Of particular focus is the

**Fundamental Theorem of Calculus**stating that if $F(x)$ is a primitive function of $f(x)$, that is $\frac{dF}{dx}=f$, then- $\int_a^bf(x)\, dx = F(b) - F(a)$.

For example, if $f(x)=1$, then $F(x)=x$, and so we can formally write

- $1 = \int_{100}^{101} f(x)\, dx = F(101)-F(100)=101 - 100$

The radiative flux of heat energy $Q$ from a warm body of temperature $T_2$ to a colder body of temperature $T_1 < T_2$ is (essentially) given by the integral

- $Q = \int_{T_1}^{T_2}f(\nu )\,d\nu$

warm body the "overkill" spectrum $f(\nu )$ above the cut-off $T_1$ for the cold body, with the overkill effectively causing the heating while the shared spectrum below cut-off has no heating effect, as explained in more detail here.

With $F(T)=\sigma T^4$ and $F$ acting as a primitive function of the Planck function $f$, the radiative flux $Q$ can now according to the Fundamental Theorem formally be expressed in the form of the Stefan-Boltzmann Law

- $Q = \sigma T_2^4 - \sigma T_1^4$

thus formally expressing the positive quantity $Q$ as the difference between two (large) numbers $\sigma T_2^4$ and $\sigma T_1^4$.

The Voodoo Physics of the Greenhouse Effect is the result of giving the mathematical identity

Note that $F(T)=\sigma T^4$ can be interpreted physically as the one-way radiative flux from a body of temperature $T$ into a background at 0 K. But the physics is missing of viewing the one-way radiative flux between two bodies as a difference between two separate fluxes into a background at 0 K.

- $\int_{T_1}^{T_2}f(\nu )\,d\nu = \sigma T_2^4 - \sigma T_1^4$

Note that $F(T)=\sigma T^4$ can be interpreted physically as the one-way radiative flux from a body of temperature $T$ into a background at 0 K. But the physics is missing of viewing the one-way radiative flux between two bodies as a difference between two separate fluxes into a background at 0 K.

A property (identity) of a mathematical model is thus freely interpreted to be real physics, in the same way as an operation of a voodoo doll is believed to be able to have a real effect on a real person.

This is nothing but Voodoo Physics, and this is the nature of the Greenhouse Effect based on Back Radiation from cold to warm underlying the CO2-alarmism so forcefully preached by IPCC with now Macron as ardent follower.

Macron, despite (or maybe thanks to) his education in French elite schools with all its mathematics, thus appears to be overwhelmed by the Fundamental Theorem of Calculus and cannot separate model from reality.

Macron, despite (or maybe thanks to) his education in French elite schools with all its mathematics, thus appears to be overwhelmed by the Fundamental Theorem of Calculus and cannot separate model from reality.

Voodoo = operation on model believed to have real effect on real person.

if the Earth surface was radiating directly to surrounding space at 0 K and not to the atmosphere at $T_1$ K.

You can buy a pyrgeometer from Kipp and Zonen and play with it as a voodoo doll believing it reports real physics if you want to sell CO2-alarmism. From the above analysis you may understand that in fact it represents a symbiosis of science and commercial industry serving CO2-alarmism by supplying fictional physics.

**PS**The Pyrgeometer is a perfect example of a voodoo doll, reporting (non-physical) back radiation $\sigma T_1^4$ from a cold atmosphere at $T_1$ to warmer Earth surface at $T_2$, by measuring $Q$ and (erroneously) viewing $\sigma T_2^4$ to be radiation from the Earth surface, asif the Earth surface was radiating directly to surrounding space at 0 K and not to the atmosphere at $T_1$ K.

You can buy a pyrgeometer from Kipp and Zonen and play with it as a voodoo doll believing it reports real physics if you want to sell CO2-alarmism. From the above analysis you may understand that in fact it represents a symbiosis of science and commercial industry serving CO2-alarmism by supplying fictional physics.

Yes, that looks like non-interpretable voodoo (for instance, the integral in (*) should have the limits 0 to \(\infty\), right?).

SvaraRaderaIsn't it better that we look at the physical origin of the blackbody radiation energy density and its relation to \(Q\) instead?

We only need to use classical thermodynamics for this.

First, we note that for radiation to be in thermal equilibrium with matter, the matter only needs to be sufficiently optically dense, so that enough interaction takes place for the radiation to equilibrate. This means that both solids, liquids, gases and even a hollow cavity can function to equilibrate the radiation, given sufficient optical depth. This is essentially the physical meaning of a blackbody.

To make the discussion meaningful, we make the ordinary assumption that the matter functions as a thermal bath for the radiation. That is that the temperature of the matter changes on a much larger timescale, so that we can treat it as having a constant temperature.

We now note that the most important property of blackbody radiation is that it is radiation in

thermal equilibriumwith matter at temperature \(T\). Using onlyclassical thermodynamics, we can then quite easily show that the energy density of such radiation is only dependent on the temperature of the matter, according to the formula\(u(T) = \beta T^4\),

where \(\beta\) is some constant. We also note that we don't need the Planck spectrum for this result, the result relies only on the facts that electromagnetic radiation carries energy and momentum which gives us a relation between the radiation pressure and the energy density.

When the radiation in thermal equilibrium escapes the matter, the energy density of the radiation maintains the same as when in equilibrium with the matter, since it no longer interacts. Looking at the enclosing area this escaping energy density can be interpreted as a radiation flux.

For incoming radiation, the situation we have is reversed. The incoming radiation is not interacting until it starts to equilibrate with the thermal bath. Because of our assumption of sufficient optical thickness the incoming radiation will eventually equilibrate.

We now get the quantity that you call \(Q\), comes naturally from ordinary energy conservation over the enclosing surface.

We note that this physical picture is a much more general one, since the incoming radiation not necessarily needs to be thermal radiation. This better reflects a real situation when two bodies at different temperatures exchange energy by radiation, black or otherwise.

I think you misunderstand the pyrgeometer.

SvaraRaderaIt clearly says in the text in the picture that it is the outgoing energy (red arrows) that is assumed to be calculated with the Stefan-Boltzmann law.

Hence the incoming is not radiation following a Stefan-Boltzmann law, so there is not a relation of the type Q = sT2^4 - sT1^4.

Ah, forgot the point :)

SvaraRaderaThe point is that the Fundamental law of calculus has completely nothing to do with the device.

It is stated that the outgoing radiation is postulated to be sigma T^4 with T the pyrgeometer/Earth temperature from which the incoming radiation is computed. But the outgoing sigma T^4 is the radiation to an environment at 0 K which is not the temperature of the atmosphere (which is what is effectively being measured). So you have not properly understood the setting.

SvaraRaderaI wonder if you read the comment by Kevin above. It clearly shows how you misunderstand the situation.

SvaraRaderaThe energy density for the radiation being in thermal equilibrium is of course only dependent on the temperature of the body it is interacting with. The ambient temperature can not be involved here, otherwise there is no thermal equilibrium. This is very reasonable since the time it takes for the radiation to reach equilibrium is much quicker than the time it takes for the body to change its temperature. The heat capacity of the radiation is many, many, many orders of magnitudes smaller then that for solids, liquids and ordinary gases (if I remember correctly there is a ratio of about 10^-12 to liquids for constant volume capacities).

The energy density for the radiation in equilibrium can easily be shown to follow a T^4 dependence, where T is the temperature of the body. When the radiation that is in thermal equilibrium leaves, it has the same energy density at that instance. So the assumption that this is a formula for radiation to a 0K background is very flawed. It must be independent of the temperature of the surroundings. How could it not??

You speak about "energy density" of something. I consider transfer of heat energy by radiation between bodies, which is similar to transfer by conduction in a dependence on temperature difference: The one-way transfer of heat energy from the inside of your house to a colder exterior by conduction depends on the temperature difference between inside and outside, and the same holds for radiation.

SvaraRaderaYou do understand that Fourier's law is a phenomenological law, right?

RaderaThe thermal conductivity \(\kappa\) in a material is a fundamentally different property than the \(\beta\) (as introduced by Kevin above) in radiative energy transport.

Would be interesting to see if you actually can explain the difference between thermal conduction and thermal radiation.

My analysis of radiative heat transfer is presented in detail at https://computationalblackbody.wordpress.com . And yes, heat transport by conduction and radiation are similar as energy transport scaling with temperature difference warm to cold.

RaderaYou are making a fundamental mistake.

RaderaAt the moment I don't have time to go into detail, but for your conclusions about energy transport by heat, following conclusions from Fourier's law, you are assuming a general validity of Fourier's law. And more specific, you are begging the question about the direction of the flow, since the assumption of one-directional flow is an pre-assumption for the validity of the Fourier law.

But Fourier's law is not true in general, it is an approximation, as been known for long. It has since way back been noted that Fourier's law is incomplete due to lack of inertial response to thermal perturbations. More specific, this makes the law violate microscopic reversibility, and based on how you model yourself, I don't think that is something you would promote violation of. When it comes to radiative exchange, between spatially separated bodies, thing would get even worse...

Lets say I have two light bulbs, one stronger (hotter) than the other, and the strong one is turned off, while the weak one is turned on. Then the weaker one radiates onto the stronger one. Then I turn on the stronger one. Then you say that the weaker one stops radiating onto the stronger one; is that correct?

SvaraRaderaWhat does ”radiate onto” mean? Transfer of heat energy? From cold to warm?

SvaraRadera