, would emit as radiative heat to a perfect vacuum at absolute zero. They are nothing but mathematical constructs. We measure the temperatures of the two opposing surfaces (in vacuum) and derive the radiative heat flux between them. Because if we want to measure the heat flux coming off the warmer surface with some kind of externally introduced sensor, then what we end up detecting is not the heat from that surface to the other one, but from that surface to … our sensor. Our sensor would have to be at the exact same temperature (and also have the exact same emissivity) as the other surface for us to get a ‘correct’ result.

What matters here is the temperatures of the opposing surfaces, on each side of the vacuum between them. There is ONLY ONE transfer of energy going on, the radiative heat flux from hot to cold.

“It is actually q(net) and has no temperature profile at all and is certainly not equal to sigmaT^4.

Plot some Planck curves using these values T = 255.1 (roughly 240 W/sqm) and T = 303.3 K and also add another column for double the column for the T = 255.1 curve.

Use wavelength and multiply each result in each column by pi.

You will quickly see that doubling every value for T = 255.1 K does not produce the 303.3 K curve.”

Rosco, in this particular discussion, blackbody radiation, Plack curves and Wien’s displacement law are all irrelevant. I specified this very carefully in the first post. This discussion deals with temperature change as a result of energy transfers between systems/regions/objects/surfaces and therefore strictly relates to the field of THERMODYNAMICS. The only thing that matters is ‘how much energy is transferred as radiative heat from the hot object to the cold object per unit time’?

Read up on the 1st Law of Thermodynamics. If more heat (energy per unit time) is coming in to a system than what goes out, then there will be a surplus piling up, some of the incoming energy will accumulate, and the system’s ‘internal energy’ [U] – and thus its ‘temperature’ [T] – will rise.

“The only way I can see whereby the higher frequency (shorter wavelengths) emissions can be induced is if there is a hotter source – I cannot see you have explained how this occurs.”

ALL the energy accumulating inside the sphere, making it warmer, is ALWAYS from its own central energy source, Rosco. NOTHING comes back from the shell. The point is only, how fast is the energy moving OUT of the sphere (to the shell) as compared to the incoming (from the source)?

“This is all a supposed justification for the “natural greenhouse effect” of Earth and how it doesn’t breach thermodynamic rules.”

You obviously haven’t read anything I’ve written. You’re fighting a gigantic straw man here, Rosco. If you seriously think that I’m one who’s goal it is to somehow justify the rGHE, I cannot but laugh!

“The surface may be heated to 303.3 K but it is the solar radiation doing it.”

Indeed. And no one ever claimed otherwise on this blog …

4. geran says:

Yeah, I was working from memory (always a bad idea) and it seemed like either the incoming energy was held constant or the sphere temp. So, I had to go back and check and you are right, Willis maintained a constant energy and allowed the temperature of the sphere to increase.

It’s been about 3 years since I’ve worked on this, but now I remember and interesting sidetrack. For example, suppose the sphere has an area of 2 square meters instead of just one. The power out to space must remain 240 Watts (Pout = Pin). So, that means the flux to space must be 120 W/m^2. Using your methodology, that then gives a shell temp of 214.4 K, and a sphere temp of 255 K. IOW, if we increase the size of the sphere, we cool it off, everything else being the same.

That should make some Warmist’s head explode!

5. Truthseeker says:

So much clarity … so little confusion …

6. Rosco says:

You claim –

“It seems to me you came here with very particular preconceptions about what you think I’m claiming.”

I simply believe what you claim is inappropriate application of the Stefan-Boltzmann equation.

I still believe you have made the same mistake I said before – you write q as a flux when it clearly isn’t – it is clearly q(net) and unique to a particular set of circumstances.

You quote my statement –

“The radiation from the sphere is actually performing work on the shell – increasing the shell’s internal energy.”

And respond –

“No it’s not. It’s heating the shell, increasing the shell’s internal energy: ΔU = Q.”

You have actually COMPLETELY AGREED with me on the main point but you object to my use of the word “work”.

But the fundamental claim I made must be completely sound because you totally agree with it.

240 X A Watts increases the internal energy of the shell to a point (temperature) where it can emit 120 X 2A Watts and this without needing to increase the temperature of the sphere.

The only objection you have to this is it means the Shell is not emitting 240 Watts externally and if that is not absolutely you demanding preservation of flux then I don’t know what else it is.

Then you claim –

“Rosco, in this particular discussion, blackbody radiation, Planck curves and Wien’s displacement law are all irrelevant. ”

Well I am very sorry to inform you your simply mistaken on this point !

Are you or are you not using the Stefan-Boltzmann equation to “prove” all of your claims ???

Do you have some other mechanism relating the radiation emission spectrum from an object at any particular temperature that is not Planck’s law ??

Because if you do then the Stefan-Boltzmann equation you inappropriately apply is obviously “irrelevant” – by your own words !

The Stefan-Boltzmann equation is explicitly defined as pi times the integral from zero to infinity of the values of the Planck equation over the particular variable chosen – often wavelength or wavenumber.

Sources for this are too many to list but SpectralCalc, Wolfram Alpha and Hyperphysics will do.

As such the Stefan-Boltzmann equation is the area under any Planck curve and provides no real description of the radiation emission other than the gross total power.

Without knowing the radiative emission spectrum one has little real knowledge.

To claim they are irrelevant is the real straw man in this discussion.

I know your algebraic manipulations are simply mistaken.

The equation you write as your starting point is q = σ(Th^4 – Tc^4) – this equation simply defines one particular case and you require knowledge of 2 of the values to calculate the third.

Let’s say Th is 303.33 K and Tc is 255.06 K then one would calculate your q as ~480 – 240 = 240.

But as either temperature changes so does q.

You use a constant q/sigma in your equation – I say this is inappropriate.

Then you write –

Th = 4√((q/σ) + Tc4) to calculate Th ???

C’mon – if Th is more than ~255 K the sphere is no longer emitting 240 X A Watts !

You are really saying the temperature of the sphere is equal to the temperature of the sphere plus the temperature of the shell ???

Expand your equation – Th = 4√((q/σ) + Tc4) = 4√(q/σ) + 4√(Tc4) –

and it simply becomes Th = Th + Tc !

(because Th (= 4√(q/σ) + Tc( = 4√(Tc4) which agrees totally with your original claim q = σT4 which is exactly T = 4√(q/σ) .

I don’t recall any actual Physics reference to anything like this.

Didn’t you notice this inconvenient truth ?

What source do you cite establishing your algebraic manipulations are actually correct and scientifically valid.

Your argument is a convoluted proof of back radiation even though you say it isn’t – your results are exactly the same.

Your logic is circular and a notational trick – not a real proof.

The author of this forgot some of the fundamental rules of the scientific method including but not limited to:-

1. No theorem can be proved by purely notational tricks.

2. The fundamental rule of mathematical sloppiness is that you are allowed to be sloppy as long as you know how to do things correctly.

• okulaer says:

Hi, Rosco.

You say: “Your argument is a convoluted proof of back radiation even though you say it isn’t – your results are exactly the same.”

It’s not. You are simply not willing to even try to understand what I’m saying, because you’re completely caught up in your rigid preconceptions of what you think I’m saying.

You say for instance: “240 X A Watts increases the internal energy of the shell to a point (temperature) where it can emit 120 X 2A Watts and this without needing to increase the temperature of the sphere.” Likewise, you say: “C’mon – if Th is more than ~255 K the sphere is no longer emitting 240 X A Watts !”

Don’t you get it, Rosco? This is YOU being stuck on a “back radiation” interpretation of the situation. This is YOU being stuck on thinking that the two σT^4 terms are each real fluxes of energy. Not me.

The 240 W/m^2 radiative heat flux from the sphere to the shell warms the shell until it emits a radiative heat flux of 240 W/m^2 to space. There is no radiative transfer of energy of ANY kind from the shell back to the sphere. Try to pay attention!

The sphere at 303.3K does NOT emit more than 240 W/m^2 worth of radiation! Only in your literal (rGHE/AGW fundamentalist) interpretation of the Stefan-Boltzmann equation does this NEED to happen. When the facing surface is warmer than 0 K, it doesn’t happen. The temperature of the opposing surface impedes the escape of energy (by radiation) from the sphere. Just like a blanket or a piece of clothing impedes the escape of energy (by convection and evaporation) from your body.

The whole bidrectional flux thing is all in YOUR mind, Rosco, not mine. Sorry to inform you.

7. Rosco says:

Th = 4√((q/σ) + Tc4)

Is obviously only valid when Tc = zero !

• Rosco says: January 13, 2015 at 1:52 am

“Th = 4√((q/σ) + Tc4) Is obviously only valid when Tc = zero !

Tc can never be zero. That Th is precisely correct for any black mass within an environment of any possible Tc.

8. Truthseeker says:

Okulaer,

I have been reading the discussion between you and Rosco and the following thought occurred to me.

You have said that there is a heat flow between the sphere and the shell. Isn’t the heat flow really between the sphere and the internal vacuum between the sphere and the shell? Isn’t a vacuum an infinite heat sink? Why would a presence of a shell at some point outside the sphere affect how the shell interacts with the vacuum that immediately surrounds it?

I think that the flaw in your logic is that you are not representing the flow correctly. I see it as the following;

Energy Source > Sphere > Internal Vacuum > Shell > External Vacuum

The radiation eminating from the sphere travels the internal vacuum unimpeded to the shell and then from the shell outwards. So the sphere does not build up heat because the interaction is between it and the internal vacuum at 0K. The effect of the shell is to temporarily delay the radiated energy from the sphere reaching the space outside the shell but that is all.

Just my 2c worth …

• okulaer says:

Hi, TruthSeeker.

I see your point. But you need to remember, even though there is no MASS inside the vacuum, there’s definitely a RADIATION FIELD. The energy is transferred as heat through this radiation field, as a result of the particular state/condition of this radiation field. The surface of the sphere is simply at all times in direct contact with the inner surface of the shell via the dynamic radiation field between them.

You can read about this concept (because, yes indeed, it is a conceptual idea, just like the bidirectional stream principle is a conceptual model of how a radiation field works) here, for instance:
http://claesjohnson.blogspot.no/2014/02/physics-illusion-2-photons-as-light.html

I will get to this in my next blog post.

• Truthseeker says:

OK, but if Tsphere goes up then doesn’t that increase the differential between Tsphere and Tshell which will increase the heat flow from the sphere and Tsphere will drop back down to the previous level?

Also since the shell is in contact with an infinite heat sink in the external vacuum will it not just shed the heat as fast as it arrives as per its internal conductivity? This means that Tshell will not increase in the way that you are suggesting.

• okulaer says:

Why would the Tsphere drop back to the previous level? When the Tsphere goes up, it increases the difference in ‘radiative heat potentials’ down to the shell back UP to the previous differential. First the differential is reduced. This is when new energy from the sphere’s nuclear energy source starts piling up again inside the sphere (because now Qin is once again larger than Qout). This raises the temp of the sphere until the differential is back to where it was, and we have reached a new dynamic equilibrium level. Tsphere increases until we reach this new steady state and then it increases no more. Until the Tshell increases even more. Then the whole cycle starts anew. Like I said, this is technically a stepwise process, but in reality it is just a smooth rise in temperature without any apparent reduction in heat output.

Hope this makes sense …

9. Bryan says:

Okulaer, good work but just to add that the same results can be obtained by considering photon flow.
Both methods, that is classical thermodynamics(your article) and bi-directional photon flow (origin Prevosts theory of exchanges) give the same results when used consistently.

The main advantage of photon flow method is to challenge the likes of scienceofdoom who will ask what happens to the returning radiation from the shell ?
‘Is it absorbed by the core?’

SoDs next question will be ;
‘Surely this means that the shell heats the core?’

The bidirectional method says
Radiation (or photon flow) from hotter to colder objects and colder to hotter occurs.
So energy transfer is bidirectional
However the hotter flow is more intense and includes higher frequency photons.
The net energy flow is the heat transfer and the spontaneous direction is always from hotter to colder.
So heat transfer is unidirectional .

Looking forward to your next article.

• okulaer says:

Yeah, I would rather drop the idea of a ‘counter-flow’ altogether. And that’s what I try to accomplish. First of all, the bidirectional idea is but a quantum-theoretical conceptual model. It has never experimentally been shown to be a ‘real thing’. The old masters knew perfectly well that it was just that, a descriptive model of reality, they never claimed otherwise. Niels Bohr, for instance, stated: “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…”

Today, however, people like SoD seem to take for granted that the concept is in fact Reality, a fact, a law of nature, established Truth. And then they argue from that basis.

I’ve explained earlier that the two-way flow model breaks down as soon as you introduce a third body to the mix, a constant energy source to the warmer object. Because in this situation you are forced to explain the extra heating that you DO observe in the warmer (and now heated) object by the addition of “back-radiated” energy from the passive cooler body opposing the warmer one.

This is why people like Rosco here get confused. He thinks that I’m stuck with the bidirectional idea. Simply because HE is. And then, whenever he sees extra warming of the central sphere, this is confirmation to him of “back radiation heating”. So in his world, he CANNOT allow extra warming of the sphere. In my world (and, in fact, in the real world), this is no problem. It is still the energy input from the source of the warmer object that does the heating. “Back radiation” doesn’t even enter the equation. There is only an opposing ‘temperature potential‘ impeding the radiative heat out from the sphere through the radiation field between the two.

Thanks, Bryan.

• Bryan says: January 13, 2015 at 9:18 am

“Okulaer, good work but just to add that the same results can be obtained by considering photon flow.
Both methods, that is classical thermodynamics(your article) and bi-directional photon flow (origin Prevosts theory of exchanges) give the same results when used consistently.”

“The main advantage of photon flow method is to challenge the likes of scienceofdoom who will ask what happens to the returning radiation from the shell ?
‘Is it absorbed by the core?’

Answer – No! No EM flux at any fequency is ever emitted in a direction of higher field strength. There is nothing to absorb. To claim so is a direct contradiction of Maxwell’s equations. There is only flux in the direction of lower field strength and always proportional to the difference in opposing field strengths. Prevost had no photons. Your photon crap never works through any dispersive media like the atmosphere. -will-

• Alex says:

“the two-way flow model breaks down as soon as you introduce a third body to the mix”

Sorry, but you are incorrect. There is not radiative transfer problem in existence where the accounting of energy fluxes from hotter to colder and colder to hotter breaks down. Perhaps you don’t understand how to solve such problems properly, but others do.

However, I have yet to see anyone be able to derive the correct equation for the heat transfer between two gray body plates without in some way explicitly accounting for the energy transferred from the colder to the hotter plates.

I’d like to see you try to do it.

10. Rosco says:

I must agree that I didn’t read your whole article carefully. I have now read it more carefully.

I’ll make some comments citing references and I ask you to provide references citing why they are wrong or confused.

You write –

“Only in the most basic state, surrounded by nothing but a perfect vacuum at absolute zero (Figure 1), will the steady-state radiative heat flux radiated isotropically from the blackbody surface of our sphere into its perfect heat sink (cold reservoir) be precisely determined – through the Stefan-Boltzmann equation – by its own steady-state temperature alone.”

What references can you cite to prove this assertion ?

I think this assertion is false.

None of the texts I read make this claim and therefore I would appreciate a reference as I am always ready to resolve my confusion.

Young and Freedman University Physics simply quotes H = AeσT^4. There is no mention that this relation is valid only in a 0 K environment ! To the contrary they also provide examples using this simple relationship to calculate fluxes.

Further, the Stefan-Boltzmann equation was derived from experimental evidence conducted on Earth at whatever ambient temperature existed at the times measurements were recorded.

It was certainly not derived from experiments in a vacuum at 0 Kelvin.

How do you explain the derivation of their equation from observational data when they had absolutley no method to resolve q(net) except the equation they were striving to derive ?

How do you explain that Planck’s equation produces an emission- versus, say, wavelength curve based solely on the object’s temperature and that the area under this curve times pi equals = σT^4 – the Stefan-Boltzmann equation ?

Does this also only apply in a vacuum at 0 K ?

Also you can go further and plot photon fluxes and observe that what Bryan says is correct –

“However the hotter flow is more intense and includes higher frequency photons.
The net energy flow is the heat transfer and the spontaneous direction is always from hotter to colder.
So heat transfer is unidirectional .”

The vast majority of scientists disagree with your assertion – “Yeah, I would rather drop the idea of a ‘counter-flow’ altogether.”

This in no way implies either point of view is correct – both could be wrong.

And finally I guess you do not believe in expensive precise infra-red thermometers – again cited in Young and Freedman – despite the evidence they seem to work perfectly fine in any ambient temperature environment ?

They simply apply the Stefan-Boltzmann equation to reconcile the emission to temperature – there are many manufacturers listed on the internet who cite the SB equation forms part of the electronic circuitry. They include an emissivity correction factor as well.

If I am wrong in what I have written above then it will be simple for you to cite reputable references highlighting my errors.

If reading these references convinces me that I am “confused” I will happily acknowledge this point.

I will post another problem I see in your discussion soon.

11. okulaer says: January 13, 2015 at 1:12 pm

“Yeah, I would rather drop the idea of a ‘counter-flow’ altogether. And that’s what I try to accomplish. First of all, the bidirectional idea is but a quantum-theoretical conceptual model. It has never experimentally been shown to be a ‘real thing’.”

Quantum Electro Dynamics correctly demonstrates that Thermal EMR flux is never in a direction to lower temperature. See my post to Bryan

Your article has a serious flaw in that you insist of a constant flux (P/m^2) rather than constant P. Any shell must have a greater radius than the sphere, so its area must also be greater. this is the concept of power flux density, which decreases as 1/r^2 for EMR.

• okulaer says:

Hi, Will.

It is not a ‘flaw’. It is just a point I have chosen to ignore in my analysis of the problem, to highlight the actual concepts I want people to focus their attention on.

Indeed, if you account for the larger surface area of the shell, the emissive power (Watts), and thus total energy per unit time (J/s), would be the same at steady state, but not its power density flux (Watts per area). You’re absolutely correct. Flux is not conserved. Energy is.

I’m not trying to hide anything here, Will. Don’t worry. I’m only trying to keep things simple.

• Kristan,
I believe you are trying to explain the “is” of this physical!

I was nitpicking to observe your response. That “Aw Shit” is acceptable to me, and I hope contributes to your explanation. Please right here, explain the ClimAstrologist misuse of “warming”. Scientifically “warming”, or “heating” is transfer of power to, and in some cases, accumulation of that power as sensible heat, expressed as an increase in temperature.

Your ClimAstrologist insist that warming must be increase in temperature. The EMR flux from the sun does little of that. Most solar power is converted to chemical biomass production, and vast latent heat of evaporation. This re- conversion of latent heat to sensible heat in the atmosphere, allows the atmosphere to be a more efficient producer of exitance flux of waste heat than the surface can possibly be. Surface temperature is just hanging about saying “do wat”.

12. Rosco says:

You claim –

“The sphere at 303.3K does NOT emit more than 240 W/m^2 worth of radiation! ”

Again – your claim is at odds with all of the references I can find. Do not think I don’t understand why you make this claim – I completely understand your algebraic manipulations – I just do not agree they are correct.

The references I can find claim the system will emit 240 W/m^2 NET radiation to space if the sphere is at 303.3 K and the shell is at 255 K –

q(net) = σ x ~303.3^4 – σ x 255^4 = ~240 net to space.

This is exactly the same as your equation Th = 4√((q/σ) + Tc4) without the re-arrangement.

There is no argument there really – I simply believe you have not made the case to justify the doubling of energy necessary to prove your claim and increase the temperature of the sphere – that is my point in a nutshell.

The reason I think you have arbitrarily doubled available energy is what I wrote before.

From your starting point of sphere emitting 240 at 255 K in isolation we add the shell which has zero internal energy.

The shell is now absorbing 240 x A Watts from the sphere and this will increase its internal energy and hence raise its temperature.

When it reaches ~214 K it will radiate about 120 x A Watts on each side of the shell because the external surface is facing a cold reservoir at 0 K. At least that is what is claimed for a shell – 120 x A watts over each surface.

120 x 2A joules per second form the 2 surfaces of the shell is equivalent to the emission from the surface of the sphere at 240 X A joules per second.

You would have to agree that seems like a balance of energy right there – surely ?

Every joule per second emitted by the sphere is required to maintain the shell at ~214 K emitting 120 x A joules per second over 2 surfaces.

I have no problem with that relationship and can see how it is balanced :-

q(net) = σ x ~255^4 – σ x 214.5^4 = ~120 net to space.

People who support Willis’ proposal insist that unless the shell is radiating at 240 x A to space energy is accumulating in the system.

Your conclusion is not dissimilar to their position though you dismiss back radiation.

I am guessing you agree with the proposal that another shell – indeed any number of shells – will increase the temperature of the sphere equivalent to an extra 240 W/sqm. per shell ?

13. Rosco says: January 14, 2015 at 2:36 am

“I must agree that I didn’t read your whole article carefully. I have now read it more carefully.”

“I’ll make some comments citing references and I ask you to provide references citing why they are wrong or confused.”

You write –

(“Only in the most basic state, surrounded by nothing but a perfect vacuum at absolute zero (Figure 1), will the steady-state radiative heat flux radiated isotropically from the blackbody surface of our sphere into its perfect heat sink (cold reservoir) be precisely determined – through the Stefan-Boltzmann equation – by its own steady-state temperature alone.”)

“What references can you cite to prove this assertion ? I think this assertion is false.”

The assertation is correct! Nether temperature within parentheses of the S-B equation can be at 0 Kelvin,as that would be rasing an asymtote to the forth power, however, each can approach 0 Kelvin as close as necessary.

“None of the texts I read make this claim and therefore I would appreciate a reference as I am always ready to resolve my confusion.”
“Young and Freedman University Physics simply quotes H = AeσT^4. There is no mention that this relation is valid only in a 0 K environment ! To the contrary they also provide examples using this simple relationship to calculate fluxes.”

“Young and Freedman University Physics” has always been totally at odds with the science of electromagnetic field theory. Please read some text on electromagnetic radiation. No EM flux at any fequency is ever emitted in a direction of higher field strength. To claim such is a direct contradiction of Maxwell’s equations. There is only flux in the direction of lower field strength and always proportional to the difference in opposing field strengths.

14. Rosco says:

My final comment – I promise.

I will await your references however.

Problem numbers 1023 and 1026 from the text “Problems and Solutions on Thermodynamics and Statistical Mechanics (Major American Universities Ph.D. Qualifying Questions and Solutions) Volume 5 Edited by Young-Kuo Lim” deal with radiation emission from a sphere surrounded by a shell.

Problem 1023 is worded thus

“Consider a black sphere of radius R at temperature T which radiates to distant black surroundings at T = 0 K.

(a) Surround the sphere with a nearby heat shield in the form of a black shell whose temperature is determined by radiative equilibrium. What is the temperature of the shell and what is the effect of the shell on the total power radiated to the surroundings ?

(b) How is the total power radiated affected by additional heat shields ?
“Note that this is a crude model of a star surrounded by a dust cloud”.”

The solution they provide is that at radiative equilibrium the temperature of the shell is T/4√2 because the radiative emission has been reduced by half.

This is the solution I proposed by considering energy conservation and not your proposal where the shell MUST equal the original temperature T with the star heating up internally !

Problem 1026 provides a comprehensive set of radiative balance equations and again shows that the radiation emitted is reduced by a factor determined by a relationship involving real radii – again it disagrees with your assertions.

I simply never see any references that support your claims !

I believe you have misapplied the Stefan-Boltzmann equation and invented an interesting text to support your assertions.

If you can provide me with any references that contradict the references I cite I will eagerly read them and if convinced they prove your assertions I will happily acknowledge this !

But for the present I have cited numerous references which do not support your assertions.

All you have provided thus far is your opinion and interpretation which you acknowledge does not confirm to accepted thermodynamics –

“Yeah, I would rather drop the idea of a ‘counter-flow’ altogether. And that’s what I try to accomplish. First of all, the bidirectional idea is but a quantum-theoretical conceptual model.”

You may well be right but you have to overcome a lot of contrary evidence.

Without reputable reference you are merely citing your interpretation.

You may continue to claim I am confused but I have cited several reputable references that disagree with your interpretations – thus I have supporting documentation versus your opinion.

• okulaer says:

Rosco, your comments are a bit unruly (expansive, winding and intricate). So I need some time to take in every little point you’re trying to make.

Meaning, I’m just writing here now to let you know that I will reply, but it will have to be later. Rather busy here in my real life at the moment 🙂

But I appreciate your continued interest. Always interesting to have one’s ideas challenged …

FYI, a couple of years back I was firmly on your side on this topic – the sphere WILL NOT HEAT! I’ve since come to realise that this approach is at odds with reality. The phenomenon is called ‘insulation’ and is a very real physical thing. It works by the same principle whether it’s through convection/evaporation, conduction or radiation – by impeding/reducing heat loss. A thermodynamic system needs to keep its Qout equal to its Qin to maintain its Q (‘net heat’) at zero and hence its ‘internal energy’ (U) and temperature (T) constant. If you increase Qin, then U and T will naturally rise. Likewise, if you reduce Qout, U and T will ALSO rise. It’s that basic. And that’s what’s happening to the sphere.

However, the reduction in Qout could NEVER come about by returning energy to the heated (and insulated) object from a cooler place, the whole “back radiation heating” absurdity. We STILL very much agree on this point. The heat source of the insulated object is ALWAYS its ONLY provider of energy. The cooler insulating layer simply reduces the OUTGOING energy from the warmer object at a certain temperature. The difference between us seems to be that I say there IS no “back radiation” in the sense of a returning FLUX/TRANSFER of energy from the cooler shell, while you (seemingly) say there IS, only it doesn’t do anything.

If you go back to the original discussion thread on this issue at Joe Postma’s website, you’ll find me as the commenter ‘Kristian’:
http://climateofsophistry.com/2013/03/08/the-fraud-of-the-aghe-part-11-quantum-mechanics-the-sheer-stupidity-of-ghe-science-on-wuwt/

Also, Rosco, try to address ‘Greg House’ about your ideas about the bidirectional principle. He basically says the same thing I do, that what we actually observe (detect) is the HEAT, not two opposing fluxes somehow making up a ‘net’ flow of energy we simply call heat. This is nothing but a mental model. It has never been proven by experiment. In fact, it cannot physically be proven experimentally. Because any sensor would only ever detect the heat, there is no way you can create a setup where you would be able to extract one individual ‘hemiflux’ from a heat transfer radiation field, separate from the other one. Even if they both DID exist. You could simply never verify their existence at all. Only assume and claim it.

In other words, in calling the actually observed heat flux the ‘net’ of two unobserved opposing fluxes you’re frankly turning reality on its head. You’re actually making up, inventing reality for yourself. It’s a bit like having a thermometer measuring a temperature of 10C and claiming this temperature is somehow the ‘net’ of two different temperatures, one at 20 and one at 0, both affecting the thermometer, maybe from each side of it, I don’t know. Well, what the thermomenter (nature, basically) actually shows you is the 10 degrees and that’s it. The rest is only in your head.

• okulaer says:

Rosco,

I’ve changed my mind. I will not address all your points. Because they amount to the same misunderstanding of what I’m trying to convey.

Rather, I will try to explain my position as clearly as possible, to finally ask you to consider your own position on this topic.

It is all quite straightforward.

You know of course the 1st Law of Thermodynamics, expressed here for a closed system: ΔU = Q – W.

There is no thermodynamic ‘work’ being done either by or on the central sphere system during our warming process, so we can confidently set W to zero. This leaves us with a very simple equation: ΔU = Q: The change in the sphere’s ‘internal energy’ [U] – and thus its temperature – is solely due to the balance between the ‘heat’ absorbed by the sphere [Qin] and the ‘heat’ emitted by the sphere [Qout]. If they are equal, then the sphere’s ‘net heat’ [Q] is zero and there is no change in ‘internal energy’: ΔU = 0. If, however, the incoming ‘heat’ exceeds the outgoing ‘heat’, then Q will be positive and so naturally will ΔU.

What you need to keep in mind here is that the sphere system is involved in two separate heat transfers at the same time. It gains energy from its nuclear energy source (its hot resevoir), while simultaneously losing energy to its surroundings (its cold reservoir). The energy gain is the Qin. The energy loss is the Qout. The balance between the two is Q:

We know that the sphere’s nuclear energy source provides a constant input of energy to the sphere, from within. It never changes. It remains equivalent to 240 W/m2. This input is effectively the Qin of the sphere. So we see that in order for the sphere to achieve a Q=0, hence a ΔU=0, its Qout needs to balance its Qin. Meaning, the steady state radiative heat flux escaping the sphere to its cold reservoir will have to be 240 W/m2 as well. If the source provides a power input to the sphere of 240 W, we understand that the total area of the sphere’s outer surface will have to be 1 m2 for the flux to be 240 W/m2.

Now, Rosco, the important thing to note here, is that if the sphere’s Qout for some reason were to drop below this steady-state flux density (240 W/m2), then its Q would once again become positive, and its ‘internal energy’ [U] – and accordingly its temperature – would grow as a result. It would grow from the energy (still constantly) supplied by its nuclear energy source (hot reservoir) piling up inside its mass, since the energy can no longer escape as fast from its outer surface as it enters from the core.

This is very basic stuff. It is simply a matter of straightforward energy budgeting.

So, enter the Stefan-Boltzmann radiative heat transfer equation for blackbodies: Q̇ / A = σ[Th4 – Tc4], or simply q = σ[Th4 – Tc4]. (Yes, we will once again simplify the situation by assuming the two opposing surfaces (sphere vs. shell) are so close that neither the inverse square law nor any difference in opposing areas come significantly into play.)

This determines the radiative heat transfer from a hot object to a cold object across a vacuum, or, if there is no opposing object, simply the radiative heat transfer from the hot object.

Now, if you look at the figure above, you will see that q, the radiative heat flux from the surface of the sphere to its surroundings, be it the vacuum of space or (later) the surrounding steel shell, equals the Qout of the sphere. The energy that escapes the sphere as ‘heat’ is the very same energy that is received by the shell as ‘heat’.

So, what happens?

With the surroundings of the sphere being nothing but a perfect vacuum at 0 K [Tc4 = 0] (no shell in place), q simply becomes 240 W/m2 (q = σTh4, q = σ2554).

With the shell up around the sphere, however, the surroundings will no longer stay at 0 K. They will warm. From having a thermal mass and absorbing the radiative heat flux from the sphere.

What, then, happens to q when Tc rises to say 100K? It is reduced: q = σ[Th4 – Tc4], q = σ[2554 – 1004], q = 234 W/m2.

Rosco, this is how the heat transfer equation works.

As the surroundings of the sphere warm, the radiative heat flux (q) escaping its surface is naturally diminished. This is the work of an insulating layer.

As q as we know equals the sphere’s Qout, we now have a situation where (the constant) Qin to the sphere from its nuclear energy source at its core (240 W/m2) is once more larger than its Qout (234 W/m2). Its Q thus goes positive again and its U (and T) will necessarily rise as a result.

It is up to you to decide, then, Rosco, whether you believe – in your bidirectional flow mindset – that it is the increased “back radiation” flux resulting from the higher Tc (the shell) opposing the sphere surface, or if it is simply a matter of less energy per unit time moving out from the surface of the sphere in the first place, causing the natural warming of the sphere to make its Qout restore the balance with its Qin.

I’m afraid you cannot avoid the further warming of the sphere, Rosco. The question is, is it returning energy from the cooler insulating layer doing the extra warming, or is it the opposing temperature potential of the insulating layer reducing the energy escaping the central object to begin with?

• Rosco says: .January 14, 2015 at 3:40 am My final comment – I promise.

“Problem numbers 1023 and 1026 from the text “Problems and Solutions on Thermodynamics and Statistical Mechanics (Major American Universities Ph.D. Qualifying Questions and Solutions) Volume 5 Edited by Young-Kuo Lim” deal with radiation emission from a sphere surrounded by a shell.”

Problem 1023 is worded thus

“Consider a black sphere of radius R at temperature T which radiates to distant black surroundings at T = 0 K.”

For any temperature T to be maintained, there must be a power source that will sustain temperature T, with emitting area 4PI R^2, 100% emissivity radiating in all directions of no opposing radiance. Just what is that power? and how is that power supplied?

“(a) Surround the sphere with a nearby heat shield in the form of a black shell whose temperature is determined by radiative equilibrium. What is the temperature of the shell and what is the effect of the shell on the total power radiated to the surroundings ?”

This entirely depends on how the power source responds to its increase in temperature. The power dissipated to surroundings must be identical to that power generated.

“(b) How is the total power radiated affected by additional heat shields ?
“Note that this is a crude model of a star surrounded by a dust cloud”.”

The power dissipated to surroundings must be identical to that power generated.

The solution they provide is that at radiative equilibrium the temperature of the shell is T/4√2 because the radiative emission has been reduced by half.

This is inherent in the property that the rethermalization of thermal EMR flux is equivalent to a 3db reduction of antenna gain, of the source, or the reduction to one half of any surface emissivity of the source. This requires an increase in temperature of the source to retain the very same power exitance.

15. Bryan says:

Rosco quotes from Young and Freedman University Physics.

This is the most widely used Physics textbbook for ist and 2nd year physics and maths students at the worlds universities .
So we have to deal with the interested intelligent readers who use this or a similar source for their physics education.

okulaer gets the same results as Willis but with a completely different interpretation.

We cannot ignore the fact that physics has moved on since Maxwell.
Quantum mechanics is used to answer for situations where classical mechanics is found wanting.

For EM radiation > 10um both classical and QM approaches give the same answers
For EM radiation < QM must increasingly be used for answers that match reality

Of course radiation is only one of four methods of heat transfer and there is justifiable reason to think that IPCC 'science' unscrupulously use the results of cavity radiation well beyond its theoretical justification.

All this to show that for a superconducting model of the Earth the solar radiation through the greenhouse effect can raise the surface temperature by about 20K .
If you do the maths it would take longer than the age of the universe to achieve this

• Rosco says:

I also quote another reference which gives 2 problems dealing specifically with concentric spheres exactly like the Steel Greenhouse proposition.

The people at WUWT urged on by Robert Brown claimed these are not the same problem because the problem did not specifically say there is an internal energy source and thus continued promoting their “pet theory” without thinking.

Unlike the authors who set these PhD qualifying problem sets – American Universities use these books covering all sciences – the people at WUWT urged on by Robert Brown did not have the intelligence to realise one simple fact in their pointless criticism of standard science –

If an object in 0 Kelvin space is at a constant temperature IT MUST have an internal energy source !!

Otherwise it could not be at a constant temperature !

Only someone who isn’t thinking straight requires that the “internal energy source” MUST be explicitly stated in the problem.

The authors of such texts do not explicitly state such obvious basic necessities because the text is aimed at PhD students not high school.

The whole construct in this post relies on 2 propositions

1. Does H = AeσT^4 ONLY apply in a 0 Kelvin vacuum environment ?

I think most – if not all -science references do not support this.

If it were true then expensive IR sensing equipment is useless as there is no way to “zero” it.

And of course it is impossible to even derive the SB equation in the first place.

2. Does q = σ x Th^4 – σ x Tc^4 represent a “real” flux subject to algebraic manipulation ?

Or is it really q(net) as stated in ALL the texts – not just Young and Freedman !

In the equation ALL the texts say the terms σ x T^4 are real fluxes.

And q is q(net) and they always state this.

If these are true then two of the fundamental pillars supporting the author’s assertions collapse.

I really don’t know – after all we all are simply students relying on accumulated knowledge – even those who create truly stunning advances.

The author of this may well be right but he needs to demonstrate that points 1. and 2. above are mistaken.

I personally don’t think they are mistaken and I don’t think the author has made his case.

My beliefs do not mean I support ANY “back radiation” heating capability – I think this claim is nonsense unless the source of any such radiation is hotter than the receiver and in this case it isn’t really “back radiation” at all is it ?

I am not sure about “photons” but they provide some useful terminology to talk about phenomena.

It is a well established fact that the photoelectric effect has a threshold energy level below which you can bombard a photoelectric material with huge quantities of energy but not one single electron is stimulated enough to be ejected and create an electric current.

If this is true then there is a threshold energy level below which the number of “photons” cannot cause increases in energy levels because unless the “photons” have a higher energy level which is determined by frequency or wavelength alone – not the number – they can only replace photons of the same energy level already radiating away and hence do not increase temperature.

Logically such a threshold level in thermodynamics is the temperature.

Almost ALL texts books say this in a roundabout way as they need to sell. None ever explicitly say this.

Climate science has so politicised and corrupted academia that any Academic who stands up and challenges the standard dogma is committing career suicide.

However I could be completely mistaken.

• Bryan says: January 14, 2015 at 10:28 am

“Rosco quotes from Young and Freedman University Physics.This is the most widely used Physics textbbook for ist and 2nd year physics and maths students at the worlds universities ”

Written by incompetent academics published via fraud, at exorbitant prices to the inocent, as a method. of promoting PRIDE, and INCOMPETENCE.

“So we have to deal with the interested intelligent readers who use this or a similar source for their physics education.”

Do we do that by also promoting the incompetence?

“okulaer gets the same results as Willis but with a completely different interpretation.”

So? Both are the result of Sloth, neither exhibit due diligence!.

“We cannot ignore the fact that physics has moved on since Maxwell. Quantum mechanics is used to answer for situations where classical mechanics is found wanting.”

Quantum mechanics, like statistical mechanics, attempts to describe “all”: in terms of Newtonian mechanics. Nothing in this fantasy concerns the electromagnetic, not even mass. If you take your auto to a quantum mechanic for repair, you will likely get back a Camel.

“For EM radiation > 10um both classical and QM approaches give the same answers
For EM radiation < QM must increasingly be used for answers that match reality"

What reality, that only in your fantasy! At all frequencies, Maxwell's equations still describe all that has ever been observed, detected, or measured.

"Of course radiation is only one of four methods of heat transfer and there is justifiable reason to think that IPCC 'science' unscrupulously use the results of cavity radiation well beyond its theoretical justification."

Please demonstrate any cavity radiative flux within an isothermic cavity.! So far there are only two methods of "heat" transfer, conduction, and convection. both require mass and specific heat. Please describe the other two power transfers that you claim, and how do they relate to "heat"? Remember minus "heat" is the conjugate of temperature!

"All this to show that for a superconducting model of the Earth the solar radiation through the greenhouse effect can raise the surface temperature by about 20K . "If you do the maths it would take longer than the age of the universe to achieve this"

If you believe in some "model", that model must theoretically show your belief. What has that to do with this physical? This physical is so very dynamic, that it can never be modeled, it simply is, internal or external to your fantasy!

16. Bryan says:

Will Janoschka says;

“What reality, that only in your fantasy! At all frequencies, Maxwell’s equations still describe all that has ever been observed, detected, or measured.”

So for you new Physics stopped around 1900.

You have a lot of catching up to do!

Rather than overload you, read up about the photoelectric effect then we can move on to some more recent work.

http://en.wikipedia.org/wiki/Photoelectric_effect

• Bryan says: January 14, 2015 at 10:06 pm
Will Janoschka says;

(“What reality, that only in your fantasy! At all frequencies, Maxwell’s equations still describe all that has ever been observed, detected, or measured.”)

”So for you new Physics stopped around 1900.You have a lot of catching up to do!”

There is no new physics, there is only vast new stupidity, all created by the insane concept of “photon”.

“Rather than overload you, read up about the photoelectric effect then we can move on to some more recent work. http://en.wikipedia.org/wiki/Photoelectric_effect

Dr. Einstein discovered the photoelectric effect. The complex conjugate of all things mechanical. Rather than the stupid integral of power over time to energy. A true conflict to force (power) over distance to work, never energy! The integral of work over time is called action (a construct). Power^2 within a time interval is called a quantum. Sometimes sufficient to create an “event”, the emission of an electron from a photocathode.

I do not know, I must severely polk at your claim of knowledge, kiddie!

• Bryan says:

Will says

“Dr. Einstein discovered the photoelectric effect. ”

Its not about what Einstein said or did not say.

The Photoelectric effect is a real physical phenomena unlike the pseudo Greenhouse Effect which is no effect at all.

Electrons will be liberated from a clean zinc plate only if the incident EM wave frequency exceeds a certain value.
Classical EM theory (Maxwell) would predict a higher intensity lower frequency wave should also work.
It does not.

Here is a long list of real physical phenomena where classical physics fails to give any explanation.

You have a lot of ground to catch up Rip Van Winkle.

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

• Sorry Kristian, But I must!!!

Bryan says: January 15, 2015 at 9:23 am

“You have a lot of ground to catch up Rip Van Winkle”

This Rip Van Winkle is not interested at all in your fantasy Bull Shit!!!

(Will says “Dr. Einstein discovered the photoelectric effect. ”)

“Its not about what Einstein said or did not say. Its about a physical effect.”

Ah! Such arrogance, Byran gets to decide what this thread is about This thread is about the the nonsense of Willis Echenbach! Perhaps you are, or are like, other deluded University lecturers such as Drs. Joel Shore, Tim Folkerts, or Robert Brown. From your comments, it is evident that your University education is but a scam and qualifies you only for the position of University lecturer, to continue the scam of Universities and textbook publishers. Perhaps with some on the job training you would qualify as janitor at the same “institutuion of higher learning”.

“The Photoelectric effect is a real physical phenomena unlike the pseudo Greenhouse Effect which is no effect at all.”

Indeed Dr. Einstein was no dummy. Today such effect is called the “work function” or “action” of the metal or other material. With low temperature semiconductor surfaces the dislocation of a lattice defect can be observed down to your fake energy of 0.05 ev (20 microns). This in no way demonstrates that an EM wavelet ever acts like a particle until such is absorbed within a particular time interval and the action becomes evident. Such evidence has no indication of anything prior to absorption.

“Electrons will be liberated from a clean zinc plate only if the incident EM wave frequency exceeds a certain value.Classical EM theory (Maxwell) would predict a higher intensity lower frequency wave should also work. It does not.”

Which of the 22 equations of Maxwel,l predict such a thing? Which of the 4 translations to vector geometry by J. Poynting predict such a thing? Have you even read and understand any one of those 26 equations? Please demonstrate your complete lack of understanding of any electromagnetic phenomena! Your claim of thermal electromagnetic flux being emitted in the direction of higher temperature (higher field strength) verify your lack of knowledge (stupidity.).

“Here is a long list of real physical phenomena where classical physics fails to give any explanation.” http://hyperphysics.phy-astr.gsu.edu/hbase/qapp.html#c3

None of those in your list show any failure of the prior, to explain the optical properties of matter..
Your Quantum mechanical POV is interesting, but remains fantasy as never observed or demonstrated! Probabilities remain probabilities. Evidence please? Produce a photon at any frequency, so the rest of us may measure your fantasy. Your “sub atomic particles” may exist. It will take QED to demonstrate that they be the complex conjugate of what we observe. -will-

17. Kristan,
I enjoy your blog. No political correctness, but much wrangling about different opinions.
More formally this would be your position in the Colosseum, with your big guy that will determine what is! Truly exciting! The rest of us are sitting back and polishing the points of the pitchforks!

18. Rosco says: January 14, 2015 at 3:40 am My final comment – I promise.
I will await your references however.

“Problem numbers 1023 and 1026 from the text “Problems and Solutions on Thermodynamics and Statistical Mechanics (Major American Universities Ph.D. Qualifying Questions and Solutions) Volume 5 Edited by Young-Kuo Lim” deal with radiation emission from a sphere surrounded by a shell. Problem 1023 is worded thus”

“Consider a black sphere of radius R at temperature T which radiates to distant black surroundings at T = 0 K.”

OK for near 0 Kelvin

“(a) Surround the sphere with a nearby heat shield in the form of a black shell whose temperature is determined by radiative equilibrium. What is the temperature of the shell and what is the effect of the shell on the total power radiated to the surroundings ?”

If your illusionary shell has precisely the same surface area as your sphere the temperature of the shell must be identical to that of the original sphere temperature. Because of the rethermalization at the shell, the new temperature of the sphere must be 2^(0.25) of that original. All flux is identical outward.

“(b) How is the total power radiated affected by additional heat shields ?
“Note that this is a crude model of a star surrounded by a dust cloud”.”

The total power radiated must be exactly the power generated else all temperatures must spontaneously readjust, to make it so. Captain Picard..

“The solution they provide is that at radiative equilibrium the temperature of the shell is T/4√2 because the radiative emission has been reduced by half.”

What total nonsense, The “effective” emissivity of the source must be reduced to one half, because of rethermalization of the shell resulting in the increased temperature of the sphere.

-snip- much more Bull Shit

19. Truthseeker says:

Okulaer,

i think I believe I have (with help) worked out why your analysis is wrong.

There is a heat flow you have not included. That is the heat flow from the Shell to the external vacuum which is effectively an infinite heat sink. The effect of this heat flow will be to continually reduce the temperature of the Shell.

The other missing component is that the energy required to increase the temperature of an object increases as the temperature increases. It takes more energy to raise the temperature of an object from 100K to 101K than it does to raise it from 0K to 1K. What this means is the energy flow from the Sphere will not be restricted in any way. Even when the two objects reach an equilibrium state, it will take all of the energy flow from the Sphere to keep the Shell at that equilibrium temperature (which will of course be lower than the temperature of the Sphere) because of the infinite heat sink that the Shell flows energy to throughout this process.

The presence of the Shell does not affect the temperature of the Sphere.

• okulaer says:

Truthseeker,

I must say, I do wonder who’s that helper of yours.

You say: “There is a heat flow you have not included. That is the heat flow from the Shell to the external vacuum which is effectively an infinite heat sink. The effect of this heat flow will be to continually reduce the temperature of the Shell.”

Have I? Have I really forgotten about it? Then what does that top arrow on the figure below represent?

You seem to have missed the whole analysis, Truthseeker. How the shell’s radiative heat flux to space grows larger as it gets warmer, thus progressively making it harder for it to warm as fast as before. Until we reach dynamic equlibrium and as much energy escapes from the outside to space as heat as what comes in as heat from the sphere on the inside, hence no more warming, no further accumulation of energy.

How did you all of a sudden miss this? It’s the whole point of my post.

You also say: “The other missing component is that the energy required to increase the temperature of an object increases as the temperature increases. It takes more energy to raise the temperature of an object from 100K to 101K than it does to raise it from 0K to 1K.”

Does it indeed? How did you figure that out? How does that work?

Listen, as long as the ‘heat capacity’ of the object remains unchanged, then raising its temperature by 1 degree will of course always require the same amount of energy, if it goes from 0 to 1K or from 100 to 101K. What does change is the time it takes for the object to warm by one full degree. It takes more time for it to go from 100 to 101K than for it to go from 0 to 1K. As long as the energy input stays constant. Why? Because the object’s simultaneous energy output is naturally larger at 100K than at 0.

“What this means is the energy flow from the Sphere will not be restricted in any way.”

Huh!? How did you get from there to here?

https://okulaer.wordpress.com/2015/01/11/to-heat-a-planetary-surface-for-dummies-part-2/comment-page-1/#comment-161

Hopefully you will understand why the sphere will have to warm with the shell insulating it and its nuclear energy source constantly supplying it with ‘new’ energy. What Rosco (and Postma) forgets is that there are two heat transfers going on at the same time, not just one. They ignore the constant input from the nuclear energy source to the sphere and then feel they can freely let the heat flux from sphere to shell reduce all the way down to zero and establish a dynamic equilibrium that way. But what this means is simply that the sphere would have an equivalent incoming heat (power, from its source) of 240 W and an outgoing heat (power, to its sink) of 0 W. This would leave the ‘net heat’ (Q) to the sphere at 240 W. None of the energy constantly entering is able to escape the sphere as heat, so all of it would necessarily accumulate inside the sphere, 240 joules per second.

That’s thermodynamics, Truthseeker. Q_in vs. Q_out. ΔU. T.

• Truthseeker says:

Okulaer,

You have made the same mistake that Willis made. You are conserving energy flux, not energy.

If the Sphere is emitting 240 W/m2, the Shell is receiving less than that as a flux because the m2 is bigger. Same energy over greater area equals less flux.

Lets put some numbers on this.

The surface area of a sphere is 4 x pi x r^2. So if the sphere has a radius of 10m then the surface area of the sphere is 4 x 3.1416 x 10 x 10 = 1256.64 m2. At 240 W/m2 the sphere has a total energy output of 301593.6 Watts (joules per second). Now let’s say that the shell is 10m from the sphere. The surface area of the shell is 4 x 3.1416 x 20 x 20 = 5026.56 m2. Dividing the 301593.6 Watts that the sphere is producing by the surface area of the shell we get an incoming flux of 60 W/m2. So your diagram is fundamentally wrong. The Shell never receives 240 W/m2. The Shell also emits 60 W/m2. So it needs the 60 W/m2 incoming flux from the Sphere just to maintain the equilibrium temperature.

Your own diagram has magically multiplied the energy involved by a factor of 4.

You seem to need more help than I do.

• okulaer says:

Truthseeker,

The area of an insulating layer is of no importance to this layer’s basic insulating ability. If the surface area of the insulating layer is larger than the surface area of the object being insulated, this doesn’t change the fact that the presence of the insulating layer around the insulated object reduces the power output (energy per unit time) of the insulated object (from having a temperature, which space hasn’t got) and hence – with a power source constantly providing the insulated object with energy from within – forces it to warm: Q_in > Q_out; naturally leads to +U; naturally leads to +T.

Trying to bring in ‘area’ to somehow invalidate my analysis just shows you haven’t understood my argument in the least. Or that you don’t want to understand …

• Truthseeker says:

Okulaer,

The internal vacuum has a much greater insulating ability than the Shell. This is because it restricts the emission of energy to radiation only which is a very poor transfer mechanism. The Shell does not ADD to this insulating ability as you call it. The temperature of the Sphere is dictated by the energy source, the mass of the Sphere and the emission characteristics of the Sphere itself. That’s it.

You talk about energy flows and then do not do the basic math to calculate them correctly.

It is you that is not understanding what is really going on.

• okulaer says:

I’m interested to know, how did you get from ‘clarity’ to such confusion, Truthseeker …?

• eilert says:

Like Truthseeker I also have a problem with the calculation of the increase of the temperature of the sphere, surrounded by a shell, which in turn emits into a vacuum. In my mind the temperature increase is only valid if the shell has its own energy source or is in contact with the sphere, but is emitting at a different, slower rate, as the sphere would have done. The shell would then become an insulator to the sphere.

The general heat flow equation: q = σ (Th^4 – Tc^4 )

For the specific case where the sphere emits into the vacuum this formula can be written as:
qsp = σ (Th^4 – 0^4) (0 = 0 kelvin temperature of vacuum)

For the case where the sphere emits into the shell and the shell into the vacuum, we have the same overall flux, since the energy to the sphere did not increase, but two fluxes, which can be written as:
qsp = qss + qs0
where qss flux from sphere to shell: qss = σ (Th^4 – Tc^4 )
and qs0 flux from shell to vacuum: qs0 = σ (Tc^4 – 0^4 )

Thus as the sphere heats the shell the flux qss decreases, but the flux qs0 from the shell to the vacuum increases. Eventually qss = 0, if the sphere and the shell are at the same temperature and the flux from the shell to the vacuum qs0 = qsp.

Is my logic correct here or did I miss something.

• eilert says:

I meant ‘the Power to the sphere did not increase’

• okulaer says:

Hi, Eilert.

I appreciate your take on this, but no, your logic is not correct. You mix two separate heat transfers into one operation. The sphere’s Q_out needs to balance its Q_in from the internal power source (just like the shell’s Q_out (to space) needs to balance its Q_in (from the sphere) … at dynamic equilibrium, not before), otherwise more energy would enter it per unit time than what escapes, and it would necessarily warm (its U would increase). Since the Q_in is always constant, then that means the Q_out must remain constant as well. If there is to be no further change in temperature.

So when the shell warms, by the Stefan-Boltzmann radiative heat transfer equation, if the temp of the sphere is kept constant, then q (the heat flux from sphere to shell) will decrease. And since the heat flux between the sphere and the shell is equal to the sphere’s Q_out, this means the sphere’s Q_out becomes smaller.

We have a situation, then, where the sphere’s Q_in from its internal power source remains the same as always, but where its Q_out to the shell is reduced: Q_in > Q_out = +Q = +U = +T. The sphere’s surface temperature MUST rise also.

It is that basic. That’s how it works when you insulate a heated object.

I will very soon publish Part 3 of this series where my position on this will (hopefully) be made clear …

Cheers 🙂

20. PTruthseeker says:
January 16, 2015 at 5:57 am

Okulaer,

“i think I believe I have (with help) worked out why your analysis is wrong.”

The analysis of Kristian, is somewhat sloppy but not incorrect. Please consider that your “heat flow” is but a power transfer.

“There is a heat flow you have not included. That is the heat flow from the Shell to the external vacuum which is effectively an infinite heat sink. The effect of this heat flow will be to continually reduce the temperature of the Shell”

The EM power transfer to space is quite limited. If that power is replaced from the sphere, no reduction of temperature of shell can occur. All power is produced from the sphere. The shell is but a place keeper at constant temperature transferring the power from the sphere to else with very low radiance.

“The other missing component is that the energy required to increase the temperature of an object increases as the temperature increases. It takes more energy to raise the temperature of an object from 100K to 101K than it does to raise it from 0K to 1K.”

So what? If all is but power, not energy, transfer at the spontaneous temperatures required for such power transfer.

‘What this means is the energy flow from the Sphere will not be restricted in any way.”

Huh! The radiative power transfer. is always limited by the radiance of the acceptor, (shell).

“Even when the two objects reach an equilibrium state, it will take all of the energy flow from the Sphere to keep the Shell at that equilibrium temperature (which will of course be lower than the temperature of the Sphere) because of the infinite heat sink that the Shell flows energy to throughout this process.”

This is correct! Please replace energy flow (rate) by power, or flux!

“The presence of the Shell does not affect the temperature of the Sphere”

What nonsense. The rethermalization of the shell must increase the temperature of the sphere
for transfer of such power or flux as per the Stefan Boltzmann equation.

21. I like the name OKulaer,
When referring to the who of your blog (God that controls nothing but this blog), Please pick Ku or Kr.
For present I shall use Kr, K (itself) is always reserved by Lord Kelvin as indicating 0K as an asymptotic value far far away from newly defined 1K. Lotsa problems there as you have experienced.
{rant} Other Post normal nonsense is claiming that sensible heat, latent heat, and electrical as the catchall “”” internal energy”””. Each has been well known as distinct for 200 years as having selective properties. Actually each should be referenced as some accumulation of something measurable but requires no integration over something orthogonal. Count the rocks in a pile of rocks for a magnitude of rocks. Time to count is nonsense, however may make counting difficult. This is called variance. Shitcan that stupid concept of integrating, “over a time interval”, that may not exist.{/rant}

Your explanation is fine, but misses your true concept, that mixing unknown symbols, with, addition, multiplication, and the raising of something to power (multiplying self by a non integer of times, God^(2^0.5) must be incorrect, please express yourself in some understandable manner
This algebraic symbolic nonsense is called “post normal science”.. If the expected result requires actual numeric result, “numbers” not symbols must be used. Even a guess is much better than a symbol,Eingineering vs scientology. Dry-lab from symbols to numbers, measure, AW shit, explain the difference! This is known as learning. Learning does not come from books, or from professors. I shall return to read Joe Postma’s blog most reluctantly!

• okulaer says:

Will, if you wonder about the concept of ‘internal energy’, read for instance my Part 1:
https://okulaer.wordpress.com/2015/01/02/to-heat-a-planetary-surface-for-dummies-part-1/

‘Internal energy’ is all the energy kept microscopically inside an object or a system, not all of it necessarily affecting its temperature directly. But only that part that does affect a system’s temperature is naturally relevant to the analysis of temperature change upon energy transfer and the resulting change in ‘internal energy’.

Transferring energy to a system as ‘heat’ or as ‘work’ basically has the exact same effect on that system, raising its temperature. Once the energy is inside the system, however, you cannot tell wherefrom or from what process it came. And all the energy transferred doesn’t necessarily translate into a rise in temperature, even though all the energy transferred indeed does translate into a rise in ‘internal energy’.

• Kristian,
I have no problem with the concept all “powers or forces” non attributable to mass or gravitational field. However it is the aggregation of all these “all” into your “internal” energy, that you claim, must be specific heat, converted to sensible heat and temperature! No storage of chemical, or electrical power need have any relation to temperature. ;

“Once the energy is inside the system, however, you cannot tell wherefrom or from what process it came. And all the energy transferred doesn’t necessarily translate into a rise in temperature, even though all the energy transferred indeed does translate into a rise in ‘internal energy’”

What nonsense, You claim the accumulation (storage), of sensible, latent, chemical, and electrical “power”. needs some amorphous and no identifiable difference between the members of such accumulation, as some “internal” energy. Please distinguish your concept as to.physical as opposed to fantasy?

Offside:
Have you determined “why” Postma’s claims are BS? Postma insists that his useful symbolic algebra describes and is identical to the arithmetic of this physical.

• okulaer says:

Will, I simply can’t grasp what you’re trying to say. You need to be a bit clearer. You introduce some chemical and electrical energy plus ‘latent heat’ into the mix that isn’t even part of the problem analysis in the first place. Electrical energy, for instance, if it doesn’t just heat directly, it will normally do some kind of work, but even so, there will always be some amount of residual warming, meaning the internal energy KE of the system powered has increased and so, accordingly, has its temperature.

Btw, the ‘internal energy’ concept [U] is not mine. It is one of the fundamentals of Thermodynamics. Like ‘heat’ [Q], like ‘work’ [W], like ‘entropy’ [S], like ‘enthalpy’ [H].

• Kristian,
“Btw, the ‘internal energy’ concept [U] is not mine. It is one of the fundamentals of Thermodynamics. Like ‘heat’ [Q], like ‘work’ [W], like ‘entropy’ [S], like ‘enthalpy’ [H].”

Indeed U is a construct created by post modern academics to confuse. It was an attempt to correct the problem of entropy [S] when radiative power exchange was introduced. Such only created many more problems for Physics students. It failed miserably. Such is not taught in engineering, where all must struggle through entropy, for understanding. Your [U] has the numerical value of [S] divided by the temperature of that [S]. Your [U] is but the specific heat of any mass times its own temperature, one of many many form of storage of power this one destroying power unless a lower temperature sink is available.. Storage of power is never equivalent work [W], which is the conversion of power to structure over time. (one nail at a time).

Will, I simply can’t grasp what you’re trying to say. You need to be a bit clearer. You introduce some chemical and electrical energy plus ‘latent heat’ into the mix that isn’t even part of the problem analysis in the first place.

Kristian, If you can even say that, that saying is the ‘problem’ to be analyzed. Some conversions of power storage to another are reversible if no power is lost. Such is a revisable process. The most irreversible process is construction [W]. Construction requires even more power to be lost to entropy, (the temporary storage. of lost power), that some claim is needed for the conservation of something), an accounting game having no significance.
Most efficient conversion of one means of power storage to another, is accompanied by some loss of that power. That power is collected in the lowest temperature mass. Come along thermal EMR and even that power is transferred in a direction of lower radiance. What is conserved?

If you can conceptualize this ‘problem’, perhaps I can be more clear. Have you determined “why” Postma’s claims are BS? Postma insists that his useful symbolic algebra describes and is identical to the arithmetic of this physical. Hint The improper mixing of vector symbols with scalar symbols, resulting in only Not A Number!

• Such is a revisable process.
Sorry Such is a reversible process. 😕 words suck

• “Construction requires even more power to be lost to entropy,”
Mia-culpa Destruction requires even more power to be lost to entropy, Damn words