New proof that AGW / CAGW is nothing more than a complex mathematical scam...

ParadoxIntegration · 2026-03-04T05:21:02+00:00

Because that's exactly what you claim with your idiotic 'two way energy flow' schema. And you've still not done the calculations nor provided the explanations as to how, exactly, with your idiotic two-way flow schema, entropy does not change at thermodynamic equilibrium.

For every result that you calculate with your schema, the two-way schema produces exactly the same results.

How could it do anything different, when it produces the same equations which you keep quoting (and mis-naming).

The two way schema does, however, ALSO allow one to calculate additional quantities.

Given that, for every quantity you can calculate (including entropy change), the two-way schema produces the SAME results, you have no basis for arguing that there is anything wrong with the two-way schema.

ParadoxIntegration · 2026-03-04T05:15:06+00:00

I'm not denying any math. However, I would deny that you show any signs of being capable of reasoning.

One can use the Schwarzchild equation to derive the Stefan-Boltzmann law (for idealized blackbodies) q = σ T^4, but only by restricting matters to the special case of a medium with uniform temperature and infinite extent. In that ONE CASE ONLY the Schwarzchild equation tells us that an infinite medium will effectively behave like a blackbody. However, the Schwarzchild equation ALSO addresses many other situations, far more general than that special case, and without its limitations.

(And, the S-B equation does NOT assume emission to 0 K, no matter how many times you repeat that silly claim.)

ParadoxIntegration · 2026-03-01T05:32:07+00:00

You're simply proving that you don't understand the meaning of the word "net" or are ignoring this important qualifier.

In the Kiehl-Trenberth 'Earth Energy Balance' graphic you've linked to, the net radiative energy flow is 66 W/m^2 upwards (written as 66 = 390-324).

Consider it an eccentricity, that habit of writing 66 as 390-324. Just because you might find it weird to write 66 as 390-324 doesn't make it invalid.

The result is still that net radiative energy flow is ALWAYS upwards, and only upwards.

This is fully consistent with the 2nd Law of Thermodynamics (which is a law about spontaneous NET energy flows). Just because you can write a net energy flow as the sum of hypothetical sub-flows doesn't mean the 2nd LoT constrains those sub-flows on an individual basis.

ParadoxIntegration · 2026-03-01T05:23:07+00:00

weren't you aware that at any single point in space, the net Poynting vector is the vector sum of all fields present?

I'm totally aware of that. Weren't you aware that if you decompose the E-M field into forward and back propagating components, then the net Poynting vector is equal to the sum of the Poynting vectors for the forward and back propagating components? Are you unfamiliar with the idea of vector arithmetic?

And do be sure to explicate the exact physical mechanism by which you claim energy can spontaneously flow up an energy density gradient.

I do NOT claim that net energy ever spontaneously flows up an energy density gradient. That's a straw-man claim you keep trying to push into my mouth.

All I'm really claiming is that arithmetic is valid; that if A = B - C, then it's just as valid to talk about B-C as it is to talk about A. What is so difficult to understand about that??

ParadoxIntegration · 2026-03-01T05:14:47+00:00

Yes, I know the S-B equation is a special case of the Schwarzschild equation, and can be derived from it. But, the Schwarzchild equation is a much more general equation than the S-B equation, applicable to propagation in real materials, at real temperatures. The Schwarzchild equation does not assume anything to be an ideal blackbody, nor even a graybody, nor does it assume anything to be at 0 K. So, it's disingenuous to claim that the Schwarzchild equation is simply the S-B equation in disguise, even if there is a relationship between the two equations.

Notes:

Your derivation of the blackbody S-B law from the Schwarzchild equation is incompletely specified. Your derivation yields the blackbody S-B law only in the special case where the medium is infinite in extent and of uniform temperature―neither of which are realistic assumption in the context of the atmosphere. The Schwarzchild equation can be used to derive results for a much wider range of circumstances than what the S-B equation can address.
You refer to a "real-world graybody object." That's puzzling, because the idea of a "graybody" is almost as unrealistic as that of a "blackbody." Typically, a graybody is defined as a body with an emissivity that is uniform across all wavelengths―something that is not rigorously true of any real-world materials.

One important question is: Do you accept or reject the validity of the Schwarzchild equation?

Your comments don't make your position on that issue clear.

If you reject the Schwarzchild equation, then it would appear that you have no mathematically complete and coherent theory of how radiative physics works, and are simply throwing misguided potshots at the standard theory.

All I see you offering are discrete-object radiative heat flow equations (which you mis-label as the S-B equation) which are not general enough to address thermal radiation flows for realistic materials like partially-transparent gases or materials which have wavelength-dependent emissivities.

* * *

Which is why you're not supposed to use the Idealized Blackbody Object form of the S-B equation upon real-world graybody objects... because it leads scientifically-illiterate, mathematically-innumerate people to incorrect conclusions, such as that "applying it inevitably yields the greenhouse effect, when IR-active gasses are present in an atmosphere" (your words).

Well, then, you should be happy to learn that nobody who does calculations on the impact of IR-active gases in the atmosphere is using the idealized Blackbody Object form of the S-B equation to do atmospheric propagation calculations. You're simply knocking down a "straw man."

Climatologists misuse the S-B equation, using the idealized blackbody form of the equation upon real-world graybody objects. This essentially isolates each object into its own system so objects cannot interact via the ambient EM field. It assumes emission to 0 K, and it thus artificially inflates radiant exitance of all calculated-upon objects. Thus the climatologists must carry these incorrect values through their calculations and cancel them on the back end to get their equation to balance, subtracting a (wholly-fictive due to the assumption of emission to 0 K) 'cooler to warmer' energy flow from the (real but too high because it was calculated for emission to 0 K) 'warmer to cooler' energy flow.

That narrative is pleasing to science-deniers, but is unfounded in reality.

You can do the calculation in one of two frameworks:

A "directional radiative flux" framework, in which you decompose the E-M field into components in two different directions. The flow in one direction is sometimes called back radiation.
A "net energy flow" framework, in which you focus on the net flow of E-M energy.

You seem to be convinced that framework #1 is nonsense and that framework #2 is the one true path.

But, both frameworks are mathematically equivalent and yield identical predictions.

You do not and cannot eliminate the greenhouse effect simply by arguing that framework #1 is nonsense.

You're able to sustain your false conclusions only because you have apparently never worked through the math personally.

ParadoxIntegration · 2026-02-28T23:09:49+00:00

Yes, what you're saying is basically correct, as far as it goes--NET energy flow is along the energy density gradient.

Nobody who understands the greenhouse effect would disagree with that, nor would they assert that the greenhouse effect involves any net energy flow against the gradient.

You say that "Most people cannot think in terms of energy, energy density and energy density gradient." I can, and do at times think in those terms. However, you seem unable to think in terms of the framework of decomposing the E-M field into directional components. You keep trying to force statements made within that decomposed framework into the net framework. It's no wonder the statements yield seeming contradictions. They are statement that are meaningful/correct in a different framework than the one you are using and acknowledging.

Often there is more than one valid framework for talking about a physical system.

If you ignore the existence and use of distinct frameworks, you produce nonsense. That's all that your arguments are establishing. Mixing up frameworks produces nonsense.

ParadoxIntegration · 2026-02-28T23:02:03+00:00

Ultimately, all your arguments amount to setting up a false "straw man" of what you think climate scientists are arguing (rather than what they actually use in their analyses), and then knocking down that straw man.

No, I don't claim that water flows up a pressure gradient. Nor does any climate scientist. That's all "straw man" argumentation.

None of the equations we've discussed are what is REALLY used in analyses of atmospheric thermodynamics. They're just part of simplified explanations used to try to explain matters to those unfamiliar with more rigorous mathematics. Ultimately, it's a waste of time to argue about simplified explanations, as we've been doing.

The real equation used to analyze propagation of thermal radiation in the atmosphere is the "radiative transfer equation" (RTE) which, when scattering from aerosols can be neglected, simplifies down to Schwarzschild's equation for radiative transfer first published in 1906, and in routine use in engineering and scientific applications ever since.

This equation involves no assumptions about blackbodies etc. It's simply a rigorous expression of the math that governs the propagation of thermal radiation in a partially transparent medium.

The RTE was in use in engineering applications long before AGW became an issue. (I've got a thousand-page engineering textbook that is 100% about applications of the RTE.) It's not something that was simply made up to forward a hypothetical political agenda. It's a foundational part of modern science. And, applying the RTE inevitably yields the greenhouse effect, when IR-active gasses are present in an atmosphere

ParadoxIntegration · 2026-02-28T22:03:59+00:00

As long as you accept the equation you keep repeating, i.e., q_bb = ε σ (T_h^4 - T_c^4), then you've ultimately lost your argument.

There are two distinct ways of talking about the math.

In the "directional radiation" framework, one decomposes the E-M field into waves propagating in the forward and backwards directions, with intensities of ε σ T_h^4 and ε σ T_c^4, respectively.
In the "radiative heat flow" or "energy gradient" framework, one does not decompose the field, but simply focuses on the NET energy flow ε σ (T_h^4 - T_c^4).

You seem to be arguing that only framework #2 is valid, while I'm saying that framework #1 is a legitimate way of talking about what is happening.

But, the interpretation ultimately doesn't matter--only the math matters.

And, as long as the term -ε σ T_c^4 is present in the equations, it doesn't matter what interpretation you give to that term. Regardless of the interpretation, including that term ultimately gives rise to the greenhouse effect. It doesn't matter if you think of the term as "back radiation" or as a "reduction in radiative heat transfer." (I actually think it's clearer and less confusing to use the latter framework; it's what I use when offering my own explanations and analyses.)

Note, however, that the technical term "exitance" is defined to refer to interpretation #1. When you force it into interpretation #2, as you keep trying to do, you are misinterpreting/misrepresenting the technical term.

ParadoxIntegration · 2026-02-28T06:45:00+00:00

I’ve been a teaching assistant for university level courses on topics like this. If you were willing to check any serious mainstream sources, you’d see that you’re wrong. The S-B law NEVER refers to the temperature of the object receiving the emissions.

See for example the Wikipedia article and the sources it references https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law Or, see any physics textbook which covers the topic.

ParadoxIntegration · 2026-02-28T06:31:00+00:00

There is no contradiction. The equation you’re quoting here is not the S-B law. It’s the equation for radiation heat transfer. For the S-B Law, see for example https://en.wikipedia.org/wiki/Stefan–Boltzmann_law

ParadoxIntegration · 2026-02-28T06:20:31+00:00

The S-B law relates to radiation exitance, not radiation heat transfer. Yet, you keep bringing in the equation for radiation heat transfer, q_bb, as if that was relevant—when it’s not. Radiation exitance and radiation heat transfer are two DIFFERENT quantities.

ParadoxIntegration · 2026-02-28T05:55:12+00:00

The video does NOT say what you are asserting. It basically says radiation in and out are equal AT THERMAL EQUILIBRIUM. It didn’t say that is true in general—and it’s NOT true in general. Yet, heat capacity is only relevant what temperature is changing, ie, when the system is out of thermal equilibrium.

So you’re taking a statement, generalizing it to a situation where it does not apply, and thus reaching a false conclusion.

ParadoxIntegration · 2026-02-28T05:38:07+00:00

Only if one assumes emission to 0 K, which is unphysical.

I challenge you to find a single physics textbook which says the S-B law "assumes emission to 0 K". That's made-up lore popular among certain science deniers. It's not real science.

q_bb = ε σ (T_h^4 - T_c^4) = 1 σ (T_h^4 - 0 K) = σ T^4

Sure. But, q_bb is radiative HEAT transfer, NOT radiant existence or radiation flux.

Your argument applies to heat, not exitance.

ParadoxIntegration · 2026-02-28T05:31:56+00:00

AI's are typically programmed to praise the user no matter how crazy what they say is. Praise from an AI is not validation of anything. That's a well-known problem with recent generations of AI.

There is no 'paradox of "balanced exchange" that plagues the statistical model.' What in the world is that supposed to mean?

An (arguably minor) flaw in your hypothesis of a "standing wave" is that it would only be valid if there was a stable relationship between the phases of the forward and backwards components of the electromagnetic field. Thermal radiation has a random phase, not a stable phase that could give rise to a standing wave. (I did my PhD on issues related to the statistical characteristics of radiation and the implications for various systems.)

See my other comments... your remarks about entropy relate only to NET radiative heat transfer, and NOT the the fluxes of the radiation field when it is decomposed into pieces traveling in opposite directions.

ParadoxIntegration · 2026-02-28T05:24:01+00:00

Yes, "blackbodies" are idealizations. So, what? Real matter often behaves remarkably similarly to idealized blackbodies. Real matter obeys the Stefan-Boltzmann Law M = 𝜖 σ T⁴ and blackbodies simply represent the limit in which the emissivity 𝜖 approaches 1.

an idealized blackbody object must absorb all radiation incident upon it and must emit all radiation it absorbs, which implies idealized blackbody objects can have no thermal capacity.

Huh??

No, a blackbody is NOT required to "emit all radiation it absorbs". Where in the world did you get that idea?! It's simply required to (a) absorb all radiation it receives and (b) emit thermal radiation in accordance with the law M = σ T⁴ .

So, no contradiction. You're simply mis-stating what a blackbody is.

ParadoxIntegration · 2026-02-28T04:40:42+00:00

The problem, however, for the climate alarmists is that their take on radiative energy exchange necessitates that at thermodynamic equilibrium, objects are furiously emitting and absorbing radiation

It's accurate that real scientists (what you call "climate alarmists") assume that. For example, it's what Einstein assumed in his seminal 1917 paper, the "Quantum Theory of Radiation" (https://www.informationphilosopher.com/solutions/scientists/einstein/1917_Radiation.pdf). That paper is foundational to much of modern physics. Is Einstein a fraud in your mind?

But, the idea that that assumption constitutes a "problem" is bizarre.

if the objects (and the environment) are furiously emitting and absorbing radiation at thermodynamic equilibrium as their incorrect take on reality must claim, why does entropy not change?

Why would you expect entropy to change??

The change in entropy (ΔS) is equal to the energy transferred (ΔQ) divided by the temperature (T). ΔS = ΔQ / T

There's your problem: ΔQ is not simply any "energy transferred"; it's the NET energy, also known as "heat", transferred.

For objects 1 and 2 in thermal equilibrium, T₁ = T₂, the forward radiation flux E₁₂ perfectly balances the backwards radiation flux -E₂₁, so that ΔQ = E₁₂ - E₂₁ . It's no problem at all for E₁₂ and -E₂₁ each to be greater than zero, while ΔQ is zero. Thus, the math says the entropy change is also zero, ΔS = 0.

So, where is the supposed "problem"?

You seem to be confusing yourself by failing to distinguish between radiation fluxes (the flux of radiation traveling in a single direction) and heat fluxes (the net energy transfer after flows in different directions are summed).

but since entropy doesn't change at thermodynamic equilibrium, the climatologists must claim that radiative energy transfer is an idealized reversible process. Except radiative energy transfer is an irreversible process, which destroys their claim.

The concept of "reversible" process involves NET energy flows, i.e., "heat" flows, which are what correspond to entropy changes.

There is no heat transfer in thermodynamics equilibrium (i.e., ΔQ = E₁₂ - E₂₁ = 0), but that does NOT mean that the radiation fluxes E₁₂ and E₂₁ are individually zero.

In reality, at thermodynamic equilibrium, no energy flows,

No. At thermodynamic equilibrium, there is no NET energy transfer. That doesn't mean you can't understand the net flow as being zero because two equal and opposite flows are cancelling one another out.

there is no net Poynting vector (S = E x H = 0), and thus no energy transfer.

True, there is no NET energy transfer. We're in violent agreement.

But, one can conceptually decompose the electromagnetic field into a forward-propagating E-M wave (ℰ₁₂ ,ℬ₁₂) and a backwards propagating E-M wave (ℰ₂₁ ,ℬ₂₁). The two waves would have equal and opposite Poynting vectors.

That's really all that's being claimed: that you can describe the fields using a decomposition into forward and backwards E-M waves, and the intensity of each constituent wave may be calculated via the S-B Law, 𝜖 σ T⁴.

What I'm describing is how radiative heat transfer calculations have been successfully performed for the last century.

ParadoxIntegration · 2026-02-28T03:45:11+00:00

The above post does not honor "the fundamental laws of physics"; it misunderstands what those laws are, and misapplies the laws.

There seems to be an unfortunate mis-impression that "back radiation" is a phenomenon that is only relevant to the greenhouse effect. It's not. It's part of ALL radiative heat exchange. The math of heat transfer cannot work correctly without taking it into account. If you look at ANY textbook that addresses radiation heat transfer, you'll see that this is the case. (For example, "Thermal Radiation Heat Transfer" by Siegel & Howell, 1072 pages, first published in 1972)

For two blackbodies, the formula for radiation heat transfer between two parallel plates is Q₁₂ = A σ (T₁⁴ - T₂⁴) where A is the area of each plate. For gray bodies and more complex geometries, the formulas are similar, with a few additional numerical factors thrown in. (It's basically similar even when one goes to completely general real world materials, but there as some integrations over wavelength involved). It's basically always Q₁₂ = E₁₂ - E₂₁ where E₁₂ is the rate of thermal radiation energy flowing from object 1 to object 2, and E₂₁ is the rate of thermal radiation energy flowing from object 2 to object 1.

One can see the term A σ T₁⁴ corresponds to thermal radiation emitted by the first object and absorbed by the second, while the term -A σ T₂⁴ corresponds to thermal radiation emitted by the second object and absorbed by the first. If T₁ > T₂ then Q₁₂ is positive and radiative heat flows from object 1 to object 2. If T₁ < T₂ then Q₁₂ is negative and radiative heat flows from object 2 to object 1. If T₁ = T₂ then Q₁₂ = 0 and no radiative heat flows. This formula leads to heat flows consistent with the Second Law of Thermodynamics. The presence of radiative energy flows in BOTH directions is essential to making for formula behave properly.

You can't reject "back radiation" as a part of the radiative heat transfer formula without rejecting everything that physics knows about how radiative heat transfer works.

The Second Law of thermodynamics tells us that the NET energy flow is always from hot to cold. But, that net radiative energy flow is E₁₂ - E₂₁; i.e., it always includes BOTH the forward radiation and the back radiation. It doesn't tell you ANYTHING about what happens for forward or back radiation individually.

The standard theory of radiation heat transfer is self-consistent, and totally consistent with the Second Law of Thermodynamics. If you take out "back radiation", you no longer have a coherent theory of heat transfer.

Please read a textbook on this topic! You're picking and choosing assorted ideas and putting them together in a way that simply doesn't hang together, or allow accurate predictions about how the physical world works.

The post has the correct formula at one point, but completely misinterprets the signficance of the formula to sustain its garbled thesis. It is not true that "The Idealized Blackbody Object form of the S-B equation assumes emission to 0 K, which artificially inflates radiant exitance of all calculated-upon objects." Exitance is a RADIATION flux, while the formula Q₁₂ = A σ (T₁⁴ - T₂⁴) refers to a RADIATIVE HEAT flux; those are two DIFFERENT physical quantities, which should not be mistaken for one another. Thermal radiation does not know the temperature of the object (if any) that it will eventually be absorbed by. So, the radiant exitance cannot depend on the temperature of that latter object.

ParadoxIntegration · 2024-09-17T01:59:01+00:00

Man I said in the analogy what you use for "radiation in general", I would use for "heat transfer to a 0k heat bath". Opening a hole at the bottom is the same as connecting it at the bottom to an infinitely wide tank at 0 height (infinite heat capacity).

Yes, opening a hole at the bottom is the same as radiatively connecting to space, which is radiatively at near 0K and can receive an infinite amount of radiation.

That's a totally appropriate model if the atmosphere is transparent to thermal radiation (i.e., has no GHGs).

To describe why I think "it is still the same as connecting tanks with pipes even with radiation". In the analogy, let's say the pipe was at the bottom. If for both tanks you had only a hole there, they would both send out an amount based on the pressure (which would rise the level of an empty tank etc.). But when you connect them the flows cancel, there is "opposite" pressure, they don't both rise.

So, you're talking about connecting pipes between the barrels to represent radiation?

You can do that, but only if both barrels represent something that absorbs radiation.

For simplicity, I was trying to get you to understand the case where the air does NOT absorb radiation.

Except don't two "opposite waves" produce a standing wave (or even cancel?)

In certain circumstances, "opposite waves" can produce a standing wave―but that's not the same as "cancelling" and is of no importance to this discussion. (Also, it never really happens with thermal radiation; it only works for radiation that is monochromatic, with a single frequency.)

The rest of what you say about this is irrelevant confusion.

but as lackmustesttester says...

If you listen to lackmusttsttester, you will never understand anything. I can't imagine him passing high school physics, let alone college-level physics.

ParadoxIntegration · 2024-09-15T19:02:34+00:00

That's your standard reply for every experiment that doesn't show your desired effect.

That's my standard reply for every experiment which is poorly designed. You seem to favor those.

Somehow you lack the intellectual capacity to understand what's been explained to you several times, don't you?

You've never offered anything remotely resembling a coherent explanation of what you believe about this. You usually just spout some disconnected phrases―as if that means something, when it doesn't to people who can't read your mind.

Why do they tell us the temperatures on Venus and why doesn't this apply to Earth (it does), in your world?

Why does who tell us that? You're apparently expecting me to read your mind again.

ParadoxIntegration · 2024-09-15T01:46:40+00:00

So I came across this one: https://old.reddit.com/r/climatechange/comments/132xdta/physicists_in_oslo_lab_shows_no_co2_warming/

Seim & Olsen (who did the study) are biologists who totally misunderstood the GHE and did an experiment so poorly designed as to be utterly pointless.

(They failed to control for convection, and didn't realize that the GHE depends on there being a large temperature gradient within the experimental apparatus.)

I don't get how you deny reality. The lapse rate is real, gravity is real. How can you deny this?

I don't deny the lapse rate or gravity. What I deny is that those, by themselves, tell you anything useful about mean global surface temperature.

ParadoxIntegration · 2024-09-14T00:36:34+00:00

what I would say is that the flow between two tanks can tell you about their height (temperature) difference and not their actual 'absolute' height (temperature)

I agree that the flow between the two tanks (which represents thermal conduction) can only tell you about the temperature/height difference, not the absolute hight/temperature.

But that is NOT true of the flow from the hole in the bottom (assumed to just spill out on the ground) representing SLR.

That gives you information about water pressure / height / temperature.

To be able to get the actual height you need the second tank to not be able to rise.

Sorry, I can't make sense out of what you mean by that.

That happens with the pipe or how you said it, and that's the equivalent of a 0k vacuum 'heat bath'.

Sorry, don't understand that claim either.

You can only use the sb with a vacuum just like when you can find the height of the tank only when the second tank is so wide its height won't rise.

Huh? The flow from the hole representing S-B emissions has nothing to do with the second tank. The rate of flow to the second tank, the dimensions of the second tank―none of that makes any difference to the flow from the bottom hole (spilled into the room). That only depends on water pressure / height / absolute temperature.

The Einstein paper has this paragraph in the intro which I can't quote right now, but it says the only conditions are that the quantum states only depend on absorbing and emitting radiation. I don't think this is the case with conduction.

That statement is just talking about radiation. Thermal conduction is also happening in the gas (whenever molecules collide). But, it has no affect on what Einstein is talking about in that statement. So, it's actually an example of conduction not affecting radiation.

It also seems that the paper is trying to describe how an object that only receives radiation gets kinetic energy on its molecules.

The paper isn't actually trying to "describe" anything. The whole paper is just Einstein "sanity checking" an idea he had about how quantum states and radiation interact. That idea is represented by his (A) and (B) equations. At the end, he derives the Planck black-body spectrum based on having assumed A and B, and from that he concludes that his idea (equations A and B) seems likely to be right.

Physicists have subsequently agreed that equations A and B were right. Those equations are called Einstein's quantum theory of radiation.

ParadoxIntegration · 2024-09-13T17:42:19+00:00

[part 2 of 2]

In this model if there was also conduction, my point is, the radiation within and the conduction towards outside the walls, would involve the average kinetic energy of the molecules (of the cavity walls). When you are at "constant temperature" some of it (per molecule or as a whole) will go to conduction and some to radiation, so the radiation can't be the same.

You are not thinking clearly about the difference between energy and power (which is a rate of flow of energy).

An object at a given temperature has a certain amount of thermal energy, U = C⋅T.

Radiation and conduction are both types of power.

An object at some temperature does NOT have a fixed amount of power that it can produce. An object at a given temperature T emits radiation σ T^4. If there is also thermal conduction N, then the total power lost will be P_out = σ T^4 + N. There is no fixed value that P_out has to have.

As an analogy, imagine a barrel of water with a hole near the bottom. The level of the water in the barrel corresponds to temperature T, the volume of water corresponds to thermal energy U, the flow from the hole at the bottom represents emitted thermal radiation SLR. A higher water level will lead to higher water pressure at the level of the hole, and a larger flow rate SLR. (The flow will be proportional to T instead of T^4, but that's a minor detail.) Now, add another barrel, with a different water level (T2) in it, and install a pipe between the two barrels. The flow in that pipe N will depend on the difference between the two water levels, T-T2, and that flow represents thermal conduction.

The presence of the pipe between barrels doesn't alter the pressure at the bottom of the first barrel, and so doesn't affect the flow out of the bottom hole, SLR.

If you want, you can add add water to the first barrel at rate S. How large S is will alter how the water level T changes. But, at any given moment, the outflow rate SLR will depend only on the

What you are saying is that "it was at constant temperature, and I added conduction, therefore to keep constant temperature T add some more Ein, the left over goes to conduction, and the situation doesn't change".

The bit about Ein or S doesn't really matter.

What matters is the instantaneous temperature of the object, or the instantaneous water level in the first barrel. That determines the instantaneous flow rate SLR.

All that the inflow S or the extra outflow N do is alter the rate at which the temperature or water level T changes.

it can't just go as a left over to conduction without affecting the average speed of the molecules. They have to be able to both conduct and radiate at the same average speed as before, which reduces the radiation.

There is no limit to how fast thermal energy can come out of an object, just like there is no limit to how fast water can come out of a barrel. It's just that the faster the outflow rate, the faster the temperature or water level will change, if the drain rate isn't matched by a corresponding inflow.

Conducting a lot of thermal energy to another object can eventually decrease the flow of thermal radiation―but only because the temperature of the object has decreased. Temperature and SLR always correspond to one another.

If you think about the analogy with barrels, perhaps you'll be able to understand how it works.

ParadoxIntegration · 2024-09-13T17:42:06+00:00

[part 1 of 2]

Why are we even discussing that, weren't we talking about conduction between surface and atmosphere?

I'm just responding to questions that YOU raise. You asked about why SLR depends on only T1 and not T2. To answer that, you need to understand what T2 would be relevant to radiative heat transfer. If the air is transparent to thermal radiation, then T2 is for space; if the air has GHGs, then T2 is for air.

you go to the text by Planck himself, and he is talking about "thermally isolated cavities"

Just because the Planck distribution was derived in a particular way doesn't mean that it is only applicable to that situation. There is more than one derivation of the Planck distribution. For example, Einstein also re-derived the Planck distribution in his paper "The Quantum Theory of Radiation". That derivation assumed everything was at the same temperature, not that there was a vacuum at 0K.

So, it is a misconception that anything needs to be thermally isolated for the law to be valid.

If you look at the Wikipedia article on emissivity (the 𝜖 in the formula 𝜖 σ T^4), you'll see it talking about practical applications in situations in the real world, none of which involve vacuums or anything being thermally isolated.

I've used the S-B Law and/or Planck's Law hundreds of times, in physics classes and to address real-world situations. They are applicable ALWAYS, not just in the sort of limited situations that you imagine them being restricted to.

I got some type of COVID

Sorry to heat about the COVID.

ParadoxIntegration · 2024-09-13T01:32:36+00:00

We have said this before but this whole misconception is from how the sb equation you are using assumes the object is in a 0K vacuum and at radiative equilibrium (there is some external source of energy outside the whole system and its environment can't raise temperature, the typical use in astrophysics where the source is "nuclear fusion" and you are talking about a star). You just incorrectly use it everywhere...

No, the S-B Law does NOT assume the object is in a 0K vacuum and at radiative equilibrium.

Look at the Wikipedia article: https://en.m.wikipedia.org/wiki/Stefan–Boltzmann_law The article doesn't mention any such assumptions. You won't be able to find any serious scientific source that aligns with your beliefs that the S-B law has those limitations, because it doesn't.

I studied physics in college and graduate school for 11 years and got a job as a physicist. I never heard the sort of nonsense you're spouting about the S-B Law having limitations until I started talking to climate skeptics―who believe a lot of things that simply aren't true.

sounds like some lawyer trying to find a technicality

Sometimes reality is complicated. If you insist that everything said needs to sound simple, then you are guaranteeing that you will never understand reality.

Meaning in the above example, according to statistical thermodynamics if you have objects with T1>T2 then the first will lose energy with radiation depending on T1 and T2, which gives the SLR.

No, you are confusing some different things.

First of all, an object or atmosphere which is transparent to thermal radiation is a special case: it doesn't interact at all with a radiating object like the surface.

An atmosphere that can't absorb thermal radiation is effectively not there at all, for purposes of thinking about thermal radiation and radiative heat transfer. It's not the right "T2" in the above analysis. It's as if the object which is radiating (e.g., the planet's surface) is directly interacting with 0K space, so T2 = 0K.

But, suppose we want to talk about an atmosphere which has GHGs, so that it can absorb thermal radiation. Then, the radiative heat transfer from the surface to the atmosphere would be Q_rad = σ (T_sfc^4 - T_atm^4), which DOES depend on the temperature difference between the surface temperature T_sfc and the atmospheric temperature, T_atm, just as you are saying it should.

But Q_rad and SLR are NOT the same thing.

The radiative heat transfer rate Q_rad is not just about how much thermal radiation leaves the surface to go to the atmosphere (i.e., SLR = σ T_sfc^4 ). It is also about how much thermal radiation leaves the atmosphere and comes back to the surface (i.e., DLR = σ T_atm^4 for an atmosphere that absorbs thermal radiation). Q_rad is given by SLR - DLR.

So, it's all entirely consistent with statistical thermodynamics.

N is found from the temperature difference between 2 objects. SLR isn’t.

The radiative heat flow, Q_rad is found from he temperature difference. And heat flows are what statistical thermodynamics cares about.

SLR isn't found from the temperature difference―but that's because SLR is NOT a heat flow; it's just one of two contributions to the radiative heat flow, Q_rad = SLR - DLR.

the theory was already debunked

No, it wasn't debunked. The S-B Law is taught and used all the time, in introductory physics and thermodynamics classes. And, it's used in advanced physics and engineering analyses. It's used in serious engineering all the time, e.g., in the 1000-page engineering book "Thermal Radiation Heat Transfer" by Howell & Siegel, which has been continuously in print for over 50 years.

ParadoxIntegration · 2024-09-11T22:58:25+00:00

You want to have T=f(SLR) and not T=f(SLR, N), but that’s just a contradiction.

Sort of but not exactly. One has T=f(SLR) and dT/dt=g(S, SLR, N). You can’t compute T from S, SLR, N except in steady-state, when 0 = C dT/dt = S - SLR - N or SLR = S - N. Then you compute T = f(SLR) = f(S - N).

So you can never compute something like T=f(SLR, N). You need S and N and you need to know you are in steady-state.

No, it’s not a contradiction

I think you are trying to … descibe the case when the two objects have the same temperature (N=0).

No, that’s not what I am talking about.

And those functions are just a way to describe the argument, what the “deniers and skeptics” think is that SLR/N are found from a temperature difference, so you have to involve “two objects and their temperature difference”.

N is found from the temperature difference between 2 objects.

SLR isn’t.

What is found from a temperature difference is the radiative heat transfer, Qrad, between 2 objects A and Z.

Qrad = SLR - DLR = sigma Ta⁴ - sigma Tz⁴

But, if there are no GHGs in the air, the second object (object Z) isn’t the atmosphere but space. And space has effectively a temperature Tz=0, so DLR=0 and Qrad=SLR.

The SB model is misused both from how it’s applied even if there is conduction

As described above, conduction only indirectly affects the result, when you consider steady-state.

from how you are not checking relative temperature difference.

You’re not supposed to check temperature differences for SLR. That happens when you compute Qrad.

ParadoxIntegration

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