Planet Mean Surface Temperature Equation: Tmean =[Φ(1-a)S (β*N*cp)¹∕ ⁴ /4σ]¹∕ ⁴

based on Planet Surface Rotational Warming Phenomenon

Stefan-Boltzmann emission law doesn’t work vice-versa !

The T = ( J /σ )¹∕ ⁴ is a mistake !

Stefan-Boltzmann emission law doesn’t work vice-versa !

The old convincement that the Stefan-Boltzmann emission law works vice-versa is based on assumption, that EM energy obeys the 1st Law of Thermodynamics (1LOT). That assumption was never verified, it was never been confirmed by experiment.

Let’s see:

The Stefan-Boltzmann emission law states:

 J = σ*Τ⁴ (W/m²) EM energy flux (1)

The mathematical ability to obtain T, for a given J led to the misfortunate believe that the Stefan-Boltzmann emission law formula can be used vise-versa:

T = ( J /σ ) ¹∕ ⁴ (K) (2) as the surface (vise-versa) radiative emission temperature “definition”.

Well, this is theoretically right for a blackbody theoretical approach. Blackbody surface behavioral property is compared with a tiny hole in a stove. The incident in the hole radiative energy vanishes inside the stove… The hole is infinitesimally smaller than the stove’s inside walls area. Thus the incident in the hole EM energy cannot escape out of the stove.

After multiple interactions with the stove’s walls, the incident in the hole the entire EM energy is transformed into heat and is, eventually, evenly dissipated and accumulated as HEAT in the stove’s inner walls…

The EM energy emitted out of the stove’s hole is then only the inside stove uniform surface temperature T dependent function

J = σ*Τ⁴ (W/m²).

But the

T = ( J /σ ) ¹∕ ⁴ (K) (2) as the irradiated surface (vise-versa) radiative emission temperature “definition”… is utterly unacceptable, because it has not a physical analogue in the real world.

That is why we should consider planet effective temperature Te = [ (1-a) S /4σ ]¹∕ ⁴ (K)

as a mathematical abstraction, which doesn't describe the real world processes.

 

Transformation of energy is inevitably accompanied by heat release, but it is not energy transfer itself…

The notion:

T = ( J /σ ) ¹∕ ⁴ (K)

is a misfortunate scientific mislead, because it is a huge mistake. The Stefan-Boltzmann emission law doesn’t work vice-versa. 

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Opponent:

"The Stefan-Boltzmann law is based on observations and is a foundation of modern physics. However it – and any odd derivation thereof – cannot be used to calculate Earth’s surface temperature."

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" S-B is used all the time to calculate an ‘effective temperature’. That is not the surface temperature – as a direct result of atmospheric radiative transfers."

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Answer:

Planet effective temperature is written as:

Te = [ (1-a) S /4σ ]¹∕ ⁴ (K)

Planet effective temperature uses the notion:

T = ( J /σ ) ¹∕ ⁴ (K)

which is utterly unacceptable, because it has not a physical analogue in the real world. That is why we should consider planet effective temperature

Te = [ (1-a) S /4σ ]¹∕ ⁴ (K)

as a mathematical abstraction.

The Stefan-Boltzmann emission law

J = σ*Τ⁴ (W/m²)

EM energy flux, doesn’t work vice-versa. 

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Opponent:

"The effective average radiating temperature is not 288 K...

 J is an average emission. (1-a)S simply estimates IR emissions based on absorbed radiation. It works because it is empirical physics – but it does not give the surface temperature."

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Answer:

It is an emission of a body at temperature T.

Opponenet:

> “(1-a)S simply estimates IR emissions based on absorbed radiation.”

Answer:

It was thought that (1-a)S is entirely absorbed, but it is an assumption. It is a very much mistaken assumption.

Opponenet:

> “It works because it is empirical physics – but it does not give the surface temperature.”

Answer:

No, it does not work that way… The Stefan-Boltzmann emission law

J = σ*Τ⁴ (W/m²)

EM energy flux, doesn’t work vice-versa.

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Opponent:

"Energy is conserved – it is either reflected or absorbed and reemitted. "

___

Answer:

It is either SW reflected or transformed into IR EM energy… and to HEAT which gets accumulated. When rotating faster and when having a higher cp, less SW energy is transformed into IR EM energy, and more SW energy is transformed to HEAT, and, therefore, more HEAT is accumulated.

Later that accumulated HEAT is also IR emitted, but when there is more HEAT accumulated and more energy emitted later in the night, the planet on average surface is warmer.

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The New equation

Tmean = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴

actually can be simplified as:

Tmean = Β*(N*cp)^(1/16)* [ Φ(1-a) S ]¹∕ ⁴

Planet effective temperature is written as:

Te = [ (1-a) S /4σ ]¹∕ ⁴ (K)

Planet effective temperature uses the notion:

T = ( J /σ ) ¹∕ ⁴ (K)

which is utterly unacceptable, because it has not a physical analogue in the real world.

That is why we should consider planet effective temperature

Te = [ (1-a) S /4σ ]¹∕ ⁴ (K)

as a mathematical abstraction.

The Stefan-Boltzmann emission law

J = σ*Τ⁴ (W/m²)

EM energy flux, doesn’t work vice-versa.

Not the entire not reflected portion of incident solar flux is transformed to HEAT. 

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When averaging solar flux

Te = [ (1-a) S /4σ ]¹∕ ⁴ (K)

we do not know how much SW solar energy is transformed to HEAT…

That is why flux cannot be averaged. it irradiates planet from one side, and the amount of SW energy transformed and accumulated to HEAT is determined by the Planet Surface Rotational Warming Phenomenon.

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The New equation considers planet in total. There is not any averaging taking place.

Tmean = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴

the

Tmean = [ (1-a) S /4σ ]¹∕ ⁴ (K)

is a partial case, when a planet has

N*cp = 1 /150

and therefore, since

β =150

(β*N*cp) = 1

and

Tmean = [ Φ (1-a) S (150 /150)¹∕ ⁴ /4σ ]¹∕ ⁴

thus

Tmean = [ (1-a) S /4σ ]¹∕ ⁴ (K)

There cannot be any planet effective temperature.

Planet average surface temperatures are strongly dependent on the Rotational Warming Phenomenon.

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Opponenet:

" 90% of incoming energy is converted to kinetic energy in oceans – measurable by thermometers. 4% is in latent heat in liquid water and vapor. The rest goes to warming land. "

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Answer:

When averaging solar flux

Te = [ (1-a) S /4σ ]¹∕ ⁴ (K)

we do not know how much SW solar energy is transformed to HEAT…

That is why flux cannot be averaged.

Solar flux irradiates planet from one side, and the amount of SW energy transformed and accumulated to HEAT is determined by the Planet Surface Rotational Warming Phenomenon.

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Opponent:

"All solar irradiance that is not reflected heats the planet – one way or another.

The sun heats oceans and land – that’s what results in IR emissions – that are recycled between surface and atmosphere when photons are not being emitted to space. "

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Answer:

When solar flux hits planet surface the process is a radiative SW EM energy interaction with matter.

When SW EM energy interacts with matter a TRANSFORMATION of energy occurs.

Transformation of energy is inevitably accompanied by heat release, but it is not energy transfer itself…

The notion:

T = ( J /σ ) ¹∕ ⁴ (K)

is a misfortunate scientific mislead, because it is a huge mistake.

The Stefan-Boltzmann emission law doesn’t work vice-versa.

These two IR emission mechanisms cannot be compared, because they are different.

Here is the discovered Rotating Planet Surface Solar irradiation Interaction-Emission New Equation:

Jemit = 4πr²σΤmean⁴ /(β*N*cp)¹∕ ⁴ (W)

And compare with the Stefan-Boltzmann emission law equation:

J = σΤe⁴ (W/m²)

– When comparing with the Stefan-Boltzmann blackbody emission – it is obvious when planet emission is considered – it is a different mechanism of emission.

Thus the equation describing IR emission from irradiated rotating planet surface is different.

-

When averaged over the entire planet surface, in order to compare with the Stefan-Boltzmann blackbody emission:

Jemit = σΤmean⁴ /(β*N*cp)¹∕ ⁴ (W/m²)

And compare with the Stefan-Boltzmann emission law equation:

J = σΤe⁴ (W/m²)

It is obvious these two IR emission mechanisms cannot be compared, because they are different.

Thus, the planet effective temperature Te formula:

Te = [ (1-a) S /4σ ]¹∕ ⁴

Which results for Earth Te =255K

cannot be compared with the planet measured average surface temperature Tmean = 288K.

When we say different, it is something else, not the same as what we used to think – it is different…

"If it is a different mechanism of emission Is it magic then?”

When solar irradiated, planet surface doesn’t behave as the inner surface of the blackbody stove…

That is why it is so much different. And planet when IR emitting do not use a hole to let the EM energy out.

Thus, there is not any magic in solar irradiated planet surface IR EM energy emission mechanism.

It is simply a different than the Stefan-Boltzmann equation describes for the blackbody emission mechanism.

It is different, not magic. When we say different, it is something else, not the same as what we used to think – it is different…

The same radiation from a warmer planet - yes, that is how it happen.

"If a black body is rotating, does it alter its radiation because of that. No."

1). Planet is not a blackbody.

2). Blackbody is not getting warmed by irradiation. Blackbody is already a warm body.

Now. you asking

"if planet rotates, does it alter its radiation because of it ?"

And you gave, all by yourself, the correct answer. You answered "No."

- Of course, when planet rotates faster, it is a warmer planet. The same radiation from a warmer planet - yes, that is how it happen.

It is the Planet Surface Rotational Warming Phenomenon which does the job! 

By the reversed Stefan-Boltzmann law what we are referring to is the “absorbing” surface.

 May 4, 2022

Opponent:

"Christos Vournas Have you ever used an IR thermometer to get a temperature reading at a distance? The instrument receives IR to a sensor.

Based upon the temperature change of the sensor based upon a reference a calculation is made using the Stefan-Boltzmann relationship of radiant energy to temperature (taking into account emissivity).

You can experimentally verify that the Stefan-Boltzmann Law works in reverse by comparing the temperature reading you get on the IR thermometer with using a conventional thermometer on the same object to see how close they match (try it with water that has a reasonable high emissivity).

That a glass of water an get a reading with an IR thermometer then use a conventional thermometer on the water and see how close they match."

Answer:

Thank you for your respond.

“…by comparing the temperature reading you get on the IR thermometer with using a conventional thermometer on the same object to see how close they match…”

You describe the IR thermometer calibration process… What IR thermometer does is to measure surface temperature depending on the surface’s IR radiation intensity…

Knowing " T ", we can calculate " J ". Or, knowing " J ", we can calculate " T ". The equation works either way, at the emitting surface.

Well, you do not use the Stefan-Boltzmann emission law in reverse here…

The Stefan-Boltzmann emission law states:

J = σ*Τ⁴ (W/m²) EM energy flux (1)

In your example you refer to the by surface the IR EM energy emission intensity.

The reversed Stefan-Boltzmann law is about the incident on the surface EM flux’s " J " ability to warm the surface in the reversed way.

By the reversed Stefan-Boltzmann law what we are referring to is the “absorbing” surface.

The equation is no longer valid (for the purpose of irradiated surface mean temperature evaluation), as the not reflected portion of incoming flux is not entirely absorbed and emitted.

Thank you for helping to clear this out.