The Knowledge of Photons

We often see the statement from climate lukewarmers and alarmists that the radiation from a cool object (such as the atmosphere) can not “know” that it is not supposed to travel to and heat up a warmer object (such as the surface), and thus, radiation from a colder object will heat up a warmer object.  That is, the colder atmosphere must heat the warmer surface.

Of course, this defies all common sense and heat transfer mathematics and thermodynamics, but alas, it is what they say.  They use this “net flow” argument, where cold heats up hot and hot heats up cold, but the “net heating” is hot heating cold since the hotter heats the colder by a larger amount.  But simply look at the logic: if cold heats hot, then as hot heats cold, the colder will heat up the hotter more, thus heating the colder more, thus heating the hotter more…ad infinitum.  So, it’s ridiculous.  And of course, heat does not flow from cold to hot in any case, and heat flow input is what is required for temperature increase.

Look at the 1st Law of Thermodynamics:

First law of thermodynamics: When energy passes, as work, as heat, or with matter, into or out from a system, its internal energy changes in accord with the law of conservation of energy. Equivalently, perpetual motion machines of the first kind are impossible.

So, internal energy in the thermal case would be thermal energy measurable as the system’s temperature.  To increase temperature when the action of work or exchange of matter are not occurring, then one requires heat.  Thus for the electromagnetic case, look at the radiative heat flow equation (for 2 parallel walls with unit emissivity say):

Q’ = A*σ*(Thot4 – Tcool4)

This defines Q’ as the heat.  There is a hot and cool term, and there is an exchange of energy between them since they are subtracting from one to the other; however, only that result after subtraction is heat.  Only Q’ is heat.  The radiation from the cool object to the hot object is not heat, and only the greater portion of the radiation from the hot object relative to the cool object is heat, and it transfers or flows only in the direction from the hot object to the cool object, from the greater power to the lesser power.

Since these are very basic laws of thermodynamics and heat transfer, then it has been totally established that a cool object doesn’t make a warmer object warmer still.  And hence the radiative greenhouse effect of climate alarm, and climate alarm which depends upon it, is false.  And we didn’t have to look at any pretty pictures to get here…simply math and logic.  If you want a different result than that, then you’re going to have to change the 1st Law of Thermodynamics.  Good luck with that.

Speed of Light

The lukewarmers and alarmists will still continue to wonder why the radiation from the cool object doesn’t warm the warmer object, and although this can be demonstrated via the above analysis based on first principles of physics, it doesn’t seem sufficient for them and they simply take their dissatisfaction with the answer as justification that they should continue to believe that cold can heat hot.  They ask, still:  How does the photon from the cold object “know” not to heat the warmer object?

It is a strange question, because we typically do not ascribe mental functions such as “knowing” to photons or other phenomena in physics.  The mechanism of what occurs is described by the mathematics, and by adherence to the laws of physics, and we don’t ask, for example, how a planet “knows” to obey gravity.  The question is of course incoherent, because we don’t ask for the phenomenon to have to “know” what it is doing – instead, the mathematics and physics explains what happens.  Thus, since it is not possible to answer how a scientific phenomenon knows how to do anything, because the question is scientifically incoherent, then apparently they believe that this gives them the freedom to believe something which the mathematics and physics otherwise doesn’t support, since the mathematics doesn’t tell us how it “knows” to be as it is.

A related question that they should ask themselves, is how a photon “knows” to travel at the speed of light with no dependence on the reference frame of measurement?  How does a photon know how to do that?  We have the equations based on the phenomenon of light speed invariance, just as we have the equations for heat flow, but in both cases, how do photons “know” how to travel at the speed of light, and how do they “know” to transfer heat only from hot to cold?  It seems that indeed, photons know how to do things.  Like behave mathematically.  We have the mathematics that describes what they do, but, why do they do what they do?  How do they know?

Photon Thought

The problem is that people don’t think like how a photon thinks nearly enough.  People think like humans, with anthropocentric biases…particularly with non-relativistic anthropocentric biases!  If you want to know why a photon does something, maybe you should imagine the universe in the way a photon would see the universe, rather than the way you see the universe.  So how would you do this?

We know how do this.  We all know about the spacetime modifications that a material object witnesses when it travels near the speed of light – time slows down and lengths contract.  Of course, a material object never can actually get to the speed of light itself.  However, a photon is at the speed of light.  So what happens to the experience of spacetime when you actually do travel at the speed of light, like a photon does?

Relativity already has the answer: the Lorentz transformations show that time stops for a photon, i.e. time is dilated to infinity, and they show that all space distance has shrunk to zero, i.e. length contraction is infinite.

So for a photon, there is no experience of spacetime at all…  At least not nearly in the way a human usually experiences it.

We think of a photon as travelling ‘from here to there’, but this is a totally anthropocentric material-based non-relativistic conception, and hence has no ontological validity whatsoever in terms of a statement of how reality functions.  You have to go with what the mathematics and physics says, not what your material-based sense-perception would make you think. Photons, experiencing spacetime as infinitely length contracted and infinitely time dilated, are totally outside the anthropocentric experience of spacetime.

From the photon’s point of view there is neither any time nor any distance between where it originates in spacetime, originating say from thermal emission from a surface, and its destination at some other surface which in human experience is at some extended distance, but in the photon’s experience for which that distance is shrunk to zero, and requiring no time to get there.

Heat of the Gaps

Willis Eschenbach’s steel greenhouse concept gets to the heart of this matter.  If we take the original sphere with some central internal energy source, then we have a basic diagram like this:

A sphere of radius r with internal power source X will have a surface flux of F, as indicated in the diagram.  Now, what happens when you add another layer on this sphere?

All that happens is that you increase the sphere’s radius – you make a bigger sphere – resulting in a reduction of the sphere’s surface flux F relative to what it was before given a fixed P.  This means a lower surface temperature.  Adding a new layer to the sphere is the same thing as simply having the original sphere with a larger radius.  A sphere with a larger radius won’t have a higher internal or exterior temperature just because it is a larger sphere, and likewise adding a layer to smaller sphere won’t cause the smaller sphere to be warmer either internally or on its exterior.  This is of course all about conductive heat energy transfer.

Now imagine that you divide a layer from the existing sphere and begin expanding this layer outwards, such as to create a vacuum gap between an inner sphere and surrounding shell, as shown in cut-out in the next diagram.  Or equally, simply bring in a shell from elsewhere to surround the sphere:The energy from the sphere, in the form of electromagnetic waves from surface thermal emission, or photons, now has to cross the gap in order for the energy to transfer to the shell.  From a photon’s point of view, that gap is length-contracted to zero distance (and in fact the entire shell and everything else in the universe is length contracted to zero).

Now in conductive transfer of heat energy we imagine a physical or mechanical propagation of energy.  Conductive transfer is, however, entirely mediated by the electromagnetic force because this transfer occurs via the electromagnetic forces of the electron clouds surrounding the constituent atoms.  This exchange of force between “touching atoms” is mediated by virtual photons.

So you see the connection here?  There is no real ontological difference between conductive transfer of heat energy and radiative transfer of heat energy; radiative transfer of heat is simply conductive transfer but occurring at what we experience as a distance. They are mediated by the exact same force of electromagnetism.  From the photon’s point of view however, there is no distance and no time required to travel that distance at all.

Therefore, since making a smaller sphere with a central power source into a bigger sphere doesn’t increase the sphere’s temperature, then putting a shell around the sphere doesn’t increase the sphere’s temperature either, because ontologically, from the point of view of the mediator of the energy, the photon, there is no distance.  So, if you want to know “why” or “how” an individual photon knows not to send heat energy backwards from a cold object to a warmer object, and why and how the radiative heat flow equation shows that heat only flows in one direction across a gap, the reason is because the gap doesn’t exist for the photon.  The quantification is a bit different, sure, with conductive transfer being proportional to the linear difference of temperature, and radiative transfer being proportional to the difference of fourth powers of temperature, but ontologically it is still the same fundamental force of physics (electromagnetism mediated by photons) involved in the transfer of heat energy.

And this is why the same laws of thermodynamics apply: there is no such thing as “back-conduction” of heat energy, just as there is no such thing as “back-radiation” of heat energy.

If you would like to read more about Ontological Mathematics, and to learn what it is and how it can be used to scientifically, mathematically, and logically explain things that anthropocentric sense-perception based materialist science has insurmountable problems with, please see the treatise in the series of philosophy books titled “The God Series“.

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49 Responses to Ontological Mathematics is the Answer to GHE-Based Climate Alarm

1. Arfur Bryant says:

Thank you, Joe.

I am shocked that Warmists and Lukewarmists alike have such a problem in understanding the difference between radiation transfer between objects and heat transfer. It is simple: there is energy flowing both ways but heat only flows one way.

The steel greenhouse concept has caused so much confusion in the way it was ‘sold’.

Arfur

2. squid2112 says:

Awesome stuff Joseph!!! … awesome!

3. Bryan says:

Excellent summary

The Willis metal greenhouse model and similar models start, like most frauds, with a simple almost throwaway remark

“The energy supply to the planet core does not mater ‘lets say its nuclear’.”
Sounds innocent enough doesn’t it ?
Hell no!
By this remark they say the second law does not matter.
Energy has quality (2nd law) as well as quantity (1st law)

A simple AAA 1.5 volt battery contains much less energy than say a litre of water at 90 degrees C

Yet by attaching the battery to a circuit with a tungsten MES lamp temperatures of 3000C are achieved
Try getting the litre of water to transform itself any higher than 90C – its impossible

Electrical energy in the battery is highest quality and has no problem in being transferred into other energy forms.
Thermal energy like the litre of hot water is low quality and the transfer into other more useful forms is strictly regulated by the second LOT.

4. Quokka says:

For the example of two parallel walls, isn’t the standard scientific understanding that the hot surface radiates σ*Thot4 W/m2
to the cold surface which is absorbed, and the cold surface radiates σ*Tcool4 W/m2 to the hot surface which is absorbed,
and that, therefore, the hot surface loses energy at the rate σ*(Thot4 – Tcool4)
W/m2, and the cold surface gains energy at the same rate?

I don’t see how this could be characterized as “cold heating hot”, since the hot surface is losing energy. Nor can I see why any sort of thermal runaway should happen. If you do the accounting correctly so that energy is conserved, then the two walls will both converge to an intermediate temperature, as they should. A thermal runaway can only happen if you break the 2nd law of thermo and add new energy to the system. I wouldn’t say that heat flows from the cold plate to the hot place, but I would say that energy does flow from from cold plate to hot plate, as it presumably must if the cold plate is radiating.

It is not obvious to me how what you say above differs from this. Or are you claiming that σ*(Thot4 – Tcool4) W/m2 radiates from the hot surface and is absorbed by the cold surface, and 0 from the cold surface to the hot surface? This would result in the same amount of heat being transferred from the hot wall to the cool wall.

If you disagree with my figures above, can you please clarify how much energy you believe is being radiated from each surface, and where it goes? (reflected, absorbed, ???)

5. Carl Allen says:

“First law of thermodynamics: When energy passes, as work, as heat, or with matter, into or out from a system, its internal energy changes in accord with the law of conservation of energy. Equivalently, perpetual motion machines of the first kind are impossible.”

Let’s remember what spawned the “greenhouse effect” in the first place. The “greenhouse effect” hypothesis presumes to explain why the Earth’s “mean global surface temperature” is some 33 °C higher than the Earth’s “effective radiating temperature”. By this definition if the Earth’s “mean global surface temperature” were equal to the Earth’s “effective radiating temperature” then the “greenhouse effect” would = 0.000.

If you consider that the “surface” of the Earth is an atmosphere (85% of which is contained within the troposphere) and you recall that the mean temperature of tropospheric air (based on the International Standard Atmosphere) is about -18 °C, then the “surface” temperature of the Earth does, in fact, equal its “effective radiating temperature”. Thus by their own definition the “greenhouse effect” = 0.000. In other words, the Earth’s “surface”, i.e., its atmosphere, cannot be shown to be retaining any excess thermal energy.

The “greenhouse effect” confusion arises because those who spawned the hypothesis chose arbitrarily to only measure the temperature of the hottest part of the troposphere—sea level air—and call that temperature the Earth’s “mean global surface temperature”. Remember that all land based weather stations are sited ~1.5 meters off of the ground and only measure the temperature of 1/11,000th of the troposphere (the one meter thick layer of air that is being measured is only one out of 11,000 meters—the average thickness of the troposphere.)

Here is where the First Law of TD comes into play. “When energy passes, as work, as heat, or with matter, into or out from a system, its internal energy changes in accord with the law of conservation of energy.”

The one-meter thick layer of air that is about 1.5 meters off of the ground whose temperature is being measured and averaged to yield what is being called the “global mean surface temperature” is a completely “open system” in that its boundaries allow both mass transfer as well as energy transfer. This means that the air whose temperature is being measured in the afternoon is not the same air whose temperature was measured earlier that morning. The morning air is long gone by the afternoon and has been replaced by air that was somewhere else in the troposphere earlier in the day.

Conversely, the “greenhouse effect” hypothesis treats the one-meter thick layer of air that is about 1.5 meters off of the ground whose temperature is being measured and averaged to yield what is being called the “global mean surface temperature” as a “closed system” that is thermally isolated except to IR radiation. Is it any wonder why they cannot figure out why that one meter layer of air is 33 °C warmer than the average temperature of the troposphere as a whole?

If you want to properly analyze why sea-level air contains as much internal energy as it does you have to factor in the “work” that is done on it prior to its arriving at sea-level. For example, the larger thermodynamic system called the Troposphere is continually overturning. Within the Troposphere we observe the existence of great cells of circulating air, the largest being the Hadley Cell in which massive quantities of air convect upwards to the tropopause at the equator driven by daily uneven solar heating. This air then moves along the tropopause to the higher latitudes both north and south and descends back to the ground at ~30 degrees north/south.

The laws of physics demand that when the air from the tropopause descends back to sea level within the down-going leg of the Hadley Cell (many ground based weather stations are sited near this latitude) the “work” done on it by its progressively increasing higher pressure surroundings converts into thermal energy within that descending air. This is the well-known “adiabatic process” that is taught in every class on meteorology and we know that the “work” done on descending air raises the internal energy of that air enough to create a temperature lapse rate of 9.8 C/km. If it were not for contravening thermodynamic forces this would raise the temperature of the air on its descent from the tropopause to sea-level some 108 C. (The average temperature of the air at the tropopause is -60 C and 11 km x 9.8C/km= 108 C.) Without contravening thermodynamic forces sea level air would be 48 C instead of the 15C (what it is now according to the International Standard Atmosphere.) Let me say that again, if it were not for contravening thermodynamic forces the adiabatic process, which is triggered by the continuous overturning of tropospheric air, would keep sea-level air very hot at about 48C.

One of the “contravening thermodynamic” forces that keeps sea-level air cooler than 48 °C is radiative heat transfer, because higher radiation from the warmer and more emissive lower atmosphere moves heat up the atmospheric column against the cooler and less emissive upper atmosphere. Extensive surface radiation data being gathered at NOAA’a SURFRAD sites around the country as well as the formula for “net radiation heat loss rate” demonstrates that the movement of heat within the atmospheric column via IR radiation is virtually always upward away from the surface. When people talk about the fact that carbon dioxide and water vapor enhance the ability of the lower atmosphere to cool this is what they are talking about.

The “greenhouse effect” hypothesis completely ignores the fact that the air whose temperature is being measured at sea level ultimately came from the tropopause where its temperature was on average about -60 °C and that it acquired enough internal energy by the “work” done on it on its way to sea level to raise its temperature up to 48 °C–the temperature that it would be if contravening thermodynamic forces did not cool it down to about 15 °C. They instead believe that sea-level air only acquires thermal energy from the IR radiation that is being emitted by the ground which isn’t enough to keep sea-level air as warm as it is. They therefore concocted the “back-radiation” theory in which a portion of the heat “trapped” in the air by “greenhouse gases” is re-radiated back down to the ground raising its temperature, all the while being completely oblivious to the entire array of thermodynamic forces that are at work within the greater thermodynamic system called the “troposphere”. Here is one such example from wikipedia:
“The greenhouse effect is a process by which thermal radiation from a planetary surface is absorbed by atmospheric greenhouse gases, and is re-radiated in all directions. Since part of this re-radiation is back towards the surface and the lower atmosphere, it results in an elevation of the average surface temperature above what it would be in the absence of the gases.”

When they say that sea-level air is warmer than what it would be if warmed by sunlight alone they are correct, because it is also being adiabatically warmed as a result of the “work” that is continually being done on descending air. If this were not the case then diesel engines, which have no spark plugs, would not work. The fuel/air mixture within a diesel piston achieves “flash point” temperature simply by the “work” done on the fuel/air mixture during the upstroke of the piston. When a parcel of air descends from the tropopause to sea-level it experiences the same thermodynamic force as the fuel/air mixture within a diesel engine on the piston’s upstroke. At the present time many people (even the Pope and his Pontifical Academy of Sciences) refuse to acknowledge that “work” done on descending air within the continually overturning troposphere warms it and are instead calling the consequential extra sea-level warmth a “greenhouse effect”.

Carl

6. Greg House says:

I like the part about photons psychology.

7. RF says:

“There is no real ontological difference between conductive transfer of heat energy and radiative transfer of heat energy; radiative transfer of heat is simply conductive transfer but occurring at what we experience as a distance. They are mediated by the exact same force of electromagnetism.”

8. Ah Tim Casey’s site…great work there which also clearly debunks the radiative greenhouse effect.

Excellent comment.

9. Greg House says:

“There is a hot and cool term, and there is an exchange of energy between them since they are subtracting from one to the other”

Not necessarily. This part should be better avoided. Firstly, two terms do not require 2 separate processes, examples are known. Secondly, I have yet to see a direct physical evidence. Thirdly, the “cold” term can be explained differently, like gain from the hot minus loss in the opposite direction, not towards hot.

10. Truthseeker says:

I think that the no back-radiation greenhouse effect message is spread wider than we may realise.

Check out this discussion over at Jo Nova’s …

11. Greg House says:

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

No wonder the “greenhouse effect” can not be detected, since it does not exist.

12. squid2112 says:

@Arfur Bryant says:
2015/05/17 at 1:16 AM

It is simple: there is energy flowing both ways but heat only flows one way.

This is because one must ask oneself, what is heat? The problem is, heat isn’t a “thing” that can move around. Heat is a “result”. A result of the energy and vibrational state of a molecule. The higher the vibrational state, the higher the temperature of the molecule, which is felt as “heat”, which is a “result” of the energy. Heat itself is not transferable, only the energy is transferable. This is why heat can only go from hot to cold. It is exactly the bicycle wheel analogy that I have given before. Only a higher energy state can increase increase the energy state of a molecule, and thus increase its heat. That is, if object A is radiating (or conducting) towards object B, object B can only increase in energy if object A is at greater energy than object B. There is NO heat transfer from A to B either, only energy is transferred, but the “result” of that increased energy transfer is increased heat for object B. It is really as simple as that. This principle holds true in all cases, conduction and radiation alike, just as Joseph describes with this article.

13. squid2112 says:

Incidentally, what I have described in my last comment explains why there is no “heat” in space, that is, space itself has no temperature. It is neither hot nor cold. And by the way, in that context, the term cold is meaningless. There is only hot. Cold can only exist in but one particular case, and that is absolute zero vibration of a molecule. Any other energy state results in heat. The 2nd law should really be thought of as, “heat can only be transferred to a lesser warm object” But again, “heat” itself cannot be transferred anyway, only the energy can transfer.

14. Truthseeker says:

Squiddy … that explanation of how energy transfers (8:04pm) is GOLD. It explains everything and especially why the “NET heat flow” argument is totally bogus.

You are right about space and also about absolute zero and hot. However we perceive things as humans and so what has a lower temperature than us is “cold” and a higher temperature is “warm” and a much higher temperature is “hot”. It is easier to explain things as “cold” and “hot” rather than “warm’ and “less warm”.

15. Truthseeker says:

Greg House says:
2015/05/17 at 5:58 PM

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

No wonder the “greenhouse effect” can not be detected, since it does not exist.
================================================================
Which is precisely what all of those different people were saying in different ways …

16. Tom says:

17. Cheers Tom.

18. HYPeRGY says:

Greg House says:2015/05/17 at 12:59 PM
I like the part about photons psychology.

“Psychophysics”….great book that one…Lux et Veritas!

Brilliant work JP, keep the math coming!

19. squid2112 says:

@Truthseeker,

Thank you very much for your kind words. I just been trying to describe these things that can paint a mental picture. I know this helps me to understand these things. For me, during the past couple of years, I have had an epiphany concerning thermodynamics, in large part and thanks to the works of people like Joseph. I fully understand how the fundamentals of thermodynamics works and can picture it clearly. And after all, it is all about fundamentals. As it relates to the AGW topic, if you haven’t got the fundamentals, then the theory is a non-starter, which it is concerning the so-called “greenhouse effect”. The so-called “greenhouse effect” completely violates the very fundamentals of thermodynamics and the laws of nature.

Agreed on the hot/cold thing. Cold is a perception. It is really a measure of relativity. That is, this object is not as warm relative to me, therefore it is colder than I. It is cold. Yeah, I understand why we say “cold”. I was just trying to point out that “technically”, there is no such thing as cold, technically speaking.

20. Damian Scott says:

“Agreed on the hot/cold thing.”
What about actual sensible heat input into the system? For example, the increase in vehicles over the last century. Surely Mankind should be warming the atmosphere purely through increased thermal emissions?

21. squid2112 says:

Damian Scott,

What is “actual sensible heat” ?

22. Yes although the energy we use isn’t 100% efficient and so loses energy to heat, the amount is negligible compared to the input from the sun across the disk cross-section. Can look up the annual human consumption of energy, and then compare that number to a calculation for annual solar energy input.

23. Mr Pettersen says:

I have a question i cant find the answer to.
If freqency and wawelength of a molecule set the energy level co2 must change the wawelenght it emitts when its heated. That will mean that co2 near the surface must emitt shorter wawelengts at higher frequency than co2 higher up. If frequency are set by other molecules in the atmosphere they can not have wawelenghts longer than the rest of the atmosphere at the same level.
Will co2 then be able to change from far infrared to near infrared depending on its hight in the atmosphere?

24. geran says:

Mr. Petterson, if it helps, the wavelength, frequency, and energy of a photon are ALL determined by the temperature of the emitter.

25. @Quokka 2015/05/17 at 7:30 AM

Heat is what is required to increase temperature. The energy from the cold source is not heat. Q’ = A*s*(Thot^4 – Tcool^4). Only Q’ is heat. The radiation from the cool object to the hot object is not heat, and only the greater portion of the radiation from the hot object relative to the cool object is heat, and it transfers or flows only in the direction from the hot object to the cool object, from the greater power to the lesser power, as heat. Thus, radiation from the cool object does not warm the warmer object, because as per the 1st Law of Thermodynamics, heat input is what is required to increase temperature, and as per the heat flow equation, no heat transfers from cold to hot.

THEY say that the radiation from the cold object must heat the warmer object; it doesn’t, heat is only the difference of the terms.

The radiative heat flow equation shows what energy is being emitted, and only what balance of that energy can manifest as heat. The cold object has no heat to send to the warmer object hence can’t raise its temperature; only the balance of the warmer object’s energy can be heat energy and cause temperature increase.

And so there’s no radiative greenhouse effect.

26. Exactly Carl!

If you consider that the “surface” of the Earth is an atmosphere (85% of which is contained within the troposphere) and you recall that the mean temperature of tropospheric air (based on the International Standard Atmosphere) is about -18 °C, then the “surface” temperature of the Earth does, in fact, equal its “effective radiating temperature”. Thus by their own definition the “greenhouse effect” = 0.000. In other words, the Earth’s “surface”, i.e., its atmosphere, cannot be shown to be retaining any excess thermal energy.

Great entire comment!

27. Mr Pettersen says:

Thank you Geran! Thats what i learned at school. But will this mean that not all co2 emitts the same wawelenghts? if they vibrate at the same frequency as the rest of the atmosphere at that level, the output must be different for every level since temperature change with altitude.

28. Greg House says:

Carl,

if you accept that bogus -18°C calculation and try to explain the discrepancy, it will end in a bogus explanation. Now, let’s start with your redefining the surface. So, the only heat source the Sun can warm your “surface” to -18°C MAXIMUM, but underneath your “surface” it is WARMER? Then, Carl, this warmer underneath would radiate more energy away that the Sun delivers. Which means creating energy out of nothing. The 33°C difference is hence nonsense.

29. geran says:

@ Mr Pettersen: The simple answer is “yes”. But remember, a CO2 molecule is not a black body. If you had a “surface” of CO2 molecules, then emissivity would need to be considered. All matter emits IR radiation, and the photon characteristics are affected by the emitter temperature.

30. Reblogged this on ajmarciniak.

31. Quokka says:

We all agree (I think) that in the two parallel plates scenario, heat flows from the hot plate to the cold one at (Thot^4 – Tcool^4) W/m^2 (for simplicity I’ll assume σ=1 and drop the W/m^2 from now on). And people keep talking about energy flowing both directions, so do you also agree that the hot plate is radiating Thot^4 (S-B law), and the cold one is radiating Tcool^4?

[JP: Quokka…I wrote the equation – here it is again: Q’ = A*s*(Thot^4 – Tcool^4).

You see two terms there. The radiation from the hot term, and the radiation from the cool term. Only Q’s is the heat, whereas each term is the power/energy from each object.]

So if, hypothetically, both plates absorb the energy coming from the other plate, then we get the right result: the hot plate loses energy at (Thot^4 – Tcool^4) and the cold one gains it at the same rate. Without any thermal runaway happening, or energy being created out of nothing (as I have seen people here claim would happen).

[JP: We don’t claim that runaway heating occurs in this correct scenario as I have already described it to you. Telling it back to me but pretending you’re the one saying it doesn’t put me on the defensive about it. The radiative GHE model and crowd says that the cold term increases the temperature of the hot term – that is their mistake and nothing to do with me, and reciting back to me what I’ve said about how it correctly works doesn’t leverage me out of the correct position to be replaced by the sophists.]

I gather you don’t like that model, but here’s the funny thing: even after reading at least a couple of dozen posts in this site, I can’t work out what your alternative is.

[JP: Q’ = A*s*(Thot^4 – Tcool^4) where only Q’ is heat and only Q’ can cause temperature increase is the model of the laws of thermodynamics, which I subscribe to. The radiative GHE sophistry has the alternative to the laws of thermodynamics, not me.]

I realize that energy isn’t heat. Things can store energy without heating up if, for example, the energy is converted into mechanical or chemical potential energy, or used to change the phase of something (e.g. melting ice to water). But in this scenario, let’s assume none of that is happening. If energy is absorbed, it is going to warm the stuff that absorbs it, because it has nowhere else to go;

[JP: And there’s your sleight of hand sophistry. Wrong. Only Q’ = heat can cause temperature increase, i.e. “warm the stuff that absorbs it”. But here you’re trying to go back to saying that all energy causes warming. The cool term does not cause the hot object to increase in temperature, i.e. “to be warmed”.]

if something emits energy, it will cool down. So, once again, because I want to compare your model with the other one: in your model, how much energy is leaving both surfaces, and where does it go – howbout much of it is absorbed/reflected/whatevered, by which surfaces?

[JP: Well you’ve set up some sophistry here about who’s model is who’s. I don’t have a model. I use the laws of thermodynamics. The radiative GHE has an alternative model to the laws of thermodynamics.

Again, here is the heat equation: Q’ = A*s*(Thot^4 – Tcool^4). It shows how much energy leaves either surface. It shows which portion of that energy will be heat and can cause warming and in which direction. The energy propagates into the cooler body and out of its exterior side relative to the warm term – in the steel greenhouse, this is the exterior of the shell, or for two parallel walls, the other side of the cool wall.]

32. Gary says:

My current understanding is that the radiation from a warmer object is greater than a colder object. The warmer object sends more photons than it receives and the colder object sends fewer photons than it receives, so the net transfer is still from hot to cold.

I saw the light bulb experiment at Watt’s Up With That. Allegedly the colder outer glass sent back radiation to the light bulb and the light bulb warmed. Forget about the outside power that warmed the entire system! LOL! Cut the power and the parlor trick … er experiment would fail. According to the Bolzman equation, if you replace a cold object (ambient air) with a warmer cold object (glass enclosure) the light bulb loses less energy. Add the the outside power, and of course it’s hotter than it would be if exposed to air.

A very rude and arrogant individual named Curt did the experiment and post. He and Eisenbach think the outer shell around the planet should have the same power output as the surface. According to them, Ein = Eout. Wrong equation! Without input from the sun, it should be U1 – Eout = U2, and U2 < U1 and so the total system cannot have more than U1, the total energy of the planet. Yet Eisenbach has a shell and planet putting out watts to correspond to 2U1! Both the first and second law of thermodynamics are violated.

Anyway, I'm done ranting. Enjoyed your article, especially your insight on the photon's perspective.

33. Gary, you saw my debunk of Watts and Curt Wilson’s silly experiment?

https://climateofsophistry.com/2013/06/05/slaying-watts-with-watts/

These guys are amateurs and fools…sophists.

34. Quokka says:

Responding to Joe in the 2014/04/22 at 8:14 AM post:

Quokka…I wrote the equation – here it is again: Q’ = A*s*(Thot^4 – Tcool^4).

You see two terms there. The radiation from the hot term, and the radiation from the cool term.

Ah, some clarification finally: I take this to mean that you agree that the amount of energy radiating form the hot surface is A*s*Thot^4.

But we know that the hot surface is only losing heat at the rate of A*s*(Thot^4 – Tcool^4). So obviously it must be getting A*s*Tcool^4 from somewhere, to make up the difference.

[JP: You are conflating energy and heat again. The hot object is transferring heat at the rate specified by the Q’ equation; however, the Q’ equation contains the energy the hot object is emitting in total, which is A*s*Thot^4.]

Where does that from? I hypothesize that it is the A*s*Tcool^4 radiated from the cold surface. If it is not that energy, then there are two problems to explain:

1) Where does the A*s*Tcool^4 come from, to stop the hot surface losing more energy than it should?

[JP: As I said, the hot surface loses all of the energy it should, in the form of its own term in the heat transfer equation A*s*Thot^4. Q’ is just the portion of A*s*Thot^4 which is acting as heat for the cooler object.]

2) What happens to the A*s*Tcool^4 radiated from the cold surface when it reaches the hot surface?

[JP: Nothing. It’s bosonic energy. Just overlaps with no effect. And it doesn’t have the higher energy frequency components required to add to the warmer object to raise its temperature. In fact since the hot object replaces all of the energy that the cool object “emits” towards it, one can think of the energy from the cool object as not actually having left the cool object at all since you can’t actually label and track that energy. All of the energy from the powered warm source then gets emitted on the other side of the cool object, and of course the context here with the radiative greenhouse effect is where only the hot object is powered, and the cool object is passive, as in the steel greenhouse – the energy from the inner sphere passes through the shell and heats the shell along the way until the shell is as warm as it can get, and at that point the energy leaving the shell on its exterior is equal to the energy leaving or being generated by the sphere. The shell can not send heat to the sphere and so the sphere doesn’t increase in temperature.]

We don’t claim that runaway heating occurs in this correct scenario as I have already described it to you.

So the “free energy oven” here was a mistake, then? You seemed to think then that absorbing back radiation would lead to runaway heating. You even explicitly said “It doesn’t matter if there’s an input source of energy or not”

[JP: Yes, backradiation thermal absorption would still lead to runaway heating. Nothing has changed here or there.]

Well you’ve set up some sophistry here about who’s model is who’s. I don’t have a model. I use the laws of thermodynamics. The radiative GHE has an alternative model to the laws of thermodynamics.

As far as I know I’m using the laws of thermo too: energy is conserved, the correct amount of heat flows from hot to cold, surfaces radiate in accordance with S-B. If you think I’m doing it wrong, please clarify how your energy flows differ from the ones I’ve suggested.

[JP: You are conflating energy and heat. A*s*Thot^4 or A*s*Tcool^4 is always the energy. Only the difference is heat. Heat flow is not conserved. Only energy is. You can only call A*s*Thot^4 or A*s*Tcool^4 “heat” when they are by themselves with absolute zero background temperature, but even then it doesn’t make sense since empty space can’t be heated, i.e., can not attain a temperature, and so those terms are more properly still just total energy emission to space.]

And repeating “Q’ = A*s*(Thot^4 – Tcool^4)” doesn’t actually help. I already know it, and it doesn’t tell me what the internal energy flows are, only what the overall net heat flow is.

[JP: The equation has the individual energy contributions right in it.]

35. Quokka says:

Responding to Joe, 23/05/2015:

But we know that the hot surface is only losing heat at the rate of A*s*(Thot^4 – Tcool^4). So obviously it must be getting A*s*Tcool^4 from somewhere, to make up the difference.

[JP: You are conflating energy and heat again. The hot object is transferring heat at the rate specified by the Q’ equation; however, the Q’ equation contains the energy the hot object is emitting in total, which is A*s*Thot^4.]

This is technobabble. I understand the distinction between heat and energy. Let’s just consider the energy flows, which must add up. If the hot surface is emitting A*s*Thot^4, but only losing A*s*(Thot^4 – Tcool^4), obviously it must (by the 1st law of thermo) also be absorbing A*s*Tcool^4 from somewhere. Where? Coincidentally, A*s*Tcool^4 just happens to be the amount of energy from the cold surface that is hitting the hot surface, but (according to you) not being absorbed by it.

2) What happens to the A*s*Tcool^4 radiated from the cold surface when it reaches the hot surface?

[JP: Nothing. It’s bosonic energy. Just overlaps with no effect. And it doesn’t have the higher energy frequency components required to add to the warmer object to raise its temperature.

I’ve seen you quote the 1st law of thermodynamics, to the effect that the internal energy of a system changes in accord to the energy passing in or out of the system. Yet here, you seem to be proposing that the hot surface ‘absorbs’ the energy, with no change in state – it doesn’t get warmer, it (presumably) doesn’t convert it into chemical energy, or mechanical energy, or matter. It just … disappears? Does it ever come back again? If so, where was it stored in the meantime?

I have a problem with this claim that “it doesn’t have the higher energy frequency components required to add to the warmer object”. I have never heard of this effect. Can you find documentation for it at a reputable source (I’d accept Wikipedia).

By contrast, I have read a few times that absorptivity always equals emissivity for any given frequency, and there are thermodynamic reasons why this must be so. Since hot surfaces emit at low frequencies (because they emit more at all frequencies than colder surfaces do), therefore they must also be able to absorb at those frequencies.

Now, if you think that’s impossible, can you find a reference that says so? And for a given temperature, what is the formula which calculates the frequency below which it can’t absorb radiation? Who discovered this formula?

And repeating “Q’ = A*s*(Thot^4 – Tcool^4)” doesn’t actually help. I already know it, and it doesn’t tell me what the internal energy flows are, only what the overall net heat flow is.

[JP: The equation has the individual energy contributions right in it.]

No. It is a single equation, so it can solve for one variable. It tells you the difference between the incoming and outgoing energy. Any values for which outgoing – incoming = Q’ would work. My proposed solution is that outgoing = A*s*Thot^4, and incoming = A*s*Tcool^4, and that gives the right answer (but only if the hot surface actually absorbs the A*s*Tcool^4, and it doesn’t mysteriously vanish). Do you have a different solution? If so, what?

So, once again (3rd time lucky, maybe): in this dead simple scenario (two parallel surfaces, one hotter than the other), what are the *energy* flows? How much is each surface emitting and absorbing, and where does that energy go?

36. I’ll just stick to your last question Quokka since it captures all your other questions. First, in an instantaneous snapshot when the temperatures aren’t equal, then of course heat will be flowing from hotter to cooler. Of course the end-result is interesting, perhaps more interesting too, because this snapshot is part of a differential equation and so with heat flowing, things will be changing and heading to some end state. Let’s agree that our scenario is for one powered wall, i.e. one wall with a power source inside (and this is the “hot” wall), and another passive wall (cold) placed nearby.

The heat flow between the walls at any snapshot in time is this:

Q’ = A*sigma*(Thot^4 – Tcool^4)

Of course, Q’ = heat energy. Equilibrium is when Q’ = 0, which means no more heat flow. At that point Thot = Tcool.

The powered wall has a uniform power P0 Joules/second per unit surface area (A = 1), and so has a surface temperature of Thot = (P0/sigma)^(1/4). For even more simplicity I’ve assumed that the powered wall is perfectly insulated on one side, so that all of its power is emitted only on one side.

The cool wall sheds no heat energy on the side facing the powered wall, because Q’ is always either into the cool wall and so the cool wall is gaining energy, or, Q’ goes to zero and so the cool wall is no longer gaining heat energy; so the cool wall never has the opportunity for a positive Q’ away from itself towards the hot wall, because it never gets warmer than the hot wall. Since positive heat input is what is required for temperature increase, the cold wall never raises temperature of the hot wall, because it never sends a positive heat input to the hot wall.

The cool wall sheds energy per unit area at sigma*Tcool^4 on its side facing away from the powered wall – this energy is gone to space. When equilibrium is reached and the heat energy received on the side facing the powered wall has completely diffused through the cool wall, then the cool wall temperature = Thot, and emits sigma*Thot^4 on the side facing away from the powered wall, which is equal to P0 from the powered wall.

And so, 1st Law of Thermo is satisfied, all energy is conserved and accounted for, and, heat energy flow is always only from hotter to cooler, with the only possible end result being that heat flow goes to zero between the walls which means that the walls have stopped changing in temperature, and are the same temperature.

You will ask what happened to the energy from the cool wall to the hot wall. Here are some things to think about: The cool wall never loses heat energy towards the hot wall; Q’ is always into the cool wall, only ending at Q’ = 0 at equilibrium. Also think of diffusion: if the cool wall touched the hot wall, it wouldn’t heat up the hot wall, and the heat from the hot wall would diffuse through the cool wall and simply make the cool wall the new surface of the hot wall. The point being that you can also ask what happened to the diffusive exchange of energy from the cool wall “towards” the hot wall – there never was any diffusive loss of energy from the cool wall to the hot wall, just like there was never any radiative loss of energy from the cool wall to the hot wall. The energy from the hot wall goes into the thermal energy content of the cool wall, and eventually diffuses to the other side of the cool wall and then radiates away from there.

37. Quokka says:

When the hot wall was non-powered you seemed to accept that yes, energy was leaving the cold wall and travelling to the hot wall. Are you saying that now that the hot wall is powered, that energy flow has stopped? You have sigma*Thot^4 travelling from the hot wall to the cold wall, and sigma*Thot^4 leaving the far side of the cold wall to space. Your energy flows add up, but only because you have no energy leaving the cold wall towards the hot wall. This would appear to break the laws of physics, which say that the cold wall should be emitting sigma*Thot^4 on both sides. On the other hand, if energy is leaving the cold wall towards the hot wall, that breaks another law of physics, because now more energy is leaving the cold wall than is coming into it.

In your 2015/05/26 at 6:53 PM comment in this post, you certainly seem to agree that energy is travelling from the cold plate to the hot plate. But this energy is not included in your energy budget above. If you do include it, how do you get everything to balance?

You say things like

The cool wall sheds no heat energy on the side facing the powered wall…

and:

there was never any radiative loss of energy from the cool wall to the hot wall

When you say things like that, can you be surprised that, as you say, “[sophists] always then ask if I deny that the cool object emits any energy at all!”. It’s extraordinarily difficult to get a handle on what you mean because you alternate between statements that seem to accept that the cold wall radiates towards the hot wall, and statements that seem to deny it.

No-one is saying that “heat energy”, in your term, goes from the cold wall to the hot wall. But the concept of heat as a net energy flow presupposes that energy has to be flowing in both directions, but that “heat” goes from the hot wall to the cold wall because the hot wall radiates more than the cold wall.

When you say that the cold wall radiates “energy”, but not “heat energy”, you make it sound as if there is something called “heat energy”, which is different from “energy” (I don’t know if that is your intention). But as I see it, both walls radiate energy; heat is just the name we give to the net flow. If we know all the energy flows, we have a complete picture of what is happening.

So I will ask once again, because it wasn’t fully answered last time : what are all the energy flows in your powered hot wall scenario?

38. Yes Quokka I make the mistake of assuming that “machines” will be able to intuit the meaning. Any energy lost by the cool wall is replaced by the hot wall, and so it doesn’t lose any energy on the side facing the hot wall. Yes, heat energy is Q’ which has the equation for it – that is its definition, and this is different than the individual energy terms. There is a clear difference in the respective equations. No heat energy, no temperature increase.

All the energy flows are and were accounted for.

39. ScottM says:

Any energy lost by the cool wall is replaced by the hot wall, and so it doesn’t lose any energy on the side facing the hot wall.

Okay. So if a portion of the energy from the hot wall is being used to replace the energy lost by the cool wall, then the other portion is available to heat the cool wall. So as long as we satisfy

Q’ = A*sigma*(Thot^4 – Tcool^4)

then all is well. We can rewrite that as

“(Portion of hot wall’s radiated power output available to heat the cool wall) = (hot wall’s radiated power output) – (radiated power from the hot wall that is used to replace power radiated by the cool wall on the side facing the hot wall)”

And the fact that it needs to be replaced means that there really is A*sigma*Tcool^4 radiated from the cool wall in the direction of the hot wall.

The cool wall sheds energy per unit area at sigma*Tcool^4 on its side facing away from the powered wall – this energy is gone to space. When equilibrium is reached and the heat energy received on the side facing the powered wall has completely diffused through the cool wall, then the cool wall temperature = Thot, and emits sigma*Thot^4 on the side facing away from the powered wall, which is equal to P0 from the powered wall.

If the equilibrium condition is defined by Thot = Tcool, then according to the equation there is no heat flow from the hot wall to the cool wall. But there is a flow of A*sigma*Tcool^4 (equal to A*sigma*Thot^4 according to the stated equilibrium condition) from the (formerly) cool wall, on the side away from the powered wall. How is the formerly cool wall getting power to replace that radiated off into space, or does it just cool off, moving *away* from the equilibrium condition? If the latter, then at what point does it stop cooling?

40. “How is the formerly cool wall getting power to replace that radiated off into space, or does it just cool off, moving *away* from the equilibrium condition? If the latter, then at what point does it stop cooling?”

The outside wall facing space gets energy from the powered heating wall. At equilibrium, the outside wall emits the same amount of energy that is supplied by the power from the heating wall.

“When equilibrium is reached and the heat energy received on the side facing the powered wall has completely diffused through the cool wall, then the cool wall temperature = Thot, and emits sigma*Thot^4 on the side facing away from the powered wall, which is equal to P0 from the powered wall.”

There is no heat flow between the powered and passive wall at equilibrium…but there is energy shared between them with the energy coming from the power source.

At equilibrium, as the cool wall emits energy to space, that energy is replaced spontaneously/instantly/at the same time from the powered hot wall.

41. ScottM says:

“The outside wall facing space gets energy from the powered heating wall.”

Since the outer wall is radiating A*sigma*Tcool^4 in both directions, then the power from the heating wall must be 2*A*sigma*Tcool^4.

To quote your article above, “There is a hot and cool term, and there is an exchange of energy between them since they are subtracting from one to the other; however, only that result after subtraction is heat.”

That paragraph acknowledges an exchange of energy, so that the cool wall is radiating towards the hot wall. As it must, otherwise we would not get the equation Q’ = A*sigma*(Thot^4-Tcool^4). We would simply get Q’ = A*sigma*Thot^4. There would not be A*sigma*Tcool^4 to subtract from the first term.

To meet the first law of thermodynamics, you need a net power flow from the first wall to the second equal to the power flow from the second wall to space, if the second wall’s temperature is to be constant. No work is being done by this power, only heating, so the net power flow from the first wall to the second is Q’. Q’ cannot be zero if the second wall is radiating to space and maintaining a constant temperature.

42. ScottM – The energy from the cool wall is immediately replaced by 1) the exact same energy frequencies and quantity of them from the warmer wall, 2) actually a greater quantity of them from the warmer wall, 3) and additionally higher frequencies from the warm wall. Therefore the cool wall does not lose energy in the direction of the warmer wall, and the warmer wall of course heats the cooler wall given the remaining difference in their emissions. The cooler wall comes to equilibrium with the warmer wall, to the same temperature as the warm wall, without the warmer wall “needing to” emit more energy than it actually produces.

The heat Q’ from the cool wall to space is A*sigma*Thot^4 once the cool wall comes to equilibrium with the warm wall. And this energy is provided by the warm wall.

The cool wall does not lose energy toward the warmer wall, but gains energy from that direction. That is your mistake.

43. ScottM says:

Okay, I was confused by whether you were saying the equilibrium was defined as the point where “the outside wall emits the same amount of energy that is supplied by the power from the heating wall”, or the point where “the cool wall temperature = Thot”. I guess one of these is the definition, and the other the consequence.

I’m still puzzled by something. You said earlier that “The cool wall sheds energy per unit area at sigma*Tcool^4 on its side facing away from the powered wall”. But that assumes the outer wall is a blackbody (emissivity = 1).

Suppose we coat the outside of the outer wall with a low-emissivity paint. To make the numbers simple, lets give it an emissivity of 1/16. The inside surface of the outer wall will remain black.

This means the outer wall must double in temperature (Tnew = 2*Tcool) in order to emit the same power it was receiving before? Because the new radiated flux is (1/16)*sigma*Tnew^4. Doubling the temperature multiplies the result of that expression by 16 to overcome the factor of 1/16 for the low emissivity. Now the power emitted to space is once again equal to the power supplied to the heating wall.

Is the new equilibrium temperature for the heating wall:

a) twice the old equilibrium temperature, so that it matches the temperature of the outer wall at equilibrium?
b) unchanged?
c) none of the above?

If the answer is (a), then the heating wall is emitting four times the power it was previously emitting. Where is this new power coming from?

If the answer is (b), then the unpowered wall is now hotter than the powered wall. Won’t that mean there is now heat flow from the unpowered wall to the powered one? If so, how can that be considered equilibrium, as it means the powered wall’s temperature will start to rise?

I’m thinking the answer must be (c). But how would you calculate it?

44. ScottM: Yes if emissivity is reduced then of course this has an effect. The entire system, hot wall and cold wall, would increase to the same new higher temperature required to emit the energy given the new lower emissivity on the outer side of the outer wall. Nothing wrong with that.

Changing the emissivity has no effect on the underlying laws and restrictions of physics, and lower emissivity is not the GHE nor does it mean that the outer wall with lower emissivity is sending energy back to the warmer wall.

The outer wall is simply the new surface of the inner wall previously by itself. So all that is happening is that the surface of emission is changing emissivity. The power still comes from the power source, and is converted to thermal energy (“temperature”) until the point at which emission equals the power input, given the ability of the surface to emit (emissivity). The power input, and the power emission, is still the same.