# Two Proofs of Planck’s Law vs Backradiation

Authors

Planck trying to convince a disbelieving Bohr that his proof of Planck’s Law is correct. On the blackboard: Maxwell’s equations.

My recent debate with Roy Spencer and Fred Singer and others about the concept of backradiation, which has come to play a fundamental role in CO2 alarmism, illustrates the danger of misunderstanding mathematics and thus misunderstanding physics.

Central in the discussion is the interpretation of Planck’s Law of blackbody radiation expressing the energy $E (T,\nu )$ radiated from a blackbody B as a function of temperature $T$ frequency $\nu$ as

• $E(T,\nu ) =\gamma T\nu^2$ ,

where $\gamma$ is a universal constant.  The question centers around the interpretation of the physics supposedly described by Planck’s Law. There are two possibilities:

1. B radiates $E$ into a surrounding S seemingly independent of the temperature of S. This allows an interpretation with R radiating into S even if S has higher temperature, which is exploited by CO2 alarmists as backradiation causing global warming.
2. B and S form a system and B radiates energy into S only if B has higher temperature than S. There is no backradiation, and no ground for CO2 alarmism.

The question is now if Planck’s Law describes 1 or 2. How can we tell? By looking at the proof of Planck’s Law, which I do below. The result of the analysis is the following:

1. In Planck’s proof B acts independent of S. It allows an interpretation that B emits into S even if S has higher temperature = backradiation, even if this was not the intention by Planck.
2. In my proof B and S form a resonating system. B emits into S only if S has lower temperature.

My conclusion is that 2. with B and S forming an interacting system is closer to actual physics, than 1. with B independent of S which can be misinterpreted as alarming backradiation.

The analysis shows the importance of judging the proof of a mathematical result when interpreting the result. Blind interpretation can lead astray. Planck did not speak about backradiation, but his proof brings in non-physical statistics and thereby opens to a misinterpretation inventing non-physical backradiation.

## Planck’s Proof: Statistics of Quanta

The derivation of Planck’s Law (of blackbody radiation) given by Planck is based on particle statistics borrowed from thermodynamics with little connection to the real physics of radiation as interaction of matter and electromagnetic wave motion.

Planck was forced into particle statistics because a radiation law based on wave motion seemed to lead to a Rayleigh-Jeans law with an “ultraviolet catastrophe” with energy exploding to infinity like $\nu^2$ with frequency $\nu$ without upper bound. By a statistical assumption that highly energetic waves are rare Planck gave physics a way to avoid the catastrophe which earned almost infinite fame and a Nobel Prize.

But Planck left a question without answer: Is it possible to derive a correct radiation law without resort to statistics, using instead classical electrodynamics described by Maxwell’s wave equations?

## New Proof: Deterministic Finite Precision Computation

In Mathematical Physics of Blackbody Radiation and Computational Blackbody Radiation I show that this is indeed possible, by replacing statistics by a much more basic physical assumption of finite precision computation. Let me here presents the essence of my argument:

I consider mathematical wave model of the form:

• $U_{tt} - U_{xx} - \gamma U_{ttt} - \delta^2U_{xxt} = f$

where the subindices indicate differentiation with respect to space $x$ and time $t$, and

1. $U_{tt} - U_{xx}$: material force from vibrating string with U displacement
2. $- \gamma U_{ttt}$ is Abraham-Lorentz (radiation reaction) force
3. $- \delta^2U_{xxt}$ is a viscous force
4. $f$ is exterior forcing,

and $\gamma$ and $\delta^2$ are (small) positive coefficients subject to a frequency dependent switch from outgoing radiation with to absorption/internal heating defined below.

This model with $\delta =0$ is essentially the starting point also for Planck, before resorting to statistics of quanta: A system of resonators $U$ in resonance with an exterior forcing $f$.

The finite precision computation is represented by the $\delta^2$ viscosity which sets a smallest scale in space of size $\delta$ by damping frequencies higher than $\frac{1}{\delta}$.

The coefficients $\gamma$ and $\delta$ are chosen so that only one is positive for a certain frequency $\nu$ in a spectral decomposition, with a switch from $\gamma > 0$ to $\delta > 0$ at $\nu = \frac{T}{h}$, where $T$ is temperature and $h$ is a fixed precision parameter representing atomic dimension in the string of atoms modeled by the wave equation.

The switch thus defines a temperature dependent cut-off frequency $\frac{T}{h}$ with the effect that only frequencies below the cut-off are emitted as radiation while frequencies above cut-off are stored as internal heat energy of the string.

The essence of the proof is the following energy balance established by a spectral analysis assuming periodicity in space and time:

• $\int \gamma U_{tt}^2dxdt + \int \delta^2U_{xt}^2dxdt = \epsilon \int f^2 dxdt$

with $\epsilon\approx 1$ is a coefficient of emissivity (= absorptivity), which we write in condensed form as

• $R + A = F$, or in a spectral decomposition $R_\nu + A_\nu = F_\nu$, or in words
• Radiation + Absorption = Forcing,

where $R = \int \gamma U_{tt}^2dxdt$, $A=\int \delta^2U_{xt}^2dxdt$ and $F=\epsilon \int f^2 dxdt$.

As just said, the switch effectively means that

• $R_\nu = F_\nu$ for $\nu < \frac{T}{h}$
• $A_\nu = F_\nu$ for $\nu > \frac{T}{h}$,

which expresses that the forcing $F_\nu$ is remitted as outgoing radiation for $\nu < \frac{T}{h}$, and is absorbed and stored as internal heat energy for $\nu > \frac{T}{h}$.

The coefficient $\epsilon$ represents emissivity = absorptivity, in accordance with Kirchhoff’s Law. An essential aspect is that $\epsilon$ is independent of $\gamma$ and $\delta$, which expresses that radiation (to up to the constant $\epsilon$) is independent of the nature of the radiating/absorbing body.

Planck’s law in the form $R+A = F$ with only one of $R$ and $A$ non-zero for each frequency is a variant of Planck’s classical law with a sharp switch instead of a continuous transition from radiation to absorption/heating.

Planck’s law in the form $R + A = F$ thus contains the classical law for radiation as $R=F$ but also the further information that absorption into internal heat energy $A=R$ only occurs for frequencies above cut-off.

The proof that $R + A = F$ is non-trival and expresses a fundamental aspect of near-resonance in systems of oscillators with small damping. The proof shows Planck’s Law $R_\nu\sim \gamma T\nu^2$ for $\nu < \frac{T}{h}$ and Kirchhoff’s Law, without resort to statistics.

The effects of Planck’s statistics in modern physics are subject to an investigation in Dr Faustus of Modern Physics: Is it true that Planck’s technique for avoiding the ultraviolet catastrophe led to a much bigger catastrophe of abandoning rationality in physics?

## Sum Up and Conclusion

Both proofs essential start from the same wave equation model but differ  as concerns the mechanism for high frequency cut-off: Planck uses statistics of quanta and I use deterministic finite precision computation.

The difference comes out as a different view of the system of blackbody B and surrounding S.

With Planck’s statististics B is detached from its surrounding and B is supposed to emit whatever statistics decides B to emit irrespective of the temperature of S.

But in my derivation B and S form an interacting system and B only emits energy into S if B has higher temperature than S.

My conclusion is that my proof with B and S forming a system is closer to physics than Planck’s proof with B acting independently of S.

What is your conclusion: System or Not System? To Interact or Not To Interact?

1. ### iceskaterfinland

I cannot see that the issue you are objecting to is something we need to call backradiation.

You are objecting to standard net radiation heat loss curves.

Ie the well estabished practical observation, that If you have a warm surface at 50C it loses heat less rapidly when exposed to a surface warmer than for example 0C or 49C.

Therefore if you place matter at 4k between an object at 298K and the coldness of space at 3k, then the object at 298K is now cooling more slowly, and if it is heated either from within the material due to a temperature gradient or externally heated it will get hotter at the surface.

There is 14.7 pounds of atmosphere above the thin surface layer occupying only one square inch. So even extremely cold temperatures in the atmosphere will reduce the cooling rate of the surface.

What actually are you arguing against? It is not clear to me.

The basic physical observations are well established

• ### iceskaterfinland

Ie I am saying that if hot objects cannot receive radiation from cold objects, you still need to account for why net heat transfer by radiation is reduced when the surface is exposed to a colder atmosphere that is warmer than the vast unwarmable cold of deep space

• ### iceskaterfinland

I think i get what you are saying now. You are saying that all heat leaving the surface can only return if it is emitted by a warmer object.

And you are basing this idea on some kind of resonant connection between masses some distance apart where by some method the emitting molecule knows the heat content of the mass it is part of, and knows the heat content of the mass of which the molecule absorbing is part of, so that if a thermometer was in both masses you would know one mass had greater total vibrational movement than the other – even though you have no idea about the temperature of the individual molecules in the mass. And you have some objection to the use of statistics to look at probabilities of different molecular vibrational states.

• ### Richard T. Fowler

“[. . .] emitting molecule knows the heat content of the mass it is part of [. . .]”

My friend, it only needs to know the temperature of the particles with which it is in informational contact with. All right?

RTF

2. ### claesjohnson

I object to the idea that there is spontaneous transfer of heat energy from a cold body to a warmer body, because that violates the 2nd law, and I prove the 2nd law from basic wave mechanics. I do not object to standard net radiation heat loss curves as they conform to my proof.

• ### iceskaterfinland

The second law as orgiinated by Clausius and by Lord Kelvin is talking about the impossibility of cooling a material to create a self sustaining heat engine that does work and is against the idea we now have of the impossibility of perpetual motion machines. Obviously by backradiation it is not possible to create a machine that can do work. back radiation is always part of a cooling process.

If we are talking about radiation we cannot meaningfully talk about large warm bodies of precisely the same temperature. Temperature is only matter in motion at atomic vibrational level. Rather we have to talk about atoms and molecules reducing their vibration energy when radiation is released. Which means they are cooler.

A warm body has both hot and cold vibrational states within it.

There are no 2nd law violations

• ### iceskaterfinland

sorry… correction. the kelvin statement is in support of the impossibility of perpetual motion machines

• ### iceskaterfinland

>>I object to the idea that there is spontaneous transfer of heat energy from a cold body to a warmer body

would it interest you to see the temperature at the surface of a hot body rising when a colder body is placed near it? where previously the hot body was exposed to a colder object. Can you account for that?

• ### iceskaterfinland

Also we cannot really say that heat energy can be transfered by radiation.

All we can say is that matter greater than absolute zero has atomic and molecular vibrational energy, and the atomic vibrations are reduced when radiation is emitted and travelling to another object and when the radiation is absorbed the absorbing atoms have greater vibrational energy.

So a heat pump cools one object and gets hotter itself and then transfers that heat to another object and becomes cooler for no violation of the second law and external work is needed to drive the no net change process.

Via radiation emission, an atom is cooled and another form of energy is simultaneously produced. and later another atom is warmed and the alternate form of energy no longer exists for a no net change process requiring no external work.

The reduction in vibrational energy and creation of radiation of equal power cannnot be considered a violation of the second law

3. ### iceskaterfinland

>>I prove the 2nd law from basic wave mechanics.

Are you objecting to quantum physics and the ideas put forward by Einstein?

Basic wave mechanics’ sounds like an approximation rather like Newtons universal theory of gravity

4. ### claesjohnson

You seem to be getting some of the message, but I do not know if you want to understand or just object.

• ### iceskaterfinland

Understanding sound fine. But not if i have to understand you but you dont want to understand me

• ### iceskaterfinland

It seems to me you are objecting to the foundation ideas about atoms and molecules and want to view mass as something that is indivisible, or at least you have some objection to viewing matter at the atomic scale when you consider individual processes such as a molecule cooling when it emits radiation where the radiation is not what we call heat

• ### iceskaterfinland

at the atomic scale a molecule can emit radiation and ‘be cooled’ and warm another molecule that is ‘heated’. No work is done.

And yet you talk about second law violations

• ### Richard T. Fowler

Are you familiar with the “first law of motion”?

RTF

5. ### iceskaterfinland

Another aspect of this is why is the sky at night not full of a uniform colour of light rather than points of light amongst blackness? One early idea was that the universe is full of dust, but it was soon realised that the dust itself would warm to the temperature of the stars eventually. It was then concluded that the universe is not infinite – which was news to me! Hence the big bang theory. Therefore the reason for the lack of light of approximately uniform colour is that there is just not much matter in space but most of the matter there is hot.

So via your theory how does the earth cool to outerspace? Space is not cold as such. It is just lots of hot stuff in a vacuum. Via your theory the earth would surely begin rising to the temperature of the vast bulk of matter in the universe.

So you need a theory on the universe also. I actually thought the expanding universe meant that light was so red shifted it became invisible but apparently that is not todays view of things?

6. ### claesjohnson

I see no difficulty with the idea of “empty space” filled with some form of “dust” at 3 K.

• ### iceskaterfinland

Yes but space is not filled with dust. It has enormously hot stars where these stars would warm all matter around them including the dust. Therefore it is reasoned the hot objects and the dust are emitting to a substantial amount of nothingness where no absorption will take place, space is not finite and there was a big bang

7. ### Tor

But if that you have written about your model is true and some light is scattered, then it seems that you are not modelling a blackbody, by definition.

“An ideal body is now defined, called a blackbody. A blackbody allows all incident radiation to pass into it (no reflected energy) and __internally absorbs all the incident radiation__ (no energy transmitted through the body). This is true of radiation for all wavelengths and for all angles of incidence. Hence the blackbody is a perfect absorber for all incident radiation.”

8. ### claesjohnson

An ideal blackbody without physical model belongs to philosophy and not physics. I consider a specific physical model as an example of a (non-ideal) blackbody.

9. ### Tor

In your opinion, is a mathematical pendulum also philosophy then?

I do not agree that idealizations have no place in physics, quite the contrary. Everything in physics are idealizations!

Physics is about describing and model reality using idealized models.

10. ### claesjohnson

A blackbody as a cavity with a little peep hole is a physical model of an ideal blackbody. This is a silly model of a real material radiating body, which is better modeled as a system of damped resonators subject to forcing.

• ### Tor

To be fair, the cavity is an approximate realization of an ideal blackbody (the type that physicists talk about, that is).

But that should be a good gauge. How well does your model fit to measurements on radiation from such a cavity? There should be more than ample data to compare with.

Further, I’m a bit curios on what your definition of an non-ideal blackbody is?

11. ### Richard T. Fowler

“Further, I’m a bit curios on what your definition of an non-ideal blackbody is?”

Well, whatever definition one chooses, it obviously includes the type of cavity the utility of which you have been defending. It is physical, thus non-ideal.

Of course, it is not a real blackbody. But such talk of “real blackbodies” is really nonsense, since by the definition of a blackbody, a “real blackbody” is not real, and thus is “not a real blackbody”.

In other words, by its definition, one can deduce that “If it exists, then it does not exist, and if it does not exist, then it can exist”. I am hard pressed to think of anything more pointless than debating such a semantic question in a serious tone as if to suggest that it is a legitimate scientific endeavor.

RTF

• ### iceskaterfinland

Something very close to a real black body might exist but at the moment we can only get within about 5% of the idea. Just as 150 years ago it was not possible to get very near the idea of absolute zero.

The fact we got closer and closer to the idea of zero and found it harder and harder to do so suggests the idea is real.

Science is founded on ideas and approximations.

Another idea being that there is no single temperature in an object but rather there is continual heating and cooling and an averaging of forces to produce what can be measured.

It makes no sense to replace these ideas, which have been found useful to make advances in Science, because you have decided the ideas are pointless on semantic grounds.

You also need to consider that mathematics do not describe reality but rather reality exists and then humans look for mathematics that can describe what is observed.

For example proving the second law of thermodynamics mathematically is not the same thing as observing the law in action experimentally.

And it is interesting that when you observe brownian movement of pollen in water and can reason that the water is vibrating to cause the movement of pollen you can then analyse the movement to conclude that some particles will be more energetic than other particles. You can then reason that statistically there will be a distribution of different temperatures or heat amounts or vibrational amounts for each molecule.

So why conclude that statistics is of no meaning to this particular idea of backradiation?

It appears the result does not fit expectations so is set aside for what does fit expectations.

12. ### claesjohnson

Einstein did not like the idea of atoms throwing dice, and neither do I.

13. ### Richard T. Fowler

“The fact we got closer and closer to the idea of zero and found it harder and harder to do so suggests the idea is real.”

Wow, so you actually believe that the harder we look without finding, the more likely it becomes that the thing exists?

And you call that approach “science”?

RTF

• ### iceskaterfinland

We found it got harder and harder to get to absolute zero but we got closer and closer. I think now we are only truelly tiny amounts from absolute zero. Ie we have arrived at something like 0.01K

• ### iceskaterfinland

And on the topic of very good/idea black body absorbers, there has been huge research going on in the area of things like infrared detection where nanotechnological devices are being built and where if they can find better absorbers they can make even more advances, so as i said, it is possible that like absolute zero, in time humans will get closer and closer to the idea of the perfect absorber.

You just managed to totally transmogrify the point i was making

• ### Richard T. Fowler

I’m glad I could be of assistance.

RTF

14. ### iceskaterfinland

Einstein i think demonstrated that a molecule is as equally like to cool via emission as it is to warm by absorption.

If you have two molecules in an object one can be cool and the other hot. Without throwing dice they can warm and cool each other alternately for ever.

You asked if modern physics supported the idea that atoms and molecules emit in all directions regardless of the properties of other atoms and molecules. And the answer as far as I am aware is absolutely yes they do. I have never heard of your kind of idea. So it must be a pretty unusual one.

However, If you do have any historical documents that post date Lord Kelvin and co, to suggest there was ever a serious debate on this topic it would be interesting to become aware of that debate.

As far as i know all that has happened over the years is that the existing ideas have been reinforced by the new discoveries.

Your idea that you can say to Roy Spencer that you are right and he is wrong seems to me to be totally at odds with all of science

• ### iceskaterfinland

Einstein also demonstrated that Newtons gravity was wrong to say that gravity acts at a distance but Newton himself thought it was absurd that gravity acted at a distance but he found that his theory enabled accurate calculations of anything observable at that time. The fact that Newton was able to realise this is a testament to the mans abilities. Your theory also acts at a distance so even Newton would say the idea is absurd. It sounds enormously far fetched to me, that a molecule 100 million light years away has some connection to me. It sounds however a nice thought as i write that though. Even the absorptions created by emissions from my parents and grandparents are somehow connected to me – as are all the absorptions from every living creature on the planet and all who ever lived before me going back hundreds if not billions of years into the past.

• ### Richard T. Fowler

As I wrote at drroyspencer.com recently, I consider Claes’ model to be an ideal one like a blackbody. If a blackbody doesn’t have to be physically real to be valid for some purposes, why does Claes’ model have to?

This is why some of Claes’ detractors in the scientific community jar me so much with their questioning of the instantaneity. (I think that’s a word.) I know they understand this concept of idealism, so I have to ask myself, “Are they just delusional, or is there something more going on here?” I know they are not stupid, and I know that this issue is simple. So the question is frankly unavoidable — because the issue keeps coming up over and over, and over.

And over. Case in point, the Air Vent.

RTF

15. ### blouis79blouis79

I understood the 2nd law was not specific about mode of energy/heat transfer. So then any consideration of radiation alone can only be valid in a vacuum. Should be easy to do some proper physics experiments in a vacuum.

In the real world, heat transfer by radiation is small in comparison with conduction and convection. Anyone playing with an IR thermometer can demonstrate presence of radiation with very little heat flow. That is what climate science is doing. Nonsense physics.

So when a system can transfer heat by conduction, convection and radiation, can we compute which modes transmit how much energy?????

• ### iceskaterfinland

blouis, all the earths incoming solar energy has to leave by radiation alone, and you only have to stand next to a warm surface to realise that radiation is a significant factor in air.

But that is not the issue here. The issue here is how much different the models will be if radiation transfer slows down, rather than does not slow down but some comes back to the sender.

Either way for the surface layer the result is the same. Higher up you need more complex analysis to see what difference it makes. Roy Spencer implied it would be the same higher up for the same other assumptions. If he is wrong that would be interesting to me and probably to him too.

• ### Richard T. Fowler

Have you ever tried measuring a temperature both over and under a heat source, at equal distance from the source? I think that that quantity of difference in temperature should approximately relate to the difference between rates of propagation by convection and radiation

RTF

• ### Richard T. Fowler

Also, what is the average density of matter in space adjacent to the Earth boundary? What is the amount of thermal energy per square meter per second that can be moved across that boundary by convection and conduction?

I bet the answer to that is profoundly higher than many here have been inclined to think it is. Especially since everyone who I’ve seen comment about this issue seems to believe it is approximately zero.

I was born in the day, but not yesterday.

RTF

• ### iceskaterfinland

Re first law of motion. Are you sure you are not thinking of the 3rd law? Otherwise your point is lost on me.

Re convection versus radiation. Convection could be enormously more important than the models consider.

However the point is what difference does it make if you slow down radiation as compared to allow it to go fast and send some back. We know the surface layer would be the same if we start from that point upwards. I dont know sufficient about the heat content and heat capacity of the atmosphere to make a judgement further up but Spencer seems to think it would be the same anyway. Much of the cold surface and cold atmosphere is currently warming much of the warmer atmosphere in the models. It sounds pretty complex and i dont know the answer. We know it is warmer at the surface when the atmosphere is colder above and water is present in the atmosphere – but this is true for either the current model or the one proposed here. Water is hugely more powerful than the other absorbers and it seems to me a model would be hopelessly wrong *constantly* if it could not correctly model the impact of radiation from water.

• ### Richard T. Fowler

Just remember, you wrote that — not me or anyone on my side. Noted for the record.

RTF

16. ### iceskaterfinland

Richard, the atmosphere ends where space begins. Space begins when you can no longer detect the atmosphere. That is something like 50,000 miles up as far as i know.

Any atoms or molecules at that height that are in space are not going to make much difference to heat transfer by collisions with the atmosphere.

17. ### iceskaterfinland

Richard, Re: action at a distance and ideal behaviours, What benefits do we get from considering such a difficult idea over one that is enormously simple to understand?

The idea also seems flawed in that it talks about heat transfer. When heat cannot be transferred. All that can happen is old heat is removed when a new energy is released and then the new energy creates some new heat, where with all knowledge currently know the speed of light is finite. Heat therefore dissapears at the atomic level.

And a foundation of current thinking is that there are vibrating atoms and molecules where such simple observations as brownian motion give us information as to the nature of the vibrations. Since these are only vibrations of particles why is it that we cannot consider these particles behave like any other group of particles that collide and transfer different amounts of energy in a random fashion so that some particles have more energy and others have almost none at all?

As a beginning point therefore it seems that some particular theory of molecular collisions is needed. But again that stretches my mind to think that humans do not already have detailed understanding of the nature of such collisions which you can suppose follow normal perfectly elastic collision ‘mechanics’?

Another basic problem is people like Lord Kelvin and others worked thru their ideas with thought experiments along the lines of maxwells demon where they knew that statistically their laws were only approximations at the atomic level.

18. ### Richard T. Fowler

Hint: Atmosphere, theoretically, is always detectable at all altitudes within the universe. If not, we would not be able to estimate the density of matter in “outer space”.

RTF

• ### iceskaterfinland

Yes, but the Earths boundary is when you can no longer tell the difference. Meaning there is bugger all up there to transfer heat by conduction where atoms have to collide to conduct.

• ### Richard T. Fowler

I’m sorry but I strongly disagree that the effect on results is negligible.

RTF

• ### iceskaterfinland

what mechanism are you proposing for meaningful heat transfer by conduction in what is essentially a vacuum? Ie what is the basis of your strong disagreement?

• ### Richard T. Fowler

“billions and billions” of particles of matter.

• ### iceskaterfinland

>>“billions and billions” of particles of matter.

What particles are you referring to? Heat is by definition atomic and molecular vibrations. Conduction has to be via those particles.

Are we quibbling over the word zero or something more substantial than that?

• ### Richard T. Fowler

Well, I think we’re quibbling over the question of whether the effect of billions and billions of particles of matter crossing back and forth across the TOA is negligible. By “negligible” in this context I mean “reasonable to neglect”.

RTF

19. ### Richard T. Fowler

Regarding “what benefits do we get [. . .]”

We don’t know until we do sufficient calculations, do we? So what are you waiting for?

(If you can’t do the calculations then fine; at least stop questioning the rationale of those who want to.)

As far as potential benefits, how about not having the world’s economy destroyed and killing hundreds of millions of people with insane “climate” policy?

RTF

• ### iceskaterfinland

if the rational for the theory is to stop insane climate modelling then you are better off tackling the models fairly flacky *assumptions* than attacking what seems to me to the bedrock of modern scientific thinking going back hundreds of years.

• ### Richard T. Fowler

Again, please familiarize yourself with the “first law of motion” and the history of its supposed “application” to problems.

RTF

20. ### iceskaterfinland

Richard, I am not involved in climate models. So noting what i say is not going to help change the course of history.

Re: First law: The velocity of a body remains constant unless the body is acted upon by an external force.

I do not know what you are getting at. Neither is it obvious to me how i find out the history involving this law and how it was used incorrectly.

• ### Richard T. Fowler

It’s all right, skater. Give it time. I see reason for hope in you.

RTF

• ### iceskaterfinland

OK it seems you are referring to the first law being wrong or only valid when referenced correctly in response to this:

“It seems to me you are objecting to the foundation ideas about atoms and molecules and want to view mass as something that is indivisible, or at least you have some objection to viewing matter at the atomic scale when you consider individual processes such as a molecule cooling when it emits radiation where the radiation is not what we call heat

At the atomic scale a molecule can emit radiation and ‘be cooled’ and warm another molecule that is ‘heated’. No work is done.

And yet you talk about second law violations.”

All I am asking is you clarify what part of that you were focusing upon when you responded

• ### Richard T. Fowler

I don’t believe the “law” is wrong. I am not positively certain it is always true in natural interactions, but I am close enough that as a rule, I always assume it. I put it in quotes because I don’t think it’s correct to call it a “law”.

My reference in this context was to the idea of no work being done.

RTF

• ### iceskaterfinland

what work is being done? if a molecule cools by emitting an equal amount of energy, the total energy is the same. And if another molecule of the same energy level as the cooled molecule absorbs the energy then this is now warmed to the original temperature of the first molecule. One got cooler and one got hotter for no change in the system

Where is the second law violation?

Anyway the second law as originally written down by Clausius and then by Kelvin is referring to the impossibility of self actuating heat engines that continue operating, rather than a machine that can cool a part of a system it exists in and then go no further.

• ### Richard T. Fowler

Are you trying to make me mad?

When did I mention the “second law”?

Are you seriously telling me that you actually forgot what I was talking about?

Because you sure don’t seem to have forgotten anything else. And I have a pretty large-size population of commentary to draw from in reaching that conclusion.

RTF

21. ### iceskaterfinland

I am trying to work out why you mentioned the first law of motion. And you said you were using it:

————-quote of you——————

My reference in this context was to the idea of no work being done

————-end quote of you————–

And previous to that i had said that it appeared when you first mentioned the first law you were responding to the following which mentions “no work” and “the second law of thermodynamics”

——————–My quote of myself————————–

“OK it seems you are referring to the first law being wrong or only valid when referenced correctly in response to this:

“It seems to me you are objecting to the foundation ideas about atoms and molecules and want to view mass as something that is indivisible, or at least you have some objection to viewing matter at the atomic scale when you consider individual processes such as a molecule cooling when it emits radiation where the radiation is not what we call heat

At the atomic scale a molecule can emit radiation and ‘be cooled’ and warm another molecule that is ‘heated’. No work is done.

And yet you talk about second law violations.”

All I am asking is you clarify what part of that you were focusing upon when you responded

———-end of my quote of myself

• ### Richard T. Fowler

What a brilliant move. You quote all that and, conveniently for you, fail to mention that you wrote this:

“Where is the second law violation? ”

which, of course, implies that I that I was stating that there was one.

I think it is now pretty obvious that you are aware that I didn’t state that. I was talking about the “first law of motion”, not, the “second law of thermodynamics”.

Can we now please have some honest discussion?

RTF

• ### iceskaterfinland

I was talking about no work being done and i was talking about the second law of thermodynamics after Claes mentioned the second law

In the text i just quoted from ***myself*** i said:

“No work is done.

And yet you talk about second law violations.”

THAT WAS BEFORE YOU MENTIONED THE FIRST LAW OF MOTION.

So once again the question is why did you mention the first law of motion?

• ### Richard T. Fowler

Hopefully you can see that I was saying that for the “second law of thermo” (2LT) to not be violated, the first law of motion (1LM) must be, because for 2LT to be inviolate, by your logic, there must be no work. But the 1LM requires that work be done. So I was not saying you are violating 2LT; I was saying that you are violating 1LM. Clear enough?

RTF

• ### iceskaterfinland

I am not sure what you are saying

Are you saying that a molecule cannot spontaneously emit a photon, and after or during emission have less kinetic energy?

I was assuming there would a collision, an atomic vibration would occur or whatever happens at that point to result in emission, and therefore the collision would no longer be the elastic collision expected without emission, and less energy would be present in the molecule. The energy would be in the photon.

• ### Richard T. Fowler

Well, a number of different issues, but the most important, I think, is that you have a wave in the aether, this constitutes a pressure gradient, and the exertion of pressure requires work. Therefore, it is impossible to assume that there is no loss with distance traveled, without violating 1LM.

I could cite at least two other problems, but I’ll let them be right now.

RTF

22. ### iceskaterfinland

Instead of saying i am dishonest just attempt to see what the nature of the misunderstanding is between us. I have no idea why you mentioned the first law of motion in the context of no work being done. However i talked about no work being done and the second law of thermodynamics…..so assumed your comment was related to that somehow.

• ### Richard T. Fowler

Excuse me, I wasn’t assuming anything. I had very good reason to believe that, and I laid all my evidence out. I left nothing to the imagination. As I should, since were dealing here with an environment poisoned and almost completely permeated with abject fraud and ubiquitous baseless attacks, including on character.

I accept that you apparently misunderstood. Hopefully you can accept the need to be a bit more careful in your own assumptions. You assumed the worst possible thing I can even imagine about what was going on in my head. Absolutely out of bounds for civil discussion, in my opinion. But no hard feelings, anyway. It looks like you are just a little confused.

RTF

• ### iceskaterfinland

??????

What did i say that was out of bounds?

You said this

—————————–

“What a brilliant move. You quote all that and, conveniently for you, fail to mention that you wrote this:

“Where is the second law violation? ”

which, of course, implies that I that I was stating that there was one.

I think it is now pretty obvious that you are aware that I didn’t state that. I was talking about the “first law of motion”, not, the “second law of thermodynamics”.

Can we now please have some honest discussion?

————————–

So you think it was a brilliant move by me……

You think it now pretty obvious i am aware you never said XYZ

You request a more honest discussion

————————-

And you say i am confused! 🙂

You just called me a liar…..

But no worries………..

Please dont blame me like that though

• ### Richard T. Fowler

Sir, please, I did not call you a liar. And please note that I retracted what concerns I had. Your latest comment can be seen to suggest that I have not. I have.

As far as your question about what you stated that was out of bounds for civil discussion, I was referring to the statement: “Where is the second law violation?” and supporting commentary.

RTF

23. ### iceskaterfinland

>>Well, a number of different issues, but the most important, I think, is that you have a wave in the aether, this constitutes a pressure gradient, and the exertion of pressure requires work. Therefore, it is impossible to assume that there is no loss with distance traveled, without violating 1LM.

But with this reasoning any isolated system would be doing work. So energy would be lost unless it can be returned/received. Also it seems to imply radiation would slow down with distance or lose amplitude

• ### Richard T. Fowler

Yes!

By the grace of God, you’re starting to understand.

RTF

24. ### iceskaterfinland

What did i say to justify you saying that i had said anything out of bounds?????

Seems to me you lost your temper and now find it hard to admit you called me a liar.

You just said I had deliberately planned to obfuscate over something or other to hide what i was up to when it was obvious what i was doing. Then you asked for a more honest discussion

Fairly obviously you want to avoid responsibility for calling me a liar with the following:

Hopefully you can accept the need to be a bit more careful in your own assumptions. You assumed the worst possible thing I can even imagine about what was going on in my head. Absolutely out of bounds for civil discussion, in my opinion. But no hard feelings, anyway. It looks like you are just a little confused.

And now later by making out that comment has something to do with our earlier conversation

• ### Richard T. Fowler

What you said to justify my statement that you said anything out of bounds for civil discussion:

“Where is the second law violation?”

That’s it.

I am not trying to get out of anything. If it’s so important to you, I acknowledge that I implied you were probably being dishonest. I was never sure, and I was very careful not to make a statement of certainty. I subsequently came to believe that you had not been dishonest, and I (in the spirit of honesty) said so.

Fair enough?

RTF

• ### iceskaterfinland

I still have no idea what generated this from you

“Hopefully you can accept the need to be a bit more careful in your own assumptions. You assumed the worst possible thing I can even imagine about what was going on in my head. Absolutely out of bounds for civil discussion, in my opinion.”

How does this:

“Where is the second law violation?” and subsequent discussion.

me assuming the worst possible thing you can even imagine about what was going on in your head that is absolutely out of bounds for civil discussion in your opinion

???????????????????

• ### Richard T. Fowler

Because I would have to be very dumb to say “first law of motion” when I was (hypothetically) trying to attack the “second law of thermodynamics.”

I think you are taking this way too hard. There is no need to keep being upset. It was a misunderstanding. These things happen, right?

RTF

• ### Richard T. Fowler

Rather, to attack the application of it.

RTF

25. ### iceskaterfinland

You entered a conversation that was not happening between me and Claes where i mentioned the second law of thermodynamics and no work being done.

You then were pretty obtuse about why you mentioned the first law. But i persisted rather than giving up even though you thought i was some modelling fraudster criminal where you felt some need to take down notes about me to be used later.

I was not saying you were dumb. You generated all of that. And that lead to you losing your temper and saying in clear terms i was deliberatingly winding you up and i was not honest.

• ### iceskaterfinland

Yes i wondered if you were mixing the first and third law of motion. Yes i must have thought you were dumb

• ### Richard T. Fowler

Whatever, my friend. No worries.

One thing, though: At no time have I thought that you are a modelling fraudster criminal. I only thought that you were defending the modelling fraudster criminals. And I stand by the assertion that you were. But I can now say that you clearly are an honest person, and I really do respect that. Thank you.

RTF

26. ### Richard T. Fowler

By the way, there is a reason to be somewhat vague in discussing these types of matters. It is because if one’s audience is made to do some of the thinking, they will feel more sense of “ownership” of any resulting insights. If they are just told what they should believe, it is easier to dismiss it because, after all, it was never “theirs” to begin with.

RTF

27. ### iceskaterfinland

Ok. Moving on then. And shaken but not stirred. In summary it seems you guys have an alternate theory of the universe so whatever i make comments upon is only coming from my ability to reason with existing paradigms and it appears that my comments are always going to be invalid in your paradigm……..

Do you though have an explanation of how an uncooled microbolometer used in a IR camera can accurately know the temperature of a cold object if emissivity is correctly set for that surface, where the camera is using the resistance of an absorber to calculate the temperature of the detector? Where importantly because of the lens it appears the camera sees more of the object than the object can see of the detector. The researchers claim the emissions are warming a thermally isolated vacuum protected detector.

• ### Richard T. Fowler

First of all, there is a huge gulf between Claes’ physics and cosmology and mine. But I support his work, because I think some of it has promise.

Secondly, I don’t think he supports my statements above about the first law of motion and its role in thermal transfer. I could be wrong, but I don’t think he does.

Thirdly, he has written about the issue you are asking about many times. I am really not too comfortable going into the details, but I can tell you that the general idea is that cooler “temps” are recorded because the radiation creates some kind of disturbance in the receptor, without causing warming.

I’m sure Claes will probably have more to say about it, so I won’t bother linking to his past writings.

RTF

• ### Richard T. Fowler

Excuse me. Without causing thermal transfer. See how hard this is?

RTF

• ### iceskaterfinland

thanks. The more i think about those cameras the more it does my head in. After all you have an ambient temp detector surrounded by huge thermal energy you would imagine, but apparently not so huge that objects two miles away can be detected using IR because the resistance of the detector changes. Sometimes i really wonder if they got this stuff from aliens! 🙂