# Radiant Energy Transfer Surface to Atmosphere

Written by Pierre R Latour PhD PE

α1 εo < α0 ε1, to disprove the greenhouse gas theory.  It explains why radiant energy flows two way. Flow from Earth’s surface absorbed by atmosphere exceeds that from atmosphere absorbed by surface.

Introduction of more atmospheric CO2 increases atmosphere’s emissivity and absorptivity and probably cools atmosphere and surface slightly.

Introduction
The Global Warming from Green House Gas (CO2) Theory, GHGT, controversy has been raging
since 1990 [1, 2]. Proponents say radiant energy transfers from colder atmosphere with CO2 to
Earths warmer surface, warming it (Fig. 1). Critics say since heat flows from higher temperature
to lower, this must be wrong and the whole GHGT a hoax.

Many global warming experts claim radiant energy is transferred between two radiators at a
rate proportional to their temperature difference (to power 4); much like thermal and
convective heat transfer. There is confusion between electromagnetic energy intensity from a
radiator (all matter emits) and energy transfer rate between two radiators (all matter absorbs,
transmits and reflects).This paper explains why this is only true in special, unrealistic cases. This allows the controversy to be resolved. More complete models have been published [3,4]

Emission
Radiant energy, em, is emitted from matter with intensity according to the Stefan-Boltzmann
equation, Ii = σ εi T4 , where Ii = radiant exitance, radiant emittance, emission intensity, w/m2,
rate per m2 , a function of wavelength i, T = surface temperature, deg K/100, εi = emissivity of
radiator, a physical property of matter, function of wavelength i. 0 < εi < 1, and σ = StefanBoltzmann fundamental constant of nature, 5.67 w/m2/(K/100)4

Many experts simplify with emissivity εi = 1 for both radiators, since emissivity is difficult to
determine. This is the black body assumption. This is a source of controversy. Real matter has εi< 1.
By integrating over all wavelengths, i, we can find a weighted average emissivity ε = ∫ εi di/I and
corresponding average emission intensity I = ∫ Ii di/i = ∫ σ εi T4 di/i = σ ε T4. Therefore I, w/m2 =
5.67 ε T4 where ε is the average emissivity of the emitter and T is emitter temperature. This is the
generalized Stefan-Boltzmann Equation for all matter radiators.

Absorption
Radiant energy is absorbed by matter with intensity that is the absorption fraction of incident
radiation from the emitter. Ji = αi I0 = αi σ ε0 T04, where Ji = absorption intensity, w/m2, rate per m2, a function of wavelength i, I0 = emitter intensity, w/m2 T0 = emitter surface temperature, deg K/100, αi = absorptivity of absorber, a physical property of matter, function of wavelength i. 0 < αi < 1 ε0 = average emissivity of emitter σ = Stefan-Boltzmann fundamental constant of nature, 5.67 w/m2/(K/100)4
The remaining incident radiation is transmitted through radiator, β, and reflected to surroundings,
γ, such that α + β + γ = 1. Like emissivity, absorptivity has a spectrum of exitance vs. wavelength unique to each atom and molecule. By integrating over all wavelengths, i, we can find a weighted average absorptivity α1 = ∫ αi di/I and average absorption intensity J = ∫ Ji di/i = ∫ σ αi ε0 T0
4 di/i = σ α1 ε0 T04. J1 = σ α1 ε0 T04 , where α is average absorptivity and J is average absorption intensity. This is the fraction of incident em from emitter 0 absorbed by absorber 1.

Energy Transfer Rate Law
The driving force for net radiant energy transfer is an intensity difference between radiators, like a temperature difference for conduction and convection. In the latter two there is a temperature gradient through matter, transmitting the molecular/atomic kinetic energy (indicated by temperature). In the former case there is an electromagnetic energy field everywhere throughout space, transmitting electromagnetic energy among radiators in all directions by an intensity difference, gradient, among them.

The radiant energy transfer from radiator 0 to radiator 1 is intensity emitted by 0 and absorbed
by 1. The energy transfer from radiator 1 to radiator 0 is the intensity emitted by 1 and
absorbed by 0. This two-way energy transfer occurs simultaneously because radiation exists at
many frequencies. The net energy transfer from 0 to 1 is the first less the second [5]

For simplicity we assume steady state from here. For unsteady state, just include energy accumulation rate, mCpdT/dt, in the energy balance, input – output for the describing differential equation of the body of interest temperature. Latour [2] gives a rigorous model of changing interacting energy flows to estimate radiator temperature changes.

This is the general equation for net radiant energy transfer between dissimilar radiators. The
first term is the intensity from radiator 0 times the fraction absorbed by radiator 1, α1, i.e. flow
from 0 to 1. The second is the intensity from radiator 1 times the fraction absorbed by 0, α0, i.e.
flow from 1 to 0.

Here we have radiant energy flowing from the colder radiator T14 = 4 to the hotter radiator T0
4 =6, because the colder has higher emissivity and lower absorptivity. (This does not prove GHGT
Fig. 1 that net energy flows down from cold atmosphere to warmer surface. Just that it is
theoretically possible with some physical property combinations.)

Energy balance of plate 0: Rate of accumulation in 0 equals input – output, where m0 Cp0 dT0/dt = Q0 – Q0,1 Q0 = thermal energy input to 0 Q0,1 = radiant energy transfer from 0 to 1 when Q0,1 > 0. Otherwise rate is from 1 to 0. m0 = mass of plate, kg Cp0 = heat capacity of plate, joule/T – g – s
t = time, s.

Rate of energy accumulation in 1 equals input – output, where m1 Cp1 dT1/dt = Q1 + Q0,1,
and where Q1 = thermal energy input to 1.
This is a pair of coupled ordinary differential equations. With appropriate T0 and T1 initial
conditions and specified input functions, Q0(t) and Q1(t), these can be solved for transient
changes in T0(t) and T1(t).
These can be combined by addition, eliminating the em terms, m0 Cp0 dT0/dt + m1 Cp1 dT1/dt =
Q0 + Q1. At steady state both derivatives are 0, and Q0 = – Q1.
Input rate = output rate = constant, naturally. One side is heated, and one side is refrigerated.
Each rate equals +- Q0,1, which need not be 0. T1 need not equal T0 at steady state when each is
constant.

Net radiant energy flows from 0 to 1; surface to atmosphere.
If radiating atmospheric CO2 increases, both ε1 and α1 increase. To determine the change in
these properties quantitively is beyond the scope of this paper.
Assume CO2 increases from 400 to 800 ppmv and atmosphere properties increase 0.1% to:
ε1 = 0.083006 * 1.001 = 0.083089
α1 = 0.402980 * 1.001 = 0.40701
Q0,1 = 0.06612 * 1.001 * 2.88 4
– 0.007263 * 1.001 * 2.554
= 0.066186 * 68.7971 – 0.007270 * 42.2825
= 4.55341 – 0.307406 = 4.24600 > 0 and > 4.24155
Both atmospheric emissivity and absorptivity increased with CO2 doubling, but emissivity
dominates. Net energy transfer rate from surface to atmosphere increased with CO2, which
would cool the surface. Connecting all interacting variables shows the same thing [2]

Conclusion
From the law of radiant energy transfer we have proven it can flow from the colder to warmer
radiator, depending on emissivity and absorptivity properties of both radiators. And we have
derived the specific physical property conditions of radiators where this can happen: the colder

When T0 > T1 and α1 εo < α0 ε1, radiant energy flows from 1 to 0, provided Q0,1 = α1 ε0 T0 4- α0 ε1T14 < 0

Using realistic average physical properties for Earth’s surface and atmosphere, we find energy
transfers from surface to atmosphere at a higher rate than the reverse with increasing CO2. This
indicates net global cooling by radiant energy transfer from increasing atmospheric CO2. This
contradicts the Green House Gas Theory.

References
1. Latour, P. R., “Undeniable and Unfalsifiable”, January 2014. https://principiascientific.org/undeniable-unfalsifiable.html/
2. Latour, P. E., “Physics Proves Cooling”, December 2014. https://principiascientific.org/physics-proves-radiating-gases-decrease-global-temperature.html/
3. Latour, P. R., “ChE Models Earth’s Temperature Response to Fuel Combustion”, July
2015. https://principia-scientific.org/che-models-earth-s-temperature-response-to-fuelcombustion/
4. Latour, P. R., “Radiation Physics Laws Give the Effect of CO2 on Earth’s Temperatures –
A Primer”, Principia Scientific International, February 2017. http://principiascientific.org/radiation-physics-laws-give-effect-co2-earths-temperatures-primer/
5. Latour, P. R., “Simultaneous Conduction and Radiation Energy Transfer”, March 2017.

Editor’s note: readers should be aware that our software does not reproduce super and subscripts correctly – the original PDF is here Radiant Energy Transfer Nov19 (2)

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• ### Squidly

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BS! … the whole “two way” garbage is complete BS! .. the surface cannot absorb ANY energy from the atmosphere because the radiation from the atmosphere is lower frequency. The surface cannot absorb any of it. There is not “two way” exchange. There is only a one way exchange. A molecule receiving radiation at a lower frequency than itself doesn’t feel a thing. It doesn’t absorb it, and it doesn’t impact it at all! .. NONE!

The only way Molecule A can receive any energy from Molecule B, is if, and ONLY if, Molecule B is of greater energy state than Molecule A .. this is a proven physical law of our universe!

So stop with the stupid “two way” exchange nonsense. It is just as nonsensical as the so-called “greenhouse effect” itself.

• ### tom0mason

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Absolutely correct Squidly,
These weather worrier can only get away with such BS because they devalue and deliberately misunderstand the influence the sun has on our weather.

These BSers try to say that atmospheric gases make our climate, when it’s the solar radiation that makes and drives the temperature differential, from equator to poles, that naturally drives the climate. There is NO observed evidence for CO2 affecting the atmospheric temperature, or the climate. If CO2 has any affect it is so meager as to be lost in the natural atmospheric thermal noise and chaotic climatic turbulence.

IT IS THE SUN THAT DRIVES THE CLIMATE NOT CO2!
These climate worriers will find out in a few short year (10 or so maybe less) as this current solar decline hits home.

• ### James McGinn

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Squidly:
BS! … the whole “two way” garbage is complete BS! .. the surface cannot absorb ANY energy from the atmosphere because the radiation from the atmosphere is lower frequency. The surface cannot absorb any of it.

James McGinn:
Nonsense. The surface does absorb energy from the atmosphere. No real scientists believes otherwise. You’ve been brainwashed by Postma.

Squidly:
There is not “two way” exchange. There is only a one way exchange. A molecule receiving radiation at a lower frequency than itself doesn’t feel a thing. It doesn’t absorb it, and it doesn’t impact it at all! .. NONE!
The only way Molecule A can receive any energy from Molecule B, is if, and ONLY if, Molecule B is of greater energy state than Molecule A .. this is a proven physical law of our universe!

James McGinn: No such physical law/principle actually exists. You are a victim of a group delusion.

All objects in the universe are emitting energy in all directions constantly. All other objects can potentially absorb and re-radiate energy from all other objects. Objects that are receiving more than they are radiating are said to be warming (net energy gain). Objects that emit more than they receive are said to be cooling (net energy loss).
Squidly, stop being a child. You know full well that a “proven physical law of our universe,” (your claim) should be incredibly easy to substantiate with direct reference to experimental evidence. Yet you provide none. Why? Because the truth is you don’t really understand it.

James McGinn / Genius / Solving Tornadoes
Do you believe in convection? Did you know that convection is to meteorologists what GHG is to alarmists?
The ‘Missing Link’ of Meteorology’s Theory of Storms
http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16329

• ### Pierre R Latour

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Squidly, You criticize my article based on your own authority; no proof, no evidence, no references.
Got any science to back it up? Where is my error; math, physics, English? You would be tossed out of any law court for a frivolous charge.
You say my paper is BS. What does BS stand for and what do you mean by it? All animals pass undigestible residue. Including you. It is natural and necessary. What is the relevance to radiant energy transfer?
Why don’t physical properties of two facing radiators have any bearing on the rate of radiant energy transfer between them? How does the warmer radiator know whether incident radiation came from a cooler radiator with higher emissivity and intensity or not? (It doesn’t; my rate law handles transfer rate precisely and accurately.) Disprove that if you can.

• ### John Harrison

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Squidly. I am sure that we have spoken of this before but you are focussing exclusively on unimolecular interactions with an IR photon. The surface is not unimolecular as you may have gathered and neither are they all in the same energy state. The surface is a multimolecular grey body therefore some of the molecules have a sufficiently low energy state to absorb IR photons. Open your mind a little further to take in the whole picture.

• ### lifeisthermal

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Exactly! A warmer body already emits a higher density of photons in the wavelengths of the colder body, because of more molecules excited to higher frequencies, at all frequencies. Absorbing those photons would, if anything, lead to “dilution” in flux density=cooling. Which is the effect of a cold body on a warm body. So, even IF absorbed, it doesn´t warm anything. But I agree, they´re not absorbed, a warm body is cooled by the transfer of heat according to dT.

• ### Norman

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Squidly

It might help your mindless ranting if you actually studied real physics. You can make endless false declarations based upon your distorted illusions. You can convince the crackpot audience on this unscientific blog. You will never convince one person who has actually studied science.

Basically you don’t know any actual physics but are so arrogant in thought process you think a powerful declaration of nonsense convinces yourself you are right.

There is certainly a two-way exchange of radiant energy. You are too ignorant to understand that the vast majority of surface molecules at room temperature are in “ground state” which means many many molecules able to absorb incoming energy.

The lunatics on this blog make the same phony claims over and over. It seems to never end. Not one of you crackpots can prove any of your phony assertions with anything but you continue to declare them. All it does is make you look incredibly stupid.

• ### John Harrison

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Norman. Harsh comments but very true nonetheless. Don’t be too disappointed if you don’t get a response from Squidly or his compatriots. They bang on about having proved that GHG back-radiation is thermodynamically impossible. All they are doing are demonstrating a surprising lack of understanding of basic physics which is getting skeptics a bad name.

• ### E.N. Tropy

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Norman. What’s in a name? In your case, one possibility is N. Moron. Methinks a ranting crackpot is calling the kettle black.

Matter can only absorb energy emanating from a higher energy source. Energy from a lower energy source will be reflected. There’s never any two-way, net exchange. The transfer is always down a thermodynamic one-way street until equilibrium. At equilibrium, both entities will continue to radiate and energy can potentially flow in both directions, but absorption will occur only if the recipient has momentarily dropped (e.g., through radiating) to a lower energy state to the sender at the instance of sending.

Return to your lunatic, CO2-global warming greenhouse asylum, with its two-way radiant energy exchange, close the door, lie down and contemplate the fundamental laws of the universe while slowly filling the greenhouse with CO2.

• ### Norman

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E.N. Tropy

Perhaps if you actually studied physics instead of pretending you know something about the topic I could give you some respect. Where do you get your declaration from “Matter can only absorb energy emanating from a higher energy source. Energy from a lower energy source will be reflected.”

You pull this out of your behind and pretend it is valid. It is just a made up belief you have. Not based upon any actual science. You are like the Flat-Earth society. You make up junk science and think you know something. This is a nuthouse full of all types of nuts. None care to learn actual physics but they are deluded enough to think they know some. You certainly don’t have a clue about science and make things up as you go.

• ### E.N. Tropy

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I have a Ph.D. in Physics. X-ray lasers.

• ### Norman

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E.N. Tropy

From your limited logic skills and poor attempts at insult I doubt you possess enough thought process to understand this. I think it is a waste of time trying to reason with a lunatic but I can try.

The graphic on this link will show you a two-way exchange of energy for solid objects.
https://en.wikipedia.org/wiki/Elastic_collision

If you scroll down a bit you will see various energy exchanges in collisions. When a slower object is moving toward a faster moving object and the two collide they exchange energy. The energy of the slower one is transferred to the faster one. An object with less energy can transfer energy to one with more energy. You have zero understanding of valid physics and it is like communicating with a cultist. You think you are correct and you can be if science were based upon the make up declarations of crackpots. Thankfully science is based upon reason, logic and empirical data.

Simple experiments can prove you wrong but your empty head will not accept the results.

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• ### Norman

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E.N. Tropy

It does not appear you have a Ph.D in physics or you would understand radiant heat transfer. There are many on this blog that pretend physics Maybe you are not one of them but I would hope you knew better than the others.

If you have a heated plate in outer space the radiant energy from a colder object will increase its temperature. It is gaining energy from the colder source as well as whatever source is heating it. There is NOT one valid physics textbook that agrees with your position and all disagree. It is well established experimentally verified physics and is used in all engineering of radiant heat transfer. It would be impossible for you to not had studied this to get your degree. So all that can be correct would be you reject established verified science in favor of the crackpot version of making up your own ideas and not having to prove any of them. What a horrible downfall for you to be caught up in the phony made up physics and reject the established verified version you learn while gaining you degree. Get out of the fake science and come back to reality. The fake stuff may be more fun because all you have to do is declare things, the real science is far more rewarding. It leads to a greater understanding of the truth. It can be proven.

• ### geran

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When Norman attacks others, he always forgets to add the disclaimer that he has no background in the relevant physics. He pounds on his keyboard, relentlessly, making big claims, but not able to convince anyone but himself. He lovies his pseudoscience.

Lacking any ability to think for himself, he actually believes he can bake a turkey with ice cubes.

Hilariious.

• ### Norman

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JDHuffman

You are still wrong. I know so much more about actual physics than you. Unlike you that pretends to know physics by finding words like Poynting Vectors and using the concept incorrectly but you never will actually read any physics. You have too much fun making up your own versions rather than study the real thing. Like all things about you you are fake. You lie about the ice cubes and turkey but continue with such lies. You are wrong about all the physics you post but you keep going. I guess you hope there are a few crackpots around you can convince. Good luck with that.

• ### geran

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When Norman attacks others, he always forgets to add the disclaimer that he has no background in the relevant physics. He pounds on his keyboard, relentlessly, making big claims, but not able to convince anyone but himself. He loves his pseudoscience.

And, as if he’s not already funny enough, he gets peoples’ names mixed up!

Hilariious.

• ### Norman

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JDHuffman

Whatever you say.

• ### Herb Rose

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Hi Pierre,
I believe in the statement “Energy from a lower source will be reflected.” should be Energy from a lower source will be re-emitted.
An object can reflect energy, absorb energy, transmit energy or combination of these three functions depending on the nature of the energy striking it and properties of the objects, with the difference being the direction and size of the energy vectors coming from the object.
In reflection the energy is directed back to the source by the surface as with a mirror, radar, or red paint and never interacts with the object’s interior molecules. In absorption the energy the energy enter the object and interacts with molecules of the object and then is emitted in all directions as energy by the object either as the same wavelength energy or different wavelength. A red piece of glass will absorb all the spectrum of light, including red, and will emit it as red light.
When energies transmitted through an object they don’t interact with the molecules on the surface or in the object and continue in the same direction. Radio waves will pass through most objects unless they encounter the right size antenna to absorb them.
If an object reflects all the energy it receives it will have the appearance of emitting that energy the same as if the object was at equilibrium with the energy and absorbing that energy then re-emitting it.
Herb

• ### E.N. Trophy

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Herb,
Good points. However, re-emission implies prior absorption, which means a temporary increase in energy state, which would be impossible if the transmitter is a lower energy state source.

• ### geran

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People that don’t understand the relevant physics, and those that want to pervert and corrupt science, always claim that a surface will absorb ALL photons. Warmists need this to be true so they can push the false narrative that the atmosphere can warm the surface. But, not all photons are always absorbed, so their false narrative becomes their false religion.

If all photons could always be absorbed that would allow baking a turkey with ice cubes.

(Right now, there are clowns reading this who will rush out to buy bags of ice for Thanksgiving! The comedy continues).

• ### Pierre R Latour

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OK Herb. While my sentence is true, yours is better.

• ### Zoe Phin

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This is a great article highlighting the fundamental conception problem of the radiative greenhouse effect.

“There is confusion between electromagnetic energy intensity from a radiator (all matter emits) and energy transfer rate between two radiators (all matter absorbs”

I would have done it a little differently.
There’s
1) Energy
2) Energy Flow (Heat)
3) Radiating Potential to 0 Kelvin

They confuse #3 with #2, and that’s why they make up nonsense like “two way energy flow” or “NET heat flow”.

There is no two way energy flows.

Energy flows from high energy to low energy, and we call that heat!

There can’t be two way energy flows because electromagnetic waves (or photons) can’t occupy the same space for them to go two ways. It just doesn’t make sense.

They confuse energy flows with radiating-potentials-to-0K.

Am I right?

• ### Squidly

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You are pretty much right. For me, one of the biggest problems is that people conflate “heat” with “energy” .. you cannot move “heat” from one place to another. “Heat” is not a transferable “thing” .. “heat” is a “result” .. IR is not “heat” .. IR is electromagnetic radiation .. “heat” is not electromagnetic radiation. “Heat” is a vibrational state of the molecule, in a sense. “Heat” is a result of the vibrational state of a molecule.

• ### lifeisthermal

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“They confuse energy flows with radiating-potentials-to-0K”

Just like the manipulation of the transfer equation in pyrgeometers does.

• ### Pierre R Latour,

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Squidly,
The case you offer is when surface absorptivity is zero, in which case my presentation agrees.
But what if surface absorptivity is not zero everywhere?
You declare two-way radiant energy transfer is BS, without proof.

• ### Squidly

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BS! … more conflation … there is no such thing as “two way transfer” .. period .. you cannot transfer energy from Molecule A to Molecule B unless (and only if) Molecule A is of greater energy state than Molecule B .. there is no exception to this rule ..period! … Conflating with “absorptivity” is absurd. It is just more sophistry nonsense.

And no Pierre, I am not talking specifically about “net” energy .. I am talking any energy (including “net”). If Molecule A is of lower energy state than Molecule B .. then Molecule B does not even see Molecule A .. Molecule B doesn’t even know Molecule A exists, unless Molecule B comes into contact (conduction) with Molecule A, and then if Molecule A is of lesser energy state than Molecule B, Molecule A actually steals energy from Molecule B, decreasing the energy state of Molecule B and raising the energy state of Molecule A. (think of your hand smacking a bicycle wheel to spin it. Very same principle. You can only speed the wheel if you smack it with greater energy than it already has, otherwise it will stay the same, or slow if you hand is lesser energy).

Me having to describe this very fundamental physical law of our universe to you is, well, embarrassing. Think about this for just one moment and ask yourself this. In the grand scheme of our universe, could our universe exist in any other way? … the answer: NO, it absolutely could not.

What I have described has to be true .. there is no other choice. We simply could not exist if it were not true. Check that again .. we could not exist otherwise !!

• ### Herb Rose

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Hi Squidly,
When you see the different phases of the moon you not only see the part of the moon illuminated by the sun but also the shaded area illuminated by light being reflected off the Earth, then reflected back to the Earth. This is an example where a object with lower energy is transferring energy (in the form of light) to an object with greater energy. All objects above absolute zero radiate energy. All objects absorb radiated energy. Objects with different energy levels will reach an equilibrium where the amount of energy received and the amount radiated do not result in a change in the energy of the objects.
The Earth was once a lot hotter than it is now. It radiated energy in all directions while at the same time it was receiving energy from the sun. The point where the Earth’s energy was in equilibrium with the energy coming from the sun was above the surface of the Earth. Since the Earth was radiating energy into space in all directions (not just at the sun) it was losing energy and the equilbrium point shifted towards the Earth’s surface. This allowed the crust to form on top of the hotter mantle and brought the Earth into equilibrium with the energy from the sun establishing the sun as the determining factor for the energy of the Earth. When there is volcanic activity or forest fires (where stored solar energy is released) the Earth radiates more energy into space than it receives from the sun, re-establishing the equilibrium. Through volcanism the Earth will continue to cool creating a thicker crust and the energy level of the sun will continue to determine our climate.
Herb

• ### geran

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Herb, you are confusing “reflection” with “absorption” and “emission”. We see the moon because it is “reflecting” visible light. It is NOT “emitting” visible light.

So, your statement “All objects absorb radiated energy.” is only applicable in certain cases, i.e., the statement is wrong more than it’s right.

• ### Herb Rose

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Hi Geran,
You’re right about the light from the moon being reflected.I should have said the light from distant stars that travel thousands of light years before reaching Earth. Do they transmit energy to the Earth even though the energy is so weak you may need a telescope to detect it?
If an object radiates energy but doesn’t absorb energy doesn’t it eventually lose all its energy and reach absolute zero and be undetectable? Or would an object at absolute zero be a perfect reflector as opposed to a black body which absorbs all energy? How would you distinguish an object reflecting all energy from an object emitting energy?
Herb

• ### geran

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Herb,

Photons from the distant stars are the same wavelength/energy as when they were emitted. It’s only the flux that gets reduced, not the energy of the individual photons. But a photon does not always imply heat transfer. For heat transfer to occur, the photon must be absorbed.

Yes, if an object emitted energy, but never absorbed energy, it would continue to drop in temperature.

An object at absolute zero would likely be a perfect absorber, until it rose in temperature. An object at very high temperature, like a star, would be a very poor absorber. Photon absorption is a function of compatibility between the surface and the arriving photon. And that compatibility is affected by temperature.

• ### Herb Rose

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Hi again Geran,
If the photon from a distant star has the same wavelength/energy as when emitted why wouldn’t the photon be absorbed like photons coming from the sun which have the same wavelength/energy?
Your comment that the statement “All object absorb radiated energy.” is mostly wrong I took to mean that there are objects that don’t absorb radiated energy.
The emission and absorption of radiated energy is done be the electric and magnetic fields of the molecules the object is made of, not by the photons striking matter.. It is the size of those molecules and the length of bonds that determine what wavelengths are absorbed, reflected, or are transmitted without being affected. It is like the molecules are tuning into a radio wave that matches their antenna and when emitting they can change those wavelengths or emit the same wavelength absorbed in all directions. Outgoing wavelengths only block incoming wavelengths if they are completely out of phase cancelling each other out as in the dark interference bands in the dual thin slit experiment.
I still believe that all objects absorb radiated energy even though that energy may not result in an increase of the objects energy. It is like adding water to a bucket with holes in it. Eventually the water leaking out will match the water being added.
Herb.

• ### Pierre R Latour,

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Herb Rose gets it. Squidly doesn’t.

• ### geran

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Herb, if two photons have the same wavelength, they are the same.

And your “belief”, “I still believe that all objects absorb radiated energy…” is easily disproved. Take a look all around you. You see objects because visible light is being reflected. It was NOT absorbed.

• ### Herb Rose

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Hi Geran,
If you see something it is because certain wavelength are being reflected no all of them. A red object is red because that wavelength is being reflected (red paint) or emitted (red glass) while other wave lengths are absorbed and converted to other wavelengths of energy.
Herb

• ### geran

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Exactly Herb, not all photons are always absorbed.

• ### Finn McCool

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Pierre
One problem I always have with similar articles is trying to pick apart the math which is buried in text.
Is it possible for you to use Latex or something similar to format the equations to make them more clear?
I am just a dumb computer coder who needs to have all assumptions and variables clearly stated, hopefully with meaningful names 🙂 This helps to replicate results of the maths on real world data.

• ### Pierre R Latour,

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PSI has trouble with superscripts and subscripts. Sorry.

• ### Zoe Phin

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Test

εσ(T₁-T₀)⁴

₀¹₂³4₅⁶₇⁸

Do you see super/sub-scripts?

• ### geran

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This posting is somewhere between confusing and wrong. Just one example of “wrong”:

The driving force for net radiant energy transfer is an intensity difference between radiators, like a temperature difference for conduction and convection.”

WRONG! The “driving force” is surface temperature. As Squidly implied, “net” has no useful significance related to the S/B Law.

• ### Squidly

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Thanks Geran !! .. I am so sick and tired of the misdirection with this so-called “net energy” sophistry bullshit. We have high energy particle colliders that have demonstrated, empirically, that the whole concept of this “net energy” is bunk. Unless you are of greater energy state (radiatively), I cannot even see you. You cannot affect me. period…

• ### lifeisthermal

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Unfortunately many books on heat transfer labels it as “net transfer” and explains it as a two way transfer. But it was never explained that way in the beginning, when the laws were discovered. And it´s important to remember that there exists no measurements of any incoming radiation at the warm body surface from the cold body, while it´s easy to measure the heat arriving at the cold body.

But you know this.

• ### Herb Rose

|

The trouble with the GHGT and all climatology is the bad data. The thermometer does not give an accurate measurement of kinetic energy. The molecules of 0 C water have more movement (kinetic energy) than the molecules of 0 C ice.
To use a thermometer correctly you must expose the same surface area to the medium being measured so the same number of molecules (mass) transfer energy to the thermometer. If to little or to much of the material expanding to record the energy is exposed, less or more mass will transfer energy and it will give an inaccurate measurement.
In the atmosphere the number of molecules (mass) transferring kinetic energy to the thermometer decreases with increasing altitude so both the velocity and mass of the molecules being measured vary. Anyone looking at the graph of temperature vs altitude in the atmosphere can plainly see that it does not show how energy flows from a source.
How can scientists who live in heated homes and eat food cooked in an oven, where oxygen and nitrogen are absorbing energy and transferring that energy to other objects, believe that the sun does not heat theses gases in the atmosphere? They accept that the sun heats theses gases in the thermosphere but then say the light penetrating the thermosphere cannot heat these same gases at lower altitudes.
What energy the atoms in a molecule absorbs does not correlate to the energy of the molecules as a whole any more than the rotational energy of the Earth correlates to its orbital energy.

• ### Squidly

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Herb, the problem with GHGT is that it violates physical laws of our universe … period!

• ### Zoe Phin

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Herb,
“The molecules of 0 C water have more movement (kinetic energy) than the molecules of 0 C ice.
To use a thermometer correctly you must expose the same surface area to the medium being measured so the same number of molecules (mass) transfer energy to the thermometer.”

I’m sorry, Herb, but that’s wrong. The definition of Kinetic Energy includes mass and velocity. It seems like you want to redefine K.E. to just be velocity.

And we define movement as translational, not vibrational. We don’t count the bonds’ vibrations.

• ### Herb Rose

|

Zoe,
The whole comment was about not incorporating the change of mass in the atmosphere. All water molecules have the same mass so any variation in kinetic energy is due to the velocity of the molecules. It seems to be that not only do you not understand gravity and thermodynamics but also you cannot read.
Since economics was your major here’s a way to profit from your education. Write to the school that gave you A s in physics and tell them if they return your tuition and expenses you will never divulge the name of the school.

• ### Zoe Phin

|

Herb,
I got a free scholarship.

You’re redefining what kinetic energy (a BULK measure). You’ve reduced K.E. to just average velocity. So why don’t you just say that: average velocity. Why destroy physics?

• ### Herb Rose

|

Zoe,
I am not the one who defines temperature as the mean kinetic energy of the molecules in the medium being measured.
The thermometer measures the total kinetic energy transferred to it by the molecules striking its measuring surface. O2 molecules and N2 molecules with the same kinetic energy have different velocities because of their different masses. In water, where the molecules have the same mass, there is a bell curve representing the velocity of the different molecules with the temperature representing the mean/average velocity of the molecules. Because liquid water molecules are in close contact with other molecules there are constant collisions where velocities are equalized forming a narrow bell curve. In a gas, where collisions are less frequent the bell curve is broader but the temperature is still the total kinetic energy being transferred to the measuring surface.
In a 100 C oven and a 100 C pot of boiling water the total kinetic energy being transferred to the measuring surface of the thermometer are the same. In the oven there are fewer molecules (less mass) transferring energy to the measuring surface so in order for them to transfer the same total kinetic energy to the same size area their velocity (kinetic energy) must be greater than the more numerous molecules in the water.It is you who believes that in the atmosphere where with increasing altitude and decreasing mass transferring energy to the same area that the decreasing temperature (total kinetic energy) means the molecules have less kinetic energy (lower velocity of identical molecules) instead of the decreasing temperature resulting from the decreasing number of molecules (less mass) transferring energy to the thermometer. It is you who maintains kinetic energy is purely from velocity.

• ### Zoe Phin

|

Herb,
“It is you …”
Evidence?

“It is you who maintains kinetic energy is purely from velocity.”

Pure projection. No, as I’ve said and as physics DEFINES: kinetic energy is a function of both mass and avelocity.

• ### James McGinn

|

Zoe:
“And we define movement as translational, not vibrational. We don’t count the bonds’ vibrations.”

Yes. It is customary to deliberately ignore, “the bonds vibrations.”

There are a lot of deliberately ignored vibrations in H2O (also ammonia.) It is these vibrations that are associated with liquid H2O’s (and liquid ammonia’s) high heat capacity.

It is because of the high elasticity of these hydrogen bonds (unlike covalent or ionic bonds) that there is movement (KE) conserved by this movement.

These facts are poorly understood and therefore, deliberately ignored/minimized by the tendency to dumb down models for public consumption.

James McGinn / Genius
Correction to The Current Model of Hydrogen Bonding in Water
http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=17448

• ### T L Winslow

|

The CO2 AGW hoax relies on claiming a “net energy flow” between atmospheric CO2 and the surface to argue that the Second Law of Thermodynamics isn’t violated by CO2 back radiation, when we all know it is.

Here’s the hoax being pushed by the DeSmog smear blog:

In its table titled “Fact/Myth/Fallacy” it states:
“2nd law talks about net flow of energy, and doesn’t forbid some flow from cool to hot.”

It’s all about the pesky persistent Planck (blackbody) Radiation Law, the Wien Displacement Law, and the Second Law of Thermodynamics:

The CO2 hoaxers’ con game is exposed by the simple observation that the Second Law states that a colder body can’t raise the TEMPERATURE of a hotter body. It says nothing about net energy flow. CO2’s Max, er, max Planck radiation wavelength is 15 microns, which corresponds to a Planck temperature of 193K (-80C), which can’t even melt an ice cube. It might warm a -81C ice cube if it isn’t in water form 🙂

Check my work here by typing 193K into the temperature field:
https://www.spectralcalc.com/blackbody_calculator/blackbody.php

You can sit in an ice cave and bask in all that 0C (10.6 micron) Planck radiation all day, but it won’t cook your meal. To raise the temperature of an object with max Planck radiation wavelength l, the radiating object must have a max Planck radiation of wavelength less than l, which requires it to have a higher Planck temperature. You can’t fool Mother Nature. Any longer wavelength radiation will either bounce off (reflect, like light) or get absorbed and reemitted at that longer wavelength,, which doesn’t warm the hotter object one gnat ‘s whisker because the Planck radiation curve already includes all longer wavelengths but only the max power wavelength determines the temperature, along with what temperature objects it can warm.

It’s worse than that, because to raise the temperature of an object the second object must not only have a hotter temperature (lower max Planck wavelength), it must have the same or greater power to switch it to a higher temperature curve. Try holding a lit match in front of a cigarette at arm’s length and seeing how it won’t light until you narrow the distance.

In short, a colder object’s radiation can’t push the Planck radiation curve to the left (shorter wavelength) and warm it up, but rather the hotter object will pull the colder object’s curve to the left, warming it up, while standing pat. Look up Doctor Dolittle’s Pushmi-Pullyu and see if 5th grader scientists can groove on it.

https://en.wikipedia.org/wiki/List_of_Doctor_Dolittle_characters

Therefore, the whole CO2 back radiation narrative of so many watts per square meter of -80C radiation from some indeterminate range of heights in the sky is a pure hoax designed to fool thermodynamics ignoramuses, which sadly includes most so-called physicists.

“The law that entropy always increases holds, I think, the supreme position among the laws of Nature. If someone points out to you that your pet theory of the universe is in disagreement with Maxwell’s equations – then so much the worse for Maxwell’s equations. If it is found to be contradicted by observation – well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation.” – Sir Arthur Stanley Eddington (1882-1944)

Read my super cool collection of Quora articles tearing the CO2 greenhouse warming theory apart and revealing that it’s not only a scientific hoax, it’s being pushed for political purposes and is the ultimate fake news. How can we grab the public’s attention without a world stage like Great Tuna, er, Greta Thunberg gets?

http://www.historyscoper.com/tlwsquoraclimatechangearticles.html

• ### Pierre R Latour

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The temperature of a body only increases when the rate or energy in all forms going in is greater than the rate of energy in all forms going out. That is the First Law of Thermodynamics; Conservation of Energy.
The rate of temperature change, dT/dt, depends on the rate IO difference and body mass & heat capacity. In particular mCpdT/dt = accumulation rate = input rate – output rate. Note no problem adding and subtracting fluxes.

• ### geran

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You have no problem adding and subtracting fluxes because you don’t understand the science. You are unknowingly championing pseudoscience.

A certain PhD (meteorology) would agree with your entire paragraph, then use it to support the GHE. And you’re tagging right along….

• ### Pierre R Latour

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Geran, Perhaps you have difficulty adding and subtracting flows because you don’t understand science or engineering? Even after I explained it with a simple example? Did you know your body adds and subtracts energy with every swallow and twitch?

• ### lifeisthermal

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But that doesn´t mean that you can decrease the outgoing energy and heat something up. Because if that was true, I could cool down the outside of the walls of my house and the Indoor temperature would rise. And that´s exactly what the GHE says, that less emission by already cold air(which means that the air is cooling) makes the heat source(surface) warmer. In the video linked below, the reactor´s emission is reduced by the water surrounding it, just like the fluid atmosphere reduces the emission by the surface, but the reactor doesn´t get warmer from that, it gets colder from the water:

• ### Pierre R Latour

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Lifeisthermal,
You are correct. Radiant energy from the colder absorbed by the hotter emitter usually does not raise hotter temperature is because hotter emits and transfers at a greater rate, usually. But not always. My paper gave the criteria.

• ### Tom Martin

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after all is said and done, if the alarmist were actually scientific types — as they say they would have done an experiment that demonstrates doubling CO2 actually raises the temp 2 or more degrees.

• ### Pierre D. Bernier

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And all references are to himself ! ! !
GAAAAAAAAAAWWWWWWWWWWWDDDDDDDDDDDDDD

• ### Pierre R Latour,

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And what pray tell is wrong with that?

• ### geran

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Dr. Latour, you are a knowledgeable Skeptic. I have seen some of your other work, and you mostly get it right. But here, you have some things wrong.

The equation that involves subtracting the fourth power of two different temperatures is NOT the Stefan-Boltzmann equation. You are trying to use that equation to somehow disprove the GHE. Doing so leads you to some misinterpretations.

The equation is invalid, at the outset. So is the GHE. This recent article, sadly, just adds confusion. I doubt that was your intention.

• ### Pierre R Latour

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Geran, I didn’t say it was. I said S-B gives intensity, a fraction of which is absorbed. The net transfer is difference of two way absorbed intensities. What law of radiant energy transfer do you propose? Light travels is all directions.

• ### geran

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Light can travel in all directions, but you’re still making the false assumption that all photons will be absorbed. You’re still trying to use the bogus “2-T’s” equation.

• ### Pierre R Latour

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Not so. I made no assumption about photons; I did not mention them. I use the fraction of incident radiation at a surface that is absorbed as absorptivity, by definition. The rest of incident radiation is transmitted and reflected. Elementary.

• ### Pierre D. Bernier

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From our very good friend Rosco at Climate of Sophistry…
https://dl.dropboxusercontent.com/s/vzjo38j2vn6p6p0/239.7%20%2B%20239.7.png
Adding and subtracting power fluxes and calculating temperatures using the SB equation does not give the same answer as performing the exact same algebra using Planck’s law.

• ### Zoe Phin

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Pierre,
Rosco uses a strawman. No one claims that adding fluxes means you stack spectral intensities on top of each other.

For example, take two bodies with same heat capacity. One is 20C, the other 10C. Suppose they both emit blackbody radiation. The result is a completely different 15C blackbody, and this has nothing to do with stacking spectral intensities on top of each other.

• ### Pierre R Latour

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Agree result at 15C is different than 10C + 20C. Know why? It is in my paper. Q = T14 – T04. The fourth power is not linear.
Ever notice every automobile in the world has two headlights since 1910? If intensities are not additive they would get by just as well with one.

• ### Matt

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Avoiding close proximity blindspots is primary and cross beam hazard and distance illumination and evaluation are drivers for two headlights.
Regards. Matt

• ### James McGinn

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Bernier:
And all references are to himself ! ! !
GAAAAAAAAAAWWWWWWWWWWWDDDDDDDDDDDDDD

Latour:
And what pray tell is wrong with that?

James McGinn:
Absolutely nothing!!! Pierre, be aware that in this group (PSI) there are a lot of groupies of conventional scientific thought. They have formed emotional attachments to oversimplified models meant to appeal to the lowest common denominator of science consumers.

James McGinn / Genius / Solving Tornadoes
http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16329&start=360#p125466
I discovered the empirical shortcomings of meteorology after I discovered them in climatology. My reasoning was very simple. Knowing that the origins of climatology are in meteorology, I reasoned that if AGW is as bad as it appears then meteorology must also have skeletons in its closet. So I did something that nobody has done before, I looked at the convection model of storm theory with scrutiny. I found numerous fatal flaws and I found that meteorologists have long ago established a tradition of ignoring these fatal flaws.
My point is that you/we cannot defeat a conversational science based on empiricism because conversational sciences are based on allegories that appeal to the base sensations of the public. The only way to defeat a conversational science is to reveal it as such to the public. And the best way to reveal it to the public is to start with meteorology since this is the spring from which it sprang (or is it sprung?). The conversational tradition is the problem and its roots are in meteorology, not climatology.

• ### Herb Rose

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Hi James,
It’s good to hear from you again.
Climate is determined by the energy received from the sun by the Earth. Weather is how that energy is redistributed around the Earth.
Herb

• ### Pierre R Latour

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Probably right. I clearly simplified by averaging emissivity and absorptivity over wavelengths. Just use Plank to get effective properties to harmonize.

• ### Pierre R Latour

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I didn’t calculate temperatures from S-B, merely took difference between radiant energy absorbed by each radiator to find the net transfer between them. Elementary. I never mentioned heat.
Nobody has derived an alternate radiant energy transfer law.

• ### geran

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The “T0^4 – T1^4” is your problem. You are attempting to subtract fluxes.

• ### Pierre R Latour

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Nothing wrong with subtracting fluxes. Accumulation of water in Lake Tahoe is incoming flow – out going flow. Why dies anyone have as problem with that? Law of Conservation of Matter! Rate of convective heat transfer through a brick is surroundings temperature – brick temperature.

• ### geran

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In general, fluxes cannot be treated with simple math. They cannot be added, subtracted, averaged, etc. The confusion comes from the fact that most people have little experience with radiative physics.

Fluxes are photons. Not all photons are the same. Photons from one surface temperature will be different than photons from a different temperature surface. That’s why flux is not a conserved quantity.

Your example of water flow would be an example of conservation of mass. Photons and fluxes are different “animals”. Different fluxes will have different spectra. Different fluxes will have different impact on a surface. Different fluxes are DIFFERENT.

• ### Pierre R Latour

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But energy is not. Agree energy fluxes have different spectra but that does not preclude adding energy fluxes. Matter does it all the time. Two dissimilar fluxes absorbed (perhaps with different absorptivity) by a brick heat or cool it by First Law I gave.

• ### geran

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No matter does NOT do it all the time. Now, you’re just making things up.

Different spectra is exactly what does preclude the fluxes simply adding.

• ### Zoe Phin

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Geran,
Water vapor in the atmosphere can readily absorb both infrared from earth and uv/light from the sun at the same time – and gain internal energy.

Water vapor doesn’t do math and compare earth to sun or sun to earth, it just readily absorbs in the frequencies it does and that’s that.

• ### Pierre R Latour

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So you deny all matter radiates em energy and absorbs em energy? Wow!!!!
The spectra of two flashlights shining on floor add intensities, becoming new beam with different spectra.

• ### Pierre R Latour

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Fluxes/flows of matter and energy are vectors. Mathematics shows how to add and subtract vectors. Navigation is based on it.

• ### lifeisthermal

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Good example. Now direct them straight towards each other. Does anyone of them increase their output in W/m^2?

They don´t. For the same reason the atmosphere doesn´t increase surface emission, at any wavelength.

• ### Pierre R Latour

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You want observational evidence intensity sets radiation rate, not temperature? Study the diagram of sunshine. Intensity is greatest at noon; so is energy transfer and surface temperature. Intensity is lower near sunset; so is energy transfer and temperature. If temperature difference with sun is greater, why isn’t energy rate? Because, unlike conduction & convection, radiation rate is driven by intensity difference, not temperature difference. Evening gets cooler. There is no violation of Second Law in my article.

• ### geran

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All wrong! The solar flux is greater at noon, than at sunset, due to the angle and the atmosphere.

I’m not even sure you understand the basics, now.

• ### Pierre R Latour

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Geran,
You are guilty again of claiming something I did not say! Reread my answer carefully. You just said what I said; then disagreed. Hard to debate a poor reader.

• ### geran

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Wrong again, Latour. But false accusations appear to be one of your tools. Your physics is incompetent and your logic is even worse.

What I was guilty of was believing you were knowledgeable. I have since corrected that false belief.

• ### Pierre R Latour

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Intensity is better term than flux.
Your first sentence is true. Second does not follow and is false.
I said solar transfer from sun to surface is less at sunset, proving transfer is less even though surface is at lower temperature than at noon and temperature difference is greater. I rest my case. I doubt you understand the basics.

• ### Tom Anderson

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There seems to be many misstatements and errors here Nobody seems to be a physicist qualified to discuss the physics of the atmosphere and climate authoritatively much less correctly. Seems it is always like on our skeptical sites.

As to apportionment of radiation to a sphere as opposed to a plane, I recommend Dr. Charles Anderson’s calculations made in cooperation with Dr. Alan Siddons at:

• ### geran

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Yes Tom, Dr. Anderson is an especially knowledgeable physicist.

He’s not a relative, is he?

• ### John Harrison

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I have been over this many times but there are some who are far too attached to their pet theories and cling to a belief that they can prove that back radiation and net flow of energy via IR photons cannot occur. The counter-argument is usually that radiation from a cold object cannot warm a surface at a higher temperature but this is a deliberate mis-direction as that is not the claim being made by reasonably intelligent scientists. The indisputable fact is that some of the IR photons which are emitted from “activated” GHG molecules will strike the surface of the Earth and some of those must be absorbed. As previously pointed out, the Earth’s surface is multimolecular. The AVERAGE temperature of the Earth’s surface is indeed higher than that of the GHG but the miriad individual molecules on the Earth’s will occupy an infinite number of energy states. Therefore those molecules which occupy the lower energy states are perfectly capable of absorbing an IR photon back-radiated from a GHG molecule. However, in the meantime the Earth’s surface, as it is at a higher AVERAGE temperature will have emitted far more IR photons that it has absorbed and so continues to cool. So undoubtedly there is energy flow in both directions giving rise to a NET flow away from the Earth’s surface as it cools. However, the rate of cooling will depend on the magnitude of the NET flow of electromagnetic energy. This is the plain fact, you can apply all the inappropriate and irrelevant equations and as many mis-directions as you wish but this truth is inescapable. Move on…

• ### geran

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Harrison, that’s an awful lot of pounding on your keyboard just to say that photons are flying in all directions.

But at least you ended up with the correct conclusion that the surface warms the atmosphere, as it cools. The heat flow is from Sun to surface to atmosphere to space.

We can move on now.

• ### John Harrison

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Geran. I note that you have avoided mentioning the implications of the above wrt back radiation reducing of the rate of cooling of the Earth’s surface. Would you care to comment on that?

• ### geran

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I didn’t avoid anything. I acknowledged your acceptance of reality:

“…in the meantime the Earth’s surface, as it is at a higher AVERAGE temperature will have emitted far more IR photons that it has absorbed and so continues to cool.”

“So undoubtedly there is energy flow in both directions giving rise to a NET flow away from the Earth’s surface as it cools.”

• ### Zoe Phin

|

John,
geran is correct.
There’s no evidence that backradiation does anything.
Pyrgeometers measure Upwelling-from-instrument IR.
And another instrument measures backradiation only because it’s hypercooled to 4K.

• ### John Harrison

|

Hi Zoe. Whether back-radiation has any significant effect is an entirely different question. Instead of trying to claim that such contravenes the laws of thermodynamics, which is so patently incorrect that it is giving skeptics like myself a bad name, it is the magnitude of the effect in reducing cooling rates of the Earth’s surface and atmosphere which should be in question. Unfortunately, Geran, Joseph, Geraint etc have found themselves in the same position as the alarmists in that they have painted themselves into a corner and fear to acknowledge that they might be in error. Instead, just like the alarmists, they are resorting to abuse and incorrect application and interpretation of the physics in order to set up a smokescreen behind which they hope to escape exposure to ridicule, This has not worked with some of the more knowledgeable physicists who have posted above.

• ### geran

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Harrison, what gives people like yourself a bad name is your corruption of science and misrepresentation of others.

Now pound on your keyboard some more. It’s fun to watch.

• ### Pierre R Latour

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John Harrison is correct.
I want to thank all commenters on their interest in my article, now available as pdf from link at bottom of article.
I knew many commentators would try to find fault because it corrects along held misconceptions, like
1. Rate of radiant energy transfer is only proportional to temp difference, independent of radiator and absorber properties. Even if it seems outlandish.
2. Radiant energy always transfers from warmer radiator (500C) with low emissivity (0.2) and high absorptivity (0.9) to a cooler radiator (499C) with high emissivity (0.9) and low absorptivity (0.3). Even if it seems outlandish.
3. Warm radiators cannot absorb radiation from cooler radiators, although they have no way of detecting such cooler radiators from afar. Which seems outlandish.
4. The notion of net radiation rate from a cooler radiator to a warmer one violates Second Law of Thermodynamics. Which seems outlandish.

• ### Pierre R Latour

|

Please omit “net” from last entry #4. Although I think it is correct.
As cooler radiator cools more and hot gets hotter, they reach steady state where each emits and absorbs at the same rate.
Remember Tom Edison worked for a decade to find a bulb filament with high enough emissivity so the temperature was not so hot and filament did not burn up.

• ### James McGinn

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A few days ago, on his website, Postma revealed that the definition of “net” that he has been assuming all of this time. It is not a scientifically concise definition but it is a definition he has adopted from economics!!!

https://climateofsophistry.com/2019/10/26/watch-/#comment-53316

This is Postma’s thermodynamically invalid definition of “net”:
1 : free from all charges or deductions: such as
a : remaining after the deduction of all charges, outlay, or loss”

Here is the correct, thermodynamically valid, definition of net:
The total sum of heating and cooling of an entity after all inputs and outputs of energy to the entity have been accounted for.

In the same thread, postma continued his rant saying, “. . . Latour is a clandestine fraud who spent years infiltrating us. Seriously, about two or three years. Once he was firmly “in”, then suddenly he comes out with some “solution” where he claims that cold can heat hot, . . .”. And then we get an indication of the depth of Postma’s paranoia driven delusion: “. . . he was quite the operation. A good several years getting in with us. Now, he’s just another Doug Cotton, or Zoe, attempting to harass me.”

So, apparently, according to Joe Postma, anybody that does not agree with Postma’s method of borrowing definitions from other disciplines and inserting them into thermodynamics is a fraud.

1. Rate of radiant energy transfer is only proportional to temp difference, independent of radiator and absorber properties. Even if it seems outlandish.

Excellent point. This is especially a problem when dealing with water because the high elasticity of the categorically distinct bonding of hydrogen bonding between water molecules and its implications have been completely misunderstood by modern science. (Herb is starting to catch on about the categorical differences, but he isn’t quite getting it. At least, not yet, IMO.)

Radiant energy always transfers from warmer radiator (500C) with low emissivity (0.2) and high absorptivity (0.9) to a cooler radiator (499C) with high emissivity (0.9) and low absorptivity (0.3). Even if it seems outlandish.
Warm radiators cannot absorb radiation from cooler radiators, although they have no way of detecting such cooler radiators from afar. Which seems outlandish.

It’s surreal that anybody believes this. Postma is most notorious for pushing this misconception.

The notion of net radiation rate from a cooler radiator to a warmer one violates Second Law of Thermodynamics. Which seems outlandish.

Way back in 2013, in response to one of Postma’s rants on this topic, I pointed this out. Since then I have continued to see many (Squidly, Postma, and many others) continue with this nonsense.

My final comment: Don’t bring nonscientific definitions to a scientific argument.

James McGinn / Genius / Solving Tornadoes
http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16329&start=165#p122167
The ‘plasma’ of my model is novel, unfamiliar and, therefore, hard to accept. But that is the case for any scientific discovery. Alfred Wegener proposed continental drift in 1912. It took geologists 50 years to warm up to the idea. Now, however, when you look at a map of the southern Atlantic ocean the congruence of the eastern and western shorelines jumps out at you. When it came to deducing the molecular composition of atmospheric vortices and arriving at the conjecture that they contained wind shear generated, rapidly spinning polymers of H2O, I feel that I had a huge advantage that allowed me to avoid a common misassumption that traps others. I knew that the sheath of the tornado must involve some kind of molecular distinction.

These principles prevented me from making the common error of casually assuming that the molecular composition of tornadoes was the same as that of air and/or moist air. I’ve encountered a number of other tornado theorists and it is very common for them to casually assume that a tornado is just fast spinning air. They don’t take into account the fact that the sheath needed to possess the ability to resist itself from casually mixing with the surrounding air molecules. In other words, my principles of entitiness allowed me to realize that tornadoes could not persist as entities if the molecules that comprise the sheath of the tornado did not possess some kind of internal resilience greater than that of just air. Otherwise the molecules in the sheath would casually mix with those outside the sheath and the tornado would not have persistence.

• ### Pierre R Latour

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James McGinn, Good statement.
To refute GHGT it is better to use a valid radiant energy transfer law than an invalid one.
I take the Postma remarks you gave to show I still have my ability to learn. He did seem to go astray a few years ago. I admired his earlier work. Then, like Michael Mann, he blocked publication of one of my articles he could not disprove, with colorful emotion.
To be clear, if properties are such that net energy flows from cold radiator with higher intensity to warm one with high absorptivity because cold intensity is greater than warm, the colder one does heat the warmer one a bit until their intensities are equal and steady state is reached. Their temperatures remain different because they are different kinds of radiators. I submit not all the surfaces in your den are at “room temperature”. I admit I have not found properties of this kind, because I am not looking for them.

• ### Herb Rose

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Hi Pierre,
I don’t recall if I tried to explain my thoughts to you before on the inaccuracy of the reading of a thermometer and kinetic energy so excuse me if I am repeating it.
If a 1 m^3 oven has a temperature of100 C and 1 m^3 pot of boiling water has a temperature of 100 C according to the thermometer the kinetic energy of the molecules in the oven is the same as the kinetic energy of the molecules in the water. A 1 m^3 oven at sea level has about 1 kg of molecules. A I m^3 pot of boiling water has 1000 kg of molecules. If the area of the thermometer measuring the temperature is the same in both containers there will be 1000 molecules in the water transferring heat to the thermometer for every molecule in the oven transferring heat. The mass transferring heat in the water will be about 1000 times the mass transferring energy in the oven, so if the same amount of kinetic energy is going to both thermometer the velocity of the molecules in the water must be 1000 ^ -2 that of the molecules in the oven. Therefore the kinetic energy of the gas molecules is greater than the kinetic energy of the water molecules and even though they have the same temperature the gas molecules will transfer energy to the water molecules.
Herb

• ### Pierre R Latour

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Hi Herb. I part company with “The mass transferring heat in the water will be about 1000 times”. I think you confuse transient with steady state. Temperature is a point property of the kinetic energy of matter. At steady state, thermal equilibrium, the temperatures of the small thermometer bulb surface and its immediate surroundings are the same, hence no thermal energy transfer either way. This by definition of steady state. This is independent of the size/mass of the medium surrounding the thermometer. No matter how big the ocean is, the measured temperature one inch below the surface at Cape Cod is what it is. Of course if matter is moving by the bulb, situation is more complicated without steady state.

• ### Herb Rose

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Hi Pierre,
I consider the steady state to be when the energy being absorbed/radiated by the bulb is equal to the energy being radiated/absorbed by the thermometer’s body. If you were to expose the tip of the thermometers bulb to ice water (0 C) only a mall amount of the mercury would be transferring heat into the water while a greater amount of mercury in the bulb and in the body of the thermometer would be absorbing heat. This would result in the mercury contracting less than if the entire bulb was submerged and the temperature reading, when the two rates equalized, will be greater than 0 C.
Thermometers are designed with a fix area to be exposed to the medium being tested and are calibrated using ice water and boiling water. If the thermometer is not used correctly it does not give an accurate reading so it is possible to get readings of boiling water greater than 100 C and ice water with a temperature lower than 0 C. The fixed area of the bulb is needed to give a constant number of molecules transferring energy to it. So how does using the thermometer in a gas (which it is not designed or calibrated for) differ from not submerging the entire bulb in a liquid?
Herb

• ### Pierre R Latour

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Herb, I think I answered that. My definition differs from yours, but key is all temperatures are at equilibrium; constant. The assumption is the bulb surroundings are at the same temperature around the bulb; usually pretty close. I realize mercury temperature in stem may not be the same. All measurements are error prone. Thermocouples work better; they are smaller.
This seems irrelevant to my paper. Temperature cannot be detected at a distance across space, even filled with matter. Radiation does not suffer than handicap; hence the different transfer rate law.

• ### Pierre R Latour

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While several take issue with my result, none have questioned my proof. None have disproved it except with authoritarian language.
I properly account for physical properties of both radiators. I proved net radiant energy transfer is from Earth’s surface to atmosphere and increasing CO2 increases it because absorptivity outweighs emissivity. This nullifies GHGT. Deniers rejoice. QED