The Sophistry of Backradiation

Written by Joseph E Postma

Another Tale of Two Versions: Sophistry finds its greatest expression and success in deceit when it can switch reference frames, goal posts, and contexts without switching the language. sophistry We’ve seen this already in the two versions of “the greenhouse effect”, where we have two physically distinct mechanisms, one of which exists in reality and the other of which is a simulacrum of that reality upon which many other lies can and have been built, and where both use the same label.

It has now become apparent that the term “backradiation” is likewise being used by the pseudoscientific, pseudomeritocratic climate propaganda establishment in a sophistical manner to generate additional cognitive dissonance and sophistry.

 Form vs. Content

There are two ways to understand the term “backradiation”. The term is sophistical in and of itself, but the two versions have a difference which the sophists exploit to create cognitive dissonance.

The first most obvious way is in terms of its content, i.e. in the terms of what it is functionally required to do as part of the mechanism of the alarmist radiative greenhouse effect from which the term itself originates.  And that function is to cause heating, is to cause a surface source of thermal radiation to become even hotter still because its own radiant thermal energy is sent back to it (i.e., backradiation) after thermal absorption and re-emission at a cooler target.  More generally, the function is that the thermal radiation from a cool object will be absorbed by and thus cause heating (i.e. temperature increase) on a warmer object.

 This is what is occurring in the diagram (right) from the University of Washington Department of Atmospheric Sciences. assumed energy balance

The concept here is of course sophistry because radiation from a cool object does not heat up a warmer object – heat doesn’t flow from cold to hot.  It is sophistry because there is no such thing as “backdiffusion” of heat energy in thermodynamics, and diffusion works via the same fundamental electromagnetic force of physics that radiative transfer is mediated by.  If there is backdiffusion heating then there will be backradiation heating, and if there is not backdiffusion heating then there will not be backradiation heating.  There is no such thing as backdiffusion heating.

The second way to use the term “backradiation” is in terms of its form, i.e., in simple direct reference to the thermal radiation that a cooler object emits (with context in the vicinity of a warmer object).  All surfaces with non-zero emissivity and non-zero temperature will emit thermal radiation, and so whether we call a surface warm, or cold, etc., it does indeed emit thermal radiation.  

This is sophistry because it is an invented term – there’s no need to call radiant energy from a cooler object “backradiation” just as there isn’t a need to call conductive energy  from a cooler object “backdiffusion”, because heat does not “backdiffuse” from a cooler object to a warmer one.  The label is of course invented in order to attach it to the idea that radiant thermal energy from a cool object must be absorbed by and thus increase the temperature of a warmer object.  

Of course, there is no concept in thermodynamics that thermal energy will or even can diffuse from a cool object to a warmer one, just because the cool object contains thermal energy; in fact, this directly contradicts the set of Laws of Thermodynamics.

Energy does not physically, by physical conductive diffusion, transfer heat from a cool object to a warmer one.  However, the sophists have exploited various anthropocentric sense-perception-based propensities of the sensing-type majority population (and essentially all of science) to confuse them about what energy might do across a “non-physical” gap it might “jump” via radiation, as discussed in a previous post.  

Of course, it doesn’t matter if the energy transfer is by physical diffusion or across a gap by radiation.  Either way, heat only transfers from hot to cold and so therefore the cold object can not increase the temperature of the warm object.  If it confuses them so as to how energy can “know” not to transfer heat radiatively from cold to hot across a gap, they should likewise be confused as to how energy can “know” not to diffuse heat conductively from cold to hot with objects in physical contact.  It is simply because they are sensing types that they accept the latter without question, but are so exploitable to confusion and sophistry with the former.

The Sophist Switch

The Slayers have always said that “backradiation doesn’t exist”.  We say that because “backdiffusion” doesn’t exist.  What it is supposed to do doesn’t exist, and what it is supposed to do is to cause a warmer object to become warmer still due to the thermal energy from a cool object.  This is only possible if heat flows from cold to hot, which it of course doesn’t. Hence backradiation doesn’t cause heating and likewise everyone already knows that backdiffusion doesn’t exist either.

At this point the sophist radiative greenhouse alarmist will switch the usage of the term “backradiation” to its form, and accuse us of denying that a cold object emits any radiation at all.  They make the sophist assumption that if you reject the notion that a cold object can heat a warmer object, then you must deny that backradiation exists in the form of radiant energy from a cooler object at all, because they implicitly assume that any radiation whatsoever can cause heating irrespective of its source temperature.

They either won’t admit to, or can’t understand, that while the thermal energy from a cool object can and does indeed exist, this energy is not capable of passing heat to a warmer object; this is true with either diffusive, or radiative, energy exchange.

If we look at either the radiative or diffusive heat flow equations (respectively)

Q’ = A*σ*(Thot4 – Tcool4)

Q’ = A*k*(Thot – Tcool)

we see that there are indeed two terms in the equation: a hot term and a cool term.  Only the greater portion of the energy can act as heat, and the greater portion is of course only available from the hotter object both in terms of quantity and more importantly in terms of quality, in terms of having higher frequency energy components that the cooler object does not contain.

Heat flow is of course all about Fourier Mathematics, and Fourier math is all about waves and the frequencies of those waves.  Higher temperature means higher frequency energy waves, and so these are higher frequencies which the lower temperatures do not contain.  Lower temperature can not increase the frequencies of a higher temperature, and thus increase that higher temperature’s temperature, because it doesn’t contain the higher frequencies required to add to the higher temperature.

heat flow between planck curves

The Follow On

It is very funny sophist behaviour to watch, because after I write out the heat flow equation for them and explain that it has two terms and that only the difference can act as heat and raise the temperature of the cool object, they always then ask if I deny that the cool object emits any energy at all!

So let’s get this straight:  Do you see the radiative heat flow equation in the next line?

Q’ = A*σ*(Thot4 – Tcool4)

The equation has two terms – a hot term and a cool term.  Since the terms are there, then it means that both objects emit energy.  Therefore it would be impossible to write and use that equation while denying that one of the energy terms doesn’t exist.  The problem for the sophists, is that they are unwilling to admit that only Q’ is heat.

The equation defines that only the difference of energy is heat.  Each term is not itself heat.  Repeat:

Each term is not itself heat.

Only the difference of the energy is the heat and this heat flows from hot to cold and can only cause the cold object to raise in temperature.  You can only call A*σ*Thot4 or A*s*Tcool4“heat” when they are by themselves with absolute zero background temperature, but even then it doesn’t make sense since empty space can’t be heated, i.e., can not attain a temperature, and so those terms are more properly still just total energy emission to space. Q’ is just the portion of A*σ*Thot4 which is acting as heat for the cooler object.

There is no two-way flow of heat which also has a “net” heat.  If there was a two-way flow of heat, then the cold object would raise the temperature of the warm object, and you just violated thermodynamics.  And think of that logic: if the cool object heats the warmer object, then as the warmer object heats the cooler object, the cooler object will cause the warmer object to warm up some more, requiring the cooler object to become warmer, resulting in the warmer object becoming warmer leading to the cooler object becoming warmer, which causes the warmer object to warm…ad infinitum.

This is a very basic logical sequence and the infinite recursion produced is there to debunk itself.

And even before that, a cold object can not send heat to a warmer object; heat does not diffuse from cold to hot and therefore heat does not radiate from cold to hot since in both cases the transfer of heat is mediated by electromagnetism.  What is so hard to understand about heat not flowing from cold to hot?!

There is two-way flow of energy; there is not two-way flow of heat.

Demo

It has now become clear that another thing which is confusing people is calling the thermal emission from an object “heat”.  This is wrong.  If we have for a single hot source

A*σ*Thot4

this term is actually its power, P = A*σ*Thot4, not heat!  This term is its energy emission, not its heat.  Heat is only the portion of the energy which can act to increase the temperature of another object, hence why heat is written as

Q’ = A*σ*(Thot4 – Tcool4)

You could only equate P with heat if it was emitting towards an object with absolute zero temperature, so that

Q’ = A*σ*(Thot4 – 04) = A*σ*Thot4 = P

Otherwise, heat is only the portion of P which is above and beyond A*σ*Tcool4.

Here is what goes wrong when you call any emission from any source “heat”.  Following on from the Gift of the Steel Greenhouse post, for an infinite plane which is perfectly insulated on its backside and is perfectly conductive, the differential equation for the temperature of a section of the plane of area A is:

1]                           mCp*dT/dt = ∑Q’i

where ‘m’ is the mass of the section, Cp is its thermal capacity, T is its temperature (and is actually T(t), a function of time, but I’ll just write T instead of T(t) for simplicity), and ∑Q’i is the sum of the “heat” flow into the section.  We start with the section having its own internal power source P0, and so

2]                           ∑Q’i = Q’0 = P0 – AσT4

Subbing that into equation 1 we get:

3]                             mCp*dT/dt = P0 – AσT4

Thermal equilibrium for the section is when the rate of change of temperature goes to zero, i.e., is when the temperature stops changing, and so the equilibrium temperature is:

4]                         mCp*dT/dt = 0 = P0 – AσT04

Solving for T0:

5]                              T0 = [P0/(Aσ)]1/4

which is a fixed constant value.

Now we bring in another plane with a small gap spacing between it and the original powered plane.  The new plane is not powered, not insulated, and it doesn’t have to be perfectly conductive.  As the powered plane raises in temperature it emits the requisite energy, and this energy heats the new plane.  And so, as the new plane is heated, it then emits its requisite energy and we must call this energy “heat”, and we must assume that it is absorbed by the original plane, and thus must add to the temperature forcing for the original plane.  

The energy re-emitted by the new plane is simply equal to the energy that it receives which generates its temperature, that energy being AσT4 from the original plane. Thus, subbing the “new heat input” to the original plane with the new plane in place, we get:

6]                           ∑Q’i = P0 – AσT4 + AσT4

Subbing that into equation 1 we get:

7]                             mCp*dT/dt = P0 – 0

which has the solution

8]                              T(t) = [P0/(mCp)]*t

which goes to infinity as t goes to infinity.  That is, it grows without bound.

This is the runaway heating problem you get when you consider any emission from an object to be heat, and that it must be absorbed as heat by any other surface irrespective of the direction of heat flow.  The solution going off to infinity shows that it is wrong, not right!

Look at the diagrams below (Sequence A, Sequence B) and try to visualize which one demonstrates the logical, thermodynamic, reality-based sequence of energy emission and heat flow, and which one doesn’t.

sequence A

sequence B

Read more by Joseph Postma at climateofsophistry.com{jcomments off}